id question solution final_answer context image modality difficulty is_multiple_answer unit answer_type error question_type subfield subject language 0 Let $P$ and $P^{\prime}$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\ell$ through $O$ the segment $\ell \cap P$ is strictly longer than the segment $\ell \cap P^{\prime}$. Is it possible that the ratio of the area of $P^{\prime}$ to the area of $P$ is greater than $1.9 ?$ ['The answer is in the affirmative: Given a positive $\\epsilon<2$, the ratio in question may indeed be greater than $2-\\epsilon$.\n\n\n\nTo show this, consider a square $A B C D$ centred at $O$, and let $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ be the reflections of $O$ in $A, B$, and $C$, respectively. Notice that, if $\\ell$ is a line through $O$, then the segments $\\ell \\cap A B C D$ and $\\ell \\cap A^{\\prime} B^{\\prime} C^{\\prime}$ have equal lengths, unless $\\ell$ is the line $A C$.\n\n\n\nNext, consider the points $M$ and $N$ on the segments $B^{\\prime} A^{\\prime}$ and $B^{\\prime} C^{\\prime}$, respectively, such that $B^{\\prime} M / B^{\\prime} A^{\\prime}=B^{\\prime} N / B^{\\prime} C^{\\prime}=(1-\\epsilon / 4)^{1 / 2}$. Finally, let $P^{\\prime}$ be the image of the convex quadrangle $B^{\\prime} M O N$ under the homothety of ratio $(1-\\epsilon / 4)^{1 / 4}$ centred at $O$. Clearly, the quadrangles $P \\equiv A B C D$ and $P^{\\prime}$ satisfy the conditions in the statement, and the ratio of the area of $P^{\\prime}$ to the area of $P$ is exactly $2-\\epsilon / 2$.\n\n\n\n'] [] Text-only Competition True Theorem proof Geometry Math English 1 "Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that: $$ x=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor $$ Prove that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$" "['We prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.\n\n\n\nLet $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,\n\n\n\n$$\n\n\\sum_{b \\in B, b \\leq c}\\left\\lfloor\\sqrt[k]{\\frac{c}{b}}\\right\\rfloor=c\n\n$$\n\n\n\nTo this end, notice that every positive integer has a unique representation as a product of an element in $B$ and a $k$ th power. Consequently, the set of all positive integers less than or equal to $c$ splits into\n\n\n\n$$\n\nC_{b}=\\left\\{x: x \\in \\mathbb{Z}_{>0}, x \\leq c, \\text { and } x / b \\text { is a } k \\text { th power }\\right\\}, \\quad b \\in B, b \\leq c\n\n$$\n\n\n\nClearly, $\\left|C_{b}\\right|=\\lfloor\\sqrt[k]{c / b}\\rfloor$, whence the desired equality.\n\n\n\nFinally, enumerate $B$ according to the natural order: $1=b_{1}b_{n-1}=a_{n-1}$ and\n\n\n\n$$\n\nb_{n}=\\sum_{i=1}^{n}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{b_{i}}}\\right\\rfloor=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{b_{i}}}\\right\\rfloor+1=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{a_{i}}}\\right\\rfloor+1\n\n$$\n\n\n\nthe definition of $a_{n}$ forces $a_{n} \\leq b_{n}$. Were $a_{n}0$ then $f_{n}(x) \\leq f_{n}(x-1)$; this means that such an $x$ cannot equal $a_{n}$. Thus $a_{j} / a_{i}$ is never the $k$ th power of an integer if $j>i$.\n\n\n\nNow we are prepared to prove by induction on $n$ that $a_{1}, a_{2}, \\ldots, a_{n}$ are exactly all $k$ thpower-free integers in $\\left[1, a_{n}\\right]$. The base case $n=1$ is trivial.\n\n\n\n\n\n\n\nAssume that all the $k$ th-power-free integers on $\\left[1, a_{n}\\right]$ are exactly $a_{1}, \\ldots, a_{n}$. Let $b$ be the least integer larger than $a_{n}$ such that $g_{n}(b)=0$. We claim that: (1) $b=a_{n+1}$; and (2) $b$ is the least $k$ th-power-free number greater than $a_{n}$.\n\n\n\nTo prove (1), notice first that all the numbers of the form $a_{j} / a_{i}$ with $1 \\leq i1$ be the greatest integer such that $y^{k} \\mid b$; then $b / y^{k}$ is $k$ th-power-free and hence $b / y^{k}=a_{i}$ for some $i \\leq n$. So $b / a_{i}$ is the $k$ th power of an integer, which contradicts the definition of $b$.\n\n\n\nThus $a_{1}, a_{2}, \\ldots$ are exactly all $k$ th-power-free positive integers; consequently all primes are contained in this sequence.']" [] Text-only Competition True Theorem proof Number Theory Math English 2 $2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exactly once. Prove that some edge has never been chosen. ['Step 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i\n\n\n\nFirst Approach. Since $\\mathrm{O}_{1} \\mathrm{O}_{2} \\mathrm{O}_{3} \\mathrm{O}_{4}$ is a parallelogram, $\\overrightarrow{F_{1} F_{2}}+\\overrightarrow{F_{3} F_{4}}=\\mathbf{0}$ and $\\overrightarrow{F_{2} F_{3}}+\\overrightarrow{F_{4} F_{1}}=\\mathbf{0}$; this still holds in the degenerate case, for if the $O_{i}$ are collinear, then they all lie on the line $T_{1} T_{4}$, and each $O_{i}$ is the midpoint of the segment $T_{i} T_{i+1}$. Consequently,\n\n\n\n$$\n\n\\begin{aligned}\n\n\\overrightarrow{P Q}-\\overrightarrow{R S}+\\overrightarrow{T U}-\\overrightarrow{V W}= & \\left(\\overrightarrow{P F_{1}}+\\overrightarrow{F_{1} F_{2}}+\\overrightarrow{F_{2} Q}\\right)-\\left(\\overrightarrow{R F_{2}}+\\overrightarrow{F_{2} F_{3}}+\\overrightarrow{F_{3} S}\\right) \\\\\n\n& +\\left(\\overrightarrow{T F_{3}}+\\overrightarrow{F_{3} F_{4}}+\\overrightarrow{F_{4} U}\\right)-\\left(\\overrightarrow{V F_{4}}+\\overrightarrow{F_{4} F_{1}}+\\overrightarrow{F_{1} W}\\right) \\\\\n\n= & \\left(\\overrightarrow{P F_{1}}-\\overrightarrow{F_{1} W}\\right)-\\left(\\overrightarrow{R F_{2}}-\\overrightarrow{F_{2} Q}\\right)+\\left(\\overrightarrow{T F_{3}}-\\overrightarrow{F_{3} S}\\right)-\\left(\\overrightarrow{V F_{4}}-\\overrightarrow{F_{4} U}\\right) \\\\\n\n& +\\left(\\overrightarrow{F_{1} F_{2}}+\\overrightarrow{F_{3} F_{4}}\\right)-\\left(\\overrightarrow{F_{2} F_{3}}+\\overrightarrow{F_{4} F_{1}}\\right)=\\mathbf{0} .\n\n\\end{aligned}\n\n$$\n\n\n\nAlternatively, but equivalently, $\\overrightarrow{P Q}+\\overrightarrow{T U}=\\overrightarrow{R S}+\\overrightarrow{V W}$, as required.\n\n\n\nSecond Approach. This is merely another way of reading the previous argument. Fix an orientation of the line $P W$, say, from $P$ towards $W$, and use a lower case letter to denote the coordinate of a point labelled by the corresponding upper case letter.\n\n\n\nSince the diagonals of a parallelogram bisect one another, $f_{1}+f_{3}=f_{2}+f_{4}$, the common value being twice the coordinate of the projection to $P W$ of the point where $O_{1} O_{3}$ and $O_{2} O_{4}$ cross; the relation clearly holds in the degenerate case as well.\n\n\n\n\n\n\n\nPlug $f_{1}=\\frac{1}{2}(p+w), f_{2}=\\frac{1}{2}(q+r), f_{3}=\\frac{1}{2}(s+t)$ and $f_{4}=\\frac{1}{2}(u+v)$ into the above equality to get $p+w+s+t=q+r+u+v$. Alternatively, but equivalently, $(q-p)+(u-t)=(s-r)+(w-v)$, that is, $P Q+T U=R Q+V W$, as required.'] [] Text-only Competition True Theorem proof Geometry Math English 4 A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4 . Prove that the number of such ways to assign the leaders is divisible by 17 . "[""Assume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.\n\n\n\nConsider the variables $x_{1}, x_{2}, \\ldots, x_{17}$ corresponding to the workers. Assign each brigade (from the $i$-th through the $j$-th worker) the polynomial $f_{i j}=x_{i}+x_{i+1}+\\cdots+x_{j}$, and form the product $f=\\prod_{1 \\leq i \\leq j \\leq 17} f_{i j}$. The number $N$ is the sum $\\Sigma(f)$ of the coefficients of all monomials $x_{1}^{\\alpha_{1}} x_{2}^{\\alpha_{2}} \\ldots x_{17}^{\\alpha_{17}}$ in the expansion of $f$, where the $\\alpha_{i}$ are all congruent to 1 modulo 4 . For any polynomial $P$, let $\\Sigma(P)$ denote the corresponding sum. From now on, all polynomials are considered with coefficients in the finite field $\\mathbb{F}_{17}$.\n\n\n\nRecall that for any positive integer $n$, and any integers $a_{1}, a_{2}, \\ldots, a_{n}$, there exist indices $i \\leq j$ such that $a_{i}+a_{i+1}+\\cdots+a_{j}$ is divisible by $n$. Consequently, $f\\left(a_{1}, a_{2}, \\ldots, a_{17}\\right)=0$ for all $a_{1}, a_{2}, \\ldots, a_{17}$ in $\\mathbb{F}_{17}$.\n\n\n\nNow, if some monomial in the expansion of $f$ is divisible by $x_{i}^{17}$, replace that $x_{i}^{17}$ by $x_{i}$; this does not alter the above overall vanishing property (by Fermat's Little Theorem), and preserves $\\Sigma(f)$. After several such changes, $f$ transforms into a polynomial $g$ whose degree in each variable does not exceed 16 , and $g\\left(a_{1}, a_{2}, \\ldots, a_{17}\\right)=0$ for all $a_{1}, a_{2}, \\ldots, a_{17}$ in $\\mathbb{F}_{17}$. For such a polynomial, an easy induction on the number of variables shows that it is identically zero. Consequently, $\\Sigma(g)=0$, so $\\Sigma(f)=0$ as well, as desired.""]" [] Text-only Competition True Theorem proof Number Theory Math English 5 Let $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$. ['Let the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ coincide, so $J_{b} J_{c} \\perp A D$.\n\n\n\nNow let $O$ be the circumcentre of the triangle $A J_{b} J_{c}$. Then $\\angle J_{b} A O=\\pi / 2-\\angle A O J_{b} / 2=$ $\\pi / 2-\\angle A J_{c} J_{b}=\\angle X A J_{c}=\\frac{1}{2} \\angle D A C$. Therefore, $\\angle B A O=\\angle B A J_{b}+\\angle J_{b} A O=\\frac{1}{2} \\angle B A D+$ $\\frac{1}{2} \\angle D A C=\\frac{1}{2} \\angle B A C$, and the conclusion follows.\n\n\n\n\n\n\n\nFig. 1'] [] Text-only Competition True Theorem proof Geometry Math English 6 " Prove that every positive integer $n$ can be written uniquely in the form $$ n=\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}} $$ where $k \geq 0$ and $0 \leq m_{1}0$, and even if $n \\leq 0$. The integer $w(n)=\\lfloor\\ell / 2\\rfloor$ is called the weight of $n$.\n\n\n\nExistence once proved, uniqueness follows from the fact that there are as many such representations as integers in the range $-2^{M}+1$ through $2^{M}$, namely, $2^{M+1}$.\n\n\n\nTo prove existence, notice that the base case $M=0$ is clear, so let $M \\geq 1$ and let $n$ be an integer in the range $-2^{M}+1$ through $2^{M}$.\n\n\n\nIf $-2^{M}+1 \\leq n \\leq-2^{M-1}$, then $1 \\leq n+2^{M} \\leq 2^{M-1}$, so $n+2^{M}=\\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}$ for some integers $k \\geq 0$ and $0 \\leq m_{1}<\\cdots\n\n\n\nRecall that $\\overrightarrow{A H}=2 \\overrightarrow{O M}$, so $\\overrightarrow{O M}=\\overrightarrow{A K}$. Hence $A O M K$ is a parallelogram, so $M K=O A=O B=$ $O C$.\n\n\n\nConsider the rotation through $\\angle(\\overrightarrow{O C}, \\overrightarrow{O B})$ about $M$. It maps $P$ to $Q$; let it map $K$ to some point $L$. Then $M K=M L=O B=O C$ and $\\angle L M K=\\angle B O C$, so the triangles $O B C$ and $M K L$ are congruent. Hence $B C=K L \\leq K Q+L Q=$ $K Q+K P$, as required.' ""Let $\\alpha=\\angle(P B, B C)=\\angle(Q C, B C)$. Since $P$ lies on the circle of diameter $O B$, $\\angle(O P, O M)=\\alpha$. Since also $Q$ lies on the circle of diameter $O C$, it immediately follows that $M P=M Q=R \\sin \\alpha$ by sine theorem in triangles $\\triangle O P M$ and $\\triangle O Q M$.\n\n\n\nBecause $P Q$ is the projection of $B C$ on line $\\ell$, it follows that $P Q=B C \\sin \\alpha$. Just like in the first solution, $K M=A O=R$ (the circumradius of triangle $\\triangle A B C$ ).\n\n\n\nNow apply Ptolemy's inequality for the quadrilateral $K P M Q: K P \\cdot M Q+K Q \\cdot M P \\geq P Q \\cdot K M$, and now substitute the relations from above, leading to\n\n\n\n$$\n\nR \\sin \\alpha(K P+K Q) \\geq R \\sin \\alpha \\cdot B C\n\n$$\n\n\n\nwhich is precisely the conclusion whenever $\\sin \\alpha \\neq 0$. The case when $\\sin \\alpha=0$ can be treated either directly, or via a limit argument."" 'Denote by $R$ and $O$ the circumradius and the circumcentre of triangle $A B C$, respectively. As in Solution 1, we see that $M K=R$.\n\n\n\nAssume now that $\\ell$ is fixed, while $A$ moves along the fixed circle $(A B C)$. Then $K$ will move along a cricle centred at $M$ with radius $R$. We must show that for each point $K$ on this circle we have $B C \\leq K P+K Q$. In doing so, we prove that the afore-mentioned circle contains an ellipse with foci at $Q$ and $P$ with distance $B C$.\n\n\n\nLet $S$ be the foot of the perpendicular from $M$ to $P Q$, it is easy to verify that $S$ is the center of the ellipse. We shall then consider it as the origin. Let $u=\\frac{B C}{2}$ and $t=\\frac{P Q}{2}$; notice that $u$ is the major semi-axis of the ellipse and $\\sqrt{u^{2}-t^{2}}$ is the minor one. Assume $X(x, y)$ is a point on this ellipse. We now need to prove $M X \\leq R$.\n\n\n\nSince $X$ is on the ellipse, we can write $(x, y)=$ $\\left(u \\cos \\theta, \\sqrt{u^{2}-t^{2}} \\sin \\theta\\right)$, for some $\\theta \\in(0,2 \\pi)$. Since $M X^{2}=x^{2}+(y+M S)^{2}$, we can expand and obtain\n\n\n\n$M X^{2}=u^{2}+M S^{2}-t^{2} \\cdot \\sin ^{2} \\theta+2 M S \\cdot \\sqrt{u^{2}-t^{2}} \\cdot \\sin \\theta$.\n\n\n\nAdd and subtract $M S^{2}\\left(u^{2}-t^{2}\\right) / t^{2}$ in order to obtain a square on the right hand side: $M X^{2}=u^{2}+$ $M S^{2}+\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}-\\left(t \\sin \\theta-\\frac{M S \\sqrt{u^{2}-t^{2}}}{t}\\right)^{2}$. It now suffices to show that $u^{2}+M S^{2}+\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}=$ $R^{2}$, since then it would immediately follow that $M X^{2} \\leq R^{2}$.\n\n\n\nApplying Pythagorean theorem in triangles $O B M$ and $O S M$, we obtain $R^{2}=u^{2}+O M^{2}$ and $O M^{2}=M S^{2}+O S^{2}$, so it remains to prove that $O S^{2}=\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}$. Let $\\alpha=\\angle(O P, B M)$, then $O S / M S=\\tan \\alpha$ and $t / u=\\cos \\alpha$, so $O S^{2}=$ $M S^{2} \\tan ^{2} \\alpha=M S^{2}\\left(\\frac{1-\\cos ^{2} \\alpha}{\\cos ^{2} \\alpha}\\right)=M S^{2} \\cdot \\frac{u^{2}-t^{2}}{t^{2}}$, which is the desired result.']" [] Text-only Competition True Theorem proof Geometry Math English 8 "Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$. Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$." "['Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\\operatorname{deg} Q \\mid \\operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically, as desired.\n\n\n\nChoose the polynomials $T(x)$ and $M(x)$ such that\n\n\n\n$$\n\nS(x)=T(Q(x))+M(x) \\tag{*}\n\n$$\n\n\n\nwhere $\\operatorname{deg} M$ is minimised; if $M=0$, then we get the desired result. For the sake of contradiction, suppose $M \\neq 0$. Then $q \\nmid m=\\operatorname{deg} M$; otherwise, $M(x)=\\beta Q(x)^{m / q}+M_{1}(x)$, where $\\beta$ is some number and $\\operatorname{deg} M_{1}<\\operatorname{deg} M$, contradicting the choice of $M$. In particular, $01$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that $\\operatorname{deg}\\left(P(x)-U(x)^{c}\\right)<(c-1) b$ and $\\operatorname{deg}(R(x)-$ $\\left.V(x)^{c}\\right)<(c-1) a$. Then $\\operatorname{deg}\\left(F(x)-U(Q(x))^{c}\\right)=$ $\\operatorname{deg}\\left(P(Q(x))-U(Q(x))^{c}\\right)<(c-1) a b d, \\operatorname{deg}(F(x)-$ $\\left.V(S(x))^{c}\\right)=\\operatorname{deg}\\left(R(S(x))-V(S(x))^{c}\\right)<(c-1) a b d$, so $\\operatorname{deg}\\left(U(Q(x))^{c}-V(S(x))^{c}\\right)=\\operatorname{deg}((F(x)-$ $\\left.\\left.V(S(x))^{c}\\right)-\\left(F(x)-U(Q(x))^{c}\\right)\\right)<(c-1) a b d$.\n\n\n\nOn the other hand, $U(Q(x))^{c}-V(S(x))^{c}=$ $(U(Q(x))-V(S(x)))\\left(U(Q(x))^{c-1}+\\cdots+\\right.$ $\\left.V(S(x))^{c-1}\\right)$.\n\n\n\nBy the preceding, the degree of the left-hand member is (strictly) less than $(c-1) a b d$ which is precisely the degree of the second factor in the righthand member. This forces $U(Q(x))=V(S(x))$, so $U(Q(x))=V(S(x))$ has degree $a b d1$. Let $V$ denote the vertex set of $\\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.\n\n\n\nCase 1: There exists a partition $V=A \\sqcup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.\n\n\n\nSince $\\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.\n\n\n\nAssume that $T$, one of the three trees, does not contain $e$. Then the graph $T \\cup\\{e\\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e^{\\prime}$ connecting $A$ and $B$; the edge $e^{\\prime}$ is also blue. Replace $e^{\\prime}$ by $e$ in $T$ to get another tree $T^{\\prime}$ with the same number of edges of each colour as in $T$, but containing $e$.\n\n\n\nPerforming such an operation to all three trees, we arrive at the situation where the three trees $T_{r}^{\\prime}$, $T_{g}^{\\prime}$, and $T_{b}^{\\prime}$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\\Gamma^{*}$, and set $r^{*}=r, g^{*}=g$, and $b^{*}=b-1$. The new graph satisfies the conditions in the statement for those new values - indeed, under the shrinking, each of the trees $T_{r}^{\\prime}, T_{g}^{\\prime}$, and $T_{b}^{\\prime}$ loses a blue edge. So $\\Gamma^{*}$ has a spanning tree with exactly $r$ red, exactly $g$ green, and exactly $b-1$ blue edges. Finally, pass back to $\\Gamma$ by restoring $e$, to obtain the a desired spanning tree in $\\Gamma$.\n\n\n\nCase 2: There is no such a partition.\n\n\n\nConsider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \\cup G \\cup B$ has the smallest number $k$ of components. If $k=1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).\n\n\n\nAssume now that $k \\geq 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \\backslash V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.\n\n\n\nLet $e^{\\prime}$ be a red edge in $C$ and set $R^{\\prime}=R \\backslash\\left\\{e^{\\prime}\\right\\} \\cup$ $\\{e\\}$. Then $\\left(R^{\\prime}, G, B\\right)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.' 'For a spanning tree $T$ in $\\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.\n\n\n\nAssume that $\\mathcal{C}$ is some collection of spanning trees in $\\Gamma$. Write\n\n\n\n$$\n\\begin{aligned}\n\nr(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} r(T), & g(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} g(T), \\\\\n\nb(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} b(T), & R(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}}(T), \\\\\n\nG(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}} g(T), & B(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}} b(T) .\n\n\\end{aligned}\n$$\n\n\n\nSay that a collection $\\mathcal{C}$ is good if $r \\in[r(\\mathcal{C}, R(\\mathcal{C})]$, $g \\in[g(\\mathcal{C}, G(\\mathcal{C})]$, and $b \\in[b(\\mathcal{C}, B(\\mathcal{C})]$. By the problem conditions, the collection of all spanning trees in $\\Gamma$ is good.\n\n\n\nFor a good collection $\\mathcal{C}$, say that an edge $e$ of $\\Gamma$ is suspicious if $e$ belongs to some tree in $\\mathcal{C}$ but not to all trees in $\\mathcal{C}$. Choose now a good collection $\\mathcal{C}$ minimizing the number of suspicious edges. If $\\mathcal{C}$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(\\mathcal{C})g$.\n\n\n\nWe now distinguish two cases.\n\n\n\nCase 1: $B(\\mathcal{C})=b$.\n\n\n\nLet $T^{0}$ be a tree in $\\mathcal{C}$ with $g\\left(T^{0}\\right)=g(\\mathcal{C}) \\leq g$. Since $G(\\mathcal{C})>g$, there exists a green edge $e$ contained in some tree in $\\mathcal{C}$ but not in $T^{0}$; clearly, $e$ is suspicious. Fix one such green edge $e$.\n\n\n\nNow, for every $T$ in $\\mathcal{C}$, define a spanning tree $T_{1}$ of $\\Gamma$ as follows. If $T$ does not contain $e$, then $T_{1}=T$; in particular, $\\left(T^{0}\\right)_{1}=T^{0}$. Otherwise, the graph $T \\backslash\\{e\\}$ falls into two components. The tree $T_{0}$ contains some edge $e^{\\prime}$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_{1}=T \\backslash\\{e\\} \\cup\\left\\{e^{\\prime}\\right\\}$.\n\n\n\nLet $\\mathcal{C}_{1}=\\left\\{T_{1}: T \\in \\mathcal{C}\\right\\}$. All edges suspicious for $\\mathcal{C}_{1}$ are also suspicious for $\\mathcal{C}$, but no tree in $\\mathcal{C}_{1}$ con- tains $e$. So the number of suspicious edges for $\\mathcal{C}_{1}$ is strictly smaller than that for $\\mathcal{C}$.\n\n\n\nWe now show that $\\mathcal{C}_{1}$ is good, reaching thereby a contradiction with the choice of $\\mathcal{C}$. For every $T$ in $\\mathcal{C}$, the tree $T_{1}$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g\\left(\\mathcal{C}_{1}\\right) \\leq g(\\mathcal{C}) \\leq g, G\\left(\\mathcal{C}_{1}\\right) \\geq G(\\mathcal{C})-1 \\geq g$, $R\\left(\\mathcal{C}_{1}\\right) \\geq R(\\mathcal{C}) \\geq r, r\\left(\\mathcal{C}_{1}\\right) \\leq r(\\mathcal{C})+1 \\leq r$, and $B\\left(\\mathcal{C}_{1}\\right) \\geq B(\\mathcal{C}) \\geq b$. Finally, we get $b\\left(T^{0}\\right) \\leq$ $B(\\mathcal{C})=b$; since $\\mathcal{C}_{1}$ contains $T^{0}$, it follows that $b\\left(\\mathcal{C}_{1}\\right) \\leq b\\left(T^{0}\\right) \\leq b$, which concludes the proof.\n\n\n\nCase 2: $B(\\mathcal{C})>b$.\n\n\n\nConsider a tree $T^{0}$ in $\\mathcal{C}$ satisfying $r\\left(T^{0}\\right)=$ $R(\\mathcal{C}) \\geq r$. Since $r(\\mathcal{C})$ $1-1 / 100$. "[""If $x \\geq 1-1 /(100 \\cdot 2016)$, then\n\n\n\n$$\n\nx^{2016} \\geq\\left(1-\\frac{1}{100 \\cdot 2016}\\right)^{2016}>1-2016 \\cdot \\frac{1}{100 \\cdot 2016}=1-\\frac{1}{100}\n\n$$\n\n\n\nby Bernoulli's inequality, whence the conclusion.\n\n\n\nIf $x<1-1 /(100 \\cdot 2016)$, then $y \\geq(1-x)^{1 / 2016}>(100 \\cdot 2016)^{-1 / 2016}$, and it is sufficient to show that the latter is greater than $1-1 / 100=99 / 100$; alternatively, but equivalently, that\n\n\n\n$$\n\n\\left(1+\\frac{1}{99}\\right)^{2016}>100 \\cdot 2016\n\n$$\n\n\n\nTo establish the latter, refer again to Bernoulli's inequality to write\n\n\n\n$$\n\n\\left(1+\\frac{1}{99}\\right)^{2016}>\\left(1+\\frac{1}{99}\\right)^{99 \\cdot 20}>\\left(1+99 \\cdot \\frac{1}{99}\\right)^{20}=2^{20}>100 \\cdot 2016 .\n\n$$""]" [] Text-only Competition True Theorem proof Algebra Math English 12 "A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the $\operatorname{arc} A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$. Prove that $R \geq r_{1}+r_{2}+r_{3}$." ['Let $\\ell_{1}$ be the tangent to $\\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\\omega_{1}$. The tangents $\\ell_{2}$ and $\\ell_{3}$ are defined similarly. The lines $\\ell_{1}$ and $\\ell_{2}, \\ell_{2}$ and $\\ell_{3}, \\ell_{3}$ and $\\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.\n\n\n\nEach of the triangles $\\Delta_{1}=\\triangle X S_{1} T_{1}, \\Delta_{2}=\\triangle T_{2} X S_{2}$, and $\\Delta_{3}=\\triangle S_{3} T_{3} X$ is similar to $\\Delta=\\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\\Delta_{i}$ and $\\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\\rho_{i}$ is the inradius of $\\Delta_{i}$, then $\\rho_{1}+\\rho_{2}+\\rho_{3}=R$.\n\n\n\nFinally, notice that $\\omega_{i}$ is interior to $\\Delta_{i}$, so $r_{i} \\leq \\rho_{i}$, and the conclusion follows by the preceding.\n\n\n\n\n\nFig. 1\n\n\n\n\n\n\n\nFig. 2'] [] Text-only Competition True Theorem proof Geometry Math English 13 "A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the $\operatorname{arc} A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$. If $R=r_{1}+r_{2}+r_{3}$, prove that the six points where the circles $\omega_{i}$ touch the diagonals $A_{1} B_{2}, A_{2} B_{3}, A_{3} B_{1}$ are concyclic." ['Let $\\ell_{1}$ be the tangent to $\\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\\omega_{1}$. The tangents $\\ell_{2}$ and $\\ell_{3}$ are defined similarly. The lines $\\ell_{1}$ and $\\ell_{2}, \\ell_{2}$ and $\\ell_{3}, \\ell_{3}$ and $\\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.\n\n\n\nEach of the triangles $\\Delta_{1}=\\triangle X S_{1} T_{1}, \\Delta_{2}=\\triangle T_{2} X S_{2}$, and $\\Delta_{3}=\\triangle S_{3} T_{3} X$ is similar to $\\Delta=\\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\\Delta_{i}$ and $\\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\\rho_{i}$ is the inradius of $\\Delta_{i}$, then $\\rho_{1}+\\rho_{2}+\\rho_{3}=R$.\n\n\n\nFinally, notice that $\\omega_{i}$ is interior to $\\Delta_{i}$, so $r_{i} \\leq \\rho_{i}$, and the conclusion follows by the preceding.\n\n\n\n\n\nFig. 1\n\n\n\n\n\n\n\nFig. 2\n\nBy above, the equality $R=r_{1}+r_{2}+r_{3}$ holds if and only if $r_{i}=\\rho_{i}$ for all $i$, which implies in turn that $\\omega_{i}$ is the incircle of $\\Delta_{i}$. Let $K_{i}, L_{i}, M_{i}$ be the points where $\\omega_{i}$ touches the sides $X S_{i}, X T_{i}, S_{i} T_{i}$, respectively. We claim that the six points $K_{i}$ and $L_{i}(i=1,2,3)$ are equidistant from $X$.\n\n\n\nClearly, $X K_{i}=X L_{i}$, and we are to prove that $X K_{2}=X L_{1}$ and $X K_{3}=X L_{2}$. By similarity, $\\angle T_{1} M_{1} L_{1}=\\angle C_{3} M_{1} M_{2}$ and $\\angle S_{2} M_{2} K_{2}=\\angle C_{3} M_{2} M_{1}$, so the points $M_{1}, M_{2}, L_{1}, K_{2}$ are collinear. Consequently, $\\angle X K_{2} L_{1}=\\angle C_{3} M_{1} M_{2}=\\angle C_{3} M_{2} M_{1}=\\angle X L_{1} K_{2}$, so $X K_{2}=X L_{1}$. Similarly, $X K_{3}=X L_{2}$.'] [] Text-only Competition True Theorem proof Geometry Math English 14 A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $\mathcal{A}$ and $\mathcal{B}$. An $\mathcal{A B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $\mathcal{A}$ and the other in $\mathcal{B}$, and such that no segments form a closed polyline. An $\mathcal{A B}$-tree is transformed into another as follows: choose three distinct segments $A_{1} B_{1}, B_{1} A_{2}$ and $A_{2} B_{2}$ in the $\mathcal{A B}$-tree such that $A_{1}$ is in $\mathcal{A}$ and $A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and remove the segment $A_{1} B_{1}$ to replace it by the segment $A_{1} B_{2}$. Given any $\mathcal{A B}$-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps. ['The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $\\mathcal{A}$ and $\\mathcal{B}$, and transforming one into another preserves the degree of any vertex in $\\mathcal{A}$, but not necessarily that of a vertex in $\\mathcal{B}$.\n\n\n\nThe idea is to devise a strict semi-invariant of the process, i.e., assign each $\\mathcal{A B}$-tree a real number strictly decreasing under a transformation. Since the number of trees on the $n$ points is finite, the conclusion follows.\n\n\n\nTo describe the assignment, consider an $\\mathcal{A B}$-tree $\\mathcal{T}=(\\mathcal{A} \\sqcup \\mathcal{B}, \\mathcal{E})$. Removal of an edge $e$ of $\\mathcal{T}$ splits the graph into exactly two components. Let $p_{\\mathcal{T}}(e)$ be the number of vertices in $\\mathcal{A}$ lying in the component of $\\mathcal{T}-e$ containing the $\\mathcal{A}$-endpoint of $e$; since $\\mathcal{T}$ is a tree, $p_{\\mathcal{T}}(e)$ counts the number of paths in $\\mathcal{T}-e$ from the $\\mathcal{A}$-endpoint of $e$ to vertices in $\\mathcal{A}$ (including the one-vertex path). Define $f(\\mathcal{T})=\\sum_{e \\in \\mathcal{E}} p_{\\mathcal{T}}(e)|e|$, where $|e|$ is the Euclidean length of $e$.\n\n\n\nWe claim that $f$ strictly decreases under a transformation. To prove this, let $\\mathcal{T}^{\\prime}$ be obtained from $\\mathcal{T}$ by a transformation involving the polyline $A_{1} B_{1} A_{2} B_{2}$; that is, $A_{1}$ and $A_{2}$ are in $\\mathcal{A}, B_{1}$ and $B_{2}$ are in $\\mathcal{B}, A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and $\\mathcal{T}^{\\prime}=\\mathcal{T}-A_{1} B_{1}+A_{1} B_{2}$. It is readily checked that $p_{\\mathcal{T}^{\\prime}}(e)=p_{\\mathcal{T}}(e)$ for every edge $e$ of $\\mathcal{T}$ different from $A_{1} B_{1}, A_{2} B_{1}$ and $A_{2} B_{2}, p_{\\mathcal{T}^{\\prime}}\\left(A_{1} B_{2}\\right)=$ $p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right), p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{1}\\right)=p_{\\mathcal{T}}\\left(A_{2} B_{1}\\right)+p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)$, and $p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B b_{2}\\right)=p_{\\mathcal{T}}\\left(A_{2} B_{2}\\right)-p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)$. Consequently,\n\n\n\n$$\n\n\\begin{aligned}\n\nf\\left(\\mathcal{T}^{\\prime}\\right)-f(\\mathcal{T})= & p_{\\mathcal{T}^{\\prime}}\\left(A_{1} B_{2}\\right) \\cdot A_{1} B_{2}+\\left(p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{1}\\right)-p_{\\mathcal{T}}\\left(A_{2} B_{1}\\right)\\right) \\cdot A_{2} B_{1}+ \\\\\n\n& \\left(p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{2}\\right)-p_{\\mathcal{T}}\\left(A_{2} B_{2}\\right)\\right) \\cdot A_{2} B_{2}-p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right) \\cdot A_{1} B_{1} \\\\\n\n= & p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)\\left(A_{1} B_{2}+A_{2} B_{1}-A_{2} B_{2}-A_{1} B_{1}\\right)<0\n\n\\end{aligned}\n\n$$'] [] Text-only Competition True Theorem proof Geometry Math English 15 "Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$. Prove that there are infinitely many positive integers $n$ such that $\lambda(n)=\lambda(n+1)=+1$" "['Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nThe Pell equation $x^{2}-6 y^{2}=1$ has infinitely many solutions in positive integers; all solutions are given by $\\left(x_{n}, y_{n}\\right)$, where $x_{n}+y_{n} \\sqrt{6}=(5+2 \\sqrt{6})^{n}$. Since $\\lambda\\left(6 y^{2}\\right)=1$ and also $\\lambda\\left(6 y^{2}+1\\right)=\\lambda\\left(x^{2}\\right)=1$, the thesis is proven.' 'Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nTake any existing pair with $\\lambda(n)=$ $\\lambda(n+1)=1$. Then $\\lambda\\left((2 n+1)^{2}-1\\right)=\\lambda\\left(4 n^{2}+4 n\\right)=\\lambda(4) \\cdot \\lambda(n)$. $\\lambda(n+1)=1$, and also $\\lambda\\left((2 n+1)^{2}\\right)=\\lambda(2 n+1)^{2}=1$, so we have built a larger $(1,1)$ pair.']" [] Text-only Competition True Theorem proof Algebra Math English 16 "Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$. Prove that there are infinitely many positive integers $n$ such that $\lambda(n)=\lambda(n+1)=-1$." "['Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nThe equation $3 x^{2}-2 y^{2}=1$ (again Pell theory) has also infinitely many solutions in positive integers, given by $\\left(x_{n}, y_{n}\\right)$, where $x_{n} \\sqrt{3}+y_{n} \\sqrt{2}=(\\sqrt{3}+\\sqrt{2})^{2 n+1}$. Since $\\lambda\\left(2 y^{2}\\right)=$ -1 and $\\lambda\\left(2 y^{2}+1\\right)=\\lambda\\left(3 x^{2}\\right)=-1$, the thesis is proven.' 'Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nAssume $(\\lambda(n-1), \\lambda(n))$ is the largest $(-1,-1)$ pair, therefore $\\lambda(n+1)=1$ and $\\lambda\\left(n^{2}+n\\right)=\\lambda(n)$. $\\lambda(n+1)=-1$, therefore again $\\lambda\\left(n^{2}+n+1\\right)=1$. But then $\\lambda\\left(n^{3}-1\\right)=\\lambda(n-1) \\cdot \\lambda\\left(n^{2}+n+1\\right)=-1$, and also $\\lambda\\left(n^{3}\\right)=$ $\\lambda(n)^{3}=-1$, so we found yet a larger such pair than the one we started with, contradiction.' 'Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nAssume the pairs of consecutive terms $(-1,-1)$ in $\\mathfrak{S}$ are finitely many. Then from some rank on we only have subsequences $(1,-1,1,1, \\ldots, 1,-1,1)$. By ""doubling"" such a subsequence (like at point ii)), we will produce\n\n\n\n$$\n\n(-1, ?, 1, ?,-1, ?,-1, ?, \\ldots, ?,-1, ?, 1, ?,-1)\n\n$$\n\n\n\nAccording with our assumption, all ?-terms ought to be 1, hence the produced subsequence is\n\n\n\n$$\n\n(-1,1,1,1,-1,1,-1,1, \\ldots, 1,-1,1,1,1,-1)\n\n$$\n\n\n\nand so the ""separating packets"" of l\'s contain either one or three terms. Now assume some far enough $(1,1,1,1)$ or $(-1,1,1,-1)$ subsequence of $\\mathfrak{S}$ were to exist. Since it lies within some ""doubled"" subsequence, it contradicts the structure described above, which thus is the only prevalent from some rank on. But then all the positions of the $(-1)$ terms will have the same parity. However though, we have $\\lambda(p)=\\lambda\\left(2 p^{2}\\right)=-1$ for all odd primes $p$, and these terms have different parity of their positions. A contradiction has been reached.']" [] Text-only Competition True Theorem proof Algebra Math English 17 Consider an integer $n \geq 2$ and write the numbers $1,2, \ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. For all integers $n \geq 2$, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves. "['The answer is in the affirmative for all $n \\geq 2$. Induct on $n$. Leaving aside the trivial case $n=2$, deal first with particular cases $n=5$ and $n=6$.\n\n\n\nIf $n=5$, remove first the pair $(2,5)$, notice that $3=|2-5|$ is already on the board, so $7=2+5$ alone is written down. Removal of the pair $(3,4)$ then leaves exactly two numbers on the board, 1 and 7 , since $|3 \\pm 4|$ are both already there.\n\n\n\nIf $n=6$, remove first the pair $(1,6)$, notice that $5=|1-6|$ is already on the board, so $7=1+6$ alone is written down. Next, remove the pair $(2,5)$ and notice that $|2 \\pm 5|$ are both already on the board, so no new number is written down. Finally, removal of the pair $(3,4)$ provides a single number to be written down, $1=|3-4|$, since $7=3+4$ is already on the board. At this stage, the process comes to an end: 1 and 7 are the two numbers left.\n\n\n\nIn the remaining cases, the problem for $n$ is brought down to the corresponding problem for $[n / 2\\rceilb$ on the board, we can replace them by $a+b$ and $a-b$, and then, performing a move on the two new numbers, by $(a+b)+(a-b)=2 a$ and $(a+b)-(a-b)=2 b$. So we can double any two numbers on the board.\n\n\n\nWe now show that, if the board contains two even numbers $a$ and $b$, we can divide them both by 2 , while keeping the other numbers unchanged. If $k$ is even, split the other numbers into pairs to multiply each pair by 2 ; then clear out the common factor 2 . If $k$ is odd, split all numbers but $a$ into pairs to multiply each by 2 ; then do the same for all numbers but $b$; finally, clear out the common factor 4.\n\n\n\nBack to the problem, if two of the numbers $a_{1}, \\ldots, a_{k}$ are even, reduce them both by 2 to get a set with a smaller sum, which is impossible. Otherwise, two numbers, say, $a_{1}\n\n\n\nUsing the Claim above, we now show that the citizens may be placed at the $3 n^{2}+3 n+1$ tile vertices.\n\n\n\nConsider any tile $T_{1}$; its vertices are at least 1 apart from each other. Moreover, let $B A C$ be a part of the boundary of some tile $T$, and let $X$ be any point of the boundary of $T$, lying outside the half-open intervals $[A, B)$ and $[A, C)$ (in this case, we say that $X$ is not adjacent to $A$ ). Then $A X \\geq \\sqrt{3} / 2$.\n\n\n\nNow consider any two tile vertices $A$ and $B$. If they are vertices of the same tile we already know $A B \\geq 1$; otherwise, the segment $A B$ crosses the boundaries of some tiles containing $A$ and $B$ at some points $X$ and $Y$ not adjacent to $A$ and $B$, respectively. Hence $A B \\geq A X+Y B \\geq \\sqrt{3}>1$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 19 Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? "['Such a tester pair exists. We may biject $\\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\\alpha, \\beta$ (which we will specify further later). Take\n\n\n\n$$\n\ng(x)=\\max (x-\\alpha, 0) \\quad \\text { and } \\quad h(x)=\\min (x+\\beta, 1)\n\n$$15\n\n\n\nSay a set $S \\subseteq[0,1]$ is invariant if $f(S) \\subseteq S$ for all functions $f$ commuting with both $g$ and $h$. Note that intersections and unions of invariant sets are invariant. Preimages of invariant sets under $g$ and $h$ are also invariant; indeed, if $S$ is invariant and, say, $T=g^{-1}(S)$, then $g(f(T))=$ $f(g(T)) \\subseteq f(S) \\subseteq S$, thus $f(T) \\subseteq T$.\n\n\n\nWe claim that (if we choose $\\alpha+\\beta<1$ ) the intervals $[0, n \\alpha-m \\beta]$ are invariant where $n$ and $m$ are nonnegative integers with $0 \\leq n \\alpha-m \\beta \\leq 1$. We prove this by induction on $m+n$.\n\n\n\nThe set $\\{0\\}$ is invariant, as for any $f$ commuting with $g$ we have $g(f(0))=f(g(0))=f(0)$, so $f(0)$ is a fixed point of $g$. This gives that $f(0)=0$, thus the induction base is established.\n\n\n\nSuppose now we have some $m, n$ such that $\\left[0, n^{\\prime} \\alpha-m^{\\prime} \\beta\\right]$ is invariant whenever $m^{\\prime}+n^{\\prime}<$ $m+n$. At least one of the numbers $(n-1) \\alpha-m \\beta$ and $n \\alpha-(m-1) \\beta$ lies in $(0,1)$. Note however that in the first case $[0, n \\alpha-m \\beta]=g^{-1}([0,(n-1) \\alpha-m \\beta])$, so $[0, n \\alpha-m \\beta]$ is invariant. In the second case $[0, n \\alpha-m \\beta]=h^{-1}([0, n \\alpha-(m-1) \\beta])$, so again $[0, n \\alpha-m \\beta]$ is invariant. This completes the induction.\n\n\n\nWe claim that if we choose $\\alpha+\\beta<1$, where $0<\\alpha \\notin \\mathbb{Q}$ and $\\beta=1 / k$ for some integer $k>1$, then all intervals $[0, \\delta]$ are invariant for $0 \\leq \\delta<1$. This occurs, as by the previous claim, for all nonnegative integers $n$ we have $[0,(n \\alpha \\bmod 1)]$ is invariant. The set of $n \\alpha \\bmod 1$ is dense in $[0,1]$, so in particular\n\n\n\n$$\n\n[0, \\delta]=\\bigcap_{(n \\alpha \\bmod 1)>\\delta}[0,(n \\alpha \\bmod 1)]\n\n$$\n\n\n\nis invariant.\n\n\n\nA similar argument establishes that $[\\delta, 1]$ is invariant, so by intersecting these $\\{\\delta\\}$ is invariant for $0<\\delta<1$. Yet we also have $\\{0\\},\\{1\\}$ both invariant, which proves $f$ to be the identity.' ""Let us agree that a sequence $\\mathbf{x}=\\left(x_{n}\\right)_{n=1,2, \\ldots}$ is cofinally non-constant if for every index $m$ there exists an index $n>m$ such that $x_{m} \\neq x_{n}$.\n\n\n\nBiject $\\mathbb{R}$ with the set of cofinally non-constant sequences of 0's and 1's, and define $g$ and $h$ by\n\n\n\n$$\n\ng(\\epsilon, \\mathbf{x})=\\left\\{\\begin{array}{ll}\n\n\\epsilon, \\mathbf{x} & \\text { if } \\epsilon=0 \\\\\n\n\\mathbf{x} & \\text { else }\n\n\\end{array} \\quad \\text { and } \\quad h(\\epsilon, \\mathbf{x})= \\begin{cases}\\epsilon, \\mathbf{x} & \\text { if } \\epsilon=1 \\\\\n\n\\mathbf{x} & \\text { else }\\end{cases}\\right.\n\n$$\n\n\n\nwhere $\\epsilon, \\mathbf{x}$ denotes the sequence formed by appending $\\mathbf{x}$ to the single-element sequence $\\epsilon$. Note that $g$ fixes precisely those sequences beginning with 0 , and $h$ fixes precisely those beginning with 1.\n\n\n\nNow assume that $f$ commutes with both $f$ and $g$. To prove that $f(\\mathbf{x})=\\mathbf{x}$ for all $\\mathbf{x}$ we show that $\\mathbf{x}$ and $f(\\mathbf{x})$ share the same first $n$ terms, by induction on $n$.\n\n\n\nThe base case $n=1$ is simple, as we have noticed above that the set of sequences beginning with a 0 is precisely the set of $g$-fixed points, so is preserved by $f$, and similarly for the set of sequences starting with 1 .\n\n\n\n\n\n\n\nSuppose that $f(\\mathbf{x})$ and $\\mathbf{x}$ agree for the first $n$ terms, whatever $\\mathbf{x}$. Consider any sequence, and write it as $\\mathbf{x}=\\epsilon, \\mathbf{y}$. Without loss of generality, we may (and will) assume that $\\epsilon=0$, so $f(\\mathbf{x})=0, \\mathbf{y}^{\\prime}$ by the base case. Yet then $f(\\mathbf{y})=f(h(\\mathbf{x}))=h(f(\\mathbf{x}))=h\\left(0, \\mathbf{y}^{\\prime}\\right)=\\mathbf{y}^{\\prime}$. Consequently, $f(\\mathbf{x})=0, f(\\mathbf{y})$, so $f(\\mathbf{x})$ and $\\mathbf{x}$ agree for the first $n+1$ terms by the inductive hypothesis.\n\n\n\nThus $f$ fixes all of cofinally non-constant sequences, and the conclusion follows."" 'We will show that there exists a tester pair of bijective functions $g$ and $h$.\n\n\n\nFirst of all, let us find out when a pair of functions is a tester pair. Let $g, h: \\mathbb{R} \\rightarrow \\mathbb{R}$ be arbitrary functions. We construct a directed graph $G_{g, h}$ with $\\mathbb{R}$ as the set of vertices, its edges being painted with two colors: for every vertex $x \\in \\mathbb{R}$, we introduce a red edge $x \\rightarrow g(x)$ and a blue edge $x \\rightarrow h(x)$.\n\n\n\nNow, assume that the function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \\in \\mathbb{R}$. This means exactly that if there exists an edge $x \\rightarrow y$, then there also exists an edge $f(x) \\rightarrow f(y)$ of the same color; that is $-f$ is an endomorphism of $G_{g, h}$.\n\n\n\nThus, a pair $(g, h)$ is a tester pair if and only if the graph $G_{g, h}$ admits no nontrivial endomorphisms. Notice that each endomorphism maps a component into a component. Thus, to construct a tester pair, it suffices to construct a continuum of components with no nontrivial endomorphisms and no homomorphisms from one to another. It can be done in many ways; below we present one of them.\n\n\n\nLet $g(x)=x+1$; the construction of $h$ is more involved. For every $x \\in[0,1)$ we define the set $S_{x}=x+\\mathbb{Z}$; the sets $S_{x}$ will be exactly the components of $G_{g, h}$. Now we will construct these components.\n\n\n\nLet us fix any $x \\in[0,1)$; let $x=0 . x_{1} x_{2} \\ldots$ be the binary representation of $x$. Define $h(x-n)=x-n+1$ for every $n>3$. Next, let $h(x-3)=x, h(x)=x-2, h(x-2)=x-1$, and $h(x-1)=x+1$ (that would be a ""marker"" which fixes a point in our component).\n\n\n\nNext, for every $i=1,2, \\ldots$, we define\n\n\n\n(1) $h(x+3 i-2)=x+3 i-1, h(x+3 i-1)=x+3 i$, and $h(x+3 i)=x+3 i+1$, if $x_{i}=0$;\n\n\n\n(2) $h(x+3 i-2)=x+3 i, h(x+3 i)=3 i-1$, and $h(x+3 i-1)=x+3 i+1$, if $x_{i}=1$.\n\n\n\nClearly, $h$ is a bijection mapping each $S_{x}$ to itself. Now we claim that the graph $G_{g, h}$ satisfies the desired conditions.\n\n\n\nConsider any homomorphism $f_{x}: S_{x} \\rightarrow S_{y}$ ( $x$ and $y$ may coincide). Since $g$ is a bijection, consideration of the red edges shows that $f_{x}(x+n)=x+n+k$ for a fixed real $k$. Next, there exists a blue edge $(x-3) \\rightarrow x$, and the only blue edge of the form $(y+m-3) \\rightarrow(y+m)$ is $(y-3) \\rightarrow y$; thus $f_{x}(x)=y$, and $k=0$.\n\n\n\nNext, if $x_{i}=0$ then there exists a blue edge $(x+3 i-2) \\rightarrow(x+3 i-1)$; then the edge $(y+3 i-2) \\rightarrow(y+3 i-1)$ also should exist, so $y_{i}=0$. Analogously, if $x_{i}=1$ then there exists a blue edge $(x+3 i-2) \\rightarrow(x+3 i)$; then the edge $(y+3 i-2) \\rightarrow(y+3 i)$ also should exist, so $y_{i}=1$. We conclude that $x=y$, and $f_{x}$ is the identity mapping, as required.']" [] Text-only Competition True Theorem proof Algebra Math English 20 Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$. The lines $A B$ and $C D$ meet at $P$, the lines $A D$ and $B C$ meet at $Q$, and the diagonals $A C$ and $B D$ meet at $R$. Let $M$ be the midpoint of the segment $P Q$, and let $K$ be the common point of the segment $M R$ and the circle $\omega$. Prove that the circumcircle of the triangle $K P Q$ and $\omega$ are tangent to one another. ['Let $O$ be the centre of $\\omega$. Notice that the points $P, Q$, and $R$ are the poles (with respect to $\\omega$ ) of the lines $Q R, R P$, and $P Q$, respectively. Hence we have $O P \\perp Q R, O Q \\perp R P$, and $O R \\perp P Q$, thus $R$ is the orthocentre of the triangle $O P Q$. Now, if $M R \\perp P Q$, then the points $P$ and $Q$ are the reflections of one another in the line $M R=M O$, and the triangle $P Q K$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\n\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $M R$, and let $U$ be the common point of the lines $O V$ and $P Q$. Since $U$ lies on the polar line of $R$ and $O U \\perp M R$, we obtain that $U$ is the pole of $M R$. Therefore, the line $U K$ is tangent to $\\omega$. Hence it is enough to prove that $U K^{2}=U P \\cdot U Q$, since this relation implies that $U K$ is also tangent to the circle $K P Q$.\n\n\n\nFrom the rectangular triangle $O K U$, we get $U K^{2}=U V \\cdot U O$. Let $\\Omega$ be the circumcircle of triangle $O P Q$, and let $R^{\\prime}$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $P Q$. It is well known that $R^{\\prime}$ is the point of $\\Omega$ opposite to $O$, hence $O R^{\\prime}$ is the diameter of $\\Omega$. Finally, since $\\angle O V R^{\\prime}=90^{\\circ}$, the point $V$ also lies on $\\Omega$, hence $U P \\cdot U Q=U V \\cdot U O=U K^{2}$, as required.\n\n\n\n'] [] Text-only Competition True Theorem proof Geometry Math English 21 Does there exist an infinite sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$ ? ['The answer is in the affirmative.\n\n\n\nThe idea is to consider a sequence of pairwise distinct primes $p_{1}, p_{2}, p_{3}, \\ldots$, cover the positive integers by a sequence of finite non-empty sets $I_{n}$ such that $I_{m}$ and $I_{n}$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_{n}=\\prod_{i \\in I_{n}} p_{i}, n=1,2,3, \\ldots$\n\n\n\nOne possible way of finding such sets is the following. For all positive integers $n$, let\n\n\n\n$$\n\n\\begin{aligned}\n\n& 2 n \\in I_{k} \\quad \\text { for all } k=n, n+3, n+5, n+7, \\ldots ; \\quad \\text { and } \\\\\n\n& 2 n-1 \\in I_{k} \\quad \\text { for all } k=n, n+2, n+4, n+6, \\ldots\n\n\\end{aligned}\n\n$$\n\n\n\nClearly, each $I_{k}$ is finite, since it contains none of the numbers greater than $2 k$. Next, the number $p_{2 n}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i}$, while the number $p_{2 n-1}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i+1}$ for $i=1,2, \\ldots$. Finally, none of the indices appears in two consecutive sets.'] [] Text-only Competition True Theorem proof Algebra Math English 22 "A finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers $$ a+b, a-b, b-a, a \times b, a / b(\text { if } b \neq 0), b / a(\text { if } a \neq 0) $$ Prove that, for every integer $n>100$, there are only finitely many integers $k \geq 0$, such that, starting from the list $$ k+1, k+2, \ldots, k+n $$ it is possible to obtain, after $n-1$ operations, the value $n$ !." ['We prove the problem statement even for all positive integer $n$.\n\n\n\nThere are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \\ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an elementary operation from two such formulas on two disjoint sets of the $x_{i}$.\n\n\n\nA straightforward induction on $n$ shows that the outcome of each such construction is a number of the form\n\n\n\n$$\n\n\\frac{\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in\\{0,1\\}} a_{\\alpha_{1}, \\ldots, \\alpha_{n}} x_{1}^{\\alpha_{1}} \\cdots x_{n}^{\\alpha_{n}}}{\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in\\{0,1\\}} b_{\\alpha_{1}, \\ldots, \\alpha_{n}} x_{1}^{\\alpha_{1}} \\cdots x_{n}^{\\alpha_{n}}},\n\\tag{*}\n$$\n\n\n\nwhere the $a_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ and $b_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ are all in the set $\\{0, \\pm 1\\}$, not all zero of course, $a_{0, \\ldots, 0}=b_{1, \\ldots, 1}=0$, and also $a_{\\alpha_{1}, \\ldots, \\alpha_{n}} \\cdot b_{\\alpha_{1}, \\ldots, \\alpha_{n}}=0$ for every set of indices.\n\n\n\nSince $\\left|a_{\\alpha_{1}, \\ldots, \\alpha_{n}}\\right| \\leq 1$, and $a_{0,0, \\ldots, 0}=0$, the absolute value of the numerator does not exceed $\\left(1+\\left|x_{1}\\right|\\right) \\cdots\\left(1+\\left|x_{n}\\right|\\right)-1$; in particular, if $c$ is an integer in the range $-n, \\ldots,-1$, and $x_{k}=c+k$, $k=1, \\ldots, n$, then the absolute value of the numerator is at most $(-c) !(n+c+1) !-1 \\leq n !-10>p_{2}$. The graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ cross at distinct points $A$ and $B$. The four tangents to $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ at $A$ and $B$ form a convex quadrilateral which has an inscribed circle. Prove that the graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ have the same axis of symmetry. "['Let $\\mathcal{A}_{i}$ and $\\mathcal{B}_{i}$ be the tangents to $\\mathcal{G}_{i}$ at $A$ and $B$, respectively, and let $C_{i}=\\mathcal{A}_{i} \\cap \\mathcal{B}_{i}$. Since $f_{1}(x)$ is convex and $f_{2}(x)$ is concave, the convex quadrangle formed by the four tangents is exactly $A C_{1} B C_{2}$.\n\n\n\nLemma. If $C A$ and $C B$ are the tangents drawn from a point $C$ to the graph $\\mathcal{G}$ of a quadratic trinomial $f(x)=p x^{2}+q x+r, A, B \\in \\mathcal{G}, A \\neq B$, then the abscissa of $C$ is the arithmetic mean of the abscissae of $A$ and $B$.\n\n\n\nProof. Assume, without loss of generality, that $C$ is at the origin, so the equations of the two tangents have the form $y=k_{a} x$ and $y=k_{b} x$. Next, the abscissae $x_{A}$ and $x_{B}$ of the tangency points $A$ and $B$, respectively, are multiple roots of the polynomials $f(x)-k_{a} x$ and $f(x)-k_{b} x$, respectively. By the Vieta theorem, $x_{A}^{2}=r / p=x_{B}^{2}$, so $x_{A}=-x_{B}$, since the case $x_{A}=x_{B}$ is ruled out by $A \\neq B$.\n\n\n\n\n\nThe Lemma shows that the line $C_{1} C_{2}$ is parallel to the $y$-axis and the points $A$ and $B$ are equidistant from this line.\n\n\n\nSuppose, if possible, that the incentre $O$ of the quadrangle $A C_{1} B C_{2}$ does not lie on the line $C_{1} C_{2}$. Assume, without loss of generality, that $O$ lies inside the triangle $A C_{1} C_{2}$ and let $A^{\\prime}$ be the reflection of $A$ in the line $C_{1} C_{2}$. Then the ray $C_{i} B$ emanating from $C_{i}$ lies inside the angle $A C_{i} A^{\\prime}$, so $B$ lies inside the quadrangle $A C_{1} A^{\\prime} C_{2}$, whence $A$ and $B$ are not equidistant from $C_{1} C_{2}$ - a contradiction.\n\n\n\nThus $O$ lies on $C_{1} C_{2}$, so the lines $A C_{i}$ and $B C_{i}$ are reflections of one another in the line $C_{1} C_{2}$, and $B=A^{\\prime}$. Hence $y_{A}=y_{B}$, and since $f_{i}(x)=y_{A}+p_{i}\\left(x-x_{A}\\right)\\left(x-x_{B}\\right)$, the line $C_{1} C_{2}$ is the axis of symmetry of both parabolas, as required.' 'Use the standard equation of a tangent to a smooth curve in the plane, to deduce that the tangents at two distinct points $A$ and $B$ on the parabola of equation $y=p x^{2}+q x+r$,\n\n\n\n\n\n\n\n$p \\neq 0$, meet at some point $C$ whose coordinates are\n\n\n\n$$\n\nx_{C}=\\frac{1}{2}\\left(x_{A}+x_{B}\\right) \\quad \\text { and } \\quad y_{C}=p x_{A} x_{B}+q \\cdot \\frac{1}{2}\\left(x_{A}+x_{B}\\right)+r\n\n$$\n\n\n\nUsage of the standard formula for Euclidean distance yields\n\n\n\n$$\n\nC A=\\frac{1}{2}\\left|x_{B}-x_{A}\\right| \\sqrt{1+\\left(2 p x_{A}+q\\right)^{2}} \\quad \\text { and } \\quad C B=\\frac{1}{2}\\left|x_{B}-x_{A}\\right| \\sqrt{1+\\left(2 p x_{B}+q\\right)^{2}},\n\n$$\n\n\n\nso, after obvious manipulations,\n\n\n\n$$\n\nC B-C A=\\frac{2 p\\left(x_{B}-x_{A}\\right)\\left|x_{B}-x_{A}\\right|\\left(p\\left(x_{A}+x_{B}\\right)+q\\right)}{\\sqrt{1+\\left(2 p x_{A}+q\\right)^{2}}+\\sqrt{1+\\left(2 p x_{B}+q\\right)^{2}}}\n\n$$\n\n\n\nNow, write the condition in the statement in the form $C_{1} B-C_{1} A=C_{2} B-C_{2} A$, apply the above formula and clear common factors to get\n\n\n\n$$\n\n\\frac{p_{1}\\left(p_{1}\\left(x_{A}+x_{B}\\right)+q_{1}\\right)}{\\sqrt{1+\\left(2 p_{1} x_{A}+q_{1}\\right)^{2}}+\\sqrt{1+\\left(2 p_{1} x_{B}+q_{1}\\right)^{2}}}=\\frac{p_{2}\\left(p_{2}\\left(x_{A}+x_{B}\\right)+q_{2}\\right)}{\\sqrt{1+\\left(2 p_{2} x_{A}+q_{2}\\right)^{2}}+\\sqrt{1+\\left(2 p_{2} x_{B}+q_{2}\\right)^{2}}}\n\n$$\n\n\n\nNext, use the fact that $x_{A}$ and $x_{B}$ are the solutions of the quadratic equation $\\left(p_{1}-p_{2}\\right) x^{2}+$ $\\left(q_{1}-q_{2}\\right) x+r_{1}-r_{2}=0$, so $x_{A}+x_{B}=-\\left(q_{1}-q_{2}\\right) /\\left(p_{1}-p_{2}\\right)$, to obtain\n\n\n\n$$\n\n\\frac{p_{1}\\left(p_{1} q_{2}-p_{2} q_{1}\\right)}{\\sqrt{1+\\left(2 p_{1} x_{A}+q_{1}\\right)^{2}}+\\sqrt{1+\\left(2 p_{1} x_{B}+q_{1}\\right)^{2}}}=\\frac{p_{2}\\left(p_{1} q_{2}-p_{2} q_{1}\\right)}{\\sqrt{1+\\left(2 p_{2} x_{A}+q_{2}\\right)^{2}}+\\sqrt{1+\\left(2 p_{2} x_{B}+q_{2}\\right)^{2}}}\n\n$$\n\n\n\nFinally, since $p_{1} p_{2}<0$ and the denominators above are both positive, the last equality forces $p_{1} q_{2}-p_{2} q_{1}=0$; that is, $q_{1} / p_{1}=q_{2} / p_{2}$, so the two parabolas have the same axis.']" [] Text-only Competition True Theorem proof Geometry Math English 24 Let $A B C D$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $A B, B C, C D$, and $D A$, respectively. It is given that the segments $P R$ and $Q S$ dissect $A B C D$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic. "[""We start with a lemma which holds even in a more general setup.\n\n\n\nLemma 1. Let $P Q R S$ be a convex quadrangle whose diagonals meet at $O$. Let $\\omega_{1}$ and $\\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\\ell$ be their radical axis. Finally, choose the points $A, B$, and $C$ outside this quadrangle so that: the point $P$ (respectively, $Q$ ) lies on the segment $A B$ (respectively, $B C$ ); and $A O \\perp P S, B O \\perp P Q$, and $C O \\perp Q R$. Then the three lines $A C, P Q$, and $\\ell$ are concurrent or parallel.\n\n\n\nProof. Assume first that the lines $P R$ and $Q S$ are not perpendicular. Let $H_{1}$ and $H_{2}$ be the orthocentres of the triangles $O S P$ and $O Q R$, respectively; notice that $H_{1}$ and $H_{2}$ do not coincide.\n\n\n\nSince $H_{1}$ is the radical centre of the circles on diameters $R S, S P$, and $P Q$, it lies on $\\ell$. Similarly, $H_{2}$ lies on $\\ell$, so the lines $H_{1} H_{2}$ and $\\ell$ coincide.\n\n\n\nThe corresponding sides of the triangles $A P H_{1}$ and $C Q H_{2}$ meet at $O, B$, and the orthocentre of the triangle $O P Q$ (which lies on $O B$ ). By Desargues' theorem, the lines $A C, P Q$ and $\\ell$ are concurrent or parallel.\n\n\n\n\n\n\n\nThe case when $P R \\perp Q S$ may be considered as a limit case, since the configuration in the statement of the lemma allows arbitrarily small perturbations. The lemma is proved.\n\n\n\nBack to the problem, let the segments $P R$ and $Q S$ cross at $O$, let $\\omega_{1}$ and $\\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\\ell$ be their radical axis. By the Lemma, the three lines $A C, \\ell$, and $P Q$ are concurrent or parallel, and similarly so are the three lines $A C, \\ell$, and $R S$. Thus, if the lines $A C$ and $\\ell$ are distinct, all four lines are concurrent or pairwise parallel.\n\n\n\nThis is clearly the case when the lines $P S$ and $Q R$ are not parallel (since $\\ell$ crosses $O A$ and $O C$ at the orthocentres of $O S P$ and $O Q R$, these orthocentres being distinct from $A$ and $C$ ). In this case, denote the concurrency point by $T$. If $T$ is not ideal, then we have $T P \\cdot T Q=T R \\cdot T S$ (as $T \\in \\ell$ ), so $P Q R S$ is cyclic. If $T$ is ideal (i.e., all four lines are parallel), then the segments $P Q$ and $R S$ have the same perpendicular bisector (namely, the line of centers of $\\omega_{1}$ and $\\omega_{2}$ ), and $P Q R S$ is cyclic again.\n\n\n\nAssume now $P S$ and $Q R$ parallel. By symmetry, $P Q$ and $R S$ may also be assumed parallel: otherwise, the preceding argument goes through after relabelling. In this case, we need to prove that the parallelogram $P Q R S$ is a rectangle.\n\n\n\n\n\n\n\n\n\n\n\nSuppose, by way of contradiction, that $O P>O Q$. Let the line through $O$ and parallel to $P Q$ meet $A B$ at $M$, and $C B$ at $N$. Since $O P>O Q$, the angle $S P Q$ is acute and the angle $P Q R$ is obtuse, so the angle $A O B$ is obtuse, the angle $B O C$ is acute, $M$ lies on the segment $A B$, and $N$ lies on the extension of the segment $B C$ beyond $C$. Therefore: $O A>O M$, since the angle $O M A$ is obtuse; $O M>O N$, since $O M: O N=K P: K Q$, where $K$ is the projection of $O$ onto $P Q$; and $O N>O C$, since the angle $O C N$ is obtuse. Consequently, $O A>O C$.\n\n\n\nSimilarly, $O R>O S$ yields $O C>O A$ : a contradiction. Consequently, $O P=O Q$ and $P Q R S$ is a rectangle. This ends the proof."" 'To begin, we establish a useful lemma.\n\n\n\nLemma 2. If $P$ is a point on the side $A B$ of a triangle $O A B$, then\n\n\n\n$$\n\n\\frac{\\sin A O P}{O B}+\\frac{\\sin P O B}{O A}=\\frac{\\sin A O B}{O P}\n\n$$\n\n\n\nProof. Let $[X Y Z]$ denote the area of a triangle $X Y Z$, to write\n\n\n\n$0=2([A O B]-[P O B]-[P O C])=O A \\cdot O B \\cdot \\sin A O B-O B \\cdot O P \\cdot \\sin P O B-O P \\cdot O A \\cdot \\sin A O P$, and divide by $O A \\cdot O B \\cdot O P$ to get the required identity.\n\n\n\nA similar statement remains valid if the point $C$ lies on the line $A B$; the proof is obtained by using signed areas and directed lengths.\n\n\n\nWe now turn to the solution. We first prove some sort of a converse statement, namely:\n\n\n\nClaim. Let $P Q R S$ be a cyclic quadrangle with $O=P R \\cap Q S$; assume that no its diagonal is perpendicular to a side. Let $\\ell_{A}, \\ell_{B}, \\ell_{C}$, and $\\ell_{D}$ be the lines through $O$ perpendicular to $S P$, $P Q, Q R$, and $R S$, respectively. Choose any point $A \\in \\ell_{A}$ and successively define $B=A P \\cap \\ell_{B}$, $C=B Q \\cap \\ell_{C}, D=C R \\cap \\ell_{D}$, and $A^{\\prime}=D S \\cap \\ell_{A}$. Then $A^{\\prime}=A$.\n\n\n\nProof. We restrict ourselves to the case when the points $A, B, C, D$, and $A^{\\prime}$ lie on $\\ell_{A}, \\ell_{B}$, $\\ell_{C}, \\ell_{D}$, and $\\ell_{A}$ on the same side of $O$ as their points of intersection with the respective sides of the quadrilateral $P Q R S$. Again, a general case is obtained by suitable consideration of directed lengths.\n\n\n\n\n\n\n\n\n\n\n\nDenote\n\n\n\n$$\n\n\\begin{gathered}\n\n\\alpha=\\angle Q P R=\\angle Q S R=\\pi / 2-\\angle P O B=\\pi / 2-\\angle D O S, \\\\\n\n\\beta=\\angle R P S=\\angle R Q S=\\pi / 2-\\angle A O P=\\pi / 2-\\angle Q O C, \\\\\n\n\\gamma=\\angle S Q P=\\angle S R P=\\pi / 2-\\angle B O Q=\\pi / 2-\\angle R O D, \\\\\n\n\\delta=\\angle P R Q=\\angle P S Q=\\pi / 2-\\angle C O R=\\pi / 2-\\angle S O A .\n\n\\end{gathered}\n\n$$\n\n\n\nBy Lemma 2 applied to the lines $A P B, P Q C, C R D$, and $D S A^{\\prime}$, we get\n\n\n\n$$\n\n\\begin{aligned}\n\n& \\frac{\\sin (\\alpha+\\beta)}{O P}=\\frac{\\cos \\alpha}{O A}+\\frac{\\cos \\beta}{O B}, \\quad \\frac{\\sin (\\beta+\\gamma)}{O Q}=\\frac{\\cos \\beta}{O B}+\\frac{\\cos \\gamma}{O C} \\\\\n\n& \\frac{\\sin (\\gamma+\\delta)}{O R}=\\frac{\\cos \\gamma}{O C}+\\frac{\\cos \\delta}{O D}, \\quad \\frac{\\sin (\\delta+\\alpha)}{O S}=\\frac{\\cos \\delta}{O D}+\\frac{\\cos \\alpha}{O A^{\\prime}}\n\n\\end{aligned}\n\n$$\n\n\n\nAdding the two equalities on the left and subtracting the two on the right, we see that the required equality $A=A^{\\prime}$ (i.e., $\\cos \\alpha / O A=\\cos \\alpha / O A^{\\prime}$, in view of $\\cos \\alpha \\neq 0$ ) is equivalent to the relation\n\n\n\n$$\n\n\\frac{\\sin Q P S}{O P}+\\frac{\\sin S R Q}{O R}=\\frac{\\sin P Q R}{O Q}+\\frac{\\sin R S P}{O S}\n\n$$\n\n\n\nLet $d$ denote the circumdiameter of $P Q R S$, so $\\sin Q P S=\\sin S R Q=Q S / d$ and $\\sin R S P=$ $\\sin P Q R=P R / d$. Thus the required relation reads\n\n\n\n$$\n\n\\frac{Q S}{O P}+\\frac{Q S}{O R}=\\frac{P R}{O S}+\\frac{P R}{O Q}, \\quad \\text { or } \\quad \\frac{Q S \\cdot P R}{O P \\cdot O R}=\\frac{P R \\cdot Q S}{O S \\cdot O Q}\n\n$$\n\n\n\nThe last relation is trivial, due again to cyclicity.\n\n\n\nFinally, it remains to derive the problem statement from our Claim. Assume that $P Q R S$ is not cyclic, e.g., that $O P \\cdot O R>O Q \\cdot O S$, where $O=P R \\cap Q S$. Mark the point $S^{\\prime}$ on the ray $O S$ so that $O P \\cdot O R=O Q \\cdot O S^{\\prime}$. Notice that no diagonal of $P Q R S$ is perpendicular to a side, so the quadrangle $P Q R S^{\\prime}$ satisfies the conditions of the claim.\n\n\n\nLet $\\ell_{A}^{\\prime}$ and $\\ell_{D}^{\\prime}$ be the lines through $O$ perpendicular to $P S^{\\prime}$ and $R S^{\\prime}$, respectively. Then $\\ell_{A}^{\\prime}$ and $\\ell_{D}^{\\prime}$ cross the segments $A P$ and $R D$, respectively, at some points $A^{\\prime}$ and $D^{\\prime}$. By the Claim, the line $A^{\\prime} D^{\\prime}$ passes through $S^{\\prime}$. This is impossible, because the segment $A^{\\prime} D^{\\prime}$ crosses the segment $O S$ at some interior point, while $S^{\\prime}$ lies on the extension of this segment. This contradiction completes the proof.']" [] Text-only Competition True Theorem proof Geometry Math English 25 Let $a, b, c, d$ be positive integers such that $a d \neq b c$ and $\operatorname{gcd}(a, b, c, d)=1$. Prove that, as $n$ runs through the positive integers, the values $\operatorname{gcd}(a n+b, c n+d)$ may achieve form the set of all positive divisors of some integer. "['We extend the problem statement by allowing $a$ and $c$ take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \\leq a \\leq c$. Let $S(a, b, c, d)=\\left\\{\\operatorname{gcd}(a n+b, c n+d): n \\in \\mathbb{Z}_{>0}\\right\\}$.\n\nNow we induct on $a$. We first deal with the inductive step, leaving the base case $a=0$ to the end of the solution. So, assume that $a>0$; we intend to find a 4 -tuple $\\left(a^{\\prime}, b^{\\prime}, c^{\\prime}, d^{\\prime}\\right)$ satisfying the requirements of the extended problem, such that $S\\left(a^{\\prime}, b^{\\prime}, c^{\\prime}, d^{\\prime}\\right)=S(a, b, c, d)$ and $0 \\leq a^{\\prime}1)$. We thus $\\operatorname{get} \\operatorname{gcd}\\left(b^{\\prime}, c\\right)=1$ and $S\\left(0, b^{\\prime}, c, d\\right)=S(0, b, c, d)$.\n\nFinally, it is easily seen that $S\\left(0, b^{\\prime}, c, d\\right)$ is the set of all positive divisors of $b^{\\prime}$. Each member of $S\\left(0, b^{\\prime}, c, d\\right)$ is clearly a divisor of $b^{\\prime}$. Conversely, if $\\delta$ is a positive divisor of $b^{\\prime}$, then $c n+d \\equiv \\delta$ $\\left(\\bmod b^{\\prime}\\right)$ for some $n$, since $b^{\\prime}$ and $c$ are coprime, so $\\delta$ is indeed a member of $S\\left(0, b^{\\prime}, c, d\\right)$.' 'For positive integers $s$ and $t$ and prime $p$, we will denote by $\\operatorname{gcd}_{p}(s, t)$ the greatest common $p$-power divisor of $s$ and $t$.\n\nClaim 1. For any positive integer $n, \\operatorname{gcd}(a n+b, c n+d) \\mid a d-b c$.\n\nProof. This is clear from the identity\n\n$$\na(c n+d)-c(a n+b)=a d-b c . \\tag{\\dagger}\n$$\n\nClaim 2. The set of values taken by $\\operatorname{gcd}(a n+b, c n+d)$ is exactly the set of values taken by the product\n\n$$\n\\prod_{p \\mid a d-b c} \\operatorname{gcd}_{p}\\left(a n_{p}+b, c n_{p}+d\\right)\n$$\n\nas the $\\left(n_{p}\\right)_{p \\mid a d-b c}$ each range over positive integers.\n\n\n\nProof. From the identity\n\n$$\n\\operatorname{gcd}(a n+b, c n+d)=\\prod_{p \\mid a d-b c} \\operatorname{gcd}_{p}(a n+b, c n+d),\n$$\n\nit is clear that every value taken by $\\operatorname{gcd}(a n+b, c n+d)$ is also a value taken by the product (with all $\\left.n_{p}=n\\right)$. Conversely, it suffices to show that, given any positive integers $\\left(n_{p}\\right)_{p \\mid a d-b c}$, there is a positive integer $n$ such that $\\operatorname{gcd}_{p}(a n+b, c n+d)=\\operatorname{gcd}_{p}\\left(a n_{p}+b, c n_{p}+d\\right)$ for each $p \\mid a d-b c$. This can be achieved by requiring that $n$ be congruent to $n_{p}$ modulo a sufficiently large ${ }^{1}$ power of $p$ (using the Chinese Remainder Theorem).\n\nUsing Claim 2, it suffices to determine the sets of values taken by $\\operatorname{gcd}_{p}(a n+b, c n+d)$ as $n$ ranges over all positive integers. There are two cases.\n\nClaim 3. If $p \\mid a, c$, then $\\operatorname{gcd}_{p}(a n+b, c n+d)=1$ for all $n$.\n\nProof. If $p \\mid a n+b, c n+d$, then we would have $p \\mid a, b, c, d$, which is not the case.\n\nClaim 4. If $p \\nmid a$ or $p \\nmid c$, then the values taken by $\\operatorname{gcd}_{p}(a n+b, c n+d)$ are exactly the $p$-power divisors of $a d-b c$.\n\nProof. Assume without loss of generality that $p \\nmid a$. Then from identity $(\\dagger)$ we have $\\operatorname{gcd}_{p}(a n+$ $b, c n+d)=\\operatorname{gcd}_{p}(a n+b, a d-b c)$. But since $p \\nmid a$, the arithmetic progression $a n+b$ takes all possible values modulo the highest $p$-power divisor of $a d-b c$, and in particular the values taken by $\\operatorname{gcd}_{p}(a n+b, a d-b c)$ are exactly the $p$-power divisors of $a d-b c$.\n\nConclusion. Using claims 2,3 and 4 , we see that the set of values taken by $\\operatorname{gcd}(a n+b, c n+d)$ is equal to the set of products of $p$-power divisors of $a d-b c$, where we only consider those primes $p$ not dividing $\\operatorname{gcd}(a, c)$. Thus the set of values of $\\operatorname{gcd}(a n+b, c n+d)$ is equal to the set of divisors of the largest factor of $a d-b c$ coprime to $\\operatorname{gcd}(a, c)$.']" [] Text-only Competition True Theorem proof Algebra Math English 26 Fix a circle $\Gamma$, a line $\ell$ tangent to $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ cross $\ell$ at $Y$ and $Z$. Prove that, as $X$ traces $\Omega$, the circle $X Y Z$ is tangent to two fixed circles. ['Assume $\\Gamma$ of unit radius and invert with respect to $\\Gamma$. No reference will be made to the original configuration, so images will be denoted by the same letters. Letting $\\Gamma$ be centred at $G$, notice that inversion in $\\Gamma$ maps tangents to $\\Gamma$ to circles of unit diameter through $G$ (hence internally tangent to $\\Gamma$ ). Under inversion, the statement reads as follows:\n\nFix a circle $\\Gamma$ of unit radius centred at $G$, a circle $\\ell$ of unit diameter through $G$, and a circle $\\Omega$ inside $\\ell$ disjoint from $\\ell$. The circles $\\eta$ and $\\zeta$ of unit diameter, through $G$ and a variable point $X$ on $\\Omega$, cross $\\ell$ again at $Y$ and $Z$, respectively. Prove that, as $X$ traces $\\Omega$, the circle $X Y Z$ is tangent to two fixed circles.\n\n\n\nSince $\\eta$ and $\\zeta$ are the reflections of the circumcircle $\\ell$ of the triangle $G Y Z$ in its sidelines $G Y$ and $G Z$, respectively, they pass through the orthocentre of this triangle. And since $\\eta$ and $\\zeta$ cross again at $X$, the latter is the orthocentre of the triangle $G Y Z$. Hence the circle $\\xi$ through $X, Y, Z$ is the reflection of $\\ell$ in the line $Y Z$; in particular, $\\xi$ is also of unit diameter.\n\nLet $O$ and $L$ be the centres of $\\Omega$ and $\\ell$, respectively, and let $R$ be the (variable) centre of $\\xi$. Let $G X \\operatorname{cross} \\xi$ again at $X^{\\prime}$; then $G$ and $X^{\\prime}$ are reflections of one another in the line $Y Z$, so $G L R X^{\\prime}$ is an isosceles trapezoid. Then $L R \\| G X$ and $\\angle(L G, G X)=\\angle\\left(G X^{\\prime}, X^{\\prime} R\\right)=\\angle(R X, X G)$, i.e., $L G \\| R X$; this means that $G L R X$ is a parallelogram, so $\\overrightarrow{X R}=\\overrightarrow{G L}$ is constant.\n\nFinally, consider the fixed point $N$ defined by $\\overrightarrow{O N}=\\overrightarrow{G L}$. Then $X R N O$ is a parallelogram, so the distance $R N=O X$ is constant. Consequently, $\\xi$ is tangent to the fixed circles centred at $N$ of radii $|1 / 2-O X|$ and $1 / 2+O X$.\n\nOne last check is needed to show that the inverse images of the two obtained circles are indeed circles and not lines. This might happen if one of them contained $G$; we show that this is impossible. Indeed, since $\\Omega$ lies inside $\\ell$, we have $O L<1 / 2-O X$, so\n\n$$\nN G=|\\overrightarrow{G L}+\\overrightarrow{L O}+\\overrightarrow{O N}|=|2 \\overrightarrow{G L}+\\overrightarrow{L O}| \\geq 2|\\overrightarrow{G L}|-|\\overrightarrow{L O}|>1-(1 / 2-O X)=1 / 2+O X\n$$\n\nthis shows that $G$ is necessarily outside the obtained circles.'] [] Text-only Competition True Theorem proof Geometry Math English 27 Let $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$. The circle $A B D$ meets the segment $A C$ again at an interior point $E$. The circle $A C D$ meets the segment $A B$ again at an interior point $F$. Let $A^{\prime}$ be the reflection of $A$ in the line $B C$. The lines $A^{\prime} C$ and $D E$ meet at $P$, and the lines $A^{\prime} B$ and $D F$ meet at $Q$. Prove that the lines $A D, B P$ and $C Q$ are concurrent (or all parallel). "['Let $\\sigma$ denote reflection in the line $B C$. Since $\\angle B D F=\\angle B A C=$ $\\angle C D E$, by concyclicity, the lines $D E$ and $D F$ are images of one another under $\\sigma$, so the lines $A C$ and $D F$ meet at $P^{\\prime}=\\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\\prime}=\\sigma(Q)$. Consequently, the lines $P Q$ and $P^{\\prime} Q^{\\prime}=\\sigma(P Q)$ meet at some (possibly ideal) point $R$ on the line $B C$.\n\n\n\nSince the pairs of lines $(C A, Q D),(A B, D P),(B C, P Q)$ meet at three collinear points, namely $P^{\\prime}, Q^{\\prime}, R$ respectively, the triangles $A B C$ and $D P Q$ are perspective, i.e., the lines $A D, B P, C Q$ are concurrent, by the Desargues theorem.\n\n\n\n\n\n\n\nFig. 1\n\n\n\n\n\n\n\nFig. 2' 'As in the first solution, $\\sigma$ denotes reflection in the line $B C$, the lines $D E$ and $D F$ are images of one another under $\\sigma$, the lines $A C$ and $D F$ meet at $P^{\\prime}=\\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\\prime}=\\sigma(Q)$.\n\n\n\nLet the line $A D$ meet the circle $A B C$ again at $M$. Letting $M^{\\prime}=\\sigma(M)$, it is sufficient to prove that the lines $D M^{\\prime}=\\sigma(A D), B P^{\\prime}=\\sigma(B P)$ and $C Q^{\\prime}=\\sigma(C Q)$ are concurrent.\n\n\n\n\n\n\n\nBegin by noticing that $\\angle\\left(B M^{\\prime}, M^{\\prime} D\\right)=-\\angle(B M, M A)=-\\angle(B C, C A)=\\angle(B F, F D)$, to infer that $M^{\\prime}$ lies on the circle $B D F$. Similarly, $M^{\\prime}$ lies on the circle $C D E$, so the line $D M^{\\prime}$ is the radical axis of the circles $B D F$ and $C D E$.\n\n\n\nSince $P^{\\prime}$ lies on the lines $A C$ and $D F$, it is the radical centre of the circles $A B C, A D C$, and $B D F$; hence the line $B P^{\\prime}$ is the radical axis of the circles $B D F$ and $A B C$. Similarly, the line $C Q^{\\prime}$ is the radical axis of the circles $C D E$ and $A B C$. So the conclusion follows: the lines $D M^{\\prime}$, $B P^{\\prime}$ and $C Q^{\\prime}$ are concurrent at the radical centre of the circles $A B C, B D F$ and $C D E$; thus the lines $D M, B P^{\\prime}$ and $C Q^{\\prime}$ are also concurrent.' 'As in the previous solutions, $\\sigma$ denotes reflection in the line $B C$. Let the lines $B E$ and $C F$ meet at $X$. Due to the circles $B D E A$ and $C D F A$, we have $\\angle X B D=$ $\\angle E A D=\\angle X F D$, so the quadrilateral $B F X D$ is cyclic; similarly, the quadrilateral $C E X D$ is cyclic. Hence $\\angle X D B=\\angle C F A=\\angle C D A$, the lines $D X$ and $D A$ are therefore images of one another under $\\sigma$, and $X^{\\prime}=\\sigma(X)$ lies on the line $A D$. Let $E^{\\prime}=\\sigma(E)$ and $F^{\\prime}=\\sigma(F)$, and apply the Pappus theorem to the hexagon $B P F^{\\prime} C Q E^{\\prime}$ to infer that $X^{\\prime}, D$, and $B P \\cap C Q$ are collinear. The conclusion follows.\n\n\n\n\n\n\n\nFig. 3']" [] Text-only Competition True Theorem proof Geometry Math English 28 Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy replaces the number $n$ on the blackboard with a number of the form $n^{k}$, where $k$ is a positive integer. Bob wins if the number on the board becomes zero. Can Amy prevent Bob's win? "[""The answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree that $S(0)=0$.\n\n\n\nNow we show that (i) on any move of hers, Amy does not increase the square-free part of the positive integer on the board; and (ii) on any move of his, Bob always can replace a positive integer $n$ with a non-negative integer $k$ with $S(k)' 'Let $Q$ be the isogonal conjugate of $P$ with respect to $\\triangle A E D$, so $\\angle(Q A, A D)=$ $\\angle(E A, A P)=\\angle(E B, B A)$ and $\\angle(Q D, D A)=\\angle(E D, D P)=\\angle(E C, C D)$. Now our aim is to prove that $Q E \\| C D$; this will yield that $\\angle(E C, C D)=\\angle(A E, E Q)=\\angle(P E, E D)$, whence $P E$ is tangent to $\\Omega$.\n\n\n\nLet $D Q$ meet $A B$ at $X$. Then we have $\\angle(X D, D A)=\\angle(E C, C D)=\\angle(E A, A B)$ and $\\angle(D A, A X)=\\angle(A B, B C)$, hence the triangles $D A X$ and $A B C$ are similar. Since $\\angle(A B, B E)=$ $\\angle(D A, A Q)$, the points $Q$ and $E$ correspond to each other in these triangles, hence $Q$ is the midpoint of $D X$. This yields that the points $Q$ and $E$ lie on the midline of the trapezoid parallel to $C D$, as desired.\n\n\n\n\n\n\n\n' 'Let $O$ be the intersection of the diagonals $A C$ and $B D$. Let $F$ be the midpoint of $B D$. Let $S$ be the second intersection point of the circumcircles of triangles $A O F$ and $D O E$. We will prove that $S D$ and $S E$ are tangent to $\\Omega$; the symmetric argument would then imply also that $S A$ and $S F$ are tangent to $\\Gamma$. Thus $S=P$ and the claimed tangency holds.\n\n\n\nWe first prove that $O S$ is parallel to $A B$ and $D C$. Compute the powers of the points $A, B$ with respect to the circumcircles of $A O F$ and $D O E$ :\n\n\n\n$$\n\n\\begin{gathered}\n\nd(A, A O F)=0, \\quad d(A, D O E)=A O \\cdot A E \\\\\n\nd(B, A O F)=B O \\cdot B F, \\quad d(B, D O E)=B O \\cdot B D=2 B O \\cdot B F\n\n\\end{gathered}\n\n$$\n\n\n\nAnd therefore\n\n\n\n$$\n\nd(B, D O E)-d(B, A O F)=B O \\cdot B F=A O \\cdot A E=d(A, D O E)-d(A, A O F)\n\n$$\n\n\n\nThus both $A$ and $B$ belong to a locus of the form\n\n\n\n$$\n\nd(X, D O E)-d(X, A O F)=\\mathrm{const}\n\n$$\n\n\n\nwhich is always a lines parallel to the radical axis of the respective circles. Since this radical axis is $O S$ by definition, it follows that $A B$ is parallel to $O S$, as claimed.\n\n\n\nNow by angle chasing in the cyclic quadrilateral $D S O E$, we find\n\n\n\n$$\n\n\\begin{aligned}\n\n\\angle(S D, D E) & =\\angle(S O, O E)=\\angle(D C, C E), \\\\\n\n\\angle(S E, E D)=\\angle(S O, O D) & =\\angle(D C, D B)=\\angle(A C, C D)=\\angle(E C, C D),\n\n\\end{aligned}\n\n$$\n\n\n\nand these angle equalities are exactly the conditions of $S D, S E$ being tanget to $\\Omega$, as claimed.']" [] Text-only Competition True Theorem proof Geometry Math English 30 "Given any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\varepsilon) v$ edges has two distinct simple cycles of equal lengths. (Recall that the notion of a simple cycle does not allow repetition of vertices in a cycle.)" "['Fix a positive real number $\\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.\n\n\n\nAssuming $\\varepsilon^{2} v \\geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of simple cycles in $G$, showing thereby that $v$ cannot be arbitrarily large, whence the conclusion.\n\n\n\nSince a simple cycle in $G$ has at most $v$ vertices, and each length class contains at most one such, $G$ has at most $v$ pairwise distinct simple cycles. This establishes the desired upper bound.\n\n\n\nFor the lower bound, consider a spanning tree for each component of $G$, and collect them all together to form a spanning forest $F$. Let $A$ be the set of edges of $F$, and let $B$ be the set of all other edges of $G$. Clearly, $|A| \\leq v-1$, so $|B| \\geq(1+\\varepsilon) v-|A| \\geq(1+\\varepsilon) v-(v-1)=\\varepsilon v+1>\\varepsilon v$.\n\n\n\nFor each edge $b$ in $B$, adjoining $b$ to $F$ produces a unique simple cycle $C_{b}$ through $b$. Let $S_{b}$ be the set of edges in $A$ along $C_{b}$. Since the $C_{b}$ have pairwise distinct lengths, $\\sum_{b \\in B}\\left|S_{b}\\right| \\geq$ $2+\\cdots+(|B|+1)=|B|(|B|+3) / 2>|B|^{2} / 2>\\varepsilon^{2} v^{2} / 2$.\n\n\n\nConsequently, some edge in $A$ lies in more than $\\varepsilon^{2} v^{2} /(2 v)=\\varepsilon^{2} v / 2$ of the $S_{b}$. Fix such an edge $a$ in $A$, and let $B^{\\prime}$ be the set of all edges $b$ in $B$ whose corresponding $S_{b}$ contain $a$, so $\\left|B^{\\prime}\\right|>\\varepsilon^{2} v / 2$.\n\n\n\nFor each 2-edge subset $\\left\\{b_{1}, b_{2}\\right\\}$ of $B^{\\prime}$, the union $C_{b_{1}} \\cup C_{b_{2}}$ of the cycles $C_{b_{1}}$ and $C_{b_{2}}$ forms a $\\theta$-graph, since their common part is a path in $F$ through $a$; and since neither of the $b_{i}$ lies along this path, $C_{b_{1}} \\cup C_{b_{2}}$ contains a third simple cycle $C_{b_{1}, b_{2}}$ through both $b_{1}$ and $b_{2}$. Finally, since $B^{\\prime} \\cap C_{b_{1}, b_{2}}=\\left\\{b_{1}, b_{2}\\right\\}$, the assignment $\\left\\{b_{1}, b_{2}\\right\\} \\mapsto C_{b_{1}, b_{2}}$ is injective, so the total number of simple cycles in $G$ is at least $\\left(\\begin{array}{c}\\left|B^{\\prime}\\right| \\\\ 2\\end{array}\\right)>\\left(\\begin{array}{c}\\varepsilon^{2} v / 2 \\\\ 2\\end{array}\\right)$. This establishes the desired lower bound and concludes the proof.' 'Recall that the girth of a graph $G$ is the minimal length of a (simple) cycle in this graph.\n\n\n\nLemma. For any fixed positive $\\delta$, a graph on $v$ vertices whose girth is at least $\\delta v$ has at most $v+o(v)$ edges.\n\n\n\nProof. Define $f(v)$ to be the maximal number $f$ such that a graph on $v$ vertices whose girth is at least $\\delta v$ may have $v+f$ edges. We are interested in the recursive estimates for $f$.\n\n\n\nLet $G$ be a graph on $v$ vertices whose gifth is at least $\\delta v$ containing $v+f(v)$ edges. If $G$ contains a leaf (i.e., a vertex of degree 1), then one may remove this vertex along with its edge, obtaining a graph with at most $v-1+f(v-1)$ edges. Thus, in this case $f(v) \\leq f(v-1)$.\n\n\n\nDefine an isolated path of length $k$ to be a sequence of vertices $v_{0}, v_{1}, \\ldots, v_{k}$, such that $v_{i}$ is connected to $v_{i+1}$, and each of the vertices $v_{1}, \\ldots, v_{k-1}$ has degree 2 (so, these vertices are connected only to their neighbors in the path). If $G$ contains an isolated path $v_{0}, \\ldots, v_{k}$ of length, say, $k>\\sqrt{v}$, then one may remove all its middle vertices $v_{1}, \\ldots, v_{k-1}$, along with all their $k$ edges. We obtain a graph on $v-k+1$ vertices with at most $(v-k+1)+f(v-k+1)$ edges. Thus, in this case $f(v) \\leq f(v-k+1)+1$.\n\n\n\nAssume now that the lengths of all isolated paths do not exceed $\\sqrt{v}$; we show that in this case $v$ is bounded from above. For that purpose, we replace each maximal isolated path by an edge between its endpoints, removing all middle vertices. We obtain a graph $H$ whose girth is at least $\\delta v / \\sqrt{v}=\\delta \\sqrt{v}$. Each vertex of $H$ has degree at least 3. By the girth condition, the neighborhood of any vertex $x$ of radius $r=\\lfloor(\\delta \\sqrt{v}-1) / 2\\rfloor$ is a tree rooted at $x$. Any vertex at level $i\\sqrt{v}$. This easily yields $f(v)=o(v)$, as desired.\n\n\n\nNow we proceed to the problem. Consider a graph on $v$ vertices containing no two simple cycles of the same length. Take its $\\lfloor\\varepsilon v / 2\\rfloor$ shortest cycles (or all its cycles, if their total number is smaller) and remove an edge from each. We get a graph of girth at least $\\varepsilon v / 2$. By the lemma, the number of edges in the obtained graph is at most $v+o(v)$, so the number of edges in the initial graph is at most $v+\\varepsilon v / 2+o(v)$, which is smaller than $(1+\\varepsilon) v$ if $v$ is large enough.']" [] Text-only Competition True Theorem proof Algebra Math English 31 "Prove that there exist two functions $$ f, g: \mathbb{R} \rightarrow \mathbb{R} $$ such that $f \circ g$ is strictly decreasing, while $g \circ f$ is strictly increasing." ['Let\n\n\n\n$$\n\n\\begin{aligned}\n\n& \\text { - } A=\\bigcup_{k \\in \\mathbb{Z}}\\left(\\left[-2^{2 k+1},-2^{2 k}\\right) \\bigcup\\left(2^{2 k}, 2^{2 k+1}\\right]\\right) ; \\\\\n\n& \\text { - } B=\\bigcup_{k \\in \\mathbb{Z}}\\left(\\left[-2^{2 k},-2^{2 k-1}\\right) \\bigcup\\left(2^{2 k-1}, 2^{2 k}\\right]\\right) .\n\n\\end{aligned}\n\n$$\n\n\n\nThus $A=2 B, B=2 A, A=-A, B=-B, A \\cap B=\\varnothing$, and finally $A \\cup B \\cup\\{0\\}=\\mathbb{R}$. Let us take\n\n\n\n$$\n\nf(x)=\\left\\{\\begin{array}{lll}\n\nx & \\text { for } & x \\in A \\\\\n\n-x & \\text { for } & x \\in B \\\\\n\n0 & \\text { for } & x=0\n\n\\end{array}\\right.\n\n$$\n\n\n\nTake $g(x)=2 f(x)$. Thus $f(g(x))=f(2 f(x))=-2 x$ and $g(f(x))=2 f(f(x))=2 x$.'] [] Text-only Competition True Theorem proof Algebra Math English 32 Let $A B C D$ be a cyclic quadrangle and let $P$ be a point on the side $A B$. The diagonal $A C$ crosses the segment $D P$ at $Q$. The parallel through $P$ to $C D$ crosses the extension of the side $B C$ beyond $B$ at $K$, and the parallel through $Q$ to $B D$ crosses the extension of the side $B C$ beyond $B$ at $L$. Prove that the circumcircles of the triangles $B K P$ and $C L Q$ are tangent. ['We show that the circles $B K P$ and $C L Q$ are tangent at the point $T$ where the line $D P$ crosses the circle $A B C D$ again.\n\nSince $B C D T$ is cyclic, we have $\\angle K B T=\\angle C D T$. Since $K P \\| C D$, we get $\\angle C D T=\\angle K P T$. Thus, $\\angle K B T=\\angle C D T=\\angle K P T$, which shows that $T$ lies on the circle $B K P$. Similarly, the equalities $\\angle L C T=\\angle B D T=\\angle L Q T$ show that $T$ also lies on the circle $C L Q$.\n\nIt remains to prove that these circles are indeed tangent at $T$. This follows from the fact that the chords $T P$ and $T Q$ in the circles $B K T P$ and $C L T Q$, respectively, both lie along the same line and subtend equal angles $\\angle T B P=\\angle T B A=\\angle T C A=\\angle T C Q$.\n\n\n\n### 分析'] [] Text-only Competition True Theorem proof Geometry Math English 33 "Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying $$ P(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} $$" "['The answer is in the negative. Comparing the degrees of both sides in $(1)$ we get $\\operatorname{deg} P=21 n$ and $\\operatorname{deg} Q=10 n$ for some positive integer $n$. Take the derivative of $(1)$ to obtain\n\n$$\nP^{\\prime} P^{8}(10 P+9)=Q^{\\prime} Q^{19}(21 Q+20)\n$$\n\nSince $\\operatorname{gcd}(10 P+9, P)=\\operatorname{gcd}(10 P+9, P+1)=1$, it follows that $\\operatorname{gcd}\\left(10 P+9, P^{9}(P+1)\\right)=1$, so $\\operatorname{gcd}(10 P+9, Q)=1$, by $(1)$. Thus $(2)$ yields $10 P+9 \\mid Q^{\\prime}(21 Q+20)$, which is impossible since $0<\\operatorname{deg}\\left(Q^{\\prime}(21 Q+20)\\right)=20 n-1<21 n=\\operatorname{deg}(10 P+9)$. A contradiction.' 'Letting $r$ and $s$ be integers such that $r \\geq 2$ and $s \\geq 2 r$, we show that if $P^{r}+P^{r-1}=$ $Q^{s}+Q^{s-1}$, then $Q$ is constant.\n\nLet $m=\\operatorname{deg} P$ and $n=\\operatorname{deg} Q$. A degree inspection in the given relation shows that $m \\geq 2 n$.\n\nWe will prove that $P(P+1)$ has at least $m+1$ distinct complex roots. Assuming this for the moment, notice that $Q$ takes on one of the values 0 or -1 at each of those roots. Since $m+1 \\geq 2 n+1$, it follows that $Q$ takes on one of the values 0 and -1 at more than $n$ distinct points, so $Q$ must be constant.\n\nFinally, we prove that $P(P+1)$ has at least $m+1$ distinct complex roots. This can be done either by referring to the Mason-Stothers theorem or directly, in terms of multiplicities of the roots in question.\n\nSince $P$ and $P+1$ are relatively prime, the Mason-Stothers theorem implies that the number of distinct roots of $P(P+1)$ is greater than $m$, hence at least $m+1$.\n\nFor a direct proof, let $z_{1}, \\ldots, z_{t}$ be the distinct complex roots of $P(P+1)$, and let $z_{k}$ have multiplicity $\\alpha_{k}, k=1, \\ldots, t$. Since $P$ and $P+1$ have no roots in common, and $P^{\\prime}=(P+1)^{\\prime}$, it follows that $P^{\\prime}$ has a root of multiplicity $\\alpha_{k}-1$ at $z_{k}$. Consequently, $m-1=\\operatorname{deg} P^{\\prime} \\geq$ $\\sum_{k=1}^{t}\\left(\\alpha_{k}-1\\right)=\\sum_{k=1}^{t} \\alpha_{k}-t=2 m-t$; that is, $t \\geq m+1$. This completes the prof.\n\n### 分析']" [] Text-only Competition True Theorem proof Algebra Math English 34 Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win? "[""The answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.\n\nWe say that two unit grid-segments form a low-left corner (or LL-corner) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An up-right corner (or UR-corner) is defined similarly. The common endpoint of two unit grid-segments at a corner is the joint of that corner.\n\nFix a vertical line on the grid and call it the midline; the unit grid-segments along the midline are called middle segments. The unit grid-segments lying to the left/right of the midline are called left/right segments. Partition all left segments into LL-corners, and all right segments into UR-corners.\n\nWe now describe Ann's strategy. Her first move consists in orienting some middle segment arbitrarily. Assume that at some stage, Bob orients some segment $s$. If $s$ is a middle segment, Ann orients any free middle segment arbitrarily. Otherwise, $s$ forms a corner in the partition with some other segment $t$. Then Ann orients $t$ so that the joint of the corner is either the source of both arrows, or the target of both. Notice that after any move of Ann's, each corner in the partition is either completely oriented or completely not oriented. This means that Ann can always make a required move.\n\nAssume that Bob wins at some stage, i.e., an oriented cycle $C$ occurs. Let $X$ be the lowest of the leftmost points of $C$, and let $Y$ be the topmost of the rightmost points of $C$. If $X$ lies (strictly) to the left of the midline, then $X$ is the joint of some corner whose segments are both oriented. But, according to Ann's strategy, they are oriented so that they cannot occur in a cycle - a contradiction. Otherwise, $Y$ lies to the right of the midline, and a similar argument applies. Thus, Bob will never win, as desired.\n\n\n### 分析""]" [] Text-only Competition True Theorem proof Combinatorics Math English 35 "Given a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of real numbers. For each $i(1 \leq i \leq n)$ define $$ d_{i}=\max \left\{a_{j}: 1 \leq j \leq i\right\}-\min \left\{a_{j}: i \leq j \leq n\right\} $$ and let $$ d=\max \left\{d_{i}: 1 \leq i \leq n\right\} $$ Prove that for arbitrary real numbers $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$, $$ \max \left\{\left|x_{i}-a_{i}\right|: 1 \leq i \leq n\right\} \geq \frac{d}{2} \tag{1} $$" ['Let $1 \\leq p \\leq q \\leq r \\leq n$ be indices for which\n\n$$\nd=d_{q}, \\quad a_{p}=\\max \\left\\{a_{j}: 1 \\leq j \\leq q\\right\\}, \\quad a_{r}=\\min \\left\\{a_{j}: q \\leq j \\leq n\\right\\}\n$$\n\nand thus $d=a_{p}-a_{r}$. (These indices are not necessarily unique.)\n\n\n\nFor arbitrary real numbers $x_{1} \\leq x_{2} \\leq \\ldots \\leq x_{n}$, consider just the two quantities $\\left|x_{p}-a_{p}\\right|$ and $\\left|x_{r}-a_{r}\\right|$. Since\n\n$$\n\\left(a_{p}-x_{p}\\right)+\\left(x_{r}-a_{r}\\right)=\\left(a_{p}-a_{r}\\right)+\\left(x_{r}-x_{p}\\right) \\geq a_{p}-a_{r}=d\n$$\n\nwe have either $a_{p}-x_{p} \\geq \\frac{d}{2}$ or $x_{r}-a_{r} \\geq \\frac{d}{2}$. Hence,\n\n$$\n\\max \\left\\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\right\\} \\geq \\max \\left\\{\\left|x_{p}-a_{p}\\right|,\\left|x_{r}-a_{r}\\right|\\right\\} \\geq \\max \\left\\{a_{p}-x_{p}, x_{r}-a_{r}\\right\\} \\geq \\frac{d}{2}\n$$'] [] Text-only Competition True Theorem proof Algebra Math English 36 "Given a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of real numbers. For each $i(1 \leq i \leq n)$ define $$ d_{i}=\max \left\{a_{j}: 1 \leq j \leq i\right\}-\min \left\{a_{j}: i \leq j \leq n\right\} $$ and let $$ d=\max \left\{d_{i}: 1 \leq i \leq n\right\} $$ Show that there exists a sequence $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$ of real numbers such that we have equality in (1)." ['Define the sequence $\\left(x_{k}\\right)$ as\n\n$$\nx_{1}=a_{1}-\\frac{d}{2}, \\quad x_{k}=\\max \\left\\{x_{k-1}, a_{k}-\\frac{d}{2}\\right\\} \\quad \\text { for } 2 \\leq k \\leq n\n$$\n\nWe show that we have equality in (1) for this sequence.\n\nBy the definition, sequence $\\left(x_{k}\\right)$ is non-decreasing and $x_{k}-a_{k} \\geq-\\frac{d}{2}$ for all $1 \\leq k \\leq n$. Next we prove that\n\n$$\nx_{k}-a_{k} \\leq \\frac{d}{2} \\quad \\text { for all } 1 \\leq k \\leq n\\tag{2}\n$$\n\nConsider an arbitrary index $1 \\leq k \\leq n$. Let $\\ell \\leq k$ be the smallest index such that $x_{k}=x_{\\ell}$. We have either $\\ell=1$, or $\\ell \\geq 2$ and $x_{\\ell}>x_{\\ell-1}$. In both cases,\n\n$$\nx_{k}=x_{\\ell}=a_{\\ell}-\\frac{d}{2}\\tag{3}\n$$\n\nSince\n\n$$\na_{\\ell}-a_{k} \\leq \\max \\left\\{a_{j}: 1 \\leq j \\leq k\\right\\}-\\min \\left\\{a_{j}: k \\leq j \\leq n\\right\\}=d_{k} \\leq d\n$$\n\nequality (3) implies\n\n$$\nx_{k}-a_{k}=a_{\\ell}-a_{k}-\\frac{d}{2} \\leq d-\\frac{d}{2}=\\frac{d}{2}\n$$\n\nWe obtained that $-\\frac{d}{2} \\leq x_{k}-a_{k} \\leq \\frac{d}{2}$ for all $1 \\leq k \\leq n$, so\n\n$$\n\\max \\left\\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\right\\} \\leq \\frac{d}{2}\n$$\n\nWe have equality because $\\left|x_{1}-a_{1}\\right|=\\frac{d}{2}$.'] [] Text-only Competition True Theorem proof Algebra Math English 37 "Let $n$ be a positive integer, and let $x$ and $y$ be positive real numbers such that $x^{n}+y^{n}=1$. Prove that $$ \left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{1}{(1-x)(1-y)} $$" "['For each real $t \\in(0,1)$,\n\n$$\n\\frac{1+t^{2}}{1+t^{4}}=\\frac{1}{t}-\\frac{(1-t)\\left(1-t^{3}\\right)}{t\\left(1+t^{4}\\right)}<\\frac{1}{t}\n$$\n\nSubstituting $t=x^{k}$ and $t=y^{k}$,\n\n$$\n0<\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}<\\sum_{k=1}^{n} \\frac{1}{x^{k}}=\\frac{1-x^{n}}{x^{n}(1-x)} \\quad \\text { and } \\quad 0<\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}<\\sum_{k=1}^{n} \\frac{1}{y^{k}}=\\frac{1-y^{n}}{y^{n}(1-y)}\n$$\n\nSince $1-y^{n}=x^{n}$ and $1-x^{n}=y^{n}$,\n\n$$\n\\frac{1-x^{n}}{x^{n}(1-x)}=\\frac{y^{n}}{x^{n}(1-x)}, \\quad \\frac{1-y^{n}}{y^{n}(1-y)}=\\frac{x^{n}}{y^{n}(1-y)}\n$$\n\nand therefore\n\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right)<\\frac{y^{n}}{x^{n}(1-x)} \\cdot \\frac{x^{n}}{y^{n}(1-y)}=\\frac{1}{(1-x)(1-y)}\n$$' 'We prove\n\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right)<\\frac{\\left(\\frac{1+\\sqrt{2}}{2} \\ln 2\\right)^{2}}{(1-x)(1-y)}<\\frac{0.7001}{(1-x)(1-y)}\\tag{1}\n$$\n\nThe idea is to estimate each term on the left-hand side with the same constant. To find the upper bound for the expression $\\frac{1+x^{2 k}}{1+x^{4 k}}$, consider the function $f(t)=\\frac{1+t}{1+t^{2}}$ in interval $(0,1)$. Since\n\n$$\nf^{\\prime}(t)=\\frac{1-2 t-t^{2}}{\\left(1+t^{2}\\right)^{2}}=\\frac{(\\sqrt{2}+1+t)(\\sqrt{2}-1-t)}{\\left(1+t^{2}\\right)^{2}},\n$$\n\nthe function increases in interval $(0, \\sqrt{2}-1]$ and decreases in $[\\sqrt{2}-1,1)$. Therefore the maximum is at point $t_{0}=\\sqrt{2}-1$ and\n\n$$\nf(t)=\\frac{1+t}{1+t^{2}} \\leq f\\left(t_{0}\\right)=\\frac{1+\\sqrt{2}}{2}=\\alpha \\text {. }\n$$\n\nApplying this to each term on the left-hand side of (1), we obtain\n\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right) \\leq n \\alpha \\cdot n \\alpha=(n \\alpha)^{2} \\tag{2}\n$$\n\nTo estimate $(1-x)(1-y)$ on the right-hand side, consider the function\n\n$$\ng(t)=\\ln \\left(1-t^{1 / n}\\right)+\\ln \\left(1-(1-t)^{1 / n}\\right) .\n$$\n\n\n\nSubstituting $s$ for $1-t$, we have\n\n$$\n-n g^{\\prime}(t)=\\frac{t^{1 / n-1}}{1-t^{1 / n}}-\\frac{s^{1 / n-1}}{1-s^{1 / n}}=\\frac{1}{s t}\\left(\\frac{(1-t) t^{1 / n}}{1-t^{1 / n}}-\\frac{(1-s) s^{1 / n}}{1-s^{1 / n}}\\right)=\\frac{h(t)-h(s)}{s t}\n$$\n\nThe function\n\n$$\nh(t)=t^{1 / n} \\frac{1-t}{1-t^{1 / n}}=\\sum_{i=1}^{n} t^{i / n}\n$$\n\nis obviously increasing for $t \\in(0,1)$, hence for these values of $t$ we have\n\n$$\ng^{\\prime}(t)>0 \\Longleftrightarrow h(t)2$, and let $a(1), a(2), \ldots$ be a sequence of nonnegative real numbers such that $$ a(m+n) \leq 2 a(m)+2 a(n) \text { for all } m, n \geq 1\tag{1} $$ and $$ a\left(2^{k}\right) \leq \frac{1}{(k+1)^{c}} \quad \text { for all } k \geq 0\tag{2} $$ Prove that the sequence $a(n)$ is bounded." "['For convenience, define $a(0)=0$; then condition (1) persists for all pairs of nonnegative indices.\n\nLemma 1. For arbitrary nonnegative indices $n_{1}, \\ldots, n_{k}$, we have\n\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq \\sum_{i=1}^{k} 2^{i} a\\left(n_{i}\\right)\\tag{3}\n$$\n\nand\n\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq 2 k \\sum_{i=1}^{k} a\\left(n_{i}\\right)\\tag{4}\n$$\n\nProof. Inequality (3) is proved by induction on $k$. The base case $k=1$ is trivial, while the induction step is provided by\n\n$$\na\\left(\\sum_{i=1}^{k+1} n_{i}\\right)=a\\left(n_{1}+\\sum_{i=2}^{k+1} n_{i}\\right) \\leq 2 a\\left(n_{1}\\right)+2 a\\left(\\sum_{i=1}^{k} n_{i+1}\\right) \\leq 2 a\\left(n_{1}\\right)+2 \\sum_{i=1}^{k} 2^{i} a\\left(n_{i+1}\\right)=\\sum_{i=1}^{k+1} 2^{i} a\\left(n_{i}\\right)\n$$\n\nTo establish (4), first the inequality\n$$\na\\left(\\sum_{i=1}^{2^{d}} n_{i}\\right) \\leq 2^{d} \\sum_{i=1}^{2^{d}} a\\left(n_{i}\\right)\n$$\ncan be proved by an obvious induction on $d$. Then, turning to (4), we find an integer $d$ such that $2^{d-1}d$, and take some positive integer $f$ such that $M_{f}>d$. Applying (3), we get\n$$\na(n)=a\\left(\\sum_{k=1}^{f} \\sum_{M_{k-1} \\leq i2$. We can assume that $s_{1} \\leq s_{2} \\leq \\cdots \\leq s_{k}$. Note that\n$$\n\\sum_{i=1}^{k-1} 2^{-s_{i}} \\leq 1-2^{-s_{k-1}}\n$$\nsince the left-hand side is a fraction with the denominator $2^{s_{k-1}}$, and this fraction is less than 1. Define $s_{k-1}^{\\prime}=s_{k-1}-1$ and $n_{k-1}^{\\prime}=n_{k-1}+n_{k}$; then we have\n$$\n\\sum_{i=1}^{k-2} 2^{-s_{i}}+2^{-s_{k-1}^{\\prime}} \\leq\\left(1-2 \\cdot 2^{-s_{k-1}}\\right)+2^{1-s_{k-1}}=1 .\n$$\nNow, the induction hypothesis can be applied to achieve\n$$\n\\begin{aligned}\na\\left(\\sum_{i=1}^{k} n_{i}\\right)=a\\left(\\sum_{i=1}^{k-2} n_{i}+n_{k-1}^{\\prime}\\right) & \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}^{\\prime}} a\\left(n_{k-1}^{\\prime}\\right) \\\\\n& \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}-1} \\cdot 2\\left(a\\left(n_{k-1}\\right)+a\\left(n_{k}\\right)\\right) \\\\\n& \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}} a\\left(n_{k-1}\\right)+2^{s_{k}} a\\left(n_{k}\\right) .\n\\end{aligned}\n$$\n\n\nLet $q=c / 2>1$. Take an arbitrary positive integer $n$ and write\n$$\nn=\\sum_{i=1}^{k} 2^{u_{i}}, \\quad 0 \\leq u_{1}1$.']" [] Text-only Competition True Theorem proof Algebra Math English 39 "Let $a_{1}, a_{2}, \ldots, a_{100}$ be nonnegative real numbers such that $a_{1}^{2}+a_{2}^{2}+\ldots+a_{100}^{2}=1$. Prove that $$ a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\ldots+a_{100}^{2} a_{1}<\frac{12}{25} $$" ['Let $S=\\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$. (As usual, we consider the indices modulo 100, e.g. we set $a_{101}=a_{1}$ and $a_{102}=a_{2}$.)\n\nApplying the Cauchy-Schwarz inequality to sequences $\\left(a_{k+1}\\right)$ and $\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)$, and then the AM-GM inequality to numbers $a_{k+1}^{2}$ and $a_{k+2}^{2}$,\n$$\n\\begin{aligned}\n(3 S)^{2} & =\\left(\\sum_{k=1}^{100} a_{k+1}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)\\right)^{2} \\leq\\left(\\sum_{k=1}^{100} a_{k+1}^{2}\\right)\\left(\\sum_{k=1}^{100}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)^{2}\\right) \\\\\n& =1 \\cdot \\sum_{k=1}^{100}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)^{2}=\\sum_{k=1}^{100}\\left(a_{k}^{4}+4 a_{k}^{2} a_{k+1} a_{k+2}+4 a_{k+1}^{2} a_{k+2}^{2}\\right) \\\\\n& \\leq \\sum_{k=1}^{100}\\left(a_{k}^{4}+2 a_{k}^{2}\\left(a_{k+1}^{2}+a_{k+2}^{2}\\right)+4 a_{k+1}^{2} a_{k+2}^{2}\\right)=\\sum_{k=1}^{100}\\left(a_{k}^{4}+6 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\\right) .\n\\end{aligned}\n$$\nApplying the trivial estimates\n$$\n\\sum_{k=1}^{100}\\left(a_{k}^{4}+2 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\\right) \\leq\\left(\\sum_{k=1}^{100} a_{k}^{2}\\right)^{2} \\quad \\text { and } \\quad \\sum_{k=1}^{100} a_{k}^{2} a_{k+1}^{2} \\leq\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}\\right)\\left(\\sum_{j=1}^{50} a_{2 j}^{2}\\right)\n$$\nwe obtain that\n$$\n(3 S)^{2} \\leq\\left(\\sum_{k=1}^{100} a_{k}^{2}\\right)^{2}+4\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}\\right)\\left(\\sum_{j=1}^{50} a_{2 j}^{2}\\right) \\leq 1+\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}+\\sum_{j=1}^{50} a_{2 j}^{2}\\right)^{2}=2,\n$$\nhence\n$$\nS \\leq \\frac{\\sqrt{2}}{3} \\approx 0.4714<\\frac{12}{25}=0.48\n$$'] [] Text-only Competition True Theorem proof Algebra Math English 40 A unit square is dissected into $n>1$ rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of some rectangle. Prove that in this dissection, there exists a rectangle having no point on the boundary of the square. "['Call the directions of the sides of the square horizontal and vertical. A horizontal or vertical line, which intersects the interior of the square but does not intersect the interior of any rectangle, will be called a splitting line. A rectangle having no point on the boundary of the square will be called an interior rectangle.\n\nSuppose, to the contrary, that there exists a dissection of the square into more than one rectangle, such that no interior rectangle and no splitting line appear. Consider such a dissection with the least possible number of rectangles. Notice that this number of rectangles is greater than 2, otherwise their common side provides a splitting line.\n\nIf there exist two rectangles having a common side, then we can replace them by their union (see Figure 1). The number of rectangles was greater than 2, so in a new dissection it is greater than 1. Clearly, in the new dissection, there is also no splitting line as well as no interior rectangle. This contradicts the choice of the original dissection.\n\nDenote the initial square by $A B C D$, with $A$ and $B$ being respectively the lower left and lower right vertices. Consider those two rectangles $a$ and $b$ containing vertices $A$ and $B$, respectively. (Note that $a \\neq b$, otherwise its top side provides a splitting line.) We can assume that the height of $a$ is not greater than that of $b$. Then consider the rectangle $c$ neighboring to the lower right corner of $a$ (it may happen that $c=b$ ). By aforementioned, the heights of $a$ and $c$ are distinct. Then two cases are possible.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nCase 1. The height of $c$ is less than that of $a$. Consider the rectangle $d$ which is adjacent to both $a$ and $c$, i. e. the one containing the angle marked in Figure 2. This rectangle has no common point with $B C$ (since $a$ is not higher than $b$ ), as well as no common point with $A B$ or with $A D$ (obviously). Then $d$ has a common point with $C D$, and its left side provides a splitting line. Contradiction.\n\nCase 2. The height of $c$ is greater than that of $a$. Analogously, consider the rectangle $d$ containing the angle marked on Figure 3. It has no common point with $A D$ (otherwise it has a common side with $a$ ), as well as no common point with $A B$ or with $B C$ (obviously). Then $d$ has a common point with $C D$. Hence its right side provides a splitting line, and we get the contradiction again.' 'Again, we suppose the contrary. Consider an arbitrary counterexample. Then we know that each rectangle is attached to at least one side of the square. Observe that a rectangle cannot be attached to two opposite sides, otherwise one of its sides lies on a splitting line.\n\nWe say that two rectangles are opposite if they are attached to opposite sides of $A B C D$. We claim that there exist two opposite rectangles having a common point.\n\nConsider the union $L$ of all rectangles attached to the left. Assume, to the contrary, that $L$ has no common point with the rectangles attached to the right. Take a polygonal line $p$ connecting the top and the bottom sides of the square and passing close from the right to the boundary of $L$ (see Figure 4). Then all its points belong to the rectangles attached either to the top or to the bottom. Moreover, the upper end-point of $p$ belongs to a rectangle attached to the top, and the lower one belongs to an other rectangle attached to the bottom. Hence, there is a point on $p$ where some rectangles attached to the top and to the bottom meet each other. So, there always exists a pair of neighboring opposite rectangles.\n\n\n\nFigure 4\n\n\n\nFigure 5\n\n\n\nFigure 6\n\nNow, take two opposite neighboring rectangles $a$ and $b$. We can assume that $a$ is attached to the left and $b$ is attached to the right. Let $X$ be their common point. If $X$ belongs to their horizontal sides (in particular, $X$ may appear to be a common vertex of $a$ and $b$ ), then these sides provide a splitting line (see Figure 5). Otherwise, $X$ lies on the vertical sides. Let $\\ell$ be the line containing these sides.\n\nSince $\\ell$ is not a splitting line, it intersects the interior of some rectangle. Let $c$ be such a rectangle, closest to $X$; we can assume that $c$ lies above $X$. Let $Y$ be the common point of $\\ell$ and the bottom side of $c$ (see Figure 6). Then $Y$ is also a vertex of two rectangles lying below $c$.\n\nSo, let $Y$ be the upper-right and upper-left corners of the rectangles $a^{\\prime}$ and $b^{\\prime}$, respectively. Then $a^{\\prime}$ and $b^{\\prime}$ are situated not lower than $a$ and $b$, respectively (it may happen that $a=a^{\\prime}$ or $b=b^{\\prime}$ ). We claim that $a^{\\prime}$ is attached to the left. If $a=a^{\\prime}$ then of course it is. If $a \\neq a^{\\prime}$ then $a^{\\prime}$ is above $a$, below $c$ and to the left from $b^{\\prime}$. Hence, it can be attached to the left only.\n\nAnalogously, $b^{\\prime}$ is attached to the right. Now, the top sides of these two rectangles pass through $Y$, hence they provide a splitting line again. This last contradiction completes the proof.']" [] Text-only Competition True Theorem proof Combinatorics Math English 41 "Let $A_{0}=\left(a_{1}, \ldots, a_{n}\right)$ be a finite sequence of real numbers. For each $k \geq 0$, from the sequence $A_{k}=\left(x_{1}, \ldots, x_{n}\right)$ we construct a new sequence $A_{k+1}$ in the following way. 1. We choose a partition $\{1, \ldots, n\}=I \cup J$, where $I$ and $J$ are two disjoint sets, such that the expression $$ \left|\sum_{i \in I} x_{i}-\sum_{j \in J} x_{j}\right| $$ attains the smallest possible value. (We allow the sets $I$ or $J$ to be empty; in this case the corresponding sum is 0 .) If there are several such partitions, one is chosen arbitrarily. 2. We set $A_{k+1}=\left(y_{1}, \ldots, y_{n}\right)$, where $y_{i}=x_{i}+1$ if $i \in I$, and $y_{i}=x_{i}-1$ if $i \in J$. Prove that for some $k$, the sequence $A_{k}$ contains an element $x$ such that $|x| \geq n / 2$." ['Lemma. Suppose that all terms of the sequence $\\left(x_{1}, \\ldots, x_{n}\\right)$ satisfy the inequality $\\left|x_{i}\\right|1)$. By the induction hypothesis there exists a splitting $\\{1, \\ldots, n-1\\}=I^{\\prime} \\cup J^{\\prime}$ such that\n$$\n\\left|\\sum_{i \\in I^{\\prime}} x_{i}-\\sum_{j \\in J^{\\prime}} x_{j}\\right|n-2 \\cdot \\frac{n}{2}=0\n$$\n\nThus we obtain $S_{q}>S_{q-1}>\\cdots>S_{p}$. This is impossible since $A_{p}=A_{q}$ and hence $S_{p}=S_{q}$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 42 In the Cartesian coordinate plane define the strip $S_{n}=\{(x, y) \mid n \leq xb$. Then $a_{1}>b_{1} \\geq 1$, so $a_{1} \\geq 3$.\n\nChoose integers $k$ and $\\ell$ such that $k a_{1}-\\ell b_{1}=1$ and therefore $k a-\\ell b=d$. Since $a_{1}$ and $b_{1}$ are odd, $k+\\ell$ is odd as well. Hence, for every integer $n$, strips $S_{n}$ and $S_{n+k a-\\ell b}=S_{n+d}$ have opposite colors. This also implies that the coloring is periodic with period $2 d$, i.e. strips $S_{n}$ and $S_{n+2 d}$ have the same color for every $n$.\n\n\n\nFigure 1\n\nWe will construct the desired rectangle $A B C D$ with $A B=C D=a$ and $B C=A D=b$ in a position such that vertex $A$ lies on the $x$-axis, and the projection of side $A B$ onto the $x$-axis is of length $2 d$ (see Figure 1). This is possible since $a=a_{1} d>2 d$. The coordinates of the vertices will have the forms\n$$\nA=(t, 0), \\quad B=\\left(t+2 d, y_{1}\\right), \\quad C=\\left(u+2 d, y_{2}\\right), \\quad D=\\left(u, y_{3}\\right)\n$$\nLet $\\varphi=\\sqrt{a_{1}^{2}-4}$. By Pythagoras' theorem,\n$$\ny_{1}=B B_{0}=\\sqrt{a^{2}-4 d^{2}}=d \\sqrt{a_{1}^{2}-4}=d \\varphi .\n$$\nSo, by the similar triangles $A D D_{0}$ and $B A B_{0}$, we have the constraint\n$$\nu-t=A D_{0}=\\frac{A D}{A B} \\cdot B B_{0}=\\frac{b d}{a} \\varphi \\tag{1}\n$$\n\nfor numbers $t$ and $u$. Computing the numbers $y_{2}$ and $y_{3}$ is not required since they have no effect to the colors.\n\nObserve that the number $\\varphi$ is irrational, because $\\varphi^{2}$ is an integer, but $\\varphi$ is not: $a_{1}>\\varphi \\geq$ $\\sqrt{a_{1}^{2}-2 a_{1}+2}>a_{1}-1$.\n\nBy the periodicity, points $A$ and $B$ have the same color; similarly, points $C$ and $D$ have the same color. Furthermore, these colors depend only on the values of $t$ and $u$. So it is sufficient to choose numbers $t$ and $u$ such that vertices $A$ and $D$ have the same color.\n\nLet $w$ be the largest positive integer such that there exist $w$ consecutive strips $S_{n_{0}}, S_{n_{0}+1}, \\ldots$, $S_{n_{0}+w-1}$ with the same color, say red. (Since $S_{n_{0}+d}$ must be blue, we have $w \\leq d$.) We will choose $t$ from the interval $\\left(n_{0}, n_{0}+w\\right)$.\n\n\n\nFigure 2\n\nConsider the interval $I=\\left(n_{0}+\\frac{b d}{a} \\varphi, n_{0}+\\frac{b d}{a} \\varphi+w\\right)$ on the $x$-axis (see Figure 2). Its length is $w$, and the end-points are irrational. Therefore, this interval intersects $w+1$ consecutive strips. Since at most $w$ consecutive strips may have the same color, interval $I$ must contain both red and blue points. Choose $u \\in I$ such that the line $x=u$ is red and set $t=u-\\frac{b d}{a} \\varphi$, according to the constraint (1). Then $t \\in\\left(n_{0}, n_{0}+w\\right)$ and $A=(t, 0)$ is red as well as $D=\\left(u, y_{3}\\right)$.\n\nHence, variables $u$ and $t$ can be set such that they provide a rectangle with four red vertices.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 43 "In a mathematical competition some competitors are friends; friendship is always mutual. Call a group of competitors a clique if each two of them are friends. The number of members in a clique is called its size. It is known that the largest size of cliques is even. Prove that the competitors can be arranged in two rooms such that the largest size of cliques in one room is the same as the largest size of cliques in the other room." ['We present an algorithm to arrange the competitors. Let the two rooms be Room A and Room B. We start with an initial arrangement, and then we modify it several times by sending one person to the other room. At any state of the algorithm, $A$ and $B$ denote the sets of the competitors in the rooms, and $c(A)$ and $c(B)$ denote the largest sizes of cliques in the rooms, respectively.\n\nStep 1. Let $M$ be one of the cliques of largest size, $|M|=2 m$. Send all members of $M$ to Room $A$ and all other competitors to Room $B$.\n\nSince $M$ is a clique of the largest size, we have $c(A)=|M| \\geq c(B)$.\n\nStep 2. While $c(A)>c(B)$, send one person from Room $A$ to Room $B$.\n\n\n\nNote that $c(A)>c(B)$ implies that Room $A$ is not empty.\n\nIn each step, $c(A)$ decreases by one and $c(B)$ increases by at most one. So at the end we have $c(A) \\leq c(B) \\leq c(A)+1$.\n\nWe also have $c(A)=|A| \\geq m$ at the end. Otherwise we would have at least $m+1$ members of $M$ in Room $B$ and at most $m-1$ in Room $A$, implying $c(B)-c(A) \\geq(m+1)-(m-1)=2$.\n\nStep 3. Let $k=c(A)$. If $c(B)=k$ then $S T O P$.\n\nIf we reached $c(A)=c(B)=k$ then we have found the desired arrangement.\n\nIn all other cases we have $c(B)=k+1$.\n\nFrom the estimate above we also know that $k=|A|=|A \\cap M| \\geq m$ and $|B \\cap M| \\leq m$.\n\nStep 4. If there exists a competitor $x \\in B \\cap M$ and a clique $C \\subset B$ such that $|C|=k+1$ and $x \\notin C$, then move $x$ to Room $A$ and STOP.\n\n\n\nAfter moving $x$ back to Room $A$, we will have $k+1$ members of $M$ in Room $A$, thus $c(A)=k+1$. Due to $x \\notin C, c(B)=|C|$ is not decreased, and after this step we have $c(A)=c(B)=k+1$.\n\n\n\nIf there is no such competitor $x$, then in Room $B$, all cliques of size $k+1$ contain $B \\cap M$ as a subset.\n\nStep 5. While $c(B)=k+1$, choose a clique $C \\subset B$ such that $|C|=k+1$ and move one member of $C \\backslash M$ to Room $A$.\n\n\n\nNote that $|C|=k+1>m \\geq|B \\cap M|$, so $C \\backslash M$ cannot be empty.\n\nEvery time we move a single person from Room $B$ to Room $A$, so $c(B)$ decreases by at most 1 . Hence, at the end of this loop we have $c(B)=k$.\n\nIn Room $A$ we have the clique $A \\cap M$ with size $|A \\cap M|=k$ thus $c(A) \\geq k$. We prove that there is no clique of larger size there. Let $Q \\subset A$ be an arbitrary clique. We show that $|Q| \\leq k$.\n\n\n\nIn Room $A$, and specially in set $Q$, there can be two types of competitors:\n\n- Some members of $M$. Since $M$ is a clique, they are friends with all members of $B \\cap M$.\n- Competitors which were moved to Room $A$ in Step 5 . Each of them has been in a clique with $B \\cap M$ so they are also friends with all members of $B \\cap M$.\n\nHence, all members of $Q$ are friends with all members of $B \\cap M$. Sets $Q$ and $B \\cap M$ are cliques themselves, so $Q \\cup(B \\cap M)$ is also a clique. Since $M$ is a clique of the largest size,\n$$\n|M| \\geq|Q \\cup(B \\cap M)|=|Q|+|B \\cap M|=|Q|+|M|-|A \\cap M|\n$$\ntherefore\n$$\n|Q| \\leq|A \\cap M|=k\n$$\nFinally, after Step 5 we have $c(A)=c(B)=k$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 44 Let $\alpha<\frac{3-\sqrt{5}}{2}$ be a positive real number. Prove that there exist positive integers $n$ and $p>\alpha \cdot 2^{n}$ for which one can select $2 p$ pairwise distinct subsets $S_{1}, \ldots, S_{p}, T_{1}, \ldots, T_{p}$ of the set $\{1,2, \ldots, n\}$ such that $S_{i} \cap T_{j} \neq \varnothing$ for all $1 \leq i, j \leq p$. ['Let $k$ and $m$ be positive integers (to be determined later) and set $n=k m$. Decompose the set $\\{1,2, \\ldots, n\\}$ into $k$ disjoint subsets, each of size $m$; denote these subsets by $A_{1}, \\ldots, A_{k}$. Define the following families of sets:\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\left\\{S \\subset\\{1,2, \\ldots, n\\}: \\forall i S \\cap A_{i} \\neq \\varnothing\\right\\} \\\\\n\\mathcal{T}_{1} & =\\left\\{T \\subset\\{1,2, \\ldots, n\\}: \\exists i A_{i} \\subset T\\right\\}, \\quad \\mathcal{T}=\\mathcal{T}_{1} \\backslash \\mathcal{S} .\n\\end{aligned}\n$$\nFor each set $T \\in \\mathcal{T} \\subset \\mathcal{T}_{1}$, there exists an index $1 \\leq i \\leq k$ such that $A_{i} \\subset T$. Then for all $S \\in \\mathcal{S}$, $S \\cap T \\supset S \\cap A_{i} \\neq \\varnothing$. Hence, each $S \\in \\mathcal{S}$ and each $T \\in \\mathcal{T}$ have at least one common element.\n\nBelow we show that the numbers $m$ and $k$ can be chosen such that $|\\mathcal{S}|,|\\mathcal{T}|>\\alpha \\cdot 2^{n}$. Then, choosing $p=\\min \\{|\\mathcal{S}|,|\\mathcal{T}|\\}$, one can select the desired $2 p$ sets $S_{1}, \\ldots, S_{p}$ and $T_{1}, \\ldots, T_{p}$ from families $\\mathcal{S}$ and $\\mathcal{T}$, respectively. Since families $\\mathcal{S}$ and $\\mathcal{T}$ are disjoint, sets $S_{i}$ and $T_{j}$ will be pairwise distinct.\n\nTo count the sets $S \\in \\mathcal{S}$, observe that each $A_{i}$ has $2^{m}-1$ nonempty subsets so we have $2^{m}-1$ choices for $S \\cap A_{i}$. These intersections uniquely determine set $S$, so\n$$\n|\\mathcal{S}|=\\left(2^{m}-1\\right)^{k}\n$$\nSimilarly, if a set $H \\subset\\{1,2, \\ldots, n\\}$ does not contain a certain set $A_{i}$ then we have $2^{m}-1$ choices for $H \\cap A_{i}$ : all subsets of $A_{i}$, except $A_{i}$ itself. Therefore, the complement of $\\mathcal{T}_{1}$ contains $\\left(2^{m}-1\\right)^{k}$ sets and\n$$\n\\left|\\mathcal{T}_{1}\\right|=2^{k m}-\\left(2^{m}-1\\right)^{k} .\n$$\nNext consider the family $\\mathcal{S} \\backslash \\mathcal{T}_{1}$. If a set $S$ intersects all $A_{i}$ but does not contain any of them, then there exists $2^{m}-2$ possible values for each $S \\cap A_{i}$ : all subsets of $A_{i}$ except $\\varnothing$ and $A_{i}$. Therefore the number of such sets $S$ is $\\left(2^{m}-2\\right)^{k}$, so\n$$\n\\left|\\mathcal{S} \\backslash \\mathcal{T}_{1}\\right|=\\left(2^{m}-2\\right)^{k}\n$$\nFrom (1), (2), and (3) we obtain\n$$\n|\\mathcal{T}|=\\left|\\mathcal{T}_{1}\\right|-\\left|\\mathcal{S} \\cap \\mathcal{T}_{1}\\right|=\\left|\\mathcal{T}_{1}\\right|-\\left(|\\mathcal{S}|-\\left|\\mathcal{S} \\backslash \\mathcal{T}_{1}\\right|\\right)=2^{k m}-2\\left(2^{m}-1\\right)^{k}+\\left(2^{m}-2\\right)^{k}\n$$\nLet $\\delta=\\frac{3-\\sqrt{5}}{2}$ and $k=k(m)=\\left[2^{m} \\log \\frac{1}{\\delta}\\right]$. Then\n$$\n\\lim _{m \\rightarrow \\infty} \\frac{|\\mathcal{S}|}{2^{k m}}=\\lim _{m \\rightarrow \\infty}\\left(1-\\frac{1}{2^{m}}\\right)^{k}=\\exp \\left(-\\lim _{m \\rightarrow \\infty} \\frac{k}{2^{m}}\\right)=\\delta\n$$\nand similarly\n$$\n\\lim _{m \\rightarrow \\infty} \\frac{|\\mathcal{T}|}{2^{k m}}=1-2 \\lim _{m \\rightarrow \\infty}\\left(1-\\frac{1}{2^{m}}\\right)^{k}+\\lim _{m \\rightarrow \\infty}\\left(1-\\frac{2}{2^{m}}\\right)^{k}=1-2 \\delta+\\delta^{2}=\\delta\n$$\nHence, if $m$ is sufficiently large then $\\frac{|\\mathcal{S}|}{2^{m k}}$ and $\\frac{|\\mathcal{T}|}{2^{m k}}$ are greater than $\\alpha$ (since $\\alpha<\\delta$ ). So $|\\mathcal{S}|,|\\mathcal{T}|>\\alpha \\cdot 2^{m k}=\\alpha \\cdot 2^{n}$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 45 "Given a convex $n$-gon $P$ in the plane. For every three vertices of $P$, consider the triangle determined by them. Call such a triangle good if all its sides are of unit length. Prove that there are not more than $\frac{2}{3} n$ good triangles." ['Consider all good triangles containing a certain vertex $A$. The other two vertices of any such triangle lie on the circle $\\omega_{A}$ with unit radius and center $A$. Since $P$ is convex, all these vertices lie on an arc of angle less than $180^{\\circ}$. Let $L_{A} R_{A}$ be the shortest such arc, oriented clockwise (see Figure 1). Each of segments $A L_{A}$ and $A R_{A}$ belongs to a unique good triangle. We say that the good triangle with side $A L_{A}$ is assigned counterclockwise to $A$, and the second one, with side $A R_{A}$, is assigned clockwise to $A$. In those cases when there is a single good triangle containing vertex $A$, this triangle is assigned to $A$ twice.\n\nThere are at most two assignments to each vertex of the polygon. (Vertices which do not belong to any good triangle have no assignment.) So the number of assignments is at most $2 n$.\n\nConsider an arbitrary good triangle $A B C$, with vertices arranged clockwise. We prove that $A B C$ is assigned to its vertices at least three times. Then, denoting the number of good triangles by $t$, we obtain that the number $K$ of all assignments is at most $2 n$, while it is not less than $3 t$. Then $3 t \\leq K \\leq 2 n$, as required.\n\nActually, we prove that triangle $A B C$ is assigned either counterclockwise to $C$ or clockwise to $B$. Then, by the cyclic symmetry of the vertices, we obtain that triangle $A B C$ is assigned either counterclockwise to $A$ or clockwise to $C$, and either counterclockwise to $B$ or clockwise to $A$, providing the claim.\n\n\nFigure 1\n\n\n\nFigure 2\n\nAssume, to the contrary, that $L_{C} \\neq A$ and $R_{B} \\neq A$. Denote by $A^{\\prime}, B^{\\prime}, C^{\\prime}$ the intersection points of circles $\\omega_{A}, \\omega_{B}$ and $\\omega_{C}$, distinct from $A, B, C$ (see Figure 2). Let $C L_{C} L_{C}^{\\prime}$ be the good triangle containing $C L_{C}$. Observe that the angle of arc $L_{C} A$ is less than $120^{\\circ}$. Then one of the points $L_{C}$ and $L_{C}^{\\prime}$ belongs to $\\operatorname{arc} B^{\\prime} A$ of $\\omega_{C}$; let this point be $X$. In the case when $L_{C}=B^{\\prime}$ and $L_{C}^{\\prime}=A$, choose $X=B^{\\prime}$.\n\nAnalogously, considering the good triangle $B R_{B}^{\\prime} R_{B}$ which contains $B R_{B}$ as an edge, we see that one of the points $R_{B}$ and $R_{B}^{\\prime}$ lies on $\\operatorname{arc} A C^{\\prime}$ of $\\omega_{B}$. Denote this point by $Y, Y \\neq A$. Then angles $X A Y, Y A B, B A C$ and $C A X$ (oriented clockwise) are not greater than $180^{\\circ}$. Hence, point $A$ lies in quadrilateral $X Y B C$ (either in its interior or on segment $X Y$ ). This is impossible, since all these five points are vertices of $P$.\n\nHence, each good triangle has at least three assignments, and the statement is proved.'] [] Text-only Competition True Theorem proof Combinatorics Math English 46 In triangle $A B C$, the angle bisector at vertex $C$ intersects the circumcircle and the perpendicular bisectors of sides $B C$ and $C A$ at points $R, P$, and $Q$, respectively. The midpoints of $B C$ and $C A$ are $S$ and $T$, respectively. Prove that triangles $R Q T$ and $R P S$ have the same area. "['. If $A C=B C$ then triangle $A B C$ is isosceles, triangles $R Q T$ and $R P S$ are symmetric about the bisector $C R$ and the statement is trivial. If $A C \\neq B C$ then it can be assumed without loss of generality that $A C\n\nDenote the circumcenter by $O$. The right triangles $C T Q$ and $C S P$ have equal angles at vertex $C$, so they are similar, $\\angle C P S=\\angle C Q T=\\angle O Q P$ and\n$$\n\\frac{Q T}{P S}=\\frac{C Q}{C P}\n$$\nLet $\\ell$ be the perpendicular bisector of chord $C R$; of course, $\\ell$ passes through the circumcenter $O$. Due to the equal angles at $P$ and $Q$, triangle $O P Q$ is isosceles with $O P=O Q$. Then line $\\ell$ is the axis of symmetry in this triangle as well. Therefore, points $P$ and $Q$ lie symmetrically on line segment $C R$,\n$$\nR P=C Q \\quad \\text { and } \\quad R Q=C P .\n$$\nTriangles $R Q T$ and $R P S$ have equal angles at vertices $Q$ and $P$, respectively. Then\n$$\n\\frac{\\operatorname{area}(R Q T)}{\\operatorname{area}(R P S)}=\\frac{\\frac{1}{2} \\cdot R Q \\cdot Q T \\cdot \\sin \\angle R Q T}{\\frac{1}{2} \\cdot R P \\cdot P S \\cdot \\sin \\angle R P S}=\\frac{R Q}{R P} \\cdot \\frac{Q T}{P S}\n$$\nSubstituting (1) and (2),\n$$\n\\frac{\\operatorname{area}(R Q T)}{\\operatorname{area}(R P S)}=\\frac{R Q}{R P} \\cdot \\frac{Q T}{P S}=\\frac{C P}{C Q} \\cdot \\frac{C Q}{C P}=1\n$$\nHence, $\\operatorname{area}(R Q T)=\\operatorname{area}(R S P)$.' '. Assume again $A C\n\nConsider the rotation around point $O$ by angle $\\gamma$. This transform moves $A$ to $R, R$ to $B$ and $Q$ to $P$; hence triangles $R Q A$ and $B P R$ are congruent and they have the same area.\n\nTriangles $R Q T$ and $R Q A$ have $R Q$ as a common side, so the ratio between their areas is\n$$\n\\frac{\\operatorname{area}(R Q T)}{\\operatorname{area}(R Q A)}=\\frac{d(T, C R)}{d(A, C R)}=\\frac{C T}{C A}=\\frac{1}{2}\n$$\n$(d(X, Y Z)$ denotes the distance between point $X$ and line $Y Z)$.\n\nIt can be obtained similarly that\n$$\n\\frac{\\operatorname{area}(R P S)}{\\operatorname{area}(B P R)}=\\frac{C S}{C B}=\\frac{1}{2}\n$$\nNow the proof can be completed as\n$$\n\\operatorname{area}(R Q T)=\\frac{1}{2} \\operatorname{area}(R Q A)=\\frac{1}{2} \\operatorname{area}(B P R)=\\operatorname{area}(R P S)\n$$']" [] Text-only Competition True Theorem proof Geometry Math English 47 Given an isosceles triangle $A B C$ with $A B=A C$. The midpoint of side $B C$ is denoted by $M$. Let $X$ be a variable point on the shorter arc $M A$ of the circumcircle of triangle $A B M$. Let $T$ be the point in the angle domain $B M A$, for which $\angle T M X=90^{\circ}$ and $T X=B X$. Prove that $\angle M T B-\angle C T M$ does not depend on $X$. "['. Let $N$ be the midpoint of segment $B T$ (see Figure 1). Line $X N$ is the axis of symmetry in the isosceles triangle $B X T$, thus $\\angle T N X=90^{\\circ}$ and $\\angle B X N=\\angle N X T$. Moreover, in triangle $B C T$, line $M N$ is the midline parallel to $C T$; hence $\\angle C T M=\\angle N M T$.\n\nDue to the right angles at points $M$ and $N$, these points lie on the circle with diameter $X T$. Therefore,\n$$\n\\angle M T B=\\angle M T N=\\angle M X N \\quad \\text { and } \\quad \\angle C T M=\\angle N M T=\\angle N X T=\\angle B X N .\n$$\nHence\n$$\n\\angle M T B-\\angle C T M=\\angle M X N-\\angle B X N=\\angle M X B=\\angle M A B\n$$\nwhich does not depend on $X$.\n\n\n\nFigure 1\n\n\n\nFigure 2' '. Let $S$ be the reflection of point $T$ over $M$ (see Figure 2). Then $X M$ is the perpendicular bisector of $T S$, hence $X B=X T=X S$, and $X$ is the circumcenter of triangle $B S T$. Moreover, $\\angle B S M=\\angle C T M$ since they are symmetrical about $M$. Then\n$$\n\\angle M T B-\\angle C T M=\\angle S T B-\\angle B S T=\\frac{\\angle S X B-\\angle B X T}{2} .\n$$\nObserve that $\\angle S X B=\\angle S X T-\\angle B X T=2 \\angle M X T-\\angle B X T$, so\n$$\n\\angle M T B-\\angle C T M=\\frac{2 \\angle M X T-2 \\angle B X T}{2}=\\angle M X B=\\angle M A B\n$$\nwhich is constant.']" [] Text-only Competition True Theorem proof Geometry Math English 48 The diagonals of a trapezoid $A B C D$ intersect at point $P$. Point $Q$ lies between the parallel lines $B C$ and $A D$ such that $\angle A Q D=\angle C Q B$, and line $C D$ separates points $P$ and $Q$. Prove that $\angle B Q P=\angle D A Q$. ['Let $t=\\frac{A D}{B C}$. Consider the homothety $h$ with center $P$ and scale $-t$. Triangles $P D A$ and $P B C$ are similar with ratio $t$, hence $h(B)=D$ and $h(C)=A$.\n\n\n\nLet $Q^{\\prime}=h(Q)$ (see Figure 1). Then points $Q, P$ and $Q^{\\prime}$ are obviously collinear. Points $Q$ and $P$ lie on the same side of $A D$, as well as on the same side of $B C$; hence $Q^{\\prime}$ and $P$ are also on the same side of $h(B C)=A D$, and therefore $Q$ and $Q^{\\prime}$ are on the same side of $A D$. Moreover, points $Q$ and $C$ are on the same side of $B D$, while $Q^{\\prime}$ and $A$ are on the opposite side (see Figure above).\n\nBy the homothety, $\\angle A Q^{\\prime} D=\\angle C Q B=\\angle A Q D$, hence quadrilateral $A Q^{\\prime} Q D$ is cyclic. Then\n$$\n\\angle D A Q=\\angle D Q^{\\prime} Q=\\angle D Q^{\\prime} P=\\angle B Q P\n$$\n(the latter equality is valid by the homothety again).'] [] Text-only Competition True Theorem proof Geometry Math English 49 Consider five points $A, B, C, D, E$ such that $A B C D$ is a parallelogram and $B C E D$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$, and let $\ell$ intersect segment $D C$ and line $B C$ at points $F$ and $G$, respectively. Suppose that $E F=E G=E C$. Prove that $\ell$ is the bisector of angle $D A B$. ['If $C F=C G$, then $\\angle F G C=\\angle G F C$, hence $\\angle G A B=\\angle G F C=\\angle F G C=\\angle F A D$, and $\\ell$ is a bisector.\n\nAssume that $C F\\sqrt{E C^{2}-L C^{2}}=L E .\n$$\nSince quadrilateral $B C E D$ is cyclic, we have $\\angle E D C=\\angle E B C$, so the right triangles $B E L$ and $D E K$ are similar. Then $K E>L E$ implies $D K>B L$, and hence\n$$\nD F=D K-K F>B L-L C=B C=A D .\n$$\nBut triangles $A D F$ and $G C F$ are similar, so we have $1>\\frac{A D}{D F}=\\frac{G C}{C F}$; this contradicts our assumption.\n\nThe case $C F>G C$ is completely similar. We consequently obtain the converse inequalities $K F>L C, K E'] [] Text-only Competition True Theorem proof Geometry Math English 50 Let $A B C$ be a fixed triangle, and let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $B C, C A, A B$, respectively. Let $P$ be a variable point on the circumcircle. Let lines $P A_{1}, P B_{1}, P C_{1}$ meet the circumcircle again at $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Assume that the points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$ are distinct, and lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ form a triangle. Prove that the area of this triangle does not depend on $P$. "[""Let $A_{0}, B_{0}, C_{0}$ be the points of intersection of the lines $A A^{\\prime}, B B^{\\prime}$ and $C C^{\\prime}$ (see Figure). We claim that area $\\left(A_{0} B_{0} C_{0}\\right)=\\frac{1}{2} \\operatorname{area}(A B C)$, hence it is constant.\n\nConsider the inscribed hexagon $A B C C^{\\prime} P A^{\\prime}$. By Pascal's theorem, the points of intersection of its opposite sides (or of their extensions) are collinear. These points are $A B \\cap C^{\\prime} P=C_{1}$, $B C \\cap P A^{\\prime}=A_{1}, C C^{\\prime} \\cap A^{\\prime} A=B_{0}$. So point $B_{0}$ lies on the midline $A_{1} C_{1}$ of triangle $A B C$. Analogously, points $A_{0}$ and $C_{0}$ lie on lines $B_{1} C_{1}$ and $A_{1} B_{1}$, respectively.\n\nLines $A C$ and $A_{1} C_{1}$ are parallel, so triangles $B_{0} C_{0} A_{1}$ and $A C_{0} B_{1}$ are similar; hence we have\n$$\n\\frac{B_{0} C_{0}}{A C_{0}}=\\frac{A_{1} C_{0}}{B_{1} C_{0}}\n$$\nAnalogously, from $B C \\| B_{1} C_{1}$ we obtain\n$$\n\\frac{A_{1} C_{0}}{B_{1} C_{0}}=\\frac{B C_{0}}{A_{0} C_{0}}\n$$\nCombining these equalities, we get\n$$\n\\frac{B_{0} C_{0}}{A C_{0}}=\\frac{B C_{0}}{A_{0} C_{0}}\n$$\nor\n$$\nA_{0} C_{0} \\cdot B_{0} C_{0}=A C_{0} \\cdot B C_{0}\n$$\nHence we have\n\n\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\frac{1}{2} A_{0} C_{0} \\cdot B_{0} C_{0} \\sin \\angle A_{0} C_{0} B_{0}=\\frac{1}{2} A C_{0} \\cdot B C_{0} \\sin \\angle A C_{0} B=\\operatorname{area}\\left(A B C_{0}\\right) \\text {. }\n$$\nSince $C_{0}$ lies on the midline, we have $d\\left(C_{0}, A B\\right)=\\frac{1}{2} d(C, A B)$ (we denote by $d(X, Y Z)$ the distance between point $X$ and line $Y Z)$. Then we obtain\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\operatorname{area}\\left(A B C_{0}\\right)=\\frac{1}{2} A B \\cdot d\\left(C_{0}, A B\\right)=\\frac{1}{4} A B \\cdot d(C, A B)=\\frac{1}{2} \\operatorname{area}(A B C)\n$$"" '. Again, we prove that $\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\frac{1}{2} \\operatorname{area}(A B C)$.\n\nWe can assume that $P$ lies on arc $A C$. Mark a point $L$ on side $A C$ such that $\\angle C B L=$ $\\angle P B A$; then $\\angle L B A=\\angle C B A-\\angle C B L=\\angle C B A-\\angle P B A=\\angle C B P$. Note also that $\\angle B A L=\\angle B A C=\\angle B P C$ and $\\angle L C B=\\angle A P B$. Hence, triangles $B A L$ and $B P C$ are similar, and so are triangles $L C B$ and $A P B$.\n\nAnalogously, mark points $K$ and $M$ respectively on the extensions of sides $C B$ and $A B$ beyond point $B$, such that $\\angle K A B=\\angle C A P$ and $\\angle B C M=\\angle P C A$. For analogous reasons, $\\angle K A C=\\angle B A P$ and $\\angle A C M=\\angle P C B$. Hence $\\triangle A B K \\sim \\triangle A P C \\sim \\triangle M B C, \\triangle A C K \\sim$ $\\triangle A P B$, and $\\triangle M A C \\sim \\triangle B P C$. From these similarities, we have $\\angle C M B=\\angle K A B=\\angle C A P$, while we have seen that $\\angle C A P=\\angle C B P=\\angle L B A$. Hence, $A K\\|B L\\| C M$.\n\n\n\n\n\nLet line $C C^{\\prime}$ intersect $B L$ at point $X$. Note that $\\angle L C X=\\angle A C C^{\\prime}=\\angle A P C^{\\prime}=\\angle A P C_{1}$, and $P C_{1}$ is a median in triangle $A P B$. Since triangles $A P B$ and $L C B$ are similar, $C X$ is a median in triangle $L C B$, and $X$ is a midpoint of $B L$. For the same reason, $A A^{\\prime}$ passes through this midpoint, so $X=B_{0}$. Analogously, $A_{0}$ and $C_{0}$ are the midpoints of $A K$ and $C M$.\n\nNow, from $A A_{0} \\| C C_{0}$, we have\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\operatorname{area}\\left(A C_{0} A_{0}\\right)-\\operatorname{area}\\left(A B_{0} A_{0}\\right)=\\operatorname{area}\\left(A C A_{0}\\right)-\\operatorname{area}\\left(A B_{0} A_{0}\\right)=\\operatorname{area}\\left(A C B_{0}\\right)\n$$\nFinally,\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\operatorname{area}\\left(A C B_{0}\\right)=\\frac{1}{2} B_{0} L \\cdot A C \\sin A L B_{0}=\\frac{1}{4} B L \\cdot A C \\sin A L B=\\frac{1}{2} \\operatorname{area}(A B C)\n$$']" [] Text-only Competition True Theorem proof Geometry Math English 51 "Given an acute triangle $A B C$ with angles $\alpha, \beta$ and $\gamma$ at vertices $A, B$ and $C$, respectively, such that $\beta>\gamma$. Point $I$ is the incenter, and $R$ is the circumradius. Point $D$ is the foot of the altitude from vertex $A$. Point $K$ lies on line $A D$ such that $A K=2 R$, and $D$ separates $A$ and $K$. Finally, lines $D I$ and $K I$ meet sides $A C$ and $B C$ at $E$ and $F$, respectively. Prove that if $I E=I F$ then $\beta \leq 3 \gamma$." "['We first prove that\n$$\n\\angle K I D=\\frac{\\beta-\\gamma}{2} \\tag{1}\n$$\neven without the assumption that $I E=I F$. Then we will show that the statement of the problem is a consequence of this fact.\n\nDenote the circumcenter by $O$. On the circumcircle, let $P$ be the point opposite to $A$, and let the angle bisector $A I$ intersect the circle again at $M$. Since $A K=A P=2 R$, triangle $A K P$ is isosceles. It is known that $\\angle B A D=\\angle C A O$, hence $\\angle D A I=\\angle B A I-\\angle B A D=\\angle C A I-$ $\\angle C A O=\\angle O A I$, and $A M$ is the bisector line in triangle $A K P$. Therefore, points $K$ and $P$ are symmetrical about $A M$, and $\\angle A M K=\\angle A M P=90^{\\circ}$. Thus, $M$ is the midpoint of $K P$, and $A M$ is the perpendicular bisector of $K P$.\n\n\n\nDenote the perpendicular feet of incenter $I$ on lines $B C, A C$, and $A D$ by $A_{1}, B_{1}$, and $T$, respectively. Quadrilateral $D A_{1} I T$ is a rectangle, hence $T D=I A_{1}=I B_{1}$.\n\nDue to the right angles at $T$ and $B_{1}$, quadrilateral $A B_{1} I T$ is cyclic. Hence $\\angle B_{1} T I=$ $\\angle B_{1} A I=\\angle C A M=\\angle B A M=\\angle B P M$ and $\\angle I B_{1} T=\\angle I A T=\\angle M A K=\\angle M A P=$ $\\angle M B P$. Therefore, triangles $B_{1} T I$ and $B P M$ are similar and $\\frac{I T}{I B_{1}}=\\frac{M P}{M B}$.\n\nIt is well-known that $M B=M C=M I$. Then right triangles $I T D$ and $K M I$ are also\n\n\n\nsimilar, because $\\frac{I T}{T D}=\\frac{I T}{I B_{1}}=\\frac{M P}{M B}=\\frac{K M}{M I}$. Hence, $\\angle K I M=\\angle I D T=\\angle I D A$, and\n$$\n\\angle K I D=\\angle M I D-\\angle K I M=(\\angle I A D+\\angle I D A)-\\angle I D A=\\angle I A D .\n$$\nFinally, from the right triangle $A D B$ we can compute\n$$\n\\angle K I D=\\angle I A D=\\angle I A B-\\angle D A B=\\frac{\\alpha}{2}-\\left(90^{\\circ}-\\beta\\right)=\\frac{\\alpha}{2}-\\frac{\\alpha+\\beta+\\gamma}{2}+\\beta=\\frac{\\beta-\\gamma}{2} .\n$$\nNow let us turn to the statement and suppose that $I E=I F$. Since $I A_{1}=I B_{1}$, the right triangles $I E B_{1}$ and $I F A_{1}$ are congruent and $\\angle I E B_{1}=\\angle I F A_{1}$. Since $\\beta>\\gamma, A_{1}$ lies in the interior of segment $C D$ and $F$ lies in the interior of $A_{1} D$. Hence, $\\angle I F C$ is acute. Then two cases are possible depending on the order of points $A, C, B_{1}$ and $E$.\n\n\nIf point $E$ lies between $C$ and $B_{1}$ then $\\angle I F C=\\angle I E A$, hence quadrilateral $C E I F$ is cyclic and $\\angle F C E=180^{\\circ}-\\angle E I F=\\angle K I D$. By (1), in this case we obtain $\\angle F C E=\\gamma=\\angle K I D=$ $\\frac{\\beta-\\gamma}{2}$ and $\\beta=3 \\gamma$.\n\nOtherwise, if point $E$ lies between $A$ and $B_{1}$, quadrilateral $C E I F$ is a deltoid such that $\\angle I E C=\\angle I F C<90^{\\circ}$. Then we have $\\angle F C E>180^{\\circ}-\\angle E I F=\\angle K I D$. Therefore, $\\angle F C E=\\gamma>\\angle K I D=\\frac{\\beta-\\gamma}{2}$ and $\\beta<3 \\gamma$.' '$$\n\\angle K I D=\\frac{\\beta-\\gamma}{2} \\tag{1}\n$$\nWe give a different proof for (1). Then the solution can be finished in the same way as above.\n\nDefine points $M$ and $P$ again; it can be proved in the same way that $A M$ is the perpendicular bisector of $K P$. Let $J$ be the center of the excircle touching side $B C$. It is well-known that points $B, C, I, J$ lie on a circle with center $M$; denote this circle by $\\omega_{1}$.\n\nLet $B^{\\prime}$ be the reflection of point $B$ about the angle bisector $A M$. By the symmetry, $B^{\\prime}$ is the second intersection point of circle $\\omega_{1}$ and line $A C$. Triangles $P B A$ and $K B^{\\prime} A$ are symmetrical\n\n\n\nwith respect to line $A M$, therefore $\\angle K B^{\\prime} A=\\angle P B A=90^{\\circ}$. By the right angles at $D$ and $B^{\\prime}$, points $K, D, B^{\\prime}, C$ are concyclic and\n$$\nA D \\cdot A K=A B^{\\prime} \\cdot A C\n$$\nFrom the cyclic quadrilateral $I J C B^{\\prime}$ we obtain $A B^{\\prime} \\cdot A C=A I \\cdot A J$ as well, therefore\n$$\nA D \\cdot A K=A B^{\\prime} \\cdot A C=A I \\cdot A J\n$$\nand points $I, J, K, D$ are also concyclic. Denote circle $I D K J$ by $\\omega_{2}$.\n\n\n\nLet $N$ be the point on circle $\\omega_{2}$ which is opposite to $K$. Since $\\angle N D K=90^{\\circ}=\\angle C D K$, point $N$ lies on line $B C$. Point $M$, being the center of circle $\\omega_{1}$, is the midpoint of segment $I J$, and $K M$ is perpendicular to $I J$. Therefore, line $K M$ is the perpendicular bisector of $I J$ and hence it passes through $N$.\n\nFrom the cyclic quadrilateral $I D K N$ we obtain\n$$\n\\angle K I D=\\angle K N D=90^{\\circ}-\\angle D K N=90^{\\circ}-\\angle A K M=\\angle M A K=\\frac{\\beta-\\gamma}{2}\n$$']" [] Text-only Competition True Theorem proof Geometry Math English 52 Point $P$ lies on side $A B$ of a convex quadrilateral $A B C D$. Let $\omega$ be the incircle of triangle $C P D$, and let $I$ be its incenter. Suppose that $\omega$ is tangent to the incircles of triangles $A P D$ and $B P C$ at points $K$ and $L$, respectively. Let lines $A C$ and $B D$ meet at $E$, and let lines $A K$ and $B L$ meet at $F$. Prove that points $E, I$, and $F$ are collinear. ['Let $\\Omega$ be the circle tangent to segment $A B$ and to rays $A D$ and $B C$; let $J$ be its center. We prove that points $E$ and $F$ lie on line $I J$.\n\n\n\nDenote the incircles of triangles $A D P$ and $B C P$ by $\\omega_{A}$ and $\\omega_{B}$. Let $h_{1}$ be the homothety with a negative scale taking $\\omega$ to $\\Omega$. Consider this homothety as the composition of two homotheties: one taking $\\omega$ to $\\omega_{A}$ (with a negative scale and center $K$ ), and another one taking $\\omega_{A}$ to $\\Omega$ (with a positive scale and center $A$ ). It is known that in such a case the three centers of homothety are collinear (this theorem is also referred to as the theorem on the three similitude centers). Hence, the center of $h_{1}$ lies on line $A K$. Analogously, it also lies on $B L$, so this center is $F$. Hence, $F$ lies on the line of centers of $\\omega$ and $\\Omega$, i. e. on $I J$ (if $I=J$, then $F=I$ as well, and the claim is obvious).\n\nConsider quadrilateral $A P C D$ and mark the equal segments of tangents to $\\omega$ and $\\omega_{A}$ (see the figure below to the left). Since circles $\\omega$ and $\\omega_{A}$ have a common point of tangency with $P D$, one can easily see that $A D+P C=A P+C D$. So, quadrilateral $A P C D$ is circumscribed; analogously, circumscribed is also quadrilateral $B C D P$. Let $\\Omega_{A}$ and $\\Omega_{B}$ respectively be their incircles.\n\n\n\n\nConsider the homothety $h_{2}$ with a positive scale taking $\\omega$ to $\\Omega$. Consider $h_{2}$ as the composition of two homotheties: taking $\\omega$ to $\\Omega_{A}$ (with a positive scale and center $C$ ), and taking $\\Omega_{A}$ to $\\Omega$ (with a positive scale and center $A$ ), respectively. So the center of $h_{2}$ lies on line $A C$. By analogous reasons, it lies also on $B D$, hence this center is $E$. Thus, $E$ also lies on the line of centers $I J$, and the claim is proved.'] [] Text-only Competition True Theorem proof Geometry Math English 53 Let $b, n>1$ be integers. Suppose that for each $k>1$ there exists an integer $a_{k}$ such that $b-a_{k}^{n}$ is divisible by $k$. Prove that $b=A^{n}$ for some integer $A$. ['Let the prime factorization of $b$ be $b=p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$, where $p_{1}, \\ldots, p_{s}$ are distinct primes. Our goal is to show that all exponents $\\alpha_{i}$ are divisible by $n$, then we can set $A=p_{1}^{\\alpha_{1} / n} \\ldots p_{s}^{\\alpha_{s} / n}$.\n\nApply the condition for $k=b^{2}$. The number $b-a_{k}^{n}$ is divisible by $b^{2}$ and hence, for each $1 \\leq i \\leq s$, it is divisible by $p_{i}^{2 \\alpha_{i}}>p_{i}^{\\alpha_{i}}$ as well. Therefore\n$$\na_{k}^{n} \\equiv b \\equiv 0 \\quad\\left(\\bmod p_{i}^{\\alpha_{i}}\\right)\n$$\nand\n$$\na_{k}^{n} \\equiv b \\not \\equiv 0 \\quad\\left(\\bmod p_{i}^{\\alpha_{i}+1}\\right)\n$$\nwhich implies that the largest power of $p_{i}$ dividing $a_{k}^{n}$ is $p_{i}^{\\alpha_{i}}$. Since $a_{k}^{n}$ is a complete $n$th power, this implies that $\\alpha_{i}$ is divisible by $n$.'] [] Text-only Competition True Theorem proof Number Theory Math English 54 "For every integer $k \geq 2$, prove that $2^{3 k}$ divides the number $$ \left(\begin{array}{c} 2^{k+1} \\ 2^{k} \end{array}\right)-\left(\begin{array}{c} 2^{k} \\ 2^{k-1} \end{array}\right) \tag{1} $$ but $2^{3 k+1}$ does not." ['We use the notation $(2 n-1) ! !=1 \\cdot 3 \\cdots(2 n-1)$ and $(2 n) ! !=2 \\cdot 4 \\cdots(2 n)=2^{n} n !$ for any positive integer $n$. Observe that $(2 n) !=(2 n) ! !(2 n-1) ! !=2^{n} n !(2 n-1) ! !$.\n\nFor any positive integer $n$ we have\n$$\n\\begin{aligned}\n& \\left(\\begin{array}{c}\n4 n \\\\\n2 n\n\\end{array}\\right)=\\frac{(4 n) !}{(2 n) !^{2}}=\\frac{2^{2 n}(2 n) !(4 n-1) ! !}{(2 n) !^{2}}=\\frac{2^{2 n}}{(2 n) !}(4 n-1) ! ! \\\\\n& \\left(\\begin{array}{c}\n2 n \\\\\nn\n\\end{array}\\right)=\\frac{1}{(2 n) !}\\left(\\frac{(2 n) !}{n !}\\right)^{2}=\\frac{1}{(2 n) !}\\left(2^{n}(2 n-1) ! !\\right)^{2}=\\frac{2^{2 n}}{(2 n) !}(2 n-1) ! !^{2} .\n\\end{aligned}\n$$\nThen expression (1) can be rewritten as follows:\n$$\n\\begin{aligned}\n\\left(\\begin{array}{c}\n2^{k+1} \\\\\n2^{k}\n\\end{array}\\right) & -\\left(\\begin{array}{c}\n2^{k} \\\\\n2^{k-1}\n\\end{array}\\right)=\\frac{2^{2^{k}}}{\\left(2^{k}\\right) !}\\left(2^{k+1}-1\\right) ! !-\\frac{2^{2^{k}}}{\\left(2^{k}\\right) !}\\left(2^{k}-1\\right) ! !^{2} \\\\\n& =\\frac{2^{2^{k}}\\left(2^{k}-1\\right) ! !}{\\left(2^{k}\\right) !} \\cdot\\left(\\left(2^{k}+1\\right)\\left(2^{k}+3\\right) \\ldots\\left(2^{k}+2^{k}-1\\right)-\\left(2^{k}-1\\right)\\left(2^{k}-3\\right) \\ldots\\left(2^{k}-2^{k}+1\\right)\\right) .\n\\end{aligned} \\tag{2}\n$$\nWe compute the exponent of 2 in the prime decomposition of each factor (the first one is a rational number but not necessarily an integer; it is not important).\n\nFirst, we show by induction on $n$ that the exponent of 2 in $\\left(2^{n}\\right)$ ! is $2^{n}-1$. The base case $n=1$ is trivial. Suppose that $\\left(2^{n}\\right) !=2^{2^{n}-1}(2 d+1)$ for some integer $d$. Then we have\n$$\n\\left(2^{n+1}\\right) !=2^{2^{n}}\\left(2^{n}\\right) !\\left(2^{n+1}-1\\right) ! !=2^{2^{n}} 2^{2^{n}-1} \\cdot(2 d+1)\\left(2^{n+1}-1\\right) ! !=2^{2^{n+1}-1} \\cdot(2 q+1)\n$$\nfor some integer $q$. This finishes the induction step.\n\nHence, the exponent of 2 in the first factor in $(2)$ is $2^{k}-\\left(2^{k}-1\\right)=1$.\n\nThe second factor in (2) can be considered as the value of the polynomial\n$$\nP(x)=(x+1)(x+3) \\ldots\\left(x+2^{k}-1\\right)-(x-1)(x-3) \\ldots\\left(x-2^{k}+1\\right) .\\tag{3}\n$$\nat $x=2^{k}$. Now we collect some information about $P(x)$.\n\nObserve that $P(-x)=-P(x)$, since $k \\geq 2$. So $P(x)$ is an odd function, and it has nonzero coefficients only at odd powers of $x$. Hence $P(x)=x^{3} Q(x)+c x$, where $Q(x)$ is a polynomial with integer coefficients.\n\nCompute the exponent of 2 in $c$. We have\n$$\n\\begin{aligned}\nc & =2\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}} \\frac{1}{2 i-1}=\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}}\\left(\\frac{1}{2 i-1}+\\frac{1}{2^{k}-2 i+1}\\right) \\\\\n& =\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}} \\frac{2^{k}}{(2 i-1)\\left(2^{k}-2 i+1\\right)}=2^{k} \\sum_{i=1}^{2^{k-1}} \\frac{\\left(2^{k}-1\\right) ! !}{(2 i-1)\\left(2^{k}-2 i+1\\right)}=2^{k} S .\n\\end{aligned}\n$$\n\nFor any integer $i=1, \\ldots, 2^{k-1}$, denote by $a_{2 i-1}$ the residue inverse to $2 i-1$ modulo $2^{k}$. Clearly, when $2 i-1$ runs through all odd residues, so does $a_{2 i-1}$, hence\n$$\n\\begin{gathered}\nS=\\sum_{i=1}^{2^{k-1}} \\frac{\\left(2^{k}-1\\right) ! !}{(2 i-1)\\left(2^{k}-2 i+1\\right)} \\equiv-\\sum_{i=1}^{2^{k-1}} \\frac{\\left(2^{k}-1\\right) ! !}{(2 i-1)^{2}} \\equiv-\\sum_{i=1}^{2^{k-1}}\\left(2^{k}-1\\right) ! ! a_{2 i-1}^{2} \\\\\n=-\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}}(2 i-1)^{2}=-\\left(2^{k}-1\\right) ! ! \\frac{2^{k-1}\\left(2^{2 k}-1\\right)}{3} \\quad\\left(\\bmod 2^{k}\\right) .\n\\end{gathered}\n$$\nTherefore, the exponent of 2 in $S$ is $k-1$, so $c=2^{k} S=2^{2 k-1}(2 t+1)$ for some integer $t$.\n\nFinally we obtain that\n$$\nP\\left(2^{k}\\right)=2^{3 k} Q\\left(2^{k}\\right)+2^{k} c=2^{3 k} Q\\left(2^{k}\\right)+2^{3 k-1}(2 t+1),\n$$\nwhich is divisible exactly by $2^{3 k-1}$. Thus, the exponent of 2 in $(2)$ is $1+(3 k-1)=3 k$.'] [] Text-only Competition True Theorem proof Number Theory Math English 55 Let $k$ be a positive integer. Prove that the number $\left(4 k^{2}-1\right)^{2}$ has a positive divisor of the form $8 k n-1$ if and only if $k$ is even. ['The statement follows from the following fact.\n\nLemma. For arbitrary positive integers $x$ and $y$, the number $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ if and only if $x=y$.\n\nProof. If $x=y$ then $4 x y-1=4 x^{2}-1$ obviously divides $\\left(4 x^{2}-1\\right)^{2}$ so it is sufficient to consider the opposite direction.\n\nCall a pair $(x, y)$ of positive integers bad if $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ but $x \\neq y$. In order to prove that bad pairs do not exist, we present two properties of them which provide an infinite descent.\n\nProperty (i). If $(x, y)$ is a bad pair and $xr$. Then $\\nu_{p}\\left(q p^{k}+r\\right)=\\nu_{p}\\left(q p^{k}\\right)+\\nu_{p}(r)$.\n\nProof. We claim that $\\operatorname{ord}_{p}\\left(q p^{k}+i\\right)=\\operatorname{ord}_{p}(i)$ for all $01$. By the construction of the sequence, $p_{i}^{n_{\\ell_{1}}}$ divides $n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}$; clearly, $p_{i}^{n_{\\ell_{1}}}>n_{\\ell_{1}}$ for all $1 \\leq i \\leq k$. Therefore the Lemma can be applied for $p=p_{i}, k=r=n_{\\ell_{1}}$ and $q p^{k}=n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}$ to obtain\n$$\nf_{i}\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f_{i}\\left(n_{\\ell_{1}}\\right)+f_{i}\\left(n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right) \\quad \\text { for all } 1 \\leq i \\leq k\n$$\nand hence\n$$\nf\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}\\right)+\\ldots+f\\left(n_{\\ell_{m}}\\right)\n$$\nby the induction hypothesis.\n\nNow consider the values $f\\left(n_{1}\\right), f\\left(n_{2}\\right), \\ldots$ There exist finitely many possible values of $f$. Hence, there exists an infinite sequence of indices $\\ell_{1}<\\ell_{2}<\\ldots$ such that $f\\left(n_{\\ell_{1}}\\right)=f\\left(n_{\\ell_{2}}\\right)=\\ldots$ and thus\n$$\nf\\left(n_{\\ell_{m+1}}+n_{\\ell_{m+2}}+\\ldots+n_{\\ell_{m+d}}\\right)=f\\left(n_{\\ell_{m+1}}\\right)+\\ldots+f\\left(n_{\\ell_{m+d}}\\right)=d \\cdot f\\left(n_{\\ell_{1}}\\right)=(\\overline{0}, \\ldots, \\overline{0})\n$$\nfor all $m$. We have found infinitely many suitable numbers.' 'We use the same Lemma and definition of the function $f$.\n\nLet $S=\\{f(n): n \\in \\mathbb{N}\\}$. Obviously, set $S$ is finite. For every $s \\in S$ choose the minimal $n_{s}$ such that $f\\left(n_{s}\\right)=s$. Denote $N=\\max _{s \\in S} n_{s}$. Moreover, let $g$ be an integer such that $p_{i}^{g}>N$ for each $i=1,2, \\ldots, k$. Let $P=\\left(p_{1} p_{2} \\ldots p_{k}\\right)^{g}$.\n\nWe claim that\n$$\n\\{f(n) \\mid n \\in[m P, m P+N]\\}=S \\tag{1}\n$$\nfor every positive integer $m$. In particular, since $(\\overline{0}, \\ldots, \\overline{0})=f(1) \\in S$, it follows that for an arbitrary $m$ there exists $n \\in[m P, m P+N]$ such that $f(n)=(\\overline{0}, \\ldots, \\overline{0})$. So there are infinitely many suitable numbers.\n\nTo prove (1), let $a_{i}=f_{i}(m P)$. Consider all numbers of the form $n_{m, s}=m P+n_{s}$ with $s=\\left(s_{1}, \\ldots, s_{k}\\right) \\in S$ (clearly, all $n_{m, s}$ belong to $[m P, m P+N]$ ). Since $n_{s} \\leq N(c-b)(c+b)>a^{2}$.\n\nNow we turn to the induction step. Let $n>1$ and put $t=\\lfloor n / 2\\rfloor\\frac{n}{2} \\cdot \\frac{9 n}{2} \\geq(n+1)^{2} \\geq i^{2}$,\n\nso this triangle is obtuse. The proof is completed.'] [] Text-only Competition True Theorem proof Algebra Math English 58 "Let $f$ be a function from the set of real numbers to itself that satisfies $$ f(x+y) \leq y f(x)+f(f(x)) \tag{1} $$ for all real numbers $x$ and $y$. Prove that $f(x)=0$ for all $x \leq 0$." "['Substituting $y=t-x$, we rewrite (11) as\n\n$$\nf(t) \\leq t f(x)-x f(x)+f(f(x)) .\n\\tag{2}\n$$\n\nConsider now some real numbers $a, b$ and use (2) with $t=f(a), x=b$ as well as with $t=f(b)$, $x=a$. We get\n\n$$\n\\begin{aligned}\n& f(f(a))-f(f(b)) \\leq f(a) f(b)-b f(b) \\\\\n& f(f(b))-f(f(a)) \\leq f(a) f(b)-a f(a)\n\\end{aligned}\n$$\n\nAdding these two inequalities yields\n\n$$\n2 f(a) f(b) \\geq a f(a)+b f(b)\n$$\n\nNow, substitute $b=2 f(a)$ to obtain $2 f(a) f(b) \\geq a f(a)+2 f(a) f(b)$, or $a f(a) \\leq 0$. So, we get\n\n$$\nf(a) \\geq 0 \\text { for all } a<0\n\\tag{3}\n$$\n\nNow suppose $f(x)>0$ for some real number $x$. From (2) we immediately get that for every $t<\\frac{x f(x)-f(f(x))}{f(x)}$ we have $f(t)<0$. This contradicts (3); therefore\n\n$$\nf(x) \\leq 0 \\text { for all real } x\n\\tag{4}\n$$\n\nand by (3) again we get $f(x)=0$ for all $x<0$.\n\nWe are left to find $f(0)$. Setting $t=x<0$ in (2) we get\n\n$$\n0 \\leq 0-0+f(0)\n$$\n\nso $f(0) \\geq 0$. Combining this with (4) we obtain $f(0)=0$.' 'We will also use the condition of the problem in form (2). For clarity we divide the argument into four steps.\n\n\n\nStep 1. We begin by proving that $f$ attains nonpositive values only. Assume that there exist some real number $z$ with $f(z)>0$. Substituting $x=z$ into (2) and setting $A=f(z)$, $B=-z f(z)-f(f(z))$ we get $f(t) \\leq A t+B$ for all real $t$. Hence, if for any positive real number $t$ we substitute $x=-t, y=t$ into (1), we get\n\n$$\n\\begin{aligned}\nf(0) & \\leq t f(-t)+f(f(-t)) \\leq t(-A t+B)+A f(-t)+B \\\\\n& \\leq-t(A t-B)+A(-A t+B)+B=-A t^{2}-\\left(A^{2}-B\\right) t+(A+1) B .\n\\end{aligned}\n$$\n\nBut surely this cannot be true if we take $t$ to be large enough. This contradiction proves that we have indeed $f(x) \\leq 0$ for all real numbers $x$. Note that for this reason (1) entails\n\n$$\nf(x+y) \\leq y f(x)\n\\tag{5}\n$$\n\nfor all real numbers $x$ and $y$.\n\nStep 2. We proceed by proving that $f$ has at least one zero. If $f(0)=0$, we are done. Otherwise, in view of Step 1 we get $f(0)<0$. Observe that (5) tells us now $f(y) \\leq y f(0)$ for all real numbers $y$. Thus we can specify a positive real number $a$ that is so large that $f(a)^{2}>-f(0)$. Put $b=f(a)$ and substitute $x=b$ and $y=-b$ into (5); we learn $-b^{2}\sqrt{2}$ and $a^{2}+b^{2}+c^{2}=3$. Prove that $$ \frac{a}{(b+c-a)^{2}}+\frac{b}{(c+a-b)^{2}}+\frac{c}{(a+b-c)^{2}} \geq \frac{3}{(a b c)^{2}} \tag{1} $$ Throughout both solutions, we denote the sums of the form $f(a, b, c)+f(b, c, a)+f(c, a, b)$ by $\sum f(a, b, c)$." "[""The condition $b+c>\\sqrt{2}$ implies $b^{2}+c^{2}>1$, so $a^{2}=3-\\left(b^{2}+c^{2}\\right)<2$, i.e. $a<\\sqrt{2}0$, and also $c+a-b>0$ and $a+b-c>0$ for similar reasons.\n\nWe will use the variant of HÖLDER's inequality\n\n$$\n\\frac{x_{1}^{p+1}}{y_{1}^{p}}+\\frac{x_{1}^{p+1}}{y_{1}^{p}}+\\ldots+\\frac{x_{n}^{p+1}}{y_{n}^{p}} \\geq \\frac{\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)^{p+1}}{\\left(y_{1}+y_{2}+\\ldots+y_{n}\\right)^{p}}\n$$\n\nwhich holds for all positive real numbers $p, x_{1}, x_{2}, \\ldots, x_{n}, y_{1}, y_{2}, \\ldots, y_{n}$. Applying it to the left-hand side of (1) with $p=2$ and $n=3$, we get\n\n$$\n\\sum \\frac{a}{(b+c-a)^{2}}=\\sum \\frac{\\left(a^{2}\\right)^{3}}{a^{5}(b+c-a)^{2}} \\geq \\frac{\\left(a^{2}+b^{2}+c^{2}\\right)^{3}}{\\left(\\sum a^{5 / 2}(b+c-a)\\right)^{2}}=\\frac{27}{\\left(\\sum a^{5 / 2}(b+c-a)\\right)^{2}}\n\\tag{2}\n$$\n\nTo estimate the denominator of the right-hand part, we use an instance of ScHUR's inequality, namely\n\n$$\n\\sum a^{3 / 2}(a-b)(a-c) \\geq 0\n$$\n\nwhich can be rewritten as\n\n$$\n\\sum a^{5 / 2}(b+c-a) \\leq a b c(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})\n$$\n\nMoreover, by the inequality between the arithmetic mean and the fourth power mean we also have\n\n$$\n\\left(\\frac{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{3}\\right)^{4} \\leq \\frac{a^{2}+b^{2}+c^{2}}{3}=1\n$$\n\ni.e., $\\sqrt{a}+\\sqrt{b}+\\sqrt{c} \\leq 3$. Hence, (2) yields\n\n$$\n\\sum \\frac{a}{(b+c-a)^{2}} \\geq \\frac{27}{(a b c(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}))^{2}} \\geq \\frac{3}{a^{2} b^{2} c^{2}}\n$$\n\nthus solving the problem."" 'we mention that all the numbers $b+c-a, a+c-b, a+b-c$ are positive. We will use only this restriction and the condition\n\n$$\na^{5}+b^{5}+c^{5} \\geq 3\n\\tag{3}\n$$\n\nwhich is weaker than the given one. Due to the symmetry we may assume that $a \\geq b \\geq c$.\n\nIn view of (3), it suffices to prove the inequality\n\n$$\n\\sum \\frac{a^{3} b^{2} c^{2}}{(b+c-a)^{2}} \\geq \\sum a^{5}\n$$\n\nor, moving all the terms into the left-hand part,\n\n$$\n\\sum \\frac{a^{3}}{(b+c-a)^{2}}\\left((b c)^{2}-(a(b+c-a))^{2}\\right) \\geq 0\n\\tag{4}\n$$\n\nNote that the signs of the expressions $(y z)^{2}-(x(y+z-x))^{2}$ and $y z-x(y+z-x)=(x-y)(x-z)$ are the same for every positive $x, y, z$ satisfying the triangle inequality. So the terms in (4) corresponding to $a$ and $c$ are nonnegative, and hence it is sufficient to prove that the sum of the terms corresponding to $a$ and $b$ is nonnegative. Equivalently, we need the relation\n\n$$\n\\frac{a^{3}}{(b+c-a)^{2}}(a-b)(a-c)(b c+a(b+c-a)) \\geq \\frac{b^{3}}{(a+c-b)^{2}}(a-b)(b-c)(a c+b(a+c-b)) .\n$$\n\nObviously, we have\n\n$$\na^{3} \\geq b^{3} \\geq 0, \\quad 0(a-b)^{2}\n$$\n\nwhich holds since $c>a-b \\geq 0$ and $a+b>a-b \\geq 0$.']" [] Text-only Competition True Theorem proof Algebra Math English 60 Suppose that 1000 students are standing in a circle. Prove that there exists an integer $k$ with $100 \leq k \leq 300$ such that in this circle there exists a contiguous group of $2 k$ students, for which the first half contains the same number of girls as the second half. ['Number the students consecutively from 1 to 1000. Let $a_{i}=1$ if the $i$ th student is a girl, and $a_{i}=0$ otherwise. We expand this notion for all integers $i$ by setting $a_{i+1000}=$ $a_{i-1000}=a_{i}$. Next, let\n\n$$\nS_{k}(i)=a_{i}+a_{i+1}+\\cdots+a_{i+k-1}\n$$\n\nNow the statement of the problem can be reformulated as follows:\n\nThere exist an integer $k$ with $100 \\leq k \\leq 300$ and an index $i$ such that $S_{k}(i)=S_{k}(i+k)$.\n\nAssume now that this statement is false. Choose an index $i$ such that $S_{100}(i)$ attains the maximal possible value. In particular, we have $S_{100}(i-100)-S_{100}(i)<0$ and $S_{100}(i)-S_{100}(i+100)>0$, for if we had an equality, then the statement would hold. This means that the function $S(j)-$ $S(j+100)$ changes sign somewhere on the segment $[i-100, i]$, so there exists some index $j \\in$ $[i-100, i-1]$ such that\n\n$$\nS_{100}(j) \\leq S_{100}(j+100)-1, \\quad \\text { but } \\quad S_{100}(j+1) \\geq S_{100}(j+101)+1\n\\tag{1}\n$$\n\nSubtracting the first inequality from the second one, we get $a_{j+100}-a_{j} \\geq a_{j+200}-a_{j+100}+2$, so\n\n$$\na_{j}=0, \\quad a_{j+100}=1, \\quad a_{j+200}=0\n$$\n\nSubstituting this into the inequalities of (1), we also obtain $S_{99}(j+1) \\leq S_{99}(j+101) \\leq S_{99}(j+1)$, which implies\n\n$$\nS_{99}(j+1)=S_{99}(j+101) .\n\\tag{2}\n$$\n\nNow let $k$ and $\\ell$ be the least positive integers such that $a_{j-k}=1$ and $a_{j+200+\\ell}=1$. By symmetry, we may assume that $k \\geq \\ell$. If $k \\geq 200$ then we have $a_{j}=a_{j-1}=\\cdots=a_{j-199}=0$, so $S_{100}(j-199)=S_{100}(j-99)=0$, which contradicts the initial assumption. Hence $\\ell \\leq k \\leq 199$. Finally, we have\n\n$$\n\\begin{gathered}\nS_{100+\\ell}(j-\\ell+1)=\\left(a_{j-\\ell+1}+\\cdots+a_{j}\\right)+S_{99}(j+1)+a_{j+100}=S_{99}(j+1)+1 \\\\\nS_{100+\\ell}(j+101)=S_{99}(j+101)+\\left(a_{j+200}+\\cdots+a_{j+200+\\ell-1}\\right)+a_{j+200+\\ell}=S_{99}(j+101)+1\n\\end{gathered}\n$$\n\nComparing with (2) we get $S_{100+\\ell}(j-\\ell+1)=S_{100+\\ell}(j+101)$ and $100+\\ell \\leq 299$, which again contradicts our assumption.'] [] Text-only Competition True Theorem proof Combinatorics Math English 61 "Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another point $Q$ of $\mathcal{S}$. The point $Q$ now takes over as the new pivot. This process continues indefinitely, with the pivot always being a point from $\mathcal{S}$. Show that for a suitable $P \in \mathcal{S}$ and a suitable starting line $\ell$ containing $P$, the resulting windmill will visit each point of $\mathcal{S}$ as a pivot infinitely often." ['Give the rotating line an orientation and distinguish its sides as the oranje side and the blue side. Notice that whenever the pivot changes from some point $T$ to another point $U$, after the change, $T$ is on the same side as $U$ was before. Therefore, the number of elements of $\\mathcal{S}$ on the oranje side and the number of those on the blue side remain the same throughout the whole process (except for those moments when the line contains two points).\n\n\nFirst consider the case that $|\\mathcal{S}|=2 n+1$ is odd. We claim that through any point $T \\in \\mathcal{S}$, there is a line that has $n$ points on each side. To see this, choose an oriented line through $T$ containing no other point of $\\mathcal{S}$ and suppose that it has $n+r$ points on its oranje side. If $r=0$ then we have established the claim, so we may assume that $r \\neq 0$. As the line rotates through $180^{\\circ}$ around $T$, the number of points of $\\mathcal{S}$ on its oranje side changes by 1 whenever the line passes through a point; after $180^{\\circ}$, the number of points on the oranje side is $n-r$. Therefore there is an intermediate stage at which the oranje side, and thus also the blue side, contains $n$ points.\n\nNow select the point $P$ arbitrarily, and choose a line through $P$ that has $n$ points of $\\mathcal{S}$ on each side to be the initial state of the windmill. We will show that during a rotation over $180^{\\circ}$, the line of the windmill visits each point of $\\mathcal{S}$ as a pivot. To see this, select any point $T$ of $\\mathcal{S}$ and select a line $\\boldsymbol{\\ell}$ through $T$ that separates $\\mathcal{S}$ into equal halves. The point $T$ is the unique point of $\\mathcal{S}$ through which a line in this direction can separate the points of $\\mathcal{S}$ into equal halves (parallel translation would disturb the balance). Therefore, when the windmill line is parallel to $\\ell$, it must be $\\ell$ itself, and so pass through $T$.\n\nNext suppose that $|\\mathcal{S}|=2 n$. Similarly to the odd case, for every $T \\in \\mathcal{S}$ there is an oriented\n\n\n\nline through $T$ with $n-1$ points on its oranje side and $n$ points on its blue side. Select such an oriented line through an arbitrary $P$ to be the initial state of the windmill.\n\nWe will now show that during a rotation over $360^{\\circ}$, the line of the windmill visits each point of $\\mathcal{S}$ as a pivot. To see this, select any point $T$ of $\\mathcal{S}$ and an oriented line $\\ell$ through $T$ that separates $\\mathcal{S}$ into two subsets with $n-1$ points on its oranje and $n$ points on its blue side. Again, parallel translation would change the numbers of points on the two sides, so when the windmill line is parallel to $\\ell$ with the same orientation, the windmill line must pass through $T$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 62 "Let $n$ be a positive integer and let $W=\ldots x_{-1} x_{0} x_{1} x_{2} \ldots$ be an infinite periodic word consisting of the letters $a$ and $b$. Suppose that the minimal period $N$ of $W$ is greater than $2^{n}$. A finite nonempty word $U$ is said to appear in $W$ if there exist indices $k \leq \ell$ such that $U=x_{k} x_{k+1} \ldots x_{\ell}$. A finite word $U$ is called ubiquitous if the four words $U a, U b, a U$, and $b U$ all appear in $W$. Prove that there are at least $n$ ubiquitous finite nonempty words." ['Throughout the solution, all the words are nonempty. For any word $R$ of length $m$, we call the number of indices $i \\in\\{1,2, \\ldots, N\\}$ for which $R$ coincides with the subword $x_{i+1} x_{i+2} \\ldots x_{i+m}$ of $W$ the multiplicity of $R$ and denote it by $\\mu(R)$. Thus a word $R$ appears in $W$ if and only if $\\mu(R)>0$. Since each occurrence of a word in $W$ is both succeeded by either the letter $a$ or the letter $b$ and similarly preceded by one of those two letters, we have\n\n$$\n\\mu(R)=\\mu(R a)+\\mu(R b)=\\mu(a R)+\\mu(b R)\n\\tag{1}\n$$\n\nfor all words $R$.\n\nWe claim that the condition that $N$ is in fact the minimal period of $W$ guarantees that each word of length $N$ has multiplicity 1 or 0 depending on whether it appears or not. Indeed, if the words $x_{i+1} x_{i+2} \\ldots x_{i+N}$ and $x_{j+1} \\ldots x_{j+N}$ are equal for some $1 \\leq i2^{n}$, at least one of the two words $a$ and $b$ has a multiplicity that is strictly larger than $2^{n-1}$.\n\nFor each $k=0,1, \\ldots, n-1$, let $U_{k}$ be a subword of $W$ whose multiplicity is strictly larger than $2^{k}$ and whose length is maximal subject to this property. Note that such a word exists in view of the two observations made in the two previous paragraphs.\n\nFix some index $k \\in\\{0,1, \\ldots, n-1\\}$. Since the word $U_{k} b$ is longer than $U_{k}$, its multiplicity can be at most $2^{k}$, so in particular $\\mu\\left(U_{k} b\\right)<\\mu\\left(U_{k}\\right)$. Therefore, the word $U_{k} a$ has to appear by (1). For a similar reason, the words $U_{k} b, a U_{k}$, and $b U_{k}$ have to appear as well. Hence, the word $U_{k}$ is ubiquitous. Moreover, if the multiplicity of $U_{k}$ were strictly greater than $2^{k+1}$, then by (11) at least one of the two words $U_{k} a$ and $U_{k} b$ would have multiplicity greater than $2^{k}$ and would thus violate the maximality condition imposed on $U_{k}$.\n\nSo we have $\\mu\\left(U_{0}\\right) \\leq 2<\\mu\\left(U_{1}\\right) \\leq 4<\\ldots \\leq 2^{n-1}<\\mu\\left(U_{n-1}\\right)$, which implies in particular that the words $U_{0}, U_{1}, \\ldots, U_{n-1}$ have to be distinct. As they have been proved to be ubiquitous as well, the problem is solved.'] [] Text-only Competition True Theorem proof Combinatorics Math English 63 Let $A B C$ be an acute triangle. Let $\omega$ be a circle whose center $L$ lies on the side $B C$. Suppose that $\omega$ is tangent to $A B$ at $B^{\prime}$ and to $A C$ at $C^{\prime}$. Suppose also that the circumcenter $O$ of the triangle $A B C$ lies on the shorter arc $B^{\prime} C^{\prime}$ of $\omega$. Prove that the circumcircle of $A B C$ and $\omega$ meet at two points. ['The point $B^{\\prime}$, being the perpendicular foot of $L$, is an interior point of side $A B$. Analogously, $C^{\\prime}$ lies in the interior of $A C$. The point $O$ is located inside the triangle $A B^{\\prime} C^{\\prime}$, hence $\\angle C O B<\\angle C^{\\prime} O B^{\\prime}$.\n\n\n\nLet $\\alpha=\\angle C A B$. The angles $\\angle C A B$ and $\\angle C^{\\prime} O B^{\\prime}$ are inscribed into the two circles with centers $O$ and $L$, respectively, so $\\angle C O B=2 \\angle C A B=2 \\alpha$ and $2 \\angle C^{\\prime} O B^{\\prime}=360^{\\circ}-\\angle C^{\\prime} L B^{\\prime}$. From the kite $A B^{\\prime} L C^{\\prime}$ we have $\\angle C^{\\prime} L B^{\\prime}=180^{\\circ}-\\angle C^{\\prime} A B^{\\prime}=180^{\\circ}-\\alpha$. Combining these, we get\n\n$$\n2 \\alpha=\\angle C O B<\\angle C^{\\prime} O B^{\\prime}=\\frac{360^{\\circ}-\\angle C^{\\prime} L B^{\\prime}}{2}=\\frac{360^{\\circ}-\\left(180^{\\circ}-\\alpha\\right)}{2}=90^{\\circ}+\\frac{\\alpha}{2}\n$$\n\nso\n\n$$\n\\alpha<60^{\\circ}\n$$\n\nLet $O^{\\prime}$ be the reflection of $O$ in the line $B C$. In the quadrilateral $A B O^{\\prime} C$ we have\n\n$$\n\\angle C O^{\\prime} B+\\angle C A B=\\angle C O B+\\angle C A B=2 \\alpha+\\alpha<180^{\\circ},\n$$\n\nso the point $O^{\\prime}$ is outside the circle $A B C$. Hence, $O$ and $O^{\\prime}$ are two points of $\\omega$ such that one of them lies inside the circumcircle, while the other one is located outside. Therefore, the two circles intersect.'] [] Text-only Competition True Theorem proof Geometry Math English 64 "Let $A_{1} A_{2} A_{3} A_{4}$ be a non-cyclic quadrilateral. Let $O_{1}$ and $r_{1}$ be the circumcenter and the circumradius of the triangle $A_{2} A_{3} A_{4}$. Define $O_{2}, O_{3}, O_{4}$ and $r_{2}, r_{3}, r_{4}$ in a similar way. Prove that $$ \frac{1}{O_{1} A_{1}^{2}-r_{1}^{2}}+\frac{1}{O_{2} A_{2}^{2}-r_{2}^{2}}+\frac{1}{O_{3} A_{3}^{2}-r_{3}^{2}}+\frac{1}{O_{4} A_{4}^{2}-r_{4}^{2}}=0 $$" "['Let $M$ be the point of intersection of the diagonals $A_{1} A_{3}$ and $A_{2} A_{4}$. On each diagonal choose a direction and let $x, y, z$, and $w$ be the signed distances from $M$ to the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$, respectively.\n\nLet $\\omega_{1}$ be the circumcircle of the triangle $A_{2} A_{3} A_{4}$ and let $B_{1}$ be the second intersection point of $\\omega_{1}$ and $A_{1} A_{3}$ (thus, $B_{1}=A_{3}$ if and only if $A_{1} A_{3}$ is tangent to $\\omega_{1}$ ). Since the expression $O_{1} A_{1}^{2}-r_{1}^{2}$ is the power of the point $A_{1}$ with respect to $\\omega_{1}$, we get\n\n$$\nO_{1} A_{1}^{2}-r_{1}^{2}=A_{1} B_{1} \\cdot A_{1} A_{3} .\n$$\n\nOn the other hand, from the equality $M B_{1} \\cdot M A_{3}=M A_{2} \\cdot M A_{4}$ we obtain $M B_{1}=y w / z$. Hence, we have\n\n$$\nO_{1} A_{1}^{2}-r_{1}^{2}=\\left(\\frac{y w}{z}-x\\right)(z-x)=\\frac{z-x}{z}(y w-x z) .\n$$\n\nSubstituting the analogous expressions into the sought sum we get\n\n$$\n\\sum_{i=1}^{4} \\frac{1}{O_{i} A_{i}^{2}-r_{i}^{2}}=\\frac{1}{y w-x z}\\left(\\frac{z}{z-x}-\\frac{w}{w-y}+\\frac{x}{x-z}-\\frac{y}{y-w}\\right)=0\n$$\n\nas desired.' 'Introduce a Cartesian coordinate system in the plane. Every circle has an equation of the form $p(x, y)=x^{2}+y^{2}+l(x, y)=0$, where $l(x, y)$ is a polynomial of degree at most 1 . For any point $A=\\left(x_{A}, y_{A}\\right)$ we have $p\\left(x_{A}, y_{A}\\right)=d^{2}-r^{2}$, where $d$ is the distance from $A$ to the center of the circle and $r$ is the radius of the circle.\n\nFor each $i$ in $\\{1,2,3,4\\}$ let $p_{i}(x, y)=x^{2}+y^{2}+l_{i}(x, y)=0$ be the equation of the circle with center $O_{i}$ and radius $r_{i}$ and let $d_{i}$ be the distance from $A_{i}$ to $O_{i}$. Consider the equation\n\n$$\n\\sum_{i=1}^{4} \\frac{p_{i}(x, y)}{d_{i}^{2}-r_{i}^{2}}=1\n\\tag{1}\n$$\n\n\n\nSince the coordinates of the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$ satisfy (1) but these four points do not lie on a circle or on an line, equation (1) defines neither a circle, nor a line. Hence, the equation is an identity and the coefficient of the quadratic term $x^{2}+y^{2}$ also has to be zero, i.e.\n\n$$\n\\sum_{i=1}^{4} \\frac{1}{d_{i}^{2}-r_{i}^{2}}=0\n$$']" [] Text-only Competition True Theorem proof Geometry Math English 65 Let $A B C D$ be a convex quadrilateral whose sides $A D$ and $B C$ are not parallel. Suppose that the circles with diameters $A B$ and $C D$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_{E}$ be the circle through the feet of the perpendiculars from $E$ to the lines $A B, B C$, and $C D$. Let $\omega_{F}$ be the circle through the feet of the perpendiculars from $F$ to the lines $C D, D A$, and $A B$. Prove that the midpoint of the segment $E F$ lies on the line through the two intersection points of $\omega_{E}$ and $\omega_{F}$. ['Denote by $P, Q, R$, and $S$ the projections of $E$ on the lines $D A, A B, B C$, and $C D$ respectively. The points $P$ and $Q$ lie on the circle with diameter $A E$, so $\\angle Q P E=\\angle Q A E$; analogously, $\\angle Q R E=\\angle Q B E$. So $\\angle Q P E+\\angle Q R E=\\angle Q A E+\\angle Q B E=90^{\\circ}$. By similar reasons, we have $\\angle S P E+\\angle S R E=90^{\\circ}$, hence we get $\\angle Q P S+\\angle Q R S=90^{\\circ}+90^{\\circ}=180^{\\circ}$, and the quadrilateral $P Q R S$ is inscribed in $\\omega_{E}$. Analogously, all four projections of $F$ onto the sides of $A B C D$ lie on $\\omega_{F}$.\n\nDenote by $K$ the meeting point of the lines $A D$ and $B C$. Due to the arguments above, there is no loss of generality in assuming that $A$ lies on segment $D K$. Suppose that $\\angle C K D \\geq 90^{\\circ}$; then the circle with diameter $C D$ covers the whole quadrilateral $A B C D$, so the points $E, F$ cannot lie inside this quadrilateral. Hence our assumption is wrong. Therefore, the lines $E P$ and $B C$ intersect at some point $P^{\\prime}$, while the lines $E R$ and $A D$ intersect at some point $R^{\\prime}$.\n\n\n\nFigure 1\n\nWe claim that the points $P^{\\prime}$ and $R^{\\prime}$ also belong to $\\omega_{E}$. Since the points $R, E, Q, B$ are concyclic, $\\angle Q R K=\\angle Q E B=90^{\\circ}-\\angle Q B E=\\angle Q A E=\\angle Q P E$. So $\\angle Q R K=\\angle Q P P^{\\prime}$, which means that the point $P^{\\prime}$ lies on $\\omega_{E}$. Analogously, $R^{\\prime}$ also lies on $\\omega_{E}$.\n\nIn the same manner, denote by $M$ and $N$ the projections of $F$ on the lines $A D$ and $B C$\n\n\n\nrespectively, and let $M^{\\prime}=F M \\cap B C, N^{\\prime}=F N \\cap A D$. By the same arguments, we obtain that the points $M^{\\prime}$ and $N^{\\prime}$ belong to $\\omega_{F}$.\n\n\n\nFigure 2\n\nNow we concentrate on Figure 2, where all unnecessary details are removed. Let $U=N N^{\\prime} \\cap$ $P P^{\\prime}, V=M M^{\\prime} \\cap R R^{\\prime}$. Due to the right angles at $N$ and $P$, the points $N, N^{\\prime}, P, P^{\\prime}$ are concyclic, so $U N \\cdot U N^{\\prime}=U P \\cdot U P^{\\prime}$ which means that $U$ belongs to the radical axis $g$ of the circles $\\omega_{E}$ and $\\omega_{F}$. Analogously, $V$ also belongs to $g$.\n\nFinally, since $E U F V$ is a parallelogram, the radical axis $U V$ of $\\omega_{E}$ and $\\omega_{F}$ bisects $E F$.'] [] Text-only Competition True Theorem proof Geometry Math English 66 Let $A B C$ be an acute triangle with circumcircle $\Omega$. Let $B_{0}$ be the midpoint of $A C$ and let $C_{0}$ be the midpoint of $A B$. Let $D$ be the foot of the altitude from $A$, and let $G$ be the centroid of the triangle $A B C$. Let $\omega$ be a circle through $B_{0}$ and $C_{0}$ that is tangent to the circle $\Omega$ at a point $X \neq A$. Prove that the points $D, G$, and $X$ are collinear. "['If $A B=A C$, then the statement is trivial. So without loss of generality we may assume $A B\n\nDenote by $T$ the second intersection point of $\\Omega$ and the line $D X$. Note that $O$ belongs to $\\Omega_{1}$. Using the circles $\\gamma$ and $\\Omega$, we find $\\angle D A T=\\angle A D X-\\angle A T D=\\frac{1}{2}\\left(360^{\\circ}-\\angle A W X\\right)-\\frac{1}{2} \\angle A O X=$ $180^{\\circ}-\\frac{1}{2}(\\angle A W X+\\angle A O X)=90^{\\circ}$. So, $A D \\perp A T$, and hence $A T \\| B C$. Thus, $A T C B$ is an isosceles trapezoid inscribed in $\\Omega$.\n\nDenote by $A_{0}$ the midpoint of $B C$, and consider the image of $A T C B$ under the homothety $h$ with center $G$ and factor $-\\frac{1}{2}$. We have $h(A)=A_{0}, h(B)=B_{0}$, and $h(C)=C_{0}$. From the\n\n\n\nsymmetry about $B_{0} C_{0}$, we have $\\angle T C B=\\angle C B A=\\angle B_{0} C_{0} A=\\angle D C_{0} B_{0}$. Using $A T \\| D A_{0}$, we conclude $h(T)=D$. Hence the points $D, G$, and $T$ are collinear, and $X$ lies on the same line.' 'We define the points $A_{0}, O$, and $W$ as in the previous solution and we concentrate on the case $A B\n\nTo complete the argument, we note that the homothety centered at $G$ sending the triangle $A B C$ to the triangle $A_{0} B_{0} C_{0}$ maps the altitude $A D$ to the altitude $A_{0} Q$. Therefore it maps $D$ to $Q$, so the points $D, G$, and $Q$ are collinear. Hence $G$ lies on $\\ell$ as well.']" [] Text-only Competition True Theorem proof Geometry Math English 67 Let $A B C$ be a triangle with incenter $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with the lines $A I$ and $B I$, respectively. The chord $D E$ meets $A C$ at a point $F$, and $B C$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $A D$ and the line through $G$ parallel to $B E$. Suppose that the tangents to $\omega$ at $A$ and at $B$ meet at a point $K$. Prove that the three lines $A E, B D$, and $K P$ are either parallel or concurrent. "['Since\n\n$$\n\\angle I A F=\\angle D A C=\\angle B A D=\\angle B E D=\\angle I E F\n$$\n\nthe quadrilateral $A I F E$ is cyclic. Denote its circumcircle by $\\omega_{1}$. Similarly, the quadrilateral $B D G I$ is cyclic; denote its circumcircle by $\\omega_{2}$.\n\nThe line $A E$ is the radical axis of $\\omega$ and $\\omega_{1}$, and the line $B D$ is the radical axis of $\\omega$ and $\\omega_{2}$. Let $t$ be the radical axis of $\\omega_{1}$ and $\\omega_{2}$. These three lines meet at the radical center of the three circles, or they are parallel to each other. We will show that $t$ is in fact the line $P K$.\n\nLet $L$ be the second intersection point of $\\omega_{1}$ and $\\omega_{2}$, so $t=I L$. (If the two circles are tangent to each other then $L=I$ and $t$ is the common tangent.)\n\n\n\nLet the line $t$ meet the circumcircles of the triangles $A B L$ and $F G L$ at $K^{\\prime} \\neq L$ and $P^{\\prime} \\neq L$, respectively. Using oriented angles we have\n\n$$\n\\angle\\left(A B, B K^{\\prime}\\right)=\\angle\\left(A L, L K^{\\prime}\\right)=\\angle(A L, L I)=\\angle(A E, E I)=\\angle(A E, E B)=\\angle(A B, B K),\n$$\n\n\n\nso $B K^{\\prime} \\| B K$. Similarly we have $A K^{\\prime} \\| A K$, and therefore $K^{\\prime}=K$. Next, we have\n\n$$\n\\angle\\left(P^{\\prime} F, F G\\right)=\\angle\\left(P^{\\prime} L, L G\\right)=\\angle(I L, L G)=\\angle(I D, D G)=\\angle(A D, D E)=\\angle(P F, F G),\n$$\n\nhence $P^{\\prime} F \\| P F$ and similarly $P^{\\prime} G \\| P G$. Therefore $P^{\\prime}=P$. This means that $t$ passes through $K$ and $P$, which finishes the proof.' ""Let $M$ be the intersection point of the tangents to $\\omega$ at $D$ and $E$, and let the lines $A E$ and $B D$ meet at $T$; if $A E$ and $B D$ are parallel, then let $T$ be their common ideal point. It is well-known that the points $K$ and $M$ lie on the line $T I$ (as a consequence of PASCAL's theorem, applied to the inscribed degenerate hexagons $A A D B B E$ and $A D D B E E$ ).\n\nThe lines $A D$ and $B E$ are the angle bisectors of the angles $\\angle C A B$ and $\\angle A B C$, respectively, so $D$ and $E$ are the midpoints of the arcs $B C$ and $C A$ of the circle $\\omega$, respectively. Hence, $D M$ is parallel to $B C$ and $E M$ is parallel to $A C$.\n\nApply PASCAL's theorem to the degenerate hexagon $C A D D E B$. By the theorem, the points $C A \\cap D E=F, A D \\cap E B=I$ and the common ideal point of lines $D M$ and $B C$ are collinear, therefore $F I$ is parallel to $B C$ and $D M$. Analogously, the line $G I$ is parallel to $A C$ and $E M$.\n\n\n\nNow consider the homothety with scale factor $-\\frac{F G}{E D}$ which takes $E$ to $G$ and $D$ to $F$. Since the triangles $E D M$ and $G F I$ have parallel sides, the homothety takes $M$ to $I$. Similarly, since the triangles $D E I$ and $F G P$ have parallel sides, the homothety takes $I$ to $P$. Hence, the points $M, I, P$ and the homothety center $H$ must lie on the same line. Therefore, the point $P$ also lies on the line $T K I M$.""]" [] Text-only Competition True Theorem proof Geometry Math English 68 Let $A B C$ be a triangle with $A B=A C$, and let $D$ be the midpoint of $A C$. The angle bisector of $\angle B A C$ intersects the circle through $D, B$, and $C$ in a point $E$ inside the triangle $A B C$. The line $B D$ intersects the circle through $A, E$, and $B$ in two points $B$ and $F$. The lines $A F$ and $B E$ meet at a point $I$, and the lines $C I$ and $B D$ meet at a point $K$. Show that $I$ is the incenter of triangle $K A B$. "[""Let $D^{\\prime}$ be the midpoint of the segment $A B$, and let $M$ be the midpoint of $B C$. By symmetry at line $A M$, the point $D^{\\prime}$ has to lie on the circle $B C D$. Since the $\\operatorname{arcs} D^{\\prime} E$ and $E D$ of that circle are equal, we have $\\angle A B I=\\angle D^{\\prime} B E=\\angle E B D=I B K$, so $I$ lies on the angle bisector of $\\angle A B K$. For this reason it suffices to prove in the sequel that the ray $A I$ bisects the angle $\\angle B A K$.\n\nFrom\n\n$$\n\\angle D F A=180^{\\circ}-\\angle B F A=180^{\\circ}-\\angle B E A=\\angle M E B=\\frac{1}{2} \\angle C E B=\\frac{1}{2} \\angle C D B\n$$\n\nwe derive $\\angle D F A=\\angle D A F$ so the triangle $A F D$ is isosceles with $A D=D F$.\n\n\n\nApplying Menelaus's theorem to the triangle $A D F$ with respect to the line $C I K$, and applying the angle bisector theorem to the triangle $A B F$, we infer\n\n$$\n1=\\frac{A C}{C D} \\cdot \\frac{D K}{K F} \\cdot \\frac{F I}{I A}=2 \\cdot \\frac{D K}{K F} \\cdot \\frac{B F}{A B}=2 \\cdot \\frac{D K}{K F} \\cdot \\frac{B F}{2 \\cdot A D}=\\frac{D K}{K F} \\cdot \\frac{B F}{A D}\n$$\n\nand therefore\n\n$$\n\\frac{B D}{A D}=\\frac{B F+F D}{A D}=\\frac{B F}{A D}+1=\\frac{K F}{D K}+1=\\frac{D F}{D K}=\\frac{A D}{D K}\n$$\n\n\n\nIt follows that the triangles $A D K$ and $B D A$ are similar, hence $\\angle D A K=\\angle A B D$. Then\n\n$$\n\\angle I A B=\\angle A F D-\\angle A B D=\\angle D A F-\\angle D A K=\\angle K A I\n$$\n\nshows that the point $K$ is indeed lying on the angle bisector of $\\angle B A K$."" 'It can be shown in the same way as in the first solution that $I$ lies on the angle bisector of $\\angle A B K$. Here we restrict ourselves to proving that $K I$ bisects $\\angle A K B$.\n\n\n\nDenote the circumcircle of triangle $B C D$ and its center by $\\omega_{1}$ and by $O_{1}$, respectively. Since the quadrilateral $A B F E$ is cyclic, we have $\\angle D F E=\\angle B A E=\\angle D A E$. By the same reason, we have $\\angle E A F=\\angle E B F=\\angle A B E=\\angle A F E$. Therefore $\\angle D A F=\\angle D F A$, and hence $D F=D A=D C$. So triangle $A F C$ is inscribed in a circle $\\omega_{2}$ with center $D$.\n\nDenote the circumcircle of triangle $A B D$ by $\\omega_{3}$, and let its center be $O_{3}$. Since the arcs $B E$ and $E C$ of circle $\\omega_{1}$ are equal, and the triangles $A D E$ and $F D E$ are congruent, we have $\\angle A O_{1} B=2 \\angle B D E=\\angle B D A$, so $O_{1}$ lies on $\\omega_{3}$. Hence $\\angle O_{3} O_{1} D=\\angle O_{3} D O_{1}$.\n\nThe line $B D$ is the radical axis of $\\omega_{1}$ and $\\omega_{3}$. Point $C$ belongs to the radical axis of $\\omega_{1}$ and $\\omega_{2}$, and $I$ also belongs to it since $A I \\cdot I F=B I \\cdot I E$. Hence $K=B D \\cap C I$ is the radical center of $\\omega_{1}$, $\\omega_{2}$, and $\\omega_{3}$, and $A K$ is the radical axis of $\\omega_{2}$ and $\\omega_{3}$. Now, the radical axes $A K, B K$ and $I K$ are perpendicular to the central lines $O_{3} D, O_{3} O_{1}$ and $O_{1} D$, respectively. By $\\angle O_{3} O_{1} D=\\angle O_{3} D O_{1}$, we get that $K I$ is the angle bisector of $\\angle A K B$.' 'Again, let $M$ be the midpoint of $B C$. As in the previous solutions, we can deduce $\\angle A B I=\\angle I B K$. We show that the point $I$ lies on the angle bisector of $\\angle K A B$.\n\nLet $G$ be the intersection point of the circles $A F C$ and $B C D$, different from $C$. The lines\n\n\n\n$C G, A F$, and $B E$ are the radical axes of the three circles $A G F C, C D B$, and $A B F E$, so $I=A F \\cap B E$ is the radical center of the three circles and $C G$ also passes through $I$.\n\n\n\nThe angle between line $D E$ and the tangent to the circle $B C D$ at $E$ is equal to $\\angle E B D=$ $\\angle E A F=\\angle A B E=\\angle A F E$. As the tangent at $E$ is perpendicular to $A M$, the line $D E$ is perpendicular to $A F$. The triangle $A F E$ is isosceles, so $D E$ is the perpendicular bisector of $A F$ and thus $A D=D F$. Hence, the point $D$ is the center of the circle $A F C$, and this circle passes through $M$ as well since $\\angle A M C=90^{\\circ}$.\n\nLet $B^{\\prime}$ be the reflection of $B$ in the point $D$, so $A B C B^{\\prime}$ is a parallelogram. Since $D C=D G$ we have $\\angle G C D=\\angle D B C=\\angle K B^{\\prime} A$. Hence, the quadrilateral $A K C B^{\\prime}$ is cyclic and thus $\\angle C A K=\\angle C B^{\\prime} K=\\angle A B D=2 \\angle M A I$. Then\n\n$$\n\\angle I A B=\\angle M A B-\\angle M A I=\\frac{1}{2} \\angle C A B-\\frac{1}{2} \\angle C A K=\\frac{1}{2} \\angle K A B\n$$\n\nand therefore $A I$ is the angle bisector of $\\angle K A B$.']" [] Text-only Competition True Theorem proof Geometry Math English 69 Let $A B C D E F$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with center $O$. Suppose that the circumcircle of triangle $A C E$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $C D$. Suppose that the perpendicular from $B$ to $D F$ intersects the line $E O$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $D E$. Prove that $D J=D L$. "['Since $\\omega$ and the circumcircle of triangle $A C E$ are concentric, the tangents from $A$, $C$, and $E$ to $\\omega$ have equal lengths; that means that $A B=B C, C D=D E$, and $E F=F A$. Moreover, we have $\\angle B C D=\\angle D E F=\\angle F A B$.\n\n\n\nConsider the rotation around point $D$ mapping $C$ to $E$; let $B^{\\prime}$ and $L^{\\prime}$ be the images of the points $B$ and $J$, respectively, under this rotation. Then one has $D J=D L^{\\prime}$ and $B^{\\prime} L^{\\prime} \\perp D E$; moreover, the triangles $B^{\\prime} E D$ and $B C D$ are congruent. Since $\\angle D E O<90^{\\circ}$, the lines $E O$ and $B^{\\prime} L^{\\prime}$ intersect at some point $K^{\\prime}$. We intend to prove that $K^{\\prime} B \\perp D F$; this would imply $K=K^{\\prime}$, therefore $L=L^{\\prime}$, which proves the problem statement.\n\nAnalogously, consider the rotation around $F$ mapping $A$ to $E$; let $B^{\\prime \\prime}$ be the image of $B$ under this rotation. Then the triangles $F A B$ and $F E B^{\\prime \\prime}$ are congruent. We have $E B^{\\prime \\prime}=A B=B C=$ $E B^{\\prime}$ and $\\angle F E B^{\\prime \\prime}=\\angle F A B=\\angle B C D=\\angle D E B^{\\prime}$, so the points $B^{\\prime}$ and $B^{\\prime \\prime}$ are symmetrical with respect to the angle bisector $E O$ of $\\angle D E F$. So, from $K^{\\prime} B^{\\prime} \\perp D E$ we get $K^{\\prime} B^{\\prime \\prime} \\perp E F$. From these two relations we obtain\n\n$$\nK^{\\prime} D^{2}-K^{\\prime} E^{2}=B^{\\prime} D^{2}-B^{\\prime} E^{2} \\text { and } K^{\\prime} E^{2}-K^{\\prime} F^{2}=B^{\\prime \\prime} E^{2}-B^{\\prime \\prime} F^{2} \\text {. }\n$$\n\nAdding these equalities and taking into account that $B^{\\prime} E=B^{\\prime \\prime} E$ we obtain\n\n$$\nK^{\\prime} D^{2}-K^{\\prime} F^{2}=B^{\\prime} D^{2}-B^{\\prime \\prime} F^{2}=B D^{2}-B F^{2},\n$$\n\n\n\nwhich means exactly that $K^{\\prime} B \\perp D F$.' 'Let us denote the points of tangency of $A B, B C, C D, D E, E F$, and $F A$ to $\\omega$ by $R, S, T, U, V$, and $W$, respectively. As in the previous solution, we mention that $A R=$ $A W=C S=C T=E U=E V$.\n\nThe reflection in the line $B O$ maps $R$ to $S$, therefore $A$ to $C$ and thus also $W$ to $T$. Hence, both lines $R S$ and $W T$ are perpendicular to $O B$, therefore they are parallel. On the other hand, the lines $U V$ and $W T$ are not parallel, since otherwise the hexagon $A B C D E F$ is symmetric with respect to the line $B O$ and the lines defining the point $K$ coincide, which contradicts the conditions of the problem. Therefore we can consider the intersection point $Z$ of $U V$ and $W T$.\n\n\n\nNext, we recall a well-known fact that the points $D, F, Z$ are collinear. Actually, $D$ is the pole of the line $U T, F$ is the pole of $V W$, and $Z=T W \\cap U V$; so all these points belong to the polar line of $T U \\cap V W$.\n\n\n\nNow, we put $O$ into the origin, and identify each point (say $X$ ) with the vector $\\overrightarrow{O X}$. So, from now on all the products of points refer to the scalar products of the corresponding vectors.\n\nSince $O K \\perp U Z$ and $O B \\perp T Z$, we have $K \\cdot(Z-U)=0=B \\cdot(Z-T)$. Next, the condition $B K \\perp D Z$ can be written as $K \\cdot(D-Z)=B \\cdot(D-Z)$. Adding these two equalities we get\n\n$$\nK \\cdot(D-U)=B \\cdot(D-T)\n$$\n\nBy symmetry, we have $D \\cdot(D-U)=D \\cdot(D-T)$. Subtracting this from the previous equation, we obtain $(K-D) \\cdot(D-U)=(B-D) \\cdot(D-T)$ and rewrite it in vector form as\n\n$$\n\\overrightarrow{D K} \\cdot \\overrightarrow{U D}=\\overrightarrow{D B} \\cdot \\overrightarrow{T D}\n$$\n\nFinally, projecting the vectors $\\overrightarrow{D K}$ and $\\overrightarrow{D B}$ onto the lines $U D$ and $T D$ respectively, we can rewrite this equality in terms of segment lengths as $D L \\cdot U D=D J \\cdot T D$, thus $D L=D J$.']" [] Text-only Competition True Theorem proof Geometry Math English 70 "Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is tangent to the circle $\omega$. To avoid a large case distinction, we will use the notion of oriented angles. Namely, for two lines $\ell$ and $m$, we denote by $\angle(\ell, m)$ the angle by which one may rotate $\ell$ anticlockwise to obtain a line parallel to $m$. Thus, all oriented angles are considered modulo $180^{\circ}$. ![]" "[""Denote by $T$ the point of tangency of $t$ and $\\omega$. Let $A^{\\prime}=t_{b} \\cap t_{c}, B^{\\prime}=t_{a} \\cap t_{c}$, $C^{\\prime}=t_{a} \\cap t_{b}$. Introduce the point $A^{\\prime \\prime}$ on $\\omega$ such that $T A=A A^{\\prime \\prime}\\left(A^{\\prime \\prime} \\neq T\\right.$ unless $T A$ is a diameter). Define the points $B^{\\prime \\prime}$ and $C^{\\prime \\prime}$ in a similar way.\n\nSince the points $C$ and $B$ are the midpoints of arcs $T C^{\\prime \\prime}$ and $T B^{\\prime \\prime}$, respectively, we have\n\n$$\n\\begin{aligned}\n\\angle\\left(t, B^{\\prime \\prime} C^{\\prime \\prime}\\right) & =\\angle\\left(t, T C^{\\prime \\prime}\\right)+\\angle\\left(T C^{\\prime \\prime}, B^{\\prime \\prime} C^{\\prime \\prime}\\right)=2 \\angle(t, T C)+2 \\angle\\left(T C^{\\prime \\prime}, B C^{\\prime \\prime}\\right) \\\\\n& =2(\\angle(t, T C)+\\angle(T C, B C))=2 \\angle(t, B C)=\\angle\\left(t, t_{a}\\right) .\n\\end{aligned}\n$$\n\nIt follows that $t_{a}$ and $B^{\\prime \\prime} C^{\\prime \\prime}$ are parallel. Similarly, $t_{b} \\| A^{\\prime \\prime} C^{\\prime \\prime}$ and $t_{c} \\| A^{\\prime \\prime} B^{\\prime \\prime}$. Thus, either the triangles $A^{\\prime} B^{\\prime} C^{\\prime}$ and $A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime}$ are homothetic, or they are translates of each other. Now we will prove that they are in fact homothetic, and that the center $K$ of the homothety belongs\n\n\n\nto $\\omega$. It would then follow that their circumcircles are also homothetic with respect to $K$ and are therefore tangent at this point, as desired.\n\nWe need the two following claims.\n\nClaim 1. The point of intersection $X$ of the lines $B^{\\prime \\prime} C$ and $B C^{\\prime \\prime}$ lies on $t_{a}$.\n\nProof. Actually, the points $X$ and $T$ are symmetric about the line $B C$, since the lines $C T$ and $C B^{\\prime \\prime}$ are symmetric about this line, as are the lines $B T$ and $B C^{\\prime \\prime}$.\n\nClaim 2. The point of intersection $I$ of the lines $B B^{\\prime}$ and $C C^{\\prime}$ lies on the circle $\\omega$.\n\nProof. We consider the case that $t$ is not parallel to the sides of $A B C$; the other cases may be regarded as limit cases. Let $D=t \\cap B C, E=t \\cap A C$, and $F=t \\cap A B$.\n\nDue to symmetry, the line $D B$ is one of the angle bisectors of the lines $B^{\\prime} D$ and $F D$; analogously, the line $F B$ is one of the angle bisectors of the lines $B^{\\prime} F$ and $D F$. So $B$ is either the incenter or one of the excenters of the triangle $B^{\\prime} D F$. In any case we have $\\angle(B D, D F)+\\angle(D F, F B)+$ $\\angle\\left(B^{\\prime} B, B^{\\prime} D\\right)=90^{\\circ}$, so\n\n$$\n\\angle\\left(B^{\\prime} B, B^{\\prime} C^{\\prime}\\right)=\\angle\\left(B^{\\prime} B, B^{\\prime} D\\right)=90^{\\circ}-\\angle(B C, D F)-\\angle(D F, B A)=90^{\\circ}-\\angle(B C, A B) .\n$$\n\nAnalogously, we get $\\angle\\left(C^{\\prime} C, B^{\\prime} C^{\\prime}\\right)=90^{\\circ}-\\angle(B C, A C)$. Hence,\n\n$$\n\\angle(B I, C I)=\\angle\\left(B^{\\prime} B, B^{\\prime} C^{\\prime}\\right)+\\angle\\left(B^{\\prime} C^{\\prime}, C^{\\prime} C\\right)=\\angle(B C, A C)-\\angle(B C, A B)=\\angle(A B, A C),\n$$\n\nwhich means exactly that the points $A, B, I, C$ are concyclic.\n\nNow we can complete the proof. Let $K$ be the second intersection point of $B^{\\prime} B^{\\prime \\prime}$ and $\\omega$. Applying PASCAL's theorem to hexagon $K B^{\\prime \\prime} C I B C^{\\prime \\prime}$ we get that the points $B^{\\prime}=K B^{\\prime \\prime} \\cap I B$ and $X=B^{\\prime \\prime} C \\cap B C^{\\prime \\prime}$ are collinear with the intersection point $S$ of $C I$ and $C^{\\prime \\prime} K$. So $S=$ $C I \\cap B^{\\prime} X=C^{\\prime}$, and the points $C^{\\prime}, C^{\\prime \\prime}, K$ are collinear. Thus $K$ is the intersection point of $B^{\\prime} B^{\\prime \\prime}$ and $C^{\\prime} C^{\\prime \\prime}$ which implies that $K$ is the center of the homothety mapping $A^{\\prime} B^{\\prime} C^{\\prime}$ to $A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime}$, and it belongs to $\\omega$."" ""Define the points $T, A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $B C, C A$, and $A B$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $A B C$, therefore the points $X, Y, Z$ are also collinear. Moreover, we have $X \\in B^{\\prime} C^{\\prime}, Y \\in C^{\\prime} A^{\\prime}$, $Z \\in A^{\\prime} B^{\\prime}$.\n\nDenote $\\alpha=\\angle(t, T C)=\\angle(B T, B C)$. Using the symmetry in the lines $A C$ and $B C$, we get\n\n$$\n\\angle(B C, B X)=\\angle(B T, B C)=\\alpha \\quad \\text { and } \\quad \\angle\\left(X C, X C^{\\prime}\\right)=\\angle(t, T C)=\\angle\\left(Y C, Y C^{\\prime}\\right)=\\alpha \\text {. }\n$$\n\nSince $\\angle\\left(X C, X C^{\\prime}\\right)=\\angle\\left(Y C, Y C^{\\prime}\\right)$, the points $X, Y, C, C^{\\prime}$ lie on some circle $\\omega_{c}$. Define the circles $\\omega_{a}$ and $\\omega_{b}$ analogously. Let $\\omega^{\\prime}$ be the circumcircle of triangle $A^{\\prime} B^{\\prime} C^{\\prime}$.\n\n\n\nNow, applying MiqueL's theorem to the four lines $A^{\\prime} B^{\\prime}, A^{\\prime} C^{\\prime}, B^{\\prime} C^{\\prime}$, and $X Y$, we obtain that the circles $\\omega^{\\prime}, \\omega_{a}, \\omega_{b}, \\omega_{c}$ intersect at some point $K$. We will show that $K$ lies on $\\omega$, and that the tangent lines to $\\omega$ and $\\omega^{\\prime}$ at this point coincide; this implies the problem statement.\n\nDue to symmetry, we have $X B=T B=Z B$, so the point $B$ is the midpoint of one of the $\\operatorname{arcs} X Z$ of circle $\\omega_{b}$. Therefore $\\angle(K B, K X)=\\angle(X Z, X B)$. Analogously, $\\angle(K X, K C)=$ $\\angle(X C, X Y)$. Adding these equalities and using the symmetry in the line $B C$ we get\n\n$$\n\\angle(K B, K C)=\\angle(X Z, X B)+\\angle(X C, X Z)=\\angle(X C, X B)=\\angle(T B, T C) .\n$$\n\nTherefore, $K$ lies on $\\omega$.\n\nNext, let $k$ be the tangent line to $\\omega$ at $K$. We have\n\n$$\n\\begin{aligned}\n\\angle\\left(k, K C^{\\prime}\\right) & =\\angle(k, K C)+\\angle\\left(K C, K C^{\\prime}\\right)=\\angle(K B, B C)+\\angle\\left(X C, X C^{\\prime}\\right) \\\\\n& =(\\angle(K B, B X)-\\angle(B C, B X))+\\alpha=\\angle\\left(K B^{\\prime}, B^{\\prime} X\\right)-\\alpha+\\alpha=\\angle\\left(K B^{\\prime}, B^{\\prime} C^{\\prime}\\right),\n\\end{aligned}\n$$\n\nwhich means exactly that $k$ is tangent to $\\omega^{\\prime}$.\n\n""]" [] Text-only Competition True Theorem proof Geometry Math English 71 For any integer $d>0$, let $f(d)$ be the smallest positive integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16$, and $f(6)=12$ ). Prove that for every integer $k \geq 0$ the number $f\left(2^{k}\right)$ divides $f\left(2^{k+1}\right)$. ['For any positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. Let $n=\\prod_{p} p^{a(p)}$ be the prime factorization of $n$ where $p$ ranges over the prime numbers, the integers $a(p)$ are nonnegative and all but finitely many $a(p)$ are zero. Then we have $d(n)=\\prod_{p}(a(p)+1)$. Thus, $d(n)$ is a power of 2 if and only if for every prime $p$ there is a nonnegative integer $b(p)$ with $a(p)=2^{b(p)}-1=1+2+2^{2}+\\cdots+2^{b(p)-1}$. We then have\n\n$$\nn=\\prod_{p} \\prod_{i=0}^{b(p)-1} p^{2^{i}}, \\quad \\text { and } \\quad d(n)=2^{k} \\quad \\text { with } \\quad k=\\sum_{p} b(p)\n$$\n\nLet $\\mathcal{S}$ be the set of all numbers of the form $p^{2^{r}}$ with $p$ prime and $r$ a nonnegative integer. Then we deduce that $d(n)$ is a power of 2 if and only if $n$ is the product of the elements of some finite subset $\\mathcal{T}$ of $\\mathcal{S}$ that satisfies the following condition: for all $t \\in \\mathcal{T}$ and $s \\in \\mathcal{S}$ with $s \\mid t$ we have $s \\in \\mathcal{T}$. Moreover, if $d(n)=2^{k}$ then the corresponding set $\\mathcal{T}$ has $k$ elements.\n\nNote that the $\\operatorname{set} \\mathcal{T}_{k}$ consisting of the smallest $k$ elements from $\\mathcal{S}$ obviously satisfies the condition above. Thus, given $k$, the smallest $n$ with $d(n)=2^{k}$ is the product of the elements of $\\mathcal{T}_{k}$. This $n$ is $f\\left(2^{k}\\right)$. Since obviously $\\mathcal{T}_{k} \\subset \\mathcal{T}_{k+1}$, it follows that $f\\left(2^{k}\\right) \\mid f\\left(2^{k+1}\\right)$.'] [] Text-only Competition True Theorem proof Number Theory Math English 72 Consider a polynomial $P(x)=\left(x+d_{1}\right)\left(x+d_{2}\right) \cdot \ldots \cdot\left(x+d_{9}\right)$, where $d_{1}, d_{2}, \ldots, d_{9}$ are nine distinct integers. Prove that there exists an integer $N$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20 . "['Note that the statement of the problem is invariant under translations of $x$; hence without loss of generality we may suppose that the numbers $d_{1}, d_{2}, \\ldots, d_{9}$ are positive.\n\nThe key observation is that there are only eight primes below 20, while $P(x)$ involves more than eight factors.\n\nWe shall prove that $N=d^{8}$ satisfies the desired property, where $d=\\max \\left\\{d_{1}, d_{2}, \\ldots, d_{9}\\right\\}$. Suppose for the sake of contradiction that there is some integer $x \\geq N$ such that $P(x)$ is composed of primes below 20 only. Then for every index $i \\in\\{1,2, \\ldots, 9\\}$ the number $x+d_{i}$ can be expressed as product of powers of the first 8 primes.\n\nSince $x+d_{i}>x \\geq d^{8}$ there is some prime power $f_{i}>d$ that divides $x+d_{i}$. Invoking the pigeonhole principle we see that there are two distinct indices $i$ and $j$ such that $f_{i}$ and $f_{j}$ are powers of the same prime number. For reasons of symmetry, we may suppose that $f_{i} \\leq f_{j}$. Now both of the numbers $x+d_{i}$ and $x+d_{j}$ are divisible by $f_{i}$ and hence so is their difference $d_{i}-d_{j}$. But as\n\n$$\n0<\\left|d_{i}-d_{j}\\right| \\leq \\max \\left(d_{i}, d_{j}\\right) \\leq dD_{i}$ the numerator of the fraction we thereby get cannot be 1 , and hence it has to be divisible by some prime number $p_{i}<20$.\n\nBy the pigeonhole principle, there are a prime number $p$ and two distinct indices $i$ and $j$ such that $p_{i}=p_{j}=p$. Let $p^{\\alpha_{i}}$ and $p^{\\alpha_{j}}$ be the greatest powers of $p$ dividing $x+d_{i}$ and $x+d_{j}$, respectively. Due to symmetry we may suppose $\\alpha_{i} \\leq \\alpha_{j}$. But now $p^{\\alpha_{i}}$ divides $d_{i}-d_{j}$ and hence also $D_{i}$, which means that all occurrences of $p$ in the numerator of the fraction $\\left(x+d_{i}\\right) / D_{i}$ cancel out, contrary to the choice of $p=p_{i}$. This contradiction proves our claim.' 'Given a nonzero integer $N$ as well as a prime number $p$ we write $v_{p}(N)$ for the exponent with which $p$ occurs in the prime factorization of $|N|$.\n\nEvidently, if the statement of the problem were not true, then there would exist an infinite sequence $\\left(x_{n}\\right)$ of positive integers tending to infinity such that for each $n \\in \\mathbb{Z}_{+}$the integer $P\\left(x_{n}\\right)$ is not divisible by any prime number $>20$. Observe that the numbers $-d_{1},-d_{2}, \\ldots,-d_{9}$ do not appear in this sequence.\n\nNow clearly there exists a prime $p_{1}<20$ for which the sequence $v_{p_{1}}\\left(x_{n}+d_{1}\\right)$ is not bounded; thinning out the sequence $\\left(x_{n}\\right)$ if necessary we may even suppose that\n\n$$\nv_{p_{1}}\\left(x_{n}+d_{1}\\right) \\longrightarrow \\infty\n$$\n\nRepeating this argument eight more times we may similarly choose primes $p_{2}, \\ldots, p_{9}<20$ and suppose that our sequence $\\left(x_{n}\\right)$ has been thinned out to such an extent that $v_{p_{i}}\\left(x_{n}+d_{i}\\right) \\longrightarrow \\infty$ holds for $i=2, \\ldots, 9$ as well. In view of the pigeonhole principle, there are distinct indices $i$ and $j$ as well as a prime $p<20$ such that $p_{i}=p_{j}=p$. Setting $k=v_{p}\\left(d_{i}-d_{j}\\right)$ there now has to be some $n$ for which both $v_{p}\\left(x_{n}+d_{i}\\right)$ and $v_{p}\\left(x_{n}+d_{j}\\right)$ are greater than $k$. But now the numbers $x_{n}+d_{i}$ and $x_{n}+d_{j}$ are divisible by $p^{k+1}$ whilst their difference $d_{i}-d_{j}$ is not - a contradiction.']" [] Text-only Competition True Theorem proof Number Theory Math English 73 Let $f$ be a function from the set of integers to the set of positive integers. Suppose that for any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove that for all integers $m, n$ with $f(m) \leq f(n)$ the number $f(n)$ is divisible by $f(m)$. "['Suppose that $x$ and $y$ are two integers with $f(x)0\n$$\n\nso $f(x-y) \\leq f(y)-f(x)f(1)$. Note that such a number exists due to the symmetry of $f$ obtained in Claim 2 .\n\n\n\nClaim 3. $f(n) \\neq f(1)$ if and only if $a \\mid n$.\n\nProof. Since $f(1)=\\cdots=f(a-1)0$, so $f(n+a) \\leq$ $f(a)-f(n)0$ this implies $M \\leq 3^{P(m+a x-b y)}-1$. But $P(m+a x-b y)$ is listed among $P(1), P(2), \\ldots, P(d)$, so\n\n$$\nM<3^{P(m+a x-b y)} \\leq 3^{\\max \\{P(1), P(2), \\ldots, P(d)\\}}\n$$\n\n\n\nwhich contradicts (1).""]" [] Text-only Competition True Theorem proof Number Theory Math English 75 "Let $p$ be an odd prime number. For every integer $a$, define the number $$ S_{a}=\frac{a}{1}+\frac{a^{2}}{2}+\cdots+\frac{a^{p-1}}{p-1} $$ Let $m$ and $n$ be integers such that $$ S_{3}+S_{4}-3 S_{2}=\frac{m}{n} $$ Prove that $p$ divides $m$." "[""For rational numbers $p_{1} / q_{1}$ and $p_{2} / q_{2}$ with the denominators $q_{1}, q_{2}$ not divisible by $p$, we write $p_{1} / q_{1} \\equiv p_{2} / q_{2}(\\bmod p)$ if the numerator $p_{1} q_{2}-p_{2} q_{1}$ of their difference is divisible by $p$.\n\nWe start with finding an explicit formula for the residue of $S_{a}$ modulo $p$. Note first that for every $k=1, \\ldots, p-1$ the number $\\left(\\begin{array}{l}p \\\\ k\\end{array}\\right)$ is divisible by $p$, and\n\n$$\n\\frac{1}{p}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right)=\\frac{(p-1)(p-2) \\cdots(p-k+1)}{k !} \\equiv \\frac{(-1) \\cdot(-2) \\cdots(-k+1)}{k !}=\\frac{(-1)^{k-1}}{k} \\quad(\\bmod p)\n$$\n\nTherefore, we have\n\n$$\nS_{a}=-\\sum_{k=1}^{p-1} \\frac{(-a)^{k}(-1)^{k-1}}{k} \\equiv-\\sum_{k=1}^{p-1}(-a)^{k} \\cdot \\frac{1}{p}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right) \\quad(\\bmod p)\n$$\n\nThe number on the right-hand side is integer. Using the binomial formula we express it as\n\n$$\n-\\sum_{k=1}^{p-1}(-a)^{k} \\cdot \\frac{1}{p}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right)=-\\frac{1}{p}\\left(-1-(-a)^{p}+\\sum_{k=0}^{p}(-a)^{k}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right)\\right)=\\frac{(a-1)^{p}-a^{p}+1}{p}\n$$\n\nsince $p$ is odd. So, we have\n\n$$\nS_{a} \\equiv \\frac{(a-1)^{p}-a^{p}+1}{p} \\quad(\\bmod p)\n$$\n\nFinally, using the obtained formula we get\n\n$$\n\\begin{aligned}\nS_{3}+S_{4}-3 S_{2} & \\equiv \\frac{\\left(2^{p}-3^{p}+1\\right)+\\left(3^{p}-4^{p}+1\\right)-3\\left(1^{p}-2^{p}+1\\right)}{p} \\\\\n& =\\frac{4 \\cdot 2^{p}-4^{p}-4}{p}=-\\frac{\\left(2^{p}-2\\right)^{2}}{p} \\quad(\\bmod p) .\n\\end{aligned}\n$$\n\nBy Fermat's theorem, $p \\mid 2^{p}-2$, so $p^{2} \\mid\\left(2^{p}-2\\right)^{2}$ and hence $S_{3}+S_{4}-3 S_{2} \\equiv 0(\\bmod p)$."" ""One may solve the problem without finding an explicit formula for $S_{a}$. It is enough to find the following property.\n\nLemma. For every integer $a$, we have $S_{a+1} \\equiv S_{-a}(\\bmod p)$.\n\nProof. We expand $S_{a+1}$ using the binomial formula as\n\n$$\nS_{a+1}=\\sum_{k=1}^{p-1} \\frac{1}{k} \\sum_{j=0}^{k}\\left(\\begin{array}{c}\nk \\\\\nj\n\\end{array}\\right) a^{j}=\\sum_{k=1}^{p-1}\\left(\\frac{1}{k}+\\sum_{j=1}^{k} a^{j} \\cdot \\frac{1}{k}\\left(\\begin{array}{c}\nk \\\\\nj\n\\end{array}\\right)\\right)=\\sum_{k=1}^{p-1} \\frac{1}{k}+\\sum_{j=1}^{p-1} a^{j} \\sum_{k=j}^{p-1} \\frac{1}{k}\\left(\\begin{array}{l}\nk \\\\\nj\n\\end{array}\\right) a^{k} .\n$$\n\nNote that $\\frac{1}{k}+\\frac{1}{p-k}=\\frac{p}{k(p-k)} \\equiv 0(\\bmod p)$ for all $1 \\leq k \\leq p-1$; hence the first sum vanishes modulo $p$. For the second sum, we use the relation $\\frac{1}{k}\\left(\\begin{array}{c}k \\\\ j\\end{array}\\right)=\\frac{1}{j}\\left(\\begin{array}{c}k-1 \\\\ j-1\\end{array}\\right)$ to obtain\n\n$$\nS_{a+1} \\equiv \\sum_{j=1}^{p-1} \\frac{a^{j}}{j} \\sum_{k=1}^{p-1}\\left(\\begin{array}{c}\nk-1 \\\\\nj-1\n\\end{array}\\right) \\quad(\\bmod p)\n$$\n\nFinally, from the relation\n\n$$\n\\sum_{k=1}^{p-1}\\left(\\begin{array}{c}\nk-1 \\\\\nj-1\n\\end{array}\\right)=\\left(\\begin{array}{c}\np-1 \\\\\nj\n\\end{array}\\right)=\\frac{(p-1)(p-2) \\ldots(p-j)}{j !} \\equiv(-1)^{j} \\quad(\\bmod p)\n$$\n\nwe obtain\n\n$$\nS_{a+1} \\equiv \\sum_{j=1}^{p-1} \\frac{a^{j}(-1)^{j}}{j !}=S_{-a}\n$$\n\nNow we turn to the problem. Using the lemma we get\n\n$$\nS_{3}-3 S_{2} \\equiv S_{-2}-3 S_{2}=\\sum_{\\substack{1 \\leq k \\leq p-1 \\\\ k \\text { is even }}} \\frac{-2 \\cdot 2^{k}}{k}+\\sum_{\\substack{1 \\leq k \\leq p-1 \\\\ k \\text { is odd }}} \\frac{-4 \\cdot 2^{k}}{k}(\\bmod p)\n\\tag{1}\n$$\n\nThe first sum in (11) expands as\n\n$$\n\\sum_{\\ell=1}^{(p-1) / 2} \\frac{-2 \\cdot 2^{2 \\ell}}{2 \\ell}=-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{4^{\\ell}}{\\ell}\n$$\n\nNext, using Fermat's theorem, we expand the second sum in (11) as\n\n$$\n-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{2^{2 \\ell+1}}{2 \\ell-1} \\equiv-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{2^{p+2 \\ell}}{p+2 \\ell-1}=-\\sum_{m=(p+1) / 2}^{p-1} \\frac{2 \\cdot 4^{m}}{2 m}=-\\sum_{m=(p+1) / 2}^{p-1} \\frac{4^{m}}{m}(\\bmod p)\n$$\n\n(here we set $m=\\ell+\\frac{p-1}{2}$ ). Hence,\n\n$$\nS_{3}-3 S_{2} \\equiv-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{4^{\\ell}}{\\ell}-\\sum_{m=(p+1) / 2}^{p-1} \\frac{4^{m}}{m}=-S_{4} \\quad(\\bmod p)\n$$""]" [] Text-only Competition True Theorem proof Number Theory Math English 76 Let $k$ be a positive integer and set $n=2^{k}+1$. Prove that $n$ is a prime number if and only if the following holds: there is a permutation $a_{1}, \ldots, a_{n-1}$ of the numbers $1,2, \ldots, n-1$ and a sequence of integers $g_{1}, g_{2}, \ldots, g_{n-1}$ such that $n$ divides $g_{i}^{a_{i}}-a_{i+1}$ for every $i \in\{1,2, \ldots, n-1\}$, where we set $a_{n}=a_{1}$. "['Let $N=\\{1,2, \\ldots, n-1\\}$. For $a, b \\in N$, we say that $b$ follows $a$ if there exists an integer $g$ such that $b \\equiv g^{a}(\\bmod n)$ and denote this property as $a \\rightarrow b$. This way we have a directed graph with $N$ as set of vertices. If $a_{1}, \\ldots, a_{n-1}$ is a permutation of $1,2, \\ldots, n-1$ such that $a_{1} \\rightarrow a_{2} \\rightarrow \\ldots \\rightarrow a_{n-1} \\rightarrow a_{1}$ then this is a Hamiltonian cycle in the graph.\n\nStep I. First consider the case when $n$ is composite. Let $n=p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$ be its prime factorization. All primes $p_{i}$ are odd.\n\nSuppose that $\\alpha_{i}>1$ for some $i$. For all integers $a, g$ with $a \\geq 2$, we have $g^{a} \\not \\equiv p_{i}\\left(\\bmod p_{i}^{2}\\right)$, because $g^{a}$ is either divisible by $p_{i}^{2}$ or it is not divisible by $p_{i}$. It follows that in any Hamiltonian cycle $p_{i}$ comes immediately after 1 . The same argument shows that $2 p_{i}$ also should come immediately after 1, which is impossible. Hence, there is no Hamiltonian cycle in the graph.\n\nNow suppose that $n$ is square-free. We have $n=p_{1} p_{2} \\ldots p_{s}>9$ and $s \\geq 2$. Assume that there exists a Hamiltonian cycle. There are $\\frac{n-1}{2}$ even numbers in this cycle, and each number which follows one of them should be a quadratic residue modulo $n$. So, there should be at least $\\frac{n-1}{2}$ nonzero quadratic residues modulo $n$. On the other hand, for each $p_{i}$ there exist exactly $\\frac{p_{i}+1}{2}$ quadratic residues modulo $p_{i}$; by the Chinese Remainder Theorem, the number of quadratic residues modulo $n$ is exactly $\\frac{p_{1}+1}{2} \\cdot \\frac{p_{2}+1}{2} \\cdot \\ldots \\cdot \\frac{p_{s}+1}{2}$, including 0 . Then we have a contradiction by\n\n$$\n\\frac{p_{1}+1}{2} \\cdot \\frac{p_{2}+1}{2} \\cdot \\ldots \\cdot \\frac{p_{s}+1}{2} \\leq \\frac{2 p_{1}}{3} \\cdot \\frac{2 p_{2}}{3} \\cdot \\ldots \\cdot \\frac{2 p_{s}}{3}=\\left(\\frac{2}{3}\\right)^{s} n \\leq \\frac{4 n}{9}<\\frac{n-1}{2}\n$$\n\nThis proves the ""if""-part of the problem.\n\nStep II. Now suppose that $n$ is prime. For any $a \\in N$, denote by $\\nu_{2}(a)$ the exponent of 2 in the prime factorization of $a$, and let $\\mu(a)=\\max \\left\\{t \\in[0, k] \\mid 2^{t} \\rightarrow a\\right\\}$.\n\nLemma. For any $a, b \\in N$, we have $a \\rightarrow b$ if and only if $\\nu_{2}(a) \\leq \\mu(b)$.\n\nProof. Let $\\ell=\\nu_{2}(a)$ and $m=\\mu(b)$.\n\nSuppose $\\ell \\leq m$. Since $b$ follows $2^{m}$, there exists some $g_{0}$ such that $b \\equiv g_{0}^{2^{m}}(\\bmod n)$. By $\\operatorname{gcd}(a, n-1)=2^{\\ell}$ there exist some integers $p$ and $q$ such that $p a-q(n-1)=2^{\\ell}$. Choosing $g=g_{0}^{2^{m-\\ell} p}$ we have $g^{a}=g_{0}^{2^{m-\\ell} p a}=g_{0}^{2^{m}+2^{m-\\ell} q(n-1)} \\equiv g_{0}^{2^{m}} \\equiv b(\\bmod n)$ by FERMAT\'s theorem. Hence, $a \\rightarrow b$.\n\nTo prove the reverse statement, suppose that $a \\rightarrow b$, so $b \\equiv g^{a}(\\bmod n)$ with some $g$. Then $b \\equiv\\left(g^{a / 2^{\\ell}}\\right)^{2^{\\ell}}$, and therefore $2^{\\ell} \\rightarrow b$. By the definition of $\\mu(b)$, we have $\\mu(b) \\geq \\ell$. The lemma is\n\n\n\nproved.\n\nNow for every $i$ with $0 \\leq i \\leq k$, let\n\n$$\n\\begin{aligned}\nA_{i} & =\\left\\{a \\in N \\mid \\nu_{2}(a)=i\\right\\}, \\\\\nB_{i} & =\\{a \\in N \\mid \\mu(a)=i\\}, \\\\\n\\text { and } C_{i} & =\\{a \\in N \\mid \\mu(a) \\geq i\\}=B_{i} \\cup B_{i+1} \\cup \\ldots \\cup B_{k} .\n\\end{aligned}\n$$\n\nWe claim that $\\left|A_{i}\\right|=\\left|B_{i}\\right|$ for all $0 \\leq i \\leq k$. Obviously we have $\\left|A_{i}\\right|=2^{k-i-1}$ for all $i=$ $0, \\ldots, k-1$, and $\\left|A_{k}\\right|=1$. Now we determine $\\left|C_{i}\\right|$. We have $\\left|C_{0}\\right|=n-1$ and by Fermat\'s theorem we also have $C_{k}=\\{1\\}$, so $\\left|C_{k}\\right|=1$. Next, notice that $C_{i+1}=\\left\\{x^{2} \\bmod n \\mid x \\in C_{i}\\right\\}$. For every $a \\in N$, the relation $x^{2} \\equiv a(\\bmod n)$ has at most two solutions in $N$. Therefore we have $2\\left|C_{i+1}\\right| \\leq\\left|C_{i}\\right|$, with the equality achieved only if for every $y \\in C_{i+1}$, there exist distinct elements $x, x^{\\prime} \\in C_{i}$ such that $x^{2} \\equiv x^{\\prime 2} \\equiv y(\\bmod n)$ (this implies $\\left.x+x^{\\prime}=n\\right)$. Now, since $2^{k}\\left|C_{k}\\right|=\\left|C_{0}\\right|$, we obtain that this equality should be achieved in each step. Hence $\\left|C_{i}\\right|=2^{k-i}$ for $0 \\leq i \\leq k$, and therefore $\\left|B_{i}\\right|=2^{k-i-1}$ for $0 \\leq i \\leq k-1$ and $\\left|B_{k}\\right|=1$.\n\nFrom the previous arguments we can see that for each $z \\in C_{i}(0 \\leq i0$. Finally, if $\\lambda=k-1$, then $C$ contains $2^{k-1}$ which is the only element of $A_{k-1}$. Since $B_{k-1}=\\left\\{2^{k}\\right\\}=A_{k}$ and $B_{k}=\\{1\\}$, the cycle $C$ contains the path $2^{k-1} \\rightarrow 2^{k} \\rightarrow 1$ and it contains an odd number again. This completes the proof of the ""only if""-part of the problem.']" [] Text-only Competition True Theorem proof Number Theory Math English 77 "Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that $$ 36 \leq 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \leq 48 . $$" "['Observe that\n\n$$\n\\begin{gathered}\n4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)-\\left(a^{4}+b^{4}+c^{4}+d^{4}\\right)=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right) \\\\\n+6\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-4(a+b+c+d)+4 \\\\\n=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right)+52\n\\end{gathered}\n$$\n\nNow, introducing $x=a-1, y=b-1, z=c-1, t=d-1$, we need to prove the inequalities\n\n$$\n16 \\geq x^{4}+y^{4}+z^{4}+t^{4} \\geq 4\n$$\n\nunder the constraint\n\n$$\nx^{2}+y^{2}+z^{2}+t^{2}=\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-2(a+b+c+d)+4=4\n\\tag{1}\n$$\n\n(we will not use the value of $x+y+z+t$ though it can be found).\n\nNow the rightmost inequality in (1) follows from the power mean inequality:\n\n$$\nx^{4}+y^{4}+z^{4}+t^{4} \\geq \\frac{\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}}{4}=4\n$$\n\nFor the other one, expanding the brackets we note that\n\n$$\n\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}=\\left(x^{4}+y^{4}+z^{4}+t^{4}\\right)+q,\n$$\n\nwhere $q$ is a nonnegative number, so\n\n$$\nx^{4}+y^{4}+z^{4}+t^{4} \\leq\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}=16\n$$\n\nand we are done.' 'First, we claim that $0 \\leq a, b, c, d \\leq 3$. Actually, we have\n\n$$\na+b+c=6-d, \\quad a^{2}+b^{2}+c^{2}=12-d^{2}\n$$\n\nhence the power mean inequality\n\n$$\na^{2}+b^{2}+c^{2} \\geq \\frac{(a+b+c)^{2}}{3}\n$$\n\nrewrites as\n\n$$\n12-d^{2} \\geq \\frac{(6-d)^{2}}{3} \\quad \\Longleftrightarrow \\quad 2 d(d-3) \\leq 0\n$$\n\n\n\nwhich implies the desired inequalities for $d$; since the conditions are symmetric, we also have the same estimate for the other variables.\n\nNow, to prove the rightmost inequality, we use the obvious inequality $x^{2}(x-2)^{2} \\geq 0$ for each real $x$; this inequality rewrites as $4 x^{3}-x^{4} \\leq 4 x^{2}$. It follows that\n\n$$\n\\left(4 a^{3}-a^{4}\\right)+\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\leq 4\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=48\n$$\n\nas desired.\n\nNow we prove the leftmost inequality in an analogous way. For each $x \\in[0,3]$, we have $(x+1)(x-1)^{2}(x-3) \\leq 0$ which is equivalent to $4 x^{3}-x^{4} \\geq 2 x^{2}+4 x-3$. This implies that $\\left(4 a^{3}-a^{4}\\right)+\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\geq 2\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)+4(a+b+c+d)-12=36$, as desired.' 'First, expanding $48=4\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)$ and applying the AM-GM inequality, we have\n\n$$\n\\begin{aligned}\na^{4}+b^{4}+c^{4}+d^{4}+48 & =\\left(a^{4}+4 a^{2}\\right)+\\left(b^{4}+4 b^{2}\\right)+\\left(c^{4}+4 c^{2}\\right)+\\left(d^{4}+4 d^{2}\\right) \\\\\n& \\geq 2\\left(\\sqrt{a^{4} \\cdot 4 a^{2}}+\\sqrt{b^{4} \\cdot 4 b^{2}}+\\sqrt{c^{4} \\cdot 4 c^{2}}+\\sqrt{d^{4} \\cdot 4 d^{2}}\\right) \\\\\n& =4\\left(\\left|a^{3}\\right|+\\left|b^{3}\\right|+\\left|c^{3}\\right|+\\left|d^{3}\\right|\\right) \\geq 4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right),\n\\end{aligned}\n$$\n\nwhich establishes the rightmost inequality.\n\nTo prove the leftmost inequality, we first show that $a, b, c, d \\in[0,3]$ as in the previous solution. Moreover, we can assume that $0 \\leq a \\leq b \\leq c \\leq d$. Then we have $a+b \\leq b+c \\leq$ $\\frac{2}{3}(b+c+d) \\leq \\frac{2}{3} \\cdot 6=4$.\n\nNext, we show that $4 b-b^{2} \\leq 4 c-c^{2}$. Actually, this inequality rewrites as $(c-b)(b+c-4) \\leq 0$, which follows from the previous estimate. The inequality $4 a-a^{2} \\leq 4 b-b^{2}$ can be proved analogously.\n\nFurther, the inequalities $a \\leq b \\leq c$ together with $4 a-a^{2} \\leq 4 b-b^{2} \\leq 4 c-c^{2}$ allow us to apply the Chebyshev inequality obtaining\n\n$$\n\\begin{aligned}\na^{2}\\left(4 a-a^{2}\\right)+b^{2}\\left(4 b-b^{2}\\right)+c^{2}\\left(4 c-c^{2}\\right) & \\geq \\frac{1}{3}\\left(a^{2}+b^{2}+c^{2}\\right)\\left(4(a+b+c)-\\left(a^{2}+b^{2}+c^{2}\\right)\\right) \\\\\n& =\\frac{\\left(12-d^{2}\\right)\\left(4(6-d)-\\left(12-d^{2}\\right)\\right)}{3} .\n\\end{aligned}\n$$\n\nThis implies that\n\n$$\n\\begin{aligned}\n\\left(4 a^{3}-a^{4}\\right) & +\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\geq \\frac{\\left(12-d^{2}\\right)\\left(d^{2}-4 d+12\\right)}{3}+4 d^{3}-d^{4} \\\\\n& =\\frac{144-48 d+16 d^{3}-4 d^{4}}{3}=36+\\frac{4}{3}(3-d)(d-1)\\left(d^{2}-3\\right) .\n\\end{aligned}\n\\tag{2}\n$$\n\nFinally, we have $d^{2} \\geq \\frac{1}{4}\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=3$ (which implies $d>1$ ); so, the expression $\\frac{4}{3}(3-d)(d-1)\\left(d^{2}-3\\right)$ in the right-hand part of $(2)$ is nonnegative, and the desired inequality is proved.']" [] Text-only Competition True Theorem proof Algebra Math English 78 A sequence $x_{1}, x_{2}, \ldots$ is defined by $x_{1}=1$ and $x_{2 k}=-x_{k}, x_{2 k-1}=(-1)^{k+1} x_{k}$ for all $k \geq 1$. Prove that $x_{1}+x_{2}+\cdots+x_{n} \geq 0$ for all $n \geq 1$. "['We start with some observations. First, from the definition of $x_{i}$ it follows that for each positive integer $k$ we have\n\n$$\nx_{4 k-3}=x_{2 k-1}=-x_{4 k-2} \\quad \\text { and } \\quad x_{4 k-1}=x_{4 k}=-x_{2 k}=x_{k} \\text {. }\n\\tag{1}\n$$\n\nHence, denoting $S_{n}=\\sum_{i=1}^{n} x_{i}$, we have\n\n$$\nS_{4 k}=\\sum_{i=1}^{k}\\left(\\left(x_{4 k-3}+x_{4 k-2}\\right)+\\left(x_{4 k-1}+x_{4 k}\\right)\\right)=\\sum_{i=1}^{k}\\left(0+2 x_{k}\\right)=2 S_{k}\\tag{2}\n$$\n$$\nS_{4 k+2}=S_{4 k}+\\left(x_{4 k+1}+x_{4 k+2}\\right)=S_{4 k}\\tag{3}\n$$\n\nObserve also that $S_{n}=\\sum_{i=1}^{n} x_{i} \\equiv \\sum_{i=1}^{n} 1=n(\\bmod 2)$.\n\nNow we prove by induction on $k$ that $S_{i} \\geq 0$ for all $i \\leq 4 k$. The base case is valid since $x_{1}=x_{3}=x_{4}=1, x_{2}=-1$. For the induction step, assume that $S_{i} \\geq 0$ for all $i \\leq 4 k$. Using the relations (1)-(3), we obtain\n\n$$\nS_{4 k+4}=2 S_{k+1} \\geq 0, \\quad S_{4 k+2}=S_{4 k} \\geq 0, \\quad S_{4 k+3}=S_{4 k+2}+x_{4 k+3}=\\frac{S_{4 k+2}+S_{4 k+4}}{2} \\geq 0\n$$\n\nSo, we are left to prove that $S_{4 k+1} \\geq 0$. If $k$ is odd, then $S_{4 k}=2 S_{k} \\geq 0$; since $k$ is odd, $S_{k}$ is odd as well, so we have $S_{4 k} \\geq 2$ and hence $S_{4 k+1}=S_{4 k}+x_{4 k+1} \\geq 1$.\n\nConversely, if $k$ is even, then we have $x_{4 k+1}=x_{2 k+1}=x_{k+1}$, hence $S_{4 k+1}=S_{4 k}+x_{4 k+1}=$ $2 S_{k}+x_{k+1}=S_{k}+S_{k+1} \\geq 0$. The step is proved.' 'We will use the notation of $S_{n}$ and the relations (1)-(3) from the previous solution.\n\nAssume the contrary and consider the minimal $n$ such that $S_{n+1}<0$; surely $n \\geq 1$, and from $S_{n} \\geq 0$ we get $S_{n}=0, x_{n+1}=-1$. Hence, we are especially interested in the set $M=\\left\\{n: S_{n}=0\\right\\}$; our aim is to prove that $x_{n+1}=1$ whenever $n \\in M$ thus coming to a contradiction.\n\nFor this purpose, we first describe the set $M$ inductively. We claim that (i) $M$ consists only of even numbers, (ii) $2 \\in M$, and (iii) for every even $n \\geq 4$ we have $n \\in M \\Longleftrightarrow[n / 4] \\in M$. Actually, (i) holds since $S_{n} \\equiv n(\\bmod 2)$, (ii) is straightforward, while (iii) follows from the relations $S_{4 k+2}=S_{4 k}=2 S_{k}$.\n\nNow, we are left to prove that $x_{n+1}=1$ if $n \\in M$. We use the induction on $n$. The base case is $n=2$, that is, the minimal element of $M$; here we have $x_{3}=1$, as desired.\n\nFor the induction step, consider some $4 \\leq n \\in M$ and let $m=[n / 4] \\in M$; then $m$ is even, and $x_{m+1}=1$ by the induction hypothesis. We prove that $x_{n+1}=x_{m+1}=1$. If $n=4 m$ then we have $x_{n+1}=x_{2 m+1}=x_{m+1}$ since $m$ is even; otherwise, $n=4 m+2$, and $x_{n+1}=-x_{2 m+2}=x_{m+1}$, as desired. The proof is complete.']" [] Text-only Competition True Theorem proof Algebra Math English 79 Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n))=f(n)+1$ and $g(f(n))=$ $g(n)+1$ hold for all positive integers. Prove that $f(n)=g(n)$ for all positive integer $n$. "['Throughout the solution, by $\\mathbb{N}$ we denote the set of all positive integers. For any function $h: \\mathbb{N} \\rightarrow \\mathbb{N}$ and for any positive integer $k$, define $h^{k}(x)=\\underbrace{h(h(\\ldots h}_{k}(x) \\ldots)$ ) (in particular, $\\left.h^{0}(x)=x\\right)$.\n\nObserve that $f\\left(g^{k}(x)\\right)=f\\left(g^{k-1}(x)\\right)+1=\\cdots=f(x)+k$ for any positive integer $k$, and similarly $g\\left(f^{k}(x)\\right)=g(x)+k$. Now let $a$ and $b$ are the minimal values attained by $f$ and $g$, respectively; say $f\\left(n_{f}\\right)=a, g\\left(n_{g}\\right)=b$. Then we have $f\\left(g^{k}\\left(n_{f}\\right)\\right)=a+k, g\\left(f^{k}\\left(n_{g}\\right)\\right)=b+k$, so the function $f$ attains all values from the set $N_{f}=\\{a, a+1, \\ldots\\}$, while $g$ attains all the values from the set $N_{g}=\\{b, b+1, \\ldots\\}$.\n\nNext, note that $f(x)=f(y)$ implies $g(x)=g(f(x))-1=g(f(y))-1=g(y)$; surely, the converse implication also holds. Now, we say that $x$ and $y$ are similar (and write $x \\sim y$ ) if $f(x)=f(y)$ (equivalently, $g(x)=g(y)$ ). For every $x \\in \\mathbb{N}$, we define $[x]=\\{y \\in \\mathbb{N}: x \\sim y\\}$; surely, $y_{1} \\sim y_{2}$ for all $y_{1}, y_{2} \\in[x]$, so $[x]=[y]$ whenever $y \\in[x]$.\n\nNow we investigate the structure of the sets $[x]$.\n\nClaim 1. Suppose that $f(x) \\sim f(y)$; then $x \\sim y$, that is, $f(x)=f(y)$. Consequently, each class $[x]$ contains at most one element from $N_{f}$, as well as at most one element from $N_{g}$.\n\nProof. If $f(x) \\sim f(y)$, then we have $g(x)=g(f(x))-1=g(f(y))-1=g(y)$, so $x \\sim y$. The second statement follows now from the sets of values of $f$ and $g$.\n\nNext, we clarify which classes do not contain large elements.\n\nClaim 2. For any $x \\in \\mathbb{N}$, we have $[x] \\subseteq\\{1,2, \\ldots, b-1\\}$ if and only if $f(x)=a$. Analogously, $[x] \\subseteq\\{1,2, \\ldots, a-1\\}$ if and only if $g(x)=b$.\n\nProof. We will prove that $[x] \\nsubseteq\\{1,2, \\ldots, b-1\\} \\Longleftrightarrow f(x)>a$; the proof of the second statement is similar.\n\nNote that $f(x)>a$ implies that there exists some $y$ satisfying $f(y)=f(x)-1$, so $f(g(y))=$ $f(y)+1=f(x)$, and hence $x \\sim g(y) \\geq b$. Conversely, if $b \\leq c \\sim x$ then $c=g(y)$ for some $y \\in \\mathbb{N}$, which in turn follows $f(x)=f(g(y))=f(y)+1 \\geq a+1$, and hence $f(x)>a$.\n\nClaim 2 implies that there exists exactly one class contained in $\\{1, \\ldots, a-1\\}$ (that is, the class $\\left[n_{g}\\right]$ ), as well as exactly one class contained in $\\{1, \\ldots, b-1\\}$ (the class $\\left[n_{f}\\right]$ ). Assume for a moment that $a \\leq b$; then $\\left[n_{g}\\right]$ is contained in $\\{1, \\ldots, b-1\\}$ as well, hence it coincides with $\\left[n_{g}\\right]$. So, we get that\n\n$$\nf(x)=a \\Longleftrightarrow g(x)=b \\Longleftrightarrow x \\sim n_{f} \\sim n_{g} .\n\\tag{1}\n$$\n\nClaim 3. $a=b$.\n\nProof. By Claim 2, we have $[a] \\neq\\left[n_{f}\\right]$, so $[a]$ should contain some element $a^{\\prime} \\geq b$ by Claim 2 again. If $a \\neq a^{\\prime}$, then $[a]$ contains two elements $\\geq a$ which is impossible by Claim 1 . Therefore, $a=a^{\\prime} \\geq b$. Similarly, $b \\geq a$.\n\nNow we are ready to prove the problem statement. First, we establish the following\n\nClaim 4. For every integer $d \\geq 0, f^{d+1}\\left(n_{f}\\right)=g^{d+1}\\left(n_{f}\\right)=a+d$.\n\nProof. Induction on $d$. For $d=0$, the statement follows from (1) and Claim 3. Next, for $d>1$ from the induction hypothesis we have $f^{d+1}\\left(n_{f}\\right)=f\\left(f^{d}\\left(n_{f}\\right)\\right)=f\\left(g^{d}\\left(n_{f}\\right)\\right)=f\\left(n_{f}\\right)+d=a+d$. The equality $g^{d+1}\\left(n_{f}\\right)=a+d$ is analogous.\n\n\n\nFinally, for each $x \\in \\mathbb{N}$, we have $f(x)=a+d$ for some $d \\geq 0$, so $f(x)=f\\left(g^{d}\\left(n_{f}\\right)\\right)$ and hence $x \\sim g^{d}\\left(n_{f}\\right)$. It follows that $g(x)=g\\left(g^{d}\\left(n_{f}\\right)\\right)=g^{d+1}\\left(n_{f}\\right)=a+d=f(x)$ by Claim 4 .' ""We start with the same observations, introducing the relation $\\sim$ and proving Claim 1 from the previous solution.\n\nNote that $f(a)>a$ since otherwise we have $f(a)=a$ and hence $g(a)=g(f(a))=g(a)+1$, which is false.\n\nClaim $2^{\\prime} . \\quad a=b$.\n\nProof. We can assume that $a \\leq b$. Since $f(a) \\geq a+1$, there exists some $x \\in \\mathbb{N}$ such that $f(a)=f(x)+1$, which is equivalent to $f(a)=f(g(x))$ and $a \\sim g(x)$. Since $g(x) \\geq b \\geq a$, by Claim 1 we have $a=g(x) \\geq b$, which together with $a \\leq b$ proves the Claim.\n\nNow, almost the same method allows to find the values $f(a)$ and $g(a)$.\n\nClaim $3^{\\prime} . \\quad f(a)=g(a)=a+1$.\n\nProof. Assume the contrary; then $f(a) \\geq a+2$, hence there exist some $x, y \\in \\mathbb{N}$ such that $f(x)=f(a)-2$ and $f(y)=g(x)($ as $g(x) \\geq a=b)$. Now we get $f(a)=f(x)+2=f\\left(g^{2}(x)\\right)$, so $a \\sim g^{2}(x) \\geq a$, and by Claim 1 we get $a=g^{2}(x)=g(f(y))=1+g(y) \\geq 1+a$; this is impossible. The equality $g(a)=a+1$ is similar.\n\nNow, we are prepared for the proof of the problem statement. First, we prove it for $n \\geq a$. Claim 4'. For each integer $x \\geq a$, we have $f(x)=g(x)=x+1$.\n\nProof. Induction on $x$. The base case $x=a$ is provided by Claim $3^{\\prime}$, while the induction step follows from $f(x+1)=f(g(x))=f(x)+1=(x+1)+1$ and the similar computation for $g(x+1)$.\n\nFinally, for an arbitrary $n \\in \\mathbb{N}$ we have $g(n) \\geq a$, so by Claim $4^{\\prime}$ we have $f(n)+1=$ $f(g(n))=g(n)+1$, hence $f(n)=g(n)$.""]" [] Text-only Competition True Theorem proof Algebra Math English 80 "Let $a_{1}, \ldots, a_{r}$ be positive real numbers. For $n>r$, we inductively define $$ a_{n}=\max _{1 \leq k \leq n-1}\left(a_{k}+a_{n-k}\right) \tag{1} $$ Prove that there exist positive integers $\ell \leq r$ and $N$ such that $a_{n}=a_{n-\ell}+a_{\ell}$ for all $n \geq N$." "['First, from the problem conditions we have that each $a_{n}(n>r)$ can be expressed as $a_{n}=a_{j_{1}}+a_{j_{2}}$ with $j_{1}, j_{2}r$ then we can proceed in the same way with $a_{j_{1}}$, and so on. Finally, we represent $a_{n}$ in a form\n\n$$\na_{n}=a_{i_{1}}+\\cdots+a_{i_{k}}\\tag{2}\n$$\n$$\n1 \\leq i_{j} \\leq r, \\quad i_{1}+\\cdots+i_{k}=n .\n\n\\tag{3}\n$$\n\nMoreover, if $a_{i_{1}}$ and $a_{i_{2}}$ are the numbers in (2) obtained on the last step, then $i_{1}+i_{2}>r$. Hence we can adjust (3) as\n\n$$\n1 \\leq i_{j} \\leq r, \\quad i_{1}+\\cdots+i_{k}=n, \\quad i_{1}+i_{2}>r\n\\tag{4}\n$$\n\nOn the other hand, suppose that the indices $i_{1}, \\ldots, i_{k}$ satisfy the conditions (4). Then, denoting $s_{j}=i_{1}+\\cdots+i_{j}$, from (1) we have\n\n$$\na_{n}=a_{s_{k}} \\geq a_{s_{k-1}}+a_{i_{k}} \\geq a_{s_{k-2}}+a_{i_{k-1}}+a_{i_{k}} \\geq \\cdots \\geq a_{i_{1}}+\\cdots+a_{i_{k}}\n$$\n\nSummarizing these observations we get the following\n\nClaim. For every $n>r$, we have\n\n$$\na_{n}=\\max \\left\\{a_{i_{1}}+\\cdots+a_{i_{k}}: \\text { the collection }\\left(i_{1}, \\ldots, i_{k}\\right) \\text { satisfies }(4)\\right\\}\n$$\n\nNow we denote\n\n$$\ns=\\max _{1 \\leq i \\leq r} \\frac{a_{i}}{i}\n$$\n\nand fix some index $\\ell \\leq r$ such that $s=\\frac{a_{\\ell}}{\\ell}$.\n\nConsider some $n \\geq r^{2} \\ell+2 r$ and choose an expansion of $a_{n}$ in the form (2), (4). Then we have $n=i_{1}+\\cdots+i_{k} \\leq r k$, so $k \\geq n / r \\geq r \\ell+2$. Suppose that none of the numbers $i_{3}, \\ldots, i_{k}$ equals $\\ell$. Then by the pigeonhole principle there is an index $1 \\leq j \\leq r$ which appears among $i_{3}, \\ldots, i_{k}$ at least $\\ell$ times, and surely $j \\neq \\ell$. Let us delete these $\\ell$ occurrences of $j$ from $\\left(i_{1}, \\ldots, i_{k}\\right)$, and add $j$ occurrences of $\\ell$ instead, obtaining a sequence $\\left(i_{1}, i_{2}, i_{3}^{\\prime}, \\ldots, i_{k^{\\prime}}^{\\prime}\\right)$ also satisfying (4). By Claim, we have\n\n$$\na_{i_{1}}+\\cdots+a_{i_{k}}=a_{n} \\geq a_{i_{1}}+a_{i_{2}}+a_{i_{3}^{\\prime}}+\\cdots+a_{i_{k^{\\prime}}^{\\prime}}\n$$\n\nor, after removing the coinciding terms, $\\ell a_{j} \\geq j a_{\\ell}$, so $\\frac{a_{\\ell}}{\\ell} \\leq \\frac{a_{j}}{j}$. By the definition of $\\ell$, this means that $\\ell a_{j}=j a_{\\ell}$, hence\n\n$$\na_{n}=a_{i_{1}}+a_{i_{2}}+a_{i_{3}^{\\prime}}+\\cdots+a_{i_{k^{\\prime}}^{\\prime}}\n$$\n\nThus, for every $n \\geq r^{2} \\ell+2 r$ we have found a representation of the form (2), (4) with $i_{j}=\\ell$ for some $j \\geq 3$. Rearranging the indices we may assume that $i_{k}=\\ell$.\n\nFinally, observe that in this representation, the indices $\\left(i_{1}, \\ldots, i_{k-1}\\right)$ satisfy the conditions (4) with $n$ replaced by $n-\\ell$. Thus, from the Claim we get\n\n$$\na_{n-\\ell}+a_{\\ell} \\geq\\left(a_{i_{1}}+\\cdots+a_{i_{k-1}}\\right)+a_{\\ell}=a_{n}\n$$\n\nwhich by (1) implies\n\n$$\na_{n}=a_{n-\\ell}+a_{\\ell} \\quad \\text { for each } n \\geq r^{2} \\ell+2 r,\n$$\n\nas desired.' ""As in the previous solution, we involve the expansion (2), (3), and we fix some index $1 \\leq \\ell \\leq r$ such that\n\n$$\n\\frac{a_{\\ell}}{\\ell}=s=\\max _{1 \\leq i \\leq r} \\frac{a_{i}}{i}\n$$\n\nNow, we introduce the sequence $\\left(b_{n}\\right)$ as $b_{n}=a_{n}-s n$; then $b_{\\ell}=0$.\n\nWe prove by induction on $n$ that $b_{n} \\leq 0$, and $\\left(b_{n}\\right)$ satisfies the same recurrence relation as $\\left(a_{n}\\right)$. The base cases $n \\leq r$ follow from the definition of $s$. Now, for $n>r$ from the induction hypothesis we have\n\n$$\nb_{n}=\\max _{1 \\leq k \\leq n-1}\\left(a_{k}+a_{n-k}\\right)-n s=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}+n s\\right)-n s=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}\\right) \\leq 0\n$$\n\nas required.\n\nNow, if $b_{k}=0$ for all $1 \\leq k \\leq r$, then $b_{n}=0$ for all $n$, hence $a_{n}=s n$, and the statement is trivial. Otherwise, define\n\n$$\nM=\\max _{1 \\leq i \\leq r}\\left|b_{i}\\right|, \\quad \\varepsilon=\\min \\left\\{\\left|b_{i}\\right|: 1 \\leq i \\leq r, b_{i}<0\\right\\}\n$$\n\nThen for $n>r$ we obtain\n\n$$\nb_{n}=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}\\right) \\geq b_{\\ell}+b_{n-\\ell}=b_{n-\\ell}\n$$\n\nSo\n\n$$\n0 \\geq b_{n} \\geq b_{n-\\ell} \\geq b_{n-2 \\ell} \\geq \\cdots \\geq-M\n$$\n\nThus, in view of the expansion (2), (3) applied to the sequence $\\left(b_{n}\\right)$, we get that each $b_{n}$ is contained in a set\n\n$$\nT=\\left\\{b_{i_{1}}+b_{i_{2}}+\\cdots+b_{i_{k}}: i_{1}, \\ldots, i_{k} \\leq r\\right\\} \\cap[-M, 0]\n$$\n\nWe claim that this set is finite. Actually, for any $x \\in T$, let $x=b_{i_{1}}+\\cdots+b_{i_{k}}\\left(i_{1}, \\ldots, i_{k} \\leq r\\right)$. Then among $b_{i_{j}}$ 's there are at most $\\frac{M}{\\varepsilon}$ nonzero terms (otherwise $x<\\frac{M}{\\varepsilon} \\cdot(-\\varepsilon)<-M$ ). Thus $x$ can be expressed in the same way with $k \\leq \\frac{M}{\\varepsilon}$, and there is only a finite number of such sums.\n\nFinally, for every $t=1,2, \\ldots, \\ell$ we get that the sequence\n\n$$\nb_{r+t}, b_{r+t+\\ell}, b_{r+t+2 \\ell}, \\ldots\n$$\n\nis non-decreasing and attains the finite number of values; therefore it is constant from some index. Thus, the sequence $\\left(b_{n}\\right)$ is periodic with period $\\ell$ from some index $N$, which means that\n\n$$\nb_{n}=b_{n-\\ell}=b_{n-\\ell}+b_{\\ell} \\quad \\text { for all } n>N+\\ell\n$$\n\nand hence\n\n$$\na_{n}=b_{n}+n s=\\left(b_{n-\\ell}+(n-\\ell) s\\right)+\\left(b_{\\ell}+\\ell s\\right)=a_{n-\\ell}+a_{\\ell} \\quad \\text { for all } n>N+\\ell,\n$$\n\nas desired.""]" [] Text-only Competition True Theorem proof Algebra Math English 81 "Given six positive numbers $a, b, c, d, e, f$ such that $a\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} . \tag{1} $$" "['We define also $\\sigma=a c+c e+a e, \\tau=b d+b f+d f$. The idea of the solution is to interpret (1) as a natural inequality on the roots of an appropriate polynomial.\n\nActually, consider the polynomial\n\n$$\n\\begin{array}{r}\nP(x)=(b+d+f)(x-a)(x-c)(x-e)+(a+c+e)(x-b)(x-d)(x-f) \\\\\n=T\\left(x^{3}-S x^{2}+\\sigma x-a c e\\right)+S\\left(x^{3}-T x^{2}+\\tau x-b d f\\right) .\n\\end{array}\n\\tag{2}\n$$\n\nSurely, $P$ is cubic with leading coefficient $S+T>0$. Moreover, we have\n\n$$\n\\begin{array}{ll}\nP(a)=S(a-b)(a-d)(a-f)<0, & P(c)=S(c-b)(c-d)(c-f)>0 \\\\\nP(e)=S(e-b)(e-d)(e-f)<0, & P(f)=T(f-a)(f-c)(f-e)>0\n\\end{array}\n$$\n\nHence, each of the intervals $(a, c),(c, e),(e, f)$ contains at least one root of $P(x)$. Since there are at most three roots at all, we obtain that there is exactly one root in each interval (denote them by $\\alpha \\in(a, c), \\beta \\in(c, e), \\gamma \\in(e, f))$. Moreover, the polynomial $P$ can be factorized as\n\n$$\nP(x)=(T+S)(x-\\alpha)(x-\\beta)(x-\\gamma) \\text {. }\n\\tag{3}\n$$\n\nEquating the coefficients in the two representations (2) and (3) of $P(x)$ provides\n\n$$\n\\alpha+\\beta+\\gamma=\\frac{2 T S}{T+S}, \\quad \\alpha \\beta+\\alpha \\gamma+\\beta \\gamma=\\frac{S \\tau+T \\sigma}{T+S}\n$$\n\nNow, since the numbers $\\alpha, \\beta, \\gamma$ are distinct, we have\n\n$$\n0<(\\alpha-\\beta)^{2}+(\\alpha-\\gamma)^{2}+(\\beta-\\gamma)^{2}=2(\\alpha+\\beta+\\gamma)^{2}-6(\\alpha \\beta+\\alpha \\gamma+\\beta \\gamma)\n$$\n\nwhich implies\n\n$$\n\\frac{4 S^{2} T^{2}}{(T+S)^{2}}=(\\alpha+\\beta+\\gamma)^{2}>3(\\alpha \\beta+\\alpha \\gamma+\\beta \\gamma)=\\frac{3(S \\tau+T \\sigma)}{T+S}\n$$\n\nor\n\n$$\n4 S^{2} T^{2}>3(T+S)(T \\sigma+S \\tau)\n$$\n\nwhich is exactly what we need.' 'Let\n\n$$\nU=\\frac{1}{2}\\left((e-a)^{2}+(c-a)^{2}+(e-c)^{2}\\right)=S^{2}-3(a c+a e+c e)\n$$\n\nand\n\n$$\nV=\\frac{1}{2}\\left((f-b)^{2}+(f-d)^{2}+(d-b)^{2}\\right)=T^{2}-3(b d+b f+d f)\n$$\n\nThen\n\n$$\n\\begin{aligned}\n& \\text { (L.H.S. })^{2}-(\\text { R.H.S. })^{2}=(2 S T)^{2}-(S+T)(S \\cdot 3(b d+b f+d f)+T \\cdot 3(a c+a e+c e))= \\\\\n& \\quad=4 S^{2} T^{2}-(S+T)\\left(S\\left(T^{2}-V\\right)+T\\left(S^{2}-U\\right)\\right)=(S+T)(S V+T U)-S T(T-S)^{2},\n\\end{aligned}\n$$\n\nand the statement is equivalent with\n\n$$\n(S+T)(S V+T U)>S T(T-S)^{2} .\\tag{4}\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n(S+T)(T U+S V) \\geq(\\sqrt{S \\cdot T U}+\\sqrt{T \\cdot S V})^{2}=S T(\\sqrt{U}+\\sqrt{V})^{2} .\n\\tag{5}\n$$\n\nEstimate the quantities $\\sqrt{U}$ and $\\sqrt{V}$ by the QM-AM inequality with the positive terms $(e-c)^{2}$ and $(d-b)^{2}$ being omitted:\n\n$$\n\\begin{aligned}\n\\sqrt{U}+\\sqrt{V} & >\\sqrt{\\frac{(e-a)^{2}+(c-a)^{2}}{2}}+\\sqrt{\\frac{(f-b)^{2}+(f-d)^{2}}{2}} \\\\\n& >\\frac{(e-a)+(c-a)}{2}+\\frac{(f-b)+(f-d)}{2}=\\left(f-\\frac{d}{2}-\\frac{b}{2}\\right)+\\left(\\frac{e}{2}+\\frac{c}{2}-a\\right) \\\\\n& =(T-S)+\\frac{3}{2}(e-d)+\\frac{3}{2}(c-b)>T-S .\n\\end{aligned}\\tag{6}\n$$\n\nThe estimates (5) and (6) prove (4) and hence the statement.' 'We define $\\sigma=a c+c e+a e, \\tau=b d+b f+d f$.\nMoreover, let $s=c+e$. Note that\n\n$$\n(c-b)(c-d)+(e-f)(e-d)+(e-f)(c-b)<0\n$$\n\nsince each summand is negative. This rewrites as\n\n$$\n\\begin{gathered}\n(b d+b f+d f)-(a c+c e+a e)<(c+e)(b+d+f-a-c-e), \\text { or } \\\\\n\\tau-\\sigma\\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} .\n$$' 'We introduce the expressions $\\sigma$ and $\\tau$ as in the previous solutions. The idea of the solution is to change the values of variables $a, \\ldots, f$ keeping the left-hand side unchanged and increasing the right-hand side; it will lead to a simpler inequality which can be proved in a direct way.\n\nNamely, we change the variables (i) keeping the (non-strict) inequalities $a \\leq b \\leq c \\leq d \\leq$ $e \\leq f$; (ii) keeping the values of sums $S$ and $T$ unchanged; and finally (iii) increasing the values of $\\sigma$ and $\\tau$. Then the left-hand side of (1) remains unchanged, while the right-hand side increases. Hence, the inequality (1) (and even a non-strict version of (1)) for the changed values would imply the same (strict) inequality for the original values.\n\nFirst, we find the sufficient conditions for (ii) and (iii) to be satisfied.\n\nLemma. Let $x, y, z>0$; denote $U(x, y, z)=x+y+z, v(x, y, z)=x y+x z+y z$. Suppose that $x^{\\prime}+y^{\\prime}=x+y$ but $|x-y| \\geq\\left|x^{\\prime}-y^{\\prime}\\right|$; then we have $U\\left(x^{\\prime}, y^{\\prime}, z\\right)=U(x, y, z)$ and $v\\left(x^{\\prime}, y^{\\prime}, z\\right) \\geq$ $v(x, y, z)$ with equality achieved only when $|x-y|=\\left|x^{\\prime}-y^{\\prime}\\right|$.\n\nProof. The first equality is obvious. For the second, we have\n\n$$\n\\begin{aligned}\nv\\left(x^{\\prime}, y^{\\prime}, z\\right)=z\\left(x^{\\prime}+y^{\\prime}\\right)+x^{\\prime} y^{\\prime} & =z\\left(x^{\\prime}+y^{\\prime}\\right)+\\frac{\\left(x^{\\prime}+y^{\\prime}\\right)^{2}-\\left(x^{\\prime}-y^{\\prime}\\right)^{2}}{4} \\\\\n& \\geq z(x+y)+\\frac{(x+y)^{2}-(x-y)^{2}}{4}=v(x, y, z),\n\\end{aligned}\n$$\n\nwith the equality achieved only for $\\left(x^{\\prime}-y^{\\prime}\\right)^{2}=(x-y)^{2} \\Longleftrightarrow\\left|x^{\\prime}-y^{\\prime}\\right|=|x-y|$, as desired.\n\nNow, we apply Lemma several times making the following changes. For each change, we denote the new values by the same letters to avoid cumbersome notations.\n\n1. Let $k=\\frac{d-c}{2}$. Replace $(b, c, d, e)$ by $(b+k, c+k, d-k, e-k)$. After the change we have $a0$, while in the latter one $S(T)=1^{3}+1^{3}+(-1)^{3}+(-1)^{3}=0$, as desired.\n\nNow we turn to the general problem. Consider a tournament $T$ with no bad companies and enumerate the players by the numbers from 1 to $n$. For every 4 players $i_{1}, i_{2}, i_{3}, i_{4}$ consider a ""sub-tournament"" $T_{i_{1} i_{2} i_{3} i_{4}}$ consisting of only these players and the games which they performed with each other. By the abovementioned, we have $S\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right) \\geq 0$. Our aim is to prove that\n\n$$\nS(T)=\\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right)\n\\tag{2}\n$$\n\nwhere the sum is taken over all 4 -tuples of distinct numbers from the set $\\{1, \\ldots, n\\}$. This way the problem statement will be established.\n\nWe interpret the number $\\left(w_{i}-\\ell_{i}\\right)^{3}$ as following. For $i \\neq j$, let $\\varepsilon_{i j}=1$ if the $i$ th player wins against the $j$ th one, and $\\varepsilon_{i j}=-1$ otherwise. Then\n\n$$\n\\left(w_{i}-\\ell_{i}\\right)^{3}=\\left(\\sum_{j \\neq i} \\varepsilon_{i j}\\right)^{3}=\\sum_{j_{1}, j_{2}, j_{3} \\neq i} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}\n$$\n\nHence,\n\n$$\nS(T)=\\sum_{i \\notin\\left\\{j_{1}, j_{2}, j_{3}\\right\\}} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}\n$$\n\nTo simplify this expression, consider all the terms in this sum where two indices are equal. If, for instance, $j_{1}=j_{2}$, then the term contains $\\varepsilon_{i j_{1}}^{2}=1$, so we can replace this term by $\\varepsilon_{i j_{3}}$. Make such replacements for each such term; obviously, after this change each term of the form $\\varepsilon_{i j_{3}}$ will appear $P(T)$ times, hence\n\n$$\nS(T)=\\sum_{\\left|\\left\\{i, j_{1}, j_{2}, j_{3}\\right\\}\\right|=4} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}+P(T) \\sum_{i \\neq j} \\varepsilon_{i j}=S_{1}(T)+P(T) S_{2}(T)\n$$\n\n\n\nWe show that $S_{2}(T)=0$ and hence $S(T)=S_{1}(T)$ for each tournament. Actually, note that $\\varepsilon_{i j}=-\\varepsilon_{j i}$, and the whole sum can be split into such pairs. Since the sum in each pair is 0 , so is $S_{2}(T)$.\n\nThus the desired equality (2) rewrites as\n\n$$\nS_{1}(T)=\\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S_{1}\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right)\n\\tag{3}\n$$\n\nNow, if all the numbers $j_{1}, j_{2}, j_{3}$ are distinct, then the set $\\left\\{i, j_{1}, j_{2}, j_{3}\\right\\}$ is contained in exactly one 4-tuple, hence the term $\\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}$ appears in the right-hand part of (3) exactly once, as well as in the left-hand part. Clearly, there are no other terms in both parts, so the equality is established.' ""Similarly to the first solution, we call the subsets of players as companies, and the $k$-element subsets will be called as $k$-companies.\n\nIn any company of the players, call a player the local champion of the company if he defeated all other members of the company. Similarly, if a player lost all his games against the others in the company then call him the local loser of the company. Obviously every company has at most one local champion and at most one local loser. By the condition of the problem, whenever a 4-company has a local loser, then this company has a local champion as well.\n\nSuppose that $k$ is some positive integer, and let us count all cases when a player is the local champion of some $k$-company. The $i$ th player won against $w_{i}$ other player. To be the local champion of a $k$-company, he must be a member of the company, and the other $k-1$ members must be chosen from those whom he defeated. Therefore, the $i$ th player is the local champion of $\\left(\\begin{array}{c}w_{i} \\\\ k-1\\end{array}\\right) k$-companies. Hence, the total number of local champions of all $k$-companies is $\\sum_{i=1}^{n}\\left(\\begin{array}{c}w_{i} \\\\ k-1\\end{array}\\right)$.\n\nSimilarly, the total number of local losers of the $k$-companies is $\\sum_{i=1}^{n}\\left(\\begin{array}{c}\\ell_{i} \\\\ k-1\\end{array}\\right)$.\n\nNow apply this for $k=2,3$ and 4 .\n\nSince every game has a winner and a loser, we have $\\sum_{i=1}^{n} w_{i}=\\sum_{i=1}^{n} \\ell_{i}=\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$, and hence\n\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)=0\n\\tag{4}\n$$\n\nIn every 3-company, either the players defeated one another in a cycle or the company has both a local champion and a local loser. Therefore, the total number of local champions and local losers in the 3 -companies is the same, $\\sum_{i=1}^{n}\\left(\\begin{array}{c}w_{i} \\\\ 2\\end{array}\\right)=\\sum_{i=1}^{n}\\left(\\begin{array}{c}\\ell_{i} \\\\ 2\\end{array}\\right)$. So we have\n\n$$\n\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n2\n\\end{array}\\right)\\right)=0\\tag{5}\n$$\n\nIn every 4-company, by the problem's condition, the number of local losers is less than or equal to the number of local champions. Then the same holds for the total numbers of local\n\n\n\nchampions and local losers in all 4-companies, so $\\sum_{i=1}^{n}\\left(\\begin{array}{c}w_{i} \\\\ 3\\end{array}\\right) \\geq \\sum_{i=1}^{n}\\left(\\begin{array}{c}\\ell_{i} \\\\ 3\\end{array}\\right)$. Hence,\n\n$$\n\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n3\n\\end{array}\\right)\\right) \\geq 0\\tag{6}\n$$\n\nNow we establish the problem statement (1) as a linear combination of (4), (5) and (6). It is easy check that\n\n$$\n(x-y)^{3}=24\\left(\\left(\\begin{array}{l}\nx \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{l}\ny \\\\\n3\n\\end{array}\\right)\\right)+24\\left(\\left(\\begin{array}{l}\nx \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{l}\ny \\\\\n2\n\\end{array}\\right)\\right)-\\left(3(x+y)^{2}-4\\right)(x-y)\n$$\n\nApply this identity to $x=w_{1}$ and $y=\\ell_{i}$. Since every player played $n-1$ games, we have $w_{i}+\\ell_{i}=n-1$, and thus\n\n$$\n\\left(w_{i}-\\ell_{i}\\right)^{3}=24\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n3\n\\end{array}\\right)\\right)+24\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n2\n\\end{array}\\right)\\right)-\\left(3(n-1)^{2}-4\\right)\\left(w_{i}-\\ell_{i}\\right)\n$$\n\nThen\n\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3}=24 \\underbrace{\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n3\n\\end{array}\\right)\\right)}_{\\geq 0}+24 \\underbrace{\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n2\n\\end{array}\\right)\\right)}_{0}-\\left(3(n-1)^{2}-4\\right) \\underbrace{\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)}_{0} \\geq 0 .\n$$""]" [] Text-only Competition True Theorem proof Combinatorics Math English 83 "Given a positive integer $k$ and other two integers $b>w>1$. There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: (i) The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then $k$ first ones (if they consist of more than one pearl) are chosen; if there are less than $k$ strings longer than 1 , then one chooses all of them. (ii) Next, one cuts each chosen string into two parts differing in length by at most one. (For instance, if there are strings of 5, 4, 4, 2 black pearls, strings of $8,4,3$ white pearls and $k=4$, then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts $(4,4),(3,2),(2,2)$ and $(2,2)$, respectively.) The process stops immediately after the step when a first isolated white pearl appears. Prove that at this stage, there will still exist a string of at least two black pearls." "['Denote the situation after the $i$ th step by $A_{i}$; hence $A_{0}$ is the initial situation, and $A_{i-1} \\rightarrow A_{i}$ is the $i$ th step. We call a string containing $m$ pearls an $m$-string; it is an $m$-w-string or a $m$-b-string if it is white or black, respectively.\n\nWe continue the process until every string consists of a single pearl. We will focus on three moments of the process: (a) the first stage $A_{s}$ when the first 1-string (no matter black or white) appears; (b) the first stage $A_{t}$ where the total number of strings is greater than $k$ (if such moment does not appear then we put $t=\\infty$ ); and (c) the first stage $A_{f}$ when all black pearls are isolated. It is sufficient to prove that in $A_{f-1}$ (or earlier), a 1-w-string appears.\n\nWe start with some easy properties of the situations under consideration. Obviously, we have $s \\leq f$. Moreover, all b-strings from $A_{f-1}$ become single pearls in the $f$ th step, hence all of them are 1 - or 2 -b-strings.\n\nNext, observe that in each step $A_{i} \\rightarrow A_{i+1}$ with $i \\leq t-1$, all $(>1)$-strings were cut since there are not more than $k$ strings at all; if, in addition, $i1$ as $s-1 \\leq \\min \\{s, t\\}$. Now, if $s=f$, then in $A_{s-1}$, there is no $1-\\mathrm{w}$-string as well as no $(>2)$-b-string. That is, $2=B_{s-1} \\geq W_{s-1} \\geq b_{s-1} \\geq w_{s-1}>1$, hence all these numbers equal 2 . This means that in $A_{s-1}$, all strings contain 2 pearls, and there are $2^{s-1}$ black and $2^{s-1}$ white strings, which means $b=2 \\cdot 2^{s-1}=w$. This contradicts the problem conditions.\n\nHence we have $s \\leq f-1$ and thus $s \\leq t$. Therefore, in the $s$ th step each string is cut into two parts. Now, if a 1-b-string appears in this step, then from $w_{s-1} \\leq b_{s-1}$ we see that a 1-w-string appears as well; so, in each case in the sth step a 1-w-string appears, while not all black pearls become single, as desired.\n\nCase 2. Now assume that $t+1 \\leq s$ and $t+2 \\leq f$. Then in $A_{t}$ we have exactly $2^{t}$ white and $2^{t}$ black strings, all being larger than 1 , and $2^{t+1}>k \\geq 2^{t}$ (the latter holds since $2^{t}$ is the total number of strings in $\\left.A_{t-1}\\right)$. Now, in the $(t+1)$ st step, exactly $k$ strings are cut, not more than $2^{t}$ of them being black; so the number of w-strings in $A_{t+1}$ is at least $2^{t}+\\left(k-2^{t}\\right)=k$. Since the number of w-strings does not decrease in our process, in $A_{f-1}$ we have at least $k$ white strings as well.\n\nFinally, in $A_{f-1}$, all b-strings are not larger than 2, and at least one 2-b-string is cut in the $f$ th step. Therefore, at most $k-1$ white strings are cut in this step, hence there exists a w-string $\\mathcal{W}$ which is not cut in the $f$ th step. On the other hand, since a 2 -b-string is cut, all $(\\geq 2)$-w-strings should also be cut in the $f$ th step; hence $\\mathcal{W}$ should be a single pearl. This is exactly what we needed.' 'We use the same notations as introduced in the first paragraph of the previous solution. We claim that at every stage, there exist a $u$-b-string and a $v$-w-string such that either\n\n(i) $u>v \\geq 1$, or\n\n(ii) $2 \\leq u \\leq v<2 u$, and there also exist $k-1$ of $(>v / 2)$-strings other than considered above.\n\nFirst, we notice that this statement implies the problem statement. Actually, in both cases (i) and (ii) we have $u>1$, so at each stage there exists a $(\\geq 2)$-b-string, and for the last stage it is exactly what we need.\n\nNow, we prove the claim by induction on the number of the stage. Obviously, for $A_{0}$ the condition (i) holds since $b>w$. Further, we suppose that the statement holds for $A_{i}$, and prove it for $A_{i+1}$. Two cases are possible.\n\nCase 1. Assume that in $A_{i}$, there are a $u$-b-string and a $v$-w-string with $u>v$. We can assume that $v$ is the length of the shortest w-string in $A_{i}$; since we are not at the final stage, we have $v \\geq 2$. Now, in the $(i+1)$ st step, two subcases may occur.\n\nSubcase 1a. Suppose that either no $u$-b-string is cut, or both some $u$-b-string and some $v$-w-string are cut. Then in $A_{i+1}$, we have either a $u$-b-string and a $(\\leq v)$-w-string (and (i) is valid), or we have a $\\lceil u / 2\\rceil$-b-string and a $\\lfloor v / 2\\rfloor$-w-string. In the latter case, from $u>v$ we get $\\lceil u / 2\\rceil>\\lfloor v / 2\\rfloor$, and (i) is valid again.\n\nSubcase 1b. Now, some $u$-b-string is cut, and no $v$-w-string is cut (and hence all the strings which are cut are longer than $v$ ). If $u^{\\prime}=\\lceil u / 2\\rceil>v$, then the condition (i) is satisfied since we have a $u^{\\prime}$-b-string and a $v$-w-string in $A_{i+1}$. Otherwise, notice that the inequality $u>v \\geq 2$ implies $u^{\\prime} \\geq 2$. Furthermore, besides a fixed $u$-b-string, other $k-1$ of $(\\geq v+1)$-strings should be cut in the $(i+1)$ st step, hence providing at least $k-1$ of $(\\geq\\lceil(v+1) / 2\\rceil)$-strings, and $\\lceil(v+1) / 2\\rceil>v / 2$. So, we can put $v^{\\prime}=v$, and we have $u^{\\prime} \\leq vv$, so each one results in a $(>v / 2)$ string. Hence in $A_{i+1}$, there exist $k \\geq k-1$ of $(>v / 2)$-strings other than the considered $u$ - and $v$-strings, and the condition (ii) is satisfied.\n\nSubcase 2c. In the remaining case, all $u$-b-strings are cut. This means that all $(\\geq u)$-strings are cut as well, hence our $v$-w-string is cut. Therefore in $A_{i+1}$ there exists a $[u / 2]$-b-string together with a $\\lfloor v / 2\\rfloor$-w-string. Now, if $u^{\\prime}=\\lceil u / 2\\rceil>\\lfloor v / 2\\rfloor=v^{\\prime}$ then the condition (i) is fulfilled. Otherwise, we have $u^{\\prime} \\leq v^{\\prime}$ $|f(W)|$; if there is no such set then we set $W=\\varnothing$. Denote $W^{\\prime}=f(W), U=V \\backslash W, U^{\\prime}=V^{\\prime} \\backslash W^{\\prime}$.\n\n\n\nBy our assumption and the Lemma condition, $|f(V)|=\\left|V^{\\prime}\\right| \\geq|V|$, hence $W \\neq V$ and $U \\neq \\varnothing$. Permuting the coordinates, we can assume that $U^{\\prime}=\\left\\{v_{i j}: 1 \\leq i \\leq \\ell\\right\\}, W^{\\prime}=\\left\\{v_{i j}: \\ell+1 \\leq i \\leq k\\right\\}$.\n\nConsider the induced subgraph $G^{\\prime}$ of $G$ on the vertices $U \\cup U^{\\prime}$. We claim that for every $X \\subset U$, we get $\\left|f(X) \\cap U^{\\prime}\\right| \\geq|X|$ (so $G^{\\prime}$ satisfies the conditions of Hall's lemma). Actually, we have $|W| \\geq|f(W)|$, so if $|X|>\\left|f(X) \\cap U^{\\prime}\\right|$ for some $X \\subset U$, then we have\n\n$$\n|W \\cup X|=|W|+|X|>|f(W)|+\\left|f(X) \\cap U^{\\prime}\\right|=\\left|f(W) \\cup\\left(f(X) \\cap U^{\\prime}\\right)\\right|=|f(W \\cup X)|\n$$\n\nThis contradicts the maximality of $|W|$.\n\nThus, applying Hall's lemma, we can assign to each $L \\in U$ some vertex $v_{i j} \\in U^{\\prime}$ so that to distinct elements of $U$, distinct vertices of $U^{\\prime}$ are assigned. In this situation, we say that $L \\in U$ corresponds to the $i$ th axis, and write $g(L)=i$. Since there are $n_{i}-1$ vertices of the form $v_{i j}$, we get that for each $1 \\leq i \\leq \\ell$, not more than $n_{i}-1$ subgrids correspond to the $i$ th axis.\n\nFinally, we are ready to present the desired point. Since $W \\neq V$, there exists a point $b=\\left(b_{1}, b_{2}, \\ldots, b_{k}\\right) \\in N \\backslash\\left(\\cup_{L \\in W} L\\right)$. On the other hand, for every $1 \\leq i \\leq \\ell$, consider any subgrid $L \\in U$ with $g(L)=i$. This means exactly that $L$ is orthogonal to the $i$ th axis, and hence all its elements have the same $i$ th coordinate $c_{L}$. Since there are at most $n_{i}-1$ such subgrids, there exists a number $0 \\leq a_{i} \\leq n_{i}-1$ which is not contained in a set $\\left\\{c_{L}: g(L)=i\\right\\}$. Choose such number for every $1 \\leq i \\leq \\ell$. Now we claim that point $a=\\left(a_{1}, \\ldots, a_{\\ell}, b_{\\ell+1}, \\ldots, b_{k}\\right)$ is not covered, hence contradicting the Lemma condition.\n\nSurely, point $a$ cannot lie in some $L \\in U$, since all the points in $L$ have $g(L)$ th coordinate $c_{L} \\neq a_{g(L)}$. On the other hand, suppose that $a \\in L$ for some $L \\in W$; recall that $b \\notin L$. But the points $a$ and $b$ differ only at first $\\ell$ coordinates, so $L$ should be orthogonal to at least one of the first $\\ell$ axes, and hence our graph contains some edge $\\left(L, v_{i j}\\right)$ for $i \\leq \\ell$. It contradicts the definition of $W^{\\prime}$. The Lemma is proved.\n\nNow we turn to the problem. Let $d_{j}$ be the step of the progression $P_{j}$. Note that since $n=$ l.c.m. $\\left(d_{1}, \\ldots, d_{s}\\right)$, for each $1 \\leq i \\leq k$ there exists an index $j(i)$ such that $p_{i}^{\\alpha_{i}} \\mid d_{j(i)}$. We assume that $n>1$; otherwise the problem statement is trivial.\n\nFor each $0 \\leq m \\leq n-1$ and $1 \\leq i \\leq k$, let $m_{i}$ be the residue of $m$ modulo $p_{i}^{\\alpha_{i}}$, and let $m_{i}=\\overline{r_{i \\alpha_{i}} \\ldots r_{i 1}}$ be the base $p_{i}$ representation of $m_{i}$ (possibly, with some leading zeroes). Now, we put into correspondence to $m$ the sequence $r(m)=\\left(r_{11}, \\ldots, r_{1 \\alpha_{1}}, r_{21}, \\ldots, r_{k \\alpha_{k}}\\right)$. Hence $r(m)$ lies in a $\\underbrace{p_{1} \\times \\cdots \\times p_{1}}_{\\alpha_{1} \\text { times }} \\times \\cdots \\times \\underbrace{p_{k} \\times \\cdots \\times p_{k}}_{\\alpha_{k} \\text { times }} \\operatorname{grid} N$.\n\nSurely, if $r(m)=r\\left(m^{\\prime}\\right)$ then $p_{i}^{\\alpha_{i}} \\mid m_{i}-m_{i}^{\\prime}$, which follows $p_{i}^{\\alpha_{i}} \\mid m-m^{\\prime}$ for all $1 \\leq i \\leq k$; consequently, $n \\mid m-m^{\\prime}$. So, when $m$ runs over the set $\\{0, \\ldots, n-1\\}$, the sequences $r(m)$ do not repeat; since $|N|=n$, this means that $r$ is a bijection between $\\{0, \\ldots, n-1\\}$ and $N$. Now we will show that for each $1 \\leq i \\leq s$, the set $L_{i}=\\left\\{r(m): m \\in P_{i}\\right\\}$ is a subgrid, and that for each axis there exists a subgrid orthogonal to this axis. Obviously, these subgrids cover $N$, and the condition (ii') follows directly from (ii). Hence the Lemma provides exactly the estimate we need.\n\nConsider some $1 \\leq j \\leq s$ and let $d_{j}=p_{1}^{\\gamma_{1}} \\ldots p_{k}^{\\gamma_{k}}$. Consider some $q \\in P_{j}$ and let $r(q)=$ $\\left(r_{11}, \\ldots, r_{k \\alpha_{k}}\\right)$. Then for an arbitrary $q^{\\prime}$, setting $r\\left(q^{\\prime}\\right)=\\left(r_{11}^{\\prime}, \\ldots, r_{k \\alpha_{k}}^{\\prime}\\right)$ we have\n\n$$\nq^{\\prime} \\in P_{j} \\quad \\Longleftrightarrow \\quad p_{i}^{\\gamma_{i}} \\mid q-q^{\\prime} \\text { for each } 1 \\leq i \\leq k \\quad \\Longleftrightarrow \\quad r_{i, t}=r_{i, t}^{\\prime} \\text { for all } t \\leq \\gamma_{i} \\text {. }\n$$\n\nHence $L_{j}=\\left\\{\\left(r_{11}^{\\prime}, \\ldots, r_{k \\alpha_{k}}^{\\prime}\\right) \\in N: r_{i, t}=r_{i, t}^{\\prime}\\right.$ for all $\\left.t \\leq \\gamma_{i}\\right\\}$ which means that $L_{j}$ is a subgrid containing $r(q)$. Moreover, in $L_{j(i)}$, all the coordinates corresponding to $p_{i}$ are fixed, so it is orthogonal to all of their axes, as desired."" ""We start with introducing some notation. For positive integer $r$, we denote $[r]=\\{1,2, \\ldots, r\\}$. Next, we say that a set of progressions $\\mathcal{P}=\\left\\{P_{1}, \\ldots, P_{s}\\right\\}$ cover $\\mathbb{Z}$ if each integer belongs to some of them; we say that this covering is minimal if no proper subset of $\\mathcal{P}$ covers $\\mathbb{Z}$. Obviously, each covering contains a minimal subcovering.\n\nNext, for a minimal covering $\\left\\{P_{1}, \\ldots, P_{s}\\right\\}$ and for every $1 \\leq i \\leq s$, let $d_{i}$ be the step of progression $P_{i}$, and $h_{i}$ be some number which is contained in $P_{i}$ but in none of the other progressions. We assume that $n>1$, otherwise the problem is trivial. This implies $d_{i}>1$, otherwise the progression $P_{i}$ covers all the numbers, and $n=1$.\n\nWe will prove a more general statement, namely the following\n\nClaim. Assume that the progressions $P_{1}, \\ldots, P_{s}$ and number $n=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}>1$ are chosen as in the problem statement. Moreover, choose some nonempty set of indices $I=\\left\\{i_{1}, \\ldots, i_{t}\\right\\} \\subseteq[k]$ and some positive integer $\\beta_{i} \\leq \\alpha_{i}$ for every $i \\in I$. Consider the set of indices\n\n$$\nT=\\left\\{j: 1 \\leq j \\leq s, \\text { and } p_{i}^{\\alpha_{i}-\\beta_{i}+1} \\mid d_{j} \\text { for some } i \\in I\\right\\} .\n$$\n\nThen\n\n$$\n|T| \\geq 1+\\sum_{i \\in I} \\beta_{i}\\left(p_{i}-1\\right)\n\\tag{2}\n$$\n\nObserve that the Claim for $I=[k]$ and $\\beta_{i}=\\alpha_{i}$ implies the problem statement, since the left-hand side in (2) is not greater than $s$. Hence, it suffices to prove the Claim.\n\n1. First, we prove the Claim assuming that all $d_{j}$ 's are prime numbers. If for some $1 \\leq i \\leq k$ we have at least $p_{i}$ progressions with the step $p_{i}$, then they do not intersect and hence cover all the integers; it means that there are no other progressions, and $n=p_{i}$; the Claim is trivial in this case.\n\nNow assume that for every $1 \\leq i \\leq k$, there are not more than $p_{i}-1$ progressions with step $p_{i}$; each such progression covers the numbers with a fixed residue modulo $p_{i}$, therefore there exists a residue $q_{i} \\bmod p_{i}$ which is not touched by these progressions. By the Chinese Remainder Theorem, there exists a number $q$ such that $q \\equiv q_{i}\\left(\\bmod p_{i}\\right)$ for all $1 \\leq i \\leq k$; this number cannot be covered by any progression with step $p_{i}$, hence it is not covered at all. A contradiction.\n\n\n\n2. Now, we assume that the general Claim is not valid, and hence we consider a counterexample $\\left\\{P_{1}, \\ldots, P_{s}\\right\\}$ for the Claim; we can choose it to be minimal in the following sense:\n\n- the number $n$ is minimal possible among all the counterexamples; of $n$.\n- the sum $\\sum_{i} d_{i}$ is minimal possible among all the counterexamples having the chosen value\n\nAs was mentioned above, not all numbers $d_{i}$ are primes; hence we can assume that $d_{1}$ is composite, say $p_{1} \\mid d_{1}$ and $d_{1}^{\\prime}=\\frac{d_{1}}{p_{1}}>1$. Consider a progression $P_{1}^{\\prime}$ having the step $d_{1}^{\\prime}$, and containing $P_{1}$. We will focus on two coverings constructed as follows.\n\n(i) Surely, the progressions $P_{1}^{\\prime}, P_{2}, \\ldots, P_{s}$ cover $\\mathbb{Z}$, though this covering in not necessarily minimal. So, choose some minimal subcovering $\\mathcal{P}^{\\prime}$ in it; surely $P_{1}^{\\prime} \\in \\mathcal{P}^{\\prime}$ since $h_{1}$ is not covered by $P_{2}, \\ldots, P_{s}$, so we may assume that $\\mathcal{P}^{\\prime}=\\left\\{P_{1}^{\\prime}, P_{2}, \\ldots, P_{s^{\\prime}}\\right\\}$ for some $s^{\\prime} \\leq s$. Furthermore, the period of the covering $\\mathcal{P}^{\\prime}$ can appear to be less than $n$; so we denote this period by\n\n$$\nn^{\\prime}=p_{1}^{\\alpha_{1}-\\sigma_{1}} \\ldots p_{k}^{\\alpha_{k}-\\sigma_{k}}=\\text { l.c.m. }\\left(d_{1}^{\\prime}, d_{2}, \\ldots, d_{s^{\\prime}}\\right)\n$$\n\nObserve that for each $P_{j} \\notin \\mathcal{P}^{\\prime}$, we have $h_{j} \\in P_{1}^{\\prime}$, otherwise $h_{j}$ would not be covered by $\\mathcal{P}$.\n\n(ii) On the other hand, each nonempty set of the form $R_{i}=P_{i} \\cap P_{1}^{\\prime}(1 \\leq i \\leq s)$ is also a progression with a step $r_{i}=$ l.c.m. $\\left(d_{i}, d_{1}^{\\prime}\\right)$, and such sets cover $P_{1}^{\\prime}$. Scaling these progressions with the ratio $1 / d_{1}^{\\prime}$, we obtain the progressions $Q_{i}$ with steps $q_{i}=r_{i} / d_{1}^{\\prime}$ which cover $\\mathbb{Z}$. Now we choose a minimal subcovering $\\mathcal{Q}$ of this covering; again we should have $Q_{1} \\in \\mathcal{Q}$ by the reasons of $h_{1}$. Now, denote the period of $\\mathcal{Q}$ by\n\n$$\nn^{\\prime \\prime}=\\text { l.c.m. }\\left\\{q_{i}: Q_{i} \\in \\mathcal{Q}\\right\\}=\\frac{\\text { l.c.m. }\\left\\{r_{i}: Q_{i} \\in \\mathcal{Q}\\right\\}}{d_{1}^{\\prime}}=\\frac{p_{1}^{\\gamma_{1}} \\ldots p_{k}^{\\gamma_{k}}}{d_{1}^{\\prime}} \\text {. }\n$$\n\nNote that if $h_{j} \\in P_{1}^{\\prime}$, then the image of $h_{j}$ under the scaling can be covered by $Q_{j}$ only; so, in this case we have $Q_{j} \\in \\mathcal{Q}$.\n\nOur aim is to find the desired number of progressions in coverings $\\mathcal{P}$ and $\\mathcal{Q}$. First, we have $n \\geq n^{\\prime}$, and the sum of the steps in $\\mathcal{P}^{\\prime}$ is less than that in $\\mathcal{P}$; hence the Claim is valid for $\\mathcal{P}^{\\prime}$. We apply it to the set of indices $I^{\\prime}=\\left\\{i \\in I: \\beta_{i}>\\sigma_{i}\\right\\}$ and the exponents $\\beta_{i}^{\\prime}=\\beta_{i}-\\sigma_{i}$; hence the set under consideration is\n\n$$\nT^{\\prime}=\\left\\{j: 1 \\leq j \\leq s^{\\prime}, \\text { and } p_{i}^{\\left(\\alpha_{i}-\\sigma_{i}\\right)-\\beta_{i}^{\\prime}+1}=p_{i}^{\\alpha_{i}-\\beta_{i}+1} \\mid d_{j} \\text { for some } i \\in I^{\\prime}\\right\\} \\subseteq T \\cap\\left[s^{\\prime}\\right] \\text {, }\n$$\n\nand we obtain that\n\n$$\n\\left|T \\cap\\left[s^{\\prime}\\right]\\right| \\geq\\left|T^{\\prime}\\right| \\geq 1+\\sum_{i \\in I^{\\prime}}\\left(\\beta_{i}-\\sigma_{i}\\right)\\left(p_{i}-1\\right)=1+\\sum_{i \\in I}\\left(\\beta_{i}-\\sigma_{i}\\right)_{+}\\left(p_{i}-1\\right)\n$$\n\nwhere $(x)_{+}=\\max \\{x, 0\\}$; the latter equality holds as for $i \\notin I^{\\prime}$ we have $\\beta_{i} \\leq \\sigma_{i}$.\n\nObserve that $x=(x-y)_{+}+\\min \\{x, y\\}$ for all $x, y$. So, if we find at least\n\n$$\nG=\\sum_{i \\in I} \\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}\\left(p_{i}-1\\right)\n$$\n\nindices in $T \\cap\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}$, then we would have\n\n$|T|=\\left|T \\cap\\left[s^{\\prime}\\right]\\right|+\\left|T \\cap\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}\\right| \\geq 1+\\sum_{i \\in I}\\left(\\left(\\beta_{i}-\\sigma_{i}\\right)_{+}+\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}\\right)\\left(p_{i}-1\\right)=1+\\sum_{i \\in I} \\beta_{i}\\left(p_{i}-1\\right)$,\n\nthus leading to a contradiction with the choice of $\\mathcal{P}$. We will find those indices among the indices of progressions in $\\mathcal{Q}$.\n\n\n\n3. Now denote $I^{\\prime \\prime}=\\left\\{i \\in I: \\sigma_{i}>0\\right\\}$ and consider some $i \\in I^{\\prime \\prime}$; then $p_{i}^{\\alpha_{i}} \\nmid n^{\\prime}.$. On the other hand, there exists an index $j(i)$ such that $p_{i}^{\\alpha_{i}} \\mid d_{j(i)}$; this means that $d_{j(i)} \\nmid n^{\\prime}$ and hence $P_{j(i)}$ cannot appear in $\\mathcal{P}^{\\prime}$, so $j(i)>s^{\\prime}$. Moreover, we have observed before that in this case $h_{j(i)} \\in P_{1}^{\\prime}$, hence $Q_{j(i)} \\in \\mathcal{Q}$. This means that $q_{j(i)} \\mid n^{\\prime \\prime}$, therefore $\\gamma_{i}=\\alpha_{i}$ for each $i \\in I^{\\prime \\prime}$ (recall here that $q_{i}=r_{i} / d_{1}^{\\prime}$ and hence $\\left.d_{j(i)}\\left|r_{j(i)}\\right| d_{1}^{\\prime} n^{\\prime \\prime}\\right)$.\n\nLet $d_{1}^{\\prime}=p_{1}^{\\tau_{1}} \\ldots p_{k}^{\\tau_{k}}$. Then $n^{\\prime \\prime}=p_{1}^{\\gamma_{1}-\\tau_{1}} \\ldots p_{k}^{\\gamma_{i}-\\tau_{i}}$. Now, if $i \\in I^{\\prime \\prime}$, then for every $\\beta$ the condition $p_{i}^{\\left(\\gamma_{i}-\\tau_{i}\\right)-\\beta+1} \\mid q_{j}$ is equivalent to $p_{i}^{\\alpha_{i}-\\beta+1} \\mid r_{j}$.\n\nNote that $n^{\\prime \\prime} \\leq n / d_{1}^{\\prime}0$. So, the set under consideration is\n\n$$\n\\begin{aligned}\nT^{\\prime \\prime} & =\\left\\{j: Q_{j} \\in \\mathcal{Q}, \\text { and } p_{i}^{\\left(\\gamma_{i}-\\tau_{i}\\right)-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid q_{j} \\text { for some } i \\in I^{\\prime \\prime}\\right\\} \\\\\n& =\\left\\{j: Q_{j} \\in \\mathcal{Q}, \\text { and } p_{i}^{\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid r_{j} \\text { for some } i \\in I^{\\prime \\prime}\\right\\},\n\\end{aligned}\n$$\n\nand we obtain $\\left|T^{\\prime \\prime}\\right| \\geq 1+G$. Finally, we claim that $T^{\\prime \\prime} \\subseteq T \\cap\\left(\\{1\\} \\cup\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}\\right)$; then we will obtain $\\left|T \\cap\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}\\right| \\geq G$, which is exactly what we need.\n\nTo prove this, consider any $j \\in T^{\\prime \\prime}$. Observe first that $\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1>\\alpha_{i}-\\sigma_{i} \\geq \\tau_{i}$, hence from $p_{i}^{\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid r_{j}=$ l.c.m. $\\left(d_{1}^{\\prime}, d_{j}\\right)$ we have $p_{i}^{\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid d_{j}$, which means that $j \\in T$. Next, the exponent of $p_{i}$ in $d_{j}$ is greater than that in $n^{\\prime}$, which means that $P_{j} \\notin \\mathcal{P}^{\\prime}$. This may appear only if $j=1$ or $j>s^{\\prime}$, as desired. This completes the proof.""]" [] Text-only Competition True Theorem proof Geometry Math English 85 Let $A B C$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $B C, C A, A B$ respectively. One of the intersection points of the line $E F$ and the circumcircle is $P$. The lines $B P$ and $D F$ meet at point $Q$. Prove that $A P=A Q$. "['The line $E F$ intersects the circumcircle at two points. Depending on the choice of $P$, there are two different cases to consider.\n\nCase 1: The point $P$ lies on the ray $E F$ (see Fig. 1).\n\nLet $\\angle C A B=\\alpha, \\angle A B C=\\beta$ and $\\angle B C A=\\gamma$. The quadrilaterals $B C E F$ and $C A F D$ are cyclic due to the right angles at $D, E$ and $F$. So,\n\n$$\n\\begin{aligned}\n& \\angle B D F=180^{\\circ}-\\angle F D C=\\angle C A F=\\alpha, \\\\\n& \\angle A F E=180^{\\circ}-\\angle E F B=\\angle B C E=\\gamma, \\\\\n& \\angle D F B=180^{\\circ}-\\angle A F D=\\angle D C A=\\gamma .\n\\end{aligned}\n$$\n\nSince $P$ lies on the arc $A B$ of the circumcircle, $\\angle P B A<\\angle B C A=\\gamma$. Hence, we have\n\n$$\n\\angle P B D+\\angle B D F=\\angle P B A+\\angle A B D+\\angle B D F<\\gamma+\\beta+\\alpha=180^{\\circ}\n$$\n\nand the point $Q$ must lie on the extensions of $B P$ and $D F$ beyond the points $P$ and $F$, respectively.\n\nFrom the cyclic quadrilateral $A P B C$ we get\n\n$$\n\\angle Q P A=180^{\\circ}-\\angle A P B=\\angle B C A=\\gamma=\\angle D F B=\\angle Q F A .\n$$\n\nHence, the quadrilateral $A Q P F$ is cyclic. Then $\\angle A Q P=180^{\\circ}-\\angle P F A=\\angle A F E=\\gamma$.\n\nWe obtained that $\\angle A Q P=\\angle Q P A=\\gamma$, so the triangle $A Q P$ is isosceles, $A P=A Q$.\n\n\n\nFig. 1\n\n\n\nFig. 2\n\n\n\nCase 2: The point $P$ lies on the ray $F E$ (see Fig. 2). In this case the point $Q$ lies inside the segment $F D$.\n\nSimilarly to the first case, we have\n\n$$\n\\angle Q P A=\\angle B C A=\\gamma=\\angle D F B=180^{\\circ}-\\angle A F Q .\n$$\n\nHence, the quadrilateral $A F Q P$ is cyclic.\n\nThen $\\angle A Q P=\\angle A F P=\\angle A F E=\\gamma=\\angle Q P A$. The triangle $A Q P$ is isosceles again, $\\angle A Q P=\\angle Q P A$ and thus $A P=A Q$.' 'For arbitrary points $X, Y$ on the circumcircle, denote by $\\overparen{X Y}$ the central angle of the $\\operatorname{arc} X Y$.\n\nLet $P$ and $P^{\\prime}$ be the two points where the line $E F$ meets the circumcircle; let $P$ lie on the $\\operatorname{arc} A B$ and let $P^{\\prime}$ lie on the $\\operatorname{arc} C A$. Let $B P$ and $B P^{\\prime}$ meet the line $D F$ and $Q$ and $Q^{\\prime}$, respectively (see Fig. 3). We will prove that $A P=A P^{\\prime}=A Q=A Q^{\\prime}$.\n\n\n\nFig. 3\n\nLike in the first solution, we have $\\angle A F E=\\angle B F P=\\angle D F B=\\angle B C A=\\gamma$ from the cyclic quadrilaterals $B C E F$ and $C A F D$.\n\nBy $\\overparen{P B}+\\overparen{P^{\\prime} A}=2 \\angle A F P^{\\prime}=2 \\gamma=2 \\angle B C A=\\overparen{A P}+\\overparen{P B}$, we have\n\n$$\n\\overparen{A P}=\\overparen{P^{\\prime} A}, \\quad \\angle P B A=\\angle A B P^{\\prime} \\quad \\text { and } \\quad A P=A P^{\\prime} \\text {. }\\tag{1}\n$$\n\nDue to $\\overparen{A P}=\\overparen{P^{\\prime} A}$, the lines $B P$ and $B Q^{\\prime}$ are symmetrical about line $A B$.\n\nSimilarly, by $\\angle B F P=\\angle Q^{\\prime} F B$, the lines $F P$ and $F Q^{\\prime}$ are symmetrical about $A B$. It follows that also the points $P$ and $P^{\\prime}$ are symmetrical to $Q^{\\prime}$ and $Q$, respectively. Therefore,\n\n$$\nA P=A Q^{\\prime} \\quad \\text { and } \\quad A P^{\\prime}=A Q .\\tag{2}\n$$\n\nThe relations (1) and (2) together prove $A P=A P^{\\prime}=A Q=A Q^{\\prime}$.']" [] Text-only Competition True Theorem proof Geometry Math English 86 Point $P$ lies inside triangle $A B C$. Lines $A P, B P, C P$ meet the circumcircle of $A B C$ again at points $K, L, M$, respectively. The tangent to the circumcircle at $C$ meets line $A B$ at $S$. Prove that $S C=S P$ if and only if $M K=M L$. "['We assume that $C A>C B$, so point $S$ lies on the ray $A B$.\n\nFrom the similar triangles $\\triangle P K M \\sim \\triangle P C A$ and $\\triangle P L M \\sim \\triangle P C B$ we get $\\frac{P M}{K M}=\\frac{P A}{C A}$ and $\\frac{L M}{P M}=\\frac{C B}{P B}$. Multiplying these two equalities, we get\n\n$$\n\\frac{L M}{K M}=\\frac{C B}{C A} \\cdot \\frac{P A}{P B}\n$$\n\nHence, the relation $M K=M L$ is equivalent to $\\frac{C B}{C A}=\\frac{P B}{P A}$.\n\nDenote by $E$ the foot of the bisector of angle $B$ in triangle $A B C$. Recall that the locus of points $X$ for which $\\frac{X A}{X B}=\\frac{C A}{C B}$ is the Apollonius circle $\\Omega$ with the center $Q$ on the line $A B$, and this circle passes through $C$ and $E$. Hence, we have $M K=M L$ if and only if $P$ lies on $\\Omega$, that is $Q P=Q C$.\n\n\n\nFig. 1\n\nNow we prove that $S=Q$, thus establishing the problem statement. We have $\\angle C E S=$ $\\angle C A E+\\angle A C E=\\angle B C S+\\angle E C B=\\angle E C S$, so $S C=S E$. Hence, the point $S$ lies on $A B$ as well as on the perpendicular bisector of $C E$ and therefore coincides with $Q$.' 'As in the previous solution, we assume that $S$ lies on the ray $A B$.\n\n1. Let $P$ be an arbitrary point inside both the circumcircle $\\omega$ of the triangle $A B C$ and the angle $A S C$, the points $K, L, M$ defined as in the problem. We claim that $S P=S C$ implies $M K=M L$.\n\nLet $E$ and $F$ be the points of intersection of the line $S P$ with $\\omega$, point $E$ lying on the segment $S P$ (see Fig. 2).\n\n\n\n\n\nFig. 2\n\nWe have $S P^{2}=S C^{2}=S A \\cdot S B$, so $\\frac{S P}{S B}=\\frac{S A}{S P}$, and hence $\\triangle P S A \\sim \\triangle B S P$. Then $\\angle B P S=\\angle S A P$. Since $2 \\angle B P S=\\overparen{B E}+\\overparen{L F}$ and $2 \\angle S A P=\\overparen{B E}+\\overparen{E K}$ we have\n\n$$\n\\overparen{L F}=\\overparen{E K} .\\tag{1}\n$$\n\nOn the other hand, from $\\angle S P C=\\angle S C P$ we have $\\overparen{E C}+\\overparen{M F}=\\overparen{E C}+\\overparen{E M}$, or\n\n$$\n\\overparen{M F}=\\overparen{E M}\\tag{2}\n$$\n\nFrom (1) and (2) we get $\\overparen{M F L}=\\overparen{M F}+\\overparen{F L}=\\overparen{M E}+\\overparen{E K}=\\overparen{M E K}$ and hence $M K=M L$. The claim is proved.\n\n2. We are left to prove the converse. So, assume that $M K=M L$, and introduce the points $E$ and $F$ as above. We have $S C^{2}=S E \\cdot S F$; hence, there exists a point $P^{\\prime}$ lying on the segment $E F$ such that $S P^{\\prime}=S C$ (see Fig. 3).\n\n\n\nFig. 3\n\n\n\nAssume that $P \\neq P^{\\prime}$. Let the lines $A P^{\\prime}, B P^{\\prime}, C P^{\\prime}$ meet $\\omega$ again at points $K^{\\prime}, L^{\\prime}, M^{\\prime}$ respectively. Now, if $P^{\\prime}$ lies on the segment $P F$ then by the first part of the solution we have $\\overparen{M^{\\prime} F L^{\\prime}}=\\overparen{M^{\\prime} E K^{\\prime}}$. On the other hand, we have $\\overparen{M F L}>\\overparen{M^{\\prime} F L^{\\prime}}=\\overparen{M^{\\prime} E K^{\\prime}}>\\overparen{M E K}$, therefore $\\overparen{M F L}>\\overparen{M E K}$ which contradicts $M K=M L$.\n\nSimilarly, if point $P^{\\prime}$ lies on the segment $E P$ then we get $\\overparen{M F L}<\\overparen{M E K}$ which is impossible. Therefore, the points $P$ and $P^{\\prime}$ coincide and hence $S P=S P^{\\prime}=S C$.' 'We present a different proof of the converse direction, that is, $M K=M L \\Rightarrow$ $S P=S C$. As in the previous solutions we assume that $C A>C B$, and the line $S P$ meets $\\omega$ at $E$ and $F$.\n\nFrom $M L=M K$ we get $\\overparen{M E K}=\\overparen{M F L}$. Now we claim that $\\overparen{M E}=\\overparen{M F}$ and $\\overparen{E K}=\\overparen{F L}$.\n\nTo the contrary, suppose first that $\\overparen{M E}>\\overparen{M F}$; then $\\overparen{E K}=\\overparen{M E K}-\\overparen{M E}<\\overparen{M F L}-\\overparen{M F}=$ $\\overparen{F L}$. Now, the inequality $\\overparen{M E}>\\overparen{M F}$ implies $2 \\angle S C M=\\overparen{E C}+\\overparen{M E}>\\overparen{E C}+\\overparen{M F}=2 \\angle S P C$ and hence $S P>S C$. On the other hand, the inequality $\\overparen{E K}<\\overparen{F L}$ implies $2 \\angle S P K=$ $\\overparen{E K}+\\overparen{A F}<\\overparen{F L}+\\overparen{A F}=2 \\angle A B L$, hence\n\n$$\n\\angle S P A=180^{\\circ}-\\angle S P K>180^{\\circ}-\\angle A B L=\\angle S B P .\n$$\n\n\n\nFig. 4\n\nConsider the point $A^{\\prime}$ on the ray $S A$ for which $\\angle S P A^{\\prime}=\\angle S B P$; in our case, this point lies on the segment $S A$ (see Fig. 4). Then $\\triangle S B P \\sim \\triangle S P A^{\\prime}$ and $S P^{2}=S B \\cdot S A^{\\prime}S C$.\n\nSimilarly, one can prove that the inequality $\\overparen{M E}<\\overparen{M F}$ is also impossible. So, we get $\\overparen{M E}=\\overparen{M F}$ and therefore $2 \\angle S C M=\\overparen{E C}+\\overparen{M E}=\\overparen{E C}+\\overparen{M F}=2 \\angle S P C$, which implies $S C=S P$.']" [] Text-only Competition True Theorem proof Geometry Math English 87 "Let $A_{1} A_{2} \ldots A_{n}$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_{1}, \ldots, P_{n}$ onto lines $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_{1}, \ldots, X_{n}$ on sides $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively, $$ \max \left\{\frac{X_{1} X_{2}}{P_{1} P_{2}}, \ldots, \frac{X_{n} X_{1}}{P_{n} P_{1}}\right\} \geq 1 $$" "['Denote $P_{n+1}=P_{1}, X_{n+1}=X_{1}, A_{n+1}=A_{1}$.\n\nLemma. Let point $Q$ lies inside $A_{1} A_{2} \\ldots A_{n}$. Then it is contained in at least one of the circumcircles of triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$.\n\nProof. If $Q$ lies in one of the triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$, the claim is obvious. Otherwise $Q$ lies inside the polygon $X_{1} X_{2} \\ldots X_{n}$ (see Fig. 1). Then we have\n\n$$\n\\begin{aligned}\n& \\left(\\angle X_{1} A_{2} X_{2}+\\angle X_{1} Q X_{2}\\right)+\\cdots+\\left(\\angle X_{n} A_{1} X_{1}+\\angle X_{n} Q X_{1}\\right) \\\\\n& \\quad \\quad=\\left(\\angle X_{1} A_{1} X_{2}+\\cdots+\\angle X_{n} A_{1} X_{1}\\right)+\\left(\\angle X_{1} Q X_{2}+\\cdots+\\angle X_{n} Q X_{1}\\right)=(n-2) \\pi+2 \\pi=n \\pi\n\\end{aligned}\n$$\n\nhence there exists an index $i$ such that $\\angle X_{i} A_{i+1} X_{i+1}+\\angle X_{i} Q X_{i+1} \\geq \\frac{\\pi n}{n}=\\pi$. Since the quadrilateral $Q X_{i} A_{i+1} X_{i+1}$ is convex, this means exactly that $Q$ is contained the circumcircle of $\\triangle X_{i} A_{i+1} X_{i+1}$, as desired.\n\nNow we turn to the solution. Applying lemma, we get that $P$ lies inside the circumcircle of triangle $X_{i} A_{i+1} X_{i+1}$ for some $i$. Consider the circumcircles $\\omega$ and $\\Omega$ of triangles $P_{i} A_{i+1} P_{i+1}$ and $X_{i} A_{i+1} X_{i+1}$ respectively (see Fig. 2); let $r$ and $R$ be their radii. Then we get $2 r=A_{i+1} P \\leq 2 R$ (since $P$ lies inside $\\Omega$ ), hence\n\n$$\nP_{i} P_{i+1}=2 r \\sin \\angle P_{i} A_{i+1} P_{i+1} \\leq 2 R \\sin \\angle X_{i} A_{i+1} X_{i+1}=X_{i} X_{i+1},\n$$\n\nQED.\n\n\n\nFig. 1\n\n\n\nFig. 2' 'we assume that all indices of points are considered modulo $n$. We will prove a bit stronger inequality, namely\n\n$$\n\\max \\left\\{\\frac{X_{1} X_{2}}{P_{1} P_{2}} \\cos \\alpha_{1}, \\ldots, \\frac{X_{n} X_{1}}{P_{n} P_{1}} \\cos \\alpha_{n}\\right\\} \\geq 1\n$$\n\nwhere $\\alpha_{i}(1 \\leq i \\leq n)$ is the angle between lines $X_{i} X_{i+1}$ and $P_{i} P_{i+1}$. We denote $\\beta_{i}=\\angle A_{i} P_{i} P_{i-1}$ and $\\gamma_{i}=\\angle A_{i+1} P_{i} P_{i+1}$ for all $1 \\leq i \\leq n$.\n\nSuppose that for some $1 \\leq i \\leq n$, point $X_{i}$ lies on the segment $A_{i} P_{i}$, while point $X_{i+1}$ lies on the segment $P_{i+1} A_{i+2}$. Then the projection of the segment $X_{i} X_{i+1}$ onto the line $P_{i} P_{i+1}$ contains segment $P_{i} P_{i+1}$, since $\\gamma_{i}$ and $\\beta_{i+1}$ are acute angles (see Fig. 3). Therefore, $X_{i} X_{i+1} \\cos \\alpha_{i} \\geq$ $P_{i} P_{i+1}$, and in this case the statement is proved.\n\nSo, the only case left is when point $X_{i}$ lies on segment $P_{i} A_{i+1}$ for all $1 \\leq i \\leq n$ (the case when each $X_{i}$ lies on segment $A_{i} P_{i}$ is completely analogous).\n\nNow, assume to the contrary that the inequality\n\n$$\nX_{i} X_{i+1} \\cos \\alpha_{i}P_{i+1} Y_{i+1}^{\\prime}$ (again since $\\gamma_{i}$ and $\\beta_{i+1}$ are acute; see Fig. 4). Hence, we have\n\n$$\nX_{i} P_{i} \\cos \\gamma_{i}>X_{i+1} P_{i+1} \\cos \\beta_{i+1}, \\quad 1 \\leq i \\leq n\n$$\n\nMultiplying these inequalities, we get\n\n$$\n\\cos \\gamma_{1} \\cos \\gamma_{2} \\cdots \\cos \\gamma_{n}>\\cos \\beta_{1} \\cos \\beta_{2} \\cdots \\cos \\beta_{n}\n\\tag{2}\n$$\n\nOn the other hand, the sines theorem applied to triangle $P P_{i} P_{i+1}$ provides\n\n$$\n\\frac{P P_{i}}{P P_{i+1}}=\\frac{\\sin \\left(\\frac{\\pi}{2}-\\beta_{i+1}\\right)}{\\sin \\left(\\frac{\\pi}{2}-\\gamma_{i}\\right)}=\\frac{\\cos \\beta_{i+1}}{\\cos \\gamma_{i}}\n$$\n\nMultiplying these equalities we get\n\n$$\n1=\\frac{\\cos \\beta_{2}}{\\cos \\gamma_{1}} \\cdot \\frac{\\cos \\beta_{3}}{\\cos \\gamma_{2}} \\cdots \\frac{\\cos \\beta_{1}}{\\cos \\gamma_{n}}\n$$\n\nwhich contradicts (2).\n\n\n\nFig. 3\n\n\n\nFig. 4']" [] Text-only Competition True Theorem proof Geometry Math English 88 Let $I$ be the incenter of a triangle $A B C$ and $\Gamma$ be its circumcircle. Let the line $A I$ intersect $\Gamma$ at a point $D \neq A$. Let $F$ and $E$ be points on side $B C$ and $\operatorname{arc} B D C$ respectively such that $\angle B A F=\angle C A E<\frac{1}{2} \angle B A C$. Finally, let $G$ be the midpoint of the segment $I F$. Prove that the lines $D G$ and $E I$ intersect on $\Gamma$. "['Let $X$ be the second point of intersection of line $E I$ with $\\Gamma$, and $L$ be the foot of the bisector of angle $B A C$. Let $G^{\\prime}$ and $T$ be the points of intersection of segment $D X$ with lines $I F$ and $A F$, respectively. We are to prove that $G=G^{\\prime}$, or $I G^{\\prime}=G^{\\prime} F$. By the Menelaus theorem applied to triangle $A I F$ and line $D X$, it means that we need the relation\n\n$$\n1=\\frac{G^{\\prime} F}{I G^{\\prime}}=\\frac{T F}{A T} \\cdot \\frac{A D}{I D}, \\quad \\text { or } \\quad \\frac{T F}{A T}=\\frac{I D}{A D}\n$$\n\nLet the line $A F$ intersect $\\Gamma$ at point $K \\neq A$ (see Fig. 1); since $\\angle B A K=\\angle C A E$ we have $\\overparen{B K}=\\overparen{C E}$, hence $K E \\| B C$. Notice that $\\angle I A T=\\angle D A K=\\angle E A D=\\angle E X D=\\angle I X T$, so the points $I, A, X, T$ are concyclic. Hence we have $\\angle I T A=\\angle I X A=\\angle E X A=\\angle E K A$, so $I T\\|K E\\| B C$. Therefore we obtain $\\frac{T F}{A T}=\\frac{I L}{A I}$.\n\nSince $C I$ is the bisector of $\\angle A C L$, we get $\\frac{I L}{A I}=\\frac{C L}{A C}$. Furthermore, $\\angle D C L=\\angle D C B=$ $\\angle D A B=\\angle C A D=\\frac{1}{2} \\angle B A C$, hence the triangles $D C L$ and $D A C$ are similar; therefore we get $\\frac{C L}{A C}=\\frac{D C}{A D}$. Finally, it is known that the midpoint $D$ of $\\operatorname{arc} B C$ is equidistant from points $I$, $B, C$, hence $\\frac{D C}{A D}=\\frac{I D}{A D}$.\n\nSummarizing all these equalities, we get\n\n$$\n\\frac{T F}{A T}=\\frac{I L}{A I}=\\frac{C L}{A C}=\\frac{D C}{A D}=\\frac{I D}{A D}\n$$\n\nas desired.\n\n\n\nFig. 1\n\n\n\nFig. 2' 'As in the previous solution, we introduce the points $X, T$ and $K$ and note that it suffice to prove the equality\n\n$$\n\\frac{T F}{A T}=\\frac{D I}{A D} \\quad \\Longleftrightarrow \\quad \\frac{T F+A T}{A T}=\\frac{D I+A D}{A D} \\quad \\Longleftrightarrow \\quad \\frac{A T}{A D}=\\frac{A F}{D I+A D} .\n$$\n\nSince $\\angle F A D=\\angle E A I$ and $\\angle T D A=\\angle X D A=\\angle X E A=\\angle I E A$, we get that the triangles $A T D$ and $A I E$ are similar, therefore $\\frac{A T}{A D}=\\frac{A I}{A E}$.\n\nNext, we also use the relation $D B=D C=D I$. Let $J$ be the point on the extension of segment $A D$ over point $D$ such that $D J=D I=D C$ (see Fig. 2). Then $\\angle D J C=$ $\\angle J C D=\\frac{1}{2}(\\pi-\\angle J D C)=\\frac{1}{2} \\angle A D C=\\frac{1}{2} \\angle A B C=\\angle A B I$. Moreover, $\\angle B A I=\\angle J A C$, hence triangles $A B I$ and $A J C$ are similar, so $\\frac{A B}{A J}=\\frac{A I}{A C}$, or $A B \\cdot A C=A J \\cdot A I=(D I+A D) \\cdot A I$.\n\nOn the other hand, we get $\\angle A B F=\\angle A B C=\\angle A E C$ and $\\angle B A F=\\angle C A E$, so triangles $A B F$ and $A E C$ are also similar, which implies $\\frac{A F}{A C}=\\frac{A B}{A E}$, or $A B \\cdot A C=A F \\cdot A E$.\n\nSummarizing we get\n\n$$\n(D I+A D) \\cdot A I=A B \\cdot A C=A F \\cdot A E \\quad \\Rightarrow \\quad \\frac{A I}{A E}=\\frac{A F}{A D+D I} \\quad \\Rightarrow \\quad \\frac{A T}{A D}=\\frac{A F}{A D+D I}\n$$\n\nas desired.']" [] Text-only Competition True Theorem proof Geometry Math English 89 Let $A B C D E$ be a convex pentagon such that $B C \| A E, A B=B C+A E$, and $\angle A B C=$ $\angle C D E$. Let $M$ be the midpoint of $C E$, and let $O$ be the circumcenter of triangle $B C D$. Given that $\angle D M O=90^{\circ}$, prove that $2 \angle B D A=\angle C D E$. "['Choose point $T$ on ray $A E$ such that $A T=A B$; then from $A E \\| B C$ we have $\\angle C B T=\\angle A T B=\\angle A B T$, so $B T$ is the bisector of $\\angle A B C$. On the other hand, we have $E T=A T-A E=A B-A E=B C$, hence quadrilateral $B C T E$ is a parallelogram, and the midpoint $M$ of its diagonal $C E$ is also the midpoint of the other diagonal $B T$.\n\nNext, let point $K$ be symmetrical to $D$ with respect to $M$. Then $O M$ is the perpendicular bisector of segment $D K$, and hence $O D=O K$, which means that point $K$ lies on the circumcircle of triangle $B C D$. Hence we have $\\angle B D C=\\angle B K C$. On the other hand, the angles $B K C$ and $T D E$ are symmetrical with respect to $M$, so $\\angle T D E=\\angle B K C=\\angle B D C$.\n\nTherefore, $\\angle B D T=\\angle B D E+\\angle E D T=\\angle B D E+\\angle B D C=\\angle C D E=\\angle A B C=180^{\\circ}-$ $\\angle B A T$. This means that the points $A, B, D, T$ are concyclic, and hence $\\angle A D B=\\angle A T B=$ $\\frac{1}{2} \\angle A B C=\\frac{1}{2} \\angle C D E$, as desired.\n' 'Let $\\angle C B D=\\alpha, \\angle B D C=\\beta, \\angle A D E=\\gamma$, and $\\angle A B C=\\angle C D E=2 \\varphi$. Then we have $\\angle A D B=2 \\varphi-\\beta-\\gamma, \\angle B C D=180^{\\circ}-\\alpha-\\beta, \\angle A E D=360^{\\circ}-\\angle B C D-\\angle C D E=$ $180^{\\circ}-2 \\varphi+\\alpha+\\beta$, and finally $\\angle D A E=180^{\\circ}-\\angle A D E-\\angle A E D=2 \\varphi-\\alpha-\\beta-\\gamma$.\n\n\nLet $N$ be the midpoint of $C D$; then $\\angle D N O=90^{\\circ}=\\angle D M O$, hence points $M, N$ lie on the circle with diameter $O D$. Now, if points $O$ and $M$ lie on the same side of $C D$, we have $\\angle D M N=\\angle D O N=\\frac{1}{2} \\angle D O C=\\alpha ;$ in the other case, we have $\\angle D M N=180^{\\circ}-\\angle D O N=\\alpha ;$\n\n\n\nso, in both cases $\\angle D M N=\\alpha$ (see Figures). Next, since $M N$ is a midline in triangle $C D E$, we have $\\angle M D E=\\angle D M N=\\alpha$ and $\\angle N D M=2 \\varphi-\\alpha$.\n\nNow we apply the sine rule to the triangles $A B D, A D E$ (twice), $B C D$ and $M N D$ obtaining\n\n$$\n\\begin{gathered}\n\\frac{A B}{A D}=\\frac{\\sin (2 \\varphi-\\beta-\\gamma)}{\\sin (2 \\varphi-\\alpha)}, \\quad \\frac{A E}{A D}=\\frac{\\sin \\gamma}{\\sin (2 \\varphi-\\alpha-\\beta)}, \\quad \\frac{D E}{A D}=\\frac{\\sin (2 \\varphi-\\alpha-\\beta-\\gamma)}{\\sin (2 \\varphi-\\alpha-\\beta)} \\\\\n\\frac{B C}{C D}=\\frac{\\sin \\beta}{\\sin \\alpha}, \\quad \\frac{C D}{D E}=\\frac{C D / 2}{D E / 2}=\\frac{N D}{N M}=\\frac{\\sin \\alpha}{\\sin (2 \\varphi-\\alpha)}\n\\end{gathered}\n$$\n\nwhich implies\n\n$$\n\\frac{B C}{A D}=\\frac{B C}{C D} \\cdot \\frac{C D}{D E} \\cdot \\frac{D E}{A D}=\\frac{\\sin \\beta \\cdot \\sin (2 \\varphi-\\alpha-\\beta-\\gamma)}{\\sin (2 \\varphi-\\alpha) \\cdot \\sin (2 \\varphi-\\alpha-\\beta)}\n$$\n\nHence, the condition $A B=A E+B C$, or equivalently $\\frac{A B}{A D}=\\frac{A E+B C}{A D}$, after multiplying by the common denominator rewrites as\n\n$$\n\\begin{aligned}\n& \\sin (2 \\varphi-\\alpha-\\beta) \\cdot \\sin (2 \\varphi-\\beta-\\gamma)=\\sin \\gamma \\cdot \\sin (2 \\varphi-\\alpha)+\\sin \\beta \\cdot \\sin (2 \\varphi-\\alpha-\\beta-\\gamma) \\\\\n& \\Longleftrightarrow \\cos (\\gamma-\\alpha)-\\cos (4 \\varphi-2 \\beta-\\alpha-\\gamma)=\\cos (2 \\varphi-\\alpha-2 \\beta-\\gamma)-\\cos (2 \\varphi+\\gamma-\\alpha) \\\\\n& \\Longleftrightarrow \\cos (\\gamma-\\alpha)+\\cos (2 \\varphi+\\gamma-\\alpha)=\\cos (2 \\varphi-\\alpha-2 \\beta-\\gamma)+\\cos (4 \\varphi-2 \\beta-\\alpha-\\gamma) \\\\\n& \\Longleftrightarrow \\cos \\varphi \\cdot \\cos (\\varphi+\\gamma-\\alpha)=\\cos \\varphi \\cdot \\cos (3 \\varphi-2 \\beta-\\alpha-\\gamma) \\\\\n& \\Longleftrightarrow \\cos \\varphi \\cdot(\\cos (\\varphi+\\gamma-\\alpha)-\\cos (3 \\varphi-2 \\beta-\\alpha-\\gamma))=0 \\\\\n& \\Longleftrightarrow \\cos \\varphi \\cdot \\sin (2 \\varphi-\\beta-\\alpha) \\cdot \\sin (\\varphi-\\beta-\\gamma)=0 .\n\\end{aligned}\n$$\n\nSince $2 \\varphi-\\beta-\\alpha=180^{\\circ}-\\angle A E D<180^{\\circ}$ and $\\varphi=\\frac{1}{2} \\angle A B C<90^{\\circ}$, it follows that $\\varphi=\\beta+\\gamma$, hence $\\angle B D A=2 \\varphi-\\beta-\\gamma=\\varphi=\\frac{1}{2} \\angle C D E$, as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 90 The vertices $X, Y, Z$ of an equilateral triangle $X Y Z$ lie respectively on the sides $B C$, $C A, A B$ of an acute-angled triangle $A B C$. Prove that the incenter of triangle $A B C$ lies inside triangle $X Y Z$. "['We will prove a stronger fact; namely, we will show that the incenter $I$ of triangle $A B C$ lies inside the incircle of triangle $X Y Z$ (and hence surely inside triangle $X Y Z$ itself). We denote by $d(U, V W)$ the distance between point $U$ and line $V W$.\n\nDenote by $O$ the incenter of $\\triangle X Y Z$ and by $r, r^{\\prime}$ and $R^{\\prime}$ the inradii of triangles $A B C, X Y Z$ and the circumradius of $X Y Z$, respectively. Then we have $R^{\\prime}=2 r^{\\prime}$, and the desired inequality is $O I \\leq r^{\\prime}$. We assume that $O \\neq I$; otherwise the claim is trivial.\n\nLet the incircle of $\\triangle A B C$ touch its sides $B C, A C, A B$ at points $A_{1}, B_{1}, C_{1}$ respectively. The lines $I A_{1}, I B_{1}, I C_{1}$ cut the plane into 6 acute angles, each one containing one of the points $A_{1}, B_{1}, C_{1}$ on its border. We may assume that $O$ lies in an angle defined by lines $I A_{1}$, $I C_{1}$ and containing point $C_{1}$ (see Fig. 1). Let $A^{\\prime}$ and $C^{\\prime}$ be the projections of $O$ onto lines $I A_{1}$ and $I C_{1}$, respectively.\n\nSince $O X=R^{\\prime}$, we have $d(O, B C) \\leq R^{\\prime}$. Since $O A^{\\prime} \\| B C$, it follows that $d\\left(A^{\\prime}, B C\\right)=$ $A^{\\prime} I+r \\leq R^{\\prime}$, or $A^{\\prime} I \\leq R^{\\prime}-r$. On the other hand, the incircle of $\\triangle X Y Z$ lies inside $\\triangle A B C$, hence $d(O, A B) \\geq r^{\\prime}$, and analogously we get $d(O, A B)=C^{\\prime} C_{1}=r-I C^{\\prime} \\geq r^{\\prime}$, or $I C^{\\prime} \\leq r-r^{\\prime}$.\n\n\n\nFig. 1\n\n\n\nFig. 2\n\nFinally, the quadrilateral $I A^{\\prime} O C^{\\prime}$ is circumscribed due to the right angles at $A^{\\prime}$ and $C^{\\prime}$ (see Fig. 2). On its circumcircle, we have $\\overparen{A^{\\prime} O C^{\\prime}}=2 \\angle A^{\\prime} I C^{\\prime}<180^{\\circ}=\\overparen{O C^{\\prime} I \\text {, hence } 180^{\\circ} \\geq}$ $\\overparen{I C^{\\prime}}>\\overparen{A^{\\prime} O}$. This means that $I C^{\\prime}>A^{\\prime} O$. Finally, we have $O I \\leq I A^{\\prime}+A^{\\prime} O90^{\\circ}$ thus leading to a contradiction.\n\nNote that $\\omega$ intersects each of the segments $X Y$ and $Y Z$ at two points; let $U, U^{\\prime}$ and $V$, $V^{\\prime}$ be the points of intersection of $\\omega$ with $X Y$ and $Y Z$, respectively $\\left(U Y>U^{\\prime} Y, V Y>V^{\\prime} Y\\right.$; see Figs. 3 and 4). Note that $60^{\\circ}=\\angle X Y Z=\\frac{1}{2}\\left(\\overparen{U V}-\\overparen{U^{\\prime} V^{\\prime}}\\right) \\leq \\frac{1}{2} \\overparen{U V}$, hence $\\overparen{U V} \\geq 120^{\\circ}$.\n\n\n\nOn the other hand, since $I$ lies in $\\triangle A Y Z$, we get $\\sqrt{U V^{\\prime}}<180^{\\circ}$, hence $\\sqrt{U A_{1} U^{\\prime}} \\leq \\sqrt{U A_{1} V^{\\prime}}<$ $180^{\\circ}-\\overparen{U V} \\leq 60^{\\circ}$.\n\nNow, two cases are possible due to the order of points $Y, B_{1}$ on segment $A C$.\n\n\n\nFig. 3\n\n\n\nFig. 4\n\nCase 1. Let point $Y$ lie on the segment $A B_{1}$ (see Fig. 3). Then we have $\\angle Y X C=$ $\\frac{1}{2}\\left(\\overparen{A_{1} U^{\\prime}}-\\overparen{A_{1} U}\\right) \\leq \\frac{1}{2} \\overparen{U A_{1} U^{\\prime}}<30^{\\circ}$; analogously, we get $\\angle X Y C \\leq \\frac{1}{2} \\overparen{U A_{1} U^{\\prime}}<30^{\\circ}$. Therefore, $\\angle Y C X=180^{\\circ}-\\angle Y X C-\\angle X Y C>120^{\\circ}$, as desired.\n\nCase 2. Now let point $Y$ lie on the segment $C B_{1}$ (see Fig. 4). Analogously, we obtain $\\angle Y X C<30^{\\circ}$. Next, $\\angle I Y X>\\angle Z Y X=60^{\\circ}$, but $\\angle I Y X<\\angle I Y B_{1}$, since $Y B_{1}$ is a tangent and $Y X$ is a secant line to circle $\\omega$ from point $Y$. Hence, we get $120^{\\circ}<\\angle I Y B_{1}+\\angle I Y X=$ $\\angle B_{1} Y X=\\angle Y X C+\\angle Y C X<30^{\\circ}+\\angle Y C X$, hence $\\angle Y C X>120^{\\circ}-30^{\\circ}=90^{\\circ}$, as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 91 "Let $a, b$ be integers, and let $P(x)=a x^{3}+b x$. For any positive integer $n$ we say that the pair $(a, b)$ is $n$-good if $n \mid P(m)-P(k)$ implies $n \mid m-k$ for all integers $m, k$. We say that $(a, b)$ is very good if $(a, b)$ is $n$-good for infinitely many positive integers $n$. Show that all 2010-good pairs are very good." ['We will show that if a pair $(a, b)$ is $2010-\\operatorname{good}$ then $(a, b)$ is $67^{i}$-good for all positive integer $i$.\n\nClaim 1. If $(a, b)$ is $2010-\\operatorname{good}$ then $(a, b)$ is 67 -good.\n\nProof. Assume that $P(m)=P(k)(\\bmod 67)$. Since 67 and 30 are coprime, there exist integers $m^{\\prime}$ and $k^{\\prime}$ such that $k^{\\prime} \\equiv k(\\bmod 67), k^{\\prime} \\equiv 0(\\bmod 30)$, and $m^{\\prime} \\equiv m(\\bmod 67), m^{\\prime} \\equiv 0$ $(\\bmod 30)$. Then we have $P\\left(m^{\\prime}\\right) \\equiv P(0) \\equiv P\\left(k^{\\prime}\\right)(\\bmod 30)$ and $P\\left(m^{\\prime}\\right) \\equiv P(m) \\equiv P(k) \\equiv P\\left(k^{\\prime}\\right)$ $(\\bmod 67)$, hence $P\\left(m^{\\prime}\\right) \\equiv P\\left(k^{\\prime}\\right)(\\bmod 2010)$. This implies $m^{\\prime} \\equiv k^{\\prime}(\\bmod 2010)$ as $(a, b)$ is 2010 -good. It follows that $m \\equiv m^{\\prime} \\equiv k^{\\prime} \\equiv k(\\bmod 67)$. Therefore, $(a, b)$ is 67 -good.\n\nClaim 2. If $(a, b)$ is 67 -good then $67 \\mid a$.\n\nProof. Suppose that $67 \\nmid a$. Consider the sets $\\left\\{a t^{2}(\\bmod 67): 0 \\leq t \\leq 33\\right\\}$ and $\\left\\{-3 a s^{2}-b\\right.$ $\\bmod 67: 0 \\leq s \\leq 33\\}$. Since $a \\not \\equiv 0(\\bmod 67)$, each of these sets has 34 elements. Hence they have at least one element in common. If $a t^{2} \\equiv-3 a s^{2}-b(\\bmod 67)$ then for $m=t \\pm s, k=\\mp 2 s$ we have\n\n$$\n\\begin{aligned}\nP(m)-P(k)=a\\left(m^{3}-k^{3}\\right)+b(m-k) & =(m-k)\\left(a\\left(m^{2}+m k+k^{2}\\right)+b\\right) \\\\\n& =(t \\pm 3 s)\\left(a t^{2}+3 a s^{2}+b\\right) \\equiv 0 \\quad(\\bmod 67)\n\\end{aligned}\n$$\n\nSince $(a, b)$ is 67 -good, we must have $m \\equiv k(\\bmod 67)$ in both cases, that is, $t \\equiv 3 s(\\bmod 67)$ and $t \\equiv-3 s(\\bmod 67)$. This means $t \\equiv s \\equiv 0(\\bmod 67)$ and $b \\equiv-3 a s^{2}-a t^{2} \\equiv 0(\\bmod 67)$. But then $67 \\mid P(7)-P(2)=67 \\cdot 5 a+5 b$ and $67 / 17-2$, contradicting that $(a, b)$ is 67 -good.\n\nClaim 3. If $(a, b)$ is $2010-\\operatorname{good}$ then $(a, b)$ is $67^{i}-\\operatorname{good}$ all $i \\geq 1$.\n\nProof. By Claim 2 , we have $67 \\mid a$. If $67 \\mid b$, then $P(x) \\equiv P(0)(\\bmod 67)$ for all $x$, contradicting that $(a, b)$ is 67 -good. Hence, $67 /\\langle b$.\n\nSuppose that $67^{i} \\mid P(m)-P(k)=(m-k)\\left(a\\left(m^{2}+m k+k^{2}\\right)+b\\right)$. Since $67 \\mid a$ and $67 \\not \\nmid b$, the second factor $a\\left(m^{2}+m k+k^{2}\\right)+b$ is coprime to 67 and hence $67^{i} \\mid m-k$. Therefore, $(a, b)$ is $67^{i}$-good.'] [] Text-only Competition True Theorem proof Number Theory Math English 92 "The rows and columns of a $2^{n} \times 2^{n}$ table are numbered from 0 to $2^{n}-1$. The cells of the table have been colored with the following property being satisfied: for each $0 \leq i, j \leq 2^{n}-1$, the $j$ th cell in the $i$ th row and the $(i+j)$ th cell in the $j$ th row have the same color. (The indices of the cells in a row are considered modulo $2^{n}$.) Prove that the maximal possible number of colors is $2^{n}$." ['Throughout the solution we denote the cells of the table by coordinate pairs; $(i, j)$ refers to the $j$ th cell in the $i$ th row.\n\nConsider the directed graph, whose vertices are the cells of the board, and the edges are the arrows $(i, j) \\rightarrow(j, i+j)$ for all $0 \\leq i, j \\leq 2^{n}-1$. From each vertex $(i, j)$, exactly one edge passes (to $\\left(j, i+j \\bmod 2^{n}\\right)$ ); conversely, to each cell $(j, k)$ exactly one edge is directed (from the cell $\\left.\\left(k-j \\bmod 2^{n}, j\\right)\\right)$. Hence, the graph splits into cycles.\n\nNow, in any coloring considered, the vertices of each cycle should have the same color by the problem condition. On the other hand, if each cycle has its own color, the obtained coloring obviously satisfies the problem conditions. Thus, the maximal possible number of colors is the same as the number of cycles, and we have to prove that this number is $2^{n}$.\n\nNext, consider any cycle $\\left(i_{1}, j_{1}\\right),\\left(i_{2}, j_{2}\\right), \\ldots$; we will describe it in other terms. Define a sequence $\\left(a_{0}, a_{1}, \\ldots\\right)$ by the relations $a_{0}=i_{1}, a_{1}=j_{1}, a_{n+1}=a_{n}+a_{n-1}$ for all $n \\geq 1$ (we say that such a sequence is a Fibonacci-type sequence). Then an obvious induction shows that $i_{k} \\equiv a_{k-1}\\left(\\bmod 2^{n}\\right), j_{k} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$. Hence we need to investigate the behavior of Fibonacci-type sequences modulo $2^{n}$.\n\nDenote by $F_{0}, F_{1}, \\ldots$ the Fibonacci numbers defined by $F_{0}=0, F_{1}=1$, and $F_{n+2}=$ $F_{n+1}+F_{n}$ for $n \\geq 0$. We also set $F_{-1}=1$ according to the recurrence relation.\n\nFor every positive integer $m$, denote by $\\nu(m)$ the exponent of 2 in the prime factorization of $m$, i.e. for which $2^{\\nu(m)} \\mid m$ but $2^{\\nu(m)+1} \\mid\\langle m$.\n\nLemma 1. For every Fibonacci-type sequence $a_{0}, a_{1}, a_{2}, \\ldots$, and every $k \\geq 0$, we have $a_{k}=$ $F_{k-1} a_{0}+F_{k} a_{1}$.\n\nProof. Apply induction on $k$. The base cases $k=0,1$ are trivial. For the step, from the induction hypothesis we get\n\n$$\na_{k+1}=a_{k}+a_{k-1}=\\left(F_{k-1} a_{0}+F_{k} a_{1}\\right)+\\left(F_{k-2} a_{0}+F_{k-1} a_{1}\\right)=F_{k} a_{0}+F_{k+1} a_{1}\n$$\n\nLemma 2. For every $m \\geq 3$,\n\n(a) we have $\\nu\\left(F_{3 \\cdot 2^{m-2}}\\right)=m$;\n\n(b) $d=3 \\cdot 2^{m-2}$ is the least positive index for which $2^{m} \\mid F_{d}$;\n\n(c) $F_{3 \\cdot 2^{m-2}+1} \\equiv 1+2^{m-1}\\left(\\bmod 2^{m}\\right)$.\n\nProof. Apply induction on $m$. In the base case $m=3$ we have $\\nu\\left(F_{3 \\cdot 2^{m-2}}\\right)=F_{6}=8$, so $\\nu\\left(F_{3 \\cdot 2^{m-2}}\\right)=\\nu(8)=3$, the preceding Fibonacci-numbers are not divisible by 8 , and indeed $F_{3 \\cdot 2^{m-2}+1}=F_{7}=13 \\equiv 1+4(\\bmod 8)$.\n\nNow suppose that $m>3$ and let $k=3 \\cdot 2^{m-3}$. By applying Lemma 1 to the Fibonacci-type sequence $F_{k}, F_{k+1}, \\ldots$ we get\n\n$$\n\\begin{gathered}\nF_{2 k}=F_{k-1} F_{k}+F_{k} F_{k+1}=\\left(F_{k+1}-F_{k}\\right) F_{k}+F_{k+1} F_{k}=2 F_{k+1} F_{k}-F_{k}^{2}, \\\\\nF_{2 k+1}=F_{k} \\cdot F_{k}+F_{k+1} \\cdot F_{k+1}=F_{k}^{2}+F_{k+1}^{2} .\n\\end{gathered}\n$$\n\nBy the induction hypothesis, $\\nu\\left(F_{k}\\right)=m-1$, and $F_{k+1}$ is odd. Therefore we get $\\nu\\left(F_{k}^{2}\\right)=$ $2(m-1)>(m-1)+1=\\nu\\left(2 F_{k} F_{k+1}\\right)$, which implies $\\nu\\left(F_{2 k}\\right)=m$, establishing statement (a).\n\n\n\nMoreover, since $F_{k+1}=1+2^{m-2}+a 2^{m-1}$ for some integer $a$, we get\n\n$$\nF_{2 k+1}=F_{k}^{2}+F_{k+1}^{2} \\equiv 0+\\left(1+2^{m-2}+a 2^{m-1}\\right)^{2} \\equiv 1+2^{m-1} \\quad\\left(\\bmod 2^{m}\\right)\n$$\n\nas desired in statement (c).\n\nWe are left to prove that $2^{m} \\nmid F_{\\ell}$ for $\\ell<2 k$. Assume the contrary. Since $2^{m-1} \\mid F_{\\ell}$, from the induction hypothesis it follows that $\\ell>k$. But then we have $F_{\\ell}=F_{k-1} F_{\\ell-k}+F_{k} F_{\\ell-k+1}$, where the second summand is divisible by $2^{m-1}$ but the first one is not (since $F_{k-1}$ is odd and $\\ell-k0$ such that $a_{k+p} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$ for all $k \\geq 0$.\n\nLemma 3. Let $A=\\left(a_{0}, a_{1}, \\ldots\\right)$ be a Fibonacci-type sequence such that $\\mu(A)=k0$ )\n\n$$\n\\left\\lfloor a_{i+1}\\right\\rfloor \\leq a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle<\\left\\lfloor a_{i}\\right\\rfloor\n$$\n\nthe sequence $\\left\\lfloor a_{i}\\right\\rfloor$ is strictly decreasing as long as its terms are in $[1, \\infty)$. Eventually there appears a number from the interval $[0,1)$ and all subsequent terms are 0 .\n\nNow pass to the more interesting situation where $a_{0}<0$; then all $a_{i} \\leq 0$. Suppose the sequence never hits 0 . Then we have $\\left\\lfloor a_{i}\\right\\rfloor \\leq-1$ for all $i$, and so\n\n$$\n1+\\left\\lfloor a_{i+1}\\right\\rfloor>a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle>\\left\\lfloor a_{i}\\right\\rfloor ;\n$$\n\nthis means that the sequence $\\left\\lfloor a_{i}\\right\\rfloor$ is nondecreasing. And since all its terms are integers from $(-\\infty,-1]$, this sequence must be constant from some term on:\n\n$$\n\\left\\lfloor a_{i}\\right\\rfloor=c \\quad \\text { for } \\quad i \\geq i_{0} ; \\quad c \\text { a negative integer. }\n$$\n\nThe defining formula becomes\n\n$$\na_{i+1}=c \\cdot\\left\\langle a_{i}\\right\\rangle=c\\left(a_{i}-c\\right)=c a_{i}-c^{2}\n$$\n\nConsider the sequence\n\n$$\nb_{i}=a_{i}-\\frac{c^{2}}{c-1} \\tag{1}\n$$\n\nIt satisfies the recursion rule\n\n$$\nb_{i+1}=a_{i+1}-\\frac{c^{2}}{c-1}=c a_{i}-c^{2}-\\frac{c^{2}}{c-1}=c b_{i} \n$$\n\nimplying\n\n$$\nb_{i}=c^{i-i_{0}} b_{i_{0}} \\quad \\text { for } \\quad i \\geq i_{0} \\tag{2}\n$$\n\nSince all the numbers $a_{i}$ (for $i \\geq i_{0}$ ) lie in $\\left[c, c+1\\right.$ ), the sequence $\\left(b_{i}\\right)$ is bounded. The equation (2) can be satisfied only if either $b_{i_{0}}=0$ or $|c|=1$, i.e., $c=-1$.\n\n\n\nIn the first case, $b_{i}=0$ for all $i \\geq i_{0}$, so that\n\n$$\na_{i}=\\frac{c^{2}}{c-1} \\quad \\text { for } \\quad i \\geq i_{0}\n$$\n\nIn the second case, $c=-1$, equations (1) and (2) say that\n\n$$\na_{i}=-\\frac{1}{2}+(-1)^{i-i_{0}} b_{i_{0}}= \\begin{cases}a_{i_{0}} & \\text { for } i=i_{0}, i_{0}+2, i_{0}+4, \\ldots \\\\ 1-a_{i_{0}} & \\text { for } i=i_{0}+1, i_{0}+3, i_{0}+5, \\ldots\\end{cases}\n$$\n\nSummarising, we see that (from some point on) the sequence $\\left(a_{i}\\right)$ either is constant or takes alternately two values from the interval $(-1,0)$. The result follows.'] [] Text-only Competition True Theorem proof Algebra Math English 94 "The sequence of real numbers $a_{0}, a_{1}, a_{2}, \ldots$ is defined recursively by $$ a_{0}=-1, \quad \sum_{k=0}^{n} \frac{a_{n-k}}{k+1}=0 \quad \text { for } \quad n \geq 1 $$ Show that $a_{n}>0$ for $n \geq 1$." ['The proof goes by induction. For $n=1$ the formula yields $a_{1}=1 / 2$. Take $n \\geq 1$, assume $a_{1}, \\ldots, a_{n}>0$ and write the recurrence formula for $n$ and $n+1$, respectively as\n\n$$\n\\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1}=0 \\quad \\text { and } \\quad \\sum_{k=0}^{n+1} \\frac{a_{k}}{n-k+2}=0\n$$\n\nSubtraction yields\n\n$$\n\\begin{aligned}\n0=(n+2) \\sum_{k=0}^{n+1} \\frac{a_{k}}{n-k+2}-(n+1) \\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1} \\\\\n=(n+2) a_{n+1}+\\sum_{k=0}^{n}\\left(\\frac{n+2}{n-k+2}-\\frac{n+1}{n-k+1}\\right) a_{k} .\n\\end{aligned}\n$$\n\nThe coefficient of $a_{0}$ vanishes, so\n\n$$\na_{n+1}=\\frac{1}{n+2} \\sum_{k=1}^{n}\\left(\\frac{n+1}{n-k+1}-\\frac{n+2}{n-k+2}\\right) a_{k}=\\frac{1}{n+2} \\sum_{k=1}^{n} \\frac{k}{(n-k+1)(n-k+2)} a_{k} .\n$$\n\nThe coefficients of $a_{1}, \\ldots, a_{n}$ are all positive. Therefore, $a_{1}, \\ldots, a_{n}>0$ implies $a_{n+1}>0$.'] [] Text-only Competition True Theorem proof Algebra Math English 95 "The sequence $c_{0}, c_{1}, \ldots, c_{n}, \ldots$ is defined by $c_{0}=1, c_{1}=0$ and $c_{n+2}=c_{n+1}+c_{n}$ for $n \geq 0$. Consider the set $S$ of ordered pairs $(x, y)$ for which there is a finite set $J$ of positive integers such that $x=\sum_{j \in J} c_{j}, y=\sum_{j \in J} c_{j-1}$. Prove that there exist real numbers $\alpha, \beta$ and $m, M$ with the following property: An ordered pair of nonnegative integers $(x, y)$ satisfies the inequality $$ m<\alpha x+\beta y1$ and $-1<\\psi<0$, this implies $\\alpha \\varphi+\\beta=0$.\n\nTo satisfy $\\alpha \\varphi+\\beta=0$, one can set for instance $\\alpha=\\psi, \\beta=1$. We now find the required $m$ and $M$ for this choice of $\\alpha$ and $\\beta$.\n\nNote first that the above displayed equation gives $c_{n} \\psi+c_{n-1}=\\psi^{n-1}, n \\geq 1$. In the sequel, we denote the pairs in $S$ by $\\left(a_{J}, b_{J}\\right)$, where $J$ is a finite subset of the set $\\mathbb{N}$ of positive integers and $a_{J}=\\sum_{j \\in J} c_{j}, b_{J}=\\sum_{j \\in J} c_{j-1}$. Since $\\psi a_{J}+b_{J}=\\sum_{j \\in J}\\left(c_{j} \\psi+c_{j-1}\\right)$, we obtain\n\n$$\n\\psi a_{J}+b_{J}=\\sum_{j \\in J} \\psi^{j-1} \\quad \\text { for each }\\left(a_{J}, b_{J}\\right) \\in S \\tag{1}\n$$\n\nOn the other hand, in view of $-1<\\psi<0$,\n\n$$\n-1=\\frac{\\psi}{1-\\psi^{2}}=\\sum_{j=0}^{\\infty} \\psi^{2 j+1}<\\sum_{j \\in J} \\psi^{j-1}<\\sum_{j=0}^{\\infty} \\psi^{2 j}=\\frac{1}{1-\\psi^{2}}=1-\\psi=\\varphi\n$$\n\nTherefore, according to (1),\n\n$$\n-1<\\psi a_{J}+b_{J}<\\varphi \\quad \\text { for each }\\left(a_{J}, b_{J}\\right) \\in S\n$$\n\nThus $m=-1$ and $M=\\varphi$ is an appropriate choice.\n\nConversely, we prove that if an ordered pair of nonnegative integers $(x, y)$ satisfies the inequality $-1<\\psi x+y<\\varphi$ then $(x, y) \\in S$.\n\n\n\nLemma. Let $x, y$ be nonnegative integers such that $-1<\\psi x+y<\\varphi$. Then there exists a subset $J$ of $\\mathbb{N}$ such that\n\n$$\n\\psi x+y=\\sum_{j \\in J} \\psi^{j-1} \\tag{2}\n$$\n\nProof. For $x=y=0$ it suffices to choose the empty subset of $\\mathbb{N}$ as $J$, so let at least one of $x, y$ be nonzero. There exist representations of $\\psi x+y$ of the form\n\n$$\n\\psi x+y=\\psi^{i_{1}}+\\cdots+\\psi^{i_{k}}\n$$\n\nwhere $i_{1} \\leq \\cdots \\leq i_{k}$ is a sequence of nonnegative integers, not necessarily distinct. For instance, we can take $x$ summands $\\psi^{1}=\\psi$ and $y$ summands $\\psi^{0}=1$. Consider all such representations of minimum length $k$ and focus on the ones for which $i_{1}$ has the minimum possible value $j_{1}$. Among them, consider the representations where $i_{2}$ has the minimum possible value $j_{2}$. Upon choosing $j_{3}, \\ldots, j_{k}$ analogously, we obtain a sequence $j_{1} \\leq \\cdots \\leq j_{k}$ which clearly satisfies $\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}$. To prove the lemma, it suffices to show that $j_{1}, \\ldots, j_{k}$ are pairwise distinct.\n\nSuppose on the contrary that $j_{r}=j_{r+1}$ for some $r=1, \\ldots, k-1$. Let us consider the case $j_{r} \\geq 2$ first. Observing that $2 \\psi^{2}=1+\\psi^{3}$, we replace $j_{r}$ and $j_{r+1}$ by $j_{r}-2$ and $j_{r}+1$, respectively. Since\n\n$$\n\\psi^{j_{r}}+\\psi^{j_{r+1}}=2 \\psi^{j_{r}}=\\psi^{j_{r}-2}\\left(1+\\psi^{3}\\right)=\\psi^{j_{r}-2}+\\psi^{j_{r}+1}\n$$\n\nthe new sequence also represents $\\psi x+y$ as needed, and the value of $i_{r}$ in it contradicts the minimum choice of $j_{r}$.\n\nLet $j_{r}=j_{r+1}=0$. Then the sum $\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}$ contains at least two summands equal to $\\psi^{0}=1$. On the other hand $j_{s} \\neq 1$ for all $s$, because the equality $1+\\psi=\\psi^{2}$ implies that a representation of minimum length cannot contain consecutive $i_{r}$ 's. It follows that\n\n$$\n\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}>2+\\psi^{3}+\\psi^{5}+\\psi^{7}+\\cdots=2-\\psi^{2}=\\varphi\n$$\n\ncontradicting the condition of the lemma.\n\nLet $j_{r}=j_{r+1}=1$; then $\\sum_{r=1}^{k} \\psi^{j_{r}}$ contains at least two summands equal to $\\psi^{1}=\\psi$. Like in the case $j_{r}=j_{r+1}=0$, we also infer that $j_{s} \\neq 0$ and $j_{s} \\neq 2$ for all $s$. Therefore\n\n$$\n\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}<2 \\psi+\\psi^{4}+\\psi^{6}+\\psi^{8}+\\cdots=2 \\psi-\\psi^{3}=-1\n$$\n\nwhich is a contradiction again. The conclusion follows.\n\nNow let the ordered pair $(x, y)$ satisfy $-1<\\psi x+y<\\varphi$; hence the lemma applies to $(x, y)$. Let $J \\subset \\mathbb{N}$ be such that (2) holds. Comparing (1) and (2), we conclude that $\\psi x+y=\\psi a_{J}+b_{J}$. Now, $x, y, a_{J}$ and $b_{J}$ are integers, and $\\psi$ is irrational. So the last equality implies $x=a_{J}$ and $y=b_{J}$. This shows that the numbers $\\alpha=\\psi, \\beta=1, m=-1, M=\\varphi$ meet the requirements.""]" [] Text-only Competition True Theorem proof Algebra Math English 96 "Prove the inequality $$ \sum_{i\\sqrt{a+b}>\\sqrt{c}$.\n\nLet $x=\\sqrt{b}+\\sqrt{c}-\\sqrt{a}, y=\\sqrt{c}+\\sqrt{a}-\\sqrt{b}$ and $z=\\sqrt{a}+\\sqrt{b}-\\sqrt{c}$. Then\n\n$b+c-a=\\left(\\frac{z+x}{2}\\right)^{2}+\\left(\\frac{x+y}{2}\\right)^{2}-\\left(\\frac{y+z}{2}\\right)^{2}=\\frac{x^{2}+x y+x z-y z}{2}=x^{2}-\\frac{1}{2}(x-y)(x-z)$\n\nand\n\n$$\n\\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}=\\sqrt{1-\\frac{(x-y)(x-z)}{2 x^{2}}} \\leq 1-\\frac{(x-y)(x-z)}{4 x^{2}}\n$$\n\napplying $\\sqrt{1+2 u} \\leq 1+u$ in the last step. Similarly we obtain\n\n$$\n\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}} \\leq 1-\\frac{(z-x)(z-y)}{4 z^{2}} \\quad \\text { and } \\quad \\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 1-\\frac{(y-z)(y-x)}{4 y^{2}}\n$$\n\nSubstituting these quantities into the statement, it is sufficient to prove that\n\n$$\n\\frac{(x-y)(x-z)}{x^{2}}+\\frac{(y-z)(y-x)}{y^{2}}+\\frac{(z-x)(z-y)}{z^{2}} \\geq 0 \\tag{1}\n$$\n\nBy symmetry we can assume $x \\leq y \\leq z$. Then\n\n$$\n\\begin{gathered}\n\\frac{(x-y)(x-z)}{x^{2}}=\\frac{(y-x)(z-x)}{x^{2}} \\geq \\frac{(y-x)(z-y)}{y^{2}}=-\\frac{(y-z)(y-x)}{y^{2}} \\\\\n\\frac{(z-x)(z-y)}{z^{2}} \\geq 0\n\\end{gathered}\n$$\n\nand (1) follows.' 'Due to the symmetry of variables, it can be assumed that $a \\geq b \\geq c$. We claim that\n\n$$\n\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 1 \\quad \\text { and } \\quad \\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}+\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}} \\leq 2\n$$\n\nThe first inequality follows from\n\n$$\n\\sqrt{a+b-c}-\\sqrt{a}=\\frac{(a+b-c)-a}{\\sqrt{a+b-c}+\\sqrt{a}} \\leq \\frac{b-c}{\\sqrt{b}+\\sqrt{c}}=\\sqrt{b}-\\sqrt{c}\n$$\n\nFor proving the second inequality, let $p=\\sqrt{a}+\\sqrt{b}$ and $q=\\sqrt{a}-\\sqrt{b}$. Then $a-b=p q$ and the inequality becomes\n\n$$\n\\frac{\\sqrt{c-p q}}{\\sqrt{c}-q}+\\frac{\\sqrt{c+p q}}{\\sqrt{c}+q} \\leq 2\n$$\n\nFrom $a \\geq b \\geq c$ we have $p \\geq 2 \\sqrt{c}$. Applying the Cauchy-Schwarz inequality,\n\n$$\n\\begin{gathered}\n\\left(\\frac{\\sqrt{c-p q}}{\\sqrt{c}-q}+\\frac{\\sqrt{c+p q}}{\\sqrt{c}+q}\\right)^{2} \\leq\\left(\\frac{c-p q}{\\sqrt{c}-q}+\\frac{c+p q}{\\sqrt{c}+q}\\right)\\left(\\frac{1}{\\sqrt{c}-q}+\\frac{1}{\\sqrt{c}+q}\\right) \\\\\n=\\frac{2\\left(c \\sqrt{c}-p q^{2}\\right)}{c-q^{2}} \\cdot \\frac{2 \\sqrt{c}}{c-q^{2}}=4 \\cdot \\frac{c^{2}-\\sqrt{c} p q^{2}}{\\left(c-q^{2}\\right)^{2}} \\leq 4 \\cdot \\frac{c^{2}-2 c q^{2}}{\\left(c-q^{2}\\right)^{2}} \\leq 4 .\n\\end{gathered}\n$$']" [] Text-only Competition True Theorem proof Algebra Math English 98 "We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: - if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$ ) are in the same state, then $L_{i}$ is switched off; - otherwise, $L_{i}$ is switched on. Initially all the lamps are off except the leftmost one which is on. Prove that there are infinitely many integers $n$ for which all the lamps will eventually be off." "[""Experiments with small $n$ lead to the guess that every $n$ of the form $2^{k}$ should be good. This is indeed the case, and more precisely: let $A_{k}$ be the $2^{k} \\times 2^{k}$ matrix whose rows represent the evolution of the system, with entries 0,1 (for off and on respectively). The top row shows the initial state $[1,0,0, \\ldots, 0]$; the bottom row shows the state after $2^{k}-1$ steps. The claim is that:\n\n$$\n\\text { The bottom row of } A_{k} \\text { is }[1,1,1, \\ldots, 1] \\text {. }\n$$\n\nThis will of course suffice because one more move then produces $[0,0,0, \\ldots, 0]$, as required.\n\nThe proof is by induction on $k$. The base $k=1$ is obvious. Assume the claim to be true for a\n\n$k \\geq 1$ and write the matrix $A_{k+1}$ in the block form $\\left(\\begin{array}{cc}A_{k} & O_{k} \\\\ B_{k} & C_{k}\\end{array}\\right)$ with four $2^{k} \\times 2^{k}$ matrices. After $m$ steps, the last 1 in a row is at position $m+1$. Therefore $O_{k}$ is the zero matrix. According to the induction hypothesis, the bottom row of $\\left[A_{k} O_{k}\\right]$ is $[1, \\ldots, 1,0, \\ldots, 0]$, with $2^{k}$ ones and $2^{k}$ zeros. The next row is thus\n\n$$\n[\\underbrace{0, \\ldots, 0}_{2^{k}-1}, 1,1, \\underbrace{0, \\ldots, 0}_{2^{k}-1}]\n$$\n\nIt is symmetric about its midpoint, and this symmetry is preserved in all subsequent rows because the procedure described in the problem statement is left/right symmetric. Thus $B_{k}$ is the mirror image of $C_{k}$. In particular, the rightmost column of $B_{k}$ is identical with the leftmost column of $C_{k}$.\n\nImagine the matrix $C_{k}$ in isolation from the rest of $A_{k+1}$. Suppose it is subject to evolution as defined in the problem: the first (leftmost) term in a row depends only on the two first terms in the preceding row, according as they are equal or not. Now embed $C_{k}$ again in $A_{k}$. The 'leftmost' terms in the rows of $C_{k}$ now have neighbours on their left side - but these neighbours are their exact copies. Consequently the actual evolution within $C_{k}$ is the same, whether or not $C_{k}$ is considered as a piece of $A_{k+1}$ or in isolation. And since the top row of $C_{k}$ is $[1,0, \\ldots, 0]$, it follows that $C_{k}$ is identical with $A_{k}$.\n\n\n\nThe bottom row of $A_{k}$ is $[1,1, \\ldots, 1]$; the same is the bottom row of $C_{k}$, hence also of $B_{k}$, which mirrors $C_{k}$. So the bottom row of $A_{k+1}$ consists of ones only and the induction is complete.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 99 "We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: - if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$ ) are in the same state, then $L_{i}$ is switched off; - otherwise, $L_{i}$ is switched on. Initially all the lamps are off except the leftmost one which is on. Prove that there are infinitely many integers $n$ for which the lamps will never be all off." ['There are many ways to produce an infinite sequence of those $n$ for which the state $[0,0, \\ldots, 0]$ will never be achieved. As an example, consider $n=2^{k}+1$ (for $k \\geq 1$ ). The evolution of the system can be represented by a matrix $\\mathcal{A}$ of width $2^{k}+1$ with infinitely many rows. The top $2^{k}$ rows form the matrix $A_{k}$ discussed above, with one column of zeros attached at its right.\n\nIn the next row we then have the vector $[0,0, \\ldots, 0,1,1]$. But this is just the second row of $\\mathcal{A}$ reversed. Subsequent rows will be mirror copies of the foregoing ones, starting from the second one. So the configuration $[1,1,0, \\ldots, 0,0]$, i.e. the second row of $\\mathcal{A}$, will reappear. Further rows will periodically repeat this pattern and there will be no row of zeros.'] [] Text-only Competition True Theorem proof Combinatorics Math English 100 "Let $S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $P$ whose vertices are in $S$, let $a(P)$ be the number of vertices of $P$, and let $b(P)$ be the number of points of $S$ which are outside $P$. Prove that for every real number $x$ $$ \sum_{P} x^{a(P)}(1-x)^{b(P)}=1 $$ where the sum is taken over all convex polygons with vertices in $S$. NB. A line segment, a point and the empty set are considered as convex polygons of 2,1 and 0 vertices, respectively." "['For each convex polygon $P$ whose vertices are in $S$, let $c(P)$ be the number of points of $S$ which are inside $P$, so that $a(P)+b(P)+c(P)=n$, the total number of points in $S$. Denoting $1-x$ by $y$,\n\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{P} x^{a(P)} y^{b(P)}(x+y)^{c(P)}=\\sum_{P} \\sum_{i=0}^{c(P)}\\left(\\begin{array}{c}\nc(P) \\\\\ni\n\\end{array}\\right) x^{a(P)+i} y^{b(P)+c(P)-i}\n$$\n\nView this expression as a homogeneous polynomial of degree $n$ in two independent variables $x, y$. In the expanded form, it is the sum of terms $x^{r} y^{n-r}(0 \\leq r \\leq n)$ multiplied by some nonnegative integer coefficients.\n\nFor a fixed $r$, the coefficient of $x^{r} y^{n-r}$ represents the number of ways of choosing a convex polygon $P$ and then choosing some of the points of $S$ inside $P$ so that the number of vertices of $P$ and the number of chosen points inside $P$ jointly add up to $r$.\n\nThis corresponds to just choosing an $r$-element subset of $S$. The correspondence is bijective because every set $T$ of points from $S$ splits in exactly one way into the union of two disjoint subsets, of which the first is the set of vertices of a convex polygon - namely, the convex hull of $T$ - and the second consists of some points inside that polygon.\n\nSo the coefficient of $x^{r} y^{n-r}$ equals $\\left(\\begin{array}{l}n \\\\ r\\end{array}\\right)$. The desired result follows:\n\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{r=0}^{n}\\left(\\begin{array}{l}\nn \\\\\nr\n\\end{array}\\right) x^{r} y^{n-r}=(x+y)^{n}=1\n$$' 'Apply induction on the number $n$ of points. The case $n=0$ is trivial. Let $n>0$ and assume the statement for less than $n$ points. Take a set $S$ of $n$ points.\n\nLet $C$ be the set of vertices of the convex hull of $S$, let $m=|C|$.\n\nLet $X \\subset C$ be an arbitrary nonempty set. For any convex polygon $P$ with vertices in the set $S \\backslash X$, we have $b(P)$ points of $S$ outside $P$. Excluding the points of $X-$ all outside $P$ - the set $S \\backslash X$ contains exactly $b(P)-|X|$ of them. Writing $1-x=y$, by the induction hypothesis\n\n$$\n\\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)-|X|}=1\n$$\n\n(where $P \\subset S \\backslash X$ means that the vertices of $P$ belong to the set $S \\backslash X$ ). Therefore\n\n$$\n\\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)}=y^{|X|}\n$$\n\nAll convex polygons appear at least once, except the convex hull $C$ itself. The convex hull adds $x^{m}$. We can use the inclusion-exclusion principle to compute the sum of the other terms:\n\n$$\n\\begin{gathered}\n\\sum_{P \\neq C} x^{a(P)} y^{b(P)}=\\sum_{k=1}^{m}(-1)^{k-1} \\sum_{|X|=k} \\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)}=\\sum_{k=1}^{m}(-1)^{k-1} \\sum_{|X|=k} y^{k} \\\\\n=\\sum_{k=1}^{m}(-1)^{k-1}\\left(\\begin{array}{c}\nm \\\\\nk\n\\end{array}\\right) y^{k}=-\\left((1-y)^{m}-1\\right)=1-x^{m}\n\\end{gathered}\n$$\n\nand then\n\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{P=C}+\\sum_{P \\neq C}=x^{m}+\\left(1-x^{m}\\right)=1\n$$']" [] Text-only Competition True Theorem proof Combinatorics Math English 101 "A cake has the form of an $n \times n$ square composed of $n^{2}$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$. Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\mathcal{B}$ than of arrangement $\mathcal{A}$. Prove that arrangement $\mathcal{B}$ can be obtained from $\mathcal{A}$ by performing a number of switches, defined as follows: A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle." ['We use capital letters to denote unit squares; $O$ is the top left corner square. For any two squares $X$ and $Y$ let $[X Y]$ be the smallest grid rectangle containing these two squares. Strawberries lie on some squares in arrangement $\\mathcal{A}$. Put a plum on each square of the target configuration $\\mathcal{B}$. For a square $X$ denote by $a(X)$ and $b(X)$ respectively the number of strawberries and the number of plums in $[O X]$. By hypothesis $a(X) \\leq b(X)$ for each $X$, with strict inequality for some $X$ (otherwise the two arrangements coincide and there is nothing to prove).\n\nThe idea is to show that by a legitimate switch one can obtain an arrangement $\\mathcal{A}^{\\prime}$ such that\n\n$$\na(X) \\leq a^{\\prime}(X) \\leq b(X) \\quad \\text { for each } X ; \\quad \\sum_{X} a(X)<\\sum_{X} a^{\\prime}(X) \\tag{1}\n$$\n\n(with $a^{\\prime}(X)$ defined analogously to $a(X)$; the sums range over all unit squares $X$ ). This will be enough because the same reasoning then applies to $\\mathcal{A}^{\\prime}$, giving rise to a new arrangement $\\mathcal{A}^{\\prime \\prime}$, and so on (induction). Since $\\sum a(X)<\\sum a^{\\prime}(X)<\\sum a^{\\prime \\prime}(X)<\\ldots$ and all these sums do not exceed $\\sum b(X)$, we eventually obtain a sum with all summands equal to the respective $b(X) \\mathrm{s}$; all strawberries will meet with plums.\n\nConsider the uppermost row in which the plum and the strawberry lie on different squares $P$ and $S$ (respectively); clearly $P$ must be situated left to $S$. In the column passing through $P$, let $T$ be the top square and $B$ the bottom square. The strawberry in that column lies below the plum (because there is no plum in that column above $P$, and the positions of strawberries and plums coincide everywhere above the row of $P$ ). Hence there is at least one strawberry in the region $[B S]$ below $[P S]$. Let $V$ be the position of the uppermost strawberry in that region.\n\n\n\n\n\nDenote by $W$ the square at the intersection of the row through $V$ with the column through $S$ and let $R$ be the square vertex-adjacent to $W$ up-left. We claim that\n\n$$\na(X)\n\nIf $\\Gamma$ is above line $\\ell$ then so is $\\Gamma+\\Delta$, which contradicts the maximum choice of $\\Delta$. If $\\Gamma$ contains cells from row $B$, observe that $\\Gamma+\\Delta$ contains $u$. Let $s$ be the size of $\\Gamma+\\Delta$. Being full of $H \\backslash\\{u\\}, \\Gamma+\\Delta$ contains $s$ holes other than $u$. But it also contains $u$, contradicting the assumption that $H$ is spread out.\n\nThe claim follows, showing that $b_{1}=w$ is an appropriate choice for $a_{1}=u$ and $a_{2}=v$. As explained above, this is enough to complete the induction.']" [] Text-only Competition True Theorem proof Combinatorics Math English 104 "Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it. Call a pair of points of the polyhedron antipodal if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes. Let $A$ be the number of antipodal pairs of vertices, and let $B$ be the number of antipodal pairs of midpoints of edges. Determine the difference $A-B$ in terms of the numbers of vertices, edges and faces." "['Denote the polyhedron by $\\Gamma$; let its vertices, edges and faces be $V_{1}, V_{2}, \\ldots, V_{n}$, $E_{1}, E_{2}, \\ldots, E_{m}$ and $F_{1}, F_{2}, \\ldots, F_{\\ell}$, respectively. Denote by $Q_{i}$ the midpoint of edge $E_{i}$.\n\nLet $S$ be the unit sphere, the set of all unit vectors in three-dimensional space. Map the boundary elements of $\\Gamma$ to some objects on $S$ as follows.\n\nFor a face $F_{i}$, let $S^{+}\\left(F_{i}\\right)$ and $S^{-}\\left(F_{i}\\right)$ be the unit normal vectors of face $F_{i}$, pointing outwards from $\\Gamma$ and inwards to $\\Gamma$, respectively. These points are diametrically opposite.\n\nFor an edge $E_{j}$, with neighbouring faces $F_{i_{1}}$ and $F_{i_{2}}$, take all support planes of $\\Gamma$ (planes which have a common point with $\\Gamma$ but do not intersect it) containing edge $E_{j}$, and let $S^{+}\\left(E_{j}\\right)$ be the set of their outward normal vectors. The set $S^{+}\\left(E_{j}\\right)$ is an arc of a great circle on $S$. Arc $S^{+}\\left(E_{j}\\right)$ is perpendicular to edge $E_{j}$ and it connects points $S^{+}\\left(F_{i_{1}}\\right)$ and $S^{+}\\left(F_{i_{2}}\\right)$.\n\nDefine also the set of inward normal vectors $S^{-}\\left(E_{i}\\right)$ which is the reflection of $S^{+}\\left(E_{i}\\right)$ across the origin.\n\nFor a vertex $V_{k}$, which is the common endpoint of edges $E_{j_{1}}, \\ldots, E_{j_{h}}$ and shared by faces $F_{i_{1}}, \\ldots, F_{i_{h}}$, take all support planes of $\\Gamma$ through point $V_{k}$ and let $S^{+}\\left(V_{k}\\right)$ be the set of their outward normal vectors. This is a region on $S$, a spherical polygon with vertices $S^{+}\\left(F_{i_{1}}\\right), \\ldots, S^{+}\\left(F_{i_{h}}\\right)$ bounded by $\\operatorname{arcs} S^{+}\\left(E_{j_{1}}\\right), \\ldots, S^{+}\\left(E_{j_{h}}\\right)$. Let $S^{-}\\left(V_{k}\\right)$ be the reflection of $S^{+}\\left(V_{k}\\right)$, the set of inward normal vectors.\n\nNote that region $S^{+}\\left(V_{k}\\right)$ is convex in the sense that it is the intersection of several half spheres.\n\n\nNow translate the conditions on $\\Gamma$ to the language of these objects.\n\n(a) Polyhedron $\\Gamma$ has no parallel edges - the great circles of $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{j}\\right)$ are different for all $i \\neq j$.\n\n(b) If an edge $E_{i}$ does not belong to a face $F_{j}$ then they are not parallel - the great circle which contains $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{i}\\right)$ does not pass through points $S^{+}\\left(F_{j}\\right)$ and $S^{-}\\left(F_{j}\\right)$.\n\n(c) Polyhedron $\\Gamma$ has no parallel faces - points $S^{+}\\left(F_{i}\\right)$ and $S^{-}\\left(F_{j}\\right)$ are pairwise distinct.\n\nThe regions $S^{+}\\left(V_{k}\\right)$, arcs $S^{+}\\left(E_{j}\\right)$ and points $S^{+}\\left(F_{i}\\right)$ provide a decomposition of the surface of the sphere. Regions $S^{-}\\left(V_{k}\\right)$, arcs $S^{-}\\left(E_{j}\\right)$ and points $S^{-}\\left(F_{i}\\right)$ provide the reflection of this decomposition. These decompositions are closely related to the problem.\n\n\n\nLemma 1. For any $1 \\leq i, j \\leq n$, regions $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right)$ overlap if and only if vertices $V_{i}$ and $V_{j}$ are antipodal.\n\nLemma 2. For any $1 \\leq i, j \\leq m$, arcs $S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect if and only if the midpoints $Q_{i}$ and $Q_{j}$ of edges $E_{i}$ and $E_{j}$ are antipodal.\n\nProof of lemma 1. First note that by properties (a,b,c) above, the two regions cannot share only a single point or an arc. They are either disjoint or they overlap.\n\nAssume that the two regions have a common interior point $u$. Let $P_{1}$ and $P_{2}$ be two parallel support planes of $\\Gamma$ through points $V_{i}$ and $V_{j}$, respectively, with normal vector $u$. By the definition of regions $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right), u$ is the inward normal vector of $P_{1}$ and the outward normal vector of $P_{2}$. Therefore polyhedron $\\Gamma$ lies between the two planes; vertices $V_{i}$ and $V_{j}$ are antipodal.\n\nTo prove the opposite direction, assume that $V_{i}$ and $V_{j}$ are antipodal. Then there exist two parallel support planes $P_{1}$ and $P_{2}$ through $V_{i}$ and $V_{j}$, respectively, such that $\\Gamma$ is between them. Let $u$ be the inward normal vector of $P_{1}$; then $u$ is the outward normal vector of $P_{2}$, therefore $u \\in S^{-}\\left(V_{i}\\right) \\cap S^{+}\\left(V_{j}\\right)$. The two regions have a common point, so they overlap.\n\nProof of lemma 2. Again, by properties (a,b) above, the endpoints of arc $S^{-}\\left(E_{i}\\right)$ cannot belong to $S^{+}\\left(E_{j}\\right)$ and vice versa. The two arcs are either disjoint or intersecting.\n\nAssume that arcs $S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect at point $u$. Let $P_{1}$ and $P_{2}$ be the two support planes through edges $E_{i}$ and $E_{j}$, respectively, with normal vector $u$. By the definition of $\\operatorname{arcs} S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$, vector $u$ points inwards from $P_{1}$ and outwards from $P_{2}$. Therefore $\\Gamma$ is between the planes. Since planes $P_{1}$ and $P_{2}$ pass through $Q_{i}$ and $Q_{j}$, these points are antipodal.\n\nFor the opposite direction, assume that points $Q_{i}$ and $Q_{j}$ are antipodal. Let $P_{1}$ and $P_{2}$ be two support planes through these points, respectively. An edge cannot intersect a support plane, therefore $E_{i}$ and $E_{j}$ lie in the planes $P_{1}$ and $P_{2}$, respectively. Let $u$ be the inward normal vector of $P_{1}$, which is also the outward normal vector of $P_{2}$. Then $u \\in S^{-}\\left(E_{i}\\right) \\cap S^{+}\\left(E_{j}\\right)$. So the two arcs are not disjoint; they therefore intersect.\n\nNow create a new decomposition of sphere $S$. Draw all $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{j}\\right)$ on sphere $S$ and put a knot at each point where two arcs meet. We have $\\ell$ knots at points $S^{+}\\left(F_{i}\\right)$ and another $\\ell$ knots at points $S^{-}\\left(F_{i}\\right)$, corresponding to the faces of $\\Gamma$; by property (c) they are different. We also have some pairs $1 \\leq i, j \\leq m$ where $\\operatorname{arcs} S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect. By Lemma 2, each antipodal pair $\\left(Q_{i}, Q_{j}\\right)$ gives rise to two such intersections; hence, the number of all intersections is $2 B$ and we have $2 \\ell+2 B$ knots in all.\n\nEach intersection knot splits two arcs, increasing the number of arcs by 2 . Since we started with $2 m$ arcs, corresponding the edges of $\\Gamma$, the number of the resulting curve segments is $2 m+4 B$.\n\nThe network of these curve segments divides the sphere into some ""new"" regions. Each new region is the intersection of some overlapping sets $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right)$. Due to the convexity, the intersection of two overlapping regions is convex and thus contiguous. By Lemma 1, each pair of overlapping regions corresponds to an antipodal vertex pair and each antipodal vertex pair gives rise to two different overlaps, which are symmetric with respect to the origin. So the number of new regions is $2 A$.\n\nThe result now follows from Euler\'s polyhedron theorem. We have $n+l=m+2$ and\n\n$$\n(2 \\ell+2 B)+2 A=(2 m+4 B)+2,\n$$\n\ntherefore\n\n$$\nA-B=m-\\ell+1=n-1 \\text {. }\n$$\n\nTherefore $A-B$ is by one less than the number of vertices of $\\Gamma$.' ""Use the same notations for the polyhedron and its vertices, edges and faces as in Solution 1. We regard points as vectors starting from the origin. Polyhedron $\\Gamma$ is regarded as a closed convex set, including its interior. In some cases the edges and faces of $\\Gamma$ are also regarded as sets of points. The symbol $\\partial$ denotes the boundary of the certain set; e.g. $\\partial \\Gamma$ is the surface of $\\Gamma$.\n\nLet $\\Delta=\\Gamma-\\Gamma=\\{U-V: U, V \\in \\Gamma\\}$ be the set of vectors between arbitrary points of $\\Gamma$. Then $\\Delta$, being the sum of two bounded convex sets, is also a bounded convex set and, by construction, it is also centrally symmetric with respect to the origin. We will prove that $\\Delta$ is also a polyhedron and express the numbers of its faces, edges and vertices in terms $n, m, \\ell, A$ and $B$.\n\nLemma 1. For points $U, V \\in \\Gamma$, point $W=U-V$ is a boundary point of $\\Delta$ if and only if $U$ and $V$ are antipodal. Moreover, for each boundary point $W \\in \\partial \\Delta$ there exists exactly one pair of points $U, V \\in \\Gamma$ such that $W=U-V$.\n\nProof. Assume first that $U$ and $V$ are antipodal points of $\\Gamma$. Let parallel support planes $P_{1}$ and $P_{2}$ pass through them such that $\\Gamma$ is in between. Consider plane $P=P_{1}-U=$ $P_{2}-V$. This plane separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. After reflecting one of the sets, e.g. $\\Gamma-V$, the sets $\\Gamma-U$ and $-\\Gamma+V$ lie in the same half space bounded by $P$. Then $(\\Gamma-U)+(-\\Gamma+V)=\\Delta-W$ lies in that half space, so $0 \\in P$ is a boundary point of the set $\\Delta-W$. Translating by $W$ we obtain that $W$ is a boundary point of $\\Delta$.\n\nTo prove the opposite direction, let $W=U-V$ be a boundary point of $\\Delta$, and let $\\Psi=$ $(\\Gamma-U) \\cap(\\Gamma-V)$. We claim that $\\Psi=\\{0\\}$. Clearly $\\Psi$ is a bounded convex set and $0 \\in \\Psi$. For any two points $X, Y \\in \\Psi$, we have $U+X, V+Y \\in \\Gamma$ and $W+(X-Y)=(U+X)-(V+Y) \\in \\Delta$. Since $W$ is a boundary point of $\\Delta$, the vector $X-Y$ cannot have the same direction as $W$. This implies that the interior of $\\Psi$ is empty. Now suppose that $\\Psi$ contains a line segment $S$. Then $S+U$ and $S+V$ are subsets of some faces or edges of $\\Gamma$ and these faces/edges are parallel to $S$. In all cases, we find two faces, two edges, or a face and an edge which are parallel, contradicting the conditions of the problem. Therefore, $\\Psi=\\{0\\}$ indeed.\n\nSince $\\Psi=(\\Gamma-U) \\cap(\\Gamma-V)$ consists of a single point, the interiors of bodies $\\Gamma-U$ and $\\Gamma-V$ are disjoint and there exists a plane $P$ which separates them. Let $u$ be the normal vector of $P$ pointing into that half space bounded by $P$ which contains $\\Gamma-U$. Consider the planes $P+U$ and $P+V$; they are support planes of $\\Gamma$, passing through $U$ and $V$, respectively. From plane $P+U$, the vector $u$ points into that half space which contains $\\Gamma$. From plane $P+V$, vector $u$ points into the opposite half space containing $\\Gamma$. Therefore, we found two proper support through points $U$ and $V$ such that $\\Gamma$ is in between.\n\nFor the uniqueness part, assume that there exist points $U_{1}, V_{1} \\in \\Gamma$ such that $U_{1}-V_{1}=U-V$. The points $U_{1}-U$ and $V_{1}-V$ lie in the sets $\\Gamma-U$ and $\\Gamma-V$ separated by $P$. Since $U_{1}-U=V_{1}-V$, this can happen only if both are in $P$; but the only such point is 0 . Therefore, $U_{1}-V_{1}=U-V$ implies $U_{1}=U$ and $V_{1}=V$. The lemma is complete.\n\nLemma 2. Let $U$ and $V$ be two antipodal points and assume that plane $P$, passing through 0 , separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. Let $\\Psi_{1}=(\\Gamma-U) \\cap P$ and $\\Psi_{2}=(\\Gamma-V) \\cap P$. Then $\\Delta \\cap(P+U-V)=\\Psi_{1}-\\Psi_{2}+U-V$.\n\nProof. The sets $\\Gamma-U$ and $-\\Gamma+V$ lie in the same closed half space bounded by $P$. Therefore, for any points $X \\in(\\Gamma-U)$ and $Y \\in(-\\Gamma+V)$, we have $X+Y \\in P$ if and only if $X, Y \\in P$. Then\n\n$(\\Delta-(U-V)) \\cap P=((\\Gamma-U)+(-\\Gamma+V)) \\cap P=((\\Gamma-U) \\cap P)+((-\\Gamma+V) \\cap P)=\\Psi_{1}-\\Psi_{2}$.\n\nNow a translation by $(U-V)$ completes the lemma.\n\n\n\nNow classify the boundary points $W=U-V$ of $\\Delta$, according to the types of points $U$ and $V$. In all cases we choose a plane $P$ through 0 which separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. We will use the notation $\\Psi_{1}=(\\Gamma-U) \\cap P$ and $\\Psi_{2}=(\\Gamma-V) \\cap P$ as well.\n\nCase 1: Both $U$ and $V$ are vertices of $\\Gamma$. Bodies $\\Gamma-U$ and $\\Gamma-V$ have a common vertex which is 0 . Choose plane $P$ in such a way that $\\Psi_{1}=\\Psi_{2}=\\{0\\}$. Then Lemma 2 yields $\\Delta \\cap(P+W)=\\{W\\}$. Therefore $P+W$ is a support plane of $\\Delta$ such that they have only one common point so no line segment exists on $\\partial \\Delta$ which would contain $W$ in its interior.\n\nSince this case occurs for antipodal vertex pairs and each pair is counted twice, the number of such boundary points on $\\Delta$ is $2 A$.\n\nCase 2: Point $U$ is an interior point of an edge $E_{i}$ and $V$ is a vertex of $\\Gamma$. Choose plane $P$ such that $\\Psi_{1}=E_{i}-U$ and $\\Psi_{2}=\\{0\\}$. By Lemma $2, \\Delta \\cap(P+W)=E_{i}-V$. Hence there exists a line segment in $\\partial \\Delta$ which contains $W$ in its interior, but there is no planar region in $\\partial \\Delta$ with the same property.\n\nWe obtain a similar result if $V$ belongs to an edge of $\\Gamma$ and $U$ is a vertex.\n\nCase 3: Points $U$ and $V$ are interior points of edges $E_{i}$ and $E_{j}$, respectively. Let $P$ be the plane of $E_{i}-U$ and $E_{j}-V$. Then $\\Psi_{1}=E_{i}-U, \\Psi_{2}=E_{j}-V$ and $\\Delta \\cap(P+W)=E_{i}-E_{j}$. Therefore point $W$ belongs to a parallelogram face on $\\partial \\Delta$.\n\nThe centre of the parallelogram is $Q_{i}-Q_{j}$, the vector between the midpoints. Therefore an edge pair $\\left(E_{i}, E_{j}\\right)$ occurs if and only if $Q_{i}$ and $Q_{j}$ are antipodal which happens $2 B$ times.\n\nCase 4: Point $U$ lies in the interior of a face $F_{i}$ and $V$ is a vertex of $\\Gamma$. The only choice for $P$ is the plane of $F_{i}-U$. Then we have $\\Psi_{1}=F_{i}-U, \\Psi_{2}=\\{0\\}$ and $\\Delta \\cap(P+W)=F_{i}-V$. This is a planar face of $\\partial \\Delta$ which is congruent to $F_{i}$.\n\nFor each face $F_{i}$, the only possible vertex $V$ is the farthest one from the plane of $F_{i}$.\n\nIf $U$ is a vertex and $V$ belongs to face $F_{i}$ then we obtain the same way that $W$ belongs to a face $-F_{i}+U$ which is also congruent to $F_{i}$. Therefore, each face of $\\Gamma$ has two copies on $\\partial \\Delta$, a translated and a reflected copy.\n\nCase 5: Point $U$ belongs to a face $F_{i}$ of $\\Gamma$ and point $V$ belongs to an edge or a face $G$. In this case objects $F_{i}$ and $G$ must be parallel which is not allowed.\n\n\nNow all points in $\\partial \\Delta$ belong to some planar polygons (cases 3 and 4 ), finitely many line segments (case 2) and points (case 1). Therefore $\\Delta$ is indeed a polyhedron. Now compute the numbers of its vertices, edges and faces.\n\nThe vertices are obtained in case 1 , their number is $2 A$.\n\nFaces are obtained in cases 3 and 4 . Case 3 generates $2 B$ parallelogram faces. Case 4 generates $2 \\ell$ faces.\n\nWe compute the number of edges of $\\Delta$ from the degrees (number of sides) of faces of $\\Gamma$. Let $d_{i}$ be the the degree of face $F_{i}$. The sum of degrees is twice as much as the number of edges, so $d_{1}+d_{2}+\\ldots+d_{l}=2 m$. The sum of degrees of faces of $\\Delta$ is $2 B \\cdot 4+2\\left(d_{1}+d_{2}+\\cdots+d_{l}\\right)=8 B+4 m$, so the number of edges on $\\Delta$ is $4 B+2 m$.\n\nApplying Euler's polyhedron theorem on $\\Gamma$ and $\\Delta$, we have $n+l=m+2$ and $2 A+(2 B+2 \\ell)=$ $(4 B+2 m)+2$. Then the conclusion follows:\n\n$$\nA-B=m-\\ell+1=n-1 \\text {. }\n$$""]" [] Text-only Competition True Theorem proof Combinatorics Math English 105 "Let $A B C$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies $$ \angle P B A+\angle P C A=\angle P B C+\angle P C B . $$ Show that $A P \geq A I$ and that equality holds if and only if $P$ coincides with $I$." ['Let $\\angle A=\\alpha, \\angle B=\\beta, \\angle C=\\gamma$. Since $\\angle P B A+\\angle P C A+\\angle P B C+\\angle P C B=\\beta+\\gamma$, the condition from the problem statement is equivalent to $\\angle P B C+\\angle P C B=(\\beta+\\gamma) / 2$, i. e. $\\angle B P C=90^{\\circ}+\\alpha / 2$.\n\nOn the other hand $\\angle B I C=180^{\\circ}-(\\beta+\\gamma) / 2=90^{\\circ}+\\alpha / 2$. Hence $\\angle B P C=\\angle B I C$, and since $P$ and $I$ are on the same side of $B C$, the points $B, C, I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\\omega$ of triangle $B C I$.\n\n\n\nLet $\\Omega$ be the circumcircle of triangle $A B C$. It is a well-known fact that the centre of $\\omega$ is the midpoint $M$ of the $\\operatorname{arc} B C$ of $\\Omega$. This is also the point where the angle bisector $A I$ intersects $\\Omega$.\n\nFrom triangle $A P M$ we have\n\n$$\nA P+P M \\geq A M=A I+I M=A I+P M\n$$\n\nTherefore $A P \\geq A I$. Equality holds if and only if $P$ lies on the line segment $A I$, which occurs if and only if $P=I$.'] [] Text-only Competition True Theorem proof Geometry Math English 106 "Let $A B C D$ be a trapezoid with parallel sides $A B>C D$. Points $K$ and $L$ lie on the line segments $A B$ and $C D$, respectively, so that $A K / K B=D L / L C$. Suppose that there are points $P$ and $Q$ on the line segment $K L$ satisfying $$ \angle A P B=\angle B C D \quad \text { and } \quad \angle C Q D=\angle A B C \text {. } $$ Prove that the points $P, Q, B$ and $C$ are concyclic." "['Because $A B \\| C D$, the relation $A K / K B=D L / L C$ readily implies that the lines $A D, B C$ and $K L$ have a common point $S$.\n\n\n\nConsider the second intersection points $X$ and $Y$ of the line $S K$ with the circles $(A B P)$ and $(C D Q)$, respectively. Since $A P B X$ is a cyclic quadrilateral and $A B \\| C D$, one has\n\n$$\n\\angle A X B=180^{\\circ}-\\angle A P B=180^{\\circ}-\\angle B C D=\\angle A B C \\text {. }\n$$\n\nThis shows that $B C$ is tangent to the circle $(A B P)$ at $B$. Likewise, $B C$ is tangent to the circle $(C D Q)$ at $C$. Therefore $S P \\cdot S X=S B^{2}$ and $S Q \\cdot S Y=S C^{2}$.\n\nLet $h$ be the homothety with centre $S$ and ratio $S C / S B$. Since $h(B)=C$, the above conclusion about tangency implies that $h$ takes circle $(A B P)$ to circle $(C D Q)$. Also, $h$ takes $A B$ to $C D$, and it easily follows that $h(P)=Y, h(X)=Q$, yielding $S P / S Y=S B / S C=S X / S Q$.\n\nEqualities $S P \\cdot S X=S B^{2}$ and $S Q / S X=S C / S B$ imply $S P \\cdot S Q=S B \\cdot S C$, which is equivalent to $P, Q, B$ and $C$ being concyclic.' ""The case where $P=Q$ is trivial. Thus assume that $P$ and $Q$ are two distinct points. As in the first solution, notice that the lines $A D, B C$ and $K L$ concur at a point $S$.\n\n\n\nLet the lines $A P$ and $D Q$ meet at $E$, and let $B P$ and $C Q$ meet at $F$. Then $\\angle E P F=\\angle B C D$ and $\\angle F Q E=\\angle A B C$ by the condition of the problem. Since the angles $B C D$ and $A B C$ add up to $180^{\\circ}$, it follows that $P E Q F$ is a cyclic quadrilateral.\n\nApplying Menelaus' theorem, first to triangle $A S P$ and line $D Q$ and then to triangle $B S P$ and line $C Q$, we have\n\n$$\n\\frac{A D}{D S} \\cdot \\frac{S Q}{Q P} \\cdot \\frac{P E}{E A}=1 \\quad \\text { and } \\quad \\frac{B C}{C S} \\cdot \\frac{S Q}{Q P} \\cdot \\frac{P F}{F B}=1\n$$\n\nThe first factors in these equations are equal, as $A B \\| C D$. Thus the last factors are also equal, which implies that $E F$ is parallel to $A B$ and $C D$. Using this and the cyclicity of $P E Q F$, we obtain\n\n$$\n\\angle B C D=\\angle B C F+\\angle F C D=\\angle B C Q+\\angle E F Q=\\angle B C Q+\\angle E P Q .\n$$\n\nOn the other hand,\n\n$$\n\\angle B C D=\\angle A P B=\\angle E P F=\\angle E P Q+\\angle Q P F,\n$$\n\nand consequently $\\angle B C Q=\\angle Q P F$. The latter angle either coincides with $\\angle Q P B$ or is supplementary to $\\angle Q P B$, depending on whether $Q$ lies between $K$ and $P$ or not. In either case it follows that $P, Q, B$ and $C$ are concyclic.""]" [] Text-only Competition True Theorem proof Geometry Math English 107 "Let $A B C D E$ be a convex pentagon such that $$ \angle B A C=\angle C A D=\angle D A E \quad \text { and } \quad \angle A B C=\angle A C D=\angle A D E \text {. } $$ The diagonals $B D$ and $C E$ meet at $P$. Prove that the line $A P$ bisects the side $C D$." "[""Let the diagonals $A C$ and $B D$ meet at $Q$, the diagonals $A D$ and $C E$ meet at $R$, and let the ray $A P$ meet the side $C D$ at $M$. We want to prove that $C M=M D$ holds.\n\n\n\nThe idea is to show that $Q$ and $R$ divide $A C$ and $A D$ in the same ratio, or more precisely\n\n$$\n\\frac{A Q}{Q C}=\\frac{A R}{R D} \\tag{1}\n$$\n\n(which is equivalent to $Q R \\| C D$ ). The given angle equalities imply that the triangles $A B C$, $A C D$ and $A D E$ are similar. We therefore have\n\n$$\n\\frac{A B}{A C}=\\frac{A C}{A D}=\\frac{A D}{A E}\n$$\n\nSince $\\angle B A D=\\angle B A C+\\angle C A D=\\angle C A D+\\angle D A E=\\angle C A E$, it follows from $A B / A C=$ $A D / A E$ that the triangles $A B D$ and $A C E$ are also similar. Their angle bisectors in $A$ are $A Q$ and $A R$, respectively, so that\n\n$$\n\\frac{A B}{A C}=\\frac{A Q}{A R}\n$$\n\nBecause $A B / A C=A C / A D$, we obtain $A Q / A R=A C / A D$, which is equivalent to (1). Now Ceva's theorem for the triangle $A C D$ yields\n\n$$\n\\frac{A Q}{Q C} \\cdot \\frac{C M}{M D} \\cdot \\frac{D R}{R A}=1\n$$\n\nIn view of (1), this reduces to $C M=M D$, which completes the proof.""]" [] Text-only Competition True Theorem proof Geometry Math English 108 A point $D$ is chosen on the side $A C$ of a triangle $A B C$ with $\angle C<\angle A<90^{\circ}$ in such a way that $B D=B A$. The incircle of $A B C$ is tangent to $A B$ and $A C$ at points $K$ and $L$, respectively. Let $J$ be the incentre of triangle $B C D$. Prove that the line $K L$ intersects the line segment $A J$ at its midpoint. ['Denote by $P$ be the common point of $A J$ and $K L$. Let the parallel to $K L$ through $J$ meet $A C$ at $M$. Then $P$ is the midpoint of $A J$ if and only if $A M=2 \\cdot A L$, which we are about to show.\n\n\n\nDenoting $\\angle B A C=2 \\alpha$, the equalities $B A=B D$ and $A K=A L$ imply $\\angle A D B=2 \\alpha$ and $\\angle A L K=90^{\\circ}-\\alpha$. Since $D J$ bisects $\\angle B D C$, we obtain $\\angle C D J=\\frac{1}{2} \\cdot\\left(180^{\\circ}-\\angle A D B\\right)=90^{\\circ}-\\alpha$. Also $\\angle D M J=\\angle A L K=90^{\\circ}-\\alpha$ since $J M \\| K L$. It follows that $J D=J M$.\n\nLet the incircle of triangle $B C D$ touch its side $C D$ at $T$. Then $J T \\perp C D$, meaning that $J T$ is the altitude to the base $D M$ of the isosceles triangle $D M J$. It now follows that $D T=M T$, and we have\n\n$$\nD M=2 \\cdot D T=B D+C D-B C \\text {. }\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\nA M & =A D+(B D+C D-B C) \\\\\n& =A D+A B+D C-B C \\\\\n& =A C+A B-B C \\\\\n& =2 \\cdot A L,\n\\end{aligned}\n$$\n\nwhich completes the proof.'] [] Text-only Competition True Theorem proof Geometry Math English 109 Circles $\omega_{1}$ and $\omega_{2}$ with centres $O_{1}$ and $O_{2}$ are externally tangent at point $D$ and internally tangent to a circle $\omega$ at points $E$ and $F$, respectively. Line $t$ is the common tangent of $\omega_{1}$ and $\omega_{2}$ at $D$. Let $A B$ be the diameter of $\omega$ perpendicular to $t$, so that $A, E$ and $O_{1}$ are on the same side of $t$. Prove that lines $A O_{1}, B O_{2}, E F$ and $t$ are concurrent. "[""Point $E$ is the centre of a homothety $h$ which takes circle $\\omega_{1}$ to circle $\\omega$. The radii $O_{1} D$ and $O B$ of these circles are parallel as both are perpendicular to line $t$. Also, $O_{1} D$ and $O B$ are on the same side of line $E O$, hence $h$ takes $O_{1} D$ to $O B$. Consequently, points $E$, $D$ and $B$ are collinear. Likewise, points $F, D$ and $A$ are collinear as well.\n\nLet lines $A E$ and $B F$ intersect at $C$. Since $A F$ and $B E$ are altitudes in triangle $A B C$, their common point $D$ is the orthocentre of this triangle. So $C D$ is perpendicular to $A B$, implying that $C$ lies on line $t$. Note that triangle $A B C$ is acute-angled. We mention the well-known fact that triangles $F E C$ and $A B C$ are similar in ratio $\\cos \\gamma$, where $\\gamma=\\angle A C B$. In addition, points $C, E, D$ and $F$ lie on the circle with diameter $C D$.\n\n\n\nLet $P$ be the common point of lines $E F$ and $t$. We are going to prove that $P$ lies on line $A O_{1}$. Denote by $N$ the second common point of circle $\\omega_{1}$ and $A C$; this is the point of $\\omega_{1}$ diametrically opposite to $D$. By Menelaus' theorem for triangle $D C N$, points $A, O_{1}$ and $P$ are collinear if and only if\n\n$$\n\\frac{C A}{A N} \\cdot \\frac{N O_{1}}{O_{1} D} \\cdot \\frac{D P}{P C}=1\n$$\n\nBecause $N O_{1}=O_{1} D$, this reduces to $C A / A N=C P / P D$. Let line $t$ meet $A B$ at $K$. Then $C A / A N=C K / K D$, so it suffices to show that\n\n$$\n\\frac{C P}{P D}=\\frac{C K}{K D}\n$$\n\nTo verify (1), consider the circumcircle $\\Omega$ of triangle $A B C$. Draw its diameter $C U$ through $C$, and let $C U$ meet $A B$ at $V$. Extend $C K$ to meet $\\Omega$ at $L$. Since $A B$ is parallel to $U L$, we have $\\angle A C U=\\angle B C L$. On the other hand $\\angle E F C=\\angle B A C, \\angle F E C=\\angle A B C$ and $E F / A B=\\cos \\gamma$, as stated above. So reflection in the bisector of $\\angle A C B$ followed by a homothety with centre $C$ and ratio $1 / \\cos \\gamma$ takes triangle $F E C$ to triangle $A B C$. Consequently, this transformation\n\n\n\ntakes $C D$ to $C U$, which implies $C P / P D=C V / V U$. Next, we have $K L=K D$, because $D$ is the orthocentre of triangle $A B C$. Hence $C K / K D=C K / K L$. Finally, $C V / V U=C K / K L$ because $A B$ is parallel to $U L$. Relation (1) follows, proving that $P$ lies on line $A O_{1}$. By symmetry, $P$ also lies on line $A O_{2}$ which completes the solution.\n\nSolution 2. We proceed as in the first solution to define a triangle $A B C$ with orthocentre $D$, in which $A F$ and $B E$ are altitudes.\n\nDenote by $M$ the midpoint of $C D$. The quadrilateral $C E D F$ is inscribed in a circle with centre $M$, hence $M C=M E=M D=M F$.\n\n\n\nConsider triangles $A B C$ and $O_{1} O_{2} M$. Lines $O_{1} O_{2}$ and $A B$ are parallel, both of them being perpendicular to line $t$. Next, $M O_{1}$ is the line of centres of circles $(C E F)$ and $\\omega_{1}$ whose common chord is $D E$. Hence $M O_{1}$ bisects $\\angle D M E$ which is the external angle at $M$ in the isosceles triangle $C E M$. It follows that $\\angle D M O_{1}=\\angle D C A$, so that $M O_{1}$ is parallel to $A C$. Likewise, $M O_{2}$ is parallel to $B C$.\n\nThus the respective sides of triangles $A B C$ and $O_{1} O_{2} M$ are parallel; in addition, these triangles are not congruent. Hence there is a homothety taking $A B C$ to $O_{1} O_{2} M$. The lines $A O_{1}$, $B O_{2}$ and $C M=t$ are concurrent at the centre $Q$ of this homothety.\n\nFinally, apply Pappus' theorem to the triples of collinear points $A, O, B$ and $O_{2}, D, O_{1}$. The theorem implies that the points $A D \\cap O O_{2}=F, A O_{1} \\cap B O_{2}=Q$ and $O O_{1} \\cap B D=E$ are collinear. In other words, line $E F$ passes through the common point $Q$ of $A O_{1}, B O_{2}$ and $t$.\n\nComment. Relation (1) from Solution 1 expresses the well-known fact that points $P$ and $K$ are harmonic conjugates with respect to points $C$ and $D$. It is also easy to justify it by direct computation. Denoting $\\angle C A B=\\alpha, \\angle A B C=\\beta$, it is straightforward to obtain $C P / P D=C K / K D=\\tan \\alpha \\tan \\beta$.""]" [] Text-only Competition True Theorem proof Geometry Math English 110 In a triangle $A B C$, let $M_{a}, M_{b}, M_{c}$ be respectively the midpoints of the sides $B C, C A$, $A B$ and $T_{a}, T_{b}, T_{c}$ be the midpoints of the $\operatorname{arcs} B C, C A, A B$ of the circumcircle of $A B C$, not containing the opposite vertices. For $i \in\{a, b, c\}$, let $\omega_{i}$ be the circle with $M_{i} T_{i}$ as diameter. Let $p_{i}$ be the common external tangent to $\omega_{j}, \omega_{k}(\{i, j, k\}=\{a, b, c\})$ such that $\omega_{i}$ lies on the opposite side of $p_{i}$ than $\omega_{j}, \omega_{k}$ do. Prove that the lines $p_{a}, p_{b}, p_{c}$ form a triangle similar to $A B C$ and find the ratio of similitude. ['Let $T_{a} T_{b}$ intersect circle $\\omega_{b}$ at $T_{b}$ and $U$, and let $T_{a} T_{c}$ intersect circle $\\omega_{c}$ at $T_{c}$ and $V$. Further, let $U X$ be the tangent to $\\omega_{b}$ at $U$, with $X$ on $A C$, and let $V Y$ be the tangent to $\\omega_{c}$ at $V$, with $Y$ on $A B$. The homothety with centre $T_{b}$ and ratio $T_{b} T_{a} / T_{b} U$ maps the circle $\\omega_{b}$ onto the circumcircle of $A B C$ and the line $U X$ onto the line tangent to the circumcircle at $T_{a}$, which is parallel to $B C$; thus $U X \\| B C$. The same is true of $V Y$, so that $U X\\|B C\\| V Y$.\n\nLet $T_{a} T_{b}$ cut $A C$ at $P$ and let $T_{a} T_{c}$ cut $A B$ at $Q$. The point $X$ lies on the hypotenuse $P M_{b}$ of the right triangle $P U M_{b}$ and is equidistant from $U$ and $M_{b}$. So $X$ is the midpoint of $M_{b} P$. Similarly $Y$ is the midpoint of $M_{c} Q$.\n\nDenote the incentre of triangle $A B C$ as usual by $I$. It is a known fact that $T_{a} I=T_{a} B$ and $T_{c} I=T_{c} B$. Therefore the points $B$ and $I$ are symmetric across $T_{a} T_{c}$, and consequently $\\angle Q I B=\\angle Q B I=\\angle I B C$. This implies that $B C$ is parallel to the line $I Q$, and likewise, to $I P$. In other words, $P Q$ is the line parallel to $B C$ passing through $I$.\n\n\n\nClearly $M_{b} M_{c} \\| B C$. So $P M_{b} M_{c} Q$ is a trapezoid and the segment $X Y$ connects the midpoints of its nonparallel sides; hence $X Y \\| B C$. This combined with the previously established relations $U X\\|B C\\| V Y$ shows that all the four points $U, X, Y, V$ lie on a line which is the common tangent to circles $\\omega_{b}, \\omega_{c}$. Since it leaves these two circles on one side and the circle $\\omega_{a}$ on the other, this line is just the line $p_{a}$ from the problem statement.\n\nLine $p_{a}$ runs midway between $I$ and $M_{b} M_{c}$. Analogous conclusions hold for the lines $p_{b}$ and $p_{c}$. So these three lines form a triangle homothetic from centre $I$ to triangle $M_{a} M_{b} M_{c}$ in ratio $1 / 2$, hence similar to $A B C$ in ratio $1 / 4$.'] [] Text-only Competition True Theorem proof Geometry Math English 111 "Let $A B C D$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that $$ \angle P A B+\angle P D C \leq 90^{\circ} \quad \text { and } \quad \angle P B A+\angle P C D \leq 90^{\circ} \text {. } $$ Prove that $A B+C D \geq B C+A D$." ['We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $A B C D$. Then:\n\n\n\n$$\n\\text{Circles $(B C T)$ and $(D A T)$ are tangent at $T$}\n\\newline\n\\text { if and only if } \\angle A D T+\\angle B C T=\\angle A T B \\text {. } \\tag{1}\n$$\n\nIndeed, if the two circles touch each other then their common tangent at $T$ intersects the segment $A B$ at a point $Z$, and so $\\angle A D T=\\angle A T Z, \\angle B C T=\\angle B T Z$, by the tangent-chord theorem. Thus $\\angle A D T+\\angle B C T=\\angle A T Z+\\angle B T Z=\\angle A T B$.\n\nAnd conversely, if $\\angle A D T+\\angle B C T=\\angle A T B$ then one can draw from $T$ a ray $T Z$ with $Z$ on $A B$ so that $\\angle A D T=\\angle A T Z, \\angle B C T=\\angle B T Z$. The first of these equalities implies that $T Z$ is tangent to the circle $(D A T)$; by the second equality, $T Z$ is tangent to the circle $(B C T)$, so the two circles are tangent at $T$.\n\n\n\nSo the equivalence (1) is settled. It will be used later on. Now pass to the actual solution. Its key idea is to introduce the circumcircles of triangles $A B P$ and $C D P$ and to consider their second intersection $Q$ (assume for the moment that they indeed meet at two distinct points $P$ and $Q$ ).\n\nSince the point $A$ lies outside the circle $(B C P)$, we have $\\angle B C P+\\angle B A P<180^{\\circ}$. Therefore the point $C$ lies outside the circle $(A B P)$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same $\\operatorname{arc} C D$ of the circle $(B C P)$.\n\n\n\n\n\nBy symmetry, $P$ and $Q$ lie on the same arc $A B$ of the circle $(A B P)$. Thus the point $Q$ lies either inside the angle $B P C$ or inside the angle $A P D$. Without loss of generality assume that $Q$ lies inside the angle $B P C$. Then\n\n$$\n\\angle A Q D=\\angle P Q A+\\angle P Q D=\\angle P B A+\\angle P C D \\leq 90^{\\circ}\\tag{2}\n$$\n\nby the condition of the problem.\n\nIn the cyclic quadrilaterals $A P Q B$ and $D P Q C$, the angles at vertices $A$ and $D$ are acute. So their angles at $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $B P C$ but in fact inside the triangle $B P C$, hence also inside the quadrilateral $A B C D$.\n\nNow an argument similar to that used in deriving (2) shows that\n\n$$\n\\angle B Q C=\\angle P A B+\\angle P D C \\leq 90^{\\circ} .\\tag{3}\n$$\n\nMoreover, since $\\angle P C Q=\\angle P D Q$, we get\n\n$$\n\\angle A D Q+\\angle B C Q=\\angle A D P+\\angle P D Q+\\angle B C P-\\angle P C Q=\\angle A D P+\\angle B C P .\n$$\n\nThe last sum is equal to $\\angle A P B$, according to the observation (1) applied to $T=P$. And because $\\angle A P B=\\angle A Q B$, we obtain\n\n$$\n\\angle A D Q+\\angle B C Q=\\angle A Q B\n$$\n\nApplying now (1) to $T=Q$ we conclude that the circles $(B C Q)$ and $(D A Q)$ are externally tangent at $Q$. (We have assumed $P \\neq Q$; but if $P=Q$ then the last conclusion holds trivially.)\n\nFinally consider the halfdiscs with diameters $B C$ and $D A$ constructed inwardly to the quadrilateral $A B C D$. They have centres at $M$ and $N$, the midpoints of $B C$ and $D A$ respectively. In view of (2) and (3), these two halfdiscs lie entirely inside the circles $(B Q C)$ and $(A Q D)$; and since these circles are tangent, the two halfdiscs cannot overlap. Hence $M N \\geq \\frac{1}{2} B C+\\frac{1}{2} D A$.\n\nOn the other hand, since $\\overrightarrow{M N}=\\frac{1}{2}(\\overrightarrow{B A}+\\overrightarrow{C D})$, we have $M N \\leq \\frac{1}{2}(A B+C D)$. Thus indeed $A B+C D \\geq B C+D A$, as claimed.'] [] Text-only Competition True Theorem proof Geometry Math English 112 Points $A_{1}, B_{1}, C_{1}$ are chosen on the sides $B C, C A, A B$ of a triangle $A B C$, respectively. The circumcircles of triangles $A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$ intersect the circumcircle of triangle $A B C$ again at points $A_{2}, B_{2}, C_{2}$, respectively $\left(A_{2} \neq A, B_{2} \neq B, C_{2} \neq C\right)$. Points $A_{3}, B_{3}, C_{3}$ are symmetric to $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of the sides $B C, C A, A B$ respectively. Prove that the triangles $A_{2} B_{2} C_{2}$ and $A_{3} B_{3} C_{3}$ are similar. ['We will work with oriented angles between lines. For two straight lines $\\ell, m$ in the plane, $\\angle(\\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\\pi$ are identified, so that\n\n$$\n\\angle(\\ell, m)=-\\angle(m, \\ell), \\quad \\angle(\\ell, m)+\\angle(m, n)=\\angle(\\ell, n)\n$$\n\nIf $\\ell$ is the line through points $K, L$ and $m$ is the line through $M, N$, one writes $\\angle(K L, M N)$ for $\\angle(\\ell, m)$; the characters $K, L$ are freely interchangeable; and so are $M, N$.\n\nThe counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K, L, M, N$ are four noncollinear points in the plane then\n\n$$\nK, L, M, N \\text { are concyclic if and only if } \\angle(K M, L M)=\\angle(K N, L N) \\text {. }\\tag{1}\n$$\n\nPassing to the solution proper, we first show that the three circles $\\left(A B_{1} C_{1}\\right),\\left(B C_{1} A_{1}\\right)$, $\\left(C A_{1} B_{1}\\right)$ have a common point. So, let $\\left(A B_{1} C_{1}\\right)$ and $\\left(B C_{1} A_{1}\\right)$ intersect at the points $C_{1}$ and $P$. Then by $(1)$\n\n$$\n\\begin{gathered}\n\\angle\\left(P A_{1}, C A_{1}\\right)=\\angle\\left(P A_{1}, B A_{1}\\right)=\\angle\\left(P C_{1}, B C_{1}\\right) \\\\\n=\\angle\\left(P C_{1}, A C_{1}\\right)=\\angle\\left(P B_{1}, A B_{1}\\right)=\\angle\\left(P B_{1}, C B_{1}\\right) .\n\\end{gathered}\n$$\n\nDenote this angle by $\\varphi$.\n\nThe equality between the outer terms shows, again by (1), that the points $A_{1}, B_{1}, P, C$ are concyclic. Thus $P$ is the common point of the three mentioned circles.\n\nFrom now on the basic property (1) will be used without explicit reference. We have\n\n$$\n\\varphi=\\angle\\left(P A_{1}, B C\\right)=\\angle\\left(P B_{1}, C A\\right)=\\angle\\left(P C_{1}, A B\\right)\\tag{2}\n$$\n\n\n\n\n\nLet lines $A_{2} P, B_{2} P, C_{2} P$ meet the circle $(A B C)$ again at $A_{4}, B_{4}, C_{4}$, respectively. As\n\n$$\n\\angle\\left(A_{4} A_{2}, A A_{2}\\right)=\\angle\\left(P A_{2}, A A_{2}\\right)=\\angle\\left(P C_{1}, A C_{1}\\right)=\\angle\\left(P C_{1}, A B\\right)=\\varphi\n$$\n\nwe see that line $A_{2} A$ is the image of line $A_{2} A_{4}$ under rotation about $A_{2}$ by the angle $\\varphi$. Hence the point $A$ is the image of $A_{4}$ under rotation by $2 \\varphi$ about $O$, the centre of $(A B C)$. The same rotation sends $B_{4}$ to $B$ and $C_{4}$ to $C$. Triangle $A B C$ is the image of $A_{4} B_{4} C_{4}$ in this map. Thus\n\n$$\n\\angle\\left(A_{4} B_{4}, A B\\right)=\\angle\\left(B_{4} C_{4}, B C\\right)=\\angle\\left(C_{4} A_{4}, C A\\right)=2 \\varphi\n$$\n\nSince the rotation by $2 \\varphi$ about $O$ takes $B_{4}$ to $B$, we have $\\angle\\left(A B_{4}, A B\\right)=\\varphi$. Hence by $(2)$\n\n$$\n\\angle\\left(A B_{4}, P C_{1}\\right)=\\angle\\left(A B_{4}, A B\\right)+\\angle\\left(A B, P C_{1}\\right)=\\varphi+(-\\varphi)=0\n$$\n\nwhich means that $A B_{4} \\| P C_{1}$.\n\n\nLet $C_{5}$ be the intersection of lines $P C_{1}$ and $A_{4} B_{4}$; define $A_{5}, B_{5}$ analogously. So $A B_{4} \\| C_{1} C_{5}$ and, by (3) and (2),\n\n$$\n\\angle\\left(A_{4} B_{4}, P C_{1}\\right)=\\angle\\left(A_{4} B_{4}, A B\\right)+\\angle\\left(A B, P C_{1}\\right)=2 \\varphi+(-\\varphi)=\\varphi ;\n$$\n\ni.e., $\\angle\\left(B_{4} C_{5}, C_{5} C_{1}\\right)=\\varphi$. This combined with $\\angle\\left(C_{5} C_{1}, C_{1} A\\right)=\\angle\\left(P C_{1}, A B\\right)=\\varphi$ (see (2)) proves that the quadrilateral $A B_{4} C_{5} C_{1}$ is an isosceles trapezoid with $A C_{1}=B_{4} C_{5}$.\n\nInterchanging the roles of $A$ and $B$ we infer that also $B C_{1}=A_{4} C_{5}$. And since $A C_{1}+B C_{1}=$ $A B=A_{4} B_{4}$, it follows that the point $C_{5}$ lies on the line segment $A_{4} B_{4}$ and partitions it into segments $A_{4} C_{5}, B_{4} C_{5}$ of lengths $B C_{1}\\left(=A C_{3}\\right)$ and $A C_{1}$ (=BC $C_{3}$ ). In other words, the rotation which maps triangle $A_{4} B_{4} C_{4}$ onto $A B C$ carries $C_{5}$ onto $C_{3}$. Likewise, it sends $A_{5}$ to $A_{3}$ and $B_{5}$ to $B_{3}$. So the triangles $A_{3} B_{3} C_{3}$ and $A_{5} B_{5} C_{5}$ are congruent. It now suffices to show that the latter is similar to $A_{2} B_{2} C_{2}$.\n\nLines $B_{4} C_{5}$ and $P C_{5}$ coincide respectively with $A_{4} B_{4}$ and $P C_{1}$. Thus by (4)\n\n$$\n\\angle\\left(B_{4} C_{5}, P C_{5}\\right)=\\varphi\n$$\n\nAnalogously (by cyclic shift) $\\varphi=\\angle\\left(C_{4} A_{5}, P A_{5}\\right)$, which rewrites as\n\n$$\n\\varphi=\\angle\\left(B_{4} A_{5}, P A_{5}\\right)\n$$\n\n\n\nThese relations imply that the points $P, B_{4}, C_{5}, A_{5}$ are concyclic. Analogously, $P, C_{4}, A_{5}, B_{5}$ and $P, A_{4}, B_{5}, C_{5}$ are concyclic quadruples. Therefore\n\n$$\n\\angle\\left(A_{5} B_{5}, C_{5} B_{5}\\right)=\\angle\\left(A_{5} B_{5}, P B_{5}\\right)+\\angle\\left(P B_{5}, C_{5} B_{5}\\right)=\\angle\\left(A_{5} C_{4}, P C_{4}\\right)+\\angle\\left(P A_{4}, C_{5} A_{4}\\right) .\n$$\n\nOn the other hand, since the points $A_{2}, B_{2}, C_{2}, A_{4}, B_{4}, C_{4}$ all lie on the circle $(A B C)$, we have\n\n$$\n\\angle\\left(A_{2} B_{2}, C_{2} B_{2}\\right)=\\angle\\left(A_{2} B_{2}, B_{4} B_{2}\\right)+\\angle\\left(B_{4} B_{2}, C_{2} B_{2}\\right)=\\angle\\left(A_{2} A_{4}, B_{4} A_{4}\\right)+\\angle\\left(B_{4} C_{4}, C_{2} C_{4}\\right)\n$$\n\nBut the lines $A_{2} A_{4}, B_{4} A_{4}, B_{4} C_{4}, C_{2} C_{4}$ coincide respectively with $P A_{4}, C_{5} A_{4}, A_{5} C_{4}, P C_{4}$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\\angle\\left(A_{5} B_{5}, C_{5} B_{5}\\right)=\\angle\\left(A_{2} B_{2}, C_{2} B_{2}\\right)$. Hence (by cyclic shift, once more) also $\\angle\\left(B_{5} C_{5}, A_{5} C_{5}\\right)=\\angle\\left(B_{2} C_{2}, A_{2} C_{2}\\right)$ and $\\angle\\left(C_{5} A_{5}, B_{5} A_{5}\\right)=\\angle\\left(C_{2} A_{2}, B_{2} A_{2}\\right)$. This means that the triangles $A_{5} B_{5} C_{5}$ and $A_{2} B_{2} C_{2}$ have their corresponding angles equal, and consequently they are similar.'] [] Text-only Competition True Theorem proof Geometry Math English 113 To each side $a$ of a convex polygon we assign the maximum area of a triangle contained in the polygon and having $a$ as one of its sides. Show that the sum of the areas assigned to all sides of the polygon is not less than twice the area of the polygon. "['Lemma. Every convex $(2 n)$-gon, of area $S$, has a side and a vertex that jointly span a triangle of area not less than $S / n$.\n\nProof. By main diagonals of the (2n)-gon we shall mean those which partition the (2n)-gon into two polygons with equally many sides. For any side $b$ of the $(2 n)$-gon denote by $\\Delta_{b}$ the triangle $A B P$ where $A, B$ are the endpoints of $b$ and $P$ is the intersection point of the main diagonals $A A^{\\prime}, B B^{\\prime}$. We claim that the union of triangles $\\Delta_{b}$, taken over all sides, covers the whole polygon.\n\nTo show this, choose any side $A B$ and consider the main diagonal $A A^{\\prime}$ as a directed segment. Let $X$ be any point in the polygon, not on any main diagonal. For definiteness, let $X$ lie on the left side of the ray $A A^{\\prime}$. Consider the sequence of main diagonals $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}, \\ldots$, where $A, B, C, \\ldots$ are consecutive vertices, situated right to $A A^{\\prime}$.\n\nThe $n$-th item in this sequence is the diagonal $A^{\\prime} A$ (i.e. $A A^{\\prime}$ reversed), having $X$ on its right side. So there are two successive vertices $K, L$ in the sequence $A, B, C, \\ldots$ before $A^{\\prime}$ such that $X$ still lies to the left of $K K^{\\prime}$ but to the right of $L L^{\\prime}$. And this means that $X$ is in the triangle $\\Delta_{\\ell^{\\prime}}, \\ell^{\\prime}=K^{\\prime} L^{\\prime}$. Analogous reasoning applies to points $X$ on the right of $A A^{\\prime}$ (points lying on main diagonals can be safely ignored). Thus indeed the triangles $\\Delta_{b}$ jointly cover the whole polygon.\n\nThe sum of their areas is no less than $S$. So we can find two opposite sides, say $b=A B$ and $b^{\\prime}=A^{\\prime} B^{\\prime}$ (with $A A^{\\prime}, B B^{\\prime}$ main diagonals) such that $\\left[\\Delta_{b}\\right]+\\left[\\Delta_{b^{\\prime}}\\right] \\geq S / n$, where $[\\cdots]$ stands for the area of a region. Let $A A^{\\prime}, B B^{\\prime}$ intersect at $P$; assume without loss of generality that $P B \\geq P B^{\\prime}$. Then\n\n$$\n\\left[A B A^{\\prime}\\right]=[A B P]+\\left[P B A^{\\prime}\\right] \\geq[A B P]+\\left[P A^{\\prime} B^{\\prime}\\right]=\\left[\\Delta_{b}\\right]+\\left[\\Delta_{b^{\\prime}}\\right] \\geq S / n\n$$\n\nproving the lemma.\n\nNow, let $\\mathcal{P}$ be any convex polygon, of area $S$, with $m$ sides $a_{1}, \\ldots, a_{m}$. Let $S_{i}$ be the area of the greatest triangle in $\\mathcal{P}$ with side $a_{i}$. Suppose, contrary to the assertion, that\n\n$$\n\\sum_{i=1}^{m} \\frac{S_{i}}{S}<2\n$$\n\nThen there exist rational numbers $q_{1}, \\ldots, q_{m}$ such that $\\sum q_{i}=2$ and $q_{i}>S_{i} / S$ for each $i$.\n\nLet $n$ be a common denominator of the $m$ fractions $q_{1}, \\ldots, q_{m}$. Write $q_{i}=k_{i} / n$; so $\\sum k_{i}=2 n$. Partition each side $a_{i}$ of $\\mathcal{P}$ into $k_{i}$ equal segments, creating a convex (2n)-gon of area $S$ (with some angles of size $180^{\\circ}$ ), to which we apply the lemma. Accordingly, this refined polygon has a side $b$ and a vertex $H$ spanning a triangle $T$ of area $[T] \\geq S / n$. If $b$ is a piece of a side $a_{i}$ of $\\mathcal{P}$, then the triangle $W$ with base $a_{i}$ and summit $H$ has area\n\n$$\n[W]=k_{i} \\cdot[T] \\geq k_{i} \\cdot S / n=q_{i} \\cdot S>S_{i}\n$$\n\nin contradiction with the definition of $S_{i}$. This ends the proof.' 'As in the first solution, we allow again angles of size $180^{\\circ}$ at some vertices of the convex polygons considered.\n\nTo each convex $n$-gon $\\mathcal{P}=A_{1} A_{2} \\ldots A_{n}$ we assign a centrally symmetric convex $(2 n)$-gon $\\mathcal{Q}$ with side vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}, 1 \\leq i \\leq n$. The construction is as follows. Attach the $2 n$ vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}$ at a common origin and label them $\\overrightarrow{\\mathbf{b}_{1}}, \\overrightarrow{\\mathbf{b}_{2}}, \\ldots, \\overrightarrow{\\mathbf{b}_{2 n}}$ in counterclockwise direction; the choice of the first vector $\\overrightarrow{\\mathbf{b}_{1}}$ is irrelevant. The order of labelling is well-defined if $\\mathcal{P}$ has neither parallel sides nor angles equal to $180^{\\circ}$. Otherwise several collinear vectors with the same direction are labelled consecutively $\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, \\overrightarrow{\\mathbf{b}_{j+r}}$. One can assume that in such cases the respective opposite vectors occur in the order $-\\overrightarrow{\\mathbf{b}_{j}},-\\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots,-\\overrightarrow{\\mathbf{b}_{j+r}}$, ensuring that $\\overrightarrow{\\mathbf{b}_{j+n}}=-\\overrightarrow{\\mathbf{b}_{j}}$ for $j=1, \\ldots, 2 n$. Indices are taken cyclically here and in similar situations below.\n\nChoose points $B_{1}, B_{2}, \\ldots, B_{2 n}$ satisfying $\\overrightarrow{B_{j} B_{j+1}}=\\overrightarrow{\\mathbf{b}_{j}}$ for $j=1, \\ldots, 2 n$. The polygonal line $\\mathcal{Q}=B_{1} B_{2} \\ldots B_{2 n}$ is closed, since $\\sum_{j=1}^{2 n} \\overrightarrow{\\mathbf{b}_{j}}=\\overrightarrow{0}$. Moreover, $\\mathcal{Q}$ is a convex $(2 n)$-gon due to the arrangement of the vectors $\\overrightarrow{\\mathbf{b}_{j}}$, possibly with $180^{\\circ}$-angles. The side vectors of $\\mathcal{Q}$ are $\\pm \\overrightarrow{A_{i} A_{i+1}}$, $1 \\leq i \\leq n$. So in particular $\\mathcal{Q}$ is centrally symmetric, because it contains as side vectors $\\overrightarrow{A_{i} A_{i+1}}$ and $-\\overrightarrow{A_{i} A_{i+1}}$ for each $i=1, \\ldots, n$. Note that $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are opposite sides of $\\mathcal{Q}$, $1 \\leq j \\leq n$. We call $\\mathcal{Q}$ the associate of $\\mathcal{P}$.\n\nLet $S_{i}$ be the maximum area of a triangle with side $A_{i} A_{i+1}$ in $\\mathcal{P}, 1 \\leq i \\leq n$. We prove that\n\n$$\n\\left[B_{1} B_{2} \\ldots B_{2 n}\\right]=2 \\sum_{i=1}^{n} S_{i} \\tag{1}\n$$\n\nand\n\n$$\n\\left[B_{1} B_{2} \\ldots B_{2 n}\\right] \\geq 4\\left[A_{1} A_{2} \\ldots A_{n}\\right] \\tag{2}\n$$\n\nIt is clear that (1) and (2) imply the conclusion of the original problem.\n\nLemma. For a side $A_{i} A_{i+1}$ of $\\mathcal{P}$, let $h_{i}$ be the maximum distance from a point of $\\mathcal{P}$ to line $A_{i} A_{i+1}$, $i=1, \\ldots, n$. Denote by $B_{j} B_{j+1}$ the side of $\\mathcal{Q}$ such that $\\overrightarrow{A_{i} A_{i+1}}=\\overrightarrow{B_{j} B_{j+1}}$. Then the distance between $B_{j} B_{j+1}$ and its opposite side in $\\mathcal{Q}$ is equal to $2 h_{i}$.\n\nProof. Choose a vertex $A_{k}$ of $\\mathcal{P}$ at distance $h_{i}$ from line $A_{i} A_{i+1}$. Let $\\mathbf{u}$ be the unit vector perpendicular to $A_{i} A_{i+1}$ and pointing inside $\\mathcal{P}$. Denoting by $\\mathbf{x} \\cdot \\mathbf{y}$ the dot product of vectors $\\mathbf{x}$ and $\\mathbf{y}$, we have\n\n$$\nh=\\mathbf{u} \\cdot \\overrightarrow{A_{i} A_{k}}=\\mathbf{u} \\cdot\\left(\\overrightarrow{A_{i} A_{i+1}}+\\cdots+\\overrightarrow{A_{k-1} A_{k}}\\right)=\\mathbf{u} \\cdot\\left(\\overrightarrow{A_{i} A_{i-1}}+\\cdots+\\overrightarrow{A_{k+1} A_{k}}\\right)\n$$\n\nIn $\\mathcal{Q}$, the distance $H_{i}$ between the opposite sides $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ is given by\n\n$$\nH_{i}=\\mathbf{u} \\cdot\\left(\\overrightarrow{B_{j} B_{j+1}}+\\cdots+\\overrightarrow{B_{j+n-1} B_{j+n}}\\right)=\\mathbf{u} \\cdot\\left(\\overrightarrow{\\mathbf{b}_{j}}+\\overrightarrow{\\mathbf{b}_{j+1}}+\\cdots+\\overrightarrow{\\mathbf{b}_{j+n-1}}\\right) .\n$$\n\nThe choice of vertex $A_{k}$ implies that the $n$ consecutive vectors $\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, \\overrightarrow{\\mathbf{b}_{j+n-1}}$ are precisely $\\overrightarrow{A_{i} A_{i+1}}, \\ldots, \\overrightarrow{A_{k-1} A_{k}}$ and $\\overrightarrow{A_{i} A_{i-1}}, \\ldots, \\overrightarrow{A_{k+1} A_{k}}$, taken in some order. This implies $H_{i}=2 h_{i}$.\n\nFor a proof of (1), apply the lemma to each side of $\\mathcal{P}$. If $O$ the centre of $\\mathcal{Q}$ then, using the notation of the lemma,\n\n$$\n\\left[B_{j} B_{j+1} O\\right]=\\left[B_{j+n} B_{j+n+1} O\\right]=\\left[A_{i} A_{i+1} A_{k}\\right]=S_{i}\n$$\n\nSummation over all sides of $\\mathcal{P}$ yields (1).\n\nSet $d(\\mathcal{P})=[\\mathcal{Q}]-4[\\mathcal{P}]$ for a convex polygon $\\mathcal{P}$ with associate $\\mathcal{Q}$. Inequality $(2)$ means that $d(\\mathcal{P}) \\geq 0$ for each convex polygon $\\mathcal{P}$. The last inequality will be proved by induction on the\n\n\n\nnumber $\\ell$ of side directions of $\\mathcal{P}$, i. e. the number of pairwise nonparallel lines each containing a side of $\\mathcal{P}$.\n\nWe choose to start the induction with $\\ell=1$ as a base case, meaning that certain degenerate polygons are allowed. More exactly, we regard as degenerate convex polygons all closed polygonal lines of the form $X_{1} X_{2} \\ldots X_{k} Y_{1} Y_{2} \\ldots Y_{m} X_{1}$, where $X_{1}, X_{2}, \\ldots, X_{k}$ are points in this order on a line segment $X_{1} Y_{1}$, and so are $Y_{m}, Y_{m-1}, \\ldots, Y_{1}$. The initial construction applies to degenerate polygons; their associates are also degenerate, and the value of $d$ is zero. For the inductive step, consider a convex polygon $\\mathcal{P}$ which determines $\\ell$ side directions, assuming that $d(\\mathcal{P}) \\geq 0$ for polygons with smaller values of $\\ell$.\n\nSuppose first that $\\mathcal{P}$ has a pair of parallel sides, i. e. sides on distinct parallel lines. Let $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$ be such a pair, and let $A_{i} A_{i+1} \\leq A_{j} A_{j+1}$. Remove from $\\mathcal{P}$ the parallelogram $R$ determined by vectors $\\overrightarrow{A_{i} A_{i+1}}$ and $\\overrightarrow{A_{i} A_{j+1}}$. Two polygons are obtained in this way. Translating one of them by vector $\\overrightarrow{A_{i} A_{i+1}}$ yields a new convex polygon $\\mathcal{P}^{\\prime}$, of area $[\\mathcal{P}]-[R]$ and with value of $\\ell$ not exceeding the one of $\\mathcal{P}$. The construction just described will be called operation A.\n\n\nThe associate of $\\mathcal{P}^{\\prime}$ is obtained from $\\mathcal{Q}$ upon decreasing the lengths of two opposite sides by an amount of $2 A_{i} A_{i+1}$. By the lemma, the distance between these opposite sides is twice the distance between $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$. Thus operation $\\mathbf{A}$ decreases $[\\mathcal{Q}]$ by the area of a parallelogram with base and respective altitude twice the ones of $R$, i. e. by $4[R]$. Hence $\\mathbf{A}$ leaves the difference $d(\\mathcal{P})=[\\mathcal{Q}]-4[\\mathcal{P}]$ unchanged.\n\nNow, if $\\mathcal{P}^{\\prime}$ also has a pair of parallel sides, apply operation $\\mathbf{A}$ to it. Keep doing so with the subsequent polygons obtained for as long as possible. Now, A decreases the number $p$ of pairs of parallel sides in $\\mathcal{P}$. Hence its repeated applications gradually reduce $p$ to 0 , and further applications of $\\mathbf{A}$ will be impossible after several steps. For clarity, let us denote by $\\mathcal{P}$ again the polygon obtained at that stage.\n\nThe inductive step is complete if $\\mathcal{P}$ is degenerate. Otherwise $\\ell>1$ and $p=0$, i. e. there are no parallel sides in $\\mathcal{P}$. Observe that then $\\ell \\geq 3$. Indeed, $\\ell=2$ means that the vertices of $\\mathcal{P}$ all lie on the boundary of a parallelogram, implying $p>0$.\n\nFurthermore, since $\\mathcal{P}$ has no parallel sides, consecutive collinear vectors in the sequence $\\left(\\overrightarrow{\\mathbf{b}_{k}}\\right)$ (if any) correspond to consecutive $180^{\\circ}$-angles in $\\mathcal{P}$. Removing the vertices of such angles, we obtain a convex polygon with the same value of $d(\\mathcal{P})$.\n\nIn summary, if operation $\\mathbf{A}$ is impossible for a nondegenerate polygon $\\mathcal{P}$, then $\\ell \\geq 3$. In addition, one may assume that $\\mathcal{P}$ has no angles of size $180^{\\circ}$.\n\nThe last two conditions then also hold for the associate $\\mathcal{Q}$ of $\\mathcal{P}$, and we perform the following construction. Since $\\ell \\geq 3$, there is a side $B_{j} B_{j+1}$ of $\\mathcal{Q}$ such that the sum of the angles at $B_{j}$ and $B_{j+1}$ is greater than $180^{\\circ}$. (Such a side exists in each convex $k$-gon for $k>4$.) Naturally, $B_{j+n} B_{j+n+1}$ is a side with the same property. Extend the pairs of sides $B_{j-1} B_{j}, B_{j+1} B_{j+2}$\n\n\n\nand $B_{j+n-1} B_{j+n}, B_{j+n+1} B_{j+n+2}$ to meet at $U$ and $V$, respectively. Let $\\mathcal{Q}^{\\prime}$ be the centrally symmetric convex 2(n+1)-gon obtained from $\\mathcal{Q}$ by inserting $U$ and $V$ into the sequence $B_{1}, \\ldots, B_{2 n}$ as new vertices between $B_{j}, B_{j+1}$ and $B_{j+n}, B_{j+n+1}$, respectively. Informally, we adjoin to $\\mathcal{Q}$ the congruent triangles $B_{j} B_{j+1} U$ and $B_{j+n} B_{j+n+1} V$. Note that $B_{j}, B_{j+1}, B_{j+n}$ and $B_{j+n+1}$ are kept as vertices of $\\mathcal{Q}^{\\prime}$, although $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are no longer its sides.\n\nLet $A_{i} A_{i+1}$ be the side of $\\mathcal{P}$ such that $\\overrightarrow{A_{i} A_{i+1}}=\\overrightarrow{B_{j} B_{j+1}}=\\overrightarrow{\\mathbf{b}_{j}}$. Consider the point $W$ such that triangle $A_{i} A_{i+1} W$ is congruent to triangle $B_{j} B_{j+1} U$ and exterior to $\\mathcal{P}$. Insert $W$ into the sequence $A_{1}, A_{2}, \\ldots, A_{n}$ as a new vertex between $A_{i}$ and $A_{i+1}$ to obtain an $(n+1)$-gon $\\mathcal{P}^{\\prime}$. We claim that $\\mathcal{P}^{\\prime}$ is convex and its associate is $\\mathcal{Q}^{\\prime}$.\n\n\nVectors $\\overrightarrow{A_{i} W}$ and $\\overrightarrow{\\mathbf{b}_{j-1}}$ are collinear and have the same direction, as well as vectors $\\overrightarrow{W A_{i+1}}$ and $\\overrightarrow{\\mathbf{b}_{j+1}}$. Since $\\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}$ are consecutive terms in the sequence $\\left(\\overrightarrow{\\mathbf{b}_{k}}\\right)$, the angle inequalities $\\angle\\left(\\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{\\mathbf{b}_{j}}\\right) \\leq \\angle\\left(\\overrightarrow{A_{i-1} A_{i}}, \\overrightarrow{\\mathbf{b}_{j}}\\right)$ and $\\angle\\left(\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}\\right) \\leq \\angle\\left(\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{A_{i+1} A_{i+2}}\\right)$ hold true. They show that $\\mathcal{P}^{\\prime}$ is a convex polygon. To construct its associate, vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}= \\pm \\overrightarrow{\\mathbf{b}_{j}}$ must be deleted from the defining sequence $\\left(\\overrightarrow{\\mathbf{b}_{k}}\\right)$ of $\\mathcal{Q}$, and the vectors $\\pm \\overrightarrow{A_{i} W}, \\pm \\overrightarrow{W A_{i+1}}$ must be inserted appropriately into it. The latter can be done as follows:\n\n$$\n\\ldots, \\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{A_{i} W}, \\overrightarrow{W A_{i+1}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots,-\\overrightarrow{\\mathbf{b}_{j-1}},-\\overrightarrow{A_{i} W},-\\overrightarrow{W A_{i+1}},-\\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots\n$$\n\nThis updated sequence produces $\\mathcal{Q}^{\\prime}$ as the associate of $\\mathcal{P}^{\\prime}$.\n\nIt follows from the construction that $\\left[\\mathcal{P}^{\\prime}\\right]=[\\mathcal{P}]+\\left[A_{i} A_{i+1} W\\right]$ and $\\left[\\mathcal{Q}^{\\prime}\\right]=[\\mathcal{Q}]+2\\left[A_{i} A_{i+1} W\\right]$. Therefore $d\\left(\\mathcal{P}^{\\prime}\\right)=d(\\mathcal{P})-2\\left[A_{i} A_{i+1} W\\right]f(n)$ infinitely often. (b) Prove that $f(n+1)f(n)$ and $d(n+1)1, d(n) \\geq 2$ holds, with equality if and only if $n$ is prime. Since $f(6)=7 / 3>2$, it follows that $f(n)>2$ holds for all $n \\geq 6$.\n\nSince there are infinitely many primes, $d(n+1)=2$ holds for infinitely many values of $n$, and for each such $n \\geq 6$ we have $d(n+1)=2\\max \\{d(1), d(2), \\ldots, d(n)\\}$ for infinitely many $n$. For all such $n$, we have $d(n+1)>f(n)$. This completes the solution.'] [] Text-only Competition True Theorem proof Number Theory Math English 116 Let $P$ be a polynomial of degree $n>1$ with integer coefficients and let $k$ be any positive integer. Consider the polynomial $Q(x)=P(P(\ldots P(P(x)) \ldots))$, with $k$ pairs of parentheses. Prove that $Q$ has no more than $n$ integer fixed points, i.e. integers satisfying the equation $Q(x)=x$. ['The claim is obvious if every integer fixed point of $Q$ is a fixed point of $P$ itself. For the sequel assume that this is not the case. Take any integer $x_{0}$ such that $Q\\left(x_{0}\\right)=x_{0}$, $P\\left(x_{0}\\right) \\neq x_{0}$ and define inductively $x_{i+1}=P\\left(x_{i}\\right)$ for $i=0,1,2, \\ldots ;$ then $x_{k}=x_{0}$.\n\nIt is evident that\n\n$$\nP(u)-P(v) \\text { is divisible by } u-v \\text { for distinct integers } u, v \\text {. }\\tag{1}\n$$\n\n(Indeed, if $P(x)=\\sum a_{i} x^{i}$ then each $a_{i}\\left(u^{i}-v^{i}\\right)$ is divisible by $u-v$.) Therefore each term in the chain of (nonzero) differences\n\n$$\nx_{0}-x_{1}, \\quad x_{1}-x_{2}, \\quad \\ldots, \\quad x_{k-1}-x_{k}, \\quad x_{k}-x_{k+1}\\tag{2}\n$$\n\nis a divisor of the next one; and since $x_{k}-x_{k+1}=x_{0}-x_{1}$, all these differences have equal absolute values. For $x_{m}=\\min \\left(x_{1}, \\ldots, x_{k}\\right)$ this means that $x_{m-1}-x_{m}=-\\left(x_{m}-x_{m+1}\\right)$. Thus $x_{m-1}=x_{m+1}\\left(\\neq x_{m}\\right)$. It follows that consecutive differences in the sequence (2) have opposite signs. Consequently, $x_{0}, x_{1}, x_{2}, \\ldots$ is an alternating sequence of two distinct values. In other words, every integer fixed point of $Q$ is a fixed point of the polynomial $P(P(x))$. Our task is to prove that there are at most $n$ such points.\n\nLet $a$ be one of them so that $b=P(a) \\neq a$ (we have assumed that such an $a$ exists); then $a=P(b)$. Take any other integer fixed point $\\alpha$ of $P(P(x))$ and let $P(\\alpha)=\\beta$, so that $P(\\beta)=\\alpha$; the numbers $\\alpha$ and $\\beta$ need not be distinct ( $\\alpha$ can be a fixed point of $P$ ), but each of $\\alpha, \\beta$ is different from each of $a, b$. Applying property (1) to the four pairs of integers $(\\alpha, a),(\\beta, b)$, $(\\alpha, b),(\\beta, a)$ we get that the numbers $\\alpha-a$ and $\\beta-b$ divide each other, and also $\\alpha-b$ and $\\beta-a$ divide each other. Consequently\n\n$$\n\\alpha-b= \\pm(\\beta-a), \\quad \\alpha-a= \\pm(\\beta-b)\\tag{3}\n$$\n\nSuppose we have a plus in both instances: $\\alpha-b=\\beta-a$ and $\\alpha-a=\\beta-b$. Subtraction yields $a-b=b-a$, a contradiction, as $a \\neq b$. Therefore at least one equality in (3) holds with a minus sign. For each of them this means that $\\alpha+\\beta=a+b$; equivalently $a+b-\\alpha-P(\\alpha)=0$.\n\nDenote $a+b$ by $C$. We have shown that every integer fixed point of $Q$ other that $a$ and $b$ is a root of the polynomial $F(x)=C-x-P(x)$. This is of course true for $a$ and $b$ as well. And since $P$ has degree $n>1$, the polynomial $F$ has the same degree, so it cannot have more than $n$ roots. Hence the result.'] [] Text-only Competition True Theorem proof Number Theory Math English 117 Prove that, for every positive integer $n$, there exists an integer $m$ such that $2^{m}+m$ is divisible by $n$. ['We will prove by induction on $d$ that, for every positive integer $N$, there exist positive integers $b_{0}, b_{1}, \\ldots, b_{d-1}$ such that, for each $i=0,1,2, \\ldots, d-1$, we have $b_{i}>N$ and\n\n$$\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d)\n$$\n\nThis yields the claim for $m=b_{0}$.\n\nThe base case $d=1$ is trivial. Take an $a>1$ and assume that the statement holds for all $d\\max \\left(2^{M}, N\\right)$ and\n\n$$\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d) \\quad \\text { for } \\quad i=0,1,2, \\ldots, d-1 \\tag{1}\n$$\n\nFor each $i=0,1, \\ldots, d-1$ consider the sequence\n\n$$\n2^{b_{i}}+b_{i}, \\quad 2^{b_{i}+k}+\\left(b_{i}+k\\right), \\ldots, 2^{b_{i}+\\left(a^{\\prime}-1\\right) k}+\\left(b_{i}+\\left(a^{\\prime}-1\\right) k\\right)\\tag{2}\n$$\n\nModulo $a$, these numbers are congruent to\n\n$$\n2^{b_{i}}+b_{i}, 2^{b_{i}}+\\left(b_{i}+k\\right), \\ldots, 2^{b_{i}}+\\left(b_{i}+\\left(a^{\\prime}-1\\right) k\\right),\n$$\n\nrespectively. The $d$ sequences contain $a^{\\prime} d=a$ numbers altogether. We shall now prove that no two of these numbers are congruent modulo $a$.\n\nSuppose that\n\n$$\n2^{b_{i}}+\\left(b_{i}+m k\\right) \\equiv 2^{b_{j}}+\\left(b_{j}+n k\\right) \\quad(\\bmod a)\\tag{3}\n$$\n\nfor some values of $i, j \\in\\{0,1, \\ldots, d-1\\}$ and $m, n \\in\\left\\{0,1, \\ldots, a^{\\prime}-1\\right\\}$. Since $d$ is a divisor of $a$, we also have\n\n$$\n2^{b_{i}}+\\left(b_{i}+m k\\right) \\equiv 2^{b_{j}}+\\left(b_{j}+n k\\right) \\quad(\\bmod d) .\n$$\n\nBecause $d$ is a divisor of $k$ and in view of $(1)$, we obtain $i \\equiv j(\\bmod d)$. As $i, j \\in\\{0,1, \\ldots, d-1\\}$, this just means that $i=j$. Substituting this into (3) yields $m k \\equiv n k(\\bmod a)$. Therefore $m k^{\\prime} \\equiv n k^{\\prime}\\left(\\bmod a^{\\prime}\\right)$; and since $a^{\\prime}$ and $k^{\\prime}$ are coprime, we get $m \\equiv n\\left(\\bmod a^{\\prime}\\right)$. Hence also $m=n$\n\nIt follows that the $a$ numbers that make up the $d$ sequences (2) satisfy all the requirements; they are certainly all greater than $N$ because we chose each $b_{i}>\\max \\left(2^{M}, N\\right)$. So the statement holds for $a$, completing the induction.'] [] Text-only Competition True Theorem proof Number Theory Math English 118 Let $n$ be an integer, and let $A$ be a subset of $\left\{0,1,2,3, \ldots, 5^{n}\right\}$ consisting of $4 n+2$ numbers. Prove that there exist $a, b, c \in A$ such that $a3 b$. "['(By contradiction) Suppose that there exist $4 n+2$ non-negative integers $x_{0}<$ $x_{1}<\\cdots5^{n} \\cdot 1\n$$' 'Denote the maximum element of $A$ by $c$. For $k=0, \\ldots, 4 n-1$ let\n\n$$\nA_{k}=\\left\\{x \\in A:\\left(1-(2 / 3)^{k}\\right) c \\leqslant x<\\left(1-(2 / 3)^{k+1}\\right) c\\right\\} \\text {. }\n$$\n\nNote that\n\n$$\n\\left(1-(2 / 3)^{4 n}\\right) c=c-(16 / 81)^{n} c>c-(1 / 5)^{n} c \\geqslant c-1 \\text {, }\n$$\n\nwhich shows that the sets $A_{0}, A_{1}, \\ldots, A_{4 n-1}$ form a partition of $A \\backslash\\{c\\}$. Since $A \\backslash\\{c\\}$ has $4 n+1$ elements, by the pigeonhole principle some set $A_{k}$ does contain at least two elements of $A \\backslash\\{c\\}$. Denote these two elements $a$ and $b$ and assume $a3 b\n$$\n\nas desired.']" [] Text-only Competition True Theorem proof Algebra Math English 119 "Show that for all real numbers $x_{1}, \ldots, x_{n}$ the following inequality holds: $$ \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\left|x_{i}-x_{j}\right|} \leqslant \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\left|x_{i}+x_{j}\right|} $$" "['If we add $t$ to all the variables then the left-hand side remains constant and the right-hand side becomes\n\n$$\nH(t):=\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{\\left|x_{i}+x_{j}+2 t\\right|}\n$$\n\nLet $T$ be large enough such that both $H(-T)$ and $H(T)$ are larger than the value $L$ of the lefthand side of the inequality we want to prove. Not necessarily distinct points $p_{i, j}:=-\\left(x_{i}+x_{j}\\right) / 2$ together with $T$ and $-T$ split the real line into segments and two rays such that on each of these segments and rays the function $H(t)$ is concave since $f(t):=\\sqrt{|\\ell+2 t|}$ is concave on both intervals $(-\\infty,-\\ell / 2]$ and $[-\\ell / 2,+\\infty)$. Let $[a, b]$ be the segment containing zero. Then concavity implies $H(0) \\geqslant \\min \\{H(a), H(b)\\}$ and, since $H( \\pm T)>L$, it suffices to prove the inequalities $H\\left(-\\left(x_{i}+x_{j}\\right) / 2\\right) \\geqslant L$, that is to prove the original inequality in the case when all numbers are shifted in such a way that two variables $x_{i}$ and $x_{j}$ add up to zero. In the following we denote the shifted variables still by $x_{i}$.\n\nIf $i=j$, i.e. $x_{i}=0$ for some index $i$, then we can remove $x_{i}$ which will decrease both sides by $2 \\sum_{k} \\sqrt{\\left|x_{k}\\right|}$. Similarly, if $x_{i}+x_{j}=0$ for distinct $i$ and $j$ we can remove both $x_{i}$ and $x_{j}$ which decreases both sides by\n\n$$\n2 \\sqrt{2\\left|x_{i}\\right|}+2 \\cdot \\sum_{k \\neq i, j}\\left(\\sqrt{\\left|x_{k}+x_{i}\\right|}+\\sqrt{\\left|x_{k}+x_{j}\\right|}\\right)\n$$\n\nIn either case we reduced our inequality to the case of smaller $n$. It remains to note that for $n=0$ and $n=1$ the inequality is trivial.' 'For real $p$ consider the integral\n\n$$\nI(p)=\\int_{0}^{\\infty} \\frac{1-\\cos (p x)}{x \\sqrt{x}} d x\n$$\n\nwhich clearly converges to a strictly positive number. By changing the variable $y=|p| x$ one notices that $I(p)=\\sqrt{|p|} I(1)$. Hence, by using the trigonometric formula $\\cos (\\alpha-\\beta)-\\cos (\\alpha+$ $\\beta)=2 \\sin \\alpha \\sin \\beta$ we obtain\n\n$\\sqrt{|a+b|}-\\sqrt{|a-b|}=\\frac{1}{I(1)} \\int_{0}^{\\infty} \\frac{\\cos ((a-b) x)-\\cos ((a+b) x)}{x \\sqrt{x}} d x=\\frac{1}{I(1)} \\int_{0}^{\\infty} \\frac{2 \\sin (a x) \\sin (b x)}{x \\sqrt{x}} d x$,\n\nfrom which our inequality immediately follows:\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{\\left|x_{i}+x_{j}\\right|}-\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{\\left|x_{i}-x_{j}\\right|}=\\frac{2}{I(1)} \\int_{0}^{\\infty} \\frac{\\left(\\sum_{i=1}^{n} \\sin \\left(x_{i} x\\right)\\right)^{2}}{x \\sqrt{x}} d x \\geqslant 0\n$$']" [] Text-only Competition True Theorem proof Algebra Math English 120 "Let $n \geqslant 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove that $$ \sum_{k=1}^{n} \frac{a_{k}}{1-a_{k}}\left(a_{1}+a_{2}+\cdots+a_{k-1}\right)^{2}<\frac{1}{3} $$" "['For all $k \\leqslant n$, let\n\n$$\ns_{k}=a_{1}+a_{2}+\\cdots+a_{k} \\quad \\text { and } \\quad b_{k}=\\frac{a_{k} s_{k-1}^{2}}{1-a_{k}}\n$$\n\nwith the convention that $s_{0}=0$. Note that $b_{k}$ is exactly a summand in the sum we need to estimate. We shall prove the inequality\n\n$$\nb_{k}<\\frac{s_{k}^{3}-s_{k-1}^{3}}{3}\n\\tag{1}\n$$\n\nIndeed, it suffices to check that\n\n$$\n\\begin{aligned}\n(1) & \\Longleftrightarrow 0<\\left(1-a_{k}\\right)\\left(\\left(s_{k-1}+a_{k}\\right)^{3}-s_{k-1}^{3}\\right)-3 a_{k} s_{k-1}^{2} \\\\\n& \\Longleftrightarrow 0<\\left(1-a_{k}\\right)\\left(3 s_{k-1}^{2}+3 s_{k-1} a_{k}+a_{k}^{2}\\right)-3 s_{k-1}^{2} \\\\\n& \\Longleftrightarrow 0<-3 a_{k} s_{k-1}^{2}+3\\left(1-a_{k}\\right) s_{k-1} a_{k}+\\left(1-a_{k}\\right) a_{k}^{2} \\\\\n& \\Longleftrightarrow 0<3\\left(1-a_{k}-s_{k-1}\\right) s_{k-1} a_{k}+\\left(1-a_{k}\\right) a_{k}^{2}\n\\end{aligned}\n$$\n\nwhich holds since $a_{k}+s_{k-1}=s_{k} \\leqslant 1$ and $a_{k} \\in(0,1)$.\n\nThus, adding inequalities (1) for $k=1, \\ldots, n$, we conclude that\n\n$$\nb_{1}+b_{2}+\\cdots+b_{n}<\\frac{s_{n}^{3}-s_{0}^{3}}{3}=\\frac{1}{3}\n$$\n\nas desired.' 'First, let us define\n\n$$\nS\\left(a_{1}, \\ldots, a_{n}\\right):=\\sum_{k=1}^{n} \\frac{a_{k}}{1-a_{k}}\\left(a_{1}+a_{2}+\\cdots+a_{k-1}\\right)^{2}\n$$\n\n\n\nFor some index $i$, denote $a_{1}+\\cdots+a_{i-1}$ by $s$. If we replace $a_{i}$ with two numbers $a_{i} / 2$ and $a_{i} / 2$, i.e. replace the tuple $\\left(a_{1}, \\ldots, a_{n}\\right)$ with $\\left(a_{1}, \\ldots, a_{i-1}, a_{i} / 2, a_{i} / 2, a_{i+1}, \\ldots, a_{n}\\right)$, the sum will increase by\n\n$$\n\\begin{aligned}\nS\\left(a_{1}, \\ldots, a_{i} / 2, a_{i} / 2, \\ldots, a_{n}\\right)-S\\left(a_{1}, \\ldots, a_{n}\\right) & =\\frac{a_{i} / 2}{1-a_{i} / 2}\\left(s^{2}+\\left(s+a_{i} / 2\\right)^{2}\\right)-\\frac{a_{i}}{1-a_{i}} s^{2} \\\\\n& =a_{i} \\frac{\\left(1-a_{i}\\right)\\left(2 s^{2}+s a_{i}+a_{i}^{2} / 4\\right)-\\left(2-a_{i}\\right) s^{2}}{\\left(2-a_{i}\\right)\\left(1-a_{i}\\right)} \\\\\n& =a_{i} \\frac{\\left(1-a_{i}-s\\right) s a_{i}+\\left(1-a_{i}\\right) a_{i}^{2} / 4}{\\left(2-a_{i}\\right)\\left(1-a_{i}\\right)},\n\\end{aligned}\n$$\n\nwhich is strictly positive. So every such replacement strictly increases the sum. By repeating this process and making maximal number in the tuple tend to zero, we keep increasing the sum which will converge to\n\n$$\n\\int_{0}^{1} x^{2} d x=\\frac{1}{3}\n$$\n\nThis completes the proof.' 'We sketch a probabilistic version of the first solution. Let $x_{1}, x_{2}, x_{3}$ be drawn uniformly and independently at random from the segment [0,1]. Let $I_{1} \\cup I_{2} \\cup \\cdots \\cup I_{n}$ be a partition of $[0,1]$ into segments of length $a_{1}, a_{2}, \\ldots, a_{n}$ in this order. Let $J_{k}:=I_{1} \\cup \\cdots \\cup I_{k-1}$ for $k \\geqslant 2$ and $J_{1}:=\\varnothing$. Then\n\n$$\n\\begin{aligned}\n\\frac{1}{3}= & \\sum_{k=1}^{n} \\mathbb{P}\\left\\{x_{1} \\geqslant x_{2}, x_{3} ; x_{1} \\in I_{k}\\right\\} \\\\\n= & \\sum_{k=1}^{n}\\left(\\mathbb{P}\\left\\{x_{1} \\in I_{k} ; x_{2}, x_{3} \\in J_{k}\\right\\}+2 \\cdot \\mathbb{P}\\left\\{x_{1} \\geqslant x_{2} ; x_{1}, x_{2} \\in I_{k} ; x_{3} \\in J_{k}\\right\\}\\right. \\\\\n& \\left.\\quad+\\mathbb{P}\\left\\{x_{1} \\geqslant x_{2}, x_{3} ; x_{1}, x_{2}, x_{3} \\in I_{k}\\right\\}\\right) \\\\\n= & \\sum_{k=1}^{n}\\left(a_{k}\\left(a_{1}+\\cdots+a_{k-1}\\right)^{2}+2 \\cdot \\frac{a_{k}^{2}}{2} \\cdot\\left(a_{1}+\\cdots+a_{k-1}\\right)+\\frac{a_{k}^{3}}{3}\\right) \\\\\n> & \\sum_{k=1}^{n}\\left(a_{k}\\left(a_{1}+\\cdots+a_{k-1}\\right)^{2}+a_{k}^{2}\\left(a_{1}+\\cdots+a_{k-1}\\right) \\cdot \\frac{a_{1}+\\cdots+a_{k-1}}{1-a_{k}}\\right)\n\\end{aligned}\n$$\n\nwhere for the last inequality we used that $1-a_{k} \\geqslant a_{1}+\\cdots+a_{k-1}$. This completes the proof since\n\n$$\na_{k}+\\frac{a_{k}^{2}}{1-a_{k}}=\\frac{a_{k}}{1-a_{k}}\n$$']" [] Text-only Competition True Theorem proof Algebra Math English 121 Let $A$ be a finite set of (not necessarily positive) integers, and let $m \geqslant 2$ be an integer. Assume that there exist non-empty subsets $B_{1}, B_{2}, B_{3}, \ldots, B_{m}$ of $A$ whose elements add up to the sums $m^{1}, m^{2}, m^{3}, \ldots, m^{m}$, respectively. Prove that $A$ contains at least $m / 2$ elements. ['Let $A=\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$. Assume that, on the contrary, $k=|A|\left(\frac{2 n}{3}\right)^{3 / 2} . $$" "[""Lemma 1.1. If $a, b, c$ are non-negative numbers such that $a b-c^{2} \\geqslant 1$, then\n\n$$\n(a+2 b)^{2} \\geqslant(b+2 c)^{2}+6\n$$\n\nProof. $(a+2 b)^{2}-(b+2 c)^{2}=(a-b)^{2}+2(b-c)^{2}+6\\left(a b-c^{2}\\right) \\geqslant 6$.\n\nLemma 1.2. $\\sqrt{1}+\\cdots+\\sqrt{n}>\\frac{2}{3} n^{3 / 2}$.\n\nProof. Bernoulli's inequality $(1+t)^{3 / 2}>1+\\frac{3}{2} t$ for $0>t \\geqslant-1$ (or, alternatively, a straightforward check) gives\n\n$$\n(k-1)^{3 / 2}=k^{3 / 2}\\left(1-\\frac{1}{k}\\right)^{3 / 2}>k^{3 / 2}\\left(1-\\frac{3}{2 k}\\right)=k^{3 / 2}-\\frac{3}{2} \\sqrt{k}\n\\tag{*}\n$$\n\nSumming up (*) over $k=1,2, \\ldots, n$ yields\n\n$$\n0>n^{3 / 2}-\\frac{3}{2}(\\sqrt{1}+\\cdots+\\sqrt{n}) .\n$$\n\nNow put $y_{i}:=2 x_{i}+x_{i+1}$ for $i=0,1, \\ldots, n$. We get $y_{0} \\geqslant 0$ and $y_{i}^{2} \\geqslant y_{i-1}^{2}+6$ for $i=1,2, \\ldots, n$ by Lemma 1.1. Thus, an easy induction on $i$ gives $y_{i} \\geqslant \\sqrt{6 i}$. Using this estimate and Lemma 1.2 we get\n\n$$\n3\\left(x_{0}+\\ldots+x_{n+1}\\right) \\geqslant y_{1}+\\ldots+y_{n} \\geqslant \\sqrt{6}(\\sqrt{1}+\\sqrt{2}+\\ldots+\\sqrt{n})>\\sqrt{6} \\cdot \\frac{2}{3} n^{3 / 2}=3\\left(\\frac{2 n}{3}\\right)^{3 / 2}\n$$"" 'Say that an index $i \\in\\{0,1, \\ldots, n+1\\}$ is $\\operatorname{good}$, if $x_{i} \\geqslant \\sqrt{\\frac{2}{3}} i$, otherwise call the index $i$ bad.\n\nLemma 2.1. There are no two consecutive bad indices.\n\nProof. Assume the contrary and consider two bad indices $j, j+1$ with minimal possible $j$. Since 0 is good, we get $j>0$, thus by minimality $j-1$ is a good index and we have\n\n$$\n\\frac{2}{3} \\sqrt{j(j+1)}>x_{j} x_{j+1} \\geqslant x_{j-1}^{2}+1 \\geqslant \\frac{2}{3}(j-1)+1=\\frac{2}{3} \\cdot \\frac{j+(j+1)}{2}\n$$\n\nthat contradicts the AM-GM inequality for numbers $j$ and $j+1$.\n\nLemma 2.2. If an index $j \\leqslant n-1$ is good, then\n\n$$\nx_{j+1}+x_{j+2} \\geqslant \\sqrt{\\frac{2}{3}}(\\sqrt{j+1}+\\sqrt{j+2})\n$$\n\nProof. We have\n\n$$\nx_{j+1}+x_{j+2} \\geqslant 2 \\sqrt{x_{j+1} x_{j+2}} \\geqslant 2 \\sqrt{x_{j}^{2}+1} \\geqslant 2 \\sqrt{\\frac{2}{3} j+1} \\geqslant \\sqrt{\\frac{2}{3} j+\\frac{2}{3}}+\\sqrt{\\frac{2}{3} j+\\frac{4}{3}},\n$$\n\nthe last inequality follows from concavity of the square root function, or, alternatively, from the AM-QM inequality for the numbers $\\sqrt{\\frac{2}{3} j+\\frac{2}{3}}$ and $\\sqrt{\\frac{2}{3} j+\\frac{4}{3}}$.\n\n\n\nLet $S_{i}=x_{1}+\\ldots+x_{i}$ and $T_{i}=\\sqrt{\\frac{2}{3}}(\\sqrt{1}+\\ldots+\\sqrt{i})$.\n\nLemma 2.3. If an index $i$ is good, then $S_{i} \\geqslant T_{i}$.\n\nProof. Induction on $i$. The base case $i=0$ is clear. Assume that the claim holds for good indices less than $i$ and prove it for a good index $i>0$.\n\nIf $i-1$ is good, then by the inductive hypothesis we get $S_{i}=S_{i-1}+x_{i} \\geqslant T_{i-1}+\\sqrt{\\frac{2}{3} i}=T_{i}$.\n\nIf $i-1$ is bad, then $i>1$, and $i-2$ is good by Lemma 2.1. Then using Lemma 2.2 and the inductive hypothesis we get\n\n$$\nS_{i}=S_{i-2}+x_{i-1}+x_{i} \\geqslant T_{i-2}+\\sqrt{\\frac{2}{3}}(\\sqrt{i-1}+\\sqrt{i})=T_{i}\n$$\n\nSince either $n$ or $n+1$ is good by Lemma 2.1, Lemma 2.3 yields in both cases $S_{n+1} \\geqslant T_{n}$']" [] Text-only Competition True Theorem proof Algebra Math English 123 Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a, b, c, d \in S$ with $\operatorname{gcd}(a, b) \neq \operatorname{gcd}(c, d)$. Prove that there exist three pairwise distinct $x, y, z \in S$ such that $\operatorname{gcd}(x, y)=\operatorname{gcd}(y, z) \neq \operatorname{gcd}(z, x)$. ['There exists $\\alpha \\in S$ so that $\\{\\operatorname{gcd}(\\alpha, s) \\mid s \\in S, s \\neq \\alpha\\}$ contains at least two elements. Since $\\alpha$ has only finitely many divisors, there is a $d \\mid \\alpha$ such that the set $B=\\{\\beta \\in$ $S \\mid \\operatorname{gcd}(\\alpha, \\beta)=d\\}$ is infinite. Pick $\\gamma \\in S$ so that $\\operatorname{gcd}(\\alpha, \\gamma) \\neq d$. Pick $\\beta_{1}, \\beta_{2} \\in B$ so that $\\operatorname{gcd}\\left(\\beta_{1}, \\gamma\\right)=\\operatorname{gcd}\\left(\\beta_{2}, \\gamma\\right)=: d^{\\prime}$. If $d=d^{\\prime}$, then $\\operatorname{gcd}\\left(\\alpha, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\gamma, \\beta_{1}\\right) \\neq \\operatorname{gcd}(\\alpha, \\gamma)$. If $d \\neq d^{\\prime}$, then either $\\operatorname{gcd}\\left(\\alpha, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\alpha, \\beta_{2}\\right)=d$ and $\\operatorname{gcd}\\left(\\beta_{1}, \\beta_{2}\\right) \\neq d \\operatorname{or} \\operatorname{gcd}\\left(\\gamma, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\gamma, \\beta_{2}\\right)=d^{\\prime}$ and $\\operatorname{gcd}\\left(\\beta_{1}, \\beta_{2}\\right) \\neq d^{\\prime}$.'] [] Text-only Competition True Theorem proof Algebra Math English 124 "A thimblerigger has 2021 thimbles numbered from 1 through 2021. The thimbles are arranged in a circle in arbitrary order. The thimblerigger performs a sequence of 2021 moves; in the $k^{\text {th }}$ move, he swaps the positions of the two thimbles adjacent to thimble $k$. Prove that there exists a value of $k$ such that, in the $k^{\text {th }}$ move, the thimblerigger swaps some thimbles $a$ and $b$ such that $ak \geqslant 1$. There are $2 n+1$ students standing in a circle. Each student $S$ has $2 k$ neighbours - namely, the $k$ students closest to $S$ on the right, and the $k$ students closest to $S$ on the left. Suppose that $n+1$ of the students are girls, and the other $n$ are boys. Prove that there is a girl with at least $k$ girls among her neighbours." "[""We replace the girls by 1's, and the boys by 0's, getting the numbers $a_{1}, a_{2}, \\ldots, a_{2 n+1}$ arranged in a circle. We extend this sequence periodically by letting $a_{2 n+1+k}=a_{k}$ for all $k \\in \\mathbb{Z}$. We get an infinite periodic sequence\n\n$$\n\\ldots, a_{1}, a_{2}, \\ldots, a_{2 n+1}, a_{1}, a_{2}, \\ldots, a_{2 n+1}, \\ldots\n$$\n\nConsider the numbers $b_{i}=a_{i}+a_{i-k-1}-1 \\in\\{-1,0,1\\}$ for all $i \\in \\mathbb{Z}$. We know that\n\n$$\nb_{m+1}+b_{m+2}+\\cdots+b_{m+2 n+1}=1 \\quad(m \\in \\mathbb{Z})\n\\tag{1}\n$$\n\nin particular, this yields that there exists some $i_{0}$ with $b_{i_{0}}=1$. Now we want to find an index $i$ such that\n\n$$\nb_{i}=1 \\quad \\text { and } \\quad b_{i+1}+b_{i+2}+\\cdots+b_{i+k} \\geqslant 0 .\n\\tag{2}\n$$\n\nThis will imply that $a_{i}=1$ and\n\n$$\n\\left(a_{i-k}+a_{i-k+1}+\\cdots+a_{i-1}\\right)+\\left(a_{i+1}+a_{i+2}+\\cdots+a_{i+k}\\right) \\geqslant k,\n$$\n\nas desired.\n\nSuppose, to the contrary, that for every index $i$ with $b_{i}=1$ the sum $b_{i+1}+b_{i+2}+\\cdots+b_{i+k}$ is negative. We start from some index $i_{0}$ with $b_{i_{0}}=1$ and construct a sequence $i_{0}, i_{1}, i_{2}, \\ldots$, where $i_{j}(j>0)$ is the smallest possible index such that $i_{j}>i_{j-1}+k$ and $b_{i_{j}}=1$. We can choose two numbers among $i_{0}, i_{1}, \\ldots, i_{2 n+1}$ which are congruent modulo $2 n+1$ (without loss of generality, we may assume that these numbers are $i_{0}$ and $i_{T}$ ).\n\nOn the one hand, for every $j$ with $0 \\leqslant j \\leqslant T-1$ we have\n\n$$\nS_{j}:=b_{i_{j}}+b_{i_{j}+1}+b_{i_{j}+2}+\\cdots+b_{i_{j+1}-1} \\leqslant b_{i_{j}}+b_{i_{j}+1}+b_{i_{j}+2}+\\cdots+b_{i_{j}+k} \\leqslant 0\n$$\n\nsince $b_{i_{j}+k+1}, \\ldots, b_{i_{j+1}-1} \\leqslant 0$. On the other hand, since $\\left(i_{T}-i_{0}\\right) \\mid(2 n+1)$, from (1) we deduce\n\n$$\nS_{0}+\\cdots+S_{T-1}=\\sum_{i=i_{0}}^{i_{T}-1} b_{i}=\\frac{i_{T}-i_{0}}{2 n+1}>0\n$$\n\nThis contradiction finishes the solution.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 127 "A hunter and an invisible rabbit play a game on an infinite square grid. First the hunter fixes a colouring of the cells with finitely many colours. The rabbit then secretly chooses a cell to start in. Every minute, the rabbit reports the colour of its current cell to the hunter, and then secretly moves to an adjacent cell that it has not visited before (two cells are adjacent if they share a side). The hunter wins if after some finite time either - the rabbit cannot move; or - the hunter can determine the cell in which the rabbit started. Decide whether there exists a winning strategy for the hunter." "[""A central idea is that several colourings $C_{1}, C_{2}, \\ldots, C_{k}$ can be merged together into a single product colouring $C_{1} \\times C_{2} \\times \\cdots \\times C_{k}$ as follows: the colours in the product colouring are ordered tuples $\\left(c_{1}, \\ldots, c_{n}\\right)$ of colours, where $c_{i}$ is a colour used in $C_{i}$, so that each cell gets a tuple consisting of its colours in the individual colourings $C_{i}$. This way, any information which can be determined from one of the individual colourings can also be determined from the product colouring.\n\nNow let the hunter merge the following colourings:\n\n- The first two colourings $C_{1}$ and $C_{2}$ allow the tracking of the horizontal and vertical movements of the rabbit.\n\nThe colouring $C_{1}$ colours the cells according to the residue of their $x$-coordinates modulo 3 , which allows to determine whether the rabbit moves left, moves right, or moves vertically. Similarly, the colouring $C_{2}$ uses the residues of the $y$-coordinates modulo 3 , which allows to determine whether the rabbit moves up, moves down, or moves horizontally.\n\n- Under the condition that the rabbit's $x$-coordinate is unbounded, colouring $C_{3}$ allows to determine the exact value of the $x$-coordinate:\n\nIn $C_{3}$, the columns are coloured white and black so that the gaps between neighboring black columns are pairwise distinct. As the rabbit's $x$-coordinate is unbounded, it will eventually visit two black cells in distinct columns. With the help of colouring $C_{1}$ the hunter can catch that moment, and determine the difference of $x$-coordinates of those two black cells, hence deducing the precise column.\n\nSymmetrically, under the condition that the rabbit's $y$-coordinate is unbounded, there is a colouring $C_{4}$ that allows the hunter to determine the exact value of the $y$-coordinate.\n\n- Finally, under the condition that the sum $x+y$ of the rabbit's coordinates is unbounded, colouring $C_{5}$ allows to determine the exact value of this sum: The diagonal lines $x+y=$ const are coloured black and white, so that the gaps between neighboring black diagonals are pairwise distinct.\n\nUnless the rabbit gets stuck, at least two of the three values $x, y$ and $x+y$ must be unbounded as the rabbit keeps moving. Hence the hunter can eventually determine two of these three values; thus he does know all three. Finally the hunter works backwards with help of the colourings $C_{1}$ and $C_{2}$ and computes the starting cell of the rabbit.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 128 "Consider a checkered $3 m \times 3 m$ square, where $m$ is an integer greater than 1 . A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F$. The frog can hop from any cell to either the next cell to the right or the next cell upwards. Some cells can be sticky, and the frog gets trapped once it hops on such a cell. A set $X$ of cells is called blocking if the frog cannot reach $F$ from $S$ when all the cells of $X$ are sticky. A blocking set is minimal if it does not contain a smaller blocking set. (a) Prove that there exists a minimal blocking set containing at least $3 m^{2}-3 m$ cells." ['In the following example the square is divided into $m$ stripes of size $3 \\times 3 \\mathrm{~m}$. It is easy to see that $X$ is a minimal blocking set. The first and the last stripe each contains $3 m-1$ cells from the set $X$; every other stripe contains $3 m-2$ cells, see Figure 1 . The total number of cells in the set $X$ is $3 m^{2}-2 m+2$.\n\n\n\nFigure 1'] [] Text-only Competition True Theorem proof Combinatorics Math English 129 "Consider a checkered $3 m \times 3 m$ square, where $m$ is an integer greater than 1 . A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F$. The frog can hop from any cell to either the next cell to the right or the next cell upwards. Some cells can be sticky, and the frog gets trapped once it hops on such a cell. A set $X$ of cells is called blocking if the frog cannot reach $F$ from $S$ when all the cells of $X$ are sticky. A blocking set is minimal if it does not contain a smaller blocking set. Prove that every minimal blocking set contains at most $3 m^{2}$ cells. Note. An example of a minimal blocking set for $m=2$ is shown below. Cells of the set $X$ are marked by letters $x$. | | | | | | $F$ | | :--- | :--- | :--- | :--- | :--- | :--- | | $x$ | $x$ | | | | | | | | $x$ | | | | | | | | $x$ | | | | | | | | $x$ | | | $S$ | | $x$ | | | |" "['For a given blocking set $X$, say that a non-sticky cell is red if the frog can reach it from $S$ via some hops without entering set $X$. We call a non-sticky cell blue if the frog can reach $F$ from that cell via hops without entering set $X$. One can regard the blue cells as those reachable from $F$ by anti-hops, i.e. moves downwards and to the left. We also colour all cells in $X$ green. It follows from the definition of the blocking set that no cell will be coloured twice. In Figure 2 we show a sample of a blocking set and the corresponding colouring.\n\nNow assume that $X$ is a minimal blocking set. We denote by $R$ (resp., $B$ and $G$ ) be the total number of red (resp., blue and green) cells.\n\nWe claim that $G \\leqslant R+1$ and $G \\leqslant B+1$. Indeed, there are at most $2 R$ possible frog hops from red cells. Every green or red cell (except for $S$ ) is accessible by such hops. Hence $2 R \\geqslant G+(R-1)$, or equivalently $G \\leqslant R+1$. In order to prove the inequality $G \\leqslant B+1$, we turn over the board and apply the similar arguments.\n\nTherefore we get $9 m^{2} \\geqslant B+R+G \\geqslant 3 G-2$, so $G \\leqslant 3 m^{2}$.\n\n\n\n\n\nFigure 2 (a)\n\n\n\nFigure 2 (b)' 'We shall use the same colouring as in the above solution. Again, assume that $X$ is a minimal blocking set.\n\nNote that any $2 \\times 2$ square cannot contain more than 2 green cells. Indeed, on Figure 3(a) the cell marked with ""?"" does not block any path, while on Figure 3(b) the cell marked with ""?"" should be coloured red and blue simultaneously. So we can split all green cells into chains consisting of three types of links shown on Figure 4 (diagonal link in the other direction is not allowed, corresponding green cells must belong to different chains). For example, there are 3 chains in Figure 2(b).\n\n\nFigure 3\n\n\n\n\nand\n\n\nFigure 4\n\n\n\nand\n\n\n\nFigure 5\n\nWe will inscribe green chains in disjoint axis-aligned rectangles so that the number of green cells in each rectangle will not exceed $1 / 3$ of the area of the rectangle. This will give us the bound $G \\leqslant 3 m^{2}$. Sometimes the rectangle will be the minimal bounding rectangle of the chain, sometimes minimal bounding rectangles will be expanded in one or two directions in order to have sufficiently large area.\n\nNote that for any two consecutive cells in the chain the colouring of some neighbouring cells is uniquely defined (see Figure 5). In particular, this observation gives a corresponding rectangle for the chains of height (or width) 1 (see Figure 6(a)). A separate green cell can be inscribed in $1 \\times 3$ or $3 \\times 1$ rectangle with one red and one blue cell, see Figure $6(\\mathrm{~b})-(\\mathrm{c})$, otherwise we get one of impossible configurations shown in Figure 3.\n\n\n\n$(a)$\n\n\n\n(b)\n\n\n\n$(c)$\n\nFigure 6\n\n\n\n(a)\n\n\n\n(b)\n\nFigure 7\n\nAny diagonal chain of length 2 is always inscribed in a $2 \\times 3$ or $3 \\times 2$ rectangle without another green cells. Indeed, one of the squares marked with ""?"" in Figure 7(a) must be red. If it is the bottom question mark, then the remaining cell in the corresponding $2 \\times 3$ rectangle must have the same colour, see Figure 7(b).\n\nA longer chain of height (or width) 2 always has a horizontal (resp., vertical) link and can be inscribed into a $3 \\times a$ rectangle. In this case we expand the minimal bounding rectangle across the long side which touches the mentioned link. On Figure 8(a) the corresponding expansion of the minimal bounding rectangle is coloured in light blue. The upper right corner cell must be also blue. Indeed it cannot be red or green. If it is not coloured in blue, see Figure 8(b), then all anti-hop paths from $F$ to ""?"" are blocked with green cells. And these green cells are surrounded by blue ones, what is impossible. In this case the green chain contains $a$ cells, which is exactly $1 / 3$ of the area of the rectangle.\n\n\n\n\n\nFigure 8 (a)\n\n\n\nFigure 8 (b)\n\nIn the remaining case the minimal bounding rectangle of the chain is of size $a \\times b$ where $a, b \\geqslant 3$. Denote by $\\ell$ the length of the chain (i.e. the number of cells in the chain).\n\nIf the chain has at least two diagonal links (see Figure 9), then $\\ell \\leqslant a+b-3 \\leqslant a b / 3$.\n\nIf the chain has only one diagonal link then $\\ell=a+b-2$. In this case the chain has horizontal and vertical end-links, and we expand the minimal bounding rectangle in two directions to get an $(a+1) \\times(b+1)$ rectangle. On Figure 10 a corresponding expansion of the minimal bounding rectangle is coloured in light red. Again the length of the chain does not exceed $1 / 3$ of the rectangle\'s area: $\\ell \\leqslant a+b-2 \\leqslant(a+1)(b+1) / 3$.\n\nOn the next step we will use the following statement: all cells in constructed rectangles are coloured red, green or blue (the cells upwards and to the right of green cells are blue; the cells downwards and to the left of green cells are red). The proof repeats the same arguments as before (see Figure 8(b).)\n\n\n\nFigure 9\n\n\n\nFigure 10\n\n\n\n(a)\n\n\n\n(b)\n\nFigure 11\n\nNote that all constructed rectangles are disjoint. Indeed, assume that two rectangles have a common cell. Using the above statement, one can see that the only such cell can be a common corner cell, as shown in Figure 11. Moreover, in this case both rectangles should be expanded, otherwise they would share a green corner cell.\n\nIf they were expanded along the same axis (see Figure 11(a)), then again the common corner cannot be coloured correctly. If they were expanded along different axes (see Figure 11(b)) then the two chains have a common point and must be connected in one chain. (These arguments work for $2 \\times 3$ and $1 \\times 3$ rectangles in a similar manner.)']" [] Text-only Competition True Theorem proof Combinatorics Math English 130 Let $A B C D$ be a parallelogram such that $A C=B C$. A point $P$ is chosen on the extension of the segment $A B$ beyond $B$. The circumcircle of the triangle $A C D$ meets the segment $P D$ again at $Q$, and the circumcircle of the triangle $A P Q$ meets the segment $P C$ again at $R$. Prove that the lines $C D, A Q$, and $B R$ are concurrent. "['Since $A C=B C=A D$, we have $\\angle A B C=\\angle B A C=\\angle A C D=\\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n\n$$\n\\angle C R A=180^{\\circ}-\\angle A R P=180^{\\circ}-\\angle A Q P=\\angle D Q A=\\angle D C A=\\angle C B A,\n$$\n\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\nIntroduce the point $X=A Q \\cap C D$; we need to prove that $B, R$ and $X$ are collinear.\n\nBy means of the circle $(A P R Q)$ we have\n\n$$\n\\angle R Q X=180^{\\circ}-\\angle A Q R=\\angle R P A=\\angle R C X\n$$\n\n(the last equality holds in view of $A B \\| C D$ ), which means that the points $C, Q, R$, and $X$ also lie on some circle $\\delta$.\n\nUsing the circles $\\delta$ and $\\gamma$ we finally obtain\n\n$$\n\\angle X R C=\\angle X Q C=180^{\\circ}-\\angle C Q A=\\angle A D C=\\angle B A C=180^{\\circ}-\\angle C R B,\n$$\n\nthat proves the desired collinearity.\n\n' ""Since $A C=B C=A D$, we have $\\angle A B C=\\angle B A C=\\angle A C D=\\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n\n$$\n\\angle C R A=180^{\\circ}-\\angle A R P=180^{\\circ}-\\angle A Q P=\\angle D Q A=\\angle D C A=\\angle C B A,\n$$\n\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\n\nLet $\\alpha$ denote the circle $(A P R Q)$. Since\n\n$$\n\\angle C A P=\\angle A C D=\\angle A Q D=180^{\\circ}-\\angle A Q P\n$$\n\nthe line $A C$ is tangent to $\\alpha$.\n\nNow, let $A D$ meet $\\alpha$ again at a point $Y$ (which necessarily lies on the extension of $D A$ beyond $A$ ). Using the circle $\\gamma$, along with the fact that $A C$ is tangent to $\\alpha$, we have\n\n$$\n\\angle A R Y=\\angle C A D=\\angle A C B=\\angle A R B\n$$\n\nso the points $Y, B$, and $R$ are collinear.\n\nApplying Pascal's theorem to the hexagon $A A Y R P Q$ (where $A A$ is regarded as the tangent to $\\alpha$ at $A$ ), we see that the points $A A \\cap R P=C, A Y \\cap P Q=D$, and $Y R \\cap Q A$ are collinear. Hence the lines $C D, A Q$, and $B R$ are concurrent."" 'Since $A C=B C=A D$, we have $\\angle A B C=\\angle B A C=\\angle A C D=\\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n\n$$\n\\angle C R A=180^{\\circ}-\\angle A R P=180^{\\circ}-\\angle A Q P=\\angle D Q A=\\angle D C A=\\angle C B A,\n$$\n\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\n\nwe introduce the point $X=A Q \\cap C D$ and aim at proving that the points $B, R$, and $X$ are collinear. we denote $\\alpha=(A P Q R)$; but now we define $Y$ to be the second meeting point of $R B$ with $\\alpha$.\n\nUsing the circle $\\alpha$ and noticing that $C D$ is tangent to $\\gamma$, we obtain\n\n$$\n\\angle R Y A=\\angle R P A=\\angle R C X=\\angle R B C .\n$$\n\nSo $A Y \\| B C$, and hence $Y$ lies on $D A$.\n\nNow the chain of equalities (1) shows also that $\\angle R Y D=\\angle R C X$, which implies that the points $C, D, Y$, and $R$ lie on some circle $\\beta$. Hence, the lines $C D, A Q$, and $Y B R$ are the pairwise radical axes of the circles $(A Q C D), \\alpha$, and $\\beta$, so those lines are concurrent.']" [] Text-only Competition True Theorem proof Geometry Math English 131 Let $A B C D$ be a convex quadrilateral circumscribed around a circle with centre $I$. Let $\omega$ be the circumcircle of the triangle $A C I$. The extensions of $B A$ and $B C$ beyond $A$ and $C$ meet $\omega$ at $X$ and $Z$, respectively. The extensions of $A D$ and $C D$ beyond $D$ meet $\omega$ at $Y$ and $T$, respectively. Prove that the perimeters of the (possibly self-intersecting) quadrilaterals $A D T X$ and $C D Y Z$ are equal. ['The point $I$ is the intersection of the external bisector of the angle $T C Z$ with the circumcircle $\\omega$ of the triangle $T C Z$, so $I$ is the midpoint of the $\\operatorname{arc} T C Z$ and $I T=I Z$. Similarly, $I$ is the midpoint of the $\\operatorname{arc} Y A X$ and $I X=I Y$. Let $O$ be the centre of $\\omega$. Then $X$ and $T$ are the reflections of $Y$ and $Z$ in $I O$, respectively. So $X T=Y Z$.\n\n\n\nLet the incircle of $A B C D$ touch $A B, B C, C D$, and $D A$ at points $P, Q, R$, and $S$, respectively.\n\nThe right triangles $I X P$ and $I Y S$ are congruent, since $I P=I S$ and $I X=I Y$. Similarly, the right triangles $I R T$ and $I Q Z$ are congruent. Therefore, $X P=Y S$ and $R T=Q Z$.\n\nDenote the perimeters of $A D T X$ and $C D Y Z$ by $P_{A D T X}$ and $P_{C D Y Z}$ respectively. Since $A S=A P, C Q=R C$, and $S D=D R$, we obtain\n\n$$\n\\begin{aligned}\nP_{A D T X}=X T+X A+A S+ & S D+D T=X T+X P+R T \\\\\n& =Y Z+Y S+Q Z=Y Z+Y D+D R+R C+C Z=P_{C D Y Z}\n\\end{aligned}\n$$\n\nas required.'] [] Text-only Competition True Theorem proof Geometry Math English 132 Let $A B C D$ be a quadrilateral inscribed in a circle $\Omega$. Let the tangent to $\Omega$ at $D$ intersect the rays $B A$ and $B C$ at points $E$ and $F$, respectively. A point $T$ is chosen inside the triangle $A B C$ so that $T E \| C D$ and $T F \| A D$. Let $K \neq D$ be a point on the segment $D F$ such that $T D=T K$. Prove that the lines $A C, D T$ and $B K$ intersect at one point. "['Let the segments $T E$ and $T F$ cross $A C$ at $P$ and $Q$, respectively. Since $P E \\| C D$ and $E D$ is tangent to the circumcircle of $A B C D$, we have\n\n$$\n\\angle E P A=\\angle D C A=\\angle E D A\n$$\n\nand so the points $A, P, D$, and $E$ lie on some circle $\\alpha$. Similarly, the points $C, Q, D$, and $F$ lie on some circle $\\gamma$.\n\nWe now want to prove that the line $D T$ is tangent to both $\\alpha$ and $\\gamma$ at $D$. Indeed, since $\\angle F C D+\\angle E A D=180^{\\circ}$, the circles $\\alpha$ and $\\gamma$ are tangent to each other at $D$. To prove that $T$ lies on their common tangent line at $D$ (i.e., on their radical axis), it suffices to check that $T P \\cdot T E=T Q \\cdot T F$, or that the quadrilateral $P E F Q$ is cyclic. This fact follows from\n\n$$\n\\angle Q F E=\\angle A D E=\\angle A P E .\n$$\n\nSince $T D=T K$, we have $\\angle T K D=\\angle T D K$. Next, as $T D$ and $D E$ are tangent to $\\alpha$ and $\\Omega$, respectively, we obtain\n\n$$\n\\angle T K D=\\angle T D K=\\angle E A D=\\angle B D E\n$$\n\nwhich implies $T K \\| B D$.\n\nNext we prove that the five points $T, P, Q, D$, and $K$ lie on some circle $\\tau$. Indeed, since $T D$ is tangent to the circle $\\alpha$ we have\n\n$$\n\\angle E P D=\\angle T D F=\\angle T K D\n$$\n\nwhich means that the point $P$ lies on the circle (TDK). Similarly, we have $Q \\in(T D K)$.\n\nFinally, we prove that $P K \\| B C$. Indeed, using the circles $\\tau$ and $\\gamma$ we conclude that\n\n$$\n\\angle P K D=\\angle P Q D=\\angle D F C\n$$\n\nwhich means that $P K \\| B C$.\n\nTriangles $T P K$ and $D C B$ have pairwise parallel sides, which implies the fact that $T D, P C$ and $K B$ are concurrent, as desired.\n\n' 'Consider the spiral similarity $\\phi$ centred at $D$ which maps $B$ to $F$. Recall that for any two points $X$ and $Y$, the triangles $D X \\phi(X)$ and $D Y \\phi(Y)$ are similar.\n\nDefine $T^{\\prime}=\\phi(E)$. Then\n\n$$\n\\angle C D F=\\angle F B D=\\angle \\phi(B) B D=\\angle \\phi(E) E D=\\angle T^{\\prime} E D,\n$$\n\nso $C D \\| T^{\\prime} E$. Using the fact that $D E$ is tangent to $(A B D)$ and then applying $\\phi$ we infer\n\n$$\n\\angle A D E=\\angle A B D=\\angle T^{\\prime} F D\n$$\n\nso $A D \\| T^{\\prime} F$; hence $T^{\\prime}$ coincides with $T$. Therefore,\n\n$$\n\\angle B D E=\\angle F D T=\\angle D K T\n$$\n\nwhence $T K \\| B D$.\n\nLet $B K \\cap T D=X, A C \\cap T D=Y$, and $A C \\cap T F=Q$. Notice that $T K \\| B D$ implies\n\n$$\n\\frac{T X}{X D}=\\frac{T K}{B D}=\\frac{T D}{B D}\n$$\n\nSo we wish to prove that $\\frac{T Y}{Y D}$ is equal to the same ratio.\n\nWe first show that $\\phi(A)=Q$. Indeed,\n\n$$\n\\angle D A \\phi(A)=\\angle D B F=\\angle D A C\n$$\n\nand so $\\phi(A) \\in A C$. Together with $\\phi(A) \\in \\phi(E B)=T F$ this yields $\\phi(A)=Q$. It follows that\n\n$$\n\\frac{TQ}{AE}=\\frac{TD}{ED}\n$$\n\n\n\n\nNow, since $T F \\| A D$ and $\\triangle E A D \\sim \\triangle E D B$, we have\n\n$$\n\\frac{T Y}{Y D}=\\frac{T Q}{A D}=\\frac{T Q}{A E} \\cdot \\frac{A E}{A D}=\\frac{T D}{E D} \\cdot \\frac{E D}{B D}=\\frac{T D}{B D}\n$$\n\nwhich completes the proof.']" [] Text-only Competition True Theorem proof Geometry Math English 133 "Let $A B C D$ be a cyclic quadrilateral whose sides have pairwise different lengths. Let $O$ be the circumcentre of $A B C D$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $B_{1}$ and $D_{1}$, respectively. Let $O_{B}$ be the centre of the circle which passes through $B$ and is tangent to $A C$ at $D_{1}$. Similarly, let $O_{D}$ be the centre of the circle which passes through $D$ and is tangent to $A C$ at $B_{1}$. Assume that $B D_{1} \| D B_{1}$. Prove that $O$ lies on the line $O_{B} O_{D}$." "['We introduce some objects and establish some preliminary facts common for all solutions below.\n\nLet $\\Omega$ denote the circle $(A B C D)$, and let $\\gamma_{B}$ and $\\gamma_{D}$ denote the two circles from the problem statement (their centres are $O_{B}$ and $O_{D}$, respectively). Clearly, all three centres $O, O_{B}$, and $O_{D}$ are distinct.\n\nAssume, without loss of generality, that $A B>B C$. Suppose that $A D>D C$, and let $H=A C \\cap B D$. Then the rays $B B_{1}$ and $D D_{1}$ lie on one side of $B D$, as they contain the midpoints of the arcs $A D C$ and $A B C$, respectively. However, if $B D_{1} \\| D B_{1}$, then $B_{1}$ and $D_{1}$ should be separated by $H$. This contradiction shows that $A D\ncentres $O_{B} O$; likewise, $D T_{D} \\perp O_{D} O$. Therefore, $O \\in O_{B} O_{D} \\Longleftrightarrow O_{B} O\\left\\|O_{D} O \\Longleftrightarrow B T_{B}\\right\\|$ $D T_{D}$, and the problem reduces to showing that\n\n$$\nB T_{B} \\| D T_{D}\n\\tag{1}\n$$\n\n\nLet the diagonals $A C$ and $B D$ cross at $H$. Consider the homothety $h$ centred at $H$ and mapping $B$ to $D$. Since $B D_{1} \\| D B_{1}$, we have $h\\left(D_{1}\\right)=B_{1}$.\n\nLet the tangents to $\\Omega$ at $B$ and $D$ meet $A C$ at $L_{B}$ and $L_{D}$, respectively. We have\n\n$$\n\\angle L_{B} B B_{1}=\\angle L_{B} B C+\\angle C B B_{1}=\\angle B A L_{B}+\\angle B_{1} B A=\\angle B B_{1} L_{B},\n$$\n\nwhich means that the triangle $L_{B} B B_{1}$ is isosceles, $L_{B} B=L_{B} B_{1}$. The powers of $L_{B}$ with respect to $\\Omega$ and $\\gamma_{D}$ are $L_{B} B^{2}$ and $L_{B} B_{1}^{2}$, respectively; so they are equal, whence $L_{B}$ lies on the radical axis $T_{D} D$ of those two circles. Similarly, $L_{D}$ lies on the radical axis $T_{B} B$ of $\\Omega$ and $\\gamma_{B}$.\n\nBy the sine rule in the triangle $B H L_{B}$, we obtain\n\n$$\n\\frac{H L_{B}}{\\sin \\angle H B L_{B}}=\\frac{B L_{B}}{\\sin \\angle B H L_{B}}=\\frac{B_{1} L_{B}}{\\sin \\angle B H L_{B}}\n\\tag{2}\n$$\n\nsimilarly,\n\n$$\n\\frac{H L_{D}}{\\sin \\angle H D L_{D}}=\\frac{D L_{D}}{\\sin \\angle D H L_{D}}=\\frac{D_{1} L_{D}}{\\sin \\angle D H L_{D}}\n\\tag{3}\n$$\n\nClearly, $\\angle B H L_{B}=\\angle D H L_{D}$. In the circle $\\Omega$, tangent lines $B L_{B}$ and $D L_{D}$ form equal angles with the chord $B D$, so $\\sin \\angle H B L_{B}=\\sin \\angle H D L_{D}$ (this equality does not depend on the picture). Thus, dividing (2) by (3) we get\n\n$$\n\\frac{H L_{B}}{H L_{D}}=\\frac{B_{1} L_{B}}{D_{1} L_{D}}, \\quad \\text { and hence } \\quad \\frac{H L_{B}}{H L_{D}}=\\frac{H L_{B}-B_{1} L_{B}}{H L_{D}-D_{1} L_{D}}=\\frac{H B_{1}}{H D_{1}}\n$$\n\nSince $h\\left(D_{1}\\right)=B_{1}$, the obtained relation yields $h\\left(L_{D}\\right)=L_{B}$, so $h$ maps the line $L_{D} B$ to $L_{B} D$, and these lines are parallel, as desired.\n\n\n\n' ""Let $B D_{1}$ and $T_{B} D_{1}$ meet $\\Omega$ again at $X_{B}$ and $Y_{B}$, respectively. Then\n\n$$\n\\angle B D_{1} C=\\angle B T_{B} D_{1}=\\angle B T_{B} Y_{B}=\\angle B X_{B} Y_{B}\n$$\n\nwhich shows that $X_{B} Y_{B} \\| A C$. Similarly, let $D B_{1}$ and $T_{D} B_{1}$ meet $\\Omega$ again at $X_{D}$ and $Y_{D}$, respectively; then $X_{D} Y_{D} \\| A C$.\n\nLet $M_{D}$ and $M_{B}$ be the midpoints of the arcs $A B C$ and $A D C$, respectively; then the points $D_{1}$ and $B_{1}$ lie on $D M_{D}$ and $B M_{B}$, respectively. Let $K$ be the midpoint of $A C$ (which lies on $M_{B} M_{D}$ ). Applying Pascal's theorem to $M_{D} D X_{D} X_{B} B M_{B}$, we obtain that the points $D_{1}=M_{D} D \\cap X_{B} B, B_{1}=D X_{D} \\cap B M_{B}$, and $X_{D} X_{B} \\cap M_{B} M_{D}$ are collinear, which means that $X_{B} X_{D}$ passes through $K$. Due to symmetry, the diagonals of an isosceles trapezoid $X_{B} Y_{B} X_{D} Y_{D}$ cross at $K$.\n\n\n\nLet $b$ and $d$ denote the distances from the lines $X_{B} Y_{B}$ and $X_{D} Y_{D}$, respectively, to $A C$. Then we get\n\n$$\n\\frac{X_{B} Y_{B}}{X_{D} Y_{D}}=\\frac{b}{d}=\\frac{D_{1} X_{B}}{B_{1} X_{D}}\n$$\n\nwhere the second equation holds in view of $D_{1} X_{B} \\| B_{1} X_{D}$. Therefore, the triangles $D_{1} X_{B} Y_{B}$ and $B_{1} X_{D} Y_{D}$ are similar. The triangles $D_{1} T_{B} B$ and $B_{1} T_{D} D$ are similar to them and hence to each other. Since $B D_{1} \\| D B_{1}$, these triangles are also homothetical. This yields $B T_{B} \\| D T_{D}$, as desired.""]" [] Text-only Competition True Theorem proof Geometry Math English 134 A point $D$ is chosen inside an acute-angled triangle $A B C$ with $A B>A C$ so that $\angle B A D=\angle D A C$. A point $E$ is constructed on the segment $A C$ so that $\angle A D E=\angle D C B$. Similarly, a point $F$ is constructed on the segment $A B$ so that $\angle A D F=\angle D B C$. A point $X$ is chosen on the line $A C$ so that $C X=B X$. Let $O_{1}$ and $O_{2}$ be the circumcentres of the triangles $A D C$ and $D X E$. Prove that the lines $B C, E F$, and $O_{1} O_{2}$ are concurrent. "['Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\\angle B A D=\\angle D A C$, the point $Q$ lies on $A D$. Then $\\angle Q B A=\\angle D B C=\\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \\cdot A B=A D \\cdot A Q=A E \\cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.\n\n\n\nLet $T$ be the intersection of $B C$ and $F E$.\n\nClaim. $T D^{2}=T B \\cdot T C=T F \\cdot T E$.\n\nProof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get\n\n$$\n\\begin{aligned}\n& \\angle B D F=\\angle A F D-\\angle A B D=\\left(180^{\\circ}-\\angle F A D-\\angle F D A\\right)-(\\angle A B C-\\angle D B C) \\\\\n= & 180^{\\circ}-\\angle F A D-\\angle A B C=180^{\\circ}-\\angle D A E-\\angle F E A=\\angle F E D+\\angle A D E=\\angle F E D+\\angle D C B,\n\\end{aligned}\n$$\n\nwhich implies the desired tangency.\n\nSince the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \\cdot T F=T B \\cdot T C$.\n\n\n\nLet $T A$ intersect the circle $(A B C)$ again at $M$. Due to the circles $(B C E F)$ and $(A M C B)$, and using the above Claim, we get $T M \\cdot T A=T F \\cdot T E=T B \\cdot T C=T D^{2}$; in particular, the points $A, M, E$, and $F$ are concyclic.\n\nUnder the inversion with centre $T$ and radius $T D$, the point $M$ maps to $A$, and $B$ maps to $C$, which implies that the circle $(M B D)$ maps to the circle $(A D C)$. Their common point $D$ lies on the circle of the inversion, so the second intersection point $K$ also lies on that circle, which means $T K=T D$. It follows that the point $T$ and the centres of the circles $(K D E)$ and $(A D C)$ lie on the perpendicular bisector of $K D$.\n\nSince the center of $(A D C)$ is $O_{1}$, it suffices to show now that the points $D, K, E$, and $X$ are concyclic (the center of the corresponding circle will be $O_{2}$ ).\n\nThe lines $B M, D K$, and $A C$ are the pairwise radical axes of the circles $(A B C M),(A C D K)$ and $(B M D K)$, so they are concurrent at some point $P$. Also, $M$ lies on the circle $(A E F)$, thus\n\n$$\n\\begin{aligned}\n\\sphericalangle(E X, X B) & =\\sphericalangle(C X, X B)=\\sphericalangle(X C, B C)+\\sphericalangle(B C, B X)=2 \\sphericalangle(A C, C B) \\\\\n& =\\sphericalangle(A C, C B)+\\sphericalangle(E F, F A)=\\sphericalangle(A M, B M)+\\sphericalangle(E M, M A)=\\sphericalangle(E M, B M),\n\\end{aligned}\n$$\n\nso the points $M, E, X$, and $B$ are concyclic. Therefore, $P E \\cdot P X=P M \\cdot P B=P K \\cdot P D$, so the points $E, K, D$, and $X$ are concyclic, as desired.\n\n\n\n' 'Common remarks. Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\\angle B A D=\\angle D A C$, the point $Q$ lies on $A D$. Then $\\angle Q B A=\\angle D B C=\\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \\cdot A B=A D \\cdot A Q=A E \\cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.\n\n\n\nLet $T$ be the intersection of $B C$ and $F E$.\n\nClaim. $T D^{2}=T B \\cdot T C=T F \\cdot T E$.\n\nProof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get\n\n$$\n\\begin{aligned}\n& \\angle B D F=\\angle A F D-\\angle A B D=\\left(180^{\\circ}-\\angle F A D-\\angle F D A\\right)-(\\angle A B C-\\angle D B C) \\\\\n= & 180^{\\circ}-\\angle F A D-\\angle A B C=180^{\\circ}-\\angle D A E-\\angle F E A=\\angle F E D+\\angle A D E=\\angle F E D+\\angle D C B,\n\\end{aligned}\n$$\n\nwhich implies the desired tangency.\n\nSince the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \\cdot T F=T B \\cdot T C$.\n\n\n\n\nWe use only the first part of the Common remarks, namely, the facts that the tuples $(C, D, Q, E)$ and $(B, C, E, F)$ are both concyclic. We also introduce the point $T=$ $B C \\cap E F$. Let the circle $(C D E)$ meet $B C$ again at $E_{1}$. Since $\\angle E_{1} C Q=\\angle D C E$, the arcs $D E$ and $Q E_{1}$ of the circle $(C D Q)$ are equal, so $D Q \\| E E_{1}$.\n\nSince $B F E C$ is cyclic, the line $A D$ forms equal angles with $B C$ and $E F$, hence so does $E E_{1}$. Therefore, the triangle $E E_{1} T$ is isosceles, $T E=T E_{1}$, and $T$ lies on the common perpendicular bisector of $E E_{1}$ and $D Q$.\n\nLet $U$ and $V$ be the centres of circles $(A D E)$ and $(C D Q E)$, respectively. Then $U O_{1}$ is the perpendicular bisector of $A D$. Moreover, the points $U, V$, and $O_{2}$ belong to the perpendicular bisector of $D E$. Since $U O_{1} \\| V T$, in order to show that $O_{1} O_{2}$ passes through $T$, it suffices to show that\n\n$$\n\\frac{O_{2} U}{O_{2} V}=\\frac{O_{1} U}{T V}\n\\tag{1}\n$$\n\nDenote angles $A, B$, and $C$ of the triangle $A B C$ by $\\alpha, \\beta$, and $\\gamma$, respectively. Projecting onto $A C$ we obtain\n\n$$\n\\frac{O_{2} U}{O_{2} V}=\\frac{(X E-A E) / 2}{(X E+E C) / 2}=\\frac{A X}{C X}=\\frac{A X}{B X}=\\frac{\\sin (\\gamma-\\beta)}{\\sin \\alpha}\n\\tag{2}\n$$\n\nThe projection of $O_{1} U$ onto $A C$ is $(A C-A E) / 2=C E / 2$; the angle between $O_{1} U$ and $A C$ is $90^{\\circ}-\\alpha / 2$, so\n\n$$\n\\frac{O_{1} U}{E C}=\\frac{1}{2 \\sin (\\alpha / 2)}\n\\tag{3}\n$$\n\nNext, we claim that $E, V, C$, and $T$ are concyclic. Indeed, the point $V$ lies on the perpendicular bisector of $C E$, as well as on the internal angle bisector of $\\angle C T F$. Therefore, $V$ coincides with the midpoint of the arc $C E$ of the circle $(T C E)$.\n\nNow we have $\\angle E V C=2 \\angle E E_{1} C=180^{\\circ}-(\\gamma-\\beta)$ and $\\angle V E T=\\angle V E_{1} T=90^{\\circ}-\\angle E_{1} E C=$ $90^{\\circ}-\\alpha / 2$. Therefore,\n\n$$\n\\frac{E C}{T V}=\\frac{\\sin \\angle E T C}{\\sin \\angle V E T}=\\frac{\\sin (\\gamma-\\beta)}{\\cos (\\alpha / 2)}\n\\tag{4}\n$$\n\n\n\n\n\nRecalling (2) and multiplying (3) and (4) we establish (1):\n\n$$\n\\frac{O_{2} U}{O_{2} V}=\\frac{\\sin (\\gamma-\\beta)}{\\sin \\alpha}=\\frac{1}{2 \\sin (\\alpha / 2)} \\cdot \\frac{\\sin (\\gamma-\\beta)}{\\cos (\\alpha / 2)}=\\frac{O_{1} U}{E C} \\cdot \\frac{E C}{T V}=\\frac{O_{1} U}{T V}\n$$' 'Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\\angle B A D=\\angle D A C$, the point $Q$ lies on $A D$. Then $\\angle Q B A=\\angle D B C=\\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \\cdot A B=A D \\cdot A Q=A E \\cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.\n\n\n\nLet $T$ be the intersection of $B C$ and $F E$.\n\nClaim. $T D^{2}=T B \\cdot T C=T F \\cdot T E$.\n\nProof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get\n\n$$\n\\begin{aligned}\n& \\angle B D F=\\angle A F D-\\angle A B D=\\left(180^{\\circ}-\\angle F A D-\\angle F D A\\right)-(\\angle A B C-\\angle D B C) \\\\\n= & 180^{\\circ}-\\angle F A D-\\angle A B C=180^{\\circ}-\\angle D A E-\\angle F E A=\\angle F E D+\\angle A D E=\\angle F E D+\\angle D C B,\n\\end{aligned}\n$$\n\nwhich implies the desired tangency.\n\nSince the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \\cdot T F=T B \\cdot T C$.\n\n\nNotice that $\\angle A Q E=\\angle Q C B$ and $\\angle A Q F=\\angle Q B C$; so, if we replace the point $D$ with $Q$ in the problem set up, the points $E, F$, and $T$ remain the same. So, by the Claim, we have $T Q^{2}=T B \\cdot T C=T D^{2}$.\n\nThus, there exists a circle $\\Gamma$ centred at $T$ and passing through $D$ and $Q$. We denote the second meeting point of the circles $\\Gamma$ and $(A D C)$ by $K$. Let the line $A C$ meet the circle $(D E K)$ again at $Y$; we intend to prove that $Y=X$. this will yield that the point $T$, as well as the centres $O_{1}$ and $O_{2}$, all lie on the perpendicular bisector of $D K$.\n\nLet $L=A D \\cap B C$. We perform an inversion centred at $C$; the images of the points will be denoted by primes, e.g., $A^{\\prime}$ is the image of $A$. We obtain the following configuration, constructed in a triangle $A^{\\prime} C L^{\\prime}$.\n\nThe points $D^{\\prime}$ and $Q^{\\prime}$ are chosen on the circumcircle $\\Omega$ of $A^{\\prime} L^{\\prime} C$ such that $\\sphericalangle\\left(L^{\\prime} C, D^{\\prime} C\\right)=$ $\\sphericalangle\\left(Q^{\\prime} C, A^{\\prime} C\\right)$, which means that $A^{\\prime} L^{\\prime} \\| D^{\\prime} Q^{\\prime}$. The lines $D^{\\prime} Q^{\\prime}$ and $A^{\\prime} C$ meet at $E^{\\prime}$.\n\nA circle $\\Gamma^{\\prime}$ centred on $C L^{\\prime}$ passes through $D^{\\prime}$ and $Q^{\\prime}$. Notice here that $B^{\\prime}$ lies on the segment $C L^{\\prime}$, and that $\\angle A^{\\prime} B^{\\prime} C=\\angle B A C=2 \\angle L A C=2 \\angle A^{\\prime} L^{\\prime} C$, so that $B^{\\prime} L^{\\prime}=B^{\\prime} A^{\\prime}$, and $B^{\\prime}$ lies on the perpendicular bisector of $A^{\\prime} L^{\\prime}$ (which coincides with that of $D^{\\prime} Q^{\\prime}$ ). All this means that $B^{\\prime}$ is the centre of $\\Gamma^{\\prime}$.\n\nFinally, $K^{\\prime}$ is the second meeting point of $A^{\\prime} D^{\\prime}$ and $\\Gamma^{\\prime}$, and $Y^{\\prime}$ is the second meeting point of the circle $\\left(D^{\\prime} K^{\\prime} E^{\\prime}\\right)$ and the line $A^{\\prime} E^{\\prime}$, We have $\\sphericalangle\\left(Y^{\\prime} K^{\\prime}, K^{\\prime} A^{\\prime}\\right)=\\sphericalangle\\left(Y^{\\prime} E^{\\prime}, E^{\\prime} D^{\\prime}\\right)=$ $\\sphericalangle\\left(Y^{\\prime} A^{\\prime}, A^{\\prime} L^{\\prime}\\right)$, so $A^{\\prime} L^{\\prime}$ is tangent to the circumcircle $\\omega$ of the triangle $Y^{\\prime} A^{\\prime} K^{\\prime}$.\n\nLet $O$ and $O^{*}$ be the centres of $\\Omega$ and $\\omega$, respectively. Then $O^{*} A^{\\prime} \\perp A^{\\prime} L^{\\prime} \\perp B^{\\prime} O$. The projections of vectors $\\overrightarrow{O^{*} A^{\\prime}}$ and $\\overrightarrow{B^{\\prime} O}$ onto $K^{\\prime} D^{\\prime}$ are equal to $\\overrightarrow{K^{\\prime} A^{\\prime}} / 2=\\overrightarrow{K^{\\prime} D^{\\prime}} / 2-\\overrightarrow{A^{\\prime} D^{\\prime}} / 2$. So $\\overrightarrow{O^{*} A^{\\prime}}=\\overrightarrow{B^{\\prime} O}$, or equivalently $\\overrightarrow{A^{\\prime} O}=\\overrightarrow{O^{*} B^{\\prime}}$. Projecting this equality onto $A^{\\prime} C$, we see that the projection of $\\overrightarrow{O^{*} B^{\\prime}}$ equals $\\overrightarrow{A^{\\prime} C} / 2$. Since $O^{*}$ is projected to the midpoint of $A^{\\prime} Y^{\\prime}$, this yields that $B^{\\prime}$ is projected to the midpoint of $C Y^{\\prime}$, i.e., $B^{\\prime} Y^{\\prime}=B^{\\prime} C$ and $\\angle B^{\\prime} Y^{\\prime} C=\\angle B^{\\prime} C Y^{\\prime}$. In the original figure, this rewrites as $\\angle C B Y=\\angle B C Y$, so $Y$ lies on the perpendicular bisector of $B C$, as desired.\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 135 Let $\omega$ be the circumcircle of a triangle $A B C$, and let $\Omega_{A}$ be its excircle which is tangent to the segment $B C$. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_{A}$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_{A}$ at $X$ and $Y$, respectively. The tangent line at $P$ to the circumcircle of the triangle $A P X$ intersects the tangent line at $Q$ to the circumcircle of the triangle $A Q Y$ at a point $R$. Prove that $A R \perp B C$. "['Let $D$ be the point of tangency of $B C$ and $\\Omega_{A}$. Let $D^{\\prime}$ be the point such that $D D^{\\prime}$ is a diameter of $\\Omega_{A}$. Let $R^{\\prime}$ be (the unique) point such that $A R^{\\prime} \\perp B C$ and $R^{\\prime} D^{\\prime} \\| B C$. We shall prove that $R^{\\prime}$ coincides with $R$.\n\nLet $P X$ intersect $A B$ and $D^{\\prime} R^{\\prime}$ at $S$ and $T$, respectively. Let $U$ be the ideal common point of the parallel lines $B C$ and $D^{\\prime} R^{\\prime}$. Note that the (degenerate) hexagon $A S X T U C$ is circumscribed around $\\Omega_{A}$, hence by the Brianchon theorem $A T, S U$, and $X C$ concur at a point which we denote by $V$. Then $V S \\| B C$. It follows that $\\sphericalangle(S V, V X)=\\sphericalangle(B C, C X)=$ $\\sphericalangle(B A, A X)$, hence $A X S V$ is cyclic. Therefore, $\\sphericalangle(P X, X A)=\\sphericalangle(S V, V A)=\\sphericalangle\\left(R^{\\prime} T, T A\\right)$. Since $\\angle A P T=\\angle A R^{\\prime} T=90^{\\circ}$, the quadrilateral $A P R^{\\prime} T$ is cyclic. Hence,\n\n$$\n\\sphericalangle(X A, A P)=90^{\\circ}-\\sphericalangle(P X, X A)=90^{\\circ}-\\sphericalangle\\left(R^{\\prime} T, T A\\right)=\\sphericalangle\\left(T A, A R^{\\prime}\\right)=\\sphericalangle\\left(T P, P R^{\\prime}\\right)\n$$\n\nIt follows that $P R^{\\prime}$ is tangent to the circle $(A P X)$.\n\nAnalogous argument shows that $Q R^{\\prime}$ is tangent to the circle $(A Q Y)$. Therefore, $R=R^{\\prime}$ and $A R \\perp B C$.\n\n' ""Let $J$ and $r$ be the center and the radius of $\\Omega_{A}$. Denote the diameter of $\\omega$ by $d$ and its center by $O$. By Euler's formula, $O J^{2}=(d / 2)^{2}+d r$, so the power of $J$ with respect to $\\omega$ equals $d r$.\n\n\n\nLet $J X$ intersect $\\omega$ again at $L$. Then $J L=d$. Let $L K$ be a diameter of $\\omega$ and let $M$ be the midpoint of $J K$. Since $J L=L K$, we have $\\angle L M K=90^{\\circ}$, so $M$ lies on $\\omega$. Let $R^{\\prime}$ be the point such that $R^{\\prime} P$ is tangent to the circle $(A P X)$ and $A R^{\\prime} \\perp B C$. Note that the line $A R^{\\prime}$ is symmetric to the line $A O$ with respect to $A J$.\n\n\nLemma. Let $M$ be the midpoint of the side $J K$ in a triangle $A J K$. Let $X$ be a point on the circle $(A M K)$ such that $\\angle J X K=90^{\\circ}$. Then there exists a point $T$ on the line $K X$ such that the triangles $A K J$ and $A J T$ are similar and equioriented.\n\nProof. Note that $M X=M K$. We construct a parallelogram $A J N K$. Let $T$ be a point on $K X$ such that $\\sphericalangle(N J, J A)=\\sphericalangle(K J, J T)$. Then\n\n$$\n\\sphericalangle(J N, N A)=\\sphericalangle(K A, A M)=\\sphericalangle(K X, X M)=\\sphericalangle(M K, K X)=\\sphericalangle(J K, K T) .\n$$\n\nSo there exists a spiral similarity with center $J$ mapping the triangle $A J N$ to the triangle $T J K$. Therefore, the triangles $N J K$ and $A J T$ are similar and equioriented. It follows that the triangles $A K J$ and $A J T$ are similar and equioriented.\n\n\n\n\n\nBack to the problem, we construct a point $T$ as in the lemma. We perform the composition $\\phi$ of inversion with centre $A$ and radius $A J$ and reflection in $A J$. It is known that every triangle $A E F$ is similar and equioriented to $A \\phi(F) \\phi(E)$.\n\nSo $\\phi(K)=T$ and $\\phi(T)=K$. Let $P^{*}=\\phi(P)$ and $R^{*}=\\phi\\left(R^{\\prime}\\right)$. Observe that $\\phi(T K)$ is a circle with diameter $A P^{*}$. Let $A A^{\\prime}$ be a diameter of $\\omega$. Then $P^{*} K \\perp A K \\perp A^{\\prime} K$, so $A^{\\prime}$ lies on $P^{*} K$. The triangles $A R^{\\prime} P$ and $A P^{*} R^{*}$ are similar and equioriented, hence\n\n$\\sphericalangle\\left(A A^{\\prime}, A^{\\prime} P^{*}\\right)=\\sphericalangle\\left(A A^{\\prime}, A^{\\prime} K\\right)=\\sphericalangle(A X, X P)=\\sphericalangle(A X, X P)=\\sphericalangle\\left(A P, P R^{\\prime}\\right)=\\sphericalangle\\left(A R^{*}, R^{*} P^{*}\\right)$,\n\nso $A, A^{\\prime}, R^{*}$, and $P^{*}$ are concyclic. Since $A^{\\prime}$ and $R^{*}$ lie on $A O$, we obtain $R^{*}=A^{\\prime}$. So $R^{\\prime}=\\phi\\left(A^{\\prime}\\right)$, and $\\phi\\left(A^{\\prime}\\right) P$ is tangent to the circle $(A P X)$.\n\nAn identical argument shows that $\\phi\\left(A^{\\prime}\\right) Q$ is tangent to the circle $(A Q Y)$. Therefore, $R=$ $\\phi\\left(A^{\\prime}\\right)$ and $A R \\perp B C$.""]" [] Text-only Competition True Theorem proof Number Theory Math English 136 Let $n \geqslant 100$ be an integer. The numbers $n, n+1, \ldots, 2 n$ are written on $n+1$ cards, one number per card. The cards are shuffled and divided into two piles. Prove that one of the piles contains two cards such that the sum of their numbers is a perfect square. ['To solve the problem it suffices to find three squares and three cards with numbers $a, b, c$ on them such that pairwise sums $a+b, b+c, a+c$ are equal to the chosen squares. By choosing the three consecutive squares $(2 k-1)^{2},(2 k)^{2},(2 k+1)^{2}$ we arrive at the triple\n\n$$\n(a, b, c)=\\left(2 k^{2}-4 k, \\quad 2 k^{2}+1, \\quad 2 k^{2}+4 k\\right)\n$$\n\nWe need a value for $k$ such that\n\n$$\nn \\leqslant 2 k^{2}-4 k, \\quad \\text { and } \\quad 2 k^{2}+4 k \\leqslant 2 n \\text {. }\n$$\n\nA concrete $k$ is suitable for all $n$ with\n\n$$\nn \\in\\left[k^{2}+2 k, 2 k^{2}-4 k+1\\right]=: I_{k} .\n$$\n\nFor $k \\geqslant 9$ the intervals $I_{k}$ and $I_{k+1}$ overlap because\n\n$$\n(k+1)^{2}+2(k+1) \\leqslant 2 k^{2}-4 k+1 .\n$$\n\nHence $I_{9} \\cup I_{10} \\cup \\ldots=[99, \\infty)$, which proves the statement for $n \\geqslant 99$.'] [] Text-only Competition True Theorem proof Number Theory Math English 137 "Prove that there are only finitely many quadruples $(a, b, c, n)$ of positive integers such that $$ n !=a^{n-1}+b^{n-1}+c^{n-1} . $$" "[""For fixed $n$ there are clearly finitely many solutions; we will show that there is no solution with $n>100$. So, assume $n>100$. By the AM-GM inequality,\n\n$$\n\\begin{aligned}\nn ! & =2 n(n-1)(n-2)(n-3) \\cdot(3 \\cdot 4 \\cdots(n-4)) \\\\\n& \\leqslant 2(n-1)^{4}\\left(\\frac{3+\\cdots+(n-4)}{n-6}\\right)^{n-6}=2(n-1)^{4}\\left(\\frac{n-1}{2}\\right)^{n-6}<\\left(\\frac{n-1}{2}\\right)^{n-1}\n\\end{aligned}\n$$\n\nthus $a, b, c<(n-1) / 2$.\n\nFor every prime $p$ and integer $m \\neq 0$, let $\\nu_{p}(m)$ denote the $p$-adic valuation of $m$; that is, the greatest non-negative integer $k$ for which $p^{k}$ divides $m$. Legendre's formula states that\n\n$$\n\\nu_{p}(n !)=\\sum_{s=1}^{\\infty}\\left\\lfloor\\frac{n}{p^{s}}\\right\\rfloor\n$$\n\nand a well-know corollary of this formula is that\n\n$$\n\\nu_{p}(n !)<\\sum_{s=1}^{\\infty} \\frac{n}{p^{s}}=\\frac{n}{p-1}\n\\tag{1}\n$$\n\nIf $n$ is odd then $a^{n-1}, b^{n-1}, c^{n-1}$ are squares, and by considering them modulo 4 we conclude that $a, b$ and $c$ must be even. Hence, $2^{n-1} \\mid n$ ! but that is impossible for odd $n$ because $\\nu_{2}(n !)=\\nu_{2}((n-1) !)a+b$. On the other hand, $p \\mid c$ implies that $p100$.""]" [] Text-only Competition True Theorem proof Number Theory Math English 138 Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of positive integers such that $a_{n+2 m}$ divides $a_{n}+a_{n+m}$ for all positive integers $n$ and $m$. Prove that this sequence is eventually periodic, i.e. there exist positive integers $N$ and $d$ such that $a_{n}=a_{n+d}$ for all $n>N$. ['We will make repeated use of the following simple observation:\n\nLemma 1. If a positive integer $d$ divides $a_{n}$ and $a_{n-m}$ for some $m$ and $n>2 m$, it also divides $a_{n-2 m}$. If $d$ divides $a_{n}$ and $a_{n-2 m}$, it also divides $a_{n-m}$.\n\nProof. Both parts are obvious since $a_{n}$ divides $a_{n-2 m}+a_{n-m}$.\n\nClaim. The sequence $\\left(a_{n}\\right)$ is bounded.\n\nProof. Suppose the contrary. Then there exist infinitely many indices $n$ such that $a_{n}$ is greater than each of the previous terms $a_{1}, a_{2}, \\ldots, a_{n-1}$. Let $a_{n}=k$ be such a term, $n>10$. For each $s<\\frac{n}{2}$ the number $a_{n}=k$ divides $a_{n-s}+a_{n-2 s}<2 k$, therefore\n\n$$\na_{n-s}+a_{n-2 s}=k \\text {. }\n$$\n\nIn particular,\n\n$$\na_{n}=a_{n-1}+a_{n-2}=a_{n-2}+a_{n-4}=a_{n-4}+a_{n-8}\n$$\n\nthat is, $a_{n-1}=a_{n-4}$ and $a_{n-2}=a_{n-8}$. It follows from Lemma 1 that $a_{n-1}$ divides $a_{n-1-3 s}$ for $3 sN$, then $a_{j}=t$ for infinitely many $j$.\n\nClearly the sequence $\\left(a_{n+N}\\right)_{n>0}$ satisfies the divisibility condition, and it is enough to prove that this sequence is eventually periodic. Thus truncating the sequence if necessary, we can assume that each number appears infinitely many times in the sequence. Let $k$ be the maximum number appearing in the sequence.\n\nLemma 2. If a positive integer $d$ divides $a_{n}$ for some $n$, then the numbers $i$ such that $d$ divides $a_{i}$ form an arithmetical progression with an odd difference.\n\nProof. Let $i_{1}1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $$ has no positive roots." "['We first prove that, for $x>0$,\n\n$$\na_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i}\n\\tag{1}\n$$\n\nwith equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$.\n\nIf $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields\n\n$$\n\\frac{(x+1)+\\overbrace{1+1+\\cdots+1}^{a_{i}-1 \\text { ones }}}{a_{i}} \\geqslant \\sqrt[a_{i}]{(x+1) \\cdot 1^{a_{i}-1}} \\Longrightarrow a_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i} .\n$$\n\nSince $x+1>1$, the inequality is strict for $a_{i}>1$.\n\nMultiplying the inequalities (1) for $i=1,2, \\ldots, n$ yields\n\n$$\n\\prod_{i=1}^{n} a_{i}(x+1)^{1 / a_{i}} \\leqslant \\prod_{i=1}^{n}\\left(x+a_{i}\\right) \\Longleftrightarrow M(x+1)^{\\sum_{i=1}^{n} 1 / a_{i}}-\\prod_{i=1}^{n}\\left(x+a_{i}\\right) \\leqslant 0 \\Longleftrightarrow P(x) \\leqslant 0\n$$\n\nwith equality iff $a_{i}=1$ for all $i \\in\\{1,2, \\ldots, n\\}$. But this implies $M=1$, which is not possible. Hence $P(x)<0$ for all $x \\in \\mathbb{R}^{+}$, and $P$ has no positive roots.' 'We will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$.\n\nIndeed, since $a_{j} \\geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \\ldots a_{n}=M>1$ ), we have $k=\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}1$, if (2) is true for a given $r2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$." "['Set $D=2017$. Denote\n\n$$\nM_{n}=\\max _{kD$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$.\n\n(i) There exist indices $p$ and $q$ such that $a_{n}=-\\left(a_{p}+a_{q}\\right)$ and $p+q=n$. Since $a_{p}, a_{q} \\leqslant M_{n}$, we have $a_{n} \\geqslant-2 M_{n}$.\n\n(ii) On the other hand, choose an index $kD$ is lucky if $m_{n} \\leqslant 2 M_{n}$. Two cases are possible.\n\nCase 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \\leqslant 2 M_{n}$ and $M_{n} \\leqslant M_{n+1} \\leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \\leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \\leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$.\n\nCase 2. Assume now that there is no lucky index, i.e., $2 M_{n}D$. Then (1) shows that for all $n>D$ we have $m_{n} \\leqslant m_{n+1} \\leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}a_{i} \\text { for each } i-2 \\ell, \\quad \\text { and } \\quad n>D\n\\tag{2}\n$$\n\nWe first show that there must be some good index $n$. By assumption, we may take an index $N$ such that $a_{N}>\\max \\{L,-2 \\ell\\}$. Choose $n$ minimally such that $a_{n}=\\max \\left\\{a_{1}, a_{2}, \\ldots, a_{N}\\right\\}$. Now, the first condition in (2) is satisfied because of the minimality of $n$, and the second and third conditions are satisfied because $a_{n} \\geqslant a_{N}>L,-2 \\ell$, and $L \\geqslant a_{i}$ for every $i$ such that $1 \\leqslant i \\leqslant D$.\n\n\n\nLet $n$ be a good index. We derive a contradiction. We have that\n\n$$\na_{n}+a_{u}+a_{v} \\leqslant 0\\tag{3}\n$$\n\nwhenever $u+v=n$.\n\nWe define the index $u$ to maximize $a_{u}$ over $1 \\leqslant u \\leqslant n-1$, and let $v=n-u$. Then, we note that $a_{u} \\geqslant a_{v}$ by the maximality of $a_{u}$.\n\nAssume first that $v \\leqslant D$. Then, we have that\n\n$$\na_{N}+2 \\ell \\leqslant 0\n$$\n\nbecause $a_{u} \\geqslant a_{v} \\geqslant \\ell$. But this contradicts our assumption that $a_{n}>-2 \\ell$ in the second criteria of (2).\n\nNow assume that $v>D$. Then, there exist some indices $w_{1}, w_{2}$ summing up to $v$ such that\n\n$$\na_{v}+a_{w_{1}}+a_{w_{2}}=0\n$$\n\nBut combining this with (3), we have\n\n$$\na_{n}+a_{u} \\leqslant a_{w_{1}}+a_{w_{2}}\n$$\n\nBecause $a_{n}>a_{u}$, we have that $\\max \\left\\{a_{w_{1}}, a_{w_{2}}\\right\\}>a_{u}$. But since each of the $w_{i}$ is less than $v$, this contradicts the maximality of $a_{u}$.']" [] Text-only Competition True Theorem proof Algebra Math English 142 "Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text { for } n=1,2, \ldots\right. $$ Prove that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to 2017 ." "['The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$.\n\nLemma. We have $a_{n} \\geqslant 1$ for all $n \\geqslant 1$.\n\nProof. Let us suppose otherwise in order to obtain a contradiction. Let\n\n$$\nn \\geqslant 1 \\text { be the smallest integer with } a_{n} \\leqslant 0 \\text {. }\n\\tag{1}\n$$\n\nNote that $n \\geqslant 2$. It follows that $a_{n-1} \\geqslant 1$ and $a_{n-2} \\geqslant 0$. Thus we cannot have $a_{n}=$ $a_{n-1} b_{n-1}+a_{n-2}$, so we must have $a_{n}=a_{n-1} b_{n-1}-a_{n-2}$. Since $a_{n} \\leqslant 0$, we have $a_{n-1} \\leqslant a_{n-2}$. Thus we have $a_{n-2} \\geqslant a_{n-1} \\geqslant a_{n}$.\n\nLet\n\n$$\nr \\text { be the smallest index with } a_{r} \\geqslant a_{r+1} \\geqslant a_{r+2} \\text {. }\n\\tag{2}\n$$\n\nThen $r \\leqslant n-2$ by the above, but also $r \\geqslant 2$ : if $b_{1}=1$, then $a_{2}=a_{1}=1$ and $a_{3}=a_{2} b_{2}+a_{1}>a_{2}$; if $b_{1}>1$, then $a_{2}=b_{1}>1=a_{1}$.\n\nBy the minimal choice (2) of $r$, it follows that $a_{r-1}0$. In order to have $a_{r+1} \\geqslant a_{r+2}$, we must have $a_{r+2}=a_{r+1} b_{r+1}-a_{r}$ so that $b_{r} \\geqslant 2$. Putting everything together, we conclude that\n\n$$\na_{r+1}=a_{r} b_{r} \\pm a_{r-1} \\geqslant 2 a_{r}-a_{r-1}=a_{r}+\\left(a_{r}-a_{r-1}\\right)>a_{r},\n$$\n\nwhich contradicts (2).\n\nTo complete the problem, we prove that $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$ by induction. The cases $n=0,1$ are given. Assume it is true for all non-negative integers strictly less than $n$, where $n \\geqslant 2$. There are two cases:\n\nCase 1: $b_{n-1}=1$.\n\nThen $a_{n+1}=a_{n} b_{n}+a_{n-1}$. By the inductive assumption one of $a_{n-1}, a_{n}$ is at least $n-1$ and the other, by the lemma, is at least 1 . Hence\n\n$$\na_{n+1}=a_{n} b_{n}+a_{n-1} \\geqslant a_{n}+a_{n-1} \\geqslant(n-1)+1=n .\n$$\n\nThus $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$, as desired.\n\nCase 2: $b_{n-1}>1$.\n\nSince we defined $b_{0}=1$ there is an index $r$ with $1 \\leqslant r \\leqslant n-1$ such that\n\n$$\nb_{n-1}, b_{n-2}, \\ldots, b_{r} \\geqslant 2 \\quad \\text { and } \\quad b_{r-1}=1\n$$\n\nWe have $a_{r+1}=a_{r} b_{r}+a_{r-1} \\geqslant 2 a_{r}+a_{r-1}$. Thus $a_{r+1}-a_{r} \\geqslant a_{r}+a_{r-1}$.\n\nNow we claim that $a_{r}+a_{r-1} \\geqslant r$. Indeed, this holds by inspection for $r=1$; for $r \\geqslant 2$, one of $a_{r}, a_{r-1}$ is at least $r-1$ by the inductive assumption, while the other, by the lemma, is at least 1 . Hence $a_{r}+a_{r-1} \\geqslant r$, as claimed, and therefore $a_{r+1}-a_{r} \\geqslant r$ by the last inequality in the previous paragraph.\n\nSince $r \\geqslant 1$ and, by the lemma, $a_{r} \\geqslant 1$, from $a_{r+1}-a_{r} \\geqslant r$ we get the following two inequalities:\n\n$$\na_{r+1} \\geqslant r+1 \\quad \\text { and } \\quad a_{r+1}>a_{r} \\text {. }\n$$\n\n\n\nNow observe that\n\n$$\na_{m}>a_{m-1} \\Longrightarrow a_{m+1}>a_{m} \\text { for } m=r+1, r+2, \\ldots, n-1\n$$\n\nsince $a_{m+1}=a_{m} b_{m}-a_{m-1} \\geqslant 2 a_{m}-a_{m-1}=a_{m}+\\left(a_{m}-a_{m-1}\\right)>a_{m}$. Thus\n\n$$\na_{n}>a_{n-1}>\\cdots>a_{r+1} \\geqslant r+1 \\Longrightarrow a_{n} \\geqslant n .\n$$\n\nSo $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$, as desired.' 'We say that an index $n>1$ is bad if $b_{n-1}=1$ and $b_{n-2}>1$; otherwise $n$ is good. The value of $b_{0}$ is irrelevant to the definition of $\\left(a_{n}\\right)$ since $a_{0}=0$; so we assume that $b_{0}>1$.\n\nLemma 1. (a) $a_{n} \\geqslant 1$ for all $n>0$.\n\n(b) If $n>1$ is good, then $a_{n}>a_{n-1}$.\n\nProof. Induction on $n$. In the base cases $n=1,2$ we have $a_{1}=1 \\geqslant 1, a_{2}=b_{1} a_{1} \\geqslant 1$, and finally $a_{2}>a_{1}$ if 2 is good, since in this case $b_{1}>1$.\n\nNow we assume that the lemma statement is proved for $n=1,2, \\ldots, k$ with $k \\geqslant 2$, and prove it for $n=k+1$. Recall that $a_{k}$ and $a_{k-1}$ are positive by the induction hypothesis.\n\nCase 1: $k$ is bad.\n\nWe have $b_{k-1}=1$, so $a_{k+1}=b_{k} a_{k}+a_{k-1} \\geqslant a_{k}+a_{k-1}>a_{k} \\geqslant 1$, as required.\n\nCase 2: $k$ is good.\n\nWe already have $a_{k}>a_{k-1} \\geqslant 1$ by the induction hypothesis. We consider three easy subcases.\n\nSubcase 2.1: $b_{k}>1$.\n\nThen $a_{k+1} \\geqslant b_{k} a_{k}-a_{k-1} \\geqslant a_{k}+\\left(a_{k}-a_{k-1}\\right)>a_{k} \\geqslant 1$.\n\nSubcase 2.2: $b_{k}=b_{k-1}=1$.\n\nThen $a_{k+1}=a_{k}+a_{k-1}>a_{k} \\geqslant 1$.\n\nSubcase 2.3: $b_{k}=1$ but $b_{k-1}>1$.\n\nThen $k+1$ is bad, and we need to prove only (a), which is trivial: $a_{k+1}=a_{k}-a_{k-1} \\geqslant 1$.\n\nSo, in all three subcases we have verified the required relations.\n\nLemma 2. Assume that $n>1$ is bad. Then there exists a $j \\in\\{1,2,3\\}$ such that $a_{n+j} \\geqslant$ $a_{n-1}+j+1$, and $a_{n+i} \\geqslant a_{n-1}+i$ for all $1 \\leqslant i0: b_{n+i-1}>1\\right\\}\n$$\n\n(possibly $m=+\\infty$ ). We claim that $j=\\min \\{m, 3\\}$ works. Again, we distinguish several cases, according to the value of $m$; in each of them we use Lemma 1 without reference.\n\nCase 1: $m=1$, so $b_{n}>1$.\n\nThen $a_{n+1} \\geqslant 2 a_{n}+a_{n-1} \\geqslant a_{n-1}+2$, as required.\n\nCase 2: $m=2$, so $b_{n}=1$ and $b_{n+1}>1$.\n\nThen we successively get\n\n$$\n\\begin{gathered}\na_{n+1}=a_{n}+a_{n-1} \\geqslant a_{n-1}+1 \\\\\na_{n+2} \\geqslant 2 a_{n+1}+a_{n} \\geqslant 2\\left(a_{n-1}+1\\right)+a_{n}=a_{n-1}+\\left(a_{n-1}+a_{n}+2\\right) \\geqslant a_{n-1}+4\n\\end{gathered}\n$$\n\nwhich is even better than we need.\n\n\n\nCase 3: $m>2$, so $b_{n}=b_{n+1}=1$.\n\nThen we successively get\n\n$$\n\\begin{gathered}\na_{n+1}=a_{n}+a_{n-1} \\geqslant a_{n-1}+1, \\quad a_{n+2}=a_{n+1}+a_{n} \\geqslant a_{n-1}+1+a_{n} \\geqslant a_{n-1}+2, \\\\\na_{n+3} \\geqslant a_{n+2}+a_{n+1} \\geqslant\\left(a_{n-1}+1\\right)+\\left(a_{n-1}+2\\right) \\geqslant a_{n-1}+4\n\\end{gathered}\n$$\n\nas required.\n\nLemmas 1(b) and 2 provide enough information to prove that $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$ for all $n$ and, moreover, that $a_{n} \\geqslant n$ often enough. Indeed, assume that we have found some $n$ with $a_{n-1} \\geqslant n-1$. If $n$ is good, then by Lemma 1(b) we have $a_{n} \\geqslant n$ as well. If $n$ is bad, then Lemma 2 yields $\\max \\left\\{a_{n+i}, a_{n+i+1}\\right\\} \\geqslant a_{n-1}+i+1 \\geqslant n+i$ for all $0 \\leqslant i0$, we have $f(x)+y=f(y)+x$. Prove that $f(x)+y \leqslant f(y)+x$ whenever $x>y$." "['Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows:\n\nFor every $x, y \\in \\mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$.\n\nThis condition may in turn be rewritten in the following form:\n$$\n\\text{If }g(x) \\neq g(y)\\text{, then the number x+y lies (non-strictly)}\\text{ between }g(x)\\text{ and }g(y). \n\\tag{*}\n$$\nNotice here that the function $g_{1}(x)=-g(-x)$ also satisfies $(*)$, since\n\n$$\n\\begin{gathered}\ng_{1}(x) \\neq g_{1}(y) \\Longrightarrow \\quad g(-x) \\neq g(-y) \\Longrightarrow \\quad-(x+y) \\text { lies between } g(-x) \\text { and } g(-y) \\\\\n\\Longrightarrow \\quad x+y \\text { lies between } g_{1}(x) \\text { and } g_{1}(y) .\n\\end{gathered}\n$$\n\nOn the other hand, the relation we need to prove reads now as\n\n$$\ng(x) \\leqslant g(y) \\quad \\text { whenever } xX$. Similarly, if $X>2 x$, then $g$ attains at most two values on $[x ; X-x)$ - namely, $X$ and, possibly, some $YX$ and hence by $(*)$ we get $X \\leqslant a+x \\leqslant g(a)$.\n\nNow, for any $b \\in(X-x ; x)$ with $g(b) \\neq X$ we similarly get $b+x \\leqslant g(b)$. Therefore, the number $a+b$ (which is smaller than each of $a+x$ and $b+x$ ) cannot lie between $g(a)$ and $g(b)$, which by $(*)$ implies that $g(a)=g(b)$. Hence $g$ may attain only two values on $(X-x ; x]$, namely $X$ and $g(a)>X$.\n\nTo prove the second claim, notice that $g_{1}(-x)=-X<2 \\cdot(-x)$, so $g_{1}$ attains at most two values on $(-X+x,-x]$, i.e., $-X$ and, possibly, some $-Y>-X$. Passing back to $g$, we get what we need.\n\nLemma 2. If $X<2 x$, then $g$ is constant on $(X-x ; x)$. Similarly, if $X>2 x$, then $g$ is constant on $(x ; X-x)$.\n\nProof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist $a, b \\in(X-x ; x)$ with $g(a) \\neq g(b)$; by Lemma 1 , we may assume that $g(a)=X$ and $Y=g(b)>X$.\n\nNotice that $\\min \\{X-a, X-b\\}>X-x$, so there exists a $u \\in(X-x ; x)$ such that $u<\\min \\{X-a, X-b\\}$. By Lemma 1, we have either $g(u)=X$ or $g(u)=Y$. In the former case, by (*) we have $X \\leqslant u+b \\leqslant Y$ which contradicts $u2 x$, then $g(a)=X$ for all $a \\in(x ; X-x)$.\n\nProof. Again, we only prove the first claim.\n\nBy Lemmas 1 and 2, this claim may be violated only if $g$ takes on a constant value $Y>X$ on $(X-x, x)$. Choose any $a, b \\in(X-x ; x)$ with $a2 a$. Applying Lemma 2 to $a$ in place of $x$, we obtain that $g$ is constant on $(a, Y-a)$. By (2) again, we have $x \\leqslant Y-bg(y)$ for some $xc$. Now, for any $x$ with $g(x)=c$, by $(*)$ we have $c \\leqslant x+y \\leqslant g(y)$, or $c-y \\leqslant x \\leqslant g(y)-y$. Since $c$ and $y$ are constant, we get what we need.\n\nIf $g(y)-c$. By the above arguments, we obtain that all the solutions of $g_{1}(-x)=-c$ are bounded, which is equivalent to what we need.\n\nAs an immediate consequence, the function $g$ takes on infinitely many values, which shows that the next claim is indeed widely applicable.\n\nClaim 2. If $g(x)g(y)$ for some $x'] [] Text-only Competition True Theorem proof Combinatorics Math English 145 Let $n$ be a positive integer. Define a chameleon to be any sequence of $3 n$ letters, with exactly $n$ occurrences of each of the letters $a, b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3 n^{2} / 2$ swaps. "['To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps.\n\nFor any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \\geqslant d(X, Z)$ for any three chameleons $X, Y$, and $Z$.\n\nLemma. Consider two chameleons\n\n$$\nP=\\underbrace{a a \\ldots a}_{n} \\underbrace{b b \\ldots b}_{n} \\underbrace{c c \\ldots c}_{n} \\text { and } Q=\\underbrace{c c \\ldots c}_{n} \\underbrace{b b \\ldots b}_{n} \\underbrace{a a \\ldots a}_{n} .\n$$\n\nThen $d(P, Q) \\geqslant 3 n^{2}$.\n\nProof. For any chameleon $X$ and any pair of distinct letters $u, v \\in\\{a, b, c\\}$, we define $f_{u, v}(X)$ to be the number of pairs of positions in $X$ such that the left one is occupied by $u$, and the right one is occupied by $v$. Define $f(X)=f_{a, b}(X)+f_{a, c}(X)+f_{b, c}(X)$. Notice that $f_{a, b}(P)=f_{a, c}(P)=f_{b, c}(P)=n^{2}$ and $f_{a, b}(Q)=f_{a, c}(Q)=f_{b, c}(Q)=0$, so $f(P)=3 n^{2}$ and $f(Q)=0$.\n\nNow consider some swap changing a chameleon $X$ to $X^{\\prime}$; say, the letters $a$ and $b$ are swapped. Then $f_{a, b}(X)$ and $f_{a, b}\\left(X^{\\prime}\\right)$ differ by exactly 1 , while $f_{a, c}(X)=f_{a, c}\\left(X^{\\prime}\\right)$ and $f_{b, c}(X)=f_{b, c}\\left(X^{\\prime}\\right)$. This yields $\\left|f(X)-f\\left(X^{\\prime}\\right)\\right|=1$, i.e., on any swap the value of $f$ changes by 1 . Hence $d(X, Y) \\geqslant$ $|f(X)-f(Y)|$ for any two chameleons $X$ and $Y$. In particular, $d(P, Q) \\geqslant|f(P)-f(Q)|=3 n^{2}$, as desired.\n\nBack to the problem, take any chameleon $X$ and notice that $d(X, P)+d(X, Q) \\geqslant d(P, Q) \\geqslant$ $3 n^{2}$ by the lemma. Consequently, $\\max \\{d(X, P), d(X, Q)\\} \\geqslant \\frac{3 n^{2}}{2}$, which establishes the problem statement.' ""In any chameleon $X$, we enumerate the positions in it from left to right by $1,2, \\ldots, 3 n$. Define $s_{c}(X)$ as the sum of positions occupied by $c$. The value of $s_{c}$ changes by at most 1 on each swap, but this fact alone does not suffice to solve the problem; so we need an improvement.\n\nFor every chameleon $X$, denote by $X_{\\bar{c}}$ the sequence obtained from $X$ by removing all $n$ letters $c$. Enumerate the positions in $X_{\\bar{c}}$ from left to right by $1,2, \\ldots, 2 n$, and define $s_{\\bar{c}, b}(X)$ as the sum of positions in $X_{\\bar{c}}$ occupied by $b$. (In other words, here we consider the positions of the $b$ 's relatively to the $a$ 's only.) Finally, denote\n\n$$\nd^{\\prime}(X, Y):=\\left|s_{c}(X)-s_{c}(Y)\\right|+\\left|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(Y)\\right| .\n$$\n\n\n\nNow consider any swap changing a chameleon $X$ to $X^{\\prime}$. If no letter $c$ is involved into this swap, then $s_{c}(X)=s_{c}\\left(X^{\\prime}\\right)$; on the other hand, exactly one letter $b$ changes its position in $X_{\\bar{c}}$, so $\\left|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}\\left(X^{\\prime}\\right)\\right|=1$. If a letter $c$ is involved into a swap, then $X_{\\bar{c}}=X_{\\bar{c}}^{\\prime}$, so $s_{\\bar{c}, b}(X)=s_{\\bar{c}, b}\\left(X^{\\prime}\\right)$ and $\\left|s_{c}(X)-s_{c}\\left(X^{\\prime}\\right)\\right|=1$. Thus, in all cases we have $d^{\\prime}\\left(X, X^{\\prime}\\right)=1$.\n\nAs in the previous solution, this means that $d(X, Y) \\geqslant d^{\\prime}(X, Y)$ for any two chameleons $X$ and $Y$. Now, for any chameleon $X$ we will indicate a chameleon $Y$ with $d^{\\prime}(X, Y) \\geqslant 3 n^{2} / 2$, thus finishing the solution.\n\nThe function $s_{c}$ attains all integer values from $1+\\cdots+n=\\frac{n(n+1)}{2}$ to $(2 n+1)+\\cdots+3 n=$ $2 n^{2}+\\frac{n(n+1)}{2}$. If $s_{c}(X) \\leqslant n^{2}+\\frac{n(n+1)}{2}$, then we put the letter $c$ into the last $n$ positions in $Y$; otherwise we put the letter $c$ into the first $n$ positions in $Y$. In either case we already have $\\left|s_{c}(X)-s_{c}(Y)\\right| \\geqslant n^{2}$.\n\nSimilarly, $s_{\\bar{c}, b}$ ranges from $\\frac{n(n+1)}{2}$ to $n^{2}+\\frac{n(n+1)}{2}$. So, if $s_{\\bar{c}, b}(X) \\leqslant \\frac{n^{2}}{2}+\\frac{n(n+1)}{2}$, then we put the letter $b$ into the last $n$ positions in $Y$ which are still free; otherwise, we put the letter $b$ into the first $n$ such positions. The remaining positions are occupied by $a$. In any case, we have $\\left|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(Y)\\right| \\geqslant \\frac{n^{2}}{2}$, thus $d^{\\prime}(X, Y) \\geqslant n^{2}+\\frac{n^{2}}{2}=\\frac{3 n^{2}}{2}$, as desired.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 146 "For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\text {th }}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \cdots)}_{B \text { appears } a \text { times }} . $$" ['For any function $g: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ and any subset $X \\subset \\mathbb{Z}_{>0}$, we define $g(X)=$ $\\{g(x): x \\in X\\}$. We have that the image of $f_{X}$ is $f_{X}\\left(\\mathbb{Z}_{>0}\\right)=\\mathbb{Z}_{>0} \\backslash X$. We now show a general lemma about the operation $*$, with the goal of showing that $*$ is associative.\n\nLemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \\circ f_{Y}$ are equal.\n\nProof. We have\n\n$f_{X * Y}\\left(\\mathbb{Z}_{>0}\\right)=\\mathbb{Z}_{>0} \\backslash(X * Y)=\\left(\\mathbb{Z}_{>0} \\backslash X\\right) \\backslash f_{X}(Y)=f_{X}\\left(\\mathbb{Z}_{>0}\\right) \\backslash f_{X}(Y)=f_{X}\\left(\\mathbb{Z}_{>0} \\backslash Y\\right)=f_{X}\\left(f_{Y}\\left(\\mathbb{Z}_{>0}\\right)\\right)$.\n\nThus, the functions $f_{X * Y}$ and $f_{X} \\circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \\circ f_{Y}$.\n\nLemma 1 implies that $*$ is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting\n\n$$\n\\begin{gathered}\n\\mathbb{Z}_{>0} \\backslash((A * B) * C)=f_{(A * B) * C}\\left(\\mathbb{Z}_{>0}\\right)=f_{A * B}\\left(f_{C}\\left(\\mathbb{Z}_{>0}\\right)\\right)=f_{A}\\left(f_{B}\\left(f_{C}\\left(\\mathbb{Z}_{>0}\\right)\\right)\\right) \\\\\n=f_{A}\\left(f_{B * C}\\left(\\mathbb{Z}_{>0}\\right)=f_{A *(B * C)}\\left(\\mathbb{Z}_{>0}\\right)=\\mathbb{Z}_{>0} \\backslash(A *(B * C))\\right.\n\\end{gathered}\n$$\n\nIn light of the associativity of $*$, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation\n\n$$\nX^{* k}=\\underbrace{X *(X * \\cdots *(X *(X * X)) \\ldots)}_{X \\text { appears } k \\text { times }} .\n$$\n\nOur goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma.\n\nLemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$.\n\nProof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \\in X \\backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that\n\n$$\nf_{X}(s)=s+\\left|X \\cap\\left\\{1,2, \\ldots, f_{X}(s)\\right\\}\\right|\n\\tag{1}\n$$\n\nSince $f_{X}(s) \\geqslant s$, we have that\n\n$$\n\\left\\{f_{X}(s)+1, f_{X}(s)+2, \\ldots\\right\\} \\cap X=\\left\\{f_{X}(s)+1, f_{X}(s)+2, \\ldots\\right\\} \\cap Y\n$$\n\nwhich, together with the assumption that $|X|=|Y|$, gives\n\n$$\n\\left|X \\cap\\left\\{1,2, \\ldots, f_{X}(s)\\right\\}\\right|=\\left|Y \\cap\\left\\{1,2, \\ldots, f_{X}(s)\\right\\}\\right|\n\\tag{2}\n$$\n\n\n\nNow consider the equation\n\n$$\nt-|Y \\cap\\{1,2, \\ldots, t\\}|=s\n$$\n\nThis equation is satisfied only when $t \\in\\left[f_{Y}(s), f_{Y}(s+1)\\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \\notin X$ and $f_{X}(s) \\geqslant s$, we have that $f_{X}(s) \\notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$.\n\nFinally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \\notin X * Y$. However, since $s \\in X$, we have $f_{Y}(s) \\in Y * X$, a contradiction.\n\nWe are now ready to finish the proof. Note first of all that $\\left|A^{* b}\\right|=a b=\\left|B^{* a}\\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2 , we have $A^{* b}=B^{* a}$, as desired.'] [] Text-only Competition True Theorem proof Combinatorics Math English 147 Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. "['Throughout the solution, we refer to $\\angle A, \\angle B, \\angle C, \\angle D$, and $\\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \\perp C I$ and, similarly, $A C \\perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \\perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \\perp B C$.\n\nLines $I B$ and $I C$ bisect $\\angle B$ and $\\angle C$, respectively. Since $I A=I C, I B=I D$, and $A B=$ $B C=C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\\angle I A B=\\angle I C B=$ $\\angle C / 2=\\angle A / 2$, so the line $I A$ bisects $\\angle A$. Similarly, the line $I D$ bisects $\\angle D$. Finally, the line $I E$ bisects $\\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon.\n\nThe sum of the internal angles in a pentagon is $540^{\\circ}$, so\n\n$$\n\\angle E=540^{\\circ}-2 \\angle A+2 \\angle B \\text {. }\n$$\n\nIn quadrilateral $A B I E$,\n\n$$\n\\begin{aligned}\n\\angle B I E & =360^{\\circ}-\\angle E A B-\\angle A B I-\\angle A E I=360^{\\circ}-\\angle A-\\frac{1}{2} \\angle B-\\frac{1}{2} \\angle E \\\\\n& =360^{\\circ}-\\angle A-\\frac{1}{2} \\angle B-\\left(270^{\\circ}-\\angle A-\\angle B\\right) \\\\\n& =90^{\\circ}+\\frac{1}{2} \\angle B=90^{\\circ}+\\angle I B C,\n\\end{aligned}\n$$\n\nwhich means that $E I \\perp B C$, completing the proof.\n\n' ""We present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$.\n\nApplying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.""]" [] Text-only Competition True Theorem proof Geometry Math English 148 Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$. "['In the circles $\\Omega$ and $\\Gamma$ we have $\\angle J R S=\\angle J I S=\\angle A R^{\\prime} S$. On the other hand, since $R A$ is tangent to $\\Omega$, we get $\\angle S J R=\\angle S R A$. So the triangles $A R R^{\\prime}$ and $S J R$ are similar, and\n\n$$\n\\frac{R^{\\prime} R}{R J}=\\frac{A R^{\\prime}}{S R}=\\frac{A R^{\\prime}}{S R^{\\prime}}\n$$\n\nThe last relation, together with $\\angle A R^{\\prime} S=\\angle J R R^{\\prime}$, yields $\\triangle A S R^{\\prime} \\sim \\triangle R^{\\prime} J R$, hence $\\angle S A R^{\\prime}=\\angle R R^{\\prime} J$. It follows that $J R^{\\prime}$ is tangent to $\\Gamma$ at $R^{\\prime}$.\n\n' '\n\nWe notice that $\\angle J R S=\\angle J I S=\\angle A R^{\\prime} S$, so we have $R J \\| A R^{\\prime}$.\n\nLet $A^{\\prime}$ be the reflection of $A$ about $S$; then $A R A^{\\prime} R^{\\prime}$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $R A^{\\prime}$.\n\nFrom $\\angle S R^{\\prime} A^{\\prime}=\\angle S R A=\\angle S J R$ we get that the points $S, J, A^{\\prime}, R^{\\prime}$ are concyclic. This proves that $\\angle S R^{\\prime} J=\\angle S A^{\\prime} J=\\angle S A^{\\prime} R=\\angle S A R^{\\prime}$, so $J R^{\\prime}$ is tangent to $\\Gamma$ at $R^{\\prime}$.']" [] Text-only Competition True Theorem proof Geometry Math English 149 Let $O$ be the circumcenter of an acute scalene triangle $A B C$. Line $O A$ intersects the altitudes of $A B C$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $P Q H$ lies on a median of triangle $A B C$. ['Suppose, without loss of generality, that $A B\n\nLet $T$ be the center of $\\omega$ and let lines $A T$ and $B C$ meet at $M$. We will take advantage of the similarity between $A B C$ and $H P Q$ and the fact that $A H$ is tangent to $\\omega$ at $H$, with $A$ on line $P Q$. Consider the corresponding tangent $A S$ to $\\Omega$, with $S \\in B C$. Then $S$ and $A$ correspond to each other in $\\triangle A B C \\sim \\triangle H P Q$, and therefore $\\angle O S M=\\angle O A T=\\angle O A M$. Hence quadrilateral $S A O M$ is cyclic, and since the tangent line $A S$ is perpendicular to $A O$, $\\angle O M S=180^{\\circ}-\\angle O A S=90^{\\circ}$. This means that $M$ is the orthogonal projection of $O$ onto $B C$, which is its midpoint. So $T$ lies on median $A M$ of triangle $A B C$.'] [] Text-only Competition True Theorem proof Geometry Math English 150 In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. "['Denote by $\\Omega$ the circle $A E F P Q$, and denote by $\\gamma$ the circle $P Q M$. Let the line $A D$ meet $\\omega$ again at $T \\neq D$. We will show that $\\gamma$ is tangent to $\\omega$ at $T$.\n\nWe first prove that points $P, Q, M, T$ are concyclic. Let $A^{\\prime}$ be the center of $\\omega$. Since $A^{\\prime} E \\perp A E$ and $A^{\\prime} F \\perp A F, A A^{\\prime}$ is a diameter in $\\Omega$. Let $N$ be the midpoint of $D T$; from $A^{\\prime} D=A^{\\prime} T$ we can see that $\\angle A^{\\prime} N A=90^{\\circ}$ and therefore $N$ also lies on the circle $\\Omega$. Now, from the power of $D$ with respect to the circles $\\gamma$ and $\\Omega$ we get\n\n$$\nD P \\cdot D Q=D A \\cdot D N=2 D M \\cdot \\frac{D T}{2}=D M \\cdot D T\n$$\n\nso $P, Q, M, T$ are concyclic.\n\nIf $E F \\| B C$, then $A B C$ is isosceles and the problem is now immediate by symmetry. Otherwise, let the tangent line to $\\omega$ at $T$ meet line $B C$ at point $R$. The tangent line segments $R D$ and $R T$ have the same length, so $A^{\\prime} R$ is the perpendicular bisector of $D T$; since $N D=N T$, $N$ lies on this perpendicular bisector.\n\nIn right triangle $A^{\\prime} R D, R D^{2}=R N \\cdot R A^{\\prime}=R P \\cdot R Q$, in which the last equality was obtained from the power of $R$ with respect to $\\Omega$. Hence $R T^{2}=R P \\cdot R Q$, which implies that $R T$ is also tangent to $\\gamma$. Because $R T$ is a common tangent to $\\omega$ and $\\gamma$, these two circles are tangent at $T$.\n\n' 'After proving that $P, Q, M, T$ are concyclic, we finish the problem in a different fashion. We only consider the case in which $E F$ and $B C$ are not parallel. Let lines $P Q$ and $E F$ meet at point $R$. Since $P Q$ and $E F$ are radical axes of $\\Omega, \\gamma$ and $\\omega, \\gamma$, respectively, $R$ is the radical center of these three circles.\n\nWith respect to the circle $\\omega$, the line $D R$ is the polar of $D$, and the line $E F$ is the polar of $A$. So the pole of line $A D T$ is $D R \\cap E F=R$, and therefore $R T$ is tangent to $\\omega$.\n\nFinally, since $T$ belongs to $\\gamma$ and $\\omega$ and $R$ is the radical center of $\\gamma, \\omega$ and $\\Omega$, line $R T$ is the radical axis of $\\gamma$ and $\\omega$, and since it is tangent to $\\omega$, it is also tangent to $\\gamma$. Because $R T$ is a common tangent to $\\omega$ and $\\gamma$, these two circles are tangent at $T$.' 'We give an alternative proof that the circles are tangent at the common point $T$. Again, we start from the fact that $P, Q, M, T$ are concyclic. Let point $O$ be the midpoint of diameter $A A^{\\prime}$. Then $M O$ is the midline of triangle $A D A^{\\prime}$, so $M O \\| A^{\\prime} D$. Since $A^{\\prime} D \\perp P Q$, $M O$ is perpendicular to $P Q$ as well.\n\nLooking at circle $\\Omega$, which has center $O, M O \\perp P Q$ implies that $M O$ is the perpendicular bisector of the chord $P Q$. Thus $M$ is the midpoint of arc $\\overparen{P Q}$ from $\\gamma$, and the tangent line $m$ to $\\gamma$ at $M$ is parallel to $P Q$.\n\n\n\nConsider the homothety with center $T$ and ratio $\\frac{T D}{T M}$. It takes $D$ to $M$, and the line $P Q$ to the line $m$. Since the circle that is tangent to a line at a given point and that goes through another given point is unique, this homothety also takes $\\omega$ (tangent to $P Q$ and going through $T$ ) to $\\gamma$ (tangent to $m$ and going through $T$ ). We conclude that $\\omega$ and $\\gamma$ are tangent at $T$.']" [] Text-only Competition True Theorem proof Geometry Math English 151 Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$. "['If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1}\n\nFinally, let $X$ be the second intersection point of $\\omega$ and the line $D E$. Since $B M$ is a diameter in $\\omega$, we have $\\angle B X M=90^{\\circ}$. Moreover,\n\n$$\n\\angle E X M=180^{\\circ}-\\angle M B E=180^{\\circ}-\\angle F B E=\\angle E D F,\n$$\n\nso $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.' 'Define point $M$ as the point opposite to $B$ on circle $\\omega$, and point $R$ as the intersection of lines $A C, A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\\angle A D C$, like in the first solution.\n\nSince $B$ is the midpoint of the arc $\\overparen{A E C}$, the line $B E R$ is the external bisector of $\\angle C E A$. Now we show that the internal angle bisectors of $\\angle A D C$ and $\\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\\angle C E A$, which is line $E M$, meet $A C$ at $S^{\\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\\angle A D C$ and $\\angle C E A$,\n\n$$\nA S: C S=A D: C D=A R: C R=A E: C E=A S^{\\prime}: C S^{\\prime}\n$$\n\nso indeed $S=S^{\\prime}$.\n\nBy $\\angle R D S=\\angle S E R=90^{\\circ}$ the points $R, S, D$ and $E$ are concyclic.\n\n\n\nNow let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\\angle E X B=\\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\\angle E D S=\\angle E R S$ in circle $S R D E$, and $\\angle E R S=\\angle E M B$ because $S R \\perp B M$ and $E R \\perp M E$. Therefore, $\\angle E X B=\\angle E M B$, so indeed, the point $X$ lies on $\\omega$.']" [] Text-only Competition True Theorem proof Geometry Math English 152 "Let $n \geqslant 3$ be an integer. Two regular $n$-gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive. (That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary from the other vertices of $\mathcal{A}$.)" "['By a polygon we always mean its interior together with its boundary.\n\nWe start with finding a regular $n$-gon $\\mathcal{C}$ which $(i)$ is inscribed into $\\mathcal{B}$ (that is, all vertices of $\\mathcal{C}$ lie on the perimeter of $\\mathcal{B}$ ); and $($ ii) is either a translation of $\\mathcal{A}$, or a homothetic image of $\\mathcal{A}$ with a positive factor.\n\nSuch a polygon may be constructed as follows. Let $O_{A}$ and $O_{B}$ be the centers of $\\mathcal{A}$ and $\\mathcal{B}$, respectively, and let $A$ be an arbitrary vertex of $\\mathcal{A}$. Let $\\overrightarrow{O_{B} C}$ be the vector co-directional to $\\overrightarrow{O_{A} A}$, with $C$ lying on the perimeter of $\\mathcal{B}$. The rotations of $C$ around $O_{B}$ by multiples of $2 \\pi / n$ form the required polygon. Indeed, it is regular, inscribed into $\\mathcal{B}$ (due to the rotational symmetry of $\\mathcal{B}$ ), and finally the translation/homothety mapping $\\overrightarrow{O_{A} A}$ to $\\overrightarrow{O_{B} C}$ maps $\\mathcal{A}$ to $\\mathcal{C}$.\n\nNow we separate two cases.\n\n\n\nConstruction of $\\mathcal{C}$\n\n\n\nCase 1: Translation\n\nCase 1: $\\mathcal{C}$ is a translation of $\\mathcal{A}$ by a vector $\\vec{v}$.\n\nDenote by $t$ the translation transform by vector $\\vec{v}$. We need to prove that the vertices of $\\mathcal{C}$ which stay in $\\mathcal{B}$ under $t$ are consecutive. To visualize the argument, we refer the plane to Cartesian coordinates so that the $x$-axis is co-directional with $\\vec{v}$. This way, the notions of right/left and top/bottom are also introduced, according to the $x$ - and $y$-coordinates, respectively.\n\nLet $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ be the top and the bottom vertices of $\\mathcal{B}$ (if several vertices are extremal, we take the rightmost of them). They split the perimeter of $\\mathcal{B}$ into the right part $\\mathcal{B}_{\\mathrm{R}}$ and the left part $\\mathcal{B}_{\\mathrm{L}}$ (the vertices $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ are assumed to lie in both parts); each part forms a connected subset of the perimeter of $\\mathcal{B}$. So the vertices of $\\mathcal{C}$ are also split into two parts $\\mathcal{C}_{\\mathrm{L}} \\subset \\mathcal{B}_{\\mathrm{L}}$ and $\\mathcal{C}_{\\mathrm{R}} \\subset \\mathcal{B}_{\\mathrm{R}}$, each of which consists of consecutive vertices.\n\nNow, all the points in $\\mathcal{B}_{\\mathrm{R}}$ (and hence in $\\mathcal{C}_{\\mathrm{R}}$ ) move out from $\\mathcal{B}$ under $t$, since they are the rightmost points of $\\mathcal{B}$ on the corresponding horizontal lines. It remains to prove that the vertices of $\\mathcal{C}_{\\mathrm{L}}$ which stay in $\\mathcal{B}$ under $t$ are consecutive.\n\nFor this purpose, let $C_{1}, C_{2}$, and $C_{3}$ be three vertices in $\\mathcal{C}_{\\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $t\\left(C_{1}\\right)$ and $t\\left(C_{3}\\right)$ lie in $\\mathcal{B}$; we need to prove that $t\\left(C_{2}\\right) \\in \\mathcal{B}$ as well. Let $A_{i}=t\\left(C_{i}\\right)$. The line through $C_{2}$ parallel to $\\vec{v}$ crosses the segment $C_{1} C_{3}$ to the right of $C_{2}$; this means that this line crosses $A_{1} A_{3}$ to the right of $A_{2}$, so $A_{2}$ lies inside the triangle $A_{1} C_{2} A_{3}$ which is contained in $\\mathcal{B}$. This yields the desired result.\n\nCase 2: $\\mathcal{C}$ is a homothetic image of $\\mathcal{A}$ centered at $X$ with factor $k>0$.\n\n\n\nDenote by $h$ the homothety mapping $\\mathcal{C}$ to $\\mathcal{A}$. We need now to prove that the vertices of $\\mathcal{C}$ which stay in $\\mathcal{B}$ after applying $h$ are consecutive. If $X \\in \\mathcal{B}$, the claim is easy. Indeed, if $k<1$, then the vertices of $\\mathcal{A}$ lie on the segments of the form $X C$ ( $C$ being a vertex of $\\mathcal{C})$ which lie in $\\mathcal{B}$. If $k>1$, then the vertices of $\\mathcal{A}$ lie on the extensions of such segments $X C$ beyond $C$, and almost all these extensions lie outside $\\mathcal{B}$. The exceptions may occur only in case when $X$ lies on the boundary of $\\mathcal{B}$, and they may cause one or two vertices of $\\mathcal{A}$ stay on the boundary of $\\mathcal{B}$. But even in this case those vertices are still consecutive.\n\nSo, from now on we assume that $X \\notin \\mathcal{B}$.\n\nNow, there are two vertices $B_{\\mathrm{T}}$ and $\\mathcal{B}_{\\mathrm{B}}$ of $\\mathcal{B}$ such that $\\mathcal{B}$ is contained in the angle $\\angle B_{\\mathrm{T}} X B_{\\mathrm{B}}$; if there are several options, say, for $B_{\\mathrm{T}}$, then we choose the farthest one from $X$ if $k>1$, and the nearest one if $k<1$. For the visualization purposes, we refer the plane to Cartesian coordinates so that the $y$-axis is co-directional with $\\overrightarrow{B_{\\mathrm{B}} B_{\\mathrm{T}}}$, and $X$ lies to the left of the line $B_{\\mathrm{T}} B_{\\mathrm{B}}$. Again, the perimeter of $\\mathcal{B}$ is split by $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ into the right part $\\mathcal{B}_{\\mathrm{R}}$ and the left part $\\mathcal{B}_{\\mathrm{L}}$, and the set of vertices of $\\mathcal{C}$ is split into two subsets $\\mathcal{C}_{\\mathrm{R}} \\subset \\mathcal{B}_{\\mathrm{R}}$ and $\\mathcal{C}_{\\mathrm{L}} \\subset \\mathcal{B}_{\\mathrm{L}}$.\n\n\n\nCase $2, X$ inside $\\mathcal{B}$\n\n\n\nSubcase 2.1: $k>1$\n\nSubcase 2.1: $k>1$.\n\nIn this subcase, all points from $\\mathcal{B}_{\\mathrm{R}}$ (and hence from $\\mathcal{C}_{\\mathrm{R}}$ ) move out from $\\mathcal{B}$ under $h$, because they are the farthest points of $\\mathcal{B}$ on the corresponding rays emanated from $X$. It remains to prove that the vertices of $\\mathcal{C}_{\\mathrm{L}}$ which stay in $\\mathcal{B}$ under $h$ are consecutive.\n\nAgain, let $C_{1}, C_{2}, C_{3}$ be three vertices in $\\mathcal{C}_{\\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $h\\left(C_{1}\\right)$ and $h\\left(C_{3}\\right)$ lie in $\\mathcal{B}$. Let $A_{i}=h\\left(C_{i}\\right)$. Then the ray $X C_{2}$ crosses the segment $C_{1} C_{3}$ beyond $C_{2}$, so this ray crosses $A_{1} A_{3}$ beyond $A_{2}$; this implies that $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\\mathcal{B}$.\n\n\n\nSubcase $2.2: k<1$\n\nSubcase 2.2: $k<1$.\n\nThis case is completely similar to the previous one. All points from $\\mathcal{B}_{\\mathrm{L}}$ (and hence from $\\mathcal{C}_{\\mathrm{L}}$ move out from $\\mathcal{B}$ under $h$, because they are the nearest points of $\\mathcal{B}$ on the corresponding\n\n\n\nrays emanated from $X$. Assume that $C_{1}, C_{2}$, and $C_{3}$ are three vertices in $\\mathcal{C}_{\\mathrm{R}}$ such that $C_{2}$ lies between $C_{1}$ and $C_{3}$, and $h\\left(C_{1}\\right)$ and $h\\left(C_{3}\\right)$ lie in $\\mathcal{B}$; let $A_{i}=h\\left(C_{i}\\right)$. Then $A_{2}$ lies on the segment $X C_{2}$, and the segments $X A_{2}$ and $A_{1} A_{3}$ cross each other. Thus $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\\mathcal{B}$.' 'By a polygon we always mean its interior together with its boundary.\n\nLet $O_{A}$ and $O_{B}$ be the centers of $\\mathcal{A}$ and $\\mathcal{B}$, respectively. Denote $[n]=\\{1,2, \\ldots, n\\}$.\n\nWe start with introducing appropriate enumerations and notations. Enumerate the sidelines of $\\mathcal{B}$ clockwise as $\\ell_{1}, \\ell_{2}, \\ldots, \\ell_{n}$. Denote by $\\mathcal{H}_{i}$ the half-plane of $\\ell_{i}$ that contains $\\mathcal{B}\\left(\\mathcal{H}_{i}\\right.$ is assumed to contain $\\ell_{i}$ ); by $B_{i}$ the midpoint of the side belonging to $\\ell_{i}$; and finally denote $\\overrightarrow{b_{i}}=\\overrightarrow{B_{i} O_{B}}$. (As usual, the numbering is cyclic modulo $n$, so $\\ell_{n+i}=\\ell_{i}$ etc.)\n\nNow, choose a vertex $A_{1}$ of $\\mathcal{A}$ such that the vector $\\overrightarrow{O_{A} A_{1}}$ points ""mostly outside $\\mathcal{H}_{1}$ ""; strictly speaking, this means that the scalar product $\\left\\langle\\overrightarrow{O_{A} A_{1}}, \\overrightarrow{b_{1}}\\right\\rangle$ is minimal. Starting from $A_{1}$, enumerate the vertices of $\\mathcal{A}$ clockwise as $A_{1}, A_{2}, \\ldots, A_{n}$; by the rotational symmetry, the choice of $A_{1}$ yields that the vector $\\overrightarrow{O_{A} A_{i}}$ points ""mostly outside $\\mathcal{H}_{i}$ "", i.e.,\n\n$$\n\\left\\langle\\overrightarrow{O_{A} A_{i}}, \\overrightarrow{b_{i}}\\right\\rangle=\\min _{j \\in[n]}\\left\\langle\\overrightarrow{O_{A} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle\n\\tag{1}\n$$\n\n\n\nEnumerations and notations\n\nWe intend to reformulate the problem in more combinatorial terms, for which purpose we introduce the following notion. Say that a subset $I \\subseteq[n]$ is connected if the elements of this set are consecutive in the cyclic order (in other words, if we join each $i$ with $i+1 \\bmod n$ by an edge, this subset is connected in the usual graph sense). Clearly, the union of two connected subsets sharing at least one element is connected too. Next, for any half-plane $\\mathcal{H}$ the indices of vertices of, say, $\\mathcal{A}$ that lie in $\\mathcal{H}$ form a connected set.\n\nTo access the problem, we denote\n\n$$\nM=\\left\\{j \\in[n]: A_{j} \\notin \\mathcal{B}\\right\\}, \\quad M_{i}=\\left\\{j \\in[n]: A_{j} \\notin \\mathcal{H}_{i}\\right\\} \\quad \\text { for } i \\in[n]\n$$\n\nWe need to prove that $[n] \\backslash M$ is connected, which is equivalent to $M$ being connected. On the other hand, since $\\mathcal{B}=\\bigcap_{i \\in[n]} \\mathcal{H}_{i}$, we have $M=\\bigcup_{i \\in[n]} M_{i}$, where the sets $M_{i}$ are easier to investigate. We will utilize the following properties of these sets; the first one holds by the definition of $M_{i}$, along with the above remark.\n\n\n\n\n\nThe sets $M_{i}$\n\nProperty 1: Each set $M_{i}$ is connected.\n\nProperty 2: If $M_{i}$ is nonempty, then $i \\in M_{i}$.\n\nProof. Indeed, we have\n\n$$\nj \\in M_{i} \\Longleftrightarrow A_{j} \\notin \\mathcal{H}_{i} \\Longleftrightarrow\\left\\langle\\overrightarrow{B_{i} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle<0 \\Longleftrightarrow\\left\\langle\\overrightarrow{O_{A} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle<\\left\\langle\\overrightarrow{O_{A} B_{i}}, \\overrightarrow{b_{i}}\\right\\rangle .\n\\tag{2}\n$$\n\nThe right-hand part of the last inequality does not depend on $j$. Therefore, if some $j$ lies in $M_{i}$, then by (1) so does $i$.\n\nIn view of Property 2, it is useful to define the set\n\n$$\nM^{\\prime}=\\left\\{i \\in[n]: i \\in M_{i}\\right\\}=\\left\\{i \\in[n]: M_{i} \\neq \\varnothing\\right\\}\n$$\n\nProperty 3: The set $M^{\\prime}$ is connected.\n\nProof. To prove this property, we proceed on with the investigation started in (2) to write\n\n$$\ni \\in M^{\\prime} \\Longleftrightarrow A_{i} \\in M_{i} \\Longleftrightarrow\\left\\langle\\overrightarrow{B_{i} A_{i}}, \\overrightarrow{b_{i}}\\right\\rangle<0 \\Longleftrightarrow\\left\\langle\\overrightarrow{O_{B} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle<\\left\\langle\\overrightarrow{O_{B} B_{i}}, \\overrightarrow{b_{i}}\\right\\rangle+\\left\\langle\\overrightarrow{A_{i} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle\n$$\n\nThe right-hand part of the obtained inequality does not depend on $i$, due to the rotational symmetry; denote its constant value by $\\mu$. Thus, $i \\in M^{\\prime}$ if and only if $\\left\\langle\\overrightarrow{O_{B} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle<\\mu$. This condition is in turn equivalent to the fact that $B_{i}$ lies in a certain (open) half-plane whose boundary line is orthogonal to $O_{B} O_{A}$; thus, it defines a connected set.\n\nNow we can finish the solution. Since $M^{\\prime} \\subseteq M$, we have\n\n$$\nM=\\bigcup_{i \\in[n]} M_{i}=M^{\\prime} \\cup \\bigcup_{i \\in[n]} M_{i}\n$$\n\nso $M$ can be obtained from $M^{\\prime}$ by adding all the sets $M_{i}$ one by one. All these sets are connected, and each nonempty $M_{i}$ contains an element of $M^{\\prime}$ (namely, $i$ ). Thus their union is also connected.']" [] Text-only Competition True Theorem proof Geometry Math English 153 A convex quadrilateral $A B C D$ has an inscribed circle with center $I$. Let $I_{a}, I_{b}, I_{c}$, and $I_{d}$ be the incenters of the triangles $D A B, A B C, B C D$, and $C D A$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common external tangents of the circles $B I_{a} I_{c}$ and $D I_{a} I_{c}$ meet at $Y$. Prove that $\angle X I Y=90^{\circ}$. ['Denote by $\\omega_{a}, \\omega_{b}, \\omega_{c}$ and $\\omega_{d}$ the circles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}, O_{b}, O_{c}$ and $O_{d}$, and let their radii be $r_{a}, r_{b}, r_{c}$ and $r_{d}$, respectively.\n\nClaim 1. $I_{b} I_{d} \\perp A C$ and $I_{a} I_{c} \\perp B D$.\n\nProof. Let the incircles of triangles $A B C$ and $A C D$ be tangent to the line $A C$ at $T$ and $T^{\\prime}$, respectively. (See the figure to the left.) We have $A T=\\frac{A B+A C-B C}{2}$ in triangle $A B C, A T^{\\prime}=$ $\\frac{A D+A C-C D}{2}$ in triangle $A C D$, and $A B-B C=A D-C D$ in quadrilateral $A B C D$, so\n\n$$\nA T=\\frac{A C+A B-B C}{2}=\\frac{A C+A D-C D}{2}=A T^{\\prime}\n$$\n\nThis shows $T=T^{\\prime}$. As an immediate consequence, $I_{b} I_{d} \\perp A C$.\n\nThe second statement can be shown analogously.\n\n\nClaim 2. The points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the lines $A I, B I, C I$ and $D I$, respectively.\n\nProof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.)\n\nNotice first that the incircles of triangles $A B C$ and $A C D$ can be obtained from the incircle of the quadrilateral $A B C D$ with homothety centers $B$ and $D$, respectively, and homothety factors less than 1 , therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $B I$ and $D I$, respectively.\n\nAs is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $A T$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\\angle I_{b} A T=\\angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $A B C$ and $A C D$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\\angle B A C$ and $\\angle C A D$, respectively. Then\n\n$$\n\\angle O_{a} A D=\\angle O_{a} A I_{d}+\\angle I_{d} A D=\\angle I_{b} A T+\\angle I_{d} A D=\\frac{1}{2} \\angle B A C+\\frac{1}{2} \\angle C A D=\\frac{1}{2} \\angle B A D\n$$\n\nso $O_{a}$ lies on the angle bisector of $\\angle B A D$, that is, on line $A I$.\n\nThe point $X$ is the external similitude center of $\\omega_{a}$ and $\\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\\omega_{a}$ and $\\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $A C$.\n\n\n\n\n\nFrom the similarity of the circles $\\omega_{a}$ and $\\omega_{c}$, from $O_{a} I_{b}=O_{a} I_{d}=O_{a} A=r_{a}$ and $O_{c} I_{b}=$ $O_{c} I_{d}=O_{c} C=r_{c}$, and from $A C \\| O_{a} O_{c}$ we can see that\n\n$$\n\\frac{O_{a} X}{O_{c} X}=\\frac{O_{a} U}{O_{c} U}=\\frac{r_{a}}{r_{c}}=\\frac{O_{a} I_{b}}{O_{c} I_{b}}=\\frac{O_{a} I_{d}}{O_{c} I_{d}}=\\frac{O_{a} A}{O_{c} C}=\\frac{O_{a} I}{O_{c} I}\n$$\n\nSo the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a}: r_{c}$. In this Apollonius circle $X U$ is a diameter, and the lines $I U$ and $I X$ are respectively the internal and external bisectors of $\\angle O_{a} I O_{c}=\\angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $U X$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $I X$ and $I U$ are the internal and external bisectors of $\\angle I_{b} I I_{d}=\\angle B I D$, respectively.\n\nRepeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $I Y$ is the internal bisector of $\\angle A I C$ and the external bisector of $\\angle B I D$. Therefore, the lines $I X$ and $I Y$ respectively are the internal and external bisectors of $\\angle B I D$, so they are perpendicular.'] [] Text-only Competition True Theorem proof Geometry Math English 154 "Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices $i \in\{0,1, \ldots, p-1\}$ have been chosen. Then the following number is computed: $$ M=a_{0}+10 \cdot a_{1}+\cdots+10^{p-1} \cdot a_{p-1}=\sum_{j=0}^{p-1} a_{j} \cdot 10^{j} $$ The goal of Eduardo is to make the number $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy." "[""We say that a player makes the move $\\left(i, a_{i}\\right)$ if he chooses the index $i$ and then the element $a_{i}$ of the set $\\{0,1,2,3,4,5,6,7,8,9\\}$ in this move.\n\nIf $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of 10 .\n\nNow assume that the prime number $p$ does not belong to $\\{2,5\\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, $\\left(10^{(p-1) / 2}\\right)^{2}=10^{p-1} \\equiv 1(\\bmod p)$, so $p \\mid\\left(10^{(p-1) / 2}\\right)^{2}-1=\\left(10^{(p-1) / 2}+1\\right)\\left(10^{(p-1) / 2}-1\\right)$. Since $p$ is prime, either $p \\mid 10^{(p-1) / 2}+1$ or $p \\mid 10^{(p-1) / 2}-1$. Thus we have two cases:\n\nCase a: $10^{(p-1) / 2} \\equiv-1(\\bmod p)$\n\nIn this case, for each move $\\left(i, a_{i}\\right)$ of Fernando, Eduardo immediately makes the move $\\left(j, a_{j}\\right)=$ $\\left(i+\\frac{p-1}{2}, a_{i}\\right)$, if $0 \\leqslant i \\leqslant \\frac{p-3}{2}$, or $\\left(j, a_{j}\\right)=\\left(i-\\frac{p-1}{2}, a_{i}\\right)$, if $\\frac{p-1}{2} \\leqslant i \\leqslant p-2$. We will have $10^{j} \\equiv-10^{i}$ $(\\bmod p)$, and so $a_{j} \\cdot 10^{j}=a_{i} \\cdot 10^{j} \\equiv-a_{i} \\cdot 10^{i}(\\bmod p)$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\\{r, r+(p-1) / 2\\}$ with $0 \\leqslant r \\leqslant(p-3) / 2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_{k} \\cdot 10^{k}$ corresponding to the already chosen pairs $\\left(k, a_{k}\\right)$ divisible by $p$, and thus wins the game.\n\nCase b: $10^{(p-1) / 2} \\equiv 1(\\bmod p)$\n\nIn this case, for each move $\\left(i, a_{i}\\right)$ of Fernando, Eduardo immediately makes the move $\\left(j, a_{j}\\right)=$ $\\left(i+\\frac{p-1}{2}, 9-a_{i}\\right)$, if $0 \\leqslant i \\leqslant \\frac{p-3}{2}$, or $\\left(j, a_{j}\\right)=\\left(i-\\frac{p-1}{2}, 9-a_{i}\\right)$, if $\\frac{p-1}{2} \\leqslant i \\leqslant p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^{j} \\equiv 10^{i}$ $(\\bmod p)$, and so $a_{j} \\cdot 10^{j}+a_{i} \\cdot 10^{i} \\equiv\\left(a_{i}+a_{j}\\right) \\cdot 10^{i}=9 \\cdot 10^{i}(\\bmod p)$. Therefore, at the end of the game, the sum of all terms $a_{k} \\cdot 10^{k}$ will be congruent to\n\n$$\n\\sum_{i=0}^{\\frac{p-3}{2}} 9 \\cdot 10^{i}=10^{(p-1) / 2}-1 \\equiv 0 \\quad(\\bmod p)\n$$\n\nand Eduardo wins the game.""]" [] Text-only Competition True Theorem proof Number Theory Math English 155 "Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for each $(x, y)$ in the set $S$. Note: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form $$ f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} . $$" "[""First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \\pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\\left(x_{1}, y_{1}\\right)$ through $\\left(x_{n}, y_{n}\\right)$. If any two of these lattice points $\\left(x_{i}, y_{i}\\right)$ and $\\left(x_{j}, y_{j}\\right)$ lie on the same line through the origin, then $\\left(x_{j}, y_{j}\\right)=\\left(-x_{i},-y_{i}\\right)$ because both of the points are irreducible. We then have $f\\left(x_{j}, y_{j}\\right)= \\pm f\\left(x_{i}, y_{i}\\right)$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points.\n\nConsider the homogenous polynomials $\\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define\n\n$$\ng_{i}(x, y)=\\prod_{j \\neq i} \\ell_{j}(x, y)\n$$\n\nThen $\\ell_{i}\\left(x_{j}, y_{j}\\right)=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}\\left(x_{j}, y_{j}\\right)=0$ for all $j \\neq i$. Define $a_{i}=g_{i}\\left(x_{i}, y_{i}\\right)$, and note that $a_{i} \\neq 0$.\n\nNote that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties:\n\n1. $g_{i}\\left(x_{j}, y_{j}\\right)=0$ if $j \\neq i$.\n2. $g_{i}\\left(x_{i}, y_{i}\\right)=a_{i}$.\n\nFor any $N \\geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}\\left(x_{i}, y_{i}\\right)=1$, which exists since $\\left(x_{i}, y_{i}\\right)$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$.\n\nWe may now reduce the problem to the following claim:\n\nClaim: For each positive integer a, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1 , such that $f_{a}(x, y) \\equiv 1(\\bmod a)$ for all relatively prime $(x, y)$.\n\nTo see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}(1 \\leqslant i \\leqslant n)$. Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial.\n\nWe prove the claim by factoring $a$. First, if $a$ is a power of a prime $\\left(a=p^{k}\\right)$, then we may choose either:\n\n- $f_{a}(x, y)=\\left(x^{p-1}+y^{p-1}\\right)^{\\phi(a)}$ if $p$ is odd;\n- $f_{a}(x, y)=\\left(x^{2}+x y+y^{2}\\right)^{\\phi(a)}$ if $p=2$.\n\nNow suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \\cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that\n\n$$\n\\frac{a}{q_{i}} F_{q_{i}}(x, y) \\equiv \\frac{a}{q_{i}}(\\bmod a)\n$$\n\nfor any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\\frac{a}{q_{i}}$ that equals 1 . Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \\equiv 1$ $(\\bmod a)$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree."" ""As in the previous solution, label the irreducible lattice points $\\left(x_{1}, y_{1}\\right), \\ldots,\\left(x_{n}, y_{n}\\right)$ and assume without loss of generality that no two of the points are collinear with the origin. We induct on $n$ to construct a homogenous polynomial $f(x, y)$ such that $f\\left(x_{i}, y_{i}\\right)=1$ for all $1 \\leqslant i \\leqslant n$.\n\nIf $n=1$ : Since $x_{1}$ and $y_{1}$ are relatively prime, there exist some integers $c, d$ such that $c x_{1}+d y_{1}=1$. Then $f(x, y)=c x+d y$ is suitable.\n\nIf $n \\geqslant 2$ : By the induction hypothesis we already have a homogeneous polynomial $g(x, y)$ with $g\\left(x_{1}, y_{1}\\right)=\\ldots=g\\left(x_{n-1}, y_{n-1}\\right)=1$. Let $j=\\operatorname{deg} g$,\n\n$$\ng_{n}(x, y)=\\prod_{k=1}^{n-1}\\left(y_{k} x-x_{k} y\\right)\n$$\n\nand $a_{n}=g_{n}\\left(x_{n}, y_{n}\\right)$. By assumption, $a_{n} \\neq 0$. Take some integers $c, d$ such that $c x_{n}+d y_{n}=1$. We will construct $f(x, y)$ in the form\n\n$$\nf(x, y)=g(x, y)^{K}-C \\cdot g_{n}(x, y) \\cdot(c x+d y)^{L}\n$$\n\nwhere $K$ and $L$ are some positive integers and $C$ is some integer. We assume that $L=K j-n+1$ so that $f$ is homogenous.\n\nDue to $g\\left(x_{1}, y_{1}\\right)=\\ldots=g\\left(x_{n-1}, y_{n-1}\\right)=1$ and $g_{n}\\left(x_{1}, y_{1}\\right)=\\ldots=g_{n}\\left(x_{n-1}, y_{n-1}\\right)=0$, the property $f\\left(x_{1}, y_{1}\\right)=\\ldots=f\\left(x_{n-1}, y_{n-1}\\right)=1$ is automatically satisfied with any choice of $K, L$, and $C$.\n\nFurthermore,\n\n$$\nf\\left(x_{n}, y_{n}\\right)=g\\left(x_{n}, y_{n}\\right)^{K}-C \\cdot g_{n}\\left(x_{n}, y_{n}\\right) \\cdot\\left(c x_{n}+d y_{n}\\right)^{L}=g\\left(x_{n}, y_{n}\\right)^{K}-C a_{n}\n$$\n\nIf we have an exponent $K$ such that $g\\left(x_{n}, y_{n}\\right)^{K} \\equiv 1\\left(\\bmod a_{n}\\right)$, then we may choose $C$ such that $f\\left(x_{n}, y_{n}\\right)=1$. We now choose such a $K$.\n\nConsider an arbitrary prime divisor $p$ of $a_{n}$. By\n\n$$\np \\mid a_{n}=g_{n}\\left(x_{n}, y_{n}\\right)=\\prod_{k=1}^{n-1}\\left(y_{k} x_{n}-x_{k} y_{n}\\right)\n$$\n\nthere is some $1 \\leqslant k0$.)""]" [] Text-only Competition True Theorem proof Number Theory Math English 156 "Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties: - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal to 1 ; - $f(a+b, b)=f(a, b)$ for any pair of relatively prime positive integers $(a, b)$. Prove that $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right) \geqslant \sqrt{2 p}-2 $$" "['Denote by $\\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \\in \\mathbb{A}$ there exists a pair $(u, v) \\in \\mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $\\left(u_{0}, v_{0}\\right)$ is one such pair, then all such pairs are of the form $(u, v)=\\left(u_{0}+k b, v_{0}-k a\\right)$, where $k \\in \\mathbb{Z}$. So there exists a unique such pair $(u, v)$ with $-b / 20$.\n\nProof. We induct on $a+b$. The base case is $a+b=2$. In this case, we have that $a=b=1$, $g(a, b)=g(1,1)=(0,1)$ and $f(1,1)=0$, so the claim holds.\n\nAssume now that $a+b>2$, and so $a \\neq b$, since $a$ and $b$ are coprime. Two cases are possible. Case 1: $a>b$.\n\nNotice that $g(a-b, b)=(u, v+u)$, since $u(a-b)+(v+u) b=1$ and $u \\in(-b / 2, b / 2]$. Thus $f(a, b)=1 \\Longleftrightarrow f(a-b, b)=1 \\Longleftrightarrow u>0$ by the induction hypothesis.\n\nCase 2: $av b \\geqslant 1-\\frac{a b}{2}, \\quad \\text { so } \\quad \\frac{1+a}{2} \\geqslant \\frac{1}{b}+\\frac{a}{2}>v \\geqslant \\frac{1}{b}-\\frac{a}{2}>-\\frac{a}{2}\n$$\n\nThus $1+a>2 v>-a$, so $a \\geqslant 2 v>-a$, hence $a / 2 \\geqslant v>-a / 2$, and thus $g(b, a)=(v, u)$.\n\nObserve that $f(a, b)=1 \\Longleftrightarrow f(b, a)=0 \\Longleftrightarrow f(b-a, a)=0$. We know from Case 1 that $g(b-a, a)=(v, u+v)$. We have $f(b-a, a)=0 \\Longleftrightarrow v \\leqslant 0$ by the inductive hypothesis. Then, since $b>a \\geqslant 1$ and $u a+v b=1$, we have $v \\leqslant 0 \\Longleftrightarrow u>0$, and we are done.\n\nThe Lemma proves that, for all $(a, b) \\in \\mathbb{A}, f(a, b)=1$ if and only if the inverse of $a$ modulo $b$, taken in $\\{1,2, \\ldots, b-1\\}$, is at most $b / 2$. Then, for any odd prime $p$ and integer $n$ such that $n \\not \\equiv 0(\\bmod p), f\\left(n^{2}, p\\right)=1$ iff the inverse of $n^{2} \\bmod p$ is less than $p / 2$. Since $\\left\\{n^{2} \\bmod p: 1 \\leqslant n \\leqslant p-1\\right\\}=\\left\\{n^{-2} \\bmod p: 1 \\leqslant n \\leqslant p-1\\right\\}$, including multiplicities (two for each quadratic residue in each set), we conclude that the desired sum is twice the number of quadratic residues that are less than $p / 2$, i.e.,\n\n$$\n\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right)=2 \\mid\\left\\{k: 1 \\leqslant k \\leqslant \\frac{p-1}{2} \\text { and } k^{2} \\bmod p<\\frac{p}{2}\\right\\} \\mid .\n\\tag{1}\n$$\n\nSince the number of perfect squares in the interval $[1, p / 2)$ is $\\lfloor\\sqrt{p / 2}\\rfloor>\\sqrt{p / 2}-1$, we conclude that\n\n$$\n\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right)>2\\left(\\sqrt{\\frac{p}{2}}-1\\right)=\\sqrt{2 p}-2\n$$' 'We provide a different proof for the Lemma. For this purpose, we use continued fractions to find $g(a, b)=(u, v)$ explicitly.\n\nThe function $f$ is completely determined on $\\mathbb{A}$ by the following\n\nClaim. Represent $a / b$ as a continued fraction; that is, let $a_{0}$ be an integer and $a_{1}, \\ldots, a_{k}$ be positive integers such that $a_{k} \\geqslant 2$ and\n\n$$\n\\frac{a}{b}=a_{0}+\\frac{1}{a_{1}+\\frac{1}{a_{2}+\\frac{1}{\\cdots+\\frac{1}{a_{k}}}}}=\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{k}\\right] .\n$$\n\nThen $f(a, b)=0 \\Longleftrightarrow k$ is even.\n\nProof. We induct on $b$. If $b=1$, then $a / b=[a]$ and $k=0$. Then, for $a \\geqslant 1$, an easy induction shows that $f(a, 1)=f(1,1)=0$.\n\nNow consider the case $b>1$. Perform the Euclidean division $a=q b+r$, with $0 \\leqslant r0$ and define $q_{-1}=0$ if necessary. Then\n\n- $q_{k}=a_{k} q_{k-1}+q_{k-2}$, and\n- $a q_{k-1}-b p_{k-1}=p_{k} q_{k-1}-q_{k} p_{k-1}=(-1)^{k-1}$.\n\nAssume that $k>0$. Then $a_{k} \\geqslant 2$, and\n\n$$\nb=q_{k}=a_{k} q_{k-1}+q_{k-2} \\geqslant a_{k} q_{k-1} \\geqslant 2 q_{k-1} \\Longrightarrow q_{k-1} \\leqslant \\frac{b}{2}\n$$\n\nwith strict inequality for $k>1$, and\n\n$$\n(-1)^{k-1} q_{k-1} a+(-1)^{k} p_{k-1} b=1\n$$\n\nNow we finish the proof of the Lemma. It is immediate for $k=0$. If $k=1$, then $(-1)^{k-1}=1$, so\n\n$$\n-b / 2<0 \\leqslant(-1)^{k-1} q_{k-1} \\leqslant b / 2 .\n$$\n\nIf $k>1$, we have $q_{k-1}0$, we find that $g(a, b)=\\left((-1)^{k-1} q_{k-1},(-1)^{k} p_{k-1}\\right)$, and so\n\n$$\nf(a, b)=1 \\Longleftrightarrow k \\text { is odd } \\Longleftrightarrow u=(-1)^{k-1} q_{k-1}>0 .\n$$']" [] Text-only Competition True Theorem proof Number Theory Math English 157 "Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1\tag{1} $$" ['We first show the following.\n\n- Claim. For any positive real numbers $x, y$ with $x y \\geqslant 1$, we have\n\n$$\n\\left(x^{2}+1\\right)\\left(y^{2}+1\\right) \\leqslant\\left(\\left(\\frac{x+y}{2}\\right)^{2}+1\\right)^{2}\\tag{2}\n$$\n\nProof. Note that $x y \\geqslant 1$ implies $\\left(\\frac{x+y}{2}\\right)^{2}-1 \\geqslant x y-1 \\geqslant 0$. We find that\n\n$$\n\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)=(x y-1)^{2}+(x+y)^{2} \\leqslant\\left(\\left(\\frac{x+y}{2}\\right)^{2}-1\\right)^{2}+(x+y)^{2}=\\left(\\left(\\frac{x+y}{2}\\right)^{2}+1\\right)^{2} .\n$$\n\nWithout loss of generality, assume $a \\geqslant b \\geqslant c$. This implies $a \\geqslant 1$. Let $d=\\frac{a+b+c}{3}$. Note that\n\n$$\na d=\\frac{a(a+b+c)}{3} \\geqslant \\frac{1+1+1}{3}=1 .\n$$\n\nThen we can apply (2) to the pair $(a, d)$ and the pair $(b, c)$. We get\n\n$$\n\\left(a^{2}+1\\right)\\left(d^{2}+1\\right)\\left(b^{2}+1\\right)\\left(c^{2}+1\\right) \\leqslant\\left(\\left(\\frac{a+d}{2}\\right)^{2}+1\\right)^{2}\\left(\\left(\\frac{b+c}{2}\\right)^{2}+1\\right)^{2} .\\tag{3}\n$$\n\nNext, from\n\n$$\n\\frac{a+d}{2} \\cdot \\frac{b+c}{2} \\geqslant \\sqrt{a d} \\cdot \\sqrt{b c} \\geqslant 1\n$$\n\nwe can apply (2) again to the pair $\\left(\\frac{a+d}{2}, \\frac{b+c}{2}\\right)$. Together with (3), we have\n\n$$\n\\left(a^{2}+1\\right)\\left(d^{2}+1\\right)\\left(b^{2}+1\\right)\\left(c^{2}+1\\right) \\leqslant\\left(\\left(\\frac{a+b+c+d}{4}\\right)^{2}+1\\right)^{4}=\\left(d^{2}+1\\right)^{4} .\n$$\n\nTherefore, $\\left(a^{2}+1\\right)\\left(b^{2}+1\\right)\\left(c^{2}+1\\right) \\leqslant\\left(d^{2}+1\\right)^{3}$, and (1) follows by taking cube root of both sides.'] [] Text-only Competition True Theorem proof Algebra Math English 158 " Prove that for every positive integer $n$, there exists a fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0\n\nWithout loss of generality, we may assume the snails start at $A_{i}$ and $A_{j}$ respectively. Let $l_{i}$ intersect $l_{j}$ at $P$. Note that there is an odd number of points between $\\operatorname{arc} A_{i} A_{j}$. Each of these points belongs to a line $l_{k}$. Such a line $l_{k}$ must intersect exactly one of\n\n\n\nthe segments $A_{i} P$ and $A_{j} P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i} P$ and $A_{j} P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i} P$ and $A_{j} P$ (not counting $P$ ) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i} P$ and $A_{j} P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 163 "Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps. Prove that if $n$ is even then there must be a moment when some two snails occupy the same intersection point no matter how Jeff places the snails." "[""We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$.\n\nFor even $n$, we consider any way that Jeff places the snails and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_{i}$ and $A_{j}$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_{i} B_{i}$ and $A_{j} B_{j}$. Then any other segment meeting one of the segments $A_{i} P$ and $A_{j} P$ must also meet the other one, and so the number of intersections on $A_{i} P$ and $A_{j} P$ are the same. This shows the snails starting from $A_{i}$ and $A_{j}$ will meet at $P$.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 164 "In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A .\tag{1 } $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent." "['Denote the common angle in (1) by $\\theta$. As $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C}=\\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. From $E A=E D$, we get\n\n$$\n\\angle A F D=\\angle A B C=90^{\\circ}+\\theta=180^{\\circ}-\\frac{1}{2} \\angle A E D .\n$$\n\nHence, $F$ lies on the circle with centre $E$ and radius $E A$. In particular, $E F=E A=E D$. As $\\angle E F A=\\angle E A F=2 \\theta=\\angle B F C$, points $B, F, E$ are collinear.\n\nAs $\\angle E D A=\\angle M A D$, we have $E D / / A M$ and hence $E, D, X$ are collinear. As $M$ is the midpoint of $C F$ and $\\angle C B F=90^{\\circ}$, we get $M F=M B$. In the isosceles triangles $E F A$ and $M F B$, we have $\\angle E F A=\\angle M F B$ and $A F=B F$. Therefore, they are congruent to each other. Then we have $B M=A E=X M$ and $B E=B F+F E=A F+F M=A M=E X$. This shows $\\triangle E M B \\cong \\triangle E M X$. As $F$ and $D$ lie on $E B$ and $E X$ respectively and $E F=E D$, we know that lines $B D$ and $X F$ are symmetric with respect to $E M$. It follows that the three lines are concurrent.\n\n' ""From $\\angle C A D=\\angle E D A$, we have $A C / / E D$. Together with $A C / / E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\\theta$. From $\\triangle A B F \\sim \\triangle A C D$, we get $\\frac{A B}{A C}=\\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. This yields $\\angle A F D=\\angle A B C=90^{\\circ}+\\theta$ and hence $\\angle F D C=90^{\\circ}$, implying that $B C D F$ is cyclic. Let $\\Gamma_{1}$ be its circumcircle.\n\nNext, from $\\triangle A B F \\sim \\triangle A D E$, we have $\\frac{A B}{A D}=\\frac{A F}{A E}$ so that $\\triangle A B D \\sim \\triangle A F E$. Therefore,\n\n$$\n\\angle A F E=\\angle A B D=\\theta+\\angle F B D=\\theta+\\angle F C D=2 \\theta=180^{\\circ}-\\angle B F A .\n$$\n\nThis implies $B, F, E$ are collinear. Note that $F$ is the incentre of triangle $D A B$. Point $E$ lies on the internal angle bisector of $\\angle D B A$ and lies on the perpendicular bisector of $A D$. It follows that $E$ lies on the circumcircle $\\Gamma_{2}$ of triangle $A B D$, and $E A=E F=E D$.\n\nAlso, since $C F$ is a diameter of $\\Gamma_{1}$ and $M$ is the midpoint of $C F, M$ is the centre of $\\Gamma_{1}$ and hence $\\angle A M D=2 \\theta=\\angle A B D$. This shows $M$ lies on $\\Gamma_{2}$. Next, $\\angle M D X=\\angle M A E=\\angle D X M$ since $A M X E$ is a parallelogram. Hence $M D=M X$ and $X$ lies on $\\Gamma_{1}$.\n\n\n\nWe now have two ways to complete the solution.\n\n- Method 1. From $E F=E A=X M$ and $E X / / F M, E F M X$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by $\\Gamma_{3}$. Since $B D, E M, F X$ are the three radical axes of $\\Gamma_{1}, \\Gamma_{2}, \\Gamma_{3}$, they must be concurrent.\n- Method 2. As $\\angle D M F=2 \\theta=\\angle B F M$, we have $D M / / E B$. Also,\n\n$$\n\\angle B F D+\\angle X B F=\\angle B F C+\\angle C F D+90^{\\circ}-\\angle C B X=2 \\theta+\\left(90^{\\circ}-\\theta\\right)+90^{\\circ}-\\theta=180^{\\circ}\n$$\n\nimplies $D F / / X B$. These show the corresponding sides of triangles $D M F$ and $B E X$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $D B, M E, F X$ meet at a point."" 'Let the common angle in (1) be $\\theta$. From $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C}=\\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. Then $\\angle A D F=\\angle A C B=90^{\\circ}-2 \\theta=90^{\\circ}-\\angle B A D$ and hence $D F \\perp A B$. As $F A=F B$, this implies $\\triangle D A B$ is isosceles with $D A=D B$. Then $F$ is the incentre of $\\triangle D A B$.\n\nNext, from $\\angle A E D=180^{\\circ}-2 \\theta=180^{\\circ}-\\angle D B A$, points $A, B, D, E$ are concyclic. Since we also have $E A=E D$, this shows $E, F, B$ are collinear and $E A=E F=E D$.\n\n\n\nNote that $C$ lies on the internal angle bisector of $\\angle B A D$ and lies on the external angle bisector of $\\angle D B A$. It follows that it is the $A$-excentre of triangle $D A B$. As $M$ is the midpoint of $C F, M$ lies on the circumcircle of triangle $D A B$ and it is the centre of the circle passing through $D, F, B, C$. By symmetry, $D E F M$ is a rhombus. Then the midpoints of $A X, E M$ and $D F$ coincide, and it follows that $D A F X$ is a parallelogram.\n\nLet $P$ be the intersection of $B D$ and $E M$, and $Q$ be the intersection of $A D$ and $B E$. From $\\angle B A C=\\angle D C A$, we know that $D C, A B, E M$ are parallel. Thus we have $\\frac{D P}{P B}=\\frac{C M}{M A}$. This is further equal to $\\frac{A E}{B E}$ since $C M=D M=D E=A E$ and $M A=B E$. From $\\triangle A E Q \\sim \\triangle B E A$, we find that\n\n$$\n\\frac{D P}{P B}=\\frac{A E}{B E}=\\frac{A Q}{B A}=\\frac{Q F}{F B}\n$$\n\nby the Angle Bisector Theorem. This implies $Q D / / F P$ and hence $F, P, X$ are collinear, as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 165 Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$. "['Let $A M$ meet $\\Gamma$ again at $Y$ and $X Y$ meet $B C$ at $D^{\\prime}$. It suffices to show $D^{\\prime}=D$. We shall apply the following fact.\n\n- Claim. For any cyclic quadrilateral $P Q R S$ whose diagonals meet at $T$, we have\n\n$$\n\\frac{Q T}{T S}=\\frac{P Q \\cdot Q R}{P S \\cdot S R}\n$$\n\nProof. We use $\\left[W_{1} W_{2} W_{3}\\right]$ to denote the area of $W_{1} W_{2} W_{3}$. Then\n\n$$\n\\frac{Q T}{T S}=\\frac{[P Q R]}{[P S R]}=\\frac{\\frac{1}{2} P Q \\cdot Q R \\sin \\angle P Q R}{\\frac{1}{2} P S \\cdot S R \\sin \\angle P S R}=\\frac{P Q \\cdot Q R}{P S \\cdot S R}\n$$\n\nApplying the Claim to $A B Y C$ and $X B Y C$ respectively, we have $1=\\frac{B M}{M C}=\\frac{A B \\cdot B Y}{A C \\cdot C Y}$ and $\\frac{B D^{\\prime}}{D^{\\prime} C}=\\frac{X B \\cdot B Y}{X C \\cdot C Y}$. These combine to give\n\n$$\n\\frac{B D^{\\prime}}{C D^{\\prime}}=\\frac{X B}{X C} \\cdot \\frac{B Y}{C Y}=\\frac{X B}{X C} \\cdot \\frac{A C}{A B}\n\\tag{1}\n$$\n\nNext, we use directed angles to find that $\\measuredangle X B F=\\measuredangle X B A=\\measuredangle X C A=\\measuredangle X C E$ and $\\measuredangle X F B=\\measuredangle X F A=\\measuredangle X E A=\\measuredangle X E C$. This shows triangles $X B F$ and $X C E$ are directly similar. In particular, we have\n\n$$\n\\frac{X B}{X C}=\\frac{B F}{C E}\n\\tag{2}\n$$\n\nIn the following, we give two ways to continue the proof.\n\n- Method 1. Here is a geometrical method. As $\\angle F I B=\\angle A I B-90^{\\circ}=\\frac{1}{2} \\angle A C B=\\angle I C B$ and $\\angle F B I=\\angle I B C$, the triangles $F B I$ and $I B C$ are similar. Analogously, triangles $E I C$ and $I B C$ are also similar. Hence, we get\n\n$$\n\\frac{F B}{I B}=\\frac{B I}{B C} \\quad \\text { and } \\quad \\frac{E C}{I C}=\\frac{I C}{B C}\n\\tag{3}\n$$\n\n\n\n\n\nNext, construct a line parallel to $B C$ and tangent to the incircle. Suppose it meets sides $A B$ and $A C$ at $B_{1}$ and $C_{1}$ respectively. Let the incircle touch $A B$ and $A C$ at $B_{2}$ and $C_{2}$ respectively. By homothety, the line $B_{1} I$ is parallel to the external angle bisector of $\\angle A B C$, and hence $\\angle B_{1} I B=90^{\\circ}$. Since $\\angle B B_{2} I=90^{\\circ}$, we get $B B_{2} \\cdot B B_{1}=B I^{2}$, and similarly $C C_{2} \\cdot C C_{1}=C I^{2}$. Hence,\n\n$$\n\\frac{B I^{2}}{C I^{2}}=\\frac{B B_{2} \\cdot B B_{1}}{C C_{2} \\cdot C C_{1}}=\\frac{B B_{1}}{C C_{1}} \\cdot \\frac{B D}{C D}=\\frac{A B}{A C} \\cdot \\frac{B D}{C D}\n\\tag{4}\n$$\n\nCombining (1), (2), (3) and (4), we conclude\n\n$$\n\\frac{B D^{\\prime}}{C D^{\\prime}}=\\frac{X B}{X C} \\cdot \\frac{A C}{A B}=\\frac{B F}{C E} \\cdot \\frac{A C}{A B}=\\frac{B I^{2}}{C I^{2}} \\cdot \\frac{A C}{A B}=\\frac{B D}{C D}\n$$\n\nso that $D^{\\prime}=D$. The result then follows.\n\n- Method 2. Let $\\beta=\\frac{1}{2} \\angle A B C$ and $\\gamma=\\frac{1}{2} \\angle A C B$. Observe that $\\angle F I B=\\angle A I B-90^{\\circ}=\\gamma$. Hence, $\\frac{B F}{F I}=\\frac{\\sin \\angle F I B}{\\sin \\angle I B F}=\\frac{\\sin \\gamma}{\\sin \\beta}$. Similarly, $\\frac{C E}{E I}=\\frac{\\sin \\beta}{\\sin \\gamma}$. As $F I=E I$, we get\n\n$$\n\\frac{B F}{C E}=\\frac{B F}{F I} \\cdot \\frac{E I}{C E}=\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}\n\\tag{5}\n$$\n\n\n\nTogether with (1) and (2), we find that\n\n$$\n\\frac{B D^{\\prime}}{C D^{\\prime}}=\\frac{A C}{A B} \\cdot\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}=\\frac{\\sin 2 \\beta}{\\sin 2 \\gamma} \\cdot\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}=\\frac{\\tan \\gamma}{\\tan \\beta}=\\frac{I D / C D}{I D / B D}=\\frac{B D}{C D}\n$$\n\nThis shows $D^{\\prime}=D$ and the result follows.' 'Let $\\omega_{A}$ be the $A$-mixtilinear incircle of triangle $A B C$. From the properties of mixtilinear incircles, $\\omega_{A}$ touches sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose $\\omega_{A}$ is tangent to $\\Gamma$ at $T$. Let $A M$ meet $\\Gamma$ again at $Y$, and let $D_{1}, T_{1}$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $B C$ respectively. It is well-known that $\\angle B A T=\\angle D_{1} A C$ so that $A, D_{1}, T_{1}$ are collinear.\n\n\n\nWe then show that $X, M, T_{1}$ are collinear. Let $R$ be the radical centre of $\\omega_{A}, \\Gamma$ and the circumcircle of triangle $A E F$. Then $R$ lies on $A X, E F$ and the tangent at $T$ to $\\Gamma$. Let $A T$ meet $\\omega_{A}$ again at $S$ and meet $E F$ at $P$. Obviously, $S F T E$ is a harmonic quadrilateral. Projecting from $T$, the pencil $(R, P ; F, E)$ is harmonic. We further project the pencil onto $\\Gamma$ from $A$, so that $X B T C$ is a harmonic quadrilateral. As $T T_{1} / / B C$, the projection from $T_{1}$ onto $B C$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $B C$, which is $M$. This shows $X, M, T_{1}$ are collinear.\n\nWe have two ways to finish the proof.\n\n- Method 1. Note that both $A Y$ and $X T_{1}$ are chords of $\\Gamma$ passing through the midpoint $M$ of the chord $B C$. By the Butterfly Theorem, $X Y$ and $A T_{1}$ cut $B C$ at a pair of symmetric points with respect to $M$, and hence $X, D, Y$ are collinear. The proof is thus complete.\n\n\n\n- Method 2. Here, we finish the proof without using the Butterfly Theorem. As DTT $D_{1}$ is an isosceles trapezoid, we have\n\n$$\n\\measuredangle Y T D=\\measuredangle Y T T_{1}+\\measuredangle T_{1} T D=\\measuredangle Y A T_{1}+\\measuredangle A D_{1} D=\\measuredangle Y M D\n$$\n\nso that $D, T, Y, M$ are concyclic. As $X, M, T_{1}$ are collinear, we have\n\n$$\n\\measuredangle A Y D=\\measuredangle M T D=\\measuredangle D_{1} T_{1} M=\\measuredangle A T_{1} X=\\measuredangle A Y X\n$$\n\nThis shows $X, D, Y$ are collinear.']" [] Text-only Competition True Theorem proof Geometry Math English 166 "Let $B=(-1,0)$ and $C=(1,0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be nice if (i) there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $T Q$ lies entirely in $S$; and (ii) for any triangle $P_{1} P_{2} P_{3}$, there exists a unique point $A$ in $S$ and a permutation $\sigma$ of the indices $\{1,2,3\}$ for which triangles $A B C$ and $P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}$ are similar. Prove that there exist two distinct nice subsets $S$ and $S^{\prime}$ of the set $\{(x, y): x \geqslant 0, y \geqslant 0\}$ such that if $A \in S$ and $A^{\prime} \in S^{\prime}$ are the unique choices of points in (ii), then the product $B A \cdot B A^{\prime}$ is a constant independent of the triangle $P_{1} P_{2} P_{3}$." ['If in the similarity of $\\triangle A B C$ and $\\triangle P_{\\sigma(1)} P_{\\sigma(2)} P_{\\sigma(3)}, B C$ corresponds to the longest side of $\\triangle P_{1} P_{2} P_{3}$, then we have $B C \\geqslant A B \\geqslant A C$. The condition $B C \\geqslant A B$ is equivalent to $(x+1)^{2}+y^{2} \\leqslant 4$, while $A B \\geqslant A C$ is trivially satisfied for any point in the first quadrant. Then we first define\n\n$$\nS=\\left\\{(x, y):(x+1)^{2}+y^{2} \\leqslant 4, x \\geqslant 0, y \\geqslant 0\\right\\}\n$$\n\nNote that $S$ is the intersection of a disk and the first quadrant, so it is bounded and convex, and we can choose any $T \\in S$ to meet condition (i). For any point $A$ in $S$, the relation $B C \\geqslant A B \\geqslant A C$ always holds. Therefore, the point $A$ in (ii) is uniquely determined, while its existence is guaranteed by the above construction.\n\n\n\nNext, if in the similarity of $\\triangle A^{\\prime} B C$ and $\\triangle P_{\\sigma(1)} P_{\\sigma(2)} P_{\\sigma(3)}, B C$ corresponds to the second longest side of $\\triangle P_{1} P_{2} P_{3}$, then we have $A^{\\prime} B \\geqslant B C \\geqslant A^{\\prime} C$. The two inequalities are equivalent to $(x+1)^{2}+y^{2} \\geqslant 4$ and $(x-1)^{2}+y^{2} \\leqslant 4$ respectively. Then we define\n\n$$\nS^{\\prime}=\\left\\{(x, y):(x+1)^{2}+y^{2} \\geqslant 4,(x-1)^{2}+y^{2} \\leqslant 4, x \\geqslant 0, y \\geqslant 0\\right\\}\n$$\n\n\n\nThe boundedness condition is satisfied while (ii) can be argued as in the previous case. For (i), note that $S^{\\prime}$ contains points inside the disk $(x-1)^{2}+y^{2} \\leqslant 4$ and outside the disk $(x+1)^{2}+y^{2} \\geqslant 4$. This shows we can take $T^{\\prime}=(1,2)$ in (i), which is the topmost point of the circle $(x-1)^{2}+y^{2}=4$.\n\nIt remains to check that the product $B A \\cdot B A^{\\prime}$ is a constant. Suppose we are given a triangle $P_{1} P_{2} P_{3}$ with $P_{1} P_{2} \\geqslant P_{2} P_{3} \\geqslant P_{3} P_{1}$. By the similarity, we have\n\n$$\nB A=B C \\cdot \\frac{P_{2} P_{3}}{P_{1} P_{2}} \\quad \\text { and } \\quad B A^{\\prime}=B C \\cdot \\frac{P_{1} P_{2}}{P_{2} P_{3}}\n$$\n\nThus $B A \\cdot B A^{\\prime}=B C^{2}=4$, which is certainly independent of the triangle $P_{1} P_{2} P_{3}$.'] [] Text-only Competition True Theorem proof Geometry Math English 167 Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$. "['\n\nLet $\\Gamma$ be the circle with centre $E$ passing through $B$ and $C$. Since $E D \\perp A C$, the point $F$ symmetric to $C$ with respect to $D$ lies on $\\Gamma$. From $\\angle D C I=\\angle I C B=\\angle C B I$, the line $D C$ is a tangent to the circumcircle of triangle $I B C$. Let $J$ be the symmetric point of $I$ with respect to $D$. Using directed lengths, from\n\n$$\nD C \\cdot D F=-D C^{2}=-D I \\cdot D B=D J \\cdot D B\n$$\n\nthe point $J$ also lies on $\\Gamma$. Let $I^{\\prime}$ be the reflection of $I$ in $A C$. Since $I J$ and $C F$ bisect each other, $C J F I$ is a parallelogram. From $\\angle F I^{\\prime} C=\\angle C I F=\\angle F J C$, we find that $I^{\\prime}$ lies on $\\Gamma$. This gives $E I^{\\prime}=E B$.\n\nNote that $A C$ is the internal angle bisector of $\\angle B D I^{\\prime}$. This shows $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$ as $D E \\perp A C$. Together with $E I^{\\prime}=E B$, it is well-known that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.' 'Let $I^{\\prime}$ be the reflection of $I$ in $A C$ and let $S$ be the intersection of $I^{\\prime} C$ and $A I$. Using directed angles, we let $\\theta=\\measuredangle A C I=\\measuredangle I C B=\\measuredangle C B I$. We have\n\n$$\n\\measuredangle I^{\\prime} S E=\\measuredangle I^{\\prime} C A+\\measuredangle C A I=\\theta+\\left(\\frac{\\pi}{2}+2 \\theta\\right)=3 \\theta+\\frac{\\pi}{2}\n$$\n\nand\n\n$$\n\\measuredangle I^{\\prime} D E=\\measuredangle I^{\\prime} D C+\\frac{\\pi}{2}=\\measuredangle C D I+\\frac{\\pi}{2}=\\measuredangle D C B+\\measuredangle C B D+\\frac{\\pi}{2}=3 \\theta+\\frac{\\pi}{2} .\n$$\n\nThis shows $I^{\\prime}, D, E, S$ are concyclic.\n\nNext, we find $\\measuredangle I^{\\prime} S B=2 \\measuredangle I^{\\prime} S E=6 \\theta$ and $\\measuredangle I^{\\prime} D B=2 \\measuredangle C D I=6 \\theta$. Therefore, $I^{\\prime}, D, B, S$ are concyclic so that $I^{\\prime}, D, E, B, S$ lie on the same circle. The result then follows.\n\n\n\n' 'Let $I^{\\prime}$ be the reflection of $I$ in $A C$, and let $D^{\\prime}$ be the second intersection of $A I$ and the circumcircle of triangle $A B D$. Since $A D^{\\prime}$ bisects $\\angle B A D$, point $D^{\\prime}$ is the midpoint of the arc $B D$ and $D D^{\\prime}=B D^{\\prime}=C D^{\\prime}$. Obviously, $A, E, D^{\\prime}$ lie on $A I$ in this order.\n\n\n\nWe find that $\\angle E D^{\\prime} D=\\angle A D^{\\prime} D=\\angle A B D=\\angle I B C=\\angle I C B$. Next, since $D^{\\prime}$ is the circumcentre of triangle $B C D$, we have $\\angle E D D^{\\prime}=90^{\\circ}-\\angle D^{\\prime} D C=\\angle C B D=\\angle I B C$. The two relations show that triangles $E D^{\\prime} D$ and $I C B$ are similar. Therefore, we have\n\n$$\n\\frac{B C}{C I^{\\prime}}=\\frac{B C}{C I}=\\frac{D D^{\\prime}}{D^{\\prime} E}=\\frac{B D^{\\prime}}{D^{\\prime} E}\n$$\n\nAlso, we get\n\n$$\n\\angle B C I^{\\prime}=\\angle B C A+\\angle A C I^{\\prime}=\\angle B C A+\\angle I C A=\\angle B C A+\\angle D B C=\\angle B D A=\\angle B D^{\\prime} E .\n$$\n\nThese show triangles $B C I^{\\prime}$ and $B D^{\\prime} E$ are similar, and hence triangles $B C D^{\\prime}$ and $B I^{\\prime} E$ are similar. As $B C D^{\\prime}$ is isosceles, we obtain $B E=I^{\\prime} E$.\n\nAs $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$ and $E I^{\\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.' 'Let $A I$ and $B I$ meet the circumcircle of triangle $A B C$ again at $A^{\\prime}$ and $B^{\\prime}$ respectively, and let $E^{\\prime}$ be the reflection of $E$ in $A C$. From\n\n$$\n\\begin{aligned}\n\\angle B^{\\prime} A E^{\\prime} & =\\angle B^{\\prime} A D-\\angle E^{\\prime} A D=\\frac{\\angle A B C}{2}-\\frac{\\angle B A C}{2}=90^{\\circ}-\\angle B A C-\\frac{\\angle A B C}{2} \\\\\n& =90^{\\circ}-\\angle B^{\\prime} D A=\\angle B^{\\prime} D E^{\\prime},\n\\end{aligned}\n$$\n\npoints $B^{\\prime}, A, D, E^{\\prime}$ are concyclic. Then\n\n$$\n\\angle D B^{\\prime} E^{\\prime}=\\angle D A E^{\\prime}=\\frac{\\angle B A C}{2}=\\angle B A A^{\\prime}=\\angle D B^{\\prime} A^{\\prime}\n$$\n\nand hence $B^{\\prime}, E^{\\prime}, A^{\\prime}$ are collinear. It is well-known that $A^{\\prime} B^{\\prime}$ is the perpendicular bisector of $C I$, so that $C E^{\\prime}=I E^{\\prime}$. Let $I^{\\prime}$ be the reflection of $I$ in $A C$. This implies $B E=C E=I^{\\prime} E$. As $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$ and $E I^{\\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.\n\n' 'Let $F$ be the intersection of $C I$ and $A B$. Clearly, $F$ and $D$ are symmetric with respect to $A I$. Let $O$ be the circumcentre of triangle $B I F$, and let $I^{\\prime}$ be the reflection of $I$ in $A C$.\n\n\n\n\n\nFrom $\\angle B F O=90^{\\circ}-\\angle F I B=\\frac{1}{2} \\angle B A C=\\angle B A I$, we get $E I / / F O$. Also, from the relation $\\angle O I B=90^{\\circ}-\\angle B F I=90^{\\circ}-\\angle C D I=\\angle I^{\\prime} I D$, we know that $O, I, I^{\\prime}$ are collinear.\n\nNote that $E D / / O I$ since both are perpendicular to $A C$. Then $\\angle F E I=\\angle D E I=\\angle O I E$. Together with $E I / / F O$, the quadrilateral $E F O I$ is an isosceles trapezoid. Therefore, we find that $\\angle D I E=\\angle F I E=\\angle O E I$ so $O E / / I D$. Then $D E O I$ is a parallelogram. Hence, we have $D I^{\\prime}=D I=E O$, which shows $D E O I^{\\prime}$ is an isosceles trapezoid. In addition, $E D=O I=O B$ and $O E / / B D$ imply $E O B D$ is another isosceles trapezoid. In particular, both $D E O I^{\\prime}$ and $E O B D$ are cyclic. This shows $B, D, E, I^{\\prime}$ are concyclic.' 'Let $I^{\\prime}$ be the reflection of $I$ in $A C$. Denote by $T$ and $M$ the projections from $I$ to sides $A B$ and $B C$ respectively. Since $B I$ is the perpendicular bisector of $T M$, we have\n\n$$\nD T=D M\\tag{1}\n$$\n\nSince $\\angle A D E=\\angle A T I=90^{\\circ}$ and $\\angle D A E=\\angle T A I$, we have $\\triangle A D E \\sim \\triangle A T I$. This shows $\\frac{A D}{A E}=\\frac{A T}{A I}=\\frac{A T}{A I^{\\prime}}$. Together with $\\angle D A T=2 \\angle D A E=\\angle E A I^{\\prime}$, this yields $\\triangle D A T \\sim \\triangle E A I^{\\prime}$. In particular, we have\n\n$$\n\\frac{D T}{E I^{\\prime}}=\\frac{A T}{A I^{\\prime}}=\\frac{A T}{A I}\n\\tag{2}\n$$\n\nObviously, the right-angled triangles $A M B$ and $A T I$ are similar. Then we get\n\n$$\n\\frac{A M}{A B}=\\frac{A T}{A I}\\tag{3}\n$$\n\nNext, from $\\triangle A M B \\sim \\triangle A T I \\sim \\triangle A D E$, we have $\\frac{A M}{A B}=\\frac{A D}{A E}$ so that $\\triangle A D M \\sim \\triangle A E B$. It follows that\n\n$$\n\\frac{D M}{E B}=\\frac{A M}{A B}\\tag{4}\n$$\n\nCombining (1), (2), (3) and (4), we get $E B=E I^{\\prime}$. As $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$, we know that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 168 Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$. "['Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $Q$ be the midpoint of $A H$ and $N$ be the nine-point centre of triangle $A B C$. It is known that $Q$ lies on the nine-point circle of triangle $A B C, N$ is the midpoint of $Q M$ and that $Q M$ is parallel to $A O$.\n\nLet the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S^{\\prime}$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have\n\n$$\n\\begin{aligned}\n\\measuredangle S D Q & =\\measuredangle Q A S=\\measuredangle X A S-\\measuredangle X A Q=\\left(\\frac{\\pi}{2}-\\measuredangle A Y X\\right)-\\measuredangle B A P=\\measuredangle C B A-\\measuredangle A Y X \\\\\n& =(\\measuredangle C B A-\\measuredangle A C B)-\\measuredangle B C A-\\measuredangle A Y X=\\measuredangle P E M-(\\measuredangle B C A+\\measuredangle A Y X) \\\\\n& =\\measuredangle P Q M-\\angle(B C, X Y)=\\frac{\\pi}{2}-\\angle\\left(S^{\\prime} Q, B C\\right)-\\angle(B C, X Y)=\\measuredangle S S^{\\prime} Q .\n\\end{aligned}\n$$\n\nThis shows $D, S^{\\prime}, S, Q$ are concyclic.\n\n\n\nLet the perpendicular from $N$ to $B C$ intersect line $S S^{\\prime}$ at $O_{1}$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_{1}$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From\n\n$$\n\\measuredangle D S^{\\prime} O_{1}=\\measuredangle D Q S=\\measuredangle S Q A=\\angle(S Q, Q A)=\\angle\\left(O D, O_{1} N\\right)=\\measuredangle D N O_{1}\n$$\n\n\n\nsince $S Q / / O D$ and $Q A / / O_{1} N$, we know that $D, O_{1}, S^{\\prime}, N$ are concyclic. Therefore, we get\n\n$$\n\\measuredangle S D S^{\\prime}=\\measuredangle S Q S^{\\prime}=\\angle\\left(S Q, Q S^{\\prime}\\right)=\\angle\\left(N D, N S^{\\prime}\\right)=\\measuredangle D N S^{\\prime}\n$$\n\nso that $S D$ is a tangent to the circle through $D, O_{1}, S^{\\prime}, N$. Then we have\n\n$$\nS S^{\\prime} \\cdot S O_{1}=S D^{2}=S X^{2}\n\\tag{1}\n$$\n\nNext, we show that $S$ and $S^{\\prime}$ are symmetric with respect to $X Y$. By the Sine Law, we have\n\n$$\n\\frac{S S^{\\prime}}{\\sin \\angle S Q S^{\\prime}}=\\frac{S Q}{\\sin \\angle S S^{\\prime} Q}=\\frac{S Q}{\\sin \\angle S D Q}=\\frac{S Q}{\\sin \\angle S A Q}=\\frac{S A}{\\sin \\angle S Q A} .\n$$\n\nIt follows that\n\n$$\nS S^{\\prime}=S A \\cdot \\frac{\\sin \\angle S Q S^{\\prime}}{\\sin \\angle S Q A}=S A \\cdot \\frac{\\sin \\angle H O A}{\\sin \\angle O H A}=S A \\cdot \\frac{A H}{A O}=S A \\cdot 2 \\cos A,\n$$\n\nwhich is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\\angle S A C<\\angle O A C$ if and only if $\\angle Y X A>\\angle C B A$. This shows $S$ and $O_{1}$ lie on different sides of $X Y$. As $S^{\\prime}$ lies on ray $S O_{1}$, it follows that $S$ and $S^{\\prime}$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S^{\\prime}$ are symmetric with respect to $X Y$.\n\nLet $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S^{\\prime}$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S^{\\prime}=2 \\frac{S X^{2}}{d}$. It follows from (1) that $d=2 S O_{1}$. As $S O_{1}$ is the perpendicular bisector of $X Y$, point $O_{1}$ is the circumcentre of triangle $X S Y$.' 'Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_{1}$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S^{\\prime}$ and $S^{\\prime \\prime}$ and define the analogous points $X^{\\prime}, Y^{\\prime}, O_{1}^{\\prime}, X^{\\prime \\prime}, Y^{\\prime \\prime} O_{1}^{\\prime \\prime}$ accordingly. Note that $S, S^{\\prime}, S^{\\prime \\prime}$ lie on the perpendicular bisector of $A D$.\n\nAs $X X^{\\prime}$ and $Y Y^{\\prime}$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X^{\\prime} Y^{\\prime}$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X^{\\prime} Y^{\\prime}$. We find that\n\n$$\n\\measuredangle S X Y=\\frac{\\pi}{2}-\\measuredangle Y A X=\\frac{\\pi}{2}-\\measuredangle Y^{\\prime} A X^{\\prime}=\\measuredangle S^{\\prime} X^{\\prime} Y^{\\prime}\n$$\n\nand similarly $\\measuredangle S Y X=\\measuredangle S^{\\prime} Y^{\\prime} X^{\\prime}$. This shows triangles $S X Y$ and $S^{\\prime} X^{\\prime} Y^{\\prime}$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_{1}$ to $S^{\\prime}, X^{\\prime}, Y^{\\prime}, O_{1}^{\\prime}$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_{1}$ to $S^{\\prime \\prime}, X^{\\prime \\prime}, Y^{\\prime \\prime}, O_{1}^{\\prime \\prime}$. From these, we see that there is a spiral similarity taking the corresponding points $S, S^{\\prime}, S^{\\prime \\prime}$ to points $O_{1}, O_{1}^{\\prime}, O_{1}^{\\prime \\prime}$. In particular, $O_{1}, O_{1}^{\\prime}, O_{1}^{\\prime \\prime}$ are collinear.\n\n\n\n\n\nIt now suffices to show that $O_{1}$ lies on the perpendicular bisector of $P M$ for two special cases.\n\nFirstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_{1}$ is the nine-point centre and $O_{1} P=O_{1} M$. Indeed, $P$ and $M$ also lies on the nine-point circle.\n\nSecondly, we take $S^{\\prime}$ to be the midpoint of $A O$. Then $X^{\\prime}$ and $Y^{\\prime}$ are the midpoints of $A B$ and $A C$ respectively. Then $X^{\\prime} Y^{\\prime} / / B C$. Clearly, $S^{\\prime}$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X^{\\prime} Y^{\\prime}$ and $P M$ coincide. Hence, we must have $O_{1}^{\\prime} P=O_{1}^{\\prime} M$.\n']" [] Text-only Competition True Theorem proof Geometry Math English 169 Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segments $B M$ and $D M$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $X E$ and $Y F$. Prove that $P Q \perp A C$. "['\n\nLet $\\omega_{1}$ be the circumcircle of triangle $A B C$. We first prove that $Y$ lies on $\\omega_{1}$. Let $Y^{\\prime}$ be the point on ray $M D$ such that $M Y^{\\prime} \\cdot M D=M A^{2}$. Then triangles $M A Y^{\\prime}$ and $M D A$ are oppositely similar. Since $M C^{2}=M A^{2}=M Y^{\\prime} \\cdot M D$, triangles $M C Y^{\\prime}$ and $M D C$ are also oppositely similar. Therefore, using directed angles, we have\n\n$$\n\\measuredangle A Y^{\\prime} C=\\measuredangle A Y^{\\prime} M+\\measuredangle M Y^{\\prime} C=\\measuredangle M A D+\\measuredangle D C M=\\measuredangle C D A=\\measuredangle A B C\n$$\n\nso that $Y^{\\prime}$ lies on $\\omega_{1}$.\n\nLet $Z$ be the intersection point of lines $B C$ and $A D$. Since $\\measuredangle P D Z=\\measuredangle P B C=\\measuredangle P B Z$, point $Z$ lies on $\\omega$. In addition, from $\\measuredangle Y^{\\prime} B Z=\\measuredangle Y^{\\prime} B C=\\measuredangle Y^{\\prime} A C=\\measuredangle Y^{\\prime} A M=\\measuredangle Y^{\\prime} D Z$, we also know that $Y^{\\prime}$ lies on $\\omega$. Note that $\\angle A D C$ is acute implies $M A \\neq M D$ so $M Y^{\\prime} \\neq M D$. Therefore, $Y^{\\prime}$ is the second intersection of $D M$ and $\\omega$. Then $Y^{\\prime}=Y$ and hence $Y$ lies on $\\omega_{1}$.\n\nNext, by the Angle Bisector Theorem and the similar triangles, we have\n\n$$\n\\frac{F A}{F C}=\\frac{A D}{C D}=\\frac{A D}{A M} \\cdot \\frac{C M}{C D}=\\frac{Y A}{Y M} \\cdot \\frac{Y M}{Y C}=\\frac{Y A}{Y C}\n$$\n\nHence, $F Y$ is the internal angle bisector of $\\angle A Y C$.\n\nLet $B^{\\prime}$ be the second intersection of the internal angle bisector of $\\angle C B A$ and $\\omega_{1}$. Then $B^{\\prime}$ is the midpoint of arc $A C$ not containing $B$. Therefore, $Y B^{\\prime}$ is the external angle bisector of $\\angle A Y C$, so that $B^{\\prime} Y \\perp F Y$.\n\n\n\nDenote by $l$ the line through $P$ parallel to $A C$. Suppose $l$ meets line $B^{\\prime} Y$ at $S$. From\n\n$$\n\\begin{aligned}\n\\measuredangle P S Y & =\\measuredangle\\left(A C, B^{\\prime} Y\\right)=\\measuredangle A C Y+\\measuredangle C Y B^{\\prime}=\\measuredangle A C Y+\\measuredangle C A B^{\\prime}=\\measuredangle A C Y+\\measuredangle B^{\\prime} C A \\\\\n& =\\measuredangle B^{\\prime} C Y=\\measuredangle B^{\\prime} B Y=\\measuredangle P B Y,\n\\end{aligned}\n$$\n\nthe point $S$ lies on $\\omega$. Similarly, the line through $X$ perpendicular to $X E$ also passes through the second intersection of $l$ and $\\omega$, which is the point $S$. From $Q Y \\perp Y S$ and $Q X \\perp X S$, point $Q$ lies on $\\omega$ and $Q S$ is a diameter of $\\omega$. Therefore, $P Q \\perp P S$ so that $P Q \\perp A C$.' 'Denote by $\\omega_{1}$ and $\\omega_{2}$ the circumcircles of triangles $A B C$ and $A D C$ respectively. Since $\\angle A B C=\\angle A D C$, we know that $\\omega_{1}$ and $\\omega_{2}$ are symmetric with respect to the midpoint $M$ of $A C$.\n\nFirstly, we show that $X$ lies on $\\omega_{2}$. Let $X_{1}$ be the second intersection of ray $M B$ and $\\omega_{2}$ and $X^{\\prime}$ be its symmetric point with respect to $M$. Then $X^{\\prime}$ lies on $\\omega_{1}$ and $X^{\\prime} A X_{1} C$ is a parallelogram. Hence, we have\n\n$$\n\\begin{aligned}\n\\measuredangle D X_{1} B & =\\measuredangle D X_{1} A+\\measuredangle A X_{1} B=\\measuredangle D C A+\\measuredangle A X_{1} X^{\\prime}=\\measuredangle D C A+\\measuredangle C X^{\\prime} X_{1} \\\\\n& =\\measuredangle D C A+\\measuredangle C A B=\\angle(C D, A B) .\n\\end{aligned}\n$$\n\n\n\nAlso, we have\n\n$$\n\\measuredangle D P B=\\measuredangle P D C+\\angle(C D, A B)+\\measuredangle A B P=\\angle(C D, A B)\n$$\n\nThese yield $\\measuredangle D X_{1} B=\\measuredangle D P B$ and hence $X_{1}$ lies on $\\omega$. It follows that $X_{1}=X$ and $X$ lies on $\\omega_{2}$. Similarly, $Y$ lies on $\\omega_{1}$.\n\n\n\nNext, we prove that $Q$ lies on $\\omega$. Suppose the perpendicular bisector of $A C$ meet $\\omega_{1}$ at $B^{\\prime}$ and $M_{1}$ and meet $\\omega_{2}$ at $D^{\\prime}$ and $M_{2}$, so that $B, M_{1}$ and $D^{\\prime}$ lie on the same side of $A C$. Note that $B^{\\prime}$ lies on the angle bisector of $\\angle A B C$ and similarly $D^{\\prime}$ lies on $D P$.\n\nIf we denote the area of $W_{1} W_{2} W_{3}$ by $\\left[W_{1} W_{2} W_{3}\\right]$, then\n\n$$\n\\frac{B A \\cdot X^{\\prime} A}{B C \\cdot X^{\\prime} C}=\\frac{\\frac{1}{2} B A \\cdot X^{\\prime} A \\sin \\angle B A X^{\\prime}}{\\frac{1}{2} B C \\cdot X^{\\prime} C \\sin \\angle B C X^{\\prime}}=\\frac{\\left[B A X^{\\prime}\\right]}{\\left[B C X^{\\prime}\\right]}=\\frac{M A}{M C}=1\n$$\n\nAs $B E$ is the angle bisector of $\\angle A B C$, we have\n\n$$\n\\frac{E A}{E C}=\\frac{B A}{B C}=\\frac{X^{\\prime} C}{X^{\\prime} A}=\\frac{X A}{X C}\n$$\n\nTherefore, $X E$ is the angle bisector of $\\angle A X C$, so that $M_{2}$ lies on the line joining $X, E, Q$. Analogously, $M_{1}, F, Q, Y$ are collinear. Thus,\n\n$$\n\\begin{aligned}\n\\measuredangle X Q Y & =\\measuredangle M_{2} Q M_{1}=\\measuredangle Q M_{2} M_{1}+\\measuredangle M_{2} M_{1} Q=\\measuredangle X M_{2} D^{\\prime}+\\measuredangle B^{\\prime} M_{1} Y \\\\\n& =\\measuredangle X D D^{\\prime}+\\measuredangle B^{\\prime} B Y=\\measuredangle X D P+\\measuredangle P B Y=\\measuredangle X B P+\\measuredangle P B Y=\\measuredangle X B Y,\n\\end{aligned}\n$$\n\nwhich implies $Q$ lies on $\\omega$.\n\nFinally, as $M_{1}$ and $M_{2}$ are symmetric with respect to $M$, the quadrilateral $X^{\\prime} M_{2} X M_{1}$ is a parallelogram. Consequently,\n\n$$\n\\measuredangle X Q P=\\measuredangle X B P=\\measuredangle X^{\\prime} B B^{\\prime}=\\measuredangle X^{\\prime} M_{1} B^{\\prime}=\\measuredangle X M_{2} M_{1}\n$$\n\nThis shows $Q P / / M_{2} M_{1}$. As $M_{2} M_{1} \\perp A C$, we get $Q P \\perp A C$.' 'We first state two results which will be needed in our proof.\n\n- Claim 1. In $\\triangle X^{\\prime} Y^{\\prime} Z^{\\prime}$ with $X^{\\prime} Y^{\\prime} \\neq X^{\\prime} Z^{\\prime}$, let $N^{\\prime}$ be the midpoint of $Y^{\\prime} Z^{\\prime}$ and $W^{\\prime}$ be the foot of internal angle bisector from $X^{\\prime}$. Then $\\tan ^{2} \\measuredangle W^{\\prime} X^{\\prime} Z^{\\prime}=\\tan \\measuredangle N^{\\prime} X^{\\prime} W^{\\prime} \\tan \\measuredangle Z^{\\prime} W^{\\prime} X^{\\prime}$.\n\nProof.\n\n\n\nWithout loss of generality, assume $X^{\\prime} Y^{\\prime}>X^{\\prime} Z^{\\prime}$. Then $W^{\\prime}$ lies between $N^{\\prime}$ and $Z^{\\prime}$. The signs of both sides agree so it suffices to establish the relation for ordinary angles. Let $\\angle W^{\\prime} X^{\\prime} Z^{\\prime}=\\alpha, \\angle N^{\\prime} X^{\\prime} W^{\\prime}=\\beta$ and $\\angle Z^{\\prime} W^{\\prime} X^{\\prime}=\\gamma$. We have\n\n$$\n\\frac{\\sin (\\gamma-\\alpha)}{\\sin (\\alpha-\\beta)}=\\frac{N^{\\prime} X^{\\prime}}{N^{\\prime} Y^{\\prime}}=\\frac{N^{\\prime} X^{\\prime}}{N^{\\prime} Z^{\\prime}}=\\frac{\\sin (\\gamma+\\alpha)}{\\sin (\\alpha+\\beta)}\n$$\n\n\n\nThis implies\n\n$$\n\\frac{\\tan \\gamma-\\tan \\alpha}{\\tan \\gamma+\\tan \\alpha}=\\frac{\\sin \\gamma \\cos \\alpha-\\cos \\gamma \\sin \\alpha}{\\sin \\gamma \\cos \\alpha+\\cos \\gamma \\sin \\alpha}=\\frac{\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta}{\\sin \\alpha \\cos \\beta+\\cos \\alpha \\sin \\beta}=\\frac{\\tan \\alpha-\\tan \\beta}{\\tan \\alpha+\\tan \\beta}\n$$\n\nExpanding and simplifying, we get the desired result $\\tan ^{2} \\alpha=\\tan \\beta \\tan \\gamma$.\n\n- Claim 2. Let $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ be a quadrilateral inscribed in circle $\\Gamma$. Let diagonals $A^{\\prime} C^{\\prime}$ and $B^{\\prime} D^{\\prime}$ meet at $E^{\\prime}$, and $F^{\\prime}$ be the intersection of lines $A^{\\prime} B^{\\prime}$ and $C^{\\prime} D^{\\prime}$. Let $M^{\\prime}$ be the midpoint of $E^{\\prime} F^{\\prime}$. Then the power of $M^{\\prime}$ with respect to $\\Gamma$ is equal to $\\left(M^{\\prime} E^{\\prime}\\right)^{2}$.\n\nProof.\n\n\n\nLet $O^{\\prime}$ be the centre of $\\Gamma$ and let $\\Gamma^{\\prime}$ be the circle with centre $M^{\\prime}$ passing through $E^{\\prime}$. Let $F_{1}$ be the inversion image of $F^{\\prime}$ with respect to $\\Gamma$. It is well-known that $E^{\\prime}$ lies on the polar of $F^{\\prime}$ with respect to $\\Gamma$. This shows $E^{\\prime} F_{1} \\perp O^{\\prime} F^{\\prime}$ and hence $F_{1}$ lies on $\\Gamma^{\\prime}$. It follows that the inversion image of $\\Gamma^{\\prime}$ with respect to $\\Gamma$ is $\\Gamma^{\\prime}$ itself. This shows $\\Gamma^{\\prime}$ is orthogonal to $\\Gamma$, and thus the power of $M^{\\prime}$ with respect to $\\Gamma$ is the square of radius of $\\Gamma^{\\prime}$, which is $\\left(M^{\\prime} E^{\\prime}\\right)^{2}$.\n\nWe return to the main problem. Let $Z$ be the intersection of lines $A D$ and $B C$, and $W$ be the intersection of lines $A B$ and $C D$. Since $\\measuredangle P D Z=\\measuredangle P B C=\\measuredangle P B Z$, point $Z$ lies on $\\omega$. Similarly, $W$ lies on $\\omega$. Applying Claim 2 to the cyclic quadrilateral $Z B D W$, we know that the power of $M$ with respect to $\\omega$ is $M A^{2}$. Hence, $M X \\cdot M B=M A^{2}$.\n\nSuppose the line through $B$ perpendicular to $B E$ meets line $A C$ at $T$. Then $B E$ and $B T$ are the angle bisectors of $\\angle C B A$. This shows $(T, E ; A, C)$ is harmonic. Thus, we have $M E \\cdot M T=M A^{2}=M X \\cdot M B$. It follows that $E, T, B, X$ are concyclic.\n\n\n\n\n\nThe result is trivial for the special case $A D=C D$ since $P, Q$ lie on the perpendicular bisector of $A C$ in that case. Similarly, the case $A B=C B$ is trivial. It remains to consider the general cases where we can apply Claim 1 in the latter part of the proof.\n\nLet the projections from $P$ and $Q$ to $A C$ be $P^{\\prime}$ and $Q^{\\prime}$ respectively. Then $P Q \\perp A C$ if and only if $P^{\\prime}=Q^{\\prime}$ if and only if $\\frac{E P^{\\prime}}{F P^{\\prime}}=\\frac{E Q^{\\prime}}{F Q^{\\prime}}$ in terms of directed lengths. Note that\n\n$$\n\\frac{E P^{\\prime}}{F P^{\\prime}}=\\frac{\\tan \\measuredangle E F P}{\\tan \\measuredangle F E P}=\\frac{\\tan \\measuredangle A F D}{\\tan \\measuredangle A E B}\n$$\n\nNext, we have $\\frac{E Q^{\\prime}}{F Q^{\\prime}}=\\frac{\\tan \\measuredangle E F Q}{\\tan \\measuredangle F E Q}$ where $\\measuredangle F E Q=\\measuredangle T E X=\\measuredangle T B X=\\frac{\\pi}{2}+\\measuredangle E B M$ and by symmetry $\\measuredangle E F Q=\\frac{\\pi}{2}+\\measuredangle F D M$. Combining all these, it suffices to show\n\n$$\n\\frac{\\tan \\measuredangle A F D}{\\tan \\measuredangle A E B}=\\frac{\\tan \\measuredangle M B E}{\\tan \\measuredangle M D F}\n$$\n\nWe now apply Claim 1 twice to get\n\n$$\n\\tan \\measuredangle A F D \\tan \\measuredangle M D F=\\tan ^{2} \\measuredangle F D C=\\tan ^{2} \\measuredangle E B A=\\tan \\measuredangle M B E \\tan \\measuredangle A E B .\n$$\n\nThe result then follows.']" [] Text-only Competition True Theorem proof Geometry Math English 170 "Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$. Prove that $P$ lies on line $O I$ where $O$ is the circumcentre of triangle $A B C$." "['Let $A^{\\prime}$ be the reflection of $A$ in $B C$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\\Gamma$ of triangle $A B C$. As triangles $A B A^{\\prime}$ and $A O C$ are isosceles with $\\angle A B A^{\\prime}=2 \\angle A B C=\\angle A O C$, they are similar to each other. Also, triangles $A B I_{A}$ and $A I C$ are similar. Therefore we have\n\n$$\n\\frac{A A^{\\prime}}{A I_{A}}=\\frac{A A^{\\prime}}{A B} \\cdot \\frac{A B}{A I_{A}}=\\frac{A C}{A O} \\cdot \\frac{A I}{A C}=\\frac{A I}{A O}\n$$\n\nTogether with $\\angle A^{\\prime} A I_{A}=\\angle I A O$, we find that triangles $A A^{\\prime} I_{A}$ and $A I O$ are similar.\n\n\n\nDenote by $P^{\\prime}$ the intersection of line $A P$ and line $O I$. Using directed angles, we have\n\n$$\n\\begin{aligned}\n\\measuredangle M A P^{\\prime} & =\\measuredangle I_{A}^{\\prime} A I_{A}=\\measuredangle I_{A}^{\\prime} A A^{\\prime}-\\measuredangle I_{A} A A^{\\prime}=\\measuredangle A A^{\\prime} I_{A}-\\measuredangle(A M, O M) \\\\\n& =\\measuredangle A I O-\\measuredangle A M O=\\measuredangle M O P^{\\prime} .\n\\end{aligned}\n$$\n\nThis shows $M, O, A, P^{\\prime}$ are concyclic.\n\n\n\nDenote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Then\n\n$$\nI P^{\\prime}=\\frac{I A \\cdot I M}{I O}=\\frac{I O^{2}-R^{2}}{I O}\n$$\n\nis independent of $A$. Hence, $B P$ also meets line $O I$ at the same point $P^{\\prime}$ so that $P^{\\prime}=P$, and $P$ lies on $O I$.' 'Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}^{\\prime} C, A I_{B} C, I_{A} B C$ and $I_{A}^{\\prime} B C$ are all similar. From $\\triangle A I_{B}^{\\prime} C \\sim \\triangle I_{A}^{\\prime} B C$, we get $\\triangle A I_{A}^{\\prime} C \\sim \\triangle I_{B}^{\\prime} B C$. From $\\measuredangle A B P=\\measuredangle I_{B}^{\\prime} B C=\\measuredangle A I_{A}^{\\prime} C$ and $\\measuredangle B A P=\\measuredangle I_{A}^{\\prime} A C$, the triangles $A B P$ and $A I_{A}^{\\prime} C$ are directly similar.\n\n\n\nConsider the inversion with centre $A$ and radius $\\sqrt{A B \\cdot A C}$ followed by the reflection in $A I$. Then $B$ and $C$ are mapped to each other, and $I$ and $I_{A}$ are mapped to each other.\n\n\n\nFrom the similar triangles obtained, we have $A P \\cdot A I_{A}^{\\prime}=A B \\cdot A C$ so that $P$ is mapped to $I_{A}^{\\prime}$ under the transformation. In addition, line $A O$ is mapped to the altitude from $A$, and hence $O$ is mapped to the reflection of $A$ in $B C$, which we call point $A^{\\prime}$. Note that $A A^{\\prime} I_{A} I_{A}^{\\prime}$ is an isosceles trapezoid, which shows it is inscribed in a circle. The preimage of this circle is a straight line, meaning that $O, I, P$ are collinear.']" [] Text-only Competition True Theorem proof Geometry Math English 171 "Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$. Let one of the tangents from $P$ to the incircle of triangle $A B C$ meet the circumcircle at points $X$ and $Y$. Show that $\angle X I Y=120^{\circ}$." "[""By Poncelet's Porism, the other tangents to the incircle of triangle $A B C$ from $X$ and $Y$ meet at a point $Z$ on $\\Gamma$. Let $T$ be the touching point of the incircle to $X Y$, and let $D$ be the midpoint of $X Y$. We have\n\n$$\n\\begin{aligned}\nO D & =I T \\cdot \\frac{O P}{I P}=r\\left(1+\\frac{O I}{I P}\\right)=r\\left(1+\\frac{O I^{2}}{O I \\cdot I P}\\right)=r\\left(1+\\frac{R^{2}-2 R r}{R^{2}-I O^{2}}\\right) \\\\\n& =r\\left(1+\\frac{R^{2}-2 R r}{2 R r}\\right)=\\frac{R}{2}=\\frac{O X}{2} .\n\\end{aligned}\n$$"" 'Denote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Note that by the above transformation, we have $\\triangle A P O \\sim \\triangle A A^{\\prime} I_{A}^{\\prime}$ and $\\triangle A A^{\\prime} I_{A} \\sim \\triangle A I O$. Therefore, we find that\n\n$$\nP O=A^{\\prime} I_{A}^{\\prime} \\cdot \\frac{A O}{A I_{A}^{\\prime}}=A I_{A} \\cdot \\frac{A O}{A^{\\prime} I_{A}}=\\frac{A I_{A}}{A^{\\prime} I_{A}} \\cdot A O=\\frac{A O}{I O} \\cdot A O\n$$\n\nThis shows $P O \\cdot I O=R^{2}$, and it follows that $P$ and $I$ are mapped to each other under the inversion with respect to the circumcircle $\\Gamma$ of triangle $A B C$. Then $P X \\cdot P Y$, which is the power of $P$ with respect to $\\Gamma$, equals $P I \\cdot P O$. This yields $X, I, O, Y$ are concyclic.\n\nLet $T$ be the touching point of the incircle to $X Y$, and let $D$ be the midpoint of $X Y$. Then\n\n$$\nO D=I T \\cdot \\frac{P O}{P I}=r \\cdot \\frac{P O}{P O-I O}=r \\cdot \\frac{R^{2}}{R^{2}-I O^{2}}=r \\cdot \\frac{R^{2}}{2 R r}=\\frac{R}{2}\n$$\n\nThis shows $\\angle D O X=60^{\\circ}$ and hence $\\angle X I Y=\\angle X O Y=120^{\\circ}$.\nThis shows $\\angle X Z Y=60^{\\circ}$ and hence $\\angle X I Y=120^{\\circ}$.']" [] Text-only Competition True Theorem proof Geometry Math English 172 "Let $A_{1}, B_{1}$ and $C_{1}$ be points on sides $B C, C A$ and $A B$ of an acute triangle $A B C$ respectively, such that $A A_{1}, B B_{1}$ and $C C_{1}$ are the internal angle bisectors of triangle $A B C$. Let $I$ be the incentre of triangle $A B C$, and $H$ be the orthocentre of triangle $A_{1} B_{1} C_{1}$. Show that $$ A H+B H+C H \geqslant A I+B I+C I \text {. } $$" ['Without loss of generality, assume $\\alpha=\\angle B A C \\leqslant \\beta=\\angle C B A \\leqslant \\gamma=\\angle A C B$. Denote by $a, b, c$ the lengths of $B C, C A, A B$ respectively. We first show that triangle $A_{1} B_{1} C_{1}$ is acute.\n\nChoose points $D$ and $E$ on side $B C$ such that $B_{1} D / / A B$ and $B_{1} E$ is the internal angle bisector of $\\angle B B_{1} C$. As $\\angle B_{1} D B=180^{\\circ}-\\beta$ is obtuse, we have $B B_{1}>B_{1} D$. Thus,\n\n$$\n\\frac{B E}{E C}=\\frac{B B_{1}}{B_{1} C}>\\frac{D B_{1}}{B_{1} C}=\\frac{B A}{A C}=\\frac{B A_{1}}{A_{1} C}\n$$\n\nTherefore, $B E>B A_{1}$ and $\\frac{1}{2} \\angle B B_{1} C=\\angle B B_{1} E>\\angle B B_{1} A_{1}$. Similarly, $\\frac{1}{2} \\angle B B_{1} A>\\angle B B_{1} C_{1}$. It follows that\n\n$$\n\\angle A_{1} B_{1} C_{1}=\\angle B B_{1} A_{1}+\\angle B B_{1} C_{1}<\\frac{1}{2}\\left(\\angle B B_{1} C+\\angle B B_{1} A\\right)=90^{\\circ}\n$$\n\nis acute. By symmetry, triangle $A_{1} B_{1} C_{1}$ is acute.\n\nLet $B B_{1}$ meet $A_{1} C_{1}$ at $F$. From $\\alpha \\leqslant \\gamma$, we get $a \\leqslant c$, which implies\n\n$$\nB A_{1}=\\frac{c a}{b+c} \\leqslant \\frac{a c}{a+b}=B C_{1}\n$$\n\nand hence $\\angle B C_{1} A_{1} \\leqslant \\angle B A_{1} C_{1}$. As $B F$ is the internal angle bisector of $\\angle A_{1} B C_{1}$, this shows $\\angle B_{1} F C_{1}=\\angle B F A_{1} \\leqslant 90^{\\circ}$. Hence, $H$ lies on the same side of $B B_{1}$ as $C_{1}$. This shows $H$ lies inside triangle $B B_{1} C_{1}$. Similarly, from $\\alpha \\leqslant \\beta$ and $\\beta \\leqslant \\gamma$, we know that $H$ lies inside triangles $C C_{1} B_{1}$ and $A A_{1} C_{1}$.\n\n\n\n\n\nAs $\\alpha \\leqslant \\beta \\leqslant \\gamma$, we have $\\alpha \\leqslant 60^{\\circ} \\leqslant \\gamma$. Then $\\angle B I C \\leqslant 120^{\\circ} \\leqslant \\angle A I B$. Firstly, suppose $\\angle A I C \\geqslant 120^{\\circ}$.\n\nRotate points $B, I, H$ through $60^{\\circ}$ about $A$ to $B^{\\prime}, I^{\\prime}, H^{\\prime}$ so that $B^{\\prime}$ and $C$ lie on different sides of $A B$. Since triangle $A I^{\\prime} I$ is equilateral, we have\n\n$$\nA I+B I+C I=I^{\\prime} I+B^{\\prime} I^{\\prime}+I C=B^{\\prime} I^{\\prime}+I^{\\prime} I+I C .\n\\tag{1}\n$$\n\nSimilarly,\n\n$$\nA H+B H+C H=H^{\\prime} H+B^{\\prime} H^{\\prime}+H C=B^{\\prime} H^{\\prime}+H^{\\prime} H+H C \\text {. }\\tag{2}\n$$\n\nAs $\\angle A I I^{\\prime}=\\angle A I^{\\prime} I=60^{\\circ}, \\angle A I^{\\prime} B^{\\prime}=\\angle A I B \\geqslant 120^{\\circ}$ and $\\angle A I C \\geqslant 120^{\\circ}$, the quadrilateral $B^{\\prime} I^{\\prime} I C$ is convex and lies on the same side of $B^{\\prime} C$ as $A$.\n\nNext, since $H$ lies inside triangle $A C C_{1}, H$ lies outside $B^{\\prime} I^{\\prime} I C$. Also, $H$ lying inside triangle $A B I$ implies $H^{\\prime}$ lies inside triangle $A B^{\\prime} I^{\\prime}$. This shows $H^{\\prime}$ lies outside $B^{\\prime} I^{\\prime} I C$ and hence the convex quadrilateral $B^{\\prime} I^{\\prime} I C$ is contained inside the quadrilateral $B^{\\prime} H^{\\prime} H C$. It follows that the perimeter of $B^{\\prime} I^{\\prime} I C$ cannot exceed the perimeter of $B^{\\prime} H^{\\prime} H C$. From (1) and (2), we conclude that\n\n$$\nA H+B H+C H \\geqslant A I+B I+C I \\text {. }\n$$\n\nFor the case $\\angle A I C<120^{\\circ}$, we can rotate $B, I, H$ through $60^{\\circ}$ about $C$ to $B^{\\prime}, I^{\\prime}, H^{\\prime}$ so that $B^{\\prime}$ and $A$ lie on different sides of $B C$. The proof is analogous to the previous case and we still get the desired inequality.'] [] Text-only Competition True Theorem proof Geometry Math English 173 "Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^{k}+m n^{l}+1$ divides $n^{k+l}-1$. Prove that - $m=1$ and $l=2 k$; or - $l \mid k$ and $m=\frac{n^{k-l}-1}{n^{l}-1}$." "['It is given that\n\n$$\nn^{k}+m n^{l}+1 \\mid n^{k+l}-1\n\\tag{1}\n$$\n\nThis implies\n\n$$\nn^{k}+m n^{l}+1 \\mid\\left(n^{k+l}-1\\right)+\\left(n^{k}+m n^{l}+1\\right)=n^{k+l}+n^{k}+m n^{l} .\\tag{2}\n$$\n\nWe have two cases to discuss.\n\n- Case 1. $l \\geqslant k$.\n\nSince $\\left(n^{k}+m n^{l}+1, n\\right)=1,(2)$ yields\n\n$$\nn^{k}+m n^{l}+1 \\mid n^{l}+m n^{l-k}+1\n$$\n\nIn particular, we get $n^{k}+m n^{l}+1 \\leqslant n^{l}+m n^{l-k}+1$. As $n \\geqslant 2$ and $k \\geqslant 1,(m-1) n^{l}$ is at least $2(m-1) n^{l-k}$. It follows that the inequality cannot hold when $m \\geqslant 2$. For $m=1$, the above divisibility becomes\n\n$$\nn^{k}+n^{l}+1 \\mid n^{l}+n^{l-k}+1\n$$\n\nNote that $n^{l}+n^{l-k}+12 m n^{l}+1>n^{l}+m n^{l-k}+1$, it follows that $n^{k}+m n^{l}+1=n^{l}+m n^{l-k}+1$, that is,\n\n$$\nm\\left(n^{l}-n^{l-k}\\right)=n^{l}-n^{k} .\n$$\n\nIf $m \\geqslant 2$, then $m\\left(n^{l}-n^{l-k}\\right) \\geqslant 2 n^{l}-2 n^{l-k} \\geqslant 2 n^{l}-n^{l}>n^{l}-n^{k}$ gives a contradiction. Hence $m=1$ and $l-k=k$, which means $m=1$ and $l=2 k$.\n\n- Case 2. $l2 n^{k}+m>n^{k}+n^{k-l}+m$, it follows that $n^{k}+m n^{l}+1=n^{k}+n^{k-l}+m$. This gives $m=\\frac{n^{k-l}-1}{n^{l}-1}$. Note that $n^{l}-1 \\mid n^{k-l}-1$ implies $l \\mid k-l$ and hence $l \\mid k$. The proof is thus complete.']" [] Text-only Competition True Theorem proof Number Theory Math English 174 "Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that $$ k=\frac{x^{2}-a}{x^{2}-y^{2}} \tag{1} $$ for some integers $x$ and $y$ with $x>\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with $0 \leqslant x<\sqrt{a}$. Prove that $A=B$." "[""We first prove the following preliminary result.\n\n- Claim. For fixed $k$, let $x, y$ be integers satisfying (1). Then the numbers $x_{1}, y_{1}$ defined by\n\n$$\nx_{1}=\\frac{1}{2}\\left(x-y+\\frac{(x-y)^{2}-4 a}{x+y}\\right), \\quad y_{1}=\\frac{1}{2}\\left(x-y-\\frac{(x-y)^{2}-4 a}{x+y}\\right)\n$$\n\nare integers and satisfy (1) (with $x, y$ replaced by $x_{1}, y_{1}$ respectively).\n\nProof. Since $x_{1}+y_{1}=x-y$ and\n\n$$\nx_{1}=\\frac{x^{2}-x y-2 a}{x+y}=-x+\\frac{2\\left(x^{2}-a\\right)}{x+y}=-x+2 k(x-y)\n$$\n\nboth $x_{1}$ and $y_{1}$ are integers. Let $u=x+y$ and $v=x-y$. The relation (1) can be rewritten as\n\n$$\nu^{2}-(4 k-2) u v+\\left(v^{2}-4 a\\right)=0 \\text {. }\n$$\n\nBy Vieta's Theorem, the number $z=\\frac{v^{2}-4 a}{u}$ satisfies\n\n$$\nv^{2}-(4 k-2) v z+\\left(z^{2}-4 a\\right)=0\n$$\n\nSince $x_{1}$ and $y_{1}$ are defined so that $v=x_{1}+y_{1}$ and $z=x_{1}-y_{1}$, we can reverse the process and verify (1) for $x_{1}, y_{1}$.\n\nWe first show that $B \\subset A$. Take any $k \\in B$ so that (1) is satisfied for some integers $x, y$ with $0 \\leqslant x<\\sqrt{a}$. Clearly, $y \\neq 0$ and we may assume $y$ is positive. Since $a$ is not a square, we have $k>1$. Hence, we get $0 \\leqslant x\\sqrt{a}\n$$\n\nThis implies $k \\in A$ and hence $B \\subset A$.\n\n\n\nNext, we shall show that $A \\subset B$. Take any $k \\in A$ so that (1) is satisfied for some integers $x, y$ with $x>\\sqrt{a}$. Again, we may assume $y$ is positive. Among all such representations of $k$, we choose the one with smallest $x+y$. Define\n\n$$\nx_{1}=\\frac{1}{2}\\left|x-y+\\frac{(x-y)^{2}-4 a}{x+y}\\right|, \\quad y_{1}=\\frac{1}{2}\\left(x-y-\\frac{(x-y)^{2}-4 a}{x+y}\\right) .\n$$\n\nBy the Claim, $x_{1}, y_{1}$ are integers satisfying (1). Since $k>1$, we get $x>y>\\sqrt{a}$. Therefore, we have $y_{1}>\\frac{4 a}{x+y}>0$ and $\\frac{4 a}{x+y}\\sqrt{a}$, we get a contradiction due to the minimality of $x+y$. Therefore, we must have $0 \\leqslant x_{1}<\\sqrt{a}$, which means $k \\in B$ so that $A \\subset B$.\n\nThe two subset relations combine to give $A=B$."" ""The relation (1) is equivalent to\n\n$$\nk y^{2}-(k-1) x^{2}=a .\n\\tag{2}\n$$\n\nMotivated by Pell's Equation, we prove the following, which is essentially the same as the Claim in"" '- Claim. If $\\left(x_{0}, y_{0}\\right)$ is a solution to (2), then $\\left((2 k-1) x_{0} \\pm 2 k y_{0},(2 k-1) y_{0} \\pm 2(k-1) x_{0}\\right)$ is also a solution to (2).\n\nProof. We check directly that\n\n$$\n\\begin{aligned}\n& k\\left((2 k-1) y_{0} \\pm 2(k-1) x_{0}\\right)^{2}-(k-1)\\left((2 k-1) x_{0} \\pm 2 k y_{0}\\right)^{2} \\\\\n= & \\left(k(2 k-1)^{2}-(k-1)(2 k)^{2}\\right) y_{0}^{2}+\\left(k(2(k-1))^{2}-(k-1)(2 k-1)^{2}\\right) x_{0}^{2} \\\\\n= & k y_{0}^{2}-(k-1) x_{0}^{2}=a\n\\end{aligned}\n$$\n\nIf (2) is satisfied for some $0 \\leqslant x<\\sqrt{a}$ and nonnegative integer $y$, then clearly (1) implies $y>x$. Also, we have $k>1$ since $a$ is not a square number. By the Claim, consider another solution to (2) defined by\n\n$$\nx_{1}=(2 k-1) x+2 k y, \\quad y_{1}=(2 k-1) y+2(k-1) x\n$$\n\nIt satisfies $x_{1} \\geqslant(2 k-1) x+2 k(x+1)=(4 k-1) x+2 k>x$. Then we can replace the old solution by a new one which has a larger value in $x$. After a finite number of replacements, we must get a solution with $x>\\sqrt{a}$. This shows $B \\subset A$.\n\nIf (2) is satisfied for some $x>\\sqrt{a}$ and nonnegative integer $y$, by the Claim we consider another solution to (2) defined by\n\n$$\nx_{1}=|(2 k-1) x-2 k y|, \\quad y_{1}=(2 k-1) y-2(k-1) x\n$$\n\n\n\nFrom (2), we get $\\sqrt{k} y>\\sqrt{k-1} x$. This implies $k y>\\sqrt{k(k-1)} x>(k-1) x$ and hence $(2 k-1) x-2 k yy$. Then it is clear that $(2 k-1) x-2 k y>-x$. These combine to give $x_{1}|x|$, and, in case of $x>\\sqrt{a}$, another solution $\\left(x_{2}, y_{2}\\right)$ with $\\left|x_{2}\\right|<|x|$.\n\nWithout loss of generality, assume $x, y \\geqslant 0$. Let $u=x+y$ and $v=x-y$. Then $u \\geqslant v$ and (1) becomes\n\n$$\nk=\\frac{(u+v)^{2}-4 a}{4 u v}\n\\tag{3}\n$$\n\nThis is the same as\n\n$$\nv^{2}+(2 u-4 k u) v+u^{2}-4 a=0\n$$\n\nLet $v_{1}=4 k u-2 u-v$. Then $u+v_{1}=4 k u-u-v \\geqslant 8 u-u-v>u+v$. By Vieta's Theorem, $v_{1}$ satisfies\n\n$$\nv_{1}^{2}+(2 u-4 k u) v_{1}+u^{2}-4 a=0\n$$\n\nThis gives $k=\\frac{\\left(u+v_{1}\\right)^{2}-4 a}{4 u v_{1}}$. As $k$ is an integer, $u+v_{1}$ must be even. Therefore, $x_{1}=\\frac{u+v_{1}}{2}$ and $y_{1}=\\frac{v_{1}-u}{2}$ are integers. By reversing the process, we can see that $\\left(x_{1}, y_{1}\\right)$ is a solution to (1), with $x_{1}=\\frac{u+v_{1}}{2}>\\frac{u+v}{2}=x \\geqslant 0$. This completes the first half of the proof.\n\nSuppose $x>\\sqrt{a}$. Then $u+v>2 \\sqrt{a}$ and (3) can be rewritten as\n\n$$\nu^{2}+(2 v-4 k v) u+v^{2}-4 a=0\n$$\n\nLet $u_{2}=4 k v-2 v-u$. By Vieta's Theorem, we have $u u_{2}=v^{2}-4 a$ and\n\n$$\nu_{2}^{2}+(2 v-4 k v) u_{2}+v^{2}-4 a=0\n\\tag{4}\n$$\n\nBy $u>0, u+v>2 \\sqrt{a}$ and (3), we have $v>0$. If $u_{2} \\geqslant 0$, then $v u_{2} \\leqslant u u_{2}=v^{2}-4 a0$ and $u_{2}+v\\nu_{p}\\left(A_{1} A_{m+1}^{2}\\right)$ for $2 \\leqslant m \\leqslant k-1$.\n\nProof. The case $m=2$ is obvious since $\\nu_{p}\\left(A_{1} A_{2}^{2}\\right) \\geqslant p^{t}>\\nu_{p}\\left(A_{1} A_{3}^{2}\\right)$ by the condition and the above assumption.\n\nSuppose $\\nu_{p}\\left(A_{1} A_{2}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{3}^{2}\\right)>\\cdots>\\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$ where $3 \\leqslant m \\leqslant k-1$. For the induction step, we apply Ptolemy's Theorem to the cyclic quadrilateral $A_{1} A_{m-1} A_{m} A_{m+1}$ to get\n\n$$\nA_{1} A_{m+1} \\times A_{m-1} A_{m}+A_{1} A_{m-1} \\times A_{m} A_{m+1}=A_{1} A_{m} \\times A_{m-1} A_{m+1}\n$$\n\nwhich can be rewritten as\n\n$$\n\\begin{aligned}\nA_{1} A_{m+1}^{2} \\times A_{m-1} A_{m}^{2}= & A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}+A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2} \\\\\n& -2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}\n\\end{aligned}\n\\tag{1}\n$$\n\nFrom this, $2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}$ is an integer. We consider the component of $p$ of each term in (1). By the inductive hypothesis, we have $\\nu_{p}\\left(A_{1} A_{m-1}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$. Also, we have $\\nu_{p}\\left(A_{m} A_{m+1}^{2}\\right) \\geqslant p^{t}>\\nu_{p}\\left(A_{m-1} A_{m+1}^{2}\\right)$. These give\n\n$$\n\\nu_{p}\\left(A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)\n\\tag{2}\n$$\n\nNext, we have $\\nu_{p}\\left(4 A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2} \\times A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)=\\nu_{p}\\left(A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}\\right)+$ $\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)>2 \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)$ from (2). This implies\n\n$$\n\\nu_{p}\\left(2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}\\right)>\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) .\n\\tag{3}\n$$\n\nCombining (1), (2) and (3), we conclude that\n\n$$\n\\nu_{p}\\left(A_{1} A_{m+1}^{2} \\times A_{m-1} A_{m}^{2}\\right)=\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) .\n$$\n\nBy $\\nu_{p}\\left(A_{m-1} A_{m}^{2}\\right) \\geqslant p^{t}>\\nu_{p}\\left(A_{m-1} A_{m+1}^{2}\\right)$, we get $\\nu_{p}\\left(A_{1} A_{m+1}^{2}\\right)<\\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$. The Claim follows by induction.\n\n\n\nFrom the Claim, we get a chain of inequalities\n\n$$\np^{t}>\\nu_{p}\\left(A_{1} A_{3}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{4}^{2}\\right)>\\cdots>\\nu_{p}\\left(A_{1} A_{k}^{2}\\right) \\geqslant p^{t}\n$$\n\nwhich yields a contradiction. Therefore, we can show by induction that $2 S$ is divisible by $n$.""]" [] Text-only Competition True Theorem proof Number Theory Math English 176 "Let $a, b, c, d$ be four real numbers such that $a \geqslant b \geqslant c \geqslant d>0$ and $a+b+c+d=1$. Prove that $$ (a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d}<1 $$" "['The weighted AM-GM inequality with weights $a, b, c, d$ gives\n\n$$\na^{a} b^{b} c^{c} d^{d} \\leqslant a \\cdot a+b \\cdot b+c \\cdot c+d \\cdot d=a^{2}+b^{2}+c^{2}+d^{2}\n$$\n\nso it suffices to prove that $(a+2 b+3 c+4 d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)<1=(a+b+c+d)^{3}$. This can be done in various ways, for example:\n\n$$\n\\begin{gathered}\n(a+b+c+d)^{3}>a^{2}(a+3 b+3 c+3 d)+b^{2}(3 a+b+3 c+3 d) \\\\\n\\quad+c^{2}(3 a+3 b+c+3 d)+d^{2}(3 a+3 b+3 c+d) \\\\\n\\geqslant\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right) \\cdot(a+2 b+3 c+4 d)\n\\end{gathered}\n$$' 'From $b \\geqslant d$ we get\n\n$$\na+2 b+3 c+4 d \\leqslant a+3 b+3 c+3 d=3-2 a .\n$$\n\nIf $a<\\frac{1}{2}$, then the statement can be proved by\n\n$$\n(a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d} \\leqslant(3-2 a) a^{a} a^{b} a^{c} a^{d}=(3-2 a) a=1-(1-a)(1-2 a)<1 .\n$$\n\nFrom now on we assume $\\frac{1}{2} \\leqslant a<1$.\n\nBy $b, c, d<1-a$ we have\n\n$$\nb^{b} c^{c} d^{d}<(1-a)^{b} \\cdot(1-a)^{c} \\cdot(1-a)^{d}=(1-a)^{1-a} .\n$$\n\nTherefore,\n\n$$\n(a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d}<(3-2 a) a^{a}(1-a)^{1-a} .\n$$\n\nFor $00,\n$$\n\nso $g$ is strictly convex on $(0,1)$.\n\nBy $g\\left(\\frac{1}{2}\\right)=\\log 2+2 \\cdot \\frac{1}{2} \\log \\frac{1}{2}=0$ and $\\lim _{x \\rightarrow 1-} g(x)=0$, we have $g(x) \\leqslant 0$ (and hence $f(x) \\leqslant 1$ ) for all $x \\in\\left[\\frac{1}{2}, 1\\right)$, and therefore\n\n$$\n(a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d}b-1$, there is a group, $\\left(V_{i}, V_{i+1}, V_{i+2}, V_{i+3}\\right)$ that contains more white than black vertices. If three are white and one is black in that group, we are done. Otherwise, if $V_{i}, V_{i+1}, V_{i+2}, V_{i+3}$ are all white then let $V_{j}$ be the first black vertex among $V_{i+4}, \\ldots, V_{4 k+1}$ (recall that $V_{4 k+1}$ is black); then $V_{j-3}, V_{j-2}$ and $V_{j-1}$ are white and $V_{j}$ is black.\n\nNow we have four consecutive vertices $V_{i}, V_{i+1}, V_{i+2}, V_{i+3}$ that form a skew-colored quadrilateral. The remaining vertices form a convex $(4 k-3)$-gon; $w-3$ of them are white and $b-1$ are black. Since $b-1 \\geqslant k-1$ and $w-3 \\geqslant(2 k+1)-3>k-1$, we can apply the Claim with $k-1$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 180 "Let $p$ be an odd prime, and put $N=\frac{1}{4}\left(p^{3}-p\right)-1$. The numbers $1,2, \ldots, N$ are painted arbitrarily in two colors, red and blue. For any positive integer $n \leqslant N$, denote by $r(n)$ the fraction of integers in $\{1,2, \ldots, n\}$ that are red. Prove that there exists a positive integer $a \in\{1,2, \ldots, p-1\}$ such that $r(n) \neq a / p$ for all $n=1,2, \ldots, N$." ['Denote by $R(n)$ the number of red numbers in $\\{1,2, \\ldots, n\\}$, i.e., $R(n)=n r(n)$. Similarly, denote by $B(n)$ and $b(n)=B(n) / n$ the number and proportion of blue numbers in $\\{1,2, \\ldots, n\\}$, respectively. Notice that $B(n)+R(n)=n$ and $b(n)+r(n)=1$. Therefore, the statement of the problem does not change after swapping the colors.\n\nArguing indirectly, for every $a \\in\\{1,2, \\ldots, p-1\\}$ choose some positive integer $n_{a}$ such that $r\\left(n_{a}\\right)=a / p$ and, hence, $R\\left(n_{a}\\right)=a n_{a} / p$. Clearly, $p \\mid n_{a}$, so that $n_{a}=p m_{a}$ for some positive integer $m_{a}$, and $R\\left(n_{a}\\right)=a m_{a}$. Without loss of generality, we assume that $m_{1}m_{p-1}$.\n\nChoose $k$ to be the smallest index satisfying $m_{k}>m_{p-1}$; by our assumptions, we have $1\\left|R_{2}\\right|$. We will find a saddle subpair $\\left(R^{\\prime}, C^{\\prime}\\right)$ of $\\left(R_{1}, C_{1}\\right)$ with $\\left|R^{\\prime}\\right| \\leqslant\\left|R_{2}\\right|$; clearly, this implies the desired statement.\n\nStep 1: We construct maps $\\rho: R_{1} \\rightarrow R_{1}$ and $\\sigma: C_{1} \\rightarrow C_{1}$ such that $\\left|\\rho\\left(R_{1}\\right)\\right| \\leqslant\\left|R_{2}\\right|$, and $a\\left(\\rho\\left(r_{1}\\right), c_{1}\\right) \\geqslant a\\left(r_{1}, \\sigma\\left(c_{1}\\right)\\right)$ for all $r_{1} \\in R_{1}$ and $c_{1} \\in C_{1}$.\n\nSince $\\left(R_{1}, C_{1}\\right)$ is a saddle pair, for each $r_{2} \\in R_{2}$ there is $r_{1} \\in R_{1}$ such that $a\\left(r_{1}, c_{1}\\right) \\geqslant a\\left(r_{2}, c_{1}\\right)$ for all $c_{1} \\in C_{1}$; denote one such an $r_{1}$ by $\\rho_{1}\\left(r_{2}\\right)$. Similarly, we define four functions\n\n$$\n\\begin{array}{llllll}\n\\rho_{1}: R_{2} \\rightarrow R_{1} & \\text { such that } & a\\left(\\rho_{1}\\left(r_{2}\\right), c_{1}\\right) \\geqslant a\\left(r_{2}, c_{1}\\right) & \\text { for all } & r_{2} \\in R_{2}, & c_{1} \\in C_{1} ; \\\\\n\\rho_{2}: R_{1} \\rightarrow R_{2} & \\text { such that } & a\\left(\\rho_{2}\\left(r_{1}\\right), c_{2}\\right) \\geqslant a\\left(r_{1}, c_{2}\\right) & \\text { for all } & r_{1} \\in R_{1}, & c_{2} \\in C_{2} ; \\\\\n\\sigma_{1}: C_{2} \\rightarrow C_{1} & \\text { such that } & a\\left(r_{1}, \\sigma_{1}\\left(c_{2}\\right)\\right) \\leqslant a\\left(r_{1}, c_{2}\\right) & \\text { for all } & r_{1} \\in R_{1}, & c_{2} \\in C_{2} ; \\\\\n\\sigma_{2}: C_{1} \\rightarrow C_{2} & \\text { such that } & a\\left(r_{2}, \\sigma_{2}\\left(c_{1}\\right)\\right) \\leqslant a\\left(r_{2}, c_{1}\\right) & \\text { for all } & r_{2} \\in R_{2}, & c_{1} \\in C_{1} .\n\\end{array}\n\\tag{1}\n$$\n\nSet now $\\rho=\\rho_{1} \\circ \\rho_{2}: R_{1} \\rightarrow R_{1}$ and $\\sigma=\\sigma_{1} \\circ \\sigma_{2}: C_{1} \\rightarrow C_{1}$. We have\n\n$$\n\\left|\\rho\\left(R_{1}\\right)\\right|=\\left|\\rho_{1}\\left(\\rho_{2}\\left(R_{1}\\right)\\right)\\right| \\leqslant\\left|\\rho_{1}\\left(R_{2}\\right)\\right| \\leqslant\\left|R_{2}\\right| .\n$$\n\nMoreover, for all $r_{1} \\in R_{1}$ and $c_{1} \\in C_{1}$, we get\n\n$$\n\\begin{aligned}\n& a\\left(\\rho\\left(r_{1}\\right), c_{1}\\right)=a\\left(\\rho_{1}\\left(\\rho_{2}\\left(r_{1}\\right)\\right), c_{1}\\right) \\geqslant a\\left(\\rho_{2}\\left(r_{1}\\right), c_{1}\\right) \\geqslant a\\left(\\rho_{2}\\left(r_{1}\\right), \\sigma_{2}\\left(c_{1}\\right)\\right) \\\\\n& \\geqslant a\\left(r_{1}, \\sigma_{2}\\left(c_{1}\\right)\\right) \\geqslant a\\left(r_{1}, \\sigma_{1}\\left(\\sigma_{2}\\left(c_{1}\\right)\\right)\\right)=a\\left(r_{1}, \\sigma\\left(c_{1}\\right)\\right)\n\\end{aligned}\n\\tag{2}\n$$\n\nas desired.\n\nStep 2: Given maps $\\rho$ and $\\sigma$, we construct a proper saddle subpair $\\left(R^{\\prime}, C^{\\prime}\\right)$ of $\\left(R_{1}, C_{1}\\right)$.\n\nThe properties of $\\rho$ and $\\sigma$ yield that\n\n$$\na\\left(\\rho^{i}\\left(r_{1}\\right), c_{1}\\right) \\geqslant a\\left(\\rho^{i-1}\\left(r_{1}\\right), \\sigma\\left(c_{1}\\right)\\right) \\geqslant \\ldots \\geqslant a\\left(r_{1}, \\sigma^{i}\\left(c_{1}\\right)\\right)\n$$\n\nfor each positive integer $i$ and all $r_{1} \\in R_{1}, c_{1} \\in C_{1}$.\n\nConsider the images $R^{i}=\\rho^{i}\\left(R_{1}\\right)$ and $C^{i}=\\sigma^{i}\\left(C_{1}\\right)$. Clearly, $R_{1}=R^{0} \\supseteq R^{1} \\supseteq R^{2} \\supseteq \\ldots$ and $C_{1}=C^{0} \\supseteq C^{1} \\supseteq C^{2} \\supseteq \\ldots$. Since both chains consist of finite sets, there is an index $n$ such that $R^{n}=R^{n+1}=\\ldots$ and $C^{n}=C^{n+1}=\\ldots$ Then $\\rho^{n}\\left(R^{n}\\right)=R^{2 n}=R^{n}$, so $\\rho^{n}$ restricted to $R^{n}$ is a bijection. Similarly, $\\sigma^{n}$ restricted to $C^{n}$ is a bijection from $C^{n}$ to itself. Therefore, there exists a positive integer $k$ such that $\\rho^{n k}$ acts identically on $R^{n}$, and $\\sigma^{n k}$ acts identically on $C^{n}$.\n\nWe claim now that $\\left(R^{n}, C^{n}\\right)$ is a saddle subpair of $\\left(R_{1}, C_{1}\\right)$, with $\\left|R^{n}\\right| \\leqslant\\left|R^{1}\\right|=\\left|\\rho\\left(R_{1}\\right)\\right| \\leqslant$ $\\left|R_{2}\\right|$, which is what we needed. To check that this is a saddle pair, take any row $r^{\\prime}$; since $\\left(R_{1}, C_{1}\\right)$ is a saddle pair, there exists $r_{1} \\in R_{1}$ such that $a\\left(r_{1}, c_{1}\\right) \\geqslant a\\left(r^{\\prime}, c_{1}\\right)$ for all $c_{1} \\in C_{1}$. Set now $r_{*}=\\rho^{n k}\\left(r_{1}\\right) \\in R^{n}$. Then, for each $c \\in C^{n}$ we have $c=\\sigma^{n k}(c)$ and hence\n\n$$\na\\left(r_{*}, c\\right)=a\\left(\\rho^{n k}\\left(r_{1}\\right), c\\right) \\geqslant a\\left(r_{1}, \\sigma^{n k}(c)\\right)=a\\left(r_{1}, c\\right) \\geqslant a\\left(r^{\\prime}, c\\right)\n$$\n\nwhich establishes condition $(i)$. Condition (ii) is checked similarly.'] [] Text-only Competition True Theorem proof Combinatorics Math English 183 "Let $A B C$ be an isosceles triangle with $B C=C A$, and let $D$ be a point inside side $A B$ such that $A D\n\nFinally, both $O$ and $M$ lie on lines $\\ell$ and $C M$, therefore $O=M$, and $\\angle A C B=90^{\\circ}$ follows.' 'Like in the first solution, we conclude that points $C, P, Q, D, F$ and the midpoint $M$ of $A B$ lie on one circle $\\omega$ with diameter $C D$, and $M$ lies on $\\ell$, the perpendicular bisector of $P Q$.\n\nLet $B F$ and $C M$ meet at $G$ and let $\\alpha=\\angle A B F$. Then, since $E$ lies on $\\ell$, and the quadrilaterals $F C B A$ and $F C P Q$ are cyclic, we have\n\n$$\n\\angle C Q P=\\angle F P Q=\\angle F C Q=\\angle F C A=\\angle F B A=\\alpha\n$$\n\nSince points $P, E, F$ are collinear, we have\n\n$$\n\\angle F E M=\\angle F E Q+\\angle Q E M=2 \\alpha+\\left(90^{\\circ}-\\alpha\\right)=90^{\\circ}+\\alpha\n$$\n\nBut $\\angle F G M=90^{\\circ}+\\alpha$, so $F E G M$ is cyclic. Hence\n\n$$\n\\angle E G C=\\angle E F M=\\angle P F M=\\angle P C M .\n$$\n\nThus $G E \\| B C$. It follows that $\\angle F A C=\\angle C B F=\\angle E G F$, so $F E G A$ is cyclic, too. Hence $\\angle A C B=\\angle A F B=\\angle A F G=180^{\\circ}-\\angle A M G=90^{\\circ}$, that completes the proof.\n\n\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 184 "Let $A B C D$ be a convex quadrilateral. Suppose that $P$ is a point in the interior of $A B C D$ such that $$ \angle P A D: \angle P B A: \angle D P A=1: 2: 3=\angle C B P: \angle B A P: \angle B P C . $$ The internal bisectors of angles $A D P$ and $P C B$ meet at a point $Q$ inside the triangle $A B P$. Prove that $A Q=B Q$." "['Let $\\varphi=\\angle P A D$ and $\\psi=\\angle C B P$; then we have $\\angle P B A=2 \\varphi, \\angle D P A=3 \\varphi$, $\\angle B A P=2 \\psi$ and $\\angle B P C=3 \\psi$. Let $X$ be the point on segment $A D$ with $\\angle X P A=\\varphi$. Then\n\n$$\n\\angle P X D=\\angle P A X+\\angle X P A=2 \\varphi=\\angle D P A-\\angle X P A=\\angle D P X\n$$\n\nIt follows that triangle $D P X$ is isosceles with $D X=D P$ and therefore the internal angle bisector of $\\angle A D P$ coincides with the perpendicular bisector of $X P$. Similarly, if $Y$ is a point on $B C$ such that $\\angle B P Y=\\psi$, then the internal angle bisector of $\\angle P C B$ coincides with the perpendicular bisector of $P Y$. Hence, we have to prove that the perpendicular bisectors of $X P$, $P Y$, and $A B$ are concurrent.\n\n\n\nNotice that\n\n$$\n\\angle A X P=180^{\\circ}-\\angle P X D=180^{\\circ}-2 \\varphi=180^{\\circ}-\\angle P B A .\n$$\n\nHence the quadrilateral $A X P B$ is cyclic; in other words, $X$ lies on the circumcircle of triangle $A P B$. Similarly, $Y$ lies on the circumcircle of triangle $A P B$. It follows that the perpendicular bisectors of $X P, P Y$, and $A B$ all pass through the center of circle $(A B Y P X)$. This finishes the proof.' 'We define the angles $\\varphi=\\angle P A D, \\psi=\\angle C B P$ and use $\\angle P B A=2 \\varphi, \\angle D P A=$ $3 \\varphi, \\angle B A P=2 \\psi$ and $\\angle B P C=3 \\psi$ again. Let $O$ be the circumcenter of $\\triangle A P B$.\n\nNotice that $\\angle A D P=180^{\\circ}-\\angle P A D-\\angle D P A=180^{\\circ}-4 \\varphi$, which, in particular, means that $4 \\varphi<180^{\\circ}$. Further, $\\angle P O A=2 \\angle P B A=4 \\varphi=180^{\\circ}-\\angle A D P$, therefore the quadrilateral $A D P O$ is cyclic. By $A O=O P$, it follows that $\\angle A D O=\\angle O D P$. Thus $D O$ is the internal bisector of $\\angle A D P$. Similarly, $C O$ is the internal bisector of $\\angle P C B$.\n\n\n\n\n\nFinally, $O$ lies on the perpendicular bisector of $A B$ as it is the circumcenter of $\\triangle A P B$. Therefore the three given lines in the problem statement concur at point $O$.']" [] Text-only Competition True Theorem proof Geometry Math English 185 Let $A B C D$ be a convex quadrilateral with $\angle A B C>90^{\circ}, \angle C D A>90^{\circ}$, and $\angle D A B=\angle B C D$. Denote by $E$ and $F$ the reflections of $A$ in lines $B C$ and $C D$, respectively. Suppose that the segments $A E$ and $A F$ meet the line $B D$ at $K$ and $L$, respectively. Prove that the circumcircles of triangles $B E K$ and $D F L$ are tangent to each other. "['Denote by $A^{\\prime}$ the reflection of $A$ in $B D$. We will show that that the quadrilaterals $A^{\\prime} B K E$ and $A^{\\prime} D L F$ are cyclic, and their circumcircles are tangent to each other at point $A^{\\prime}$.\n\nFrom the symmetry about line $B C$ we have $\\angle B E K=\\angle B A K$, while from the symmetry in $B D$ we have $\\angle B A K=\\angle B A^{\\prime} K$. Hence $\\angle B E K=\\angle B A^{\\prime} K$, which implies that the quadrilateral $A^{\\prime} B K E$ is cyclic. Similarly, the quadrilateral $A^{\\prime} D L F$ is also cyclic.\n\n\n\nFor showing that circles $A^{\\prime} B K E$ and $A^{\\prime} D L F$ are tangent it suffices to prove that\n\n$$\n\\angle A^{\\prime} K B+\\angle A^{\\prime} L D=\\angle B A^{\\prime} D .\n$$\n\nIndeed, by $A K \\perp B C, A L \\perp C D$, and again the symmetry in $B D$ we have\n\n$$\n\\angle A^{\\prime} K B+\\angle A^{\\prime} L D=180^{\\circ}-\\angle K A^{\\prime} L=180^{\\circ}-\\angle K A L=\\angle B C D=\\angle B A D=\\angle B A^{\\prime} D,\n$$\n\nas required.' 'Note that $\\angle K A L=180^{\\circ}-\\angle B C D$, since $A K$ and $A L$ are perpendicular to $B C$ and $C D$, respectively. Reflect both circles $(B E K)$ and $(D F L)$ in $B D$. Since $\\angle K E B=\\angle K A B$ and $\\angle D F L=\\angle D A L$, the images are the circles $(K A B)$ and $(L A D)$, respectively; so they meet at $A$. We need to prove that those two reflections are tangent at $A$.\n\nFor this purpose, we observe that\n\n$$\n\\angle A K B+\\angle A L D=180^{\\circ}-\\angle K A L=\\angle B C D=\\angle B A D .\n$$\n\nThus, there exists a ray $A P$ inside angle $\\angle B A D$ such that $\\angle B A P=\\angle A K B$ and $\\angle D A P=$ $\\angle D L A$. Hence the line $A P$ is a common tangent to the circles $(K A B)$ and $(L A D)$, as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 186 "In the plane, there are $n \geqslant 6$ pairwise disjoint disks $D_{1}, D_{2}, \ldots, D_{n}$ with radii $R_{1} \geqslant R_{2} \geqslant \ldots \geqslant R_{n}$. For every $i=1,2, \ldots, n$, a point $P_{i}$ is chosen in disk $D_{i}$. Let $O$ be an arbitrary point in the plane. Prove that $$ O P_{1}+O P_{2}+\ldots+O P_{n} \geqslant R_{6}+R_{7}+\ldots+R_{n} $$" ['We will make use of the following lemma.\n\nLemma. Let $D_{1}, \\ldots, D_{6}$ be disjoint disks in the plane with radii $R_{1}, \\ldots, R_{6}$. Let $P_{i}$ be a point in $D_{i}$, and let $O$ be an arbitrary point. Then there exist indices $i$ and $j$ such that $O P_{i} \\geqslant R_{j}$.\n\nProof. Let $O_{i}$ be the center of $D_{i}$. Consider six rays $O O_{1}, \\ldots, O O_{6}$ (if $O=O_{i}$, then the ray $O O_{i}$ may be assumed to have an arbitrary direction). These rays partition the plane into six angles (one of which may be non-convex) whose measures sum up to $360^{\\circ}$; hence one of the angles, say $\\angle O_{i} O O_{j}$, has measure at most $60^{\\circ}$. Then $O_{i} O_{j}$ cannot be the unique largest side in (possibly degenerate) triangle $O O_{i} O_{j}$, so, without loss of generality, $O O_{i} \\geqslant O_{i} O_{j} \\geqslant R_{i}+R_{j}$. Therefore, $O P_{i} \\geqslant O O_{i}-R_{i} \\geqslant\\left(R_{i}+R_{j}\\right)-R_{i}=R_{j}$, as desired.\n\nNow we prove the required inequality by induction on $n \\geqslant 5$. The base case $n=5$ is trivial. For the inductive step, apply the Lemma to the six largest disks, in order to find indices $i$ and $j$ such that $1 \\leqslant i, j \\leqslant 6$ and $O P_{i} \\geqslant R_{j} \\geqslant R_{6}$. Removing $D_{i}$ from the configuration and applying the inductive hypothesis, we get\n\n$$\n\\sum_{k \\neq i} O P_{k} \\geqslant \\sum_{\\ell \\geqslant 7} R_{\\ell}\n$$\n\nAdding up this inequality with $O P_{i} \\geqslant R_{6}$ we establish the inductive step.'] [] Text-only Competition True Theorem proof Geometry Math English 187 Let $A B C D$ be a cyclic quadrilateral with no two sides parallel. Let $K, L, M$, and $N$ be points lying on sides $A B, B C, C D$, and $D A$, respectively, such that $K L M N$ is a rhombus with $K L \| A C$ and $L M \| B D$. Let $\omega_{1}, \omega_{2}, \omega_{3}$, and $\omega_{4}$ be the incircles of triangles $A N K$, $B K L, C L M$, and $D M N$, respectively. Prove that the internal common tangents to $\omega_{1}$ and $\omega_{3}$ and the internal common tangents to $\omega_{2}$ and $\omega_{4}$ are concurrent. "['Let $I_{i}$ be the center of $\\omega_{i}$, and let $r_{i}$ be its radius for $i=1,2,3,4$. Denote by $T_{1}$ and $T_{3}$ the points of tangency of $\\omega_{1}$ and $\\omega_{3}$ with $N K$ and $L M$, respectively. Suppose that the internal common tangents to $\\omega_{1}$ and $\\omega_{3}$ meet at point $S$, which is the center of homothety $h$ with negative ratio (namely, with ratio $-\\frac{r_{3}}{r_{1}}$ ) mapping $\\omega_{1}$ to $\\omega_{3}$. This homothety takes $T_{1}$ to $T_{3}$ (since the tangents to $\\omega_{1}$ and $\\omega_{3}$ at $T_{1}$ to $T_{3}$ are parallel), hence $S$ is a point on the segment $T_{1} T_{3}$ with $T_{1} S: S T_{3}=r_{1}: r_{3}$.\n\nConstruct segments $S_{1} S_{3} \\| K L$ and $S_{2} S_{4} \\| L M$ through $S$ with $S_{1} \\in N K, S_{2} \\in K L$, $S_{3} \\in L M$, and $S_{4} \\in M N$. Note that $h$ takes $S_{1}$ to $S_{3}$, hence $I_{1} S_{1} \\| I_{3} S_{3}$, and $S_{1} S: S S_{3}=r_{1}: r_{3}$. We will prove that $S_{2} S: S S_{4}=r_{2}: r_{4}$ or, equivalently, $K S_{1}: S_{1} N=r_{2}: r_{4}$. This will yield the problem statement; indeed, applying similar arguments to the intersection point $S^{\\prime}$ of the internal common tangents to $\\omega_{2}$ and $\\omega_{4}$, we see that $S^{\\prime}$ satisfies similar relations, and there is a unique point inside $K L M N$ satisfying them. Therefore, $S^{\\prime}=S$.\n\n\nFurther, denote by $I_{A}, I_{B}, I_{C}, I_{D}$ and $r_{A}, r_{B}, r_{C}, r_{D}$ the incenters and inradii of triangles $D A B, A B C, B C D$, and $C D A$, respectively. One can shift triangle $C L M$ by $\\overrightarrow{L K}$ to glue it with triangle $A K N$ into a quadrilateral $A K C^{\\prime} N$ similar to $A B C D$. In particular, this shows that $r_{1}: r_{3}=r_{A}: r_{C}$; similarly, $r_{2}: r_{4}=r_{B}: r_{D}$. Moreover, the same shift takes $S_{3}$ to $S_{1}$, and it also takes $I_{3}$ to the incenter $I_{3}^{\\prime}$ of triangle $K C^{\\prime} N$. Since $I_{1} S_{1} \\| I_{3} S_{3}$, the points $I_{1}, S_{1}, I_{3}^{\\prime}$ are collinear. Thus, in order to complete the solution, it suffices to apply the following Lemma to quadrilateral $A K C^{\\prime} N$.\n\nLemma 1. Let $A B C D$ be a cyclic quadrilateral, and define $I_{A}, I_{C}, r_{B}$, and $r_{D}$ as above. Let $I_{A} I_{C}$ meet $B D$ at $X$; then $B X: X D=r_{B}: r_{D}$.\n\nProof. Consider an inversion centered at $X$; the images under that inversion will be denoted by primes, e.g., $A^{\\prime}$ is the image of $A$.\n\nBy properties of inversion, we have\n\n$$\n\\angle I_{C}^{\\prime} I_{A}^{\\prime} D^{\\prime}=\\angle X I_{A}^{\\prime} D^{\\prime}=\\angle X D I_{A}=\\angle B D A / 2=\\angle B C A / 2=\\angle A C I_{B} .\n$$\n\nWe obtain $\\angle I_{A}^{\\prime} I_{C}^{\\prime} D^{\\prime}=\\angle C A I_{B}$ likewise; therefore, $\\triangle I_{C}^{\\prime} I_{A}^{\\prime} D^{\\prime} \\sim \\triangle A C I_{B}$. In the same manner, we get $\\triangle I_{C}^{\\prime} I_{A}^{\\prime} B^{\\prime} \\sim \\triangle A C I_{D}$, hence the quadrilaterals $I_{C}^{\\prime} B^{\\prime} I_{A}^{\\prime} D^{\\prime}$ and $A I_{D} C I_{B}$ are also similar. But the diagonals $A C$ and $I_{B} I_{D}$ of quadrilateral $A I_{D} C I_{B}$ meet at a point $Y$ such that $I_{B} Y$ :\n\n\n\n$Y I_{D}=r_{B}: r_{D}$. By similarity, we get $D^{\\prime} X: B^{\\prime} X=r_{B}: r_{D}$ and hence $B X: X D=D^{\\prime} X:$ $B^{\\prime} X=r_{B}: r_{D}$.' ""This solution is based on the following general Lemma.\n\nLemma 2. Let $E$ and $F$ be distinct points, and let $\\omega_{i}, i=1,2,3,4$, be circles lying in the same halfplane with respect to $E F$. For distinct indices $i, j \\in\\{1,2,3,4\\}$, denote by $O_{i j}^{+}$ (respectively, $O_{i j}^{-}$) the center of homothety with positive (respectively, negative) ratio taking $\\omega_{i}$ to $\\omega_{j}$. Suppose that $E=O_{12}^{+}=O_{34}^{+}$and $F=O_{23}^{+}=O_{41}^{+}$. Then $O_{13}^{-}=O_{24}^{-}$.\n\nProof. Applying Monge's theorem to triples of circles $\\omega_{1}, \\omega_{2}, \\omega_{4}$ and $\\omega_{1}, \\omega_{3}, \\omega_{4}$, we get that both points $O_{24}^{-}$and $O_{13}^{-}$lie on line $E O_{14}^{-}$. Notice that this line is distinct from $E F$. Similarly we obtain that both points $O_{24}^{-}$and $O_{13}^{-}$lie on $\\mathrm{FO}_{34}^{-}$. Since the lines $E O_{14}^{-}$and $\\mathrm{FO}_{34}^{-}$are distinct, both points coincide with the meeting point of those lines.\n\n\n\n\nTurning back to the problem, let $A B$ intersect $C D$ at $E$ and let $B C$ intersect $D A$ at $F$. Assume, without loss of generality, that $B$ lies on segments $A E$ and $C F$. We will show that the points $E$ and $F$, and the circles $\\omega_{i}$ satisfy the conditions of Lemma 2, so the problem statement follows. In the sequel, we use the notation of $O_{i j}^{ \\pm}$from the statement of Lemma 2, applied to circles $\\omega_{1}, \\omega_{2}, \\omega_{3}$, and $\\omega_{4}$.\n\nUsing the relations $\\triangle E C A \\sim \\triangle E B D, K N \\| B D$, and $M N \\| A C$. we get\n\n$$\n\\frac{A N}{N D}=\\frac{A N}{A D} \\cdot \\frac{A D}{N D}=\\frac{K N}{B D} \\cdot \\frac{A C}{N M}=\\frac{A C}{B D}=\\frac{A E}{E D}\n$$\n\nTherefore, by the angle bisector theorem, point $N$ lies on the internal angle bisector of $\\angle A E D$. We prove similarly that $L$ also lies on that bisector, and that the points $K$ and $M$ lie on the internal angle bisector of $\\angle A F B$.\n\nSince $K L M N$ is a rhombus, points $K$ and $M$ are symmetric in line $E L N$. Hence, the convex quadrilateral determined by the lines $E K, E M, K L$, and $M L$ is a kite, and therefore it has an incircle $\\omega_{0}$. Applying Monge's theorem to $\\omega_{0}, \\omega_{2}$, and $\\omega_{3}$, we get that $O_{23}^{+}$lies on $K M$. On the other hand, $O_{23}^{+}$lies on $B C$, as $B C$ is an external common tangent to $\\omega_{2}$ and $\\omega_{3}$. It follows that $F=O_{23}^{+}$. Similarly, $E=O_{12}^{+}=O_{34}^{+}$, and $F=O_{41}^{+}$.""]" [] Text-only Competition True Theorem proof Geometry Math English 188 Let $I$ and $I_{A}$ be the incenter and the $A$-excenter of an acute-angled triangle $A B C$ with $A B\n\nTo show that $\\omega$ and $\\Omega$ are tangent at $T$, let $\\ell$ be the tangent to $\\omega$ at $T$, so that $\\sphericalangle\\left(T I_{A}, \\ell\\right)=$ $\\sphericalangle\\left(E I_{A}, E T\\right)$. Using circles $(B D E T)$ and $\\left(B I C I_{A}\\right)$, we get\n\n$$\n\\sphericalangle\\left(E I_{A}, E T\\right)=\\sphericalangle(E B, E T)=\\sphericalangle(D B, D T) .\n$$\n\nTherefore,\n\n$$\n\\sphericalangle(T I, \\ell)=90^{\\circ}+\\sphericalangle\\left(T I_{A}, \\ell\\right)=90^{\\circ}+\\sphericalangle(D B, D T)=\\sphericalangle(D I, D T)\n$$\n\nwhich shows that $\\ell$ is tangent to $\\Omega$ at $T$.' 'We use the notation of circles $\\Gamma, \\omega$, and $\\Omega$ as in the previous solution.\n\nLet $L$ be the point opposite to $I$ in circle $\\Omega$. Then $\\angle I A L=\\angle I D L=90^{\\circ}$, which means that $L$ is the foot of the external bisector of $\\angle A$ in triangle $A B C$. Let $L I$ cross $\\Gamma$ again at $M$.\n\n\n\nLet $T$ be the foot of the perpendicular from $I$ onto $I_{A} L$. Then $T$ is the second intersection point of $\\Gamma$ and $\\Gamma$. We will show that $T$ is the desired tangency point.\n\nFirst, we show that $T$ lies on circle $\\omega$. Notice that\n\n$$\n\\sphericalangle(L T, L M)=\\sphericalangle(A T, A I) \\text { and } \\quad \\sphericalangle(M T, M L)=\\sphericalangle(M T, M I)=\\sphericalangle\\left(I_{A} T, I_{A} I\\right) \\text {, }\n$$\n\nwhich shows that triangles $T M L$ and $T I_{A} A$ are similar and equioriented. So there exists a rotational homothety $\\tau$ mapping $T M L$ to $T I_{A} A$.\n\nSince $\\sphericalangle(M L, L D)=\\sphericalangle(A I, A D)$, we get $\\tau(B C)=A D$. Next, since\n\n$$\n\\sphericalangle(M B, M L)=\\sphericalangle(M B, M I)=\\sphericalangle\\left(I_{A} B, I_{A} I\\right)=\\sphericalangle\\left(I_{A} E, I_{A} A\\right)\n$$\n\nwe get $\\tau(B)=E$. Similarly, $\\tau(C)=F$. Since the points $M, B, C$, and $T$ are concyclic, so are their $\\tau$-images, which means that $T$ lies on $\\omega=\\tau(\\Gamma)$.\n\n\n\nFinally, since $\\tau(L)=A$ and $\\tau(B)=E$, triangles $A T L$ and $E T B$ are similar so that\n\n$$\n\\sphericalangle(A T, A L)=\\sphericalangle(E T, E B)=\\sphericalangle\\left(E I_{A}, E T\\right) .\n$$\n\nThis means that the tangents to $\\Omega$ and $\\omega$ at $T$ make the same angle with the line $I_{A} T L$, so the circles are indeed tangent at $T$.' 'We also use the notation of circles $\\omega$, and $\\Omega$ from the previous solutions.\n\nPerform an inversion centered at $D$. The images of the points will be denoted by primes, e.g., $A^{\\prime}$ is the image of $A$.\n\nFor convenience, we use the notation $\\angle B I D=\\beta, \\angle C I D=\\gamma$, and $\\alpha=180^{\\circ}-\\beta-\\gamma=$ $90^{\\circ}-\\angle B A I$. We start with computing angles appearing after inversion. We get\n\n$$\n\\begin{gathered}\n\\angle D B^{\\prime} I^{\\prime}=\\beta, \\quad \\angle D C^{\\prime} I^{\\prime}=\\gamma, \\quad \\text { and hence } \\angle B^{\\prime} I^{\\prime} C^{\\prime}=\\alpha ; \\\\\n\\angle E^{\\prime} I_{A}^{\\prime} F^{\\prime}=\\angle E^{\\prime} I_{A}^{\\prime} D-\\angle F^{\\prime} I_{A}^{\\prime} D=\\angle I_{A} E D-\\angle I_{A} F D=\\angle E I_{A} F=180^{\\circ}-\\alpha .\n\\end{gathered}\n$$\n\nNext, we have\n\n$$\n\\angle A^{\\prime} E^{\\prime} B^{\\prime}=\\angle D E^{\\prime} B^{\\prime}=\\angle D B E=\\beta=90^{\\circ}-\\frac{\\angle D B A}{2}=90^{\\circ}-\\frac{\\angle E^{\\prime} A^{\\prime} B^{\\prime}}{2},\n$$\n\nwhich yields that triangle $A^{\\prime} B^{\\prime} E^{\\prime}$ is isosceles with $A^{\\prime} B^{\\prime}=A^{\\prime} E^{\\prime}$. Similarly, $A^{\\prime} F^{\\prime}=A^{\\prime} C^{\\prime}$.\n\nFinally, we get\n\n$$\n\\begin{aligned}\n\\angle A^{\\prime} B^{\\prime} I^{\\prime}=\\angle I^{\\prime} B^{\\prime} D-\\angle A^{\\prime} B^{\\prime} D=\\beta- & \\angle B A D=\\beta-\\left(90^{\\circ}-\\alpha\\right)+\\angle I A D \\\\\n& =\\angle I C D+\\angle I A D=\\angle C^{\\prime} I^{\\prime} D+\\angle A^{\\prime} I^{\\prime} D=\\angle C^{\\prime} I^{\\prime} A^{\\prime}\n\\end{aligned}\n$$\n\nsimilarly, $\\angle A^{\\prime} C^{\\prime} I^{\\prime}=\\angle A^{\\prime} I^{\\prime} B^{\\prime}$, so that triangles $A^{\\prime} B^{\\prime} I^{\\prime}$ and $A^{\\prime} I^{\\prime} C^{\\prime}$ are similar. Therefore, $A^{\\prime} I^{2}=A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{\\prime}$.\n\nRecall that we need to prove the tangency of line $A^{\\prime} I^{\\prime}=\\Omega^{\\prime}$ with circle $\\left(E^{\\prime} F^{\\prime} I_{A}^{\\prime}\\right)=\\omega^{\\prime}$. A desired tangency point $T^{\\prime}$ must satisfy $A^{\\prime} T^{\\prime 2}=A^{\\prime} E^{\\prime} \\cdot A^{\\prime} F^{\\prime}$; the relations obtained above yield\n\n$$\nA^{\\prime} E^{\\prime} \\cdot A^{\\prime} F^{\\prime}=A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{\\prime}=A^{\\prime} I^{\\prime 2},\n$$\n\nso that $T^{\\prime}$ should be symmetric to $I^{\\prime}$ with respect to $A^{\\prime}$.\n\nThus, let us define a point $T^{\\prime}$ as the reflection of $I^{\\prime}$ in $A^{\\prime}$, and show that $T^{\\prime}$ lies on circle $\\Omega^{\\prime}$; the equalities above will then imply that $A^{\\prime} T^{\\prime}$ is tangent to $\\Omega^{\\prime}$, as desired.\n\n\nThe property that triangles $B^{\\prime} A^{\\prime} I^{\\prime}$ and $I^{\\prime} A^{\\prime} C^{\\prime}$ are similar means that quadrilateral $B^{\\prime} I^{\\prime} C^{\\prime} T^{\\prime}$ is harmonic. Indeed, let $C^{*}$ be the reflection of $C^{\\prime}$ in the perpendicular bisector of $I^{\\prime} T^{\\prime}$; then $C^{*}$ lies on $B^{\\prime} A^{\\prime}$ by $\\angle B^{\\prime} A^{\\prime} I^{\\prime}=\\angle A^{\\prime} I^{\\prime} C^{\\prime}=\\angle T^{\\prime} I^{\\prime} C^{*}$, and then $C^{*}$ lies on circle $\\left(I^{\\prime} B^{\\prime} T^{\\prime}\\right)$ since $A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{*}=A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{\\prime}=A^{\\prime} I^{\\prime 2}=A^{\\prime} I^{\\prime} \\cdot A^{\\prime} T^{\\prime}$. Therefore, $C^{\\prime}$ also lies on that circle (and the circle is $\\left.\\left(B^{\\prime} I^{\\prime} C^{\\prime}\\right)=\\Gamma^{\\prime}\\right)$. Moreover, $B^{\\prime} C^{*}$ is a median in triangle $B^{\\prime} I^{\\prime} T^{\\prime}$, so $B^{\\prime} C^{\\prime}$ is its symmedian, which establishes harmonicity.\n\nNow we have $\\angle A^{\\prime} B^{\\prime} T^{\\prime}=\\angle I^{\\prime} B^{\\prime} C^{\\prime}=\\beta=\\angle A^{\\prime} B^{\\prime} E^{\\prime}$; which shows that $E^{\\prime}$ lies on $B^{\\prime} T^{\\prime}$. Similarly, $F^{\\prime}$ lies on $C^{\\prime} T^{\\prime}$. Hence, $\\angle E^{\\prime} T^{\\prime} F^{\\prime}=\\angle B^{\\prime} I^{\\prime} C^{\\prime}=180^{\\circ}-\\angle E^{\\prime} I_{A}^{\\prime} F^{\\prime}$, which establishes $T^{\\prime} \\in \\omega^{\\prime}$.']" [] Text-only Competition True Theorem proof Geometry Math English 189 "Let $P$ be a point on the circumcircle of an acute-angled triangle $A B C$. Let $D$, $E$, and $F$ be the reflections of $P$ in the midlines of triangle $A B C$ parallel to $B C, C A$, and $A B$, respectively. Denote by $\omega_{A}, \omega_{B}$, and $\omega_{C}$ the circumcircles of triangles $A D P, B E P$, and $C F P$, respectively. Denote by $\omega$ the circumcircle of the triangle formed by the perpendicular bisectors of segments $A D, B E$ and $C F$. Show that $\omega_{A}, \omega_{B}, \omega_{C}$, and $\omega$ have a common point." "[""Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes in triangle $A B C$, and let $m_{A}, m_{B}$, and $m_{C}$ be the midlines parallel to $B C, C A$, and $A B$, respectively. We always denote by $\\sphericalangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\\circ}$.\n\nStep 1: Circles $\\omega_{A}, \\omega_{B}$, and $\\omega_{C}$ share a common point $Q$ different from $P$.\n\nNotice that $m_{A}$ is the perpendicular bisector of $P D$, so $\\omega_{A}$ is symmetric with respect to $m_{A}$. Since $A$ and $A_{1}$ are also symmetric to each other in $m_{A}$, the point $A_{1}$ lies on $\\omega_{A}$. Similarly, $B_{1}$ and $C_{1}$ lie on $\\omega_{B}$ and $\\omega_{C}$, respectively.\n\nLet $H$ be the orthocenter of $\\triangle A B C$. Quadrilaterals $A B A_{1} B_{1}$ and $B C B_{1} C_{1}$ are cyclic, so $A H \\cdot H A_{1}=B H \\cdot H B_{1}=C H \\cdot H C_{1}$. This means that $H$ lies on pairwise radical axes of $\\omega_{A}$, $\\omega_{B}$, and $\\omega_{C}$. Point $P$ also lies on those radical axes; hence the three circles have a common radical axis $\\ell=P H$, and the second meeting point $Q$ of $\\ell$ with $\\omega_{A}$ is the second common point of the three circles. Notice here that $H$ lies inside all three circles, hence $Q \\neq P$.\n\n\nStep 2: Point $Q$ lies on $\\omega$.\n\nLet $p_{A}, p_{B}$, and $p_{C}$ denote the perpendicular bisectors of $A D, B E$, and $C F$, respectively; denote by $\\Delta$ the triangle formed by those perpendicular bisectors. By Simson's theorem, in order to show that $Q$ lies on the circumcircle $\\omega$ of $\\Delta$, it suffices to prove that the projections of $Q$ onto the sidelines $p_{A}, p_{B}$, and $p_{C}$ are collinear. Alternatively, but equivalently, it suffices to prove that the reflections $Q_{A}, Q_{B}$, and $Q_{C}$ of $Q$ in those lines, respectively, are collinear. In fact, we will show that four points $P, Q_{A}, Q_{B}$, and $Q_{C}$ are collinear.\n\nSince $p_{A}$ is the common perpendicular bisector of $A D$ and $Q Q_{A}$, the point $Q_{A}$ lies on $\\omega_{A}$, and, moreover, $\\sphericalangle\\left(D A, D Q_{A}\\right)=\\sphericalangle(A Q, A D)$. Therefore,\n\n$$\n\\sphericalangle\\left(P A, P Q_{A}\\right)=\\sphericalangle\\left(D A, D Q_{A}\\right)=\\sphericalangle(A Q, A D)=\\sphericalangle(P Q, P D)=\\sphericalangle(P Q, B C)+90^{\\circ}\n$$\n\n\n\nSimilarly, we get $\\sphericalangle\\left(P B, P Q_{B}\\right)=\\sphericalangle(P Q, C A)+90^{\\circ}$. Therefore,\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(P Q_{A}, P Q_{B}\\right)=\\sphericalangle\\left(P Q_{A}, P A\\right) & +\\sphericalangle(P A, P B)+\\sphericalangle\\left(P B, P Q_{B}\\right) \\\\\n& =\\sphericalangle(B C, P Q)+90^{\\circ}+\\sphericalangle(C A, C B)+\\sphericalangle(P Q, C A)+90^{\\circ}=0,\n\\end{aligned}\n$$\n\nwhich shows that $P, Q_{A}$, and $Q_{B}$ are collinear. Similarly, $Q_{C}$ also lies on $P Q_{A}$.""]" [] Text-only Competition True Theorem proof Geometry Math English 190 "Let $\Gamma$ and $I$ be the circumcircle and the incenter of an acute-angled triangle $A B C$. Two circles $\omega_{B}$ and $\omega_{C}$ passing through $B$ and $C$, respectively, are tangent at $I$. Let $\omega_{B}$ meet the shorter arc $A B$ of $\Gamma$ and segment $A B$ again at $P$ and $M$, respectively. Similarly, let $\omega_{C}$ meet the shorter arc $A C$ of $\Gamma$ and segment $A C$ again at $Q$ and $N$, respectively. The rays $P M$ and $Q N$ meet at $X$, and the tangents to $\omega_{B}$ and $\omega_{C}$ at $B$ and $C$, respectively, meet at $Y$. Prove that the points $A, X$, and $Y$ are collinear." "['Let $A I, B I$, and $C I$ meet $\\Gamma$ again at $D, E$, and $F$, respectively. Let $\\ell$ be the common tangent to $\\omega_{B}$ and $\\omega_{C}$ at $I$. We always denote by $\\sphericalangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\\circ}$.\n\nStep 1: We show that $Y$ lies on $\\Gamma$.\n\nRecall that any chord of a circle makes complementary directed angles with the tangents to the circle at its endpoints. Hence,\n\n$$\n\\sphericalangle(B Y, B I)+\\sphericalangle(C I, C Y)=\\sphericalangle(I B, \\ell)+\\sphericalangle(\\ell, I C)=\\sphericalangle(I B, I C) .\n$$\n\nTherefore,\n\n$$\n\\begin{array}{r}\n\\sphericalangle(B Y, B A)+\\sphericalangle(C A, C Y)=\\sphericalangle(B I, B A)+\\sphericalangle(B Y, B I)+\\sphericalangle(C I, C Y)+\\sphericalangle(C A, C I) \\\\\n=\\sphericalangle(B C, B I)+\\sphericalangle(I B, I C)+\\sphericalangle(C I, C B)=0,\n\\end{array}\n$$\n\nwhich yields $Y \\in \\Gamma$.\n\n\n\nStep 2: We show that $X=\\ell \\cap E F$.\n\nLet $X_{*}=\\ell \\cap E F$. To prove our claim, it suffices to show that $X_{*}$ lies on both $P M$ and $Q N$; this will yield $X_{*}=X$. Due to symmetry, it suffices to show $X_{*} \\in Q N$.\n\nNotice that\n\n$$\n\\sphericalangle\\left(I X_{*}, I Q\\right)=\\sphericalangle(C I, C Q)=\\sphericalangle(C F, C Q)=\\sphericalangle(E F, E Q)=\\sphericalangle\\left(E X_{*}, E Q\\right) ;\n$$\n\n\n\ntherefore, the points $X_{*}, I, Q$, and $E$ are concyclic (if $Q=E$, then the direction of $E Q$ is supposed to be the direction of a tangent to $\\Gamma$ at $Q$; in this case, the equality means that the circle $\\left(X_{*} I Q\\right)$ is tangent to $\\Gamma$ at $\\left.Q\\right)$. Then we have\n\n$$\n\\sphericalangle\\left(Q X_{*}, Q I\\right)=\\sphericalangle\\left(E X_{*}, E I\\right)=\\sphericalangle(E F, E B)=\\sphericalangle(C A, C F)=\\sphericalangle(C N, C I)=\\sphericalangle(Q N, Q I) \\text {, }\n$$\n\nwhich shows that $X_{*} \\in Q N$.\n\nStep 3: We finally show that $A, X$, and $Y$ are collinear.\n\nRecall that $I$ is the orthocenter of triangle $D E F$, and $A$ is symmetric to $I$ with respect to $E F$. Therefore,\n\n$$\n\\sphericalangle(A X, A E)=\\sphericalangle(I E, I X)=\\sphericalangle(B I, \\ell)=\\sphericalangle(B Y, B I)=\\sphericalangle(B Y, B E)=\\sphericalangle(A Y, A E) \\text {, }\n$$\n\nwhich yields the desired collinearity.' 'Perform an inversion centered at $I$; the images of the points are denoted by primes, e.g., $A^{\\prime}$ is the image of $A$.\n\nOn the inverted figure, $I$ and $\\Gamma^{\\prime}$ are the orthocenter and the circumcircle of triangle $A^{\\prime} B^{\\prime} C^{\\prime}$, respectively. The points $P^{\\prime}$ and $Q^{\\prime}$ lie on $\\Gamma^{\\prime}$ such that $B^{\\prime} P^{\\prime} \\| C^{\\prime} Q^{\\prime}$ (since $B^{\\prime} P^{\\prime}=\\omega_{B}^{\\prime}$ and $C^{\\prime} Q^{\\prime}=\\omega_{C}^{\\prime}$ ). The points $M^{\\prime}$ and $N^{\\prime}$ are the second intersections of lines $B^{\\prime} P^{\\prime}$ and $C^{\\prime} Q^{\\prime}$ with the circumcircles $\\gamma_{B}$ and $\\gamma_{C}$ of triangles $A^{\\prime} I B^{\\prime}$ and $A^{\\prime} I C^{\\prime}$, respectively. Notice here that $\\gamma_{C}$ is obtained from $\\gamma_{B}$ by the translation at $\\overrightarrow{B^{\\prime} C^{\\prime}}$; the same translation maps line $B^{\\prime} P^{\\prime}$ to $C^{\\prime} Q^{\\prime}$, and hence $M^{\\prime}$ to $N^{\\prime}$. In other words, $B^{\\prime} M^{\\prime} N^{\\prime} C^{\\prime}$ is a parallelogram, and $P^{\\prime} Q^{\\prime}$ partitions it into two isosceles trapezoids.\n\nPoint $X^{\\prime}$ is the second intersection point of circles $\\left(I P^{\\prime} M^{\\prime}\\right)$ and $\\left(I Q^{\\prime} N^{\\prime}\\right)$ that is - the reflection of $I$ in their line of centers. But the centers lie on the common perpendicular bisector $p$ of $P^{\\prime} M^{\\prime}$ and $Q^{\\prime} N^{\\prime}$, so $p$ is that line of centers. Hence, $I X^{\\prime} \\| B^{\\prime} P^{\\prime}$, as both lines are perpendicular to $p$.\n\nFinally, the point $Y$ satisfies $\\sphericalangle(B Y, B I)=\\sphericalangle(P B, P I)$ and $\\sphericalangle(C Y, C I)=\\sphericalangle(Q C, Q I)$, which yields $\\sphericalangle\\left(Y^{\\prime} B^{\\prime}, Y^{\\prime} I\\right)=\\sphericalangle\\left(B^{\\prime} P^{\\prime}, B^{\\prime} I\\right)$ and $\\sphericalangle\\left(Y^{\\prime} C^{\\prime}, Y^{\\prime} I\\right)=\\sphericalangle\\left(C^{\\prime} Q^{\\prime}, C^{\\prime} I\\right)$. Therefore,\n\n$$\n\\sphericalangle\\left(Y^{\\prime} B^{\\prime}, Y^{\\prime} C^{\\prime}\\right)=\\sphericalangle\\left(B^{\\prime} P^{\\prime}, B^{\\prime} I\\right)+\\sphericalangle\\left(C^{\\prime} I, C^{\\prime} Q^{\\prime}\\right)=\\sphericalangle\\left(C^{\\prime} I, B^{\\prime} I\\right)=\\sphericalangle\\left(A^{\\prime} B^{\\prime}, A^{\\prime} C^{\\prime}\\right)\n$$\n\nwhich shows that $Y^{\\prime} \\in \\Gamma^{\\prime}$.\n\nIn congruent circles $\\Gamma^{\\prime}$ and $\\gamma_{B}$, the chords $A^{\\prime} P^{\\prime}$ and $A^{\\prime} M^{\\prime}$ subtend the same angle $\\angle A^{\\prime} B^{\\prime} P^{\\prime}$; therefore, $A^{\\prime} P^{\\prime}=A^{\\prime} M^{\\prime}$, and hence $A^{\\prime} \\in p$. This yields $A^{\\prime} X^{\\prime}=A^{\\prime} I$, and hence $\\sphericalangle\\left(I A^{\\prime}, I X^{\\prime}\\right)=$ $\\sphericalangle\\left(X^{\\prime} I, X^{\\prime} A^{\\prime}\\right)$.\n\nFinally, we have\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(Y^{\\prime} I, Y^{\\prime} A^{\\prime}\\right) & =\\sphericalangle\\left(Y^{\\prime} I, Y^{\\prime} B^{\\prime}\\right)+\\sphericalangle\\left(Y^{\\prime} B^{\\prime}, Y^{\\prime} A^{\\prime}\\right) \\\\\n& =\\sphericalangle\\left(B^{\\prime} I, B^{\\prime} P^{\\prime}\\right)+\\sphericalangle\\left(I A^{\\prime}, I B^{\\prime}\\right)=\\sphericalangle\\left(I A^{\\prime}, B^{\\prime} P^{\\prime}\\right)=\\sphericalangle\\left(I A^{\\prime}, I X^{\\prime}\\right)=\\sphericalangle\\left(X^{\\prime} I, X^{\\prime} A^{\\prime}\\right),\n\\end{aligned}\n$$\n\nwhich yields that the points $A^{\\prime}, X^{\\prime}, Y^{\\prime}$, and $I$ are concyclic. This means exactly that $A, X$, and $Y$ are collinear.\n\n\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 191 "Prove that there exists a positive constant $c$ such that the following statement is true: Assume that $n$ is an integer with $n \geqslant 2$, and let $\mathcal{S}$ be a set of $n$ points in the plane such that the distance between any two distinct points in $\mathcal{S}$ is at least 1 . Then there is a line $\ell$ separating $\mathcal{S}$ such that the distance from any point of $\mathcal{S}$ to $\ell$ is at least $c n^{-1 / 3}$. (A line $\ell$ separates a point set $\mathcal{S}$ if some segment joining two points in $\mathcal{S}$ crosses $\ell$.)" ['We prove that the desired statement is true with $c=\\frac{1}{8}$. Set $\\delta=\\frac{1}{8} n^{-1 / 3}$. For any line $\\ell$ and any point $X$, let $X_{\\ell}$ denote the projection of $X$ to $\\ell$; a similar notation applies to sets of points.\n\nSuppose that, for some line $\\ell$, the set $\\mathcal{S}_{\\ell}$ contains two adjacent points $X$ and $Y$ with $X Y=2 d$. Then the line perpendicular to $\\ell$ and passing through the midpoint of segment $X Y$ separates $\\mathcal{S}$, and all points in $\\mathcal{S}$ are at least $d$ apart from $\\ell$. Thus, if $d \\geqslant \\delta$, then a desired line has been found. For the sake of contradiction, we assume that no such points exist, in any projection.\n\nChoose two points $A$ and $B$ in $\\mathcal{S}$ with the maximal distance $M=A B$ (i.e., $A B$ is a diameter of $\\mathcal{S})$; by the problem condition, $M \\geqslant 1$. Denote by $\\ell$ the line $A B$. The set $\\mathcal{S}$ is contained in the intersection of two disks $D_{A}$ and $D_{B}$ of radius $M$ centered at $A$ and $B$, respectively. Hence, the projection $\\mathcal{S}_{\\ell}$ is contained in the segment $A B$. Moreover, the points in $\\mathcal{S}_{\\ell}$ divide that segment into at most $n-1$ parts, each of length less than $2 \\delta$. Therefore,\n\n$$\nM\n\nChoose a point $H$ on segment $A B$ with $A H=\\frac{1}{2}$. Let $P$ be a strip between the lines $a$ and $h$ perpendicular to $A B$ and passing through $A$ and $H$, respectively; we assume that $P$ contains its boundary, which consists of lines $a$ and $h$. Set $\\mathcal{T}=P \\cap \\mathcal{S}$ and let $t=|\\mathcal{T}|$. By our assumption, segment $A H$ contains at least $\\left\\lceil\\frac{1}{2}:(2 \\delta)\\right\\rceil$ points of $S_{\\ell}$, which yields\n\n$$\nt \\geqslant \\frac{1}{4 \\delta}\\tag{2}\n$$\n\nNotice that $\\mathcal{T}$ is contained in $Q=P \\cap D_{B}$. The set $Q$ is a circular segment, and its projection $Q_{a}$ is a line segment of length\n\n$$\n2 \\sqrt{M^{2}-\\left(M-\\frac{1}{2}\\right)^{2}}<2 \\sqrt{M}\n$$\n\nOn the other hand, for any two points $X, Y \\in \\mathcal{T}$, we have $X Y \\geqslant 1$ and $X_{\\ell} Y_{\\ell} \\leqslant \\frac{1}{2}$, so $X_{a} Y_{a}=$ $\\sqrt{X Y^{2}-X_{\\ell} Y_{\\ell}^{2}} \\geqslant \\frac{\\sqrt{3}}{2}$. To summarize, $t$ points constituting $\\mathcal{T}_{a}$ lie on the segment of length less than $2 \\sqrt{M}$, and are at least $\\frac{\\sqrt{3}}{2}$ apart from each other. This yields $2 \\sqrt{M}>(t-1) \\frac{\\sqrt{3}}{2}$, or\n\n$$\nt<1+\\frac{4 \\sqrt{M}}{\\sqrt{3}}<4 \\sqrt{M}\\tag{3}\n$$\n\n\n\nas $M \\geqslant 1$.\n\nCombining the estimates (1), (2), and (3), we finally obtain\n\n$$\n\\frac{1}{4 \\delta} \\leqslant t<4 \\sqrt{M}<4 \\sqrt{2 n \\delta}, \\quad \\text { or } \\quad 512 n \\delta^{3}>1\n$$\n\nwhich does not hold for the chosen value of $\\delta$.'] [] Text-only Competition True Theorem proof Number Theory Math English 192 Given a positive integer $k$, show that there exists a prime $p$ such that one can choose distinct integers $a_{1}, a_{2}, \ldots, a_{k+3} \in\{1,2, \ldots, p-1\}$ such that $p$ divides $a_{i} a_{i+1} a_{i+2} a_{i+3}-i$ for all $i=1,2, \ldots, k$. ['First we choose distinct positive rational numbers $r_{1}, \\ldots, r_{k+3}$ such that\n\n$$\nr_{i} r_{i+1} r_{i+2} r_{i+3}=i \\text { for } 1 \\leqslant i \\leqslant k .\n$$\n\nLet $r_{1}=x, r_{2}=y, r_{3}=z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_{4}=\\frac{1}{r_{1} r_{2} r_{3}}$ and $r_{i+4}=\\frac{i+1}{i} r_{i}$. It follows that if $r_{i}$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \\equiv 1(\\bmod 4)$, by $y$ for $i \\equiv 2(\\bmod 4)$, by $z$ for $i \\equiv 3(\\bmod 4)$ and by none for $i \\equiv 0(\\bmod 4)$. Notice that $r_{i}3$ dividing a number of the form $x^{2}-x+1$ with integer $x$ there are two unconnected islands in $p$-Landia.\n\nFor brevity's sake, when a bridge connects the islands numbered $m$ and $n$, we shall speak simply that it connects $m$ and $n$.\n\nA bridge connects $m$ and $n$ if $n \\equiv m^{2}+1(\\bmod p)$ or $m \\equiv n^{2}+1(\\bmod p)$. If $m^{2}+1 \\equiv n$ $(\\bmod p)$, we draw an arrow starting at $m$ on the bridge connecting $m$ and $n$. Clearly only one arrow starts at $m$ if $m^{2}+1 \\not \\equiv m(\\bmod p)$, and no arrows otherwise. The total number of bridges does not exceed the total number of arrows.\n\nSuppose $x^{2}-x+1 \\equiv 0(\\bmod p)$. We may assume that $1 \\leqslant x \\leqslant p$; then there is no arrow starting at $x$. Since $(1-x)^{2}-(1-x)+1=x^{2}-x+1,(p+1-x)^{2}+1 \\equiv(p+1-x)(\\bmod p)$, and there is also no arrow starting at $p+1-x$. If $x=p+1-x$, that is, $x=\\frac{p+1}{2}$, then $4\\left(x^{2}-x+1\\right)=p^{2}+3$ and therefore $x^{2}-x+1$ is not divisible by $p$. Thus the islands $x$ and $p+1-x$ are different, and no arrows start at either of them. It follows that the total number of bridges in $p$-Landia does not exceed $p-2$.\n\nLet $1,2, \\ldots, p$ be the vertices of a graph $G_{p}$, where an edge connects $m$ and $n$ if and only if there is a bridge between $m$ and $n$. The number of vertices of $G_{p}$ is $p$ and the number of edges is less than $p-1$. This means that the graph is not connected, which means that there are two islands not connected by a chain of bridges.\n\nIt remains to prove that there are infinitely many primes $p$ dividing $x^{2}-x+1$ for some integer $x$. Let $p_{1}, p_{2}, \\ldots, p_{k}$ be any finite set of such primes. The number $\\left(p_{1} p_{2} \\cdot \\ldots \\cdot p_{k}\\right)^{2}-p_{1} p_{2} \\cdot \\ldots \\cdot p_{k}+1$ is greater than 1 and not divisible by any $p_{i}$; therefore it has another prime divisor with the required property."" 'One can show, by using only arithmetical methods, that for infinitely many $p$, the kingdom of $p$-Ladia contains two islands connected to no other island, except for each other.\n\nLet arrows between islands have the same meaning as in the previous solution. Suppose that positive $a3$. It follows that $a b \\equiv a(1-a) \\equiv 1$ $(\\bmod p)$. If an arrow goes from $t$ to $a$, then $t$ must satisfy the congruence $t^{2}+1 \\equiv a \\equiv a^{2}+1$ $(\\bmod p)$; the only such $t \\neq a$ is $p-a$. Similarly, the only arrow going to $b$ goes from $p-b$. If one of the numbers $p-a$ and $p-b$, say, $p-a$, is not at the end of any arrow, the pair $a, p-a$ is not connected with the rest of the islands. This is true if at least one of the congruences $x^{2}+1 \\equiv-a, x^{2}+1 \\equiv-b$ has no solutions, that is, either $-a-1$ or $-b-1$ is a quadratic non-residue modulo $p$.\n\nNote that $x^{2}-x+1 \\equiv x^{2}-(a+b) x+a b \\equiv(x-a)(x-b)(\\bmod p)$. Substituting $x=-1$ we get $(-1-a)(-1-b) \\equiv 3(\\bmod p)$. If 3 is a quadratic non-residue modulo $p$, so is one of the numbers $-1-a$ and $-1-b$.\n\nThus it is enough to find infinitely many primes $p>3$ dividing $x^{2}-x+1$ for some integer $x$ and such that 3 is a quadratic non-residue modulo $p$.\n\nIf $x^{2}-x+1 \\equiv 0(\\bmod p)$ then $(2 x-1)^{2} \\equiv-3(\\bmod p)$, that is, -3 is a quadratic residue modulo $p$, so 3 is a quadratic non-residue if and only if -1 is also a non-residue, in other words, $p \\equiv-1(\\bmod 4)$.\n\nSimilarly to the first solution, let $p_{1}, \\ldots, p_{k}$ be primes congruent to -1 modulo 4 and dividing numbers of the form $x^{2}-x+1$. The number $\\left(2 p_{1} \\cdot \\ldots \\cdot p_{k}\\right)^{2}-2 p_{1} \\cdot \\ldots \\cdot p_{k}+1$ is\n\n\n\nnot divisible by any $p_{i}$ and is congruent to -1 modulo 4 , therefore, it has some prime divisor $p \\equiv-1(\\bmod 4)$ which has the required properties.']" [] Text-only Competition True Theorem proof Number Theory Math English 194 Let $\mathcal{S}$ be a set consisting of $n \geqslant 3$ positive integers, none of which is a sum of two other distinct members of $\mathcal{S}$. Prove that the elements of $\mathcal{S}$ may be ordered as $a_{1}, a_{2}, \ldots, a_{n}$ so that $a_{i}$ does not divide $a_{i-1}+a_{i+1}$ for all $i=2,3, \ldots, n-1$. "['In all solutions, we call a set $\\mathcal{S}$ of positive integers good if no its element is a sum of two other distinct members of $\\mathcal{S}$. We will use the following simple observation.\n\nObservation A. If $a, b$, and $c$ are three distinct elements of a good set $\\mathcal{S}$ with $b>a, c$, then $b \\nmid a+c$. Otherwise, since $b \\neq a+c$, we would have $b \\leqslant(a+c) / 2<\\max \\{a, c\\}$.\n\nWe prove the following stronger statement.\n\nClaim. Let $\\mathcal{S}$ be a good set consisting of $n \\geqslant 2$ positive integers. Then the elements of $\\mathcal{S}$ may be ordered as $a_{1}, a_{2}, \\ldots, a_{n}$ so that $a_{i} \\nmid a_{i-1}+a_{i+1}$ and $a_{i} \\nmid a_{i-1}-a_{i+1}$, for all $i=2,3, \\ldots, n-1$.\n\nProof. Say that the ordering $a_{1}, \\ldots, a_{n}$ of $\\mathcal{S}$ is nice if it satisfies the required property.\n\nWe proceed by induction on $n$. The base case $n=2$ is trivial, as there are no restrictions on the ordering.\n\nTo perform the step of induction, suppose that $n \\geqslant 3$. Let $a=\\max \\mathcal{S}$, and set $\\mathcal{T}=\\mathcal{S} \\backslash\\{a\\}$. Use the inductive hypothesis to find a nice ordering $b_{1}, \\ldots, b_{n-1}$ of $\\mathcal{T}$. We will show that $a$ may be inserted into this sequence so as to reach a nice ordering of $\\mathcal{S}$. In other words, we will show that there exists a $j \\in\\{1,2, \\ldots, n\\}$ such that the ordering\n\n$$\nN_{j}=\\left(b_{1}, \\ldots, b_{j-1}, a, b_{j}, b_{j+1}, \\ldots, b_{n-1}\\right)\n$$\n\nis nice.\n\nAssume that, for some $j$, the ordering $N_{j}$ is not nice, so that some element $x$ in it divides either the sum or the difference of two adjacent ones. This did not happen in the ordering of $\\mathcal{T}$, hence $x \\in\\left\\{b_{j-1}, a, b_{j}\\right\\}$ (if, say, $b_{j-1}$ does not exist, then $x \\in\\left\\{a, b_{j}\\right\\}$; a similar agreement is applied hereafter). But the case $x=a$ is impossible: $a$ cannot divide $b_{j-1}-b_{j}$, since $0<\\left|b_{j-1}-b_{j}\\right|a, c$, then $b \\nmid a+c$. Otherwise, since $b \\neq a+c$, we would have $b \\leqslant(a+c) / 2<\\max \\{a, c\\}$.\n\n\nWe again prove a stronger statement.\n\nClaim. Let $\\mathcal{S}$ be an arbitrary set of $n \\geqslant 3$ positive integers. Then its elements can be ordered as $a_{1}, \\ldots, a_{n}$ so that, if $a_{i} \\mid a_{i-1}+a_{i+1}$, then $a_{i}=\\max \\mathcal{S}$.\n\nThe claim easily implies what we need to prove, due to Observation A.\n\nTo prove the Claim, introduce the function $f$ which assigns to any two elements $a, b \\in \\mathcal{S}$ with $aa_{j+1}>\\ldots>$ $a_{n} ;$ and\n\n(ii) $f$-avoidance: If $a0$ and $a<0$. Indeed, since $\\sum_{i=1}^{2019} u_{i}^{2}=1$, the variables $u_{i}$ cannot be all zero, and, since $\\sum_{i=1}^{2019} u_{i}=0$, the nonzero elements cannot be all positive or all negative.\n\nLet $P=\\left\\{i: u_{i}>0\\right\\}$ and $N=\\left\\{i: u_{i} \\leqslant 0\\right\\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\\sum_{i=1}^{2019} u_{i}=0$ we have $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i \\in P} u_{i}-\\sum_{i \\in N}\\left|u_{i}\\right|$, so\n\n$$\n\\sum_{i \\in P} u_{i}=\\sum_{i \\in N}\\left|u_{i}\\right|\n\\tag{1}\n$$\n\nAfter this preparation, estimate the sum of squares of the positive and nonpositive elements as follows:\n\n$$\n\\sum_{i \\in P} u_{i}^{2} \\leqslant \\sum_{i \\in P} b u_{i}=b \\sum_{i \\in P} u_{i}=b \\sum_{i \\in N}\\left|u_{i}\\right| \\leqslant b \\sum_{i \\in N}|a|=-n a b\n\\tag{2}\n$$\n$$\n\\sum_{i \\in N} u_{i}^{2} \\leqslant \\sum_{i \\in N}|a| \\cdot\\left|u_{i}\\right|=|a| \\sum_{i \\in N}\\left|u_{i}\\right|=|a| \\sum_{i \\in P} u_{i} \\leqslant|a| \\sum_{i \\in P} b=-p a b .\n\\tag{3}\n$$\n\nThe sum of these estimates is\n\n$$\n1=\\sum_{i=1}^{2019} u_{i}^{2}=\\sum_{i \\in P} u_{i}^{2}+\\sum_{i \\in N} u_{i}^{2} \\leqslant-(p+n) a b=-2019 a b\n$$\n\nthat proves $a b \\leqslant \\frac{-1}{2019}$.' 'As in the previous solution we conclude that $a<0$ and $b>0$.\n\nFor every index $i$, the number $u_{i}$ is a convex combination of $a$ and $b$, so\n\n$$\nu_{i}=x_{i} a+y_{i} b \\quad \\text { with some weights } 0 \\leqslant x_{i}, y_{i} \\leqslant 1 \\text {, with } x_{i}+y_{i}=1 \\text {. }\n$$\n\nLet $X=\\sum_{i=1}^{2019} x_{i}$ and $Y=\\sum_{i=1}^{2019} y_{i}$. From $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i=1}^{2019}\\left(x_{i} a+y_{i} b\\right)=-|a| X+b Y$, we get\n\n$$\n|a| X=b Y\n\\tag{4}\n$$\n\n\n\nFrom $\\sum_{i=1}^{2019}\\left(x_{i}+y_{i}\\right)=2019$ we have\n\n$$\nX+Y=2019\n\\tag{5}\n$$\n\nThe system of linear equations $(4,5)$ has a unique solution:\n\n$$\nX=\\frac{2019 b}{|a|+b}, \\quad Y=\\frac{2019|a|}{|a|+b}\n$$\n\nNow apply the following estimate to every $u_{i}^{2}$ in their sum:\n\n$$\nu_{i}^{2}=x_{i}^{2} a^{2}+2 x_{i} y_{i} a b+y_{i}^{2} b^{2} \\leqslant x_{i} a^{2}+y_{i} b^{2}\n$$\n\nwe obtain that\n\n$$\n1=\\sum_{i=1}^{2019} u_{i}^{2} \\leqslant \\sum_{i=1}^{2019}\\left(x_{i} a^{2}+y_{i} b^{2}\\right)=X a^{2}+Y b^{2}=\\frac{2019 b}{|a|+b}|a|^{2}+\\frac{2019|a|}{|a|+b} b^{2}=2019|a| b=-2019 a b\n$$\n\nHence, $a b \\leqslant \\frac{-1}{2019}$.']" [] Text-only Competition True Theorem proof Algebra Math English 196 "Let $n \geqslant 3$ be a positive integer and let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of $$ \left|1-\sum_{i \in X} a_{i}\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ with sum equal to 2 such that $$ \sum_{i \in X} b_{i}=1 $$" "['we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that\n\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|\n$$\n\nso we may exchange $X$ and $X^{c}$ where convenient. Let\n\n$$\n\\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i}\n$$\n\nand note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$.\n\nIf we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that\n\n$$\n\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}\n$$\n\nthen we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 .\n\n\nWithout loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, and we may assume strict inequality as otherwise $b_{i}=a_{i}$ works. Also, $X$ clearly cannot be empty.\n\nIf $n \\in X$, add $\\Delta$ to $a_{n}$, producing a sequence of $c_{i}$ with $\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}$, and then scale as described above to make the sum equal to 2 . Otherwise, there is some $k$ with $k \\in X$ and $k+1 \\in X^{c}$. Let $\\delta=a_{k+1}-a_{k}$.\n\n- If $\\delta>\\Delta$, add $\\Delta$ to $a_{k}$ and then scale.\n- If $\\delta<\\Delta$, then considering $X \\cup\\{k+1\\} \\backslash\\{k\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising.\n- If $\\delta=\\Delta$, choose any $j \\neq k, k+1$ (possible since $n \\geqslant 3$ ), and any $\\epsilon$ less than the least of $a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $j \\in X$ then add $\\Delta-\\epsilon$ to $a_{k}$ and $\\epsilon$ to $a_{j}$, then scale; otherwise, add $\\Delta$ to $a_{k}$ and $\\epsilon / 2$ to $a_{k+1}$, and subtract $\\epsilon / 2$ from $a_{j}$, then scale.' 'we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that\n\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|\n$$\n\nso we may exchange $X$ and $X^{c}$ where convenient. Let\n\n$$\n\\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i}\n$$\n\nand note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$.\n\nIf we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that\n\n$$\n\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}\n$$\n\nthen we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 .\n\nwithout loss of generality, assume $\\sum_{i \\in X} a_{i}<1$.\n\nSuppose there exists $1 \\leqslant j \\leqslant n-1$ such that $j \\in X$ but $j+1 \\in X^{c}$. Then $a_{j+1}-a_{j} \\geqslant \\Delta$, because otherwise considering $X \\cup\\{j+1\\} \\backslash\\{j\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising.\n\nIf $a_{j+1}-a_{j}>\\Delta$, put\n\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{i}, & \\text { otherwise }\\end{cases}\n$$\n\nIf $a_{j+1}-a_{j}=\\Delta$, choose any $\\epsilon$ less than the least of $\\Delta / 2, a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $|X| \\geqslant 2$, choose $k \\in X$ with $k \\neq j$, and put\n\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2-\\epsilon, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{k}+\\epsilon, & \\text { if } i=k \\\\ a_{i}, & \\text { otherwise. }\\end{cases}\n$$\n\nOtherwise, $\\left|X^{c}\\right| \\geqslant 2$, so choose $k \\in X^{c}$ with $k \\neq j+1$, and put\n\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2+\\epsilon, & \\text { if } i=j+1 \\\\ a_{k}-\\epsilon, & \\text { if } i=k ; \\\\ a_{i}, & \\text { otherwise }\\end{cases}\n$$\n\nIf there is no $1 \\leqslant j \\leqslant n$ such that $j \\in X$ but $j+1 \\in X^{c}$, there must be some $1\\Delta$, as otherwise considering $X \\cup\\{1\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. Now put\n\n$$\nb_{i}= \\begin{cases}a_{1}-\\Delta / 2, & \\text { if } i=1 \\\\ a_{n}+\\Delta / 2, & \\text { if } i=n \\\\ a_{i}, & \\text { otherwise }\\end{cases}\n$$' 'we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that\n\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|\n$$\n\nso we may exchange $X$ and $X^{c}$ where convenient. Let\n\n$$\n\\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i}\n$$\n\nand note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$.\n\nIf we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that\n\n$$\n\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}\n$$\n\nthen we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 .\n\n\nWithout loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, so $\\Delta \\geqslant 0$. If $\\Delta=0$ we can take $b_{i}=a_{i}$, so now assume that $\\Delta>0$.\n\nSuppose that there is some $k \\leqslant n$ such that $|X \\cap[k, n]|>\\left|X^{c} \\cap[k, n]\\right|$. If we choose the largest such $k$ then $|X \\cap[k, n]|-\\left|X^{c} \\cap[k, n]\\right|=1$. We can now find the required sequence $\\left(b_{i}\\right)$ by starting with $c_{i}=a_{i}$ for $i\\frac{n-k+1}{2}$ and $|X \\cap[\\ell, n]|<\\frac{n-\\ell+1}{2}$.\n\nWe now construct our sequence $\\left(b_{i}\\right)$ using this claim. Let $k$ and $\\ell$ be the greatest values satisfying the claim, and without loss of generality suppose $k=n$ and $\\ell0\n$$\n\nPartition the indices into sets $P, Q, R$, and $S$ such that\n\n$$\n\\begin{array}{ll}\nP=\\left\\{i \\mid a_{i} \\leqslant-1\\right\\} & R=\\left\\{i \\mid 00\n$$\n\nPartition the indices into sets $P, Q, R$, and $S$ such that\n\n$$\n\\begin{array}{ll}\nP=\\left\\{i \\mid a_{i} \\leqslant-1\\right\\} & R=\\left\\{i \\mid 01$ ). Therefore,\n\n$$\nt_{+}+t_{-} \\leqslant \\frac{p^{2}+s^{2}}{2}+p q+r s+p r+p s+q s=\\frac{(p+q+r+s)^{2}}{2}-\\frac{(q+r)^{2}}{2}=-\\frac{(q+r)^{2}}{2} \\leqslant 0 .\n$$\n\nIf $A$ is not empty and $p=s=0$, then there must exist $i \\in Q, j \\in R$ with $\\left|a_{i}-a_{j}\\right|>1$, and hence the earlier equality conditions cannot both occur.']" [] Text-only Competition True Theorem proof Algebra Math English 198 "Let $x_{1}, x_{2}, \ldots, x_{n}$ be different real numbers. Prove that $$ \sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}= \begin{cases}0, & \text { if } n \text { is even } \\ 1, & \text { if } n \text { is odd }\end{cases} $$" "['Since both sides of the identity are rational functions, it suffices to prove it when all $x_{i} \\notin\\{ \\pm 1\\}$. Define\n\n$$\nf(t)=\\prod_{i=1}^{n}\\left(1-x_{i} t\\right)\n$$\n\nand note that\n\n$$\nf\\left(x_{i}\\right)=\\left(1-x_{i}^{2}\\right) \\prod_{j \\neq i} 1-x_{i} x_{j}\n$$\n\nUsing the nodes $+1,-1, x_{1}, \\ldots, x_{n}$, the Lagrange interpolation formula gives us the following expression for $f$ :\n\n$$\n\\sum_{i=1}^{n} f\\left(x_{i}\\right) \\frac{(x-1)(x+1)}{\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)} \\prod_{j \\neq i} \\frac{x-x_{j}}{x_{i}-x_{j}}+f(1) \\frac{x+1}{1+1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}}+f(-1) \\frac{x-1}{-1-1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}}\n$$\n\nThe coefficient of $t^{n+1}$ in $f(t)$ is 0 , since $f$ has degree $n$. The coefficient of $t^{n+1}$ in the above expression of $f$ is\n\n$$\n\\begin{aligned}\n0 & =\\sum_{1 \\leqslant i \\leqslant n} \\frac{f\\left(x_{i}\\right)}{\\prod_{j \\neq i}\\left(x_{i}-x_{j}\\right) \\cdot\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)}+\\frac{f(1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(1-x_{j}\\right) \\cdot(1+1)}+\\frac{f(-1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(-1-x_{j}\\right) \\cdot(-1-1)} \\\\\n& =-G\\left(x_{1}, \\ldots, x_{n}\\right)+\\frac{1}{2}+\\frac{(-1)^{n+1}}{2} .\n\\end{aligned}\n$$' 'Observe that $G$ is symmetric in the variables $x_{1}, \\ldots, x_{n}$. Define $V=\\prod_{i0$. By Step 2, each of its monomials $\\mu x^{a} y^{b} z^{c}$ of the highest total degree satisfies $a, b \\geqslant c$. Applying other weak symmetries, we obtain $a, c \\geqslant b$ and $b, c \\geqslant a$; therefore, $P$ has a unique leading monomial of the form $\\mu(x y z)^{c}$. The polynomial $P_{0}(x, y, z)=P(x, y, z)-\\mu\\left(x y z-x^{2}-y^{2}-z^{2}\\right)^{c}$ has smaller total degree. Since $P_{0}$ is symmetric, it is representable as a polynomial function of $x y z-x^{2}-y^{2}-z^{2}$. Then $P$ is also of this form, completing the inductive step.' 'We will rely on the well-known identity\n\n$$\n\\cos ^{2} u+\\cos ^{2} v+\\cos ^{2} w-2 \\cos u \\cos v \\cos w-1=0 \\quad \\text { whenever } u+v+w=0 \\text {. }\n\\tag{2.1}\n$$\n\nClaim 1. The polynomial $P(x, y, z)$ is constant on the surface\n\n$$\n\\mathfrak{S}=\\{(2 \\cos u, 2 \\cos v, 2 \\cos w): u+v+w=0\\}\n$$\n\nProof. Notice that for $x=2 \\cos u, y=2 \\cos v, z=2 \\cos w$, the Vieta jumps $x \\mapsto y z-x$, $y \\mapsto z x-y, z \\mapsto x y-z$ in $(*)$ replace $(u, v, w)$ by $(v-w,-v, w),(u, w-u,-w)$ and $(-u, v, u-v)$, respectively. For example, for the first type of jump we have\n\n$$\ny z-x=4 \\cos v \\cos w-2 \\cos u=2 \\cos (v+w)+2 \\cos (v-w)-2 \\cos u=2 \\cos (v-w) \\text {. }\n$$\n\nDefine $G(u, v, w)=P(2 \\cos u, 2 \\cos v, 2 \\cos w)$. For $u+v+w=0$, the jumps give\n\n$$\n\\begin{aligned}\nG(u, v, w) & =G(v-w,-v, w)=G(w-v,-v,(v-w)-(-v))=G(-u-2 v,-v, 2 v-w) \\\\\n& =G(u+2 v, v, w-2 v)\n\\end{aligned}\n$$\n\nBy induction,\n\n$$\nG(u, v, w)=G(u+2 k v, v, w-2 k v) \\quad(k \\in \\mathbb{Z}) .\n\\tag{2.2}\n$$\n\nSimilarly,\n\n$$\nG(u, v, w)=G(u, v-2 \\ell u, w+2 \\ell u) \\quad(\\ell \\in \\mathbb{Z})\n\\tag{2.3}\n$$\n\nAnd, of course, we have\n\n$$\nG(u, v, w)=G(u+2 p \\pi, v+2 q \\pi, w-2(p+q) \\pi) \\quad(p, q \\in \\mathbb{Z}) .\n\\tag{2.4}\n$$\n\nTake two nonzero real numbers $u, v$ such that $u, v$ and $\\pi$ are linearly independent over $\\mathbb{Q}$. By combining $(2.2-2.4)$, we can see that $G$ is constant on a dense subset of the plane $u+v+w=0$. By continuity, $G$ is constant on the entire plane and therefore $P$ is constant on $\\mathfrak{S}$.\n\nClaim 2. The polynomial $T(x, y, z)=x^{2}+y^{2}+z^{2}-x y z-4$ divides $P(x, y, z)-P(2,2,2)$. Proof. By dividing $P$ by $T$ with remainders, there exist some polynomials $R(x, y, z), A(y, z)$ and $B(y, z)$ such that\n\n$$\nP(x, y, z)-P(2,2,2)=T(x, y, z) \\cdot R(x, y, z)+A(y, z) x+B(y, z)\n\\tag{2.5}\n$$\n\nOn the surface $\\mathfrak{S}$ the LHS of $(2.5)$ is zero by Claim 1 (since $(2,2,2) \\in \\mathfrak{S}$ ) and $T=0$ by (2.1). Hence, $A(y, z) x+B(y, z)$ vanishes on $\\mathfrak{S}$.\n\nNotice that for every $y=2 \\cos v$ and $z=2 \\cos w$ with $\\frac{\\pi}{3}0$ we have\n\n$$\nS_{k}=S_{k-1} \\cup\\left(x_{k}+S_{k-1}\\right)\n$$\n\n(where $\\left(x_{k}+S_{k-1}\\right)$ denotes $\\left\\{x_{k}+s: s \\in S_{k-1}\\right\\}$ ), so it suffices to prove that $x_{k} \\leqslant \\sum_{j\\sum_{j(n+1-k)(k+1)+k-1\n$$\n\nThis rearranges to $n>k(n+1-k)$, which is false for $1 \\leqslant k \\leqslant n$, giving the desired contradiction.']" [] Text-only Competition True Theorem proof Combinatorics Math English 204 "On a certain social network, there are 2019 users, some pairs of which are friends, where friendship is a symmetric relation. Initially, there are 1010 people with 1009 friends each and 1009 people with 1010 friends each. However, the friendships are rather unstable, so events of the following kind may happen repeatedly, one at a time: Let $A, B$, and $C$ be people such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends; then $B$ and $C$ become friends, but $A$ is no longer friends with them. Prove that, regardless of the initial friendships, there exists a sequence of such events after which each user is friends with at most one other user." "['The problem has an obvious rephrasing in terms of graph theory. One is given a graph $G$ with 2019 vertices, 1010 of which have degree 1009 and 1009 of which have degree 1010. One is allowed to perform operations on $G$ of the following kind:\n\nSuppose that vertex $A$ is adjacent to two distinct vertices $B$ and $C$ which are not adjacent to each other. Then one may remove the edges $A B$ and $A C$ from $G$ and add the edge $B C$ into $G$.\n\nCall such an operation a refriending. One wants to prove that, via a sequence of such refriendings, one can reach a graph which is a disjoint union of single edges and vertices.\n\n\nNote that the given graph is connected, since the total degree of any two vertices is at least 2018 and hence they are either adjacent or have at least one neighbour in common. Hence the given graph satisfies the following condition:\n$$\n\\text{Every connected component of $G$ with at least three vertices is not complete and has a vertex of odd degree.}\n\\tag{1}\n$$\nWe will show that if a graph $G$ satisfies condition (1) and has a vertex of degree at least 2 , then there is a refriending on $G$ that preserves condition (1). Since refriendings decrease the total number of edges of $G$, by using a sequence of such refriendings, we must reach a graph $G$ with maximal degree at most 1 , so we are done.\n\n\n\n\n\nPick a vertex $A$ of degree at least 2 in a connected component $G^{\\prime}$ of $G$. Since no component of $G$ with at least three vertices is complete we may assume that not all of the neighbours of $A$ are adjacent to one another. (For example, pick a maximal complete subgraph $K$ of $G^{\\prime}$. Some vertex $A$ of $K$ has a neighbour outside $K$, and this neighbour is not adjacent to every vertex of $K$ by maximality.) Removing $A$ from $G$ splits $G^{\\prime}$ into smaller connected components $G_{1}, \\ldots, G_{k}$ (possibly with $k=1$ ), to each of which $A$ is connected by at least one edge. We divide into several cases.\n\nCase 1: $k \\geqslant 2$ and $A$ is connected to some $G_{i}$ by at least two edges.\n\nChoose a vertex $B$ of $G_{i}$ adjacent to $A$, and a vertex $C$ in another component $G_{j}$ adjacent to $A$. The vertices $B$ and $C$ are not adjacent, and hence removing edges $A B$ and $A C$ and adding in edge $B C$ does not disconnect $G^{\\prime}$. It is easy to see that this preserves the condition, since the refriending does not change the parity of the degrees of vertices.\n\nCase 2: $k \\geqslant 2$ and $A$ is connected to each $G_{i}$ by exactly one edge.\n\nConsider the induced subgraph on any $G_{i}$ and the vertex $A$. The vertex $A$ has degree 1 in this subgraph; since the number of odd-degree vertices of a graph is always even, we see that $G_{i}$ has a vertex of odd degree (in $G$ ). Thus if we let $B$ and $C$ be any distinct neighbours of $A$, then removing edges $A B$ and $A C$ and adding in edge $B C$ preserves the above condition: the refriending creates two new components, and if either of these components has at least three vertices, then it cannot be complete and must contain a vertex of odd degree (since each $G_{i}$ does).\n\nCase 3: $k=1$ and $A$ is connected to $G_{1}$ by at least three edges.\n\nBy assumption, $A$ has two neighbours $B$ and $C$ which are not adjacent to one another. Removing edges $A B$ and $A C$ and adding in edge $B C$ does not disconnect $G^{\\prime}$. We are then done as in Case 1.\n\nCase 4: $k=1$ and $A$ is connected to $G_{1}$ by exactly two edges.\n\nLet $B$ and $C$ be the two neighbours of $A$, which are not adjacent. Removing edges $A B$ and $A C$ and adding in edge $B C$ results in two new components: one consisting of a single vertex; and the other containing a vertex of odd degree. We are done unless this second component would be a complete graph on at least 3 vertices. But in this case, $G_{1}$ would be a complete graph minus the single edge $B C$, and hence has at least 4 vertices since $G^{\\prime}$ is not a 4-cycle. If we let $D$ be a third vertex of $G_{1}$, then removing edges $B A$ and $B D$ and adding in edge $A D$ does not disconnect $G^{\\prime}$. We are then done as in Case 1.\n\n' 'The problem has an obvious rephrasing in terms of graph theory. One is given a graph $G$ with 2019 vertices, 1010 of which have degree 1009 and 1009 of which have degree 1010. One is allowed to perform operations on $G$ of the following kind:\n\nSuppose that vertex $A$ is adjacent to two distinct vertices $B$ and $C$ which are not adjacent to each other. Then one may remove the edges $A B$ and $A C$ from $G$ and add the edge $B C$ into $G$.\n\nCall such an operation a refriending. One wants to prove that, via a sequence of such refriendings, one can reach a graph which is a disjoint union of single edges and vertices.\n\n\nnote that a refriending preserves the property that a graph has a vertex of odd degree and (trivially) the property that it is not complete; note also that our initial graph is connected. We describe an algorithm to reduce our initial graph to a graph of maximal degree at most 1 , proceeding in two steps.\n\nStep 1: There exists a sequence of refriendings reducing the graph to a tree.\n\nProof. Since the number of edges decreases with each refriending, it suffices to prove the following: as long as the graph contains a cycle, there exists a refriending such that the resulting graph is still connected. We will show that the graph in fact contains a cycle $Z$ and vertices $A, B, C$ such that $A$ and $B$ are adjacent in the cycle $Z, C$ is not in $Z$, and is adjacent to $A$ but not $B$. Removing edges $A B$ and $A C$ and adding in edge $B C$ keeps the graph connected, so we are done.\n\n\n\nTo find this cycle $Z$ and vertices $A, B, C$, we pursue one of two strategies. If the graph contains a triangle, we consider a largest complete subgraph $K$, which thus contains at least three vertices. Since the graph itself is not complete, there is a vertex $C$ not in $K$ connected to a vertex $A$ of $K$. By maximality of $K$, there is a vertex $B$ of $K$ not connected to $C$, and hence we are done by choosing a cycle $Z$ in $K$ through the edge $A B$.\n\n\n\nIf the graph is triangle-free, we consider instead a smallest cycle $Z$. This cycle cannot be Hamiltonian (i.e. it cannot pass through every vertex of the graph), since otherwise by minimality the graph would then have no other edges, and hence would have even degree at every vertex. We may thus choose a vertex $C$ not in $Z$ adjacent to a vertex $A$ of $Z$. Since the graph is triangle-free, it is not adjacent to any neighbour $B$ of $A$ in $Z$, and we are done.\n\nStep 2: Any tree may be reduced to a disjoint union of single edges and vertices by a sequence of refriendings.\n\nProof. The refriending preserves the property of being acyclic. Hence, after applying a sequence of refriendings, we arrive at an acyclic graph in which it is impossible to perform any further refriendings. The maximal degree of any such graph is 1: if it had a vertex $A$ with two neighbours $B, C$, then $B$ and $C$ would necessarily be nonadjacent since the graph is cycle-free, and so a refriending would be possible. Thus we reach a graph with maximal degree at most 1 as desired.']" [] Text-only Competition True Theorem proof Combinatorics Math English 205 Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\angle A_{1} A_{2} A_{3}, \angle A_{2} A_{3} A_{4}, \ldots, \angle A_{2 n-2} A_{2 n-1} A_{2 n}, \angle A_{2 n-1} A_{2 n} A_{1}$, $\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group. "['Let $\\ell$ be a line separating the points into two groups ( $L$ and $R$ ) with $n$ points in each. Label the points $A_{1}, A_{2}, \\ldots, A_{2 n}$ so that $L=\\left\\{A_{1}, A_{3}, \\ldots, A_{2 n-1}\\right\\}$. We claim that this labelling works.\n\nTake a line $s=A_{2 n} A_{1}$.\n\n(a) Rotate $s$ around $A_{1}$ until it passes through $A_{2}$; the rotation is performed in a direction such that $s$ is never parallel to $\\ell$.\n\n(b) Then rotate the new $s$ around $A_{2}$ until it passes through $A_{3}$ in a similar manner.\n\n(c) Perform $2 n-2$ more such steps, after which $s$ returns to its initial position.\n\nThe total (directed) rotation angle $\\Theta$ of $s$ is clearly a multiple of $180^{\\circ}$. On the other hand, $s$ was never parallel to $\\ell$, which is possible only if $\\Theta=0$. Now it remains to partition all the $2 n$ angles into those where $s$ is rotated anticlockwise, and the others.' 'When tracing a cyclic path through the $A_{i}$ in order, with straight line segments between consecutive points, let $\\theta_{i}$ be the exterior angle at $A_{i}$, with a sign convention that it is positive if the path turns left and negative if the path turns right. Then $\\sum_{i=1}^{2 n} \\theta_{i}=360 k^{\\circ}$ for some integer $k$. Let $\\phi_{i}=\\angle A_{i-1} A_{i} A_{i+1}$ (indices $\\bmod 2 n$ ), defined as in the problem; thus $\\phi_{i}=180^{\\circ}-\\left|\\theta_{i}\\right|$.\n\nLet $L$ be the set of $i$ for which the path turns left at $A_{i}$ and let $R$ be the set for which it turns right. Then $S=\\sum_{i \\in L} \\phi_{i}-\\sum_{i \\in R} \\phi_{i}=(180(|L|-|R|)-360 k)^{\\circ}$, which is a multiple of $360^{\\circ}$ since the number of points is even. We will show that the points can be labelled such that $S=0$, in which case $L$ and $R$ satisfy the required condition of the problem.\n\nNote that the value of $S$ is defined for a slightly larger class of configurations: it is OK for two points to coincide, as long as they are not consecutive, and OK for three points to be collinear, as long as $A_{i}, A_{i+1}$ and $A_{i+2}$ do not appear on a line in that order. In what follows it will be convenient, although not strictly necessary, to consider such configurations.\n\nConsider how $S$ changes if a single one of the $A_{i}$ is moved along some straight-line path (not passing through any $A_{j}$ and not lying on any line $A_{j} A_{k}$, but possibly crossing such lines). Because $S$ is a multiple of $360^{\\circ}$, and the angles change continuously, $S$ can only change when a point moves between $R$ and $L$. Furthermore, if $\\phi_{j}=0$ when $A_{j}$ moves between $R$ and $L, S$ is unchanged; it only changes if $\\phi_{j}=180^{\\circ}$ when $A_{j}$ moves between those sets.\n\nFor any starting choice of points, we will now construct a new configuration, with labels such that $S=0$, that can be perturbed into the original one without any $\\phi_{i}$ passing through $180^{\\circ}$, so that $S=0$ for the original configuration with those labels as well.\n\nTake some line such that there are $n$ points on each side of that line. The new configuration has $n$ copies of a single point on each side of the line, and a path that alternates between\n\n\n\nsides of the line; all angles are 0 , so this configuration has $S=0$. Perturbing the points into their original positions, while keeping each point on its side of the line, no angle $\\phi_{i}$ can pass through $180^{\\circ}$, because no straight line can go from one side of the line to the other and back. So the perturbation process leaves $S=0$.' 'First, let $\\ell$ be a line in the plane such that there are $n$ points on one side and the other $n$ points on the other side. For convenience, assume $\\ell$ is horizontal (otherwise, we can rotate the plane). Then we can use the terms ""above"", ""below"", ""left"" and ""right"" in the usual way. We denote the $n$ points above the line in an arbitrary order as $P_{1}, P_{2}, \\ldots, P_{n}$, and the $n$ points below the line as $Q_{1}, Q_{2}, \\ldots, Q_{n}$.\n\nIf we connect $P_{i}$ and $Q_{j}$ with a line segment, the line segment will intersect with the line $\\ell$. Denote the intersection as $I_{i j}$. If $P_{i}$ is connected to $Q_{j}$ and $Q_{k}$, where $jk$, then the sign of $\\angle Q_{j} P_{i} Q_{k}$ is taken to be the same as for $\\angle Q_{k} P_{i} Q_{j}$.\n\nSimilarly, we can define the sign of $\\angle P_{j} Q_{i} P_{k}$ with $j\n\nSimilarly, we have\n\n$$\n\\angle P_{i_{1}} Q_{k} P_{i_{3}}=\\angle P_{i_{1}} Q_{k} P_{i_{2}}+\\angle P_{i_{2}} Q_{k} P_{i_{3}},\n\\tag{2}\n$$\n\nfor all points $Q_{k}, P_{i_{1}}, P_{i_{2}}$ and $P_{i_{3}}$, with $i_{1}\n\nThen $\\angle P_{1}$ and $\\angle Q_{1}$ are positive, $\\angle P_{2}$ and $\\angle Q_{2}$ are negative, and we have\n\n$$\n\\left|\\angle P_{1}\\right|+\\left|\\angle Q_{1}\\right|=\\left|\\angle P_{2}\\right|+\\left|\\angle Q_{2}\\right| .\n$$\n\nWith signed measures, we have\n\n$$\n\\angle P_{1}+\\angle Q_{1}+\\angle P_{2}+\\angle Q_{2}=0\n\\tag{3}\n$$\n\nIf we switch the labels of $P_{1}$ and $P_{2}$, we have the following picture:\n\n\n\nSwitching labels $P_{1}$ and $P_{2}$ has the effect of flipping the sign of all four angles (as well as swapping the magnitudes on the relabelled points); that is, the new values of $\\left(\\angle P_{1}, \\angle P_{2}, \\angle Q_{1}, \\angle Q_{2}\\right.$ ) equal the old values of $\\left(-\\angle P_{2},-\\angle P_{1},-\\angle Q_{1},-\\angle Q_{2}\\right)$. Consequently, equation (3) still holds. Similarly, when switching the labels of $Q_{1}$ and $Q_{2}$, or both the $P$ \'s and the $Q$ \'s, equation (3) still holds.\n\n\n\nThe remaining subcase of $n=2$ is that one point lies inside the triangle formed by the other three. We have the following picture.\n\n\n\nWe have\n\n$$\n\\left|\\angle P_{1}\\right|+\\left|\\angle Q_{1}\\right|+\\left|\\angle Q_{2}\\right|=\\left|\\angle P_{2}\\right|\n$$\n\nand equation (3) holds.\n\nAgain, switching the labels for $P$ \'s or the $Q$ \'s will not affect the validity of equation (3). Also, if the point lying inside the triangle of the other three is one of the $Q$ \'s rather than the $P$ \'s, the result still holds, since our sign convention is preserved when we relabel $Q$ \'s as $P$ \'s and vice-versa and reflect across $\\ell$.\n\nWe have completed the proof of the claim for $n=2$.\n\nAssume the claim holds for $n=k$, and we wish to prove it for $n=k+1$. Suppose we are given our $2(k+1)$ points. First ignore $P_{k+1}$ and $Q_{k+1}$, and form $2 k$ angles from $P_{1}, \\ldots, P_{k}$, $Q_{1}, \\ldots, Q_{k}$ as in the $n=k$ case. By the induction hypothesis we have\n\n$$\n\\sum_{i=1}^{k}\\left(\\angle P_{i}+\\angle Q_{i}\\right)=0\n$$\n\nWhen we add in the two points $P_{k+1}$ and $Q_{k+1}$, this changes our angles as follows:\n\n- the angle at $P_{k}$ changes from $\\angle Q_{k-1} P_{k} Q_{k}$ to $\\angle Q_{k-1} P_{k} Q_{k+1}$;\n- the angle at $Q_{k}$ changes from $\\angle P_{k-1} Q_{k} P_{k}$ to $\\angle P_{k-1} Q_{k} P_{k+1}$;\n- two new angles $\\angle Q_{k} P_{k+1} Q_{k+1}$ and $\\angle P_{k} Q_{k+1} P_{k+1}$ are added.\n\nWe need to prove the changes have no impact on the total sum. In other words, we need to prove\n\n$$\n\\left(\\angle Q_{k-1} P_{k} Q_{k+1}-\\angle Q_{k-1} P_{k} Q_{k}\\right)+\\left(\\angle P_{k-1} Q_{k} P_{k+1}-\\angle P_{k-1} Q_{k} P_{k}\\right)+\\left(\\angle P_{k+1}+\\angle Q_{k+1}\\right)=0\n$$\n\nIn fact, from equations (1) and (2), we have\n\n$$\n\\angle Q_{k-1} P_{k} Q_{k+1}-\\angle Q_{k-1} P_{k} Q_{k}=\\angle Q_{k} P_{k} Q_{k+1}\n$$\n\nand\n\n$$\n\\angle P_{k-1} Q_{k} P_{k+1}-\\angle P_{k-1} Q_{k} P_{k}=\\angle P_{k} Q_{k} P_{k+1} .\n$$\n\nTherefore, the left hand side of equation (4) becomes $\\angle Q_{k} P_{k} Q_{k+1}+\\angle P_{k} Q_{k} P_{k+1}+\\angle Q_{k} P_{k+1} Q_{k+1}+$ $\\angle P_{k} Q_{k+1} P_{k+1}$, which equals 0 , simply by applying the $n=2$ case of the claim. This completes the induction.' 'We shall think instead of the problem as asking us to assign a weight \\pm 1 to each angle, such that the weighted sum of all the angles is zero.\n\nGiven an ordering $A_{1}, \\ldots, A_{2 n}$ of the points, we shall assign weights according to the following recipe: walk in order from point to point, and assign the left turns +1 and the right turns -1 . This is the same weighting as in' '3, and as in that solution, the weighted sum is a multiple of $360^{\\circ}$.\n\nWe now aim to show the following:\n\nLemma. Transposing any two consecutive points in the ordering changes the weighted sum by $\\pm 360^{\\circ}$ or 0 .\n\nKnowing that, we can conclude quickly: if the ordering $A_{1}, \\ldots, A_{2 n}$ has weighted angle sum $360 k^{\\circ}$, then the ordering $A_{2 n}, \\ldots, A_{1}$ has weighted angle sum $-360 k^{\\circ}$ (since the angles are the same, but left turns and right turns are exchanged). We can reverse the ordering of $A_{1}$, $\\ldots, A_{2 n}$ by a sequence of transpositions of consecutive points, and in doing so the weighted angle sum must become zero somewhere along the way.\n\nWe now prove that lemma:\n\nProof. Transposing two points amounts to taking a section $A_{k} A_{k+1} A_{k+2} A_{k+3}$ as depicted, reversing the central line segment $A_{k+1} A_{k+2}$, and replacing its two neighbours with the dotted lines.\n\n\nFigure 1: Transposing two consecutive vertices: before (left) and afterwards (right)\n\nIn each triangle, we alter the sum by $\\pm 180^{\\circ}$. Indeed, using (anticlockwise) directed angles modulo $360^{\\circ}$, we either add or subtract all three angles of each triangle.\n\nHence both triangles together alter the sum by $\\pm 180 \\pm 180^{\\circ}$, which is $\\pm 360^{\\circ}$ or 0 .']" [] Text-only Competition True Theorem proof Combinatorics Math English 206 "Alice has a map of Wonderland, a country consisting of $n \geqslant 2$ towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be ""one way"" only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns. Alice wants to know whether there is at least one town in Wonderland with at most one outgoing road. Prove that she can always find out by asking at most $4 n$ questions." "[""We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ).\n\nPhase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of generality, suppose that the King of Hearts' answer is that the road goes from $A$ to $B$.\n\nAt the end of this phase, Alice has asked 1 question.\n\nPhase 2. During this phase there is a single (variable) town $T$ that is known to have at least one incoming road but not yet known to have any outgoing roads. Initially, $T$ is $B$. Alice does the following $n-2$ times: she picks a town $X$ she has not asked about before, and asks the direction of the road between $T$ and $X$. If it is from $X$ to $T, T$ is unchanged; if it is from $T$ to $X, X$ becomes the new choice of town $T$, as the previous $T$ is now known to have an outgoing road.\n\nAt the end of this phase, Alice has asked a total of $n-1$ questions. The final town $T$ is not yet known to have any outgoing roads, while every other town has exactly one outgoing road known. The undirected graph of roads whose directions are known is a tree.\n\nPhase 3. During this phase, Alice asks about the directions of all roads between $T$ and another town she has not previously asked about, stopping if she finds two outgoing roads from $T$. This phase involves at most $n-2$ questions. If she does not find two outgoing roads from $T$, she has answered her original question with at most $2 n-3 \\leqslant 4 n-7$ questions, so in what follows we suppose that she does find two outgoing roads, asking a total of $k$ questions in this phase, where $2 \\leqslant k \\leqslant n-2$ (and thus $n \\geqslant 4$ for what follows).\n\nFor every question where the road goes towards $T$, the town at the other end is removed from $S$ (as it already had one outgoing road known), while the last question resulted in $T$ being removed from $S$. So at the end of this phase, $|S|=n-k+1$, while a total of $n+k-1$ questions have been asked. Furthermore, the undirected graph of roads within $S$ whose directions are known contains no cycles (as $T$ is no longer a member of $S$, all questions asked in this phase involved $T$ and the graph was a tree before this phase started). Every town in $S$ has exactly one outgoing road known (not necessarily to another town in $S$ ).\n\nPhase 4. During this phase, Alice repeatedly picks any pair of towns in $S$ for which she does not know the direction of the road between them. Because every town in $S$ has exactly one outgoing road known, this always results in the removal of one of those two towns from $S$. Because there are no cycles in the graph of roads of known direction within $S$, this can continue until there are at most 2 towns left in $S$.\n\nIf it ends with $t$ towns left, $n-k+1-t$ questions were asked in this phase, so a total of $2 n-t$ questions have been asked.\n\nPhase 5. During this phase, Alice asks about all the roads from the remaining towns in $S$ that she has not previously asked about. She has definitely already asked about any road between those towns (if $t=2$ ). She must also have asked in one of the first two phases about\n\n\n\nat least one other road involving one of those towns (as those phases resulted in a tree with $n>2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase.\n\nAt the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \\leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 207 Let $A B C$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $A B$ and $A C$ again at points $D$ and $E$ respectively, and intersects segment $B C$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $B D F$ at $F$ and the tangent to circle $C E G$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $A T$ is parallel to $B C$. ['Notice that $\\angle T F B=\\angle F D A$ because $F T$ is tangent to circle $B D F$, and moreover $\\angle F D A=\\angle C G A$ because quadrilateral $A D F G$ is cyclic. Similarly, $\\angle T G B=\\angle G E C$ because $G T$ is tangent to circle $C E G$, and $\\angle G E C=\\angle C F A$. Hence,\n\n$$\n\\angle T F B=\\angle C G A \\text { and } \\angle T G B=\\angle C F A \\text {. }\n\\tag{1}\n$$\n\n\n\nTriangles $F G A$ and $G F T$ have a common side $F G$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $A T$ is parallel to line $B F G C$.'] [] Text-only Competition True Theorem proof Geometry Math English 208 Let $A B C$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $B C, C A$, and $A B$, respectively. Denote by $\omega_{B}$ and $\omega_{C}$ the incircles of triangles $B D F$ and $C D E$, and let these circles be tangent to segments $D F$ and $D E$ at $M$ and $N$, respectively. Let line $M N$ meet circles $\omega_{B}$ and $\omega_{C}$ again at $P \neq M$ and $Q \neq N$, respectively. Prove that $M P=N Q$. ['Denote the centres of $\\omega_{B}$ and $\\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $B C$ be tangent to the two circles at $T$ and $U$, respectively.\n\n\n\nFrom the cyclic quadrilaterals $A F D C$ and $A B D E$ we have\n\n$$\n\\angle M D O_{B}=\\frac{1}{2} \\angle F D B=\\frac{1}{2} \\angle B A C=\\frac{1}{2} \\angle C D E=\\angle O_{C} D N\n$$\n\nso the right-angled triangles $D M O_{B}$ and $D N O_{C}$ are similar. The ratio of similarity between the two triangles is\n\n$$\n\\frac{D N}{D M}=\\frac{O_{C} N}{O_{B} M}=\\frac{r_{C}}{r_{B}}\n$$\n\nLet $\\varphi=\\angle D M N$ and $\\psi=\\angle M N D$. The lines $F M$ and $E N$ are tangent to $\\omega_{B}$ and $\\omega_{C}$, respectively, so\n\n$$\n\\angle M T P=\\angle F M P=\\angle D M N=\\varphi \\quad \\text { and } \\quad \\angle Q U N=\\angle Q N E=\\angle M N D=\\psi .\n$$\n\n(It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $D M T$ or $D U N$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $M T P$ and $Q U N$.)\n\nIn the circles $\\omega_{B}$ and $\\omega_{C}$ the lengths of chords $M P$ and $N Q$ are\n\n$$\nM P=2 r_{B} \\cdot \\sin \\angle M T P=2 r_{B} \\cdot \\sin \\varphi \\quad \\text { and } \\quad N Q=2 r_{C} \\cdot \\sin \\angle Q U N=2 r_{C} \\cdot \\sin \\psi\n$$\n\nBy applying the sine rule to triangle $D N M$ we get\n\n$$\n\\frac{D N}{D M}=\\frac{\\sin \\angle D M N}{\\sin \\angle M N D}=\\frac{\\sin \\varphi}{\\sin \\psi}\n$$\n\nFinally, putting the above observations together, we get\n\n$$\n\\frac{M P}{N Q}=\\frac{2 r_{B} \\sin \\varphi}{2 r_{C} \\sin \\psi}=\\frac{r_{B}}{r_{C}} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{D M}{D N} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{\\sin \\psi}{\\sin \\varphi} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=1\n$$\n\nso $M P=N Q$ as required.'] [] Text-only Competition True Theorem proof Geometry Math English 209 In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\angle P P_{1} C=\angle B A C$. Similarly, on ray $Q A_{1}$, beyond $A_{1}$, let $Q_{1}$ be a point so that $\angle C Q_{1} Q=\angle C B A$. Show that points $P, Q, P_{1}$, and $Q_{1}$ are concyclic. "['Throughout the solution we use oriented angles.\n\nLet rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By\n\n$$\n\\angle Q P A_{2}=\\angle B A A_{2}=\\angle B B_{2} A_{2}=\\angle Q B_{2} A_{2},\n$$\n\npoints $P, Q, A_{2}, B_{2}$ are concyclic; denote the circle passing through these points by $\\omega$. We shall prove that $P_{1}$ and $Q_{1}$ also lie on $\\omega$.\n\n\n\nBy\n\n$$\n\\angle C A_{2} A_{1}=\\angle C A_{2} A=\\angle C B A=\\angle C Q_{1} Q=\\angle C Q_{1} A_{1},\n$$\n\npoints $C, Q_{1}, A_{2}, A_{1}$ are also concyclic. From that we get\n\n$$\n\\angle Q Q_{1} A_{2}=\\angle A_{1} Q_{1} A_{2}=\\angle A_{1} C A_{2}=\\angle B C A_{2}=\\angle B A A_{2}=\\angle Q P A_{2},\n$$\n\nso $Q_{1}$ lies on $\\omega$.\n\nIt follows similarly that $P_{1}$ lies on $\\omega$.' ""First consider the case when lines $P P_{1}$ and $Q Q_{1}$ intersect each other at some point $R$.\n\nLet line $P Q$ meet the sides $A C$ and $B C$ at $E$ and $F$, respectively. Then\n\n$$\n\\angle P P_{1} C=\\angle B A C=\\angle P E C,\n$$\n\nso points $C, E, P, P_{1}$ lie on a circle; denote that circle by $\\omega_{P}$. It follows analogously that points $C, F, Q, Q_{1}$ lie on another circle; denote it by $\\omega_{Q}$.\n\nLet $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem to the lines $A A_{1} P$ and $B B_{1} Q$ provides that points $C=A B_{1} \\cap B A_{1}, R=A_{1} Q \\cap B_{1} P$ and $T=A Q \\cap B P$ are collinear.\n\nLet line $R C T$ meet $P Q$ and $A B$ at $S$ and $U$, respectively. From $A B \\| P Q$ we obtain\n\n$$\n\\frac{S P}{S Q}=\\frac{U B}{U A}=\\frac{S F}{S E}\n$$\n\nSO\n\n$$\nS P \\cdot S E=S Q \\cdot S F\n$$\n\n\n\n\n\nSo, point $S$ has equal powers with respect to $\\omega_{P}$ and $\\omega_{Q}$, hence line $R C S$ is their radical axis; then $R$ also has equal powers to the circles, so $R P \\cdot R P_{1}=R Q \\cdot R Q_{1}$, proving that points $P, P_{1}, Q, Q_{1}$ are indeed concyclic.\n\nNow consider the case when $P P_{1}$ and $Q Q_{1}$ are parallel. Like in the previous case, let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem again to the lines $A A_{1} P$ and $B B_{1} Q$, in this limit case it shows that line $C T$ is parallel to $P P_{1}$ and $Q Q_{1}$.\n\nLet line $C T$ meet $P Q$ and $A B$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $S P \\cdot S E=S Q \\cdot S F$, so $S$ lies on the radical axis between $\\omega_{P}$ and $\\omega_{Q}$.\n\n\n\nLine $C S T$, that is the radical axis between $\\omega_{P}$ and $\\omega_{Q}$, is perpendicular to the line $\\ell$ of centres of $\\omega_{P}$ and $\\omega_{Q}$. Hence, the chords $P P_{1}$ and $Q Q_{1}$ are perpendicular to $\\ell$. So the quadrilateral $P P_{1} Q_{1} Q$ is an isosceles trapezium with symmetry axis $\\ell$, and hence is cyclic.""]" [] Text-only Competition True Theorem proof Geometry Math English 210 Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$. "['Since\n\n$$\n\\angle A P B+\\angle B P C+\\angle C P A=2 \\pi=(\\pi-\\angle A C B)+(\\pi-\\angle B A C)+(\\pi-\\angle C B A),\n$$\n\nat least one of the following inequalities holds:\n\n$$\n\\angle A P B \\geqslant \\pi-\\angle A C B, \\quad \\angle B P C \\geqslant \\pi-\\angle B A C, \\quad \\angle C P A \\geqslant \\pi-\\angle C B A .\n$$\n\nWithout loss of generality, we assume that $\\angle B P C \\geqslant \\pi-\\angle B A C$. We have $\\angle B P C>\\angle B A C$ because $P$ is inside $\\triangle A B C$. So $\\angle B P C \\geqslant \\max (\\angle B A C, \\pi-\\angle B A C)$ and hence\n\n$$\n\\sin \\angle B P C \\leqslant \\sin \\angle B A C .\n\\tag{*}\n$$\n\nLet the rays $A P, B P$, and $C P$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\\frac{P B_{1}}{B_{1} B_{3}}$ and $\\frac{P C_{1}}{C_{1} C_{3}}$ is at least 1 , which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\\Omega$.\n\nBecause $A, B, C, B_{3}$ lie on a circle, the triangles $C B_{1} B_{3}$ and $B B_{1} A$ are similar, so\n\n$$\n\\frac{C B_{1}}{B_{1} B_{3}}=\\frac{B B_{1}}{B_{1} A}\n$$\n\nApplying the sine rule we obtain\n\n$$\n\\frac{P B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{C B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{B B_{1}}{B_{1} A}=\\frac{\\sin \\angle A C P}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle P B A}\n$$\n\nSimilarly,\n\n$$\n\\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin \\angle P B A}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle A C P}\n$$\n\nMultiplying these two equations we get\n\n$$\n\\frac{P B_{1}}{B_{1} B_{3}} \\cdot \\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin ^{2} \\angle B A C}{\\sin ^{2} \\angle B P C} \\geqslant 1\n$$\n\nusing (*), which yields the desired conclusion.' 'Let the rays $A P, B P$, and $C P$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. \nAssume for the sake of contradiction that $A_{2}, B_{2}$, and $C_{2}$ all lie strictly inside circle $A B C$. It follows that $P A_{1}\n\nLet $Y$ be the foot of the bisector of $\\angle B_{3} C P$ in $\\triangle P C B_{3}$. Since $P C_{1}0$ and $\\alpha+\\beta+\\gamma=1$. Then\n\n$$\nA_{1}=\\frac{\\beta B+\\gamma C}{\\beta+\\gamma}=\\frac{1}{1-\\alpha} P-\\frac{\\alpha}{1-\\alpha} A\n$$\n\nso\n\n$$\nA_{2}=2 A_{1}-P=\\frac{1+\\alpha}{1-\\alpha} P-\\frac{2 \\alpha}{1-\\alpha} A\n$$\n\nHence\n\n$$\n\\left|A_{2}\\right|^{2}=\\left(\\frac{1+\\alpha}{1-\\alpha}\\right)^{2}|P|^{2}+\\left(\\frac{2 \\alpha}{1-\\alpha}\\right)^{2}|A|^{2}-\\frac{4 \\alpha(1+\\alpha)}{(1-\\alpha)^{2}} A \\cdot P \\text {. }\n$$\n\nUsing $|A|^{2}=1$ we obtain\n\n$$\n\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}=\\frac{1+\\alpha}{2}|P|^{2}+\\frac{2 \\alpha^{2}}{1+\\alpha}-2 \\alpha A \\cdot P\n\\tag{1}\n$$\n\nLikewise\n\n$$\n\\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}=\\frac{1+\\beta}{2}|P|^{2}+\\frac{2 \\beta^{2}}{1+\\beta}-2 \\beta B \\cdot P\n\\tag{2}\n$$\n\nand\n\n$$\n\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2}=\\frac{1+\\gamma}{2}|P|^{2}+\\frac{2 \\gamma^{2}}{1+\\gamma}-2 \\gamma C \\cdot P\n\\tag{3}\n$$\n\nSumming (1), (2) and (3) we obtain on the LHS the positive linear combination\n\n$$\n\\operatorname{LHS}=\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2}\n$$\n\nand on the RHS the quantity\n\n$$\n\\left(\\frac{1+\\alpha}{2}+\\frac{1+\\beta}{2}+\\frac{1+\\gamma}{2}\\right)|P|^{2}+\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right)-2(\\alpha A \\cdot P+\\beta B \\cdot P+\\gamma C \\cdot P) .\n$$\n\nThe first term is $2|P|^{2}$ and the last term is $-2 P \\cdot P$, so\n\n$$\n\\begin{aligned}\n\\text { RHS } & =\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right) \\\\\n& =\\frac{3 \\alpha-1}{2}+\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{3 \\beta-1}{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{3 \\gamma-1}{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\\\\n& =\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} .\n\\end{aligned}\n$$\n\nHere we used the fact that\n\n$$\n\\frac{3 \\alpha-1}{2}+\\frac{3 \\beta-1}{2}+\\frac{3 \\gamma-1}{2}=0 .\n$$\n\nWe have shown that a linear combination of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ must be at least 1 , as required.']" [] Text-only Competition True Theorem proof Geometry Math English 211 Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\angle E D C \neq 2 \cdot \angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if area $(B C D)+\operatorname{area}(A D E)=$ $\operatorname{area}(A B D)+\operatorname{area}(A B P)$. "['Let $P^{\\prime}$ be the reflection of $P$ across line $A B$, and let $M$ and $N$ be the midpoints of $P^{\\prime} E$ and $P^{\\prime} C$ respectively. Convexity ensures that $P^{\\prime}$ is distinct from both $E$ and $C$, and hence from both $M$ and $N$. We claim that both the area condition and the collinearity condition in the problem are equivalent to the condition that the (possibly degenerate) right-angled triangles $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar (equivalently, $A P^{\\prime} E$ and $B P^{\\prime} C$ are directly similar).\n\n\n\nFor the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\\angle A F M=\\angle B F N$ of signed angles modulo $\\pi$. By concyclicity of $A P^{\\prime} F M$ and $B F P^{\\prime} N$, this is equivalent to $\\angle A P^{\\prime} M=\\angle B P^{\\prime} N$, which occurs if and only if $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar.\n\n\n\nFor the other equivalence with the area condition, we have the equality of signed areas area $(A B D)+\\operatorname{area}(A B P)=\\operatorname{area}\\left(A P^{\\prime} B D\\right)=\\operatorname{area}\\left(A P^{\\prime} D\\right)+\\operatorname{area}\\left(B D P^{\\prime}\\right)$. Using the identity area $(A D E)-\\operatorname{area}\\left(A P^{\\prime} D\\right)=\\operatorname{area}(A D E)+\\operatorname{area}\\left(A D P^{\\prime}\\right)=2 \\operatorname{area}(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality\n\n$$\n\\operatorname{area}(D A M)=\\operatorname{area}(D B N)\n$$\n\nNow note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\\prime} E$ and $P^{\\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as\n\n$$\nM A \\cdot G D=N B \\cdot H D \\text {. }\n$$\n\n(In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\\prime} E$ from $P^{\\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\\prime} E$ at an angle $\\pi / 2$ clockwise from the oriented segment $P^{\\prime} E$ - we adopt the analogous conventions at $B$.)\n\n\n\n\n\nTo relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\\prime} M$ and $B P^{\\prime} N$, we use the following calculation.\n\nClaim. Let $\\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\\prime}$ with respect to $\\Gamma$. Then we have the equality\n\n$$\nG D \\cdot P^{\\prime} M=H D \\cdot P^{\\prime} N=\\frac{1}{4} h \\neq 0\n$$\n\nProof. Firstly, we have $h \\neq 0$, since otherwise $P^{\\prime}$ would lie on $\\Gamma$, and hence the internal angle bisectors of $\\angle E D P^{\\prime}$ and $\\angle P^{\\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\\angle E D C \\neq 2 \\cdot \\angle A D B$ given in the question.\n\nNext, let $E^{\\prime}$ denote the second point of intersection of $P^{\\prime} E$ with $\\Gamma$, and let $E^{\\prime \\prime}$ denote the point on $\\Gamma$ diametrically opposite $E^{\\prime}$, so that $E^{\\prime \\prime} E$ is perpendicular to $P^{\\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\\prime} E$ and $E E^{\\prime \\prime}$ of the right-angled triangle $P^{\\prime} E E^{\\prime \\prime}$; it follows that $G$ is the midpoint of $P^{\\prime} E^{\\prime \\prime}$. Since $D$ is the midpoint of $E^{\\prime} E^{\\prime \\prime}$, we have that $G D=\\frac{1}{2} P^{\\prime} E^{\\prime}$. Since $P^{\\prime} M=\\frac{1}{2} P^{\\prime} E$, we have $G D \\cdot P^{\\prime} M=\\frac{1}{4} P^{\\prime} E^{\\prime} \\cdot P^{\\prime} E=\\frac{1}{4} h$. The other equality $H D \\cdot P^{\\prime} N$ follows by exactly the same argument.\n\n\n\nFrom this claim, we see that the area condition is equivalent to the equality\n\n$$\n\\left(M A: P^{\\prime} M\\right)=\\left(N B: P^{\\prime} N\\right)\n$$\n\nof ratios of signed lengths, which is equivalent to direct similarity of $A P^{\\prime} M$ and $B P^{\\prime} N$, as desired.' 'Along the perpendicular bisector of $C E$, define the linear function\n\n$$\nf(X)=\\operatorname{area}(B C X)+\\operatorname{area}(A X E)-\\operatorname{area}(A B X)-\\operatorname{area}(A B P)\n$$\n\nwhere, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D)=0$.\n\n\n\nLet $P^{\\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\\prime}$ does not lie on the line $C E$. To see this, we let $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\\prime}$, so that $P$ is the orthogonal projection of $P^{\\prime}$ onto $A^{\\prime \\prime} B^{\\prime \\prime}$. Since $A$ lies on the perpendicular bisector of $P^{\\prime} E$, the triangle $A^{\\prime \\prime} E P^{\\prime}$ is right-angled at $E$ (and $B^{\\prime \\prime} C P^{\\prime}$ similarly). If $P^{\\prime}$ were to lie on $C E$, then the lines $A^{\\prime \\prime} E$ and $B^{\\prime \\prime} C$ would be perpendicular to $C E$ and $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\\prime \\prime} B^{\\prime \\prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\\prime}$ on $C E$.\n\nWe thus let $O$ be the centre of the circle $C E P^{\\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\\prime}$ and $C P^{\\prime}$, respectively, so\n\n$$\n\\begin{aligned}\n\\operatorname{area}(B C O)+\\operatorname{area}(A O E) & =\\operatorname{area}\\left(O P^{\\prime} B\\right)+\\operatorname{area}\\left(P^{\\prime} O A\\right)=\\operatorname{area}\\left(P^{\\prime} B O A\\right) \\\\\n& =\\operatorname{area}(A B O)+\\operatorname{area}\\left(B A P^{\\prime}\\right)=\\operatorname{area}(A B O)+\\operatorname{area}(A B P)\n\\end{aligned}\n$$\n\nand hence $f(O)=0$.\n\nNotice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\\angle E O C=2 \\cdot \\angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D)=0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear.\n\n\nIn the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides\n\n$$\n\\operatorname{area}(B C T)+\\operatorname{area}(A T E)=\\operatorname{area}(P B T A)=\\operatorname{area}(A B T)-\\operatorname{area}(A B P)\n$$\n\n$$\nf(T)=-2 \\operatorname{area}(A B P) \\neq 0 .\n$$\n\n\n\nHence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \\neq 0$.\n\nIn the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes.\n\n\n\nLet $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\\prime}$ and $B^{\\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2 , then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\\prime}$ and $B^{\\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so\n\n$$\nf(Q)=(\\operatorname{area}(B C Q)+\\operatorname{area}(A Q E))-(\\operatorname{area}(A B Q)-\\operatorname{area}(B A P))=0-0=0\n$$\n\nThe points $O$ and $Q$ are distinct. To see this, consider the circle $\\omega$ centred on $Q$ with $P^{\\prime}$ on the circumference; since triangle $P P^{\\prime} Q$ is right-angled at $P^{\\prime}$, it follows that $P$ lies outside $\\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\\omega$, so that $\\omega$ is not the circle $C E P^{\\prime}$ and $Q \\neq O$.\n\nSince $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D)=0$ as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 212 Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle D P A+\angle A Q D=\angle Q I P$. "[""Let $N$ and $M$ be the midpoints of the $\\operatorname{arcs} \\overparen{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\\angle F A E=\\angle B A C=\\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\\varphi$ (dilation in case of $A B=A C$ ) that moves $A F I E$ to $N B M C$. The directed angle in which $\\varphi$ changes directions is $\\angle(A F, N B)$, same as $\\angle(A P, N P)$ and $\\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P=E F \\cap A P$ and $Q=E F \\cap A Q$ are mapped to points $B C \\cap N P$ and $B C \\cap N Q$, respectively. Denote these points by $P^{\\prime}$ and $Q^{\\prime}$, respectively.\n\n\n\nLet $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D=L$ then line $B C$ is tangent to circle $P Q D$ ). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have\n\n$$\n\\frac{B D}{D C}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=-\\frac{B Z}{Z C}\n$$\n\nso the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \\cdot Z C=Z D \\cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \\cdot Z L=Z B \\cdot Z C=Z P \\cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim.\n\nBy $\\angle M P P^{\\prime}=\\angle M Q Q^{\\prime}=\\angle M L P^{\\prime}=\\angle M L Q^{\\prime}=90^{\\circ}$, the quadrilaterals $M L P P^{\\prime}$ and $M L Q Q^{\\prime}$ are cyclic. Then the problem statement follows by\n\n$$\n\\begin{aligned}\n\\angle D P A+\\angle A Q D & =360^{\\circ}-\\angle P A Q-\\angle Q D P=360^{\\circ}-\\angle P N Q-\\angle Q L P \\\\\n& =\\angle L P N+\\angle N Q L=\\angle P^{\\prime} M L+\\angle L M Q^{\\prime}=\\angle P^{\\prime} M Q^{\\prime}=\\angle P I Q .\n\\end{aligned}\n$$"" 'Define the point $M$ and the same spiral similarity $\\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$.\n\nBy $\\varphi(F)=B$ and $\\varphi(E)=C$ the triangles $S B F$ and $S C E$ are similar, so we have\n\n$$\n\\frac{S B}{S C}=\\frac{B F}{C E}=\\frac{B D}{C D}\n$$\n\nBy the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\\angle B S C$ and hence passes through $M$.\n\nLet $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\\varphi(K)=\\varphi(F E \\cap S I)=B C \\cap S M=D$. By $\\varphi(I)=M$, we have $K D \\| I M$.\n\n\n\nWe claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\\angle A M L=\\angle A S L=\\angle A S I=90^{\\circ}$, we have $M L \\| P Q$. Hence, $\\overparen{P L}=\\overparen{M Q}$ and therefore $\\angle P S L=\\angle M S Q=\\angle D S Q$. By $\\angle Q P S=\\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally,\n\n$$\n\\frac{S P}{S I}=\\frac{S P}{S K} \\cdot \\frac{S K}{S I}=\\frac{S M}{S Q} \\cdot \\frac{S D}{S M}=\\frac{S D}{S Q}\n$$\n\nshows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously.\n\nNow the problem statement can be proved by\n\n$$\n\\angle D P A+\\angle A Q D=\\angle D P S+\\angle S Q D=\\angle Q I S+\\angle S I P=\\angle Q I P\n$$' 'Denote the circumcircle of triangle $A B C$ by $\\Gamma$, and let rays $P D$ and $Q D$ meet $\\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \\perp I P$ and $A V \\perp I Q$. Then the problem statement will follow as\n\n$$\n\\angle D P A+\\angle A Q D=\\angle V U A+\\angle A V U=180^{\\circ}-\\angle U A V=\\angle Q I P .\n$$\n\nLet $M$ be the midpoint of arc $\\overparen{B U V C}$ and let $N$ be the midpoint of $\\operatorname{arc} \\overparen{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively.\n\nwe observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively.\n\nCombine the inversion with projecting $\\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively.\n\n\n\nThe combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \\perp I F$, $A C \\perp I E$ and $A N \\perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \\perp I P$ and $A V \\perp I Q$, and hence completes the solution.']" [] Text-only Competition True Theorem proof Geometry Math English 213 "The incircle $\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\omega$ again at $R$. Line $A R$ meets $\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. $\angle(a, b)$ denotes the directed angle between lines $a$ and $b$, measured modulo $\pi$." "['Step 1. The external bisector of $\\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\\omega$. Since $A N \\cdot A I=A E^{2}=A R \\cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R=I P$, the line $N I$ is the external bisector of $\\angle P N R$, so $P N$ meets $\\omega$ again at the point symmetric to $R$ with respect to $A N$ - i.e. at $K$.\n\nLet $D N$ cross $\\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\\omega$ meet on the polar line of the intersection of the diagonals with respect to $\\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear.\n\n\nStep 2. Let $\\Gamma$ be the circumcircle of $\\triangle B I C$. Notice that\n\n$$\n\\begin{aligned}\n\\angle(B Q, Q C)=\\angle & (B Q, Q P)+\\angle(P Q, Q C)=\\angle(B F, F P)+\\angle(P E, E C) \\\\\n& =\\angle(E F, E P)+\\angle(F P, F E)=\\angle(F P, E P)=\\angle(D F, D E)=\\angle(B I, I C),\n\\end{aligned}\n$$\n\nso $Q$ lies on $\\Gamma$. Let $Q P$ meet $\\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\\angle(B I, I T)=\\angle(B Q, Q T)=\\angle(B F, F P)=\\angle(F K, K P)$. Note $F D \\perp F K$ and $F D \\perp B I$ so $F K \\| B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I=I K$, the line $I T$ crosses $D N$ at its midpoint $M$.\n\nStep 3. Let $F^{\\prime}$ and $E^{\\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\\prime} \\cdot E^{\\prime} F=D E^{\\prime 2}=$ $B E^{\\prime} \\cdot E^{\\prime} I$, the point $E^{\\prime}$ lies on the radical axis of $\\omega$ and $\\Gamma$; the same holds for $F^{\\prime}$. Therefore, this radical axis is $E^{\\prime} F^{\\prime}$, and it passes through $M$. Thus $I M \\cdot M T=D M \\cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\\angle(D S, S T)=\\angle(D I, I T)=\\angle(D K, K P)=\\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired.\n\n\n\n' 'Namely, we introduce the points $K, L, N$, and $S$, and show that the triples $(P, N, K)$ and $(P, S, L)$ are collinear. We conclude that $K$ and $R$ are symmetric in $A I$, and reduce the problem statement to showing that $P, Q$, and $S$ are collinear.\n\nStep 1. Let $A R$ meet the circumcircle $\\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\\Omega$ again at the midpoint $M^{\\prime}$ of arc $B A C$.\n\nStep 2. Denote the circles $B F P$ and $C E P$ by $\\Omega_{B}$ and $\\Omega_{C}$, respectively. Let $\\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have\n\n$$\n\\angle(U B, B F)=\\angle(U P, P F)=\\angle(R P, P F)=\\angle(R F, F A),\n$$\n\nso $U B \\| R F$.\n\n\n\nNext, we show that the points $B, I, U$, and $X$ are concyclic. Since\n\n$$\n\\angle(U B, U X)=\\angle(R F, R X)=\\angle(A F, A R)+\\angle(F R, F A)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F),\n$$\n\nit suffices to prove $\\angle(I B, I X)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F)$, or $\\angle\\left(I B, M^{\\prime} B\\right)=\\angle(D R, D F)$. But both angles equal $\\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\\prime}$ is the midpoint of $\\operatorname{arc} B A C$ of $\\Omega$.)\n\nIt follows now from circles $B U I X$ and $B P U F Y$ that\n\n$$\n\\begin{aligned}\n\\angle(I U, U B)=\\angle(I X, B X)=\\angle\\left(M^{\\prime} X, B X\\right)= & \\frac{\\pi-\\angle A}{2} \\\\\n& =\\angle(E F, A F)=\\angle(Y F, B F)=\\angle(Y U, B U),\n\\end{aligned}\n$$\n\nso the points $Y, U$, and $I$ are collinear.\n\nLet $E F$ meet $B C$ at $W$. We have\n\n$$\n\\angle(I Y, Y W)=\\angle(U Y, F Y)=\\angle(U B, F B)=\\angle(R F, A F)=\\angle(C I, C W)\n$$\n\nso the points $W, Y, I$, and $C$ are concyclic.\n\n\n\nSimilarly, if $V$ and $Z$ are the second meeting points of $\\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples $(C, V, I, X)$ and $(B, I, Z, W)$ are both concyclic.\n\nStep 3. Let $Q^{\\prime}=C Y \\cap B Z$. We will show that $Q^{\\prime}=Q$.\n\nFirst of all, we have\n\n$$\n\\begin{aligned}\n& \\angle\\left(Q^{\\prime} Y, Q^{\\prime} B\\right)=\\angle(C Y, Z B)=\\angle(C Y, Z Y)+\\angle(Z Y, B Z) \\\\\n& \\quad=\\angle(C I, I W)+\\angle(I W, I B)=\\angle(C I, I B)=\\frac{\\pi-\\angle A}{2}=\\angle(F Y, F B)\n\\end{aligned}\n$$\n\nso $Q^{\\prime} \\in \\Omega_{B}$. Similarly, $Q^{\\prime} \\in \\Omega_{C}$. Thus $Q^{\\prime} \\in \\Omega_{B} \\cap \\Omega_{C}=\\{P, Q\\}$ and it remains to prove that $Q^{\\prime} \\neq P$. If we had $Q^{\\prime}=P$, we would have $\\angle(P Y, P Z)=\\angle\\left(Q^{\\prime} Y, Q^{\\prime} Z\\right)=\\angle(I C, I B)$. This would imply\n\n$$\n\\angle(P Y, Y F)+\\angle(E Z, Z P)=\\angle(P Y, P Z)=\\angle(I C, I B)=\\angle(P E, P F)\n$$\n\nso circles $\\Omega_{B}$ and $\\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\\prime}=Q$.\n\n\n\nStep 4. Now we are ready to show that $P, Q$, and $S$ are collinear.\n\nNotice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\\omega$, so $W$ is the pole of $A D$. Hence, $W I \\perp A D$. Since $C I \\perp D E$, this yields $\\angle(I C, W I)=\\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\\triangle D E F$, so $\\angle(D E, D A)=\\angle(D N, D F)=\\angle(D S, D F)$. Therefore,\n\n$$\n\\begin{aligned}\n\\angle(P S, P F)=\\angle(D S, D F)=\\angle(D E, D A)= & \\angle(I C, I W) \\\\\n& =\\angle(Y C, Y W)=\\angle(Y Q, Y F)=\\angle(P Q, P F)\n\\end{aligned}\n$$\n\nwhich yields the desired collinearity.']" [] Text-only Competition True Theorem proof Geometry Math English 214 "Let $\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$. Suppose that for any point $X$, and for any three lines $\ell_{1}, \ell_{2}, \ell_{3}$ passing through $X$, the points $f\left(\ell_{1}\right), f\left(\ell_{2}\right), f\left(\ell_{3}\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\ell)=P$ for any line $\ell$ passing through $P$." "[""We provide a complete characterisation of the functions satisfying the given condition.\n\nWrite $\\angle\\left(\\ell_{1}, \\ell_{2}\\right)$ for the directed angle modulo $180^{\\circ}$ between the lines $\\ell_{1}$ and $\\ell_{2}$. Given a point $P$ and an angle $\\alpha \\in\\left(0,180^{\\circ}\\right)$, for each line $\\ell$, let $\\ell^{\\prime}$ be the line through $P$ satisfying $\\angle\\left(\\ell^{\\prime}, \\ell\\right)=\\alpha$, and let $h_{P, \\alpha}(\\ell)$ be the intersection point of $\\ell$ and $\\ell^{\\prime}$. We will prove that there is some pair $(P, \\alpha)$ such that $f$ and $h_{P, \\alpha}$ are the same function. Then $P$ is the unique point in the problem statement.\n\nGiven an angle $\\alpha$ and a point $P$, let a line $\\ell$ be called $(P, \\alpha)$-good if $f(\\ell)=h_{P, \\alpha}(\\ell)$. Let a point $X \\neq P$ be called $(P, \\alpha)$-good if the circle $g(X)$ passes through $P$ and some point $Y \\neq P, X$ on $g(X)$ satisfies $\\angle(P Y, Y X)=\\alpha$. It follows from this definition that if $X$ is $(P, \\alpha)$ good then every point $Y \\neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \\alpha}(X Y)=Y$ for every $Y \\in g(X)$. Equivalently, $f(\\ell) \\in\\left\\{X, h_{P, \\alpha}(\\ell)\\right\\}$ for each line $\\ell$ passing through $X$. This shows the following lemma.\n\nLemma 1. If $X$ is $(P, \\alpha)$-good and $\\ell$ is a line passing through $X$ then either $f(\\ell)=X$ or $\\ell$ is $(P, \\alpha)$-good.\n\nLemma 2. If $X$ and $Y$ are different $(P, \\alpha)$-good points, then line $X Y$ is $(P, \\alpha)$-good.\n\nProof. If $X Y$ is not $(P, \\alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \\neq Y$.\n\nLemma 3. If $\\ell_{1}$ and $\\ell_{2}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then either $f\\left(\\ell_{1}\\right)=X$ or $f\\left(\\ell_{2}\\right)=X$ or $X$ is $(P, \\alpha)$-good.\n\nProof. If $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right) \\neq X$, then $g(X)$ is the circumcircle of $X, f\\left(\\ell_{1}\\right)$ and $f\\left(\\ell_{2}\\right)$. Since $\\ell_{1}$ and $\\ell_{2}$ are $(P, \\alpha)$-good lines, the angles\n\n$$\n\\angle\\left(\\operatorname{Pf}\\left(\\ell_{1}\\right), f\\left(\\ell_{1}\\right) X\\right)=\\angle\\left(\\operatorname{Pf}\\left(\\ell_{2}\\right), f\\left(\\ell_{2}\\right) X\\right)=\\alpha\n$$\n\nso $P$ lies on $g(X)$. Hence, $X$ is $(P, \\alpha)$-good.\n\nLemma 4. If $\\ell_{1}, \\ell_{2}$ and $\\ell_{3}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then $X$ is $(P, \\alpha)$-good.\n\nProof. This follows from the previous Lemma since at most one of the three lines $\\ell_{i}$ can satisfy $f\\left(\\ell_{i}\\right)=X$ as the three lines are all $(P, \\alpha)$-good.\n\nLemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\\alpha$ such that $A, B$ and $C$ are $(P, \\alpha) \\operatorname{good}$ points and $A B, B C$ and $C A$ are $(P, \\alpha)$-good lines.\n\n\n\n\n\nProof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \\neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles\n\n$$\n\\angle(P D, D C)=\\angle(P F, F B)=\\angle(P E, E A)=\\alpha\n$$\n\nfor some angle $\\alpha$. Hence, lines $A B, B C$ and $C A$ are all $(P, \\alpha)$-good, so by Lemma $3, A, B$ and $C$ are $(P, \\alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.)\n\nConsider the set $\\Omega$ of all points $(x, y)$ with integer coordinates $1 \\leqslant x, y \\leqslant 1000$, and consider the set $L_{\\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\\Omega}$. For each line $\\ell$ in $L_{\\Omega}$ we colour the point $f(\\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5 , so these three vertices are all $(P, \\alpha)$-good for some point $P$ and angle $\\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \\alpha)$-good for this particular pair $(P, \\alpha)$. Now by Lemma 1 , the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that $(m+1, n+1)$ is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points $(m, n+2)$ and $(m+2, n)$ then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\\Omega_{10}$ are good.\n\nNow we will use this to prove that every point $S \\neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\\ell$ through $S$ intersects at most 10 points in $\\Omega_{10}$. Hence, there are at least eight lines $\\ell$ through $S$ which contain a point $Q$ in $\\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\\ell) \\neq Q$. Hence, by Lemma 1, the line $\\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4 , the point $S$ is good. Hence, every point $S \\neq P$ is good, so by Lemma 2 , every line is good. In particular, every line $\\ell$ passing through $P$ is good, and therefore satisfies $f(\\ell)=P$, as required."" ""Note that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \\in g(X) \\cap g(Y) \\cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$.\n\nLemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \\notin\\{X, Y\\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$.\n\nProof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram below)\n\n\n\n\n\nNow pick any line $\\ell$ and any six different points $Y_{1}, \\ldots, Y_{6}$ on $\\ell \\backslash\\{f(\\ell)\\}$. Pick a point $X$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \\ldots, Y_{4}$ do not lie on $g(X)$, so that $f\\left(X Y_{i}\\right) \\notin\\left\\{X, Y_{i}\\right\\}$ for $1 \\leqslant i \\leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \\leqslant i0$, there is a point $P_{\\epsilon}$ with $g\\left(P_{\\epsilon}\\right)$ of radius at most $\\epsilon$. Then there is a point $P$ with the given property.\n\nProof. Consider a sequence $\\epsilon_{i}=2^{-i}$ and corresponding points $P_{\\epsilon_{i}}$. Because the two circles $g\\left(P_{\\epsilon_{i}}\\right)$ and $g\\left(P_{\\epsilon_{j}}\\right)$ meet, the distance between $P_{\\epsilon_{i}}$ and $P_{\\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\\sum_{i} \\epsilon_{i}$ converges, these points converge to some point $P$. For all $\\epsilon>0$, the point $P$ has distance at most $2 \\epsilon$ from $P_{\\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \\epsilon$ from $P_{\\epsilon}$, so distance at most $4 \\epsilon$ from $P$. A circle that passes distance at most $4 \\epsilon$ from $P$ for all $\\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property.\n\nLemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property.\n\nProof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2 .\n\nLemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\\left(B_{i}\\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property.\n\nProof. Consider some line $\\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\\left(B_{i}\\right)$. Fix some direction along that line, and let $X_{\\epsilon}$ be the point on $\\ell$ that has distance $\\epsilon$ from $A$ in that direction. In what follows we consider only those $\\epsilon$ for which $X_{\\epsilon}$ does not lie on any $g\\left(B_{i}\\right)$ (this restriction excludes only finitely many possible values of $\\epsilon$ ).\n\nConsider the circle $g\\left(X_{\\epsilon}\\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\\epsilon \\rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\\left(X_{\\epsilon}\\right)$ for any $\\epsilon$ in that sequence.\n\n\n\nThen $f\\left(X_{\\epsilon} B_{1}\\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\\epsilon} B_{1}$ with $g\\left(B_{1}\\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\\epsilon}, f\\left(X_{\\epsilon} B_{1}\\right)$ and $f\\left(X_{\\epsilon} B_{2}\\right)$. As $\\epsilon \\rightarrow 0$, the angle at $X_{\\epsilon}$ tends to $\\angle B_{1} A B_{2}$ or $180^{\\circ}-\\angle B_{1} A B_{2}$, which is not 0 or $180^{\\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \\epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\\left(X_{\\epsilon}\\right)$, is also bounded above by a constant multiple of $\\epsilon$, and so the result follows by Lemma 2 .\n\nLemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property.\n\nProof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.\n\nCase 1: D does not lie on that line:\n\nIn this case, consider a sequence of $X_{\\epsilon}$ at distance $\\epsilon$ from $D$, tending to $D$ along some line that is not a tangent to either circle, but perturbed slightly (by at most $\\epsilon^{2}$ ) to ensure that no three of the points $A, B$ and $X_{\\epsilon}$ are collinear and no four are concyclic.\n\nConsider the points $f\\left(X_{\\epsilon} A\\right)$ and $f\\left(X_{\\epsilon} B\\right)$, and the circles $g\\left(X_{\\epsilon}\\right)$ on which they lie. The point $f\\left(X_{\\epsilon} A\\right)$ might be either $A$ or the other intersection of $X_{\\epsilon} A$ with the circle $g(A)$, and the same applies for $B$. If, for some sequence of $\\epsilon \\rightarrow 0$, both those points are the other point of intersection, the same argument as in the proof of Lemma 4 applies to find arbitrarily small circles. Otherwise, we have either infinitely many of those circles passing through $A$, or infinitely many passing through $B$; without loss of generality, suppose infinitely many through $A$.\n\nWe now show we can find five points $B_{i}$ satisfying the conditions of Lemma 4 (together with $A$ ). Let $B_{1}$ be any of the $X_{\\epsilon}$ for which $g\\left(X_{\\epsilon}\\right)$ passes through $A$. Then repeat the following four times, for $2 \\leqslant i \\leqslant 5$.\n\nConsider some line $\\ell=X_{\\epsilon} A$ (different from those considered for previous $i$ ) that is not tangent to any of the $g\\left(B_{j}\\right)$ for $j\n\nSo, Case 2.1 can occur for only finitely many points $X$.\n\nCase 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.\n\nClearly, $R_{1} \\neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\\ell$ only at $X$ and $Q$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P X R_{i}\\right)$. Let $R_{i} Q$ meet $g\\left(R_{i}\\right)$ again at $S_{i}$; then $S_{i} \\neq Q$, as $g\\left(R_{i}\\right)$ meets $\\ell$ only at $P$ and $X$. Then $f\\left(R_{i} Q\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases.\n\n\n\nSubcase 2.2.1: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=S_{2}$; so $S_{1}, S_{2} \\in g(Q)$.\n\nIn this case we have $0=\\angle\\left(R_{1} X, X P\\right)+\\angle\\left(X P, R_{2} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} R_{2}\\right)=$ $\\angle\\left(Q S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} Q\\right)$, which shows $P \\in g(Q)$.\n\nSubcase 2.2.2: $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=R_{2}$; so $R_{1}, R_{2} \\in g(Q)$.\n\nThis can happen for at most four positions of $X$ - namely, at the intersections of $\\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \\in g(Q) \\cap p_{i}$.\n\n\n\nSubcase 2.2.3: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=R_{2}$ (the case $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=S_{2}$ is similar).\n\nIn this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\\angle\\left(Q S_{1}, S_{1} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} X\\right)=$ $\\angle\\left(R_{1} P, P X\\right)=\\angle\\left(p_{1}, \\ell\\right) \\neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \\rightarrow \\ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction.\n\nThus Case 2.2 also occurs for only finitely many points $X$, as desired.\n\nStep 3: We show that $f(P Q)=P$, as desired.\n\nThe argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \\in \\ell \\backslash\\{P, Q\\}$, which is absurd.\n\nCase 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$.\n\nLet $t(Q) \\operatorname{cross} p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X$ cross $g(R)$ again at $S$. Then $f(R X) \\in\\{R, S\\} \\cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$.\n\n\n\nSubcase 3.1.1: $S=f(R X) \\in g(X)$.\n\nWe have $\\angle(t(X), Q X)=\\angle(S X, S Q)=\\angle(S R, S Q)=\\angle(P R, P Q)=\\angle\\left(p_{2}, \\ell\\right)$. Hence $t(X) \\| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\\prime}$, so $g\\left(R^{\\prime}\\right)=\\left(P R^{\\prime} X\\right)$, and let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$; notice that $S^{\\prime} \\neq Q$. Excluding one position of $X$, we may assume that $R^{\\prime} \\notin g(Q)$, so $R^{\\prime} \\neq f\\left(R^{\\prime} Q\\right)$. Therefore, $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then, as in Case 2.1, we get $\\angle(t(Q), P Q)=\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(p_{2}, \\ell\\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible.\n\nSubcase 3.1.2: $R=f(R X) \\in g(X)$.\n\nIn this case, we have $\\angle(t(X), \\ell)=\\angle(R X, R Q)=\\angle\\left(R X, p_{1}\\right)$. Again, let $R^{\\prime}=t(X) \\cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\\left(R^{\\prime}\\right)=\\left(R^{\\prime} X P\\right)$; let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$. Due to $\\angle\\left(R^{\\prime} X, X R\\right)=\\angle(Q X, Q R)=\\angle\\left(\\ell, p_{1}\\right), R^{\\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\\prime} \\notin g(Q)$. Therefore, for those positions we have $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then $\\angle\\left(R X, p_{1}\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(R^{\\prime} S^{\\prime}, S^{\\prime} P\\right)=$ $\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well.\n\nThus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction.\n\n\n\nCase 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.\n\nBy Step $2, R_{1} \\neq R_{2}$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P Q R_{i}\\right)$. Let $R_{i} X$ meet $g\\left(R_{i}\\right)$ at $S_{i}$; then $S_{i} \\neq X$. Then $f\\left(R_{i} X\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases.\n\n\n\nSubcase 3.2.1: $f\\left(R_{1} X\\right)=S_{1}$ and $f\\left(R_{2} X\\right)=S_{2}$, so $S_{1}, S_{2} \\in g(X)$.\n\nAs in Subcase 2.2.1, we have $0=\\angle\\left(R_{1} Q, Q P\\right)+\\angle\\left(Q P, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} X\\right)$, which shows $P \\in g(X)$. But $X, Q \\in g(X)$ as well, so $g(X)$ meets $\\ell$ at three distinct points, which is absurd.\n\nSubcase 3.2.2: $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=R_{2}$, so $R_{1}, R_{2} \\in g(X)$.\n\nNow three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible.\n\nSubcase 3.2.3: $f\\left(R_{1} X\\right)=S_{1}, f\\left(R_{2} X\\right)=R_{2}$ (the case $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=S_{2}$ is similar).\n\nWe have $\\angle\\left(X R_{2}, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} P, P Q\\right)=\\angle\\left(p_{1}, \\ell\\right)$, so this case can occur for a unique position of $X$.\n\nThus, in Case 3.2, there is only a unique position of $X$, again yielding the required contradiction.\n\n\n### 分析\nThe condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\\ell$ passing through $X$, the point $f(\\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\\ell)$ other than $X$; for such points, an arbitrary choice is made.)\n\nIf there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.']" [] Text-only Competition True Theorem proof Geometry Math English 215 Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\left(\begin{array}{c}a n \\ b\end{array}\right)-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime. "['For $p$ a prime and $n$ a nonzero integer, we write $v_{p}(n)$ for the $p$-adic valuation of $n$ : the largest integer $t$ such that $p^{t} \\mid n$.\n\nWe first show that $b$ is $a$-good if and only if $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$.\n\nTo start with, the condition that $a n+1 \\mid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)-1$ can be rewritten as saying that\n\n$$\n\\frac{a n(a n-1) \\cdots(a n-b+1)}{b !} \\equiv 1 \\quad(\\bmod a n+1)\n\\tag{1}\n$$\n\nSuppose, on the one hand, there is a prime $p \\leqslant b$ with $p \\nmid a$. Take $t=v_{p}(b !)$. Then there exist positive integers $c$ such that $a c \\equiv 1\\left(\\bmod p^{t+1}\\right)$. If we take $c$ big enough, and then take $n=(p-1) c$, then $a n=a(p-1) c \\equiv p-1\\left(\\bmod p^{t+1}\\right)$ and $a n \\geqslant b$. Since $p \\leqslant b$, one of the terms of the numerator $a n(a n-1) \\cdots(a n-b+1)$ is $a n-p+1$, which is divisible by $p^{t+1}$. Hence the $p$-adic valuation of the numerator is at least $t+1$, but that of the denominator is exactly $t$. This means that $p \\mid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$, so $p \\nmid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)-1$. As $p \\mid a n+1$, we get that $a n+1 \\nmid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$, so $b$ is not $a$-good.\n\nOn the other hand, if for all primes $p \\leqslant b$ we have $p \\mid a$, then every factor of $b$ ! is coprime to $a n+1$, and hence invertible modulo $a n+1$ : hence $b$ ! is also invertible modulo $a n+1$. Then equation (1) reduces to:\n\n$$\na n(a n-1) \\cdots(a n-b+1) \\equiv b ! \\quad(\\bmod a n+1)\n$$\n\nHowever, we can rewrite the left-hand side as follows:\n\n$$\na n(a n-1) \\cdots(a n-b+1) \\equiv(-1)(-2) \\cdots(-b) \\equiv(-1)^{b} b ! \\quad(\\bmod a n+1)\n$$\n\nProvided that $a n>1$, if $b$ is even we deduce $(-1)^{b} b ! \\equiv b$ ! as needed. On the other hand, if $b$ is odd, and we take $a n+1>2(b !)$, then we will not have $(-1)^{b} b ! \\equiv b$ !, so $b$ is not $a$-good. This completes the claim.\n\nTo conclude from here, suppose that $b$ is $a$-good, but $b+2$ is not. Then $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$, but there is a prime $q \\leqslant b+2$ for which $q \\nmid a$ : so $q=b+1$ or $q=b+2$. We cannot have $q=b+2$, as that is even too, so we have $q=b+1$ : in other words, $b+1$ is prime.' ""We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p \\mid a$ for all primes $p \\leqslant b$. We do this with Lucas' theorem.\n\nSuppose that we have $p \\leqslant b$ with $p \\nmid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p \\leqslant b$. Suppose it is the $p^{t}$ digit for $t \\geqslant 1$.\n\nNow, as $n$ varies over the integers, $a n+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $a n>b$ ) such that the $p^{0}$ digit of an is $p-1$ (so $p \\mid a n+1)$ and the $p^{t}$ digit of an is 0 . Consequently, $p \\mid a n+1$ but $p \\mid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$ (by Lucas' theorem) so $p \\nmid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)-1$. Thus $b$ is not $a$-good.\n\nNow we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $a n+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\\left(\\begin{array}{c}a n \\\\ b+2\\end{array}\\right)$ and $\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$ is $(b+1)(b+2) /(a n-b)(a n-b-1)$. We know that there must be a choice of $a n+1$ such that the former binomial coefficient is 1 modulo $a n+1$ but the latter is not, which means that the given ratio must not be $1 \\bmod a n+1$. If $b+1$ and $b+2$ are both coprime to $a n+1$ then\n\n\n\nthe ratio is 1 , so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.\n\nHowever, we can observe that $b$ must be even by insisting that $a n+1$ is prime (which is possible by Dirichlet's theorem) and hence $\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right) \\equiv(-1)^{b}=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.""]" [] Text-only Competition True Theorem proof Number Theory Math English 216 Let $H=\left\{\lfloor i \sqrt{2}\rfloor: i \in \mathbb{Z}_{>0}\right\}=\{1,2,4,5,7, \ldots\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \subset\{1,2, \ldots, n\}$ satisfies $|A| \geqslant C \sqrt{n}$, then there exist $a, b \in A$ such that $a-b \in H$. (Here $\mathbb{Z}_{>0}$ is the set of positive integers, and $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z$.) "['we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|1-\\frac{1}{\\sqrt{2}}\n\\tag{1}\n$$\n\nTo see why, observe that $n \\in H$ if and only if $00}$. In other words, $01$ since $a_{i}-a_{1} \\notin H$. Furthermore, we must have $\\left\\{a_{i} / \\sqrt{2}\\right\\}<\\left\\{a_{j} / \\sqrt{2}\\right\\}$ whenever $i1 / \\sqrt{2}>$ $1-1 / \\sqrt{2}$, contradicting (1).\n\nNow, we have a sequence $0=a_{1}\\frac{1}{2 d \\sqrt{2}}\n\\tag{2}\n$$\n\nTo see why this is the case, let $h=\\lfloor d / \\sqrt{2}\\rfloor$, so $\\{d / \\sqrt{2}\\}=d / \\sqrt{2}-h$. Then\n\n$$\n\\left\\{\\frac{d}{\\sqrt{2}}\\right\\}\\left(\\frac{d}{\\sqrt{2}}+h\\right)=\\frac{d^{2}-2 h^{2}}{2} \\geqslant \\frac{1}{2}\n$$\n\nsince the numerator is a positive integer. Because $d / \\sqrt{2}+h<2 d / \\sqrt{2}$, inequality (2) follows.\n\nLet $d_{i}=a_{i+1}-a_{i}$, for $1 \\leqslant i\\sum_{i}\\left\\{\\frac{d_{i}}{\\sqrt{2}}\\right\\}>\\frac{1}{2 \\sqrt{2}} \\sum_{i} \\frac{1}{d_{i}} \\geqslant \\frac{(k-1)^{2}}{2 \\sqrt{2}} \\frac{1}{\\sum_{i} d_{i}}>\\frac{(k-1)^{2}}{2 \\sqrt{2}} \\cdot \\frac{1}{n}\n\\tag{3}\n$$\n\nHere, the first inequality holds because $\\left\\{a_{k} / \\sqrt{2}\\right\\}<1-1 / \\sqrt{2}$, the second follows from (2), the third follows from an easy application of the AM-HM inequality (or Cauchy-Schwarz), and the fourth follows from the fact that $\\sum_{i} d_{i}=a_{k}k-1\n$$\n\nwhich provides the required bound on $k$.' 'we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|0}\\right\\}$ is the complementary Beatty sequence to $H$ (in other words, $H$ and $J$ are disjoint with $H \\cup J=\\mathbb{Z}_{>0}$ ). Write $A=\\left\\{a_{1}0}$.\n\nFor any $j>i$, we have $a_{j}-a_{i}=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. Because $a_{j}-a_{i} \\in J$, we also have $a_{j}-a_{i}=\\lfloor\\alpha t\\rfloor$ for some positive integer $t$. Thus, $\\lfloor\\alpha t\\rfloor=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. The right hand side must equal either $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$ or $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor-1$, the latter of which is not a member of $J$ as $\\alpha>2$. Therefore, $t=b_{j}-b_{i}$ and so we have $\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor=\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$.\n\nFor $1 \\leqslant i1 /(2 d \\sqrt{2})$ for positive integers $d)$ proves that $1>\\left((k-1)^{2} /(2 \\sqrt{2})\\right)(\\alpha / n)$, which again rearranges to give\n\n$$\n\\sqrt{2 \\sqrt{2}-2} \\cdot \\sqrt{n}>k-1\n$$' 'we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|0} \\backslash H$, so all differences between elements of $A$ are in $J$. We start by making the following observation. Suppose we have a set $B \\subseteq\\{1,2, \\ldots, n\\}$ such that all of the differences between elements of $B$ are in $H$. Then $|A| \\cdot|B| \\leqslant 2 n$.\n\nTo see why, observe that any two sums of the form $a+b$ with $a \\in A, b \\in B$ are different; otherwise, we would have $a_{1}+b_{1}=a_{2}+b_{2}$, and so $\\left|a_{1}-a_{2}\\right|=\\left|b_{2}-b_{1}\\right|$. However, then the left hand side is in $J$ whereas the right hand side is in $H$. Thus, $\\{a+b: a \\in A, b \\in B\\}$ is a set of size $|A| \\cdot|B|$ all of whose elements are no greater than $2 n$, yielding the claimed inequality.\n\nWith this in mind, it suffices to construct a set $B$, all of whose differences are in $H$ and whose size is at least $C^{\\prime} \\sqrt{n}$ for some constant $C^{\\prime}>0$.\n\nTo do so, we will use well-known facts about the negative Pell equation $X^{2}-2 Y^{2}=-1$; in particular, that there are infinitely many solutions and the values of $X$ are given by the recurrence $X_{1}=1, X_{2}=7$ and $X_{m}=6 X_{m-1}-X_{m-2}$. Therefore, we may choose $X$ to be a solution with $\\sqrt{n} / 6\\left(\\frac{X}{\\sqrt{2}}-Y\\right) \\geqslant \\frac{-3}{\\sqrt{2 n}}\n$$\n\nfrom which it follows that $\\{X / \\sqrt{2}\\}>1-(3 / \\sqrt{2 n})$. Combined with (1), this shows that all differences between elements of $B$ are in $H$.' 'we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|\\sqrt{Y}$ elements; if that subsequence is decreasing, we may reflect (using (4) or (5)) to ensure that it is increasing. Call the subsequence $\\left\\{y_{1} \\sqrt{2}\\right\\},\\left\\{y_{2} \\sqrt{2}\\right\\}, \\ldots,\\left\\{y_{t} \\sqrt{2}\\right\\}$ for $t>\\sqrt{Y}$.\n\nNow, set $B=\\left\\{\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor: 1 \\leqslant i \\leqslant t\\right\\}$. We have $\\left\\lfloor y_{j} \\sqrt{2}\\right\\rfloor-\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor=\\left\\lfloor\\left(y_{j}-y_{i}\\right) \\sqrt{2}\\right\\rfloor$ for $i\\sqrt{Y}>\\sqrt{n} / \\sqrt{3 \\sqrt{2}}$, this is the required set.']" [] Text-only Competition True Theorem proof Number Theory Math English 217 Prove that there is a constant $c>0$ and infinitely many positive integers $n$ with the following property: there are infinitely many positive integers that cannot be expressed as the sum of fewer than $c n \log (n)$ pairwise coprime $n^{\text {th }}$ powers. "['Suppose, for an integer $n$, that we can find another integer $N$ satisfying the following property:\n\n$$\nn\\text{ is divisible by }\\varphi\\left(p^{e}\\right)\\text{ for every prime power }p^{e} \\text{exactly dividing }N.\n\\tag{1}\n$$\n\nThis property ensures that all $n^{\\text {th }}$ powers are congruent to 0 or 1 modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\\text {th }}$ powers is congruent to $m$ or $m-1$ modulo $p^{e}$, since at most one of the $n^{\\text {th }}$ powers is divisible by $p$. Thus, if $k$ denotes the number of distinct prime factors of $N$, we find by the Chinese Remainder Theorem at most $2^{k} m$ residue classes modulo $N$ which are sums of at most $m$ pairwise coprime $n^{\\text {th }}$ powers. In particular, if $N>2^{k} m$ then there are infinitely many positive integers not expressible as a sum of at most $m$ pairwise coprime $n^{\\text {th }}$ powers.\n\nIt thus suffices to prove that there are arbitrarily large pairs $(n, N)$ of integers satisfying $(1)$ such that\n\n$$\nN>c \\cdot 2^{k} n \\log (n)\n$$\n\nfor some positive constant $c$.\n\nWe construct such pairs as follows. Fix a positive integer $t$ and choose (distinct) prime numbers $p \\mid 2^{2^{t-1}}+1$ and $q \\mid 2^{2^{t}}+1$; we set $N=p q$. It is well-known that $2^{t} \\mid p-1$ and $2^{t+1} \\mid q-1$, hence\n\n$$\nn=\\frac{(p-1)(q-1)}{2^{t}}\n$$\n\nis an integer and the pair $(n, N)$ satisfies $(1)$.\n\nEstimating the size of $N$ and $n$ is now straightforward. We have\n\n$$\n\\log _{2}(n) \\leqslant 2^{t-1}+2^{t}-t<2^{t+1}<2 \\cdot \\frac{N}{n}\n$$\n\nwhich rearranges to\n\n$$\nN>\\frac{1}{8} \\cdot 2^{2} n \\log _{2}(n)\n$$\n\nand so we are done if we choose $c<\\frac{1}{8 \\log (2)} \\approx 0.18$.' 'we seek arbitrarily large pairs of integers $n$ and $N$ satisfying ( $\\dagger$ ) such that $N>c 2^{k} n \\log (n)$.\n\nThis time, to construct such pairs, we fix an integer $x \\geqslant 4$, set $N$ to be the lowest common multiple of $1,2, \\ldots, 2 x$, and set $n$ to be twice the lowest common multiple of $1,2, \\ldots, x$. The pair $(n, N)$ does indeed satisfy the condition, since if $p^{e}$ is a prime power divisor of $N$ then $\\frac{\\varphi\\left(p^{e}\\right)}{2} \\leqslant x$ is a factor of $\\frac{n}{2}=\\operatorname{lcm}_{r \\leqslant x}(r)$.\n\nNow $2 N / n$ is the product of all primes having a power lying in the interval $(x, 2 x]$, and hence $2 N / n>x^{\\pi(2 x)-\\pi(x)}$. Thus for sufficiently large $x$ we have\n\n$$\n\\log \\left(\\frac{2 N}{2^{\\pi(2 x)} n}\\right)>(\\pi(2 x)-\\pi(x)) \\log (x)-\\log (2) \\pi(2 x) \\sim x\n$$\n\nusing the Prime Number Theorem $\\pi(t) \\sim t / \\log (t)$.\n\nOn the other hand, $n$ is a product of at most $\\pi(x)$ prime powers less than or equal to $x$, and so we have the upper bound\n\n$$\n\\log (n) \\leqslant \\pi(x) \\log (x) \\sim x\n$$\n\nagain by the Prime Number Theorem. Combined with the above inequality, we find that for any $\\epsilon>0$, the inequality\n\n$$\n\\log \\left(\\frac{N}{2^{\\pi(2 x)} n}\\right)>(1-\\epsilon) \\log (n)\n$$\n\nholds for sufficiently large $x$. Rearranging this shows that\n\n$$\nN>2^{\\pi(2 x)} n^{2-\\epsilon}>2^{\\pi(2 x)} n \\log (n)\n$$\n\nfor all sufficiently large $x$ and we are done.']" [] Text-only Competition True Theorem proof Number Theory Math English 218 "Let $a$ and $b$ be two positive integers. Prove that the integer $$ a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil $$ is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to $z$.)" "['Arguing indirectly, assume that\n\n$$\na^{2}+\\left\\lceil\\frac{4 a^{2}}{b}\\right\\rceil=(a+k)^{2}, \\quad \\text { or } \\quad\\left\\lceil\\frac{(2 a)^{2}}{b}\\right\\rceil=(2 a+k) k\n$$\n\nClearly, $k \\geqslant 1$. In other words, the equation\n\n$$\n\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k\n\\tag{1}\n$$\n\nhas a positive integer solution $(c, k)$, with an even value of $c$.\n\nChoose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From\n\n$$\n\\frac{c^{2}}{b}>\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil-1=c k+k^{2}-1 \\geqslant c k\n$$\n\nand\n\n$$\n\\frac{(c-k)(c+k)}{b}<\\frac{c^{2}}{b} \\leqslant\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k\n$$\n\nit can be seen that $c>b k>c-k$, so\n\n$$\nc=k b+r \\quad \\text { with some } 00$ and $0a$, so\n\n$$\n\\begin{aligned}\n& c^{2}-1(n-1)+2 .\n$$\n\nThat completes the solution.'] [] Text-only Competition True Theorem proof Algebra Math English 220 "Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014) \end{aligned} $$ are permutations of each other. Prove that there exist distinct block-similar polynomials of degree $n+1$." ['For convenience, we set $k=2015=2 \\ell+1$.\n\nConsider the following polynomials of degree $n+1$ :\n\n$$\nP(x)=\\prod_{i=0}^{n}(x-i k) \\quad \\text { and } \\quad Q(x)=\\prod_{i=0}^{n}(x-i k-1)\n$$\n\nSince $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\\cdots=P(n k)$, these polynomials are block-similar (and distinct).'] [] Text-only Competition True Theorem proof Algebra Math English 221 "Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014) \end{aligned} $$ are permutations of each other. Prove that there do not exist distinct block-similar polynomials of degree $n$." "['For convenience, we set $k=2015=2 \\ell+1$.\n\nFor every polynomial $F(x)$ and every nonnegative integer $m$, define $\\Sigma_{F}(m)=$ $\\sum_{i=1}^{m} F(i)$; in particular, $\\Sigma_{F}(0)=0$. It is well-known that for every nonnegative integer $d$ the $\\operatorname{sum} \\sum_{i=1}^{m} i^{d}$ is a polynomial in $m$ of degree $d+1$. Thus $\\Sigma_{F}$ may also be regarded as a real polynomial of degree $\\operatorname{deg} F+1$ (with the exception that if $F=0$, then $\\Sigma_{F}=0$ as well). This allows us to consider the values of $\\Sigma_{F}$ at all real points (where the initial definition does not apply).\n\nAssume for the sake of contradiction that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Then both polynomials $\\Sigma_{P-Q}(x)$ and $\\Sigma_{P^{2}-Q^{2}}(x)$ have roots at the points $0, k, 2 k, \\ldots, n k$. This motivates the following lemma, where we use the special polynomial\n\n$$\nT(x)=\\prod_{i=0}^{n}(x-i k)\n$$\n\nLemma. Assume that $F(x)$ is a nonzero polynomial such that $0, k, 2 k, \\ldots, n k$ are among the roots of the polynomial $\\Sigma_{F}(x)$. Then $\\operatorname{deg} F \\geqslant n$, and there exists a polynomial $G(x)$ such that $\\operatorname{deg} G=\\operatorname{deg} F-n$ and $F(x)=T(x) G(x)-T(x-1) G(x-1)$.\n\nProof. If $\\operatorname{deg} Fp^{+}$, then $S\\left(p^{-}\\right)=P\\left(p^{-}\\right)+Q\\left(p^{-}\\right) \\leqslant Q\\left(p^{+}\\right)+P\\left(p^{+}\\right)=S\\left(p^{+}\\right)$, so our claim holds.\n\nWe now show that the remaining case $p^{-}$ $Q\\left(p^{+}\\right)$. Then, like in Step 1 , we have $R\\left(p^{-}\\right) \\leqslant 0, R\\left(p^{+}\\right)>0$, and $R(i k) \\leqslant 0$, so $R(x)$ has a root in each of the intervals $\\left[p^{-}, p^{+}\\right)$and $\\left(p^{+}, i k\\right]$. This contradicts the result of Step 1.\n\nWe are left only with the case $p^{-}\\ell_{j}>r_{i}$.\n\nClearly, there is no town which can sweep $T_{n}$ away from the right. Then we may choose the leftmost town $T_{k}$ which cannot be swept away from the right. One can observe now that no town $T_{i}$ with $i>k$ may sweep away some town $T_{j}$ with $jm$. As we have already observed, $p$ cannot be greater than $k$. On the other hand, $T_{m}$ cannot sweep $T_{p}$ away, so a fortiori it cannot sweep $T_{k}$ away.\n\nClaim 2. Any town $T_{m}$ with $m \\neq k$ can be swept away by some other town.\n\n\n\nProof. If $mk$.\n\nLet $T_{p}$ be a town among $T_{k}, T_{k+1}, \\ldots, T_{m-1}$ having the largest right bulldozer. We claim that $T_{p}$ can sweep $T_{m}$ away. If this is not the case, then $r_{p}<\\ell_{q}$ for some $q$ with $pm \geqslant N$." "['We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \\neq n+a_{n}$ for $m \\neq n$, each positive integer receives at most one arrow. There are some positive integers, such as 1 , that receive no arrows; these will be referred to as starting points in the sequel. When one starts at any of the starting points and keeps following the arrows, one is led to an infinite path, called its ray, that visits a strictly increasing sequence of positive integers. Since the length of any arrow is at most 2015, such a ray, say with starting point $s$, meets every interval of the form $[n, n+2014]$ with $n \\geqslant s$ at least once.\n\nSuppose for the sake of contradiction that there would be at least 2016 starting points. Then we could take an integer $n$ that is larger than the first 2016 starting points. But now the interval $[n, n+2014]$ must be met by at least 2016 rays in distinct points, which is absurd. We have thereby shown that the number $b$ of starting points satisfies $1 \\leqslant b \\leqslant 2015$. Let $N$ denote any integer that is larger than all starting points. We contend that $b$ and $N$ are as required.\n\nTo see this, let any two integers $m$ and $n$ with $n>m \\geqslant N$ be given. The sum $\\sum_{i=m+1}^{n} a_{i}$ gives the total length of the arrows emerging from $m+1, \\ldots, n$. Taken together, these arrows form $b$ subpaths of our rays, some of which may be empty. Now on each ray we look at the first number that is larger than $m$; let $x_{1}, \\ldots, x_{b}$ denote these numbers, and let $y_{1}, \\ldots, y_{b}$ enumerate in corresponding order the numbers defined similarly with respect to $n$. Then the list of differences $y_{1}-x_{1}, \\ldots, y_{b}-x_{b}$ consists of the lengths of these paths and possibly some zeros corresponding to empty paths. Consequently, we obtain\n\n$$\n\\sum_{i=m+1}^{n} a_{i}=\\sum_{j=1}^{b}\\left(y_{j}-x_{j}\\right)\n$$\n\nwhence\n\n$$\n\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)=\\sum_{j=1}^{b}\\left(y_{j}-n\\right)-\\sum_{j=1}^{b}\\left(x_{j}-m\\right)\n$$\n\nNow each of the $b$ rays meets the interval $[m+1, m+2015]$ at some point and thus $x_{1}-$ $m, \\ldots, x_{b}-m$ are $b$ distinct members of the set $\\{1,2, \\ldots, 2015\\}$. Moreover, since $m+1$ is not a starting point, it must belong to some ray; so 1 has to appear among these numbers, wherefore\n\n$$\n1+\\sum_{j=1}^{b-1}(j+1) \\leqslant \\sum_{j=1}^{b}\\left(x_{j}-m\\right) \\leqslant 1+\\sum_{j=1}^{b-1}(2016-b+j)\n$$\n\nThe same argument applied to $n$ and $y_{1}, \\ldots, y_{b}$ yields\n\n$$\n1+\\sum_{j=1}^{b-1}(j+1) \\leqslant \\sum_{j=1}^{b}\\left(y_{j}-n\\right) \\leqslant 1+\\sum_{j=1}^{b-1}(2016-b+j)\n$$\n\n\n\nSo altogether we get\n\n$$\n\\begin{aligned}\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| & \\leqslant \\sum_{j=1}^{b-1}((2016-b+j)-(j+1))=(b-1)(2015-b) \\\\\n& \\leqslant\\left(\\frac{(b-1)+(2015-b)}{2}\\right)^{2}=1007^{2},\n\\end{aligned}\n$$\n\nas desired.' 'Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have\n\n$$\nn+1 \\leqslant s_{n} \\leqslant n+2015\n$$\n\nfor all $n \\in \\mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \\ldots$ are distinct. We shall investigate the set\n\n$$\nM=\\mathbb{Z}_{>0} \\backslash\\left\\{s_{1}, s_{2}, \\ldots\\right\\}\n$$\n\nClaim. At most 2015 numbers belong to $M$.\n\nProof. Otherwise let $m_{1}m \\geqslant N$. Plugging $r=n$ and $r=m$ into (3) and subtracting the estimates that result, we deduce\n\n$$\n\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)=\\sum C_{n}-\\sum C_{m}\n$$\n\n\n\nSince $C_{n}$ and $C_{m}$ are subsets of $\\{1,2, \\ldots, 2014\\}$ with $\\left|C_{n}\\right|=\\left|C_{m}\\right|=b-1$, it is clear that the absolute value of the right-hand side of the above inequality attains its largest possible value if either $C_{m}=\\{1,2, \\ldots, b-1\\}$ and $C_{n}=\\{2016-b, \\ldots, 2014\\}$, or the other way around. In these two cases we have\n\n$$\n\\left|\\sum C_{n}-\\sum C_{m}\\right|=(b-1)(2015-b)\n$$\n\nso in the general case we find\n\n$$\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| \\leqslant(b-1)(2015-b) \\leqslant\\left(\\frac{(b-1)+(2015-b)}{2}\\right)^{2}=1007^{2}\n$$\n\nas desired.']" [] Text-only Competition True Theorem proof Combinatorics Math English 224 Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean. ['Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\\mathbb{Z}_{>0}$ denote the set of all positive integers.\n\nSuppose, to the contrary, that there exist only finitely many positive integers that are not clean. Therefore, there exists a positive integer $N$ such that every integer $n>N$ has exactly one odd representation.\n\nClearly, $S$ is infinite. We now claim the following properties of odd and even representations.\n\nProperty 1. Any positive integer $n$ has at most one odd and at most one even representation. Proof. We first show that every integer $n$ has at most one even representation. Since $S$ is infinite, there exists $x \\in S$ such that $x>\\max \\{n, N\\}$. Then, the number $n+x$ must be clean, and $x$ does not appear in any even representation of $n$. If $n$ has more than one even representation, then we obtain two distinct odd representations of $n+x$ by adding $x$ to the even representations of $n$, which is impossible. Therefore, $n$ can have at most one even representation.\n\nSimilarly, there exist two distinct elements $y, z \\in S$ such that $y, z>\\max \\{n, N\\}$. If $n$ has more than one odd representation, then we obtain two distinct odd representations of $n+y+z$ by adding $y$ and $z$ to the odd representations of $n$. This is again a contradiction.\n\nProperty 2. Fix $s \\in S$. Suppose that a number $n>N$ has no even representation. Then $n+2$ as has an even representation containing $s$ for all integers $a \\geqslant 1$.\n\nProof. It is sufficient to prove the following statement: If $n$ has no even representation without $s$, then $n+2 s$ has an even representation containing $s$ (and hence no even representation without $s$ by Property 1).\n\nNotice that the odd representation of $n+s$ does not contain $s$; otherwise, we have an even representation of $n$ without $s$. Then, adding $s$ to this odd representation of $n+s$, we get that $n+2 s$ has an even representation containing $s$, as desired.\n\nProperty 3. Every sufficiently large integer has an even representation.\n\nProof. Fix any $s \\in S$, and let $r$ be an arbitrary element in $\\{1,2, \\ldots, 2 s\\}$. Then, Property 2 implies that the set $Z_{r}=\\{r+2 a s: a \\geqslant 0\\}$ contains at most one number exceeding $N$ with no even representation. Therefore, $Z_{r}$ contains finitely many positive integers with no even representation, and so does $\\mathbb{Z}_{>0}=\\bigcup_{r=1}^{2 s} Z_{r}$.\n\nIn view of Properties 1 and 3, we may assume that $N$ is chosen such that every $n>N$ has exactly one odd and exactly one even representation. In particular, each element $s>N$ of $S$ has an even representation.\n\nProperty 4. For any $s, t \\in S$ with $NN$. Then, Property 4 implies that for every $i>k$ the even representation of $s_{i}$ contains all the numbers $s_{k}, s_{k+1}, \\ldots, s_{i-1}$. Therefore,\n\n$$\ns_{i}=s_{k}+s_{k+1}+\\cdots+s_{i-1}+R_{i}=\\sigma_{i-1}-\\sigma_{k-1}+R_{i},\\tag{1}\n$$\n\nwhere $R_{i}$ is a sum of some of $s_{1}, \\ldots, s_{k-1}$. In particular, $0 \\leqslant R_{i} \\leqslant s_{1}+\\cdots+s_{k-1}=\\sigma_{k-1}$.\n\n\n\nLet $j_{0}$ be an integer satisfying $j_{0}>k$ and $\\sigma_{j_{0}}>2 \\sigma_{k-1}$. Then (1) shows that, for every $j>j_{0}$,\n\n$$\ns_{j+1} \\geqslant \\sigma_{j}-\\sigma_{k-1}>\\sigma_{j} / 2\\tag{2}\n$$\n\nNext, let $p>j_{0}$ be an index such that $R_{p}=\\min _{i>j_{0}} R_{i}$. Then,\n\n$$\ns_{p+1}=s_{k}+s_{k+1}+\\cdots+s_{p}+R_{p+1}=\\left(s_{p}-R_{p}\\right)+s_{p}+R_{p+1} \\geqslant 2 s_{p}\n$$\n\nTherefore, there is no element of $S$ larger than $s_{p}$ but smaller than $2 s_{p}$. It follows that the even representation $\\tau$ of $2 s_{p}$ does not contain any element larger than $s_{p}$. On the other hand, inequality (2) yields $2 s_{p}>s_{1}+\\cdots+s_{p-1}$, so $\\tau$ must contain a term larger than $s_{p-1}$. Thus, it must contain $s_{p}$. After removing $s_{p}$ from $\\tau$, we have that $s_{p}$ has an odd representation not containing $s_{p}$, which contradicts Property 1 since $s_{p}$ itself also forms an odd representation of $s_{p}$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 225 In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3 , and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that it is possible to partition the company into 11 parts so that no two enemies are in the same part. "['Let $G=(V, E)$ be a graph where $V$ is the set of people in the company and $E$ is the set of the enemy pairs - the edges of the graph. In this language, partitioning into 11 disjoint enemy-free subsets means properly coloring the vertices of this graph with 11 colors.\n\nWe will prove the following more general statement.\n\nClaim. Let $G$ be a graph with chromatic number $k \\geqslant 3$. Then $G$ contains at least $2^{k-1}-k$ unsociable groups.\n\nRecall that the chromatic number of $G$ is the least $k$ such that a proper coloring\n\n$$\nV=V_{1} \\sqcup \\cdots \\sqcup V_{k}\n\\tag{1}\n$$\n\nexists. In view of $2^{11}-12>2015$, the claim implies the problem statement.\n\nLet $G$ be a graph with chromatic number $k$. We say that a proper coloring (1) of $G$ is leximinimal, if the $k$-tuple $\\left(\\left|V_{1}\\right|,\\left|V_{2}\\right|, \\ldots,\\left|V_{k}\\right|\\right)$ is lexicographically minimal; in other words, the following conditions are satisfied: the number $n_{1}=\\left|V_{1}\\right|$ is minimal; the number $n_{2}=\\left|V_{2}\\right|$ is minimal, subject to the previously chosen value of $n_{1} ; \\ldots$; the number $n_{k-1}=\\left|V_{k-1}\\right|$ is minimal, subject to the previously chosen values of $n_{1}, \\ldots, n_{k-2}$.\n\nThe following lemma is the core of the proof.\n\nLemma 1. Suppose that $G=(V, E)$ is a graph with odd chromatic number $k \\geqslant 3$, and let (1) be one of its leximinimal colorings. Then $G$ contains an odd cycle which visits all color classes $V_{1}, V_{2}, \\ldots, V_{k}$.\n\nProof of Lemma 1. Let us call a cycle colorful if it visits all color classes.\n\nDue to the definition of the chromatic number, $V_{1}$ is nonempty. Choose an arbitrary vertex $v \\in V_{1}$. We construct a colorful odd cycle that has only one vertex in $V_{1}$, and this vertex is $v$.\n\nWe draw a subgraph of $G$ as follows. Place $v$ in the center, and arrange the sets $V_{2}, V_{3}, \\ldots, V_{k}$ in counterclockwise circular order around it. For convenience, let $V_{k+1}=V_{2}$. We will draw arrows to add direction to some edges of $G$, and mark the vertices these arrows point to. First we draw arrows from $v$ to all its neighbors in $V_{2}$, and mark all those neighbors. If some vertex $u \\in V_{i}$ with $i \\in\\{2,3, \\ldots, k\\}$ is already marked, we draw arrows from $u$ to all its neighbors in $V_{i+1}$ which are not marked yet, and we mark all of them. We proceed doing this as long as it is possible. The process of marking is exemplified in Figure 1.\n\nNotice that by the rules of our process, in the final state, marked vertices in $V_{i}$ cannot have unmarked neighbors in $V_{i+1}$. Moreover, $v$ is connected to all marked vertices by directed paths.\n\nNow move each marked vertex to the next color class in circular order (see an example in Figure 3). In view of the arguments above, the obtained coloring $V_{1} \\sqcup W_{2} \\sqcup \\cdots \\sqcup W_{k}$ is proper. Notice that $v$ has a neighbor $w \\in W_{2}$, because otherwise\n\n$$\n\\left(V_{1} \\backslash\\{v\\}\\right) \\sqcup\\left(W_{2} \\cup\\{v\\}\\right) \\sqcup W_{3} \\sqcup \\cdots \\sqcup W_{k}\n$$\n\nwould be a proper coloring lexicographically smaller than (1). If $w$ was unmarked, i.e., $w$ was an element of $V_{2}$, then it would be marked at the beginning of the process and thus moved to $V_{3}$, which did not happen. Therefore, $w$ is marked and $w \\in V_{k}$.\n\n\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSince $w$ is marked, there exists a directed path from $v$ to $w$. This path moves through the sets $V_{2}, \\ldots, V_{k}$ in circular order, so the number of edges in it is divisible by $k-1$ and thus even. Closing this path by the edge $w \\rightarrow v$, we get a colorful odd cycle, as required.\n\nProof of the claim. Let us choose a leximinimal coloring (1) of $G$. For every set $C \\subseteq\\{1,2, \\ldots, k\\}$ such that $|C|$ is odd and greater than 1 , we will provide an odd cycle visiting exactly those color classes whose indices are listed in the set $C$. This property ensures that we have different cycles for different choices of $C$, and it proves the claim because there are $2^{k-1}-k$ choices for the set $C$.\n\nLet $V_{C}=\\bigcup_{c \\in C} V_{c}$, and let $G_{C}$ be the induced subgraph of $G$ on the vertex set $V_{C}$. We also have the induced coloring of $V_{C}$ with $|C|$ colors; this coloring is of course proper. Notice further that the induced coloring is leximinimal: if we had a lexicographically smaller coloring $\\left(W_{c}\\right)_{c \\in C}$ of $G_{C}$, then these classes, together the original color classes $V_{i}$ for $i \\notin C$, would provide a proper coloring which is lexicographically smaller than (1). Hence Lemma 1, applied to the subgraph $G_{C}$ and its leximinimal coloring $\\left(V_{c}\\right)_{c \\in C}$, provides an odd cycle that visits exactly those color classes that are listed in the set $C$.' ""We say that a graph is critical if deleting any vertex from the graph decreases the graph's chromatic number. Obviously every graph contains a critical induced subgraph with the same chromatic number.\n\nLemma 2. Suppose that $G=(V, E)$ is a critical graph with chromatic number $k \\geqslant 3$. Then every vertex $v$ of $G$ is contained in at least $2^{k-2}-1$ unsociable groups.\n\nProof. For every set $X \\subseteq V$, denote by $n(X)$ the number of neighbors of $v$ in the set $X$.\n\nSince $G$ is critical, there exists a proper coloring of $G \\backslash\\{v\\}$ with $k-1$ colors, so there exists a proper coloring $V=V_{1} \\sqcup V_{2} \\sqcup \\cdots \\sqcup V_{k}$ of $G$ such that $V_{1}=\\{v\\}$. Among such colorings, take one for which the sequence $\\left(n\\left(V_{2}\\right), n\\left(V_{3}\\right), \\ldots, n\\left(V_{k}\\right)\\right)$ is lexicographically minimal. Clearly, $n\\left(V_{i}\\right)>0$ for every $i=2,3, \\ldots, k$; otherwise $V_{2} \\sqcup \\ldots \\sqcup V_{i-1} \\sqcup\\left(V_{i} \\cup V_{1}\\right) \\sqcup V_{i+1} \\sqcup \\ldots V_{k}$ would be a proper coloring of $G$ with $k-1$ colors.\n\nWe claim that for every $C \\subseteq\\{2,3, \\ldots, k\\}$ with $|C| \\geqslant 2$ being even, $G$ contains an unsociable group so that the set of its members' colors is precisely $C \\cup\\{1\\}$. Since the number of such sets $C$ is $2^{k-2}-1$, this proves the lemma. Denote the elements of $C$ by $c_{1}, \\ldots, c_{2 \\ell}$ in increasing order. For brevity, let $U_{i}=V_{c_{i}}$. Denote by $N_{i}$ the set of neighbors of $v$ in $U_{i}$.\n\n\n\nWe show that for every $i=1, \\ldots, 2 \\ell-1$ and $x \\in N_{i}$, the subgraph induced by $U_{i} \\cup U_{i+1}$ contains a path that connects $x$ with another point in $N_{i+1}$. For the sake of contradiction, suppose that no such path exists. Let $S$ be the set of vertices that lie in the connected component of $x$ in the subgraph induced by $U_{i} \\cup U_{i+1}$, and let $P=U_{i} \\cap S$, and $Q=U_{i+1} \\cap S$ (see Figure 3). Since $x$ is separated from $N_{i+1}$, the sets $Q$ and $N_{i+1}$ are disjoint. So, if we re-color $G$ by replacing $U_{i}$ and $U_{i+1}$ by $\\left(U_{i} \\cup Q\\right) \\backslash P$ and $\\left(U_{i+1} \\cup P\\right) \\backslash Q$, respectively, we obtain a proper coloring such that $n\\left(U_{i}\\right)=n\\left(V_{c_{i}}\\right)$ is decreased and only $n\\left(U_{i+1}\\right)=n\\left(V_{c_{i+1}}\\right)$ is increased. That contradicts the lexicographical minimality of $\\left(n\\left(V_{2}\\right), n\\left(V_{3}\\right), \\ldots, n\\left(V_{k}\\right)\\right)$.\n\n\n\nFigure 3\n\nNext, we build a path through $U_{1}, U_{2}, \\ldots, U_{2 \\ell}$ as follows. Let the starting point of the path be an arbitrary vertex $v_{1}$ in the set $N_{1}$. For $i \\leqslant 2 \\ell-1$, if the vertex $v_{i} \\in N_{i}$ is already defined, connect $v_{i}$ to some vertex in $N_{i+1}$ in the subgraph induced by $U_{i} \\cup U_{i+1}$, and add these edges to the path. Denote the new endpoint of the path by $v_{i+1}$; by the construction we have $v_{i+1} \\in N_{i+1}$ again, so the process can be continued. At the end we have a path that starts at $v_{1} \\in N_{1}$ and ends at some $v_{2 \\ell} \\in N_{2 \\ell}$. Moreover, all edges in this path connect vertices in neighboring classes: if a vertex of the path lies in $U_{i}$, then the next vertex lies in $U_{i+1}$ or $U_{i-1}$. Notice that the path is not necessary simple, so take a minimal subpath of it. The minimal subpath is simple and connects the same endpoints $v_{1}$ and $v_{2 \\ell}$. The property that every edge steps to a neighboring color class (i.e., from $U_{i}$ to $U_{i+1}$ or $U_{i-1}$ ) is preserved. So the resulting path also visits all of $U_{1}, \\ldots, U_{2 \\ell}$, and its length must be odd. Closing the path with the edges $v v_{1}$ and $v_{2 \\ell} v$ we obtain the desired odd cycle (see Figure 4).\n\n\n\nFigure 4\n\nNow we prove the claim by induction on $k \\geqslant 3$. The base case $k=3$ holds by applying Lemma 2 to a critical subgraph. For the induction step, let $G_{0}$ be a critical $k$-chromatic subgraph of $G$, and let $v$ be an arbitrary vertex of $G_{0}$. By Lemma $2, G_{0}$ has at least $2^{k-2}-1$ unsociable groups containing $v$. On the other hand, the graph $G_{0} \\backslash\\{v\\}$ has chromatic number $k-1$, so it contains at least $2^{k-2}-(k-1)$ unsociable groups by the induction hypothesis. Altogether, this gives $2^{k-2}-1+2^{k-2}-(k-1)=2^{k-1}-k$ distinct unsociable groups in $G_{0}$ (and thus in $G)$.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 226 Let $A B C$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $A B G H$ is a parallelogram. Let $I$ be the point on the line $G H$ such that $A C$ bisects $H I$. Suppose that the line $A C$ intersects the circumcircle of the triangle $G C I$ at $C$ and $J$. Prove that $I J=A H$. "['Since $H G \\| A B$ and $B G \\| A H$, we have $B G \\perp B C$ and $C H \\perp G H$. Therefore, the quadrilateral $B G C H$ is cyclic. Since $H$ is the orthocenter of the triangle $A B C$, we have $\\angle H A C=90^{\\circ}-\\angle A C B=\\angle C B H$. Using that $B G C H$ and $C G J I$ are cyclic quadrilaterals, we get\n\n$$\n\\angle C J I=\\angle C G H=\\angle C B H=\\angle H A C .\n$$\n\nLet $M$ be the intersection of $A C$ and $G H$, and let $D \\neq A$ be the point on the line $A C$ such that $A H=H D$. Then $\\angle M J I=\\angle H A C=\\angle M D H$.\n\nSince $\\angle M J I=\\angle M D H, \\angle I M J=\\angle H M D$, and $I M=M H$, the triangles $I M J$ and $H M D$ are congruent, and thus $I J=H D=A H$.\n\n' 'Obtain $\\angle C G H=\\angle H A C$ as in the previous solution. In the parallelogram $A B G H$ we have $\\angle B A H=\\angle H G B$. It follows that\n\n$$\n\\angle H M C=\\angle B A C=\\angle B A H+\\angle H A C=\\angle H G B+\\angle C G H=\\angle C G B .\n$$\n\nSo the right triangles $C M H$ and $C G B$ are similar. Also, in the circumcircle of triangle $G C I$ we have similar triangles $M I J$ and $M C G$. Therefore,\n\n$$\n\\frac{I J}{C G}=\\frac{M I}{M C}=\\frac{M H}{M C}=\\frac{G B}{G C}=\\frac{A H}{C G}\n$$\n\nHence $I J=A H$.']" [] Text-only Competition True Theorem proof Geometry Math English 227 Let $A B C$ be a triangle inscribed into a circle $\Omega$ with center $O$. A circle $\Gamma$ with center $A$ meets the side $B C$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\Gamma$ and $\Omega$. We assume that $F$ lies on the arc $A B$ of $\Omega$ not containing $C$, and $G$ lies on the arc $A C$ of $\Omega$ not containing $B$. The circumcircles of the triangles $B D F$ and $C E G$ meet the sides $A B$ and $A C$ again at $K$ and $L$, respectively. Suppose that the lines $F K$ and $G L$ are distinct and intersect at $X$. Prove that the points $A, X$, and $O$ are collinear. "['It suffices to prove that the lines $F K$ and $G L$ are symmetric about $A O$. Now the segments $A F$ and $A G$, being chords of $\\Omega$ with the same length, are clearly symmetric with respect to $A O$. Hence it is enough to show\n\n$$\n\\angle K F A=\\angle A G L \\text {. }\n\\tag{1}\n$$\n\nLet us denote the circumcircles of $B D F$ and $C E G$ by $\\omega_{B}$ and $\\omega_{C}$, respectively. To prove (1), we start from\n\n$$\n\\angle K F A=\\angle D F G+\\angle G F A-\\angle D F K .\n$$\n\nIn view of the circles $\\omega_{B}, \\Gamma$, and $\\Omega$, this may be rewritten as\n\n$$\n\\angle K F A=\\angle C E G+\\angle G B A-\\angle D B K=\\angle C E G-\\angle C B G .\n$$\n\nDue to the circles $\\omega_{C}$ and $\\Omega$, we obtain $\\angle K F A=\\angle C L G-\\angle C A G=\\angle A G L$. Thereby the problem is solved.\n\n\n\nFigure 1' 'Again, we denote the circumcircle of $B D K F$ by $\\omega_{B}$. In addition, we set $\\alpha=$ $\\angle B A C, \\varphi=\\angle A B F$, and $\\psi=\\angle E D A=\\angle A E D$ (see Figure 2). Notice that $A F=A G$ entails $\\varphi=\\angle G C A$, so all three of $\\alpha, \\varphi$, and $\\psi$ respect the ""symmetry"" between $B$ and $C$ of our configuration. Again, we reduce our task to proving (1).\n\nThis time, we start from\n\n$$\n2 \\angle K F A=2(\\angle D F A-\\angle D F K) .\n$$\n\nSince the triangle $A F D$ is isosceles, we have\n\n$$\n\\angle D F A=\\angle A D F=\\angle E D F-\\psi=\\angle B F D+\\angle E B F-\\psi .\n$$\n\n\n\nMoreover, because of the circle $\\omega_{B}$ we have $\\angle D F K=\\angle C B A$. Altogether, this yields\n\n$$\n2 \\angle K F A=\\angle D F A+(\\angle B F D+\\angle E B F-\\psi)-2 \\angle C B A,\n$$\n\nwhich simplifies to\n\n$$\n2 \\angle K F A=\\angle B F A+\\varphi-\\psi-\\angle C B A .\n$$\n\nNow the quadrilateral $A F B C$ is cyclic, so this entails $2 \\angle K F A=\\alpha+\\varphi-\\psi$.\n\nDue to the ""symmetry"" between $B$ and $C$ alluded to above, this argument also shows that $2 \\angle A G L=\\alpha+\\varphi-\\psi$. This concludes the proof of (1).\n\n\n\nFigure 2']" [] Text-only Competition True Theorem proof Geometry Math English 228 Let $A B C$ be a triangle with $\angle C=90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $C B H$ so that $C H$ bisects $A D$. Let $P$ be the intersection point of the lines $B D$ and $C H$. Let $\omega$ be the semicircle with diameter $B D$ that meets the segment $C B$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $C Q$ and $A D$ meet on $\omega$. "['Let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 1). Since $P H \\| D K$, we have\n\n$$\n\\frac{P D}{P B}=\\frac{H K}{H B}=\\frac{A H}{H B}\n\\tag{1}\n$$\n\nLet $L$ be the projection of $Q$ onto $D B$. Since $P Q$ is tangent to $\\omega$ and $\\angle D Q B=\\angle B L Q=$ $90^{\\circ}$, we have $\\angle P Q D=\\angle Q B P=\\angle D Q L$. Therefore, $Q D$ and $Q B$ are respectively the internal and the external bisectors of $\\angle P Q L$. By the angle bisector theorem, we obtain\n\n$$\n\\frac{P D}{D L}=\\frac{P Q}{Q L}=\\frac{P B}{B L}\n\\tag{2}\n$$\n\nThe relations (1) and (2) yield $\\frac{A H}{H B}=\\frac{P D}{P B}=\\frac{D L}{L B}$. So, the spiral similarity $\\tau$ centered at $B$ and sending $A$ to $D$ maps $H$ to $L$. Moreover, $\\tau$ sends the semicircle with diameter $A B$ passing through $C$ to $\\omega$. Due to $C H \\perp A B$ and $Q L \\perp D B$, it follows that $\\tau(C)=Q$.\n\nHence, the triangles $A B D$ and $C B Q$ are similar, so $\\angle A D B=\\angle C Q B$. This means that the lines $A D$ and $C Q$ meet at some point $T$, and this point satisfies $\\angle B D T=\\angle B Q T$. Therefore, $T$ lies on $\\omega$, as needed.\n\n\n\nFigure 1\n\n\n\nFigure 2' 'Let $\\Gamma$ be the circumcircle of $A B C$, and let $A D$ meet $\\omega$ at $T$. Then $\\angle A T B=$ $\\angle A C B=90^{\\circ}$, so $T$ lies on $\\Gamma$ as well. As in the previous solution, let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 2).\n\nOur goal now is to prove that the points $C, Q$, and $T$ are collinear. Let $C T$ meet $\\omega$ again at $Q^{\\prime}$. Then, it suffices to show that $P Q^{\\prime}$ is tangent to $\\omega$, or that $\\angle P Q^{\\prime} D=\\angle Q^{\\prime} B D$.\n\nSince the quadrilateral $B D Q^{\\prime} T$ is cyclic and the triangles $A H C$ and $K H C$ are congruent, we have $\\angle Q^{\\prime} B D=\\angle Q^{\\prime} T D=\\angle C T A=\\angle C B A=\\angle A C H=\\angle H C K$. Hence, the right triangles $C H K$ and $B Q^{\\prime} D$ are similar. This implies that $\\frac{H K}{C K}=\\frac{Q^{\\prime} D}{B D}$, and thus $H K \\cdot B D=C K \\cdot Q^{\\prime} D$.\n\nNotice that $P H \\| D K$; therefore, we have $\\frac{P D}{B D}=\\frac{H K}{B K}$, and so $P D \\cdot B K=H K \\cdot B D$. Consequently, $P D \\cdot B K=H K \\cdot B D=C K \\cdot Q^{\\prime} D$, which yields $\\frac{P D}{Q^{\\prime} D}=\\frac{C K}{B K}$.\n\nSince $\\angle C K A=\\angle K A C=\\angle B D Q^{\\prime}$, the triangles $C K B$ and $P D Q^{\\prime}$ are similar, so $\\angle P Q^{\\prime} D=$ $\\angle C B A=\\angle Q^{\\prime} B D$, as required.' ""Introduce the points $T$ and $Q^{\\prime}$ as in the previous solution. Note that $T$ lies on the circumcircle of $A B C$. Here we present yet another proof that $P Q^{\\prime}$ is tangent to $\\omega$.\n\nLet $\\Omega$ be the circle completing the semicircle $\\omega$. Construct a point $F$ symmetric to $C$ with respect to $A B$. Let $S \\neq T$ be the second intersection point of $F T$ and $\\Omega$ (see Figure 4).\n\n\n\nFigure 4\n\nSince $A C=A F$, we have $\\angle D K C=\\angle H C K=\\angle C B A=\\angle C T A=\\angle D T S=180^{\\circ}-$ $\\angle S K D$. Thus, the points $C, K$, and $S$ are collinear. Notice also that $\\angle Q^{\\prime} K D=\\angle Q^{\\prime} T D=$ $\\angle H C K=\\angle K F H=180^{\\circ}-\\angle D K F$. This implies that the points $F, K$, and $Q^{\\prime}$ are collinear.\n\nApplying PASCAL's theorem to the degenerate hexagon $K Q^{\\prime} Q^{\\prime} T S S$, we get that the tangents to $\\Omega$ passing through $Q^{\\prime}$ and $S$ intersect on $C F$. The relation $\\angle Q^{\\prime} T D=\\angle D T S$ yields that $Q^{\\prime}$ and $S$ are symmetric with respect to $B D$. Therefore, the two tangents also intersect on $B D$. Thus, the two tangents pass through $P$. Hence, $P Q^{\\prime}$ is tangent to $\\omega$, as needed.""]" [] Text-only Competition True Theorem proof Geometry Math English 229 Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$. "['We may assume that $C A>C B$. Observe that $H^{\\prime}$ and $I^{\\prime}$ lie inside the segments $C F$ and $C G$, respectively. Therefore, $M$ lies outside $\\triangle A B C$ (see Figure 1).\n\nDue to the powers of points $A$ and $B$ with respect to the circle $\\Gamma$, we have\n\n$$\nC H^{\\prime} \\cdot C A=A H \\cdot A C=A D^{2}=B D^{2}=B I \\cdot B C=C I^{\\prime} \\cdot C B .\n$$\n\nTherefore, $C H^{\\prime} \\cdot C F=C I^{\\prime} \\cdot C G$. Hence, the quadrilateral $H^{\\prime} I^{\\prime} G F$ is cyclic, and so $\\angle I^{\\prime} H^{\\prime} C=$ $\\angle C G F$.\n\nLet $D F$ and $D G$ meet $\\Gamma$ again at $R$ and $S$, respectively. We claim that the points $R$ and $S$ lie on the line $H^{\\prime} I^{\\prime}$.\n\nObserve that $F H^{\\prime} \\cdot F A=F H \\cdot F C=F R \\cdot F D$. Thus, the quadrilateral $A D H^{\\prime} R$ is cyclic, and hence $\\angle R H^{\\prime} F=\\angle F D A=\\angle C G F=\\angle I^{\\prime} H^{\\prime} C$. Therefore, the points $R, H^{\\prime}$, and $I^{\\prime}$ are collinear. Similarly, the points $S, H^{\\prime}$, and $I^{\\prime}$ are also collinear, and so all the points $R, H^{\\prime}, Q, I^{\\prime}, S$, and $M$ are all collinear.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nThen, $\\angle R S D=\\angle R D A=\\angle D F G$. Hence, the quadrilateral $R S G F$ is cyclic (see Figure 2). Therefore, $M H^{\\prime} \\cdot M I^{\\prime}=M F \\cdot M G=M R \\cdot M S=M P \\cdot M C$. Thus, the quadrilateral $C P I^{\\prime} H^{\\prime}$ is also cyclic. Let $\\omega$ be its circumcircle.\n\nNotice that $\\angle H^{\\prime} C Q=\\angle S D C=\\angle S R C$ and $\\angle Q C I^{\\prime}=\\angle C D R=\\angle C S R$. Hence, $\\triangle C H^{\\prime} Q \\sim \\triangle R C Q$ and $\\triangle C I^{\\prime} Q \\sim \\triangle S C Q$, and therefore $Q H^{\\prime} \\cdot Q R=Q C^{2}=Q I^{\\prime} \\cdot Q S$.\n\nWe apply the inversion with center $Q$ and radius $Q C$. Observe that the points $R, C$, and $S$ are mapped to $H^{\\prime}, C$, and $I^{\\prime}$, respectively. Therefore, the circumcircle $\\Gamma$ of $\\triangle R C S$ is mapped to the circumcircle $\\omega$ of $\\triangle H^{\\prime} C I^{\\prime}$. Since $P$ and $C$ belong to both circles and the point $C$ is preserved by the inversion, we have that $P$ is also mapped to itself. We then get $Q P^{2}=Q C^{2}$. Hence, $Q P=Q C$.' ""Let $X=H I \\cap A B$, and let the tangent to $\\Gamma$ at $C$ meet $A B$ at $Y$. Let $X C$ meet $\\Gamma$ again at $X^{\\prime}$ (see Figure 3). Projecting from $C, X$, and $C$ again, we have $(X, A ; D, B)=$ $\\left(X^{\\prime}, H ; D, I\\right)=(C, I ; D, H)=(Y, B ; D, A)$. Since $A$ and $B$ are symmetric about $D$, it follows that $X$ and $Y$ are also symmetric about $D$.\n\nNow, Menelaus' theorem applied to $\\triangle A B C$ with the line $H I X$ yields\n\n$$\n1=\\frac{C H}{H A} \\cdot \\frac{B I}{I C} \\cdot \\frac{A X}{X B}=\\frac{A H^{\\prime}}{H^{\\prime} C} \\cdot \\frac{C I^{\\prime}}{I^{\\prime} B} \\cdot \\frac{B Y}{Y A}\n$$\n\nBy the converse of Menelaus' theorem applied to $\\triangle A B C$ with points $H^{\\prime}, I^{\\prime}, Y$, we get that the points $H^{\\prime}, I^{\\prime}, Y$ are collinear.\n\n\n\nFigure 3\n\nLet $T$ be the midpoint of $C D$, and let $O$ be the center of $\\Gamma$. Let $C M$ meet $T Y$ at $N$. To avoid confusion, we clean some superfluous details out of the picture (see Figure 4).\n\nLet $V=M T \\cap C Y$. Since $M T \\| Y D$ and $D T=T C$, we get $C V=V Y$. Then CEvA's theorem applied to $\\triangle C T Y$ with the point $M$ yields\n\n$$\n1=\\frac{T Q}{Q C} \\cdot \\frac{C V}{V Y} \\cdot \\frac{Y N}{N T}=\\frac{T Q}{Q C} \\cdot \\frac{Y N}{N T}\n$$\n\nTherefore, $\\frac{T Q}{Q C}=\\frac{T N}{N Y}$. So, $N Q \\| C Y$, and thus $N Q \\perp O C$.\n\nNote that the points $O, N, T$, and $Y$ are collinear. Therefore, $C Q \\perp O N$. So, $Q$ is the orthocenter of $\\triangle O C N$, and hence $O Q \\perp C P$. Thus, $Q$ lies on the perpendicular bisector of $C P$, and therefore $C Q=Q P$, as required.\n\n\n\nFigure 4""]" [] Text-only Competition True Theorem proof Geometry Math English 230 Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other. "['Let $A^{\\prime}$ be the point diametrically opposite to $A$ on $\\Gamma$. Since $\\angle A Q A^{\\prime}=90^{\\circ}$ and $\\angle A Q H=90^{\\circ}$, the points $Q, H$, and $A^{\\prime}$ are collinear. Similarly, if $Q^{\\prime}$ denotes the point on $\\Gamma$ diametrically opposite to $Q$, then $K, H$, and $Q^{\\prime}$ are collinear. Let the line $A H F$ intersect $\\Gamma$ again at $E$; it is known that $M$ is the midpoint of the segment $H A^{\\prime}$ and that $F$ is the midpoint of $H E$. Let $J$ be the midpoint of $H Q^{\\prime}$.\n\nConsider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\\angle H K T=\\angle H Q K$ and we are to prove that $\\angle M K T=\\angle C F K$. Thus it remains to show that $\\angle H Q K=\\angle C F K+\\angle H K M$. Due to $\\angle H Q K=90^{\\circ}-\\angle Q^{\\prime} H A^{\\prime}$ and $\\angle C F K=90^{\\circ}-\\angle K F A$, this means the same as $\\angle Q^{\\prime} H A^{\\prime}=$ $\\angle K F A-\\angle H K M$. Now, since the triangles $K H E$ and $A H Q^{\\prime}$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\\angle K F A=\\angle H J A$, and analogously one may obtain $\\angle H K M=\\angle J Q H$. Thereby our task is reduced to verifying\n\n$$\n\\angle Q^{\\prime} H A^{\\prime}=\\angle H J A-\\angle J Q H .\n$$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nTo avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\\angle Q^{\\prime} H A^{\\prime}=\\angle J Q H+\\angle H J Q$ and $\\angle H J A=\\angle Q J A+\\angle H J Q$, we just have to show that $2 \\angle J Q H=\\angle Q J A$. To this end, it suffices to remark that $A Q A^{\\prime} Q^{\\prime}$ is a rectangle and that $J$, being defined to be the midpoint of $H Q^{\\prime}$, has to lie on the mid parallel of $Q A^{\\prime}$ and $Q^{\\prime} A$.' 'We define the points $A^{\\prime}$ and $E$ and prove that the ray $M H$ passes through $Q$ in the same way as in the first solution. Notice that the points $A^{\\prime}$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A^{\\prime}$ is the second intersection of the line $M H$ with $\\Gamma$, and $E$ is the point on $\\Gamma$ with the property $\\angle H E A^{\\prime}=90^{\\circ}$ (see Figure 3).\n\nIn the circles $K Q H$ and $E A^{\\prime} H$, the line segments $H Q$ and $H A^{\\prime}$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A^{\\prime} H$. Their pairwise radical axes are the lines $Q K$, $A^{\\prime} E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\\angle Q K H=$\n\n\n\n\n\nFigure 3\n\n$\\angle H E A^{\\prime}=90^{\\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K=S E=S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$.\n\nThe circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have\n\n$$\nS M \\cdot S F=S H^{2}=S K^{2}\n$$\n\nSo, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.']" [] Text-only Competition True Theorem proof Geometry Math English 231 Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other. "[""Denote by $\\gamma_{A}, \\gamma_{B}, \\gamma_{C}$, and $\\gamma_{D}$ the incircles of the quadrilaterals $A P O S, B Q O P$, $C R O Q$, and $D S O R$, respectively.\n\nWe start with proving that the quadrilateral $A B C D$ also has an incircle which will be referred to as $\\Omega$. Denote the points of tangency as in Figure 1. It is well-known that $Q Q_{1}=O O_{1}$ (if $B C \\| P R$, this is obvious; otherwise, one may regard the two circles involved as the incircle and an excircle of the triangle formed by the lines $O Q, P R$, and $B C$ ). Similarly, $O O_{1}=P P_{1}$. Hence we have $Q Q_{1}=P P_{1}$. The other equalities of segment lengths marked in Figure 1 can be proved analogously. These equalities, together with $A P_{1}=A S_{1}$ and similar ones, yield $A B+C D=A D+B C$, as required.\n\n\n\nFigure 1\n\nNext, let us draw the lines parallel to $Q S$ through $P$ and $R$, and also draw the lines parallel to $P R$ through $Q$ and $S$. These lines form a parallelogram; denote its vertices by $A^{\\prime}, B^{\\prime}, C^{\\prime}$, and $D^{\\prime}$ as shown in Figure 2.\n\nSince the quadrilateral $A P O S$ has an incircle, we have $A P-A S=O P-O S=A^{\\prime} S-A^{\\prime} P$. It is well-known that in this case there also exists a circle $\\omega_{A}$ tangent to the four rays $A P$, $A S, A^{\\prime} P$, and $A^{\\prime} S$. It is worth mentioning here that in case when, say, the lines $A B$ and $A^{\\prime} B^{\\prime}$ coincide, the circle $\\omega_{A}$ is just tangent to $A B$ at $P$. We introduce the circles $\\omega_{B}, \\omega_{C}$, and $\\omega_{D}$ in a similar manner.\n\nAssume that the radii of the circles $\\omega_{A}$ and $\\omega_{C}$ are different. Let $X$ be the center of the homothety having a positive scale factor and mapping $\\omega_{A}$ to $\\omega_{C}$.\n\nNow, Monge's theorem applied to the circles $\\omega_{A}, \\Omega$, and $\\omega_{C}$ shows that the points $A, C$, and $X$ are collinear. Applying the same theorem to the circles $\\omega_{A}, \\omega_{B}$, and $\\omega_{C}$, we see that the points $P, Q$, and $X$ are also collinear. Similarly, the points $R, S$, and $X$ are collinear, as required.\n\nIf the radii of $\\omega_{A}$ and $\\omega_{C}$ are equal but these circles do not coincide, then the degenerate version of the same theorem yields that the three lines $A C, P Q$, and $R S$ are parallel to the line of centers of $\\omega_{A}$ and $\\omega_{C}$.\n\nFinally, we need to say a few words about the case when $\\omega_{A}$ and $\\omega_{C}$ coincide (and thus they also coincide with $\\Omega, \\omega_{B}$, and $\\omega_{D}$ ). It may be regarded as the limit case in the following manner.\n\n\n\n\n\nFigure 2\n\nLet us fix the positions of $A, P, O$, and $S$ (thus we also fix the circles $\\omega_{A}, \\gamma_{A}, \\gamma_{B}$, and $\\gamma_{D}$ ). Now we vary the circle $\\gamma_{C}$ inscribed into $\\angle Q O R$; for each of its positions, one may reconstruct the lines $B C$ and $C D$ as the external common tangents to $\\gamma_{B}, \\gamma_{C}$ and $\\gamma_{C}, \\gamma_{D}$ different from $P R$ and $Q S$, respectively. After such variation, the circle $\\Omega$ changes, so the result obtained above may be applied."" ""Applying Menelaus' theorem to $\\triangle A B C$ with the line $P Q$ and to $\\triangle A C D$ with the line $R S$, we see that the line $A C$ meets $P Q$ and $R S$ at the same point (possibly at infinity) if and only if\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=1\n\\tag{1}\n$$\n\nSo, it suffices to prove (1).\n\nWe start with the following result.\n\nLemma 1. Let $E F G H$ be a circumscribed quadrilateral, and let $M$ be its incenter. Then\n\n$$\n\\frac{E F \\cdot F G}{G H \\cdot H E}=\\frac{F M^{2}}{H M^{2}}\n$$\n\nProof. Notice that $\\angle E M H+\\angle G M F=\\angle F M E+\\angle H M G=180^{\\circ}, \\angle F G M=\\angle M G H$, and $\\angle H E M=\\angle M E F$ (see Figure 3). By the law of sines, we get\n\n$$\n\\frac{E F}{F M} \\cdot \\frac{F G}{F M}=\\frac{\\sin \\angle F M E \\cdot \\sin \\angle G M F}{\\sin \\angle M E F \\cdot \\sin \\angle F G M}=\\frac{\\sin \\angle H M G \\cdot \\sin \\angle E M H}{\\sin \\angle M G H \\cdot \\sin \\angle H E M}=\\frac{G H}{H M} \\cdot \\frac{H E}{H M} .\n$$\n\n\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nWe denote by $I, J, K$, and $L$ the incenters of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$, respectively. Applying Lemma 1 to these four quadrilaterals we get\n\n$$\n\\frac{A P \\cdot P O}{O S \\cdot S A} \\cdot \\frac{B Q \\cdot Q O}{O P \\cdot P B} \\cdot \\frac{C R \\cdot R O}{O Q \\cdot Q C} \\cdot \\frac{D S \\cdot S O}{O R \\cdot R D}=\\frac{P I^{2}}{S I^{2}} \\cdot \\frac{Q J^{2}}{P J^{2}} \\cdot \\frac{R K^{2}}{Q K^{2}} \\cdot \\frac{S L^{2}}{R L^{2}}\n$$\n\nwhich reduces to\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=\\frac{P I^{2}}{P J^{2}} \\cdot \\frac{Q J^{2}}{Q K^{2}} \\cdot \\frac{R K^{2}}{R L^{2}} \\cdot \\frac{S L^{2}}{S I^{2}}\n\\tag{2}\n$$\n\nNext, we have $\\angle I P J=\\angle J O I=90^{\\circ}$, and the line $O P$ separates $I$ and $J$ (see Figure 4). This means that the quadrilateral $I P J O$ is cyclic. Similarly, we get that the quadrilateral $J Q K O$ is cyclic with $\\angle J Q K=90^{\\circ}$. Thus, $\\angle Q K J=\\angle Q O J=\\angle J O P=\\angle J I P$. Hence, the right triangles $I P J$ and $K Q J$ are similar. Therefore, $\\frac{P I}{P J}=\\frac{Q K}{Q J}$. Likewise, we obtain $\\frac{R K}{R L}=\\frac{S I}{S L}$. These two equations together with (2) yield (1)."" ""We present another approach for showing this equation:\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=1 \\tag{1}\n$$\n\nLemma 2. Let $E F G H$ and $E^{\\prime} F^{\\prime} G^{\\prime} H^{\\prime}$ be circumscribed quadrilaterals such that $\\angle E+\\angle E^{\\prime}=$ $\\angle F+\\angle F^{\\prime}=\\angle G+\\angle G^{\\prime}=\\angle H+\\angle H^{\\prime}=180^{\\circ}$. Then\n\n$$\n\\frac{E F \\cdot G H}{F G \\cdot H E}=\\frac{E^{\\prime} F^{\\prime} \\cdot G^{\\prime} H^{\\prime}}{F^{\\prime} G^{\\prime} \\cdot H^{\\prime} E^{\\prime}}\n$$\n\nProof. Let $M$ and $M^{\\prime}$ be the incenters of $E F G H$ and $E^{\\prime} F^{\\prime} G^{\\prime} H^{\\prime}$, respectively. We use the notation $[X Y Z]$ for the area of a triangle $X Y Z$.\n\nTaking into account the relation $\\angle F M E+\\angle F^{\\prime} M^{\\prime} E^{\\prime}=180^{\\circ}$ together with the analogous ones, we get\n\n$$\n\\begin{aligned}\n\\frac{E F \\cdot G H}{F G \\cdot H E} & =\\frac{[M E F] \\cdot[M G H]}{[M F G] \\cdot[M H E]}=\\frac{M E \\cdot M F \\cdot \\sin \\angle F M E \\cdot M G \\cdot M H \\cdot \\sin \\angle H M G}{M F \\cdot M G \\cdot \\sin \\angle G M F \\cdot M H \\cdot M E \\cdot \\sin \\angle E M H} \\\\\n& =\\frac{M^{\\prime} E^{\\prime} \\cdot M^{\\prime} F^{\\prime} \\cdot \\sin \\angle F^{\\prime} M^{\\prime} E^{\\prime} \\cdot M^{\\prime} G^{\\prime} \\cdot M^{\\prime} H^{\\prime} \\cdot \\sin \\angle H^{\\prime} M^{\\prime} G^{\\prime}}{M^{\\prime} F^{\\prime} \\cdot M^{\\prime} G^{\\prime} \\cdot \\sin \\angle G^{\\prime} M^{\\prime} F^{\\prime} \\cdot M^{\\prime} H^{\\prime} \\cdot M^{\\prime} E^{\\prime} \\cdot \\sin \\angle E^{\\prime} M^{\\prime} H^{\\prime}}=\\frac{E^{\\prime} F^{\\prime} \\cdot G^{\\prime} H^{\\prime}}{F^{\\prime} G^{\\prime} \\cdot H^{\\prime} E^{\\prime}}\n\\end{aligned}\n$$\n\n\n\nFigure 6\n\nDenote by $h$ the homothety centered at $O$ that maps the incircle of $C R O Q$ to the incircle of $A P O S$. Let $Q^{\\prime}=h(Q), C^{\\prime}=h(C), R^{\\prime}=h(R), O^{\\prime}=O, S^{\\prime}=S, A^{\\prime}=A$, and $P^{\\prime}=P$. Furthermore, define $B^{\\prime}=A^{\\prime} P^{\\prime} \\cap C^{\\prime} Q^{\\prime}$ and $D^{\\prime}=A^{\\prime} S^{\\prime} \\cap C^{\\prime} R^{\\prime}$ as shown in Figure 6. Then\n\n$$\n\\frac{A P \\cdot O S}{P O \\cdot S A}=\\frac{A^{\\prime} P^{\\prime} \\cdot O^{\\prime} S^{\\prime}}{P^{\\prime} O^{\\prime} \\cdot S^{\\prime} A^{\\prime}}\n$$\n\nholds trivially. We also have\n\n$$\n\\frac{C R \\cdot O Q}{R O \\cdot Q C}=\\frac{C^{\\prime} R^{\\prime} \\cdot O^{\\prime} Q^{\\prime}}{R^{\\prime} O^{\\prime} \\cdot Q^{\\prime} C^{\\prime}}\n$$\n\nby the similarity of the quadrilaterals $C R O Q$ and $C^{\\prime} R^{\\prime} O^{\\prime} Q^{\\prime}$.\n\n\n\nNext, consider the circumscribed quadrilaterals $B Q O P$ and $B^{\\prime} Q^{\\prime} O^{\\prime} P^{\\prime}$ whose incenters lie on different sides of the quadrilaterals' shared side line $O P=O^{\\prime} P^{\\prime}$. Observe that $B Q \\| B^{\\prime} Q^{\\prime}$ and that $B^{\\prime}$ and $Q^{\\prime}$ lie on the lines $B P$ and $Q O$, respectively. It is now easy to see that the two quadrilaterals satisfy the hypotheses of Lemma 2 . Thus, we deduce\n\n$$\n\\frac{B Q \\cdot O P}{Q O \\cdot P B}=\\frac{B^{\\prime} Q^{\\prime} \\cdot O^{\\prime} P^{\\prime}}{Q^{\\prime} O^{\\prime} \\cdot P^{\\prime} B^{\\prime}}\n$$\n\nSimilarly, we get\n\n$$\n\\frac{D S \\cdot O R}{S O \\cdot R D}=\\frac{D^{\\prime} S^{\\prime} \\cdot O^{\\prime} R^{\\prime}}{S^{\\prime} O^{\\prime} \\cdot R^{\\prime} D^{\\prime}}\n$$\n\nMultiplying these four equations, we obtain\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=\\frac{A^{\\prime} P^{\\prime}}{P^{\\prime} B^{\\prime}} \\cdot \\frac{B^{\\prime} Q^{\\prime}}{Q^{\\prime} C^{\\prime}} \\cdot \\frac{C^{\\prime} R^{\\prime}}{R^{\\prime} D^{\\prime}} \\cdot \n\\frac{D^{\\prime} S^{\\prime}}{S^{\\prime} A^{\\prime}}\n\\tag{3}\n$$\n\nFinally, we apply BRianchon's theorem to the circumscribed hexagon $A^{\\prime} P^{\\prime} R^{\\prime} C^{\\prime} Q^{\\prime} S^{\\prime}$ and deduce that the lines $A^{\\prime} C^{\\prime}, P^{\\prime} Q^{\\prime}$, and $R^{\\prime} S^{\\prime}$ are either concurrent or parallel to each other. So, by Menelaus' theorem, we obtain\n\n$$\n\\frac{A^{\\prime} P^{\\prime}}{P^{\\prime} B^{\\prime}} \\cdot \\frac{B^{\\prime} Q^{\\prime}}{Q^{\\prime} C^{\\prime}} \\cdot \\frac{C^{\\prime} R^{\\prime}}{R^{\\prime} D^{\\prime}} \\cdot \\frac{D^{\\prime} S^{\\prime}}{S^{\\prime} A^{\\prime}}=1\n$$\n\nThis equation together with (3) yield (1).""]" [] Text-only Competition True Theorem proof Geometry Math English 232 "A triangulation of a convex polygon $\Pi$ is a partitioning of $\Pi$ into triangles by diagonals having no common points other than the vertices of the polygon. We say that a triangulation is a Thaiangulation if all triangles in it have the same area. Prove that any two different Thaiangulations of a convex polygon $\Pi$ differ by exactly two triangles. (In other words, prove that it is possible to replace one pair of triangles in the first Thaiangulation with a different pair of triangles so as to obtain the second Thaiangulation.)" ['We denote by $[S]$ the area of a polygon $S$.\n\nRecall that each triangulation of a convex $n$-gon has exactly $n-2$ triangles. This means that all triangles in any two Thaiangulations of a convex polygon $\\Pi$ have the same area.\n\nLet $\\mathcal{T}$ be a triangulation of a convex polygon $\\Pi$. If four vertices $A, B, C$, and $D$ of $\\Pi$ form a parallelogram, and $\\mathcal{T}$ contains two triangles whose union is this parallelogram, then we say that $\\mathcal{T}$ contains parallelogram $A B C D$. Notice here that if two Thaiangulations $\\mathcal{T}_{1}$ and $\\mathcal{T}_{2}$ of $\\Pi$ differ by two triangles, then the union of these triangles is a quadrilateral each of whose diagonals bisects its area, i.e., a parallelogram.\n\nWe start with proving two properties of triangulations.\n\nLemma 1. A triangulation of a convex polygon $\\Pi$ cannot contain two parallelograms.\n\nProof. Arguing indirectly, assume that $P_{1}$ and $P_{2}$ are two parallelograms contained in some triangulation $\\mathcal{T}$. If they have a common triangle in $\\mathcal{T}$, then we may assume that $P_{1}$ consists of triangles $A B C$ and $A D C$ of $\\mathcal{T}$, while $P_{2}$ consists of triangles $A D C$ and $C D E$ (see Figure 1). But then $B C\\|A D\\| C E$, so the three vertices $B, C$, and $E$ of $\\Pi$ are collinear, which is absurd.\n\nAssume now that $P_{1}$ and $P_{2}$ contain no common triangle. Let $P_{1}=A B C D$. The sides $A B$, $B C, C D$, and $D A$ partition $\\Pi$ into several parts, and $P_{2}$ is contained in one of them; we may assume that this part is cut off from $P_{1}$ by $A D$. Then one may label the vertices of $P_{2}$ by $X$, $Y, Z$, and $T$ so that the polygon $A B C D X Y Z T$ is convex (see Figure 2; it may happen that $D=X$ and/or $T=A$, but still this polygon has at least six vertices). But the sum of the external angles of this polygon at $B, C, Y$, and $Z$ is already $360^{\\circ}$, which is impossible. A final contradiction.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nLemma 2. Every triangle in a Thaiangulation $\\mathcal{T}$ of $\\Pi$ contains a side of $\\Pi$.\n\nProof. Let $A B C$ be a triangle in $\\mathcal{T}$. Apply an affine transform such that $A B C$ maps to an equilateral triangle; let $A^{\\prime} B^{\\prime} C^{\\prime}$ be the image of this triangle, and $\\Pi^{\\prime}$ be the image of $\\Pi$. Clearly, $\\mathcal{T}$ maps into a Thaiangulation $\\mathcal{T}^{\\prime}$ of $\\Pi^{\\prime}$.\n\nAssume that none of the sides of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is a side of $\\Pi^{\\prime}$. Then $\\mathcal{T}^{\\prime}$ contains some other triangles with these sides, say, $A^{\\prime} B^{\\prime} Z, C^{\\prime} A^{\\prime} Y$, and $B^{\\prime} C^{\\prime} X$; notice that $A^{\\prime} Z B^{\\prime} X C^{\\prime} Y$ is a convex hexagon (see Figure 3). The sum of its external angles at $X, Y$, and $Z$ is less than $360^{\\circ}$. So one of these angles (say, at $Z$ ) is less than $120^{\\circ}$, hence $\\angle A^{\\prime} Z B^{\\prime}>60^{\\circ}$. Then $Z$ lies on a circular arc subtended by $A^{\\prime} B^{\\prime}$ and having angular measure less than $240^{\\circ}$; consequently, the altitude $Z H$ of $\\triangle A^{\\prime} B^{\\prime} Z$ is less than $\\sqrt{3} A^{\\prime} B^{\\prime} / 2$. Thus $\\left[A^{\\prime} B^{\\prime} Z\\right]<\\left[A^{\\prime} B^{\\prime} C^{\\prime}\\right]$, and $\\mathcal{T}^{\\prime}$ is not a Thaiangulation. A contradiction.\n\n\n\nNow we pass to the solution. We say that a triangle in a triangulation of $\\Pi$ is an ear if it contains two sides of $\\Pi$. Note that each triangulation of a polygon contains some ear.\n\nArguing indirectly, we choose a convex polygon $\\Pi$ with the least possible number of sides such that some two Thaiangulations $\\mathcal{T}_{1}$ and $\\mathcal{T}_{2}$ of $\\Pi$ violate the problem statement (thus $\\Pi$ has at least five sides). Consider now any ear $A B C$ in $\\mathcal{T}_{1}$, with $A C$ being a diagonal of $\\Pi$. If $\\mathcal{T}_{2}$ also contains $\\triangle A B C$, then one may cut $\\triangle A B C$ off from $\\Pi$, getting a polygon with a smaller number of sides which also violates the problem statement. This is impossible; thus $\\mathcal{T}_{2}$ does not contain $\\triangle A B C$.\n\nNext, $\\mathcal{T}_{1}$ contains also another triangle with side $A C$, say $\\triangle A C D$. By Lemma 2, this triangle contains a side of $\\Pi$, so $D$ is adjacent to either $A$ or $C$ on the boundary of $\\Pi$. We may assume that $D$ is adjacent to $C$.\n\nAssume that $\\mathcal{T}_{2}$ does not contain the triangle $B C D$. Then it contains two different triangles $B C X$ and $C D Y$ (possibly, with $X=Y$ ); since these triangles have no common interior points, the polygon $A B C D Y X$ is convex (see Figure 4). But, since $[A B C]=[B C X]=$ $[A C D]=[C D Y]$, we get $A X \\| B C$ and $A Y \\| C D$ which is impossible. Thus $\\mathcal{T}_{2}$ contains $\\triangle B C D$.\n\nTherefore, $[A B D]=[A B C]+[A C D]-[B C D]=[A B C]$, and $A B C D$ is a parallelogram contained in $\\mathcal{T}_{1}$. Let $\\mathcal{T}^{\\prime}$ be the Thaiangulation of $\\Pi$ obtained from $\\mathcal{T}_{1}$ by replacing the diagonal $A C$ with $B D$; then $\\mathcal{T}^{\\prime}$ is distinct from $\\mathcal{T}_{2}$ (otherwise $\\mathcal{T}_{1}$ and $\\mathcal{T}_{2}$ would differ by two triangles). Moreover, $\\mathcal{T}^{\\prime}$ shares a common ear $B C D$ with $\\mathcal{T}_{2}$. As above, cutting this ear away we obtain that $\\mathcal{T}_{2}$ and $\\mathcal{T}^{\\prime}$ differ by two triangles forming a parallelogram different from $A B C D$. Thus $\\mathcal{T}^{\\prime}$ contains two parallelograms, which contradicts Lemma 1.\n\n\n\nFigure 4\n\n\n\nFigure 5'] [] Text-only Competition True Theorem proof Geometry Math English 233 Let $a$ and $b$ be positive integers such that $a ! b !$ is a multiple of $a !+b !$. Prove that $3 a \geqslant 2 b+2$. "['If $a>b$, we immediately get $3 a \\geqslant 2 b+2$. In the case $a=b$, the required inequality is equivalent to $a \\geqslant 2$, which can be checked easily since $(a, b)=(1,1)$ does not satisfy $a !+b ! \\mid a ! b !$. We now assume $aa !$, which is impossible. We observe that $c ! \\mid M$ since $M$ is a product of $c$ consecutive integers. Thus $\\operatorname{gcd}(1+M, c !)=1$, which implies\n\n$$\n1+M \\mid \\frac{a !}{c !}=(c+1)(c+2) \\cdots a\n\\tag{1}\n$$\n\nIf $a \\leqslant 2 c$, then $\\frac{a !}{c !}$ is a product of $a-c \\leqslant c$ integers not exceeding $a$ whereas $M$ is a product of $c$ integers exceeding $a$. Therefore, $1+M>\\frac{a !}{c !}$, which is a contradiction.\n\nIt remains to exclude the case $a=2 c+1$. Since $a+1=2(c+1)$, we have $c+1 \\mid M$. Hence, we can deduce from (1) that $1+M \\mid(c+2)(c+3) \\cdots a$. Now $(c+2)(c+3) \\cdots a$ is a product of $a-c-1=c$ integers not exceeding $a$; thus it is smaller than $1+M$. Again, we arrive at a contradiction.' 'we may assume that $aa+c$; otherwise, $a+1 \\leqslant 2 c+2 \\leqslant 2 p \\leqslant a+c$ so $p \\mid N-1$, again impossible. Thus, we have $p \\in\\left(\\frac{a+c}{2}, a\\right]$, and $p^{2} \\nmid(a+c)$ ! since $2 p>a+c$. Therefore, $p^{2} \\nmid N$ as well.\n\nIf $a \\leqslant c+2$, then the interval $\\left(\\frac{a+c}{2}, a\\right]$ contains at most one integer and hence at most one prime number, which has to be $a$. Since $p^{2} \\nmid N$, we must have $N=p=a$ or $N=1$, which is absurd since $N>a \\geqslant 1$. Thus, we have $a \\geqslant c+3$, and so $\\frac{a+c+1}{2} \\geqslant c+2$. It follows that $p$ lies in the interval $[c+2, a]$.\n\nThus, every prime appearing in the prime factorization of $N$ lies in the interval $[c+2, a]$, and its exponent is exactly 1. So we must have $N \\mid(c+2)(c+3) \\cdots a$. However, $(c+2)(c+3) \\cdots a$ is a product of $a-c-1 \\leqslant c$ numbers not exceeding $a$, so it is less than $N$. This is a contradiction.']" [] Text-only Competition True Theorem proof Number Theory Math English 234 Let $m$ and $n$ be positive integers such that $m>n$. Define $x_{k}=(m+k) /(n+k)$ for $k=$ $1,2, \ldots, n+1$. Prove that if all the numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers, then $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime. ['Assume that $x_{1}, x_{2}, \\ldots, x_{n+1}$ are integers. Define the integers\n\n$$\na_{k}=x_{k}-1=\\frac{m+k}{n+k}-1=\\frac{m-n}{n+k}>0\n$$\n\nfor $k=1,2, \\ldots, n+1$.\n\nLet $P=x_{1} x_{2} \\cdots x_{n+1}-1$. We need to prove that $P$ is divisible by an odd prime, or in other words, that $P$ is not a power of 2 . To this end, we investigate the powers of 2 dividing the numbers $a_{k}$.\n\nLet $2^{d}$ be the largest power of 2 dividing $m-n$, and let $2^{c}$ be the largest power of 2 not exceeding $2 n+1$. Then $2 n+1 \\leqslant 2^{c+1}-1$, and so $n+1 \\leqslant 2^{c}$. We conclude that $2^{c}$ is one of the numbers $n+1, n+2, \\ldots, 2 n+1$, and that it is the only multiple of $2^{c}$ appearing among these numbers. Let $\\ell$ be such that $n+\\ell=2^{c}$. Since $\\frac{m-n}{n+\\ell}$ is an integer, we have $d \\geqslant c$. Therefore, $2^{d-c+1} \\nmid a_{\\ell}=\\frac{m-n}{n+\\ell}$, while $2^{d-c+1} \\mid a_{k}$ for all $k \\in\\{1, \\ldots, n+1\\} \\backslash\\{\\ell\\}$.\n\nComputing modulo $2^{d-c+1}$, we get\n\n$$\nP=\\left(a_{1}+1\\right)\\left(a_{2}+1\\right) \\cdots\\left(a_{n+1}+1\\right)-1 \\equiv\\left(a_{\\ell}+1\\right) \\cdot 1^{n}-1 \\equiv a_{\\ell} \\not \\equiv 0 \\quad\\left(\\bmod 2^{d-c+1}\\right) .\n$$\n\nTherefore, $2^{d-c+1} \\nmid P$.\n\nOn the other hand, for any $k \\in\\{1, \\ldots, n+1\\} \\backslash\\{\\ell\\}$, we have $2^{d-c+1} \\mid a_{k}$. So $P \\geqslant a_{k} \\geqslant 2^{d-c+1}$, and it follows that $P$ is not a power of 2 .'] [] Text-only Competition True Theorem proof Number Theory Math English 235 "Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\left(a_{n}\right)$ is eventually periodic; in other words, there exist integers $N \geqslant 0$ and $t>0$ such that $a_{n+t}=a_{n}$ for all $n \geqslant N$." "['Let $s_{n}=a_{n}+b_{n}$. Notice that if $a_{n} \\mid b_{n}$, then $a_{n+1}=a_{n}+1, b_{n+1}=b_{n}-1$ and $s_{n+1}=s_{n}$. So, $a_{n}$ increases by 1 and $s_{n}$ does not change until the first index is reached with $a_{n} \\nmid s_{n}$. Define\n\n$$\nW_{n}=\\left\\{m \\in \\mathbb{Z}_{>0}: m \\geqslant a_{n} \\text { and } m \\nmid s_{n}\\right\\} \\quad \\text { and } \\quad w_{n}=\\min W_{n}\n$$\n\nClaim 1. The sequence $\\left(w_{n}\\right)$ is non-increasing.\n\nProof. If $a_{n} \\mid b_{n}$ then $a_{n+1}=a_{n}+1$. Due to $a_{n} \\mid s_{n}$, we have $a_{n} \\notin W_{n}$. Moreover $s_{n+1}=s_{n}$; therefore, $W_{n+1}=W_{n}$ and $w_{n+1}=w_{n}$.\n\nOtherwise, if $a_{n} \\nmid b_{n}$, then $a_{n} \\nmid s_{n}$, so $a_{n} \\in W_{n}$ and thus $w_{n}=a_{n}$. We show that $a_{n} \\in W_{n+1}$; this implies $w_{n+1} \\leqslant a_{n}=w_{n}$. By the definition of $W_{n+1}$, we need that $a_{n} \\geqslant a_{n+1}$ and $a_{n} \\nmid s_{n+1}$. The first relation holds because of $\\operatorname{gcd}\\left(a_{n}, b_{n}\\right)0}$ denote the set of positive integers. Consider a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$. For any $m, n \in \mathbb{Z}_{>0}$ we write $f^{n}(m)=\underbrace{f(f(\ldots f}_{n}(m) \ldots))$. Suppose that $f$ has the following two properties: (i) If $m, n \in \mathbb{Z}_{>0}$, then $\frac{f^{n}(m)-m}{n} \in \mathbb{Z}_{>0}$; (ii) The set $\mathbb{Z}_{>0} \backslash\left\{f(n) \mid n \in \mathbb{Z}_{>0}\right\}$ is finite. Prove that the sequence $f(1)-1, f(2)-2, f(3)-3, \ldots$ is periodic." "['We split the solution into three steps. In the first of them, we show that the function $f$ is injective and explain how this leads to a useful visualization of $f$. Then comes the second step, in which most of the work happens: its goal is to show that for any $n \\in \\mathbb{Z}_{>0}$ the sequence $n, f(n), f^{2}(n), \\ldots$ is an arithmetic progression. Finally, in the third step we put everything together, thus solving the problem.\n\nStep 1. We commence by checking that $f$ is injective. For this purpose, we consider any $m, k \\in \\mathbb{Z}_{>0}$ with $f(m)=f(k)$. By $(i)$, every positive integer $n$ has the property that\n\n$$\n\\frac{k-m}{n}=\\frac{f^{n}(m)-m}{n}-\\frac{f^{n}(k)-k}{n}\n$$\n\nis a difference of two integers and thus integral as well. But for $n=|k-m|+1$ this is only possible if $k=m$. Thereby, the injectivity of $f$ is established.\n\nNow recall that due to condition $(i i)$ there are finitely many positive integers $a_{1}, \\ldots, a_{k}$ such that $\\mathbb{Z}_{>0}$ is the disjoint union of $\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$ and $\\left\\{f(n) \\mid n \\in \\mathbb{Z}_{>0}\\right\\}$. Notice that by plugging $n=1$ into condition $(i)$ we get $f(m)>m$ for all $m \\in \\mathbb{Z}_{>0}$.\n\nWe contend that every positive integer $n$ may be expressed uniquely in the form $n=f^{j}\\left(a_{i}\\right)$ for some $j \\geqslant 0$ and $i \\in\\{1, \\ldots, k\\}$. The uniqueness follows from the injectivity of $f$. The existence can be proved by induction on $n$ in the following way. If $n \\in\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$, then we may take $j=0$; otherwise there is some $n^{\\prime}0$; and $T=1$ and $A=0$ if $t=0$. For every integer $n \\geqslant A$, the interval $\\Delta_{n}=[n+1, n+T]$ contains exactly $T / T_{i}$\n\n\n\nelements of the $i^{\\text {th }}$ row $(1 \\leqslant i \\leqslant t)$. Therefore, the number of elements from the last $(k-t)$ rows of the Table contained in $\\Delta_{n}$ does not depend on $n \\geqslant A$. It is not possible that none of these intervals $\\Delta_{n}$ contains an element from the $k-t$ last rows, because infinitely many numbers appear in these rows. It follows that for each $n \\geqslant A$ the interval $\\Delta_{n}$ contains at least one member from these rows.\n\nThis yields that for every positive integer $d$, the interval $[A+1, A+(d+1)(k-t) T]$ contains at least $(d+1)(k-t)$ elements from the last $k-t$ rows; therefore, there exists an index $x$ with $t+1 \\leqslant x \\leqslant k$, possibly depending on $d$, such that our interval contains at least $d+1$ elements from the $x^{\\text {th }}$ row. In this situation we have\n\n$$\nf^{d}\\left(a_{x}\\right) \\leqslant A+(d+1)(k-t) T .\n$$\n\nFinally, since there are finitely many possibilities for $x$, there exists an index $x \\geqslant t+1$ such that the set\n\n$$\nX=\\left\\{d \\in \\mathbb{Z}_{>0} \\mid f^{d}\\left(a_{x}\\right) \\leqslant A+(d+1)(k-t) T\\right\\}\n$$\n\nis infinite. Thereby we have found the ""dense row"" promised above.\n\nBy assumption $(i)$, for every $d \\in X$ the number\n\n$$\n\\beta_{d}=\\frac{f^{d}\\left(a_{x}\\right)-a_{x}}{d}\n$$\n\nis a positive integer not exceeding\n\n$$\n\\frac{A+(d+1)(k-t) T}{d} \\leqslant \\frac{A d+2 d(k-t) T}{d}=A+2(k-t) T\n$$\n\nThis leaves us with finitely many choices for $\\beta_{d}$, which means that there exists a number $T_{x}$ such that the set\n\n$$\nY=\\left\\{d \\in X \\mid \\beta_{d}=T_{x}\\right\\}\n$$\n\nis infinite. Notice that we have $f^{d}\\left(a_{x}\\right)=a_{x}+d \\cdot T_{x}$ for all $d \\in Y$.\n\nNow we are prepared to prove that the numbers in the $x^{\\text {th }}$ row form an arithmetic progression, thus coming to a contradiction with our assumption. Let us fix any positive integer $j$. Since the set $Y$ is infinite, we can choose a number $y \\in Y$ such that $y-j>\\left|f^{j}\\left(a_{x}\\right)-\\left(a_{x}+j T_{x}\\right)\\right|$. Notice that both numbers\n\n$$\nf^{y}\\left(a_{x}\\right)-f^{j}\\left(a_{x}\\right)=f^{y-j}\\left(f^{j}\\left(a_{x}\\right)\\right)-f^{j}\\left(a_{x}\\right) \\quad \\text { and } \\quad f^{y}\\left(a_{x}\\right)-\\left(a_{x}+j T_{x}\\right)=(y-j) T_{x}\n$$\n\nare divisible by $y-j$. Thus, the difference between these numbers is also divisible by $y-j$. Since the absolute value of this difference is less than $y-j$, it has to vanish, so we get $f^{j}\\left(a_{x}\\right)=$ $a_{x}+j \\cdot T_{x}$.\n\nHence, it is indeed true that all rows of the Table are arithmetic progressions.\n\nStep 3. Keeping the above notation in force, we denote the step of the $i^{\\text {th }}$ row of the table by $T_{i}$.\n\n\n\n$$\nT=\\operatorname{lcm}\\left(T_{1}, \\ldots, T_{k}\\right)\n$$\n\nTo see this, let any $n \\in \\mathbb{Z}_{>0}$ be given and denote the index of the row in which it appears in the Table by $i$. Then we have $f^{j}(n)=n+j \\cdot T_{i}$ for all $j \\in \\mathbb{Z}_{>0}$, and thus indeed\n\n$$\nf(n+T)-f(n)=f^{1+T / T_{i}}(n)-f(n)=\\left(n+T+T_{i}\\right)-\\left(n+T_{i}\\right)=T .\n$$\n\nThis concludes the solution.']" [] Text-only Competition True Theorem proof Number Theory Math English 237 "Let $z_{0}0$.\n\nNotice that\n\n$$\nn z_{n+1}-\\left(z_{0}+z_{1}+\\cdots+z_{n}\\right)=(n+1) z_{n+1}-\\left(z_{0}+z_{1}+\\cdots+z_{n}+z_{n+1}\\right)=-d_{n+1}\n$$\n\nso the second inequality in (1) is equivalent to $d_{n+1} \\leqslant 0$. Therefore, we have to prove that there is a unique index $n \\geqslant 1$ that satisfies $d_{n}>0 \\geqslant d_{n+1}$.\n\nBy its definition the sequence $d_{1}, d_{2}, \\ldots$ consists of integers and we have\n\n$$\nd_{1}=\\left(z_{0}+z_{1}\\right)-1 \\cdot z_{1}=z_{0}>0\n$$\n\nFrom\n\n$d_{n+1}-d_{n}=\\left(\\left(z_{0}+\\cdots+z_{n}+z_{n+1}\\right)-(n+1) z_{n+1}\\right)-\\left(\\left(z_{0}+\\cdots+z_{n}\\right)-n z_{n}\\right)=n\\left(z_{n}-z_{n+1}\\right)<0$\n\nwe can see that $d_{n+1}d_{2}>\\ldots$ of integers such that its first element $d_{1}$ is positive. The sequence must drop below 0 at some point, and thus there is a unique index $n$, that is the index of the last positive term, satisfying $d_{n}>0 \\geqslant d_{n+1}$.'] [] Text-only Competition True Theorem proof Algebra Math English 238 "Define the function $f:(0,1) \rightarrow(0,1)$ by $$ f(x)= \begin{cases}x+\frac{1}{2} & \text { if } x<\frac{1}{2} \\ x^{2} & \text { if } x \geqslant \frac{1}{2}\end{cases} $$ Let $a$ and $b$ be two real numbers such that $00$. Show that there exists a positive integer $n$ such that $$ \left(a_{n}-a_{n-1}\right)\left(b_{n}-b_{n-1}\right)<0 . $$" ['Note that\n\n$$\nf(x)-x=\\frac{1}{2}>0\n$$\n\nif $x<\\frac{1}{2}$ and\n\n$$\nf(x)-x=x^{2}-x<0\n$$\n\nif $x \\geqslant \\frac{1}{2}$. So if we consider $(0,1)$ as being divided into the two subintervals $I_{1}=\\left(0, \\frac{1}{2}\\right)$ and $I_{2}=\\left[\\frac{1}{2}, 1\\right)$, the inequality\n\n$$\n\\left(a_{n}-a_{n-1}\\right)\\left(b_{n}-b_{n-1}\\right)=\\left(f\\left(a_{n-1}\\right)-a_{n-1}\\right)\\left(f\\left(b_{n-1}\\right)-b_{n-1}\\right)<0\n$$\n\nholds if and only if $a_{n-1}$ and $b_{n-1}$ lie in distinct subintervals.\n\nLet us now assume, to the contrary, that $a_{k}$ and $b_{k}$ always lie in the same subinterval. Consider the distance $d_{k}=\\left|a_{k}-b_{k}\\right|$. If both $a_{k}$ and $b_{k}$ lie in $I_{1}$, then\n\n$$\nd_{k+1}=\\left|a_{k+1}-b_{k+1}\\right|=\\left|a_{k}+\\frac{1}{2}-b_{k}-\\frac{1}{2}\\right|=d_{k}\n$$\n\nIf, on the other hand, $a_{k}$ and $b_{k}$ both lie in $I_{2}$, then $\\min \\left(a_{k}, b_{k}\\right) \\geqslant \\frac{1}{2}$ and $\\max \\left(a_{k}, b_{k}\\right)=$ $\\min \\left(a_{k}, b_{k}\\right)+d_{k} \\geqslant \\frac{1}{2}+d_{k}$, which implies\n\n$$\nd_{k+1}=\\left|a_{k+1}-b_{k+1}\\right|=\\left|a_{k}^{2}-b_{k}^{2}\\right|=\\left|\\left(a_{k}-b_{k}\\right)\\left(a_{k}+b_{k}\\right)\\right| \\geqslant\\left|a_{k}-b_{k}\\right|\\left(\\frac{1}{2}+\\frac{1}{2}+d_{k}\\right)=d_{k}\\left(1+d_{k}\\right) \\geqslant d_{k}\n$$\n\nThis means that the difference $d_{k}$ is non-decreasing, and in particular $d_{k} \\geqslant d_{0}>0$ for all $k$.\n\nWe can even say more. If $a_{k}$ and $b_{k}$ lie in $I_{2}$, then\n\n$$\nd_{k+2} \\geqslant d_{k+1} \\geqslant d_{k}\\left(1+d_{k}\\right) \\geqslant d_{k}\\left(1+d_{0}\\right)\n$$\n\nIf $a_{k}$ and $b_{k}$ both lie in $I_{1}$, then $a_{k+1}$ and $b_{k+1}$ both lie in $I_{2}$, and so we have\n\n$$\nd_{k+2} \\geqslant d_{k+1}\\left(1+d_{k+1}\\right) \\geqslant d_{k+1}\\left(1+d_{0}\\right)=d_{k}\\left(1+d_{0}\\right)\n$$\n\nIn either case, $d_{k+2} \\geqslant d_{k}\\left(1+d_{0}\\right)$, and inductively we get\n\n$$\nd_{2 m} \\geqslant d_{0}\\left(1+d_{0}\\right)^{m}\n$$\n\nFor sufficiently large $m$, the right-hand side is greater than 1 , but since $a_{2 m}, b_{2 m}$ both lie in $(0,1)$, we must have $d_{2 m}<1$, a contradiction.\n\nThus there must be a positive integer $n$ such that $a_{n-1}$ and $b_{n-1}$ do not lie in the same subinterval, which proves the desired statement.'] [] Text-only Competition True Theorem proof Algebra Math English 239 Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect $R$ into at least $n+1$ smaller rectangles. "['Let $k$ be the number of rectangles in the dissection. The set of all points that are corners of one of the rectangles can be divided into three disjoint subsets:\n\n- $A$, which consists of the four corners of the original rectangle $R$, each of which is the corner of exactly one of the smaller rectangles,\n- $B$, which contains points where exactly two of the rectangles have a common corner (T-junctions, see the figure below),\n- $C$, which contains points where four of the rectangles have a common corner (crossings, see the figure below).\n\n\nFigure 1: A T-junction and a crossing\n\nWe denote the number of points in $B$ by $b$ and the number of points in $C$ by $c$. Since each of the $k$ rectangles has exactly four corners, we get\n\n$$\n4 k=4+2 b+4 c\n$$\n\nIt follows that $2 b \\leqslant 4 k-4$, so $b \\leqslant 2 k-2$.\n\nEach of the $n$ given points has to lie on a side of one of the smaller rectangles (but not of the original rectangle $R$ ). If we extend this side as far as possible along borders between rectangles, we obtain a line segment whose ends are T-junctions. Note that every point in $B$ can only be an endpoint of at most one such segment containing one of the given points, since it is stated that no two of them lie on a common line parallel to the sides of $R$. This means that\n\n$$\nb \\geqslant 2 n \\text {. }\n$$\n\nCombining our two inequalities for $b$, we get\n\n$$\n2 k-2 \\geqslant b \\geqslant 2 n\n$$\n\nthus $k \\geqslant n+1$, which is what we wanted to prove.' ""Let $k$ denote the number of rectangles. In the following, we refer to the directions of the sides of $R$ as 'horizontal' and 'vertical' respectively. Our goal is to prove the inequality $k \\geqslant n+1$ for fixed $n$. Equivalently, we can prove the inequality $n \\leqslant k-1$ for each $k$, which will be done by induction on $k$. For $k=1$, the statement is trivial.\n\nNow assume that $k>1$. If none of the line segments that form the borders between the rectangles is horizontal, then we have $k-1$ vertical segments dividing $R$ into $k$ rectangles. On each of them, there can only be one of the $n$ points, so $n \\leqslant k-1$, which is exactly what we want to prove.\n\nOtherwise, consider the lowest horizontal line $h$ that contains one or more of these line segments. Let $R^{\\prime}$ be the rectangle that results when everything that lies below $h$ is removed from $R$ (see the example in the figure below).\n\nThe rectangles that lie entirely below $h$ form blocks of rectangles separated by vertical line segments. Suppose there are $r$ blocks and $k_{i}$ rectangles in the $i^{\\text {th }}$ block. The left and right border of each block has to extend further upwards beyond $h$. Thus we can move any points that lie on these borders upwards, so that they now lie inside $R^{\\prime}$. This can be done without violating the conditions, one only needs to make sure that they do not get to lie on a common horizontal line with one of the other given points.\n\nAll other borders between rectangles in the $i^{\\text {th }}$ block have to lie entirely below $h$. There are $k_{i}-1$ such line segments, each of which can contain at most one of the given points. Finally, there can be one point that lies on $h$. All other points have to lie in $R^{\\prime}$ (after moving some of them as explained in the previous paragraph).\n\n\n\nFigure 2: Illustration of the inductive argument\n\nWe see that $R^{\\prime}$ is divided into $k-\\sum_{i=1}^{r} k_{i}$ rectangles. Applying the induction hypothesis to $R^{\\prime}$, we find that there are at most\n\n$$\n\\left(k-\\sum_{i=1}^{r} k_{i}\\right)-1+\\sum_{i=1}^{r}\\left(k_{i}-1\\right)+1=k-r\n$$\n\npoints. Since $r \\geqslant 1$, this means that $n \\leqslant k-1$, which completes our induction.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 240 We have $2^{m}$ sheets of paper, with the number 1 written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a+b$ on both sheets. Prove that after $m 2^{m-1}$ steps, the sum of the numbers on all the sheets is at least $4^{m}$. ['Let $P_{k}$ be the product of the numbers on the sheets after $k$ steps.\n\nSuppose that in the $(k+1)^{\\text {th }}$ step the numbers $a$ and $b$ are replaced by $a+b$. In the product, the number $a b$ is replaced by $(a+b)^{2}$, and the other factors do not change. Since $(a+b)^{2} \\geqslant 4 a b$, we see that $P_{k+1} \\geqslant 4 P_{k}$. Starting with $P_{0}=1$, a straightforward induction yields\n\n$$\nP_{k} \\geqslant 4^{k}\n$$\n\nfor all integers $k \\geqslant 0$; in particular\n\n$$\nP_{m \\cdot 2^{m-1}} \\geqslant 4^{m \\cdot 2^{m-1}}=\\left(2^{m}\\right)^{2^{m}}\n$$\n\nso by the AM-GM inequality, the sum of the numbers written on the sheets after $m 2^{m-1}$ steps is at least\n\n$$\n2^{m} \\cdot \\sqrt[2^{m}]{P_{m \\cdot 2^{m-1}}} \\geqslant 2^{m} \\cdot 2^{m}=4^{m}\n$$'] [] Text-only Competition True Theorem proof Combinatorics Math English 241 Consider $n \geqslant 3$ lines in the plane such that no two lines are parallel and no three have a common point. These lines divide the plane into polygonal regions; let $\mathcal{F}$ be the set of regions having finite area. Prove that it is possible to colour $\lceil\sqrt{n / 2}\rceil$ of the lines blue in such a way that no region in $\mathcal{F}$ has a completely blue boundary. (For a real number $x,\lceil x\rceil$ denotes the least integer which is not smaller than $x$.) ['Let $L$ be the given set of lines. Choose a maximal (by inclusion) subset $B \\subseteq L$ such that when we colour the lines of $B$ blue, no region in $\\mathcal{F}$ has a completely blue boundary. Let $|B|=k$. We claim that $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$.\n\nLet us colour all the lines of $L \\backslash B$ red. Call a point blue if it is the intersection of two blue lines. Then there are $\\left(\\begin{array}{l}k \\\\ 2\\end{array}\\right)$ blue points.\n\nNow consider any red line $\\ell$. By the maximality of $B$, there exists at least one region $A \\in \\mathcal{F}$ whose only red side lies on $\\ell$. Since $A$ has at least three sides, it must have at least one blue vertex. Let us take one such vertex and associate it to $\\ell$.\n\nSince each blue point belongs to four regions (some of which may be unbounded), it is associated to at most four red lines. Thus the total number of red lines is at most $4\\left(\\begin{array}{l}k \\\\ 2\\end{array}\\right)$. On the other hand, this number is $n-k$, so\n\n$$\nn-k \\leqslant 2 k(k-1), \\quad \\text { thus } n \\leqslant 2 k^{2}-k \\leqslant 2 k^{2} \\text {, }\n$$\n\nand finally $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$, which gives the desired result.'] [] Text-only Competition True Theorem proof Combinatorics Math English 242 Let $M$ be a set of $n \geqslant 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $A B$ and $C D$ sharing a common interior point and replace them by the segments $A C$ and $B D$ if none of them is present at this moment. Prove that it is impossible to perform $n^{3} / 4$ or more such moves. ['A line is said to be red if it contains two points of $M$. As no three points of $M$ are collinear, each red line determines a unique pair of points of $M$. Moreover, there are precisely $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)<\\frac{n^{2}}{2}$ red lines. By the value of a segment we mean the number of red lines intersecting it in its interior, and the value of a set of segments is defined to be the sum of the values of its elements. We will prove that $(i)$ the value of the initial set of segments is smaller than $n^{3} / 2$ and that (ii) each step decreases the value of the set of segments present by at least 2 . Since such a value can never be negative, these two assertions imply the statement of the problem.\n\nTo show $(i)$ we just need to observe that each segment has a value that is smaller than $n^{2} / 2$. Thus the combined value of the $n$ initial segments is indeed below $n \\cdot n^{2} / 2=n^{3} / 2$.\n\nIt remains to establish (ii). Suppose that at some moment we have two segments $A B$ and $C D$ sharing an interior point $S$, and that at the next moment we have the two segments $A C$ and $B D$ instead. Let $X_{A B}$ denote the set of red lines intersecting the segment $A B$ in its interior and let the sets $X_{A C}, X_{B D}$, and $X_{C D}$ be defined similarly. We are to prove that $\\left|X_{A C}\\right|+\\left|X_{B D}\\right|+2 \\leqslant\\left|X_{A B}\\right|+\\left|X_{C D}\\right|$.\n\nAs a first step in this direction, we claim that\n\n$$\n\\left|X_{A C} \\cup X_{B D}\\right|+2 \\leqslant\\left|X_{A B} \\cup X_{C D}\\right|\n\\tag{1}\n$$\n\nIndeed, if $g$ is a red line intersecting, e.g. the segment $A C$ in its interior, then it has to intersect the triangle $A C S$ once again, either in the interior of its side $A S$, or in the interior of its side $C S$, or at $S$, meaning that it belongs to $X_{A B}$ or to $X_{C D}$ (see Figure 1). Moreover, the red lines $A B$ and $C D$ contribute to $X_{A B} \\cup X_{C D}$ but not to $X_{A C} \\cup X_{B D}$. Thereby (1) is proved.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nSimilarly but more easily one obtains\n\n$$\n\\left|X_{A C} \\cap X_{B D}\\right| \\leqslant\\left|X_{A B} \\cap X_{C D}\\right|\n\\tag{2}\n$$\n\nIndeed, a red line $h$ appearing in $X_{A C} \\cap X_{B D}$ belongs, for similar reasons as above, also to $X_{A B} \\cap X_{C D}$. To make the argument precise, one may just distinguish the cases $S \\in h$ (see Figure 2) and $S \\notin h$ (see Figure 3). Thereby (2) is proved.\n\nAdding (1) and (2) we obtain the desired conclusion, thus completing the solution of this problem.'] [] Text-only Competition True Theorem proof Combinatorics Math English 243 "There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding along the current circle until he reaches an intersection with another circle. Then he continues his journey on this new circle and also changes the direction of moving, i.e. from clockwise to anticlockwise or vice versa. Suppose that Turbo's path entirely covers all circles. Prove that $n$ must be odd." "['Replace every cross (i.e. intersection of two circles) by two small circle arcs that indicate the direction in which the snail should leave the cross (see Figure 1.1). Notice that the placement of the small arcs does not depend on the direction of moving on the curves; no matter which direction the snail is moving on the circle arcs, he will follow the same curves (see Figure 1.2). In this way we have a set of curves, that are the possible paths of the snail. Call these curves snail orbits or just orbits. Every snail orbit is a simple closed curve that has no intersection with any other orbit.\n\n\n\nFigure 1.1\n\n\n\nFigure 1.2\n\nWe prove the following, more general statement.\n\n(*) In any configuration of $n$ circles such that no two of them are tangent, the number of snail orbits has the same parity as the number $n$. (Note that it is not assumed that all circle pairs intersect.)\n\nThis immediately solves the problem.\n\nLet us introduce the following operation that will be called flipping a cross. At a cross, remove the two small arcs of the orbits, and replace them by the other two arcs. Hence, when the snail arrives at a flipped cross, he will continue on the other circle as before, but he will preserve the orientation in which he goes along the circle arcs (see Figure 2).\n\n\nFigure 2\n\nConsider what happens to the number of orbits when a cross is flipped. Denote by $a, b, c$, and $d$ the four arcs that meet at the cross such that $a$ and $b$ belong to the same circle. Before the flipping $a$ and $b$ were connected to $c$ and $d$, respectively, and after the flipping $a$ and $b$ are connected to $d$ and $c$, respectively.\n\nThe orbits passing through the cross are closed curves, so each of the arcs $a, b, c$, and $d$ is connected to another one by orbits outside the cross. We distinguish three cases.\n\nCase 1: $a$ is connected to $b$ and $c$ is connected to $d$ by the orbits outside the cross (see Figure 3.1).\n\n\n\nWe show that this case is impossible. Remove the two small arcs at the cross, connect $a$ to $b$, and connect $c$ to $d$ at the cross. Let $\\gamma$ be the new closed curve containing $a$ and $b$, and let $\\delta$ be the new curve that connects $c$ and $d$. These two curves intersect at the cross. So one of $c$ and $d$ is inside $\\gamma$ and the other one is outside $\\gamma$. Then the two closed curves have to meet at least one more time, but this is a contradiction, since no orbit can intersect itself.\n\n\n\nFigure 3.1\n\n\n\nFigure 3.2\n\n\nFigure 3.3\n\nCase 2: $a$ is connected to $c$ and $b$ is connected to $d$ (see Figure 3.2).\n\nBefore the flipping $a$ and $c$ belong to one orbit and $b$ and $d$ belong to another orbit. Flipping the cross merges the two orbits into a single orbit. Hence, the number of orbits decreases by 1 .\n\nCase 3: $a$ is connected to $d$ and $b$ is connected to $c$ (see Figure 3.3).\n\nBefore the flipping the $\\operatorname{arcs} a, b, c$, and $d$ belong to a single orbit. Flipping the cross splits that orbit in two. The number of orbits increases by 1 .\n\nAs can be seen, every flipping decreases or increases the number of orbits by one, thus changes its parity.\n\nNow flip every cross, one by one. Since every pair of circles has 0 or 2 intersections, the number of crosses is even. Therefore, when all crosses have been flipped, the original parity of the number of orbits is restored. So it is sufficient to prove $(*)$ for the new configuration, where all crosses are flipped. Of course also in this new configuration the (modified) orbits are simple closed curves not intersecting each other.\n\nOrient the orbits in such a way that the snail always moves anticlockwise along the circle arcs. Figure 4 shows the same circles as in Figure 1 after flipping all crosses and adding orientation. (Note that this orientation may be different from the orientation of the orbit as a planar curve; the orientation of every orbit may be negative as well as positive, like the middle orbit in Figure 4.) If the snail moves around an orbit, the total angle change in his moving direction, the total curvature, is either $+2 \\pi$ or $-2 \\pi$, depending on the orientation of the orbit. Let $P$ and $N$ be the number of orbits with positive and negative orientation, respectively. Then the total curvature of all orbits is $(P-N) \\cdot 2 \\pi$.\n\n\n\nFigure 4\n\n\n\nFigure 5\n\nDouble-count the total curvature of all orbits. Along every circle the total curvature is $2 \\pi$. At every cross, the two turnings make two changes with some angles having the same absolute value but opposite signs, as depicted in Figure 5. So the changes in the direction at the crosses cancel out. Hence, the total curvature is $n \\cdot 2 \\pi$.\n\nNow we have $(P-N) \\cdot 2 \\pi=n \\cdot 2 \\pi$, so $P-N=n$. The number of (modified) orbits is $P+N$, that has a same parity as $P-N=n$.' 'We present a different proof of $(*)$.\n\nWe perform a sequence of small modification steps on the configuration of the circles in such a way that at the end they have no intersection at all (see Figure 6.1). We use two kinds of local changes to the structure of the orbits (see Figure 6.2):\n\n- Type-1 step: An arc of a circle is moved over an arc of another circle; such a step creates or removes two intersections.\n- Type-2 step: An arc of a circle is moved through the intersection of two other circles.\n\n\n\nFigure 6.1\n\n\n\nype-1\n\n\n\nFigure 6.2\n\nWe assume that in every step only one circle is moved, and that this circle is moved over at most one arc or intersection point of other circles.\n\nWe will show that the parity of the number of orbits does not change in any step. As every circle becomes a separate orbit at the end of the procedure, this fact proves (*).\n\nConsider what happens to the number of orbits when a Type- 1 step is performed. The two intersection points are created or removed in a small neighbourhood. Denote some points of the two circles where they enter or leave this neighbourhood by $a, b, c$, and $d$ in this order around the neighbourhood; let $a$ and $b$ belong to one circle and let $c$ and $d$ belong to the other circle. The two circle arcs may have the same or opposite orientations. Moreover, the four end-points of the two arcs are connected by the other parts of the orbits. This can happen in two ways without intersection: either $a$ is connected to $d$ and $b$ is connected to $c$, or $a$ is connected to $b$ and $c$ is connected to $d$. Altogether we have four cases, as shown in Figure 7.\n\n\nFigure 7\n\nWe can see that the number of orbits is changed by -2 or +2 in the leftmost case when the arcs have the same orientation, $a$ is connected to $d$, and $b$ is connected to $c$. In the other three cases the number of orbits is not changed. Hence, Type-1 steps do not change the parity of the number of orbits.\n\nNow consider a Type-2 step. The three circles enclose a small, triangular region; by the step, this triangle is replaced by another triangle. Again, the modification of the orbits is done in some small neighbourhood; the structure does not change outside. Each side of the triangle shaped region can be convex or concave; the number of concave sides can be $0,1,2$ or 3 , so there are 4 possible arrangements of the orbits inside the neighbourhood, as shown in Figure 8.\n\n\n\n\n\nFigure 8\n\nDenote the points where the three circles enter or leave the neighbourhood by $a, b, c, d$, $e$, and $f$ in this order around the neighbourhood. As can be seen in Figure 8, there are only two essentially different cases; either $a, c, e$ are connected to $b, d, f$, respectively, or $a, c, e$ are connected to $f, b, d$, respectively. The step either preserves the set of connections or switches to the other arrangement. Obviously, in the earlier case the number of orbits is not changed; therefore we have to consider only the latter case.\n\nThe points $a, b, c, d, e$, and $f$ are connected by the orbits outside, without intersection. If $a$ was connected to $c$, say, then this orbit would isolate $b$, so this is impossible. Hence, each of $a, b, c, d, e$ and $f$ must be connected either to one of its neighbours or to the opposite point. If say $a$ is connected to $d$, then this orbit separates $b$ and $c$ from $e$ and $f$, therefore $b$ must be connected to $c$ and $e$ must be connected to $f$. Altogether there are only two cases and their reverses: either each point is connected to one of its neighbours or two opposite points are connected and the the remaining neigh boring pairs are connected to each other. See Figure 9.\n\n\nFigure 9\n\nWe can see that if only neighbouring points are connected, then the number of orbits is changed by +2 or -2 . If two opposite points are connected ( $a$ and $d$ in the figure), then the orbits are re-arranged, but their number is unchanged. Hence, Type-2 steps also preserve the parity. This completes the proof of $(*)$.' ""Like in the previous solutions, we do not need all circle pairs to intersect but we assume that the circles form a connected set. Denote by $\\mathcal{C}$ and $\\mathcal{P}$ the sets of circles and their intersection points, respectively.\n\nThe circles divide the plane into several simply connected, bounded regions and one unbounded region. Denote the set of these regions by $\\mathcal{R}$. We say that an intersection point or a region is odd or even if it is contained inside an odd or even number of circles, respectively. Let $\\mathcal{P}_{\\text {odd }}$ and $\\mathcal{R}_{\\text {odd }}$ be the sets of odd intersection points and odd regions, respectively.\n\nClaim.\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right|-\\left|\\mathcal{P}_{\\text {odd }}\\right| \\equiv n \\quad(\\bmod 2)\n\\tag{1}\n$$\n\nProof. For each circle $c \\in \\mathcal{C}$, denote by $R_{c}, P_{c}$, and $X_{c}$ the number of regions inside $c$, the number of intersection points inside $c$, and the number of circles intersecting $c$, respectively. The circles divide each other into several arcs; denote by $A_{c}$ the number of such arcs inside $c$. By double counting the regions and intersection points inside the circles we get\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right| \\equiv \\sum_{c \\in \\mathcal{C}} R_{c} \\quad(\\bmod 2) \\quad \\text { and } \\quad\\left|\\mathcal{P}_{\\text {odd }}\\right| \\equiv \\sum_{c \\in \\mathcal{C}} P_{c} \\quad(\\bmod 2)\n$$\n\n\n\nFor each circle $c$, apply EULER's polyhedron theorem to the (simply connected) regions in $c$. There are $2 X_{c}$ intersection points on $c$; they divide the circle into $2 X_{c}$ arcs. The polyhedron theorem yields $\\left(R_{c}+1\\right)+\\left(P_{c}+2 X_{c}\\right)=\\left(A_{c}+2 X_{c}\\right)+2$, considering the exterior of $c$ as a single region. Therefore,\n\n$$\nR_{c}+P_{c}=A_{c}+1 .\n\\tag{2}\n$$\n\nMoreover, we have four arcs starting from every interior points inside $c$ and a single arc starting into the interior from each intersection point on the circle. By double-counting the end-points of the interior arcs we get $2 A_{c}=4 P_{c}+2 X_{c}$, so\n\n$$\nA_{c}=2 P_{c}+X_{c} .\n\\tag{3}\n$$\n\nThe relations (2) and (3) together yield\n\n$$\nR_{c}-P_{c}=X_{c}+1\n\\tag{4}\n$$\n\nBy summing up (4) for all circles we obtain\n\n$$\n\\sum_{c \\in \\mathcal{C}} R_{c}-\\sum_{c \\in \\mathcal{C}} P_{c}=\\sum_{c \\in \\mathcal{C}} X_{c}+|\\mathcal{C}|\n$$\n\nwhich yields\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right|-\\left|\\mathcal{P}_{\\text {odd }}\\right| \\equiv \\sum_{c \\in \\mathcal{C}} X_{c}+n \\quad(\\bmod 2)\n\\tag{5}\n$$\n\nNotice that in $\\sum_{c \\in \\mathcal{C}} X_{c}$ each intersecting circle pair is counted twice, i.e., for both circles in the pair, so\n\n$$\n\\sum_{c \\in \\mathcal{C}} X_{c} \\equiv 0 \\quad(\\bmod 2)\n$$\n\nwhich finishes the proof of the Claim.\n\nNow insert the same small arcs at the intersections as in the first solution, and suppose that there is a single snail orbit $b$.\n\nFirst we show that the odd regions are inside the curve $b$, while the even regions are outside. Take a region $r \\in \\mathcal{R}$ and a point $x$ in its interior, and draw a ray $y, \\operatorname{starting}$ from $x$, that does not pass through any intersection point of the circles and is neither tangent to any of the circles. As is well-known, $x$ is inside the curve $b$ if and only if $y$ intersects $b$ an odd number of times (see Figure 10). Notice that if an arbitrary circle $c$ contains $x$ in its interior, then $c$ intersects $y$ at a single point; otherwise, if $x$ is outside $c$, then $c$ has 2 or 0 intersections with $y$. Therefore, $y$ intersects $b$ an odd number of times if and only if $x$ is contained in an odd number of circles, so if and only if $r$ is odd.\n\n\n\nFigure 10\n\nNow consider an intersection point $p$ of two circles $c_{1}$ and $c_{2}$ and a small neighbourhood around $p$. Suppose that $p$ is contained inside $k$ circles.\n\n\n\nWe have four regions that meet at $p$. Let $r_{1}$ be the region that lies outside both $c_{1}$ and $c_{2}$, let $r_{2}$ be the region that lies inside both $c_{1}$ and $c_{2}$, and let $r_{3}$ and $r_{4}$ be the two remaining regions, each lying inside exactly one of $c_{1}$ and $c_{2}$. The region $r_{1}$ is contained inside the same $k$ circles as $p$; the region $r_{2}$ is contained also by $c_{1}$ and $c_{2}$, so by $k+2$ circles in total; each of the regions $r_{3}$ and $r_{4}$ is contained inside $k+1$ circles. After the small arcs have been inserted at $p$, the regions $r_{1}$ and $r_{2}$ get connected, and the regions $r_{3}$ and $r_{4}$ remain separated at $p$ (see Figure 11). If $p$ is an odd point, then $r_{1}$ and $r_{2}$ are odd, so two odd regions are connected at $p$. Otherwise, if $p$ is even, then we have two even regions connected at $p$.\n\n\n\nFigure 11\n\n\n\nFigure 12\n\nConsider the system of odd regions and their connections at the odd points as a graph. In this graph the odd regions are the vertices, and each odd point establishes an edge that connects two vertices (see Figure 12). As $b$ is a single closed curve, this graph is connected and contains no cycle, so the graph is a tree. Then the number of vertices must be by one greater than the number of edges, so\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right|-\\left|\\mathcal{P}_{\\text {odd }}\\right|=1\n\\tag{6}\n$$\n\nThe relations (1) and (6) together prove that $n$ must be odd.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 244 The points $P$ and $Q$ are chosen on the side $B C$ of an acute-angled triangle $A B C$ so that $\angle P A B=\angle A C B$ and $\angle Q A C=\angle C B A$. The points $M$ and $N$ are taken on the rays $A P$ and $A Q$, respectively, so that $A P=P M$ and $A Q=Q N$. Prove that the lines $B M$ and $C N$ intersect on the circumcircle of the triangle $A B C$. "['Denote by $S$ the intersection point of the lines $B M$ and $C N$. Let moreover $\\beta=\\angle Q A C=\\angle C B A$ and $\\gamma=\\angle P A B=\\angle A C B$. From these equalities it follows that the triangles $A B P$ and $C A Q$ are similar (see Figure 1). Therefore we obtain\n\n$$\n\\frac{B P}{P M}=\\frac{B P}{P A}=\\frac{A Q}{Q C}=\\frac{N Q}{Q C}\n$$\n\nMoreover,\n\n$$\n\\angle B P M=\\beta+\\gamma=\\angle C Q N\n$$\n\nHence the triangles $B P M$ and $N Q C$ are similar. This gives $\\angle B M P=\\angle N C Q$, so the triangles $B P M$ and $B S C$ are also similar. Thus we get\n\n$$\n\\angle C S B=\\angle B P M=\\beta+\\gamma=180^{\\circ}-\\angle B A C\n$$\n\nwhich completes the solution.\n\n\n\nFigure 1\n\n\n\nFigure 2' ""As in the previous solution, denote by $S$ the intersection point of the lines $B M$ and $N C$. Let moreover the circumcircle of the triangle $A B C$ intersect the lines $A P$ and $A Q$ again at $K$ and $L$, respectively (see Figure 2).\n\nNote that $\\angle L B C=\\angle L A C=\\angle C B A$ and similarly $\\angle K C B=\\angle K A B=\\angle B C A$. It implies that the lines $B L$ and $C K$ meet at a point $X$, being symmetric to the point $A$ with respect to the line $B C$. Since $A P=P M$ and $A Q=Q N$, it follows that $X$ lies on the line $M N$. Therefore, using PASCAL's theorem for the hexagon $A L B S C K$, we infer that $S$ lies on the circumcircle of the triangle $A B C$, which finishes the proof.""]" [] Text-only Competition True Theorem proof Geometry Math English 245 Let $A B C$ be a triangle. The points $K, L$, and $M$ lie on the segments $B C, C A$, and $A B$, respectively, such that the lines $A K, B L$, and $C M$ intersect in a common point. Prove that it is possible to choose two of the triangles $A L M, B M K$, and $C K L$ whose inradii sum up to at least the inradius of the triangle $A B C$. "[""Denote\n\n$$\na=\\frac{B K}{K C}, \\quad b=\\frac{C L}{L A}, \\quad c=\\frac{A M}{M B}\n$$\n\nBy CEvA's theorem, $a b c=1$, so we may, without loss of generality, assume that $a \\geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than 1. Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first component not less than 1 and the second one not greater than 1 . Without loss of generality, assume that $1 \\leqslant a$ and $b \\leqslant 1$.\n\nTherefore, we obtain $b c \\leqslant 1$ and $1 \\leqslant c a$, or equivalently\n\n$$\n\\frac{A M}{M B} \\leqslant \\frac{L A}{C L} \\quad \\text { and } \\quad \\frac{M B}{A M} \\leqslant \\frac{B K}{K C}\n$$\n\nThe first inequality implies that the line passing through $M$ and parallel to $B C$ intersects the segment $A L$ at a point $X$ (see Figure 1). Therefore the inradius of the triangle $A L M$ is not less than the inradius $r_{1}$ of triangle $A M X$.\n\nSimilarly, the line passing through $M$ and parallel to $A C$ intersects the segment $B K$ at a point $Y$, so the inradius of the triangle $B M K$ is not less than the inradius $r_{2}$ of the triangle $B M Y$. Thus, to complete our solution, it is enough to show that $r_{1}+r_{2} \\geqslant r$, where $r$ is the inradius of the triangle $A B C$. We prove that in fact $r_{1}+r_{2}=r$.\n\n\n\nFigure 1\n\nSince $M X \\| B C$, the dilation with centre $A$ that takes $M$ to $B$ takes the incircle of the triangle $A M X$ to the incircle of the triangle $A B C$. Therefore\n\n$$\n\\frac{r_{1}}{r}=\\frac{A M}{A B}, \\quad \\text { and similarly } \\quad \\frac{r_{2}}{r}=\\frac{M B}{A B}\n$$\n\nAdding these equalities gives $r_{1}+r_{2}=r$, as required.""]" [] Text-only Competition True Theorem proof Geometry Math English 246 Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $A B C$ with $A B>B C$. The angle bisector of $\angle A B C$ intersects $\Omega$ at $M \neq B$. Let $\Gamma$ be the circle with diameter $B M$. The angle bisectors of $\angle A O B$ and $\angle B O C$ intersect $\Gamma$ at points $P$ and $Q$, respectively. The point $R$ is chosen on the line $P Q$ so that $B R=M R$. Prove that $B R \| A C$. (Here we always assume that an angle bisector is a ray.) ['Let $K$ be the midpoint of $B M$, i.e., the centre of $\\Gamma$. Notice that $A B \\neq B C$ implies $K \\neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.\n\nLet $N$ be the second point of intersection of $\\Gamma$ with the line $O M$. Since $B M$ is a diameter of $\\Gamma$, the lines $B N$ and $A C$ are both perpendicular to $O M$. Hence $B N \\| A C$, and it suffices to prove that $B N$ passes through $R$. Our plan for doing this is to interpret the lines $B N, O K$, and $P Q$ as the radical axes of three appropriate circles.\n\nLet $\\omega$ be the circle with diameter $B O$. Since $\\angle B N O=\\angle B K O=90^{\\circ}$, the points $N$ and $K$ lie on $\\omega$.\n\nNext we show that the points $O, K, P$, and $Q$ are concyclic. To this end, let $D$ and $E$ be the midpoints of $B C$ and $A B$, respectively. Clearly, $D$ and $E$ lie on the rays $O Q$ and $O P$, respectively. By our assumptions about the triangle $A B C$, the points $B, E, O, K$, and $D$ lie in this order on $\\omega$. It follows that $\\angle E O R=\\angle E B K=\\angle K B D=\\angle K O D$, so the line $K O$ externally bisects the angle $P O Q$. Since the point $K$ is the centre of $\\Gamma$, it also lies on the perpendicular bisector of $P Q$. So $K$ coincides with the midpoint of the arc $P O Q$ of the circumcircle $\\gamma$ of triangle $P O Q$.\n\nThus the lines $O K, B N$, and $P Q$ are pairwise radical axes of the circles $\\omega, \\gamma$, and $\\Gamma$. Hence they are concurrent at $R$, as required.\n\n'] [] Text-only Competition True Theorem proof Geometry Math English 247 Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circumcircles of the triangles $A M P$ and $B M C$. Prove that as $P$ varies, the point $Q$ lies on a fixed circle. "[""Throughout the solution, we denote by $\\sphericalangle(a, b)$ the directed angle between the lines $a$ and $b$.\n\nLet $D$ be the point on the segment $A B$ such that $B D=\\lambda \\cdot B A$. We will show that either $Q=D$, or $\\sphericalangle(D Q, Q B)=\\sphericalangle(A B, B C)$; this would mean that the point $Q$ varies over the constant circle through $D$ tangent to $B C$ at $B$, as required.\n\nDenote the circumcircles of the triangles $A M P$ and $B M C$ by $\\omega_{A}$ and $\\omega_{B}$, respectively. The lines $A P, B C$, and $M Q$ are pairwise radical axes of the circles $\\Gamma, \\omega_{A}$, and $\\omega_{B}$, thus either they are parallel, or they share a common point $X$.\n\nAssume that these lines are parallel (see Figure 1). Then the segments $A P, Q M$, and $B C$ have a common perpendicular bisector; the reflection in this bisector maps the segment $C P$ to $B A$, and maps $M$ to $Q$. Therefore, in this case $Q$ lies on $A B$, and $B Q / A B=C M / C P=$ $B D / A B$; so we have $Q=D$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nNow assume that the lines $A P, Q M$, and $B C$ are concurrent at some point $X$ (see Figure 2). Notice that the points $A, B, Q$, and $X$ lie on a common circle $\\Omega$ by MiQUEL's theorem applied to the triangle $X P C$. Let us denote by $Y$ the symmetric image of $X$ about the perpendicular bisector of $A B$. Clearly, $Y$ lies on $\\Omega$, and the triangles $Y A B$ and $\\triangle X B A$ are congruent. Moreover, the triangle $X P C$ is similar to the triangle $X B A$, so it is also similar to the triangle $Y A B$.\n\nNext, the points $D$ and $M$ correspond to each other in similar triangles $Y A B$ and $X P C$, since $B D / B A=C M / C P=\\lambda$. Moreover, the triangles $Y A B$ and $X P C$ are equi-oriented, so $\\sphericalangle(M X, X P)=\\sphericalangle(D Y, Y A)$. On the other hand, since the points $A, Q, X$, and $Y$ lie on $\\Omega$, we have $\\sphericalangle(Q Y, Y A)=\\sphericalangle(M X, X P)$. Therefore, $\\sphericalangle(Q Y, Y A)=\\sphericalangle(D Y, Y A)$, so the points $Y, D$, and $Q$ are collinear.\n\nFinally, we have $\\sphericalangle(D Q, Q B)=\\sphericalangle(Y Q, Q B)=\\sphericalangle(Y A, A B)=\\sphericalangle(A B, B X)=\\sphericalangle(A B, B C)$, as desired."" 'As in the previous solution, we introduce the radical centre $X=A P \\cap B C \\cap M Q$ of the circles $\\omega_{A}, \\omega_{B}$, and $\\Gamma$. Next, we also notice that the points $A, Q, B$, and $X$ lie on a common circle $\\Omega$.\n\nIf the point $P$ lies on the $\\operatorname{arc} B A C$ of $\\Gamma$, then the point $X$ is outside $\\Gamma$, thus the point $Q$ belongs to the ray $X M$, and therefore the points $P, A$, and $Q$ lie on the same side of $B C$. Otherwise, if $P$ lies on the $\\operatorname{arc} B C$ not containing $A$, then $X$ lies inside $\\Gamma$, so $M$ and $Q$ lie on different sides of $B C$; thus again $Q$ and $A$ lie on the same side of $B C$. So, in each case the points $Q$ and $A$ lie on the same side of $B C$.\n\n\n\nFigure 3\n\nNow we prove that the ratio\n\n$$\n\\frac{Q B}{\\sin \\angle Q B C}=\\frac{Q B}{Q X} \\cdot \\frac{Q X}{\\sin \\angle Q B X}\n$$\n\nis constant. Since the points $A, Q, B$, and $X$ are concyclic, we have\n\n$$\n\\frac{Q X}{\\sin \\angle Q B X}=\\frac{A X}{\\sin \\angle A B C}\n$$\n\nNext, since the points $B, Q, M$, and $C$ are concyclic, the triangles $X B Q$ and $X M C$ are similar, so\n\n$$\n\\frac{Q B}{Q X}=\\frac{C M}{C X}=\\lambda \\cdot \\frac{C P}{C X}\n$$\n\nAnalogously, the triangles $X C P$ and $X A B$ are also similar, so\n\n$$\n\\frac{C P}{C X}=\\frac{A B}{A X}\n$$\n\nTherefore, we obtain\n\n$$\n\\frac{Q B}{\\sin \\angle Q B C}=\\lambda \\cdot \\frac{A B}{A X} \\cdot \\frac{A X}{\\sin \\angle A B C}=\\lambda \\cdot \\frac{A B}{\\sin \\angle A B C}\n$$\n\nso this ratio is indeed constant. Thus the circle passing through $Q$ and tangent to $B C$ at $B$ is also constant, and $Q$ varies over this fixed circle.' ""Let us perform an inversion centred at $C$. Denote by $X^{\\prime}$ the image of a point $X$ under this inversion.\n\nThe circle $\\Gamma$ maps to the line $\\Gamma^{\\prime}$ passing through the constant points $A^{\\prime}$ and $B^{\\prime}$, and containing the variable point $P^{\\prime}$. By the problem condition, the point $M$ varies over the circle $\\gamma$ which is the homothetic image of $\\Gamma$ with centre $C$ and coefficient $\\lambda$. Thus $M^{\\prime}$ varies over the constant line $\\gamma^{\\prime} \\| A^{\\prime} B^{\\prime}$ which is the homothetic image of $A^{\\prime} B^{\\prime}$ with centre $C$ and coefficient $1 / \\lambda$, and $M=\\gamma^{\\prime} \\cap C P^{\\prime}$. Next, the circumcircles $\\omega_{A}$ and $\\omega_{B}$ of the triangles $A M P$ and $B M C$ map to the circumcircle $\\omega_{A}^{\\prime}$ of the triangle $A^{\\prime} M^{\\prime} P^{\\prime}$ and to the line $B^{\\prime} M^{\\prime}$, respectively; the point $Q$ thus maps to the second point of intersection of $B^{\\prime} M^{\\prime}$ with $\\omega_{A}^{\\prime}$ (see Figure 4).\n\n\n\nFigure 4\n\nLet $J$ be the (constant) common point of the lines $\\gamma^{\\prime}$ and $C A^{\\prime}$, and let $\\ell$ be the (constant) line through $J$ parallel to $C B^{\\prime}$. Let $V$ be the common point of the lines $\\ell$ and $B^{\\prime} M^{\\prime}$. Applying PAppus' theorem to the triples $\\left(C, J, A^{\\prime}\\right)$ and $\\left(V, B^{\\prime}, M^{\\prime}\\right)$ we get that the points $C B^{\\prime} \\cap J V$, $J M^{\\prime} \\cap A^{\\prime} B^{\\prime}$, and $C M^{\\prime} \\cap A^{\\prime} V$ are collinear. The first two of these points are ideal, hence so is the third, which means that $C M^{\\prime} \\| A^{\\prime} V$.\n\nNow we have $\\sphericalangle\\left(Q^{\\prime} A^{\\prime}, A^{\\prime} P^{\\prime}\\right)=\\sphericalangle\\left(Q^{\\prime} M^{\\prime}, M^{\\prime} P^{\\prime}\\right)=\\angle\\left(V M^{\\prime}, A^{\\prime} V\\right)$, which means that the triangles $B^{\\prime} Q^{\\prime} A^{\\prime}$ and $B^{\\prime} A^{\\prime} V$ are similar, and $\\left(B^{\\prime} A^{\\prime}\\right)^{2}=B^{\\prime} Q^{\\prime} \\cdot B^{\\prime} V$. Thus $Q^{\\prime}$ is the image of $V$ under the second (fixed) inversion with centre $B^{\\prime}$ and radius $B^{\\prime} A^{\\prime}$. Since $V$ varies over the constant line $\\ell, Q^{\\prime}$ varies over some constant circle $\\Theta$. Thus, applying the first inversion back we get that $Q$ also varies over some fixed circle.\n\nOne should notice that this last circle is not a line; otherwise $\\Theta$ would contain $C$, and thus $\\ell$ would contain the image of $C$ under the second inversion. This is impossible, since $C B^{\\prime} \\| \\ell$.""]" [] Text-only Competition True Theorem proof Geometry Math English 248 "Let $A B C D$ be a convex quadrilateral with $\angle B=\angle D=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $B D$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle $S C T$ and $$ \angle S H C-\angle B S C=90^{\circ}, \quad \angle T H C-\angle D T C=90^{\circ} . $$ Prove that the circumcircle of triangle $S H T$ is tangent to the line $B D$." ['Let the line passing through $C$ and perpendicular to the line $S C$ intersect the line $A B$ at $Q$ (see Figure 1). Then\n\n$$\n\\angle S Q C=90^{\\circ}-\\angle B S C=180^{\\circ}-\\angle S H C\n$$\n\nwhich implies that the points $C, H, S$, and $Q$ lie on a common circle. Moreover, since $S Q$ is a diameter of this circle, we infer that the circumcentre $K$ of triangle $S H C$ lies on the line $A B$. Similarly, we prove that the circumcentre $L$ of triangle $C H T$ lies on the line $A D$.\n\n\n\nFigure 1\n\nIn order to prove that the circumcircle of triangle $S H T$ is tangent to $B D$, it suffices to show that the perpendicular bisectors of $H S$ and $H T$ intersect on the line $A H$. However, these two perpendicular bisectors coincide with the angle bisectors of angles $A K H$ and $A L H$. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that\n\n$$\n\\frac{A K}{K H}=\\frac{A L}{L H}\\tag{1}\n$$\n\nWe present two proofs of this equality.\n\nFirst proof. Let the lines $K L$ and $H C$ intersect at $M$ (see Figure 2). Since $K H=K C$ and $L H=L C$, the points $H$ and $C$ are symmetric to each other with respect to the line $K L$. Therefore $M$ is the midpoint of $H C$. Denote by $O$ the circumcentre of quadrilateral $A B C D$. Then $O$ is the midpoint of $A C$. Therefore we have $O M \\| A H$ and hence $O M \\perp B D$. This together with the equality $O B=O D$ implies that $O M$ is the perpendicular bisector of $B D$ and therefore $B M=D M$.\n\nSince $C M \\perp K L$, the points $B, C, M$, and $K$ lie on a common circle with diameter $K C$. Similarly, the points $L, C, M$, and $D$ lie on a circle with diameter $L C$. Thus, using the sine law, we obtain\n\n$$\n\\frac{A K}{A L}=\\frac{\\sin \\angle A L K}{\\sin \\angle A K L}=\\frac{D M}{C L} \\cdot \\frac{C K}{B M}=\\frac{C K}{C L}=\\frac{K H}{L H}\n$$\n\n\n\nwhich finishes the proof of (1).\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nSecond proof. If the points $A, H$, and $C$ are collinear, then $A K=A L$ and $K H=L H$, so the equality (1) follows. Assume therefore that the points $A, H$, and $C$ do not lie in a line and consider the circle $\\omega$ passing through them (see Figure 3). Since the quadrilateral $A B C D$ is cyclic,\n\n$$\n\\angle B A C=\\angle B D C=90^{\\circ}-\\angle A D H=\\angle H A D .\n$$\n\nLet $N \\neq A$ be the intersection point of the circle $\\omega$ and the angle bisector of $\\angle C A H$. Then $A N$ is also the angle bisector of $\\angle B A D$. Since $H$ and $C$ are symmetric to each other with respect to the line $K L$ and $H N=N C$, it follows that both $N$ and the centre of $\\omega$ lie on the line $K L$. This means that the circle $\\omega$ is an Apollonius circle of the points $K$ and $L$. This immediately yields (1).'] [] Text-only Competition True Theorem proof Geometry Math English 249 "Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ at $S$ and $T$, respectively. We call the pair $(E, F)$ interesting, if the quadrilateral $K S A T$ is cyclic. Suppose that the pairs $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ are interesting. Prove that $$ \frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C} $$" "[""For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting.\n\nLet $E F K$ be an interesting triangle. Firstly, we prove that $\\angle K E F=\\angle K F E=\\angle A$, which also means that the circumcircle $\\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $K F$.\n\nDenote by $\\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $A M$ intersect the line $S T$ and the circle $\\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1).\n\nSince $E F \\| T S$ and $M$ is the midpoint of $E F, N$ is the midpoint of $S T$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $S T$, we have $\\angle K N S=\\angle M N S=$ $\\angle L N T$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $S T$. Therefore $K L \\| S T$.\n\nLet $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $E F$, and we may assume that it lies on the ray $M F$. One has\n\n$$\n\\angle K G E=\\angle K N S=\\angle S N M=\\angle K L A=180^{\\circ}-\\angle K S A\n$$\n\n(if $K=L$, then the angle $K L A$ is understood to be the angle between $A L$ and the tangent to $\\omega$ at $L$ ). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $K S G T$ is a parallelogram, we obtain $\\angle K E F=\\angle K S G=180^{\\circ}-\\angle T K S=\\angle A$. Since $K E=K F$, we also have $\\angle K F E=\\angle K E F=\\angle A$.\n\nAfter having proved this fact, one may finish the solution by different methods.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nFirst method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.\n\n\n\nLemma. Let $A B C$ be an arbitrary triangle. Choose two points $E_{1}$ and $E_{2}$ on the side $A C$, two points $F_{1}$ and $F_{2}$ on the side $A B$, and two points $K_{1}$ and $K_{2}$ on the side $B C$, in a way that the triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ are similar. Then the six circumcircles of the triangles $A E_{i} F_{i}$, $B F_{i} K_{i}$, and $C E_{i} K_{i}(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_{1} F_{1} K_{1}$ to the triangle $E_{2} F_{2} K_{2}$.\n\nProof. Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $A E_{i} F_{i}, B F_{i} K_{i}$, and $C K_{i} E_{i}$ have a common point $Z_{i}$ by MiqueL's theorem. Moreover, we have\n\n$\\sphericalangle\\left(Z_{i} F_{i}, Z_{i} E_{i}\\right)=\\sphericalangle(A B, C A), \\quad \\sphericalangle\\left(Z_{i} K_{i}, Z_{i} F_{i}\\right)=\\sphericalangle(B C, A B), \\quad \\sphericalangle\\left(Z_{i} E_{i}, Z_{i} K_{i}\\right)=\\sphericalangle(C A, B C)$.\n\nThis yields that the points $Z_{1}$ and $Z_{2}$ correspond to each other in similar triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.\n\nFinally, in order to show that $Z_{1}=Z_{2}$, one may notice that $\\sphericalangle\\left(A B, A Z_{1}\\right)=\\sphericalangle\\left(E_{1} F_{1}, E_{1} Z_{1}\\right)=$ $\\sphericalangle\\left(E_{2} F_{2}, E_{2} Z_{2}\\right)=\\sphericalangle\\left(A B, A Z_{2}\\right)$ (see Figure 2$)$. Similarly, one has $\\sphericalangle\\left(B C, B Z_{1}\\right)=\\sphericalangle\\left(B C, B Z_{2}\\right)$ and $\\sphericalangle\\left(C A, C Z_{1}\\right)=\\sphericalangle\\left(C A, C Z_{2}\\right)$. This yields $Z_{1}=Z_{2}$.\n\nNow, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively, and let $R$ be the midpoint of $B C$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $B C P Q$. Thus we obtain $\\angle A P Q=\\angle B$ and $\\angle R P C=\\angle C$, which yields $\\angle Q P R=\\angle A$. Similarly, we show that $\\angle P Q R=\\angle A$. Thus, all interesting triangles are similar to the triangle $P Q R$.\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nDenote now by $Z$ the common point of the circumcircles of $A P Q, B Q R$, and $C P R$. Let $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ be two interesting triangles. By the lemma, $Z$ is the centre of any spiral similarity taking one of the triangles $E_{1} F_{1} K_{1}, E_{2} F_{2} K_{2}$, and $P Q R$ to some other of them. Therefore the triangles $Z E_{1} E_{2}$ and $Z F_{1} F_{2}$ are similar, as well as the triangles $Z E_{1} F_{1}$ and $Z P Q$. Hence\n\n$$\n\\frac{E_{1} E_{2}}{F_{1} F_{2}}=\\frac{Z E_{1}}{Z F_{1}}=\\frac{Z P}{Z Q}\n$$\n\nMoreover, the equalities $\\angle A Z Q=\\angle A P Q=\\angle A B C=180^{\\circ}-\\angle Q Z R$ show that the point $Z$ lies on the line $A R$ (see Figure 4). Therefore the triangles $A Z P$ and $A C R$ are similar, as well as the triangles $A Z Q$ and $A B R$. This yields\n\n$$\n\\frac{Z P}{Z Q}=\\frac{Z P}{R C} \\cdot \\frac{R B}{Z Q}=\\frac{A Z}{A C} \\cdot \\frac{A B}{A Z}=\\frac{A B}{A C},\n$$\n\nwhich completes the solution.\n\n\n\nSecond method. Now we will start from the fact that $\\omega_{1}$ is tangent to the lines $K E$ and $K F$ (see Figure 5$)$. We prove that if $(E, F)$ is an interesting pair, then\n\n$$\n\\frac{A E}{A B}+\\frac{A F}{A C}=2 \\cos \\angle A\n\\tag{1}\n$$\n\nLet $Y$ be the intersection point of the segments $B E$ and $C F$. The points $B, K$, and $C$ are collinear, hence applying PASCAL's theorem to the degenerated hexagon $A F F Y E E$, we infer that $Y$ lies on the circle $\\omega_{1}$.\n\nDenote by $Z$ the second intersection point of the circumcircle of the triangle $B F Y$ with the line $B C$ (see Figure 6). By Miquel's theorem, the points $C, Z, Y$, and $E$ are concyclic. Therefore we obtain\n\n$$\nB F \\cdot A B+C E \\cdot A C=B Y \\cdot B E+C Y \\cdot C F=B Z \\cdot B C+C Z \\cdot B C=B C^{2}\n$$\n\nOn the other hand, $B C^{2}=A B^{2}+A C^{2}-2 A B \\cdot A C \\cos \\angle A$, by the cosine law. Hence\n\n$$\n(A B-A F) \\cdot A B+(A C-A E) \\cdot A C=A B^{2}+A C^{2}-2 A B \\cdot A C \\cos \\angle A\n$$\n\nwhich simplifies to the desired equality (1).\n\nLet now $\\left(E_{1}, F_{1}\\right)$ and $\\left(E_{2}, F_{2}\\right)$ be two interesting pairs of points. Then we get\n\n$$\n\\frac{A E_{1}}{A B}+\\frac{A F_{1}}{A C}=\\frac{A E_{2}}{A B}+\\frac{A F_{2}}{A C}\n$$\n\nwhich gives the desired result.\n\n\n\nFigure 5\n\n\n\nFigure 6\n\nThird method. Again, we make use of the fact that all interesting triangles are similar (and equi-oriented). Let us put the picture onto a complex plane such that $A$ is at the origin, and identify each point with the corresponding complex number.\n\nLet $E F K$ be any interesting triangle. The equalities $\\angle K E F=\\angle K F E=\\angle A$ yield that the ratio $\\nu=\\frac{K-E}{F-E}$ is the same for all interesting triangles. This in turn means that the numbers $E$, $F$, and $K$ satisfy the linear equation\n\n$$\nK=\\mu E+\\nu F, \\quad \\text { where } \\quad \\mu=1-\\nu\n\\tag{2}\n$$\n\n\n\nNow let us choose the points $X$ and $Y$ on the rays $A B$ and $A C$, respectively, so that $\\angle C X A=\\angle A Y B=\\angle A=\\angle K E F$ (see Figure 7). Then each of the triangles $A X C$ and $Y A B$ is similar to any interesting triangle, which also means that\n\n$$\nC=\\mu A+\\nu X=\\nu X \\quad \\text { and } \\quad B=\\mu Y+\\nu A=\\mu Y\n\\tag{3}\n$$\n\nMoreover, one has $X / Y=\\overline{C / B}$.\n\nSince the points $E, F$, and $K$ lie on $A C, A B$, and $B C$, respectively, one gets\n\n$$\nE=\\rho Y, \\quad F=\\sigma X, \\quad \\text { and } \\quad K=\\lambda B+(1-\\lambda) C\n$$\n\nfor some real $\\rho, \\sigma$, and $\\lambda$. In view of (3), the equation (2) now reads $\\lambda B+(1-\\lambda) C=K=$ $\\mu E+\\nu F=\\rho B+\\sigma C$, or\n\n$$\n(\\lambda-\\rho) B=(\\sigma+\\lambda-1) C .\n$$\n\nSince the nonzero complex numbers $B$ and $C$ have different arguments, the coefficients in the brackets vanish, so $\\rho=\\lambda$ and $\\sigma=1-\\lambda$. Therefore,\n\n$$\n\\frac{E}{Y}+\\frac{F}{X}=\\rho+\\sigma=1\n\\tag{4}\n$$\n\nNow, if $\\left(E_{1}, F_{1}\\right)$ and $\\left(E_{2}, F_{2}\\right)$ are two distinct interesting pairs, one may apply (4) to both pairs. Subtracting, we get\n\n$$\n\\frac{E_{1}-E_{2}}{Y}=\\frac{F_{2}-F_{1}}{X}, \\quad \\text { so } \\quad \\frac{E_{1}-E_{2}}{F_{2}-F_{1}}=\\frac{Y}{X}=\\frac{\\bar{B}}{\\bar{C}}\n$$\n\nTaking absolute values provides the required result.\n\n\n\nFigure 7"" ""Let $(E, F)$ be an interesting pair. This time we prove that\n\n$$\n\\frac{A M}{A K}=\\cos \\angle A\n\\tag{5}\n$$\n\nWe introduce the circle $\\omega$ passing through the points $K, S$, $A$, and $T$, together with the points $N$ and $L$ at which the line $A M$ intersect the line $S T$ and the circle $\\omega$ for the second time, respectively. Let moreover $O$ be the centre of $\\omega$ (see Figures 8 and 9). we note that $N$ is the midpoint of $S T$ and show that $K L \\| S T$, which implies $\\angle F A M=\\angle E A K$.\n\n\n\nFigure 8\n\n\n\nFigure 9\n\nSuppose now that $K \\neq L$ (see Figure 8). Then $K L \\| S T$, and consequently the lines $K M$ and $K L$ are perpendicular. It implies that the lines $L O$ and $K M$ meet at a point $X$ lying on the circle $\\omega$. Since the lines $O N$ and $X M$ are both perpendicular to the line $S T$, they are parallel to each other, and hence $\\angle L O N=\\angle L X K=\\angle M A K$. On the other hand, $\\angle O L N=\\angle M K A$, so we infer that triangles $N O L$ and $M A K$ are similar. This yields\n\n$$\n\\frac{A M}{A K}=\\frac{O N}{O L}=\\frac{O N}{O T}=\\cos \\angle T O N=\\cos \\angle A\n$$\n\nIf, on the other hand, $K=L$, then the points $A, M, N$, and $K$ lie on a common line, and this line is the perpendicular bisector of $S T$ (see Figure 9). This implies that $A K$ is a diameter of $\\omega$, which yields $A M=2 O K-2 N K=2 O N$. So also in this case we obtain\n\n$$\n\\frac{A M}{A K}=\\frac{2 O N}{2 O T}=\\cos \\angle T O N=\\cos \\angle A\n$$\n\nThus (5) is proved.\n\nLet $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively (see Figure 10). We claim that the point $M$ lies on the line $P Q$. Consider now the composition of the dilatation with factor $\\cos \\angle A$ and centre $A$, and the reflection with respect to the angle bisector of $\\angle B A C$. This transformation is a similarity that takes $B, C$, and $K$ to $P, Q$, and $M$, respectively. Since $K$ lies on the line $B C$, the point $M$ lies on the line $P Q$.\n\n\n\n\n\nFigure 10\n\nSuppose that $E \\neq P$. Then also $F \\neq Q$, and by Menelaus' theorem, we obtain\n\n$$\n\\frac{A Q}{F Q} \\cdot \\frac{F M}{E M} \\cdot \\frac{E P}{A P}=1\n$$\n\nUsing the similarity of the triangles $A P Q$ and $A B C$, we infer that\n\n$$\n\\frac{E P}{F Q}=\\frac{A P}{A Q}=\\frac{A B}{A C}, \\quad \\text { and hence } \\quad \\frac{E P}{A B}=\\frac{F Q}{A C}\n$$\n\nThe last equality holds obviously also in case $E=P$, because then $F=Q$. Moreover, since the line $P Q$ intersects the segment $E F$, we infer that the point $E$ lies on the segment $A P$ if and only if the point $F$ lies outside of the segment $A Q$.\n\nLet now $\\left(E_{1}, F_{1}\\right)$ and $\\left(E_{2}, F_{2}\\right)$ be two interesting pairs. Then we obtain\n\n$$\n\\frac{E_{1} P}{A B}=\\frac{F_{1} Q}{A C} \\quad \\text { and } \\quad \\frac{E_{2} P}{A B}=\\frac{F_{2} Q}{A C}\n$$\n\nIf $P$ lies between the points $E_{1}$ and $E_{2}$, we add the equalities above, otherwise we subtract them. In any case we obtain\n\n$$\n\\frac{E_{1} E_{2}}{A B}=\\frac{F_{1} F_{2}}{A C}\n$$\n\nwhich completes the solution.""]" [] Text-only Competition True Theorem proof Geometry Math English 250 Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ intersect $A V$ at $X$, and let the line passing through $V$ and parallel to $A I$ intersect $A B$ at $Y$. Let $W$ and $Z$ be the midpoints of $A X$ and $B C$, respectively. Prove that if the points $I, X$, and $Y$ are collinear, then the points $I, W$, and $Z$ are also collinear. "['We start with some general observations. Set $\\alpha=\\angle A / 2, \\beta=\\angle B / 2, \\gamma=\\angle C / 2$. Then obviously $\\alpha+\\beta+\\gamma=90^{\\circ}$. Since $\\angle U I C=90^{\\circ}$, we obtain $\\angle I U C=\\alpha+\\beta$. Therefore $\\angle B I V=\\angle I U C-\\angle I B C=\\alpha=\\angle B A I=\\angle B Y V$, which implies that the points $B, Y, I$, and $V$ lie on a common circle (see Figure 1).\n\nAssume now that the points $I, X$ and $Y$ are collinear. We prove that $\\angle Y I A=90^{\\circ}$.\n\nLet the line $X U$ intersect $A B$ at $N$. Since the lines $A I, U X$, and $V Y$ are parallel, we get\n\n$$\n\\frac{N X}{A I}=\\frac{Y N}{Y A}=\\frac{V U}{V I}=\\frac{X U}{A I}\n$$\n\nimplying $N X=X U$. Moreover, $\\angle B I U=\\alpha=\\angle B N U$. This implies that the quadrilateral $B U I N$ is cyclic, and since $B I$ is the angle bisector of $\\angle U B N$, we infer that $N I=U I$. Thus in the isosceles triangle $N I U$, the point $X$ is the midpoint of the base $N U$. This gives $\\angle I X N=90^{\\circ}$, i.e., $\\angle Y I A=90^{\\circ}$.\n\n\n\nFigure 1\n\nLet $S$ be the midpoint of the segment $V C$. Let moreover $T$ be the intersection point of the lines $A X$ and $S I$, and set $x=\\angle B A V=\\angle B C V$. Since $\\angle C I A=90^{\\circ}+\\beta$ and $S I=S C$, we obtain\n\n$$\n\\angle T I A=180^{\\circ}-\\angle A I S=90^{\\circ}-\\beta-\\angle C I S=90^{\\circ}-\\beta-\\gamma-x=\\alpha-x=\\angle T A I,\n$$\n\nwhich implies that $T I=T A$. Therefore, since $\\angle X I A=90^{\\circ}$, the point $T$ is the midpoint of $A X$, i.e., $T=W$.\n\nTo complete our solution, it remains to show that the intersection point of the lines $I S$ and $B C$ coincide with the midpoint of the segment $B C$. But since $S$ is the midpoint of the segment $V C$, it suffices to show that the lines $B V$ and $I S$ are parallel.\n\n\n\nSince the quadrilateral $B Y I V$ is cyclic, $\\angle V B I=\\angle V Y I=\\angle Y I A=90^{\\circ}$. This implies that $B V$ is the external angle bisector of the angle $A B C$, which yields $\\angle V A C=\\angle V C A$. Therefore $2 \\alpha-x=2 \\gamma+x$, which gives $\\alpha=\\gamma+x$. Hence $\\angle S C I=\\alpha$, so $\\angle V S I=2 \\alpha$.\n\nOn the other hand, $\\angle B V C=180^{\\circ}-\\angle B A C=180^{\\circ}-2 \\alpha$, which implies that the lines $B V$ and $I S$ are parallel. This completes the solution.' 'As in' '1, we first prove that the points $B, Y, I, V$ lie on a common circle and $\\angle Y I A=90^{\\circ}$. The remaining part of the solution is based on the following lemma, which holds true for any triangle $A B C$, not necessarily with the property that $I, X, Y$ are collinear.\n\nLemma. Let $A B C$ be the triangle inscribed in a circle $\\Gamma$ and let $I$ be its incentre. Assume that the line passing through $I$ and perpendicular to the line $A I$ intersects the side $A B$ at the point $Y$. Let the circumcircle of the triangle $B Y I$ intersect the circle $\\Gamma$ for the second time at $V$, and let the excircle of the triangle $A B C$ opposite to the vertex $A$ be tangent to the side $B C$ at $E$. Then\n\n$$\n\\angle B A V=\\angle C A E .\n$$\n\nProof. Let $\\rho$ be the composition of the inversion with centre $A$ and radius $\\sqrt{A B \\cdot A C}$, and the symmetry with respect to $A I$. Clearly, $\\rho$ interchanges $B$ and $C$.\n\nLet $J$ be the excentre of the triangle $A B C$ opposite to $A$ (see Figure 2). Then we have $\\angle J A C=\\angle B A I$ and $\\angle J C A=90^{\\circ}+\\gamma=\\angle B I A$, so the triangles $A C J$ and $A I B$ are similar, and therefore $A B \\cdot A C=A I \\cdot A J$. This means that $\\rho$ interchanges $I$ and $J$. Moreover, since $Y$ lies on $A B$ and $\\angle A I Y=90^{\\circ}$, the point $Y^{\\prime}=\\rho(Y)$ lies on $A C$, and $\\angle J Y^{\\prime} A=90^{\\circ}$. Thus $\\rho$ maps the circumcircle $\\gamma$ of the triangle $B Y I$ to a circle $\\gamma^{\\prime}$ with diameter $J C$.\n\nFinally, since $V$ lies on both $\\Gamma$ and $\\gamma$, the point $V^{\\prime}=\\rho(V)$ lies on the line $\\rho(\\Gamma)=A B$ as well as on $\\gamma^{\\prime}$, which in turn means that $V^{\\prime}=E$. This implies the desired result.\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nNow we turn to the solution of the problem.\n\nAssume that the incircle $\\omega_{1}$ of the triangle $A B C$ is tangent to $B C$ at $D$, and let the excircle $\\omega_{2}$ of the triangle $A B C$ opposite to the vertex $A$ touch the side $B C$ at $E$ (see Figure 3 ). The homothety with centre $A$ that takes $\\omega_{2}$ to $\\omega_{1}$ takes the point $E$ to some point $F$, and the\n\n\n\ntangent to $\\omega_{1}$ at $F$ is parallel to $B C$. Therefore $D F$ is a diameter of $\\omega_{1}$. Moreover, $Z$ is the midpoint of $D E$. This implies that the lines $I Z$ and $F E$ are parallel.\n\nLet $K=Y I \\cap A E$. Since $\\angle Y I A=90^{\\circ}$, the lemma yields that $I$ is the midpoint of $X K$. This implies that the segments $I W$ and $A K$ are parallel. Therefore, the points $W, I$ and $Z$ are collinear.']" [] Text-only Competition True Theorem proof Geometry Math English 251 A coin is called a Cape Town coin if its value is $1 / n$ for some positive integer $n$. Given a collection of Cape Town coins of total value at most $99+\frac{1}{2}$, prove that it is possible to split this collection into at most 100 groups each of total value at most 1 . "['We will show that for every positive integer $N$ any collection of Cape Town coins of total value at most $N-\\frac{1}{2}$ can be split into $N$ groups each of total value at most 1 . The problem statement is a particular case for $N=100$.\n\nWe start with some preparations. If several given coins together have a total value also of the form $\\frac{1}{k}$ for a positive integer $k$, then we may merge them into one new coin. Clearly, if the resulting collection can be split in the required way then the initial collection can also be split.\n\nAfter each such merging, the total number of coins decreases, thus at some moment we come to a situation when no more merging is possible. At this moment, for every even $k$ there is at most one coin of value $\\frac{1}{k}$ (otherwise two such coins may be merged), and for every odd $k>1$ there are at most $k-1$ coins of value $\\frac{1}{k}$ (otherwise $k$ such coins may also be merged).\n\nNow, clearly, each coin of value 1 should form a single group; if there are $d$ such coins then we may remove them from the collection and replace $N$ by $N-d$. So from now on we may assume that there are no coins of value 1 .\n\nFinally, we may split all the coins in the following way. For each $k=1,2, \\ldots, N$ we put all the coins of values $\\frac{1}{2 k-1}$ and $\\frac{1}{2 k}$ into a group $G_{k}$; the total value of $G_{k}$ does not exceed\n\n$$\n(2 k-2) \\cdot \\frac{1}{2 k-1}+\\frac{1}{2 k}<1\n$$\n\nIt remains to distribute the ""small"" coins of values which are less than $\\frac{1}{2 N}$; we will add them one by one. In each step, take any remaining small coin. The total value of coins in the groups at this moment is at most $N-\\frac{1}{2}$, so there exists a group of total value at most $\\frac{1}{N}\\left(N-\\frac{1}{2}\\right)=1-\\frac{1}{2 N}$; thus it is possible to put our small coin into this group. Acting so, we will finally distribute all the coins.']" [] Text-only Competition True Theorem proof Number Theory Math English 252 "Let $n>1$ be a given integer. Prove that infinitely many terms of the sequence $\left(a_{k}\right)_{k \geqslant 1}$, defined by $$ a_{k}=\left\lfloor\frac{n^{k}}{k}\right\rfloor $$ are odd. (For a real number $x,\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.)" "[""If $n$ is odd, let $k=n^{m}$ for $m=1,2, \\ldots$ Then $a_{k}=n^{n^{m}-m}$, which is odd for each $m$.\n\nHenceforth, assume that $n$ is even, say $n=2 t$ for some integer $t \\geqslant 1$. Then, for any $m \\geqslant 2$, the integer $n^{2^{m}}-2^{m}=2^{m}\\left(2^{2^{m}-m} \\cdot t^{2^{m}}-1\\right)$ has an odd prime divisor $p$, since $2^{m}-m>1$. Then, for $k=p \\cdot 2^{m}$, we have\n\n$$\nn^{k}=\\left(n^{2^{m}}\\right)^{p} \\equiv\\left(2^{m}\\right)^{p}=\\left(2^{p}\\right)^{m} \\equiv 2^{m}\n$$\n\nwhere the congruences are taken modulo $p\\left(\\right.$ recall that $2^{p} \\equiv 2(\\bmod p)$, by FERMAT's little theorem). Also, from $n^{k}-2^{m}m$ ). Note that for different values of $m$, we get different values of $k$, due to the different powers of 2 in the prime factorisation of $k$."" 'Treat the (trivial) case when $n$ is odd.\nNow assume that $n$ is even and $n>2$. Let $p$ be a prime divisor of $n-1$.\n\nProceed by induction on $i$ to prove that $p^{i+1}$ is a divisor of $n^{p^{i}}-1$ for every $i \\geqslant 0$. The case $i=0$ is true by the way in which $p$ is chosen. Suppose the result is true for some $i \\geqslant 0$. The factorisation\n\n$$\nn^{p^{i+1}}-1=\\left(n^{p^{i}}-1\\right)\\left[n^{p^{i}(p-1)}+n^{p^{i}(p-2)}+\\cdots+n^{p^{i}}+1\\right]\n$$\n\ntogether with the fact that each of the $p$ terms between the square brackets is congruent to 1 modulo $p$, implies that the result is also true for $i+1$.\n\nHence $\\left\\lfloor\\frac{n^{p^{i}}}{p^{i}}\\right\\rfloor=\\frac{n^{p^{i}}-1}{p^{i}}$, an odd integer for each $i \\geqslant 1$.\n\nFinally, we consider the case $n=2$. We observe that $3 \\cdot 4^{i}$ is a divisor of $2^{3 \\cdot 4^{i}}-4^{i}$ for every $i \\geqslant 1$ : Trivially, $4^{i}$ is a divisor of $2^{3 \\cdot 4^{i}}-4^{i}$, since $3 \\cdot 4^{i}>2 i$. Furthermore, since $2^{3 \\cdot 4^{i}}$ and $4^{i}$ are both congruent to 1 modulo 3 , we have $3 \\mid 2^{3 \\cdot 4^{i}}-4^{i}$. Hence, $\\left\\lfloor\\frac{2^{3 \\cdot 4^{i}}}{3 \\cdot 4^{i}}\\right\\rfloor=\\frac{2^{3 \\cdot 4^{i}}-4^{i}}{3 \\cdot 4^{i}}=\\frac{2^{3 \\cdot 4^{i}-2 i}-1}{3}$, which is odd for every $i \\geqslant 1$.' 'Treat the (trivial) case when $n$ is odd.\nLet $n$ be even, and let $p$ be a prime divisor of $n+1$. Define the sequence $\\left(a_{i}\\right)_{i \\geqslant 1}$ by\n\n$$\na_{i}=\\min \\left\\{a \\in \\mathbb{Z}_{>0}: 2^{i} \\text { divides } a p+1\\right\\}\n$$\n\nRecall that there exists $a$ with $1 \\leqslant a<2^{i}$ such that $a p \\equiv-1\\left(\\bmod 2^{i}\\right)$, so each $a_{i}$ satisfies $1 \\leqslant a_{i}<2^{i}$. This implies that $a_{i} p+1

x_{n}\\tag{2}\n$$\n\nholds for all integers $n \\geqslant 0$, it is also strictly increasing. Since $x_{n+1}$ is by (1) coprime to $c$ for any $n \\geqslant 0$, it suffices to prove that for each $n \\geqslant 2$ there exists a prime number $p$ dividing $x_{n}$ but none of the numbers $x_{1}, \\ldots, x_{n-1}$. Let us begin by establishing three preliminary claims.\n\nClaim 1. If $i \\equiv j(\\bmod m)$ holds for some integers $i, j \\geqslant 0$ and $m \\geqslant 1$, then $x_{i} \\equiv x_{j}\\left(\\bmod x_{m}\\right)$ holds as well.\n\nProof. Evidently, it suffices to show $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}\\right)$ for all integers $i \\geqslant 0$ and $m \\geqslant 1$. For this purpose we may argue for fixed $m$ by induction on $i$ using $x_{0}=0$ in the base case $i=0$. Now, if we have $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}\\right)$ for some integer $i$, then the recursive equation (1) yields\n\n$$\nx_{i+m+1} \\equiv c^{2}\\left(x_{i+m}^{3}-4 x_{i+m}^{2}+5 x_{i+m}\\right)+1 \\equiv c^{2}\\left(x_{i}^{3}-4 x_{i}^{2}+5 x_{i}\\right)+1 \\equiv x_{i+1} \\quad\\left(\\bmod x_{m}\\right),\n$$\n\nwhich completes the induction.\n\nClaim 2. If the integers $i, j \\geqslant 2$ and $m \\geqslant 1$ satisfy $i \\equiv j(\\bmod m)$, then $x_{i} \\equiv x_{j}\\left(\\bmod x_{m}^{2}\\right)$ holds as well.\n\nProof. Again it suffices to prove $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}^{2}\\right)$ for all integers $i \\geqslant 2$ and $m \\geqslant 1$. As above, we proceed for fixed $m$ by induction on $i$. The induction step is again easy using (1), but this time the base case $i=2$ requires some calculation. Set $L=5 c^{2}$. By (1) we have $x_{m+1} \\equiv L x_{m}+1\\left(\\bmod x_{m}^{2}\\right)$, and hence\n\n$$\n\\begin{aligned}\nx_{m+1}^{3}-4 x_{m+1}^{2}+5 x_{m+1} & \\equiv\\left(L x_{m}+1\\right)^{3}-4\\left(L x_{m}+1\\right)^{2}+5\\left(L x_{m}+1\\right) \\\\\n& \\equiv\\left(3 L x_{m}+1\\right)-4\\left(2 L x_{m}+1\\right)+5\\left(L x_{m}+1\\right) \\equiv 2 \\quad\\left(\\bmod x_{m}^{2}\\right),\n\\end{aligned}\n$$\n\nwhich in turn gives indeed $x_{m+2} \\equiv 2 c^{2}+1 \\equiv x_{2}\\left(\\bmod x_{m}^{2}\\right)$.\n\nClaim 3. For each integer $n \\geqslant 2$, we have $x_{n}>x_{1} \\cdot x_{2} \\cdots x_{n-2}$.\n\nProof. The cases $n=2$ and $n=3$ are clear. Arguing inductively, we assume now that the claim holds for some $n \\geqslant 3$. Recall that $x_{2} \\geqslant 3$, so by monotonicity and (2) we get $x_{n} \\geqslant x_{3} \\geqslant x_{2}\\left(x_{2}-2\\right)^{2}+x_{2}+1 \\geqslant 7$. It follows that\n\n$$\nx_{n+1}>x_{n}^{3}-4 x_{n}^{2}+5 x_{n}>7 x_{n}^{2}-4 x_{n}^{2}>x_{n}^{2}>x_{n} x_{n-1},\n$$\n\nwhich by the induction hypothesis yields $x_{n+1}>x_{1} \\cdot x_{2} \\cdots x_{n-1}$, as desired.\n\n\n\nNow we direct our attention to the problem itself: let any integer $n \\geqslant 2$ be given. By Claim 3 there exists a prime number $p$ appearing with a higher exponent in the prime factorisation of $x_{n}$ than in the prime factorisation of $x_{1} \\cdots x_{n-2}$. In particular, $p \\mid x_{n}$, and it suffices to prove that $p$ divides none of $x_{1}, \\ldots, x_{n-1}$.\n\nOtherwise let $k \\in\\{1, \\ldots, n-1\\}$ be minimal such that $p$ divides $x_{k}$. Since $x_{n-1}$ and $x_{n}$ are coprime by (1) and $x_{1}=1$, we actually have $2 \\leqslant k \\leqslant n-2$. Write $n=q k+r$ with some integers $q \\geqslant 0$ and $0 \\leqslant r1$, so there exists a prime $p$ with $v_{p}(N)>0$. Since $N$ is a fraction of two odd numbers, $p$ is odd.\n\nBy our lemma,\n\n$$\n01$ and $a_{n+2}>1$. Since $\\left(a_{n+1}\\right)^{2}-1 \\leqslant\\left(1-a_{n}\\right)\\left(a_{n+2}-1\\right)$, we deduce that $0<1-a_{n}<1<1+a_{n+2}$, thus $\\left(a_{n+1}\\right)^{2}-1<\\left(a_{n+2}+1\\right)\\left(a_{n+2}-1\\right)=\\left(a_{n+2}\\right)^{2}-1$.\n\nOn the other hand, $\\left(a_{n+2}\\right)^{2}-1 \\leqslant\\left(1-a_{n+3}\\right)\\left(a_{n+1}-1\\right)<\\left(1+a_{n+1}\\right)\\left(a_{n+1}-1\\right)=\\left(a_{n+1}\\right)^{2}-1$, a contradiction. We have shown that we cannot have two consecutive terms, except maybe $a_{1}$ and $a_{2}$, strictly greater than 1 .\n\nFinally, suppose $a_{2022}>1$. This implies that $a_{2021} \\leqslant 1$ and $a_{2023} \\leqslant 1$. Therefore $0<$ $\\left(a_{2022}\\right)^{2}-1 \\leqslant\\left(1-a_{2021}\\right)\\left(a_{2023}-1\\right) \\leqslant 0$, a contradiction. We conclude that $a_{2022} \\leqslant 1$.'] [] Text-only Competition True Theorem proof Algebra Math English 257 "Let $n \geqslant 3$ be an integer, and let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers in the interval $[0,1]$. Let $s=x_{1}+x_{2}+\ldots+x_{n}$, and assume that $s \geqslant 3$. Prove that there exist integers $i$ and $j$ with $1 \leqslant i2^{s-3} $$" "[""Let $1 \\leqslant a\\max \\left\\{\\frac{100^{n-1} a_{n-1}}{\\left(10^{\\frac{1}{n-1}}-9^{\\frac{1}{n-1}}\\right)^{n-1}}, \\frac{a_{n-1}}{9} 10^{n-1}, \\frac{a_{n-1}}{9}\\left(10 a_{n-1}\\right)^{n-1}, \\ldots, \\frac{a_{n-1}}{9}\\left(10 a_{0}\\right)^{n-1}\\right\\} .\n$$\n\nAs a direct consequence of $10^{t}$ being bigger than the first quantity listed in the above set, we get that the interval\n\n$$\nI=\\left[\\left(\\frac{9}{a_{n-1}} 10^{t}\\right)^{\\frac{1}{n-1}},\\left(\\frac{1}{a_{n-1}} 10^{t+1}\\right)^{\\frac{1}{n-1}}\\right)\n$$\n\ncontains at least 100 consecutive positive integers.\n\nLet $X$ be a positive integer in $I$ such that $X$ is congruent to $1 \\bmod 100$. Since $X \\in I$ we have\n\n$$\n9 \\cdot 10^{t} \\leqslant a_{n-1} X^{n-1}<10^{t+1}\n$$\n\nthus the first digit (from the left) of $a_{n-1} X^{n-1}$ must be 9 .\n\nNext, we observe that $a_{n-1}\\left(10 a_{i}\\right)^{n-1}<9 \\cdot 10^{t} \\leqslant a_{n-1} X^{n-1}$, thus $10 a_{i}10^{\\alpha i} a_{n-1} X^{n-1}>$ $10^{\\alpha i} a_{i} X^{i}$, the terms $a_{i} 10^{\\alpha i} X^{i}$ do not interact when added; in particular, there is no carryover caused by addition. Thus we have $s\\left(P\\left(10^{\\alpha} X\\right)\\right)=s\\left(X^{n}\\right)+s\\left(a_{n-1} X^{n-1}\\right)+\\cdots+s\\left(a_{0}\\right)$.\n\nWe now look at $P\\left(10^{\\alpha-1} X\\right)=10^{(\\alpha-1) n} X^{n}+a_{n-1} 10^{(\\alpha-1)(n-1)} X^{n-1}+\\cdots+a_{0}$. Firstly, if $i10^{(\\alpha-1) i} a_{n-1} X^{n-1} \\geqslant 10^{(\\alpha-1) i+1} a_{i} X^{i}$, thus all terms $10^{(\\alpha-1) i} a_{i} X^{i}$ for $0 \\leqslant i \\leqslant n-1$ come in 'blocks', exactly as in the previous case.\n\nFinally, $10^{(\\alpha-1) n+1}>10^{(\\alpha-1)(n-1)} a_{n-1} X^{n-1} \\geqslant 10^{(\\alpha-1) n}$, thus $10^{(\\alpha-1)(n-1)} a_{n-1} X^{n-1}$ has exactly $(\\alpha-1) n+1$ digits, and its first digit is 9 , as established above. On the other hand, $10^{(\\alpha-1) n} X^{n}$ has exactly $(\\alpha-1) n$ zeros, followed by $01($ as $X$ is $1 \\bmod 100)$. Therefore, when we add the terms, the 9 and 1 turn into 0 , the 0 turns into 1 , and nothing else is affected.\n\nPutting everything together, we obtain\n\n$$\ns\\left(P\\left(10^{\\alpha-1} X\\right)\\right)=s\\left(X^{n}\\right)+s\\left(a_{n-1} X^{n-1}\\right)+\\cdots+s\\left(a_{0}\\right)-9=s\\left(P\\left(10^{\\alpha} X\\right)\\right)-9\n$$\n\nthus $s\\left(P\\left(10^{\\alpha} X\\right)\\right)$ and $s\\left(P\\left(10^{\\alpha-1} X\\right)\\right)$ have different parities, as claimed.""]" [] Text-only Competition True Theorem proof Algebra Math English 259 "For a positive integer $n$, an $n$-sequence is a sequence $\left(a_{0}, \ldots, a_{n}\right)$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \leqslant n$, then $a_{i}+a_{j} \leqslant n$ and $a_{a_{i}+a_{j}}=a_{i+j}$. Let $f(n)$ be the number of $n$-sequences. Prove that there exist positive real numbers $c_{1}, c_{2}$ and $\lambda$ such that $$ c_{1} \lambda^{n}k$ for some $i$, and small if no such $i$ exists. For now we will assume that $\\left(a_{i}\\right)$ is not the identity sequence (in other words, that $a_{i} \\neq i$ for some $i$ ).\n\nLemma 1. If $a_{r}=a_{s}$ and $r, sr$, and let $d$ be the minimum positive integer such that $a_{r+d}=a_{r}$. Then\n\n1. The subsequence $\\left(a_{r}, a_{r+1}, \\ldots, a_{n}\\right)$ is periodic with minimal period $d$. That is, for $uv_{0}$. Thus $u_{0}=v_{0}$, so $d \\mid u-v$.\n2. If $r=0$ there is nothing to prove. Otherwise $a_{0}=a_{2 a_{0}}$ so $2 a_{0}=0$. Then we have $a_{a_{i}}=a_{i}$ for all $i$, so $a_{i}=i$ for $ik+1$. We show that $a_{i} \\leqslant k$ for all $i$ by induction. Note that Lemma 2 already establishes this for $i \\leqslant k$. We must have $d \\mid a_{d / 2}$ and $a_{d / 2} \\leqslant kk$, if $a_{j} \\leqslant k$ for $jk+1$ and $a_{i}=i$ for all $0 \\leqslant ik$.\n\nWe already have $a_{i}=i$ for $id$, this means that $a_{i}=i$ for $i \\leqslant k$.\n\nFinally, one can show inductively that $a_{i}=i$ for $kc_{1} \\lambda^{n}$ for some $c_{1}$, we note that\n\n$$\nf(n)>g\\left(k+1,\\left\\lfloor\\frac{k+1}{3}\\right\\rfloor\\right) \\geqslant 3^{\\lfloor(k+1) / 3\\rfloor}>3^{n / 6}-1\n$$\n\nTo show that $f(n)\sum_{x \in X_{i}} f(x) \text { for all } i \neq k $$ Prove that the number of nice functions is at least $n^{n}$." ['For a subset $Y \\subseteq X$, we write $f(Y)$ for $\\sum_{y \\in Y} f(y)$. Note that a function $f: X \\rightarrow$ $\\{1, \\ldots, n+1\\}$ is nice, if and only if $f\\left(X_{i}\\right)$ is maximized by a unique index $i \\in\\{1, \\ldots, m\\}$.\n\nWe will first investigate the set $\\mathcal{F}$ of functions $f: X \\rightarrow\\{1, \\ldots, n\\}$; note that $|\\mathcal{F}|=n^{n}$. For every function $f \\in \\mathcal{F}$, define a corresponding function $f^{+}: X \\rightarrow\\{1,2, \\ldots, n+1\\}$ in the following way: Pick some set $X_{l}$ that maximizes the value $f\\left(X_{l}\\right)$.\n\n- For all $x \\in X_{l}$, define $f^{+}(x)=f(x)+1$.\n- For all $x \\in X \\backslash X_{l}$, define $f^{+}(x)=f(x)$.\n\nClaim. The resulting function $f^{+}$is nice.\n\nProof. Note that $f^{+}\\left(X_{i}\\right)=f\\left(X_{i}\\right)+\\left|X_{i} \\cap X_{l}\\right|$ holds for all $X_{i}$. We show that $f^{+}\\left(X_{i}\\right)$ is maximized at the unique index $i=l$. Hence consider some arbitrary index $j \\neq l$. Then $X_{l} \\subset X_{j}$ is impossible, as this would imply $f\\left(X_{j}\\right)>f\\left(X_{l}\\right)$ and thereby contradict the choice of set $X_{l}$; this in particular yields $\\left|X_{l}\\right|>\\left|X_{j} \\cap X_{l}\\right|$.\n\n$$\nf^{+}\\left(X_{l}\\right)=f\\left(X_{l}\\right)+\\left|X_{l}\\right| \\geqslant f\\left(X_{j}\\right)+\\left|X_{l}\\right|>f\\left(X_{j}\\right)+\\left|X_{j} \\cap X_{l}\\right|=f^{+}\\left(X_{j}\\right)\n$$\n\nThe first inequality follows since $X_{l}$ was chosen to maximize the value $f\\left(X_{l}\\right)$. The second (strict) inequality follows from $\\left|X_{l}\\right|>\\left|X_{j} \\cap X_{l}\\right|$ as observed above. This completes the proof of the claim.\n\nNext observe that function $f$ can be uniquely reconstructed from $f^{+}$: the claim yields that $f^{+}$has a unique maximizer $X_{l}$, and by decreasing the value of $f^{+}$on $X_{l}$ by 1 , we get we can fully determine the values of $f$. As each of the $n^{n}$ functions $f \\in \\mathcal{F}$ yields a (unique) corresponding nice function $f^{+}: X \\rightarrow\\{1,2, \\ldots, n+1\\}$, the proof is complete.'] [] Text-only Competition True Theorem proof Combinatorics Math English 261 Let $A B C D E$ be a convex pentagon such that $B C=D E$. Assume there is a point $T$ inside $A B C D E$ with $T B=T D, T C=T E$ and $\angle T B A=\angle A E T$. Let lines $C D$ and $C T$ intersect line $A B$ at points $P$ and $Q$, respectively, and let lines $C D$ and $D T$ intersect line $A E$ at points $R$ and $S$, respectively. Assume that points $P, B, A, Q$ and $R, E, A, S$ respectively, are collinear and occur on their lines in this order. Prove that the points $P, S, Q, R$ are concyclic. "['By the conditions we have $B C=D E, C T=E T$ and $T B=T D$, so the triangles $T B C$ and $T D E$ are congruent, in particular $\\angle B T C=\\angle D T E$.\n\nIn triangles $T B Q$ and $T E S$ we have $\\angle T B Q=\\angle S E T$ and $\\angle Q T B=180^{\\circ}-\\angle B T C=180^{\\circ}-$ $\\angle D T E=\\angle E T S$, so these triangles are similar to each other. It follows that $\\angle T S E=\\angle B Q T$ and\n\n$$\n\\frac{T D}{T Q}=\\frac{T B}{T Q}=\\frac{T E}{T S}=\\frac{T C}{T S}\n$$\n\nBy rearranging this relation we get $T D \\cdot T S=T C \\cdot T Q$, so $C, D, Q$ and $S$ are concyclic. (Alternatively, we can get $\\angle C Q D=\\angle C S D$ from the similar triangles $T C S$ and $T D Q$.) Hence, $\\angle D C Q=\\angle D S Q$.\n\nFinally, from the angles of triangle $C Q P$ we get\n\n$$\n\\angle R P Q=\\angle R C Q-\\angle P Q C=\\angle D S Q-\\angle D S R=\\angle R S Q\n$$\n\nwhich proves that $P, Q, R$ and $S$ are concyclic.\n\n' 'As in the previous solution, we note that triangles $T B C$ and $T D E$ are congruent. Denote the intersection point of $D T$ and $B A$ by $V$, and the intersection point of $C T$ and $E A$ by $W$. From triangles $B C Q$ and $D E S$ we then have\n\n$$\n\\begin{aligned}\n\\angle V S W & =\\angle D S E=180^{\\circ}-\\angle S E D-\\angle E D S=180^{\\circ}-\\angle A E T-\\angle T E D-\\angle E D T \\\\\n& =180^{\\circ}-\\angle T B A-\\angle T C B-\\angle C B T=180^{\\circ}-\\angle Q C B-\\angle C B Q=\\angle B Q C=\\angle V Q W,\n\\end{aligned}\n$$\n\nmeaning that $V S Q W$ is cyclic, and in particular $\\angle W V Q=\\angle W S Q$. Since\n\n$$\n\\angle V T B=180^{\\circ}-\\angle B T C-\\angle C T D=180^{\\circ}-\\angle C T D-\\angle D T E=\\angle E T W\n$$\n\nand $\\angle T B V=\\angle W E T$ by assumption, we have that the triangles $V T B$ and $W T E$ are similar, hence\n\n$$\n\\frac{V T}{W T}=\\frac{B T}{E T}=\\frac{D T}{C T}\n$$\n\n\n\nThus $C D \\| V W$, and angle chasing yields\n\n$$\n\\angle R P Q=\\angle W V Q=\\angle W S Q=\\angle R S Q\n$$\n\nconcluding the proof.']" [] Text-only Competition True Theorem proof Geometry Math English 262 "In the acute-angled triangle $A B C$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $A F$. The lines through $P$ parallel to $A C$ and $A B$ meet $B C$ at $D$ and $E$, respectively. Points $X \neq A$ and $Y \neq A$ lie on the circles $A B D$ and $A C E$, respectively, such that $D A=D X$ and $E A=E Y$. Prove that $B, C, X$ and $Y$ are concyclic." ['Let $A^{\\prime}$ be the intersection of lines $B X$ and $C Y$. By power of a point, it suffices to prove that $A^{\\prime} B \\cdot A^{\\prime} X=A^{\\prime} C \\cdot A^{\\prime} Y$, or, equivalently, that $A^{\\prime}$ lies on the radical axis of the circles $A B D X$ and $A C E Y$.\n\nFrom $D A=D X$ it follows that in circle $A B D X$, point $D$ bisects of one of the arcs $A X$. Therefore, depending on the order of points, the line $B C$ is either the internal or external bisector of $\\angle A B X$. In both cases, line $B X$ is the reflection of $B A$ in line $B D C$. Analogously, line $C Y$ is the reflection of $C A$ in line $B C$; we can see that $A^{\\prime}$ is the reflection of $A$ in line $B C$, so $A, F$ and $A^{\\prime}$ are collinear.\n\nBy $P D \\| A C$ and $P E \\| A B$ we have $\\frac{F D}{F C}=\\frac{F P}{F A}=\\frac{F E}{F B}$, hence $F D \\cdot F B=F E \\cdot F C$. So, point $F$ has equal powers with respect to circles $A B D X$ and $A C E Y$.\n\nPoint $A$, being a common point of the two circles, is another point with equal powers, so the radical axis of circles $A B D X$ and $A C E Y$ is the altitude $A F$ that passes through $A^{\\prime}$.\n\n'] [] Text-only Competition True Theorem proof Geometry Math English 263 Let $A B C D$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $A C$ is tangent to the circle $A D Q$, and the line $B D$ is tangent to the circle $B C P$. Let $M$ and $N$ be the midpoints of $B C$ and $A D$, respectively. Prove that the following three lines are concurrent: line $C D$, the tangent of circle $A N Q$ at point $A$, and the tangent to circle $B M P$ at point $B$. "['We first prove that triangles $A D Q$ and $C D B$ are similar. Since $A B C D$ is cyclic, we have $\\angle D A Q=\\angle D C B$. By the tangency of $A C$ to the circle $A Q D$ we also have $\\angle C B D=\\angle C A D=\\angle A Q D$. The claimed similarity is proven.\n\nLet $R$ be the midpoint of $C D$. Points $N$ and $R$ correspond in the proven similarity, and so $\\angle Q N A=\\angle B R C$.\n\n\n\nLet $K$ be the second common point of line $C D$ with circle $A B R$ (i.e., if $C D$ intersects circle $A B R$, then $K \\neq R$ is the other intersection; otherwise, if $C D$ is tangent to $C D$, then $K=R$ ). In both cases, we have $\\angle B A K=\\angle B R C=\\angle Q N A$; that indicates that $A K$ is tangent to circle $A N Q$. It can be showed analogously that $B K$ is tangent to circle $B M P$.' 'We present a second solution, without using the condition that $A B C D$ is cyclic. Again, $M$ and $N$ can be any points on lines $B C$ and $A D$ such that $B M: M C=D N: N A$.\n\nLet $A B$ and $C D$ meet at $T$ (if $A B \\| C D$ then $T$ is their common ideal point). Let $C D$ meet the tangent to the circle $A N Q$ at $A$, and the tangent to the circle $B M P$ at $B$ at points $K_{1}$ and $K_{2}$, respectively.\n\n\n\nLet $I$ and $J$ be the ideal points of $A D$ and $B C$, respectively. Notice that the pencils $\\left(A D, A C, A T, A K_{1}\\right)$ and $(Q A, Q D, Q I, Q N)$ of lines are congruent, because $\\angle K_{1} A D=\\angle A Q N$, $\\angle C A D=\\angle A Q D$ and $\\angle I A T=\\angle I Q T$. Hence,\n\n$$\n\\left(D, C ; T, K_{1}\\right)=\\left(A D, A C ; A T, A K_{1}\\right)=(Q A, Q D ; Q I, Q N)=(A, D ; I, N)=\\frac{D N}{N A}\n$$\n\nIt can be obtained analogously that\n\n$$\n\\left(D, C ; T, K_{2}\\right)=\\left(B D, B C ; B T, B K_{2}\\right)=(P C, P B ; P J, P M)=(C, B ; I, N)=\\frac{B M}{M C}\n$$\n\nFrom $B M: M C=D N: D A$ we get $\\left(D, C ; T, K_{1}\\right)=\\left(D, C ; T, K_{2}\\right)$ and hence $K_{1}=K_{2}$.']" [] Text-only Competition True Theorem proof Geometry Math English 264 "Let $A B C$ be an acute-angled triangle with $A C>A B$, let $O$ be its circumcentre, and let $D$ be a point on the segment $B C$. The line through $D$ perpendicular to $B C$ intersects the lines $A O, A C$ and $A B$ at $W, X$ and $Y$, respectively. The circumcircles of triangles $A X Y$ and $A B C$ intersect again at $Z \neq A$. Prove that if $O W=O D$, then $D Z$ is tangent to the circle $A X Y$." "['Let $A O$ intersect $B C$ at $E$. As $E D W$ is a right-angled triangle and $O$ is on $W E$, the condition $O W=O D$ means $O$ is the circumcentre of this triangle. So $O D=O E$ which establishes that $D, E$ are reflections in the perpendicular bisector of $B C$.\n\nNow observe:\n\n$$\n180^{\\circ}-\\angle D X Z=\\angle Z X Y=\\angle Z A Y=\\angle Z C D \\text {, }\n$$\n\nwhich shows $C D X Z$ is cyclic.\n\n\n\nWe next show that $A Z \\| B C$. To do this, introduce point $Z^{\\prime}$ on circle $A B C$ such that $A Z^{\\prime} \\| B C$. By the previous result, it suffices to prove that $C D X Z^{\\prime}$ is cyclic. Notice that triangles $B A E$ and $C Z^{\\prime} D$ are reflections in the perpendicular bisector of $B C$. Using this and that $A, O, E$ are collinear:\n\n$$\n\\angle D Z^{\\prime} C=\\angle B A E=\\angle B A O=90^{\\circ}-\\frac{1}{2} \\angle A O B=90^{\\circ}-\\angle C=\\angle D X C,\n$$\n\nso $D X Z^{\\prime} C$ is cyclic, giving $Z \\equiv Z^{\\prime}$ as desired.\n\nUsing $A Z \\| B C$ and $C D X Z$ cyclic we get:\n\n$$\n\\angle A Z D=\\angle C D Z=\\angle C X Z=\\angle A Y Z\n$$\n\nwhich by the converse of alternate segment theorem shows $D Z$ is tangent to circle $A X Y$.' 'Notice that point $Z$ is the Miquel-point of lines $A C, B C, B A$ and $D Y$; then $B, D, Z, Y$ and $C, D, X, Y$ are concyclic. Moreover, $Z$ is the centre of the spiral similarity that maps $B C$ to $Y X$.\n\nBy $B C \\perp Y X$, the angle of that similarity is $90^{\\circ}$; hence the circles $A B C Z$ and $A X Y Z$ are perpendicular, therefore the radius $O Z$ in circle $A B C Z$ is tangent to circle $A X Y Z$.\n\n\n\nBy $O W=O D$, the triangle $O W D$ is isosceles, and\n\n$$\n\\angle Z O A=2 \\angle Z B A=2 \\angle Z B Y=2 \\angle Z D Y=\\angle O D W+\\angle D W O\n$$\n\nso $D$ lies on line $Z O$ that is tangent to circle $A X Y$.']" [] Text-only Competition True Theorem proof Geometry Math English 265 "Let $A B C$ be a triangle, and let $\ell_{1}$ and $\ell_{2}$ be two parallel lines. For $i=1,2$, let $\ell_{i}$ meet the lines $B C, C A$, and $A B$ at $X_{i}, Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $B C$, the line through $Y_{i}$ perpendicular to $C A$, and finally the line through $Z_{i}$ perpendicular to $A B$, determine a non-degenerate triangle $\Delta_{i}$. Show that the circumcircles of $\Delta_{1}$ and $\Delta_{2}$ are tangent to each other." "['Throughout the solutions, $\\sphericalangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\\circ}$.\n\nLet the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, and denote the circumcircle of $\\Delta_{i}$ by $\\omega_{i}$.\n\nIn triangles $D_{1} Y_{1} Z_{1}$ and $D_{2} Y_{2} Z_{2}$ we have $Y_{1} Z_{1} \\| Y_{2} Z_{2}$ because they are parts of $\\ell_{1}$ and $\\ell_{2}$. Moreover, $D_{1} Y_{1} \\| D_{2} Y_{2}$ are perpendicular to $A C$ and $D_{1} Z_{1} \\| D_{2} Z_{2}$ are perpendicular to $A B$, so the two triangles are homothetic and their homothetic centre is $Y_{1} Y_{2} \\cap Z_{1} Z_{2}=A$. Hence, line $D_{1} D_{2}$ passes through $A$. Analogously, line $E_{1} E_{2}$ passes through $B$ and $F_{1} F_{2}$ passes through $C$.\n\n\n\nThe corresponding sides of $\\Delta_{1}$ and $\\Delta_{2}$ are parallel, because they are perpendicular to the respective sides of triangle $A B C$. Hence, $\\Delta_{1}$ and $\\Delta_{2}$ are either homothetic, or they can be translated to each other. Using that $B, X_{2}, Z_{2}$ and $E_{2}$ are concyclic, $C, X_{2}, Y_{2}$ and $F_{2}$ are concyclic, $Z_{2} E_{2} \\perp A B$ and $Y_{2}, F_{2} \\perp A C$ we can calculate\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(E_{1} E_{2}, F_{1} F_{2}\\right) & =\\sphericalangle\\left(E_{1} E_{2}, X_{1} X_{2}\\right)+\\sphericalangle\\left(X_{1} X_{2}, F_{1} F_{2}\\right)=\\sphericalangle\\left(B E_{2}, B X_{2}\\right)+\\sphericalangle\\left(C X_{2}, C F_{2}\\right) \\\\\n& =\\sphericalangle\\left(Z_{2} E_{2}, Z_{2} X_{2}\\right)+\\sphericalangle\\left(Y_{2} X_{2}, Y_{2} F_{2}\\right)=\\sphericalangle\\left(Z_{2} E_{2}, \\ell_{2}\\right)+\\sphericalangle\\left(\\ell_{2}, Y_{2} F_{2}\\right) \\\\\n& =\\sphericalangle\\left(Z_{2} E_{2}, Y_{2} F_{2}\\right)=\\sphericalangle(A B, A C) \\not \\equiv 0,\n\\end{aligned}\n\\tag{1}\n$$\n\nand conclude that lines $E_{1} E_{2}$ and $F_{1} F_{2}$ are not parallel. Hence, $\\Delta_{1}$ and $\\Delta_{2}$ are homothetic; the lines $D_{1} D_{2}, E_{1} E_{2}$, and $F_{1} F_{2}$ are concurrent at the homothetic centre of the two triangles. Denote this homothetic centre by $H$.\n\nFor $i=1,2$, using (1), and that $A, Y_{i}, Z_{i}$ and $D_{i}$ are concyclic,\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(H E_{i}, H F_{i}\\right) & =\\sphericalangle\\left(E_{1} E_{2}, F_{1} F_{2}\\right)=\\sphericalangle(A B, A C) \\\\\n& =\\sphericalangle\\left(A Z_{i}, A Y_{i}\\right)=\\sphericalangle\\left(D_{i} Z_{i}, D_{i} Y_{i}\\right)=\\sphericalangle\\left(D_{i} E_{i}, D_{i} F_{i}\\right),\n\\end{aligned}\n$$\n\n\n\nso $H$ lies on circle $\\omega_{i}$.\n\nThe same homothety that maps $\\Delta_{1}$ to $\\Delta_{2}$, sends $\\omega_{1}$ to $\\omega_{2}$ as well. Point $H$, that is the centre of the homothety, is a common point of the two circles, That finishes proving that $\\omega_{1}$ and $\\omega_{2}$ are tangent to each other.' ""As in the first solution, let the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X_{i}, Y_{i}$ and $Z_{i}$, respectively. In the same way we conclude that $\\left(A, D_{1}, D_{2}\\right),\\left(B, E_{1}, E_{2}\\right)$ and $\\left(C, F_{1}, F_{2}\\right)$ are collinear.\n\nThe corresponding sides of triangles $A B C$ and $D_{i} E_{i} F_{i}$ are perpendicular to each other. Hence, there is a spiral similarity with rotation $\\pm 90^{\\circ}$ that maps $A B C$ to $D_{i} E_{i} F_{i}$; let $M_{i}$ be the centre of that similarity. Hence, $\\sphericalangle\\left(M_{i} A, M_{i} D_{i}\\right)=\\sphericalangle\\left(M_{i} B, M_{i} E_{i}\\right)=\\sphericalangle\\left(M_{i} C, M_{i} F_{i}\\right)=90^{\\circ}$. The circle with diameter $A D_{i}$ passes through $M_{i}, Y_{i}, Z_{i}$, so $M_{i}, A, Y_{i}, Z_{i}, D_{i}$ are concyclic; analogously $\\left(M_{i}, B, X_{i}, Z_{i}, E_{i}\\right)$ and $\\left(M_{i}, C, X_{i}, Y_{i}, F_{i}\\right)$ are concyclic.\n\nBy applying Desargues' theorem to triangles $A B C$ and $D_{i} E_{i} F_{i}$ we conclude that the lines $A D_{i}, B E_{i}$ and $B F_{i}$ are concurrent; let their intersection be $H$. Since $\\left(A, D_{1} . D_{2}\\right),\\left(B, E_{1} \\cdot E_{2}\\right)$ and $\\left(C, F_{1} . F_{2}\\right)$ are collinear, we obtain the same point $H$ for $i=1$ and $i=2$.\n\n\n\nBy $\\sphericalangle(C B, C H)=\\sphericalangle\\left(C X_{i}, C F_{i}\\right)=\\sphericalangle\\left(Y_{i} X_{i}, Y_{i} F_{i}\\right)=\\sphericalangle\\left(Y_{i} Z_{i}, Y_{i} D_{i}\\right)=\\sphericalangle\\left(A Z_{i}, A D_{i}\\right)=$ $\\sphericalangle(A B, A H)$, point $H$ lies on circle $A B C$.\n\nAnalogously, from $\\sphericalangle\\left(F_{i} D_{i}, F_{i} H\\right)=\\sphericalangle\\left(F_{i} Y_{i}, F_{i} C\\right)=\\sphericalangle\\left(X_{i} Y_{i}, X_{i} C\\right)=\\sphericalangle\\left(X_{i} Z_{i}, X_{i} B\\right)=$ $\\sphericalangle\\left(E_{i} Z_{i}, E_{i} B\\right)=\\sphericalangle\\left(E_{i} D_{i}, E_{i} H\\right)$, we can see that point $H$ lies on circle $D_{i} E_{i} F_{i}$ as well. Therefore, circles $A B C$ and $D_{i} E_{i} F_{i}$ intersect at point $H$.\n\nThe spiral similarity moves the circle $A B C$ to circle $D_{i} E_{i} F_{i}$, so the two circles are perpendicular. Hence, both circles $D_{1} E_{1} F_{1}$ and $D_{2} E_{2} F_{2}$ are tangent to the radius of circle $A B C$ at $H$.""]" [] Text-only Competition True Theorem proof Geometry Math English 266 In an acute-angled triangle $A B C$, point $H$ is the foot of the altitude from $A$. Let $P$ be a moving point such that the bisectors $k$ and $\ell$ of angles $P B C$ and $P C B$, respectively, intersect each other on the line segment $A H$. Let $k$ and $A C$ meet at $E$, let $\ell$ and $A B$ meet at $F$, and let $E F$ and $A H$ meet at $Q$. Prove that, as $P$ varies, the line $P Q$ passes through a fixed point. "[""Let the reflections of the line $B C$ with respect to the lines $A B$ and $A C$ intersect at point $K$. We will prove that $P, Q$ and $K$ are collinear, so $K$ is the common point of the varying line $P Q$.\n\nLet lines $B E$ and $C F$ intersect at $I$. For every point $O$ and $d>0$, denote by $(O, d)$ the circle centred at $O$ with radius $d$, and define $\\omega_{I}=(I, I H)$ and $\\omega_{A}=(A, A H)$. Let $\\omega_{K}$ and $\\omega_{P}$ be the incircle of triangle $K B C$ and the $P$-excircle of triangle $P B C$, respectively.\n\nSince $I H \\perp B C$ and $A H \\perp B C$, the circles $\\omega_{A}$ and $\\omega_{I}$ are tangent to each other at $H$. So, $H$ is the external homothetic centre of $\\omega_{A}$ and $\\omega_{I}$. From the complete quadrangle $B C E F$ we have $(A, I ; Q, H)=-1$, therefore $Q$ is the internal homothetic centre of $\\omega_{A}$ and $\\omega_{I}$. Since $B A$ and $C A$ are the external bisectors of angles $\\angle K B C$ and $\\angle K C B$, circle $\\omega_{A}$ is the $K$-excircle in triangle $B K C$. Hence, $K$ is the external homothetic centre of $\\omega_{A}$ and $\\omega_{K}$. Also it is clear that $P$ is the external homothetic centre of $\\omega_{I}$ and $\\omega_{P}$. Let point $T$ be the tangency point of $\\omega_{P}$ and $B C$, and let $T^{\\prime}$ be the tangency point of $\\omega_{K}$ and $B C$. Since $\\omega_{I}$ is the incircle and $\\omega_{P}$ is the $P$-excircle of $P B C, T C=B H$ and since $\\omega_{K}$ is the incircle and $\\omega_{A}$ is the $K$-excircle of $K B C, T^{\\prime} C=B H$. Therefore $T C=T^{\\prime} C$ and $T \\equiv T^{\\prime}$. It yields that $\\omega_{K}$ and $\\omega_{P}$ are tangent to each other at $T$.\n\n\n\nLet point $S$ be the internal homothetic centre of $\\omega_{A}$ and $\\omega_{P}$, and let $S^{\\prime}$ be the internal homothetic centre of $\\omega_{I}$ and $\\omega_{K}$. It's obvious that $S$ and $S^{\\prime}$ lie on $B C$. We claim that $S \\equiv S^{\\prime}$. To prove our claim, let $r_{A}, r_{I}, r_{P}$, and $r_{K}$ be the radii of $\\omega_{A}, \\omega_{I}, \\omega_{P}$ and $\\omega_{k}$, respectively.\n\nIt is well known that if the sides of a triangle are $a, b, c$, its semiperimeter is $s=(a+b+c) / 2$, and the radii of the incircle and the $a$-excircle are $r$ and $r_{a}$, respectively, then $r \\cdot r_{a}=(s-b)(s-c)$. Applying this fact to triangle $P B C$ we get $r_{I} \\cdot r_{P}=B H \\cdot C H$. The same fact in triangle $K C B$\n\n\n\nyields $r_{K} \\cdot r_{A}=C T \\cdot B T$. Since $B H=C T$ and $B T=C H$, from these two we get\n\n$$\n\\frac{H S}{S T}=\\frac{r_{A}}{r_{P}}=\\frac{r_{I}}{r_{K}}=\\frac{H S^{\\prime}}{S^{\\prime} T}\n$$\n\nso $S=S^{\\prime}$ indeed.\n\nFinally, by applying the generalised Monge's theorem to the circles $\\omega_{A}, \\omega_{I}$, and $\\omega_{K}$ (with two pairs of internal and one pair of external common tangents), we can see that points $Q$, $S$, and $K$ are collinear. Similarly one can show that $Q, S$ and $P$ are collinear, and the result follows."" ""Again, let $B E$ and $C F$ meet at $I$, that is the incentre in triangle $B C P$; then $P I$ is the third angle bisector. From the tangent segments of the incircle we have $B P-C P=$ $B H-C H$; hence, the possible points $P$ lie on a branch of a hyperbola $\\mathcal{H}$ with foci $B, C$, and $H$ is a vertex of $\\mathcal{H}$. Since $P I$ bisects the angle between the radii $B P$ and $C P$ of the hyperbola, line $P I$ is tangent to $\\mathcal{H}$.\n\n\n\nLet $K$ be the second intersection of $P Q$ and $\\mathcal{H}$, we will show that $A K$ is tangent to $\\mathcal{H}$ at $K$; this property determines $K$.\n\nLet $G=K I \\cap A P$ and $M=P I \\cap A K$. From the complete quadrangle $B C E F$ we can see that $(H, Q ; I, A)$ is harmonic, so in the complete quadrangle $A P I K$, point $H$ lies on line $G M$.\n\nConsider triangle $A I M$. Its side $A I$ is tangent to $\\mathcal{H}$ at $H$, the side $I M$ is tangent to $\\mathcal{H}$ at $P$, and $K$ is a common point of the third side $A M$ and the hyperbola such that the lines $A P$, $I K$ and $M H$ are concurrent at the generalised Gergonne-point $G$. It follows that the third side, $A M$ is also tangent to $\\mathcal{H}$ at $K$.\n\n(Alternatively, in the last step we can apply the converse of Brianchon's theorem to the degenerate hexagon $A H I P M K$. By the theorem there is a conic section $\\mathcal{H}^{\\prime}$ such that lines $A I, I M$ and $M A$ are tangent to $\\mathcal{H}^{\\prime}$ at $H, P$ and $K$, respectively. But the three points $H, K$ and $P$, together with the tangents at $H$ and $P$ uniquely determine $\\mathcal{H}^{\\prime}$, so indeed $\\mathcal{H}^{\\prime}=\\mathcal{H}$.)""]" [] Text-only Competition True Theorem proof Geometry Math English 267 Let $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ be two triangles having the same circumcircle $\omega$, and the same orthocentre $H$. Let $\Omega$ be the circumcircle of the triangle determined by the lines $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$. Prove that $H$, the centre of $\omega$, and the centre of $\Omega$ are collinear. "[""In what follows, $\\sphericalangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\\circ}$. Denote by $O$ the centre of $\\omega$. In any triangle, the homothety with ratio $-\\frac{1}{2}$ centred at the centroid of the triangle takes the vertices to the midpoints of the opposite sides and it takes the orthocentre to the circumcentre. Therefore the triangles $A B C$ and $A^{\\prime} B^{\\prime} C^{\\prime}$ share the same centroid $G$ and the midpoints of their sides lie on a circle $\\rho$ with centre on $O H$. We will prove that $\\omega, \\Omega$, and $\\rho$ are coaxial, so in particular it follows that their centres are collinear on $O H$.\n\nLet $D=B B^{\\prime} \\cap C C^{\\prime}, E=C C^{\\prime} \\cap A A^{\\prime}, F=A A^{\\prime} \\cap B B^{\\prime}, S=B C^{\\prime} \\cap B^{\\prime} C$, and $T=B C \\cap B^{\\prime} C^{\\prime}$. Since $D, S$, and $T$ are the intersections of opposite sides and of the diagonals in the quadrilateral $B B^{\\prime} C C^{\\prime}$ inscribed in $\\omega$, by Brocard's theorem triangle $D S T$ is self-polar with respect to $\\omega$, i.e. each vertex is the pole of the opposite side. We apply this in two ways.\n\n\n\nFirst, from $D$ being the pole of $S T$ it follows that the inverse $D^{*}$ of $D$ with respect to $\\omega$ is the projection of $D$ onto $S T$. In particular, $D^{*}$ lies on the circle with diameter $S D$. If $N$ denotes the midpoint of $S D$ and $R$ the radius of $\\omega$, then the power of $O$ with respect to this circle is $O N^{2}-N D^{2}=O D \\cdot O D^{*}=R^{2}$. By rearranging, we see that $N D^{2}$ is the power of $N$ with respect to $\\omega$.\n\nSecond, from $T$ being the pole of $S D$ it follows that $O T$ is perpendicular to $S D$. Let $M$ and $M^{\\prime}$ denote the midpoints of $B C$ and $B^{\\prime} C^{\\prime}$. Then since $O M \\perp B C$ and $O M^{\\prime} \\perp B^{\\prime} C^{\\prime}$ it follows that $O M M^{\\prime} T$ is cyclic and\n\n$$\n\\sphericalangle(S D, B C)=\\sphericalangle(O T, O M)=\\sphericalangle\\left(B^{\\prime} C^{\\prime}, M M^{\\prime}\\right) .\n$$\n\nFrom $B B^{\\prime} C C^{\\prime}$ being cyclic we also have $\\sphericalangle\\left(B C, B B^{\\prime}\\right)=\\sphericalangle\\left(C C^{\\prime}, B^{\\prime} C^{\\prime}\\right)$, hence we obtain\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(S D, B B^{\\prime}\\right) & =\\sphericalangle(S D, B C)+\\sphericalangle\\left(B C, B B^{\\prime}\\right) \\\\\n& =\\sphericalangle\\left(B^{\\prime} C^{\\prime}, M M^{\\prime}\\right)+\\sphericalangle\\left(C C^{\\prime}, B^{\\prime} C^{\\prime}\\right)=\\sphericalangle\\left(C C^{\\prime}, M M^{\\prime}\\right) .\n\\end{aligned}\n$$\n\nNow from the homothety mentioned in the beginning, we know that $M M^{\\prime}$ is parallel to $A A^{\\prime}$, hence the above implies that $\\sphericalangle\\left(S D, B B^{\\prime}\\right)=\\sphericalangle\\left(C C^{\\prime}, A A^{\\prime}\\right)$, which shows that $\\Omega$ is tangent to $S D$ at $D$. In particular, $N D^{2}$ is also the power of $N$ with respect to $\\Omega$.\n\n\n\nAdditionally, from $B B^{\\prime} C C^{\\prime}$ being cyclic it follows that triangles $D B C$ and $D C^{\\prime} B^{\\prime}$ are inversely similar, so $\\sphericalangle\\left(B B^{\\prime}, D M^{\\prime}\\right)=\\sphericalangle\\left(D M, C C^{\\prime}\\right)$. This yields\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(S D, D M^{\\prime}\\right) & =\\sphericalangle\\left(S D, B B^{\\prime}\\right)+\\sphericalangle\\left(B B^{\\prime}, D M^{\\prime}\\right) \\\\\n& =\\sphericalangle\\left(C C^{\\prime}, M M^{\\prime}\\right)+\\sphericalangle\\left(D M, C C^{\\prime}\\right)=\\sphericalangle\\left(D M, M M^{\\prime}\\right),\n\\end{aligned}\n$$\n\nwhich shows that the circle $D M M^{\\prime}$ is also tangent to $S D$. Since $N, M$, and $M^{\\prime}$ are collinear on the Newton-Gauss line of the complete quadrilateral determined by the lines $B B^{\\prime}, C C^{\\prime}, B C^{\\prime}$, and $B^{\\prime} C$, it follows that $N D^{2}=N M \\cdot N M^{\\prime}$. Hence $N$ has the same power with respect to $\\omega$, $\\Omega$, and $\\rho$.\n\nBy the same arguments there exist points on the tangents to $\\Omega$ at $E$ and $F$ which have the same power with respect to $\\omega, \\Omega$, and $\\rho$. The tangents to a given circle at three distinct points cannot be concurrent, hence we obtain at least two distinct points with the same power with respect to $\\omega, \\Omega$, and $\\rho$. Hence the three circles are coaxial, as desired.""]" [] Text-only Competition True Theorem proof Geometry Math English 268 "Let $A A^{\prime} B C C^{\prime} B^{\prime}$ be a convex cyclic hexagon such that $A C$ is tangent to the incircle of the triangle $A^{\prime} B^{\prime} C^{\prime}$, and $A^{\prime} C^{\prime}$ is tangent to the incircle of the triangle $A B C$. Let the lines $A B$ and $A^{\prime} B^{\prime}$ meet at $X$ and let the lines $B C$ and $B^{\prime} C^{\prime}$ meet at $Y$. Prove that if $X B Y B^{\prime}$ is a convex quadrilateral, then it has an incircle." "[""Denote by $\\omega$ and $\\omega^{\\prime}$ the incircles of $\\triangle A B C$ and $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ and let $I$ and $I^{\\prime}$ be the centres of these circles. Let $N$ and $N^{\\prime}$ be the second intersections of $B I$ and $B^{\\prime} I^{\\prime}$ with $\\Omega$, the circumcircle of $A^{\\prime} B C C^{\\prime} B^{\\prime} A$, and let $O$ be the centre of $\\Omega$. Note that $O N \\perp A C, O N^{\\prime} \\perp A^{\\prime} C^{\\prime}$ and $O N=O N^{\\prime}$ so $N N^{\\prime}$ is parallel to the angle bisector $I I^{\\prime}$ of $A C$ and $A^{\\prime} C^{\\prime}$. Thus $I I^{\\prime} \\| N N^{\\prime}$ which is antiparallel to $B B^{\\prime}$ with respect to $B I$ and $B^{\\prime} I^{\\prime}$. Therefore $B, I, I^{\\prime}, B^{\\prime}$ are concyclic.\n\n\n\nFurther define $P$ as the intersection of $A C$ and $A^{\\prime} C^{\\prime}$ and $M$ as the antipode of $N^{\\prime}$ in $\\Omega$. Consider the circle $\\Gamma_{1}$ with centre $N$ and radius $N A=N C$ and the circle $\\Gamma_{2}$ with centre $M$ and radius $M A^{\\prime}=M C^{\\prime}$. Their radical axis passes through $P$ and is perpendicular to $M N \\perp N N^{\\prime} \\| I P$, so $I$ lies on their radical axis. Therefore, since $I$ lies on $\\Gamma_{1}$, it must also lie on $\\Gamma_{2}$. Thus, if we define $Z$ as the second intersection of $M I$ with $\\Omega$, we have that $I$ is the incentre of triangle $Z A^{\\prime} C^{\\prime}$. (Note that the point $Z$ can also be constructed directly via Poncelet's porism.)\n\nConsider the incircle $\\omega_{c}$ with centre $I_{c}$ of triangle $C^{\\prime} B^{\\prime} Z$. Note that $\\angle Z I C^{\\prime}=90^{\\circ}+$ $\\frac{1}{2} \\angle Z A^{\\prime} C^{\\prime}=90^{\\circ}+\\frac{1}{2} \\angle Z B^{\\prime} C^{\\prime}=\\angle Z I_{c} C^{\\prime}$, so $Z, I, I_{c}, C^{\\prime}$ are concyclic. Similarly $B^{\\prime}, I^{\\prime}, I_{c}, C^{\\prime}$ are concyclic.\n\nThe external centre of dilation from $\\omega$ to $\\omega_{c}$ is the intersection of $I I_{c}$ and $C^{\\prime} Z$ ( $D$ in the picture), that is the radical centre of circles $\\Omega, C^{\\prime} I_{c} I Z$ and $I I^{\\prime} I_{c}$. Similarly, the external centre of dilation from $\\omega^{\\prime}$ to $\\omega_{c}$ is the intersection of $I^{\\prime} I_{c}$ and $B^{\\prime} C^{\\prime}$ ( $D^{\\prime}$ in the picture), that is the radical centre of circles $\\Omega, B^{\\prime} I^{\\prime} I_{c} C^{\\prime}$ and $I I^{\\prime} I_{c}$. Therefore the Monge line of $\\omega, \\omega^{\\prime}$ and $\\omega_{c}$ is line $D D^{\\prime}$, and the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$ coincide. Hence the external centre $T$ of dilation from $\\omega$ to $\\omega^{\\prime}$ is also on the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$.\n\n\n\n\n\nNow since $B, I, I^{\\prime}, B^{\\prime}$ are concyclic, the intersection $T^{\\prime}$ of $B B^{\\prime}$ and $I I^{\\prime}$ is on the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$. Thus $T^{\\prime}=T$ and $T$ lies on line $B B^{\\prime}$. Finally, construct a circle $\\Omega_{0}$ tangent to $A^{\\prime} B^{\\prime}, B^{\\prime} C^{\\prime}, A B$ on the same side of these lines as $\\omega^{\\prime}$. The centre of dilation from $\\omega^{\\prime}$ to $\\Omega_{0}$ is $B^{\\prime}$, so by Monge's theorem the external centre of dilation from $\\Omega_{0}$ to $\\omega$ must be on the line $T B B^{\\prime}$. However, it is on line $A B$, so it must be $B$ and $B C$ must be tangent to $\\Omega_{0}$ as desired.\n\n""]" [] Text-only Competition True Theorem proof Geometry Math English 269 "For each $1 \leqslant i \leqslant 9$ and $T \in \mathbb{N}$, define $d_{i}(T)$ to be the total number of times the digit $i$ appears when all the multiples of 1829 between 1 and $T$ inclusive are written out in base 10 . Show that there are infinitely many $T \in \mathbb{N}$ such that there are precisely two distinct values among $d_{1}(T), d_{2}(T), \ldots, d_{9}(T)$." ['Let $n:=1829$. First, we choose some $k$ such that $n \\mid 10^{k}-1$. For instance, any multiple of $\\varphi(n)$ would work since $n$ is coprime to 10 . We will show that either $T=10^{k}-1$ or $T=10^{k}-2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrary large.\n\nFor this it suffices to show that $\\#\\left\\{d_{i}\\left(10^{k}-1\\right): 1 \\leqslant i \\leqslant 9\\right\\} \\leqslant 2$. Indeed, if\n$$\n\\#\\left\\{d_{i}\\left(10^{k}-1\\right): 1 \\leqslant i \\leqslant 9\\right\\}=1\n$$\nthen, since $10^{k}-1$ which consists of all nines is a multiple of $n$, we have\n$$\nd_{i}\\left(10^{k}-2\\right)=d_{i}\\left(10^{k}-1\\right) \\text { for } i \\in\\{1, \\ldots, 8\\} \\text {, and } d_{9}\\left(10^{k}-2\\right)0$ independent of the set $Q$ such that for any positive integer $N>100$, the number of special integers in $[1, N]$ is at least $c N$. (For example, if $Q=\{3,7\}$, then $p(42)=3, q(42)=2, p(63)=3, q(63)=3, p(2022)=3$, $q(2022)=1$.)" ['Let us call two positive integers $m, n$ friends if $p(m)+p(n)$ and $q(m)+q(n)$ are both even integers. We start by noting that the pairs $(p(k), q(k))$ modulo 2 can take at most 4 different values; thus, among any five different positive integers there are two which are friends.\n\nIn addition, both functions $p$ and $q$ satisfy $f(a b)=f(a)+f(b)$ for any $a, b$. Therefore, if $m$ and $n$ are divisible by $d$, then both $p$ and $q$ satisfy the equality $f(m)+f(n)=f(m / d)+$ $f(n / d)+2 f(d)$. This implies that $m, n$ are friends if and only if $m / d, n / d$ are friends.\n\nLet us call a set of integers $\\left\\{n_{1}, n_{2}, \\ldots, n_{5}\\right\\}$ an interesting set if for any indexes $i, j$, the difference $d_{i j}=\\left|n_{i}-n_{j}\\right|$ divides both $n_{i}$ and $n_{j}$. We claim that if elements of an interesting set are all positive, then we can obtain a special integer. Indeed, if we were able to construct such a set, then there would be a pair of integers $\\left\\{n_{i}, n_{j}\\right\\}$ which are friends, according to the first observation. Additionally, the second observation yields that the quotients $n_{i} / d_{i j}, n_{j} / d_{i j}$ form a pair of friends, which happen to be consecutive integers, thus giving a special integer as desired.\n\nIn order to construct a family of interesting sets, we can start by observing that the set $\\{0,6,8,9,12\\}$ is an interesting set. Using that $72=2^{3} \\cdot 3^{2}$ is the least common multiple of all pairwise differences in this set, we obtain a family of interesting sets by considering\n$$\n\\{72 k, 72 k+6,72 k+8,72 k+9,72 k+12\\}\n$$\nfor any $k \\geqslant 1$. If we consider the quotients (of these numbers by the appropriate differences), then we obtain that the set\n$$\nS_{k}=\\{6 k, 8 k, 9 k, 12 k, 12 k+1,18 k+2,24 k+2,24 k+3,36 k+3,72 k+8\\}\n$$\nhas at least one special integer. In particular, the interval $[1,100 k]$ contains the sets $S_{1}, S_{2}, \\ldots, S_{k}$, each of which has a special number. Any special number can be contained in at most ten sets $S_{k}$, from where we conclude that the number of special integers in $[1,100 k]$ is at least $k / 10$.\n\nFinally, let $N=100 k+r$, with $k \\geqslant 1$ and $0 \\leqslant r<100$, so that we have $N<100(k+1) \\leqslant$ $200 k$. Then the number of special integers in $[1, N]$ is at least $k / 10>N / 2000$, as we wanted to prove.'] [] Text-only Competition True Theorem proof Number Theory Math English 271 Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (modulo rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbors is of the form $x^{2}+x+k$ for some positive integer $x$. "[""Let us allow the value $x=0$ as well; we prove the same statement under this more general constraint. Obviously that implies the statement with the original conditions.\n\nCall a pair $\\{p, q\\}$ of primes with $p \\neq q$ special if $p q=x^{2}+x+k$ for some nonnegative integer $x$. The following claim is the key mechanism of the problem:\n\nClaim.\n\n(a) For every prime $r$, there are at most two primes less than $r$ forming a special pair with $r$.\n\n(b) If such $p$ and $q$ exist, then $\\{p, q\\}$ is itself special.\n\nWe present two proofs of the claim.\n\nProof 1. We are interested in integers $1 \\leqslant x\\sqrt{m}$ choices for $x$ and $y$, so there are more than $m$ possible pairs $(x, y)$. Hence, two of these sums are congruent modulo $m$ : $5^{k+1} x_{1}+3^{k+1} y_{1} \\equiv 5^{k+1} x_{2}+3^{k+1} y_{2}(\\bmod m)$.\n\nNow choose $a=x_{1}-x_{2}$ and $b=y_{1}-y_{2}$; at least one of $a, b$ is nonzero, and\n$$\n5^{k+1} a+3^{k+1} b \\equiv 0 \\quad(\\bmod m), \\quad|a|,|b| \\leqslant \\sqrt{m}\n$$\nFrom\n$$\n0 \\equiv\\left(5^{k+1} a\\right)^{2}-\\left(3^{k+1} b\\right)^{2}=5^{n+1} a^{2}-3^{n+1} b^{2} \\equiv 5 \\cdot 3^{n} a^{2}-3^{n+1} b^{2}=3^{n}\\left(5 a^{2}-3 b^{2}\\right) \\quad(\\bmod m)\n$$\nwe can see that $\\left|5 a^{2}-3 b^{2}\\right|$ is a multiple of $m$. Since at least one of $a$ and $b$ is nonzero, $5 a^{2} \\neq 3 b^{2}$. Hence, by the choice of $a, b$, we have $0<\\left|5 a^{2}-3 b^{2}\\right| \\leqslant \\max \\left(5 a^{2}, 3 b^{2}\\right) \\leqslant 5 m$. That shows that $m_{1} \\leqslant 5 m$\n\nII. Next, we show that $m_{1}$ cannot be divisible by 2,3 and 5 . Since $m_{1}$ equals either $\\left|5 a^{2}-3 b^{2}\\right|$ or $\\left|a^{2}-15 b^{2}\\right|$ with some integers $a, b$, we have six cases to check. In all six cases, we will get a contradiction by presenting another multiple of $m$, smaller than $m_{1}$.\n\n- If $5 \\mid m_{1}$ and $m_{1}=\\left|5 a^{2}-3 b^{2}\\right|$, then $5 \\mid b$ and $\\left|a^{2}-15\\left(\\frac{b}{5}\\right)^{2}\\right|=\\frac{m_{1}}{5}2 x_{n}$ for some $n$, then $\\sum_{x \\in F} 1 / x0$, at least one of $a_{1}, \\ldots, a_{n}$ is positive; without loss of generality suppose $a_{1} \\geqslant 1$.\n\nConsider the polynomials $F_{1}=\\Delta_{1} F$ and $G_{1}=\\Delta G$. On the $\\operatorname{grid}\\left\\{0, \\ldots, a_{1}-1\\right\\} \\times\\left\\{0, \\ldots, a_{2}\\right\\} \\times$ $\\ldots \\times\\left\\{0, \\ldots, a_{n}\\right\\}$ we have\n\n$$\n\\begin{aligned}\nF_{1}\\left(x_{1}, \\ldots, x_{n}\\right) & =F\\left(x_{1}+1, x_{2}, \\ldots, x_{n}\\right)-F\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)= \\\\\n& =G\\left(x_{1}+\\ldots+x_{n}+1\\right)-G\\left(x_{1}+\\ldots+x_{n}\\right)=G_{1}\\left(x_{1}+\\ldots+x_{n}\\right) .\n\\end{aligned}\n$$\n\nSince $G$ is nonconstant, we have $\\operatorname{deg} G_{1}=\\operatorname{deg} G-1 \\leqslant\\left(a_{1}-1\\right)+a_{2}+\\ldots+a_{n}$. Therefore we can apply the induction hypothesis to $F_{1}$ and $G_{1}$ and conclude that $F_{1}$ is not the zero polynomial and $\\operatorname{deg} F_{1} \\geqslant \\operatorname{deg} G_{1}$. Hence, $\\operatorname{deg} F \\geqslant \\operatorname{deg} F_{1}+1 \\geqslant \\operatorname{deg} G_{1}+1=\\operatorname{deg} G$. That finishes the proof.\n\nTo prove the problem statement, take the unique polynomial $g(x)$ so that $g(x)=\\left\\lfloor\\frac{x}{m}\\right\\rfloor$ for $x \\in\\{0,1, \\ldots, n(m-1)\\}$ and $\\operatorname{deg} g \\leqslant n(m-1)$. Notice that precisely $n(m-1)+1$ values of $g$ are prescribed, so $g(x)$ indeed exists and is unique. Notice further that the constraints $g(0)=g(1)=0$ and $g(m)=1$ together enforce $\\operatorname{deg} g \\geqslant 2$.\n\nBy applying the lemma to $a_{1}=\\ldots=a_{n}=m-1$ and the polynomials $f$ and $g$, we achieve $\\operatorname{deg} f \\geqslant \\operatorname{deg} g$. Hence we just need a suitable lower bound on $\\operatorname{deg} g$.\n\nConsider the polynomial $h(x)=g(x+m)-g(x)-1$. The degree of $g(x+m)-g(x)$ is $\\operatorname{deg} g-1 \\geqslant 1$, so $\\operatorname{deg} h=\\operatorname{deg} g-1 \\geqslant 1$, and therefore $h$ cannot be the zero polynomial. On the other hand, $h$ vanishes at the points $0,1, \\ldots, n(m-1)-m$, so $h$ has at least $(n-1)(m-1)$ roots. Hence,\n\n$$\n\\operatorname{deg} f \\geqslant \\operatorname{deg} g=\\operatorname{deg} h+1 \\geqslant(n-1)(m-1)+1 \\geqslant n\n$$'] [] Text-only Competition True Theorem proof Algebra Math English 275 Let $n \geqslant 3$ be an integer. Prove that there exists a set $S$ of $2 n$ positive integers satisfying the following property: For every $m=2,3, \ldots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$. ['We show that one of possible examples is the set\n\n$$\nS=\\left\\{1 \\cdot 3^{k}, 2 \\cdot 3^{k}: k=1,2, \\ldots, n-1\\right\\} \\cup\\left\\{1, \\frac{3^{n}+9}{2}-1\\right\\}\n$$\n\nIt is readily verified that all the numbers listed above are distinct (notice that the last two are not divisible by 3 ).\n\nThe sum of elements in $S$ is\n\n$$\n\\Sigma=1+\\left(\\frac{3^{n}+9}{2}-1\\right)+\\sum_{k=1}^{n-1}\\left(1 \\cdot 3^{k}+2 \\cdot 3^{k}\\right)=\\frac{3^{n}+9}{2}+\\sum_{k=1}^{n-1} 3^{k+1}=\\frac{3^{n}+9}{2}+\\frac{3^{n+1}-9}{2}=2 \\cdot 3^{n}\n$$\n\nHence, in order to show that this set satisfies the problem requirements, it suffices to present, for every $m=2,3, \\ldots, n$, an $m$-element subset $A_{m} \\subset S$ whose sum of elements equals $3^{n}$.\n\nSuch a subset is\n\n$$\nA_{m}=\\left\\{2 \\cdot 3^{k}: k=n-m+1, n-m+2, \\ldots, n-1\\right\\} \\cup\\left\\{1 \\cdot 3^{n-m+1}\\right\\} .\n$$\n\nClearly, $\\left|A_{m}\\right|=m$. The sum of elements in $A_{m}$ is\n\n$$\n3^{n-m+1}+\\sum_{k=n-m+1}^{n-1} 2 \\cdot 3^{k}=3^{n-m+1}+\\frac{2 \\cdot 3^{n}-2 \\cdot 3^{n-m+1}}{2}=3^{n}\n$$\n\nas required.'] [] Text-only Competition True Theorem proof Combinatorics Math English 276 "Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n+1$ squares in a row, numbered 0 to $n$ from left to right. Initially, $n$ stones are put into square 0, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of those stones and moves it to the right by at most $k$ squares (the stone should stay within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than $$ \left\lceil\frac{n}{1}\right\rceil+\left\lceil\frac{n}{2}\right\rceil+\left\lceil\frac{n}{3}\right\rceil+\cdots+\left\lceil\frac{n}{n}\right\rceil $$ turns. (As usual, $\lceil x\rceil$ stands for the least integer not smaller than $x$.)" ['The stones are indistinguishable, and all have the same origin and the same final position. So, at any turn we can prescribe which stone from the chosen square to move. We do it in the following manner. Number the stones from 1 to $n$. At any turn, after choosing a square, Sisyphus moves the stone with the largest number from this square.\n\nThis way, when stone $k$ is moved from some square, that square contains not more than $k$ stones (since all their numbers are at most $k$ ). Therefore, stone $k$ is moved by at most $k$ squares at each turn. Since the total shift of the stone is exactly $n$, at least $\\lceil n / k\\rceil$ moves of stone $k$ should have been made, for every $k=1,2, \\ldots, n$.\n\nBy summing up over all $k=1,2, \\ldots, n$, we get the required estimate.'] [] Text-only Competition True Theorem proof Combinatorics Math English 277 "Let $a$ and $b$ be distinct positive integers. The following infinite process takes place on an initially empty board. (i) If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by $a$ and the other by $b$. (ii) If no such pair exists, we write down two times the number 0. Prove that, no matter how we make the choices in $(i)$, operation $(i i)$ will be performed only finitely many times." "['We may assume $\\operatorname{gcd}(a, b)=1$; otherwise we work in the same way with multiples of $d=\\operatorname{gcd}(a, b)$.\n\nSuppose that after $N$ moves of type (ii) and some moves of type $(i)$ we have to add two new zeros. For each integer $k$, denote by $f(k)$ the number of times that the number $k$ appeared on the board up to this moment. Then $f(0)=2 N$ and $f(k)=0$ for $k<0$. Since the board contains at most one $k-a$, every second occurrence of $k-a$ on the board produced, at some moment, an occurrence of $k$; the same stands for $k-b$. Therefore,\n\n$$\nf(k)=\\left\\lfloor\\frac{f(k-a)}{2}\\right\\rfloor+\\left\\lfloor\\frac{f(k-b)}{2}\\right\\rfloor\n\\tag{1}\n$$\n\nyielding\n\n$$\nf(k) \\geqslant \\frac{f(k-a)+f(k-b)}{2}-1\n\\tag{2}\n$$\n\nSince $\\operatorname{gcd}(a, b)=1$, every integer $x>a b-a-b$ is expressible in the form $x=s a+t b$, with integer $s, t \\geqslant 0$.\n\nWe will prove by induction on $s+t$ that if $x=s a+b t$, with $s, t$ nonnegative integers, then\n\n$$\nf(x)>\\frac{f(0)}{2^{s+t}}-2\n\\tag{3}\n$$\n\nThe base case $s+t=0$ is trivial. Assume now that (3) is true for $s+t=v$. Then, if $s+t=v+1$ and $x=s a+t b$, at least one of the numbers $s$ and $t$ - say $s-$ is positive, hence by (2),\n\n$$\nf(x)=f(s a+t b) \\geqslant \\frac{f((s-1) a+t b)}{2}-1>\\frac{1}{2}\\left(\\frac{f(0)}{2^{s+t-1}}-2\\right)-1=\\frac{f(0)}{2^{s+t}}-2 .\n$$\n\nAssume now that we must perform moves of type (ii) ad infinitum. Take $n=a b-a-b$ and suppose $b>a$. Since each of the numbers $n+1, n+2, \\ldots, n+b$ can be expressed in the form $s a+t b$, with $0 \\leqslant s \\leqslant b$ and $0 \\leqslant t \\leqslant a$, after moves of type (ii) have been performed $2^{a+b+1}$ times and we have to add a new pair of zeros, each $f(n+k), k=1,2, \\ldots, b$, is at least 2 . In this case (1) yields inductively $f(n+k) \\geqslant 2$ for all $k \\geqslant 1$. But this is absurd: after a finite number of moves, $f$ cannot attain nonzero values at infinitely many points.' 'We start by showing that the result of the process in the problem does not depend on the way the operations are performed. For that purpose, it is convenient to modify the process a bit.\n\nClaim 1. Suppose that the board initially contains a finite number of nonnegative integers, and one starts performing type (i) moves only. Assume that one had applied $k$ moves which led to a final arrangement where no more type $(i)$ moves are possible. Then, if one starts from the same initial arrangement, performing type (i) moves in an arbitrary fashion, then the process will necessarily stop at the same final arrangement\n\nProof. Throughout this proof, all moves are supposed to be of type $(i)$.\n\nInduct on $k$; the base case $k=0$ is trivial, since no moves are possible. Assume now that $k \\geqslant 1$. Fix some canonical process, consisting of $k$ moves $M_{1}, M_{2}, \\ldots, M_{k}$, and reaching the final arrangement $A$. Consider any sample process $m_{1}, m_{2}, \\ldots$ starting with the same initial arrangement and proceeding as long as possible; clearly, it contains at least one move. We need to show that this process stops at $A$.\n\nLet move $m_{1}$ consist in replacing two copies of $x$ with $x+a$ and $x+b$. If move $M_{1}$ does the same, we may apply the induction hypothesis to the arrangement appearing after $m_{1}$. Otherwise, the canonical process should still contain at least one move consisting in replacing $(x, x) \\mapsto(x+a, x+b)$, because the initial arrangement contains at least two copies of $x$, while the final one contains at most one such.\n\nLet $M_{i}$ be the first such move. Since the copies of $x$ are indistinguishable and no other copy of $x$ disappeared before $M_{i}$ in the canonical process, the moves in this process can be permuted as $M_{i}, M_{1}, \\ldots, M_{i-1}, M_{i+1}, \\ldots, M_{k}$, without affecting the final arrangement. Now it suffices to perform the move $m_{1}=M_{i}$ and apply the induction hypothesis as above.\n\nClaim 2. Consider any process starting from the empty board, which involved exactly $n$ moves of type (ii) and led to a final arrangement where all the numbers are distinct. Assume that one starts with the board containing $2 n$ zeroes (as if $n$ moves of type (ii) were made in the beginning), applying type (i) moves in an arbitrary way. Then this process will reach the same final arrangement.\n\nProof. Starting with the board with $2 n$ zeros, one may indeed model the first process mentioned in the statement of the claim, omitting the type (ii) moves. This way, one reaches the same final arrangement. Now, Claim 1 yields that this final arrangement will be obtained when type (i) moves are applied arbitrarily.\n\nClaim 2 allows now to reformulate the problem statement as follows: There exists an integer $n$ such that, starting from $2 n$ zeroes, one may apply type (i) moves indefinitely.\n\nIn order to prove this, we start with an obvious induction on $s+t=k \\geqslant 1$ to show that if we start with $2^{s+t}$ zeros, then we can get simultaneously on the board, at some point, each of the numbers $s a+t b$, with $s+t=k$.\n\nSuppose now that $a\n\nNow $\\angle F B D=\\angle F D B$ gives $\\overparen{A F}=\\overparen{B F}+\\overparen{A K}=\\overparen{B F}+\\overparen{A L}$, hence $\\overparen{B F}=\\overparen{L F}$. In a similar way, we get $\\overparen{C G}=\\overparen{G K}$. This yields\n\n$$\n\\angle(A P, F G)=\\frac{\\overparen{A F}+\\overparen{P G}}{2}=\\frac{\\overparen{A L}+\\overparen{L F}+\\overparen{P C}+\\overparen{C G}}{2}=\\frac{\\overparen{K L}+\\overparen{L B}+\\overparen{B C}+\\overparen{C K}}{4}=90^{\\circ}\n$$' 'Let $Z=A B \\cap F G, T=A C \\cap F G$. It suffices to prove that $\\angle A T Z=\\angle A Z T$.\n\nLet $X$ be the point for which $F X A D$ is a parallelogram. Then\n\n$$\n\\angle F X A=\\angle F D A=180^{\\circ}-\\angle F D B=180^{\\circ}-\\angle F B D \\text {, }\n$$\n\nwhere in the last equality we used that $F D=F B$. It follows that the quadrilateral $B F X A$ is cyclic, so $X$ lies on $\\Gamma$.\n\n\n\n\n\nAnalogously, if $Y$ is the point for which $G Y A E$ is a parallelogram, then $Y$ lies on $\\Gamma$. So the quadrilateral $X F G Y$ is cyclic and $F X=A D=A E=G Y$, hence $X F G Y$ is an isosceles trapezoid.\n\nNow, by $X F \\| A Z$ and $Y G \\| A T$, it follows that $\\angle A T Z=\\angle Y G F=\\angle X F G=\\angle A Z T$.' 'As in the first solution, we prove that $F G \\perp A P$, where $P$ is the midpoint of the small arc $\\overparen{B C}$.\n\nLet $O$ be the circumcentre of the triangle $A B C$, and let $M$ and $N$ be the midpoints of the small arcs $\\overparen{A B}$ and $\\overparen{A C}$, respectively. Then $O M$ and $O N$ are the perpendicular bisectors of $A B$ and $A C$, respectively.\n\n\n\nThe distance $d$ between $O M$ and the perpendicular bisector of $B D$ is $\\frac{1}{2} A B-\\frac{1}{2} B D=\\frac{1}{2} A D$, hence it is equal to the distance between $O N$ and the perpendicular bisector of $C E$.\n\nThis shows that the isosceles trapezoid determined by the diameter $\\delta$ of $\\Gamma$ through $M$ and the chord parallel to $\\delta$ through $F$ is congruent to the isosceles trapezoid determined by the diameter $\\delta^{\\prime}$ of $\\Gamma$ through $N$ and the chord parallel to $\\delta^{\\prime}$ through $G$. Therefore $M F=N G$, yielding $M N \\| F G$.\n\nNow\n\n$$\n\\angle(M N, A P)=\\frac{1}{2}(\\overparen{A M}+\\overparen{P C}+\\overparen{C N})=\\frac{1}{4}(\\overparen{A B}+\\overparen{B C}+\\overparen{C A})=90^{\\circ}\n$$\n\nhence $M N \\perp A P$, and the conclusion follows.']" [] Text-only Competition True Theorem proof Geometry Math English 280 Let $A B C$ be a triangle with $A B=A C$, and let $M$ be the midpoint of $B C$. Let $P$ be a point such that $P B

\n\nNow let $Z$ be the common point of $A M$ and the perpendicular through $Y$ to $P C$ (notice that $Z$ lies on to the ray $A M$ beyond $M$ ). We have $\\angle P A Z=\\angle P Y Z=90^{\\circ}$. Thus the points $P, A, Y$, and $Z$ are concyclic.\n\nSince $\\angle C M Z=\\angle C Y Z=90^{\\circ}$, the quadrilateral $C Y Z M$ is cyclic, hence $\\angle C Z M=$ $\\angle C Y M$. By the condition in the statement, $\\angle C Y M=\\angle B X M$, and, by symmetry in $Z M$, $\\angle C Z M=\\angle B Z M$. Therefore, $\\angle B X M=\\angle B Z M$. It follows that the points $B, X, Z$, and $M$ are concyclic, hence $\\angle B X Z=180^{\\circ}-\\angle B M Z=90^{\\circ}$.\n\nFinally, we have $\\angle P X Z=\\angle P Y Z=\\angle P A Z=90^{\\circ}$, hence the five points $P, A, X, Y, Z$ are concyclic. In particular, the quadrilateral $A P X Y$ is cyclic, as required.'] [] Text-only Competition True Theorem proof Geometry Math English 281 A point $T$ is chosen inside a triangle $A B C$. Let $A_{1}, B_{1}$, and $C_{1}$ be the reflections of $T$ in $B C, C A$, and $A B$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_{1} B_{1} C_{1}$. The lines $A_{1} T, B_{1} T$, and $C_{1} T$ meet $\Omega$ again at $A_{2}, B_{2}$, and $C_{2}$, respectively. Prove that the lines $A A_{2}, B B_{2}$, and $C C_{2}$ are concurrent on $\Omega$. ['By $\\sphericalangle(\\ell, n)$ we always mean the directed angle of the lines $\\ell$ and $n$, taken modulo $180^{\\circ}$.\n\nLet $C C_{2}$ meet $\\Omega$ again at $K$ (as usual, if $C C_{2}$ is tangent to $\\Omega$, we set $T=C_{2}$ ). We show that the line $B B_{2}$ contains $K$; similarly, $A A_{2}$ will also pass through $K$. For this purpose, it suffices to prove that\n\n$$\n\\sphericalangle\\left(C_{2} C, C_{2} A_{1}\\right)=\\sphericalangle\\left(B_{2} B, B_{2} A_{1}\\right) .\n\\tag{1}\n$$\n\nBy the problem condition, $C B$ and $C A$ are the perpendicular bisectors of $T A_{1}$ and $T B_{1}$, respectively. Hence, $C$ is the circumcentre of the triangle $A_{1} T B_{1}$. Therefore,\n\n$$\n\\sphericalangle\\left(C A_{1}, C B\\right)=\\sphericalangle(C B, C T)=\\sphericalangle\\left(B_{1} A_{1}, B_{1} T\\right)=\\sphericalangle\\left(B_{1} A_{1}, B_{1} B_{2}\\right) .\n$$\n\nIn circle $\\Omega$ we have $\\sphericalangle\\left(B_{1} A_{1}, B_{1} B_{2}\\right)=\\sphericalangle\\left(C_{2} A_{1}, C_{2} B_{2}\\right)$. Thus,\n\n$$\n\\sphericalangle\\left(C A_{1}, C B\\right)=\\sphericalangle\\left(B_{1} A_{1}, B_{1} B_{2}\\right)=\\sphericalangle\\left(C_{2} A_{1}, C_{2} B_{2}\\right) .\n\\tag{2}\n$$\n\nSimilarly, we get\n\n$$\n\\sphericalangle\\left(B A_{1}, B C\\right)=\\sphericalangle\\left(C_{1} A_{1}, C_{1} C_{2}\\right)=\\sphericalangle\\left(B_{2} A_{1}, B_{2} C_{2}\\right) .\n\\tag{3}\n$$\n\nThe two obtained relations yield that the triangles $A_{1} B C$ and $A_{1} B_{2} C_{2}$ are similar and equioriented, hence\n\n$$\n\\frac{A_{1} B_{2}}{A_{1} B}=\\frac{A_{1} C_{2}}{A_{1} C} \\quad \\text { and } \\quad \\sphericalangle\\left(A_{1} B, A_{1} C\\right)=\\sphericalangle\\left(A_{1} B_{2}, A_{1} C_{2}\\right) \\text {. }\n$$\n\nThe second equality may be rewritten as $\\sphericalangle\\left(A_{1} B, A_{1} B_{2}\\right)=\\sphericalangle\\left(A_{1} C, A_{1} C_{2}\\right)$, so the triangles $A_{1} B B_{2}$ and $A_{1} C C_{2}$ are also similar and equioriented. This establishes (1).\n\n'] [] Text-only Competition True Theorem proof Geometry Math English 282 "Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark) Preamble. Let $X=y \cap z, Y=x \cap z, Z=x \cap y$ and let $\Omega$ denote the circumcircle of the triangle $X Y Z$. Denote by $X_{0}, Y_{0}$, and $Z_{0}$ the second intersection points of $A I, B I$ and $C I$, respectively, with $\omega$. It is known that $Y_{0} Z_{0}$ is the perpendicular bisector of $A I, Z_{0} X_{0}$ is the perpendicular bisector of $B I$, and $X_{0} Y_{0}$ is the perpendicular bisector of $C I$. In particular, the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$ are homothetic, because their corresponding sides are parallel. The solutions below mostly exploit the following approach. Consider the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$, or some other pair of homothetic triangles $\Delta$ and $\delta$ inscribed into $\Omega$ and $\omega$, respectively. In order to prove that $\Omega$ and $\omega$ are tangent, it suffices to show that the centre $T$ of the homothety taking $\Delta$ to $\delta$ lies on $\omega$ (or $\Omega$ ), or, in other words, to show that $\Delta$ and $\delta$ are perspective (i.e., the lines joining corresponding vertices are concurrent), with their perspector lying on $\omega$ (or $\Omega$ ). We use directed angles throughout all the solutions." "['Claim 1. The reflections $\\ell_{a}, \\ell_{b}$ and $\\ell_{c}$ of the line $\\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\\omega$.\n\n\n\nProof. Notice that $\\sphericalangle\\left(\\ell_{b}, \\ell_{c}\\right)=\\sphericalangle\\left(\\ell_{b}, \\ell\\right)+\\sphericalangle\\left(\\ell, \\ell_{c}\\right)=2 \\sphericalangle(y, \\ell)+2 \\sphericalangle(\\ell, z)=2 \\sphericalangle(y, z)$. But $y \\perp B I$ and $z \\perp C I$ implies $\\sphericalangle(y, z)=\\sphericalangle(B I, I C)$, so, since $2 \\sphericalangle(B I, I C)=\\sphericalangle(B A, A C)$, we obtain\n\n$$\n\\sphericalangle\\left(\\ell_{b}, \\ell_{c}\\right)=\\sphericalangle(B A, A C) .\n\\tag{1}\n$$\n\nSince $A$ is the reflection of $D$ in $x, A$ belongs to $\\ell_{a}$; similarly, $B$ belongs to $\\ell_{b}$. Then (1) shows that the common point $T^{\\prime}$ of $\\ell_{a}$ and $\\ell_{b}$ lies on $\\omega$; similarly, the common point $T^{\\prime \\prime}$ of $\\ell_{c}$ and $\\ell_{b}$ lies on $\\omega$.\n\nIf $B \\notin \\ell_{a}$ and $B \\notin \\ell_{c}$, then $T^{\\prime}$ and $T^{\\prime \\prime}$ are the second point of intersection of $\\ell_{b}$ and $\\omega$, hence they coincide. Otherwise, if, say, $B \\in \\ell_{c}$, then $\\ell_{c}=B C$, so $\\sphericalangle(B A, A C)=\\sphericalangle\\left(\\ell_{b}, \\ell_{c}\\right)=\\sphericalangle\\left(\\ell_{b}, B C\\right)$, which shows that $\\ell_{b}$ is tangent at $B$ to $\\omega$ and $T^{\\prime}=T^{\\prime \\prime}=B$. So $T^{\\prime}$ and $T^{\\prime \\prime}$ coincide in all the cases, and the conclusion of the claim follows.\n\n\n\nNow we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\\ell_{b}, D_{c}$ lies on $\\ell_{c}$, and\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(D_{b} X, X D_{c}\\right) & =\\sphericalangle\\left(D_{b} X, D X\\right)+\\sphericalangle\\left(D X, X D_{c}\\right)=2 \\sphericalangle(y, D X)+2 \\sphericalangle(D X, z)=2 \\sphericalangle(y, z) \\\\\n& =\\sphericalangle(B A, A C)=\\sphericalangle(B T, T C),\n\\end{aligned}\n$$\n\nhence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\\prime}$ yields $\\sphericalangle\\left(D_{b} D_{c}, D_{c} X\\right)=90^{\\circ}+\\sphericalangle\\left(D_{b} D_{c}^{\\prime}, D_{c}^{\\prime} X\\right)=90^{\\circ}+\\sphericalangle\\left(D_{b} D, D D_{c}\\right)$. Therefore,\n\n$$\n\\begin{gathered}\n\\sphericalangle\\left(\\ell_{b}, X T\\right)=\\sphericalangle\\left(D_{b} T, X T\\right)=\\sphericalangle\\left(D_{b} D_{c}, D_{c} X\\right)=90^{\\circ}+\\sphericalangle\\left(D_{b} D, D D_{c}\\right) \\\\\n=90^{\\circ}+\\sphericalangle(B I, I C)=\\sphericalangle(B A, A I)=\\sphericalangle\\left(B A, A X_{0}\\right)=\\sphericalangle\\left(B T, T X_{0}\\right)=\\sphericalangle\\left(\\ell_{b}, X_{0} T\\right),\n\\end{gathered}\n$$\n\nso the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows.' 'As mentioned in the preamble, it is sufficient to prove that the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\\omega$. Thus, it suffices to prove that $\\sphericalangle\\left(T X_{0}, T Y_{0}\\right)=$ $\\sphericalangle\\left(Z_{0} X_{0}, Z_{0} Y_{0}\\right)$, or, equivalently, $\\sphericalangle\\left(X X_{0}, Y Y_{0}\\right)=\\sphericalangle\\left(Z_{0} X_{0}, Z_{0} Y_{0}\\right)$.\n\nRecall that $Y Z$ and $Y_{0} Z_{0}$ are the perpendicular bisectors of $A D$ and $A I$, respectively. Then, the vector $\\vec{x}$ perpendicular to $Y Z$ and shifting the line $Y_{0} Z_{0}$ to $Y Z$ is equal to $\\frac{1}{2} \\overrightarrow{I D}$. Define the shifting vectors $\\vec{y}=\\frac{1}{2} \\overrightarrow{I E}, \\vec{z}=\\frac{1}{2} \\overrightarrow{I F}$ similarly. Consider now the triangle $U V W$ formed by the perpendiculars to $A I, B I$, and $C I$ through $D, E$, and $F$, respectively (see figure below). This is another triangle whose sides are parallel to the corresponding sides of $X Y Z$.\n\nClaim 2. $\\overrightarrow{I U}=2 \\overrightarrow{X_{0} X}, \\overrightarrow{I V}=2 \\overrightarrow{Y_{0} Y}, \\overrightarrow{I W}=2 \\overrightarrow{Z_{0} Z}$.\n\nProof. We prove one of the relations, the other proofs being similar. To prove the equality of two vectors it suffices to project them onto two non-parallel axes and check that their projections are equal.\n\nThe projection of $\\overrightarrow{X_{0} X}$ onto $I B$ equals $\\vec{y}$, while the projection of $\\overrightarrow{I U}$ onto $I B$ is $\\overrightarrow{I E}=2 \\vec{y}$. The projections onto the other axis $I C$ are $\\vec{z}$ and $\\overrightarrow{I F}=2 \\vec{z}$. Then $\\overrightarrow{I U}=2 \\overrightarrow{X_{0} X}$ follows.\n\nNotice that the line $\\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$; thus $U, V, W$, and $I$ are concyclic. It follows from Claim 2 that $\\sphericalangle\\left(X X_{0}, Y Y_{0}\\right)=\\sphericalangle(I U, I V)=$ $\\sphericalangle(W U, W V)=\\sphericalangle\\left(Z_{0} X_{0}, Z_{0} Y_{0}\\right)$, and we are done.\n\n' 'Let $I_{a}, I_{b}$, and $I_{c}$ be the excentres of triangle $A B C$ corresponding to $A, B$, and $C$, respectively. Also, let $u, v$, and $w$ be the lines through $D, E$, and $F$ which are perpendicular to $A I, B I$, and $C I$, respectively, and let $U V W$ be the triangle determined by these lines, where $u=V W, v=U W$ and $w=U V$ (see figure above).\n\nNotice that the line $u$ is the reflection of $I_{b} I_{c}$ in the line $x$, because $u, x$, and $I_{b} I_{c}$ are perpendicular to $A D$ and $x$ is the perpendicular bisector of $A D$. Likewise, $v$ and $I_{a} I_{c}$ are reflections of each other in $y$, while $w$ and $I_{a} I_{b}$ are reflections of each other in $z$. It follows that $X, Y$, and $Z$ are the midpoints of $U I_{a}, V I_{b}$ and $W I_{c}$, respectively, and that the triangles $U V W$, $X Y Z$ and $I_{a} I_{b} I_{c}$ are either translates of each other or homothetic with a common homothety centre.\n\nConstruct the points $T$ and $S$ such that the quadrilaterals $U V I W, X Y T Z$ and $I_{a} I_{b} S I_{c}$ are homothetic. Then $T$ is the midpoint of $I S$. Moreover, note that $\\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$, hence $I$ belongs to the circumcircle of the triangle $U V W$, therefore $T$ belongs to $\\Omega$.\n\n\n\nConsider now the homothety or translation $h_{1}$ that maps $X Y Z T$ to $I_{a} I_{b} I_{c} S$ and the homothety $h_{2}$ with centre $I$ and factor $\\frac{1}{2}$. Furthermore, let $h=h_{2} \\circ h_{1}$. The transform $h$ can be a homothety or a translation, and\n\n$$\nh(T)=h_{2}\\left(h_{1}(T)\\right)=h_{2}(S)=T\n$$\n\nhence $T$ is a fixed point of $h$. So, $h$ is a homothety with centre $T$. Note that $h_{2}$ maps the excentres $I_{a}, I_{b}, I_{c}$ to $X_{0}, Y_{0}, Z_{0}$ defined in the preamble. Thus the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\\Omega$, and this completes the proof.']" [] Text-only Competition True Theorem proof Geometry Math English 283 A convex quadrilateral $A B C D$ satisfies $A B \cdot C D=B C \cdot D A$. A point $X$ is chosen inside the quadrilateral so that $\angle X A B=\angle X C D$ and $\angle X B C=\angle X D A$. Prove that $\angle A X B+$ $\angle C X D=180^{\circ}$. "['Let $B^{\\prime}$ be the reflection of $B$ in the internal angle bisector of $\\angle A X C$, so that $\\angle A X B^{\\prime}=\\angle C X B$ and $\\angle C X B^{\\prime}=\\angle A X B$. If $X, D$, and $B^{\\prime}$ are collinear, then we are done. Now assume the contrary.\n\nOn the ray $X B^{\\prime}$ take a point $E$ such that $X E \\cdot X B=X A \\cdot X C$, so that $\\triangle A X E \\sim$ $\\triangle B X C$ and $\\triangle C X E \\sim \\triangle B X A$. We have $\\angle X C E+\\angle X C D=\\angle X B A+\\angle X A B<180^{\\circ}$ and $\\angle X A E+\\angle X A D=\\angle X D A+\\angle X A D<180^{\\circ}$, which proves that $X$ lies inside the angles $\\angle E C D$ and $\\angle E A D$ of the quadrilateral $E A D C$. Moreover, $X$ lies in the interior of exactly one of the two triangles $E A D, E C D$ (and in the exterior of the other).\n\n\n\nThe similarities mentioned above imply $X A \\cdot B C=X B \\cdot A E$ and $X B \\cdot C E=X C \\cdot A B$. Multiplying these equalities with the given equality $A B \\cdot C D=B C \\cdot D A$, we obtain $X A \\cdot C D$. $C E=X C \\cdot A D \\cdot A E$, or, equivalently,\n\n$$\n\\frac{X A \\cdot D E}{A D \\cdot A E}=\\frac{X C \\cdot D E}{C D \\cdot C E}\n\\tag{*}\n$$\n\nLemma. Let $P Q R$ be a triangle, and let $X$ be a point in the interior of the angle $Q P R$ such that $\\angle Q P X=\\angle P R X$. Then $\\frac{P X \\cdot Q R}{P Q \\cdot P R}<1$ if and only if $X$ lies in the interior of the triangle $P Q R$. Proof. The locus of points $X$ with $\\angle Q P X=\\angle P R X$ lying inside the angle $Q P R$ is an arc $\\alpha$ of the circle $\\gamma$ through $R$ tangent to $P Q$ at $P$. Let $\\gamma$ intersect the line $Q R$ again at $Y$ (if $\\gamma$ is tangent to $Q R$, then set $Y=R$ ). The similarity $\\triangle Q P Y \\sim \\triangle Q R P$ yields $P Y=\\frac{P Q \\cdot P R}{Q R}$. Now it suffices to show that $P X

P Y$ for $X \\in \\overparen{Y R}$.\n\n\n\nCase 2: $Y$ lies on the ray $Q R$ beyond $R$ (see the right figure below).\n\nIn this case the whole arc $\\alpha$ lies inside triangle $P Q R$, and between $m$ and $P Q$, thus $P X<$ $P Y$ for all $X \\in \\alpha$.\n\n\nApplying the Lemma (to $\\triangle E A D$ with the point $X$, and to $\\triangle E C D$ with the point $X$ ), we obtain that exactly one of two expressions $\\frac{X A \\cdot D E}{A D \\cdot A E}$ and $\\frac{X C \\cdot D E}{C D \\cdot C E}$ is less than 1, which contradicts $(*)$.' 'The solution consists of two parts. In Part 1 we show that it suffices to prove that\n\n$$\n\\frac{X B}{X D}=\\frac{A B}{C D}\n$$\n\nand\n\n$$\n\\frac{X A}{X C}=\\frac{D A}{B C}\n$$\n\nIn Part 2 we establish these equalities.\n\nPart 1. Using the sine law and applying (1) we obtain\n\n$$\n\\frac{\\sin \\angle A X B}{\\sin \\angle X A B}=\\frac{A B}{X B}=\\frac{C D}{X D}=\\frac{\\sin \\angle C X D}{\\sin \\angle X C D}\n$$\n\nso $\\sin \\angle A X B=\\sin \\angle C X D$ by the problem conditions. Similarly, (2) yields $\\sin \\angle D X A=$ $\\sin \\angle B X C$. If at least one of the pairs $(\\angle A X B, \\angle C X D)$ and $(\\angle B X C, \\angle D X A)$ consists of supplementary angles, then we are done. Otherwise, $\\angle A X B=\\angle C X D$ and $\\angle D X A=\\angle B X C$. In this case $X=A C \\cap B D$, and the problem conditions yield that $A B C D$ is a parallelogram and hence a rhombus. In this last case the claim also holds.\n\nPart 2. To prove the desired equality (1), invert $A B C D$ at centre $X$ with unit radius; the images of points are denoted by primes.\n\nWe have\n\n$$\n\\angle A^{\\prime} B^{\\prime} C^{\\prime}=\\angle X B^{\\prime} A^{\\prime}+\\angle X B^{\\prime} C^{\\prime}=\\angle X A B+\\angle X C B=\\angle X C D+\\angle X C B=\\angle B C D .\n$$\n\n\n\nSimilarly, the corresponding angles of quadrilaterals $A B C D$ and $D^{\\prime} A^{\\prime} B^{\\prime} C^{\\prime}$ are equal.\n\nMoreover, we have\n\n$$\nA^{\\prime} B^{\\prime} \\cdot C^{\\prime} D^{\\prime}=\\frac{A B}{X A \\cdot X B} \\cdot \\frac{C D}{X C \\cdot X D}=\\frac{B C}{X B \\cdot X C} \\cdot \\frac{D A}{X D \\cdot D A}=B^{\\prime} C^{\\prime} \\cdot D^{\\prime} A^{\\prime}\n$$\n\n\n\nNow we need the following Lemma.\n\nLemma. Assume that the corresponding angles of convex quadrilaterals $X Y Z T$ and $X^{\\prime} Y^{\\prime} Z^{\\prime} T^{\\prime}$ are equal, and that $X Y \\cdot Z T=Y Z \\cdot T X$ and $X^{\\prime} Y^{\\prime} \\cdot Z^{\\prime} T^{\\prime}=Y^{\\prime} Z^{\\prime} \\cdot T^{\\prime} X^{\\prime}$. Then the two quadrilaterals are similar.\n\nProof. Take the quadrilateral $X Y Z_{1} T_{1}$ similar to $X^{\\prime} Y^{\\prime} Z^{\\prime} T^{\\prime}$ and sharing the side $X Y$ with $X Y Z T$, such that $Z_{1}$ and $T_{1}$ lie on the rays $Y Z$ and $X T$, respectively, and $Z_{1} T_{1} \\| Z T$. We need to prove that $Z_{1}=Z$ and $T_{1}=T$. Assume the contrary. Without loss of generality, $T X>X T_{1}$. Let segments $X Z$ and $Z_{1} T_{1}$ intersect at $U$. We have\n\n$$\n\\frac{T_{1} X}{T_{1} Z_{1}}<\\frac{T_{1} X}{T_{1} U}=\\frac{T X}{Z T}=\\frac{X Y}{Y Z}<\\frac{X Y}{Y Z_{1}}\n$$\n\nthus $T_{1} X \\cdot Y Z_{1}\n\nIt follows from the Lemma that the quadrilaterals $A B C D$ and $D^{\\prime} A^{\\prime} B^{\\prime} C^{\\prime}$ are similar, hence\n\n$$\n\\frac{B C}{A B}=\\frac{A^{\\prime} B^{\\prime}}{D^{\\prime} A^{\\prime}}=\\frac{A B}{X A \\cdot X B} \\cdot \\frac{X D \\cdot X A}{D A}=\\frac{A B}{A D} \\cdot \\frac{X D}{X B}\n$$\n\nand therefore\n\n$$\n\\frac{X B}{X D}=\\frac{A B^{2}}{B C \\cdot A D}=\\frac{A B^{2}}{A B \\cdot C D}=\\frac{A B}{C D}\n$$\n\nWe obtain (1), as desired; (2) is proved similarly.']" [] Text-only Competition True Theorem proof Geometry Math English 284 Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\ell_{A}, \ell_{B}$, and $\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$ is tangent to the line $O P$. ['As usual, we denote the directed angle between the lines $a$ and $b$ by $\\sphericalangle(a, b)$. We frequently use the fact that $a_{1} \\perp a_{2}$ and $b_{1} \\perp b_{2}$ yield $\\sphericalangle\\left(a_{1}, b_{1}\\right)=\\sphericalangle\\left(a_{2}, b_{2}\\right)$.\n\nLet the lines $\\ell_{B}$ and $\\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.\n\n\n\nClaim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic.\n\nProof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have\n\n$$\n\\sphericalangle\\left(O_{A} P, O_{C} P\\right)=\\sphericalangle\\left(O_{C} O, O_{A} O\\right)=\\sphericalangle(C P, A P)=\\sphericalangle(C B, A B)=\\sphericalangle\\left(O_{A} L_{B}, O_{C} L_{B}\\right) .\n$$\n\nDenote the circle through $L_{B}, P, O_{A}$, and $O_{C}$ by $\\omega_{B}$. Define the circles $\\omega_{A}$ and $\\omega_{C}$ similarly. Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$.\n\nProof. From cyclic quadruples of points in the circles $\\omega_{B}$ and $\\omega_{C}$, we have\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(L_{C} L_{A}, L_{C} P\\right) & =\\sphericalangle\\left(L_{C} O_{B}, L_{C} P\\right)=\\sphericalangle\\left(O_{A} O_{B}, O_{A} P\\right) \\\\\n& =\\sphericalangle\\left(O_{A} O_{C}, O_{A} P\\right)=\\sphericalangle\\left(L_{B} O_{C}, L_{B} P\\right)=\\sphericalangle\\left(L_{B} L_{A}, L_{B} P\\right) .\n\\end{aligned}\n$$\n\nClaim 3. The points $P, L_{C}$, and $C$ are collinear.\n\nProof. We have $\\sphericalangle\\left(P L_{C}, L_{C} L_{A}\\right)=\\sphericalangle\\left(P L_{C}, L_{C} O_{B}\\right)=\\sphericalangle\\left(P O_{A}, O_{A} O_{B}\\right)$. Further, since $O_{A}$ is the centre of the circle $A O P, \\sphericalangle\\left(P O_{A}, O_{A} O_{B}\\right)=\\sphericalangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \\sphericalangle(P A, A O)=\\pi / 2-\\sphericalangle(C A, C P)=\\sphericalangle\\left(C P, L_{C} L_{A}\\right)$. We obtain $\\sphericalangle\\left(P L_{C}, L_{C} L_{A}\\right)=$ $\\sphericalangle\\left(C P, L_{C} L_{A}\\right)$, which shows that $P \\in C L_{C}$.\n\n\n\nSimilarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that\n\n$$\n\\sphericalangle\\left(O P, P L_{A}\\right)=\\sphericalangle(P A, A O)=\\sphericalangle\\left(P L_{C}, L_{C} L_{A}\\right),\n$$\n\nwhich means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$.'] [] Text-only Competition True Theorem proof Geometry Math English 285 "Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied: (i) Each number in the table is congruent to 1 modulo $n$; (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$. Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$." "['Let $A_{i, j}$ be the entry in the $i^{\\text {th }}$ row and the $j^{\\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that\n\n$$\n\\sum_{i=1}^{n} R_{i} \\equiv(n-1)+P \\quad\\left(\\bmod n^{4}\\right)\n\\tag{1}\n$$\n\nDue to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion.\n\nBy condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence\n\n$R_{i}=\\prod_{j=1}^{n}\\left(1+a_{i, j}\\right)=1+\\sum_{j=1}^{n} a_{i, j}+\\sum_{1 \\leqslant j_{1}3$, so\n\n$$\nS_{n-1}=2^{n}+2^{\\lceil n / 2\\rceil}+2^{\\lfloor n / 2\\rfloor}-3>2^{n}+2^{\\lfloor n / 2\\rfloor}=a_{n} .\n$$\n\nAlso notice that $S_{n-1}-a_{n}=2^{[n / 2]}-3a_{n}$. Denote $c=S_{n-1}-b$; then $S_{n-1}-a_{n}2^{t}-3$.\n\nProof. The inequality follows from $t \\geqslant 3$. In order to prove the equivalence, we apply Claim 1 twice in the following manner.\n\nFirst, since $S_{2 t-3}-a_{2 t-2}=2^{t-1}-3<2^{t}-3<2^{2 t-2}+2^{t-1}=a_{2 t-2}$, by Claim 1 we have $2^{t}-3 \\sim S_{2 t-3}-\\left(2^{t}-3\\right)=2^{2 t-2}$.\n\nSecond, since $S_{4 t-7}-a_{4 t-6}=2^{2 t-3}-3<2^{2 t-2}<2^{4 t-6}+2^{2 t-3}=a_{4 t-6}$, by Claim 1 we have $2^{2 t-2} \\sim S_{4 t-7}-2^{2 t-2}=2^{4 t-6}-3$.\n\nTherefore, $2^{t}-3 \\sim 2^{2 t-2} \\sim 2^{4 t-6}-3$, as required.\n\nNow it is easy to find the required numbers. Indeed, the number $2^{3}-3=5=a_{0}+a_{1}$ is representable, so Claim 3 provides an infinite sequence of representable numbers\n\n$$\n2^{3}-3 \\sim 2^{6}-3 \\sim 2^{18}-3 \\sim \\cdots \\sim 2^{t}-3 \\sim 2^{4 t-6}-3 \\sim \\cdots\n$$\n\nOn the other hand, the number $2^{7}-3=125$ is non-representable (since by Claim 1 we have $125 \\sim S_{6}-125=24 \\sim S_{4}-24=17 \\sim S_{3}-17=4$ which is clearly non-representable). So Claim 3 provides an infinite sequence of non-representable numbers\n\n$$\n2^{7}-3 \\sim 2^{22}-3 \\sim 2^{82}-3 \\sim \\cdots \\sim 2^{t}-3 \\sim 2^{4 t-6}-3 \\sim \\cdots\n$$' 'We keep the notion of representability and the notation $S_{n}$ from the previous solution. We say that an index $n$ is good if $a_{n}$ writes as a sum of smaller terms from the sequence $a_{0}, a_{1}, \\ldots$ Otherwise we say it is bad. We must prove that there are infinitely many good indices, as well as infinitely many bad ones.\n\nLemma 1. If $m \\geqslant 0$ is an integer, then $4^{m}$ is representable if and only if either of $2 m+1$ and $2 m+2$ is good.\n\nProof. The case $m=0$ is obvious, so we may assume that $m \\geqslant 1$. Let $n=2 m+1$ or $2 m+2$. Then $n \\geqslant 3$. We notice that\n\n$$\nS_{n-1}4^{s}$.\n\nProof. We have $2^{4 k-2}s$.\n\nNow $4^{2}=a_{2}+a_{3}$ is representable, whereas $4^{6}=4096$ is not. Indeed, note that $4^{6}=2^{12}v_{p}\\left(a_{n}\\right)$, then $v_{p}\\left(a_{n} / a_{n+1}\\right)<0$, while $v_{p}\\left(\\left(a_{n+1}-a_{n}\\right) / a_{1}\\right) \\geqslant 0$, so $(*)$ is not integer again. Thus, $v_{p}\\left(a_{1}\\right) \\leqslant v_{p}\\left(a_{n+1}\\right) \\leqslant v_{p}\\left(a_{n}\\right)$.\n\nThe above arguments can now be applied successively to indices $n+1, n+2, \\ldots$, showing that all the indices greater than $n$ are large, and the sequence $v_{p}\\left(a_{n}\\right), v_{p}\\left(a_{n+1}\\right), v_{p}\\left(a_{n+2}\\right), \\ldots$ is nonincreasing - hence eventually constant.\n\nCase 2: There is no large index.\n\nWe have $v_{p}\\left(a_{1}\\right)>v_{p}\\left(a_{n}\\right)$ for all $n \\geqslant k$. If we had $v_{p}\\left(a_{n+1}\\right)1$ which divides all values of $f$. "['For every positive integer $m$, define $S_{m}=\\{n: m \\mid f(n)\\}$.\n\nLemma. If the set $S_{m}$ is infinite, then $S_{m}=\\{d, 2 d, 3 d, \\ldots\\}=d \\cdot \\mathbb{Z}_{>0}$ for some positive integer $d$. Proof. Let $d=\\min S_{m}$; the definition of $S_{m}$ yields $m \\mid f(d)$.\n\nWhenever $n \\in S_{m}$ and $n>d$, we have $m|f(n)| f(n-d)+f(d)$, so $m \\mid f(n-d)$ and therefore $n-d \\in S_{m}$. Let $r \\leqslant d$ be the least positive integer with $n \\equiv r(\\bmod d)$; repeating the same step, we can see that $n-d, n-2 d, \\ldots, r \\in S_{m}$. By the minimality of $d$, this shows $r=d$ and therefore $d \\mid n$.\n\nStarting from an arbitrarily large element of $S_{m}$, the process above reaches all multiples of $d$; so they all are elements of $S_{m}$.\n\nThe solution for the problem will be split into two cases.\n\nCase 1: The function $f$ is bounded.\n\nCall a prime $p$ frequent if the set $S_{p}$ is infinite, i.e., if $p$ divides $f(n)$ for infinitely many positive integers $n$; otherwise call $p$ sporadic. Since the function $f$ is bounded, there are only a finite number of primes that divide at least one $f(n)$; so altogether there are finitely many numbers $n$ such that $f(n)$ has a sporadic prime divisor. Let $N$ be a positive integer, greater than all those numbers $n$.\n\nLet $p_{1}, \\ldots, p_{k}$ be the frequent primes. By the lemma we have $S_{p_{i}}=d_{i} \\cdot \\mathbb{Z}_{>0}$ for some $d_{i}$. Consider the number\n\n$$\nn=N d_{1} d_{2} \\cdots d_{k}+1\n$$\n\nDue to $n>N$, all prime divisors of $f(n)$ are frequent primes. Let $p_{i}$ be any frequent prime divisor of $f(n)$. Then $n \\in S_{p_{i}}$, and therefore $d_{i} \\mid n$. But $n \\equiv 1\\left(\\bmod d_{i}\\right)$, which means $d_{i}=1$. Hence $S_{p_{i}}=1 \\cdot \\mathbb{Z}_{>0}=\\mathbb{Z}_{>0}$ and therefore $p_{i}$ is a common divisor of all values $f(n)$.\n\nCase 2: $f$ is unbounded.\n\nWe prove that $f(1)$ divides all $f(n)$.\n\nLet $a=f(1)$. Since $1 \\in S_{a}$, by the lemma it suffices to prove that $S_{a}$ is an infinite set.\n\nCall a positive integer $p$ a peak if $f(p)>\\max (f(1), \\ldots, f(p-1))$. Since $f$ is not bounded, there are infinitely many peaks. Let $1=p_{1}0}$ are coprime then $\\operatorname{gcd}(f(a), f(b)) \\mid f(1)$. In particular, if $a, b \\geqslant n_{0}$ are coprime then $f(a)$ and $f(b)$ are coprime.\n\nProof. Let $d=\\operatorname{gcd}(f(a), f(b))$. We can replicate Euclid's algorithm. Formally, apply induction on $a+b$. If $a=1$ or $b=1$ then we already have $d \\mid f(1)$.\n\nWithout loss of generality, suppose $1C$ (that is possible, because there are arbitrarily long gaps between the primes). Then we establish a contradiction\n\n$$\np_{N+1} \\leqslant \\max \\left(f(1), f\\left(q_{1}\\right), \\ldots, f\\left(q_{N}\\right)\\right)<\\max \\left(1+C, q_{1}+C, \\ldots, q_{N}+C\\right)=p_{N}+C0$ and $c, d$ are coprime.\n\nWe will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \\leqslant i \\leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\\left(b_{i}\\right)5 n\n$$\n\na contradiction.\n\nClaim 3. For every $0 \\leqslant k \\leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \\ldots, b_{k+30}$, at least $\\varphi(30)=8$ are divisible by $d$.\n\nProof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \\cdot(3-1) \\cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \\mid b_{i}$ for every uncovered index $i$.\n\nClaim 4. $|\\Delta|<\\frac{20}{n-2}$ and $d>\\frac{n-2}{20}$.\n\nProof. From the sequence (1), remove all fractions with $b_{n}<\\frac{n}{2}$, There remain at least $\\frac{n}{2}$ fractions, and they cannot exceed $\\frac{5 n}{n / 2}=10$. So we have at least $\\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\\frac{10}{n / 2-1}=\\frac{20}{n-2}$.\n\nThe second inequality follows from $\\frac{1}{d} \\leqslant \\frac{|c|}{d}=|\\Delta|$.\n\nNow we have everything to get the final contradiction. By Claim 3, we have $d \\mid b_{i}$ for at least $\\left\\lfloor\\frac{n}{30}\\right\\rfloor \\cdot 8$ indices $i$. By Claim 4, we have $d \\geqslant \\frac{n-2}{20}$. Therefore,\n\n$$\n5 n \\geqslant \\max \\left\\{b_{i}: d \\mid b_{i}\\right\\} \\geqslant\\left(\\left\\lfloor\\frac{n}{30}\\right\\rfloor \\cdot 8\\right) \\cdot d>\\left(\\frac{n}{30}-1\\right) \\cdot 8 \\cdot \\frac{n-2}{20}>5 n .\n$$'] [] Text-only Competition True Theorem proof Number Theory Math English 290 "Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=v_{1}=1$, and $$ u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text { for } k=1, \ldots, n-1 $$ Prove that $u_{n}=v_{n}$." "['We prove by induction on $k$ that\n\n$$\nu_{k}=\\sum_{\\substack{0i_{1}>\\ldots>i_{t}>n-k \\\\ i_{j}-i_{j+1} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}}\n\\tag{2}\n$$\n\nFor $k=n$ the expressions (1) and (2) coincide, so indeed $u_{n}=v_{n}$.' 'Define recursively a sequence of multivariate polynomials by\n\n$$\nP_{0}=P_{1}=1, \\quad P_{k+1}\\left(x_{1}, \\ldots, x_{k}\\right)=P_{k}\\left(x_{1}, \\ldots, x_{k-1}\\right)+x_{k} P_{k-1}\\left(x_{1}, \\ldots, x_{k-2}\\right),\n$$\n\nso $P_{n}$ is a polynomial in $n-1$ variables for each $n \\geqslant 1$. Two easy inductive arguments show that\n\n$$\nu_{n}=P_{n}\\left(a_{1}, \\ldots, a_{n-1}\\right), \\quad v_{n}=P_{n}\\left(a_{n-1}, \\ldots, a_{1}\\right)\n$$\n\n\n\nso we need to prove $P_{n}\\left(x_{1}, \\ldots, x_{n-1}\\right)=P_{n}\\left(x_{n-1}, \\ldots, x_{1}\\right)$ for every positive integer $n$. The cases $n=1,2$ are trivial, and the cases $n=3,4$ follow from $P_{3}(x, y)=1+x+y$ and $P_{4}(x, y, z)=$ $1+x+y+z+x z$.\n\nNow we proceed by induction, assuming that $n \\geqslant 5$ and the claim hold for all smaller cases. Using $F(a, b)$ as an abbreviation for $P_{|a-b|+1}\\left(x_{a}, \\ldots, x_{b}\\right)$ (where the indices $a, \\ldots, b$ can be either in increasing or decreasing order),\n\n$$\n\\begin{aligned}\nF(n, 1) & =F(n, 2)+x_{1} F(n, 3)=F(2, n)+x_{1} F(3, n) \\\\\n& =\\left(F(2, n-1)+x_{n} F(2, n-2)\\right)+x_{1}\\left(F(3, n-1)+x_{n} F(3, n-2)\\right) \\\\\n& =\\left(F(n-1,2)+x_{1} F(n-1,3)\\right)+x_{n}\\left(F(n-2,2)+x_{1} F(n-2,3)\\right) \\\\\n& =F(n-1,1)+x_{n} F(n-2,1)=F(1, n-1)+x_{n} F(1, n-2) \\\\\n& =F(1, n),\n\\end{aligned}\n$$\n\nas we wished to show.' 'Using matrix notation, we can rewrite the recurrence relation as\n\n$$\n\\left(\\begin{array}{c}\nu_{k+1} \\\\\nu_{k+1}-u_{k}\n\\end{array}\\right)=\\left(\\begin{array}{c}\nu_{k}+a_{k} u_{k-1} \\\\\na_{k} u_{k-1}\n\\end{array}\\right)=\\left(\\begin{array}{cc}\n1+a_{k} & -a_{k} \\\\\na_{k} & -a_{k}\n\\end{array}\\right)\\left(\\begin{array}{c}\nu_{k} \\\\\nu_{k}-u_{k-1}\n\\end{array}\\right)\n$$\n\nfor $1 \\leqslant k \\leqslant n-1$, and similarly\n\n$$\n\\left(v_{k+1} ; v_{k}-v_{k+1}\\right)=\\left(v_{k}+a_{n-k} v_{k-1} ;-a_{n-k} v_{k-1}\\right)=\\left(v_{k} ; v_{k-1}-v_{k}\\right)\\left(\\begin{array}{cc}\n1+a_{n-k} & -a_{n-k} \\\\\na_{n-k} & -a_{n-k}\n\\end{array}\\right)\n$$\n\nfor $1 \\leqslant k \\leqslant n-1$. Hence, introducing the $2 \\times 2$ matrices $A_{k}=\\left(\\begin{array}{cc}1+a_{k} & -a_{k} \\\\ a_{k} & -a_{k}\\end{array}\\right)$ we have\n\n$$\n\\left(\\begin{array}{c}\nu_{k+1} \\\\\nu_{k+1}-u_{k}\n\\end{array}\\right)=A_{k}\\left(\\begin{array}{c}\nu_{k} \\\\\nu_{k}-u_{k-1}\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(v_{k+1} ; v_{k}-v_{k+1}\\right)=\\left(v_{k} ; v_{k-1}-v_{k}\\right) A_{n-k} .\n$$\n\nfor $1 \\leqslant k \\leqslant n-1$. Since $\\left(\\begin{array}{c}u_{1} \\\\ u_{1}-u_{0}\\end{array}\\right)=\\left(\\begin{array}{l}1 \\\\ 0\\end{array}\\right)$ and $\\left(v_{1} ; v_{0}-v_{1}\\right)=(1 ; 0)$, we get\n\n$$\n\\left(\\begin{array}{c}\nu_{n} \\\\\nu_{n}-u_{n-1}\n\\end{array}\\right)=A_{n-1} A_{n-2} \\cdots A_{1} \\cdot\\left(\\begin{array}{l}\n1 \\\\\n0\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(v_{n} ; v_{n-1}-v_{n}\\right)=(1 ; 0) \\cdot A_{n-1} A_{n-2} \\cdots A_{1} \\text {. }\n$$\n\nIt follows that\n\n$$\n\\left(u_{n}\\right)=(1 ; 0)\\left(\\begin{array}{c}\nu_{n} \\\\\nu_{n}-u_{n-1}\n\\end{array}\\right)=(1 ; 0) \\cdot A_{n-1} A_{n-2} \\cdots A_{1} \\cdot\\left(\\begin{array}{l}\n1 \\\\\n0\n\\end{array}\\right)=\\left(v_{n} ; v_{n-1}-v_{n}\\right)\\left(\\begin{array}{l}\n1 \\\\\n0\n\\end{array}\\right)=\\left(v_{n}\\right) .\n$$']" [] Text-only Competition True Theorem proof Algebra Math English 291 "Prove that in any set of 2000 distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d$, such that $$ \left|\frac{a-b}{c-d}-1\right|<\frac{1}{100000} $$" ['For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \\leqslant D_{2} \\leqslant \\cdots \\leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m=n(n-1) / 2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{1}=1$. Let $D_{1}=1=y-x$ for $x, y \\in S$. Evidently $D_{m}=v-u$ is the difference between the largest element $v$ and the smallest element $u$ of $S$.\n\nIf $D_{i+1} / D_{i}<1+10^{-5}$ for some $i=1,2, \\ldots, m-1$ then the required inequality holds, because $0 \\leqslant D_{i+1} / D_{i}-1<10^{-5}$. Otherwise, the reverse inequality\n\n$$\n\\frac{D_{i+1}}{D_{i}} \\geqslant 1+\\frac{1}{10^{5}}\n$$\n\nholds for each $i=1,2, \\ldots, m-1$, and therefore\n\n$$\nv-u=D_{m}=\\frac{D_{m}}{D_{1}}=\\frac{D_{m}}{D_{m-1}} \\cdots \\frac{D_{3}}{D_{2}} \\cdot \\frac{D_{2}}{D_{1}} \\geqslant\\left(1+\\frac{1}{10^{5}}\\right)^{m-1} .\n$$\n\nFrom $m-1=n(n-1) / 2-1=1000 \\cdot 1999-1>19 \\cdot 10^{5}$, together with the fact that for all $n \\geqslant 1$, $\\left(1+\\frac{1}{n}\\right)^{n} \\geqslant 1+\\left(\\begin{array}{l}n \\\\ 1\\end{array}\\right) \\cdot \\frac{1}{n}=2$, we get\n\n$$\n\\left(1+\\frac{1}{10^{5}}\\right)^{19 \\cdot 10^{5}}=\\left(\\left(1+\\frac{1}{10^{5}}\\right)^{10^{5}}\\right)^{19} \\geqslant 2^{19}=2^{9} \\cdot 2^{10}>500 \\cdot 1000>2 \\cdot 10^{5},\n$$\n\nand so $v-u=D_{m}>2 \\cdot 10^{5}$.\n\nSince the distance of $x$ to at least one of the numbers $u, v$ is at least $(u-v) / 2>10^{5}$, we have\n\n$$\n|x-z|>10^{5} .\n$$\n\nfor some $z \\in\\{u, v\\}$. Since $y-x=1$, we have either $z>y>x$ (if $z=v$ ) or $y>x>z$ (if $z=u$ ). If $z>y>x$, selecting $a=z, b=y, c=z$ and $d=x$ (so that $b \\neq d$ ), we obtain\n\n$$\n\\left|\\frac{a-b}{c-d}-1\\right|=\\left|\\frac{z-y}{z-x}-1\\right|=\\left|\\frac{x-y}{z-x}\\right|=\\frac{1}{z-x}<10^{-5}\n$$\n\nOtherwise, if $y>x>z$, we may choose $a=y, b=z, c=x$ and $d=z$ (so that $a \\neq c$ ), and obtain\n\n$$\n\\left|\\frac{a-b}{c-d}-1\\right|=\\left|\\frac{y-z}{x-z}-1\\right|=\\left|\\frac{y-x}{x-z}\\right|=\\frac{1}{x-z}<10^{-5}\n$$\n\nThe desired result follows.'] [] Text-only Competition True Theorem proof Algebra Math English 292 "Let $\mathbb{Q}_{>0}$ be the set of positive rational numbers. Let $f: \mathbb{Q}_{>0} \rightarrow \mathbb{R}$ be a function satisfying the conditions $$ f(x) f(y) \geqslant f(x y) \tag{1} $$ $$ f(x+y) \geqslant f(x)+f(y) \tag{2} $$ for all $x, y \in \mathbb{Q}_{>0}$. Given that $f(a)=a$ for some rational $a>1$, prove that $f(x)=x$ for all $x \in \mathbb{Q}_{>0}$." ['Denote by $\\mathbb{Z}_{>0}$ the set of positive integers.\n\nPlugging $x=1, y=a$ into (1) we get $f(1) \\geqslant 1$. Next, by an easy induction on $n$ we get from (2) that\n\n$$\nf(n x) \\geqslant n f(x) \\text { for all } n \\in \\mathbb{Z}_{>0} \\text { and } x \\in \\mathbb{Q}_{>0} \\text {. }\n\\tag{3}\n$$\n\nIn particular, we have\n\n$$\nf(n) \\geqslant n f(1) \\geqslant n \\text { for all } n \\in \\mathbb{Z}_{>0}\n\\tag{4}\n$$\n\nFrom (1) again we have $f(m / n) f(n) \\geqslant f(m)$, so $f(q)>0$ for all $q \\in \\mathbb{Q}_{>0}$.\n\nNow, (2) implies that $f$ is strictly increasing; this fact together with (4) yields\n\n$$\nf(x) \\geqslant f(\\lfloor x\\rfloor) \\geqslant\\lfloor x\\rfloor>x-1 \\quad \\text { for all } x \\geqslant 1\n$$\n\nBy an easy induction we get from (1) that $f(x)^{n} \\geqslant f\\left(x^{n}\\right)$, so\n\n$$\nf(x)^{n} \\geqslant f\\left(x^{n}\\right)>x^{n}-1 \\quad \\Longrightarrow \\quad f(x) \\geqslant \\sqrt[n]{x^{n}-1} \\quad \\text { for all } x>1 \\text { and } n \\in \\mathbb{Z}_{>0}\n$$\n\nThis yields\n\n$$\nf(x) \\geqslant x \\text { for every } x>1 \\text {. }\\tag{5}\n$$\n\n(Indeed, if $x>y>1$ then $x^{n}-y^{n}=(x-y)\\left(x^{n-1}+x^{n-2} y+\\cdots+y^{n}\\right)>n(x-y)$, so for a large $n$ we have $x^{n}-1>y^{n}$ and thus $f(x)>y$.)\n\nNow, (1) and (5) give $a^{n}=f(a)^{n} \\geqslant f\\left(a^{n}\\right) \\geqslant a^{n}$, so $f\\left(a^{n}\\right)=a^{n}$. Now, for $x>1$ let us choose $n \\in \\mathbb{Z}_{>0}$ such that $a^{n}-x>1$. Then by (2) and (5) we get\n\n$$\na^{n}=f\\left(a^{n}\\right) \\geqslant f(x)+f\\left(a^{n}-x\\right) \\geqslant x+\\left(a^{n}-x\\right)=a^{n}\n$$\n\nand therefore $f(x)=x$ for $x>1$. Finally, for every $x \\in \\mathbb{Q}_{>0}$ and every $n \\in \\mathbb{Z}_{>0}$, from (1) and (3) we get\n\n$$\nn f(x)=f(n) f(x) \\geqslant f(n x) \\geqslant n f(x)\n$$\n\nwhich gives $f(n x)=n f(x)$. Therefore $f(m / n)=f(m) / n=m / n$ for all $m, n \\in \\mathbb{Z}_{>0}$.'] [] Text-only Competition True Theorem proof Algebra Math English 293 "Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \ldots$ by defining $a_{n+i}=a_{i}$ for all $i \geqslant 1$. If $$ a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n \tag{1} $$ and $$ a_{a_{i}} \leqslant n+i-1 \quad \text { for } i=1,2, \ldots, n \tag{2} $$ prove that $$ a_{1}+\cdots+a_{n} \leqslant n^{2} $$" "['First, we claim that\n\n$$\na_{i} \\leqslant n+i-1 \\quad \\text { for } i=1,2, \\ldots, n\n\\tag{3}\n$$\n\nAssume contrariwise that $i$ is the smallest counterexample. From $a_{n} \\geqslant a_{n-1} \\geqslant \\cdots \\geqslant a_{i} \\geqslant n+i$ and $a_{a_{i}} \\leqslant n+i-1$, taking into account the periodicity of our sequence, it follows that\n\n$$\na_{i} \\text { cannot be congruent to } i, i+1, \\ldots, n-1 \\text {, or } n(\\bmod n) \\text {. }\n\\tag{4}\n$$\n\nThus our assumption that $a_{i} \\geqslant n+i$ implies the stronger statement that $a_{i} \\geqslant 2 n+1$, which by $a_{1}+n \\geqslant a_{n} \\geqslant a_{i}$ gives $a_{1} \\geqslant n+1$. The minimality of $i$ then yields $i=1$, and (4) becomes contradictory. This establishes our first claim.\n\nIn particular we now know that $a_{1} \\leqslant n$. If $a_{n} \\leqslant n$, then $a_{1} \\leqslant \\cdots \\leqslant \\cdots a_{n} \\leqslant n$ and the desired inequality holds trivially. Otherwise, consider the number $t$ with $1 \\leqslant t \\leqslant n-1$ such that\n\n$$\na_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{t} \\leqslant na_{a_{i}}$ ) belongs to $\\left\\{a_{i}+1, \\ldots, n\\right\\}$, and for this reason $b_{i} \\leqslant n-a_{i}$.\n\nIt follows from the definition of the $b_{i}$ and (5) that\n\n$$\na_{t+1}+\\ldots+a_{n} \\leqslant n(n-t)+b_{1}+\\ldots+b_{t} .\n$$\n\nAdding $a_{1}+\\ldots+a_{t}$ to both sides and using that $a_{i}+b_{i} \\leqslant n$ for $1 \\leqslant i \\leqslant t$, we get\n\n$$\na_{1}+a_{2}+\\cdots+a_{n} \\leqslant n(n-t)+n t=n^{2}\n$$\n\nas we wished to prove.' 'In the first quadrant of an infinite grid, consider the increasing ""staircase"" obtained by shading in dark the bottom $a_{i}$ cells of the $i$ th column for $1 \\leqslant i \\leqslant n$. We will prove that there are at most $n^{2}$ dark cells.\n\nTo do it, consider the $n \\times n$ square $S$ in the first quadrant with a vertex at the origin. Also consider the $n \\times n$ square directly to the left of $S$. Starting from its lower left corner, shade in light the leftmost $a_{j}$ cells of the $j$ th row for $1 \\leqslant j \\leqslant n$. Equivalently, the light shading is obtained by reflecting the dark shading across the line $x=y$ and translating it $n$ units to the left. The figure below illustrates this construction for the sequence $6,6,6,7,7,7,8,12,12,14$.\n\n\n\nWe claim that there is no cell in $S$ which is both dark and light. Assume, contrariwise, that there is such a cell in column $i$. Consider the highest dark cell in column $i$ which is inside $S$. Since it is above a light cell and inside $S$, it must be light as well. There are two cases:\n\nCase 1. $a_{i} \\leqslant n$\n\nIf $a_{i} \\leqslant n$ then this dark and light cell is $\\left(i, a_{i}\\right)$, as highlighted in the figure. However, this is the $(n+i)$-th cell in row $a_{i}$, and we only shaded $a_{a_{i}}1)$. Then one may perform a sequence of operations resulting in a graph which admits a proper coloring in $n-1$ colors.\n\nProof. Let us apply repeatedly operation (i) to any appropriate vertices while it is possible. Since the number of vertices decreases, this process finally results in a graph where all the degrees are even. Surely this graph also admits a proper coloring in $n$ colors $1, \\ldots, n$; let us fix this coloring.\n\nNow apply the operation (ii) to this graph. A proper coloring of the resulting graph in $n$ colors still exists: one may preserve the colors of the original vertices and color the vertex $I^{\\prime}$ in a color $k+1(\\bmod n)$ if the vertex $I$ has color $k$. Then two connected original vertices still have different colors, and so do their two connected copies. On the other hand, the vertices $I$ and $I^{\\prime}$ have different colors since $n>1$.\n\nAll the degrees of the vertices in the resulting graph are odd, so one may apply operation $(i)$ to delete consecutively all the vertices of color $n$ one by one; no two of them are connected by an edge, so their degrees do not change during the process. Thus, we obtain a graph admitting a proper coloring in $n-1$ colors, as required. The lemma is proved.\n\nNow, assume that a graph $G$ has $n$ vertices; then it admits a proper coloring in $n$ colors. Applying repeatedly the lemma we finally obtain a graph admitting a proper coloring in one color, that is - a graph with no edges, as required.' 'Again, we will use the graph language.\n\nI. We start with the following observation.\n\nLemma. Assume that a graph $G$ contains an isolated vertex $A$, and a graph $G^{\\circ}$ is obtained from $G$ by deleting this vertex. Then, if one can apply a sequence of operations which makes a graph with no edges from $G^{\\circ}$, then such a sequence also exists for $G$.\n\nProof. Consider any operation applicable to $G^{\\circ}$ resulting in a graph $G_{1}^{\\circ}$; then there exists a sequence of operations applicable to $G$ and resulting in a graph $G_{1}$ differing from $G_{1}^{\\circ}$ by an addition of an isolated vertex $A$. Indeed, if this operation is of type $(i)$, then one may simply repeat it in $G$.\n\n\n\nOtherwise, the operation is of type (ii), and one may apply it to $G$ and then delete the vertex $A^{\\prime}$ (it will have degree 1).\n\nThus one may change the process for $G^{\\circ}$ into a corresponding process for $G$ step by step.\n\nIn view of this lemma, if at some moment a graph contains some isolated vertex, then we may simply delete it; let us call this operation (iii).\n\nII. Let $V=\\left\\{A_{1}^{0}, \\ldots, A_{n}^{0}\\right\\}$ be the vertices of the initial graph. Let us describe which graphs can appear during our operations. Assume that operation (ii) was applied $m$ times. If these were the only operations applied, then the resulting graph $G_{n}^{m}$ has the set of vertices which can be enumerated as\n\n$$\nV_{n}^{m}=\\left\\{A_{i}^{j}: 1 \\leqslant i \\leqslant n, 0 \\leqslant j \\leqslant 2^{m}-1\\right\\},\n$$\n\nwhere $A_{i}^{0}$ is the common ""ancestor"" of all the vertices $A_{i}^{j}$, and the binary expansion of $j$ (adjoined with some zeroes at the left to have $m$ digits) ""keeps the history"" of this vertex: the $d$ th digit from the right is 0 if at the $d$ th doubling the ancestor of $A_{i}^{j}$ was in the original part, and this digit is 1 if it was in the copy.\n\nNext, the two vertices $A_{i}^{j}$ and $A_{k}^{\\ell}$ in $G_{n}^{m}$ are connected with an edge exactly if either (1) $j=\\ell$ and there was an edge between $A_{i}^{0}$ and $A_{k}^{0}$ (so these vertices appeared at the same application of operation (ii)); or (2) $i=k$ and the binary expansions of $j$ and $\\ell$ differ in exactly one digit (so their ancestors became connected as a copy and the original vertex at some application of $(i i))$.\n\nNow, if some operations $(i)$ were applied during the process, then simply some vertices in $G_{n}^{m}$ disappeared. So, in any case the resulting graph is some induced subgraph of $G_{n}^{m}$.\n\nIII. Finally, we will show that from each (not necessarily induced) subgraph of $G_{n}^{m}$ one can obtain a graph with no vertices by applying operations $(i),($ ii) and (iii). We proceed by induction on $n$; the base case $n=0$ is trivial.\n\nFor the induction step, let us show how to apply several operations so as to obtain a graph containing no vertices of the form $A_{n}^{j}$ for $j \\in \\mathbb{Z}$. We will do this in three steps.\n\nStep 1. We apply repeatedly operation $(i)$ to any appropriate vertices while it is possible. In the resulting graph, all vertices have even degrees.\n\nStep 2. Apply operation (ii) obtaining a subgraph of $G_{n}^{m+1}$ with all degrees being odd. In this graph, we delete one by one all the vertices $A_{n}^{j}$ where the sum of the binary digits of $j$ is even; it is possible since there are no edges between such vertices, so all their degrees remain odd. After that, we delete all isolated vertices.\n\nStep 3. Finally, consider any remaining vertex $A_{n}^{j}$ (then the sum of digits of $j$ is odd). If its degree is odd, then we simply delete it. Otherwise, since $A_{n}^{j}$ is not isolated, we consider any vertex adjacent to it. It has the form $A_{k}^{j}$ for some $k3$. Finally, in the sequence $d(a, y), d\\left(a, b_{y}\\right), d\\left(a, c_{y}\\right), d\\left(a, d_{y}\\right)$ the neighboring terms differ by at most 1 , the first term is less than 3 , and the last one is greater than 3 ; thus there exists one which is equal to 3 , as required.'] [] Text-only Competition True Theorem proof Combinatorics Math English 297 "Let $n \geqslant 2$ be an integer. Consider all circular arrangements of the numbers $0,1, \ldots, n$; the $n+1$ rotations of an arrangement are considered to be equal. A circular arrangement is called beautiful if, for any four distinct numbers $0 \leqslant a, b, c, d \leqslant n$ with $a+c=b+d$, the chord joining numbers $a$ and $c$ does not intersect the chord joining numbers $b$ and $d$. Let $M$ be the number of beautiful arrangements of $0,1, \ldots, n$. Let $N$ be the number of pairs $(x, y)$ of positive integers such that $x+y \leqslant n$ and $\operatorname{gcd}(x, y)=1$. Prove that $$ M=N+1 $$" "['Given a circular arrangement of $[0, n]=\\{0,1, \\ldots, n\\}$, we define a $k$-chord to be a (possibly degenerate) chord whose (possibly equal) endpoints add up to $k$. We say that three chords of a circle are aligned if one of them separates the other two. Say that $m \\geqslant 3$ chords are aligned if any three of them are aligned. For instance, in Figure 1, $A, B$, and $C$ are aligned, while $B, C$, and $D$ are not.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nClaim. In a beautiful arrangement, the $k$-chords are aligned for any integer $k$.\n\nProof. We proceed by induction. For $n \\leqslant 3$ the statement is trivial. Now let $n \\geqslant 4$, and proceed by contradiction. Consider a beautiful arrangement $S$ where the three $k$-chords $A, B, C$ are not aligned. If $n$ is not among the endpoints of $A, B$, and $C$, then by deleting $n$ from $S$ we obtain a beautiful arrangement $S \\backslash\\{n\\}$ of $[0, n-1]$, where $A, B$, and $C$ are aligned by the induction hypothesis. Similarly, if 0 is not among these endpoints, then deleting 0 and decreasing all the numbers by 1 gives a beautiful arrangement $S \\backslash\\{0\\}$ where $A, B$, and $C$ are aligned. Therefore both 0 and $n$ are among the endpoints of these segments. If $x$ and $y$ are their respective partners, we have $n \\geqslant 0+x=k=n+y \\geqslant n$. Thus 0 and $n$ are the endpoints of one of the chords; say it is $C$.\n\nLet $D$ be the chord formed by the numbers $u$ and $v$ which are adjacent to 0 and $n$ and on the same side of $C$ as $A$ and $B$, as shown in Figure 2. Set $t=u+v$. If we had $t=n$, the $n-$ chords $A$, $B$, and $D$ would not be aligned in the beautiful arrangement $S \\backslash\\{0, n\\}$, contradicting the induction hypothesis. If $tn$ is equivalent to the case $t0$ is very small.\n\n\n\nFigure 3\n\nIf $0 \\leqslant a, b, c, d \\leqslant n$ satisfy $a+c=b+d$, then $a \\alpha+c \\alpha=b \\alpha+d \\alpha$, so the chord from $a$ to $c$ is parallel to the chord from $b$ to $d$ in $A(\\alpha)$. Hence in a cyclic arrangement all $k$-chords are parallel. In particular every cyclic arrangement is beautiful.\n\nNext we show that there are exactly $N+1$ distinct cyclic arrangements. To see this, let us see how $A(\\alpha)$ changes as we increase $\\alpha$ from 0 to 1 . The order of points $p$ and $q$ changes precisely when we cross a value $\\alpha=f$ such that $\\{p f\\}=\\{q f\\}$; this can only happen if $f$ is one of the $N$ fractions $f_{1}, \\ldots, f_{N}$. Therefore there are at most $N+1$ different cyclic arrangements.\n\nTo show they are all distinct, recall that $f_{i}=a_{i} / b_{i}$ and let $\\epsilon>0$ be a very small number. In the arrangement $A\\left(f_{i}+\\epsilon\\right)$, point $k$ lands at $\\frac{k a_{i}\\left(\\bmod b_{i}\\right)}{b_{i}}+k \\epsilon$. Therefore the points are grouped into $b_{i}$ clusters next to the points $0, \\frac{1}{b_{i}}, \\ldots, \\frac{b_{i}-1}{b_{i}}$ of the circle. The cluster following $\\frac{k}{b_{i}}$ contains the numbers congruent to $k a_{i}^{-1}$ modulo $b_{i}$, listed clockwise in increasing order. It follows that the first number after 0 in $A\\left(f_{i}+\\epsilon\\right)$ is $b_{i}$, and the first number after 0 which is less than $b_{i}$ is $a_{i}^{-1}\\left(\\bmod b_{i}\\right)$, which uniquely determines $a_{i}$. In this way we can recover $f_{i}$ from the cyclic arrangement. Note also that $A\\left(f_{i}+\\epsilon\\right)$ is not the trivial arrangement where we list $0,1, \\ldots, n$ in order clockwise. It follows that the $N+1$ cyclic arrangements $A(\\epsilon), A\\left(f_{1}+\\epsilon\\right), \\ldots, A\\left(f_{N}+\\epsilon\\right)$ are distinct.\n\nLet us record an observation which will be useful later:\n\n$$\n\\text { if } f_{i}<\\alpha\n\nFigure 4\n\nIn $A_{n}(\\alpha)$ the chord from $n-1$ to $x$ is parallel and adjacent to the chord from $n$ to $x-1$, so $n-1$ is between $x-1$ and $x$ in clockwise order, as shown in Figure 4. Similarly, $n-1$ is between $y$ and $y-1$. Therefore $x, y, x-1, n-1$, and $y-1$ occur in this order in $A_{n}(\\alpha)$ and hence in $A$ (possibly with $y=x-1$ or $x=y-1$ ).\n\nNow, $A$ may only differ from $A_{n}(\\alpha)$ in the location of $n$. In $A$, since the chord from $n-1$ to $x$ and the chord from $n$ to $x-1$ do not intersect, $n$ is between $x$ and $n-1$. Similarly, $n$ is between $n-1$ and $y$. Then $n$ must be between $x$ and $y$ and $A=A_{n}(\\alpha)$. Therefore $A$ is cyclic as desired.\n\nCase 2. There is exactly one $i$ with $\\frac{p_{1}}{q_{1}}<\\frac{i}{n}<\\frac{p_{2}}{q_{2}}$.\n\nIn this case there exist two cyclic arrangements $A_{n}\\left(\\alpha_{1}\\right)$ and $A_{n}\\left(\\alpha_{2}\\right)$ of the numbers $0, \\ldots, n$ extending $A_{n-1}(\\alpha)$, where $\\frac{p_{1}}{q_{1}}<\\alpha_{1}<\\frac{i}{n}$ and $\\frac{i}{n}<\\alpha_{2}<\\frac{p_{2}}{q_{2}}$. In $A_{n-1}(\\alpha), 0$ is the only number between $q_{2}$ and $q_{1}$ by (2). For the same reason, $n$ is between $q_{2}$ and 0 in $A_{n}\\left(\\alpha_{1}\\right)$, and between 0 and $q_{1}$ in $A_{n}\\left(\\alpha_{2}\\right)$.\n\nLetting $x=q_{2}$ and $y=q_{1}$, the argument of Case 1 tells us that $n$ must be between $x$ and $y$ in $A$. Therefore $A$ must equal $A_{n}\\left(\\alpha_{1}\\right)$ or $A_{n}\\left(\\alpha_{2}\\right)$, and therefore it is cyclic.\n\nThis concludes the proof that every beautiful arrangement is cyclic. It follows that there are exactly $N+1$ beautiful arrangements of $[0, n]$ as we wished to show.']" [] Text-only Competition True Theorem proof Combinatorics Math English 298 "Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Player $B$ then picks an integer $k$ and blackens the interval from $k / 2^{m}$ to $(k+1) / 2^{m}$ (some parts of this interval may have been blackened before). The goal of player $A$ is to reach a situation where the pot is empty and the interval $[0,1]$ is not completely blackened. Prove that there doesn't exist a strategy for player $A$ to win in a finite number of moves." ['We will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty.\n\nAt the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between 0 and $x_{r}$ has already been blackened; for completeness we define $x_{1}=0$. Let $m$ be the integer picked by player $A$ in this round; we define an integer $y_{r}$ by\n\n$$\n\\frac{y_{r}}{2^{m}} \\leqslant x_{r}<\\frac{y_{r}+1}{2^{m}}\n$$\n\nNote that $I_{0}^{r}=\\left[y_{r} / 2^{m},\\left(y_{r}+1\\right) / 2^{m}\\right]$ is the leftmost interval that may be painted in round $r$ and that still contains some uncolored point.\n\nPlayer $B$ now looks at the next interval $I_{1}^{r}=\\left[\\left(y_{r}+1\\right) / 2^{m},\\left(y_{r}+2\\right) / 2^{m}\\right]$. If $I_{1}^{r}$ still contains an uncolored point, then player $B$ blackens the interval $I_{1}^{r}$; otherwise he blackens the interval $I_{0}^{r}$. We make the convention that, at the beginning of the game, the interval $[1,2]$ is already blackened; thus, if $y_{r}+1=2^{m}$, then $B$ blackens $I_{0}^{r}$.\n\nOur aim is to estimate the amount of ink used after each round. Firstly, we will prove by induction that, if before $r$ th round the segment $[0,1]$ is not completely colored, then, before this move,\n\n(i) the amount of ink used for the segment $\\left[0, x_{r}\\right]$ is at most $3 x_{r}$; and\n\n(ii) for every $m, B$ has blackened at most one interval of length $1 / 2^{m}$ to the right of $x_{r}$.\n\nObviously, these conditions are satisfied for $r=0$. Now assume that they were satisfied before the $r$ th move, and consider the situation after this move; let $m$ be the number $A$ has picked at this move.\n\nIf $B$ has blackened the interval $I_{1}^{r}$ at this move, then $x_{r+1}=x_{r}$, and $(i)$ holds by the induction hypothesis. Next, had $B$ blackened before the $r$ th move any interval of length $1 / 2^{m}$ to the right of $x_{r}$, this interval would necessarily coincide with $I_{1}^{r}$. By our strategy, this cannot happen. So, condition (ii) also remains valid.\n\nAssume now that $B$ has blackened the interval $I_{0}^{r}$ at the $r$ th move, but the interval $[0,1]$ still contains uncolored parts (which means that $I_{1}^{r}$ is contained in $[0,1]$ ). Then condition (ii) clearly remains true, and we need to check $(i)$ only. In our case, the intervals $I_{0}^{r}$ and $I_{1}^{r}$ are completely colored after the $r$ th move, so $x_{r+1}$ either reaches the right endpoint of $I_{1}$ or moves even further to the right. So, $x_{r+1}=x_{r}+\\alpha$ for some $\\alpha>1 / 2^{m}$.\n\nNext, any interval blackened by $B$ before the $r$ th move which intersects $\\left(x_{r}, x_{r+1}\\right)$ should be contained in $\\left[x_{r}, x_{r+1}\\right]$; by (ii), all such intervals have different lengths not exceeding $1 / 2^{m}$, so the total amount of ink used for them is less than $2 / 2^{m}$. Thus, the amount of ink used for the segment $\\left[0, x_{r+1}\\right]$ does not exceed the sum of $2 / 2^{m}, 3 x_{r}$ (used for $\\left[0, x_{r}\\right]$ ), and $1 / 2^{m}$ used for the\n\n\n\nsegment $I_{0}^{r}$. In total it gives at most $3\\left(x_{r}+1 / 2^{m}\\right)<3\\left(x_{r}+\\alpha\\right)=3 x_{r+1}$. Thus condition $(i)$ is also verified in this case. The claim is proved.\n\nFinally, we can perform the desired estimation. Consider any situation in the game, say after the $(r-1)$ st move; assume that the segment $[0,1]$ is not completely black. By $(i i)$, in the segment $\\left[x_{r}, 1\\right]$ player $B$ has colored several segments of different lengths; all these lengths are negative powers of 2 not exceeding $1-x_{r}$; thus the total amount of ink used for this interval is at most $2\\left(1-x_{r}\\right)$. Using $(i)$, we obtain that the total amount of ink used is at most $3 x_{r}+2\\left(1-x_{r}\\right)<3$. Thus the pot is not empty, and therefore $A$ never wins.'] [] Text-only Competition True Theorem proof Combinatorics Math English 299 Let $A B C$ be an acute-angled triangle with orthocenter $H$, and let $W$ be a point on side $B C$. Denote by $M$ and $N$ the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_{1}$ the circumcircle of $B W N$, and let $X$ be the point on $\omega_{1}$ which is diametrically opposite to $W$. Analogously, denote by $\omega_{2}$ the circumcircle of $C W M$, and let $Y$ be the point on $\omega_{2}$ which is diametrically opposite to $W$. Prove that $X, Y$ and $H$ are collinear. ['Let $L$ be the foot of the altitude from $A$, and let $Z$ be the second intersection point of circles $\\omega_{1}$ and $\\omega_{2}$, other than $W$. We show that $X, Y, Z$ and $H$ lie on the same line.\n\nDue to $\\angle B N C=\\angle B M C=90^{\\circ}$, the points $B, C, N$ and $M$ are concyclic; denote their circle by $\\omega_{3}$. Observe that the line $W Z$ is the radical axis of $\\omega_{1}$ and $\\omega_{2}$; similarly, $B N$ is the radical axis of $\\omega_{1}$ and $\\omega_{3}$, and $C M$ is the radical axis of $\\omega_{2}$ and $\\omega_{3}$. Hence $A=B N \\cap C M$ is the radical center of the three circles, and therefore $W Z$ passes through $A$.\n\nSince $W X$ and $W Y$ are diameters in $\\omega_{1}$ and $\\omega_{2}$, respectively, we have $\\angle W Z X=\\angle W Z Y=90^{\\circ}$, so the points $X$ and $Y$ lie on the line through $Z$, perpendicular to $W Z$.\n\n\n\nThe quadrilateral $B L H N$ is cyclic, because it has two opposite right angles. From the power of $A$ with respect to the circles $\\omega_{1}$ and $B L H N$ we find $A L \\cdot A H=A B \\cdot A N=A W \\cdot A Z$. If $H$ lies on the line $A W$ then this implies $H=Z$ immediately. Otherwise, by $\\frac{A Z}{A H}=\\frac{A L}{A W}$ the triangles $A H Z$ and $A W L$ are similar. Then $\\angle H Z A=\\angle W L A=90^{\\circ}$, so the point $H$ also lies on the line $X Y Z$.'] [] Text-only Competition True Theorem proof Geometry Math English 300 Let $\omega$ be the circumcircle of a triangle $A B C$. Denote by $M$ and $N$ the midpoints of the sides $A B$ and $A C$, respectively, and denote by $T$ the midpoint of the $\operatorname{arc} B C$ of $\omega$ not containing $A$. The circumcircles of the triangles $A M T$ and $A N T$ intersect the perpendicular bisectors of $A C$ and $A B$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $A B C$. The lines $M N$ and $X Y$ intersect at $K$. Prove that $K A=K T$. "['Let $O$ be the center of $\\omega$, thus $O=M Y \\cap N X$. Let $\\ell$ be the perpendicular bisector of $A T$ (it also passes through $O$ ). Denote by $r$ the operation of reflection about $\\ell$. Since $A T$ is the angle bisector of $\\angle B A C$, the line $r(A B)$ is parallel to $A C$. Since $O M \\perp A B$ and $O N \\perp A C$, this means that the line $r(O M)$ is parallel to the line $O N$ and passes through $O$, so $r(O M)=O N$. Finally, the circumcircle $\\gamma$ of the triangle $A M T$ is symmetric about $\\ell$, so $r(\\gamma)=\\gamma$. Thus the point $M$ maps to the common point of $O N$ with the $\\operatorname{arc} A M T$ of $\\gamma$ - that is, $r(M)=X$.\n\nSimilarly, $r(N)=Y$. Thus, we get $r(M N)=X Y$, and the common point $K$ of $M N$ nd $X Y$ lies on $\\ell$. This means exactly that $K A=K T$.\n\n' 'Let $L$ be the second common point of the line $A C$ with the circumcircle $\\gamma$ of the triangle $A M T$. From the cyclic quadrilaterals $A B T C$ and $A M T L$ we get $\\angle B T C=180^{\\circ}-$ $\\angle B A C=\\angle M T L$, which implies $\\angle B T M=\\angle C T L$. Since $A T$ is an angle bisector in these quadrilaterals, we have $B T=T C$ and $M T=T L$. Thus the triangles $B T M$ and $C T L$ are congruent, so $C L=B M=A M$.\n\nLet $X^{\\prime}$ be the common point of the line $N X$ with the external bisector of $\\angle B A C$; notice that it lies outside the triangle $A B C$. Then we have $\\angle T A X^{\\prime}=90^{\\circ}$ and $X^{\\prime} A=X^{\\prime} C$, so we get $\\angle X^{\\prime} A M=90^{\\circ}+\\angle B A C / 2=180^{\\circ}-\\angle X^{\\prime} A C=180^{\\circ}-\\angle X^{\\prime} C A=\\angle X^{\\prime} C L$. Thus the triangles $X^{\\prime} A M$ and $X^{\\prime} C L$ are congruent, and therefore\n\n$$\n\\angle M X^{\\prime} L=\\angle A X^{\\prime} C+\\left(\\angle C X^{\\prime} L-\\angle A X^{\\prime} M\\right)=\\angle A X^{\\prime} C=180^{\\circ}-2 \\angle X^{\\prime} A C=\\angle B A C=\\angle M A L .\n$$\n\nThis means that $X^{\\prime}$ lies on $\\gamma$.\n\nThus we have $\\angle T X N=\\angle T X X^{\\prime}=\\angle T A X^{\\prime}=90^{\\circ}$, so $T X \\| A C$. Then $\\angle X T A=\\angle T A C=$ $\\angle T A M$, so the cyclic quadrilateral $M A T X$ is an isosceles trapezoid. Similarly, $N A T Y$ is an isosceles trapezoid, so again the lines $M N$ and $X Y$ are the reflections of each other about the perpendicular bisector of $A T$. Thus $K$ belongs to this perpendicular bisector.\n\n\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 301 Let $A B C$ be a triangle with $\angle B>\angle C$. Let $P$ and $Q$ be two different points on line $A C$ such that $\angle P B A=\angle Q B A=\angle A C B$ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $B Q$ for which $P D=P B$. Let the ray $A D$ intersect the circle $A B C$ at $R \neq A$. Prove that $Q B=Q R$. "['Denote by $\\omega$ the circumcircle of the triangle $A B C$, and let $\\angle A C B=\\gamma$. Note that the condition $\\gamma<\\angle C B A$ implies $\\gamma<90^{\\circ}$. Since $\\angle P B A=\\gamma$, the line $P B$ is tangent to $\\omega$, so $P A \\cdot P C=P B^{2}=P D^{2}$. By $\\frac{P A}{P D}=\\frac{P D}{P C}$ the triangles $P A D$ and $P D C$ are similar, and $\\angle A D P=\\angle D C P$.\n\nNext, since $\\angle A B Q=\\angle A C B$, the triangles $A B C$ and $A Q B$ are also similar. Then $\\angle A Q B=$ $\\angle A B C=\\angle A R C$, which means that the points $D, R, C$, and $Q$ are concyclic. Therefore $\\angle D R Q=$ $\\angle D C Q=\\angle A D P$.\n\n\n\nFigure 1\n\nNow from $\\angle A R B=\\angle A C B=\\gamma$ and $\\angle P D B=\\angle P B D=2 \\gamma$ we get\n\n$$\n\\angle Q B R=\\angle A D B-\\angle A R B=\\angle A D P+\\angle P D B-\\angle A R B=\\angle D R Q+\\gamma=\\angle Q R B\n$$\n\nso the triangle $Q R B$ is isosceles, which yields $Q B=Q R$.' 'Again, denote by $\\omega$ the circumcircle of the triangle $A B C$. Denote $\\angle A C B=\\gamma$. Since $\\angle P B A=\\gamma$, the line $P B$ is tangent to $\\omega$.\n\nLet $E$ be the second intersection point of $B Q$ with $\\omega$. If $V^{\\prime}$ is any point on the ray $C E$ beyond $E$, then $\\angle B E V^{\\prime}=180^{\\circ}-\\angle B E C=180^{\\circ}-\\angle B A C=\\angle P A B$; together with $\\angle A B Q=$ $\\angle P B A$ this shows firstly, that the rays $B A$ and $C E$ intersect at some point $V$, and secondly that the triangle $V E B$ is similar to the triangle $P A B$. Thus we have $\\angle B V E=\\angle B P A$. Next, $\\angle A E V=\\angle B E V-\\gamma=\\angle P A B-\\angle A B Q=\\angle A Q B$; so the triangles $P B Q$ and $V A E$ are also similar.\n\nLet $P H$ be an altitude in the isosceles triangle $P B D$; then $B H=H D$. Let $G$ be the intersection point of $P H$ and $A B$. By the symmetry with respect to $P H$, we have $\\angle B D G=\\angle D B G=\\gamma=$ $\\angle B E A$; thus $D G \\| A E$ and hence $\\frac{B G}{G A}=\\frac{B D}{D E}$. Thus the points $G$ and $D$ correspond to each other in the similar triangles $P A B$ and $V E B$, so $\\angle D V B=\\angle G P B=90^{\\circ}-\\angle P B Q=90^{\\circ}-\\angle V A E$. Thus $V D \\perp A E$.\n\n\n\nLet $T$ be the common point of $V D$ and $A E$, and let $D S$ be an altitude in the triangle $B D R$. The points $S$ and $T$ are the feet of corresponding altitudes in the similar triangles $A D E$ and $B D R$, so $\\frac{B S}{S R}=\\frac{A T}{T E}$. On the other hand, the points $T$ and $H$ are feet of corresponding altitudes in the similar triangles $V A E$ and $P B Q$, so $\\frac{A T}{T E}=\\frac{B H}{H Q}$. Thus $\\frac{B S}{S R}=\\frac{A T}{T E}=\\frac{B H}{H Q}$, and the triangles $B H S$ and $B Q R$ are similar.\n\nFinally, $S H$ is a median in the right-angled triangle $S B D$; so $B H=H S$, and hence $B Q=Q R$.\n\n\n\nFigure 2' 'Denote by $\\omega$ and $O$ the circumcircle of the triangle $A B C$ and its center, respectively. From the condition $\\angle P B A=\\angle B C A$ we know that $B P$ is tangent to $\\omega$.\n\nLet $E$ be the second point of intersection of $\\omega$ and $B D$. Due to the isosceles triangle $B D P$, the tangent of $\\omega$ at $E$ is parallel to $D P$ and consequently it intersects $B P$ at some point $L$. Of course, $P D \\| L E$. Let $M$ be the midpoint of $B E$, and let $H$ be the midpoint of $B R$. Notice that $\\angle A E B=\\angle A C B=\\angle A B Q=\\angle A B E$, so $A$ lies on the perpendicular bisector of $B E$; thus the points $L, A, M$, and $O$ are collinear. Let $\\omega_{1}$ be the circle with diameter $B O$. Let $Q^{\\prime}=H O \\cap B E$; since $H O$ is the perpendicular bisector of $B R$, the statement of the problem is equivalent to $Q^{\\prime}=Q$.\n\nConsider the following sequence of projections (see Fig. 3).\n\n1. Project the line $B E$ to the line $L B$ through the center $A$. (This maps $Q$ to $P$.)\n2. Project the line $L B$ to $B E$ in parallel direction with $L E .(P \\mapsto D$.\n3. Project the line $B E$ to the circle $\\omega$ through its point $A$. $(D \\mapsto R$.)\n4. Scale $\\omega$ by the ratio $\\frac{1}{2}$ from the point $B$ to the circle $\\omega_{1} .(R \\mapsto H$.\n5. Project $\\omega_{1}$ to the line $B E$ through its point $O$. $\\left(H \\mapsto Q^{\\prime}\\right.$.)\n\nWe prove that the composition of these transforms, which maps the line $B E$ to itself, is the identity. To achieve this, it suffices to show three fixed points. An obvious fixed point is $B$ which is fixed by all the transformations above. Another fixed point is $M$, its path being $M \\mapsto L \\mapsto$ $E \\mapsto E \\mapsto M \\mapsto M$.\n\n\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nIn order to show a third fixed point, draw a line parallel with $L E$ through $A$; let that line intersect $B E, L B$ and $\\omega$ at $X, Y$ and $Z \\neq A$, respectively (see Fig. 4). We show that $X$ is a fixed point. The images of $X$ at the first three transformations are $X \\mapsto Y \\mapsto X \\mapsto Z$. From $\\angle X B Z=\\angle E A Z=\\angle A E L=\\angle L B A=\\angle B Z X$ we can see that the triangle $X B Z$ is isosceles. Let $U$ be the midpoint of $B Z$; then the last two transformations do $Z \\mapsto U \\mapsto X$, and the point $X$ is fixed.']" [] Text-only Competition True Theorem proof Geometry Math English 302 Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. "['we denote $\\theta=\\angle A-\\angle D=\\angle C-\\angle F=\\angle E-\\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nLet $x=A B=D E, y=C D=F A, z=E F=B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P, E F A Q$, and $A B C R$ are parallelograms. We compute\n\n$$\n\\begin{aligned}\n\\angle P E Q & =\\angle F E Q+\\angle D E P-\\angle E=\\left(180^{\\circ}-\\angle F\\right)+\\left(180^{\\circ}-\\angle D\\right)-\\angle E \\\\\n& =360^{\\circ}-\\angle D-\\angle E-\\angle F=\\frac{1}{2}(\\angle A+\\angle B+\\angle C-\\angle D-\\angle E-\\angle F)=\\theta / 2\n\\end{aligned}\n$$\n\nSimilarly, $\\angle Q A R=\\angle R C P=\\theta / 2$.\n\n\n\nIf $\\theta=0$, since $\\triangle R C P$ is isosceles, $R=P$. Therefore $A B\\|R C=P C\\| E D$, so $A B D E$ is a parallelogram. Similarly, $B C E F$ and $C D F A$ are parallelograms. It follows that $A D, B E$ and $C F$ meet at their common midpoint.\n\nNow assume $\\theta>0$. Since $\\triangle P E Q, \\triangle Q A R$, and $\\triangle R C P$ are isosceles and have the same angle at the apex, we have $\\triangle P E Q \\sim \\triangle Q A R \\sim \\triangle R C P$ with ratios of similarity $y: z: x$. Thus\n\n$$\n\\triangle P Q R \\text { is similar to the triangle with sidelengths } y, z, \\text { and } x \\text {. }\n\\tag{1}\n$$\n\nNext, notice that\n\n$$\n\\frac{R Q}{Q P}=\\frac{z}{y}=\\frac{R A}{A F}\n$$\n\nand, using directed angles between rays,\n\n$$\n\\begin{aligned}\n\\sphericalangle(R Q, Q P) & =\\sphericalangle(R Q, Q E)+\\sphericalangle(Q E, Q P) \\\\\n& =\\sphericalangle(R Q, Q E)+\\sphericalangle(R A, R Q)=\\sphericalangle(R A, Q E)=\\sphericalangle(R A, A F) .\n\\end{aligned}\n$$\n\nThus $\\triangle P Q R \\sim \\triangle F A R$. Since $F A=y$ and $A R=z,(1)$ then implies that $F R=x$. Similarly $F P=x$. Therefore $C R F P$ is a rhombus.\n\nWe conclude that $C F$ is the perpendicular bisector of $P R$. Similarly, $B E$ is the perpendicular bisector of $P Q$ and $A D$ is the perpendicular bisector of $Q R$. It follows that $A D, B E$, and $C F$ are concurrent at the circumcenter of $P Q R$.' 'we denote $\\theta=\\angle A-\\angle D=\\angle C-\\angle F=\\angle E-\\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nLet $X=C D \\cap E F, Y=E F \\cap A B, Z=A B \\cap C D, X^{\\prime}=F A \\cap B C, Y^{\\prime}=$ $B C \\cap D E$, and $Z^{\\prime}=D E \\cap F A$. From $\\angle A+\\angle B+\\angle C=360^{\\circ}+\\theta / 2$ we get $\\angle A+\\angle B>180^{\\circ}$ and $\\angle B+\\angle C>180^{\\circ}$, so $Z$ and $X^{\\prime}$ are respectively on the opposite sides of $B C$ and $A B$ from the hexagon. Similar conclusions hold for $X, Y, Y^{\\prime}$, and $Z^{\\prime}$. Then\n\n$$\n\\angle Y Z X=\\angle B+\\angle C-180^{\\circ}=\\angle E+\\angle F-180^{\\circ}=\\angle Y^{\\prime} Z^{\\prime} X^{\\prime},\n$$\n\nand similarly $\\angle Z X Y=\\angle Z^{\\prime} X^{\\prime} Y^{\\prime}$ and $\\angle X Y Z=\\angle X^{\\prime} Y^{\\prime} Z^{\\prime}$, so $\\triangle X Y Z \\sim \\triangle X^{\\prime} Y^{\\prime} Z^{\\prime}$. Thus there is a rotation $R$ which sends $\\triangle X Y Z$ to a triangle with sides parallel to $\\triangle X^{\\prime} Y^{\\prime} Z^{\\prime}$. Since $A B=D E$ we have $R(\\overrightarrow{A B})=\\overrightarrow{D E}$. Similarly, $R(\\overrightarrow{C D})=\\overrightarrow{F A}$ and $R(\\overrightarrow{E F})=\\overrightarrow{B C}$. Therefore\n\n$$\n\\overrightarrow{0}=\\overrightarrow{A B}+\\overrightarrow{B C}+\\overrightarrow{C D}+\\overrightarrow{D E}+\\overrightarrow{E F}+\\overrightarrow{F A}=(\\overrightarrow{A B}+\\overrightarrow{C D}+\\overrightarrow{E F})+R(\\overrightarrow{A B}+\\overrightarrow{C D}+\\overrightarrow{E F})\n$$\n\nIf $R$ is a rotation by $180^{\\circ}$, then any two opposite sides of our hexagon are equal and parallel, so the three diagonals meet at their common midpoint. Otherwise, we must have\n\n$$\n\\overrightarrow{A B}+\\overrightarrow{C D}+\\overrightarrow{E F}=\\overrightarrow{0}\n$$\n\nor else we would have two vectors with different directions whose sum is $\\overrightarrow{0}$.\n\n\n\nThis allows us to consider a triangle $L M N$ with $\\overrightarrow{L M}=\\overrightarrow{E F}, \\overrightarrow{M N}=\\overrightarrow{A B}$, and $\\overrightarrow{N L}=\\overrightarrow{C D}$. Let $O$ be the circumcenter of $\\triangle L M N$ and consider the points $O_{1}, O_{2}, O_{3}$ such that $\\triangle A O_{1} B, \\triangle C O_{2} D$, and $\\triangle E O_{3} F$ are translations of $\\triangle M O N, \\triangle N O L$, and $\\triangle L O M$, respectively. Since $F O_{3}$ and $A O_{1}$ are translations of $M O$, quadrilateral $A F O_{3} O_{1}$ is a parallelogram and $O_{3} O_{1}=F A=C D=N L$. Similarly, $O_{1} O_{2}=L M$ and $O_{2} O_{3}=M N$. Therefore $\\triangle O_{1} O_{2} O_{3} \\cong \\triangle L M N$. Moreover, by means of the rotation $R$ one may check that these triangles have the same orientation.\n\nLet $T$ be the circumcenter of $\\triangle O_{1} O_{2} O_{3}$. We claim that $A D, B E$, and $C F$ meet at $T$. Let us show that $C, T$, and $F$ are collinear. Notice that $C O_{2}=O_{2} T=T O_{3}=O_{3} F$ since they are all equal to the circumradius of $\\triangle L M N$. Therefore $\\triangle T O_{3} F$ and $\\triangle C_{2} T$ are isosceles. Using directed angles between rays again, we get\n\n$$\n\\sphericalangle\\left(T F, T O_{3}\\right)=\\sphericalangle\\left(F O_{3}, F T\\right) \\quad \\text { and } \\quad \\sphericalangle\\left(T O_{2}, T C\\right)=\\sphericalangle\\left(C T, C O_{2}\\right) \\text {. }\n\\tag{2}\n$$\n\nAlso, $T$ and $O$ are the circumcenters of the congruent triangles $\\triangle O_{1} O_{2} O_{3}$ and $\\triangle L M N$ so we have $\\sphericalangle\\left(T O_{3}, T O_{2}\\right)=\\sphericalangle(O N, O M)$. Since $C O_{2}$ and $F_{3}$ are translations of $N O$ and $M O$ respectively, this implies\n\n$$\n\\sphericalangle\\left(T O_{3}, T O_{2}\\right)=\\sphericalangle\\left(C O_{2}, F O_{3}\\right) .\n\\tag{3}\n$$\n\n\n\nAdding the three equations in (2) and (3) gives\n\n$$\n\\sphericalangle(T F, T C)=\\sphericalangle(C T, F T)=-\\sphericalangle(T F, T C)\n$$\n\nwhich implies that $T$ is on $C F$. Analogous arguments show that it is on $A D$ and $B E$ also. The desired result follows.' 'we denote $\\theta=\\angle A-\\angle D=\\angle C-\\angle F=\\angle E-\\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nPlace the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A, B, C, D, E, F$ represent the corresponding complex numbers. Also consider the complex numbers $a, b, c, a^{\\prime}, b^{\\prime}, c^{\\prime}$ given by $B-A=a, D-C=b, F-E=c, E-D=a^{\\prime}$, $A-F=b^{\\prime}$, and $C-B=c^{\\prime}$. Let $k=|a| /|b|$. From $a / b^{\\prime}=-k e^{i \\angle A}$ and $a^{\\prime} / b=-k e^{i \\angle D}$ we get that $\\left(a^{\\prime} / a\\right)\\left(b^{\\prime} / b\\right)=e^{-i \\theta}$ and similarly $\\left(b^{\\prime} / b\\right)\\left(c^{\\prime} / c\\right)=e^{-i \\theta}$ and $\\left(c^{\\prime} / c\\right)\\left(a^{\\prime} / a\\right)=e^{-i \\theta}$. It follows that $a^{\\prime}=a r$, $b^{\\prime}=b r$, and $c^{\\prime}=c r$ for a complex number $r$ with $|r|=1$, as shown below.\n\n\n\nWe have\n\n$$\n0=a+c r+b+a r+c+b r=(a+b+c)(1+r) \\text {. }\n$$\n\nIf $r=-1$, then the hexagon is centrally symmetric and its diagonals intersect at its center of symmetry. Otherwise\n\n$$\na+b+c=0\n$$\n\nTherefore\n\n$$\nA=0, \\quad B=a, \\quad C=a+c r, \\quad D=c(r-1), \\quad E=-b r-c, \\quad F=-b r\n$$\n\nNow consider a point $W$ on $A D$ given by the complex number $c(r-1) \\lambda$, where $\\lambda$ is a real number with $0<\\lambda<1$. Since $D \\neq A$, we have $r \\neq 1$, so we can define $s=1 /(r-1)$. From $r \\bar{r}=|r|^{2}=1$ we get\n\n$$\n1+s=\\frac{r}{r-1}=\\frac{r}{r-r \\bar{r}}=\\frac{1}{1-\\bar{r}}=-\\bar{s}\n$$\n\nNow,\n\n$$\n\\begin{aligned}\nW \\text { is on } B E & \\Longleftrightarrow c(r-1) \\lambda-a\\|a-(-b r-c)=b(r-1) \\Longleftrightarrow c \\lambda-a s\\| b \\\\\n& \\Longleftrightarrow-a \\lambda-b \\lambda-a s\\|b \\Longleftrightarrow a(\\lambda+s)\\| b .\n\\end{aligned}\n$$\n\nOne easily checks that $r \\neq \\pm 1$ implies that $\\lambda+s \\neq 0$ since $s$ is not real. On the other hand,\n\n$$\n\\begin{aligned}\nW \\text { on } C F & \\Longleftrightarrow c(r-1) \\lambda+b r\\|-b r-(a+c r)=a(r-1) \\Longleftrightarrow c \\lambda+b(1+s)\\| a \\\\\n& \\Longleftrightarrow-a \\lambda-b \\lambda-b \\bar{s}\\|a \\Longleftrightarrow b(\\lambda+\\bar{s})\\| a \\Longleftrightarrow b \\| a(\\lambda+s),\n\\end{aligned}\n$$\n\nwhere in the last step we use that $(\\lambda+s)(\\lambda+\\bar{s})=|\\lambda+s|^{2} \\in \\mathbb{R}_{>0}$. We conclude that $A D \\cap B E=$ $C F \\cap B E$, and the desired result follows.']" [] Text-only Competition True Theorem proof Geometry Math English 303 Let the excircle of the triangle $A B C$ lying opposite to $A$ touch its side $B C$ at the point $A_{1}$. Define the points $B_{1}$ and $C_{1}$ analogously. Suppose that the circumcentre of the triangle $A_{1} B_{1} C_{1}$ lies on the circumcircle of the triangle $A B C$. Prove that the triangle $A B C$ is right-angled. "['Denote the circumcircles of the triangles $A B C$ and $A_{1} B_{1} C_{1}$ by $\\Omega$ and $\\Gamma$, respectively. Denote the midpoint of the $\\operatorname{arc} C B$ of $\\Omega$ containing $A$ by $A_{0}$, and define $B_{0}$ as well as $C_{0}$ analogously. By our hypothesis the centre $Q$ of $\\Gamma$ lies on $\\Omega$.\n\nLemma. One has $A_{0} B_{1}=A_{0} C_{1}$. Moreover, the points $A, A_{0}, B_{1}$, and $C_{1}$ are concyclic. Finally, the points $A$ and $A_{0}$ lie on the same side of $B_{1} C_{1}$. Similar statements hold for $B$ and $C$.\n\nProof. Let us consider the case $A=A_{0}$ first. Then the triangle $A B C$ is isosceles at $A$, which implies $A B_{1}=A C_{1}$ while the remaining assertions of the Lemma are obvious. So let us suppose $A \\neq A_{0}$ from now on.\n\nBy the definition of $A_{0}$, we have $A_{0} B=A_{0} C$. It is also well known and easy to show that $B C_{1}=$ $C B_{1}$. Next, we have $\\angle C_{1} B A_{0}=\\angle A B A_{0}=\\angle A C A_{0}=\\angle B_{1} C A_{0}$. Hence the triangles $A_{0} B C_{1}$ and $A_{0} C B_{1}$ are congruent. This implies $A_{0} C_{1}=A_{0} B_{1}$, establishing the first part of the Lemma. It also follows that $\\angle A_{0} C_{1} A=\\angle A_{0} B_{1} A$, as these are exterior angles at the corresponding vertices $C_{1}$ and $B_{1}$ of the congruent triangles $A_{0} B C_{1}$ and $A_{0} C B_{1}$. For that reason the points $A, A_{0}, B_{1}$, and $C_{1}$ are indeed the vertices of some cyclic quadrilateral two opposite sides of which are $A A_{0}$ and $B_{1} C_{1}$.\n\nNow we turn to the solution. Evidently the points $A_{1}, B_{1}$, and $C_{1}$ lie interior to some semicircle arc of $\\Gamma$, so the triangle $A_{1} B_{1} C_{1}$ is obtuse-angled. Without loss of generality, we will assume that its angle at $B_{1}$ is obtuse. Thus $Q$ and $B_{1}$ lie on different sides of $A_{1} C_{1}$; obviously, the same holds for the points $B$ and $B_{1}$. So, the points $Q$ and $B$ are on the same side of $A_{1} C_{1}$.\n\nNotice that the perpendicular bisector of $A_{1} C_{1}$ intersects $\\Omega$ at two points lying on different sides of $A_{1} C_{1}$. By the first statement from the Lemma, both points $B_{0}$ and $Q$ are among these points of intersection; since they share the same side of $A_{1} C_{1}$, they coincide (see Figure 1).\n\n\n\nFigure 1\n\n\n\nNow, by the first part of the Lemma again, the lines $Q A_{0}$ and $Q C_{0}$ are the perpendicular bisectors of $B_{1} C_{1}$ and $A_{1} B_{1}$, respectively. Thus\n\n$$\n\\angle C_{1} B_{0} A_{1}=\\angle C_{1} B_{0} B_{1}+\\angle B_{1} B_{0} A_{1}=2 \\angle A_{0} B_{0} B_{1}+2 \\angle B_{1} B_{0} C_{0}=2 \\angle A_{0} B_{0} C_{0}=180^{\\circ}-\\angle A B C,\n$$\n\nrecalling that $A_{0}$ and $C_{0}$ are the midpoints of the $\\operatorname{arcs} C B$ and $B A$, respectively.\n\nOn the other hand, by the second part of the Lemma we have\n\n$$\n\\angle C_{1} B_{0} A_{1}=\\angle C_{1} B A_{1}=\\angle A B C .\n$$\n\nFrom the last two equalities, we get $\\angle A B C=90^{\\circ}$, whereby the problem is solved.' 'Let $Q$ again denote the centre of the circumcircle of the triangle $A_{1} B_{1} C_{1}$, that lies on the circumcircle $\\Omega$ of the triangle $A B C$. We first consider the case where $Q$ coincides with one of the vertices of $A B C$, say $Q=B$. Then $B C_{1}=B A_{1}$ and consequently the triangle $A B C$ is isosceles at $B$. Moreover we have $B C_{1}=B_{1} C$ in any triangle, and hence $B B_{1}=B C_{1}=B_{1} C$; similarly, $B B_{1}=B_{1} A$. It follows that $B_{1}$ is the centre of $\\Omega$ and that the triangle $A B C$ has a right angle at $B$.\n\nSo from now on we may suppose $Q \\notin\\{A, B, C\\}$. We start with the following well known fact. Lemma. Let $X Y Z$ and $X^{\\prime} Y^{\\prime} Z^{\\prime}$ be two triangles with $X Y=X^{\\prime} Y^{\\prime}$ and $Y Z=Y^{\\prime} Z^{\\prime}$.\n\n(i) If $X Z \\neq X^{\\prime} Z^{\\prime}$ and $\\angle Y Z X=\\angle Y^{\\prime} Z^{\\prime} X^{\\prime}$, then $\\angle Z X Y+\\angle Z^{\\prime} X^{\\prime} Y^{\\prime}=180^{\\circ}$.\n\n(ii) If $\\angle Y Z X+\\angle X^{\\prime} Z^{\\prime} Y^{\\prime}=180^{\\circ}$, then $\\angle Z X Y=\\angle Y^{\\prime} X^{\\prime} Z^{\\prime}$.\n\nProof. For both parts, we may move the triangle $X Y Z$ through the plane until $Y=Y^{\\prime}$ and $Z=Z^{\\prime}$. Possibly after reflecting one of the two triangles about $Y Z$, we may also suppose that $X$ and $X^{\\prime}$ lie on the same side of $Y Z$ if we are in case $(i)$ and on different sides if we are in case (ii). In both cases, the points $X, Z$, and $X^{\\prime}$ are collinear due to the angle condition (see Fig. 2). Moreover we have $X \\neq X^{\\prime}$, because in case $(i)$ we assumed $X Z \\neq X^{\\prime} Z^{\\prime}$ and in case $(i i)$ these points even lie on different sides of $Y Z$. Thus the triangle $X X^{\\prime} Y$ is isosceles at $Y$. The claim now follows by considering the equal angles at its base.\n\n\n\nFigure $2(i)$\n\n\n\nFigure $2($ ii $)$\n\nRelabeling the vertices of the triangle $A B C$ if necessary we may suppose that $Q$ lies in the interior of the arc $A B$ of $\\Omega$ not containing $C$. We will sometimes use tacitly that the six triangles $Q B A_{1}, Q A_{1} C, Q C B_{1}, Q B_{1} A, Q C_{1} A$, and $Q B C_{1}$ have the same orientation.\n\nAs $Q$ cannot be the circumcentre of the triangle $A B C$, it is impossible that $Q A=Q B=Q C$ and thus we may also suppose that $Q C \\neq Q B$. Now the above Lemma $(i)$ is applicable to the triangles $Q B_{1} C$ and $Q C_{1} B$, since $Q B_{1}=Q C_{1}$ and $B_{1} C=C_{1} B$, while $\\angle B_{1} C Q=\\angle C_{1} B Q$ holds as both angles appear over the same side of the chord $Q A$ in $\\Omega$ (see Fig. 3). So we get\n\n$$\n\\angle C Q B_{1}+\\angle B Q C_{1}=180^{\\circ}\n\\tag{1}\n$$\n\n\n\nWe claim that $Q C=Q A$. To see this, let us assume for the sake of a contradiction that $Q C \\neq Q A$. Then arguing similarly as before but now with the triangles $Q A_{1} C$ and $Q C_{1} A$ we get\n\n$$\n\\angle A_{1} Q C+\\angle C_{1} Q A=180^{\\circ} .\n$$\n\nAdding this equation to (1), we get $\\angle A_{1} Q B_{1}+\\angle B Q A=360^{\\circ}$, which is absurd as both summands lie in the interval $\\left(0^{\\circ}, 180^{\\circ}\\right)$.\n\nThis proves $Q C=Q A$; so the triangles $Q A_{1} C$ and $Q C_{1} A$ are congruent their sides being equal, which in turn yields\n\n$$\n\\angle A_{1} Q C=\\angle C_{1} Q A .\n\\tag{2}\n$$\n\nFinally our Lemma ( $i i$ i) is applicable to the triangles $Q A_{1} B$ and $Q B_{1} A$. Indeed we have $Q A_{1}=Q B_{1}$ and $A_{1} B=B_{1} A$ as usual, and the angle condition $\\angle A_{1} B Q+\\angle Q A B_{1}=180^{\\circ}$ holds as $A$ and $B$ lie on different sides of the chord $Q C$ in $\\Omega$. Consequently we have\n\n$$\n\\angle B Q A_{1}=\\angle B_{1} Q A .\n\\tag{3}\n$$\n\nFrom (1) and (3) we get\n\n$$\n\\left(\\angle B_{1} Q C+\\angle B_{1} Q A\\right)+\\left(\\angle C_{1} Q B-\\angle B Q A_{1}\\right)=180^{\\circ},\n$$\n\ni.e. $\\angle C Q A+\\angle A_{1} Q C_{1}=180^{\\circ}$. In light of (2) this may be rewritten as $2 \\angle C Q A=180^{\\circ}$ and as $Q$ lies on $\\Omega$ this implies that the triangle $A B C$ has a right angle at $B$.\n\n\n\nFigure 3']" [] Text-only Competition True Theorem proof Geometry Math English 304 "Prove that for any pair of positive integers $k$ and $n$ there exist $k$ positive integers $m_{1}, m_{2}, \ldots, m_{k}$ such that $$ 1+\frac{2^{k}-1}{n}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{k}}\right) $$" "['We proceed by induction on $k$. For $k=1$ the statement is trivial. Assuming we have proved it for $k=j-1$, we now prove it for $k=j$.\n\nCase 1. $n=2 t-1$ for some positive integer $t$.\n\nObserve that\n\n$$\n1+\\frac{2^{j}-1}{2 t-1}=\\frac{2\\left(t+2^{j-1}-1\\right)}{2 t} \\cdot \\frac{2 t}{2 t-1}=\\left(1+\\frac{2^{j-1}-1}{t}\\right)\\left(1+\\frac{1}{2 t-1}\\right) .\n$$\n\nBy the induction hypothesis we can find $m_{1}, \\ldots, m_{j-1}$ such that\n\n$$\n1+\\frac{2^{j-1}-1}{t}=\\left(1+\\frac{1}{m_{1}}\\right)\\left(1+\\frac{1}{m_{2}}\\right) \\cdots\\left(1+\\frac{1}{m_{j-1}}\\right)\n$$\n\nso setting $m_{j}=2 t-1$ gives the desired expression.\n\nCase 2. $n=2 t$ for some positive integer $t$.\n\nNow we have\n\n$$\n1+\\frac{2^{j}-1}{2 t}=\\frac{2 t+2^{j}-1}{2 t+2^{j}-2} \\cdot \\frac{2 t+2^{j}-2}{2 t}=\\left(1+\\frac{1}{2 t+2^{j}-2}\\right)\\left(1+\\frac{2^{j-1}-1}{t}\\right),\n$$\n\nnoting that $2 t+2^{j}-2>0$. Again, we use that\n\n$$\n1+\\frac{2^{j-1}-1}{t}=\\left(1+\\frac{1}{m_{1}}\\right)\\left(1+\\frac{1}{m_{2}}\\right) \\cdots\\left(1+\\frac{1}{m_{j-1}}\\right)\n$$\n\nSetting $m_{j}=2 t+2^{j}-2$ then gives the desired expression.' 'Consider the base 2 expansions of the residues of $n-1$ and $-n$ modulo $2^{k}$ :\n\n$$\n\\begin{aligned}\n& n-1 \\equiv 2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{r}}\\left(\\bmod 2^{k}\\right) \\quad \\text { where } 0 \\leqslant a_{1}q_{n-1} \\text { and } q_{n}>q_{n+1}\\right\\}\n$$\n\nis infinite, since for each $n \\in S$ one has\n\n$$\np_{n}=\\max \\left\\{q_{n}, q_{n-1}\\right\\}=q_{n}=\\max \\left\\{q_{n}, q_{n+1}\\right\\}=p_{n+1}\n$$\n\nSuppose on the contrary that $S$ is finite. Since $q_{2}=7<13=q_{3}$ and $q_{3}=13>7=q_{4}$, the set $S$ is non-empty. Since it is finite, we can consider its largest element, say $m$.\n\nNote that it is impossible that $q_{m}>q_{m+1}>q_{m+2}>\\ldots$ because all these numbers are positive integers, so there exists a $k \\geqslant m$ such that $q_{k}q_{\\ell+1}$. By the minimality of $\\ell$ we have $q_{\\ell-1}k \\geqslant m$, this contradicts the maximality of $m$, and hence $S$ is indeed infinite.'] [] Text-only Competition True Theorem proof Number Theory Math English 306 Prove that there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k>N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square. "['Assume that $a_{1}, a_{2}, a_{3}, \\ldots$ is such a sequence. For each positive integer $k$, let $y_{k}=$ $\\overline{a_{k} a_{k-1} \\ldots a_{1}}$. By the assumption, for each $k>N$ there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$.\n\nI. For every $n$, let $5^{\\gamma_{n}}$ be the greatest power of 5 dividing $x_{n}$. Let us show first that $2 \\gamma_{n} \\geqslant n$ for every positive integer $n>N$.\n\nAssume, to the contrary, that there exists a positive integer $n>N$ such that $2 \\gamma_{n}N$.\n\nII. Consider now any integer $k>\\max \\{N / 2,2\\}$. Since $2 \\gamma_{2 k+1} \\geqslant 2 k+1$ and $2 \\gamma_{2 k+2} \\geqslant 2 k+2$, we have $\\gamma_{2 k+1} \\geqslant k+1$ and $\\gamma_{2 k+2} \\geqslant k+1$. So, from $y_{2 k+2}=a_{2 k+2} \\cdot 10^{2 k+1}+y_{2 k+1}$ we obtain $5^{2 k+2} \\mid y_{2 k+2}-y_{2 k+1}=a_{2 k+2} \\cdot 10^{2 k+1}$ and thus $5 \\mid a_{2 k+2}$, which implies $a_{2 k+2}=5$. Therefore,\n\n$$\n\\left(x_{2 k+2}-x_{2 k+1}\\right)\\left(x_{2 k+2}+x_{2 k+1}\\right)=x_{2 k+2}^{2}-x_{2 k+1}^{2}=y_{2 k+2}-y_{2 k+1}=5 \\cdot 10^{2 k+1}=2^{2 k+1} \\cdot 5^{2 k+2} .\n$$\n\nSetting $A_{k}=x_{2 k+2} / 5^{k+1}$ and $B_{k}=x_{2 k+1} / 5^{k+1}$, which are integers, we obtain\n\n$$\n\\left(A_{k}-B_{k}\\right)\\left(A_{k}+B_{k}\\right)=2^{2 k+1} .\n\\tag{1}\n$$\n\nBoth $A_{k}$ and $B_{k}$ are odd, since otherwise $y_{2 k+2}$ or $y_{2 k+1}$ would be a multiple of 10 which is false by $a_{1} \\neq 0$; so one of the numbers $A_{k}-B_{k}$ and $A_{k}+B_{k}$ is not divisible by 4 . Therefore (1) yields $A_{k}-B_{k}=2$ and $A_{k}+B_{k}=2^{2 k}$, hence $A_{k}=2^{2 k-1}+1$ and thus\n\n$$\nx_{2 k+2}=5^{k+1} A_{k}=10^{k+1} \\cdot 2^{k-2}+5^{k+1}>10^{k+1},\n$$\n\nsince $k \\geqslant 2$. This implies that $y_{2 k+2}>10^{2 k+2}$ which contradicts the fact that $y_{2 k+2}$ contains $2 k+2$ digits. The desired result follows.' ""Again, we assume that a sequence $a_{1}, a_{2}, a_{3}, \\ldots$ satisfies the problem conditions, introduce the numbers $x_{k}$ and $y_{k}$ as in the previous solution, and notice that\n\n$$\ny_{k+1}-y_{k}=\\left(x_{k+1}-x_{k}\\right)\\left(x_{k+1}+x_{k}\\right)=10^{k} a_{k+1}\n\\tag{2}\n$$\n\nfor all $k>N$. Consider any such $k$. Since $a_{1} \\neq 0$, the numbers $x_{k}$ and $x_{k+1}$ are not multiples of 10 , and therefore the numbers $p_{k}=x_{k+1}-x_{k}$ and $q_{k}=x_{k+1}+x_{k}$ cannot be simultaneously multiples of 20 , and hence one of them is not divisible either by 4 or by 5 . In view of (2), this means that the other one is divisible by either $5^{k}$ or by $2^{k-1}$. Notice also that $p_{k}$ and $q_{k}$ have the same parity, so both are even.\n\nOn the other hand, we have $x_{k+1}^{2}=x_{k}^{2}+10^{k} a_{k+1} \\geqslant x_{k}^{2}+10^{k}>2 x_{k}^{2}$, so $x_{k+1} / x_{k}>\\sqrt{2}$, which implies that\n\n$$\n1<\\frac{q_{k}}{p_{k}}=1+\\frac{2}{x_{k+1} / x_{k}-1}<1+\\frac{2}{\\sqrt{2}-1}<6\n\\tag{3}\n$$\n\nThus, if one of the numbers $p_{k}$ and $q_{k}$ is divisible by $5^{k}$, then we have\n\n$$\n10^{k+1}>10^{k} a_{k+1}=p_{k} q_{k} \\geqslant \\frac{\\left(5^{k}\\right)^{2}}{6}\n$$\n\nand hence $(5 / 2)^{k}<60$ which is false for sufficiently large $k$. So, assuming that $k$ is large, we get that $2^{k-1}$ divides one of the numbers $p_{k}$ and $q_{k}$. Hence\n\n$$\n\\left\\{p_{k}, q_{k}\\right\\}=\\left\\{2^{k-1} \\cdot 5^{r_{k}} b_{k}, 2 \\cdot 5^{k-r_{k}} c_{k}\\right\\} \\quad \\text { with nonnegative integers } b_{k}, c_{k}, r_{k} \\text { such that } b_{k} c_{k}=a_{k+1} \\text {. }\n$$\n\nMoreover, from (3) we get\n\n$$\n6>\\frac{2^{k-1} \\cdot 5^{r_{k}} b_{k}}{2 \\cdot 5^{k-r_{k}} c_{k}} \\geqslant \\frac{1}{36} \\cdot\\left(\\frac{2}{5}\\right)^{k} \\cdot 5^{2 r_{k}} \\quad \\text { and } \\quad 6>\\frac{2 \\cdot 5^{k-r_{k}} c_{k}}{2^{k-1} \\cdot 5^{r_{k}} b_{k}} \\geqslant \\frac{4}{9} \\cdot\\left(\\frac{5}{2}\\right)^{k} \\cdot 5^{-2 r_{k}}\n$$\n\nSo\n\n$$\n\\alpha k+c_{1}c_{1} .\n\\tag{4}\n$$\n\nConsequently, for $C=c_{2}-c_{1}+1-\\alpha>0$ we have\n\n$$\n(k+1)-r_{k+1} \\leqslant k-r_{k}+C\n\\tag{5}\n$$\n\nNext, we will use the following easy lemma.\n\nLemma. Let $s$ be a positive integer. Then $5^{s+2^{s}} \\equiv 5^{s}\\left(\\bmod 10^{s}\\right)$.\n\nProof. Euler's theorem gives $5^{2^{s}} \\equiv 1\\left(\\bmod 2^{s}\\right)$, so $5^{s+2^{s}}-5^{s}=5^{s}\\left(5^{2^{s}}-1\\right)$ is divisible by $2^{s}$ and $5^{s}$.\n\nNow, for every large $k$ we have\n\n$$\nx_{k+1}=\\frac{p_{k}+q_{k}}{2}=5^{r_{k}} \\cdot 2^{k-2} b_{k}+5^{k-r_{k}} c_{k} \\equiv 5^{k-r_{k}} c_{k} \\quad\\left(\\bmod 10^{r_{k}}\\right)\n\\tag{6}\n$$\n\nsince $r_{k} \\leqslant k-2$ by $(4)$; hence $y_{k+1} \\equiv 5^{2\\left(k-r_{k}\\right)} c_{k}^{2}\\left(\\bmod 10^{r_{k}}\\right)$. Let us consider some large integer $s$, and choose the minimal $k$ such that $2\\left(k-r_{k}\\right) \\geqslant s+2^{s}$; it exists by (4). Set $d=2\\left(k-r_{k}\\right)-\\left(s+2^{s}\\right)$. By (4) we have $2^{s}<2\\left(k-r_{k}\\right)<\\left(\\frac{2}{\\alpha}-2\\right) r_{k}-\\frac{2 c_{1}}{\\alpha}$; if $s$ is large this implies $r_{k}>s$, so (6) also holds modulo $10^{s}$. Then (6) and the lemma give\n\n$$\ny_{k+1} \\equiv 5^{2\\left(k-r_{k}\\right)} c_{k}^{2}=5^{s+2^{s}} \\cdot 5^{d} c_{k}^{2} \\equiv 5^{s} \\cdot 5^{d} c_{k}^{2} \\quad\\left(\\bmod 10^{s}\\right)\n\\tag{7}\n$$\n\nBy (5) and the minimality of $k$ we have $d \\leqslant 2 C$, so $5^{d} c_{k}^{2} \\leqslant 5^{2 C} \\cdot 81=D$. Using $5^{4}<10^{3}$ we obtain\n\n$$\n5^{s} \\cdot 5^{d} c_{k}^{2}<10^{3 s / 4} D<10^{s-1}\n$$\n\nfor sufficiently large $s$. This, together with (7), shows that the sth digit from the right in $y_{k+1}$, which is $a_{s}$, is zero. This contradicts the problem condition.""]" [] Text-only Competition True Theorem proof Number Theory Math English 307 "Fix an integer $k \geqslant 2$. Two players, called Ana and Banana, play the following game of numbers: Initially, some integer $n \geqslant k$ gets written on the blackboard. Then they take moves in turn, with Ana beginning. A player making a move erases the number $m$ just written on the blackboard and replaces it by some number $m^{\prime}$ with $k \leqslant m^{\prime}m \\geqslant k$ and $\\operatorname{gcd}(m, n)=1$, then $n$ itself is bad, for Ana has the following winning strategy in the game with initial number $n$ : She proceeds by first playing $m$ and then using Banana's strategy for the game with starting number $m$.\n\nOtherwise, if some integer $n \\geqslant k$ has the property that every integer $m$ with $n>m \\geqslant k$ and $\\operatorname{gcd}(m, n)=1$ is bad, then $n$ is good. Indeed, if Ana can make a first move at all in the game with initial number $n$, then she leaves it in a position where the first player has a winning strategy, so that Banana can defeat her.\n\nIn particular, this implies that any two good numbers have a non-trivial common divisor. Also, $k$ itself is good.\n\nFor brevity, we say that $n \\longrightarrow x$ is a move if $n$ and $x$ are two coprime integers with $n>x \\geqslant k$.\n\nClaim 1. If $n$ is good and $n^{\\prime}$ is a multiple of $n$, then $n^{\\prime}$ is also good.\n\nProof. If $n^{\\prime}$ were bad, there would have to be some move $n^{\\prime} \\longrightarrow x$, where $x$ is good. As $n^{\\prime}$ is a multiple of $n$ this implies that the two good numbers $n$ and $x$ are coprime, which is absurd.\n\nClaim 2. If $r$ and $s$ denote two positive integers for which $r s \\geqslant k$ is bad, then $r^{2} s$ is also bad. Proof. Since $r s$ is bad, there is a move $r s \\longrightarrow x$ for some good $x$. Evidently $x$ is coprime to $r^{2} s$ as well, and hence the move $r^{2} s \\longrightarrow x$ shows that $r^{2} s$ is indeed bad.\n\nClaim 3. If $p>k$ is prime and $n \\geqslant k$ is bad, then $n p$ is also bad.\n\nProof. Otherwise we choose a counterexample with $n$ being as small as possible. In particular, $n p$ is good. Since $n$ is bad, there is a move $n \\longrightarrow x$ for some good $x$. Now $n p \\longrightarrow x$ cannot be a valid move, which tells us that $x$ has to be divisible by $p$. So we can write $x=p^{r} y$, where $r$ and $y$ denote some positive integers, the latter of which is not divisible by $p$.\n\nNote that $y=1$ is impossible, for then we would have $x=p^{r}$ and the move $x \\longrightarrow k$ would establish that $x$ is bad. In view of this, there is a least power $y^{\\alpha}$ of $y$ that is at least as large as $k$. Since the numbers $n p$ and $y^{\\alpha}$ are coprime and the former is good, the latter has to be bad. Moreover, the minimality of $\\alpha$ implies $y^{\\alpha}k$, but now we get the same contradiction using Claim 3 instead of Claim 2. Thereby the problem is solved."" 'We use the same analysis of the game of numbers as in the first five paragraphs of the first solution. Let us call a prime number $p$ small in case $p \\leqslant k$ and big otherwise. We again call two integers similar if their sets of small prime factors coincide.\n\nClaim 4. For each integer $b \\geqslant k$ having some small prime factor, there exists an integer $x$ similar to it with $b \\geqslant x \\geqslant k$ and having no big prime factors.\n\nProof. Unless $b$ has a big prime factor we may simply choose $x=b$. Now let $p$ and $q$ denote a small and a big prime factor of $b$, respectively. Let $a$ be the product of all small prime factors of $b$. Further define $n$ to be the least non-negative integer for which the number $x=p^{n} a$ is at least as large as $k$. It suffices to show that $b>x$. This is clear in case $n=0$, so let us assume $n>0$ from now on. Then we have $x

2 m$. Such a pair is necessarily a child of $(c, d-c)$, and thus a descendant of some pair $\\left(c, d^{\\prime}\\right)$ with $m\\{(b+k a) \\nu\\}=\\left\\{\\left(b+k_{0} a\\right) \\nu\\right\\}+\\left(k-k_{0}\\right)\\{a \\nu\\}$ for all $k>k_{0}$, which is absurd.\n\nSimilarly, one can prove that $S_{m}$ contains finitely many pairs $(c, d)$ with $c>2 m$, thus finitely many elements at all.\n\nWe are now prepared for proving the following crucial lemma.\n\nLemma. Consider any pair $(a, b)$ with $f(a, b) \\neq m$. Then the number $g(a, b)$ of its $m$-excellent descendants is equal to the number $h(a, b)$ of ways to represent the number $t=m-f(a, b)$ as $t=k a+\\ell b$ with $k$ and $\\ell$ being some nonnegative integers.\n\nProof. We proceed by induction on the number $N$ of descendants of $(a, b)$ in $S_{m}$. If $N=0$ then clearly $g(a, b)=0$. Assume that $h(a, b)>0$; without loss of generality, we have $a \\leqslant b$. Then, clearly, $m-f(a, b) \\geqslant a$, so $f(a, b+a) \\leqslant f(a, b)+a \\leqslant m$ and $a \\leqslant m$, hence $(a, b+a) \\in S_{m}$ which is impossible. Thus in the base case we have $g(a, b)=h(a, b)=0$, as desired.\n\nNow let $N>0$. Assume that $f(a+b, b)=f(a, b)+b$ and $f(a, b+a)=f(a, b)$ (the other case is similar). If $f(a, b)+b \\neq m$, then by the induction hypothesis we have\n\n$$\ng(a, b)=g(a+b, b)+g(a, b+a)=h(a+b, b)+h(a, b+a)\n$$\n\nNotice that both pairs $(a+b, b)$ and $(a, b+a)$ are descendants of $(a, b)$ and thus each of them has strictly less descendants in $S_{m}$ than $(a, b)$ does.\n\nNext, each one of the $h(a+b, b)$ representations of $m-f(a+b, b)=m-b-f(a, b)$ as the sum $k^{\\prime}(a+b)+\\ell^{\\prime} b$ provides the representation $m-f(a, b)=k a+\\ell b$ with $k=k^{\\prime}1} d=\\sum_{d \\mid m} d\n$$\n\nas required.'] [] Text-only Competition True Theorem proof Number Theory Math English 309 "Let $a, b, c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that $$ \frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leq \frac{3}{16} . $$" "['For positive real numbers $x, y, z$, from the arithmetic-geometric-mean inequality,\n\n$$\n2 x+y+z=(x+y)+(x+z) \\geq 2 \\sqrt{(x+y)(x+z)}\n$$\n\nwe obtain\n\n$$\n\\frac{1}{(2 x+y+z)^{2}} \\leq \\frac{1}{4(x+y)(x+z)}\n$$\n\nApplying this to the left-hand side terms of the inequality to prove, we get\n\n$$\n\\begin{aligned}\n\\frac{1}{(2 a+b+c)^{2}} & +\\frac{1}{(2 b+c+a)^{2}}+\\frac{1}{(2 c+a+b)^{2}} \\\\\n& \\leq \\frac{1}{4(a+b)(a+c)}+\\frac{1}{4(b+c)(b+a)}+\\frac{1}{4(c+a)(c+b)} \\\\\n& =\\frac{(b+c)+(c+a)+(a+b)}{4(a+b)(b+c)(c+a)}=\\frac{a+b+c}{2(a+b)(b+c)(c+a)} .\n\\end{aligned}\n\\tag{1}\n$$\n\nA second application of the inequality of the arithmetic-geometric mean yields\n\n$$\na^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b \\geq 6 a b c\n$$\n\nor, equivalently,\n\n$$\n9(a+b)(b+c)(c+a) \\geq 8(a+b+c)(a b+b c+c a)\n\\tag{2}\n$$\n\nThe supposition $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=a+b+c$ can be written as\n\n$$\na b+b c+c a=a b c(a+b+c) .\n\\tag{3}\n$$\n\nApplying the arithmetic-geometric-mean inequality $x^{2} y^{2}+x^{2} z^{2} \\geq 2 x^{2} y z$ thrice, we get\n\n$$\na^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\geq a^{2} b c+a b^{2} c+a b c^{2}\n$$\n\nwhich is equivalent to\n\n$$\n(a b+b c+c a)^{2} \\geq 3 a b c(a+b+c) .\n\\tag{4}\n$$\n\n\n\nCombining (1), (2), (3), and (4), we will finish the proof:\n\n$$\n\\begin{aligned}\n\\frac{a+b+c}{2(a+b)(b+c)(c+a)} & =\\frac{(a+b+c)(a b+b c+c a)}{2(a+b)(b+c)(c+a)} \\cdot \\frac{a b+b c+c a}{a b c(a+b+c)} \\cdot \\frac{a b c(a+b+c)}{(a b+b c+c a)^{2}} \\\\\n& \\leq \\frac{9}{2 \\cdot 8} \\cdot 1 \\cdot \\frac{1}{3}=\\frac{3}{16}\n\\end{aligned}\n$$' ""Equivalently, we prove the homogenized inequality\n\n$$\n\\frac{(a+b+c)^{2}}{(2 a+b+c)^{2}}+\\frac{(a+b+c)^{2}}{(a+2 b+c)^{2}}+\\frac{(a+b+c)^{2}}{(a+b+2 c)^{2}} \\leq \\frac{3}{16}(a+b+c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\n$$\n\nfor all positive real numbers $a, b, c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a, b, c>0$, fulfilling this condition, the inequality\n\n$$\n\\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}}+\\frac{1}{(1+c)^{2}} \\leq \\frac{3}{16}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\n\\tag{5}\n$$\n\nApplying JENSEN's inequality to the function $f(x)=\\frac{x}{(1+x)^{2}}$, which is concave for $0 \\leq x \\leq 2$ and increasing for $0 \\leq x \\leq 1$, we obtain\n\n$$\n\\alpha \\frac{a}{(1+a)^{2}}+\\beta \\frac{b}{(1+b)^{2}}+\\gamma \\frac{c}{(1+c)^{2}} \\leq(\\alpha+\\beta+\\gamma) \\frac{A}{(1+A)^{2}}, \\quad \\text { where } \\quad A=\\frac{\\alpha a+\\beta b+\\gamma c}{\\alpha+\\beta+\\gamma}\n$$\n\nChoosing $\\alpha=\\frac{1}{a}, \\beta=\\frac{1}{b}$, and $\\gamma=\\frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality\n\n$$\nA=\\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}} \\leq \\frac{a+b+c}{3}=\\frac{1}{3}<1\n$$\n\nFinally we prove (5):\n\n$$\n\\begin{aligned}\n\\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}}+\\frac{1}{(1+c)^{2}} & \\leq\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\frac{A}{(1+A)^{2}} \\\\\n& \\leq\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\frac{\\frac{1}{3}}{\\left(1+\\frac{1}{3}\\right)^{2}}=\\frac{3}{16}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) .\n\\end{aligned}\n$$""]" [] Text-only Competition True Theorem proof Algebra Math English 310 "Let $a, b, c$ be positive real numbers such that $a b+b c+c a \leq 3 a b c$. Prove that $$ \sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3 \leq \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}) . $$" ['Starting with the terms of the right-hand side, the quadratic-arithmetic-mean inequality yields\n\n$$\n\\begin{aligned}\n\\sqrt{2} \\sqrt{a+b} & =2 \\sqrt{\\frac{a b}{a+b}} \\sqrt{\\frac{1}{2}\\left(2+\\frac{a^{2}+b^{2}}{a b}\\right)} \\\\\n& \\geq 2 \\sqrt{\\frac{a b}{a+b}} \\cdot \\frac{1}{2}\\left(\\sqrt{2}+\\sqrt{\\frac{a^{2}+b^{2}}{a b}}\\right)=\\sqrt{\\frac{2 a b}{a+b}}+\\sqrt{\\frac{a^{2}+b^{2}}{a+b}}\n\\end{aligned}\n$$\n\nand, analogously,\n\n$$\n\\sqrt{2} \\sqrt{b+c} \\geq \\sqrt{\\frac{2 b c}{b+c}}+\\sqrt{\\frac{b^{2}+c^{2}}{b+c}}, \\quad \\sqrt{2} \\sqrt{c+a} \\geq \\sqrt{\\frac{2 c a}{c+a}}+\\sqrt{\\frac{c^{2}+a^{2}}{c+a}}\n$$\n\nApplying the inequality between the arithmetic mean and the squared harmonic mean will finish the proof:\n\n$$\n\\sqrt{\\frac{2 a b}{a+b}}+\\sqrt{\\frac{2 b c}{b+c}}+\\sqrt{\\frac{2 c a}{c+a}} \\geq 3 \\cdot \\sqrt{\\frac{3}{\\sqrt{\\frac{a+b}{2 a b}}^{2}}+\\sqrt{\\frac{b+c}{2 b c}}+\\sqrt{\\frac{c+a}{2 c a}}^{2}}=3 \\cdot \\sqrt{\\frac{3 a b c}{a b+b c+c a}} \\geq 3\n$$'] [] Text-only Competition True Theorem proof Algebra Math English 311 "Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that $$ f(x-f(y))>y f(x)+x . $$" "['Assume that\n\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text { for all real } x, y\n\\tag{1}\n$$\n\nLet $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n\n$$\nf(y) \\leq y+a \\text { for all real } y\n\\tag{2}\n$$\n\nSetting $x=f(y)$ in (1) yields in view of (2)\n\n$$\na=f(0) \\leq y f(f(y))+f(y) \\leq y f(f(y))+y+a .\n$$\n\nThis implies $0 \\leq y(f(f(y))+1)$ and thus\n\n$$\nf(f(y)) \\geq-1 \\text { for all } y>0 \\text {. }\n\\tag{3}\n$$\n\nFrom (2) and (3) we obtain $-1 \\leq f(f(y)) \\leq f(y)+a$ for all $y>0$, so\n\n$$\nf(y) \\geq-a-1 \\text { for all } y>0\n\\tag{4}\n$$\n\nNow we show that\n\n$$\nf(x) \\leq 0 \\text { for all real } x \\text {. }\n\\tag{5}\n$$\n\nAssume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that\n\n$$\ny0\n$$\n\nand with (1) and (4) we obtain\n\n$$\ny f(x)+x \\geq f(x-f(y)) \\geq-a-1,\n$$\n\nwhence\n\n$$\ny \\geq \\frac{-a-x-1}{f(x)}\n$$\n\ncontrary to our choice of $y$. Thereby, we have established (5).\n\nSetting $x=0$ in (5) leads to $a=f(0) \\leq 0$ and (2) then yields\n\n$$\nf(x) \\leq x \\quad \\text { for all real } x\n\\tag{6}\n$$\n\nNow choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (1), (5) and\n\n\n\n(6) we obtain\n\n$$\nf(-1)=f(x-f(y)) \\leq y f(x)+x=y f(f(y)-1)+f(y)-1 \\leq y(f(y)-1)-1 \\leq-y-1,\n$$\n\ni.e. $y \\leq-f(-1)-1$, a contradiction to the choice of $y$.' 'Assume that\n\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text { for all real } x, y\n\\tag{7}\n$$\n\nLet $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n\n$$\nf(y) \\leq y+a \\text { for all real } y\n\\tag{8}\n$$\n\nNow we show that\n\n$$\nf(z) \\geq 0 \\text { for all } z \\geq 1 \\text {. }\n\\tag{9}\n$$\n\nLet $z \\geq 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives\n\n$$\nf(w)-z f(w+b) \\leq w+b \\quad \\text { for all real } w\n\\tag{10}\n$$\n\nApplying to $w, w+b, \\ldots, w+(n-1) b$, where $n=1,2, \\ldots$, leads to\n\n$$\n\\begin{gathered}\nf(w)-z^{n} f(w+n b)=(f(w)-z f(w+b))+z(f(w+b)-z f(w+2 b)) \\\\\n+\\cdots+z^{n-1}(f(w+(n-1) b)-z f(w+n b)) \\\\\n\\leq(w+b)+z(w+2 b)+\\cdots+z^{n-1}(w+n b) .\n\\end{gathered}\n$$\n\nFrom (8) we obtain\n\n$$\nf(w+n b) \\leq w+n b+a\n$$\n\nand, thus, we have for all positive integers $n$\n\n$$\nf(w) \\leq\\left(1+z+\\cdots+z^{n-1}+z^{n}\\right) w+\\left(1+2 z+\\cdots+n z^{n-1}+n z^{n}\\right) b+z^{n} a .\n\\tag{11}\n$$\n\nWith $w=0$ we get\n\n$$\na \\leq\\left(1+2 z+\\cdots+n z^{n-1}+n z^{n}\\right) b+a z^{n} .\n\\tag{12}\n$$\n\nIn view of the assumption $b<0$ we find some $n$ such that\n\n$$\na>(n b+a) z^{n}\n\\tag{13}\n$$\n\nbecause the right hand side tends to $-\\infty$ as $n \\rightarrow \\infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality\n\n$$\nf(1)>0 \\text {. }\n\\tag{14}\n$$\n\nIndeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to\n\n$$\nf(-1) \\leq-(n+1)+a\n$$\n\nwhich is false if $n$ is sufficiently large.\n\nTo complete the proof we set $t=\\min \\{-a,-2 / f(1)\\}$. Setting $x=1$ and $y=t$ in (7) gives\n\n$$\nf(1-f(t)) \\leq t f(1)+1 \\leq-2+1=-1 .\n\\tag{15}\n$$\n\n\n\nOn the other hand, by (8) and the choice of $t$ we have $f(t) \\leq t+a \\leq 0$ and hence $1-f(t) \\geq 1$. The inequality (9) yields\n\n$$\nf(1-f(t)) \\geq 0\n$$\n\nwhich contradicts (15).']" [] Text-only Competition True Theorem proof Algebra Math English 312 "Suppose that $s_{1}, s_{2}, s_{3}, \ldots$ is a strictly increasing sequence of positive integers such that the subsequences $$ s_{s_{1}}, s_{s_{2}}, s_{s_{3}}, \ldots \quad \text { and } \quad s_{s_{1}+1}, s_{s_{2}+1}, s_{s_{3}+1}, \ldots $$ are both arithmetic progressions. Prove that $s_{1}, s_{2}, s_{3}, \ldots$ is itself an arithmetic progression." "['Let $D$ be the common difference of the progression $s_{s_{1}}, s_{s_{2}}, \\ldots$. Let for $n=$ $1,2, \\ldots$\n\n$$\nd_{n}=s_{n+1}-s_{n} \\text {. }\n$$\n\nWe have to prove that $d_{n}$ is constant. First we show that the numbers $d_{n}$ are bounded. Indeed, by supposition $d_{n} \\geq 1$ for all $n$. Thus, we have for all $n$\n\n$$\nd_{n}=s_{n+1}-s_{n} \\leq d_{s_{n}}+d_{s_{n}+1}+\\cdots+d_{s_{n+1}-1}=s_{s_{n+1}}-s_{s_{n}}=D .\n$$\n\nThe boundedness implies that there exist\n\n$$\nm=\\min \\left\\{d_{n}: n=1,2, \\ldots\\right\\} \\quad \\text { and } \\quad M=\\max \\left\\{d_{n}: n=1,2, \\ldots\\right\\} \\text {. }\n$$\n\nIt suffices to show that $m=M$. Assume that $mn$ if $d_{n}=n$, because in the case $s_{n}=n$ we would have $m=d_{n}=d_{s_{n}}=M$ in contradiction to the assumption $mn$. Consequently, there is a strictly increasing sequence $n_{1}, n_{2}, \\ldots$ such that\n\n$$\nd_{s_{n_{1}}}=M, \\quad d_{s_{n_{2}}}=m, \\quad d_{s_{n_{3}}}=M, \\quad d_{s_{n_{4}}}=m, \\quad \\ldots\n$$\n\nThe sequence $d_{s_{1}}, d_{s_{2}}, \\ldots$ is the sequence of pairwise differences of $s_{s_{1}+1}, s_{s_{2}+1}, \\ldots$ and $s_{s_{1}}, s_{s_{2}}, \\ldots$, hence also an arithmetic progression. Thus $m=M$' '$\\quad$ Let the integers $D$ and $E$ be the common differences of the progressions $s_{s_{1}}, s_{s_{2}}, \\ldots$ and $s_{s_{1}+1}, s_{s_{2}+1}, \\ldots$, respectively. Let briefly $A=s_{s_{1}}-D$ and $B=s_{s_{1}+1}-E$. Then, for all positive integers $n$,\n\n$$\ns_{s_{n}}=A+n D, \\quad s_{s_{n}+1}=B+n E .\n$$\n\nSince the sequence $s_{1}, s_{2}, \\ldots$ is strictly increasing, we have for all positive integers $n$\n\n$$\ns_{s_{n}}s_{n}+m$. Then $\\left.m(m+1) \\leq m\\left(s_{n+1}-s_{n}\\right) \\leq s_{s_{n+1}}-s_{s_{n}}=(A+(n+1) D)-(A+n D)\\right)=D=m(B-A)=m^{2}$, a contradiction. Hence $s_{1}, s_{2}, \\ldots$ is an arithmetic progression with common difference $m$.']" [] Text-only Competition True Theorem proof Algebra Math English 313 "Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity $$ f(x f(x+y))=f(y f(x))+x^{2} . $$" "['It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions.\n\nLet $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0) \\neq 0$. For any real $t$, substituting $(x, y)=\\left(0, \\frac{t}{f(0)}\\right)$ into the given functional equation, we obtain\n\n$$\nf(0)=f(t)\n\\tag{1}\n$$\n\ncontradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x, y)=(t, 0)$ and $(x, y)=(t,-t)$ into the given equation, we get\n\n$$\nf(t f(t))=f(0)+t^{2}=t^{2},\n$$\n\nand\n\n$$\nf(t f(0))=f(-t f(t))+t^{2}\n$$\n\nrespectively. Therefore, we conclude that\n\n$$\nf(t f(t))=t^{2}, \\quad f(-t f(t))=-t^{2}, \\quad \\text { for every real } t\n\\tag{2}\n$$\n\nConsequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(t f(t))=t^{2}$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$.\n\nWe next show that for any real number $s$,\n\n$$\nf(-s)=-f(s) \\text {. }\n\\tag{3}\n$$\n\nThis is clear if $f(s)=0$. Suppose now $f(s)<0$, then we can find a number $t$ for which $f(s)=-t^{2}$. As $t \\neq 0$ implies $f(t) \\neq 0$, we can also find number $a$ such that $a f(t)=s$. Substituting $(x, y)=(t, a)$ into the given equation, we get\n\n$$\nf(t f(t+a))=f(a f(t))+t^{2}=f(s)+t^{2}=0\n$$\n\nand therefore, $t f(t+a)=0$, which implies $t+a=0$, and hence $s=-t f(t)$. Consequently, $f(-s)=f(t f(t))=t^{2}=-\\left(-t^{2}\\right)=-f(s)$ holds in this case.\n\nFinally, suppose $f(s)>0$ holds. Then there exists a real number $t \\neq 0$ for which $f(s)=t^{2}$. Choose a number $a$ such that $t f(a)=s$. Substituting $(x, y)=(t, a-t)$ into the given equation, we get $f(s)=f(t f(a))=f((a-t) f(t))+t^{2}=f((a-t) f(t))+f(s)$. So we have $f((a-t) f(t))=0$, from which we conclude that $(a-t) f(t)=0$. Since $f(t) \\neq 0$, we get $a=t$ so that $s=t f(t)$ and thus we see $f(-s)=f(-t f(t))=-t^{2}=-f(s)$ holds in this case also. This observation finishes the proof of (3).\n\nBy substituting $(x, y)=(s, t),(x, y)=(t,-s-t)$ and $(x, y)=(-s-t, s)$ into the given equation,\n\n\n\nwe obtain\n\n$$\n\\begin{array}{r}\nf(s f(s+t)))=f(t f(s))+s^{2} \\\\\nf(t f(-s))=f((-s-t) f(t))+t^{2}\n\\end{array}\n$$\n\nand\n\n$$\nf((-s-t) f(-t))=f(s f(-s-t))+(s+t)^{2},\n$$\n\nrespectively. Using the fact that $f(-x)=-f(x)$ holds for all $x$ to rewrite the second and the third equation, and rearranging the terms, we obtain\n\n$$\n\\begin{aligned}\nf(t f(s))-f(s f(s+t)) & =-s^{2} \\\\\nf(t f(s))-f((s+t) f(t)) & =-t^{2} \\\\\nf((s+t) f(t))+f(s f(s+t)) & =(s+t)^{2} .\n\\end{aligned}\n$$\n\nAdding up these three equations now yields $2 f(t f(s))=2 t s$, and therefore, we conclude that $f(t f(s))=t s$ holds for every pair of real numbers $s, t$. By fixing $s$ so that $f(s)=1$, we obtain $f(x)=s x$. In view of the given equation, we see that $s= \\pm 1$. It is easy to check that both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation, so these are the desired solutions.' 'It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions.\n\nLet $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0) \\neq 0$. For any real $t$, substituting $(x, y)=\\left(0, \\frac{t}{f(0)}\\right)$ into the given functional equation, we obtain\n\n$$\nf(0)=f(t)\n\\tag{1}\n$$\n\ncontradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x, y)=(t, 0)$ and $(x, y)=(t,-t)$ into the given equation, we get\n\n$$\nf(t f(t))=f(0)+t^{2}=t^{2},\n$$\n\nand\n\n$$\nf(t f(0))=f(-t f(t))+t^{2}\n$$\n\nrespectively. Therefore, we conclude that\n\n$$\nf(t f(t))=t^{2}, \\quad f(-t f(t))=-t^{2}, \\quad \\text { for every real } t\n\\tag{2}\n$$\n\nConsequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(t f(t))=t^{2}$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$.\n\nWe next show that for any real number $s$,\n\n$$\nf(-s)=-f(s) \\text {. }\n\\tag{3}\n$$\n\nThis is clear if $f(s)=0$.\n\nNow we prove that $f$ is injective. For this purpose, let us assume that $f(r)=f(s)$ for some $r \\neq s$. Then, by $(2)$\n\n$$\nr^{2}=f(r f(r))=f(r f(s))=f((s-r) f(r))+r^{2},\n$$\n\nwhere the last statement follows from the given functional equation with $x=r$ and $y=s-r$. Hence, $h=(s-r) f(r)$ satisfies $f(h)=0$ which implies $h^{2}=f(h f(h))=f(0)=0$, i.e., $h=0$. Then, by $s \\neq r$ we have $f(r)=0$ which implies $r=0$, and finally $f(s)=f(r)=f(0)=0$. Analogously, it follows that $s=0$ which gives the contradiction $r=s$.\n\nTo prove $|f(1)|=1$ we apply (2) with $t=1$ and also with $t=f(1)$ and obtain $f(f(1))=1$ and $(f(1))^{2}=f(f(1) \\cdot f(f(1)))=f(f(1))=1$.\n\nNow we choose $\\eta \\in\\{-1,1\\}$ with $f(1)=\\eta$. Using that $f$ is odd and the given equation with $x=1, y=z$ (second equality) and with $x=-1, y=z+2$ (fourth equality) we obtain\n\n$$\n\\begin{aligned}\n& f(z)+2 \\eta=\\eta(f(z \\eta)+2)=\\eta(f(f(z+1))+1)=\\eta(-f(-f(z+1))+1) \\\\\n& \\quad=-\\eta f((z+2) f(-1))=-\\eta f((z+2)(-\\eta))=\\eta f((z+2) \\eta)=f(z+2) .\n\\end{aligned}\n$$\n\nHence,\n\n$$\nf(z+2 \\eta)=\\eta f(\\eta z+2)=\\eta(f(\\eta z)+2 \\eta)=f(z)+2 .\n$$\n\nUsing this argument twice we obtain\n\n$$\nf(z+4 \\eta)=f(z+2 \\eta)+2=f(z)+4 .\n$$\n\nSubstituting $z=2 f(x)$ we have\n\n$$\nf(2 f(x))+4=f(2 f(x)+4 \\eta)=f(2 f(x+2)),\n$$\n\n\n\nwhere the last equality follows from (4). Applying the given functional equation we proceed to\n\n$$\nf(2 f(x+2))=f(x f(2))+4=f(2 \\eta x)+4\n$$\n\nwhere the last equality follows again from (4) with $z=0$, i.e., $f(2)=2 \\eta$. Finally, $f(2 f(x))=$ $f(2 \\eta x)$ and by injectivity of $f$ we get $2 f(x)=2 \\eta x$ and hence the two solutions.']" [] Text-only Competition True Theorem proof Algebra Math English 314 "Consider 2009 cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. Does the game necessarily end?" ['We interpret a card showing black as the digit 0 and a card showing gold as the digit 1. Thus each position of the 2009 cards, read from left to right, corresponds bijectively to a nonnegative integer written in binary notation of 2009 digits, where leading zeros are allowed. Each move decreases this integer, so the game must end.'] [] Text-only Competition True Theorem proof Combinatorics Math English 315 "Consider 2009 cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. Does there exist a winning strategy for the starting player?" ['We show that there is no winning strategy for the starting player. We label the cards from right to left by $1, \\ldots, 2009$ and consider the set $S$ of cards with labels $50 i, i=1,2, \\ldots, 40$. Let $g_{n}$ be the number of cards from $S$ showing gold after $n$ moves. Obviously, $g_{0}=40$. Moreover, $\\left|g_{n}-g_{n+1}\\right|=1$ as long as the play goes on. Thus, after an odd number of moves, the nonstarting player finds a card from $S$ showing gold and hence can make a move. Consequently, this player always wins.'] [] Text-only Competition True Theorem proof Combinatorics Math English 316 "Let $n$ be a positive integer. Given a sequence $\varepsilon_{1}, \ldots, \varepsilon_{n-1}$ with $\varepsilon_{i}=0$ or $\varepsilon_{i}=1$ for each $i=1, \ldots, n-1$, the sequences $a_{0}, \ldots, a_{n}$ and $b_{0}, \ldots, b_{n}$ are constructed by the following rules: $$ \begin{gathered} a_{0}=b_{0}=1, \quad a_{1}=b_{1}=7, \\ a_{i+1}=\left\{\begin{array}{ll} 2 a_{i-1}+3 a_{i}, & \text { if } \varepsilon_{i}=0, \\ 3 a_{i-1}+a_{i}, & \text { if } \varepsilon_{i}=1, \end{array} \text { for each } i=1, \ldots, n-1,\right. \\ b_{i+1}=\left\{\begin{array}{ll} 2 b_{i-1}+3 b_{i}, & \text { if } \varepsilon_{n-i}=0, \\ 3 b_{i-1}+b_{i}, & \text { if } \varepsilon_{n-i}=1, \end{array} \text { for each } i=1, \ldots, n-1 .\right. \end{gathered} $$ Prove that $a_{n}=b_{n}$." ['For a binary word $w=\\sigma_{1} \\ldots \\sigma_{n}$ of length $n$ and a letter $\\sigma \\in\\{0,1\\}$ let $w \\sigma=$ $\\sigma_{1} \\ldots \\sigma_{n} \\sigma$ and $\\sigma w=\\sigma \\sigma_{1} \\ldots \\sigma_{n}$. Moreover let $\\bar{w}=\\sigma_{n} \\ldots \\sigma_{1}$ and let $\\emptyset$ be the empty word (of length 0 and with $\\bar{\\emptyset}=\\emptyset)$. Let $(u, v)$ be a pair of two real numbers. For binary words $w$ we define recursively the numbers $(u, v)^{w}$ as follows:\n\n$$\n\\begin{array}{r}\n(u, v)^{\\emptyset}=v, \\quad(u, v)^{0}=2 u+3 v, \\quad(u, v)^{1}=3 u+v \\\\\n(u, v)^{w \\sigma \\varepsilon}= \\begin{cases}2(u, v)^{w}+3(u, v)^{w \\sigma}, & \\text { if } \\varepsilon=0 \\\\\n3(u, v)^{w}+(u, v)^{w \\sigma}, & \\text { if } \\varepsilon=1\\end{cases}\n\\end{array}\n$$\n\nIt easily follows by induction on the length of $w$ that for all real numbers $u_{1}, v_{1}, u_{2}, v_{2}, \\lambda_{1}$ and $\\lambda_{2}$\n\n$$\n\\left(\\lambda_{1} u_{1}+\\lambda_{2} u_{2}, \\lambda_{1} v_{1}+\\lambda_{2} v_{2}\\right)^{w}=\\lambda_{1}\\left(u_{1}, v_{1}\\right)^{w}+\\lambda_{2}\\left(u_{2}, v_{2}\\right)^{w}\n\\tag{1}\n$$\n\nand that for $\\varepsilon \\in\\{0,1\\}$\n\n$$\n(u, v)^{\\varepsilon w}=\\left(v,(u, v)^{\\varepsilon}\\right)^{w} .\n\\tag{2}\n$$\n\nObviously, for $n \\geq 1$ and $w=\\varepsilon_{1} \\ldots \\varepsilon_{n-1}$, we have $a_{n}=(1,7)^{w}$ and $b_{n}=(1,7)^{\\bar{w}}$. Thus it is sufficient to prove that\n\n$$\n(1,7)^{w}=(1,7)^{\\bar{w}}\n\\tag{3}\n$$\n\nfor each binary word $w$. We proceed by induction on the length of $w$. The assertion is obvious if $w$ has length 0 or 1 . Now let $w \\sigma \\varepsilon$ be a binary word of length $n \\geq 2$ and suppose that the assertion is true for all binary words of length at most $n-1$.\n\nNote that $(2,1)^{\\sigma}=7=(1,7)^{\\emptyset}$ for $\\sigma \\in\\{0,1\\},(1,7)^{0}=23$, and $(1,7)^{1}=10$.\n\nFirst let $\\varepsilon=0$. Then in view of the induction hypothesis and the equalities (1) and (2), we obtain\n\n$$\n\\begin{aligned}\n(1,7)^{w \\sigma 0}=2(1,7)^{w}+3(1,7)^{w \\sigma}=2(1,7)^{\\bar{w}}+3(1,7)^{\\sigma \\bar{w}}=2(2,1)^{\\sigma \\bar{w}}+ & 3(1,7)^{\\sigma \\bar{w}} \\\\\n& =(7,23)^{\\sigma \\bar{w}}=(1,7)^{0 \\sigma \\bar{w}}\n\\end{aligned}\n$$\n\n\n\nNow let $\\varepsilon=1$. Analogously, we obtain\n\n$$\n\\begin{aligned}\n&(1,7)^{w \\sigma 1}=3(1,7)^{w}+(1,7)^{w \\sigma}=3(1,7)^{\\bar{w}}+(1,7)^{\\sigma \\bar{w}}=3(2,1)^{\\sigma \\bar{w}}+(1,7)^{\\sigma \\bar{w}} \\\\\n&=(7,10)^{\\sigma \\bar{w}}=(1,7)^{1 \\sigma \\bar{w}}\n\\end{aligned}\n$$\n\nThus the induction step is complete, (3) and hence also $a_{n}=b_{n}$ are proved.'] [] Text-only Competition True Theorem proof Combinatorics Math English 317 For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1 . Determine the smallest possible sum of rectangle perimeters in such a partition. "[""For a $k \\times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{i j}$ in row $i$ and column $j$ has vertices $(i-1, j-1),(i-$ $1, j),(i, j-1),(i, j)$, where $i, j \\in\\{1, \\ldots, k\\}$. Without loss of generality assume that the cells $C_{i i}$, $i=1, \\ldots, k$, form a separate rectangle. Then we may consider the boards $B_{k}=\\bigcup_{10$, JENSEN's inequality immediately shows that the minimum of the right hand sight of (1) is attained for $i=k / 2$. Hence the total perimeter of the optimal partition of $B_{k}$ is at least $2 k+2 k / 2 \\log _{2} k / 2+2(k / 2) \\log _{2}(k / 2)=D_{k}$."" ""We start as in Solution 1 and present another proof that $m 2^{m+1}$ is a lower bound for the total perimeter of a partition of $B_{2^{m}}$ into $n$ rectangles. Let briefly $M=2^{m}$. For $1 \\leq i \\leq M$, let $r_{i}$ denote the number of rectangles in the partition that cover some cell from row $i$ and let $c_{j}$ be the number of rectangles that cover some cell from column $j$. Note that the total perimeter $p$ of all rectangles in the partition is\n\n$$\np=2\\left(\\sum_{i=1}^{M} r_{i}+\\sum_{i=1}^{M} c_{i}\\right)\n$$\n\nNo rectangle can simultaneously cover cells from row $i$ and from column $i$ since otherwise it would also cover the cell $C_{i i}$. We classify subsets $S$ of rectangles of the partition as follows. We say that $S$ is of type $i, 1 \\leq i \\leq M$, if $S$ contains all $r_{i}$ rectangles that cover some cell from row $i$, but none of the $c_{i}$ rectangles that cover some cell from column $i$. Altogether there are $2^{n-r_{i}-c_{i}}$ subsets of type $i$. Now we show that no subset $S$ can be simultaneously of type $i$ and of type $j$ if $i \\neq j$. Assume the contrary and let without loss of generality $i1$. By condition $\\left(2^{\\prime}\\right)$, after the Stepmother has distributed her water we have $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \\leq \\frac{5}{2}$. Therefore,\n\n$$\n\\left(y_{0}+y_{2}\\right)+\\left(y_{1}+y_{3}\\right)+\\left(y_{2}+y_{4}\\right)+\\left(y_{3}+y_{0}\\right)+\\left(y_{4}+y_{1}\\right)=2\\left(y_{0}+y_{1}+y_{2}+y_{3}+y_{4}\\right) \\leq 5\n$$\n\nand hence there is a pair of non-neighboring buckets which is not critical, say $\\left(B_{0}, B_{2}\\right)$. Now, if both of the pairs $\\left(B_{3}, B_{0}\\right)$ and $\\left(B_{2}, B_{4}\\right)$ are critical, we must have $y_{1}<\\frac{1}{2}$ and Cinderella can empty the buckets $B_{3}$ and $B_{4}$. This clearly leaves no critical pair of buckets and the total contents of all the buckets is then $y_{1}+\\left(y_{0}+y_{2}\\right) \\leq \\frac{3}{2}$. Therefore, conditions $\\left(1^{\\prime}\\right)$ and $\\left(2^{\\prime}\\right)$ are fulfilled.\n\nNow suppose that without loss of generality the pair $\\left(B_{3}, B_{0}\\right)$ is not critical. If in this case $y_{0} \\leq \\frac{1}{2}$, then one of the inequalities $y_{0}+y_{1}+y_{2} \\leq \\frac{3}{2}$ and $y_{0}+y_{3}+y_{4} \\leq \\frac{3}{2}$ must hold. But then Cinderella can empty $B_{3}$ and $B_{4}$ or $B_{1}$ and $B_{2}$, respectively and clearly fulfill the conditions.\n\nFinally consider the case $y_{0}>\\frac{1}{2}$. By $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \\leq \\frac{5}{2}$, at least one of the pairs $\\left(B_{1}, B_{3}\\right)$ and $\\left(B_{2}, B_{4}\\right)$ is not critical. Without loss of generality let this be the pair $\\left(B_{1}, B_{3}\\right)$. Since the pair $\\left(B_{3}, B_{0}\\right)$ is not critical and $y_{0}>\\frac{1}{2}$, we must have $y_{3} \\leq \\frac{1}{2}$. But then, as before, Cinderella can maintain the two conditions at the beginning of the next round by either emptying $B_{1}$ and $B_{2}$ or $B_{4}$ and $B_{0}$.']" [] Text-only Competition True Theorem proof Combinatorics Math English 319 " A grasshopper jumps along the real axis. He starts at point 0 and makes 2009 jumps to the right with lengths $1,2, \ldots, 2009$ in an arbitrary order. Let $M$ be a set of 2008 positive integers less than $1005 \cdot 2009$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$." ['We construct the set of landing points of the grasshopper.\n\nCase 1. $M$ does not contain numbers divisible by 2009.\n\nWe fix the numbers $2009 k$ as landing points, $k=1,2, \\ldots, 1005$. Consider the open intervals $I_{k}=(2009(k-1), 2009 k), k=1,2, \\ldots, 1005$. We show that we can choose exactly one point outside of $M$ as a landing point in 1004 of these intervals such that all lengths from 1 to 2009 are realized. Since there remains one interval without a chosen point, the length 2009 indeed will appear. Each interval has length 2009, hence a new landing point in an interval yields with a length $d$ also the length $2009-d$. Thus it is enough to implement only the lengths from $D=\\{1,2, \\ldots, 1004\\}$. We will do this in a greedy way. Let $n_{k}, k=1,2, \\ldots, 1005$, be the number of elements of $M$ that belong to the interval $I_{k}$. We order these numbers in a decreasing way, so let $p_{1}, p_{2}, \\ldots, p_{1005}$ be a permutation of $\\{1,2, \\ldots, 1005\\}$ such that $n_{p_{1}} \\geq n_{p_{2}} \\geq \\cdots \\geq n_{p_{1005}}$. In $I_{p_{1}}$ we do not choose a landing point. Assume that landing points have already been chosen in the intervals $I_{p_{2}}, \\ldots, I_{p_{m}}$ and the lengths $d_{2}, \\ldots, d_{m}$ from $D$ are realized, $m=1, \\ldots, 1004$. We show that there is some $d \\in D \\backslash\\left\\{d_{2}, \\ldots, d_{m}\\right\\}$ that can be implemented with a new landing point in $I_{p_{m+1}}$. Assume the contrary. Then the $1004-(m-1)$ other lengths are obstructed by the $n_{p_{m+1}}$ points of $M$ in $I_{p_{m+1}}$. Each length $d$ can be realized by two landing points, namely $2009\\left(p_{m+1}-1\\right)+d$ and $2009 p_{m+1}-d$, hence\n\n$$\nn_{p_{m+1}} \\geq 2(1005-m)\n\\tag{1}\n$$\n\nMoreover, since $|M|=2008=n_{1}+\\cdots+n_{1005}$,\n\n$$\n2008 \\geq n_{p_{1}}+n_{p_{2}}+\\cdots+n_{p_{m+1}} \\geq(m+1) n_{p_{m+1}}\n\\tag{2}\n$$\n\nConsequently, by (1) and (2),\n\n$$\n2008 \\geq 2(m+1)(1005-m)\n$$\n\nThe right hand side of the last inequality obviously attains its minimum for $m=1004$ and this minimum value is greater than 2008 , a contradiction.\n\nCase 2. $M$ does contain a number $\\mu$ divisible by 2009.\n\nBy the pigeonhole principle there exists some $r \\in\\{1, \\ldots, 2008\\}$ such that $M$ does not contain numbers with remainder $r$ modulo 2009. We fix the numbers $2009(k-1)+r$ as landing points, $k=1,2, \\ldots, 1005$. Moreover, $1005 \\cdot 2009$ is a landing point. Consider the open intervals\n\n\n\n$I_{k}=(2009(k-1)+r, 2009 k+r), k=1,2, \\ldots, 1004$. Analogously to Case 1 , it is enough to show that we can choose in 1003 of these intervals exactly one landing point outside of $M \\backslash\\{\\mu\\}$ such that each of the lengths of $D=\\{1,2, \\ldots, 1004\\} \\backslash\\{r\\}$ are implemented. Note that $r$ and $2009-r$ are realized by the first and last jump and that choosing $\\mu$ would realize these two differences again. Let $n_{k}, k=1,2, \\ldots, 1004$, be the number of elements of $M \\backslash\\{\\mu\\}$ that belong to the interval $I_{k}$ and $p_{1}, p_{2}, \\ldots, p_{1004}$ be a permutation of $\\{1,2, \\ldots, 1004\\}$ such that $n_{p_{1}} \\geq n_{p_{2}} \\geq \\cdots \\geq n_{p_{1004}}$. With the same reasoning as in Case 1 we can verify that a greedy choice of the landing points in $I_{p_{2}}, I_{p_{3}}, \\ldots, I_{p_{1004}}$ is possible. We only have to replace (1) by\n\n$$\nn_{p_{m+1}} \\geq 2(1004-m)\n$$\n\n( $D$ has one element less) and (2) by\n\n$$\n2007 \\geq n_{p_{1}}+n_{p_{2}}+\\cdots+n_{p_{m+1}} \\geq(m+1) n_{p_{m+1}}\n$$'] [] Text-only Competition True Theorem proof Combinatorics Math English 320 " Let $n$ be a nonnegative integer. A grasshopper jumps along the real axis. He starts at point 0 and makes $n+1$ jumps to the right with pairwise different positive integral lengths $a_{1}, a_{2}, \ldots, a_{n+1}$ in an arbitrary order. Let $M$ be a set of $n$ positive integers in the interval $(0, s)$, where $s=a_{1}+a_{2}+\cdots+a_{n+1}$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$." ['First of all we remark that the statement in the problem implies a strengthening of itself: Instead of $|M|=n$ it is sufficient to suppose that $|M \\cap(0, s-\\bar{a}]| \\leq n$, where $\\bar{a}=\\min \\left\\{a_{1}, a_{2}, \\ldots, a_{n+1}\\right\\}$. This fact will be used in the proof.\n\nWe prove the statement by induction on $n$. The case $n=0$ is obvious. Let $n>0$ and let the assertion be true for all nonnegative integers less than $n$. Moreover let $a_{1}, a_{2}, \\ldots, a_{n+1}, s$ and $M$ be given as in the problem. Without loss of generality we may assume that $a_{n+1}1$. If $T_{\\bar{k}-1} \\in M$, then (4) follows immediately by the minimality of $\\bar{k}$. If $T_{\\bar{k}-1} \\notin M$, by the smoothness of $\\bar{k}-1$, we obtain a situation as in Claim 1 with $m=\\bar{k}-1$ provided that $\\left|M \\cap\\left(0, T_{\\bar{k}-1}\\right]\\right| \\geq \\bar{k}-1$. Hence, we may even restrict ourselves to $\\left|M \\cap\\left(0, T_{\\bar{k}-1}\\right]\\right| \\leq \\bar{k}-2$ in this case and Claim 3 is proved.\n\nChoose an integer $v \\geq 0$ with $\\left|M \\cap\\left(0, T_{\\bar{k}}\\right)\\right|=\\bar{k}+v$. Let $r_{1}>r_{2}>\\cdots>r_{l}$ be exactly those indices $r$ from $\\{\\bar{k}+1, \\bar{k}+2, \\ldots, n+1\\}$ for which $T_{\\bar{k}}+a_{r} \\notin M$. Then\n\n$$\nn=|M|=\\left|M \\cap\\left(0, T_{\\bar{k}}\\right)\\right|+1+\\left|M \\cap\\left(T_{\\bar{k}}, s\\right)\\right| \\geq \\bar{k}+v+1+(n+1-\\bar{k}-l)\n$$\n\nand consequently $l \\geq v+2$. Note that\n\n$T_{\\bar{k}}+a_{r_{1}}-a_{1}f_{i}(\\varepsilon)$ for all $i=0, \\ldots, k$.\n\nWe will show the essential fact:\n\nFact 3. $f_{0}(n)>f_{0}(h(n))$.\n\nThen the empty string will necessarily be reached after a finite number of applications of $h$. But starting from a string without leading zeros, $\\varepsilon$ can only be reached via the strings $1 \\rightarrow 00 \\rightarrow 0 \\rightarrow \\varepsilon$. Hence also the number 1 will appear after a finite number of applications of $h$.\n\nProof of Fact 3. If the last digit $r$ of $n$ is 0 , then we write $n=x_{0} 0 \\ldots 0 x_{m-1} 0 \\varepsilon$ where the $x_{i}$ do not contain the digit 0 . Then $h(n)=x_{0} 0 \\ldots 0 x_{m-1}$ and $f_{0}(n)-f_{0}(h(n))=f_{0}(\\varepsilon)>0$.\n\nSo let the last digit $r$ of $n$ be at least 1. Let $L=y k$ and $R=z r$ be the corresponding left and right parts where $y$ is some string, $k \\leq r-1$ and the string $z$ consists only of digits not less\n\n\n\nthan $r$. Then $n=y k z r$ and $h(n)=y k z(r-1) z(r-1)$. Let $d(y)$ be the smallest digit of $y$. We consider two cases which do not exclude each other.\n\nCase 1. $d(y) \\geq k$.\n\nThen\n\n$$\nf_{k}(n)-f_{k}(h(n))=f_{k}(z r)-f_{k}(z(r-1) z(r-1)) .\n$$\n\nIn view of Fact 1 this difference is positive if and only if\n\n$$\nf_{r-1}(z r)-f_{r-1}(z(r-1) z(r-1))>0 .\n$$\n\nWe have, using Fact 2,\n\n$$\nf_{r-1}(z r)=4^{f_{r}(z r)}=4^{f_{r}(z)+4^{f_{r+1}(\\varepsilon)}} \\geq 4 \\cdot 4^{f_{r}(z)}>4^{f_{r}(z)}+4^{f_{r}(z)}+4^{f_{r}(\\varepsilon)}=f_{r-1}(z(r-1) z(r-1)) .\n$$\n\nHere we use the additional definition $f_{10}(\\varepsilon)=0$ if $r=9$. Consequently, $f_{k}(n)-f_{k}(h(n))>0$ and according to Fact $1, f_{0}(n)-f_{0}(h(n))>0$.\n\nCase 2. $d(y) \\leq k$.\n\nWe prove by induction on $d(y)=k, k-1, \\ldots, 0$ that $f_{i}(n)-f_{i}(h(n))>0$ for all $i=0, \\ldots, d(y)$. By Fact 1, it suffices to do so for $i=d(y)$. The initialization $d(y)=k$ was already treated in Case 1. Let $t=d(y)0 .\n$$\n\nThus the inequality $f_{d(y)}(n)-f_{d(y)}(h(n))>0$ is established and from Fact 1 it follows that $f_{0}(n)-f_{0}(h(n))>0$.' 'We identify integers $n \\geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from 0 to 9. Moreover, let us define that the empty string, $\\varepsilon$, is being mapped to the empty string. In the following all functions map the set of strings into the set of strings. For two functions $f$ and $g$ let $g \\circ f$ be defined by $(g \\circ f)(x)=g(f(x))$ for all strings $x$ and let, for non-negative integers $n, f^{n}$ denote the $n$-fold application of $f$. For any string $x$ let $s(x)$ be the smallest digit of $x$, and for the empty string let $s(\\varepsilon)=\\infty$. We define nine functions $g_{1}, \\ldots, g_{9}$ as follows: Let $k \\in\\{1, \\ldots, 9\\}$ and let $x$ be a string. If $x=\\varepsilon$ then $g_{k}(x)=\\varepsilon$. Otherwise, write $x$ in the form $x=y z r$ where $y$ is either the empty string or ends with a digit smaller than $k, s(z) \\geq k$ and $r$ is the rightmost digit of $x$. Then $g_{k}(x)=z r$.\n\nLemma 1. We have $g_{k} \\circ h=g_{k} \\circ h \\circ g_{k}$ for all $k=1, \\ldots, 9$.\n\nProof of Lemma 1. Let $x=y z r$ be as in the definition of $g_{k}$. If $y=\\varepsilon$, then $g_{k}(x)=x$, whence\n\n$$\ng_{k}(h(x))=g_{k}\\left(h\\left(g_{k}(x)\\right) .\\right.\n\\tag{1}\n$$\n\nSo let $y \\neq \\varepsilon$.\n\nCase 1. $z$ contains a digit smaller than $r$.\n\nLet $z=u a v$ where $a0\\end{cases}\n$$\n\n\n\nand\n\n$$\nh\\left(g_{k}(x)\\right)=h(z r)=h(u a v r)= \\begin{cases}u a v & \\text { if } r=0, \\\\ \\operatorname{uav}(r-1) v(r-1) & \\text { if } r>0 .\\end{cases}\n$$\n\nSince $y$ ends with a digit smaller than $k$, (1) is obviously true.\n\nCase 2. $z$ does not contain a digit smaller than $r$.\n\nLet $y=u v$ where $u$ is either the empty string or ends with a digit smaller than $r$ and $s(v) \\geq r$. We have\n\n$$\nh(x)= \\begin{cases}u v z & \\text { if } r=0 \\\\ u v z(r-1) v z(r-1) & \\text { if } r>0\\end{cases}\n$$\n\nand\n\n$$\nh\\left(g_{k}(x)\\right)=h(z r)= \\begin{cases}z & \\text { if } r=0 \\\\ z(r-1) z(r-1) & \\text { if } r>0\\end{cases}\n$$\n\nRecall that $y$ and hence $v$ ends with a digit smaller than $k$, but all digits of $v$ are at least $r$. Now if $r>k$, then $v=\\varepsilon$, whence the terminal digit of $u$ is smaller than $k$, which entails\n\n$$\ng_{k}(h(x))=z(r-1) z(r-1)=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right) .\n$$\n\nIf $r \\leq k$, then\n\n$$\ng_{k}(h(x))=z(r-1)=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right),\n$$\n\nso that in both cases (1) is true. Thus Lemma 1 is proved.\n\nLemma 2. Let $k \\in\\{1, \\ldots, 9\\}$, let $x$ be a non-empty string and let $n$ be a positive integer. If $h^{n}(x)=\\varepsilon$ then $\\left(g_{k} \\circ h\\right)^{n}(x)=\\varepsilon$.\n\nProof of Lemma 2. We proceed by induction on $n$. If $n=1$ we have\n\n$$\n\\varepsilon=h(x)=g_{k}(h(x))=\\left(g_{k} \\circ h\\right)(x) .\n$$\n\nNow consider the step from $n-1$ to $n$ where $n \\geq 2$. Let $h^{n}(x)=\\varepsilon$ and let $y=h(x)$. Then $h^{n-1}(y)=\\varepsilon$ and by the induction hypothesis $\\left(g_{k} \\circ h\\right)^{n-1}(y)=\\varepsilon$. In view of Lemma 1,\n\n$$\n\\begin{aligned}\n\\varepsilon=\\left(g_{k} \\circ h\\right)^{n-2}( & \\left.\\left(g_{k} \\circ h\\right)(y)\\right)=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}(h(y))\\right. \\\\\n& =\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}\\left(h\\left(g_{k}(y)\\right)\\right)=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}\\left(h\\left(g_{k}(h(x))\\right)\\right)=\\left(g_{k} \\circ h\\right)^{n}(x) .\\right.\\right.\n\\end{aligned}\n$$\n\nThus the induction step is complete and Lemma 2 is proved.\n\nWe say that the non-empty string $x$ terminates if $h^{n}(x)=\\varepsilon$ for some non-negative integer $n$.\n\nLemma 3. Let $x=y z r$ where $s(y) \\geq k, s(z) \\geq k, y$ ends with the digit $k$ and $z$ is possibly empty. If $y$ and $z r$ terminate then also $x$ terminates.\n\nProof of Lemma 3. Suppose that $y$ and $z r$ terminate. We proceed by induction on $k$. Let $k=0$. Obviously, $h(y w)=y h(w)$ for any non-empty string $w$. Let $h^{n}(z r)=\\epsilon$. It follows easily by induction on $m$ that $h^{m}(y z r)=y h^{m}(z r)$ for $m=1, \\ldots, n$. Consequently, $h^{n}(y z r)=y$. Since $y$ terminates, also $x=y z r$ terminates.\n\nNow let the assertion be true for all nonnegative integers less than $k$ and let us prove it for $k$ where $k \\geq 1$. It turns out that it is sufficient to prove that $y g_{k}(h(z r))$ terminates. Indeed:\n\nCase 1. $r=0$.\n\nThen $h(y z r)=y z=y g_{k}(h(z r))$.\n\nCase 2. $0k$.\n\nThen $h(y z r)=y h(z r)=y g_{k}(h(z r))$.\n\nNote that $y g_{k}(h(z r))$ has the form $y z^{\\prime} r^{\\prime}$ where $s\\left(z^{\\prime}\\right) \\geq k$. By the same arguments it is sufficient to prove that $y g_{k}\\left(h\\left(z^{\\prime} r^{\\prime}\\right)\\right)=y\\left(g_{k} \\circ h\\right)^{2}(z r)$ terminates and, by induction, that $y\\left(g_{k} \\circ h\\right)^{m}(z r)$ terminates for some positive integer $m$. In view of Lemma 2 there is some $m$ such that $\\left(g_{k} \\circ\\right.$ $h)^{m}(z r)=\\epsilon$, so $x=y z r$ terminates if $y$ terminates. Thus Lemma 3 is proved.\n\nNow assume that there is some string $x$ that does not terminate. We choose $x$ minimal. If $x \\geq 10$, we can write $x$ in the form $x=y z r$ of Lemma 3 and by this lemma $x$ terminates since $y$ and $z r$ are smaller than $x$. If $x \\leq 9$, then $h(x)=(x-1)(x-1)$ and $h(x)$ terminates again by Lemma 3 and the minimal choice of $x$.' 'We commence by introducing some terminology. Instead of integers, we will consider the set $S$ of all strings consisting of the digits $0,1, \\ldots, 9$, including the empty string $\\epsilon$. If $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ is a nonempty string, we let $\\rho(a)=a_{n}$ denote the terminal digit of $a$ and $\\lambda(a)$ be the string with the last digit removed. We also define $\\lambda(\\epsilon)=\\epsilon$ and denote the set of non-negative integers by $\\mathbb{N}_{0}$.\n\nNow let $k \\in\\{0,1,2, \\ldots, 9\\}$ denote any digit. We define a function $f_{k}: S \\longrightarrow S$ on the set of strings: First, if the terminal digit of $n$ belongs to $\\{0,1, \\ldots, k\\}$, then $f_{k}(n)$ is obtained from $n$ by deleting this terminal digit, i.e $f_{k}(n)=\\lambda(n)$. Secondly, if the terminal digit of $n$ belongs to $\\{k+1, \\ldots, 9\\}$, then $f_{k}(n)$ is obtained from $n$ by the process described in the problem. We also define $f_{k}(\\epsilon)=\\epsilon$. Note that up to the definition for integers $n \\leq 1$, the function $f_{0}$ coincides with the function $h$ in the problem, through interpreting integers as digit strings. The argument will be roughly as follows. We begin by introducing a straightforward generalization of our claim about $f_{0}$. Then it will be easy to see that $f_{9}$ has all these stronger properties, which means that is suffices to show for $k \\in\\{0,1, \\ldots, 8\\}$ that $f_{k}$ possesses these properties provided that $f_{k+1}$ does.\n\nWe continue to use $k$ to denote any digit. The operation $f_{k}$ is said to be separating, if the followings holds: Whenever $a$ is an initial segment of $b$, there is some $N \\in \\mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. The following two notions only apply to the case where $f_{k}$ is indeed separating, otherwise they remain undefined. For every $a \\in S$ we denote the least $N \\in \\mathbb{N}_{0}$ for which $f_{k}^{N}(a)=\\epsilon$ occurs by $g_{k}(a)$ (because $\\epsilon$ is an initial segment of $a$, such an $N$ exists if $f_{k}$ is separating). If for every two strings $a$ and $b$ such that $a$ is a terminal segment of $b$ one has $g_{k}(a) \\leq g_{k}(b)$, we say that $f_{k}$ is coherent. In case that $f_{k}$ is separating and coherent we call the digit $k$ seductive.\n\nAs $f_{9}(a)=\\lambda(a)$ for all $a$, it is obvious that 9 is seductive. Hence in order to show that 0 is seductive, which clearly implies the statement of the problem, it suffices to take any $k \\in\\{0,1, \\ldots, 8\\}$ such that $k+1$ is seductive and to prove that $k$ has to be seductive as well. Note that in doing so, we have the function $g_{k+1}$ at our disposal. We have to establish two things and we begin with\n\nStep 1. $f_{k}$ is separating.\n\nBefore embarking on the proof of this, we record a useful observation which is easily proved by induction on $M$.\n\n\n\nClaim 1. For any strings $A, B$ and any positive integer $M$ such that $f_{k}^{M-1}(B) \\neq \\epsilon$, we have\n\n$$\nf_{k}^{M}(A k B)=A k f_{k}^{M}(B)\n$$\n\nNow we call a pair $(a, b)$ of strings wicked provided that $a$ is an initial segment of $b$, but there is no $N \\in \\mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. We need to show that there are none, so assume that there were such pairs. Choose a wicked pair $(a, b)$ for which $g_{k+1}(b)$ attains its minimal possible value. Obviously $b \\neq \\epsilon$ for any wicked pair $(a, b)$. Let $z$ denote the terminal digit of $b$. Observe that $a \\neq b$, which means that $a$ is also an initial segment of $\\lambda(b)$. To facilitate the construction of the eventual contradiction, we prove\n\nClaim 2. There cannot be an $N \\in \\mathbb{N}_{0}$ such that\n\n$$\nf_{k}^{N}(b)=\\lambda(b)\n$$\n\nProof of Claim 2. For suppose that such an $N$ existed. Because $g_{k+1}(\\lambda(b))k+1$ is impossible: Set $B=f_{k}(b)$. Then also $f_{k+1}(b)=B$, but $g_{k+1}(B)g_{k}(b)$. Observe that if $f_{k}$ was incoherent, which we shall assume from now on, then such pairs existed. Now among all aggressive pairs we choose one, say $(a, b)$, for which $g_{k}(b)$ attains its least possible value. Obviously $f_{k}(a)$ cannot be injectible into $f_{k}(b)$, for otherwise the pair $\\left(f_{k}(a), f_{k}(b)\\right)$ was aggressive and contradicted our choice of $(a, b)$. Let $\\left(A_{1}, A_{2}, \\ldots, A_{m}\\right)$ and $\\left(B_{1}, B_{2}, \\ldots, B_{n}\\right)$ be the decompositions of $a$ and $b$ and take a function $H:\\{1,2, \\ldots, m\\} \\longrightarrow\\{1,2, \\ldots, n\\}$ exemplifying that $a$ is indeed injectible into $b$. If we had $H(m)\n\nFigure 1' 'Again, denote by $K, L, M$ the midpoints of segments $B P, C Q$, and $P Q$, respectively. Let $O, S, T$ be the circumcenters of triangles $A B C, K L M$, and $A P Q$, respectively (see Figure 2). Note that $M K$ and $L M$ are the midlines in triangles $B P Q$ and $C P Q$, respectively, so $\\overrightarrow{M K}=\\frac{1}{2} \\overrightarrow{Q B}$ and $\\overrightarrow{M L}=\\frac{1}{2} \\overrightarrow{P C}$. Denote by $\\operatorname{pr}_{l}(\\vec{v})$ the projection of vector $\\vec{v}$ onto line $l$. Then $\\operatorname{pr}_{A B}(\\overrightarrow{O T})=\\operatorname{pr}_{A B}(\\overrightarrow{O A}-\\overrightarrow{T A})=\\frac{1}{2} \\overrightarrow{B A}-\\frac{1}{2} \\overrightarrow{Q A}=\\frac{1}{2} \\overrightarrow{B Q}=\\overrightarrow{K M}$ and $\\operatorname{pr}_{A B}(\\overrightarrow{S M})=\\operatorname{pr}_{M K}(\\overrightarrow{S M})=$ $\\frac{1}{2} \\overrightarrow{K M}=\\frac{1}{2} \\operatorname{pr}_{A B}(\\overrightarrow{O T})$. Analogously we get $\\operatorname{pr}_{C A}(\\overrightarrow{S M})=\\frac{1}{2} \\operatorname{pr}_{C A}(\\overrightarrow{O T})$. Since $A B$ and $C A$ are not parallel, this implies that $\\overrightarrow{S M}=\\frac{1}{2} \\overrightarrow{O T}$.\n\n\n\nFigure 2\n\nNow, since the circle $k$ touches $P Q$ at $M$, we get $S M \\perp P Q$, hence $O T \\perp P Q$. Since $T$ is equidistant from $P$ and $Q$, the line $O T$ is a perpendicular bisector of segment $P Q$, and hence $O$ is equidistant from $P$ and $Q$ which finishes the proof.']" [] Text-only Competition True Theorem proof Geometry Math English 323 "Let $A B C$ be a triangle. The incircle of $A B C$ touches the sides $A B$ and $A C$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $B Y$ and $C Z$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $B C Y R$ and $B C S Z$ are parallelograms. Prove that $G R=G S$." "['Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $A B C$. Let $k$ and $k_{a}$ touch the side $B C$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $A B$ and $A C$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parallelogram are of equal length, that points of contact of the excircle and incircle to a side of a triangle lie symmetric with respect to the midpoint of this side and that segments on two tangents to a circle defined by the points of contact and their point of intersection have the same length. So we conclude\n\n$$\n\\begin{gathered}\nZ P=Z B+B P=X B+B T=B X+C X=Z S \\text { and } \\\\\nC Q=C T=B X=B Z=C S .\n\\end{gathered}\n$$\n\n\n\nSo for each of the points $Z, C$, their distances to $S$ equal the length of a tangent segment from this point to $k_{a}$. It is well-known, that all points with this property lie on the line $Z C$, which is the radical axis of $S$ and $k_{a}$. Similar arguments yield that $B Y$ is the radical axis of $R$ and $k_{a}$. So the point of intersection of $Z C$ and $B Y$, which is $G$ by definition, is the radical center of $R, S$ and $k_{a}$, from which the claim $G R=G S$ follows immediately.' ""Denote $x=A Z=A Y, y=B Z=B X, z=C X=C Y, p=Z G, q=G C$. Several lengthy calculations (MENELAOs' theorem in triangle $A Z C$, law of Cosines in triangles $A B C$ and $A Z C$ and STEWART's theorem in triangle $Z C S$ ) give four equations for $p, q, \\cos \\alpha$\n\n\n\nand $G S$ in terms of $x, y$, and $z$ that can be resolved for $G S$. The result is symmetric in $y$ and $z$, so $G R=G S$. More in detail this means:\n\nThe line $B Y$ intersects the sides of triangle $A Z C$, so MEnELaOS' theorem yields $\\frac{p}{q} \\cdot \\frac{z}{x} \\cdot \\frac{x+y}{y}=1$, hence\n\n$$\n\\frac{p}{q}=\\frac{x y}{y z+z x}\n\\tag{1}\n$$\n\nSince we only want to show that the term for $G S$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N=x y+y z+z x$. So (1) implies\n\n$$\n\\frac{p}{p+q}=\\frac{x y}{x y+y z+z x}=\\frac{x y}{N} \\quad \\text { and } \\quad \\frac{q}{p+q}=\\frac{y z+z x}{x y+y z+z x}=\\frac{y z+z x}{N}\n\\tag{2}\n$$\n\nNow the law of Cosines in triangle $A B C$ yields\n\n$$\n\\cos \\alpha=\\frac{(x+y)^{2}+(x+z)^{2}-(y+z)^{2}}{2(x+y)(x+z)}=\\frac{2 x^{2}+2 x y+2 x z-2 y z}{2(x+y)(x+z)}=1-\\frac{2 y z}{(x+y)(x+z)}\n$$\n\nWe use this result to apply the law of Cosines in triangle $A Z C$ :\n\n$$\n\\begin{aligned}\n(p+q)^{2} & =x^{2}+(x+z)^{2}-2 x(x+z) \\cos \\alpha \\\\\n& =x^{2}+(x+z)^{2}-2 x(x+z) \\cdot\\left(1-\\frac{2 y z}{(x+y)(x+z)}\\right) \\\\\n& =z^{2}+\\frac{4 x y z}{x+y} .\n\\end{aligned}\n\\tag{3}\n$$\n\nNow in triangle $Z C S$ the segment $G S$ is a cevian, so with STEWART's theorem we have $p y^{2}+q(y+z)^{2}=(p+q)\\left(G S^{2}+p q\\right)$, hence\n\n$$\nG S^{2}=\\frac{p}{p+q} \\cdot y^{2}+\\frac{q}{p+q} \\cdot(y+z)^{2}-\\frac{p}{p+q} \\cdot \\frac{q}{p+q} \\cdot(p+q)^{2}\n$$\n\nReplacing the $p$ 's and $q$ 's herein by (2) and (3) yields\n\n$$\n\\begin{aligned}\nG S^{2} & =\\frac{x y}{N} y^{2}+\\frac{y z+z x}{N}(y+z)^{2}-\\frac{x y}{N} \\cdot \\frac{y z+z x}{N} \\cdot\\left(z^{2}+\\frac{4 x y z}{x+y}\\right) \\\\\n& =\\frac{x y^{3}}{N}+\\underbrace{\\frac{y z(y+z)^{2}}{N}}_{M_{1}}+\\frac{z x(y+z)^{2}}{N}-\\frac{x y z^{3}(x+y)}{N^{2}}-\\underbrace{\\frac{4 x^{2} y^{2} z^{2}}{N^{2}}}_{M_{2}} \\\\\n& =\\frac{x y^{3}+z x(y+z)^{2}}{N}-\\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\\\\n& =\\underbrace{\\frac{x\\left(y^{3}+y^{2} z+y z^{2}+z^{3}\\right)}{N}}_{M_{3}}+\\frac{x y z^{2} N}{N^{2}}-\\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\\\\n& =\\frac{x^{2} y^{2} z^{2}+x y^{2} z^{3}+x^{2} y z^{3}-x^{2} y z^{3}-x y^{2} z^{3}}{N^{2}}+M_{1}-M_{2}+M_{3} \\\\\n& =\\frac{x^{2} y^{2} z^{2}}{N^{2}}+M_{1}-M_{2}+M_{3},\n\\end{aligned}\n$$\n\na term that is symmetric in $y$ and $z$, indeed.""]" [] Text-only Competition True Theorem proof Geometry Math English 324 Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$. "['It suffices to show that $\\angle H E F=\\angle H G E$ (see Figure 1), since in circle $E G H$ the angle over the chord $E H$ at $G$ equals the angle between the tangent at $E$ and $E H$.\n\nFirst, $\\angle B A D=180^{\\circ}-\\angle D C B=\\angle F C D$. Since triangles $F A B$ and $F C D$ have also a common interior angle at $F$, they are similar.\n\n\n\nFigure 1\n\nDenote by $\\mathcal{T}$ the transformation consisting of a reflection at the bisector of $\\angle D F C$ followed by a dilation with center $F$ and factor of $\\frac{F A}{F C}$. Then $\\mathcal{T}$ maps $F$ to $F, C$ to $A, D$ to $B$, and $H$ to $G$. To see this, note that $\\triangle F C A \\sim \\triangle F D B$, so $\\frac{F A}{F C}=\\frac{F B}{F D}$. Moreover, as $\\angle A D B=\\angle A C B$, the image of the line $D E$ under $\\mathcal{T}$ is parallel to $A C$ (and passes through $B$ ) and similarly the image of $C E$ is parallel to $D B$ and passes through $A$. Hence $E$ is mapped to the point $X$ which is the fourth vertex of the parallelogram $B E A X$. Thus, in particular $\\angle H E F=\\angle F X G$.\n\nAs $G$ is the midpoint of the diagonal $A B$ of the parallelogram $B E A X$, it is also the midpoint of $E X$. In particular, $E, G, X$ are collinear, and $E X=2 \\cdot E G$.\n\nDenote by $Y$ the fourth vertex of the parallelogram $D E C Y$. By an analogous reasoning as before, it follows that $\\mathcal{T}$ maps $Y$ to $E$, thus $E, H, Y$ are collinear with $E Y=2 \\cdot E H$. Therefore, by the intercept theorem, $H G \\| X Y$.\n\nFrom the construction of $\\mathcal{T}$ it is clear that the lines $F X$ and $F E$ are symmetric with respect to the bisector of $\\angle D F C$, as are $F Y$ and $F E$. Thus, $F, X, Y$ are collinear, which together with $H G \\| X Y$ implies $\\angle F X E=\\angle H G E$. This completes the proof.' 'We use the following\n\nLemma (Gauß). Let $A B C D$ be a quadrilateral. Let $A B$ and $C D$ intersect at $P$, and $B C$ and $D A$ intersect at $Q$. Then the midpoints $K, L, M$ of $A C, B D$, and $P Q$, respectively, are collinear.\n\nProof: Let us consider the points $Z$ that fulfill the equation\n\n$$\n(A B Z)+(C D Z)=(B C Z)+(D A Z),\n\\tag{1}\n$$\n\nwhere $(R S T)$ denotes the oriented area of the triangle $R S T$ (see Figure 2).\n\n\n\nFigure 2\n\nAs (1) is linear in $Z$, it can either characterize a line, or be contradictory, or be trivially fulfilled for all $Z$ in the plane. If (1) was fulfilled for all $Z$, then it would hold for $Z=A, Z=B$, which gives $(C D A)=(B C A),(C D B)=(D A B)$, respectively, i.e. the diagonals of $A B C D$ would bisect each other, thus $A B C D$ would be a parallelogram. This contradicts the hypothesis that $A D$ and $B C$ intersect. Since $E, F, G$ fulfill (1), it is the equation of a line which completes the proof of the lemma.\n\nNow consider the parallelograms $E A X B$ and $E C Y D$ (see Figure 1). Then $G, H$ are the midpoints of $E X, E Y$, respectively. Let $M$ be the midpoint of $E F$. By applying the Lemma to the (re-entrant) quadrilateral $A D B C$, it is evident that $G, H$, and $M$ are collinear. A dilation by a factor of 2 with center $E$ shows that $X, Y, F$ are collinear. Since $A X \\| D E$ and $B X \\| C E$, we have pairwise equal interior angles in the quadrilaterals $F D E C$ and $F B X A$. Since we have also $\\angle E B A=\\angle D C A=\\angle C D Y$, the quadrilaterals are similar. Thus, $\\angle F X A=\\angle C E F$.\n\nClearly the parallelograms $E C Y D$ and $E B X A$ are similar, too, thus $\\angle E X A=\\angle C E Y$. Consequently, $\\angle F X E=\\angle F X A-\\angle E X A=\\angle C E F-\\angle C E Y=\\angle Y E F$. By the converse of the tangent-chord angle theorem $E F$ is tangent to the circle $X E Y$. A dilation by a factor of $\\frac{1}{2}$ completes the proof.' 'As in Solution 2, G, H, M are proven to be collinear. It suffices to show that $M E^{2}=M G \\cdot M H$. If $\\boldsymbol{p}=\\overrightarrow{O P}$ denotes the vector from circumcenter $O$ to point $P$, the claim becomes\n\n$$\n\\left(\\frac{\\boldsymbol{e}-\\boldsymbol{f}}{2}\\right)^{2}=\\left(\\frac{\\boldsymbol{e}+\\boldsymbol{f}}{2}-\\frac{\\boldsymbol{a}+\\boldsymbol{b}}{2}\\right)\\left(\\frac{\\boldsymbol{e}+\\boldsymbol{f}}{2}-\\frac{\\boldsymbol{c}+\\boldsymbol{d}}{2}\\right)\n$$\n\nor equivalently\n\n$$\n4 \\boldsymbol{e} \\boldsymbol{f}-(\\boldsymbol{e}+\\boldsymbol{f})(\\boldsymbol{a}+\\boldsymbol{b}+\\boldsymbol{c}+\\boldsymbol{d})+(\\boldsymbol{a}+\\boldsymbol{b})(\\boldsymbol{c}+\\boldsymbol{d})=0\n\\tag{2}\n$$\n\nWith $R$ as the circumradius of $A B C D$, we obtain for the powers $\\mathcal{P}(E)$ and $\\mathcal{P}(F)$ of $E$ and $F$, respectively, with respect to the circumcircle\n\n$$\n\\begin{aligned}\n& \\mathcal{P}(E)=(\\boldsymbol{e}-\\boldsymbol{a})(\\boldsymbol{e}-\\boldsymbol{c})=(\\boldsymbol{e}-\\boldsymbol{b})(\\boldsymbol{e}-\\boldsymbol{d})=\\boldsymbol{e}^{2}-R^{2} \\\\\n& \\mathcal{P}(F)=(\\boldsymbol{f}-\\boldsymbol{a})(\\boldsymbol{f}-\\boldsymbol{d})=(\\boldsymbol{f}-\\boldsymbol{b})(\\boldsymbol{f}-\\boldsymbol{c})=\\boldsymbol{f}^{2}-R^{2}\n\\end{aligned}\n$$\n\nhence\n\n$$\n(\\boldsymbol{e}-\\boldsymbol{a})(\\boldsymbol{e}-\\boldsymbol{c})=\\boldsymbol{e}^{2}-R^{2}\n\\tag{3}\n$$\n$$\n(\\boldsymbol{e}-\\boldsymbol{b})(\\boldsymbol{e}-\\boldsymbol{d})=\\boldsymbol{e}^{2}-R^{2}\n\\tag{4}\n$$\n$$\n(\\boldsymbol{f}-\\boldsymbol{a})(\\boldsymbol{f}-\\boldsymbol{d})=\\boldsymbol{f}^{2}-R^{2}\n\\tag{5}\n$$\n$$\n(\\boldsymbol{f}-\\boldsymbol{b})(\\boldsymbol{f}-\\boldsymbol{c})=\\boldsymbol{f}^{2}-R^{2}\n\\tag{6}\n$$\n\nSince $F$ lies on the polar to $E$ with respect to the circumcircle, we have\n\n$$\n4 \\boldsymbol{e} \\boldsymbol{f}=4 R^{2}\n\\tag{7}\n$$\n\nAdding up (3) to (7) yields (2), as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 325 "Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have $$ \frac{|R|}{|P|} \leq \sqrt{2} $$ where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively." "['We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $\\left|R_{1}\\right| \\leq \\sqrt{2}|P|$ and $\\left|R_{3}\\right| \\leq \\sqrt{2}|P|$ holds (see Figure 1).\n\nFirst we will construct a parallelogram $R_{1} \\supseteq P$ with the property that the midpoints of the sides of $R_{1}$ are points of the boundary of $P$.\n\nChoose two points $A$ and $B$ of $P$ such that the triangle $O A B$ has maximal area. Let $a$ be the line through $A$ parallel to $O B$ and $b$ the line through $B$ parallel to $O A$. Let $A^{\\prime}, B^{\\prime}, a^{\\prime}$ and $b^{\\prime}$ be the points or lines, that are symmetric to $A, B, a$ and $b$, respectively, with respect to $O$. Now let $R_{1}$ be the parallelogram defined by $a, b, a^{\\prime}$ and $b^{\\prime}$.\n\n\n\nFigure 1\n\nObviously, $A$ and $B$ are located on the boundary of the polygon $P$, and $A, B, A^{\\prime}$ and $B^{\\prime}$ are midpoints of the sides of $R_{1}$. We note that $P \\subseteq R_{1}$. Otherwise, there would be a point $Z \\in P$ but $Z \\notin R_{1}$, i.e., one of the lines $a, b, a^{\\prime}$ or $b^{\\prime}$ were between $O$ and $Z$. If it is $a$, we have $|O Z B|>|O A B|$, which is contradictory to the choice of $A$ and $B$. If it is one of the lines $b, a^{\\prime}$ or $b^{\\prime}$ almost identical arguments lead to a similar contradiction.\n\nLet $R_{2}$ be the parallelogram $A B A^{\\prime} B^{\\prime}$. Since $A$ and $B$ are points of $P$, segment $A B \\subset P$ and so $R_{2} \\subset R_{1}$. Since $A, B, A^{\\prime}$ and $B^{\\prime}$ are midpoints of the sides of $R_{1}$, an easy argument yields\n\n$$\n\\left|R_{1}\\right|=2 \\cdot\\left|R_{2}\\right|\n\\tag{1}\n$$\n\nLet $R_{3}$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $A B$ and $B A^{\\prime}$. Obviously $R_{2} \\subset R_{3}$ and every side of $R_{3}$ contains at least one point of the boundary of $P$. Denote by $C$ the intersection point of $a$ and $b$, by $X$ the intersection point of $A B$ and $O C$, and by $X^{\\prime}$ the intersection point of $X C$ and the boundary of $R_{3}$. In a similar way denote by $D$\n\n\n\nthe intersection point of $b$ and $a^{\\prime}$, by $Y$ the intersection point of $A^{\\prime} B$ and $O D$, and by $Y^{\\prime}$ the intersection point of $Y D$ and the boundary of $R_{3}$.\n\nNote that $O C=2 \\cdot O X$ and $O D=2 \\cdot O Y$, so there exist real numbers $x$ and $y$ with $1 \\leq x, y \\leq 2$ and $O X^{\\prime}=x \\cdot O X$ and $O Y^{\\prime}=y \\cdot O Y$. Corresponding sides of $R_{3}$ and $R_{2}$ are parallel which yields\n\n$$\n\\left|R_{3}\\right|=x y \\cdot\\left|R_{2}\\right| .\n\\tag{2}\n$$\n\nThe side of $R_{3}$ containing $X^{\\prime}$ contains at least one point $X^{*}$ of $P$; due to the convexity of $P$ we have $A X^{*} B \\subset P$. Since this side of the parallelogram $R_{3}$ is parallel to $A B$ we have $\\left|A X^{*} B\\right|=\\left|A X^{\\prime} B\\right|$, so $\\left|O A X^{\\prime} B\\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A$. In a similar way we conclude that $\\left|O B^{\\prime} Y^{\\prime} A^{\\prime}\\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A^{\\prime}$. Putting things together we have $\\left|O A X^{\\prime} B\\right|=x \\cdot|O A B|,\\left|O B D A^{\\prime}\\right|=y \\cdot\\left|O B A^{\\prime}\\right|$. Since $|O A B|=\\left|O B A^{\\prime}\\right|$, we conclude that $|P| \\geq 2 \\cdot\\left|A X^{\\prime} B Y^{\\prime} A^{\\prime}\\right|=2 \\cdot\\left(x \\cdot|O A B|+y \\cdot\\left|O B A^{\\prime}\\right|\\right)=4 \\cdot \\frac{x+y}{2} \\cdot|O A B|=\\frac{x+y}{2} \\cdot R_{2}$; this is in short\n\n$$\n\\frac{x+y}{2} \\cdot\\left|R_{2}\\right| \\leq|P| \\text {. }\n\\tag{3}\n$$\n\nSince all numbers concerned are positive, we can combine (1)-(3). Using the arithmeticgeometric-mean inequality we obtain\n\n$$\n\\left|R_{1}\\right| \\cdot\\left|R_{3}\\right|=2 \\cdot\\left|R_{2}\\right| \\cdot x y \\cdot\\left|R_{2}\\right| \\leq 2 \\cdot\\left|R_{2}\\right|^{2}\\left(\\frac{x+y}{2}\\right)^{2} \\leq 2 \\cdot|P|^{2}\n$$\n\nThis implies immediately the desired result $\\left|R_{1}\\right| \\leq \\sqrt{2} \\cdot|P|$ or $\\left|R_{3}\\right| \\leq \\sqrt{2} \\cdot|P|$.' 'We construct the parallelograms $R_{1}, R_{2}$ and $R_{3}$ in the same way as in Solution 1 and will show that $\\frac{\\left|R_{1}\\right|}{|P|} \\leq \\sqrt{2}$ or $\\frac{\\left|R_{3}\\right|}{|P|} \\leq \\sqrt{2}$.\n\n\n\nFigure 2\n\nRecall that affine one-to-one maps of the plane preserve the ratio of areas of subsets of the plane. On the other hand, every parallelogram can be transformed with an affine map onto a square. It follows that without loss of generality we may assume that $R_{1}$ is a square (see Figure 2).\n\nThen $R_{2}$, whose vertices are the midpoints of the sides of $R_{1}$, is a square too, and $R_{3}$, whose sides are parallel to the diagonals of $R_{1}$, is a rectangle.\n\nLet $a>0, b \\geq 0$ and $c \\geq 0$ be the distances introduced in Figure 2. Then $\\left|R_{1}\\right|=2 a^{2}$ and\n\n\n\n$\\left|R_{3}\\right|=(a+2 b)(a+2 c)$.\n\nPoints $A, A^{\\prime}, B$ and $B^{\\prime}$ are in the convex polygon $P$. Hence the square $A B A^{\\prime} B^{\\prime}$ is a subset of $P$. Moreover, each of the sides of the rectangle $R_{3}$ contains a point of $P$, otherwise $R_{3}$ would not be minimal. It follows that\n\n$$\n|P| \\geq a^{2}+2 \\cdot \\frac{a b}{2}+2 \\cdot \\frac{a c}{2}=a(a+b+c)\n$$\n\nNow assume that both $\\frac{\\left|R_{1}\\right|}{|P|}>\\sqrt{2}$ and $\\frac{\\left|R_{3}\\right|}{|P|}>\\sqrt{2}$, then\n\n$$\n2 a^{2}=\\left|R_{1}\\right|>\\sqrt{2} \\cdot|P| \\geq \\sqrt{2} \\cdot a(a+b+c)\n$$\n\nand\n\n$$\n(a+2 b)(a+2 c)=\\left|R_{3}\\right|>\\sqrt{2} \\cdot|P| \\geq \\sqrt{2} \\cdot a(a+b+c)\n$$\n\nAll numbers concerned are positive, so after multiplying these inequalities we get\n\n$$\n2 a^{2}(a+2 b)(a+2 c)>2 a^{2}(a+b+c)^{2} .\n$$\n\nBut the arithmetic-geometric-mean inequality implies the contradictory result\n\n$$\n2 a^{2}(a+2 b)(a+2 c) \\leq 2 a^{2}\\left(\\frac{(a+2 b)+(a+2 c)}{2}\\right)^{2}=2 a^{2}(a+b+c)^{2}\n$$\n\nHence $\\frac{\\left|R_{1}\\right|}{|P|} \\leq \\sqrt{2}$ or $\\frac{\\left|R_{3}\\right|}{|P|} \\leq \\sqrt{2}$, as desired.']" [] Text-only Competition True Theorem proof Geometry Math English 326 Let the sides $A D$ and $B C$ of the quadrilateral $A B C D$ (such that $A B$ is not parallel to $C D$ ) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $A B P$ and $D C P$, respectively. Denote the midpoints of segments $O_{1} H_{1}$ and $O_{2} H_{2}$ by $E_{1}$ and $E_{2}$, respectively. Prove that the perpendicular from $E_{1}$ on $C D$, the perpendicular from $E_{2}$ on $A B$ and the line $H_{1} H_{2}$ are concurrent. "[""We keep triangle $A B P$ fixed and move the line $C D$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $\\mathrm{O}_{2}, \\mathrm{H}_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$ on $A B$ moves uniformly. Obviously, the points $O_{1}, H_{1}, E_{1}$ and the perpendicular from $E_{1}$ on $C D$ do not move at all. Hence, the intersection point $S$ of these two perpendiculars moves uniformly. Since $H_{1}$ does not move, while $H_{2}$ and $S$ move uniformly along parallel lines (both are perpendicular to $C D$ ), it is sufficient to prove their collinearity for two different positions of $C D$.\n\n\n\nFigure 1\n\nLet $C D$ pass through either point $A$ or point $B$. Note that by hypothesis these two cases are different. We will consider the case $A \\in C D$, i.e. $A=D$. So we have to show that the perpendiculars from $E_{1}$ on $A C$ and from $E_{2}$ on $A B$ intersect on the altitude $A H$ of triangle $A B C$ (see Figure 2).\n\n\n\n\n\nFigure 2\n\nTo this end, we consider the midpoints $A_{1}, B_{1}, C_{1}$ of $B C, C A, A B$, respectively. As $E_{1}$ is the center of FEUERBACH's circle (nine-point circle) of $\\triangle A B P$, we have $E_{1} C_{1}=E_{1} H$. Similarly, $E_{2} B_{1}=E_{2} H$. Note further that a point $X$ lies on the perpendicular from $E_{1}$ on $A_{1} C_{1}$ if and only if\n\n$$\nX C_{1}^{2}-X A_{1}^{2}=E_{1} C_{1}^{2}-E_{1} A_{1}^{2}\n$$\n\nSimilarly, the perpendicular from $E_{2}$ on $A_{1} B_{1}$ is characterized by\n\n$$\nX A_{1}^{2}-X B_{1}^{2}=E_{2} A_{1}^{2}-E_{2} B_{1}^{2}\n$$\n\nThe line $H_{1} H_{2}$, which is perpendicular to $B_{1} C_{1}$ and contains $A$, is given by\n\n$$\nX B_{1}^{2}-X C_{1}^{2}=A B_{1}^{2}-A C_{1}^{2}\n$$\n\nThe three lines are concurrent if and only if\n\n$$\n\\begin{aligned}\n0 & =X C_{1}^{2}-X A_{1}^{2}+X A_{1}^{2}-X B_{1}^{2}+X B_{1}^{2}-X C_{1}^{2} \\\\\n& =E_{1} C_{1}^{2}-E_{1} A_{1}^{2}+E_{2} A_{1}^{2}-E_{2} B_{1}^{2}+A B_{1}^{2}-A C_{1}^{2} \\\\\n& =-E_{1} A_{1}^{2}+E_{2} A_{1}^{2}+E_{1} H^{2}-E_{2} H^{2}+A B_{1}^{2}-A C_{1}^{2}\n\\end{aligned}\n$$\n\ni.e. it suffices to show that\n\n$$\nE_{1} A_{1}^{2}-E_{2} A_{1}^{2}-E_{1} H^{2}+E_{2} H^{2}=\\frac{A C^{2}-A B^{2}}{4}\n$$\n\nWe have\n\n$$\n\\frac{A C^{2}-A B^{2}}{4}=\\frac{H C^{2}-H B^{2}}{4}=\\frac{(H C+H B)(H C-H B)}{4}=\\frac{H A_{1} \\cdot B C}{2}\n$$\n\nLet $F_{1}, F_{2}$ be the projections of $E_{1}, E_{2}$ on $B C$. Obviously, these are the midpoints of $H P_{1}$,\n\n\n\n$H P_{2}$, where $P_{1}, P_{2}$ are the midpoints of $P B$ and $P C$ respectively. Then\n\n$$\n\\begin{aligned}\n& E_{1} A_{1}^{2}-E_{2} A_{1}^{2}-E_{1} H^{2}+E_{2} H^{2} \\\\\n& =F_{1} A_{1}^{2}-F_{1} H^{2}-F_{2} A_{1}^{2}+F_{2} H^{2} \\\\\n& =\\left(F_{1} A_{1}-F_{1} H\\right)\\left(F_{1} A_{1}+F_{1} H\\right)-\\left(F_{2} A_{1}-F_{2} H\\right)\\left(F_{2} A_{1}+F_{2} H\\right) \\\\\n& =A_{1} H \\cdot\\left(A_{1} P_{1}-A_{1} P_{2}\\right) \\\\\n& =\\frac{A_{1} H \\cdot B C}{2} \\\\\n& =\\frac{A C^{2}-A B^{2}}{4}\n\\end{aligned}\n$$\n\nwhich proves the claim."" 'Let the perpendicular from $E_{1}$ on $C D$ meet $P H_{1}$ at $X$, and the perpendicular from $E_{2}$ on $A B$ meet $P H_{2}$ at $Y$ (see Figure 3). Let $\\varphi$ be the intersection angle of $A B$ and $C D$. Denote by $M, N$ the midpoints of $P H_{1}, P H_{2}$ respectively.\n\n\n\nFigure 3\n\nWe will prove now that triangles $E_{1} X M$ and $E_{2} Y N$ have equal angles at $E_{1}, E_{2}$, and supplementary angles at $X, Y$.\n\nIn the following, angles are understood as oriented, and equalities of angles modulo $180^{\\circ}$.\n\nLet $\\alpha=\\angle H_{2} P D, \\psi=\\angle D P C, \\beta=\\angle C P H_{1}$. Then $\\alpha+\\psi+\\beta=\\varphi, \\angle E_{1} X H_{1}=\\angle H_{2} Y E_{2}=\\varphi$, thus $\\angle M X E_{1}+\\angle N Y E_{2}=180^{\\circ}$.\n\nBy considering the FEUERBACH circle of $\\triangle A B P$ whose center is $E_{1}$ and which goes through $M$, we have $\\angle E_{1} M H_{1}=\\psi+2 \\beta$. Analogous considerations with the FEUERBACH circle of $\\triangle D C P$ yield $\\angle H_{2} N E_{2}=\\psi+2 \\alpha$. Hence indeed $\\angle X E_{1} M=\\varphi-(\\psi+2 \\beta)=(\\psi+2 \\alpha)-\\varphi=\\angle Y E_{2} N$.\n\nIt follows now that\n\n$$\n\\frac{X M}{M E_{1}}=\\frac{Y N}{N E_{2}}\n$$\n\nFurthermore, $M E_{1}$ is half the circumradius of $\\triangle A B P$, while $P H_{1}$ is the distance of $P$ to the orthocenter of that triangle, which is twice the circumradius times the cosine of $\\psi$. Together\n\n\n\nwith analogous reasoning for $\\triangle D C P$ we have\n\n$$\n\\frac{M E_{1}}{P H_{1}}=\\frac{1}{4 \\cos \\psi}=\\frac{N E_{2}}{P H_{2}}\n$$\n\nBy multiplication,\n\n$$\n\\frac{X M}{P H_{1}}=\\frac{Y N}{P H_{2}}\n$$\n\nand therefore\n\n$$\n\\frac{P X}{X H_{1}}=\\frac{H_{2} Y}{Y P}\n$$\n\nLet $E_{1} X, E_{2} Y$ meet $H_{1} H_{2}$ in $R, S$ respectively.\n\nApplying the intercept theorem to the parallels $E_{1} X, P H_{2}$ and center $H_{1}$ gives\n\n$$\n\\frac{H_{2} R}{R H_{1}}=\\frac{P X}{X H_{1}}\n$$\n\nwhile with parallels $E_{2} Y, P H_{1}$ and center $H_{2}$ we obtain\n\n$$\n\\frac{H_{2} S}{S H_{1}}=\\frac{H_{2} Y}{Y P}\n$$\n\nCombination of the last three equalities yields that $R$ and $S$ coincide.']" [] Text-only Competition True Theorem proof Geometry Math English 327 Let $A B C$ be a triangle with incenter $I$ and let $X, Y$ and $Z$ be the incenters of the triangles $B I C, C I A$ and $A I B$, respectively. Let the triangle $X Y Z$ be equilateral. Prove that $A B C$ is equilateral too. ['$A Z, A I$ and $A Y$ divide $\\angle B A C$ into four equal angles; denote them by $\\alpha$. In the same way we have four equal angles $\\beta$ at $B$ and four equal angles $\\gamma$ at $C$. Obviously $\\alpha+\\beta+\\gamma=\\frac{180^{\\circ}}{4}=45^{\\circ}$; and $0^{\\circ}<\\alpha, \\beta, \\gamma<45^{\\circ}$.\n\n\n\nEasy calculations in various triangles yield $\\angle B I C=180^{\\circ}-2 \\beta-2 \\gamma=180^{\\circ}-\\left(90^{\\circ}-2 \\alpha\\right)=$ $90^{\\circ}+2 \\alpha$, hence (for $X$ is the incenter of triangle $B C I$, so $I X$ bisects $\\angle B I C$ ) we have $\\angle X I C=$ $\\angle B I X=\\frac{1}{2} \\angle B I C=45^{\\circ}+\\alpha$ and with similar aguments $\\angle C I Y=\\angle Y I A=45^{\\circ}+\\beta$ and $\\angle A I Z=\\angle Z I B=45^{\\circ}+\\gamma$. Furthermore, we have $\\angle X I Y=\\angle X I C+\\angle C I Y=\\left(45^{\\circ}+\\alpha\\right)+$ $\\left(45^{\\circ}+\\beta\\right)=135^{\\circ}-\\gamma, \\angle Y I Z=135^{\\circ}-\\alpha$, and $\\angle Z I X=135^{\\circ}-\\beta$.\n\nNow we calculate the lengths of $I X, I Y$ and $I Z$ in terms of $\\alpha, \\beta$ and $\\gamma$. The perpendicular from $I$ on $C X$ has length $I X \\cdot \\sin \\angle C X I=I X \\cdot \\sin \\left(90^{\\circ}+\\beta\\right)=I X \\cdot \\cos \\beta$. But $C I$ bisects $\\angle Y C X$, so the perpendicular from $I$ on $C Y$ has the same length, and we conclude\n\n$$\nI X \\cdot \\cos \\beta=I Y \\cdot \\cos \\alpha\n$$\n\nTo make calculations easier we choose a length unit that makes $I X=\\cos \\alpha$. Then $I Y=\\cos \\beta$ and with similar arguments $I Z=\\cos \\gamma$.\n\nSince $X Y Z$ is equilateral we have $Z X=Z Y$. The law of Cosines in triangles $X Y I, Y Z I$ yields\n\n$$\n\\begin{aligned}\n& Z X^{2}=Z Y^{2} \\\\\n\\Longrightarrow & I Z^{2}+I X^{2}-2 \\cdot I Z \\cdot I X \\cdot \\cos \\angle Z I X=I Z^{2}+I Y^{2}-2 \\cdot I Z \\cdot I Y \\cdot \\cos \\angle Y I Z \\\\\n\\Longrightarrow & I X^{2}-I Y^{2}=2 \\cdot I Z \\cdot(I X \\cdot \\cos \\angle Z I X-I Y \\cdot \\cos \\angle Y I Z) \\\\\n\\Longrightarrow & \\underbrace{\\cos ^{2} \\alpha-\\cos ^{2} \\beta}_{\\text {L.H.S. }}=\\underbrace{2 \\cdot \\cos \\gamma \\cdot\\left(\\cos \\alpha \\cdot \\cos \\left(135^{\\circ}-\\beta\\right)-\\cos \\beta \\cdot \\cos \\left(135^{\\circ}-\\alpha\\right)\\right)}_{\\text {R.H.S. }} .\n\\end{aligned}\n$$\n\nA transformation of the left-hand side (L.H.S.) yields\n\n$$\n\\begin{aligned}\n\\text { L.H.S. } & =\\cos ^{2} \\alpha \\cdot\\left(\\sin ^{2} \\beta+\\cos ^{2} \\beta\\right)-\\cos ^{2} \\beta \\cdot\\left(\\sin ^{2} \\alpha+\\cos ^{2} \\alpha\\right) \\\\\n& =\\cos ^{2} \\alpha \\cdot \\sin ^{2} \\beta-\\cos ^{2} \\beta \\cdot \\sin ^{2} \\alpha\n\\end{aligned}\n$$\n\n\n\n$$\n\\begin{aligned}\n& =(\\cos \\alpha \\cdot \\sin \\beta+\\cos \\beta \\cdot \\sin \\alpha) \\cdot(\\cos \\alpha \\cdot \\sin \\beta-\\cos \\beta \\cdot \\sin \\alpha) \\\\\n& =\\sin (\\beta+\\alpha) \\cdot \\sin (\\beta-\\alpha)=\\sin \\left(45^{\\circ}-\\gamma\\right) \\cdot \\sin (\\beta-\\alpha)\n\\end{aligned}\n$$\n\nwhereas a transformation of the right-hand side (R.H.S.) leads to\n\n$$\n\\begin{aligned}\n\\text { R.H.S. } & =2 \\cdot \\cos \\gamma \\cdot\\left(\\cos \\alpha \\cdot\\left(-\\cos \\left(45^{\\circ}+\\beta\\right)\\right)-\\cos \\beta \\cdot\\left(-\\cos \\left(45^{\\circ}+\\alpha\\right)\\right)\\right) \\\\\n& =2 \\cdot \\frac{\\sqrt{2}}{2} \\cdot \\cos \\gamma \\cdot(\\cos \\alpha \\cdot(\\sin \\beta-\\cos \\beta)+\\cos \\beta \\cdot(\\cos \\alpha-\\sin \\alpha)) \\\\\n& =\\sqrt{2} \\cdot \\cos \\gamma \\cdot(\\cos \\alpha \\cdot \\sin \\beta-\\cos \\beta \\cdot \\sin \\alpha) \\\\\n& =\\sqrt{2} \\cdot \\cos \\gamma \\cdot \\sin (\\beta-\\alpha)\n\\end{aligned}\n$$\n\nEquating L.H.S. and R.H.S. we obtain\n\n$$\n\\begin{aligned}\n& \\sin \\left(45^{\\circ}-\\gamma\\right) \\cdot \\sin (\\beta-\\alpha)=\\sqrt{2} \\cdot \\cos \\gamma \\cdot \\sin (\\beta-\\alpha) \\\\\n\\Longrightarrow & \\sin (\\beta-\\alpha) \\cdot\\left(\\sqrt{2} \\cdot \\cos \\gamma-\\sin \\left(45^{\\circ}-\\gamma\\right)\\right)=0 \\\\\n\\Longrightarrow & \\alpha=\\beta \\quad \\text { or } \\quad \\sqrt{2} \\cdot \\cos \\gamma=\\sin \\left(45^{\\circ}-\\gamma\\right) .\n\\end{aligned}\n$$\n\nBut $\\gamma<45^{\\circ}$; so $\\sqrt{2} \\cdot \\cos \\gamma>\\cos \\gamma>\\cos 45^{\\circ}=\\sin 45^{\\circ}>\\sin \\left(45^{\\circ}-\\gamma\\right)$. This leaves $\\alpha=\\beta$. With similar reasoning we have $\\alpha=\\gamma$, which means triangle $A B C$ must be equilateral.'] [] Text-only Competition True Theorem proof Geometry Math English 328 Let $A B C D$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $B C$ in $M$ and the line $C D$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$, respectively. Show that the orthocenter of $\triangle I_{1} I_{2} I_{3}$ lies on $g$. "['Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $A B M, M N C$, and $N D A$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $C B$ is also tangent to $k_{3}$.\n\n\n\nFigure 1\n\nTo this end, let $X$ denote the point of intersection of $g$ and $h$. Then $A B C X$ and $A B C D$ are circumscribed quadrilaterals, whence\n\n$$\nC D-C X=(A B+C D)-(A B+C X)=(B C+A D)-(B C+A X)=A D-A X\n$$\n\ni.e.\n\n$$\nA X+C D=C X+A D\n$$\n\nwhich in turn reveals that the quadrilateral $A X C D$ is also circumscribed. Thus $h$ touches indeed the circle $k_{3}$.\n\nMoreover, we find that $\\angle I_{3} C I_{1}=\\angle I_{3} C X+\\angle X C I_{1}=\\frac{1}{2}(\\angle D C X+\\angle X C B)=\\frac{1}{2} \\angle D C B=$ $\\frac{1}{2}\\left(180^{\\circ}-\\angle M C N\\right)=180^{\\circ}-\\angle M I_{2} N=\\angle I_{3} I_{2} I_{1}$, from which we conclude that $C, I_{1}, I_{2}, I_{3}$ are concyclic.\n\nLet now $L_{1}$ and $L_{3}$ be the reflection points of $C$ with respect to the lines $I_{2} I_{3}$ and $I_{1} I_{2}$ respectively. Since $I_{1} I_{2}$ is the angle bisector of $\\angle N M C$, it follows that $L_{3}$ lies on $g$. By analogous reasoning, $L_{1}$ lies on $g$.\n\nLet $H$ be the orthocenter of $\\triangle I_{1} I_{2} I_{3}$. We have $\\angle I_{2} L_{3} I_{1}=\\angle I_{1} C I_{2}=\\angle I_{1} I_{3} I_{2}=180^{\\circ}-\\angle I_{1} H I_{2}$, which entails that the quadrilateral $I_{2} H I_{1} L_{3}$ is cyclic. Analogously, $I_{3} H L_{1} I_{2}$ is cyclic.\n\n\n\nThen, working with oriented angles modulo $180^{\\circ}$, we have\n\n$$\n\\angle L_{3} H I_{2}=\\angle L_{3} I_{1} I_{2}=\\angle I_{2} I_{1} C=\\angle I_{2} I_{3} C=\\angle L_{1} I_{3} I_{2}=\\angle L_{1} H I_{2},\n$$\n\nwhence $L_{1}, L_{3}$, and $H$ are collinear. By $L_{1} \\neq L_{3}$, the claim follows.' 'We start by proving that $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic.\n\n\n\nFigure 2\n\nTo this end, notice first that $I_{2}, M, I_{1}$ are collinear, as are $N, I_{2}, I_{3}$ (see Figure 2). Denote by $\\alpha, \\beta, \\gamma, \\delta$ the internal angles of $A B C D$. By considerations in triangle $C M N$, it follows that $\\angle I_{3} I_{2} I_{1}=\\frac{\\gamma}{2}$. We will show that $\\angle I_{3} C I_{1}=\\frac{\\gamma}{2}$, too. Denote by $I$ the incenter of $A B C D$. Clearly, $I_{1} \\in B I, I_{3} \\in D I, \\angle I_{1} A I_{3}=\\frac{\\alpha}{2}$.\n\nUsing the abbreviation $[X, Y Z]$ for the distance from point $X$ to the line $Y Z$, we have because of $\\angle B A I_{1}=\\angle I A I_{3}$ and $\\angle I_{1} A I=\\angle I_{3} A D$ that\n\n$$\n\\frac{\\left[I_{1}, A B\\right]}{\\left[I_{1}, A I\\right]}=\\frac{\\left[I_{3}, A I\\right]}{\\left[I_{3}, A D\\right]}\n$$\n\nFurthermore, consideration of the angle sums in $A I B, B I C, C I D$ and $D I A$ implies $\\angle A I B+$ $\\angle C I D=\\angle B I C+\\angle D I A=180^{\\circ}$, from which we see\n\n$$\n\\frac{\\left[I_{1}, A I\\right]}{\\left[I_{3}, C I\\right]}=\\frac{I_{1} I}{I_{3} I}=\\frac{\\left[I_{1}, C I\\right]}{\\left[I_{3}, A I\\right]}\n$$\n\nBecause of $\\left[I_{1}, A B\\right]=\\left[I_{1}, B C\\right],\\left[I_{3}, A D\\right]=\\left[I_{3}, C D\\right]$, multiplication yields\n\n$$\n\\frac{\\left[I_{1}, B C\\right]}{\\left[I_{3}, C I\\right]}=\\frac{\\left[I_{1}, C I\\right]}{\\left[I_{3}, C D\\right]}\n$$\n\nBy $\\angle D C I=\\angle I C B=\\gamma / 2$ it follows that $\\angle I_{1} C B=\\angle I_{3} C I$ which concludes the proof of the\n\n\n\nabove statement.\n\nLet the perpendicular from $I_{1}$ on $I_{2} I_{3}$ intersect $g$ at $Z$. Then $\\angle M I_{1} Z=90^{\\circ}-\\angle I_{3} I_{2} I_{1}=$ $90^{\\circ}-\\gamma / 2=\\angle M C I_{2}$. Since we have also $\\angle Z M I_{1}=\\angle I_{2} M C$, triangles $M Z I_{1}$ and $M I_{2} C$ are similar. From this one easily proves that also $M I_{2} Z$ and $M C I_{1}$ are similar. Because $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic, $\\angle M Z I_{2}=\\angle M I_{1} C=\\angle N I_{3} C$, thus $N I_{2} Z$ and $N C I_{3}$ are similar, hence $N C I_{2}$ and $N I_{3} Z$ are similar. We conclude $\\angle Z I_{3} I_{2}=\\angle I_{2} C N=90^{\\circ}-\\gamma / 2$, hence $I_{1} I_{2} \\perp Z I_{3}$. This completes the proof.']" [] Text-only Competition True Theorem proof Geometry Math English 329 "A social club has $n$ members. They have the membership numbers $1,2, \ldots, n$, respectively. From time to time members send presents to other members, including items they have already received as presents from other members. In order to avoid the embarrassing situation that a member might receive a present that he or she has sent to other members, the club adds the following rule to its statutes at one of its annual general meetings: ""A member with membership number $a$ is permitted to send a present to a member with membership number $b$ if and only if $a(b-1)$ is a multiple of $n . ""$ Prove that, if each member follows this rule, none will receive a present from another member that he or she has already sent to other members. Alternative formulation: Let $G$ be a directed graph with $n$ vertices $v_{1}, v_{2}, \ldots, v_{n}$, such that there is an edge going from $v_{a}$ to $v_{b}$ if and only if $a$ and $b$ are distinct and $a(b-1)$ is a multiple of $n$. Prove that this graph does not contain a directed cycle." "['Suppose there is an edge from $v_{i}$ to $v_{j}$. Then $i(j-1)=i j-i=k n$ for some integer $k$, which implies $i=i j-k n$. If $\\operatorname{gcd}(i, n)=d$ and $\\operatorname{gcd}(j, n)=e$, then $e$ divides $i j-k n=i$ and thus $e$ also divides $d$. Hence, if there is an edge from $v_{i}$ to $v_{j}$, then $\\operatorname{gcd}(j, n) \\mid \\operatorname{gcd}(i, n)$.\n\nIf there is a cycle in $G$, say $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$, then we have\n\n$$\n\\operatorname{gcd}\\left(i_{1}, n\\right)\\left|\\operatorname{gcd}\\left(i_{r}, n\\right)\\right| \\operatorname{gcd}\\left(i_{r-1}, n\\right)|\\ldots| \\operatorname{gcd}\\left(i_{2}, n\\right) \\mid \\operatorname{gcd}\\left(i_{1}, n\\right)\n$$\n\nwhich implies that all these greatest common divisors must be equal, say be equal to $t$.\n\nNow we pick any of the $i_{k}$, without loss of generality let it be $i_{1}$. Then $i_{r}\\left(i_{1}-1\\right)$ is a multiple of $n$ and hence also (by dividing by $t$ ), $i_{1}-1$ is a multiple of $\\frac{n}{t}$. Since $i_{1}$ and $i_{1}-1$ are relatively prime, also $t$ and $\\frac{n}{t}$ are relatively prime. So, by the Chinese remainder theorem, the value of $i_{1}$ is uniquely determined modulo $n=t \\cdot \\frac{n}{t}$ by the value of $t$. But, as $i_{1}$ was chosen arbitrarily among the $i_{k}$, this implies that all the $i_{k}$ have to be equal, a contradiction.' 'If $a, b, c$ are integers such that $a b-a$ and $b c-b$ are multiples of $n$, then also $a c-a=a(b c-b)+(a b-a)-(a b-a) c$ is a multiple of $n$. This implies that if there is an edge from $v_{a}$ to $v_{b}$ and an edge from $v_{b}$ to $v_{c}$, then there also must be an edge from $v_{a}$ to $v_{c}$. Therefore, if there are any cycles at all, the smallest cycle must have length 2 . But suppose the vertices $v_{a}$ and $v_{b}$ form such a cycle, i. e., $a b-a$ and $a b-b$ are both multiples of $n$. Then $a-b$ is also a multiple of $n$, which can only happen if $a=b$, which is impossible.' 'Suppose there was a cycle $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$. Then $i_{1}\\left(i_{2}-1\\right)$ is a multiple of $n$, i. e., $i_{1} \\equiv i_{1} i_{2} \\bmod n$. Continuing in this manner, we get $i_{1} \\equiv i_{1} i_{2} \\equiv$ $i_{1} i_{2} i_{3} \\equiv i_{1} i_{2} i_{3} \\ldots i_{r} \\bmod n$. But the same holds for all $i_{k}$, i. e., $i_{k} \\equiv i_{1} i_{2} i_{3} \\ldots i_{r} \\bmod n$. Hence $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\bmod n$, which means $i_{1}=i_{2}=\\cdots=i_{r}$, a contradiction.' 'Let $n=k$ be the smallest value of $n$ for which the corresponding graph has a cycle. We show that $k$ is a prime power.\n\nIf $k$ is not a prime power, it can be written as a product $k=d e$ of relatively prime integers greater than 1. Reducing all the numbers modulo $d$ yields a single vertex or a cycle in the corresponding graph on $d$ vertices, because if $a(b-1) \\equiv 0 \\bmod k$ then this equation also holds modulo $d$. But since the graph on $d$ vertices has no cycles, by the minimality of $k$, we must have that all the indices of the cycle are congruent modulo $d$. The same holds modulo $e$ and hence also modulo $k=d e$. But then all the indices are equal, which is a contradiction.\n\nThus $k$ must be a prime power $k=p^{m}$. There are no edges ending at $v_{k}$, so $v_{k}$ is not contained in any cycle. All edges not starting at $v_{k}$ end at a vertex belonging to a non-multiple of $p$, and all edges starting at a non-multiple of $p$ must end at $v_{1}$. But there is no edge starting at $v_{1}$. Hence there is no cycle.' 'Suppose there was a cycle $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$. Let $q=p^{m}$ be a prime power dividing $n$. We claim that either $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\equiv 0 \\bmod q$ or $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\equiv$ $1 \\bmod q$.\n\nSuppose that there is an $i_{s}$ not divisible by $q$. Then, as $i_{s}\\left(i_{s+1}-1\\right)$ is a multiple of $q, i_{s+1} \\equiv$ $1 \\bmod p$. Similarly, we conclude $i_{s+2} \\equiv 1 \\bmod p$ and so on. So none of the labels is divisible by $p$, but since $i_{s}\\left(i_{s+1}-1\\right)$ is a multiple of $q=p^{m}$ for all $s$, all $i_{s+1}$ are congruent to 1 modulo $q$. This proves the claim.\n\nNow, as all the labels are congruent modulo all the prime powers dividing $n$, they must all be equal by the Chinese remainder theorem. This is a contradiction.']" [] Text-only Competition True Theorem proof Geometry Math English 330 "A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$. Prove that there exist distinct positive integers $a$ and $b$ such that all the numbers $P(1), P(2)$, $\ldots, P(50)$ are balanced." ['Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \\equiv f(n)+f(m) \\bmod 2$ for all positive integers $n, m$.\n\nNow for each positive integer $n$ consider the binary sequence $(f(n+1), f(n+2), \\ldots, f(n+$ $50)$ ). As there are only $2^{50}$ different such sequences there are two different positive integers $a$ and $b$ such that\n\n$$\n(f(a+1), f(a+2), \\ldots, f(a+50))=(f(b+1), f(b+2), \\ldots, f(b+50)) \\text {. }\n$$\n\nBut this implies that for the polynomial $P(x)=(x+a)(x+b)$ all the numbers $P(1), P(2)$, $\\ldots, P(50)$ are balanced, since for all $1 \\leq k \\leq 50$ we have $f(P(k)) \\equiv f(a+k)+f(b+k) \\equiv$ $2 f(a+k) \\equiv 0 \\bmod 2$.'] [] Text-only Competition True Theorem proof Number Theory Math English 331 "A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$. Prove that if $P(n)$ is balanced for all positive integers $n$, then $a=b$." ['Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \\equiv f(n)+f(m) \\bmod 2$ for all positive integers $n, m$.\n\nNow suppose $P(n)$ is balanced for all positive integers $n$ and $av_{p_{i}}(f(1))$ for all $i=1,2, \\ldots, m$, e.g. $a=\\left(p_{1} p_{2} \\ldots p_{m}\\right)^{\\alpha}$ with $\\alpha$ sufficiently large. Pick any such $a$. The condition of the problem then yields $a \\mid(f(a+1)-f(1))$. Assume $f(a+1) \\neq f(1)$. Then we must have $v_{p_{i}}(f(a+1)) \\neq$ $v_{p_{i}}(f(1))$ for at least one $i$. This yields $v_{p_{i}}(f(a+1)-f(1))=\\min \\left\\{v_{p_{i}}(f(a+1)), v_{p_{i}}(f(1))\\right\\} \\leq$ $v_{p_{1}}(f(1))p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1} \\cdot(f(r)+r)-r \\\\\n& \\geq p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1}+(f(r)+r)-r \\\\\n& >p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{m}^{\\alpha_{m}}+f(r) \\\\\n& \\geq|f(M)-f(r)| .\n\\end{aligned}\n$$\n\nBut since $M-r$ divides $f(M)-f(r)$ this can only be true if $f(r)=f(M)=f(1)$, which contradicts the choice of $r$.']" [] Text-only Competition True Theorem proof Number Theory Math English 333 Let $P(x)$ be a non-constant polynomial with integer coefficients. Prove that there is no function $T$ from the set of integers into the set of integers such that the number of integers $x$ with $T^{n}(x)=x$ is equal to $P(n)$ for every $n \geq 1$, where $T^{n}$ denotes the $n$-fold application of $T$. ['Assume there is a polynomial $P$ of degree at least 1 with the desired property for a given function $T$. Let $A(n)$ denote the set of all $x \\in \\mathbb{Z}$ such that $T^{n}(x)=x$ and let $B(n)$ denote the set of all $x \\in \\mathbb{Z}$ for which $T^{n}(x)=x$ and $T^{k}(x) \\neq x$ for all $1 \\leq k\\sqrt{a b}, b^{\\ell_{0}}>\\sqrt{a b}$, we define the polynomial\n\n$$\nP(x)=\\prod_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1}\\left(a^{k} b^{\\ell} x-\\sqrt{a b}\\right)=: \\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x^{i}\n$$\n\nwith integer coefficients $d_{i}$. By our assumption, the zeros\n\n$$\n\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}, \\quad k=0, \\ldots, k_{0}-1, \\quad \\ell=0, \\ldots, \\ell_{0}-1\n$$\n\nof $P$ are pairwise distinct.\n\nFurthermore, we consider the integer sequence\n\n$$\ny_{n}=\\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x_{n+i}, \\quad n=1,2, \\ldots\n\\tag{3}\n$$\n\nBy the theory of linear recursions, we obtain\n\n$$\ny_{n}=\\sum_{\\substack{k, \\ell \\geq 0 \\\\ k \\geq k_{0} \\text { or } \\ell \\geq \\ell_{0}}} e_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}, \\quad n=1,2, \\ldots,\n\\tag{4}\n$$\n\nwith real numbers $e_{k, \\ell}$. We have\n\n$$\n\\left|y_{n}\\right| \\leq \\sum_{\\substack{k, \\ell \\geq 0 \\\\ k \\geq k_{0} \\text { or } \\ell \\geq \\ell_{0}}}\\left|e_{k, \\ell}\\right|\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}=: M_{n}\n$$\n\n\n\nBecause the series in (4) is obtained by a finite linear combination of the absolutely convergent series (1), we conclude that in particular $M_{1}<\\infty$. Since\n\n$$\n\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}} \\leq \\lambda:=\\max \\left\\{\\frac{\\sqrt{a b}}{a^{k_{0}}}, \\frac{\\sqrt{a b}}{b^{\\ell_{0}}}\\right\\} \\quad \\text { for all } k, \\ell \\geq 0 \\text { such that } k \\geq k_{0} \\text { or } \\ell \\geq \\ell_{0}\n$$\n\nwe get the estimates $M_{n+1} \\leq \\lambda M_{n}, n=1,2, \\ldots$ Our choice of $k_{0}$ and $\\ell_{0}$ ensures $\\lambda<1$, which implies $M_{n} \\rightarrow 0$ and consequently $y_{n} \\rightarrow 0$ as $n \\rightarrow \\infty$. It follows that $y_{n}=0$ for all sufficiently large $n$.\n\nSo, equation (3) reduces to $\\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x_{n+i}=0$.\n\nUsing the theory of linear recursions again, for sufficiently large $n$ we have\n\n$$\nx_{n}=\\sum_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1} f_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}\n$$\n\nfor certain real numbers $f_{k, \\ell}$.\n\nComparing with (2), we see that $f_{k, \\ell}=c_{k, \\ell}$ for all $k, \\ell \\geq 0$ with $kj_{0}$. But this means that\n\n$$\n\\left(1-x^{\\mu}\\right)^{\\frac{1}{2}}\\left(1-x^{\\nu}\\right)^{\\frac{1}{2}}=\\sum_{j=0}^{j_{0}} g_{j} x^{j}\n$$\n\nfor all real numbers $x \\in(0,1)$. Squaring, we see that\n\n$$\n\\left(1-x^{\\mu}\\right)\\left(1-x^{\\nu}\\right)\n$$\n\nis the square of a polynomial in $x$. In particular, all its zeros are of order at least 2 , which implies $\\mu=\\nu$ by looking at roots of unity. So we obtain $\\mu=\\nu=1$, i. e., $a=b$, a contradiction.' 'We set $a^{2}=A, b^{2}=B$, and $z_{n}=\\sqrt{\\left(A^{n}-1\\right)\\left(B^{n}-1\\right)}$. Let us assume that $z_{n}$ is an integer for $n=1,2, \\ldots$ Without loss of generality, we may suppose that $b0$ as $z_{n}>0$. As before, one obtains\n\n$$\n\\begin{aligned}\n& A^{n} B^{n}-A^{n}-B^{n}+1=z_{n}^{2} \\\\\n& =\\left\\{\\delta_{0}(a b)^{n}-\\delta_{1}\\left(\\frac{a}{b}\\right)^{n}-\\delta_{2}\\left(\\frac{a}{b^{3}}\\right)^{n}-\\cdots-\\delta_{k}\\left(\\frac{a}{b^{2 k-1}}\\right)^{n}\\right\\}^{2} \\\\\n& =\\delta_{0}^{2} A^{n} B^{n}-2 \\delta_{0} \\delta_{1} A^{n}-\\sum_{i=2}^{i=k}\\left(2 \\delta_{0} \\delta_{i}-\\sum_{j=1}^{j=i-1} \\delta_{j} \\delta_{i-j}\\right)\\left(\\frac{A}{B^{i-1}}\\right)^{n}+O\\left(\\frac{A}{B^{k}}\\right)^{n} .\n\\end{aligned}\n$$\n\nEasy asymptotic calculations yield $\\delta_{0}=1, \\delta_{1}=\\frac{1}{2}, \\delta_{i}=\\frac{1}{2} \\sum_{j=1}^{j=i-1} \\delta_{j} \\delta_{i-j}$ for $i=2,3, \\ldots, k-2$, and then $a=b^{k-1}$. It follows that $k>2$ and there is some $P \\in \\mathbb{Q}[X]$ for which $(X-1)\\left(X^{k-1}-1\\right)=$ $P(X)^{2}$. But this cannot occur, for instance as $X^{k-1}-1$ has no double zeros. Thus our\n\n\n\nassumption that $z_{n}$ was an integer for $n=1,2, \\ldots$ turned out to be wrong, which solves the problem.']" [] Text-only Competition True Theorem proof Number Theory Math English 336 " Prove the inequality $$ \frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geq 1 $$ for real numbers $x, y, z \neq 1$ satisfying the condition $x y z=1$." "['We start with the substitution\n\n$$\n\\frac{x}{x-1}=a, \\quad \\frac{y}{y-1}=b, \\quad \\frac{z}{z-1}=c, \\quad \\text { i.e., } \\quad x=\\frac{a}{a-1}, \\quad y=\\frac{b}{b-1}, \\quad z=\\frac{c}{c-1}\n$$\n\nThe inequality to be proved reads $a^{2}+b^{2}+c^{2} \\geq 1$. The new variables are subject to the constraints $a, b, c \\neq 1$ and the following one coming from the condition $x y z=1$,\n\n$$\n(a-1)(b-1)(c-1)=a b c .\n$$\n\nThis is successively equivalent to\n\n$$\n\\begin{aligned}\na+b+c-1 & =a b+b c+c a, \\\\\n2(a+b+c-1) & =(a+b+c)^{2}-\\left(a^{2}+b^{2}+c^{2}\\right), \\\\\na^{2}+b^{2}+c^{2}-2 & =(a+b+c)^{2}-2(a+b+c), \\\\\na^{2}+b^{2}+c^{2}-1 & =(a+b+c-1)^{2} .\n\\end{aligned}\n$$\n\nThus indeed $a^{2}+b^{2}+c^{2} \\geq 1$, as desired.' 'Without changing variables, just setting $z=1 / x y$ and clearing fractions, the proposed inequality takes the form\n\n$$\n(x y-1)^{2}\\left(x^{2}(y-1)^{2}+y^{2}(x-1)^{2}\\right)+(x-1)^{2}(y-1)^{2} \\geq(x-1)^{2}(y-1)^{2}(x y-1)^{2} .\n$$\n\nWith the notation $p=x+y, q=x y$ this becomes, after lengthy routine manipulation and a lot of cancellation\n\n$$\nq^{4}-6 q^{3}+2 p q^{2}+9 q^{2}-6 p q+p^{2} \\geq 0\n$$\n\nIt is not hard to notice that the expression on the left is just $\\left(q^{2}-3 q+p\\right)^{2}$, hence nonnegative.\n\n(Without introducing $p$ and $q$, one is of course led with some more work to the same expression, just written in terms of $x$ and $y$; but then it is not that easy to see that it is a square.)']" [] Text-only Competition True Theorem proof Algebra Math English 337 " Show that there are infinitely many triples of rational numbers $x, y, z$ for which this inequality turns into equality." "['From the equation $a^{2}+b^{2}+c^{2}-1=(a+b+c-1)^{2}$ we see that the proposed inequality becomes an equality if and only if both sums $a^{2}+b^{2}+c^{2}$ and $a+b+c$ have value 1 . The first of them is equal to $(a+b+c)^{2}-2(a b+b c+c a)$. So the instances of equality are described by the system of two equations\n\n$$\na+b+c=1, \\quad a b+b c+c a=0\n$$\n\nplus the constraint $a, b, c \\neq 1$. Elimination of $c$ leads to $a^{2}+a b+b^{2}=a+b$, which we regard as a quadratic equation in $b$,\n\n$$\nb^{2}+(a-1) b+a(a-1)=0\n$$\n\nwith discriminant\n\n$$\n\\Delta=(a-1)^{2}-4 a(a-1)=(1-a)(1+3 a)\n$$\n\nWe are looking for rational triples $(a, b, c)$; it will suffice to have $a$ rational such that $1-a$ and $1+3 a$ are both squares of rational numbers (then $\\Delta$ will be so too). Set $a=k / m$. We want $m-k$ and $m+3 k$ to be squares of integers. This is achieved for instance by taking $m=k^{2}-k+1$ (clearly nonzero); then $m-k=(k-1)^{2}, m+3 k=(k+1)^{2}$. Note that distinct integers $k$ yield distinct values of $a=k / m$.\n\nAnd thus, if $k$ is any integer and $m=k^{2}-k+1, a=k / m$ then $\\Delta=\\left(k^{2}-1\\right)^{2} / m^{2}$ and the quadratic equation has rational roots $b=\\left(m-k \\pm k^{2} \\mp 1\\right) /(2 m)$. Choose e.g. the larger root,\n\n$$\nb=\\frac{m-k+k^{2}-1}{2 m}=\\frac{m+(m-2)}{2 m}=\\frac{m-1}{m} .\n$$\n\n\n\nComputing $c$ from $a+b+c=1$ then gives $c=(1-k) / m$. The condition $a, b, c \\neq 1$ eliminates only $k=0$ and $k=1$. Thus, as $k$ varies over integers greater than 1 , we obtain an infinite family of rational triples $(a, b, c)$-and coming back to the original variables $(x=a /(a-1)$ etc. $)$-an infinite family of rational triples $(x, y, z)$ with the needed property. (A short calculation shows that the resulting triples are $x=-k /(k-1)^{2}, y=k-k^{2}, z=(k-1) / k^{2}$; but the proof was complete without listing them.)' 'To have equality, one needs $q^{2}-3 q+p=0$. Note that $x$ and $y$ are the roots of the quadratic trinomial (in a formal variable $t$ ): $t^{2}-p t+q$. When $q^{2}-3 q+p=0$, the discriminant equals\n\n$$\n\\delta=p^{2}-4 q=\\left(3 q-q^{2}\\right)^{2}-4 q=q(q-1)^{2}(q-4)\n$$\n\nNow it suffices to have both $q$ and $q-4$ squares of rational numbers (then $p=3 q-q^{2}$ and $\\sqrt{\\delta}$ are also rational, and so are the roots of the trinomial). On setting $q=(n / m)^{2}=4+(l / m)^{2}$ the requirement becomes $4 m^{2}+l^{2}=n^{2}$ (with $l, m, n$ being integers). This is just the Pythagorean equation, known to have infinitely many integer solutions.']" [] Text-only Competition True Theorem proof Algebra Math English 338 "For an integer $m$, denote by $t(m)$ the unique number in $\{1,2,3\}$ such that $m+t(m)$ is a multiple of 3 . A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies $f(-1)=0, f(0)=1, f(1)=-1$ and $$ f\left(2^{n}+m\right)=f\left(2^{n}-t(m)\right)-f(m) \text { for all integers } m, n \geq 0 \text { with } 2^{n}>m \text {. } $$ Prove that $f(3 p) \geq 0$ holds for all integers $p \geq 0$." ['The given conditions determine $f$ uniquely on the positive integers. The signs of $f(1), f(2), \\ldots$ seem to change quite erratically. However values of the form $f\\left(2^{n}-t(m)\\right)$ are sufficient to compute directly any functional value. Indeed, let $n>0$ have base 2 representation $n=2^{a_{0}}+2^{a_{1}}+\\cdots+2^{a_{k}}, a_{0}>a_{1}>\\cdots>a_{k} \\geq 0$, and let $n_{j}=2^{a_{j}}+2^{a_{j-1}}+\\cdots+2^{a_{k}}, j=0, \\ldots, k$. Repeated applications of the recurrence show that $f(n)$ is an alternating sum of the quantities $f\\left(2^{a_{j}}-t\\left(n_{j+1}\\right)\\right)$ plus $(-1)^{k+1}$. (The exact formula is not needed for our proof.)\n\nSo we focus attention on the values $f\\left(2^{n}-1\\right), f\\left(2^{n}-2\\right)$ and $f\\left(2^{n}-3\\right)$. Six cases arise; more specifically,\n\n$t\\left(2^{2 k}-3\\right)=2, t\\left(2^{2 k}-2\\right)=1, t\\left(2^{2 k}-1\\right)=3, t\\left(2^{2 k+1}-3\\right)=1, t\\left(2^{2 k+1}-2\\right)=3, t\\left(2^{2 k+1}-1\\right)=2$.\n\nClaim. For all integers $k \\geq 0$ the following equalities hold:\n\n$$\n\\begin{array}{lll}\nf\\left(2^{2 k+1}-3\\right)=0, & f\\left(2^{2 k+1}-2\\right)=3^{k}, & f\\left(2^{2 k+1}-1\\right)=-3^{k}, \\\\\nf\\left(2^{2 k+2}-3\\right)=-3^{k}, & f\\left(2^{2 k+2}-2\\right)=-3^{k}, & f\\left(2^{2 k+2}-1\\right)=2 \\cdot 3^{k} .\n\\end{array}\n$$\n\nProof. By induction on $k$. The base $k=0$ comes down to checking that $f(2)=-1$ and $f(3)=2$; the given values $f(-1)=0, f(0)=1, f(1)=-1$ are also needed. Suppose the claim holds for $k-1$. For $f\\left(2^{2 k+1}-t(m)\\right)$, the recurrence formula and the induction hypothesis yield\n\n$$\n\\begin{aligned}\n& f\\left(2^{2 k+1}-3\\right)=f\\left(2^{2 k}+\\left(2^{2 k}-3\\right)\\right)=f\\left(2^{2 k}-2\\right)-f\\left(2^{2 k}-3\\right)=-3^{k-1}+3^{k-1}=0, \\\\\n& f\\left(2^{2 k+1}-2\\right)=f\\left(2^{2 k}+\\left(2^{2 k}-2\\right)\\right)=f\\left(2^{2 k}-1\\right)-f\\left(2^{2 k}-2\\right)=2 \\cdot 3^{k-1}+3^{k-1}=3^{k}, \\\\\n& f\\left(2^{2 k+1}-1\\right)=f\\left(2^{2 k}+\\left(2^{2 k}-1\\right)\\right)=f\\left(2^{2 k}-3\\right)-f\\left(2^{2 k}-1\\right)=-3^{k-1}-2 \\cdot 3^{k-1}=-3^{k} .\n\\end{aligned}\n$$\n\nFor $f\\left(2^{2 k+2}-t(m)\\right)$ we use the three equalities just established:\n\n$$\n\\begin{aligned}\n& f\\left(2^{2 k+2}-3\\right)=f\\left(2^{2 k+1}+\\left(2^{2 k+1}-3\\right)\\right)=f\\left(2^{2 k+1}-1\\right)-f\\left(2^{2 k+1}-3\\right)=-3^{k}-0=-3^{k}, \\\\\n& f\\left(2^{2 k+2}-2\\right)=f\\left(2^{2 k+1}+\\left(2^{2 k+1}-2\\right)\\right)=f\\left(2^{2 k+1}-3\\right)-f\\left(2^{2 k}-2\\right)=0-3^{k}=-3^{k}, \\\\\n& f\\left(2^{2 k+2}-1\\right)=f\\left(2^{2 k+1}+\\left(2^{2 k+1}-1\\right)\\right)=f\\left(2^{2 k+1}-2\\right)-f\\left(2^{2 k+1}-1\\right)=3^{k}+3^{k}=2 \\cdot 3^{k} .\n\\end{aligned}\n$$\n\nThe claim follows.\n\nA closer look at the six cases shows that $f\\left(2^{n}-t(m)\\right) \\geq 3^{(n-1) / 2}$ if $2^{n}-t(m)$ is divisible by 3 , and $f\\left(2^{n}-t(m)\\right) \\leq 0$ otherwise. On the other hand, note that $2^{n}-t(m)$ is divisible by 3 if and only if $2^{n}+m$ is. Therefore, for all nonnegative integers $m$ and $n$,\n\n(i) $f\\left(2^{n}-t(m)\\right) \\geq 3^{(n-1) / 2}$ if $2^{n}+m$ is divisible by 3 ;\n\n(ii) $f\\left(2^{n}-t(m)\\right) \\leq 0$ if $2^{n}+m$ is not divisible by 3 .\n\nOne more (direct) consequence of the claim is that $\\left|f\\left(2^{n}-t(m)\\right)\\right| \\leq \\frac{2}{3} \\cdot 3^{n / 2}$ for all $m, n \\geq 0$.\n\nThe last inequality enables us to find an upper bound for $|f(m)|$ for $m$ less than a given power of 2 . We prove by induction on $n$ that $|f(m)| \\leq 3^{n / 2}$ holds true for all integers $m, n \\geq 0$ with $2^{n}>m$.\n\n\n\nThe base $n=0$ is clear as $f(0)=1$. For the inductive step from $n$ to $n+1$, let $m$ and $n$ satisfy $2^{n+1}>m$. If $m<2^{n}$, we are done by the inductive hypothesis. If $m \\geq 2^{n}$ then $m=2^{n}+k$ where $2^{n}>k \\geq 0$. Now, by $\\left|f\\left(2^{n}-t(k)\\right)\\right| \\leq \\frac{2}{3} \\cdot 3^{n / 2}$ and the inductive assumption,\n\n$$\n|f(m)|=\\left|f\\left(2^{n}-t(k)\\right)-f(k)\\right| \\leq\\left|f\\left(2^{n}-t(k)\\right)\\right|+|f(k)| \\leq \\frac{2}{3} \\cdot 3^{n / 2}+3^{n / 2}<3^{(n+1) / 2} .\n$$\n\nThe induction is complete.\n\nWe proceed to prove that $f(3 p) \\geq 0$ for all integers $p \\geq 0$. Since $3 p$ is not a power of 2 , its binary expansion contains at least two summands. Hence one can write $3 p=2^{a}+2^{b}+c$ where $a>b$ and $2^{b}>c \\geq 0$. Applying the recurrence formula twice yields\n\n$$\nf(3 p)=f\\left(2^{a}+2^{b}+c\\right)=f\\left(2^{a}-t\\left(2^{b}+c\\right)\\right)-f\\left(2^{b}-t(c)\\right)+f(c) .\n$$\n\nSince $2^{a}+2^{b}+c$ is divisible by 3 , we have $f\\left(2^{a}-t\\left(2^{b}+c\\right)\\right) \\geq 3^{(a-1) / 2}$ by (i). Since $2^{b}+c$ is not divisible by 3 , we have $f\\left(2^{b}-t(c)\\right) \\leq 0$ by (ii). Finally $|f(c)| \\leq 3^{b / 2}$ as $2^{b}>c \\geq 0$, so that $f(c) \\geq-3^{b / 2}$. Therefore $f(3 p) \\geq 3^{(a-1) / 2}-3^{b / 2}$ which is nonnegative because $a>b$.'] [] Text-only Competition True Theorem proof Algebra Math English 339 "Let $a, b, c, d$ be positive real numbers such that $$ a b c d=1 \quad \text { and } \quad a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} $$ Prove that $$ a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} $$" ['We show that if $a b c d=1$, the sum $a+b+c+d$ cannot exceed a certain weighted mean of the expressions $\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}$ and $\\frac{b}{a}+\\frac{c}{b}+\\frac{d}{c}+\\frac{a}{d}$.\n\nBy applying the AM-GM inequality to the numbers $\\frac{a}{b}, \\frac{a}{b}, \\frac{b}{c}$ and $\\frac{a}{d}$, we obtain\n\n$$\na=\\sqrt[4]{\\frac{a^{4}}{a b c d}}=\\sqrt[4]{\\frac{a}{b} \\cdot \\frac{a}{b} \\cdot \\frac{b}{c} \\cdot \\frac{a}{d}} \\leq \\frac{1}{4}\\left(\\frac{a}{b}+\\frac{a}{b}+\\frac{b}{c}+\\frac{a}{d}\\right)\n$$\n\nAnalogously,\n\n$$\nb \\leq \\frac{1}{4}\\left(\\frac{b}{c}+\\frac{b}{c}+\\frac{c}{d}+\\frac{b}{a}\\right), \\quad c \\leq \\frac{1}{4}\\left(\\frac{c}{d}+\\frac{c}{d}+\\frac{d}{a}+\\frac{c}{b}\\right) \\quad \\text { and } \\quad d \\leq \\frac{1}{4}\\left(\\frac{d}{a}+\\frac{d}{a}+\\frac{a}{b}+\\frac{d}{c}\\right) .\n$$\n\nSumming up these estimates yields\n\n$$\na+b+c+d \\leq \\frac{3}{4}\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}\\right)+\\frac{1}{4}\\left(\\frac{b}{a}+\\frac{c}{b}+\\frac{d}{c}+\\frac{a}{d}\\right)\n$$\n\nIn particular, if $a+b+c+d>\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}$ then $a+b+c+d<\\frac{b}{a}+\\frac{c}{b}+\\frac{d}{c}+\\frac{a}{d}$.'] [] Text-only Competition True Theorem proof Algebra Math English 340 "Let $f: \mathbb{R} \rightarrow \mathbb{N}$ be a function which satisfies $$ f\left(x+\frac{1}{f(y)}\right)=f\left(y+\frac{1}{f(x)}\right) \quad \text { for all } x, y \in \mathbb{R}\tag{1} $$ Prove that there is a positive integer which is not a value of $f$." ['Suppose that the statement is false and $f(\\mathbb{R})=\\mathbb{N}$. We prove several properties of the function $f$ in order to reach a contradiction.\n\nTo start with, observe that one can assume $f(0)=1$. Indeed, let $a \\in \\mathbb{R}$ be such that $f(a)=1$, and consider the function $g(x)=f(x+a)$. By substituting $x+a$ and $y+a$ for $x$ and $y$ in (1), we have\n\n$$\ng\\left(x+\\frac{1}{g(y)}\\right)=f\\left(x+a+\\frac{1}{f(y+a)}\\right)=f\\left(y+a+\\frac{1}{f(x+a)}\\right)=g\\left(y+\\frac{1}{g(x)}\\right) .\n$$\n\nSo $g$ satisfies the functional equation (1), with the additional property $g(0)=1$. Also, $g$ and $f$ have the same set of values: $g(\\mathbb{R})=f(\\mathbb{R})=\\mathbb{N}$. Henceforth we assume $f(0)=1$.\n\nClaim 1. For an arbitrary fixed $c \\in \\mathbb{R}$ we have $\\left\\{f\\left(c+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\mathbb{N}$.\n\nProof. Equation (1) and $f(\\mathbb{R})=\\mathbb{N}$ imply\n\n$f(\\mathbb{R})=\\left\\{f\\left(x+\\frac{1}{f(c)}\\right): x \\in \\mathbb{R}\\right\\}=\\left\\{f\\left(c+\\frac{1}{f(x)}\\right): x \\in \\mathbb{R}\\right\\} \\subset\\left\\{f\\left(c+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\} \\subset f(\\mathbb{R})$.\n\nThe claim follows.\n\nWe will use Claim 1 in the special cases $c=0$ and $c=1 / 3$ :\n\n$$\n\\left\\{f\\left(\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\left\\{f\\left(\\frac{1}{3}+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\mathbb{N}\\tag{2}\n$$\n\nClaim 2. If $f(u)=f(v)$ for some $u, v \\in \\mathbb{R}$ then $f(u+q)=f(v+q)$ for all nonnegative rational $q$. Furthermore, if $f(q)=1$ for some nonnegative rational $q$ then $f(k q)=1$ for all $k \\in \\mathbb{N}$.\n\nProof. For all $x \\in \\mathbb{R}$ we have by (1)\n\n$$\nf\\left(u+\\frac{1}{f(x)}\\right)=f\\left(x+\\frac{1}{f(u)}\\right)=f\\left(x+\\frac{1}{f(v)}\\right)=f\\left(v+\\frac{1}{f(x)}\\right) .\n$$\n\nSince $f(x)$ attains all positive integer values, this yields $f(u+1 / n)=f(v+1 / n)$ for all $n \\in \\mathbb{N}$. Let $q=k / n$ be a positive rational number. Then $k$ repetitions of the last step yield\n\n$$\nf(u+q)=f\\left(u+\\frac{k}{n}\\right)=f\\left(v+\\frac{k}{n}\\right)=f(v+q)\n$$\n\nNow let $f(q)=1$ for some nonnegative rational $q$, and let $k \\in \\mathbb{N}$. As $f(0)=1$, the previous conclusion yields successively $f(q)=f(2 q), f(2 q)=f(3 q), \\ldots, f((k-1) q)=f(k q)$, as needed.\n\nClaim 3. The equality $f(q)=f(q+1)$ holds for all nonnegative rational $q$.\n\nProof. Let $m$ be a positive integer such that $f(1 / m)=1$. Such an $m$ exists by (2). Applying the second statement of Claim 2 with $q=1 / m$ and $k=m$ yields $f(1)=1$.\n\nGiven that $f(0)=f(1)=1$, the first statement of Claim 2 implies $f(q)=f(q+1)$ for all nonnegative rational $q$.\n\n\n\nClaim 4. The equality $f\\left(\\frac{1}{n}\\right)=n$ holds for every $n \\in \\mathbb{N}$.\n\nProof. For a nonnegative rational $q$ we set $x=q, y=0$ in (1) and use Claim 3 to obtain\n\n$$\nf\\left(\\frac{1}{f(q)}\\right)=f\\left(q+\\frac{1}{f(0)}\\right)=f(q+1)=f(q)\n$$\n\nBy (2), for each $n \\in \\mathbb{N}$ there exists a $k \\in \\mathbb{N}$ such that $f(1 / k)=n$. Applying the last equation with $q=1 / k$, we have\n\n$$\nn=f\\left(\\frac{1}{k}\\right)=f\\left(\\frac{1}{f(1 / k)}\\right)=f\\left(\\frac{1}{n}\\right)\n$$\n\nNow we are ready to obtain a contradiction. Let $n \\in \\mathbb{N}$ be such that $f(1 / 3+1 / n)=1$. Such an $n$ exists by (2). Let $1 / 3+1 / n=s / t$, where $s, t \\in \\mathbb{N}$ are coprime. Observe that $t>1$ as $1 / 3+1 / n$ is not an integer. Choose $k, l \\in \\mathbb{N}$ so that that $k s-l t=1$.\n\nBecause $f(0)=f(s / t)=1$, Claim 2 implies $f(k s / t)=1$. Now $f(k s / t)=f(1 / t+l)$; on the other hand $f(1 / t+l)=f(1 / t)$ by $l$ successive applications of Claim 3 . Finally, $f(1 / t)=t$ by Claim 4, leading to the impossible $t=1$. The solution is complete.'] [] Text-only Competition True Theorem proof Algebra Math English 341 "Prove that for any four positive real numbers $a, b, c, d$ the inequality $$ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geq 0 $$ holds. Determine all cases of equality." "['Denote the four terms by\n\n$$\nA=\\frac{(a-b)(a-c)}{a+b+c}, \\quad B=\\frac{(b-c)(b-d)}{b+c+d}, \\quad C=\\frac{(c-d)(c-a)}{c+d+a}, \\quad D=\\frac{(d-a)(d-b)}{d+a+b}\n$$\n\nThe expression $2 A$ splits into two summands as follows,\n\n$$\n2 A=A^{\\prime}+A^{\\prime \\prime} \\quad \\text { where } \\quad A^{\\prime}=\\frac{(a-c)^{2}}{a+b+c}, \\quad A^{\\prime \\prime}=\\frac{(a-c)(a-2 b+c)}{a+b+c}\n$$\n\nthis is easily verified. We analogously represent $2 B=B^{\\prime}+B^{\\prime \\prime}, 2 C=C^{\\prime}+C^{\\prime \\prime}, 2 B=D^{\\prime}+D^{\\prime \\prime}$ and examine each of the sums $A^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime}$ and $A^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}$ separately.\n\nWrite $s=a+b+c+d$; the denominators become $s-d, s-a, s-b, s-c$. By the CauchySchwarz inequality,\n\n$$\n\\begin{aligned}\n& \\left(\\frac{|a-c|}{\\sqrt{s-d}} \\cdot \\sqrt{s-d}+\\frac{|b-d|}{\\sqrt{s-a}} \\cdot \\sqrt{s-a}+\\frac{|c-a|}{\\sqrt{s-b}} \\cdot \\sqrt{s-b}+\\frac{|d-b|}{\\sqrt{s-c}} \\cdot \\sqrt{s-c}\\right)^{2} \\\\\n& \\quad \\leq\\left(\\frac{(a-c)^{2}}{s-d}+\\frac{(b-d)^{2}}{s-a}+\\frac{(c-a)^{2}}{s-b}+\\frac{(d-b)^{2}}{s-c}\\right)(4 s-s)=3 s\\left(A^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime}\\right) .\n\\end{aligned}\n$$\n\nHence\n\n$$\nA^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime} \\geq \\frac{(2|a-c|+2|b-d|)^{2}}{3 s} \\geq \\frac{16 \\cdot|a-c| \\cdot|b-d|}{3 s}\\tag{1}\n$$\n\nNext we estimate the absolute value of the other sum. We couple $A^{\\prime \\prime}$ with $C^{\\prime \\prime}$ to obtain\n\n$$\n\\begin{aligned}\nA^{\\prime \\prime}+C^{\\prime \\prime} & =\\frac{(a-c)(a+c-2 b)}{s-d}+\\frac{(c-a)(c+a-2 d)}{s-b} \\\\\n& =\\frac{(a-c)(a+c-2 b)(s-b)+(c-a)(c+a-2 d)(s-d)}{(s-d)(s-b)} \\\\\n& =\\frac{(a-c)(-2 b(s-b)-b(a+c)+2 d(s-d)+d(a+c))}{s(a+c)+b d} \\\\\n& =\\frac{3(a-c)(d-b)(a+c)}{M}, \\quad \\text { with } \\quad M=s(a+c)+b d .\n\\end{aligned}\n$$\n\nHence by cyclic shift\n\n$$\nB^{\\prime \\prime}+D^{\\prime \\prime}=\\frac{3(b-d)(a-c)(b+d)}{N}, \\quad \\text { with } \\quad N=s(b+d)+c a\n$$\n\nThus\n\n$$\nA^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}=3(a-c)(b-d)\\left(\\frac{b+d}{N}-\\frac{a+c}{M}\\right)=\\frac{3(a-c)(b-d) W}{M N}\\tag{2}\n$$\n\nwhere\n\n$$\nW=(b+d) M-(a+c) N=b d(b+d)-a c(a+c)\\tag{3}\n$$\n\n\n\nNote that\n\n$$\nM N>(a c(a+c)+b d(b+d)) s \\geq|W| \\cdot s\\tag{4}\n$$\n\nNow (2) and (4) yield\n\n$$\n\\left|A^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}\\right| \\leq \\frac{3 \\cdot|a-c| \\cdot|b-d|}{s}\\tag{5}\n$$\n\nCombined with (1) this results in\n\n$$\n\\begin{aligned}\n& 2(A+B+C+D)=\\left(A^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime}\\right)+\\left(A^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}\\right) \\\\\n& \\quad \\geq \\frac{16 \\cdot|a-c| \\cdot|b-d|}{3 s}-\\frac{3 \\cdot|a-c| \\cdot|b-d|}{s}=\\frac{7 \\cdot|a-c| \\cdot|b-d|}{3(a+b+c+d)} \\geq 0\n\\end{aligned}\n$$\n\nThis is the required inequality. From the last line we see that equality can be achieved only if either $a=c$ or $b=d$. Since we also need equality in (1), this implies that actually $a=c$ and $b=d$ must hold simultaneously, which is obviously also a sufficient condition.' 'We keep the notations $A, B, C, D, s$, and also $M, N, W$ from the preceding solution; the definitions of $M, N, W$ and relations (3), (4) in that solution did not depend on the foregoing considerations. Starting from\n\n$$\n2 A=\\frac{(a-c)^{2}+3(a+c)(a-c)}{a+b+c}-2 a+2 c\n$$\n\nwe get\n\n$$\n\\begin{aligned}\n2(A & +C)=(a-c)^{2}\\left(\\frac{1}{s-d}+\\frac{1}{s-b}\\right)+3(a+c)(a-c)\\left(\\frac{1}{s-d}-\\frac{1}{s-b}\\right) \\\\\n& =(a-c)^{2} \\frac{2 s-b-d}{M}+3(a+c)(a-c) \\cdot \\frac{d-b}{M}=\\frac{p(a-c)^{2}-3(a+c)(a-c)(b-d)}{M}\n\\end{aligned}\n$$\n\nwhere $p=2 s-b-d=s+a+c$. Similarly, writing $q=s+b+d$ we have\n\n$$\n2(B+D)=\\frac{q(b-d)^{2}-3(b+d)(b-d)(c-a)}{N}\n$$\n\nspecific grouping of terms in the numerators has its aim. Note that $p q>2 s^{2}$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$,\n\n$$\n2(A+B+C+D)=\\frac{p(a-c)^{2}}{M}+\\frac{3(a-c)(b-d) W}{M N}+\\frac{q(b-d)^{2}}{N}\n$$\n\nwith $W$ defined by (3).\n\nSubstitution $x=(a-c) / M, y=(b-d) / N$ brings the required inequality to the form\n\n$$\n2(A+B+C+D)=M p x^{2}+3 W x y+N q y^{2} \\geq 0\\tag{6}\n$$\n\nIt will be enough to verify that the discriminant $\\Delta=9 W^{2}-4 M N p q$ of the quadratic trinomial $M p t^{2}+3 W t+N q$ is negative; on setting $t=x / y$ one then gets (6). The first inequality in (4) together with $p q>2 s^{2}$ imply $4 M N p q>8 s^{3}(a c(a+c)+b d(b+d))$. Since\n\n$$\n(a+c) s^{3}>(a+c)^{4} \\geq 4 a c(a+c)^{2} \\quad \\text { and likewise } \\quad(b+d) s^{3}>4 b d(b+d)^{2}\n$$\n\nthe estimate continues as follows,\n\n$$\n4 M N p q>8\\left(4(a c)^{2}(a+c)^{2}+4(b d)^{2}(b+d)^{2}\\right)>32(b d(b+d)-a c(a+c))^{2}=32 W^{2} \\geq 9 W^{2}\n$$\n\nThus indeed $\\Delta<0$. The desired inequality (6) hence results. It becomes an equality if and only if $x=y=0$; equivalently, if and only if $a=c$ and simultaneously $b=d$.' '$$\n\\begin{gathered}\n(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\\\\n=((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)= \\\\\n=\\left(a^{2}+a d-b^{2}-b d\\right)\\left(a^{2}+a d-c^{2}-c d\\right)(b+c+d)= \\\\\n=\\left(a^{4}+2 a^{3} d-a^{2} b^{2}-a^{2} b d-a^{2} c^{2}-a^{2} c d+a^{2} d^{2}-a b^{2} d-a b d^{2}-a c^{2} d-a c d^{2}+b^{2} c^{2}+b^{2} c d+b c^{2} d+b c d^{2}\\right)(b+c+d)= \\\\\n=a^{4} b+a^{4} c+a^{4} d+\\left(b^{3} c^{2}+a^{2} d^{3}\\right)-a^{2} c^{3}+\\left(2 a^{3} d^{2}-b^{3} a^{2}+c^{3} b^{2}\\right)+ \\\\\n+\\left(b^{3} c d-c^{3} d a-d^{3} a b\\right)+\\left(2 a^{3} b d+c^{3} d b-d^{3} a c\\right)+\\left(2 a^{3} c d-b^{3} d a+d^{3} b c\\right) \\\\\n+\\left(-a^{2} b^{2} c+3 b^{2} c^{2} d-2 a c^{2} d^{2}\\right)+\\left(-2 a^{2} b^{2} d+2 b c^{2} d^{2}\\right)+\\left(-a^{2} b c^{2}-2 a^{2} c^{2} d-2 a b^{2} d^{2}+2 b^{2} c d^{2}\\right)+ \\\\\n+\\left(-2 a^{2} b c d-a b^{2} c d-a b c^{2} d-2 a b c d^{2}\\right)\n\\end{gathered}\n$$\n\nIntroducing the notation $S_{x y z w}=\\sum_{c y c} a^{x} b^{y} c^{z} d^{w}$, one can write\n\n$$\n\\begin{gathered}\n\\sum_{c y c}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\\\\n=S_{4100}+S_{4010}+S_{4001}+2 S_{3200}-S_{3020}+2 S_{3002}-S_{3110}+2 S_{3101}+2 S_{3011}-3 S_{2120}-6 S_{2111}= \\\\\n+\\left(S_{4100}+S_{4001}+\\frac{1}{2} S_{3110}+\\frac{1}{2} S_{3011}-3 S_{2120}\\right)+ \\\\\n+\\left(S_{4010}-S_{3020}-\\frac{3}{2} S_{3110}+\\frac{3}{2} S_{3011}+\\frac{9}{16} S_{2210}+\\frac{9}{16} S_{2201}-\\frac{9}{8} S_{2111}\\right)+ \\\\\n+\\frac{9}{16}\\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\\right)+\\frac{23}{16}\\left(S_{3200}-2 S_{3101}+S_{3002}\\right)+\\frac{39}{8}\\left(S_{3101}-S_{2111}\\right),\n\\end{gathered}\n$$\n\nwhere the expressions\n\n$$\n\\begin{gathered}\nS_{4100}+S_{4001}+\\frac{1}{2} S_{3110}+\\frac{1}{2} S_{3011}-3 S_{2120}=\\sum_{c y c}\\left(a^{4} b+b c^{4}+\\frac{1}{2} a^{3} b c+\\frac{1}{2} a b c^{3}-3 a^{2} b c^{2}\\right), \\\\\nS_{4010}-S_{3020}-\\frac{3}{2} S_{3110}+\\frac{3}{2} S_{3011}+\\frac{9}{16} S_{2210}+\\frac{9}{16} S_{2201}-\\frac{9}{8} S_{2111}=\\sum_{c y c} a^{2} c\\left(a-c-\\frac{3}{4} b+\\frac{3}{4} d\\right)^{2}, \\\\\nS_{3200}-S_{2210}-S_{2201}+S_{3002}=\\sum_{c y c} b^{2}\\left(a^{3}-a^{2} c-a c^{2}+c^{3}\\right)=\\sum_{c y c} b^{2}(a+c)(a-c)^{2}, \\\\\nS_{3200}-2 S_{3101}+S_{3002}=\\sum_{c y c} a^{3}(b-d)^{2} \\quad \\text { and } \\quad S_{3101}-S_{2111}=\\frac{1}{3} \\sum_{c y c} b d\\left(2 a^{3}+c^{3}-3 a^{2} c\\right)\n\\end{gathered}\n$$\n\nare all nonnegative.']" [] Text-only Competition True Theorem proof Algebra Math English 342 "For every positive integer $n$ determine the number of permutations $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the set $\{1,2, \ldots, n\}$ with the following property: $$ 2\left(a_{1}+a_{2}+\cdots+a_{k}\right) \quad \text { is divisible by } k \text { for } k=1,2, \ldots, n \text {. } $$" ['For each $n$ let $F_{n}$ be the number of permutations of $\\{1,2, \\ldots, n\\}$ with the required property; call them nice. For $n=1,2,3$ every permutation is nice, so $F_{1}=1, F_{2}=2, F_{3}=6$.\n\nTake an $n>3$ and consider any nice permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of $\\{1,2, \\ldots, n\\}$. Then $n-1$ must be a divisor of the number\n\n$$\n\\begin{aligned}\n2\\left(a_{1}+a_{2}+\\cdots\\right. & \\left.+a_{n-1}\\right)=2\\left((1+2+\\cdots+n)-a_{n}\\right) \\\\\n& =n(n+1)-2 a_{n}=(n+2)(n-1)+\\left(2-2 a_{n}\\right) .\n\\end{aligned}\n$$\n\nSo $2 a_{n}-2$ must be divisible by $n-1$, hence equal to 0 or $n-1$ or $2 n-2$. This means that\n\n$$\na_{n}=1 \\quad \\text { or } \\quad a_{n}=\\frac{n+1}{2} \\quad \\text { or } \\quad a_{n}=n \\text {. }\n$$\n\nSuppose that $a_{n}=(n+1) / 2$. Since the permutation is nice, taking $k=n-2$ we get that $n-2$ has to be a divisor of\n\n$$\n\\begin{aligned}\n2\\left(a_{1}+a_{2}+\\cdots+a_{n-2}\\right) & =2\\left((1+2+\\cdots+n)-a_{n}-a_{n-1}\\right) \\\\\n& =n(n+1)-(n+1)-2 a_{n-1}=(n+2)(n-2)+\\left(3-2 a_{n-1}\\right) .\n\\end{aligned}\n$$\n\nSo $2 a_{n-1}-3$ should be divisible by $n-2$, hence equal to 0 or $n-2$ or $2 n-4$. Obviously 0 and $2 n-4$ are excluded because $2 a_{n-1}-3$ is odd. The remaining possibility $\\left(2 a_{n-1}-3=n-2\\right)$ leads to $a_{n-1}=(n+1) / 2=a_{n}$, which also cannot hold. This eliminates $(n+1) / 2$ as a possible value of $a_{n}$. Consequently $a_{n}=1$ or $a_{n}=n$.\n\nIf $a_{n}=n$ then $\\left(a_{1}, a_{2}, \\ldots, a_{n-1}\\right)$ is a nice permutation of $\\{1,2, \\ldots, n-1\\}$. There are $F_{n-1}$ such permutations. Attaching $n$ to any one of them at the end creates a nice permutation of $\\{1,2, \\ldots, n\\}$.\n\nIf $a_{n}=1$ then $\\left(a_{1}-1, a_{2}-1, \\ldots, a_{n-1}-1\\right)$ is a permutation of $\\{1,2, \\ldots, n-1\\}$. It is also nice because the number\n\n$$\n2\\left(\\left(a_{1}-1\\right)+\\cdots+\\left(a_{k}-1\\right)\\right)=2\\left(a_{1}+\\cdots+a_{k}\\right)-2 k\n$$\n\nis divisible by $k$, for any $k \\leq n-1$. And again, any one of the $F_{n-1}$ nice permutations $\\left(b_{1}, b_{2}, \\ldots, b_{n-1}\\right)$ of $\\{1,2, \\ldots, n-1\\}$ gives rise to a nice permutation of $\\{1,2, \\ldots, n\\}$ whose last term is 1 , namely $\\left(b_{1}+1, b_{2}+1, \\ldots, b_{n-1}+1,1\\right)$.\n\nThe bijective correspondences established in both cases show that there are $F_{n-1}$ nice permutations of $\\{1,2, \\ldots, n\\}$ with the last term 1 and also $F_{n-1}$ nice permutations of $\\{1,2, \\ldots, n\\}$ with the last term $n$. Hence follows the recurrence $F_{n}=2 F_{n-1}$. With the base value $F_{3}=6$ this gives the outcome formula $F_{n}=3 \\cdot 2^{n-2}$ for $n \\geq 3$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 343 "Let $S=\left\{x_{1}, x_{2}, \ldots, x_{k+\ell}\right\}$ be a $(k+\ell)$-element set of real numbers contained in the interval $[0,1] ; k$ and $\ell$ are positive integers. A $k$-element subset $A \subset S$ is called nice if $$ \left|\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{\ell} \sum_{x_{j} \in S \backslash A} x_{j}\right| \leq \frac{k+\ell}{2 k \ell} $$ Prove that the number of nice subsets is at least $\frac{2}{k+\ell}\left(\begin{array}{c}k+\ell \\ k\end{array}\right)$." ['For a $k$-element subset $A \\subset S$, let $f(A)=\\frac{1}{k} \\sum_{x_{i} \\in A} x_{i}-\\frac{1}{\\ell} \\sum_{x_{j} \\in S \\backslash A} x_{j}$. Denote $\\frac{k+\\ell}{2 k \\ell}=d$. By definition a subset $A$ is nice if $|f(A)| \\leq d$.\n\nTo each permutation $\\left(y_{1}, y_{2}, \\ldots, y_{k+\\ell}\\right)$ of the set $S=\\left\\{x_{1}, x_{2}, \\ldots, x_{k+\\ell}\\right\\}$ we assign $k+\\ell$ subsets of $S$ with $k$ elements each, namely $A_{i}=\\left\\{y_{i}, y_{i+1}, \\ldots, y_{i+k-1}\\right\\}, i=1,2, \\ldots, k+\\ell$. Indices are taken modulo $k+\\ell$ here and henceforth. In other words, if $y_{1}, y_{2}, \\ldots, y_{k+\\ell}$ are arranged around a circle in this order, the sets in question are all possible blocks of $k$ consecutive elements.\n\nClaim. At least two nice sets are assigned to every permutation of $S$.\n\nProof. Adjacent sets $A_{i}$ and $A_{i+1}$ differ only by the elements $y_{i}$ and $y_{i+k}, i=1, \\ldots, k+\\ell$. By the definition of $f$, and because $y_{i}, y_{i+k} \\in[0,1]$,\n\n$$\n\\left|f\\left(A_{i+1}\\right)-f\\left(A_{i}\\right)\\right|=\\left|\\left(\\frac{1}{k}+\\frac{1}{\\ell}\\right)\\left(y_{i+k}-y_{i}\\right)\\right| \\leq \\frac{1}{k}+\\frac{1}{\\ell}=2 d\n$$\n\nEach element $y_{i} \\in S$ belongs to exactly $k$ of the sets $A_{1}, \\ldots, A_{k+\\ell}$. Hence in $k$ of the expressions $f\\left(A_{1}\\right), \\ldots, f\\left(A_{k+\\ell}\\right)$ the coefficient of $y_{i}$ is $1 / k$; in the remaining $\\ell$ expressions, its coefficient is $-1 / \\ell$. So the contribution of $y_{i}$ to the sum of all $f\\left(A_{i}\\right)$ equals $k \\cdot 1 / k-\\ell \\cdot 1 / \\ell=0$. Since this holds for all $i$, it follows that $f\\left(A_{1}\\right)+\\cdots+f\\left(A_{k+\\ell}\\right)=0$.\n\nIf $f\\left(A_{p}\\right)=\\min f\\left(A_{i}\\right), f\\left(A_{q}\\right)=\\max f\\left(A_{i}\\right)$, we obtain in particular $f\\left(A_{p}\\right) \\leq 0, f\\left(A_{q}\\right) \\geq 0$. Let $pq$ is analogous; and the claim is true for $p=q$ as $f\\left(A_{i}\\right)=0$ for all $i$ ).\n\nWe are ready to prove that at least two of the sets $A_{1}, \\ldots, A_{k+\\ell}$ are nice. The interval $[-d, d]$ has length $2 d$, and we saw that adjacent numbers in the circular arrangement $f\\left(A_{1}\\right), \\ldots, f\\left(A_{k+\\ell}\\right)$ differ by at most $2 d$. Suppose that $f\\left(A_{p}\\right)<-d$ and $f\\left(A_{q}\\right)>d$. Then one of the numbers $f\\left(A_{p+1}\\right), \\ldots, f\\left(A_{q-1}\\right)$ lies in $[-d, d]$, and also one of the numbers $f\\left(A_{q+1}\\right), \\ldots, f\\left(A_{p-1}\\right)$ lies there. Consequently, one of the sets $A_{p+1}, \\ldots, A_{q-1}$ is nice, as well as one of the sets $A_{q+1}, \\ldots, A_{p-1}$. If $-d \\leq f\\left(A_{p}\\right)$ and $f\\left(A_{q}\\right) \\leq d$ then $A_{p}$ and $A_{q}$ are nice.\n\nLet now $f\\left(A_{p}\\right)<-d$ and $f\\left(A_{q}\\right) \\leq d$. Then $f\\left(A_{p}\\right)+f\\left(A_{q}\\right)<0$, and since $\\sum f\\left(A_{i}\\right)=0$, there is an $r \\neq q$ such that $f\\left(A_{r}\\right)>0$. We have $02\\left(u_{m-1}-u_{1}\\right)$ for all $m \\geq 3$.\n\nIndeed, assume that $u_{m}-u_{1} \\leq 2\\left(u_{m-1}-u_{1}\\right)$ holds for some $m \\geq 3$. This inequality can be written as $2\\left(u_{m}-u_{m-1}\\right) \\leq u_{m}-u_{1}$. Take the unique $k$ such that $2^{k} \\leq u_{m}-u_{1}<2^{k+1}$. Then $2\\left(u_{m}-u_{m-1}\\right) \\leq u_{m}-u_{1}<2^{k+1}$ yields $u_{m}-u_{m-1}<2^{k}$. However the elements $z=u_{m}, x=u_{1}$, $y=u_{m-1}$ of $S_{a}$ then satisfy $z-y<2^{k}$ and $z-x \\geq 2^{k}$, so that $z=u_{m}$ is $k$-good to $S_{a}$.\n\nThus each term of the sequence $u_{2}-u_{1}, u_{3}-u_{1}, \\ldots, u_{p}-u_{1}$ is more than twice the previous one. Hence $u_{p}-u_{1}>2^{p-1}\\left(u_{2}-u_{1}\\right) \\geq 2^{p-1}$. But $u_{p} \\in\\left\\{1,2,3, \\ldots, 2^{n+1}\\right\\}$, so that $u_{p} \\leq 2^{n+1}$. This yields $p-1 \\leq n$, i. e. $p \\leq n+1$.\n\nIn other words, each set $S_{a}$ contains at most $n+1$ elements that are not good to it.\n\nTo summarize the conclusions, mark with red all elements in the sets $S_{a}$ that are good to the respective set, and with blue the ones that are not good. Then the total number of red elements, counting multiplicities, is at most $n \\cdot 2^{n+1}$ (each $z \\in A$ can be marked red in at most $n$ sets). The total number of blue elements is at most $(n+1) 2^{n}$ (each set $S_{a}$ contains at most $n+1$ blue elements). Therefore the sum of cardinalities of $S_{1}, S_{2}, \\ldots, S_{2^{n}}$ does not exceed $(3 n+1) 2^{n}$. By averaging, the smallest set has at most $3 n+1$ elements.' 'We show that one of the sets $S_{a}$ has at most $2 n+1$ elements. In the sequel $|\\cdot|$ denotes the cardinality of a (finite) set.\n\nClaim. For $n \\geq 2$, suppose that $k$ subsets $S_{1}, \\ldots, S_{k}$ of $\\left\\{1,2, \\ldots, 2^{n}\\right\\}$ (not necessarily different) satisfy the condition of the problem. Then\n\n$$\n\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}\n$$\n\nProof. Observe that if the sets $S_{i}(1 \\leq i \\leq k)$ satisfy the condition then so do their arbitrary subsets $T_{i}(1 \\leq i \\leq k)$. The condition also holds for the sets $t+S_{i}=\\left\\{t+x \\mid x \\in S_{i}\\right\\}$ where $t$ is arbitrary.\n\nNote also that a set may occur more than once among $S_{1}, \\ldots, S_{k}$ only if its cardinality is less than 3, in which case its contribution to the sum $\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-n\\right.$ ) is nonpositive (as $n \\geq 2$ ).\n\nThe proof is by induction on $n$. In the base case $n=2$ we have subsets $S_{i}$ of $\\{1,2,3,4\\}$. Only the ones of cardinality 3 and 4 need to be considered by the remark above; each one of\n\n\n\nthem occurs at most once among $S_{1}, \\ldots, S_{k}$. If $S_{i}=\\{1,2,3,4\\}$ for some $i$ then no $S_{j}$ is a 3 -element subset in view of the condition, hence $\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-2\\right) \\leq 2$. By the condition again, it is impossible that $S_{i}=\\{1,3,4\\}$ and $S_{j}=\\{2,3,4\\}$ for some $i, j$. So if $\\left|S_{i}\\right| \\leq 3$ for all $i$ then at most 3 summands $\\left|S_{i}\\right|-2$ are positive, corresponding to 3 -element subsets. This implies $\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-2\\right) \\leq 3$, therefore the conclusion is true for $n=2$.\n\nSuppose that the claim holds for some $n \\geq 2$, and let the sets $S_{1}, \\ldots, S_{k} \\subseteq\\left\\{1,2, \\ldots, 2^{n+1}\\right\\}$ satisfy the given property. Denote $U_{i}=S_{i} \\cap\\left\\{1,2, \\ldots, 2^{n}\\right\\}, V_{i}=S_{i} \\cap\\left\\{2^{n}+1, \\ldots, 2^{n+1}\\right\\}$. Let\n\n$$\nI=\\left\\{i|1 \\leq i \\leq k,| U_{i} \\mid \\neq 0\\right\\}, \\quad J=\\{1, \\ldots, k\\} \\backslash I\n$$\n\nThe sets $S_{j}$ with $j \\in J$ are all contained in $\\left\\{2^{n}+1, \\ldots, 2^{n+1}\\right\\}$, so the induction hypothesis applies to their translates $-2^{n}+S_{j}$ which have the same cardinalities. Consequently, this gives $\\sum_{j \\in J}\\left(\\left|S_{j}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}$, so that\n\n$$\n\\sum_{j \\in J}\\left(\\left|S_{j}\\right|-(n+1)\\right) \\leq \\sum_{j \\in J}\\left(\\left|S_{j}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}\\tag{1}\n$$\n\nFor $i \\in I$, denote by $v_{i}$ the least element of $V_{i}$. Observe that if $V_{a}$ and $V_{b}$ intersect, with $a2^{n}$, which implies $z=v_{a}$.\n\nIt follows that if the element $v_{i}$ is removed from each $V_{i}$, a family of pairwise disjoint sets $W_{i}=V_{i} \\backslash\\left\\{v_{i}\\right\\}$ is obtained, $i \\in I$ (we assume $W_{i}=\\emptyset$ if $V_{i}=\\emptyset$ ). As $W_{i} \\subseteq\\left\\{2^{n}+1, \\ldots, 2^{n+1}\\right\\}$ for all $i$, we infer that $\\sum_{i \\in I}\\left|W_{i}\\right| \\leq 2^{n}$. Therefore $\\sum_{i \\in I}\\left(\\left|V_{i}\\right|-1\\right) \\leq \\sum_{i \\in I}\\left|W_{i}\\right| \\leq 2^{n}$.\n\nOn the other hand, the induction hypothesis applies directly to the sets $U_{i}, i \\in I$, so that $\\sum_{i \\in \\mathcal{I}}\\left(\\left|U_{i}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}$. In summary,\n\n$$\n\\sum_{i \\in I}\\left(\\left|S_{i}\\right|-(n+1)\\right)=\\sum_{i \\in I}\\left(\\left|U_{i}\\right|-n\\right)+\\sum_{i \\in I}\\left(\\left|V_{i}\\right|-1\\right) \\leq(2 n-1) 2^{n-2}+2^{n}\\tag{2}\n$$\n\nThe estimates (1) and (2) are sufficient to complete the inductive step:\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-(n+1)\\right) & =\\sum_{i \\in I}\\left(\\left|S_{i}\\right|-(n+1)\\right)+\\sum_{j \\in J}\\left(\\left|S_{j}\\right|-(n+1)\\right) \\\\\n& \\leq(2 n-1) 2^{n-2}+2^{n}+(2 n-1) 2^{n-2}=(2 n+1) 2^{n-1}\n\\end{aligned}\n$$\n\nReturning to the problem, consider $k=2^{n}$ subsets $S_{1}, S_{2}, \\ldots, S_{2^{n}}$ of $\\left\\{1,2,3, \\ldots, 2^{n+1}\\right\\}$. If they satisfy the given condition, the claim implies $\\sum_{i=1}^{2^{n}}\\left(\\left|S_{i}\\right|-(n+1)\\right) \\leq(2 n+1) 2^{n-1}$. By averaging again, we see that the smallest set has at most $2 n+1$ elements.']" [] Text-only Competition True Theorem proof Combinatorics Math English 345 In an acute-angled triangle $A B C$, point $H$ is the orthocentre and $A_{0}, B_{0}, C_{0}$ are the midpoints of the sides $B C, C A, A B$, respectively. Consider three circles passing through $H: \omega_{a}$ around $A_{0}, \omega_{b}$ around $B_{0}$ and $\omega_{c}$ around $C_{0}$. The circle $\omega_{a}$ intersects the line $B C$ at $A_{1}$ and $A_{2} ; \omega_{b}$ intersects $C A$ at $B_{1}$ and $B_{2} ; \omega_{c}$ intersects $A B$ at $C_{1}$ and $C_{2}$. Show that the points $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle. "['The perpendicular bisectors of the segments $A_{1} A_{2}, B_{1} B_{2}, C_{1} C_{2}$ are also the perpendicular bisectors of $B C, C A, A B$. So they meet at $O$, the circumcentre of $A B C$. Thus $O$ is the only point that can possibly be the centre of the desired circle.\n\nFrom the right triangle $O A_{0} A_{1}$ we get\n\n$$\nO A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2} .\\tag{1}\n$$\n\nLet $K$ be the midpoint of $A H$ and let $L$ be the midpoint of $C H$. Since $A_{0}$ and $B_{0}$ are the midpoints of $B C$ and $C A$, we see that $A_{0} L \\| B H$ and $B_{0} L \\| A H$. Thus the segments $A_{0} L$ and $B_{0} L$ are perpendicular to $A C$ and $B C$, hence parallel to $O B_{0}$ and $O A_{0}$, respectively. Consequently $O A_{0} L B_{0}$ is a parallelogram, so that $O A_{0}$ and $B_{0} L$ are equal and parallel. Also, the midline $B_{0} L$ of triangle $A H C$ is equal and parallel to $A K$ and $K H$.\n\nIt follows that $A K A_{0} O$ and $H A_{0} O K$ are parallelograms. The first one gives $A_{0} K=O A=R$, where $R$ is the circumradius of $A B C$. From the second one we obtain\n\n$$\n2\\left(O A_{0}^{2}+A_{0} H^{2}\\right)=O H^{2}+A_{0} K^{2}=O H^{2}+R^{2} .\\tag{2}\n$$\n\n(In a parallelogram, the sum of squares of the diagonals equals the sum of squares of the sides).\n\nFrom (1) and (2) we get $O A_{1}^{2}=\\left(O H^{2}+R^{2}\\right) / 2$. By symmetry, the same holds for the distances $O A_{2}, O B_{1}, O B_{2}, O C_{1}$ and $O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ all lie on a circle with centre at $O$ and radius $\\left(O H^{2}+R^{2}\\right) / 2$.\n\n' 'We are going to show again that the circumcentre $O$ is equidistant from the six points in question.\n\nLet $A^{\\prime}$ be the second intersection point of $\\omega_{b}$ and $\\omega_{c}$. The line $B_{0} C_{0}$, which is the line of centers of circles $\\omega_{b}$ and $\\omega_{c}$, is a midline in triangle $A B C$, parallel to $B C$ and perpendicular to the altitude $A H$. The points $A^{\\prime}$ and $H$ are symmetric with respect to the line of centers. Therefore $A^{\\prime}$ lies on the line $A H$.\n\nFrom the two circles $\\omega_{b}$ and $\\omega_{c}$ we obtain $A C_{1} \\cdot A C_{2}=A A^{\\prime} \\cdot A H=A B_{1} \\cdot A B_{2}$. So the quadrilateral $B_{1} B_{2} C_{1} C_{2}$ is cyclic. The perpendicular bisectors of the sides $B_{1} B_{2}$ and $C_{1} C_{2}$ meet at $O$. Hence $O$ is the circumcentre of $B_{1} B_{2} C_{1} C_{2}$ and so $O B_{1}=O B_{2}=O C_{1}=O C_{2}$.\n\nAnalogous arguments yield $O A_{1}=O A_{2}=O B_{1}=O B_{2}$ and $O A_{1}=O A_{2}=O C_{1}=O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle centred at $O$.\n\n' 'Let again $O$ and $R$ be the circumcentre and circumradius. Consider the vectors\n\n$$\n\\overrightarrow{O A}=\\mathbf{a}, \\quad \\overrightarrow{O B}=\\mathbf{b}, \\quad \\overrightarrow{O C}=\\mathbf{c}, \\quad \\text { where } \\quad \\mathbf{a}^{2}=\\mathbf{b}^{2}=\\mathbf{c}^{2}=R^{2}\n$$\n\nIt is well known that $\\overrightarrow{O H}=\\mathbf{a}+\\mathbf{b}+\\mathbf{c}$. Accordingly,\n\n$$\n\\overrightarrow{A_{0} H}=\\overrightarrow{O H}-\\overrightarrow{O A_{0}}=(\\mathbf{a}+\\mathbf{b}+\\mathbf{c})-\\frac{\\mathbf{b}+\\mathbf{c}}{2}=\\frac{2 \\mathbf{a}+\\mathbf{b}+\\mathbf{c}}{2}\n$$\n\nand\n\n$$\n\\begin{gathered}\nO A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2}=\\left(\\frac{\\mathbf{b}+\\mathbf{c}}{2}\\right)^{2}+\\left(\\frac{2 \\mathbf{a}+\\mathbf{b}+\\mathbf{c}}{2}\\right)^{2} \\\\\n=\\frac{1}{4}\\left(\\mathbf{b}^{2}+2 \\mathbf{b} \\mathbf{c}+\\mathbf{c}^{2}\\right)+\\frac{1}{4}\\left(4 \\mathbf{a}^{2}+4 \\mathbf{a} \\mathbf{b}+4 \\mathbf{a} \\mathbf{c}+\\mathbf{b}^{2}+2 \\mathbf{b} \\mathbf{c}+\\mathbf{c}^{2}\\right)=2 R^{2}+(\\mathbf{a b}+\\mathbf{a c}+\\mathbf{b c}) ;\n\\end{gathered}\n$$\n\nhere $\\mathbf{a b}, \\mathbf{b c}$, etc. denote dot products of vectors. We get the same for the distances $O A_{2}, O B_{1}$, $O B_{2}, O C_{1}$ and $O C_{2}$.']" [] Text-only Competition True Theorem proof Geometry Math English 346 "Given trapezoid $A B C D$ with parallel sides $A B$ and $C D$, assume that there exist points $E$ on line $B C$ outside segment $B C$, and $F$ inside segment $A D$, such that $\angle D A E=\angle C B F$. Denote by $I$ the point of intersection of $C D$ and $E F$, and by $J$ the point of intersection of $A B$ and $E F$. Let $K$ be the midpoint of segment $E F$; assume it does not lie on line $A B$. Prove that $I$ belongs to the circumcircle of $A B K$ if and only if $K$ belongs to the circumcircle of $C D J$." ['Assume that the disposition of points is as in the diagram.\n\nSince $\\angle E B F=180^{\\circ}-\\angle C B F=180^{\\circ}-\\angle E A F$ by hypothesis, the quadrilateral $A E B F$ is cyclic. Hence $A J \\cdot J B=F J \\cdot J E$. In view of this equality, $I$ belongs to the circumcircle of $A B K$ if and only if $I J \\cdot J K=F J \\cdot J E$. Expressing $I J=I F+F J, J E=F E-F J$, and $J K=\\frac{1}{2} F E-F J$, we find that $I$ belongs to the circumcircle of $A B K$ if and only if\n\n$$\nF J=\\frac{I F \\cdot F E}{2 I F+F E}\n$$\n\nSince $A E B F$ is cyclic and $A B, C D$ are parallel, $\\angle F E C=\\angle F A B=180^{\\circ}-\\angle C D F$. Then $C D F E$ is also cyclic, yielding $I D \\cdot I C=I F \\cdot I E$. It follows that $K$ belongs to the circumcircle of $C D J$ if and only if $I J \\cdot I K=I F \\cdot I E$. Expressing $I J=I F+F J, I K=I F+\\frac{1}{2} F E$, and $I E=I F+F E$, we find that $K$ is on the circumcircle of $C D J$ if and only if\n\n$$\nF J=\\frac{I F \\cdot F E}{2 I F+F E}\n$$\n\nThe conclusion follows.\n\n'] [] Text-only Competition True Theorem proof Geometry Math English 347 Let $A B C D$ be a convex quadrilateral and let $P$ and $Q$ be points in $A B C D$ such that $P Q D A$ and $Q P B C$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $P Q$ such that $\angle P A E=\angle Q D E$ and $\angle P B E=\angle Q C E$. Show that the quadrilateral $A B C D$ is cyclic. "['Let $F$ be the point on the line $A D$ such that $E F \\| P A$. By hypothesis, the quadrilateral $P Q D A$ is cyclic. So if $F$ lies between $A$ and $D$ then $\\angle E F D=\\angle P A D=180^{\\circ}-\\angle E Q D$; the points $F$ and $Q$ are on distinct sides of the line $D E$ and we infer that $E F D Q$ is a cyclic quadrilateral. And if $D$ lies between $A$ and $F$ then a similar argument shows that $\\angle E F D=\\angle E Q D$; but now the points $F$ and $Q$ lie on the same side of $D E$, so that $E D F Q$ is a cyclic quadrilateral.\n\nIn either case we obtain the equality $\\angle E F Q=\\angle E D Q=\\angle P A E$ which implies that $F Q \\| A E$. So the triangles $E F Q$ and $P A E$ are either homothetic or parallel-congruent. More specifically, triangle $E F Q$ is the image of $P A E$ under the mapping $f$ which carries the points $P, E$ respectively to $E, Q$ and is either a homothety or translation by a vector. Note that $f$ is uniquely determined by these conditions and the position of the points $P, E, Q$ alone.\n\nLet now $G$ be the point on the line $B C$ such that $E G \\| P B$. The same reasoning as above applies to points $B, C$ in place of $A, D$, implying that the triangle $E G Q$ is the image of $P B E$ under the same mapping $f$. So $f$ sends the four points $A, P, B, E$ respectively to $F, E, G, Q$.\n\nIf $P E \\neq Q E$, so that $f$ is a homothety with a centre $X$, then the lines $A F, P E, B G$-i.e. the lines $A D, P Q, B C$-are concurrent at $X$. And since $P Q D A$ and $Q P B C$ are cyclic quadrilaterals, the equalities $X A \\cdot X D=X P \\cdot X Q=X B \\cdot X C$ hold, showing that the quadrilateral $A B C D$ is cyclic.\n\nFinally, if $P E=Q E$, so that $f$ is a translation, then $A D\\|P Q\\| B C$. Thus $P Q D A$ and $Q P B C$ are isosceles trapezoids. Then also $A B C D$ is an isosceles trapezoid, hence a cyclic quadrilateral.\n\n' 'Here is another way to reach the conclusion that the lines $A D, B C$ and $P Q$ are either concurrent or parallel. From the cyclic quadrilateral $P Q D A$ we get\n\n$$\n\\angle P A D=180^{\\circ}-\\angle P Q D=\\angle Q D E+\\angle Q E D=\\angle P A E+\\angle Q E D .\n$$\n\n\n\nHence $\\angle Q E D=\\angle P A D-\\angle P A E=\\angle E A D$. This in view of the tangent-chord theorem means that the circumcircle of triangle $E A D$ is tangent to the line $P Q$ at $E$. Analogously, the circumcircle of triangle $E B C$ is tangent to $P Q$ at $E$.\n\nSuppose that the line $A D$ intersects $P Q$ at $X$. Since $X E$ is tangent to the circle $(E A D)$, $X E^{2}=X A \\cdot X D$. Also, $X A \\cdot X D=X P \\cdot X Q$ because $P, Q, D, A$ lie on a circle. Therefore $X E^{2}=X P \\cdot X Q$.\n\nIt is not hard to see that this equation determines the position of the point $X$ on the line $P Q$ uniquely. Thus, if $B C$ also cuts $P Q$, say at $Y$, then the analogous equation for $Y$ yields $X=Y$, meaning that the three lines indeed concur. In this case, as well as in the case where $A D\\|P Q\\| B C$, the concluding argument is the same as in the first solution.\n\nIt remains to eliminate the possibility that e.g. $A D$ meets $P Q$ at $X$ while $B C \\| P Q$. Indeed, $Q P B C$ would then be an isosceles trapezoid and the angle equality $\\angle P B E=\\angle Q C E$ would force that $E$ is the midpoint of $P Q$. So the length of $X E$, which is the geometric mean of the lengths of $X P$ and $X Q$, should also be their arithmetic mean -impossible, as $X P \\neq X Q$. The proof is now complete.']" [] Text-only Competition True Theorem proof Geometry Math English 348 In an acute triangle $A B C$ segments $B E$ and $C F$ are altitudes. Two circles passing through the points $A$ and $F$ are tangent to the line $B C$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $P E$ and $Q F$ intersect on the circumcircle of triangle $A E F$. "['To approach the desired result we need some information about the slopes of the lines $P E$ and $Q F$; this information is provided by formulas (1) and (2) which we derive below.\n\nThe tangents $B P$ and $B Q$ to the two circles passing through $A$ and $F$ are equal, as $B P^{2}=B A \\cdot B F=B Q^{2}$. Consider the altitude $A D$ of triangle $A B C$ and its orthocentre $H$. From the cyclic quadrilaterals $C D F A$ and $C D H E$ we get $B A \\cdot B F=B C \\cdot B D=B E \\cdot B H$. Thus $B P^{2}=B E \\cdot B H$, or $B P / B H=B E / B P$, implying that the triangles $B P H$ and $B E P$ are similar. Hence\n\n$$\n\\angle B P E=\\angle B H P \\text {. }\\tag{1}\n$$\n\nThe point $P$ lies between $D$ and $C$; this follows from the equality $B P^{2}=B C \\cdot B D$. In view of this equality, and because $B P=B Q$,\n\n$$\nD P \\cdot D Q=(B P-B D) \\cdot(B P+B D)=B P^{2}-B D^{2}=B D \\cdot(B C-B D)=B D \\cdot D C\n$$\n\nAlso $A D \\cdot D H=B D \\cdot D C$, as is seen from the similar triangles $B D H$ and $A D C$. Combining these equalities we obtain $A D \\cdot D H=D P \\cdot D Q$. Therefore $D H / D P=D Q / D A$, showing that the triangles $H D P$ and $Q D A$ are similar. Hence $\\angle H P D=\\angle Q A D$, which can be rewritten as $\\angle B P H=\\angle B A D+\\angle B A Q$. And since $B Q$ is tangent to the circumcircle of triangle $F A Q$,\n\n$$\n\\angle B Q F=\\angle B A Q=\\angle B P H-\\angle B A D .\\tag{2}\n$$\n\nFrom (1) and (2) we deduce\n\n$$\n\\begin{aligned}\n& \\angle B P E+\\angle B Q F=(\\angle B H P+\\angle B P H)-\\angle B A D=\\left(180^{\\circ}-\\angle P B H\\right)-\\angle B A D \\\\\n& =\\left(90^{\\circ}+\\angle B C A\\right)-\\left(90^{\\circ}-\\angle A B C\\right)=\\angle B C A+\\angle A B C=180^{\\circ}-\\angle C A B .\n\\end{aligned}\n$$\n\nThus $\\angle B P E+\\angle B Q F<180^{\\circ}$, which means that the rays $P E$ and $Q F$ meet. Let $S$ be the point of intersection. Then $\\angle P S Q=180^{\\circ}-(\\angle B P E+\\angle B Q F)=\\angle C A B=\\angle E A F$.\n\nIf $S$ lies between $P$ and $E$ then $\\angle P S Q=180^{\\circ}-\\angle E S F$; and if $E$ lies between $P$ and $S$ then $\\angle P S Q=\\angle E S F$. In either case the equality $\\angle P S Q=\\angle E A F$ which we have obtained means that $S$ lies on the circumcircle of triangle $A E F$.\n\n' ""Let $H$ be the orthocentre of triangle $A B C$ and let $\\omega$ be the circle with diameter $A H$, passing through $E$ and $F$. Introduce the points of intersection of $\\omega$ with the following lines emanating from $P: P A \\cap \\omega=\\{A, U\\}, P H \\cap \\omega=\\{H, V\\}, P E \\cap \\omega=\\{E, S\\}$. The altitudes of triangle $A H P$ are contained in the lines $A V, H U, B C$, meeting at its orthocentre $Q^{\\prime}$.\n\nBy Pascal's theorem applied to the (tied) hexagon $A E S F H V$, the points $A E \\cap F H=C$, $E S \\cap H V=P$ and $S F \\cap V A$ are collinear, so $F S$ passes through $Q^{\\prime}$.\n\nDenote by $\\omega_{1}$ and $\\omega_{2}$ the circles with diameters $B C$ and $P Q^{\\prime}$, respectively. Let $D$ be the foot of the altitude from $A$ in triangle $A B C$. Suppose that $A D$ meets the circles $\\omega_{1}$ and $\\omega_{2}$ at the respective points $K$ and $L$.\n\nSince $H$ is the orthocentre of $A B C$, the triangles $B D H$ and $A D C$ are similar, and so $D A \\cdot D H=D B \\cdot D C=D K^{2}$; the last equality holds because $B K C$ is a right triangle. Since $H$ is the orthocentre also in triangle $A Q^{\\prime} P$, we analogously have $D L^{2}=D A \\cdot D H$. Therefore $D K=D L$ and $K=L$.\n\nAlso, $B D \\cdot B C=B A \\cdot B F$, from the similar triangles $A B D, C B F$. In the right triangle $B K C$ we have $B K^{2}=B D \\cdot B C$. Hence, and because $B A \\cdot B F=B P^{2}=B Q^{2}$ (by the definition of $P$ and $Q$ in the problem statement), we obtain $B K=B P=B Q$. It follows that $B$ is the centre of $\\omega_{2}$ and hence $Q^{\\prime}=Q$. So the lines $P E$ and $Q F$ meet at the point $S$ lying on the circumcircle of triangle $A E F$.\n\n""]" [] Text-only Competition True Theorem proof Geometry Math English 349 "Let $k$ and $n$ be integers with $0 \leq k \leq n-2$. Consider a set $L$ of $n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $I$ the set of intersection points of lines in $L$. Let $O$ be a point in the plane not lying on any line of $L$. A point $X \in I$ is colored red if the open line segment $O X$ intersects at most $k$ lines in $L$. Prove that $I$ contains at least $\frac{1}{2}(k+1)(k+2)$ red points." ['There are at least $\\frac{1}{2}(k+1)(k+2)$ points in the intersection set $I$ in view of the condition $n \\geq k+2$.\n\nFor each point $P \\in I$, define its order as the number of lines that intersect the open line segment $O P$. By definition, $P$ is red if its order is at most $k$. Note that there is always at least one point $X \\in I$ of order 0 . Indeed, the lines in $L$ divide the plane into regions, bounded or not, and $O$ belongs to one of them. Clearly any corner of this region is a point of $I$ with order 0.\n\nClaim. Suppose that two points $P, Q \\in I$ lie on the same line of $L$, and no other line of $L$ intersects the open line segment $P Q$. Then the orders of $P$ and $Q$ differ by at most 1 .\n\nProof. Let $P$ and $Q$ have orders $p$ and $q$, respectively, with $p \\geq q$. Consider triangle $O P Q$. Now $p$ equals the number of lines in $L$ that intersect the interior of side $O P$. None of these lines intersects the interior of side $P Q$, and at most one can pass through $Q$. All remaining lines must intersect the interior of side $O Q$, implying that $q \\geq p-1$. The conclusion follows.\n\nWe prove the main result by induction on $k$. The base $k=0$ is clear since there is a point of order 0 which is red. Assuming the statement true for $k-1$, we pass on to the inductive step. Select a point $P \\in I$ of order 0 , and consider one of the lines $\\ell \\in L$ that pass through $P$. There are $n-1$ intersection points on $\\ell$, one of which is $P$. Out of the remaining $n-2$ points, the $k$ closest to $P$ have orders not exceeding $k$ by the Claim. It follows that there are at least $k+1$ red points on $\\ell$.\n\nLet us now consider the situation with $\\ell$ removed (together with all intersection points it contains). By hypothesis of induction, there are at least $\\frac{1}{2} k(k+1)$ points of order not exceeding $k-1$ in the resulting configuration. Restoring $\\ell$ back produces at most one new intersection point on each line segment joining any of these points to $O$, so their order is at most $k$ in the original configuration. The total number of points with order not exceeding $k$ is therefore at least $(k+1)+\\frac{1}{2} k(k+1)=\\frac{1}{2}(k+1)(k+2)$. This completes the proof.'] [] Text-only Competition True Theorem proof Geometry Math English 350 "There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that $$ \angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ} \tag{1} $$ if and only if the diagonals $A C$ and $B D$ are perpendicular." "['For a point $P$ in $A B C D$ which satisfies (1), let $K, L, M, N$ be the feet of perpendiculars from $P$ to lines $A B, B C, C D, D A$, respectively. Note that $K, L, M, N$ are interior to the sides as all angles in (1) are acute. The cyclic quadrilaterals $A K P N$ and $D N P M$ give\n\n$$\n\\angle P A B+\\angle P D C=\\angle P N K+\\angle P N M=\\angle K N M .\n$$\n\nAnalogously, $\\angle P B C+\\angle P A D=\\angle L K N$ and $\\angle P C D+\\angle P B A=\\angle M L K$. Hence the equalities (1) imply $\\angle K N M=\\angle L K N=\\angle M L K=90^{\\circ}$, so that $K L M N$ is a rectangle. The converse also holds true, provided that $K, L, M, N$ are interior to sides $A B, B C, C D, D A$.\n\n(i) Suppose that there exists a point $P$ in $A B C D$ such that $K L M N$ is a rectangle. We show that $A C$ and $B D$ are parallel to the respective sides of $K L M N$.\n\nLet $O_{A}$ and $O_{C}$ be the circumcentres of the cyclic quadrilaterals $A K P N$ and $C M P L$. Line $O_{A} O_{C}$ is the common perpendicular bisector of $L M$ and $K N$, therefore $O_{A} O_{C}$ is parallel to $K L$ and $M N$. On the other hand, $O_{A} O_{C}$ is the midline in the triangle $A C P$ that is parallel to $A C$. Therefore the diagonal $A C$ is parallel to the sides $K L$ and $M N$ of the rectangle. Likewise, $B D$ is parallel to $K N$ and $L M$. Hence $A C$ and $B D$ are perpendicular.\n\n\n\n(ii) Suppose that $A C$ and $B D$ are perpendicular and meet at $R$. If $A B C D$ is a rhombus, $P$ can be chosen to be its centre. So assume that $A B C D$ is not a rhombus, and let $B R' 'For a point $P$ distinct from $A, B, C, D$, let circles $(A P D)$ and $(B P C)$ intersect again at $Q(Q=P$ if the circles are tangent). Next, let circles $(A Q B)$ and $(C Q D)$ intersect again at $R$. We show that if $P$ lies in $A B C D$ and satisfies (1) then $A C$ and $B D$ intersect at $R$ and are perpendicular; the converse is also true. It is convenient to use directed angles. Let $\\measuredangle(U V, X Y)$ denote the angle of counterclockwise rotation that makes line $U V$ parallel to line $X Y$. Recall that four noncollinear points $U, V, X, Y$ are concyclic if and only if $\\measuredangle(U X, V X)=\\measuredangle(U Y, V Y)$.\n\nThe definitions of points $P, Q$ and $R$ imply\n\n$$\n\\begin{aligned}\n\\measuredangle(A R, B R) & =\\measuredangle(A Q, B Q)=\\measuredangle(A Q, P Q)+\\measuredangle(P Q, B Q)=\\measuredangle(A D, P D)+\\measuredangle(P C, B C), \\\\\n\\measuredangle(C R, D R) & \\measuredangle(C Q, D Q)=\\measuredangle(C Q, P Q)+\\measuredangle(P Q, D Q)=\\measuredangle(C B, P B)+\\measuredangle(P A, D A), \\\\\n\\measuredangle(B R, C R) & =\\measuredangle(B R, R Q)+\\measuredangle(R Q, C R)=\\measuredangle(B A, A Q)+\\measuredangle(D Q, C D) \\\\\n& =\\measuredangle(B A, A P)+\\measuredangle(A P, A Q)+\\measuredangle(D Q, D P)+\\measuredangle(D P, C D) \\\\\n& =\\measuredangle(B A, A P)+\\measuredangle(D P, C D) .\n\\end{aligned}\n$$\n\nObserve that the whole construction is reversible. One may start with point $R$, define $Q$ as the second intersection of circles $(A R B)$ and $(C R D)$, and then define $P$ as the second intersection of circles $(A Q D)$ and $(B Q C)$. The equalities above will still hold true.\n\n\n\nAssume in addition that $P$ is interior to $A B C D$. Then\n\n$$\n\\begin{gathered}\n\\measuredangle(A D, P D)=\\angle P D A, \\measuredangle(P C, B C)=\\angle P C B, \\measuredangle(C B, P B)=\\angle P B C, \\measuredangle(P A, D A)=\\angle P A D \\\\\n\\measuredangle(B A, A P)=\\angle P A B, \\measuredangle(D P, C D)=\\angle P D C .\n\\end{gathered}\n$$\n\n(i) Suppose that $P$ lies in $A B C D$ and satisfies (1). Then $\\measuredangle(A R, B R)=\\angle P D A+\\angle P C B=90^{\\circ}$ and similarly $\\measuredangle(B R, C R)=\\measuredangle(C R, D R)=90^{\\circ}$. It follows that $R$ is the common point of lines $A C$ and $B D$, and that these lines are perpendicular.\n\n(ii) Suppose that $A C$ and $B D$ are perpendicular and intersect at $R$. We show that the point $P$ defined by the reverse construction (starting with $R$ and ending with $P$ ) lies in $A B C D$. This is enough to finish the solution, because then the angle equalities above will imply (1).\n\nOne can assume that $Q$, the second common point of circles $(A B R)$ and $(C D R)$, lies in $\\angle A R D$. Then in fact $Q$ lies in triangle $A D R$ as angles $A Q R$ and $D Q R$ are obtuse. Hence $\\angle A Q D$ is obtuse, too, so that $B$ and $C$ are outside circle $(A D Q)(\\angle A B D$ and $\\angle A C D$ are acute).\n\nNow $\\angle C A B+\\angle C D B=\\angle B Q R+\\angle C Q R=\\angle C Q B$ implies $\\angle C A B<\\angle C Q B$ and $\\angle C D B<$ $\\angle C Q B$. Hence $A$ and $D$ are outside circle $(B C Q)$. In conclusion, the second common point $P$ of circles $(A D Q)$ and $(B C Q)$ lies on their arcs $A D Q$ and $B C Q$.\n\nWe can assume that $P$ lies in $\\angle C Q D$. Since\n\n$$\n\\begin{gathered}\n\\angle Q P C+\\angle Q P D=\\left(180^{\\circ}-\\angle Q B C\\right)+\\left(180^{\\circ}-\\angle Q A D\\right)= \\\\\n=360^{\\circ}-(\\angle R B C+\\angle Q B R)-(\\angle R A D-\\angle Q A R)=360^{\\circ}-\\angle R B C-\\angle R A D>180^{\\circ},\n\\end{gathered}\n$$\n\npoint $P$ lies in triangle $C D Q$, and hence in $A B C D$. The proof is complete.\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 351 Let $A B C D$ be a convex quadrilateral with $A B \neq B C$. Denote by $\omega_{1}$ and $\omega_{2}$ the incircles of triangles $A B C$ and $A D C$. Suppose that there exists a circle $\omega$ inscribed in angle $A B C$, tangent to the extensions of line segments $A D$ and $C D$. Prove that the common external tangents of $\omega_{1}$ and $\omega_{2}$ intersect on $\omega$. "['The proof below is based on two known facts.\n\nLemma 1. Given a convex quadrilateral $A B C D$, suppose that there exists a circle which is inscribed in angle $A B C$ and tangent to the extensions of line segments $A D$ and $C D$. Then $A B+A D=C B+C D$.\n\nProof. The circle in question is tangent to each of the lines $A B, B C, C D, D A$, and the respective points of tangency $K, L, M, N$ are located as with circle $\\omega$ in the figure. Then\n\n$$\nA B+A D=(B K-A K)+(A N-D N), \\quad C B+C D=(B L-C L)+(C M-D M) .\n$$\n\nAlso $B K=B L, D N=D M, A K=A N, C L=C M$ by equalities of tangents. It follows that $A B+A D=C B+C D$.\n\n\n\nFor brevity, in the sequel we write ""excircle $A C$ "" for the excircle of a triangle with side $A C$ which is tangent to line segment $A C$ and the extensions of the other two sides.\n\nLemma 2. The incircle of triangle $A B C$ is tangent to its side $A C$ at $P$. Let $P P^{\\prime}$ be the diameter of the incircle through $P$, and let line $B P^{\\prime}$ intersect $A C$ at $Q$. Then $Q$ is the point of tangency of side $A C$ and excircle $A C$.\n\nProof. Let the tangent at $P^{\\prime}$ to the incircle $\\omega_{1}$ meet $B A$ and $B C$ at $A^{\\prime}$ and $C^{\\prime}$. Now $\\omega_{1}$ is the excircle $A^{\\prime} C^{\\prime}$ of triangle $A^{\\prime} B C^{\\prime}$, and it touches side $A^{\\prime} C^{\\prime}$ at $P^{\\prime}$. Since $A^{\\prime} C^{\\prime} \\| A C$, the homothety with centre $B$ and ratio $B Q / B P^{\\prime}$ takes $\\omega_{1}$ to the excircle $A C$ of triangle $A B C$. Because this homothety takes $P^{\\prime}$ to $Q$, the lemma follows.\n\n\n\nRecall also that if the incircle of a triangle touches its side $A C$ at $P$, then the tangency point $Q$ of the same side and excircle $A C$ is the unique point on line segment $A C$ such that $A P=C Q$.\n\nWe pass on to the main proof. Let $\\omega_{1}$ and $\\omega_{2}$ touch $A C$ at $P$ and $Q$, respectively; then $A P=(A C+A B-B C) / 2, C Q=(C A+C D-A D) / 2$. Since $A B-B C=C D-A D$ by Lemma 1, we obtain $A P=C Q$. It follows that in triangle $A B C$ side $A C$ and excircle $A C$ are tangent at $Q$. Likewise, in triangle $A D C$ side $A C$ and excircle $A C$ are tangent at $P$. Note that $P \\neq Q$ as $A B \\neq B C$.\n\nLet $P P^{\\prime}$ and $Q Q^{\\prime}$ be the diameters perpendicular to $A C$ of $\\omega_{1}$ and $\\omega_{2}$, respectively. Then Lemma 2 shows that points $B, P^{\\prime}$ and $Q$ are collinear, and so are points $D, Q^{\\prime}$ and $P$.\n\nConsider the diameter of $\\omega$ perpendicular to $A C$ and denote by $T$ its endpoint that is closer to $A C$. The homothety with centre $B$ and ratio $B T / B P^{\\prime}$ takes $\\omega_{1}$ to $\\omega$. Hence $B, P^{\\prime}$ and $T$ are collinear. Similarly, $D, Q^{\\prime}$ and $T$ are collinear since the homothety with centre $D$ and ratio $-D T / D Q^{\\prime}$ takes $\\omega_{2}$ to $\\omega$.\n\nWe infer that points $T, P^{\\prime}$ and $Q$ are collinear, as well as $T, Q^{\\prime}$ and $P$. Since $P P^{\\prime} \\| Q Q^{\\prime}$, line segments $P P^{\\prime}$ and $Q Q^{\\prime}$ are then homothetic with centre $T$. The same holds true for circles $\\omega_{1}$ and $\\omega_{2}$ because they have $P P^{\\prime}$ and $Q Q^{\\prime}$ as diameters. Moreover, it is immediate that $T$ lies on the same side of line $P P^{\\prime}$ as $Q$ and $Q^{\\prime}$, hence the ratio of homothety is positive. In particular $\\omega_{1}$ and $\\omega_{2}$ are not congruent.\n\nIn summary, $T$ is the centre of a homothety with positive ratio that takes circle $\\omega_{1}$ to circle $\\omega_{2}$. This completes the solution, since the only point with the mentioned property is the intersection of the the common external tangents of $\\omega_{1}$ and $\\omega_{2}$.']" [] Text-only Competition True Theorem proof Geometry Math English 352 "Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a, b, c$ are integers (not necessarily positive) satisfying the equations $$ a^{n}+p b=b^{n}+p c=c^{n}+p a, $$ then $a=b=c$." "['If two of $a, b, c$ are equal, it is immediate that all the three are equal. So we may assume that $a \\neq b \\neq c \\neq a$. Subtracting the equations we get $a^{n}-b^{n}=-p(b-c)$ and two cyclic copies of this equation, which upon multiplication yield\n\n$$\n\\frac{a^{n}-b^{n}}{a-b} \\cdot \\frac{b^{n}-c^{n}}{b-c} \\cdot \\frac{c^{n}-a^{n}}{c-a}=-p^{3}\n\\tag{1}\n$$\n\nIf $n$ is odd then the differences $a^{n}-b^{n}$ and $a-b$ have the same sign and the product on the left is positive, while $-p^{3}$ is negative. So $n$ must be even.\n\nLet $d$ be the greatest common divisor of the three differences $a-b, b-c, c-a$, so that $a-b=d u, \\quad b-c=d v, \\quad c-a=d w ; \\quad \\operatorname{gcd}(u, v, w)=1, u+v+w=0$.\n\nFrom $a^{n}-b^{n}=-p(b-c)$ we see that $(a-b) \\mid p(b-c)$, i.e., $u \\mid p v$; and cyclically $v|p w, w| p u$. As $\\operatorname{gcd}(u, v, w)=1$ and $u+v+w=0$, at most one of $u, v, w$ can be divisible by $p$. Supposing that the prime $p$ does not divide any one of them, we get $u|v, v| w, w \\mid u$, whence $|u|=|v|=|w|=1$; but this quarrels with $u+v+w=0$.\n\nThus $p$ must divide exactly one of these numbers. Let e.g. $p \\mid u$ and write $u=p u_{1}$. Now we obtain, similarly as before, $u_{1}|v, v| w, w \\mid u_{1}$ so that $\\left|u_{1}\\right|=|v|=|w|=1$. The equation $p u_{1}+v+w=0$ forces that the prime $p$ must be even; i.e. $p=2$. Hence $v+w=-2 u_{1}= \\pm 2$, implying $v=w(= \\pm 1)$ and $u=-2 v$. Consequently $a-b=-2(b-c)$.\n\nKnowing that $n$ is even, say $n=2 k$, we rewrite the equation $a^{n}-b^{n}=-p(b-c)$ with $p=2$ in the form\n\n$$\n\\left(a^{k}+b^{k}\\right)\\left(a^{k}-b^{k}\\right)=-2(b-c)=a-b .\n$$\n\nThe second factor on the left is divisible by $a-b$, so the first factor $\\left(a^{k}+b^{k}\\right)$ must be \\pm 1 . Then exactly one of $a$ and $b$ must be odd; yet $a-b=-2(b-c)$ is even. Contradiction ends the proof.' 'The beginning is as in the first solution. Assuming that $a, b, c$ are not all equal, hence are all distinct, we derive equation (1) with the conclusion that $n$ is even. Write $n=2 k$.\n\nSuppose that $p$ is odd. Then the integer\n\n$$\n\\frac{a^{n}-b^{n}}{a-b}=a^{n-1}+a^{n-2} b+\\cdots+b^{n-1},\n$$\n\n\n\nwhich is a factor in (1), must be odd as well. This sum of $n=2 k$ summands is odd only if $a$ and $b$ have different parities. The same conclusion holding for $b, c$ and for $c$, $a$, we get that $a, b, c, a$ alternate in their parities, which is clearly impossible.\n\nThus $p=2$. The original system shows that $a, b, c$ must be of the same parity. So we may divide (1) by $p^{3}$, i.e. $2^{3}$, to obtain the following product of six integer factors:\n\n$$\n\\frac{a^{k}+b^{k}}{2} \\cdot \\frac{a^{k}-b^{k}}{a-b} \\cdot \\frac{b^{k}+c^{k}}{2} \\cdot \\frac{b^{k}-c^{k}}{b-c} \\cdot \\frac{c^{k}+a^{k}}{2} \\cdot \\frac{c^{k}-a^{k}}{c-a}=-1\n\\tag{2}\n$$\n\nEach one of the factors must be equal to \\pm 1 . In particular, $a^{k}+b^{k}= \\pm 2$. If $k$ is even, this becomes $a^{k}+b^{k}=2$ and yields $|a|=|b|=1$, whence $a^{k}-b^{k}=0$, contradicting (2).\n\nLet now $k$ be odd. Then the sum $a^{k}+b^{k}$, with value \\pm 2 , has $a+b$ as a factor. Since $a$ and $b$ are of the same parity, this means that $a+b= \\pm 2$; and cyclically, $b+c= \\pm 2, c+a= \\pm 2$. In some two of these equations the signs must coincide, hence some two of $a, b, c$ are equal. This is the desired contradiction.']" [] Text-only Competition True Theorem proof Number Theory Math English 353 Let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct positive integers, $n \geq 3$. Prove that there exist distinct indices $i$ and $j$ such that $a_{i}+a_{j}$ does not divide any of the numbers $3 a_{1}, 3 a_{2}, \ldots, 3 a_{n}$. "[""Without loss of generality, let $0a_{n-1}\\right)$. Thus $3 a_{n-1}=a_{n}+a_{n-1}$, i. e. $a_{n}=2 a_{n-1}$.\n\nSimilarly, if $j=n$ then $3 a_{n}=k\\left(a_{n}+a_{n-1}\\right)$ for some integer $k$, and only $k=2$ is possible. Hence $a_{n}=2 a_{n-1}$ holds true in both cases remaining, $j=n-1$ and $j=n$.\n\nNow $a_{n}=2 a_{n-1}$ implies that the sum $a_{n-1}+a_{1}$ is strictly between $a_{n} / 2$ and $a_{n}$. But $a_{n-1}$ and $a_{1}$ are distinct as $n \\geq 3$, so it follows from the above that $a_{n-1}+a_{1}$ divides $a_{n}$. This provides the desired contradiction.""]" [] Text-only Competition True Theorem proof Number Theory Math English 354 Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $\operatorname{gcd}\left(a_{i}, a_{i+1}\right)>a_{i-1}$. Prove that $a_{n} \geq 2^{n}$ for all $n \geq 0$. ['Since $a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, the sequence is strictly increasing. In particular $a_{0} \\geq 1, a_{1} \\geq 2$. For each $i \\geq 1$ we also have $a_{i+1}-a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, and consequently $a_{i+1} \\geq a_{i}+a_{i-1}+1$. Hence $a_{2} \\geq 4$ and $a_{3} \\geq 7$. The equality $a_{3}=7$ would force equalities in the previous estimates, leading to $\\operatorname{gcd}\\left(a_{2}, a_{3}\\right)=\\operatorname{gcd}(4,7)>a_{1}=2$, which is false. Thus $a_{3} \\geq 8$; the result is valid for $n=0,1,2,3$. These are the base cases for a proof by induction.\n\nTake an $n \\geq 3$ and assume that $a_{i} \\geq 2^{i}$ for $i=0,1, \\ldots, n$. We must show that $a_{n+1} \\geq 2^{n+1}$. Let $\\operatorname{gcd}\\left(a_{n}, a_{n+1}\\right)=d$. We know that $d>a_{n-1}$. The induction claim is reached immediately in the following cases:\n\n$$\n\\begin{array}{ll}\n\\text { if } \\quad a_{n+1} \\geq 4 d & \\text { then } a_{n+1}>4 a_{n-1} \\geq 4 \\cdot 2^{n-1}=2^{n+1} \\\\\n\\text { if } \\quad a_{n} \\geq 3 d & \\text { then } a_{n+1} \\geq a_{n}+d \\geq 4 d>4 a_{n-1} \\geq 4 \\cdot 2^{n-1}=2^{n+1} \\text {; } \\\\\n\\text { if } a_{n}=d \\quad \\text { then } a_{n+1} \\geq a_{n}+d=2 a_{n} \\geq 2 \\cdot 2^{n}=2^{n+1}\n\\end{array}\n$$\n\nThe only remaining possibility is that $a_{n}=2 d$ and $a_{n+1}=3 d$, which we assume for the sequel. So $a_{n+1}=\\frac{3}{2} a_{n}$.\n\nLet now $\\operatorname{gcd}\\left(a_{n-1}, a_{n}\\right)=d^{\\prime}$; then $d^{\\prime}>a_{n-2}$. Write $a_{n}=m d^{\\prime} \\quad(m$ an integer $)$. Keeping in mind that $d^{\\prime} \\leq a_{n-1}9 a_{n-2} \\geq 9 \\cdot 2^{n-2}>2^{n+1} ;$\n\nif $3 \\leq m \\leq 4$ then $a_{n-1}<\\frac{1}{2} \\cdot 4 d^{\\prime}$, and hence $a_{n-1}=d^{\\prime}$,\n\n$$\na_{n+1}=\\frac{3}{2} m a_{n-1} \\geq \\frac{3}{2} \\cdot 3 a_{n-1} \\geq \\frac{9}{2} \\cdot 2^{n-1}>2^{n+1} \\text {. }\n$$\n\nSo we are left with the case $m=5$, which means that $a_{n}=5 d^{\\prime}, a_{n+1}=\\frac{15}{2} d^{\\prime}, a_{n-1}a_{n-3}$. Because $d^{\\prime \\prime}$ is a divisor of $a_{n-1}$, hence also of $2 d^{\\prime}$, we may write $2 d^{\\prime}=m^{\\prime} d^{\\prime \\prime}$ ( $m^{\\prime}$ an integer). Since $d^{\\prime \\prime} \\leq a_{n-2}\\frac{75}{4} a_{n-3} \\geq \\frac{75}{4} \\cdot 2^{n-3}>2^{n+1}\n$$\n\n$$\n\\text { if } 3 \\leq m^{\\prime} \\leq 4 \\text { then } a_{n-2}<\\frac{1}{2} \\cdot 4 d^{\\prime \\prime} \\text {, and hence } a_{n-2}=d^{\\prime \\prime} \\text {, }\n$$\n\n$$\na_{n+1}=\\frac{15}{4} m^{\\prime} a_{n-2} \\geq \\frac{15}{4} \\cdot 3 a_{n-2} \\geq \\frac{45}{4} \\cdot 2^{n-2}>2^{n+1}\n$$\n\nBoth of them have produced the induction claim. But now there are no cases left. Induction is complete; the inequality $a_{n} \\geq 2^{n}$ holds for all $n$.'] [] Text-only Competition True Theorem proof Number Theory Math English 355 "Let $n$ be a positive integer. Show that the numbers $$ \left(\begin{array}{c} 2^{n}-1 \\ 0 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 1 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 2 \end{array}\right), \quad \cdots, \quad\left(\begin{array}{c} 2^{n}-1 \\ 2^{n-1}-1 \end{array}\right) $$ are congruent modulo $2^{n}$ to $1,3,5, \ldots, 2^{n}-1$ in some order." "['It is well-known that all these numbers are odd. So the assertion that their remainders $\\left(\\bmod 2^{n}\\right)$ make up a permutation of $\\left\\{1,3, \\ldots, 2^{n}-1\\right\\}$ is equivalent just to saying that these remainders are all distinct. We begin by showing that\n\n$$\n\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k\n\\end{array}\\right)+\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k+1\n\\end{array}\\right) \\equiv 0 \\quad\\left(\\bmod 2^{n}\\right) \\quad \\text { and } \\quad\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k\n\\end{array}\\right) \\equiv(-1)^{k}\\left(\\begin{array}{c}\n2^{n-1}-1 \\\\\nk\n\\end{array}\\right) \\quad\\left(\\bmod 2^{n}\\right)\n\\tag{1}\n$$\n\nThe first relation is immediate, as the sum on the left is equal to $\\left(\\begin{array}{c}2^{n} \\\\ 2 k+1\\end{array}\\right)=\\frac{2^{n}}{2 k+1}\\left(\\begin{array}{c}2^{n}-1 \\\\ 2 k\\end{array}\\right)$, hence is divisible by $2^{n}$. The second relation:\n\n$$\n\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k\n\\end{array}\\right)=\\prod_{j=1}^{2 k} \\frac{2^{n}-j}{j}=\\prod_{i=1}^{k} \\frac{2^{n}-(2 i-1)}{2 i-1} \\cdot \\prod_{i=1}^{k} \\frac{2^{n-1}-i}{i} \\equiv(-1)^{k}\\left(\\begin{array}{c}\n2^{n-1}-1 \\\\\nk\n\\end{array}\\right) \\quad\\left(\\bmod 2^{n}\\right) .\n$$\n\nThis prepares ground for a proof of the required result by induction on $n$. The base case $n=1$ is obvious. Assume the assertion is true for $n-1$ and pass to $n$, denoting $a_{k}=\\left(\\begin{array}{c}2^{n-1}-1 \\\\ k\\end{array}\\right)$, $b_{m}=\\left(\\begin{array}{c}2^{n}-1 \\\\ m\\end{array}\\right)$. The induction hypothesis is that all the numbers $a_{k}\\left(0 \\leq k<2^{n-2}\\right)$ are distinct $\\left(\\bmod 2^{n-1}\\right)$; the claim is that all the numbers $b_{m}\\left(0 \\leq m<2^{n-1}\\right)$ are distinct $\\left(\\bmod 2^{n}\\right)$.\n\nThe congruence relations (1) are restated as\n\n$$\nb_{2 k} \\equiv(-1)^{k} a_{k} \\equiv-b_{2 k+1} \\quad\\left(\\bmod 2^{n}\\right)\n\\tag{2}\n$$\n\nShifting the exponent in the first relation of (1) from $n$ to $n-1$ we also have the congruence $a_{2 i+1} \\equiv-a_{2 i}\\left(\\bmod 2^{n-1}\\right)$. We hence conclude:\n$$\n\\text{If, for some j, }k<2^{n-2}, a_{k} \\equiv-a_{j}\\left(\\bmod 2^{n-1}\\right)\\text{, then }\\{j, k\\}=\\{2 i, 2 i+1\\}\\text{ for some i.}\\tag{3}\n$$\n\nThis is so because in the sequence $\\left(a_{k}: k<2^{n-2}\\right)$ each term $a_{j}$ is complemented to $0\\left(\\bmod 2^{n-1}\\right)$ by only one other term $a_{k}$, according to the induction hypothesis.\n\nFrom $(2)$ we see that $b_{4 i} \\equiv a_{2 i}$ and $b_{4 i+3} \\equiv a_{2 i+1}\\left(\\bmod 2^{n}\\right)$. Let\n\n$$\nM=\\left\\{m: 0 \\leq m<2^{n-1}, m \\equiv 0 \\text { or } 3(\\bmod 4)\\right\\}, \\quad L=\\left\\{l: 0 \\leq l<2^{n-1}, l \\equiv 1 \\text { or } 2(\\bmod 4)\\right\\}\n$$\n\nThe last two congruences take on the unified form\n\n$$\nb_{m} \\equiv a_{\\lfloor m / 2\\rfloor} \\quad\\left(\\bmod 2^{n}\\right) \\quad \\text { for all } \\quad m \\in M\\tag{4}\n$$\n\nThus all the numbers $b_{m}$ for $m \\in M$ are distinct $\\left(\\bmod 2^{n}\\right)$ because so are the numbers $a_{k}$ (they are distinct $\\left(\\bmod 2^{n-1}\\right)$, hence also $\\left(\\bmod 2^{n}\\right)$ ).\n\nEvery $l \\in L$ is paired with a unique $m \\in M$ into a pair of the form $\\{2 k, 2 k+1\\}$. So (2) implies that also all the $b_{l}$ for $l \\in L$ are distinct $\\left(\\bmod 2^{n}\\right)$. It remains to eliminate the possibility that $b_{m} \\equiv b_{l}\\left(\\bmod 2^{n}\\right)$ for some $m \\in M, l \\in L$.\n\nSuppose that such a situation occurs. Let $m^{\\prime} \\in M$ be such that $\\left\\{m^{\\prime}, l\\right\\}$ is a pair of the form $\\{2 k, 2 k+1\\}$, so that $(\\operatorname{see}(2)) b_{m^{\\prime}} \\equiv-b_{l}\\left(\\bmod 2^{n}\\right)$. Hence $b_{m^{\\prime}} \\equiv-b_{m}\\left(\\bmod 2^{n}\\right)$. Since both $m^{\\prime}$ and $m$ are in $M$, we have by (4) $b_{m^{\\prime}} \\equiv a_{j}, b_{m} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$ for $j=\\left\\lfloor m^{\\prime} / 2\\right\\rfloor, k=\\lfloor m / 2\\rfloor$.\n\nThen $a_{j} \\equiv-a_{k}\\left(\\bmod 2^{n}\\right)$. Thus, according to (3), $j=2 i, k=2 i+1$ for some $i$ (or vice versa). The equality $a_{2 i+1} \\equiv-a_{2 i}\\left(\\bmod 2^{n}\\right)$ now means that $\\left(\\begin{array}{c}2^{n-1}-1 \\\\ 2 i\\end{array}\\right)+\\left(\\begin{array}{c}2^{n-1}-1 \\\\ 2 i+1\\end{array}\\right) \\equiv 0\\left(\\bmod 2^{n}\\right)$. However, the sum on the left is equal to $\\left(\\begin{array}{l}2^{n-1} \\\\ 2 i+1\\end{array}\\right)$. A number of this form cannot be divisible by $2^{n}$. This is a contradiction which concludes the induction step and proves the result.' 'We again proceed by induction, writing for brevity $N=2^{n-1}$ and keeping notation $a_{k}=\\left(\\begin{array}{c}N-1 \\\\ k\\end{array}\\right), b_{m}=\\left(\\begin{array}{c}2 N-1 \\\\ m\\end{array}\\right)$. Assume that the result holds for the sequence $\\left(a_{0}, a_{1}, a_{2}, \\ldots, a_{N / 2-1}\\right)$. In view of the symmetry $a_{N-1-k}=a_{k}$ this sequence is a permutation of $\\left(a_{0}, a_{2}, a_{4}, \\ldots, a_{N-2}\\right)$. So the induction hypothesis says that this latter sequence, taken $(\\bmod N)$, is a permutation of $(1,3,5, \\ldots, N-1)$. Similarly, the induction claim is that $\\left(b_{0}, b_{2}, b_{4}, \\ldots, b_{2 N-2}\\right)$, taken $(\\bmod 2 N)$, is a permutation of $(1,3,5, \\ldots, 2 N-1)$.\n\nIn place of the congruence relations (2) we now use the following ones,\n\n$$\nb_{4 i} \\equiv a_{2 i} \\quad(\\bmod N) \\quad \\text { and } \\quad b_{4 i+2} \\equiv b_{4 i}+N \\quad(\\bmod 2 N)\\tag{5}\n$$\n\nGiven this, the conclusion is immediate: the first formula of (5) together with the induction hypothesis tells us that $\\left(b_{0}, b_{4}, b_{8}, \\ldots, b_{2 N-4}\\right)(\\bmod N)$ is a permutation of $(1,3,5, \\ldots, N-1)$. Then the second formula of $(5)$ shows that $\\left(b_{2}, b_{6}, b_{10}, \\ldots, b_{2 N-2}\\right)(\\bmod N)$ is exactly the same permutation; moreover, this formula distinguishes $(\\bmod 2 N)$ each $b_{4 i}$ from $b_{4 i+2}$.\n\nConsequently, these two sequences combined represent $(\\bmod 2 N)$ a permutation of the sequence $(1,3,5, \\ldots, N-1, N+1, N+3, N+5, \\ldots, N+N-1)$, and this is precisely the induction claim.\n\nNow we prove formulas (5); we begin with the second one. Since $b_{m+1}=b_{m} \\cdot \\frac{2 N-m-1}{m+1}$,\n\n$$\nb_{4 i+2}=b_{4 i} \\cdot \\frac{2 N-4 i-1}{4 i+1} \\cdot \\frac{2 N-4 i-2}{4 i+2}=b_{4 i} \\cdot \\frac{2 N-4 i-1}{4 i+1} \\cdot \\frac{N-2 i-1}{2 i+1} .\n$$\n\nThe desired congruence $b_{4 i+2} \\equiv b_{4 i}+N$ may be multiplied by the odd number $(4 i+1)(2 i+1)$, giving rise to a chain of successively equivalent congruences:\n\n$$\n\\begin{aligned}\nb_{4 i}(2 N-4 i-1)(N-2 i-1) & \\equiv\\left(b_{4 i}+N\\right)(4 i+1)(2 i+1) & & (\\bmod 2 N) \\\\\nb_{4 i}(2 i+1-N) & \\equiv\\left(b_{4 i}+N\\right)(2 i+1) & & (\\bmod 2 N) \\\\\n\\left(b_{4 i}+2 i+1\\right) N & \\equiv 0 & & (\\bmod 2 N)\n\\end{aligned}\n$$\n\nand the last one is satisfied, as $b_{4 i}$ is odd. This settles the second relation in (5).\n\nThe first one is proved by induction on $i$. It holds for $i=0$. Assume $b_{4 i} \\equiv a_{2 i}(\\bmod 2 N)$ and consider $i+1$ :\n\n$$\nb_{4 i+4}=b_{4 i+2} \\cdot \\frac{2 N-4 i-3}{4 i+3} \\cdot \\frac{2 N-4 i-4}{4 i+4} ; \\quad a_{2 i+2}=a_{2 i} \\cdot \\frac{N-2 i-1}{2 i+1} \\cdot \\frac{N-2 i-2}{2 i+2}\n$$\n\nBoth expressions have the fraction $\\frac{N-2 i-2}{2 i+2}$ as the last factor. Since $2 i+22 n+\\sqrt{10 n}$.\n\nLet $n=(p-1) / 2-\\ell$ where $\\ell \\geq 0$. Then $n^{2} \\equiv-1(\\bmod p)$ gives\n\n$$\n\\left(\\frac{p-1}{2}-\\ell\\right)^{2} \\equiv-1 \\quad(\\bmod p) \\quad \\text { or } \\quad(2 \\ell+1)^{2}+4 \\equiv 0 \\quad(\\bmod p)\n$$\n\nThus $(2 \\ell+1)^{2}+4=r p$ for some $r \\geq 0$. As $(2 \\ell+1)^{2} \\equiv 1 \\equiv p(\\bmod 8)$, we have $r \\equiv 5(\\bmod 8)$, so that $r \\geq 5$. Hence $(2 \\ell+1)^{2}+4 \\geq 5 p$, implying $\\ell \\geq(\\sqrt{5 p-4}-1) / 2$. Set $\\sqrt{5 p-4}=u$ for clarity; then $\\ell \\geq(u-1) / 2$. Therefore\n\n$$\nn=\\frac{p-1}{2}-\\ell \\leq \\frac{1}{2}(p-u)\n$$\n\nCombined with $p=\\left(u^{2}+4\\right) / 5$, this leads to $u^{2}-5 u-10 n+4 \\geq 0$. Solving this quadratic inequality with respect to $u \\geq 0$ gives $u \\geq(5+\\sqrt{40 n+9}) / 2$. So the estimate $n \\leq(p-u) / 2$ leads to\n\n$$\np \\geq 2 n+u \\geq 2 n+\\frac{1}{2}(5+\\sqrt{40 n+9})>2 n+\\sqrt{10 n}\n$$\n\nSince there are infinitely many primes of the form $8 k+1$, it follows easily that there are also infinitely many $n$ with the stated property.'] [] Text-only Competition True Theorem proof Number Theory Math English 357 "Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively. Here $X+Y$ denotes the set $\{x+y \mid x \in X, y \in Y\}$, for $X, Y \subseteq \mathbb{Z}$ and $X, Y \subseteq \mathbb{Q}$. a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint? b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint?" "['a) The residue classes modulo 3 yield such a partition:\n\n$$\nA=\\{3 k \\mid k \\in \\mathbb{Z}\\}, \\quad B=\\{3 k+1 \\mid k \\in \\mathbb{Z}\\}, \\quad C=\\{3 k+2 \\mid k \\in \\mathbb{Z}\\}\n$$\n\nb) The answer is no. Suppose that $\\mathbb{Q}$ can be partitioned into non-empty subsets $A, B, C$ as stated. Note that for all $a \\in A, b \\in B, c \\in C$ one has\n\n$$\na+b-c \\in C, \\quad b+c-a \\in A, \\quad c+a-b \\in B\\tag{1}\n$$\n\nIndeed $a+b-c \\notin A$ as $(A+B) \\cap(A+C)=\\emptyset$, and similarly $a+b-c \\notin B$, hence $a+b-c \\in C$. The other two relations follow by symmetry. Hence $A+B \\subset C+C, B+C \\subset A+A, C+A \\subset B+B$.\n\nThe opposite inclusions also hold. Let $a, a^{\\prime} \\in A$ and $b \\in B, c \\in C$ be arbitrary. By (1) $a^{\\prime}+c-b \\in B$, and since $a \\in A, c \\in C$, we use (1) again to obtain\n\n$$\na+a^{\\prime}-b=a+\\left(a^{\\prime}+c-b\\right)-c \\in C .\n$$\n\nSo $A+A \\subset B+C$ and likewise $B+B \\subset C+A, C+C \\subset A+B$. In summary\n\n$$\nB+C=A+A, \\quad C+A=B+B, \\quad A+B=C+C .\n$$\n\nFurthermore suppose that $0 \\in A$ without loss of generality. Then $B=\\{0\\}+B \\subset A+B$ and $C=\\{0\\}+C \\subset A+C$. So, since $B+C$ is disjoint with $A+B$ and $A+C$, it is also disjoint with $B$ and $C$. Hence $B+C$ is contained in $\\mathbb{Z} \\backslash(B \\cup C)=A$. Because $B+C=A+A$, we obtain $A+A \\subset A$. On the other hand $A=\\{0\\}+A \\subset A+A$, implying $A=A+A=B+C$.\n\nTherefore $A+B+C=A+A+A=A$, and now $B+B=C+A$ and $C+C=A+B$ yield $B+B+B=A+B+C=A, C+C+C=A+B+C=A$. In particular if $r \\in \\mathbb{Q}=A \\cup B \\cup C$ is arbitrary then $3 r \\in A$.\n\nHowever such a conclusion is impossible. Take any $b \\in B(B \\neq \\emptyset)$ and let $r=b / 3 \\in \\mathbb{Q}$. Then $b=3 r \\in A$ which is a contradiction.' 'We prove that the example for $\\mathbb{Z}$ from the first solution is unique, and then use this fact to solve part $b)$.\n\nLet $\\mathbb{Z}=A \\cup B \\cup C$ be a partition of $\\mathbb{Z}$ with $A, B, C \\neq \\emptyset$ and $A+B, B+C, C+A$ disjoint. We need the relations (1) which clearly hold for $\\mathbb{Z}$. Fix two consecutive integers from different sets, say $b \\in B$ and $c=b+1 \\in C$. For every $a \\in A$ we have, in view of (1), $a-1=a+b-c \\in C$ and $a+1=a+c-b \\in B$. So every $a \\in A$ is preceded by a number from $C$ and followed by a number from $B$.\n\nIn particular there are pairs of the form $c, c+1$ with $c \\in C, c+1 \\in A$. For such a pair and any $b \\in B$ analogous reasoning shows that each $b \\in B$ is preceded by a number from $A$ and followed by a number from $C$. There are also pairs $b, b-1$ with $b \\in B, b-1 \\in A$. We use them in a similar way to prove that each $c \\in C$ is preceded by a number from $B$ and followed by a number from $A$.\n\nBy putting the observations together we infer that $A, B, C$ are the three congruence classes modulo 3. Observe that all multiples of 3 are in the set of the partition that contains 0 .\n\n\n\nNow we turn to part b). Suppose that there is a partition of $\\mathbb{Q}$ with the given properties. Choose three rationals $r_{i}=p_{i} / q_{i}$ from the three sets $A, B, C, i=1,2,3$, and set $N=3 q_{1} q_{2} q_{3}$.\n\nLet $S \\subset \\mathbb{Q}$ be the set of fractions with denominators $N$ (irreducible or not). It is obtained through multiplication of every integer by the constant $1 / N$, hence closed under sums and differences. Moreover, if we identify each $k \\in \\mathbb{Z}$ with $k / N \\in S$ then $S$ is essentially the set $\\mathbb{Z}$ with respect to addition. The numbers $r_{i}$ belong to $S$ because\n\n$$\nr_{1}=\\frac{3 p_{1} q_{2} q_{3}}{N}, \\quad r_{2}=\\frac{3 p_{2} q_{3} q_{1}}{N}, \\quad r_{3}=\\frac{3 p_{3} q_{1} q_{2}}{N}\n$$\n\nThe partition $\\mathbb{Q}=A \\cup B \\cup C$ of $\\mathbb{Q}$ induces a partition $S=A^{\\prime} \\cup B^{\\prime} \\cup C^{\\prime}$ of $S$, with $A^{\\prime}=A \\cap S$, $B^{\\prime}=B \\cap S, C^{\\prime}=C \\cap S$. Clearly $A^{\\prime}+B^{\\prime}, B^{\\prime}+C^{\\prime}, C^{\\prime}+A^{\\prime}$ are disjoint, so this partition has the properties we consider.\n\nBy the uniqueness of the example for $\\mathbb{Z}$ the sets $A^{\\prime}, B^{\\prime}, C^{\\prime}$ are the congruence classes modulo 3 , multiplied by $1 / N$. Also all multiples of $3 / N$ are in the same set, $A^{\\prime}, B^{\\prime}$ or $C^{\\prime}$. This holds for $r_{1}, r_{2}, r_{3}$ in particular as they are all multiples of $3 / N$. However $r_{1}, r_{2}, r_{3}$ are in different sets $A^{\\prime}, B^{\\prime}, C^{\\prime}$ since they were chosen from different sets $A, B, C$. The contradiction ends the proof.']" [] Text-only Competition True Theorem proof Algebra Math English 358 "Let $a_{2}, \ldots, a_{n}$ be $n-1$ positive real numbers, where $n \geq 3$, such that $a_{2} a_{3} \cdots a_{n}=1$. Prove that $$ \left(1+a_{2}\right)^{2}\left(1+a_{3}\right)^{3} \cdots\left(1+a_{n}\right)^{n}>n^{n} . $$" ['The substitution $a_{2}=\\frac{x_{2}}{x_{1}}, a_{3}=\\frac{x_{3}}{x_{2}}, \\ldots, a_{n}=\\frac{x_{1}}{x_{n-1}}$ transforms the original problem into the inequality\n\n$$\n\\left(x_{1}+x_{2}\\right)^{2}\\left(x_{2}+x_{3}\\right)^{3} \\cdots\\left(x_{n-1}+x_{1}\\right)^{n}>n^{n} x_{1}^{2} x_{2}^{3} \\cdots x_{n-1}^{n}\\tag{*}\n$$\n\nfor all $x_{1}, \\ldots, x_{n-1}>0$. To prove this, we use the AM-GM inequality for each factor of the left-hand side as follows:\n\n$$\n\\begin{array}{rlcll}\n\\left(x_{1}+x_{2}\\right)^{2} & & & & \\geq 2^{2} x_{1} x_{2} \\\\\n\\left(x_{2}+x_{3}\\right)^{3} & = & \\left(2\\left(\\frac{x_{2}}{2}\\right)+x_{3}\\right)^{3} & & \\geq 3^{3}\\left(\\frac{x_{2}}{2}\\right)^{2} x_{3} \\\\\n\\left(x_{3}+x_{4}\\right)^{4} & = & \\left(3\\left(\\frac{x_{3}}{3}\\right)+x_{4}\\right)^{4} & & \\geq 4^{4}\\left(\\frac{x_{3}}{3}\\right)^{3} x_{4} \\\\\n& \\vdots & \\vdots && \\vdots \\\\\n\\left(x_{n-1}+x_{1}\\right)^{n} & = & \\left((n-1)\\left(\\frac{x_{n-1}}{n-1}\\right)+x_{1}\\right)^{n} && \\geq n^{n}\\left(\\frac{x_{n-1}}{n-1}\\right)^{n-1} x_{1}\n\\end{array}\n$$\n\nMultiplying these inequalities together gives $(*)$, with inequality sign $\\geq$ instead of $>$. However for the equality to occur it is necessary that $x_{1}=x_{2}, x_{2}=2 x_{3}, \\ldots, x_{n-1}=(n-1) x_{1}$, implying $x_{1}=(n-1) ! x_{1}$. This is impossible since $x_{1}>0$ and $n \\geq 3$. Therefore the inequality is strict.'] [] Text-only Competition True Theorem proof Algebra Math English 359 Let $f$ and $g$ be two nonzero polynomials with integer coefficients and $\operatorname{deg} f>\operatorname{deg} g$. Suppose that for infinitely many primes $p$ the polynomial $p f+g$ has a rational root. Prove that $f$ has a rational root. "['Since $\\operatorname{deg} f>\\operatorname{deg} g$, we have $|g(x) / f(x)|<1$ for sufficiently large $x$; more precisely, there is a real number $R$ such that $|g(x) / f(x)|<1$ for all $x$ with $|x|>R$. Then for all such $x$ and all primes $p$ we have\n\n$$\n|p f(x)+g(x)| \\geq|f(x)|\\left(p-\\frac{|g(x)|}{|f(x)|}\\right)>0\n$$\n\nHence all real roots of the polynomials $p f+g$ lie in the interval $[-R, R]$.\n\nLet $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{0}$ and $g(x)=b_{m} x^{m}+b_{m-1} x^{m-1}+\\cdots+b_{0}$ where $n>m, a_{n} \\neq 0$ and $b_{m} \\neq 0$. Upon replacing $f(x)$ and $g(x)$ by $a_{n}^{n-1} f\\left(x / a_{n}\\right)$ and $a_{n}^{n-1} g\\left(x / a_{n}\\right)$ respectively, we reduce the problem to the case $a_{n}=1$. In other words one can assume that $f$ is monic. Then the leading coefficient of $p f+g$ is $p$, and if $r=u / v$ is a rational root of $p f+g$ with $(u, v)=1$ and $v>0$, then either $v=1$ or $v=p$.\n\nFirst consider the case when $v=1$ infinitely many times. If $v=1$ then $|u| \\leq R$, so there are only finitely many possibilities for the integer $u$. Therefore there exist distinct primes $p$ and $q$ for which we have the same value of $u$. Then the polynomials $p f+g$ and $q f+g$ share this root, implying $f(u)=g(u)=0$. So in this case $f$ and $g$ have an integer root in common.\n\nNow suppose that $v=p$ infinitely many times. By comparing the exponent of $p$ in the denominators of $p f(u / p)$ and $g(u / p)$ we get $m=n-1$ and $p f(u / p)+g(u / p)=0$ reduces to an equation of the form\n\n$$\n\\left(u^{n}+a_{n-1} p u^{n-1}+\\ldots+a_{0} p^{n}\\right)+\\left(b_{n-1} u^{n-1}+b_{n-2} p u^{n-2}+\\ldots+b_{0} p^{n-1}\\right)=0 .\n$$\n\nThe equation above implies that $u^{n}+b_{n-1} u^{n-1}$ is divisible by $p$ and hence, since $(u, p)=1$, we have $u+b_{n-1}=p k$ with some integer $k$. On the other hand all roots of $p f+g$ lie in the interval $[-R, R]$, so that\n\n$$\n\\begin{gathered}\n\\frac{\\left|p k-b_{n-1}\\right|}{p}=\\frac{|u|}{p}0$ such that $f^{2 k}(n)=n+k$ is in $S$. Clearly $f$ maps $S$ into itself; moreover $f$ is injective on $S$. Indeed if $f^{i}(1)=f^{j}(1)$ with $i \\neq j$ then the values $f^{m}(1)$ start repeating periodically from some point on, and $S$ would be finite.\n\nDefine $g: S \\rightarrow S$ by $g(n)=f^{2 k_{n}}(n)=n+k_{n}$. We prove that $g$ is injective too. Suppose that $g(a)=g(b)$ with $ak_{b}$. So, since $f$ is injective on $S$, we obtain\n\n$$\nf^{2\\left(k_{a}-k_{b}\\right)}(a)=b=a+\\left(k_{a}-k_{b}\\right)\n$$\n\nHowever this contradicts the minimality of $k_{a}$ as $0n$ for $n \\in S$, so $T$ is non-empty. For each $t \\in T$ denote $C_{t}=\\left\\{t, g(t), g^{2}(t), \\ldots\\right\\}$; call $C_{t}$ the chain starting at $t$. Observe that distinct chains are disjoint because $g$ is injective. Each $n \\in S \\backslash T$ has the form $n=g\\left(n^{\\prime}\\right)$ with $n^{\\prime}k$, i. e. $k_{n}>k$. In conclusion $k_{1}, k_{2}, \\ldots$ is unbounded.'] [] Text-only Competition True Theorem proof Algebra Math English 361 "We say that a function $f: \mathbb{R}^{k} \rightarrow \mathbb{R}$ is a metapolynomial if, for some positive integers $m$ and $n$, it can be represented in the form $$ f\left(x_{1}, \ldots, x_{k}\right)=\max _{i=1, \ldots, m} \min _{j=1, \ldots, n} P_{i, j}\left(x_{1}, \ldots, x_{k}\right) $$ where $P_{i, j}$ are multivariate polynomials. Prove that the product of two metapolynomials is also a metapolynomial." ['We use the notation $f(x)=f\\left(x_{1}, \\ldots, x_{k}\\right)$ for $x=\\left(x_{1}, \\ldots, x_{k}\\right)$ and $[m]=\\{1,2, \\ldots, m\\}$. Observe that if a metapolynomial $f(x)$ admits a representation like the one in the statement for certain positive integers $m$ and $n$, then they can be replaced by any $m^{\\prime} \\geq m$ and $n^{\\prime} \\geq n$. For instance, if we want to replace $m$ by $m+1$ then it is enough to define $P_{m+1, j}(x)=P_{m, j}(x)$ and note that repeating elements of a set do not change its maximum nor its minimum. So one can assume that any two metapolynomials are defined with the same $m$ and $n$. We reserve letters $P$ and $Q$ for polynomials, so every function called $P, P_{i, j}, Q, Q_{i, j}, \\ldots$ is a polynomial function.\n\nWe start with a lemma that is useful to change expressions of the form $\\min \\max f_{i, j}$ to ones of the form $\\max \\min g_{i, j}$.\n\nLemma. Let $\\left\\{a_{i, j}\\right\\}$ be real numbers, for all $i \\in[m]$ and $j \\in[n]$. Then\n\n$$\n\\min _{i \\in[m]} \\max _{j \\in[n]} a_{i, j}=\\max _{j_{1}, \\ldots, j_{m} \\in[n]} \\min _{i \\in[m]} a_{i, j_{i}}\n$$\n\nwhere the max in the right-hand side is over all vectors $\\left(j_{1}, \\ldots, j_{m}\\right)$ with $j_{1}, \\ldots, j_{m} \\in[n]$.\n\nProof. We can assume for all $i$ that $a_{i, n}=\\max \\left\\{a_{i, 1}, \\ldots, a_{i, n}\\right\\}$ and $a_{m, n}=\\min \\left\\{a_{1, n}, \\ldots, a_{m, n}\\right\\}$. The left-hand side is $=a_{m, n}$ and hence we need to prove the same for the right-hand side. If $\\left(j_{1}, j_{2}, \\ldots, j_{m}\\right)=(n, n, \\ldots, n)$ then $\\min \\left\\{a_{1, j_{1}}, \\ldots, a_{m, j_{m}}\\right\\}=\\min \\left\\{a_{1, n}, \\ldots, a_{m, n}\\right\\}=a_{m, n}$ which implies that the right-hand side is $\\geq a_{m, n}$. It remains to prove the opposite inequality and this is equivalent to $\\min \\left\\{a_{1, j_{1}}, \\ldots, a_{m, j_{m}}\\right\\} \\leq a_{m, n}$ for all possible $\\left(j_{1}, j_{2}, \\ldots, j_{m}\\right)$. This is true because $\\min \\left\\{a_{1, j_{1}}, \\ldots, a_{m, j_{m}}\\right\\} \\leq a_{m, j_{m}} \\leq a_{m, n}$.\n\nWe need to show that the family $\\mathcal{M}$ of metapolynomials is closed under multiplication, but it turns out easier to prove more: that it is also closed under addition, maxima and minima.\n\nFirst we prove the assertions about the maxima and the minima. If $f_{1}, \\ldots, f_{r}$ are metapolynomials, assume them defined with the same $m$ and $n$. Then\n\n$$\nf=\\max \\left\\{f_{1}, \\ldots, f_{r}\\right\\}=\\max \\left\\{\\max _{i \\in[m]} \\min _{j \\in[n]} P_{i, j}^{1}, \\ldots, \\max _{i \\in[m]} \\min _{j \\in[n]} P_{i, j}^{r}\\right\\}=\\max _{s \\in[r], i \\in[m]} \\min _{j \\in[n]} P_{i, j}^{s}\n$$\n\nIt follows that $f=\\max \\left\\{f_{1}, \\ldots, f_{r}\\right\\}$ is a metapolynomial. The same argument works for the minima, but first we have to replace $\\min \\max$ by $\\max \\min$, and this is done via the lemma.\n\nAnother property we need is that if $f=\\max \\min P_{i, j}$ is a metapolynomial then so is $-f$. Indeed, $-f=\\min \\left(-\\min P_{i, j}\\right)=\\min \\max P_{i, j}$.\n\nTo prove $\\mathcal{M}$ is closed under addition let $f=\\max \\min P_{i, j}$ and $g=\\max \\min Q_{i, j}$. Then\n\n$$\n\\begin{gathered}\nf(x)+g(x)=\\max _{i \\in[m]} \\min _{j \\in[n]} P_{i, j}(x)+\\max _{i \\in[m]} \\min _{j \\in[n]} Q_{i, j}(x) \\\\\n=\\max _{i_{1}, i_{2} \\in[m]}\\left(\\min _{j \\in[n]} P_{i_{1}, j}(x)+\\min _{j \\in[n]} Q_{i_{2}, j}(x)\\right)=\\max _{i_{1}, i_{2} \\in[m]} \\min _{j_{1}, j_{2} \\in[n]}\\left(P_{i_{1}, j_{1}}(x)+Q_{i_{2}, j_{2}}(x)\\right),\n\\end{gathered}\n$$\n\nand hence $f(x)+g(x)$ is a metapolynomial.\n\nWe proved that $\\mathcal{M}$ is closed under sums, maxima and minima, in particular any function that can be expressed by sums, max, min, polynomials or even metapolynomials is in $\\mathcal{M}$.\n\nWe would like to proceed with multiplication along the same lines like with addition, but there is an essential difference. In general the product of the maxima of two sets is not equal\n\n\n\nto the maximum of the product of the sets. We need to deal with the fact that $ay$ and $x$ is to the left of $y$, and replaces the pair $(x, y)$ by either $(y+1, x)$ or $(x-1, x)$. Prove that she can perform only finitely many such iterations. "['Note first that the allowed operation does not change the maximum $M$ of the initial sequence. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be the numbers obtained at some point of the process. Consider the sum\n\n$$\nS=a_{1}+2 a_{2}+\\cdots+n a_{n} .\n$$\n\nWe claim that $S$ increases by a positive integer amount with every operation. Let the operation replace the pair $\\left(a_{i}, a_{i+1}\\right)$ by a pair $\\left(c, a_{i}\\right)$, where $a_{i}>a_{i+1}$ and $c=a_{i+1}+1$ or $c=a_{i}-1$. Then the new and the old value of $S$ differ by $d=\\left(i c+(i+1) a_{i}\\right)-\\left(i a_{i}+(i+1) a_{i+1}\\right)=a_{i}-a_{i+1}+i\\left(c-a_{i+1}\\right)$. The integer $d$ is positive since $a_{i}-a_{i+1} \\geq 1$ and $c-a_{i+1} \\geq 0$.\n\nOn the other hand $S \\leq(1+2+\\cdots+n) M$ as $a_{i} \\leq M$ for all $i=1, \\ldots, n$. Since $S$ increases by at least 1 at each step and never exceeds the constant $(1+2+\\cdots+n) M$, the process stops after a finite number of iterations.' 'Like in the first solution note that the operations do not change the maximum $M$ of the initial sequence. Now consider the reverse lexicographical order for $n$-tuples of integers. We say that $\\left(x_{1}, \\ldots, x_{n}\\right)<\\left(y_{1}, \\ldots, y_{n}\\right)$ if $x_{n}y$ and $y \\leq a \\leq x$, we see that $s_{i}$ decreases by at least 1 . This concludes the proof.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 363 "The columns and the rows of a $3 n \times 3 n$ square board are numbered $1,2, \ldots, 3 n$. Every square $(x, y)$ with $1 \leq x, y \leq 3 n$ is colored asparagus, byzantium or citrine according as the modulo 3 remainder of $x+y$ is 0,1 or 2 respectively. One token colored asparagus, byzantium or citrine is placed on each square, so that there are $3 n^{2}$ tokens of each color. Suppose that one can permute the tokens so that each token is moved to a distance of at most $d$ from its original position, each asparagus token replaces a byzantium token, each byzantium token replaces a citrine token, and each citrine token replaces an asparagus token. Prove that it is possible to permute the tokens so that each token is moved to a distance of at most $d+2$ from its original position, and each square contains a token with the same color as the square." "[""Without loss of generality it suffices to prove that the A-tokens can be moved to distinct $\\mathrm{A}$-squares in such a way that each $\\mathrm{A}$-token is moved to a distance at most $d+2$ from its original place. This means we need a perfect matching between the $3 n^{2} \\mathrm{~A}$-squares and the $3 n^{2} \\mathrm{~A}$-tokens such that the distance in each pair of the matching is at most $d+2$.\n\nTo find the matching, we construct a bipartite graph. The A-squares will be the vertices in one class of the graph; the vertices in the other class will be the A-tokens.\n\nSplit the board into $3 \\times 1$ horizontal triminos; then each trimino contains exactly one Asquare. Take a permutation $\\pi$ of the tokens which moves A-tokens to B-tokens, B-tokens to C-tokens, and C-tokens to A-tokens, in each case to a distance at most $d$. For each A-square $S$, and for each A-token $T$, connect $S$ and $T$ by an edge if $T, \\pi(T)$ or $\\pi^{-1}(T)$ is on the trimino containing $S$. We allow multiple edges; it is even possible that the same square and the same token are connected with three edges. Obviously the lengths of the edges in the graph do not exceed $d+2$. By length of an edge we mean the distance between the A-square and the A-token it connects.\n\nEach A-token $T$ is connected with the three A-squares whose triminos contain $T, \\pi(T)$ and $\\pi^{-1}(T)$. Therefore in the graph all tokens are of degree 3 . We show that the same is true for the A-squares. Let $S$ be an arbitrary A-square, and let $T_{1}, T_{2}, T_{3}$ be the three tokens on the trimino containing $S$. For $i=1,2,3$, if $T_{i}$ is an A-token, then $S$ is connected with $T_{i}$; if $T_{i}$ is a B-token then $S$ is connected with $\\pi^{-1}\\left(T_{i}\\right)$; finally, if $T_{i}$ is a C-token then $S$ is connected with $\\pi\\left(T_{i}\\right)$. Hence in the graph the A-squares also are of degree 3.\n\nSince the A-squares are of degree 3 , from every set $\\mathcal{S}$ of A-squares exactly $3|\\mathcal{S}|$ edges start. These edges end in at least $|\\mathcal{S}|$ tokens because the A-tokens also are of degree 3. Hence every set $\\mathcal{S}$ of A-squares has at least $|\\mathcal{S}|$ neighbors among the A-tokens.\n\nTherefore, by HALL's marriage theorem, the graph contains a perfect matching between the two vertex classes. So there is a perfect matching between the A-squares and A-tokens with edges no longer than $d+2$. It follows that the tokens can be permuted as specified in the problem statement.""]" [] Text-only Competition True Theorem proof Combinatorics Math English 364 "Let $k$ and $n$ be fixed positive integers. In the liar's guessing game, Amy chooses integers $x$ and $N$ with $1 \leq x \leq N$. She tells Ben what $N$ is, but not what $x$ is. Ben may then repeatedly ask Amy whether $x \in S$ for arbitrary sets $S$ of integers. Amy will always answer with yes or no, but she might lie. The only restriction is that she can lie at most $k$ times in a row. After he has asked as many questions as he wants, Ben must specify a set of at most $n$ positive integers. If $x$ is in this set he wins; otherwise, he loses. Prove that: If $n \geq 2^{k}$ then Ben can always win." "['Consider an answer $A \\in\\{y e s, n o\\}$ to a question of the kind ""Is $x$ in the set $S$ ?"" We say that $A$ is inconsistent with a number $i$ if $A=y e s$ and $i \\notin S$, or if $A=n o$ and $i \\in S$. Observe that an answer inconsistent with the target number $x$ is a lie.\n\nSuppose that Ben has determined a set $T$ of size $m$ that contains $x$. This is true initially with $m=N$ and $T=\\{1,2, \\ldots, N\\}$. For $m>2^{k}$ we show how Ben can find a number $y \\in T$ that is different from $x$. By performing this step repeatedly he can reduce $T$ to be of size $2^{k} \\leq n$ and thus win.\n\nSince only the size $m>2^{k}$ of $T$ is relevant, assume that $T=\\left\\{0,1, \\ldots, 2^{k}, \\ldots, m-1\\right\\}$. Ben begins by asking repeatedly whether $x$ is $2^{k}$. If Amy answers no $k+1$ times in a row, one of these answers is truthful, and so $x \\neq 2^{k}$. Otherwise Ben stops asking about $2^{k}$ at the first answer yes. He then asks, for each $i=1, \\ldots, k$, if the binary representation of $x$ has a 0 in the $i$ th digit. Regardless of what the $k$ answers are, they are all inconsistent with a certain number $y \\in\\left\\{0,1, \\ldots, 2^{k}-1\\right\\}$. The preceding answer yes about $2^{k}$ is also inconsistent with $y$. Hence $y \\neq x$. Otherwise the last $k+1$ answers are not truthful, which is impossible.\n\nEither way, Ben finds a number in $T$ that is different from $x$, and the claim is proven.']" [] Text-only Competition True Theorem proof Combinatorics Math English 365 "Let $k$ and $n$ be fixed positive integers. In the liar's guessing game, Amy chooses integers $x$ and $N$ with $1 \leq x \leq N$. She tells Ben what $N$ is, but not what $x$ is. Ben may then repeatedly ask Amy whether $x \in S$ for arbitrary sets $S$ of integers. Amy will always answer with yes or no, but she might lie. The only restriction is that she can lie at most $k$ times in a row. After he has asked as many questions as he wants, Ben must specify a set of at most $n$ positive integers. If $x$ is in this set he wins; otherwise, he loses. Prove that: For sufficiently large $k$ there exist $n \geq 1.99^{k}$ such that Ben cannot guarantee a win." "['Consider an answer $A \\in\\{y e s, n o\\}$ to a question of the kind ""Is $x$ in the set $S$ ?"" We say that $A$ is inconsistent with a number $i$ if $A=y e s$ and $i \\notin S$, or if $A=n o$ and $i \\in S$. Observe that an answer inconsistent with the target number $x$ is a lie.\n\nWe prove that if $1<\\lambda<2$ and $n=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1$ then Ben cannot guarantee a win. To complete the proof, then it suffices to take $\\lambda$ such that $1.99<\\lambda<2$ and $k$ large enough so that\n\n$$\nn=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1 \\geq 1.99^{k}\n$$\n\nConsider the following strategy for Amy. First she chooses $N=n+1$ and $x \\in\\{1,2, \\ldots, n+1\\}$ arbitrarily. After every answer of hers Amy determines, for each $i=1,2, \\ldots, n+1$, the number $m_{i}$ of consecutive answers she has given by that point that are inconsistent with $i$. To decide on her next answer, she then uses the quantity\n\n$$\n\\phi=\\sum_{i=1}^{n+1} \\lambda^{m_{i}}\n$$\n\nNo matter what Ben\'s next question is, Amy chooses the answer which minimizes $\\phi$.\n\nWe claim that with this strategy $\\phi$ will always stay less than $\\lambda^{k+1}$. Consequently no exponent $m_{i}$ in $\\phi$ will ever exceed $k$, hence Amy will never give more than $k$ consecutive answers inconsistent with some $i$. In particular this applies to the target number $x$, so she will never lie more than $k$ times in a row. Thus, given the claim, Amy\'s strategy is legal. Since the strategy does not depend on $x$ in any way, Ben can make no deductions about $x$, and therefore he cannot guarantee a win.\n\nIt remains to show that $\\phi<\\lambda^{k+1}$ at all times. Initially each $m_{i}$ is 0 , so this condition holds in the beginning due to $1<\\lambda<2$ and $n=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1$. Suppose that $\\phi<\\lambda^{k+1}$ at some point, and Ben has just asked if $x \\in S$ for some set $S$. According as Amy answers yes or no, the new value of $\\phi$ becomes\n\n$$\n\\phi_{1}=\\sum_{i \\in S} 1+\\sum_{i \\notin S} \\lambda^{m_{i}+1} \\quad \\text { or } \\quad \\phi_{2}=\\sum_{i \\in S} \\lambda^{m_{i}+1}+\\sum_{i \\notin S} 1\n$$\n\n\n\nSince Amy chooses the option minimizing $\\phi$, the new $\\phi$ will equal $\\min \\left(\\phi_{1}, \\phi_{2}\\right)$. Now we have\n\n$$\n\\min \\left(\\phi_{1}, \\phi_{2}\\right) \\leq \\frac{1}{2}\\left(\\phi_{1}+\\phi_{2}\\right)=\\frac{1}{2}\\left(\\sum_{i \\in S}\\left(1+\\lambda^{m_{i}+1}\\right)+\\sum_{i \\notin S}\\left(\\lambda^{m_{i}+1}+1\\right)\\right)=\\frac{1}{2}(\\lambda \\phi+n+1)\n$$\n\nBecause $\\phi<\\lambda^{k+1}$, the assumptions $\\lambda<2$ and $n=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1$ lead to\n\n$$\n\\min \\left(\\phi_{1}, \\phi_{2}\\right)<\\frac{1}{2}\\left(\\lambda^{k+2}+(2-\\lambda) \\lambda^{k+1}\\right)=\\lambda^{k+1}\n$$\n\nThe claim follows, which completes the solution.']" [] Text-only Competition True Theorem proof Combinatorics Math English 366 There are given $2^{500}$ points on a circle labeled $1,2, \ldots, 2^{500}$ in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal. ['The proof is based on the following general fact.\n\nLemma. In a graph $G$ each vertex $v$ has degree $d_{v}$. Then $G$ contains an independent set $S$ of vertices such that $|S| \\geq f(G)$ where\n\n$$\nf(G)=\\sum_{v \\in G} \\frac{1}{d_{v}+1}\n$$\n\nProof. Induction on $n=|G|$. The base $n=1$ is clear. For the inductive step choose a vertex $v_{0}$ in $G$ of minimum degree $d$. Delete $v_{0}$ and all of its neighbors $v_{1}, \\ldots, v_{d}$ and also all edges with endpoints $v_{0}, v_{1}, \\ldots, v_{d}$. This gives a new graph $G^{\\prime}$. By the inductive assumption $G^{\\prime}$ contains an independent set $S^{\\prime}$ of vertices such that $\\left|S^{\\prime}\\right| \\geq f\\left(G^{\\prime}\\right)$. Since no vertex in $S^{\\prime}$ is a neighbor of $v_{0}$ in $G$, the set $S=S^{\\prime} \\cup\\left\\{v_{0}\\right\\}$ is independent in $G$.\n\nLet $d_{v}^{\\prime}$ be the degree of a vertex $v$ in $G^{\\prime}$. Clearly $d_{v}^{\\prime} \\leq d_{v}$ for every such vertex $v$, and also $d_{v_{i}} \\geq d$ for all $i=0,1, \\ldots, d$ by the minimal choice of $v_{0}$. Therefore\n\n$$\nf\\left(G^{\\prime}\\right)=\\sum_{v \\in G^{\\prime}} \\frac{1}{d_{v}^{\\prime}+1} \\geq \\sum_{v \\in G^{\\prime}} \\frac{1}{d_{v}+1}=f(G)-\\sum_{i=0}^{d} \\frac{1}{d_{v_{i}}+1} \\geq f(G)-\\frac{d+1}{d+1}=f(G)-1 .\n$$\n\nHence $|S|=\\left|S^{\\prime}\\right|+1 \\geq f\\left(G^{\\prime}\\right)+1 \\geq f(G)$, and the induction is complete.\n\nWe pass on to our problem. For clarity denote $n=2^{499}$ and draw all chords determined by the given $2 n$ points. Color each chord with one of the colors $3,4, \\ldots, 4 n-1$ according to the sum of the numbers at its endpoints. Chords with a common endpoint have different colors. For each color $c$ consider the following graph $G_{c}$. Its vertices are the chords of color $c$, and two chords are neighbors in $G_{c}$ if they intersect. Let $f\\left(G_{c}\\right)$ have the same meaning as in the lemma for all graphs $G_{c}$.\n\nEvery chord $\\ell$ divides the circle into two arcs, and one of them contains $m(\\ell) \\leq n-1$ given points. (In particular $m(\\ell)=0$ if $\\ell$ joins two consecutive points.) For each $i=0,1, \\ldots, n-2$ there are $2 n$ chords $\\ell$ with $m(\\ell)=i$. Such a chord has degree at most $i$ in the respective graph. Indeed let $A_{1}, \\ldots, A_{i}$ be all points on either arc determined by a chord $\\ell$ with $m(\\ell)=i$ and color $c$. Every $A_{j}$ is an endpoint of at most 1 chord colored $c, j=1, \\ldots, i$. Hence at most $i$ chords of color $c$ intersect $\\ell$.\n\nIt follows that for each $i=0,1, \\ldots, n-2$ the $2 n$ chords $\\ell$ with $m(\\ell)=i$ contribute at least $\\frac{2 n}{i+1}$ to the sum $\\sum_{c} f\\left(G_{c}\\right)$. Summation over $i=0,1, \\ldots, n-2$ gives\n\n$$\n\\sum_{c} f\\left(G_{c}\\right) \\geq 2 n \\sum_{i=1}^{n-1} \\frac{1}{i}\n$$\n\nBecause there are $4 n-3$ colors in all, averaging yields a color $c$ such that\n\n$$\nf\\left(G_{c}\\right) \\geq \\frac{2 n}{4 n-3} \\sum_{i=1}^{n-1} \\frac{1}{i}>\\frac{1}{2} \\sum_{i=1}^{n-1} \\frac{1}{i}\n$$\n\nBy the lemma there are at least $\\frac{1}{2} \\sum_{i=1}^{n-1} \\frac{1}{i}$ pairwise disjoint chords of color $c$, i. e. with the same sum $c$ of the pairs of numbers at their endpoints. It remains to show that $\\frac{1}{2} \\sum_{i=1}^{n-1} \\frac{1}{i} \\geq 100$ for $n=2^{499}$. Indeed we have\n\n$$\n\\sum_{i=1}^{n-1} \\frac{1}{i}>\\sum_{i=1}^{2^{400}} \\frac{1}{i}=1+\\sum_{k=1}^{400} \\sum_{i=2^{k-1+1}}^{2^{k}} \\frac{1}{i}>1+\\sum_{k=1}^{400} \\frac{2^{k-1}}{2^{k}}=201>200\n$$\n\nThis completes the solution.'] [] Text-only Competition True Theorem proof Combinatorics Math English 367 In the triangle $A B C$ the point $J$ is the center of the excircle opposite to $A$. This excircle is tangent to the side $B C$ at $M$, and to the lines $A B$ and $A C$ at $K$ and $L$ respectively. The lines $L M$ and $B J$ meet at $F$, and the lines $K M$ and $C J$ meet at $G$. Let $S$ be the point of intersection of the lines $A F$ and $B C$, and let $T$ be the point of intersection of the lines $A G$ and $B C$. Prove that $M$ is the midpoint of $S T$. ['Let $\\alpha=\\angle C A B, \\beta=\\angle A B C$ and $\\gamma=\\angle B C A$. The line $A J$ is the bisector of $\\angle C A B$, so $\\angle J A K=\\angle J A L=\\frac{\\alpha}{2}$. By $\\angle A K J=\\angle A L J=90^{\\circ}$ the points $K$ and $L$ lie on the circle $\\omega$ with diameter $A J$.\n\nThe triangle $K B M$ is isosceles as $B K$ and $B M$ are tangents to the excircle. Since $B J$ is the bisector of $\\angle K B M$, we have $\\angle M B J=90^{\\circ}-\\frac{\\beta}{2}$ and $\\angle B M K=\\frac{\\beta}{2}$. Likewise $\\angle M C J=90^{\\circ}-\\frac{\\gamma}{2}$ and $\\angle C M L=\\frac{\\gamma}{2}$. Also $\\angle B M F=\\angle C M L$, therefore\n\n$$\n\\angle L F J=\\angle M B J-\\angle B M F=\\left(90^{\\circ}-\\frac{\\beta}{2}\\right)-\\frac{\\gamma}{2}=\\frac{\\alpha}{2}=\\angle L A J\n$$\n\nHence $F$ lies on the circle $\\omega$. (By the angle computation, $F$ and $A$ are on the same side of $B C$.) Analogously, $G$ also lies on $\\omega$. Since $A J$ is a diameter of $\\omega$, we obtain $\\angle A F J=\\angle A G J=90^{\\circ}$.\n\n\n\nThe lines $A B$ and $B C$ are symmetric with respect to the external bisector $B F$. Because $A F \\perp B F$ and $K M \\perp B F$, the segments $S M$ and $A K$ are symmetric with respect to $B F$, hence $S M=A K$. By symmetry $T M=A L$. Since $A K$ and $A L$ are equal as tangents to the excircle, it follows that $S M=T M$, and the proof is complete.'] [] Text-only Competition True Theorem proof Geometry Math English 368 Let $A B C D$ be a cyclic quadrilateral whose diagonals $A C$ and $B D$ meet at $E$. The extensions of the sides $A D$ and $B C$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $E C G D$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $A D$. Prove that $D, H, F, G$ are concyclic. ['We show first that the triangles $F D G$ and $F B E$ are similar. Since $A B C D$ is cyclic, the triangles $E A B$ and $E D C$ are similar, as well as $F A B$ and $F C D$. The parallelogram $E C G D$ yields $G D=E C$ and $\\angle C D G=\\angle D C E$; also $\\angle D C E=\\angle D C A=\\angle D B A$ by inscribed angles. Therefore\n\n$$\n\\begin{gathered}\n\\angle F D G=\\angle F D C+\\angle C D G=\\angle F B A+\\angle A B D=\\angle F B E, \\\\\n\\frac{G D}{E B}=\\frac{C E}{E B}=\\frac{C D}{A B}=\\frac{F D}{F B}\n\\end{gathered}\n$$\n\nIt follows that $F D G$ and $F B E$ are similar, and so $\\angle F G D=\\angle F E B$.\n\n\n\nSince $H$ is the reflection of $E$ with respect to $F D$, we conclude that\n\n$$\n\\angle F H D=\\angle F E D=180^{\\circ}-\\angle F E B=180^{\\circ}-\\angle F G D .\n$$\n\nThis proves that $D, H, F, G$ are concyclic.'] [] Text-only Competition True Theorem proof Geometry Math English 369 In an acute triangle $A B C$ the points $D, E$ and $F$ are the feet of the altitudes through $A$, $B$ and $C$ respectively. The incenters of the triangles $A E F$ and $B D F$ are $I_{1}$ and $I_{2}$ respectively; the circumcenters of the triangles $A C I_{1}$ and $B C I_{2}$ are $O_{1}$ and $O_{2}$ respectively. Prove that $I_{1} I_{2}$ and $O_{1} O_{2}$ are parallel. ['Let $\\angle C A B=\\alpha, \\angle A B C=\\beta, \\angle B C A=\\gamma$. We start by showing that $A, B, I_{1}$ and $I_{2}$ are concyclic. Since $A I_{1}$ and $B I_{2}$ bisect $\\angle C A B$ and $\\angle A B C$, their extensions beyond $I_{1}$ and $I_{2}$ meet at the incenter $I$ of the triangle. The points $E$ and $F$ are on the circle with diameter $B C$, so $\\angle A E F=\\angle A B C$ and $\\angle A F E=\\angle A C B$. Hence the triangles $A E F$ and $A B C$ are similar with ratio of similitude $\\frac{A E}{A B}=\\cos \\alpha$. Because $I_{1}$ and $I$ are their incenters, we obtain $I_{1} A=I A \\cos \\alpha$ and $I I_{1}=I A-I_{1} A=2 I A \\sin ^{2} \\frac{\\alpha}{2}$. By symmetry $I I_{2}=2 I B \\sin ^{2} \\frac{\\beta}{2}$. The law of sines in the triangle $A B I$ gives $I A \\sin \\frac{\\alpha}{2}=I B \\sin \\frac{\\beta}{2}$. Hence\n\n$$\nI I_{1} \\cdot I A=2\\left(I A \\sin \\frac{\\alpha}{2}\\right)^{2}=2\\left(I B \\sin \\frac{\\beta}{2}\\right)^{2}=I I_{2} \\cdot I B .\n$$\n\nTherefore $A, B, I_{1}$ and $I_{2}$ are concyclic, as claimed.\n\n\n\nIn addition $I I_{1} \\cdot I A=I I_{2} \\cdot I B$ implies that $I$ has the same power with respect to the circles $\\left(A C I_{1}\\right),\\left(B C I_{2}\\right)$ and $\\left(A B I_{1} I_{2}\\right)$. Then $C I$ is the radical axis of $\\left(A C I_{1}\\right)$ and $\\left(B C I_{2}\\right)$; in particular $C I$ is perpendicular to the line of centers $O_{1} O_{2}$.\n\nNow it suffices to prove that $C I \\perp I_{1} I_{2}$. Let $C I$ meet $I_{1} I_{2}$ at $Q$, then it is enough to check that $\\angle I I_{1} Q+\\angle I_{1} I Q=90^{\\circ}$. Since $\\angle I_{1} I Q$ is external for the triangle $A C I$, we have\n\n$$\n\\angle I I_{1} Q+\\angle I_{1} I Q=\\angle I I_{1} Q+(\\angle A C I+\\angle C A I)=\\angle I I_{1} I_{2}+\\angle A C I+\\angle C A I .\n$$\n\nIt remains to note that $\\angle I I_{1} I_{2}=\\frac{\\beta}{2}$ from the cyclic quadrilateral $A B I_{1} I_{2}$, and $\\angle A C I=\\frac{\\gamma}{2}$, $\\angle C A I=\\frac{\\alpha}{2}$. Therefore $\\angle I I_{1} Q+\\angle I_{1} I Q=\\frac{\\alpha}{2}+\\frac{\\beta}{2}+\\frac{\\gamma}{2}=90^{\\circ}$, completing the proof.'] [] Text-only Competition True Theorem proof Geometry Math English 370 Let $A B C$ be a triangle with $A B \neq A C$ and circumcenter $O$. The bisector of $\angle B A C$ intersects $B C$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $B C$. The lines through $D$ and $E$ perpendicular to $B C$ intersect the lines $A O$ and $A D$ at $X$ and $Y$ respectively. Prove that the quadrilateral $B X C Y$ is cyclic. ['The bisector of $\\angle B A C$ and the perpendicular bisector of $B C$ meet at $P$, the midpoint of the minor arc $\\widehat{B C}$ (they are different lines as $A B \\neq A C$ ). In particular $O P$ is perpendicular to $B C$ and intersects it at $M$, the midpoint of $B C$.\n\nDenote by $Y^{\\prime}$ the reflexion of $Y$ with respect to $O P$. Since $\\angle B Y C=\\angle B Y^{\\prime} C$, it suffices to prove that $B X C Y^{\\prime}$ is cyclic.\n\n\n\nWe have\n\n$$\n\\angle X A P=\\angle O P A=\\angle E Y P .\n$$\n\nThe first equality holds because $O A=O P$, and the second one because $E Y$ and $O P$ are both perpendicular to $B C$ and hence parallel. But $\\left\\{Y, Y^{\\prime}\\right\\}$ and $\\{E, D\\}$ are pairs of symmetric points with respect to $O P$, it follows that $\\angle E Y P=\\angle D Y^{\\prime} P$ and hence\n\n$$\n\\angle X A P=\\angle D Y^{\\prime} P=\\angle X Y^{\\prime} P .\n$$\n\nThe last equation implies that $X A Y^{\\prime} P$ is cyclic. By the powers of $D$ with respect to the circles $\\left(X A Y^{\\prime} P\\right)$ and $(A B P C)$ we obtain\n\n$$\nX D \\cdot D Y^{\\prime}=A D \\cdot D P=B D \\cdot D C\n$$\n\nIt follows that $B X C Y^{\\prime}$ is cyclic, as desired.'] [] Text-only Competition True Theorem proof Geometry Math English 371 Let $A B C$ be a triangle with $\angle B C A=90^{\circ}$, and let $C_{0}$ be the foot of the altitude from $C$. Choose a point $X$ in the interior of the segment $C C_{0}$, and let $K, L$ be the points on the segments $A X, B X$ for which $B K=B C$ and $A L=A C$ respectively. Denote by $M$ the intersection of $A L$ and $B K$. Show that $M K=M L$. ['Let $C^{\\prime}$ be the reflection of $C$ in the line $A B$, and let $\\omega_{1}$ and $\\omega_{2}$ be the circles with centers $A$ and $B$, passing through $L$ and $K$ respectively. Since $A C^{\\prime}=A C=A L$ and $B C^{\\prime}=B C=B K$, both $\\omega_{1}$ and $\\omega_{2}$ pass through $C$ and $C^{\\prime}$. By $\\angle B C A=90^{\\circ}, A C$ is tangent to $\\omega_{2}$ at $C$, and $B C$ is tangent to $\\omega_{1}$ at $C$. Let $K_{1} \\neq K$ be the second intersection of $A X$ and $\\omega_{2}$, and let $L_{1} \\neq L$ be the second intersection of $B X$ and $\\omega_{1}$.\n\n\n\nBy the powers of $X$ with respect to $\\omega_{2}$ and $\\omega_{1}$,\n\n$$\nX K \\cdot X K_{1}=X C \\cdot X C^{\\prime}=X L \\cdot X L_{1}\n$$\n\nso the points $K_{1}, L, K, L_{1}$ lie on a circle $\\omega_{3}$.\n\nThe power of $A$ with respect to $\\omega_{2}$ gives\n\n$$\nA L^{2}=A C^{2}=A K \\cdot A K_{1},\n$$\n\nindicating that $A L$ is tangent to $\\omega_{3}$ at $L$. Analogously, $B K$ is tangent to $\\omega_{3}$ at $K$. Hence $M K$ and $M L$ are the two tangents from $M$ to $\\omega_{3}$ and therefore $M K=M L$.'] [] Text-only Competition True Theorem proof Geometry Math English 372 Let $A B C$ be a triangle with circumcenter $O$ and incenter $I$. The points $D, E$ and $F$ on the sides $B C, C A$ and $A B$ respectively are such that $B D+B F=C A$ and $C D+C E=A B$. The circumcircles of the triangles $B F D$ and $C D E$ intersect at $P \neq D$. Prove that $O P=O I$. "[""By MiqueL's theorem the circles $(A E F)=\\omega_{A},(B F D)=\\omega_{B}$ and $(C D E)=\\omega_{C}$ have a common point, for arbitrary points $D, E$ and $F$ on $B C, C A$ and $A B$. So $\\omega_{A}$ passes through the common point $P \\neq D$ of $\\omega_{B}$ and $\\omega_{C}$.\n\nLet $\\omega_{A}, \\omega_{B}$ and $\\omega_{C}$ meet the bisectors $A I, B I$ and $C I$ at $A \\neq A^{\\prime}, B \\neq B^{\\prime}$ and $C \\neq C^{\\prime}$ respectively. The key observation is that $A^{\\prime}, B^{\\prime}$ and $C^{\\prime}$ do not depend on the particular choice of $D, E$ and $F$, provided that $B D+B F=C A, C D+C E=A B$ and $A E+A F=B C$ hold true (the last equality follows from the other two). For a proof we need the following fact.\n\nLemma. Given is an angle with vertex $A$ and measure $\\alpha$. A circle $\\omega$ through $A$ intersects the angle bisector at $L$ and sides of the angle at $X$ and $Y$. Then $A X+A Y=2 A L \\cos \\frac{\\alpha}{2}$.\n\nProof. Note that $L$ is the midpoint of arc $\\widehat{X L Y}$ in $\\omega$ and set $X L=Y L=u, X Y=v$. By PtolemY's theorem $A X \\cdot Y L+A Y \\cdot X L=A L \\cdot X Y$, which rewrites as $(A X+A Y) u=A L \\cdot v$. Since $\\angle L X Y=\\frac{\\alpha}{2}$ and $\\angle X L Y=180^{\\circ}-\\alpha$, we have $v=2 \\cos \\frac{\\alpha}{2} u$ by the law of sines, and the claim follows.\n\n\n\nApply the lemma to $\\angle B A C=\\alpha$ and the circle $\\omega=\\omega_{A}$, which intersects $A I$ at $A^{\\prime}$. This gives $2 A A^{\\prime} \\cos \\frac{\\alpha}{2}=A E+A F=B C$; by symmetry analogous relations hold for $B B^{\\prime}$ and $C C^{\\prime}$. It follows that $A^{\\prime}, B^{\\prime}$ and $C^{\\prime}$ are independent of the choice of $D, E$ and $F$, as stated.\n\nWe use the lemma two more times with $\\angle B A C=\\alpha$. Let $\\omega$ be the circle with diameter $A I$. Then $X$ and $Y$ are the tangency points of the incircle of $A B C$ with $A B$ and $A C$, and hence $A X=A Y=\\frac{1}{2}(A B+A C-B C)$. So the lemma yields $2 A I \\cos \\frac{\\alpha}{2}=A B+A C-B C$. Next, if $\\omega$ is the circumcircle of $A B C$ and $A I$ intersects $\\omega$ at $M \\neq A$ then $\\{X, Y\\}=\\{B, C\\}$, and so $2 A M \\cos \\frac{\\alpha}{2}=A B+A C$ by the lemma. To summarize,\n\n$$\n2 A A^{\\prime} \\cos \\frac{\\alpha}{2}=B C, \\quad 2 A I \\cos \\frac{\\alpha}{2}=A B+A C-B C, \\quad 2 A M \\cos \\frac{\\alpha}{2}=A B+A C\n\\tag{*}\n$$\n\nThese equalities imply $A A^{\\prime}+A I=A M$, hence the segments $A M$ and $I A^{\\prime}$ have a common midpoint. It follows that $I$ and $A^{\\prime}$ are equidistant from the circumcenter $O$. By symmetry $O I=O A^{\\prime}=O B^{\\prime}=O C^{\\prime}$, so $I, A^{\\prime}, B^{\\prime}, C^{\\prime}$ are on a circle centered at $O$.\n\nTo prove $O P=O I$, now it suffices to show that $I, A^{\\prime}, B^{\\prime}, C^{\\prime}$ and $P$ are concyclic. Clearly one can assume $P \\neq I, A^{\\prime}, B^{\\prime}, C^{\\prime}$.\n\nWe use oriented angles to avoid heavy case distinction. The oriented angle between the lines $l$ and $m$ is denoted by $\\angle(l, m)$. We have $\\angle(l, m)=-\\angle(m, l)$ and $\\angle(l, m)+\\angle(m, n)=\\angle(l, n)$ for arbitrary lines $l, m$ and $n$. Four distinct non-collinear points $U, V, X, Y$ are concyclic if and only if $\\angle(U X, V X)=\\angle(U Y, V Y)$.\n\n\n\n\n\nSuppose for the moment that $A^{\\prime}, B^{\\prime}, P, I$ are distinct and noncollinear; then it is enough to check the equality $\\angle\\left(A^{\\prime} P, B^{\\prime} P\\right)=\\angle\\left(A^{\\prime} I, B^{\\prime} I\\right)$. Because $A, F, P, A^{\\prime}$ are on the circle $\\omega_{A}$, we have $\\angle\\left(A^{\\prime} P, F P\\right)=\\angle\\left(A^{\\prime} A, F A\\right)=\\angle\\left(A^{\\prime} I, A B\\right)$. Likewise $\\angle\\left(B^{\\prime} P, F P\\right)=\\angle\\left(B^{\\prime} I, A B\\right)$. Therefore\n\n$$\n\\angle\\left(A^{\\prime} P, B^{\\prime} P\\right)=\\angle\\left(A^{\\prime} P, F P\\right)+\\angle\\left(F P, B^{\\prime} P\\right)=\\angle\\left(A^{\\prime} I, A B\\right)-\\angle\\left(B^{\\prime} I, A B\\right)=\\angle\\left(A^{\\prime} I, B^{\\prime} I\\right) .\n$$\n\nHere we assumed that $P \\neq F$. If $P=F$ then $P \\neq D, E$ and the conclusion follows similarly (use $\\angle\\left(A^{\\prime} F, B^{\\prime} F\\right)=\\angle\\left(A^{\\prime} F, E F\\right)+\\angle(E F, D F)+\\angle\\left(D F, B^{\\prime} F\\right)$ and inscribed angles in $\\left.\\omega_{A}, \\omega_{B}, \\omega_{C}\\right)$.\n\nThere is no loss of generality in assuming $A^{\\prime}, B^{\\prime}, P, I$ distinct and noncollinear. If $A B C$ is an equilateral triangle then the equalities (*) imply that $A^{\\prime}, B^{\\prime}, C^{\\prime}, I, O$ and $P$ coincide, so $O P=O I$. Otherwise at most one of $A^{\\prime}, B^{\\prime}, C^{\\prime}$ coincides with $I$. If say $C^{\\prime}=I$ then $O I \\perp C I$ by the previous reasoning. It follows that $A^{\\prime}, B^{\\prime} \\neq I$ and hence $A^{\\prime} \\neq B^{\\prime}$. Finally $A^{\\prime}, B^{\\prime}$ and $I$ are noncollinear because $I, A^{\\prime}, B^{\\prime}, C^{\\prime}$ are concyclic.""]" [] Text-only Competition True Theorem proof Geometry Math English 373 Let $A B C$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $B C, C A, A B$ intersect $\ell$ at the points $X, Y, Z$ different from $P$. Prove that the circumcircles of the triangles $A X P, B Y P$ and $C Z P$ have a common point different from $P$ or are mutually tangent at $P$. "[""Let $\\omega_{A}, \\omega_{B}, \\omega_{C}$ and $\\omega$ be the circumcircles of triangles $A X P, B Y P, C Z P$ and $A B C$ respectively. The strategy of the proof is to construct a point $Q$ with the same power with respect to the four circles. Then each of $P$ and $Q$ has the same power with respect to $\\omega_{A}, \\omega_{B}, \\omega_{C}$ and hence the three circles are coaxial. In other words they have another common point $P^{\\prime}$ or the three of them are tangent at $P$.\n\nWe first give a description of the point $Q$. Let $A^{\\prime} \\neq A$ be the second intersection of $\\omega$ and $\\omega_{A}$; define $B^{\\prime}$ and $C^{\\prime}$ analogously. We claim that $A A^{\\prime}, B B^{\\prime}$ and $C C^{\\prime}$ have a common point. Once this claim is established, the point just constructed will be on the radical axes of the three pairs of circles $\\left\\{\\omega, \\omega_{A}\\right\\},\\left\\{\\omega, \\omega_{B}\\right\\},\\left\\{\\omega, \\omega_{C}\\right\\}$. Hence it will have the same power with respect to $\\omega, \\omega_{A}, \\omega_{B}, \\omega_{C}$.\n\n\n\nWe proceed to prove that $A A^{\\prime}, B B^{\\prime}$ and $C C^{\\prime}$ intersect at one point. Let $r$ be the circumradius of triangle $A B C$. Define the points $X^{\\prime}, Y^{\\prime}, Z^{\\prime}$ as the intersections of $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ with $\\ell$. Observe that $X^{\\prime}, Y^{\\prime}, Z^{\\prime}$ do exist. If $A A^{\\prime}$ is parallel to $\\ell$ then $\\omega_{A}$ is tangent to $\\ell$; hence $X=P$ which is a contradiction. Similarly, $B B^{\\prime}$ and $C C^{\\prime}$ are not parallel to $\\ell$.\n\nFrom the powers of the point $X^{\\prime}$ with respect to the circles $\\omega_{A}$ and $\\omega$ we get\n\n$$\nX^{\\prime} P \\cdot\\left(X^{\\prime} P+P X\\right)=X^{\\prime} P \\cdot X^{\\prime} X=X^{\\prime} A^{\\prime} \\cdot X^{\\prime} A=X^{\\prime} O^{2}-r^{2}\n$$\n\nhence\n\n$$\nX^{\\prime} P \\cdot P X=X^{\\prime} O^{2}-r^{2}-X^{\\prime} P^{2}=O P^{2}-r^{2}\n$$\n\nWe argue analogously for the points $Y^{\\prime}$ and $Z^{\\prime}$, obtaining\n\n$$\nX^{\\prime} P \\cdot P X=Y^{\\prime} P \\cdot P Y=Z^{\\prime} P \\cdot P Z=O P^{2}-r^{2}=k^{2} .\n\\tag{1}\n$$\n\nIn these computations all segments are regarded as directed segments. We keep the same convention for the sequel.\n\nWe prove that the lines $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ intersect at one point by CEvA's theorem. To avoid distracting remarks we interpret everything projectively, i. e. whenever two lines are parallel they meet at a point on the line at infinity.\n\n\n\nLet $U, V, W$ be the intersections of $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ with $B C, C A, A B$ respectively. The idea is that although it is difficult to calculate the ratio $\\frac{B U}{C U}$, it is easier to deal with the cross-ratio $\\frac{B U}{C U} / \\frac{B X}{C X}$ because we can send it to the line $\\ell$. With this in mind we apply MENELAUs' theorem to the triangle $A B C$ and obtain $\\frac{B X}{C X} \\cdot \\frac{C Y}{A Y} \\cdot \\frac{A Z}{B Z}=1$. Hence Ceva's ratio can be expressed as\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=\\frac{B U}{C U} / \\frac{B X}{C X} \\cdot \\frac{C V}{A V} / \\frac{C Y}{A Y} \\cdot \\frac{A W}{B W} / \\frac{A Z}{B Z}\n$$\n\n\n\nProject the line $B C$ to $\\ell$ from $A$. The cross-ratio between $B C$ and $U X$ equals the cross-ratio between $Z Y$ and $X^{\\prime} X$. Repeating the same argument with the lines $C A$ and $A B$ gives\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=\\frac{Z X^{\\prime}}{Y X^{\\prime}} / \\frac{Z X}{Y X} \\cdot \\frac{X Y^{\\prime}}{Z Y^{\\prime}} / \\frac{X Y}{Z Y} \\cdot \\frac{Y Z^{\\prime}}{X Z^{\\prime}} / \\frac{Y Z}{X Z}\n$$\n\nand hence\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=(-1) \\cdot \\frac{Z X^{\\prime}}{Y X^{\\prime}} \\cdot \\frac{X Y^{\\prime}}{Z Y^{\\prime}} \\cdot \\frac{Y Z^{\\prime}}{X Z^{\\prime}}\n$$\n\nThe equations (1) reduce the problem to a straightforward computation on the line $\\ell$. For instance, the transformation $t \\mapsto-k^{2} / t$ preserves cross-ratio and interchanges the points $X, Y, Z$ with the points $X^{\\prime}, Y^{\\prime}, Z^{\\prime}$. Then\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=(-1) \\cdot \\frac{Z X^{\\prime}}{Y X^{\\prime}} / \\frac{Z Z^{\\prime}}{Y Z^{\\prime}} \\cdot \\frac{X Y^{\\prime}}{Z Y^{\\prime}} / \\frac{X Z^{\\prime}}{Z Z^{\\prime}}=-1\n$$\n\nWe proved that CEva's ratio equals -1 , so $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ intersect at one point $Q$."" 'First we prove that there is an inversion in space that takes $\\ell$ and $\\omega$ to parallel circles on a sphere. Let $Q R$ be the diameter of $\\omega$ whose extension beyond $Q$ passes through $P$. Let $\\Pi$ be the plane carrying our objects. In space, choose a point $O$ such that the line $Q O$ is perpendicular to $\\Pi$ and $\\angle P O R=90^{\\circ}$, and apply an inversion with pole $O$ (the radius of the inversion does not matter). For any object $\\mathcal{T}$ denote by $\\mathcal{T}^{\\prime}$ the image of $\\mathcal{T}$ under this inversion.\n\nThe inversion takes the plane $\\Pi$ to a sphere $\\Pi^{\\prime}$. The lines in $\\Pi$ are taken to circles through $O$, and the circles in $\\Pi$ also are taken to circles on $\\Pi^{\\prime}$.\n\n\n\nSince the line $\\ell$ and the circle $\\omega$ are perpendicular to the plane $O P Q$, the circles $\\ell^{\\prime}$ and $\\omega^{\\prime}$ also are perpendicular to this plane. Hence, the planes of the circles $\\ell^{\\prime}$ and $\\omega^{\\prime}$ are parallel.\n\nNow consider the circles $A^{\\prime} X^{\\prime} P^{\\prime}, B^{\\prime} Y^{\\prime} P^{\\prime}$ and $C^{\\prime} Z^{\\prime} P^{\\prime}$. We want to prove that either they have a common point (on $\\Pi^{\\prime}$ ), different from $P^{\\prime}$, or they are tangent to each other.\n\n\n\nThe point $X^{\\prime}$ is the second intersection of the circles $B^{\\prime} C^{\\prime} O$ and $\\ell^{\\prime}$, other than $O$. Hence, the lines $O X^{\\prime}$ and $B^{\\prime} C^{\\prime}$ are coplanar. Moreover, they lie in the parallel planes of $\\ell^{\\prime}$ and $\\omega^{\\prime}$. Therefore, $O X^{\\prime}$ and $B^{\\prime} C^{\\prime}$ are parallel. Analogously, $O Y^{\\prime}$ and $O Z^{\\prime}$ are parallel to $A^{\\prime} C^{\\prime}$ and $A^{\\prime} B^{\\prime}$.\n\nLet $A_{1}$ be the second intersection of the circles $A^{\\prime} X^{\\prime} P^{\\prime}$ and $\\omega^{\\prime}$, other than $A^{\\prime}$. The segments $A^{\\prime} A_{1}$ and $P^{\\prime} X^{\\prime}$ are coplanar, and therefore parallel. Now we know that $B^{\\prime} C^{\\prime}$ and $A^{\\prime} A_{1}$ are parallel to $O X^{\\prime}$ and $X^{\\prime} P^{\\prime}$ respectively, but these two segments are perpendicular because $O P^{\\prime}$ is a diameter in $\\ell^{\\prime}$. We found that $A^{\\prime} A_{1}$ and $B^{\\prime} C^{\\prime}$ are perpendicular, hence $A^{\\prime} A_{1}$ is the altitude in the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$, starting from $A$.\n\nAnalogously, let $B_{1}$ and $C_{1}$ be the second intersections of $\\omega^{\\prime}$ with the circles $B^{\\prime} P^{\\prime} Y^{\\prime}$ and $C^{\\prime} P^{\\prime} Z^{\\prime}$, other than $B^{\\prime}$ and $C^{\\prime}$ respectively. Then $B^{\\prime} B_{1}$ and $C^{\\prime} C_{1}$ are the other two altitudes in the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$.\n\n\n\nLet $H$ be the orthocenter of the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$. Let $W$ be the second intersection of the line $P^{\\prime} H$ with the sphere $\\Pi^{\\prime}$, other than $P^{\\prime}$. The point $W$ lies on the sphere $\\Pi^{\\prime}$, in the plane of the circle $A^{\\prime} P^{\\prime} X^{\\prime}$, so $W$ lies on the circle $A^{\\prime} P^{\\prime} X^{\\prime}$. Similarly, $W$ lies on the circles $B^{\\prime} P^{\\prime} Y^{\\prime}$ and $C^{\\prime} P^{\\prime} Z^{\\prime}$ as well; indeed $W$ is the second common point of the three circles.\n\nIf the line $P^{\\prime} H$ is tangent to the sphere then $W$ coincides with $P^{\\prime}$, and $P^{\\prime} H$ is the common tangent of the three circles.']" [] Text-only Competition True Theorem proof Geometry Math English 374 "An integer $a$ is called friendly if the equation $\left(m^{2}+n\right)\left(n^{2}+m\right)=a(m-n)^{3}$ has a solution over the positive integers. Prove that there are at least 500 friendly integers in the set $\{1,2, \ldots, 2012\}$." ['Every $a$ of the form $a=4 k-3$ with $k \\geq 2$ is friendly. Indeed the numbers $m=2 k-1>0$ and $n=k-1>0$ satisfy the given equation with $a=4 k-3$ :\n\n$$\n\\left(m^{2}+n\\right)\\left(n^{2}+m\\right)=\\left((2 k-1)^{2}+(k-1)\\right)\\left((k-1)^{2}+(2 k-1)\\right)=(4 k-3) k^{3}=a(m-n)^{3} .\n$$\n\nHence $5,9, \\ldots, 2009$ are friendly and so $\\{1,2, \\ldots, 2012\\}$ contains at least 502 friendly numbers.'] [] Text-only Competition True Theorem proof Number Theory Math English 375 "An integer $a$ is called friendly if the equation $\left(m^{2}+n\right)\left(n^{2}+m\right)=a(m-n)^{3}$ has a solution over the positive integers. Decide whether $a=2$ is friendly." ['We show that $a=2$ is not friendly. Consider the equation with $a=2$ and rewrite its left-hand side as a difference of squares:\n\n$$\n\\frac{1}{4}\\left(\\left(m^{2}+n+n^{2}+m\\right)^{2}-\\left(m^{2}+n-n^{2}-m\\right)^{2}\\right)=2(m-n)^{3}\n$$\n\nSince $m^{2}+n-n^{2}-m=(m-n)(m+n-1)$, we can further reformulate the equation as\n\n$$\n\\left(m^{2}+n+n^{2}+m\\right)^{2}=(m-n)^{2}\\left(8(m-n)+(m+n-1)^{2}\\right) .\n$$\n\nIt follows that $8(m-n)+(m+n-1)^{2}$ is a perfect square. Clearly $m>n$, hence there is an integer $s \\geq 1$ such that\n\n$$\n(m+n-1+2 s)^{2}=8(m-n)+(m+n-1)^{2} .\n$$\n\nSubtracting the squares gives $s(m+n-1+s)=2(m-n)$. Since $m+n-1+s>m-n$, we conclude that $s<2$. Therefore the only possibility is $s=1$ and $m=3 n$. However then the left-hand side of the given equation (with $a=2$ ) is greater than $m^{3}=27 n^{3}$, whereas its right-hand side equals $16 n^{3}$. The contradiction proves that $a=2$ is not friendly.'] [] Text-only Competition True Theorem proof Number Theory Math English 376 Let $x$ and $y$ be positive integers. If $x^{2^{n}}-1$ is divisible by $2^{n} y+1$ for every positive integer $n$, prove that $x=1$. "[""First we prove the following fact: For every positive integer $y$ there exist infinitely many primes $p \\equiv 3(\\bmod 4)$ such that $p$ divides some number of the form $2^{n} y+1$.\n\nClearly it is enough to consider the case $y$ odd. Let\n\n$$\n2 y+1=p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}}\n$$\n\nbe the prime factorization of $2 y+1$. Suppose on the contrary that there are finitely many primes $p_{r+1}, \\ldots, p_{r+s} \\equiv 3(\\bmod 4)$ that divide some number of the form $2^{n} y+1$ but do not divide $2 y+1$.\n\nWe want to find an $n$ such that $p_{i}^{e_{i}} \\| 2^{n} y+1$ for $1 \\leq i \\leq r$ and $p_{i} \\nmid 2^{n} y+1$ for $r+1 \\leq i \\leq r+s$. For this it suffices to take\n\n$$\nn=1+\\varphi\\left(p_{1}^{e_{1}+1} \\cdots p_{r}^{e_{r}+1} p_{r+1}^{1} \\cdots p_{r+s}^{1}\\right)\n$$\n\nbecause then\n\n$$\n2^{n} y+1 \\equiv 2 y+1 \\quad\\left(\\bmod p_{1}^{e_{1}+1} \\cdots p_{r}^{e_{r}+1} p_{r+1}^{1} \\cdots p_{r+s}^{1}\\right)\n$$\n\nThe last congruence means that $p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}$ divide exactly $2^{n} y+1$ and no prime $p_{r+1}, \\ldots, p_{r+s}$ divides $2^{n} y+1$. It follows that the prime factorization of $2^{n} y+1$ consists of the prime powers $p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}$ and powers of primes $\\equiv 1(\\bmod 4)$. Because $y$ is odd, we obtain\n\n$$\n2^{n} y+1 \\equiv p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}} \\equiv 2 y+1 \\equiv 3 \\quad(\\bmod 4)\n$$\n\nThis is a contradiction since $n>1$, and so $2^{n} y+1 \\equiv 1(\\bmod 4)$.\n\nNow we proceed to the problem. If $p$ is a prime divisor of $2^{n} y+1$ the problem statement implies that $x^{d} \\equiv 1(\\bmod p)$ for $d=2^{n}$. By FERMAT's little theorem the same congruence holds for $d=p-1$, so it must also hold for $d=\\left(2^{n}, p-1\\right)$. For $p \\equiv 3(\\bmod 4)$ we have $\\left(2^{n}, p-1\\right)=2$, therefore in this case $x^{2} \\equiv 1(\\bmod p)$.\n\nIn summary, we proved that every prime $p \\equiv 3(\\bmod 4)$ that divides some number of the form $2^{n} y+1$ also divides $x^{2}-1$. This is possible only if $x=1$, otherwise by the above $x^{2}-1$ would be a positive integer with infinitely many prime factors.""]" [] Text-only Competition True Theorem proof Number Theory Math English 377 Prove that for every prime $p>100$ and every integer $r$ there exist two integers $a$ and $b$ such that $p$ divides $a^{2}+b^{5}-r$. "['Throughout the solution, all congruence relations are meant modulo $p$.\n\nFix $p$, and let $\\mathcal{P}=\\{0,1, \\ldots, p-1\\}$ be the set of residue classes modulo $p$. For every $r \\in \\mathcal{P}$, let $S_{r}=\\left\\{(a, b) \\in \\mathcal{P} \\times \\mathcal{P}: a^{2}+b^{5} \\equiv r\\right\\}$, and let $s_{r}=\\left|S_{r}\\right|$. Our aim is to prove $s_{r}>0$ for all $r \\in \\mathcal{P}$.\n\nWe will use the well-known fact that for every residue class $r \\in \\mathcal{P}$ and every positive integer $k$, there are at most $k$ values $x \\in \\mathcal{P}$ such that $x^{k} \\equiv r$.\n\nLemma. Let $N$ be the number of quadruples $(a, b, c, d) \\in \\mathcal{P}^{4}$ for which $a^{2}+b^{5} \\equiv c^{2}+d^{5}$. Then\n\n$$\nN=\\sum_{r \\in \\mathcal{P}} s_{r}^{2}\\tag{a}\n$$\n\nand\n\n$$\nN \\leq p\\left(p^{2}+4 p-4\\right)\\tag{b}\n$$\n\nProof. (a) For each residue class $r$ there exist exactly $s_{r}$ pairs $(a, b)$ with $a^{2}+b^{5} \\equiv r$ and $s_{r}$ pairs $(c, d)$ with $c^{2}+d^{5} \\equiv r$. So there are $s_{r}^{2}$ quadruples with $a^{2}+b^{5} \\equiv c^{2}+d^{5} \\equiv r$. Taking the sum over all $r \\in \\mathcal{P}$, the statement follows.\n\n(b) Choose an arbitrary pair $(b, d) \\in \\mathcal{P}$ and look for the possible values of $a, c$.\n\n1. Suppose that $b^{5} \\equiv d^{5}$, and let $k$ be the number of such pairs $(b, d)$. The value $b$ can be chosen in $p$ different ways. For $b \\equiv 0$ only $d=0$ has this property; for the nonzero values of $b$ there are at most 5 possible values for $d$. So we have $k \\leq 1+5(p-1)=5 p-4$.\n\nThe values $a$ and $c$ must satisfy $a^{2} \\equiv c^{2}$, so $a \\equiv \\pm c$, and there are exactly $2 p-1$ such pairs $(a, c)$.\n\n2. Now suppose $b^{5} \\not \\equiv d^{5}$. In this case $a$ and $c$ must be distinct. By $(a-c)(a+c)=d^{5}-b^{5}$, the value of $a-c$ uniquely determines $a+c$ and thus $a$ and $c$ as well. Hence, there are $p-1$ suitable pairs $(a, c)$.\n\nThus, for each of the $k$ pairs $(b, d)$ with $b^{5} \\equiv d^{5}$ there are $2 p-1$ pairs $(a, c)$, and for each of the other $p^{2}-k$ pairs $(b, d)$ there are $p-1$ pairs $(a, c)$. Hence,\n\n$$\nN=k(2 p-1)+\\left(p^{2}-k\\right)(p-1)=p^{2}(p-1)+k p \\leq p^{2}(p-1)+(5 p-4) p=p\\left(p^{2}+4 p-4\\right)\n$$\n\nTo prove the statement of the problem, suppose that $S_{r}=\\emptyset$ for some $r \\in \\mathcal{P}$; obviously $r \\not \\equiv 0$. Let $T=\\left\\{x^{10}: x \\in \\mathcal{P} \\backslash\\{0\\}\\right\\}$ be the set of nonzero 10th powers modulo $p$. Since each residue class is the 10 th power of at most 10 elements in $\\mathcal{P}$, we have $|T| \\geq \\frac{p-1}{10} \\geq 4$ by $p>100$.\n\nFor every $t \\in T$, we have $S_{t r}=\\emptyset$. Indeed, if $(x, y) \\in S_{t r}$ and $t \\equiv z^{10}$ then\n\n$$\n\\left(z^{-5} x\\right)^{2}+\\left(z^{-2} y\\right)^{5} \\equiv t^{-1}\\left(x^{2}+y^{5}\\right) \\equiv r\n$$\n\nso $\\left(z^{-5} x, z^{-2} y\\right) \\in S_{r}$. So, there are at least $\\frac{p-1}{10} \\geq 4$ empty sets among $S_{1}, \\ldots, S_{p-1}$, and there are at most $p-4$ nonzero values among $s_{0}, s_{2}, \\ldots, s_{p-1}$. Then by the AM-QM inequality we obtain\n\n$$\nN=\\sum_{r \\in \\mathcal{P} \\backslash r T} s_{r}^{2} \\geq \\frac{1}{p-4}\\left(\\sum_{r \\in \\mathcal{P} \\backslash r T} s_{r}\\right)^{2}=\\frac{|\\mathcal{P} \\times \\mathcal{P}|^{2}}{p-4}=\\frac{p^{4}}{p-4}>p\\left(p^{2}+4 p-4\\right),\n$$\n\nwhich is impossible by the lemma.' 'If $5 \\nmid p-1$, then all modulo $p$ residue classes are complete fifth powers and the statement is trivial. So assume that $p=10 k+1$ where $k \\geq 10$. Let $g$ be a primitive root modulo $p$.\n\nWe will use the following facts:\n\n(F1) If some residue class $x$ is not quadratic then $x^{(p-1) / 2} \\equiv-1(\\bmod p)$.\n\n(F2) For every integer $d$, as a simple corollary of the summation formula for geometric progressions,\n\n$$\n\\sum_{i=0}^{2 k-1} g^{5 d i} \\equiv\\left\\{\\begin{array}{ll}\n2 k & \\text { if } 2 k \\mid d \\\\\n0 & \\text { if } 2 k \\nmid x d\n\\end{array} \\quad(\\bmod p)\\right.\n$$\n\nSuppose that, contrary to the statement, some modulo $p$ residue class $r$ cannot be expressed as $a^{2}+b^{5}$. Of course $r \\not \\equiv 0(\\bmod p)$. By $(\\mathrm{F} 1)$ we have $\\left(r-b^{5}\\right)^{(p-1) / 2}=\\left(r-b^{5}\\right)^{5 k} \\equiv-1(\\bmod p)$ for all residue classes $b$.\n\nFor $t=1,2 \\ldots, k-1$ consider the sums\n\n$$\nS(t)=\\sum_{i=0}^{2 k-1}\\left(r-g^{5 i}\\right)^{5 k} g^{5 t i}\n$$\n\nBy the indirect assumption and (F2),\n\n$$\nS(t)=\\sum_{i=0}^{2 k-1}\\left(r-\\left(g^{i}\\right)^{5}\\right)^{5 k} g^{5 t i} \\equiv \\sum_{i=0}^{2 k-1}(-1) g^{5 t i} \\equiv-\\sum_{i=0}^{2 k-1} g^{5 t i} \\equiv 0 \\quad(\\bmod p)\n$$\n\nbecause $2 k$ cannot divide $t$.\n\nOn the other hand, by the binomial theorem,\n\n$$\n\\begin{aligned}\nS(t) & =\\sum_{i=0}^{2 k-1}\\left(\\sum_{j=0}^{5 k}\\left(\\begin{array}{c}\n5 k \\\\\nj\n\\end{array}\\right) r^{5 k-j}\\left(-g^{5 i}\\right)^{j}\\right) g^{5 t i}=\\sum_{j=0}^{5 k}(-1)^{j}\\left(\\begin{array}{c}\n5 k \\\\\nj\n\\end{array}\\right) r^{5 k-j}\\left(\\sum_{i=0}^{2 k-1} g^{5(j+t) i}\\right) \\equiv \\\\\n& \\equiv \\sum_{j=0}^{5 k}(-1)^{j}\\left(\\begin{array}{c}\n5 k \\\\\nj\n\\end{array}\\right) r^{5 k-j}\\left\\{\\begin{array}{ll}\n2 k & \\text { if } 2 k \\mid j+t \\\\\n0 & \\text { if } 2 k \\nmid j+t\n\\end{array} \\quad(\\bmod p) .\\right.\n\\end{aligned}\n$$\n\nSince $1 \\leq j+t<6 k$, the number $2 k$ divides $j+t$ only for $j=2 k-t$ and $j=4 k-t$. Hence,\n\n$$\n\\begin{gathered}\n0 \\equiv S(t) \\equiv(-1)^{t}\\left(\\left(\\begin{array}{c}\n5 k \\\\\n2 k-t\n\\end{array}\\right) r^{3 k+t}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-t\n\\end{array}\\right) r^{k+t}\\right) \\cdot 2 k \\quad(\\bmod p) \\\\\n\\left(\\begin{array}{c}\n5 k \\\\\n2 k-t\n\\end{array}\\right) r^{2 k}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-t\n\\end{array}\\right) \\equiv 0 \\quad(\\bmod p) .\n\\end{gathered}\n$$\n\nTaking this for $t=1,2$ and eliminating $r$, we get\n\n$$\n\\begin{aligned}\n0 & \\equiv\\left(\\begin{array}{c}\n5 k \\\\\n2 k-2\n\\end{array}\\right)\\left(\\left(\\begin{array}{c}\n5 k \\\\\n2 k-1\n\\end{array}\\right) r^{2 k}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-1\n\\end{array}\\right)\\right)-\\left(\\begin{array}{c}\n5 k \\\\\n2 k-1\n\\end{array}\\right)\\left(\\left(\\begin{array}{c}\n5 k \\\\\n2 k-2\n\\end{array}\\right) r^{2 k}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-2\n\\end{array}\\right)\\right) \\\\\n& =\\left(\\begin{array}{c}\n5 k \\\\\n2 k-2\n\\end{array}\\right)\\left(\\begin{array}{c}\n5 k \\\\\n4 k-1\n\\end{array}\\right)-\\left(\\begin{array}{c}\n5 k \\\\\n2 k-1\n\\end{array}\\right)\\left(\\begin{array}{c}\n5 k \\\\\n4 k-2\n\\end{array}\\right) \\\\\n& =\\frac{(5 k) !^{2}}{(2 k-1) !(3 k+2) !(4 k-1) !(k+2) !}((2 k-1)(k+2)-(3 k+2)(4 k-1)) \\\\\n& =\\frac{-(5 k) !^{2} \\cdot 2 k(5 k+1)}{(2 k-1) !(3 k+2) !(4 k-1) !(k+2) !}(\\bmod p) .\n\\end{aligned}\n$$\n\nBut in the last expression none of the numbers is divisible by $p=10 k+1$, a contradiction.']" [] Text-only Competition True Theorem proof Number Theory Math English 378 According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulous for a positive integer $m$, then they are all fantabulous. Is the number $2021^{2021}$ fantabulous? "['Consider the sequence of positive integers $m, 3 m, 6 m+1,12 m+3,4 m+1,2 m$. Since each number in the sequence is fantabulous if and only if the next one is, we deduce that $m$ is fantabulous if and only if $2 m$ is fantabulous.\n\nCombined with the fact that $m$ is fantabulous if and only if $2 m+1$ is fantabulous, this implies that\n\n$m>1$ is fantabulous if and only if $f(m)=\\left[\\frac{m}{2}\\right]$ is fantabulous. We can apply $f$ sufficiently many times to any positive integer $n$ to conclude that $n$ is fantabulous if and only if 1 is fantabulous. Therefore, the fact that 2021 is fantabulous implies that 1 is fantabulous, which in turn implies that $2021^{2021}$ is fantabulous.' 'Let $m>1$ be a fantabulous number. Note that at least one of the following four cases must hold.\n\n- Case 1. The number $m$ is odd;\n\nWe have $m=2 a+1$ for some positive integer $a$, so $a\n\nProof 1:\nLet $H$ be the orthocenter of the triangle $A B P$. Then\n\n$$\n\\angle B H P=180^{\\circ}-\\angle B A C=180^{\\circ}-\\angle B C P .\n$$\n\nSo $B H P C$ is cyclic. Then we get\n\n$$\nT K \\cdot T A=T C \\cdot T B=T P \\cdot T H .\n$$\n\nSo, $A H P K$ is also cyclic. But then\n\n$$\n\\angle A K P=180^{\\circ}-\\angle A H P=\\angle A B P .\n$$\n\n\n\n\nProof 2:\nConsider the symmetric points $B^{\\prime}$ and $C^{\\prime}$ of $B$ and $C$, respectively, with respect to the line $P T$. It is clear that\n\n$$\n\\mathrm{TC}^{\\prime} \\cdot T B^{\\prime}=T C \\cdot T B=T K \\cdot T A\n$$\n\nSo $B^{\\prime} C^{\\prime} K A$ is cyclic. Also, because of the symmetry we have\n\n$$\n\\angle P C^{\\prime} B^{\\prime}=\\angle P C B=\\angle P A B .\n$$\n\nSo $B^{\\prime} C^{\\prime} P A$ is also cyclic. Therefore, the points $B^{\\prime}, C^{\\prime}, K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have\n\n$$\n\\angle P K A=\\angle P B^{\\prime} A=\\angle P B A .\n$$\n\nSo, lemma is proved and now return to the problem.\n\n\n\nLet $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M^{\\prime}$ and $N^{\\prime}$ the intersection points of the circumcircle of the triangle $H E F$ with the segments $E C$ and $F B$, respectively. We are going to show that $M=M^{\\prime}$ and $N=N^{\\prime}$ and it will prove the points $E, F, N, M$ lie on a common circle.\n\nOf course, $A$ is an orthocenter of the triangle $B C H$. Therefore $\\angle B H A=\\angle B C A, \\angle C H A=\\angle C B A$ and $\\angle H B A=\\angle H C A$. Thus\n\n$$\n\\angle H E F=\\angle H B A+\\angle E A B=\\angle H C A+\\angle F A C=\\angle H F E .\n$$\n\nSo, the triangle $H E F$ is isosceles, $H E=H F$.\n\nBy using lemma, we get\n\n$$\n\\angle A M^{\\prime} E=\\angle A H E=\\angle A C B,\n$$\n\nand\n\n$$\n\\angle A N^{\\prime} F=\\angle A H F=\\angle A B C \\text {. }\n$$\n\n\n\nTherefore $M=M^{\\prime}$ and $N=N^{\\prime}$ and we are done.' 'Let $X, Y$ be projections of $B$ on $A C$, and $C$ on $A B$, respectively. Let $\\omega$ be circumcircle of $B X Y C$. Let $Z$ be intersection of $E C$ and $\\omega$ and $D$ be projection of $E$ on $B A$.\n\n$$\n\\angle M A C=\\angle A M E-\\angle M C A=\\angle X C B-\\angle X C E=\\angle Z C B=\\angle Z X B\n$$\n\nSince $B X Y C$ is cyclic $\\angle A C Y=\\angle X B A$, and since DEXA is cyclic\n\n$$\n\\angle E X D=\\angle E A D=\\angle F A C \\text {. }\n$$\n\nTherefore, we get that the quadrangles $B Z X D$ and $C M A F$ are similar. Hence $\\angle F M C=\\angle D Z B$. Since $Z E D B$ is cyclic,\n\n$$\n\\angle D Z B=\\angle D E B=\\angle X A B .\n$$\n\nThus $\\angle F M C=\\angle X A B$. Similarly, $\\angle E N B=\\angle Y A C$. We get that $\\angle F M C=\\angle E N B$ and it implies that the points $E, F, N, M$ lie on a circle.\n\n']" [] Text-only Competition True Theorem proof Geometry Math English 380 Let $A B C$ be a triangle with incentre Iand let $D$ be an arbitrary point on the side $B C$. Let the line through $D$ perpendicular to $B I$ intersect $C I$ at $E$. Let the line through $D$ perpendicular to $C I$ intersect $B I$ at $F$. Prove that the reflection of $A$ in the line $E F$ lies on the line $B C$. ['Let us consider the case when I lies inside of triangle EFD. For the other cases the proof is almost the same only with the slight difference.\n\nWe are going to prove that the intersection point of the circumcircles of $A E C$ and $A F B$ (denote it by $T$ ) lies on the line $B C$ and this point is the symmetric point of $A$ with respect to $E F$. First of all we prove that $A E I F$ is cyclic, which implies that $T$ lies on the line $B C$, because\n\n$$\n\\angle A T B+\\angle A T C=\\angle A F B+\\angle A E C=\\angle A F I+\\angle A E I=180^{\\circ} .\n$$\n\nDenote by $N$ an intersection point of the lines $D F$ and $A C$. Of course $N$ is the symmetric point with respect to $C I$. Thus, $\\angle I N A=\\angle I D B$. Also,\n\n$$\n\\angle I F D=\\angle N D C-\\angle I B C=90^{\\circ}-\\angle I C B-\\angle I B C=\\angle I A N .\n$$\n\nSo, we get that $A, I, N$ and $F$ lie on a common circle. Therefore, we have $\\angle A F I=\\angle I N A=\\angle I D B$. Analogously, $\\angle A E I=\\angle I D C$ and we have\n\n$$\n\\angle A F I+\\angle A E I=\\angle I D B+\\angle I D C\n$$\n\nSo $\\angle A F I+\\angle A E I=180^{\\circ}$, thus $A E I F$ is cyclic and $T$ lies on the line $B C$.\n\n\n\nBecause $E C$ bisects the angle $A C B$ and $A E T C$ is cyclic we get $E A=E T$. Because of the similar reasons we have $F A=F T$. Therefore $T$ is the symmetric point of $A$ with respect to the line $E F$ and it lies on the line $B C$.\n\n\n\n# Solution 2. \n\nLike to the first solution, consider the case when I lies inside of triangle EFD. we need to prove that $A E I F$ is cyclic. The finish of the proof is the same.\n\n\n\nfirst note that $\\triangle F D B \\sim \\triangle A I B$, because $\\angle F B D=\\angle A B I$, and\n\n$$\n\\angle B F D=\\angle F D C-\\angle I B C=90^{\\circ}-\\angle I C D-\\angle I B C=\\angle I A B .\n$$\n\nBecause of the similarity we have $\\frac{A B}{A F}=\\frac{B I}{B D}$. This equality of the length ratios with $\\angle I B D=\\angle A B F$ implies that $\\triangle A B F \\sim \\triangle I B D$. Therefore, we have $\\angle I D B=\\angle A F B$. Analogously, we can get $\\angle I D C=\\angle A E C$, thus\n\n$$\n\\angle A F I+\\angle A E I=\\angle I D B+\\angle I D C=180^{\\circ} .\n$$\n\nSo, $A E I F$ is cyclic and we are done.'] [] Text-only Competition True Theorem proof Geometry Math English 381 "Does there exist a nonnegative integer $a$ for which the equation $$ \left\lfloor\frac{m}{1}\right\rfloor+\left\lfloor\frac{m}{2}\right\rfloor+\left\lfloor\frac{m}{3}\right\rfloor+\cdots+\left\lfloor\frac{m}{m}\right\rfloor=n^{2}+a $$ has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers? (The expression $\lfloor x\rfloor$ denotes the integer part(or floor) of the real number $x$. Thus $\lfloor\sqrt{2}\rfloor=1,\lfloor\pi\rfloor=\left\lfloor\frac{22}{7}\right\rfloor=$ $3,\lfloor 42\rfloor=42$ and $\lfloor 0\rfloor=0)$" ['Denote the equation from the statement by (1). The left hand side of (1) depends only on $m$, and will throughout be denoted by $L(m)$. Fix an integer $q>10^{7}$ and note that for $m=q^{3}$\n\n\n\n$$\nL\\left(q^{3}\\right)=\\sum_{k=1}^{q^{3}}\\left[\\frac{q^{3}}{k}\\right] \\leq \\sum_{k=1}^{q^{3}} \\frac{q^{3}}{k} \\leq q^{3} \\cdot \\sum_{k=1}^{q^{3}} \\frac{1}{k} \\leq q^{3} \\cdot q=q^{4} \\tag{2}\n$$\n\nIndeed, the first inequality results from $[x] \\leq x$. The second inequality can be seen (for instance) as follows. We divide the terms in the sum $\\sum_{k=1}^{q^{3}} \\frac{1}{k}$ into several groups: For $j \\geq 0$, the $j$-th group contains the $2^{j}$ consecutive terms $\\frac{1}{2^{j}}, \\ldots, \\frac{1}{2^{j+1}-1}$. Since every term in the $j$-th group is bounded by $\\frac{1}{2^{j}}$, the overall contribution of the $j$-th group to the sum is at most 1 . Since the first $q$ groups together would contain $2^{q}-1>q^{3}$ terms, the number of groups does not exceed $q$, and hence the value of the sum under consideration is indeed bounded by $q$.\n\nCall an integer $m$ special, if it satisfies $1 \\leq L(m) \\leq q^{4}$. Denote by $g(m) \\geq 1$ the largest integer whose square is bounded by $L(m)$; in other words $g^{2}(m) \\leq L(m)<(g(m)+1)^{2}$. Note that $g(m) \\leq$ $q^{2}$ for all special $m$, which implies\n\n$$\n0 \\leq L(m)-g^{2}(m)<(g(m)+1)^{2}-g^{2}(m)=2 g(m)+1 \\leq 2 q^{2}+1 . \\tag{3}\n$$\n\nFinally, we do some counting. Inequality (2) and the monotonicity of $L(m)$ imply that there exist at least $q^{3}$ special integers. Because of (3), every special integer $m$ has $0 \\leq L(m)-g^{2}(m) \\leq 2 q^{2}+1$. By averaging, at least $\\frac{q^{3}}{2 q^{2}+2}>10^{6}$ special integers must yield the same value $L(m)-g^{2}(m)$. This frequently occurring value is our choice for $\\alpha$, which yields more than $10^{6}$ solutions $(m, g(m))$ to equation (1).'] [] Text-only Competition True Theorem proof Number Theory Math English 382 Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle. "['Note that $N$ is also the midpoint of $P S$. From right-angled triangles $P A S$ and $C Q R$ we obtain $\\angle A N P=$ $2 \\angle A S P, \\angle C N Q=2 \\angle C R Q$, hence $\\angle A N C=\\angle A N P+\\angle C N Q=2(\\angle A S P+\\angle C R Q)=2(\\angle R S D+\\angle D R S)=$ $2 \\angle A D C$.\n\nSimilarly, using right-angled triangles $B A D$ and $B C D$, we obtain $\\angle A M C=2 \\angle A D C$.\n\nThus $\\angle A M C=\\angle A N C$, and the required statement follows.\n\n' 'In this proof we show that we have $\\angle N C M=\\angle N A M$ instead. From right-angled triangles $B C D$ and $Q C R$ we get $\\angle D R S=\\angle C R Q=\\angle R C N$ and $\\angle B D C=\\angle D C M$. Hence $\\angle N C M=\\angle D C M-\\angle R C N$. From rightangled triangle $A P S$ we get $\\angle P S A=\\angle S A N$. From right-angled triangle $B A D$ we have $\\angle M A D=\\angle B D A$. Moreover, $\\angle B D A=\\angle D R S+\\angle R S D-\\angle R D B$.\n\nTherefore $\\angle N A M=\\angle N A S-\\angle M A D=\\angle C D B-\\angle D R S=\\angle N C M$, and the required statement follows.' 'As $N$ is also the midpoint of $P S$, we can shrink triangle $A P S$ to a triangle $A_{0} Q R$ (where $P$ is sent to $Q$ and $S$ is sent to $R$ ). Then $A_{0}, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking the line $A_{0} R$ is parallel to the line $A D$. Therefore $\\angle C N A=\\angle C N A_{0}=2 \\angle C R A_{0}=2 \\angle C D A=\\angle C M A$. The required statement follows.']" [] Text-only Competition True Theorem proof Geometry Math English 383 There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time? "['We show that this is not possible.\n\nThe lines divide the plane into disjoint regions. We claim that there exists an alternating 2-coloring of these regions, that is each region can be colored in black or white, such that if two regions share a line segment, they have a different color. We show this inductively.\n\nIf there are no lines, this is obvious. Consider now an arrangement of $n$ lines in the plane and an alternating 2 -coloring of the regions. If we add a line $g$, we can simply switch the colors of all regions in one of the half planes divided by $g$ from white to black and vice versa. Any line segment not in $g$ will still be between two regions of different color. Any line segment in $g$ cuts a region determined by the $n$ lines in two, and since we switched colors on one side of $g$ this segment will also lie between two regions of different color.\n\nNow without loss of generality we may assume, that Turbo starts on a line segment with a white region on her left and a black one on her right. At any intersection, if she turns right, she will keep the black tile to her right. If she turns left, she will keep the white tile to her left. Thus wherever she goes, she will always have a white tile on her left and a black tile on her right. As a consequence, she can cross every line segment only in the direction where she has a white tile to her left, and never the opposite direction where she would have a black tile to the left.' ""Suppose the assumption is true.\n\nLet's label each segment in the snail's path with $\\mathbf{L}$ or $\\mathbf{R}$ depending on the direction that Turbo chose at the start point of this segment (one segment can have several labels if it has been visited several times).\n\nConsider the first segment that has been visited twice in different directions, name this segment $s_{1}$. Assume without loss of generality that it is labeled with $\\mathbf{L}$. Then next segment must be labeled with $\\mathbf{R}$, name this one $s_{2}$.\n\nLet's look at the label which $s_{1}$ can get on the second visit. If it gets $\\mathbf{L}$ then the previous segment in the path must be $s_{2}$. But in this case $s_{1}$ is not the first segment that has been visited twice in different directions because $s_{2}$ has been visited earlier. So the second label of $s_{1}$ must be $\\mathbf{R}$, and Turbo must have come from the opposite side of $s_{2}$.\n\n\n\n\n\nSince Turbo alters directions at each point, labels in her path also alter. And because two labels of $s_{1}$ are different, the number of visited segments between these two visits must be even.\n\nNow let's make the following observation: each segment in the path corresponds to exactly one line, and its previous and next segments are on opposite sides of this line.\n\n\n\nAgain consider the path between two visits of $s_{1}$.\n\nEach line intersecting this path must be crossed an even number of times because Turbo has to return to the initial side of each line. Therefore, an even number of segments of Turbo's path are contained on each of these lines. But the line containing $s_{1}$ must be crossed an odd number times. Since each crossing corresponds to exactly one segment in the path, the number of segments must be odd.\n\nHere we get the contradiction. Therefore, the assumption is false."" 'Suppose that the snail always slides slightly to the right of the line segments on her path. When turning to the right, she does not cross any line, whereas when turning to the left, she crosses exactly two lines. This means that at any time of her journey, she has crossed an even number of lines.\n\nAssuming that at some point she slides along a segment for the second time, but in the opposite direction, we argue that she needs to cross an odd number of lines. Let $\\ell$ be the line on which the revisit happens. In order to get to the other side of $\\ell$, the snail has to cross $\\ell$ an odd number of times. To visit the same segment of $\\ell$, she must cross every other line an even number of times.' 'Let us color in red all intersection points of the given lines and let us choose one of two possible directions on each segment (draw an arrow on each segment). Consider a red point $R$ where two given lines $a$ and $b$ meet, and the four segments $a_{1}, a_{2}, b_{1}, b_{2}$ with endpoint $R$ (so that $a_{i} \\subset a, b_{j} \\subset b$ ). $R$ is called a saddle if on $a_{1}, a_{2}$ the arrows go out of $R$ while on $b_{1}, b_{2}$ the arrows enter $R$, or visa versa, on $b_{1}, b_{2}$ the arrows go out of $R$ while on $a_{1}, a_{2}$ the arrows enter $R$. The set of arrows (chosen on all segments) is said to be good if all red points are saddles. It is sufficient to prove that there exists a good set of arrows. Indeed, if initially Turbo is moving along (or opposite) the arrow, then this condition holds after she turns at a red point.\n\nThe given lines cut the plane into regions. Further we need the following property of the good set of arrows (this property directly follows from the definition): the boundary of any bounded region is a directed cycle of arrows; the boundary of any unbounded region is a directed chain of arrows.\n\nWe construct a good set of arrows by induction on $n$ with trivial base $n=1$. Now erase one of $n$ given lines and assume we have a good set of arrows for remaining $n-1$ lines. Now restore the $n$-th line $\\ell$, assume that $\\ell$ is horizontal. Denote by $A_{1}, \\ldots, A_{n-1}$ all new red points on $\\ell$ from the left to the right. Each of $A_{i}$ belongs to some old segment $m_{i}$ of the line $\\ell_{i}$. Let us call $A_{i}$ ascending if the arrow on $m_{i}$ goes up, and descending if the arrow on $m_{i}$ goes down. Consider the region containing the segment $A_{i} A_{i+1}$. By the property, $A_{i}$ and $A_{i+1}$ can not be both ascending or both descending. Thus we can choose arrows on all pieces of $\\ell$ so that each arrow goes from a descending to an ascending vertex.\n\n\n\nEach of points $A_{i}$ cuts $m_{i}$ into two new pieces; the direction of new pieces supposed to be the same as on $m_{i}$. Now simultaneously change the direction of arrows on all pieces below the line $\\ell$. It is easy to see that $A_{1}, \\ldots, A_{n-1}$ become saddles, while the other red points remain saddles. This completes the induction step.']" [] Text-only Competition True Theorem proof Combinatorics Math English 384 "Let $n \geq 1$ be an integer and let $t_{1}2$ elements $t_{1}<\\ldots1$ we distinguish the two cases $t_{1}>1$ and $t_{1}=1$. If $t_{1}>1$ there exists, by the induction hypothesis, a group $A$ of size $t_{n}$ that satisfies the conditions of the problem for $t_{1}^{\\prime}=t_{1}-1, \\ldots, t_{n}^{\\prime}=t_{n}-1$. Now add a new person to $A$ and let him/her play against everyone from $A$. The new group will be of size $t_{n}+1$ and there exists a person which has played $t$ games if and only if there exists a person that has played $t-1$ games within $A$, i.e. if and only if $t \\in\\left\\{t_{1}, \\ldots, t_{n}\\right\\}$. Hence the conditions of the problem are satisfied.\n\nIf $t_{1}=1$ there exists, by the induction hypothesis, a group $B$ of size $t_{n-1}$ that satisfies the conditions of the problem for $t_{1}^{\\prime}=t_{2}-1, \\ldots, t_{n-2}^{\\prime}=t_{n-1}-1$. Now add a new person $P$ and let him/her play with everyone from group $B$ and a group $C$ of size $t_{n}-t_{n-1}>0$ and let them play with $P$. The new group will be of size $t_{n-1}+1+\\left(t_{n}-t_{n-1}\\right)+1=t_{n}+1$. Since person $P$ has played against everyone he will have played $t_{n}$ games. The people in $C$ will have played $1=t_{1}$ games. There exists a person in $B$ that has played $t$ games if and only if there exist a person in $B$ that has played $t-1$ games within $B$, i.e. if and only if $t \\in\\left\\{t_{2}, \\ldots, t_{n-1}\\right\\}$. Hence the conditions of the problem are satisfied.']" [] Text-only Competition True Theorem proof Combinatorics Math English 385 "Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that $$ \left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} . $$ Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple. There are exactly $n$ factors in the product on the left hand side." "['Notice that for odd integers $n>2$, the tuple $(1,1, \\ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.\n\nLemma 0.1. If an expensive $n$-tuple exists for some $n \\geq 4$, then also an expensive $n-2$-tuple.\n\nProof. In what follows all indices are considered modulo $n$. Let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities\n\n$$\n\\begin{gathered}\na_{t-1}+a_{t} \\leq 2 a_{t}<2\\left(a_{t}+a_{t+1}\\right) \\\\\na_{t}+a_{t+1} \\leq 2 a_{t}<2\\left(a_{t-1}+a_{t}\\right) .\n\\end{gathered}\n$$\n\nSince both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and $(2)$\n\n$$\na_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}\n$$\n\nfor some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.\n\nConsider now the $n-2$-tuple $\\left(b_{1}, \\ldots, b_{n-2}\\right)$ obtained by removing $a_{t}$ and $a_{t+1}$ from $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$. By what we just said we have\n\n$$\n\\prod_{i=1}^{n-2}\\left(b_{i}+b_{i+1}\\right)=\\frac{\\prod_{i=1}^{n}\\left(a_{i}+a_{i+1}\\right)}{\\left(a_{t-1}+a_{t}\\right)\\left(a_{t}+a_{t+1}\\right)}=2^{2(k-r)-1}\n$$\n\nand hence $\\left(b_{1}, \\ldots, b_{n-2}\\right)$ is again expensive.\n\nFrom the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2 -tuple i.e.\n\n$$\n\\left(a_{1}+a_{2}\\right)^{2}=2^{2 k-1}\n$$\n\nfor some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.\n\nWe prove this by induction. From above we saw that 1 belongs to an expensive $n$-tuple. Assume now that all odd positive integers less that $2^{k}$ belong to an expensive $n$-tuple, for some $k \\geq 1$. Hence for any odd $r<2^{k}$ there is an integer $n$ and an expensive $n$-tuple $\\left(a_{1}, \\ldots, r, \\ldots, a_{n}\\right)$. We notice that then also $\\left(a_{1}, \\ldots, r, 2^{k+1}-\\right.$ $\\left.r, r, \\ldots, a_{n}\\right)$ is expensive. Since $2^{k+1}-r$ can take all odd values between $2^{k}$ and $2^{k+1}$ the induction step is complete.' 'Consider the operators $T_{ \\pm}(n)=2 n \\pm 1$. We claim that for every odd integer $m$ there is an integer $r$ and signs $\\epsilon_{i} \\in\\{+,-\\}$ for $i=1, \\ldots r$ such that $T_{\\epsilon_{r}} \\circ \\cdots \\circ T_{\\epsilon_{1}}(1)=m$. This is certainly true for $m=1$ and for $m>1$ we find that $m=T_{-}((m+1) / 2)$ if $m \\equiv 1 \\bmod (4)$ and $m=T_{+}((m-1) / 2)$ if $m \\equiv 3 \\bmod (4)$. Note that both $(m+1) / 2$ and $(m-1) / 2$ are odd integers in their respective cases and $(m-1) / 2 \\leq(m+1) / 21$. Therefore iterating the procedure will eventually terminate in one.\n\nFor the construction it is most convenient to set $n=2 l+1$ and label the tuple $\\left(a_{-l}, a_{-l+1}, \\ldots, a_{l}\\right)$. For $m=1$ we have the expensive tuple $(1,1,1)$. For $m>1$ we will define operators $T_{ \\pm}$on expensive tuples with the condition $a_{-l}=a_{l}=1$ that give rise to a new expensive tuple $\\left(b_{-l^{\\prime}}, \\ldots, b_{l^{\\prime}}\\right)$ with $b_{-l^{\\prime}}=b_{l^{\\prime}}=1$ and $b_{0}=T_{ \\pm} a_{0}$. It is then clear that $T_{\\epsilon_{r}} \\circ \\cdots \\circ T_{\\epsilon_{1}}((1,1,1))$ is an expensive tuple containing $m$. We define $T_{ \\pm}$as follows: set $l^{\\prime}=l+1$ and $b_{-l^{\\prime}}=b_{l^{\\prime}}=1$ and $b_{i}=T_{ \\pm(-1)^{i}}\\left(a_{i}\\right)$ for $i=-l, \\ldots, l$. Here we identify + with +1 and - with -1 . We are left to prove that the new tuple is indeed expensive. If $\\pm(-1)^{l}=-1$, then $\\prod\\left(b_{i}+b_{i+1}\\right)=4 \\cdot 2^{2 l} \\prod\\left(a_{i}+a_{i+1}\\right)$, and if $\\pm(-1)^{l}=+1$, then $\\prod\\left(b_{i}+b_{i+1}\\right)=4 \\cdot 2^{2 l+2} \\prod\\left(a_{i}+a_{i+1}\\right)$. In both cases we end up with an expensive tuple again.']" [] Text-only Competition True Theorem proof Algebra Math English 386 "Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point. The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side." "['Let $H$ denote the orthocenter of $A B C$, and let $e$ denote its Euler line. Let $e_{1}, e_{2}, e_{3}$ denote the respective reflections of $e$ in $B C, C A, A B$. The proof naturally divides into two parts: we first show that pairwise intersections of the circles in question correspond to pairwise intersections of $e_{1}, e_{2}, e_{3}$, and then prove that $e_{1}, e_{2}, e_{3}$ intersect in a single point on the circumcircle of $A B C$.\n\n\n\n\n\nNow consider for example the circumcircles of $O_{1} O_{2} C$ and $G_{1} G_{2} C$. By construction, it is clear that $\\angle O_{2} C O_{1}=\\angle G_{2} C G_{1}=2 \\angle A C B$. Let $G_{1} O_{1}$ and $G_{2} O_{2}$ meet at $X$, and let $e$ meet $e_{1}, e_{2}$ at $E_{1}, E_{2}$, respectively, as shown in the diagram below. Chasing angles,\n\n$$\n\\begin{aligned}\n\\angle G_{2} X G_{1}=\\angle O_{2} X O_{1} & =\\angle E_{2} X E_{1}=180^{\\circ}-\\angle E_{1} E_{2} X-\\angle X E_{1} E_{2} \\\\\n& =180^{\\circ}-2 \\angle E_{1} E_{2} C-\\left(\\angle C E_{1} E_{2}-\\angle C E_{1} X\\right) .\n\\end{aligned}\n$$\n\nBut $\\angle C E_{1} X=\\angle B E_{1} O_{1}=\\angle B E_{1} O=180^{\\circ}-\\angle C E_{1} E_{2}$, and thus\n\n$$\n\\angle G_{2} X G_{1}=\\angle O_{2} X O_{1}=2\\left(180^{\\circ}-\\angle E_{1} E_{2} C-\\angle C E_{1} E_{2}\\right)=2 \\angle A C B .\n$$\n\n\n\nIt follows from this that $X$ lies on the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$. In other words, the second point of intersection of the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$ is the intersection of $e_{1}$ and $e_{2}$. Similarly, the circumcircles of $G_{1} G_{3} B$ and $O_{1} O_{3} B$ meet again at the intersection of $e_{1}$ and $e_{3}$, and those of $G_{2} G_{3} A$ and $O_{2} O_{3} A$ meet again at the intersection of $e_{2}$ and $e_{3}$.\n\n\n\n\n\nIt thus remains to show that $e_{1}, e_{2}, e_{3}$ are concurrent, and intersect on the circumcircle of $A B C$. Let $e$ meet the circumcircles of the triangles $B C H, A C H, A B H$ at $X_{1}, X_{2}, X_{3}$, respectively. It is well known that the reflections of $H$ in the sides of $A B C$ lie on the circumcircle of $A B C$. For this reason, the circumcircles of $B C H, A C H, A B H$ have the same radius as the circumcircle of $A B C$, and hence the reflections $X_{1}^{\\prime}, X_{2}^{\\prime}, X_{3}^{\\prime}$ of $X_{1}, X_{2}, X_{3}$ in the sides $[B C],[C A],[A B]$ lie on the circumcircle of $A B C$. By definition, $X_{1}^{\\prime}, X_{2}^{\\prime}, X_{3}^{\\prime}$ lie on $e_{1}, e_{2}, e_{3}$, respectively. It thus remains to show that they coincide.\n\nTo show that, for example, $X_{1}^{\\prime}=X_{2}^{\\prime}$, it will be sufficient to show that $\\angle X_{2} A C=\\angle X_{1}^{\\prime} A C$, since we have already shown that $X_{1}^{\\prime}$ and $X_{2}^{\\prime}$ lie on the circumcircle of $A B C$. But, chasing angles in the diagram above,\n\n$$\n\\begin{aligned}\n\\angle X_{1}^{\\prime} A C & =\\angle X_{1}^{\\prime} A B-\\angle B A C=\\left(180^{\\circ}-\\angle X_{1}^{\\prime} C B\\right)-\\angle B A C \\\\\n& =180^{\\circ}-\\angle X_{1} C B-\\angle B A C=\\left(180^{\\circ}-\\angle B A C\\right)-\\angle X_{1} H B \\\\\n& =\\angle B H C-\\angle X_{1} H B=\\angle X_{2} H C=\\angle X_{2} A C,\n\\end{aligned}\n$$\n\nwhere we have used the fact that $B H C X_{1}$ and $A H X_{2} C$ are cyclic by construction, and the fact that $\\angle B H C=180^{\\circ}-\\angle B A C$. This shows that $X_{1}^{\\prime}=X_{2}^{\\prime}$. Similarly, $X_{1}^{\\prime}=X_{3}^{\\prime}$, which completes the proof.' 'The proof consists of two parts. First, we show that if $P$ is any point inside the triangle $A B C$ and $P_{1}, P_{2}, P_{3}$ are its reflections in the sides $B C, C A, A B$, then the circumcircles of the triangles $P_{1} P_{2} C, P_{1} P_{3} B, P_{2} P_{3} A$ intersect in a point $T_{P}$ on the circumcircle of the triangle $A B C$. In the second part, we show that $T_{G}$ coincides with $T_{O}$.\n\nNow let $P$ be any point inside the triangle $A B C$ and let $P_{1}, P_{2}, P_{3}$ be the reflections in the sides as above. Let $T_{P}$ be the second intersection of the circumcircles of the triangles $P_{1} P_{2} C$ and $A B C$. We want to show that $T_{P}$ lies on the circumcircles of the triangles $P_{1} P_{3} B$ and $P_{2} P_{3} A$.\n\n\n\nBy construction, we have $P_{1} C=P_{2} C$, hence\n\n$$\n\\angle C P_{1} P_{2}=90^{\\circ}-\\frac{1}{2} \\angle P_{2} C P_{1}=90^{\\circ}-\\angle A C B .\n$$\n\nSimilarly, $\\angle P_{2} P_{3} A=90^{\\circ}-\\angle B A C$. This gives us\n\n$$\n\\begin{aligned}\n\\angle P_{2} T_{P} A & =\\angle C T_{P} A-\\angle C T_{P} P_{2}=\\angle C B A-\\angle C P_{1} P_{2} \\\\\n& =\\angle C B A-90^{\\circ}+\\angle A C B=90^{\\circ}-\\angle B A C=\\angle P_{2} P_{3} A,\n\\end{aligned}\n$$\n\nso $T_{P}$ lies on the circumcircle of the triangle $P_{2} P_{3} A$. Similarly, $T_{P}$ lies on the circumcircle of the triangle $P_{1} P_{3} B$ which completes the first part.\n\nNote that if $P_{2}$ is given, then $T_{P}$ is the unique point on the circumcircle of the triangle $A B C$ with $\\angle C T_{P} P_{2}=90^{\\circ}-\\angle A C B$. In the second part, we will use this as follows: If we can find a point $T$ on the circumcircle of the triangle $A B C$ with $\\angle C T G_{2}=\\angle C T O_{2}=90^{\\circ}-\\angle A C B$, then $T=T_{G}=T_{O}$ and we are done.\n\nLet $H$ be the orthocenter of the triangle $A B C$ and let $H_{2}$ be the reflection in the side $A C$. It is known that $\\mathrm{H}_{2}$ lies on the circumcircle of the triangle $A B C . G, O, H$ lie on the Euler line, so $G_{2}, \\mathrm{O}_{2}, \\mathrm{H}_{2}$ are collinear as well. Let $T$ be the second intersection of $G_{2} H_{2}$ and the circumcircle of the triangle $A B C$. We can now complete the proof by seeing that\n\n$$\n\\angle C T G_{2}=\\angle C T O_{2}=\\angle C T H_{2}=\\angle C B H_{2}=90^{\\circ}-\\angle A C B .\n$$\n\n' 'For every point $P$, let $p$ denote the corresponding complex number. Set $O$ to be the origin, so $o=0$, and without loss of generality we can assume that $a, b$ and $c$ lie on the unit circle. Then the centroid can be expressed as $g=\\frac{a+b+c}{3}$.\n\nThe segments $o o_{1}$ and $b c$ have a common midpoint, so $o_{1}+o=b+c$, and then $o_{1}=b+c$. Similarly $o_{2}=a+c$ and $o_{3}=a+b$. In order to compute $g_{1}$, define $y$ to be the projection of $g$ onto $b c$. Since $b$ and $c$ are on the unit circle, it is well known that $y$ can be expressed as\n\n$$\ny=\\frac{1}{2}(b+c+g-b c \\bar{g})\n$$\n\nBy using $\\bar{a}=\\frac{1}{a}, \\bar{b}=\\frac{1}{b}$ and $\\bar{c}=\\frac{1}{c}$ (points on the unit circle), we obtain\n\n$$\ng_{1}=b+c-\\frac{a b+b c+c a}{3 a}\n$$\n\nSimilarly, we get $g_{2}=a+c-\\frac{a b+b c+c a}{3 b}$ and $g_{3}=a+b-\\frac{a b+b c+c a}{3 c}$.\n\n1) Proof that circumcircles of triangles $a b c, o_{1} o_{2} c, o_{1} o_{3} b$ and $o_{2} o_{3} a$ have common point. Let $x$ be the point of intersection of circumcircles of triangles $o_{1} o_{2} c$ and $a b c(x \\neq c)$. We know that $x, o_{1}, o_{2}$ and $c$ are concyclic if and only if $\\frac{x-c}{o_{1}-c}: \\frac{x-o_{2}}{o_{1}-o_{2}}$ is real number, which is equivalent to\n\n$$\n\\frac{x-c}{\\bar{x}-\\bar{c}} \\cdot \\frac{o_{1}-o_{2}}{\\overline{o_{1}}-\\overline{o_{2}}}=\\frac{o_{1}-c}{\\overline{o_{1}}-\\bar{c}} \\cdot \\frac{x-o_{2}}{\\bar{x}-\\overline{o_{2}}}\n\\tag{1}\n$$\n\nSince $x$ and $c$ are on the unit circle $\\frac{x-c}{\\bar{x} - \\bar{c}} = -xc$. Also, $\\frac{o_1-o_2}{\\bar{o_1} - \\bar{o_2}} = \\frac{b-a}{\\bar{b} - \\bar{a}}=-ab$ and $\\frac{o_{1}-c}{\\overline{o_{1}}-\\bar{c}}=\\frac{b}{\\bar{b}}=b^{2}$. Since $\\bar{x}=\\frac{1}{x}$, from (1) and previous relations, we have:\n\n$$\nx=\\frac{a b+b c+c a}{a+b+c} \\text {. }\n$$\n\nThis formula is symmetric, so we conclude that $x$ also belongs to circumcircles of $o_{1} o_{3} b$ and $\\mathrm{o}_{2} \\mathrm{o}_{3} a$.\n\n2) Proof that $x$ belongs to circumcircles of $g_{1} g_{2} c, g_{1} g_{3} b$ and $g_{2} g_{3} a$.\n\nBecause of symmetry, it is enough to prove that $x$ belongs to circumcircle of $g_{1} g_{2} c$, i.e. to prove the following:\n\n$$\n\\frac{x-c}{\\bar{x}-\\bar{c}} \\cdot \\frac{g_{1}-g_{2}}{\\overline{g_{1}}-\\overline{g_{2}}}=\\frac{g_{1}-c}{\\overline{g_{1}}-\\bar{c}} \\cdot \\frac{x-g_{2}}{\\bar{x}-\\overline{g_{2}}} \\tag{2}\n$$\n\nEasy computations give that\n\n$$\ng_{1}-g_{2}=(b-a) \\frac{2 a b-b c-a c}{3 a b}, \\quad \\overline{g_{1}}-\\overline{g_{2}}=(\\bar{b}-\\bar{a}) \\frac{2-\\frac{a}{c}-\\frac{b}{c}}{3}\n$$\n\nand then\n\n$$\n\\frac{g_{1}-g_{2}}{\\overline{g_{1}}-\\overline{g_{2}}}=\\frac{c(b c+a c-2 a b)}{2 c-a-b}\n$$\n\n\n\nOn the other hand we have\n\n$$\ng_{1}-c=\\frac{2 a b-b c-a c}{3 a}, \\quad \\bar{g}_{1}-\\bar{c}=\\frac{2 c-a-b}{3 b c}\n$$\n\nThis implies\n\n$$\n\\frac{g_{1}-c}{{\\bar{g_{1}}-\\bar{c}}}=\\frac{2 a b-b c-a c}{2 c-a-b} \\cdot \\frac{b c}{a} .\n$$\n\nThen (2) is equivalent to\n\n$$\n\\begin{aligned}\n-x c \\cdot \\frac{c(b c+a c-2 a b)}{2 c-a-b}=\\frac{2 a b-b c-a c}{2 c-a-b} \\cdot \\frac{b c}{a} \\cdot \\frac{x-g_{2}}{\\bar{x}-\\overline{g_{2}}} \\\\\n\\Longleftrightarrow x c a\\left(\\bar{x}-\\overline{g_{2}}\\right)=b\\left(x-g_{2}\\right),\n\\end{aligned}\n$$\n\nwhich is also equivalent to\n\n$\\frac{a b+b c+c a}{a+b+c} \\cdot c a\\left(\\frac{a+b+c}{a b+b c+c a}-\\frac{1}{a}-\\frac{1}{c}+\\frac{a+b+c}{3 a c}\\right)=b \\cdot\\left(\\frac{a b+b c+c a}{a+b+c}-a-c+\\frac{a b+b c+c a}{3 b}\\right)$.\n\nThe last equality can easily be verified, which implies that $x$ belongs to circumcircle of triangle $g_{1} g_{2} c$. This concludes our proof.']" [] Text-only Competition True Theorem proof Geometry Math English 387 "Let $n$ be an odd positive integer, and let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers. Show that $$ \min _{i=1, \ldots, n}\left(x_{i}^{2}+x_{i+1}^{2}\right) \leq \max _{j=1, \ldots, n}\left(2 x_{j} x_{j+1}\right) $$ where $x_{n+1}=x_{1}$." ['In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1}-x_{k}, k=1, \\ldots, n$. Since $n$ is odd, there exists an index $j$ such that $\\left(x_{j+1}-x_{j}\\right)\\left(x_{j+2}-x_{j+1}\\right) \\geq 0$. Without loss of generality, we may and will assume both factors non-negative, so $x_{j} \\leq x_{j+1} \\leq x_{j+2}$. Consequently,\n\n$$\n\\min _{k=1, \\ldots, n}\\left(x_{k}^{2}+x_{k+1}^{2}\\right) \\leq x_{j}^{2}+x_{j+1}^{2} \\leq 2 x_{j+1}^{2} \\leq 2 x_{j+1} x_{j+2} \\leq \\max _{k=1, \\ldots, n}\\left(2 x_{k} x_{k+1}\\right)\n$$'] [] Text-only Competition True Theorem proof Algebra Math English 388 Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$. "['Let the line $X_{k} T_{k}$ and $\\omega$ meet again at $X_{k}^{\\prime}, k=1,2$, and notice that the tangent $t_{k}$ to $\\omega_{k}$ at $X_{k}$ and the tangent $t_{k}^{\\prime}$ to $\\omega$ at $X_{k}^{\\prime}$ are parallel. Since the $\\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so the $t_{k}^{\\prime}$ are parallel, and consequently the points $X_{1}^{\\prime}$ and $X_{2}^{\\prime}$ coincide (they are not antipodal, since they both lie on the same side of the line $T_{1} T_{2}$ ). The conclusion follows.' 'The circle $\\omega$ is the image of $\\omega_{k}$ under a homothety $h_{k}$ centred at $T_{k}, k=1,2$. The tangent to $\\omega$ at $X_{k}^{\\prime}=h_{k}\\left(X_{k}\\right)$ is therefore parallel to the tangent $t_{k}$ to $\\omega_{k}$ at $X_{k}$. Since the $\\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so $X_{1}^{\\prime}=X_{2}^{\\prime}$; and since the points $X_{k}, T_{k}$ and $X_{k}^{\\prime}$ are collinear, the conclusion follows.\n' 'Invert from $X_{1}$ and use an asterisk to denote images under this inversion. Notice that $\\omega_{k}^{*}$ is the tangent from $X_{2}^{*}$ to $\\omega^{*}$ at $T_{k}^{*}$, and the pole $X_{1}$ lies on the bisectrix of the angle formed by the $\\omega_{k}^{*}$, not containing $\\omega^{*}$. Letting $X_{1} T_{1}^{*}$ and $\\omega^{*}$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_{1} X_{2}^{*} T_{2}^{*}$, and the conclusion follows.']" [] Text-only Competition True Theorem proof Geometry Math English 389 Let $S$ be the set of all positive integers $n$ such that $n^{4}$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$. Prove that there are infinitely many elements of $S$ of each of the forms $7 m, 7 m+1,7 m+2,7 m+5,7 m+6$ and no elements of $S$ of the form $7 m+3$ or $7 m+4$, where $m$ is an integer. ['The conclusion is a consequence of the lemma below which actually provides a recursive description of $S$. The proof of the lemma is at the end of the solution.\n\nLemma. The fourth power of a positive integer $n$ has a divisor in the range $n^{2}+1, n^{2}+2, \\ldots, n^{2}+2 n$ if and only if at least one of the numbers $2 n^{2}+1$ and $12 n^{2}+9$ is a perfect square.\n\nConsequently, a positive integer $n$ is a member of $S$ if and only if $m^{2}-$ $2 n^{2}=1$ or $m^{2}-12 n^{2}=9$ for some positive integer $m$.\n\nThe former is a Pell equation whose solutions are $\\left(m_{1}, n_{1}\\right)=(3,2)$ and\n\n$$\n\\left(m_{k+1}, n_{k+1}\\right)=\\left(3 m_{k}+4 n_{k}, 2 m_{k}+3 n_{k}\\right), \\quad k=1,2,3, \\ldots\n$$\n\nIn what follows, all congruences are modulo 7. Iteration shows that $\\left(m_{k+3}, n_{k+3}\\right) \\equiv\\left(m_{k}, n_{k}\\right)$. Since $\\left(m_{1}, n_{1}\\right) \\equiv(3,2),\\left(m_{2}, n_{2}\\right) \\equiv(3,-2)$, and\n\n\n\n$\\left(m_{3}, n_{3}\\right) \\equiv(1,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and \\pm 2 modulo 7 .\n\nThe other equation is easily transformed into a Pell equation, $m^{\\prime 2}-$ $12 n^{\\prime 2}=1$, by noticing that $m$ and $n$ are both divisible by 3 , say $m=3 m^{\\prime}$ and $n=3 n^{\\prime}$. In this case, the solutions are $\\left(m_{1}, n_{1}\\right)=(21,6)$ and\n\n$$\n\\left(m_{k+1}, n_{k+1}\\right)=\\left(7 m_{k}+24 n_{k}, 2 m_{k}+7 n_{k}\\right), \\quad k=1,2,3, \\ldots\n$$\n\nThis time iteration shows that $\\left(m_{k+4}, n_{k+4}\\right) \\equiv\\left(m_{k}, n_{k}\\right)$. Since $\\left(m_{1}, n_{1}\\right) \\equiv$ $(0,-1),\\left(m_{2}, n_{2}\\right) \\equiv(-3,0),\\left(m_{3}, n_{3}\\right) \\equiv(0,1)$, and $\\left(m_{4}, n_{4}\\right) \\equiv(3,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and $\\pm 1$ modulo 7 .\n\nFinally, since the $n_{k}$ from the two sets of formulae exhaust $S$, by the preceding no integer in the residue classes $\\pm 3$ modulo 7 is a member of $S$.\n\nWe now turn to the proof of the lemma. Let $n$ be a member of $S$, and let $d=n^{2}+m$ be a divisor of $n^{4}$ in the range $n^{2}+1, n^{2}+2, \\ldots, n^{2}+2 n$, so $1 \\leq m \\leq 2 n$. Consideration of the square of $n^{2}=d-m$ shows $m^{2}$ divisible by $d$, so $m^{2} / d$ is a positive integer. Since $n^{2}' 'We prove that the hexagon $A B C D E F$ is the intersection of the equilateral triangles $P Q R$ and $X Y Z$.\n\nLet $d(S, A B)$ denote the signed distance from the point $S$ to the line $A B$, where the negative sign is taken if $A B$ separates $S$ and the hexagon. We define similarly the other distances $(d(S, B C)$, etc). Since $M \\in a$, we have $d(M, Z X)=d(M, Q R)$. In the same way, we have $d(M, X Y)=d(M, R P)$ and $d(M, Y Z)=d(M, P Q)$. Therefore $d(M, Z X)+d(M, X Y)+$ $d(M, Y Z)=d(M, Q R)+d(M, R P)+d(M, P Q)$.\n\nWe now use of the following well-known lemma (which can be easily proved using areas) to deduce that triangles $P Q R$ and $X Y Z$ are congruent.\n\nLemma. The sum of the signed distances from any point to the sidelines of an equilateral triangle (where the signs are taken such that all distances are positive inside the triangle) is constant and equals the length of the altitude.\n\nFor $N=b \\cap d$ we now find $d(N, Z X)=d(N, R P)$ and $d(N, X Y)=d(N, P Q)$. Using again the lemma for the point $N$, we get $d(N, Z X)+d(N, X Y)+d(N, Y Z)=d(N, Q R)+d(N, R P)+$ $d(N, P Q)$. Therefore $d(N, Y Z)=d(N, Q R)$, thus $N \\in f$.\n\nRemark. Instead of using the lemma, it is possible to use some equivalent observation in terms of signed areas.' 'We use the same notations as in Solution A. We will show that $a, c$ and $e$ are concurrent if and only if\n\n$$\nA B+C D+E F=B C+D E+F A,\n$$\n\nwhich clearly implies the problem statement by symmetry.\n\nLet $\\vec{a}$ be the vector of unit length parallel to $a$ directed from $A$ towards the interior of the hexagon. We define analogously $\\vec{b}$, etc. The angle conditions imply that opposite bisectors of the hexagon are parallel, so we have $\\vec{a}\\|\\vec{d}, \\vec{b}\\| \\vec{e}$ and $\\vec{c} \\| \\vec{f}$. Moreover, as in the previous solutions, we know that $\\vec{a}, \\vec{c}$ and $\\vec{e}$ make angles of $120^{\\circ}$ with each other. Let $M_{A}=c \\cap e, M_{C}=e \\cap a$ and $M_{E}=a \\cap c$. Then $M_{A}, M_{C}, M_{E}$ form an equilateral triangle with side length denoted by $s$. Note that the case $s=0$ is equivalent to $a, c$ and $e$ being concurrent.\n\nProjecting $\\overrightarrow{M_{E}} A+\\overrightarrow{A B}=\\overrightarrow{M_{E}} B=\\overrightarrow{M_{E} C}+\\overrightarrow{C B}$ onto $\\vec{e}=-\\vec{b}$, we obtain\n\n$$\n\\overrightarrow{A B} \\cdot \\vec{b}-\\overrightarrow{C B} \\cdot \\vec{b}=\\overrightarrow{M_{E} C} \\cdot \\vec{b}-\\overrightarrow{M_{E}} A \\cdot \\vec{b}=\\overrightarrow{M_{E} A} \\cdot \\vec{e}-\\overrightarrow{M_{E} C} \\cdot \\vec{e}\n$$\n\n\n\nWriting $\\varphi=\\frac{1}{2} \\angle B=\\frac{1}{2} \\angle D=\\frac{1}{2} \\angle F$, we know that $\\overrightarrow{A B} \\cdot \\vec{b}=-A B \\cdot \\cos (\\varphi)$, and similarly $\\overrightarrow{C B} \\cdot \\vec{b}=$ $-C B \\cdot \\cos (\\varphi)$. Because $M_{E} A$ and $M_{E} C$ intersect $e$ at $120^{\\circ}$ angles, we have $\\overrightarrow{M_{E}} A \\cdot \\vec{e}=\\frac{1}{2} M_{E} A$ and $\\overrightarrow{M_{E} C} \\cdot \\vec{e}=\\frac{1}{2} M_{E} C$. We conclude that\n\n$$\n2 \\cos (\\varphi)(A B-C B)=M_{E} C-M_{E} A\n$$\n\nAdding the analogous equalities $2 \\cos (\\varphi)(C D-E D)=M_{A} E-M_{A} C$ and $2 \\cos (\\varphi)(E F-A F)=$ $M_{C} A-M_{C} E$, we obtain\n\n$2 \\cos (\\varphi)(A B+C D+E F-C B-E D-A F)=M_{E} C-M_{E} A+M_{A} E-M_{A} C+M_{C} A-M_{C} E$.\n\nBecause $M_{A}, M_{C}$ and $M_{E}$ form an equilateral triangle with side length $s$, we have $M_{E} C-$ $M_{A} C= \\pm s, M_{C} A-M_{E} A= \\pm s$, and $M_{A} E-M_{C} E= \\pm s$. Therefore, the right hand side $M_{E} C-M_{E} A+M_{A} E-M_{A} C+M_{C} A-M_{C} E$ equals $\\pm s \\pm s \\pm s$, which (irrespective of the choices of the \\pm -signs) is 0 if and only if $s=0$. $\\operatorname{Because} \\cos (\\varphi) \\neq 0$, we conclude that\n\n$$\nA B+C D+E F=C B+E D+A F \\Longleftrightarrow s=0 \\Longleftrightarrow a, c, e \\text { concurrent, }\n$$\n\nas desired.\n\nRemark. Equalities used in the solution could appear in different forms, in particular, in terms of signed lengths.\n\nRemark. Similar solutions could be obtained by projecting onto the line perpendicular to $b$ instead of $b$.' ""We use the the same notations as in previous solutions and the fact that $a \\| d$, $b \\| e$ and $c \\| f$ make angles of $120^{\\circ}$. Also, we may assume that $E$ and $C$ are not symmetric in $a$ (if they are, the entire figure is symmetric and the conclusion is immediate).\n\nWe consider two mappings: the first one $s: a \\rightarrow B C \\rightarrow d$ sending $A^{\\prime} \\mapsto B^{\\prime} \\mapsto S$ is defined such that $A^{\\prime} B^{\\prime} \\| A B$ and $B^{\\prime} S \\| b$, and the second one $t: a \\rightarrow E F \\rightarrow d$ sending $A^{\\prime} \\mapsto F^{\\prime} \\mapsto T$ is defined such that $A^{\\prime} F^{\\prime} \\| A F$ and $F^{\\prime} T \\| f$. Both maps are affine linear since they are compositions of affine tranformations. We will prove that they coincide by finding two distinct points $A^{\\prime}, A^{\\prime \\prime} \\in a$ for which $s\\left(A^{\\prime}\\right)=t\\left(A^{\\prime}\\right)$ and $s\\left(A^{\\prime \\prime}\\right)=t\\left(A^{\\prime \\prime}\\right)$. Then we will obtain that $s(A)=t(A)$, which by construction implies that the bisectors of $\\angle B, \\angle D$ and $\\angle F$ are concurrent.\n\nWe will choose $A^{\\prime}$ to be the reflection of $C$ in $e$ and $A^{\\prime \\prime}$ to be the reflection of $E$ in $c$. They are distinct since otherwise $C$ and $E$ would be symmetric in $a$. Applying the above maps $a \\rightarrow B C$ and $a \\rightarrow E F$ to $A^{\\prime}$, we get points $B^{\\prime}$ and $F^{\\prime}$ such that $A^{\\prime} B^{\\prime} C D E F^{\\prime}$ satisfies the problem statement. However, this hexagon is symmetric in $e$, hence the bisectors of $\\angle B^{\\prime}, \\angle D, \\angle F^{\\prime}$ are concurrent and $s\\left(A^{\\prime}\\right)=t\\left(A^{\\prime}\\right)$. The same reasoning yields $s\\left(A^{\\prime \\prime}\\right)=t\\left(A^{\\prime \\prime}\\right)$, which finishes the solution.\n\nRemark. This solution is based on the fact that that two specific affine linear maps coincide. Here it was proved by exhibiting two points where they coincide. One could prove it in another way, exhibiting one such point and proving that the 'slopes' are equal.\n\nRemark. There are similar solutions where claims and proofs could be presented in more 'elementary' terms. For example, an elementary reformulation of the 'slopes' being equal is: if $b^{\\prime}$ passes through $B^{\\prime}$ parallel to $b$, and $f^{\\prime}$ passes through $F^{\\prime}$ parallel to $f$, then the line through $b \\cap f$ and $b^{\\prime} \\cap f^{\\prime}$ is parallel to $a$ (which is parallel to $d$ ).\n\nRemark. Circles $\\omega_{a}, \\omega_{c}$ and $\\omega_{e}$ could be helpful in some other solutions. In particular, the movement of $A$ along $a$ in current solution is equivalent to varying $r_{a}$."" 'We use the same notations as in previous solutions.\n\nSince the sum of the angles of a convex hexagon is $720^{\\circ}$, from the angle conditions we get $\\angle B+\\angle C=720^{\\circ} / 3=240^{\\circ}$. From $\\angle B+\\angle C=240^{\\circ}$ it follows that the angle between $c$ and $b$ equals $60^{\\circ}$. The same is analogously true for other pairs of bisectors of neighboring angles.\n\nConsider the points $O_{a} \\in a, O_{c} \\in c, O_{e} \\in e$, each at the same distance $d^{\\prime}$ from $M$, where $d^{\\prime}>\\max \\{M A, M C, M E\\}$, and such that the rays $A O_{a}, C O_{c}, E O_{e}$ point out of the hexagon. By construcion, $O_{a}$ and $O_{c}$ are symmetrical in $e$, hence $O_{a} O_{c} \\perp b$. Similarly, $O_{c} O_{e} \\perp d, O_{e} O_{a} \\perp f$. Thus it suffices to prove that perpendiculars from $B, D, F$ to the sidelines of $\\triangle O_{a} O_{c} O_{e}$ are concurrent. By a well-known criteria, this condition is equivalent to equality\n\n$$\nO_{a} B^{2}-O_{c} B^{2}+O_{c} D^{2}-O_{e} D^{2}+O_{e} F^{2}-O_{a} F^{2}=0 \\tag{*}\n$$\n\nTo prove $(*)$ consider a circle $\\omega_{a}$ centered at $O_{a}$ and tangent to $A B$ and $A F$ and define circles $\\omega_{c}$ and $\\omega_{e}$ in the same way. Rewrite $O_{a} B^{2}$ as $r_{a}^{2}+B_{a} B^{2}$, where $r_{a}$ is the radius of $\\omega_{a}$, and $B_{a}$ is the touch point of $\\omega_{a}$ with $A B$. Using similar notation for the other tangent points, transform $(*)$ into\n\n$$\nB_{a} B^{2}-B_{c} B^{2}+D_{c} D^{2}-D_{e} D^{2}+F_{e} F^{2}-F_{a} F^{2}=0 . \\tag{**}\n$$\n\nFurthermore, $\\angle O_{c} O_{a} B_{a}=\\angle M O_{a} B_{a}+\\angle O_{c} O_{a} M=\\left(90^{\\circ}-\\varphi\\right)+30^{\\circ}=120^{\\circ}-\\varphi$, where $\\varphi=$ $\\frac{1}{2} \\angle A$. (Note that $\\varphi>30^{\\circ}$, since $A B C D E F$ is convex.) By analogous arguments, $\\angle O_{a} O_{c} B_{c}=$ $\\angle O_{e} O_{c} D_{c}=\\angle O_{c} O_{e} D_{e}=\\angle O_{a} O_{e} F_{e}=\\angle O_{e} O_{a} F_{a}=120^{\\circ}-\\varphi$. It follows that rays $O_{a} B_{a}$ and $O_{c} B_{c}$ (being symmetrical in $e$ ) intersect at $U_{e} \\in e$ forming an isosceles triangle $\\triangle O_{a} U_{e} O_{c}$. Similarly define $\\triangle O_{c} U_{a} O_{e}$ and $\\triangle O_{e} U_{c} O_{a}$. These triangles are congruent (equal bases and corresponding angles). Therefore we have $O_{a} U_{c}=U_{c} O_{e}=O_{e} U_{a}=U_{a} O_{c}=O_{c} U_{e}=U_{e} O_{a}$. Moreover, we also have $B_{a} U_{e}=O_{a} U_{e}-r_{a}=O_{a} U_{c}-r_{a}=F_{a} U_{c}=x$, and thus similarly $D_{c} U_{a}=B_{c} U_{e}=y$, $F_{e} U_{c}=D_{e} U_{a}=z$.\n\nNow from quadrilateral $B B_{a} U_{e} B_{c}$ with two opposite right angles $B_{a} B^{2}-B_{c} B^{2}=B_{c} U_{e}^{2}-B_{a} U_{e}^{2}=$ $y^{2}-x^{2}$. Similarly $D_{c} D^{2}-D_{e} D^{2}=D_{e} U_{a}^{2}-D_{c} U_{a}^{2}=z^{2}-y^{2}$ and $F_{e} F^{2}-F_{a} F^{2}=F_{a} U_{c}^{2}-F_{e} U_{c}^{2}=$ $x^{2}-z^{2}$. Finally, we substitute this into $(* *)$, and the claim is proved.']" [] Text-only Competition True Theorem proof Geometry Math English 391 "A permutation of the integers $1,2, \ldots, m$ is called fresh if there exists no positive integer $k90^{\circ}$. The circumcircle $\Gamma$ of $A B C$ has radius $R$. There is a point $P$ in the interior of the line segment $A B$ such that $P B=P C$ and the length of $P A$ is $R$. The perpendicular bisector of $P B$ intersects $\Gamma$ at the points $D$ and $E$. Prove that $P$ is the incentre of triangle $C D E$." "['The angle bisector of $\\angle E C D$ intersects the circumcircle of $C D E$ (which is $\\Gamma$ ) at the midpoint $M$ of arc $D B E$. It is well-known that the incentre is the intersection of the angle bisector segment $C M$ and the circle with centre at $M$ and passing through $D, E$. We will verify this property for $P$.\n\n\n\nBy the conditions we have $A P=O A=O B=O C=O D=O E=R$. Both lines $O M$ and $A P B$ are perpendicular to $E D$, therefore $A P \\| O M$; in the quadrilateral $A O M P$ we have $A P=O A=A M=R$ and $A P \\| O M$, so $A O M P$ is a rhombus and its fourth side is $P M=R$. In the convex quadrilateral $O M B P$ we have $O M \\| P B$, so $O M B P$ is a symmetric trapezoid; the perpendicular bisector of its bases $A O$ and $P B$ coincide. From this symmetry we obtain $M D=O D=R$ and $M E=O E=R$. (Note that the triangles $O E M$ ad $O M D$ are equilateral.) We already have $M P=M D=M E=R$, so $P$ indeed lies on the circle with center $M$ and passing through $D, E$. (Notice that this circle is the reflection of $\\Gamma$ about $D E$.)\n\nFrom $P B=P C$ and $O B=O C$ we know that $B$ and $C$ are symmetrical about $O P$; from the rhombus $A O M P$ we find that $A$ and $M$ are also symmetrical about $O P$. By reflecting the collinear points $B, P, A$ (with $P$ lying in the middle) we get that $C, P, M$ are collinear (and $P$ is in the middle). Hence, $P$ lies on the line segments $C M$.' 'Let $X$ be the second intersection of $C P$ with $\\Gamma$. Using the power of the point $P$ in the circle $\\Gamma$ and the fact that $P B=P C$, we find that $P X=P A=R$. The quadrilateral $A O X P$ has four sides of equal length, so it is a rhombus and in particular $O X$ is parallel to $A P$. This proves that $O X B P$ is a trapezoid, and because the diagonals $P X$ and $O B$ have equal length, this is even an isosceles trapezoid. Because of that, $D E$ is not only the perpendicular bisector of $P B$, but also of $O X$.\n\n\n\nIn particular we have $X D=X P=X O=X E=R$, which proves that $X$ is the middle of the arc $D E$ and $P$ belongs to the circle with center $X$ going through $D$ and $E$. These properties, together with the fact that $C, P, X$ are collinear, determine uniquely the incenter of $C D E$.' 'Let $Y$ be the circumcenter of triangle $B P C$. Then from $Y B=Y P$ it follows that $Y$ lies on $D E$ (we assume $D$ lies in between $Y$ and $E$ ), and from $Y B=Y C$ it follows that $Y$ lies on $O P$, where $O$ is the center of $\\Gamma$.\n\nFrom $\\angle A O C=2 \\angle A B C=\\angle A P C$ (because $\\angle P B C=\\angle P C B)$ we deduce that $A O P C$ is a cyclic quadrilateral, and from $A P=R$ it follows that $A O P C$ is an isosceles trapezoid. We now find that $\\angle Y C P=\\angle Y P C=180^{\\circ}-\\angle O P C=180^{\\circ}-\\angle A C P$, so $Y$ lies on $A C$.\n\n\n\nPower of a point gives $Y O \\cdot Y P=Y C \\cdot Y A=Y D \\cdot Y E$, so $D, P, O$ and $E$ are concyclic. It follows that $2 \\angle D A E=\\angle D O E=\\angle D P E=\\angle D B E=180^{\\circ}-\\angle D A E$, so $\\angle D A E=60^{\\circ}$. We can now finish the proof by angle chasing.\n\nFrom $A B \\perp D E$ we have $\\angle A O D+\\angle B O E=180^{\\circ}$ and from $\\angle D O E=2 \\angle D A E=120^{\\circ}$ it follows that $\\angle B O D+\\angle B O E=120^{\\circ}$. It follows that $\\angle A O D-\\angle B O D=180^{\\circ}-120^{\\circ}=60^{\\circ}$. Let $\\angle O A B=\\angle O B A=2 \\beta$; then $\\angle A O D+\\angle B O D=\\angle A O B=180^{\\circ}-4 \\beta$. Together with $\\angle A O D-\\angle B O D=60^{\\circ}$, this yields $\\angle A O D=120^{\\circ}-2 \\beta$ and $\\angle B O D=60^{\\circ}-2 \\beta$. We now find $\\angle A E D=\\frac{1}{2} \\angle A O D=60^{\\circ}-\\beta$, which together with $\\angle D A E=60^{\\circ}$ yields $\\angle A D E=60^{\\circ}+\\beta$. From the isosceles trapezoid $A O P C$ we have $\\angle C D A=\\angle C B A=\\frac{1}{2} \\angle C P A=\\frac{1}{2} \\angle P A O=\\beta$, so $\\angle C D E=\\angle C D A+\\angle A D E=\\beta+60^{\\circ}+\\beta=60^{\\circ}+2 \\beta$.\n\nFrom $\\angle B O D=60^{\\circ}-2 \\beta$ we deduce that $\\angle B E D=30^{\\circ}-\\beta$; together with $\\angle D B E=120^{\\circ}$ this yields $\\angle E D B=30^{\\circ}+\\beta$. We now see that $\\angle P D E=\\angle B D E=30^{\\circ}+\\beta=\\frac{1}{2} \\angle C D E$, so $P$ is on the angle bisector of $\\angle C D E$. Similarly, $P$ lies on the angle bisector of $\\angle C E D$, so $P$ is the incenter of $C D E$.' 'We draw the lines $D P$ and $E P$ and let $D^{\\prime}$ resp. $E^{\\prime}$ be the second intersection point with $\\Gamma$.\n\n\n\nThe triangles $A P D^{\\prime}$ and $D P B$ are similar, and the triangles $A P E^{\\prime}$ and $E P B$ are also similar, hence they are all isosceles and it follows that $E^{\\prime}, O, P, D^{\\prime}$ lie on a circle with center $A$. In particular $A O D^{\\prime}$ and $A O E^{\\prime}$ are equilateral triangles. Angle chasing gives\n\n$$\n\\begin{gathered}\n\\angle C D P=\\angle C D D^{\\prime}=\\frac{1}{2} \\angle C O D^{\\prime}=\\frac{1}{2}\\left(60^{\\circ}+\\angle C O A\\right) \\\\\n\\angle E D P=\\angle E D D^{\\prime}=\\angle E E^{\\prime} D^{\\prime}=\\angle P E^{\\prime} D^{\\prime}=\\frac{1}{2} \\angle P A D^{\\prime}=\\frac{1}{2}\\left(60^{\\circ}+\\angle P A O\\right)\n\\end{gathered}\n$$\n\nSimilarly we prove $\\angle C E P=\\frac{1}{2}\\left(60^{\\circ}-\\angle C O A\\right)$ and $\\angle D E P=\\frac{1}{2}\\left(60^{\\circ}-\\angle P A O\\right)$ so if we can prove that $\\angle C O A=\\angle P A O$, we will have proven that $P$ belongs to the angle bisector of $\\angle C E D$ and to the angle bisector of $\\angle C D E$, which is enough to prove that $P$ is the incenter of the triangle $C D E$.\n\nLet $\\beta=\\angle A B C=\\angle P C B$. We have $\\angle A P C=2 \\beta$ and $\\angle A O C=2 \\beta$, so $A O P C$ is an inscribed quadrilateral. Moreover, since the diagonals $A P$ and $C O$ have equal length, this is actually an isosceles trapezoid, and hence $\\angle P A O=\\angle C P A=\\angle C O A$ which concludes the proof.' 'Without loss of generality we assume that $D$ and $C$ are in the same half-plane regarding line $A B$.\n\nSince $P C=B P$ and $A B C D$ is inscribed quadrilateral we have $\\angle P C B=\\angle C B P=\\angle C E A=\\alpha$. As in the other solutions, $A O P C$ is an isosceles trapezoid and $2 \\alpha=\\angle C P A=\\angle P C O$\n\n\n\nLet $K$ be intersection of $E O$ and $\\Gamma$. Then $\\angle K D E=90^{\\circ}, A B \\| D K$ and $K D B A$ is isosceles trapezoid. We obtain $D P=B D=A K$, which implies that $D P A K$ is a parallelogram and hence $D K=P A=R=O D=O K$. We see that $D O K$ is an equilateral triangle. Then $\\angle E C D=\\angle E K D=60^{\\circ}$.\n\nFurther we prove that $P C$ bisects $\\angle E C D$ using $\\angle D C B=\\angle D K B=\\angle K D A$ (from isosceles trapeziod $D K A B$ ) and that $\\angle K E C=\\angle O E C=\\angle O C E$ (from isosceles triangle $O C E)$ :\n\n$$\n\\angle D C P=\\angle D C B+\\angle B C P=\\angle K D A+\\alpha=\\angle K E C+\\angle C E A+\\alpha=\\angle O C E+2 \\alpha=\\angle P C E\n$$\n\nFurther by $\\angle D C P=\\angle P C E=\\frac{1}{2} \\angle E C D=30^{\\circ}$ distances between $P$ and sides $\\triangle C D E$ are $\\frac{1}{2} P C=\\frac{1}{2} P B$ (as ratio between cathetus and hypotenuse in right triangle with angles $60^{\\circ}$ and $30^{\\circ}$ ). So, we have found incentre.' 'Assume $A B$ is parallel to the horizontal axis, and that $\\Gamma$ is the unit circle. Write $f(\\theta)$ for the point $(\\cos (\\theta), \\sin (\\theta))$ on $\\Gamma$. As in Solution $\\mathrm{C}$, assume that $\\angle A O B=180^{\\circ}-4 \\beta$; then we can take $B=f(2 \\beta)$ and $A=f\\left(180^{\\circ}-2 \\beta\\right)$. We observe that $A O P C$ is an isosceles trapezoid, which we use to deduce that $\\angle A B C=\\frac{1}{2} \\angle A P C=\\frac{1}{2} \\angle O A B=\\beta$. We now know that $C=f\\left(180^{\\circ}-4 \\beta\\right)$.\n\nThe point $P$ lies on $A B$ with $A P=R=1$, so $P=\\left(\\cos \\left(180^{\\circ}-2 \\beta\\right)+1, \\sin (2 \\beta)\\right)=(1-$ $\\cos (2 \\beta), \\sin (2 \\beta))$. The midpoint of $B P$ therefore has coordinates $\\left(\\frac{1}{2}, \\sin (2 \\beta)\\right)$, so $D$ and $E$ have $x$-coordinate $\\frac{1}{2}$. Without loss of generality, we take $D=f\\left(60^{\\circ}\\right)$ and $E=f\\left(-60^{\\circ}\\right)$.\n\nWe have now obtained coordinates for all points in the problem, with one free parameter $(\\beta)$. To show that $P$ is the incenter of $C D E$, we will show that $P$ lies on the bisector of $\\angle C D E$; analogously, one can show that $P$ lies on the bisector of angle $C E D$. The bisector of angle $C D E$ passes through the midpoint $M$ of the arc $C E$ not containing $D E$; because $C=f\\left(180^{\\circ}-4 \\beta\\right)$ and $E=f\\left(-60^{\\circ}\\right)$, we have $M=f\\left(240^{\\circ}-2 \\beta\\right)$.\n\nIt remains to show that $P=(1-\\cos (2 \\beta), \\sin (2 \\beta))$ lies on the line connecting the points $D=$ $\\left(\\cos \\left(60^{\\circ}\\right), \\sin \\left(60^{\\circ}\\right)\\right)$ and $M=\\left(\\cos \\left(240^{\\circ}-2 \\beta\\right), \\sin \\left(240^{\\circ}-2 \\beta\\right)\\right)=\\left(-\\cos \\left(60^{\\circ}-2 \\beta\\right),-\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right)$. The equation for the line $D M$ is\n\n$\\left(Y+\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right)\\left(\\cos \\left(60^{\\circ}\\right)+\\cos \\left(60^{\\circ}-2 \\beta\\right)\\right)=\\left(\\sin \\left(60^{\\circ}\\right)+\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right)\\left(X+\\cos \\left(60^{\\circ}-2 \\beta\\right)\\right)$,\n\nwhich, using the fact that $\\cos \\left(60^{\\circ}\\right)+\\cos \\left(60^{\\circ}-2 \\beta\\right)=2 \\cos (\\beta) \\cos \\left(60^{\\circ}-\\beta\\right)$ and $\\sin \\left(60^{\\circ}\\right)+\\sin \\left(60^{\\circ}-\\right.$ $2 \\beta)=2 \\cos (\\beta) \\sin \\left(60^{\\circ}-\\beta\\right)$, simplifies to\n\n$$\n\\left(Y+\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right) \\cos \\left(60^{\\circ}-\\beta\\right)=\\left(X+\\cos \\left(60^{\\circ}-2 \\beta\\right)\\right) \\sin \\left(60^{\\circ}-\\beta\\right) \\text {. }\n$$\n\nBecause $\\cos \\left(60^{\\circ}-2 \\beta\\right) \\sin \\left(60^{\\circ}-\\beta\\right)-\\sin \\left(60^{\\circ}-2 \\beta\\right) \\cos \\left(60^{\\circ}-\\beta\\right)=\\sin (\\beta)$, this equation further simplifies to\n\n$$\nY \\cos \\left(60^{\\circ}-\\beta\\right)-X \\sin \\left(60^{\\circ}-\\beta\\right)=\\sin (\\beta) .\n$$\n\n\n\nPlugging in the coordinates of $P$, i.e., $X=1-\\cos (2 \\beta)$ and $Y=\\sin (2 \\beta)$, shows that $P$ is on this line: for this choice of $X$ and $Y$, the left hand side equals $\\sin \\left(60^{\\circ}+\\beta\\right)-\\sin \\left(60^{\\circ}-\\beta\\right)=$ $2 \\cos \\left(60^{\\circ}\\right) \\sin (\\beta)$, which is indeed equal to $\\sin (\\beta)$. So $P$ lies on the bisector $D M$ of $\\angle C D E$, as desired.' 'Let $\\Gamma$ be the complex unit circle and let $A B$ be parallel with the real line and $0<\\varphi=\\arg b<\\frac{\\pi}{2}$. Then\n\n$$\n|b|=1, \\quad a=-\\bar{b}, \\quad p=a+1=1-\\bar{b}\n$$\n\nFrom $\\operatorname{Re} d=\\operatorname{Re} e=\\operatorname{Re} \\frac{p+b}{2}=\\frac{1}{2}$ we get that $d=\\frac{1}{2}+\\frac{\\sqrt{3}}{2} i$ and $e=\\frac{1}{2}-\\frac{\\sqrt{3}}{2} i$ are conjugate 6th roots of unity; $d^{3}=e^{3}=-1, d+e=1, d^{2}=-e, e^{2}=-d$ etc.\n\nPoint $C$ is the reflection of $B$ in line $O P$. From $\\arg p=\\arg (1-\\bar{b})=\\frac{1}{2}(\\pi-\\varphi)$, we can get $\\arg c=2 \\arg p-\\arg b=\\pi-2 \\varphi$, so $c=-\\bar{b}^{2}$.\n\nNow we can verify that $E P$ bisects $\\angle C E D$. This happens if and only if $(p-e)^{2}(\\bar{c}-\\bar{e})(\\bar{d}-\\bar{e})$ is real. Since $\\bar{d}-\\bar{e}=-\\sqrt{3} i$, this is equivalent with $\\operatorname{Re}\\left[(p-e)^{2}(\\bar{c}-\\bar{e})\\right]=0$. Here\n\n$$\n\\begin{aligned}\n(p-e)^{2}(\\bar{c}-\\bar{e}) & =(1-\\bar{b}-e)^{2}\\left(-b^{2}-d\\right)=(d-\\bar{b})^{2}\\left(-b^{2}-d\\right) \\\\\n& =-|b|^{4}+2 d|b|^{2} b-d^{2} b^{2}-d \\bar{b}^{2}+2 d^{2} \\bar{b}-d^{3} \\\\\n& =-1-2 d b+\\bar{d} b^{2}-d \\bar{b}^{2}-2 \\overline{d b}+1 \\\\\n& =-2(d b-\\overline{d b})+\\left(\\bar{d} b^{2}-d \\bar{b}^{2}\\right)\n\\end{aligned}\n$$\n\nwhose real part is zero. It can be proved similarly that $D P$ bisects $\\angle E D C$.']" [] Text-only Competition True Theorem proof Geometry Math English 393 "Let $A B C$ be a triangle with $C A=C B$ and $\angle A C B=120^{\circ}$, and let $M$ be the midpoint of $A B$. Let $P$ be a variable point on the circumcircle of $A B C$, and let $Q$ be the point on the segment $C P$ such that $Q P=2 Q C$. It is given that the line through $P$ and perpendicular to $A B$ intersects the line $M Q$ at a unique point $N$. Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$." "['Let $O$ be the circumcenter of $A B C$. From the assumption that $\\angle A C B=120^{\\circ}$ it follows that $M$ is the midpoint of $C O$.\n\nLet $\\omega$ denote the circle with center in $C$ and radius $C O$. This circle in the image of the circumcircle of $A B C$ through the translation that sends $O$ to $C$. We claim that $N$ lies on $\\omega$.\n\n\nLet us consider the triangles $Q N P$ and $Q M C$. The angles in $Q$ are equal. Since $N P$ is parallel to $M C$ (both lines are perpendicular to $A B$ ), it turns out that $\\angle Q N P=\\angle Q M C$, and hence the two triangles are similar. Since $Q P=2 Q C$, it follows that\n\n$$\nN P=2 M C=C O\n$$\n\nwhich proves that $N$ lies on $\\omega$.' 'Let $M^{\\prime}$ denote the symmetric of $M$ with respect to $O$.\n\nLet us consider the quadrilateral $M M^{\\prime} P N$. The lines $M M^{\\prime}$ and $N P$ are parallel by construction. Also the lines $P M^{\\prime}$ and $N M$ are parallel (homothety from $C$ with coefficient 3). It follows that $M M^{\\prime} P N$ is a parallelogram, and hence $P N=M M^{\\prime}=O C$.' 'There are many computation approaches to this problem. For example, we can set Cartesian coordinates so that\n\n$$\nA=\\left(-\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right), \\quad B=\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right), \\quad C=(0,1), \\quad M=\\left(0, \\frac{1}{2}\\right)\n$$\n\nSetting $P=(a, b)$, we obtain that $Q=(a / 3,(2+b) / 3)$. The equation of the line through $P$ and perpendicular to $A B$ is $x=a$. The equation of the line $M Q$ (if $a \\neq 0$ ) is\n\n$$\ny-\\frac{1}{2}=\\frac{x}{a}\\left(\\frac{1}{2}+b\\right)\n$$\n\nThe intersection of the two lines is therefore\n\n$$\nN=(a, 1+b)=P+(0,1)\n$$\n\nThis shows that the map $P \\rightarrow N$ in the translation by the vector $(0,1)$. This result is independent of the position of $P$ (provided that $a \\neq 0$, because otherwise $N$ is not well-defined).\n\nWhen $P$ lies on the circumcircle of $A B C$, with the exception of the two points with $a=0$, then necessarily $N$ lies on the translated circle (which is the circle with center in $C$ and radius 1).']" [] Text-only Competition True Theorem proof Geometry Math English 394 "Consider the set $$ A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\} $$ Prove that every integer $x \geq 2$ can be written as the product of one or more elements of $A$, which are not necessarily different." ['Every integer $x \\geq 2$ can be written as the telescopic product of $x-1$ elements of $A$ as\n\n$$\nx=\\left(1+\\frac{1}{x-1}\\right) \\cdot\\left(1+\\frac{1}{x-2}\\right) \\cdot \\ldots \\cdot\\left(1+\\frac{1}{2}\\right) \\cdot\\left(1+\\frac{1}{1}\\right),\n$$\n\nwhich is enough to establish the solution.'] [] Text-only Competition True Theorem proof Algebra Math English 395 "Consider the set $$ A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\} $$ For every integer $x \geq 2$, let $f(x)$ denote the minimum integer such that $x$ can be written as the product of $f(x)$ elements of $A$, which are not necessarily different. Prove that there exist infinitely many pairs $(x, y)$ of integers with $x \geq 2, y \geq 2$, and $$ f(x y)4 k$. Since $y>2^{4 k-1}$, from (Q2.1) we already know that $f(y) \\geq 4 k$, and hence we just need to exclude that $f(y)=4 k$. Let us assume that we can represent $y$ in the form $a_{1} \\cdot \\ldots \\cdot a_{4 k}$ with every $a_{i}$ in $A$. At least one of the $a_{i}$ is not 2 (otherwise the product would be a power of 2 , while $y$ is odd), and hence it is less than or equal to $3 / 2$. It follows that\n\n$$\na_{1} \\cdot \\ldots \\cdot a_{4 k} \\leq 2^{4 k-1} \\cdot \\frac{3}{2}=15 \\cdot \\frac{2^{4 k-2}}{5}<\\frac{2^{4 k+2}}{5}2 k$. Since $y>2^{2 k-1}$ (for $k \\geq 1$ ), from (Q2.1) we already know that $f(y) \\geq 2 k$, and hence we just need to exclude that $f(y)=2 k$. Let us assume that we can represent $y$ in the form $a_{1} \\cdot \\ldots \\cdot a_{2 k}$. At least one of the factors is not 2 , and hence it is less than or equal to $3 / 2$. Thus when $k$ is large enough it follows that\n\n$$\na_{1} \\cdot \\ldots \\cdot a_{2 k} \\leq 2^{2 k-1} \\cdot \\frac{3}{2}=\\frac{3}{4} \\cdot 2^{2 k}<2^{2 k}-2^{k}7$, and any other product of at most three elements of $A$ does not exceed $2^{2} \\cdot \\frac{3}{2}=6<7$. On the other hand, $f(49) \\leq 7$ since $49=2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot \\frac{3}{2} \\cdot \\frac{49}{48}$.\n\nSuppose by contradiction that there exist only finitely many pairs $(x, y)$ that satisfy $f(x y)<$ $f(x)+f(y)$. This implies that there exists $M$ large enough so that whenever $a>M$ or $b>M$ holds we have $f(a b)=f(a)+f(b)$ (indeed, it is clear that the reverse inequality $f(a b) \\leq f(a)+f(b)$ is always satisfied).\n\nNow take any pair $(x, y)$ that satisfies $f(x y)M$ be any integer. We obtain\n\n$$\nf(n)+f(x y)=f(n x y)=f(n x)+f(y)=f(n)+f(x)+f(y),\n$$\n\nwhich contradicts $f(x y)\n\nLet us show now that at most $2^{n}-n-1$ moves are possible. To this end, let us identify a line of contestants with a permutation $\\sigma$ of $\\{1, \\ldots, n\\}$. To each permutation we associate the set of reverse pairs\n\n$$\nR(\\sigma):=\\{(i, j): 1 \\leq i\\sigma(j)\\}\n$$\n\nand the nonnegative integer\n\n$$\nW(\\sigma):=\\sum_{(i, j) \\in R(\\sigma)} 2^{i}\n$$\n\nwhich we call the total weight of the permutation. We claim that the total weight decreases after any move of the contestants. Indeed, let us assume that $C_{i}$ moves forward in the queue, let $\\sigma$ be the permutation before the move, and let $\\sigma^{\\prime}$ denote the permutation after the move. Since $C_{i}$ jumps over exactly $i$ contestants, necessarily she jumps over at least one contestant $C_{j}$ with index\n\n\n\n$j>i$. This means that the pair $(i, j)$ is reverse with respect to $\\sigma$ but not with respect to $\\sigma^{\\prime}$, and this yields a reduction of $2^{i}$ in the total weight. On the other hand, the move by $C_{i}$ can create new reverse pairs of the form $(k, i)$ with $ki$.\n\nLet us show now that the process ends after a finite number of moves. Let us assume that this is not the case. Then at least one contestant moves infinitely many times. Let $i_{0}$ be the largest index such that $C_{i_{0}}$ moves infinitely many times. Then necessarily $C_{i_{0}}$ jumps infinitely many times over some fixed $C_{j_{0}}$ with $j_{0}>i_{0}$. On the other hand, we know that $C_{j_{0}}$ makes only a finite number of moves, and therefore she can precede $C_{i_{0}}$ in the line only a finite number of times, which is absurd.\n\nIn order to estimate from above the maximal number of moves, we show that the contestant $C_{i}$ can make at most $2^{n-i}-1$ moves. Indeed, let us argue by ""backward extended induction"". To begin with, we observe that the estimate is trivially true for $C_{n}$ because she has no legal move.\n\nLet us assume now that the estimate has been proved for $C_{i}, C_{i+1}, \\ldots, C_{n}$, and let us prove it for $C_{i-1}$. When $C_{i-1}$ moves, at least one contestant $C_{j}$ with $j>i-1$ must precede her in the line. The initial configuration can provide at most $n-i$ contestants with larger index in front of $C_{i-1}$, which means at most $n-i$ moves for $C_{i-1}$. All other moves are possible only if some contestant in the range $C_{i}, C_{i+1}, \\ldots, C_{n}$ jumps over $C_{i-1}$ during her moves. As a consequence, the total number of moves of $C_{i-1}$ is at most\n\n$$\nn-i+\\sum_{k=i}^{n}\\left(2^{n-k}-1\\right)=2^{n-i+1}-1\n$$\n\nSumming over all indices we obtain that\n\n$$\n\\sum_{i=1}^{n}\\left(2^{n-i}-1\\right)=2^{n}-n-1\n$$\n\nwhich gives an estimate for the total number of moves.\n\n\nComment In every move of the example, the moving contestant jumps over exactly one contestant with larger index (and as a consequence over all contestants with smaller index).']" [] Text-only Competition True Theorem proof Combinatorics Math English 397 "A domino is a $1 \times 2$ or $2 \times 1$ tile. Let $n \geq 3$ be an integer. Dominoes are placed on an $n \times n$ board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some $k \geq 1$ such that each row and each column has a value of $k$. Prove that a balanced configuration exists for every $n \geq 3$, and find the minimum number of dominoes needed in such a configuration." ['The minimal number of dominoes required in a balanced configuration is $2 n / 3$ if $n$ is a multiple of 3 , and $2 n$ otherwise.\n\nIn order to show that this number is necessary, we count in two different ways the number of elements of the set $S$ of all pairs $(\\ell, d)$, where $\\ell$ is a row or a column of the board, and $d$ is a domino that covers at least one cell of that row or column. On the one hand, since each row or column intersects the same number $k$ of dominoes, the set $S$ has $2 n k$ elements. On the other hand, since each domino intersects 3 rows/columns, the set $S$ has $3 D$ elements, where $D$ is the total number of dominoes on the board. This leads to the equality\n\n$$\n2 n k=3 D \\text {. }\n$$\n\nIf $n$ is a multiple of 3 , from the trivial inequality $k \\geq 1$ we obtain that $D \\geq 2 n / 3$. If $n$ is not a multiple of 3 , then $k$ is a multiple of 3 , which means that $k \\geq 3$ and hence $D \\geq 2 n$.\n\nNow we need to exhibit a balanced configuration with this number of dominoes. The following diagram shows a balanced configuration with $n=3$ and $k=1$.\n\n\n\nIf $n$ is any multiple of 3 , we can obtain a balanced configuration with $k=1$ by using $n / 3$ of these $3 \\times 3$ blocks along the principal diagonal of the board.\n\nThe following diagrams show balanced configurations with $k=3$ and $n \\in\\{4,5,6,7\\}$.\n\n\nAny $n \\geq 8$ can be written in the form $4 A+r$ where $A$ is a positive integer and $r \\in\\{4,5,6,7\\}$. Therefore, we can obtain a balanced configuration with $n \\geq 8$ and $k=3$ by using one block with size $r \\times r$, and $A$ blocks with size $4 \\times 4$ along the principal diagonal of the board. In particular, this construction covers all the cases where $n$ is not a multiple of 3 .'] [] Text-only Competition True Theorem proof Combinatorics Math English 398 "Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$. Prove that $\angle A B P=\angle Q B C$." "['Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\\Omega$ and $\\Gamma$, and let $U$ be the intersection of $\\Omega$ and $A B$.\n\n\n\nThe proof can be divided in two steps:\n\n1. Proving that $M P \\cdot M Q=M B^{2}$.\n\nIt is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\\Omega$ to $\\Gamma$ sends $U$ to the point of $\\Gamma$ where the tangent to $\\Gamma$ is parallel to $A B$, and this point is $M$ ), and\n\n$$\nM V \\cdot M U=M A^{2}=M B^{2} .\n$$\n\nThis follows from the similitude between the triangles $\\triangle M A V$ and $\\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \\cdot M Y$ is independent of the choice of the line.\n\nComputing the power of $M$ with respect to $\\Omega$ we obtain that\n\n$$\nM P \\cdot M Q=M U \\cdot M V=M B^{2}\n$$\n\n2. Conclude the proof given that $M P \\cdot M Q=M B^{2}$.\n\nThe relation $M P \\cdot M Q=M B^{2}$ in turn implies that triangle $\\triangle M B P$ is similar to triangle $\\triangle M Q B$, and in particular $\\angle M B P=\\angle M Q B$. Keeping into account that $\\angle M C B=$ $\\angle M B A$, we finally conclude that\n\n$$\n\\angle Q B C=\\angle M Q B-\\angle M C B=\\angle M B P-\\angle M B A=\\angle P B A,\n$$\n\nas required.' 'Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\\Omega$ and $\\Gamma$, and let $U$ be the intersection of $\\Omega$ and $A B$.\n\n\n\nThe proof can be divided in two steps:\n\n1. Proving that $M P \\cdot M Q=M B^{2}$.\n\nLet us consider the inversion with respect to circle with center $M$ and radius $M A=M B$. This inversion switches $A B$ and $\\Gamma$, and fixes the line passing through $M, U, V$. As a consequence, it keeps $\\Omega$ fixed, and therefore it switches $P$ and $Q$. This is because they are the intersections between the fixed line $M C$ and $\\Omega$, and the only fixed point on the segment $M C$ is its intersection with the inversion circle (thus $P$ and $Q$ are switched). This implies that $M P \\cdot M Q=M B^{2}$.\n\n2. Conclude the proof given that $M P \\cdot M Q=M B^{2}$.\n\nThe relation $M P \\cdot M Q=M B^{2}$ in turn implies that triangle $\\triangle M B P$ is similar to triangle $\\triangle M Q B$, and in particular $\\angle M B P=\\angle M Q B$. Keeping into account that $\\angle M C B=$ $\\angle M B A$, we finally conclude that\n\n$$\n\\angle Q B C=\\angle M Q B-\\angle M C B=\\angle M B P-\\angle M B A=\\angle P B A,\n$$\n\nas required.' 'Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\\Omega$ and $\\Gamma$, and let $U$ be the intersection of $\\Omega$ and $A B$.\n\n\n\nThe proof can be divided in two steps:\n\n1. Proving that $M P \\cdot M Q=M B^{2}$.\n\nIt is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\\Omega$ to $\\Gamma$ sends $U$ to the point of $\\Gamma$ where the tangent to $\\Gamma$ is parallel to $A B$, and this point is $M$ ), and\n\n$$\nM V \\cdot M U=M A^{2}=M B^{2} .\n$$\n\nThis follows from the similitude between the triangles $\\triangle M A V$ and $\\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \\cdot M Y$ is independent of the choice of the line.\n\nComputing the power of $M$ with respect to $\\Omega$ we obtain that\n\n$$\nM P \\cdot M Q=M U \\cdot M V=M B^{2}\n$$\n\n2. Conclude the proof given that $M P \\cdot M Q=M B^{2}$.\n\nLet $I$ and $J$ be the incenter and the $C$-excenter of $\\triangle A B C$ respectively. It is well-known that $M A=M I=M J$, therefore the relation $M P \\cdot M Q=M A^{2}$ implies that $(P, Q, I, J)=-1$.\n\nNow observe that $\\angle I B J=90^{\\circ}$, thus $B I$ is the angle bisector of $\\angle P B Q$ as it is well-known from the theory of harmonic pencils, and this leads easily to the conclusion.' 'Let $D$ denote the intersection of $A B$ and $C M$. Let us consider an inversion with respect to $B$, and let us use primes to denote corresponding points in the transformed diagram, with the gentlemen agreement that $B^{\\prime}=B$.\n\n\n\nSince inversion preserves angles, it turns out that\n\n$$\n\\angle A^{\\prime} B^{\\prime} M^{\\prime}=\\angle A^{\\prime} M^{\\prime} B^{\\prime}=\\angle A C B,\n$$\n\nand in particular triangle $A^{\\prime} B^{\\prime} M^{\\prime}$ is isosceles with basis $B^{\\prime} M^{\\prime}$.\n\nThe image of $C M$ is the circumcircle of $B^{\\prime} C^{\\prime} M^{\\prime}$, which we denote by $\\omega^{\\prime}$. It follows that the centers of both $\\omega^{\\prime}$ and the image $\\Omega^{\\prime}$ of $\\Omega$ lie on the perpendicular bisector of $B^{\\prime} M^{\\prime}$. Therefore, the whole transformed diagram is symmetric with respect to the perpendicular bisector of $B^{\\prime} M^{\\prime}$, and in particular the $\\operatorname{arcs} D^{\\prime} P^{\\prime}$ and $Q^{\\prime} C^{\\prime}$ of $\\omega^{\\prime}$ are equal.\n\nThis is enough to conclude that $\\angle D^{\\prime} B^{\\prime} P^{\\prime}=\\angle Q^{\\prime} B^{\\prime} C^{\\prime}$, which implies the conclusion.']" [] Text-only Competition True Theorem proof Geometry Math English 399 " Prove that for every real number $t$ such that $0(1+t) s_{i}$ for every $i \\in\\{1, \\ldots, n-1\\}$, then by induction we obtain that\n\n$$\ns_{n}>(1+t)^{n-1} s_{1}\n$$\n\nAs a consequence, from (Q6.1) it follows that\n\n$$\n\\left|s_{1}\\right|=s_{1}<\\frac{1}{(1+t)^{n-1}} \\cdot s_{n} \\leq t s_{n}\n$$\n\nand therefore the required inequality is satisfied with $x=s_{1}, y=s_{n}$, and $m=0$.""]" [] Text-only Competition True Theorem proof Algebra Math English 400 " Determine whether for every real number $t$ such that $0t y $$ for every pair of different elements $x$ and $y$ of $S$ and every positive integer $m$ (i.e. $m>0$ )." "['We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form\n\n$$\n\\left|\\frac{x}{y}-m\\right|>t\n$$\n\nThis is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.\n\nNow we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{n}}>t \\quad \\forall n \\geq 1 \\tag{Q6.2}\n$$\n\nand such that for every $j>i$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{j}}<\\frac{1}{2} \\quad \\text { and } \\quad t<\\left\\{\\frac{s_{j}}{s_{i}}\\right\\}<\\frac{1}{2} \\tag{Q6.3}\n$$\n\nwhere $\\{\\alpha\\}$ denotes the fractional part of $\\alpha$. This is enough to show that the set $S:=\\left\\{s_{n}: n \\geq 1\\right\\}$ has the required property.\n\nTo this end, we consider the sequence defined recursively by\n\n$$\ns_{n+1}=\\frac{\\left(s_{1} \\cdot \\ldots \\cdot s_{n}\\right)^{2}+1}{2}\n$$\n\nwith $s_{1}$ large enough. An easy induction shows that this is an increasing sequence of odd positive integers. For every $i \\in\\{1, \\ldots, n\\}$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{n+1}} \\leq \\frac{2}{s_{i}} \\leq \\frac{2}{s_{1}}<\\frac{1}{2}\n$$\n\nbecause $s_{1}$ is large enough, which proves the first relation in (Q6.3). Moreover, it turns out that\n\n$$\n\\frac{s_{n+1}}{s_{i}}=\\frac{\\left(s_{1} \\cdot \\ldots \\cdot s_{n}\\right)^{2}}{2 s_{i}}+\\frac{1}{2 s_{i}}\n$$\n\nThe first term is a positive integer plus $1 / 2$, from which it follows that the distance of $s_{n+1} / s_{i}$ from the positive integers is greater than or equal to\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{i}} \\geq \\frac{1}{2}-\\frac{1}{2 s_{1}}\n$$\n\nwhich is greater than $t$ if $s_{1}$ is large enough. This proves the second relation in (Q6.3).' 'We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form\n\n$$\n\\left|\\frac{x}{y}-m\\right|>t\n$$\n\nThis is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.\n\nNow we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{n}}>t \\quad \\forall n \\geq 1 \\tag{Q6.2}\n$$\n\nand such that for every $j>i$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{j}}<\\frac{1}{2} \\quad \\text { and } \\quad t<\\left\\{\\frac{s_{j}}{s_{i}}\\right\\}<\\frac{1}{2} \\tag{Q6.3}\n$$\n\nwhere $\\{\\alpha\\}$ denotes the fractional part of $\\alpha$. This is enough to show that the set $S:=\\left\\{s_{n}: n \\geq 1\\right\\}$ has the required property.\n\nWe produce an increasing sequence $s_{n}$ of odd and coprime positive integers that satisfies (Q6.3) every $j>i$. As in the previous solution, this is enough to conclude.\n\nWe argue by induction. To begin with, we choose $s_{1}$ to be any odd integer satisfying the inequality in (Q6.2). Let us assume now that $s_{1}, \\ldots, s_{n}$ have already been chosen, and let us choose $s_{n+1}$ in such a way that\n\n$$\ns_{n+1} \\equiv \\frac{s_{i}-1}{2} \\quad\\left(\\bmod s_{i}\\right) \\quad \\forall i \\in\\{1, \\ldots, n\\}\n$$\n\n\n\nWe can solve this system because the previously chosen integers are odd and coprime. Moreover, any solution of this system is coprime with $s_{1}, \\ldots, s_{n}$. Indeed, for every $1 \\leq i \\leq n$ it turns out that\n\n$$\ns_{n+1}=\\frac{s_{i}-1}{2}+k_{i} s_{i}\n$$\n\nfor some positive integer $k_{i}$. Therefore, any prime $p$ that divides both $s_{n+1}$ and $s_{i}$ divides also $\\left(2 k_{i}+1\\right) s_{i}-2 s_{n+1}=1$, which is absurd. Finally, we observe that we can assume that $s_{n+1}$ is odd and large enough. In this way we can guarantee that\n\n$$\n\\frac{s_{i}}{s_{n+1}}<\\frac{1}{2} \\quad \\forall i \\in\\{1, \\ldots, n\\}\n$$\n\nwhich is the first requirement in (Q6.3), and\n\n$$\nk_{i}+tt\n$$\n\nThis is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.\n\nNow we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{n}}>t \\quad \\forall n \\geq 1 \\tag{Q6.2}\n$$\n\nand such that for every $j>i$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{j}}<\\frac{1}{2} \\quad \\text { and } \\quad t<\\left\\{\\frac{s_{j}}{s_{i}}\\right\\}<\\frac{1}{2} \\tag{Q6.3}\n$$\n\nwhere $\\{\\alpha\\}$ denotes the fractional part of $\\alpha$. This is enough to show that the set $S:=\\left\\{s_{n}: n \\geq 1\\right\\}$ has the required property.\n\nAgain we produce an increasing sequence $s_{n}$ of positive integers that satisfies (Q6.3) every $j>i$.\n\nTo this end, for every positive integer $x$, we define its security region\n\n$$\nS(x):=\\bigcup_{n \\geq 1}\\left((n+t) x,\\left(n+\\frac{1}{2}\\right) x\\right)\n$$\n\nThe security region $S(x)$ is a periodic countable union of intervals of length $\\left(\\frac{1}{2}-t\\right) x$, whose left-hand or right-hand endpoints form an arithmetic sequence. It has the property that\n\n$$\nt<\\left\\{\\frac{y}{x}\\right\\}<\\frac{1}{2} \\quad \\forall y \\in S(x)\n$$\n\nNow we prove by induction that we can choose a sequence $s_{n}$ of positive integers satisfying (Q6.3) and in addition the fact that every interval of the security region $S\\left(s_{n}\\right)$ contains at least one interval of $S\\left(s_{n-1}\\right)$.\n\nTo begin with, we choose $s_{1}$ large enough so that the length of the intervals of $S\\left(s_{1}\\right)$ is larger than 1. This guarantees that any interval of $S\\left(s_{1}\\right)$ contains at least a positive integer. Now let us choose a positive integer $s_{2} \\in S\\left(s_{1}\\right)$ that is large enough. This guarantees that $s_{1} / s_{2}$ is small enough, that the fractional part of $s_{2} / s_{1}$ is in $(t, 1 / 2)$, and that every interval of the security region $S\\left(s_{2}\\right)$ contains at least one interval of $S\\left(s_{1}\\right)$, and hence at least one positive integer.\n\nLet us now assume that $s_{1}, \\ldots, s_{n}$ have been already chosen with the required properties. We know that every interval of $S\\left(s_{n}\\right)$ contains at least one interval of $S\\left(s_{n-1}\\right)$, which in turn contains an interval in $S\\left(s_{n-2}\\right)$, and so on up to $S\\left(s_{1}\\right)$. As a consequence, we can choose a large enough positive integer $s_{n+1}$ that lies in $S\\left(s_{k}\\right)$ for every $k \\in\\{1, \\ldots, n\\}$. Since $s_{n+1}$ is large enough, we are sure that\n\n$$\n\\frac{s_{k}}{s_{n+1}}\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$. "['Let $S$ be the intersection point of $B C$ and the angle bisector of $\\angle B A D$, and let $T$ be the intersection point of $B C$ and the angle bisector of $\\angle B X C$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S=T$.\n\nFirstly denote by $M$ the middle of arc $A B$ of the circumcenter of $A B C$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $M A=M I=M B$ ); let it intersect $B C$ at $B$ and $S^{\\prime}$. Since $\\angle B A C>\\angle C B A$ it is easy to check that $S^{\\prime}$ lies on side $B C$. Denoting the angles in $A B C$ by $\\alpha, \\beta, \\gamma$ we get\n\n$$\n\\angle B A D=\\angle B A C-\\angle D A C=\\alpha-\\beta .\n$$\n\nMoreover since $\\angle M B C=\\angle M B A+\\angle A B C=\\frac{\\gamma}{2}+\\beta$, then\n\n$$\n\\angle B M S^{\\prime}=180^{\\circ}-2 \\angle M B C=180^{\\circ}-\\gamma-2 \\beta=\\alpha-\\beta .\n$$\n\nIt follows that $\\angle B A S^{\\prime}=2 \\angle B M S^{\\prime}=2 \\angle B A D$ which gives us $S=S^{\\prime}$.\n\n\nSecondly let $N$ be the middle of arc $B C$ of the circumcenter of $A B C$ which does not contain $A$. From $\\angle B A C>\\angle C B A$ we conclude that $X$ lies on the arc $A B$ of circumcircle of $A B C$ not containing $C$. Obviously both $A I$ and $X T$ are passing through $N$. Since $\\angle N B T=\\frac{\\alpha}{2}=\\angle B X N$ we obtain $\\triangle N B T \\sim \\triangle N X B$, therefore\n\n$$\nN T \\cdot N X=N B^{2}=N I^{2} .\n$$\n\nIt follows that $\\triangle N T I \\sim \\triangle N I X$. Keeping in mind that $\\angle N B C=\\angle N A C=\\angle I X A$ we get\n\n$$\n\\angle T I N=\\angle I X N=\\angle N X A-\\angle I X A=\\angle N B A-\\angle N B C=\\angle T B A .\n$$\n\nIt means that $A, I, B, T$ are concyclic which ends the proof.' ""Let $\\angle B A C=\\alpha, \\angle A B C=\\beta, \\angle B C A=\\gamma, \\angle A C X=\\phi$. Denote by $W_{1}$ and $W_{2}$ the intersections of segment $B C$ with the angle bisectors of $\\angle B X C$ and $\\angle B A D$ respectively. Then $B W_{1} / W_{1} C=B X / X C$ and $B W_{2} / W_{2} D=B A / A D$. We shall show that $B W_{1}=B W_{2}$.\n\n\n\nSince $\\angle D A C=\\angle C B A$, triangles $A D C$ and $B A C$ are similar and therefore\n\n$$\n\\frac{D C}{A C}=\\frac{A C}{B C}\n$$\n\nBy the Law of sines\n\n$$\n\\frac{B W_{2}}{W_{2} D}=\\frac{B A}{A D}=\\frac{B C}{A C}=\\frac{\\sin \\alpha}{\\sin \\beta}\n$$\n\nConsequently\n\n$$\n\\begin{gathered}\n\\frac{B D}{B W_{2}}=\\frac{W_{2} D}{B W_{2}}+1=\\frac{\\sin \\beta}{\\sin \\alpha}+1 \\\\\n\\frac{B C}{B W_{2}}=\\frac{B C}{B D} \\cdot \\frac{B D}{B W_{2}}=\\frac{1}{1-D C / B C} \\cdot \\frac{B D}{B W_{2}}=\\frac{1}{1-A C^{2} / B C^{2}} \\cdot \\frac{B D}{B W_{2}}= \\\\\n\\frac{\\sin ^{2} \\alpha}{\\sin ^{2} \\alpha-\\sin ^{2} \\beta} \\cdot \\frac{\\sin \\beta+\\sin \\alpha}{\\sin \\alpha}=\\frac{\\sin \\alpha}{\\sin \\alpha-\\sin \\beta} .\n\\end{gathered}\n$$\n\nNote that $A X B C$ is cyclic and so $\\angle B X C=\\angle B A C=\\alpha$. Hence, $\\angle X B C=180^{\\circ}-$ $\\angle B X C-\\angle B C X=180^{\\circ}-\\alpha-\\phi$. By the Law of sines for the triangle $B X C$, we have\n\n$$\n\\begin{gathered}\n\\frac{B C}{W_{1} B}=\\frac{W_{1} C}{W_{1} B}+1=\\frac{C X}{B X}+1=\\frac{\\sin \\angle C B X}{\\sin \\phi}+1= \\\\\n\\frac{\\sin (\\alpha+\\phi)}{\\sin \\phi}+1=\\sin \\alpha \\cot \\phi+\\cos \\alpha+1 .\n\\end{gathered}\n$$\n\nSo, it's enough to prove that\n\n$$\n\\frac{\\sin \\alpha}{\\sin \\alpha-\\sin \\beta}=\\sin \\alpha \\cot \\phi+\\cos \\alpha\n$$\n\nSince $A C$ is tangent to the circle $A I X$, we have $\\angle A X I=\\angle I A C=\\alpha / 2$. Moreover $\\angle X A I=\\angle X A B+\\angle B A I=\\phi+\\alpha / 2$ and $\\angle X I A=180^{\\circ}-\\angle X A I-\\angle A X I=180^{\\circ}-\\alpha-\\phi$. Applying the Law of sines again $X A C, X A I, I A C$ we obtain\n\n$$\n\\begin{gathered}\n\\frac{A X}{\\sin (\\alpha+\\phi)}=\\frac{A I}{\\sin \\alpha / 2}, \\\\\n\\frac{A X}{\\sin (\\gamma-\\phi)}=\\frac{A C}{\\sin \\angle A X C}=\\frac{A C}{\\sin \\beta}, \\\\\n\\frac{A I}{\\sin \\gamma / 2}=\\frac{A C}{\\sin (\\alpha / 2+\\gamma / 2)} .\n\\end{gathered}\n$$\n\nCombining the last three equalities we end up with\n\n$$\n\\begin{aligned}\n& \\frac{\\sin (\\gamma-\\phi)}{\\sin (\\alpha+\\phi)}=\\frac{A I}{A C} \\cdot \\frac{\\sin \\beta}{\\sin \\alpha / 2}=\\frac{\\sin \\beta}{\\sin \\alpha / 2} \\cdot \\frac{\\sin \\gamma / 2}{\\sin (\\alpha / 2+\\gamma / 2)} \\\\\n& \\frac{\\sin (\\gamma-\\phi)}{\\sin (\\alpha+\\phi)}=\\frac{\\sin \\gamma \\cot \\phi-\\cos \\gamma}{\\sin \\alpha \\cot \\phi+\\cos \\alpha}=\\frac{2 \\sin \\beta / 2 \\sin \\gamma / 2}{\\sin \\alpha / 2}\n\\end{aligned}\n$$\n\n\n\n$$\n\\frac{\\sin \\alpha \\sin \\gamma \\cot \\phi-\\sin \\alpha \\cos \\gamma}{\\sin \\gamma \\sin \\alpha \\cot \\phi+\\sin \\gamma \\cos \\alpha}=\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2}{\\cos \\gamma / 2}\n$$\n\nSubtracting 1 from both sides yields\n\n$$\n\\begin{gathered}\n\\frac{-\\sin \\alpha \\cos \\gamma-\\sin \\gamma \\cos \\alpha}{\\sin \\gamma \\sin \\alpha \\cot \\phi+\\sin \\gamma \\cos \\alpha}=\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2}{\\cos \\gamma / 2}-1= \\\\\n\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2-\\sin (\\alpha / 2+\\beta / 2)}{\\cos \\gamma / 2}=\\frac{\\sin \\beta / 2 \\cos \\alpha / 2-\\sin \\alpha / 2 \\cos \\beta / 2}{\\cos \\gamma / 2}, \\\\\n\\frac{-\\sin (\\alpha+\\gamma)}{\\sin \\gamma \\sin \\alpha \\cot \\phi+\\sin \\gamma \\cos \\alpha}=\\frac{\\sin (\\beta / 2-\\alpha / 2)}{\\cos \\gamma / 2}, \\\\\n\\frac{-\\sin \\beta}{\\sin \\alpha \\cot \\phi+\\cos \\alpha}=2 \\sin \\gamma / 2 \\sin (\\beta / 2-\\alpha / 2)= \\\\\n2 \\cos (\\beta / 2+\\alpha / 2) \\sin (\\beta / 2-\\alpha / 2)=\\sin \\beta-\\sin \\alpha,\n\\end{gathered}\n$$\n\nand the result follows. We are left to note that none of the denominators can vanish."" 'We first note that\n\n$$\n\\angle B A D=\\angle B A C-\\angle D A C=\\angle A-\\angle B .\n$$\n\nLet $C X$ and $A D$ meet at $K$. Then $\\angle C X A=\\angle A B C=\\angle K A C$. Also, we have $\\angle I X A=$ $\\angle A / 2$, since $\\omega$ is tangent to $A C$ at $A$. Therefore,\n\n$$\n\\angle D A I=|\\angle B-\\angle A / 2|=|\\angle K X A-\\angle I X A|=\\angle K X I\n$$\n\n(the absolute value depends on whether $\\angle B \\geq \\angle A / 2$ or not) which means that $X K I A$ is cyclic, i.e. $K$ lies also on $\\omega$.\n\nLet $I K$ meet $B C$ at $E$. (If $\\angle B=\\angle A / 2$, then $I K$ degenerates to the tangent line to $\\omega$ at I.) Note that $B E I A$ is cyclic, because\n\n$$\n\\angle E I A=180^{\\circ}-\\angle K X A=180^{\\circ}-\\angle A B E .\n$$\n\nWe have $\\angle E K A=180^{\\circ}-\\angle A X I=180^{\\circ}-\\angle A / 2$ and $\\angle A E I=\\angle A B I=\\angle B / 2$. Hence\n\n$$\n\\begin{aligned}\n\\angle E A K & =180^{\\circ}-\\angle E K A-\\angle A E I \\\\\n& =180^{\\circ}-\\left(180^{\\circ}-\\angle A / 2\\right)-\\angle B / 2 \\\\\n& =(\\angle A-\\angle B) / 2 \\\\\n& =\\angle B A D / 2 .\n\\end{aligned}\n$$\n\nThis means that $A E$ is the angle bisector of $\\angle B A D$. Next, let $M$ be the point of intersection of $A E$ and $B I$. Then\n\n$$\n\\angle E M I=180^{\\circ}-\\angle B / 2-\\angle B A D / 2=180^{\\circ}-\\angle A / 2 \\text {, }\n$$\n\n\n\nand so, its supplement is\n\n$$\n\\angle A M I=\\angle A / 2=\\angle A X I,\n$$\n\nso $X, M, K, I, A$ all lie on $\\omega$. Next, we have\n\n$$\n\\begin{aligned}\n\\angle X M A & =\\angle X K A \\\\\n& =180^{\\circ}-\\angle A D C-\\angle X C B \\\\\n& =180^{\\circ}-\\angle A-\\angle X C B \\\\\n& =\\angle B+\\angle X C A \\\\\n& =\\angle B+\\angle X B A \\\\\n& =\\angle X B E,\n\\end{aligned}\n$$\n\nand so $X, B, E, M$ are concyclic. Hence\n\n$$\n\\begin{aligned}\n\\angle E X C & =\\angle E X M+\\angle M X C \\\\\n& =\\angle M B E+\\angle M A K \\\\\n& =\\angle B / 2+\\angle B A D / 2 \\\\\n& =\\angle A / 2 \\\\\n& =\\angle B X C / 2 .\n\\end{aligned}\n$$\n\nThis means that $X E$ is the angle bisector of $\\angle B X C$ and so we are done!\n\n\n\n' 'It is $\\angle A B D=\\angle D A C$, and so $\\overline{A C}$ is tangent to the circumcircle of $\\triangle B A D$ at $A$. Hence $C A^{2}=C D \\cdot C B$.\n\n\n\n\n\nTriangle $\\triangle A B C$ is similar to triangle $\\triangle C A D$, because $\\angle C$ is a common angle and $\\angle C A D=\\angle A B C$, and so $\\angle A D C=\\angle B A C=2 \\varphi$.\n\nLet $Q$ be the point of intersection of $\\overline{A D}$ and $\\overline{C X}$. Since $\\angle B X C=\\angle B A C=2 \\varphi$, it follows that $B D Q X$ is cyclic. Therefore, $C D \\cdot C B=C Q \\cdot C X=C A^{2}$ which implies that $Q$ lies on $\\omega$.\n\nNext let $P$ be the point of intersection of $\\overline{A D}$ with the circumcircle of triangle $\\triangle A B C$. Then $\\angle P B C=\\angle P A C=\\angle A B C=\\angle A P C$ yielding $C A=C P$. So, let $T$ be on the side $\\overline{B C}$ such that $C T=C A=C P$. Then\n\n$$\n\\angle T A D=\\angle T A C-\\angle D A C=\\left(90^{\\circ}-\\frac{\\angle C}{2}\\right)-\\angle B=\\frac{\\angle A-\\angle B}{2}=\\frac{\\angle B A D}{2}\n$$\n\nthat is, line $\\overline{A T}$ is the angle bisector of $\\angle B A D$. We want to show that $\\overline{X T}$ is the angle bisector of $\\angle B X C$. To this end, it suffices to show that $\\angle T X C=\\varphi$.\n\nIt is $C T^{2}=C A^{2}=C Q \\cdot C X$, and so $\\overline{C T}$ is tangent to the circumcircle of $\\triangle X T Q$ at $T$. Since $\\angle T X Q=\\angle Q T C$ and $\\angle Q D C=2 \\varphi$, it suffices to show that $\\angle T Q D=\\varphi$, or, in other words, that $I, Q$, and $T$ are collinear.\n\nLet $T^{\\prime}$ is the point of intersection of $\\overline{I Q}$ and $\\overline{B C}$. Then $\\triangle A I C$ is congruent to $\\triangle T^{\\prime} I C$, since they share $\\overline{C I}$ as a common side, $\\angle A C I=\\angle T^{\\prime} C I$, and\n\n$$\n\\angle I T^{\\prime} D=2 \\varphi-\\angle T^{\\prime} Q D=2 \\varphi-\\angle I Q A=2 \\varphi-\\angle I X A=\\varphi=\\angle I A C .\n$$\n\nTherefore, $C T^{\\prime}=C A=C T$, which means that $T$ coincides with $T^{\\prime}$ and completes the proof.' 'Let $G$ be the point of intersection of $\\overline{A D}$ and $\\overline{C X}$. Since the quadrilateral $A X B C$ is cyclic, it is $\\angle A X C=\\angle A B C$.\n\n\n\n\n\nLet the line $\\overline{A D}$ meet $\\omega$ at $K$. Then it is $\\angle A X K=\\angle C A D=\\angle A B C$, because the angle that is formed by a chord and a tangent to the circle at an endpoint of the chord equals the inscribed angle to that chord. Therefore, $\\angle A X K=\\angle A X C=\\angle A X G$. This means that the point $G$ coincides with the point $K$ and so $G$ belongs to the circle $\\omega$.\n\nLet $E$ be the point of intersection of the angle bisector of $\\angle D A B$ with $\\overline{B C}$. It suffices to show that\n\n$$\n\\frac{C E}{B E}=\\frac{X C}{X B}\n$$\n\nLet $F$ be the second point of intersection of $\\omega$ with $\\overline{A B}$. Then we have $\\angle I A F=\\frac{\\angle C A B}{2}=$ $\\angle I X F$, where $I$ is the incenter of $\\triangle A B C$, because $\\angle I A F$ and $\\angle I X F$ are inscribed in the same arc of $\\omega$. Thus $\\triangle A I F$ is isosceles with $A I=I F$. Since $I$ is the incenter of $\\triangle A B C$, we have $A F=2(s-a)$, where $s=(a+b+c) / 2$ is the semiperimeter of $\\triangle A B C$. Also, it is $C E=A C=b$ because in triangle $\\triangle A C E$, we have\n\n$$\n\\begin{aligned}\n\\angle A E C & =\\angle A B C+\\angle B A E \\\\\n& =\\angle A B C+\\frac{\\angle B A D}{2} \\\\\n& =\\angle A B C+\\frac{\\angle B A C-\\angle A B C}{2} \\\\\n& =90^{\\circ}-\\frac{\\angle A C E}{2},\n\\end{aligned}\n$$\n\nand so $\\angle C A E=180^{\\circ}-\\angle A E C-\\angle A C E=90^{\\circ}-\\frac{\\angle A C E}{2}=\\angle A E C$. Hence\n\n$$\nB F=B A-A F=c-2(s-a)=a-b=C B-C E=B E \\text {. }\n$$\n\nMoreover, triangle $\\triangle C A X$ is similar to triangle $\\triangle B F X$, because $\\angle A C X=\\angle F B X$ and\n\n$$\n\\angle X F B=\\angle X A F+\\angle A X F=\\angle X A F+\\angle C A F=\\angle C A X .\n$$\n\nTherefore\n\n$$\n\\frac{C E}{B E}=\\frac{A C}{B F}=\\frac{X C}{X B}\n$$\n\nas desired. The proof is complete.' ""Let $\\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $Y$ be the second point of intersection of the circle $\\omega$ with the line $A D$. Let $L$ be the intersection of $B C$ with the angle bisector of $\\angle B A D$. We will prove $\\angle L X C=$ $1 / 2 \\angle B A C=1 / 2 \\angle B X C$.\n\nWe will refer to the angles of $\\triangle A B C$ as $\\angle A, \\angle B, \\angle C$. Thus $\\angle B A D=\\angle A-\\angle B$.\n\nOn the circumcircle of $\\triangle A B C$, we have $\\angle A X C=\\angle A B C=\\angle C A D$, and since $A C$ is tangent to $\\omega$, we have $\\angle C A D=\\angle C A Y=\\angle A X Y$. Hence $C, X, Y$ are collinear.\n\nAlso note that $\\triangle C A L$ is isosceles with $\\angle C A L=\\angle C L A=\\frac{1}{2}(\\angle B A D)+\\angle A B C=\\frac{1}{2}(\\angle A+$ $\\angle B)$ hence $A C=C L$. Moreover, $C I$ is angle bisector to $\\angle A C L$ so it's the symmetry axis for the triangle, hence $\\angle I L C=\\angle I A C=1 / 2 \\angle A$ and $\\angle A L I=\\angle L I A=1 / 2 \\angle B$. Since $A C$ is tangent to $\\omega$, we have $\\angle A Y I=\\angle I A C=1 / 2 \\angle A=\\angle L A Y+\\angle A L I$. Hence $L, Y, I$ are collinear.\n\nSince $A C$ is tangent to $\\omega$, we have $\\triangle C A Y \\sim \\triangle C X A$ hence $C A^{2}=C X \\cdot C Y$. However we proved $C A=C L$ hence $C L^{2}=C X \\cdot C Y$. Hence $\\triangle C L Y \\sim \\triangle C X L$ and hence $\\angle C X L=\\angle C L Y=\\angle C A I=1 / 2 \\angle A$.\n\n"" ""Let $M$ be the midpoint of the arc $B C$. Let $\\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $N$ be the second point of intersection of $\\omega$ with $A B$ and $L$ the intersection of $B C$ with the angle bisector of $\\angle B A D$. We know $\\frac{D L}{L B}=\\frac{A D}{A B}$ and want to prove $\\frac{X B}{X C}=\\frac{L B}{L C}$.\n\nFirst note that $\\triangle C A L$ is isosceles with $\\angle C A L=\\angle C L A=\\frac{1}{2}(\\angle B A D)+\\angle A B C$ hence $A C=C L$ and $\\frac{L B}{L C}=\\frac{L B}{A C}$.\n\nNow we calculate $\\frac{X B}{X C}$ :\n\nComparing angles on the circles $\\omega$ and the circumcircle of $\\triangle A B C$ we get $\\triangle X I N \\sim$ $\\triangle X M B$ and hence also $\\triangle X I M \\sim \\triangle X N B$ (having equal angles at $X$ and proportional adjoint sides). Hence $\\frac{X B}{X M}=\\frac{N B}{I M}$.\n\nAlso comparing angles on the circles $\\omega$ and the circumcircle of $\\triangle A B C$ and using the tangent $A C$ we get $\\triangle X A I \\sim \\triangle X C M$ and hence also $\\triangle X A C \\sim \\triangle X I M$. Hence $\\frac{X C}{X M}=$ $\\frac{A C}{I M}$.\n\nComparing the last two equations we get $\\frac{X B}{X C}=\\frac{N B}{A C}$. Comparing with $\\frac{L B}{L C}=\\frac{L B}{A C}$, it remains to prove $N B=L B$.\n\n\n\n\n\nWe prove $\\triangle I N B \\equiv \\triangle I L B$ as follows:\n\nFirst, we note that $I$ is the circumcentre of $\\triangle A L N$. Indeed, $C I$ is angle bisector in the isosceles triangle $A C L$ so it's perpendicular bisector for $A L$. As well, $\\triangle I A N$ is isosceles with $\\angle I N A=\\angle C A I=\\angle I A B$ hence $I$ is also on the perpendicular bisector of $A N$.\n\nHence $I N=I L$ and also $\\angle N I L=2 \\angle N A L=\\angle A-\\angle B=2 \\angle N I B$ (the last angle is calculated using that the exterior angle of $\\triangle N I B$ is $\\angle I N A=\\angle A / 2$. Hence $\\angle N I B=$ $\\angle L I B$ and $\\triangle I N B \\equiv \\triangle I L B$ by SAS."" 'Let $M, N$ be the midpoints of arcs $B C, B A$ of the circumcircle $A B C$, respectively. Let $Y$ be the second intersection of $A D$ and circle $A B C$. Let $E$ be the incenter of triangle $A B Y$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $Y N$ (immediate), $B C$ (since $\\angle C B Y=\\angle C A Y=\\angle C A D=\\angle A B C)$ and the angle bisector of $\\angle D A B$; so the question reduces to showing that $E$ is also on $X M$, which is the angle bisector of $\\angle C X B$.\n\nWe claim that the three lines $C X, A D Y, I E$ are concurrent at a point $D^{\\prime}$. We will complete the proof using this fact, and the proof will appear at the end (and see the solution by HEL5 for an alternative proof of this fact).\n\nTo show that $X E M$ are collinear, we construct a projective transformation which projects $M$ to $X$ through center $E$. We produce it as a composition of three other projections. Let $O$ be the intersection of lines $A D^{\\prime} D Y$ and $C I N$. Projecting the points $Y N C M$ on the circle $A B C$ through the (concyclic) point $A$ to the line $C N$ yields the points $O N C I$. Projecting these points through $E$ to the line $A Y$ yields $O Y D D^{\\prime}$ (here we use the facts that $D^{\\prime}$ lies on $I E$ and $A Y$ ). Projecting these points to the circle $A B C$ through $C$ yields $N Y B X$ (here we use the fact that $D^{\\prime}$ lies on $C X$ ). Composing, we observe that we found a projection of the circle $A B C$ to itself sending $Y N C M$ to $N Y B X$. Since the projection of the circle through $E$ also sends $Y N C$ to $N Y B$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.\n\n\n\n\n\nLet $B^{\\prime}, D^{\\prime}$ be the intersections of $A B, A D$ with the circle $A X I$, respectively. We wish to show that this $D^{\\prime}$ is the concurrency point defined above, i.e. that $C D^{\\prime} X$ and $I D^{\\prime} E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $A B^{\\prime} E$.\n\nConsider the inversion with center $C$ and radius $C A$. The circles $A X I$ and $A B D$ are tangent to $C A$ at $A$ (the former by definition, the latter since $\\angle C A D=\\angle A B C$ ), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $C A B$ to the line $A D$. Thus $X$, which is the second intersection of circles $A B C$ and $A X I$, is sent by the inversion to the second intersection of $A D$ and circle $A X I$, which is $D^{\\prime}$. In particular $C D^{\\prime} X$ are collinear.\n\nIn the circle $A I B^{\\prime}, A I$ is the angle bisector of $B^{\\prime} A$ and the tangent at $A$, so $I$ is the midpoint of the $\\operatorname{arc} A B^{\\prime}$, and in particular $A I=I B^{\\prime}$. By angle chasing, we find that $A C E$ is an isosceles triangle:\n\n$\\angle C A E=\\angle C A D+\\angle D A E=\\angle A B C+\\angle E A B=\\angle A B E+\\angle E A B=\\angle A E B=\\angle A E C$,\n\nthus the angle bisector $C I$ is the perpendicular bisector of $A E$ and $A I=I E$. Thus $I$ is the circumcenter of $A B^{\\prime} E$.\n\nWe can now show that $I D^{\\prime} E$ are collinear by angle chasing:\n\n$$\n\\angle E I B^{\\prime}=2 \\angle E A B^{\\prime}=2 \\angle E A B=\\angle D A B=\\angle D^{\\prime} A B^{\\prime}=\\angle D^{\\prime} I B^{\\prime} .\n$$' 'Let $W$ be the midpoint of $\\operatorname{arc} B C$, let $D^{\\prime}$ be the second intersection point of $A D$ and the circle $A B C$. Let $P$ be the intersection of the angle bisector $X W$ of $\\angle C X B$ with $B C$; we wish to prove that $A P$ is the angle bisector of $D A B$. Denote $\\alpha=\\frac{\\angle C A B}{2}, \\beta=\\angle A B C$.\n\nLet $M$ be the intersection of $A D$ and $X C$. Angle chasing finds:\n\n$$\n\\begin{aligned}\n\\angle M X I & =\\angle A X I-\\angle A X M=\\angle C A I-\\angle A X C=\\angle C A I-\\angle A B C=\\alpha-\\beta \\\\\n& =\\angle C A I-\\angle C A D=\\angle D A I=\\angle M A I\n\\end{aligned}\n$$\n\nAnd in particular $M$ is on $\\omega$. By angle chasing we find\n\n$$\n\\angle X I A=\\angle I X A+\\angle X A I=\\angle I C A+\\angle X A I=\\angle X A C=\\angle X B C=\\angle X B P\n$$\n\nand $\\angle P X B=\\alpha=\\angle C A I=\\angle A X I$, and it follows that $\\triangle X I A \\sim \\triangle X B P$. Let $S$ be the second intersection point of the cirumcircles of $X I A$ and $X B P$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $I S B$ and $A S P$ are collinear.\n\n\n\nLet $L$ be the second intersection of $\\omega$ and $A B$. We want to prove that $A S P$ is the angle bisector of $\\angle D A B=\\angle M A L$, i.e. that $S$ is the midpoint of the $\\operatorname{arc} M L$ of $\\omega$. And this follows easily from chasing angular arc lengths in $\\omega$ :\n\n$$\n\\begin{aligned}\n& \\overparen{A I}=\\angle C A I=\\alpha \\\\\n& \\overparen{I L}=\\angle I A L=\\alpha \\\\\n& \\overparen{M I}=\\angle M X I=\\alpha-\\beta \\\\\n& \\overparen{A I}-\\overparen{S L}=\\angle A B I=\\frac{\\beta}{2}\n\\end{aligned}\n$$\n\nAnd thus\n\n$$\n\\overparen{M L}=\\overparen{M I}+\\overparen{I L}=2 \\alpha-\\beta=2\\left(\\overparen{A I}-\\frac{\\beta}{2}\\right)=2 \\overparen{S L}\n$$\n\n' 'Let $E$ be the intersection of the bisector of $\\angle B A D$ and $B C$, and $N$ be the middle point of arc $B C$ of the circumcircle of $A B C$. Then it suffices to show that $E$ is on line $X N$.\n\nWe consider the inversion at $A$. Let $P^{*}$ be the image of a point denoted by $P$. Then $A, B^{*}, C^{*}, E^{*}$ are concyclic, $X^{*}, B^{*}, C^{*}$ are colinear, and $X^{*} I^{*}$ and $A C^{*}$ are parallel. Now it suffices to show that $A, X^{*}, E^{*}, N^{*}$ are concyclic. Let $Y$ be the intersection of $B^{*} C^{*}$ and $A E^{*}$. Then, by the power of a point, we get\n\n$$\n\\begin{aligned}\nA, X^{*}, E^{*}, N^{*} \\text { are concyclic } & \\Longleftrightarrow Y X^{*} \\cdot Y N^{*}=Y A \\cdot Y E^{*} \\\\\n& \\Longleftrightarrow Y X^{*} \\cdot Y N^{*}=Y B^{*} \\cdot Y C^{*} \\\\\n& \\left(A, B^{*}, C^{*}, E^{*} \\text { are concyclic }\\right)\n\\end{aligned}\n$$\n\nHere, by the property of inversion, we have\n\n$$\n\\angle A I^{*} B^{*}=\\angle A B I=\\frac{1}{2} \\angle A B C=\\frac{1}{2} \\angle C^{*} A D^{*} \\text {. }\n$$\n\n\n\n\n\nDefine $Q, R$ as described in the figure, and we get by simple angle chasing\n\n$$\n\\angle Q A I^{*}=\\angle Q I^{*} A, \\quad \\angle R A I^{*}=\\angle B^{*} I^{*} A\n$$\n\nEspecially, $B^{*} R$ and $A I^{*}$ are parallel, so that we have\n\n$$\n\\frac{Y B^{*}}{Y N^{*}}=\\frac{Y R}{Y A}=\\frac{Y X^{*}}{Y C^{*}}\n$$\n\nand the proof is completed.']" [] Text-only Competition True Theorem proof Geometry Math English 402 Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$. "['Let $Q X, P Y$ be tangent to the incircle of $A B C$, where $X, Y$ lie on the incircle and do not lie on $A C, A B$. Denote $\\angle B A C=\\alpha, \\angle C B A=\\beta, \\angle A C B=\\gamma$.\n\nSince $A I$ is tangent to the circumcircle of $C Q I$ we get $\\angle Q I A=\\angle Q C I=\\frac{\\gamma}{2}$. Thus\n\n$$\n\\angle I Q C=\\angle I A Q+\\angle Q I A=\\frac{\\alpha}{2}+\\frac{\\gamma}{2} .\n$$\n\nBy the definition of $X$ we have $\\angle I Q C=\\angle X Q I$, therefore\n\n$$\n\\angle A Q X=180^{\\circ}-\\angle X Q C=180^{\\circ}-\\alpha-\\gamma=\\beta .\n$$\n\nSimilarly one can prove that $\\angle A P Y=\\gamma$. This means that $Q, P, X, Y$ are collinear which leads us to the conclusion that $X=Y$ and $Q P$ is tangent to the incircle at $X$.\n' 'By the power of a point we have\n\n$$\nA D \\cdot A C=A I^{2}=A P \\cdot A B, \\quad \\text { which means that } \\frac{A Q}{A P}=\\frac{A B}{A C}\n$$\n\nand therefore triangles $A D P, A B C$ are similar. Let $J$ be the incenter of $A Q P$. We obtain\n\n$$\n\\angle J P Q=\\angle I C B=\\angle Q C I=\\angle Q I J,\n$$\n\nthus $J, P, I, Q$ are concyclic. Let $S$ be the intersection of $A I$ and $B C$. It follows that\n\n$$\n\\angle I Q P=\\angle I J P=\\angle S I C=\\angle I Q C .\n$$\n\nThis means that $I Q$ is the angle bisector of $\\angle C Q P$, so $Q P$ is indeed tangent to the incircle of $A B C$.' 'Like before, notice that $A Q \\cdot A C=A P \\cdot A B=A I^{2}$. Consider the positive inversion $\\Psi$ with center $A$ and power $A I^{2}$. This maps $P$ to $B$ (and vice-versa), $Q$ to $C$ (and vice-versa), and keeps the incenter $I$ fixed. The problem statement will follow from the fact that the image of the incircle of triangle $A B C$ under $\\Psi$ is the so-called mixtilinear incircle of $A B C$, which is defined to be the circle tangent to the lines $A B, A C$, and the circumcircle of $A B C$. Indeed, since the image of the line $Q P$ is the circumcircle of $A B C$, and inversion preserves tangencies, this implies that $Q P$ is tangent to the incircle of $A B C$.\n\nWe justify the claim as follows: let $\\gamma$ be the incircle of $A B C$ and let $\\Gamma_{A}$ be the $A$-mixtilinear incircle of $A B C$. Let $K$ and $L$ be the tangency points of $\\gamma$ with the sides $A B$ and $A C$, and let $U$ and $V$ be the tangency points of $\\Gamma_{A}$ with the sides $A B$ and $A C$, respectively. It is well-known that the incenter $I$ is the midpoint of segment $U V$. In particular, since also $A I \\perp U V$, this implies that $A U=A V=\\frac{A I}{\\cos \\frac{A}{2}}$. Note that $A K=A L=A I \\cdot \\cos \\frac{A}{2}$. Therefore, $A U \\cdot A K=A V \\cdot A L=A I^{2}$, which means that $U$ and $V$ are the images of $K$ and $L$ under $\\Psi$. Since $\\Gamma_{A}$ is the unique circle simultaneously tangent to $A B$ at $U$ and to $A C$ at $V$, it follows that the image of $\\gamma$ under $\\Psi$ must be precisely $\\Gamma_{A}$, as claimed.' 'From the power of a point theorem, we have\n\n$$\nA P \\cdot A B=A I^{2}=A Q \\cdot A C .\n$$\n\nHence $P B C Q$ is cyclic, and so, $\\angle A P Q=\\angle B C A$. Let $K$ be the circumcenter of $\\triangle B I P$ and let $L$ be the circumcenter of $\\triangle Q I C$. Then $\\overline{K L}$ is perpendicular to $\\overline{A I}$ at $I$.\n\nLet $N$ be the point of intersection of line $\\overline{K L}$ with $\\overline{A B}$. Then in the right triangle $\\triangle N I A$, we have $\\angle A N I=90^{\\circ}-\\frac{\\angle B A C}{2}$ and from the external angle theorem for triangle $\\triangle B N I$, we have $\\angle A N I=\\frac{\\angle A B C}{2}+\\angle N I B$. Hence\n\n$$\n\\angle N I B=\\angle A N I-\\frac{\\angle A B C}{2}=\\left(90^{\\circ}-\\frac{\\angle B A C}{2}\\right)-\\frac{\\angle A B C}{2}=\\frac{\\angle B C A}{2} .\n$$\n\nSince $M I$ is tangent to the circumcircle of $\\triangle B I P$ at $I$, we have\n\n$$\n\\angle B P I=\\angle B I M=\\angle N I M-\\angle N I B=90^{\\circ}-\\frac{\\angle B C A}{2} .\n$$\n\nAlso, since $\\angle A P Q=\\angle B C A$, we have\n\n$\\angle Q P I=180^{\\circ}-\\angle A P Q-\\angle B P I=180^{\\circ}-\\angle B C A-\\left(90^{\\circ}-\\frac{\\angle B C A}{2}\\right)=90^{\\circ}-\\frac{\\angle B C A}{2}$,\n\nas well. Hence $I$ lies on the angle bisector of $\\angle B P Q$, and so it is equidistant from its sides $\\overline{P Q}$ and $\\overline{P B}$. Therefore, the distance of $I$ from $\\overline{P Q}$ equals the inradius of $\\triangle A B C$, as desired.\n\n\n\n' 'Let $D$ be the point of intersection of $\\overline{A I}$ and $\\overline{B C}$ and let $R$ be the point of intersection of $\\overline{A I}$ and $\\overline{P Q}$. We have $\\angle R I P=\\angle P B I=\\frac{\\angle B}{2}$, $\\angle R I Q=\\angle I C Q=\\frac{\\angle C}{2}, \\angle I Q C=\\angle D I C=x$ and $\\angle B P I=\\angle B I D=\\varphi$, since $\\overline{A I}$ is tangent to both circles.\n\n\n\nFrom the angle bisector theorem, we have\n\n$$\n\\frac{R Q}{R P}=\\frac{A Q}{A P} \\quad \\text { and } \\quad \\frac{A C}{A B}=\\frac{D C}{B D}\n、\n$$\n\n\n\nSince $\\overline{A I}$ is tangent to both circles at $I$, we have $A I^{2}=A Q \\cdot A C$ and $A I^{2}=A P \\cdot A B$. Therefore,\n\n$$\n\\frac{R Q}{R P} \\cdot \\frac{D C}{B D}=\\frac{A Q \\cdot A C}{A B \\cdot A P}=1\n\\tag{1}\n$$\n\nFrom the sine law in triangles $\\triangle Q R I$ and $\\triangle P R I$, it follows that $\\frac{R Q}{\\sin \\frac{\\angle C}{2}}=\\frac{R I}{\\sin y}$ and $\\frac{R P}{\\sin \\frac{\\angle B}{2}}=\\frac{R I}{\\sin \\omega}$, respectively. Hence\n\n$$\n\\frac{R Q}{R P} \\cdot \\frac{\\sin \\frac{\\angle B}{2}}{\\sin \\frac{\\angle C}{2}}=\\frac{\\sin \\omega}{\\sin y}\n\\tag{2}\n$$\n\nSimilarly, from the sine law in triangles $\\triangle I D C$ and $\\triangle I D B$, it is $\\frac{D C}{\\sin x}=\\frac{I D}{\\sin \\frac{\\angle C}{2}}$ and $\\frac{B D}{\\sin \\varphi}=\\frac{I D}{\\sin \\frac{\\angle B}{2}}$, and so\n\n$$\n\\frac{D C}{B D} \\cdot \\frac{\\sin \\varphi}{\\sin x}=\\frac{\\sin \\frac{\\angle B}{2}}{\\sin \\frac{\\angle C}{2}}\n\\tag{3}\n$$\n\nBy multiplying equations (2) with (3), we obtain $\\frac{R Q}{R P} \\cdot \\frac{D C}{B D} \\cdot \\frac{\\sin \\varphi}{\\sin x}=\\frac{\\sin \\omega}{\\sin y}$, which combined with (1) and cross-multiplying yields\n\n$$\n\\sin \\varphi \\cdot \\sin y=\\sin \\omega \\cdot \\sin x\n\\tag{4}\n$$\n\nLet $\\theta=90^{\\circ}+\\frac{\\angle A}{2}$. Since $I$ is the incenter of $\\triangle A B C$, we have $x=90^{\\circ}+\\frac{\\angle A}{2}-\\varphi=\\theta-\\phi$. Also, in triangle $\\triangle P I Q$, we see that $\\omega+y+\\frac{\\angle B}{2}+\\frac{\\angle C}{2}=180^{\\circ}$, and so $y=\\theta-\\omega$.\n\nTherefore, equation (4) yields\n\n$$\n\\sin \\varphi \\cdot \\sin (\\theta-\\omega)=\\sin \\omega \\cdot \\sin (\\theta-\\varphi)\n$$\n\nor\n\n$$\n\\frac{1}{2}(\\cos (\\varphi-\\theta+\\omega)-\\cos (\\varphi+\\theta-\\omega))=\\frac{1}{2}(\\cos (\\omega-\\theta+\\varphi)-\\cos (\\omega+\\theta-\\varphi)),\n$$\n\nwhich is equivalent to\n\n$$\n\\cos (\\varphi+\\theta-\\omega)=\\cos (\\omega+\\theta-\\varphi) .\n$$\n\nSo\n\n$$\n\\varphi+\\theta-\\omega=2 k \\cdot 180^{\\circ} \\pm(\\omega+\\theta-\\varphi), \\quad(k \\in \\mathbb{Z} .)\n$$\n\nIf $\\varphi+\\theta-\\omega=2 k \\cdot 180^{\\circ}+(\\omega+\\theta-\\varphi)$, then $2(\\varphi-\\omega)=2 k \\cdot 180^{\\circ}$, with $|\\varphi-\\omega|<180^{\\circ}$ forcing $k=0$ and $\\varphi=\\omega$. If $\\varphi+\\theta-\\omega=2 k \\cdot 180^{\\circ}-(\\omega+\\theta-\\varphi)$, then $2 \\theta=2 k \\cdot 180^{\\circ}$, which contradicts the fact that $0^{\\circ}<\\theta<180^{\\circ}$. Hence $\\varphi=\\omega$, and so $P I$ is the angle bisector of $\\angle Q P B$.\n\nTherefore the distance of $I$ from $\\overline{P Q}$ is the same with the distance of $I$ from $A B$, which is equal to the inradius of $\\triangle A B C$. Consequently, $\\overline{P Q}$ is tangent to the incircle of $\\triangle A B C$.']" [] Text-only Competition True Theorem proof Geometry Math English 403 "Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions: 1. $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$; 2. the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and 3. $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$. (Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)" "['We define the $b_{i}$ recursively by letting $b_{i}$ be the smallest integer such that $b_{i} \\geq a_{i}$ and such that $b_{i}$ is not congruent to any of $b_{1}, \\ldots, b_{i-1}$ modulo $n$. Then $b_{i}-a_{i} \\leq i-1$, since of the $i$ consecutive integers $a_{i}, a_{i}+1, \\ldots, a_{i}+i-1$, at most $i-1$ are congruent to one of $b_{1}, \\ldots, b_{i-1}$ modulo $n$. Since all $b_{i}$ are distinct modulo $n$, we have $\\sum_{i=1}^{n} b_{i} \\equiv \\sum_{i=1}^{n}(i-1)=\\frac{1}{2} n(n-1)$ modulo $n$, so $n$ divides $\\sum_{i=1}^{n} b_{i}-\\frac{1}{2} n(n-1)$. Moreover, we have $\\sum_{i=1}^{n} b_{i}-\\sum_{i=1}^{n} a_{i} \\leq \\sum_{i=1}^{n}(i-1)=\\frac{1}{2} n(n-1)$, hence $\\sum_{i=1}^{n} b_{i}-\\frac{1}{2} n(n-1) \\leq \\sum_{i=1}^{n}$. As the left hand side is divisible by $n$, we have\n\n$$\n\\frac{1}{n}\\left(\\sum_{i=1}^{n} b_{i}-\\frac{1}{2} n(n-1)\\right) \\leq\\left[\\frac{1}{n} \\sum_{i=1}^{n} a_{i}\\right]\n$$\n\nwhich we can rewrite as\n\n$$\n\\sum_{i=1}^{n} b_{i} \\leq n\\left(\\frac{n-1}{2}+\\left[\\frac{1}{n} \\sum_{i=1}^{n} a_{i}\\right]\\right)\n$$\n\nas required.' ""Note that the problem is invariant under each of the following operations:\n\n- adding a multiple of $n$ to some $a_{i}$ (and the corresponding $b_{i}$ );\n- adding the same integer to all $a_{i}$ (and all $b_{i}$ );\n- permuting the index set $1,2, \\ldots, n$.\n\nWe may therefore remove the restriction that our $a_{i}$ and $b_{i}$ be positive.\n\nFor each congruence class $\\bar{k}$ modulo $n(\\bar{k}=\\overline{0}, \\ldots, \\overline{n-1})$, let $h(k)$ be the number of $i$ such that $a_{i}$ belongs to $\\bar{k}$. We will now show that the problem is solved if we can find a $t \\in \\mathbb{Z}$ such that\n\n$$\n\\begin{aligned}\nh(t) & \\geq 1 \\\\\nh(t)+h(t+1) & \\geq 2 \\\\\nh(t)+h(t+1)+h(t+2) & \\geq 3\n\\end{aligned}\n$$\n\nIndeed, these inequalities guarantee the existence of elements $a_{i_{1}} \\in \\bar{t}, a_{i_{2}} \\in \\bar{t} \\cup \\overline{t+1}$, $a_{i_{3}} \\in \\bar{t} \\cup \\overline{t+1} \\cup \\overline{t+2}$, et cetera, where all $i_{k}$ are different. Subtracting appropriate multiples of $n$ and reordering our elements, we may assume $a_{1}=t, a_{2} \\in\\{t, t+1\\}$, $a_{3} \\in\\{t, t+1, t+2\\}$, et cetera. Finally subtracting $t$ from the complete sequence, we may assume $a_{1}=0, a_{2} \\in\\{0,1\\}, a_{3} \\in\\{0,1,2\\}$ et cetera. Now simply setting $b_{i}=i-1$ for all $i$ suffices, since $a_{i} \\leq b_{i}$ for all $i$, the $b_{i}$ are all different modulo $n$, and\n\n$$\n\\sum_{i=1}^{n} b_{i}=\\frac{n(n-1)}{2} \\leq \\frac{n(n-1)}{2}+n\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right]\n$$\n\nPut $x_{i}=h(i)-1$ for all $i=0, \\ldots, n-1$. Note that $x_{i} \\geq-1$, because $h(i) \\geq 0$. If we have $x_{i} \\geq 0$ for all $i=0, \\ldots, n-1$, then taking $t=0$ completes the proof. Otherwise, we can pick some index $j$ such that $x_{j}=-1$. Let $y_{i}=x_{i}$ where $i=0, \\ldots, j-1, j+1, \\ldots, n-1$ and $y_{j}=0$. For sequence $\\left\\{y_{i}\\right\\}$ we have\n\n$$\n\\sum_{i=0}^{n-1} y_{i}=\\sum_{i=0}^{n-1} x_{i}+1=\\sum_{i=0}^{n-1} h(i)-n+1=1\n$$\n\nso from Raney's lemma there exists index $k$ such that $\\sum_{i=k}^{k+j} y_{i}>0$ for all $j=0, \\ldots, n-1$ where $y_{n+j}=y_{j}$ for $j=0, \\ldots, k-1$. Taking $t=k$ we will have\n\n$$\n\\sum_{t=k}^{k+i} h(t)-(i+1)=\\sum_{t=k}^{k+i} x(t) \\geq \\sum_{t=k}^{k+i} y(t)-1 \\geq 0\n$$\n\nfor all $i=0, \\ldots, n-1$ and we are done."" 'Choose a random permutation $c_{1}, \\ldots, c_{n}$ of the integers $1,2, \\ldots, n$. Let $b_{i}=a_{i}+f\\left(c_{i}-a_{i}\\right)$, where $f(x) \\in\\{0, \\ldots, n-1\\}$ denotes a remainder of $x$ modulo $n$. Observe, that for such defined sequence the first two conditions hold. The expected value of $B:=b_{1}+\\ldots+b_{n}$ is easily seen to be equal to $a_{1}+\\ldots+a_{n}+n(n-1) / 2$. Indeed, for each $i$ the random number $c_{i}-a_{i}$ has uniform distribution modulo $n$, thus the expected value of $f\\left(c_{i}-a_{i}\\right)$ is $(0+\\ldots+(n-1)) / n=(n-1) / 2$. Therefore we may find such $c$ that $B \\leq a_{1}+\\ldots+a_{n}+n(n-1) / 2$. But $B-n(n-1) / 2$ is divisible by $n$ and therefore $B \\leq n\\left[\\left(a_{1}+\\ldots+a_{n}\\right) / n\\right]+n(n-1) / 2$ as needed.' ""We will prove the required statement for all sequences of non-negative integers $a_{i}$ by induction on $n$.\n\nCase $n=1$ is obvious, just set $b_{1}=a_{1}$.\n\nNow suppose that the statement is true for some $n \\geq 1$; we shall prove it for $n+1$.\n\nFirst note that, by subtracting a multiple of $n+1$ to each $a_{i}$ and possibly rearranging indices we can reduce the problem to the case where $0 \\leq a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n} \\leq a_{n+1}<$ $n+1$.\n\nNow, by the induction hypothesis there exists a sequence $d_{1}, d_{2}, \\ldots, d_{n}$ which satisfies the properties required by the statement in relation to the numbers $a_{1}, \\ldots, a_{n}$. Set $I=\\{i \\mid 1 \\leq$ $i \\leq n$ and $\\left.d_{i} \\bmod n \\geq a_{i}\\right\\}$ and construct $b_{i}$, for $i=1, \\ldots, n+1$, as follows:\n\n$$\nb_{i}=\\left\\{\\begin{array}{l}\nd_{i} \\bmod n, \\text { when } i \\in I, \\\\\nn+1+\\left(d_{i} \\bmod n\\right), \\text { when } i \\in\\{1, \\ldots, n\\} \\backslash I, \\\\\nn, \\text { for } i=n+1\n\\end{array}\\right.\n$$\n\n\n\nNow, $a_{i} \\leq d_{i} \\bmod n \\leq b_{i}$ for $i \\in I$, while for $i \\notin I$ we have $a_{i} \\leq n \\leq b_{i}$. Thus the sequence $\\left(b_{i}\\right)_{i=1}^{n+1}$ satisfies the first condition from the problem statement.\n\nBy the induction hypothesis, the numbers $d_{i} \\bmod n$ are distinct for $i \\in\\{1, \\ldots, n\\}$, so the values $b_{i} \\bmod (n+1)$ are distinct elements of $\\{0, \\ldots, n-1\\}$ for $i \\in\\{1, \\ldots, n\\}$. Since $b_{n+1}=n$, the second condition is also satisfied.\n\nDenote $k=|I|$. We have\n\n$$\n\\begin{gathered}\n\\sum_{i=1}^{n+1} b_{i}=\\sum_{i=1}^{n} b_{i}+n=\\sum_{i=1}^{n} d_{i} \\bmod n+(n-k)(n+1)+n= \\\\\n\\frac{n(n+1)}{2}+(n-k)(n+1)\n\\end{gathered}\n$$\n\nhence we need to show that\n\n$$\n\\frac{n(n+1)}{2}+(n-k)(n+1) \\leq \\frac{n(n+1)}{2}+(n+1)\\left[\\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\\right]\n$$\n\nequivalently, that\n\n$$\nn-k \\leq\\left[\\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\\right]\n$$\n\nNext, from the induction hypothesis we have\n\n$$\n\\begin{gathered}\n\\frac{n(n-1)}{2}+n\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right] \\geq \\sum_{i=1}^{n} d_{i}=\\sum_{i \\in I} d_{i}+\\sum_{i \\notin I} d_{i} \\geq \\\\\n\\sum_{i \\in I} d_{i} \\bmod n+\\sum_{i \\notin I}\\left(n+d_{i} \\bmod n\\right)=\\frac{n(n-1)}{2}+(n-k) n\n\\end{gathered}\n$$\n\nor\n\n$$\nn-k \\leq\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right]\n$$\n\nThus, it's enough to show that\n\n$$\n\\frac{\\sum_{i=1}^{n} a_{i}}{n} \\leq \\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\n$$\n\nbecause then\n\n$$\nn-k \\leq\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right] \\leq\\left[\\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\\right]\n$$\n\nBut the required inequality is equivalent to $\\sum_{i=1}^{n} a_{i} \\leq n a_{n+1}$, which is obvious."" ""We can assume that all $a_{i} \\in\\{0,1, \\ldots, n-1\\}$, as we can deduct $n$ from both $a_{i}$ and $b_{i}$ for arbitrary $i$ without violating any of the three conditions from the problem statement. We shall also assume that $a_{1} \\leq \\ldots \\leq a_{n}$.\n\nNow let us provide an algorithm for constructing $b_{1}, \\ldots, b_{n}$.\n\n\n\nWe start at step 1 by choosing $f(1)$ to be the maximum $i$ in $\\{1, \\ldots, n\\}$ such that $a_{i} \\leq n-1$, that is $f(1)=n$. We set $b_{f(1)}=n-1$.\n\nHaving performed steps 1 through $j$, at step $j+1$ we set $f(j+1)$ to be the maximum $i$ in $\\{1, \\ldots, n\\} \\backslash\\{f(1), \\ldots, f(j)\\}$ such that $a_{i} \\leq n-j-1$, if such an index exists. If it does, we set $b_{f(j+1)}=n-j-1$. If there is no such index, then we define $T=j$ and assign to the terms $b_{i}$, where $i \\notin f(\\{1, \\ldots, j\\})$, the values $n, n+1 \\ldots, 2 n-j-1$, in any order, thus concluding the run of our algorithm.\n\nNotice that the sequence $\\left(b_{i}\\right)_{i=1}^{n}$ satisfies the first and second required conditions by construction. We wish to show that it also satisfies the third.\n\nNotice that, since the values chosen for the $b_{i}$ 's are those from $n-T$ to $2 n-T-1$, we have\n\n$$\n\\sum_{i=1}^{n} b_{i}=\\frac{n(n-1)}{2}+(n-T) n\n$$\n\nIt therefore suffices to show that\n\n$$\n\\left[\\frac{a_{1}+\\ldots+a_{n}}{n}\\right] \\geq n-T\n$$\n\nor (since the RHS is obviously an integer) $a_{1}+\\ldots+a_{n} \\geq(n-T) n$.\n\nFirst, we show that there exists $1 \\leq i \\leq T$ such that $n-i=b_{f(i)}=a_{f(i)}$.\n\nIndeed, this is true if $a_{n}=n-1$, so we may suppose $a_{n}t$ and in fact one can show $T=t+f(t+1)$ by proceeding inductively and using the fact that $t$ is the last time for which $a_{f(t)}=b_{f(t)}$.\n\nNow we get that, since $a_{f(t+1)+1} \\geq n-t$, then $\\sum_{i} a_{i} \\geq(n-t)(n-f(t+1))=(n-T+f(t+$ 1)) $(n-f(t+1))=n(n-T)+n f(t+1)-f(t+1)(n-T+f(t+1))=n(n-T)+t f(t+1) \\geq$ $n(n-T)$.""]" [] Text-only Competition True Theorem proof Number Theory Math English 404 "On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either (i) one of the 4038 endpoints of a chord; or (ii) an intersection point of at least two chords. Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive). Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference. Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 . (A chord is a line segment joining two different points on a circle.)" "['First we prove the following:\n\nLemma: if we color all of the points white or black, then the number of white-black edges, which we denote $E_{W B}$, is equal modulo 2 to the number of white (or black) points on the circumference, which we denote $C_{W}$, resp. $C_{B}$.\n\nObserve that changing the colour of any interior point does not change the parity of $E_{W B}$, as each interior point has even degree, so it suffices to show the statement holds when all interior points are black. But then $E_{W B}=C_{W}$ so certainly the parities are equal.\n\nNow returning to the original problem, assume that no two adjacent vertex labels differ by a multiple of three, and three-colour the vertices according to the residue class of the labels modulo 3. Let $E_{01}$ denote the number of edges between 0 -vertices and 1-vertices, and $C_{0}$ denote the number of 0 -vertices on the boundary, and so on.\n\nThen, consider the two-coloring obtained by combining the 1-vertices and 2-vertices. By applying the lemma, we see that $E_{01}+E_{02} \\equiv C_{0} \\bmod 2$.\n\n$$\n\\text { Similarly } E_{01}+E_{12} \\equiv C_{1}, \\quad \\text { and } E_{02}+E_{12} \\equiv C_{2}, \\quad \\bmod 2 \\text {. }\n$$\n\nUsing the fact that $C_{0}=C_{1}=2019$ and $C_{2}=0$, we deduce that either $E_{02}$ and $E_{12}$ are even and $E_{01}$ is odd; or $E_{02}$ and $E_{12}$ are odd and $E_{01}$ is even.\n\nBut if the edge labels are the first $N$ non-negative integers, then $E_{01}=E_{12}$ unless $N \\equiv 0$ modulo 3 , in which case $E_{01}=E_{02}$. So however Alina chooses the vertex labels, it is not possible that the multiset of edge labels is $\\{0, \\ldots, N\\}$.\n\nHence in fact two vertex labels must differ by a multiple of 3 .' 'As before, colour vertices based on their label modulo 3.\n\nSuppose this gives a valid 3-colouring of the graph with 2019 0s and 2019 1s on the circumference. Identify pairs of 0-labelled vertices and pairs of 1-labelled vertices on the circumference, with one 0 and one 1 left over. The resulting graph has even degrees except these two leaves. So the connected component $\\mathcal{C}$ containing these leaves has an Eulerian path, and any other component has an Eulerian cycle.\n\nLet $E_{01}^{*}$ denote the number of edges between 0 -vertices and 1-vertices in $\\mathcal{C}$, and let $E_{01}^{\\prime}$ denote the number of such edges in the other components, and so on. By studying whether a given vertex has label congruent to 0 modulo 3 or not as we go along the Eulerian path in $\\mathcal{C}$, we find $E_{01}^{*}+E_{02}^{*}$ is odd, and similarly $E_{01}^{*}+E_{12}^{*}$ is odd. Since neither start nor end vertex is a 2 -vertex, $E_{02}^{*}+E_{12}^{*}$ must be even.\n\nApplying the same argument for the Eulerian cycle in each other component and adding up, we find that $E_{01}^{\\prime}+E_{02}^{\\prime}, E_{01}^{\\prime}+E_{12}^{\\prime}, E_{02}^{\\prime}+E_{12}^{\\prime}$ are all even. So, again we find $E_{01}+E_{02}$, $E_{01}+E_{12}$ are odd, and $E_{02}+E_{12}$ is even, and we finish as in the original solution.']" [] Text-only Competition True Theorem proof Combinatorics Math English 405 Let $A B C$ be an acute-angled triangle in which $B C\n\nWe show that $T$ and $H$ are both on the angle bisector $\\ell$ of $\\angle B S C$.\n\nWe first prove that $H \\in \\ell$. The altitude $C H$ in triangle $A B C$ is also the altitude in isosceles triangle $P B C$ with $C P=C B$. Therefore, $C H$ is also the angle bisector of $\\angle P C B$ and hence also of $\\angle S C B$. Analogously, $B H$ is the angle bisector of $\\angle S B C$. We conclude that $H$, as the intersection of two angle bisectors of $\\triangle B S C$, is also on the third angle bisector, which is $\\ell$.\n\nWe now prove that $T \\in \\ell$.\n\nVariant 1. In the isosceles triangles $\\triangle B C P$ and $\\triangle C B Q$ we see that $\\angle B C P=180^{\\circ}-2 \\angle B$ and $\\angle C B Q=180^{\\circ}-2 \\angle C$. This yields $\\angle P S Q=\\angle B S C=180^{\\circ}-\\left(180^{\\circ}-2 \\angle B\\right)-\\left(180^{\\circ}-2 \\angle C\\right)=$ $180^{\\circ}-2 \\angle A$. Furthermore, $\\angle P T Q=2 \\angle P A Q=2 \\angle A$ ( $T$ being circumcentre of $\\left.\\triangle A P Q\\right)$. Now $\\angle P T Q+\\angle P S Q=180^{\\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\\angle P S T=\\angle P Q T=\\angle Q P T=\\angle Q S T$, so $T$ is on the angle bisector $\\angle P S Q$, which is also $\\ell$.\n\nWe conclude that $T, S$ and $H$ are collinear.\n\nVariant 2. Let $R$ be the second intersection of $B Q$ and $\\odot A P Q$.\n\n$A P R Q$ is cyclic quadrilateral, so $\\angle P R S=\\angle A, \\angle A P R=\\angle B Q C=\\angle C$. On the other hand $\\angle B P C=\\angle B$. Therefore $\\angle R P S=180^{\\circ}-\\angle B-\\angle C=\\angle A$. Hence, the triangle $P R S$ is isosceles with $S P=S R$; then $\\ell$ is the perpendicular bisector of the chord $P R$ in the circle that passes through $T$.\n\nNote that if $B Q$ is tangent to $\\odot A P Q$, then $C P$ is also tangent to $\\odot A P Q$ and triangle $A B C$ is isosceles, so $T, S$ and $H$ lie on the altitude from $A$.\n\n\n\n' '\n\nVariant 1. In the isosceles triangles $\\triangle B C P$ and $\\triangle C B Q$ we see that $\\angle B C P=180^{\\circ}-2 \\angle B$ and $\\angle C B Q=180^{\\circ}-2 \\angle C$. This yields $\\angle P S Q=\\angle B S C=180^{\\circ}-\\left(180^{\\circ}-2 \\angle B\\right)-\\left(180^{\\circ}-2 \\angle C\\right)=$ $180^{\\circ}-2 \\angle A$. Furthermore, $\\angle P T Q=2 \\angle P A Q=2 \\angle A$ ( $T$ being circumcentre of $\\left.\\triangle A P Q\\right)$. Now $\\angle P T Q+\\angle P S Q=180^{\\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\\angle P S T=\\angle P Q T=\\angle Q P T=\\angle Q S T$, so $T$ is on the angle bisector $\\angle P S Q$, which is also $\\ell$.\nFrom above, we see that $\\angle P S Q=180^{\\circ}-2 \\angle A$, so $\\angle C S Q=2 \\angle A$. From the cyclic quadrilateral $A E H D$ (with $E$ and $D$ feet of the altitudes $C H$ and $B H$ ) we see that $\\angle D H C=\\angle D A E=\\angle A$. Since $B H$ is the perpendicular bisector of $C Q$, we have $\\angle D H Q=\\angle A$ as well, so $\\angle C H Q=2 \\angle A$. From $\\angle C H Q=2 \\angle A=\\angle C S Q$, we see $C H S Q$ is a cyclic quadrilateral. This means $\\angle Q H S=\\angle Q C S$.\n\nSince triangles $P T Q$ and $C H Q$ are both isosceles with apex $2 \\angle A$, we get $\\triangle P T Q \\sim \\triangle C H Q$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\\triangle Q T H \\sim \\triangle Q P C$. This means that $\\angle Q H T=\\angle Q C P$. Combining this with $\\angle Q H S=\\angle Q C S$, we see that $\\angle Q H T=\\angle Q C P=\\angle Q C S=\\angle Q H S$. So $\\angle Q H T=\\angle Q H S$, which means that $T, S$ and $H$ are collinear.' '\n\nLet us draw a parallel $f$ to $B C$ through $A$. Let $B^{\\prime}=B Q \\cap f, C^{\\prime}=C P \\cap f$. Then $A Q B^{\\prime} \\sim C Q B$ and $A P C^{\\prime} \\sim B P C$, therefore both $Q A B^{\\prime}$ and $A P C^{\\prime}$ will be isosceles.\n\nAlso, $B C S \\sim B^{\\prime} C^{\\prime} S$ with respect to the similarity with center $S$, therefore if we take the image of the line $B H$ (which is the perpendicular bisector of the segment $C Q$ ) through this transformation, it will go through the point $B^{\\prime}$, and be perpendicular to $A Q$ (as the image of $C Q$ is parallel to $C Q$ ). As $A Q B^{\\prime}$ is isosceles, this line is the perpendicular bisector of the segment $A Q$. This means it goes through $T$, the circumcenter of $A P Q$. Similarly on the other side the image of $C H$ also goes through $T$. This means that the image of $H$ with respect to the similarity though $S$ will be $T$, so $T, S, H$ are collinear.']" [] Text-only Competition True Theorem proof Geometry Math English 406 "An infinite sequence of positive integers $a_{1}, a_{2}, \ldots$ is called good if (1) $a_{1}$ is a perfect square, and (2) for any integer $n \geq 2, a_{n}$ is the smallest positive integer such that $$ n a_{1}+(n-1) a_{2}+\ldots+2 a_{n-1}+a_{n} $$ is a perfect square. Prove that for any good sequence $a_{1}, a_{2}, \ldots$, there exists a positive integer $k$ such that $a_{n}=a_{k}$ for all integers $n \geq k$." "['Define the following auxiliary sequences:\n\n$$\n\\begin{aligned}\n& b_{1}=a_{1}, \\quad b_{n}=a_{1}+a_{2}+\\cdots+a_{n} \\\\\n& c_{1}=b_{1}, \\quad c_{n}=b_{1}+b_{2}+\\cdots+b_{n} .\n\\end{aligned}\n$$\n\nObserve that\n\n$$\n\\begin{gathered}\nc_{n}-c_{n-1}=b_{n} \\\\\nb_{n}-b_{n-1}=\\left(c_{n}-c_{n-1}\\right)-\\left(c_{n-1}-c_{n-2}\\right)=a_{n}\n\\end{gathered}\n$$\n\nTherefore,\n\n$$\nc_{n}=\\left(2 c_{n-1}-c_{n-2}\\right)+a_{n}\n$$\n\nis the smallest square greater than $2 c_{n-1}-c_{n-2}$.\n\nClaim. $\\sqrt{c_{n}}-\\sqrt{c_{n-1}} \\leq \\sqrt{c_{n-1}}-\\sqrt{c_{n-2}}$\n\nProof 1. Let $c_{n}=m_{n}^{2}, c_{n-1}=m_{n-1}^{2}, c_{n-2}=m_{n-2}^{2}$, where the value of $c_{n}$ is momentarily permitted to exceed its minimum. Any value is permitted for which\n\n$$\nb_{n}=c_{n}-c_{n-1}>b_{n-1}=c_{n-1}-c_{n-2}\n$$\n\nFactoring,\n\n$$\n\\left(m_{n}-m_{n-1}\\right)\\left(m_{n}+m_{n-1}\\right)>\\left(m_{n-1}-m_{n-2}\\right)\\left(m_{n-1}+m_{n-2}\\right) .\n$$\n\nIn the case that $m_{n}-m_{n-1}=m_{n-1}-m_{n-2}$, this inequality clearly holds, as $m_{n}+m_{n-1}>m_{n-1}+m_{n-2}$. Thus, the minimal possible value of $m_{n}$ satisfies the claimed inequality.\n\nProof 2. Denote $c_{n-1}=x^{2}, c_{n-2}=(x-d)^{2}$. Then\n\n$$\n2 c_{n-1}-c_{n-2}=2 x^{2}-(x-d)^{2}=x^{2}+2 d x-d^{2}=(x+d)^{2}-2 d^{2}<(x+d)^{2} .\n$$\n\nIt follows that $c_{n} \\leq(x+d)^{2}$.\n\nAnd so the sequence of positive integers $\\sqrt{c_{n}}-\\sqrt{c_{n-1}}$ is decreasing. Any such sequence is eventually constant.\n\nAs a corollary, $c_{n}=(x+n d)^{2}$ for sufficiently large $n$, with fixed integers $x$ and $d$. Along the lines above, it becomes clear that $a_{n}=c_{n}-2 c_{n-1}+c_{n-2}=2 d^{2}$, so the sequence $\\left(a_{n}\\right)$ is constant.' 'We write:\n\n$$\ns_{n}^{2}=S_{n}=a_{1}+\\left(a_{1}+a_{2}\\right)+\\ldots+\\left(a_{1}+\\ldots+a_{n}\\right)\n$$\n\nSo, setting $b_{n}:=a_{1}+\\ldots+a_{n}$, we have $S_{n}=b_{1}+b_{2}+\\ldots+b_{n}$ and, in particular $S_{n+1}=S_{n}+b_{n+1}$. Now, we study the quantity $S_{n} b_{n}=+b_{1}+b_{2}+\\ldots+b_{n}+b_{n}$ in two different ways. Since $b_{n+1}$ is the smallest integer strictly greater than $b_{n}$ such that $b_{1}+\\ldots+b_{n}+b_{n+1}$ is a perfect square, we must have\n\n$$\nb_{1}+b_{2}+\\ldots+b_{n}+b_{n} \\geq\\left(s_{n+1}-1\\right)^{2}\n$$\n\nHowever, we also have\n\n$$\nb_{1}+b_{2}+\\ldots+b_{n}+b_{n}=S_{n}+b_{n}=2 S_{n}-S_{n-1} \\text {. }\n$$\n\nCombining, we obtain\n\n$$\ns_{n}^{2} \\geq \\frac{s_{n-1}^{2}+\\left(s_{n+1}-1\\right)^{2}}{2}>\\left(\\frac{s_{n-1}+s_{n+1}-1}{2}\\right)^{2}\n$$\n\nwhere the final inequality is strict since the sequence $\\left(s_{k}\\right)$ is strictly increasing. Taking a square root, and noting that all the $\\left(s_{n}\\right)$ are integers, one obtains $s_{n+1}-s_{n} \\leq s_{n}-s_{n-1}$.\n\nNow we focus on the sequence $d_{n}=s_{n+1}-s_{n}$. $\\left(s_{k}\\right)$ is strictly increasing thus $\\left(d_{k}\\right)$ is positive. However we proved that $d_{n+1} \\leq d_{n}$, so the sequence $\\left(d_{k}\\right)$ is eventually constant, so eventually $s_{n}=b n+c$ and $S_{n}=(b n+c)^{2}$ with some numbers $b, c$; then\n\n$$\na_{n+2}=S_{n+2}-2 S_{n+1}+S_{n}=(b(n+2)+c)^{2}-2(b(n+1)+c)^{2}+(b n+c)^{2}=2 b^{2} .\n$$']" [] Text-only Competition True Theorem proof Algebra Math English 407 "Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct. Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle." "['Let $\\Omega$ be the circumcircle of the quadrilateral $A B C D$ and let $r$ be its radius.\n\nFirst, notice that the points $X, Y, Z, W$ are concyclic. Indeed, using oriented (modulo $180^{\\circ}$ ) angles,\n\n$$\n\\angle(X W, X Y)+\\angle(Z Y, Z W)=\\angle(X A, X B)+\\angle(Z C, Z D)=-\\frac{\\angle A+\\angle B}{2}-\\frac{\\angle C+\\angle D}{2}=0 .\n$$\n\nLet $\\omega$ be the circle passing through these four points. Our goal is to prove that $O \\in \\omega$ occurs if and only if $P \\in \\omega$.\n\nNext, we rule out the case when $A B C D$ is a trapezium. Suppose that if $A B C D$ is an isosceles trapezium; without loss of generality say $A B \\| C D$. By symmetry, points $X, Z, O$ and $P$ lie on the symmetry axis (that is, the common perpendicular bisector of $A B$ and $C D$ ), therefore they are collinear. By the conditions of the problem, these four points are distinct and $X$ and $Z$ are on $\\omega$, so neither $O$, nor $P$ can lie on $\\omega$; therefore the problem statement is obvious. From now on we assume that $A B \\nparallel C D$ and $B C \\nparallel A D$.\n\n\n\n\n\nLet $A B$ and $C D$ meet at $Q$, and let $B C$ and $A D$ meet at $R$. Without loss of generality, suppose that $B$ lies between $A$ and $Q$, and $D$ lies between $A$ and $R$. Now we show that line $Q R$ is the radical axis between circles $\\Omega$ and $\\omega$.\n\nPoint $W$ is the intersection point of the bisectors of $\\angle A$ and $\\angle D$, so in triangle $A D Q$, point $W$ is the incentre. Similary, $B Y$ and $C Y$ are the external bisectors of $\\angle Q B C$ and $\\angle B C Q$, so in triangle $B C Q$, point $Y$ is the excentre opposite to $Q$. Hence, both $Y$ and $W$ lie on the bisector of $\\angle D Q A$.\n\nBy $\\angle D Q A=180^{\\circ}-\\angle D-\\angle A=\\angle B-\\angle A$ we get\n\n$$\n\\angle B Y Q=\\angle Y B A-\\angle Y Q B=\\frac{\\angle B}{2}-\\frac{\\angle D Q A}{2}=\\frac{\\angle B}{2}-\\frac{\\angle B-\\angle A}{2}=\\frac{\\angle A}{2}=\\angle B A W\n$$\n\nso the points $A, B, Y, W$ are concyclic, and therefore $Q A \\cdot Q B=Q Y \\cdot Q W$; hence, $Q$ has equal power with respect to $\\Omega$ and $\\omega$. It can be seen similarly that $R$ has equal power with respect to the circles $\\Omega$ and $\\omega$. Hence, the radical axis of $\\Omega$ and $\\omega$ is line $Q R$.\n\nLet lines $O P$ and $Q R$ meet at point $S$. It is well-known that with respect to circle $A B C D$, the diagonal triangle $P Q R$ is autopolar. As consequences we have $O P \\perp Q R$, and the points $P$ and $S$ are symmetric to circle $A B C D$, so $O S \\cdot O P=r^{2}$.\n\nNotice that $P$ and $O$ lie inside $\\Omega$, so the polar line $Q R$ lies entirely outside, so $S$ is different from $P$ and $O$. Moreover,\n\n$$\nS O \\cdot S P=O S \\cdot(O S-O P)=O S^{2}-O S \\cdot O P=S O^{2}-r^{2}\n$$\n\nso $S O \\cdot S P$ is equal to the power of $S$ with respect to $\\Omega$. Since $S$ lies on the radical axis $Q R$, it has equal power with respect to the two circles; therefore, $S O \\cdot S P$ is equal to the power of $S$ with respect to $\\omega$. From this, it follows that $O \\in \\omega$ if and only if $P \\in \\omega$.' 'We will prove that the points $X, Y, Z, W, O, P$ lie on a conic section $\\Gamma$. Five distinct points uniquely determine a conic section, therefore\n\n$X, Y, Z, W, O$ are concyclic $\\Leftrightarrow \\Gamma$ is a circle $\\Leftrightarrow X, Y, Z, W, P$ are concyclic.\n\n\n\n\n\nLet $\\alpha=\\angle A C B=\\angle A D B, \\beta=\\angle B D C=\\angle B A C, \\gamma=\\angle C A D=\\angle C B D$ and $\\delta=\\angle D B A=$ $\\angle D C A$, where of course $\\alpha+\\beta+\\gamma+\\delta=180^{\\circ}$. For every point $\\xi$ in the plane, let $a(\\xi), b(\\xi), c(\\xi)$ and $d(\\xi)$ be the signed distances between $\\xi$ and the lines $A B, B C, C D$ and $D A$, respectively, such that the quadrilateral lies on the positive sides of these lines.\n\nThe equations of the bisectors of $\\angle A X W, \\angle B X Y, \\angle C Y Z$ and $\\angle D Z W$ are $a(\\xi)-d(\\xi)=0$, $b(\\xi)-a(\\xi)=0, c(\\xi)-b(\\xi)=0$ and $d(\\xi)-c(\\xi)=0$, respectively. Notice that the points $X, Y, Z, W$ satisfy the quadratic equations $(a(\\xi)-d(\\xi))(b(\\xi)-c(\\xi))=0$ and $(a(\\xi)-b(\\xi))(c(\\xi)-d(\\xi))=0$, so\n\n$$\na(\\xi) b(\\xi)+c(\\xi) d(\\xi)=a(\\xi) c(\\xi)+b(\\xi) d(\\xi)=a(\\xi) d(\\xi)+b(\\xi) c(\\xi) \\quad \\text { for } \\xi=X, Y, Z, W \\tag{1}\n$$\n\nWithout loss of generality, suppose that $A B C D$ is inscribed in a unit circle. Then\n\n$$\na(O)=\\cos \\alpha, \\quad b(O)=\\cos \\beta, \\quad c(O)=\\cos \\gamma, \\quad d(O)=\\cos \\delta \\tag{2a}\n$$\n\nBy $\\frac{a(P)}{b(P)}=\\frac{B P \\cdot \\sin \\delta}{B P \\cdot \\sin \\gamma}=\\frac{1 / \\sin \\gamma}{1 / \\sin \\delta}$ and the analogous relations we can see that\n\n$$\na(P)=\\frac{k}{\\sin \\gamma}, \\quad b(P)=\\frac{k}{\\sin \\delta}, \\quad c(P)=\\frac{k}{\\sin \\alpha}, \\quad d(P)=\\frac{k}{\\sin \\beta} \\tag{2b}\n$$\n\nwith some positive number $k$.\n\nNow let $s=\\frac{\\sin \\alpha \\sin \\beta \\sin \\gamma \\sin \\delta}{k^{2}}$; then, by $(2 a)$ and $(2 b)$,\n\n$$\n\\begin{aligned}\na(O) b(O)-s \\cdot a(P) b(P) & =\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta=\\cos (\\alpha+\\beta) \\\\\nc(O) d(O)-s \\cdot c(P) d(P) & =\\cos (\\gamma+\\delta)=-\\cos (\\alpha+\\beta) \\\\\na(O) b(O)+c(O) d(O) & =s \\cdot(a(P) b(P)+c(P) d(P))\n\\end{aligned} \\tag{3a}\n$$\n\nanalogous calculation provides\n\n$$\n\\begin{aligned}\n& a(O) c(O)+b(O) d(O)=s \\cdot(a(P) c(P)+b(P) d(P))\n\\end{aligned} \\tag{3b}\n$$\n$$\n\\begin{aligned}\n& a(O) d(O)+b(O) c(O)=s \\cdot(a(P) d(P)+b(P) c(P))\n\\end{aligned} \\tag{3c}\n$$\n\nIn order to find the equation of the curve $\\Gamma$, choose real numbers $u, v, w$, not all zero, such that\n\n$$\n\\begin{gathered}\nu+v+w=0\n\\end{gathered}\n\\tag{4}\n$$\n$$\n\\begin{gathered}\nu \\cdot(a(P) b(P)+c(P) d(P))+v \\cdot(a(P) c(P)+b(P) d(P))+w \\cdot(a(P) d(P)+b(P) c(P))=0\n\\end{gathered} \\tag{5}\n$$\n\nThis is always possible, because this a system of two homogeneous linear equations with three variables. Then the equation of $\\Gamma$ will be\n\n$$\nf(\\xi)=u \\cdot(a(\\xi) b(\\xi)+c(\\xi) d(\\xi))+v \\cdot(a(\\xi) c(\\xi)+b(\\xi) d(\\xi))+w \\cdot(a(\\xi) d(\\xi)+b(\\xi) c(\\xi))=0 \\tag{*}\n$$\n\nAs can be seen, $f(X)=f(Y)=f(Z)=f(W)=0$ follows from (1) and (4), $f(P)=0$ follows from (5), then $f(O)=s \\cdot f(P)=0$ follows from $(3 a-3 c)$. So, the points $X, Y, Z, W, O, P$ all satisfy the equation $f(\\xi)=0$.\n\nNotice that $f(B)=u \\cdot c(B) d(B)$ and $f(A)=w \\cdot b(A) c(A)$; since at least one of $u$ and $w$ is nonzero, either $A$ or $B$ does not satisfy $(*)$. Therefore, the equation cannot degenerate to an identity.\n\nHence, the equation $f(\\xi)=0$ is at most quadratic, it is not an identity, but satisfied by $X, Y, Z, W, O, P$, so these six points lie on a conic section.' ""Let $A^{\\prime}, B^{\\prime}, C^{\\prime}$ and $D^{\\prime}$ be the second intersection of the angle bisectors of $\\angle D A C$, $\\angle A B C, \\angle B C D$ and $\\angle C D A$ with the circumcircle of $A B C D$, respectively. Notice that $A^{\\prime}$ and $C^{\\prime}$ are the midpoints of the two arcs of $B D$ and $B^{\\prime}$ and $D^{\\prime}$ are the midpoints of arcs corresponding to $A C$ so the lines $A^{\\prime} C^{\\prime}$ and $B^{\\prime} D^{\\prime}$ intersect in $O$.\n\n\n\nWe will prove that the points $X, Y, Z, W, P$ and $O$ lie on a conic section. This is a more general statement because five points uniquely determine a conic sections so if $X, Y, Z, W$ and $P$ lie on a circle then this circle is the conic section so $O$ also lies on this circle. In the same way if $X, Y, Z$, $W$ and $O$ are on a circle then it also contains $P$.\n\nLet $E$ be the intersection of the lines $A^{\\prime} B$ and $C D^{\\prime}$. Let $F$ be the intersection of the lines $A^{\\prime} D$ and $B^{\\prime} C$. Finally, let $G$ be the intesection of $X Y$ and $Z W$.\n\nNow we use Pascal's theorem five times:\n\n1. From $B D D^{\\prime} C A A^{\\prime}$ we see that the points $P, W$ and $E$ are collinear.\n2. From $B B^{\\prime} D^{\\prime} C C^{\\prime} A^{\\prime}$ we see that the points $Y, O$ and $E$ are collinear.\n3. From $B^{\\prime} B D A^{\\prime} A C$ we see that the points $X, P$ and $F$ are collinear.\n4. From $B^{\\prime} D^{\\prime} D A^{\\prime} C^{\\prime} C$ we see that the points $O, Z$ and $F$ are collinear.\n5. From $D^{\\prime} C B^{\\prime} B A^{\\prime} D$ we see that the points $E, F$ and $G$ are collinear.\n\nSo the opposite sides of the hexagon $P X Y O Z W$ intersect at points $F, G$ and $E$, which are collinear, so by the converse of Pascal's theorem the points $X, Y, Z, W, P$ and $O$ indeed lie on a conic section.""]" [] Text-only Competition True Theorem proof Geometry Math English 408 "The side $B C$ of the triangle $A B C$ is extended beyond $C$ to $D$ so that $C D=B C$. The side $C A$ is extended beyond $A$ to $E$ so that $A E=2 C A$. Prove that if $A D=B E$, then the triangle $A B C$ is right-angled." "['Define $F$ so that $A B F D$ is a parallelogram. Then $E, A, C, F$ are collinear (as diagonals of a parallelogram bisect each other) and $B F=A D=B E$. Further, $A$ is the midpoint of $E F$, since $A F=2 A C$, and thus $A B$ is an altitude of the isosceles triangle $E B F$ with apex $B$. Therefore $A B \\perp A C$.\n\n' 'Notice that $A$ is the centroid of triangle $B D E$, since $C$ is the midpoint of $[B D]$ and $A E=2 C A$. Let $M$ be the midpoint of $[B E]$. Then $M, A, D$ lie on a line, and further, $A M=\\frac{1}{2} A D=\\frac{1}{2} B E$. This implies that $\\angle E A B=90^{\\circ}$.\n\n\n\n' 'Let $P$ be the midpoint $[A E]$. Since $C$ is the midpoint of $[B D]$, and, moreover, $A C=E P$, we have\n\n$$\n[A C D]=[A B C]=[E B P] .\n$$\n\nBut $A D=B E$, and, as mentioned previously, $A C=E P$, so this implies that\n\n$$\n\\angle B E P=\\angle C A D \\text { or } \\angle B E P=180^{\\circ}-\\angle C A D \\text {. }\n$$\n\nBut $\\angle B E P<\\angle B A C$ and $\\angle B A C+\\angle C A D=\\angle B A D<180^{\\circ}$, so we must be in the first case, i.e. $\\angle B E P=\\angle C A D$. It follows that triangles $B E P$ and $D A C$ are congruent, and thus $\\angle B P A=\\angle A C B$. But $A P=A C$, so $B A$ is a median of the isosceles triangle $B C P$. Thus $A B \\perp P C$, completing the proof.\n\n' 'Write $\\beta=\\angle E C B$, and let $x=A C, y=B C=C D, z=B E=A D$. Notice that $E C=3 x$. Then, using the cosine theorem,\n\n$$\n\\begin{array}{ll}\nz^{2}=x^{2}+y^{2}+2 x y \\cos \\beta & \\text { in triangle } A C D \\\\\nz^{2}=9 x^{2}+y^{2}-6 x y \\cos \\beta & \\text { in triangle } B C E\n\\end{array}\n$$\n\nHence $4 z^{2}=12 x^{2}+4 y^{2}$ or $z^{2}-y^{2}=3 x^{2}$. Let $H$ be the foot of the perpendicular through $B$ to $A C$, and write $h=B H$. Then\n\n$$\ny^{2}-h^{2}=C H^{2}, z^{2}-h^{2}=E H^{2} .\n$$\n\n\n\nHence $z^{2}-y^{2}=E H^{2}-C H^{2}$. Substituting from the above,\n\n$$\nE H^{2}-C H^{2}=3 x^{2}=E A^{2}-C A^{2}\n$$\n\nThus $H=A$, and hence the triangle $A B C$ is right-angled at $A$.' 'Writing $a=B C, b=C A, c=A B$, we have\n\n$$\n\\begin{array}{rlrl}\na^{2} & =b^{2}+c^{2}-2 b c \\cos \\angle A & & \\text { in triangle } A B C ;\\\\\nc^{2} & =a^{2}+b^{2}-2 a b \\cos \\angle C & & \\text { in triangle } A B C ;\\\\\nE B^{2} & =4 b^{2}+c^{2}+4 b c \\cos \\angle A & & \\text { in triangle } A E B ;\\\\\nA D^{2} & =a^{2}+b^{2}+2 a b \\cos \\angle C \\quad & & \\text { in triangle } A C D ;\n\\end{array}\n$$\n\nThus\n\n$$\n\\begin{aligned}\n6 b^{2}+3 c^{2}-2 a^{2} & =4 b^{2}+c^{2}+4 b c \\cos \\angle A=E B^{2}=A D^{2} \\\\\n& =a^{2}+b^{2}+2 a b \\cos \\angle C=2 a^{2}+2 b^{2}-c^{2}\n\\end{aligned}\n$$\n\nwhich gives $a^{2}=b^{2}+c^{2}$. Therefore $\\angle B A C$ is a right angle by the converse of the theorem of Pythagoras.' 'Let $\\overrightarrow{A C}=\\vec{x}$ and $\\overrightarrow{A B}=\\vec{y}$. Now $\\overrightarrow{A D}=2 \\vec{x}-\\vec{y}$ and $\\overrightarrow{E B}=2 \\vec{x}+\\vec{y}$. Then\n\n$$\n\\overrightarrow{B E} \\cdot \\overrightarrow{B E}=\\overrightarrow{A D} \\cdot \\overrightarrow{A D} \\quad \\Longleftrightarrow \\quad(2 \\vec{x}+\\vec{y})^{2}=(2 \\vec{x}-\\vec{y})^{2} \\Longleftrightarrow \\vec{x} \\cdot \\vec{y}=0\n$$\n\nand thus $A C \\perp A B$, whence triangle $A B C$ is right-angled at $A$.' 'Let $a, b, c, d$, $e$ denote the complex co-ordinates of the points $A, B, C, D$, $E$ and take the unit circle to be the circumcircle of $A B C$. We have\n\n$$\nd=b+2(c-b)=2 c-b \\quad \\text { and } \\quad e=c+3(a-c)=3 a-2 c \\text {. }\n$$\n\nThus $b-e=(d-a)+2(b-a)$, and hence\n\n$$\n\\begin{aligned}\nB E=A D & \\Longleftrightarrow(b-e)(\\overline{b-e})=(d-a)(\\overline{d-a}) \\\\\n& \\Longleftrightarrow 2(d-a)(\\overline{b-a})+2(\\overline{d-a})(b-a)+4(b-a)(\\overline{b-a})=0 \\\\\n& \\Longleftrightarrow 2(d-a)(a-b)+2(\\overline{d-a})(b-a) a b+4(b-a)(a-b)=0 \\\\\n& \\Longleftrightarrow(d-a)-(\\overline{d-a}) a b+2(b-a)=0 \\\\\n& \\Longleftrightarrow 2 c-b-a-2 \\bar{c} a b+a+b+2(b-a)=0 \\\\\n& \\Longleftrightarrow c^{2}-a b+b c-a c=0 \\\\\n& \\Longleftrightarrow(b+c)(c-a)=0,\n\\end{aligned}\n$$\n\nimplying $c=-b$ and that triangle $A B C$ is right-angled at $A$.' 'We use areal co-ordinates with reference to the triangle $A B C$. Recall that if $\\left(x_{1}, y_{1}, z_{1}\\right)$ and $\\left(x_{2}, y_{2}, z_{2}\\right)$ are points in the plane, then the square of the distance between these two points is $-a^{2} v w-b^{2} w u-c^{2} u v$, where $(u, v, w)=\\left(x_{1}-x_{2}, y_{1}-y_{2}, z_{1}-z_{2}\\right)$.\n\nIn our case $A=(1,0,0), B=(0,1,0), C=(0,0,1)$, so $E=(3,0,2)$ and, introducing point $F$ as in the first solution, $F=(-1,0,2)$. Then\n\n$$\nB E^{2}=A D^{2} \\Longleftrightarrow-2 a^{2}+6 b^{2}+3 c^{2}=2 a^{2}+2 b^{2}-c^{2},\n$$\n\nand thus $a^{2}=b^{2}+c^{2}$. Therefore $\\angle B A C$ is a right angle by the converse of the theorem of Pythagoras.']" [] Text-only Competition True Theorem proof Geometry Math English 409 "Let $n$ be a positive integer. Prove that there exists a set $S$ of $6 n$ pairwise different positive integers, such that the least common multiple of any two elements of $S$ is no larger than $32 n^{2}$." ['Let the set $A$ consist of the $4 n$ integers $1,2, \\ldots, 4 n$ and let the set $B$ consist of the $2 n$ even integers $4 n+2,4 n+4, \\ldots, 8 n$. We claim that the $6 n$-element set $S=A \\cup B$ has the desired property.\n\nIndeed, the least common multiple of two (even) elements of $B$ is no larger than $8 n \\cdot(8 n / 2)=32 n^{2}$, and the least common multiple of some element of $A$ and some element of $A \\cup B$ is at most their product, which is at most $4 n \\cdot 8 n=32 n^{2}$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 410 "Let $n$ be a positive integer. Prove that every set $T$ of $6 n$ pairwise different positive integers contains two elements the least common multiple of which is larger than $9 n^{2}$." "['We prove the following lemma: ""If a set $U$ contains $m+1$ integers, where $m \\geqslant 2$, that are all not less than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$.""\n\nLet the elements of $U$ be $u_{1}>u_{2}>\\cdots>u_{m+1} \\geqslant m$. Note that $1 / u_{1} \\leqslant 1 / u_{i} \\leqslant 1 / m$ for $1 \\leqslant i \\leqslant m+1$. We partition the interval $\\left[1 / u_{1} ; 1 / m\\right]$ into $m$ subintervals of equal length. By the pigeonhole principle, there exist indices $i, j$ with $1 \\leqslant im^{2}$, completing the proof of the lemma.\n\nApplying the lemma with $m=3 n$ to the $3 n+1$ largest elements of $T$, which are all not less than $3 n$, we arrive at the desired statement.\n\nA Variant. Alternatively, for part (b), we prove the following lemma: ""If a set $U$ contains $m \\geqslant 2$ integers that all are greater than $m$, then some two of its elements have least common multiple strictly larger than $\\mathrm{m}^{2} . ""$\n\nLet $u_{1}>u_{2}>\\cdots>u_{m}$ be the elements of $U$. Since $u_{m}>m=m^{2} / m$, there exists a smallest index $k$ such that $u_{k}>m^{2} / k$. If $k=1$, then $u_{1}>m^{2}$, and the least common multiple of $u_{1}$ and $u_{2}$ is strictly larger than $m^{2}$. So let us suppose $k>1$ from now on, so that we have $u_{k}>m^{2} / k$ and $u_{k-1} \\leqslant m^{2} /(k-1)$. The greatest common divisor $d$ of $u_{k-1}$ and $u_{k}$ satisfies\n\n$$\nd \\leqslant u_{k-1}-u_{k}<\\frac{m^{2}}{k-1}-\\frac{m^{2}}{k}=\\frac{m^{2}}{(k-1) k}\n$$\n\n\n\nThis implies $m^{2} /(d k)>k-1$ and $u_{k} / d>k-1$, and hence $u_{k} / d \\geqslant k$. But then the least common multiple of $u_{k-1}$ and $u_{k}$ equals\n\n$$\n\\frac{u_{k-1} u_{k}}{d} \\geqslant u_{k} \\cdot \\frac{u_{k}}{d}>\\frac{m^{2}}{k} \\cdot k=m^{2}\n$$\n\nand the proof of the lemma is complete.\n\nIf we remove the $3 n$ smallest elements from set $T$ and apply the lemma with $m=3 n$ to the remaining elements, we arrive at the desired statement.']" [] Text-only Competition True Theorem proof Combinatorics Math English 411 "Let $\Omega$ be the circumcircle of the triangle $A B C$. The circle $\omega$ is tangent to the sides $A C$ and $B C$, and it is internally tangent to $\Omega$ at the point $P$. A line parallel to $A B$ and intersecting the interior of triangle $A B C$ is tangent to $\omega$ at $Q$. Prove that $\angle A C P=\angle Q C B$." "['Assume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively and let $P E, P F, P Q$ meet $\\Omega$ at $K, L, M$, respectively. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$, and consider the homethety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Now $K$ is the image of $E$ under $\\mathscr{H}$, and $E I \\perp A C$. Hence $O K \\perp A C$, and thus $K$ is the midpoint of the arc $C A$. Similarly, $L$ is the midpoint of the $\\operatorname{arc} B C$ and $M$ is the midpoint of the arc $B A$. It follows that arcs $L M$ and $C K$ are equal, because\n\n$$\n\\begin{aligned}\n\\overparen{B M}=\\overparen{M A} & \\Longrightarrow \\overparen{B L}+\\overparen{L M}=\\overparen{M K}+\\overparen{K A} \\Longrightarrow \\overparen{L C}+\\overparen{L M}=\\overparen{M K}+\\overparen{C K} \\\\\n& \\Longrightarrow 2 \\overparen{L M}+\\overparen{M C}=\\overparen{M C}+2 \\overparen{C K} \\Longrightarrow \\overparen{L M}=\\overparen{C K}\n\\end{aligned}\n$$\n\nThus arcs $F Q$ and $D E$ of $\\omega$ are equal, too, where $D$ is the intersection of $C P$ with $\\omega$. Since $C E$ and $C F$ are tangents to $\\omega$, this implies that $\\angle D E C=\\angle C F Q$. Further, $C E=C F$, and thus triangles $C E D$ and $C F Q$ are congruent. In particular, $\\angle E C D=\\angle Q C F$, as required.\n\n' 'Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. Observe that $C I$ is the angle bisector of angle $\\angle C$, because $\\omega$ is tangent to $A C$ and $B C$. Consider the homethety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Let $M$ be the image of $Q$ under $\\mathscr{H}$. By construction, $I Q \\perp A B$, so $O M \\perp A B$. Thus the diameter $O M$ of $\\Omega$ passes through the midpoint of the $\\operatorname{arc} A B$ of $\\Omega$, which also lies on the angle bisector $C I$. This implies that $\\angle I C M=90^{\\circ}$. We next show that $P, I, Q, C$ lie on a circle. Notice that\n\n$$\n\\begin{aligned}\n\\angle P Q I & =90^{\\circ}-\\frac{1}{2} \\angle Q I P=90^{\\circ}-\\frac{1}{2} \\angle M O P=90^{\\circ}-\\left(180^{\\circ}-\\angle P C M\\right) \\\\\n& =(\\angle P C I+\\angle I C M)-90^{\\circ}=\\angle P C I .\n\\end{aligned}\n$$\n\nHence $P, I, Q, C$ lie on a circle. But $P I=I Q$, so $C I$ is the angle bisector of $\\angle P C Q$. Since $C I$ is also the angle bisector of angle $\\angle C$, it follows that $\\angle A C P=\\angle Q C B$, as required.\n\n' 'Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. Let $D$ be the second point of intersection of $C P$ with $\\omega$, and let $\\ell$ denote the tangent to $\\omega$ at $D$, which meets $A C$ at $S$. Hence $I D \\perp \\ell$. By construction, $P, I, O$ lie one a line, and hence the isosceles triangles $P I D$ and $P O C$ are similar. In particular, it follows that $O C \\perp \\ell$, so $C$ is the midpoint of the arc of $\\Omega$ defined by the points of intersection of $\\ell$ with $\\Omega$. It is easy to see that this implies that\n\n$$\n\\angle D S C=\\angle A B C .\n$$\n\nUnder reflection in the angle bisector $C I$ of $\\angle C, \\ell$ is thus mapped to a tangent to $\\omega$ parallel to $A B$ and intersecting the interior of $A B C$, since $\\omega$ is mapped to itself under this reflection. In particular, $D$ is mapped to $Q$, and thus $\\angle Q C B=\\angle A C D$, as required.\n\n' 'Let the tangent to $\\omega$ at $Q$ meet $A C$ and $B C$ at $X$ and $Y$, respectively. Then $A C / X C=B C / Y C$, and thus there is a radius $r$ such that $r^{2}=A C \\cdot Y C=B C \\cdot X C$. Let $\\Gamma$ denote the circle with centre $C$ and radius $r$, and consider the inversion $\\mathscr{I}$ in the circle $\\Gamma$. Under $\\mathscr{I}$,\n\n$A \\longmapsto A^{\\prime}$, the point on the ray $C A$ satisfying $C A^{\\prime}=C Y$;\n\n$B \\longmapsto B^{\\prime}$, the point on the ray $C B$ satisfying $C B^{\\prime}=C X$;\n\n$\\Omega \\longmapsto$ the line $A^{\\prime} B^{\\prime} ;$\n\n$\\omega \\longmapsto \\omega^{\\prime}$, the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$;\n\n$P \\longmapsto P^{\\prime}$, the point where $\\omega^{\\prime}$ touches $A^{\\prime} B^{\\prime}$;\n\nIn particular, $\\omega^{\\prime}$, the image of $\\omega$, is a circle tangent to $A C, B C$ and $A^{\\prime} B^{\\prime}$, so it is either the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$, or the incircle of $C A^{\\prime} B^{\\prime}$. Let $\\omega$ be tangent to $B C$ at $F$, and let $F^{\\prime}$ be the image of $F$ under $\\mathscr{I}$. Then $C F \\cdot C F^{\\prime}=B C \\cdot X C$. Now $C FC X=C B^{\\prime}$. Hence $\\omega^{\\prime}$ cannot be the incircle, so $\\omega^{\\prime}$ is indeed the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$.\n\nNow note that $\\omega$ is the excircle of $C X Y$ opposite $C$. The reflection about the angle bisector of $\\angle C$ maps $X$ to $B^{\\prime}, Y$ to $A^{\\prime}$. It thus maps the triangle $C X Y$ to $C B^{\\prime} A^{\\prime}, \\omega$ to $\\omega^{\\prime}$ and, finally, $Q$ to $P^{\\prime}$. It follows that $\\angle A C P=\\angle A C P^{\\prime}=\\angle Q C B$, as required.\n\n' 'Let $r$ be the radius such that $r^{2}=A C \\cdot B C$. Let $\\mathscr{J}$ denote the composition of the inversion $\\mathscr{I}$ in the circle of centre $C$ and radius $r$, followed by the reflection in the angle bisector of $\\angle C$. Under $\\mathscr{J}$,\n\n$A \\longmapsto B, B \\mapsto A ;$\n\n$\\Omega \\longmapsto$ the line $A B$;\n\n$\\omega \\longmapsto \\omega^{\\prime}$, the excircle of $A B C$ opposite the vertex $C$;\n\n$P \\longmapsto Q^{\\prime}$, the point where $\\omega^{\\prime}$ touches $A B$;\n\nIn particular, note that the image $\\omega^{\\prime}$ of $\\omega$ under $\\mathscr{J}$ is a circle tangent to $A C, B C$ and $A B$, so it is either the incircle of $A B C$, or the excircle opposite vertex $C$. Observe that $r \\geqslant \\min \\{A C, B C\\}$, so the image of the points of tangency of $\\omega$ must lie outside $A B C$, and thus $\\omega^{\\prime}$ cannot be the incircle. Thus $\\omega^{\\prime}$ is the excircle opposite vertex $C$ as claimed. Further, the point of tangency $P$ is mapped to $Q^{\\prime}$.\n\nNow, since the line $C P$ is mapped to itself under the inversion $\\mathscr{I}$, and mapped onto $C Q^{\\prime}$ under $\\mathscr{J}, C P$ and $C Q^{\\prime}$ are images of each other under reflection in the angle bisector of $\\angle C$. But $C, Q, Q^{\\prime}$ lie on a line for there is a homothety with centre $C$ that maps $\\omega$ onto the excircle $\\omega^{\\prime}$. This completes the proof.\n\n' '$\\quad$ Assume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively. Assume that $C P$ meets $\\omega$ at $D$. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. To set up a solution in the complex plane, we take the circle $\\omega$ as the unit circle centered at the origin of the complex plane, and let $P O$ be the real axis with $o>0$, where we use the convention that lowercase letters denote complex coordinates of corresponding points in the plane denoted by uppercase letters.\n\n\n\nNow, a point $Z$ on the circle $\\Omega$ satisfies\n\n$$\n|z-o|^{2}=(o+1)^{2} \\Longleftrightarrow \\quad z z^{*}-o\\left(z+z^{*}\\right)-2 o-1=0\n$$\n\nThe triangle $A B C$ is defined by the points $E$ and $F$ on $\\omega$, the intersection $C$ of the corresponding tangents lying on $\\Omega$. Thus $c=2 e f /(e+f)$, and further\n\n$$\n|c-o|^{2}=(o+1)^{2} \\Longleftrightarrow \\quad c c^{*}-o\\left(c+c^{*}\\right)-2 o-1=0\n\\tag{1}\n$$\n\nand this is the equality defining $o$. The points $A$ and $B$ are the second intersection points of $\\Omega$ with the tangents to $\\omega$ at $E$ and $F$ respectively. A point $Z$ on the tangent through $E$ is given by $z=2 e-e^{2} z^{*}$, and thus $A$ and $C$ satisfy\n\n$$\n\\begin{aligned}\n\\left(2 e-e^{2} z^{*}\\right) z^{*} & -o\\left(2 e-e^{2} z^{*}+z^{*}\\right)-2 o-1=0 \\\\\n& \\Longleftrightarrow \\quad-e^{2} z^{* 2}+\\left(2 e+o e^{2}-o\\right) z^{*}-(2 e o+2 o+1)=0 \\\\\n& \\Longleftrightarrow \\quad z^{* 2}-\\left(2 e^{*}+o-o e^{* 2}\\right) z^{*}+\\left(2 e^{*} o+2 o e^{* 2}+e^{* 2}\\right)=0\n\\end{aligned}\n$$\n\nsince $|e|=1$. Thus\n\n$$\na^{*}+c^{*}=2 e^{*}+o-o e^{* 2} \\quad \\Longrightarrow \\quad a^{*}=\\frac{2 e^{*} f}{e+f}+o\\left(1-e^{* 2}\\right)\n$$\n\nand similarly\n\n$$\nb^{*}=\\frac{2 f^{*} e}{f+e}+o\\left(1-f^{* 2}\\right)\n$$\n\nThen\n\n$$\n\\begin{aligned}\nb^{*}-a^{*} & =\\frac{2\\left(e f^{*}-e^{*} f\\right)}{e+f}+o\\left(e^{* 2}-f^{* 2}\\right) \\\\\n& =\\frac{2 e f\\left(f^{* 2}-e^{* 2}\\right)}{e+f}+o\\left(e^{* 2}-f^{* 2}\\right) \\\\\n& =\\left(f^{* 2}-e^{* 2}\\right)\\left(\\frac{2 e f}{e+f}-o\\right)=\\left(f^{* 2}-e^{* 2}\\right)(c-o) .\n\\end{aligned}\n$$\n\nNow let $Z$ be a point on the tangent to $\\omega$ parallel to $A B$ passing through $Q$. Then\n\n$$\nz=2 q-q^{2} z^{*} \\quad \\Longleftrightarrow \\quad z-q=q-q^{2} z^{*}=-q^{2}\\left(z^{*}-q^{*}\\right)\n$$\n\nfor $|q|=1$, and thus\n\n$$\n\\frac{b-a}{b^{*}-a^{*}}=\\frac{z-q}{z^{*}-q^{*}}=\\frac{-q^{2}\\left(z^{*}-q^{*}\\right)}{z^{*}-q^{*}}=-q^{2}\n$$\n\n\n\nIt follows that\n\n$$\n\\begin{aligned}\nq^{2} & =-\\frac{b-a}{b^{*}-a^{*}}=-\\frac{\\left(f^{2}-e^{2}\\right)\\left(c^{*}-o\\right)}{\\left(f^{* 2}-e^{* 2}\\right)(c-o)}=e^{2} f^{2} \\frac{c^{*}-o}{c-o} \\\\\n& =e^{2} f^{2} \\frac{\\left(c^{*}-o\\right)^{2}}{|c-o|^{2}}=e^{2} f^{2} \\frac{\\left(c^{*}-o\\right)^{2}}{(1+o)^{2}}\n\\end{aligned}\n$$\n\nwhere we have used (1). In particular,\n\n$$\nq=e f \\frac{c^{*}-o}{1+o}\n$$\n\nwhere the choice of sign is to be justified a posteriori. Further, the point $D$ satisfies\n\n$$\n-d p=\\frac{d-p}{d^{*}-p^{*}}=\\frac{c-p}{c^{*}-p^{*}} \\quad \\Longrightarrow \\quad d=-\\frac{c-p}{c^{*} p-1}=\\frac{c+1}{c^{*}+1}\n$$\n\nusing $p=-1$ to obtain the final equality.\n\nNow, it suffices to show that (i) $D Q \\| E F \\perp C I$ and (ii) the midpoint of $[D Q]$ is on $C I$. The desired equality then follows by symmetry with respect to the angle bisector of the angle $\\angle A C B$. Notice that (i) is equivalent with\n\n$$\n\\frac{d-q}{d^{*}-q^{*}}=\\frac{e-f}{e^{*}-f^{*}} \\Longleftrightarrow d q=e f\n$$\n\nfor $[D Q]$ and $[E F]$ are chords of $\\omega$. But\n\n$$\n\\begin{aligned}\nd q=e f & \\Longleftrightarrow \\frac{c+1}{c^{*}+1} e f \\frac{c^{*}-o}{1+o}=e f \\quad \\Longleftrightarrow \\quad(c+1)\\left(c^{*}-o\\right)=\\left(c^{*}+1\\right)(1+o) \\\\\n& \\Longleftrightarrow \\quad c c^{*}-o\\left(c+c^{*}\\right)-2 o-1=0 .\n\\end{aligned}\n$$\n\nThe last equality is precisely the defining relation for $o$, (1). This proves (i). Further, the midpoint of $[D Q]$ is $\\frac{1}{2}(d+q)$, so it remains to check that\n\n$$\nd q=\\frac{d+q}{d^{*}+q^{*}}=\\frac{c}{c^{*}}=e f\n$$\n\nwhere the first equality expresses that $[D Q]$ is a chord of $\\omega$ (obviously) containing its midpoint, the second equality expresses the alignment of the midpoint of $[D Q], C$ and $I$, and the third equality follows from the expression for $c$. But we have just shown that $d q=e f$. This proves (ii), justifies the choice of sign for $q$ a posteriori, and thus completes the solution of the problem.']" [] Text-only Competition True Theorem proof Geometry Math English 412 "Snow White and the Seven Dwarves are living in their house in the forest. On each of 16 consecutive days, some of the dwarves worked in the diamond mine while the remaining dwarves collected berries in the forest. No dwarf performed both types of work on the same day. On any two different (not necessarily consecutive) days, at least three dwarves each performed both types of work. Further, on the first day, all seven dwarves worked in the diamond mine. Prove that, on one of these 16 days, all seven dwarves were collecting berries." "['We define $V$ as the set of all 128 vectors of length 7 with entries in $\\{0,1\\}$. Every such vector encodes the work schedule of a single day: if the $i$-th entry is 0 then the $i$-th dwarf works in the mine, and if this entry is 1 then the $i$-th dwarf collects berries. The 16 working days correspond to 16 vectors $d_{1}, \\ldots, d_{16}$ in $V$, which we will call dayvectors. The condition imposed on any pair of distinct days means that any two distinct day-vectors $d_{i}$ and $d_{j}$ differ in at least three positions.\n\nWe say that a vector $x \\in V$ covers some vector $y \\in V$, if $x$ and $y$ differ in at most one position; note that every vector in $V$ covers exactly eight vectors. For each of the 16 day-vectors $d_{i}$ we define $B_{i} \\subset V$ as the set of the eight vectors that are covered by $d_{i}$. As, for $i \\neq j$, the day-vectors $d_{i}$ and $d_{j}$ differ in at least three positions, their corresponding sets $B_{i}$ and $B_{j}$ are disjoint. As the sets $B_{1}, \\ldots, B_{16}$ together contain $16 \\cdot 8=128=|V|$ distinct elements, they form a partition of $V$; in other words, every vector in $V$ is covered by precisely one day-vector.\n\nThe weight of a vector $v \\in V$ is defined as the number of 1-entries in $v$. For $k=0,1, \\ldots, 7$, the set $V$ contains $\\left(\\begin{array}{l}7 \\\\ k\\end{array}\\right)$ vectors of weight $k$. Let us analyse the 16 dayvectors $d_{1}, \\ldots, d_{16}$ by their weights, and let us discuss how the vectors in $V$ are covered by them.\n\n1. As all seven dwarves work in the diamond mine on the first day, the first day-vector is $d_{1}=(0000000)$. This day-vector covers all vectors in $V$ with weight 0 or 1.\n2. No day-vector can have weight 2 , as otherwise it would differ from $d_{1}$ in at most two positions. Hence each of the $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)=21$ vectors of weight 2 must be covered by some day-vector of weight 3 . As every vector of weight 3 covers three vectors of weight 2 , exactly $21 / 3=7$ day-vectors have weight 3 .\n3. How are the $\\left(\\begin{array}{l}7 \\\\ 3\\end{array}\\right)=35$ vectors of weight 3 covered by the day-vectors? Seven of them are day-vectors, and the remaining 28 ones must be covered by day-vectors of weight 4 . As every vector of weight 4 covers four vectors of weight 3 , exactly $28 / 4=7$ day-vectors have weight 4.\n\n\nTo summarize, one day-vector has weight 0 , seven have weight 3 , and seven have weight 4 . None of these 15 day-vectors covers any vector of weight 6 or 7 , so that the eight heavyweight vectors in $V$ must be covered by the only remaining day-vector; and this remaining vector must be (1111111). On the day corresponding to (1111111) all seven dwarves are collecting berries, and that is what we wanted to show.' 'If a dwarf $X$ performs the same type of work on three days $D_{1}, D_{2}, D_{3}$, then we say that this triple of days is monotonous for $X$. We claim that the following configuration cannot occur: There are three dwarves $X_{1}, X_{2}, X_{3}$ and three days $D_{1}, D_{2}$, $D_{3}$, such that the triple $\\left(D_{1}, D_{2}, D_{3}\\right)$ is monotonous for each of the dwarves $X_{1}, X_{2}, X_{3}$.\n\n(Proof: Suppose that such a configuration occurs. Then among the remaining dwarves there exist three dwarves $Y_{1}, Y_{2}, Y_{3}$ that performed both types of work on day $D_{1}$ and on day $D_{2}$; without loss of generality these three dwarves worked in the mine on day $D_{1}$ and collected berries on day $D_{2}$. On day $D_{3}$, two of $Y_{1}, Y_{2}, Y_{3}$ performed the same type of work, and without loss of generality $Y_{1}$ and $Y_{2}$ worked in the mine. But then on days $D_{1}$ and $D_{3}$, each of the five dwarves $X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}$ performed only one type of work; this is in contradiction with the problem statement.)\n\nNext we consider some fixed triple $X_{1}, X_{2}, X_{3}$ of dwarves. There are eight possible working schedules for $X_{1}, X_{2}, X_{3}$ (like mine-mine-mine, mine-mine-berries, mine-berriesmine, etc). As the above forbidden configuration does not occur, each of these eight working schedules must occur on exactly two of the sixteen days. In particular this implies that every dwarf worked exactly eight times in the mine and exactly eight times in the forest.\n\nFor $0 \\leqslant k \\leqslant 7$ we denote by $d(k)$ the number of days on which exactly $k$ dwarves were collecting berries. Since on the first day all seven dwarves were in the mine, on each of the remaining days at least three dwarves collected berries. This yields $d(0)=1$ and $d(1)=d(2)=0$. We assume, for the sake of contradiction, that $d(7)=0$ and hence\n\n$$\nd(3)+d(4)+d(5)+d(6)=15 \\tag{1}\n$$\n\nAs every dwarf collected berries exactly eight times, we get that, further,\n\n$$\n3 d(3)+4 d(4)+5 d(5)+6 d(6)=7 \\cdot 8=56 .\n\\tag{2}\n$$\n\nNext, let us count the number $q$ of quadruples $\\left(X_{1}, X_{2}, X_{3}, D\\right)$ for which $X_{1}, X_{2}, X_{3}$ are three pairwise distinct dwarves that all collected berries on day $D$. As there are $7 \\cdot 6 \\cdot 5=210$ triples of pairwise distinct dwarves, and as every working schedule for three fixed dwarves occurs on exactly two days, we get $q=420$. As every day on which $k$ dwarves collect berries contributes $k(k-1)(k-2)$ such quadruples, we also have\n\n$$\n3 \\cdot 2 \\cdot 1 \\cdot d(3)+4 \\cdot 3 \\cdot 2 \\cdot d(4)+5 \\cdot 4 \\cdot 3 \\cdot d(5)+6 \\cdot 5 \\cdot 4 \\cdot d(6)=q=420\n$$\n\n\n\nwhich simplifies to\n\n$$\nd(3)+4 d(4)+10 d(5)+20 d(6)=70 \\tag{3}\n$$\n\nFinally, we count the number $r$ of quadruples $\\left(X_{1}, X_{2}, X_{3}, D\\right)$ for which $X_{1}, X_{2}, X_{3}$ are three pairwise distinct dwarves that all worked in the mine on day $D$. Similarly as above we see that $r=420$ and that\n\n$$\n7 \\cdot 6 \\cdot 5 \\cdot d(0)+4 \\cdot 3 \\cdot 2 \\cdot d(3)+3 \\cdot 2 \\cdot 1 \\cdot d(4)=r=420 \n$$\n\nwhich simplifies to\n\n$$\n4 d(3)+d(4)=35 \\tag{4}\n$$\n\nMultiplying (1) by -40 , multiplying (2) by 10 , multiplying (3) by -1 , multiplying (4) by 4 , and then adding up the four resulting equations yields $5 d(3)=30$ and hence $d(3)=6$. Then (4) yields $d(4)=11$. As $d(3)+d(4)=17$, the total number of days cannot be 16 . We have reached the desired contradiction.']" [] Text-only Competition True Theorem proof Combinatorics Math English 413 Let $\triangle A B C$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\angle A B C$ intersects $C D$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle A D E$ again at $F$. If $\angle A D F=45^{\circ}$, show that $C F$ is tangent to $\omega$. "['\nSince $\\angle C D F=90^{\\circ}-45^{\\circ}=45^{\\circ}$, the line $D F$ bisects $\\angle C D A$, and so $F$ lies on the perpendicular bisector of segment $A E$, which meets $A B$ at $G$. Let $\\angle A B C=2 \\beta$. Since $A D E F$ is cyclic, $\\angle A F E=90^{\\circ}$, and hence $\\angle F A E=45^{\\circ}$. Further, as $B F$ bisects $\\angle A B C$, we have $\\angle F A B=90^{\\circ}-\\beta$, and thus\n\n$$\n\\angle E A B=\\angle A E G=45^{\\circ}-\\beta, \\quad \\text { and } \\quad \\angle A E D=45^{\\circ}+\\beta \\text {, }\n$$\n\nso $\\angle G E D=2 \\beta$. This implies that right-angled triangles $\\triangle E D G$ and $\\triangle B D C$ are similar, and so we have $|G D| /|C D|=|D E| /|D B|$. Thus the right-angled triangles $\\triangle D E B$ and $\\triangle D G C$ are similar, whence $\\angle G C D=\\angle D B E=\\beta$. But $\\angle D F E=\\angle D A E=45^{\\circ}-$ $\\beta$, then $\\angle G F D=45^{\\circ}-\\angle D F E=\\beta$. Hence $G D C F$ is cyclic, so $\\angle G F C=90^{\\circ}$, whence $C F$ is perpendicular to the radius $F G$ of $\\omega$.\n\nIt follows that $C F$ is a tangent to $\\omega$, as required.' 'As $\\angle A D F=45^{\\circ}$ line $D F$ is an exterior bisector of $\\angle C D B$. Since $B F$ bisects $\\angle D B C$ line $C F$ is an exterior bisector of $\\angle B C D$. Let $\\angle A B C=2 \\beta$, so $\\angle E C F=(\\angle D B C+\\angle C D B) / 2=45^{\\circ}+\\beta$. Hence $\\angle C F E=180^{\\circ}-\\angle E C F-\\angle B C E-\\angle E B C=180^{\\circ}-\\left(45^{\\circ}+\\beta+90^{\\circ}-2 \\beta+\\beta\\right)=45^{\\circ}$. It follows that $\\angle F D C=\\angle C F E$, then $C F$ is tangent to $\\omega$.' 'Note that $A E$ is diameter of circumcircle of $\\triangle A B C$ since $\\angle C D F=90^{\\circ}$. From $\\angle A E F=\\angle A D F=45^{\\circ}$ it follows that triangle $\\triangle A F E$ is right-angled and isosceles. Without loss of generality, let points $A, E$ and $F$ have coordinates $(-1,0),(1,0)$ and $(0,1)$ respectively. Points $F, E$, $B$ are collinear, hence $B$ have coordinates $(b, 1-b)$ for some $b \\neq-1$. Let point $C^{\\prime}$ be intersection of line tangent to circumcircle of $\\triangle A F E$ at $F$ with line $E D$. Thus $C^{\\prime}$ have coordinates $(c, 1)$ and from $\\overline{C^{\\prime} E} \\perp \\overline{A B}$ we get $c=2 b /(b+1)$. Now vector $\\overline{B C^{\\prime}}=(2 b /(b+1)-b, b)=b /(b+1) \\cdot(1-b, b+1)$, vector $\\overline{B F}=(-b, b)=(-1,1) \\cdot b$ and vector $\\overline{B A}=(-(b+1),-(1-b))$. Its clear that $(1-b, b+1)$ and $(-(b+1),-(1-b))$ are symmetric with respect to $\\overline{F E}=(-1,1)$, hence $B F$ bisects $\\angle C^{\\prime} B A$ and $C^{\\prime}=C$ which completes the proof.' '\n\nAgain $F$ lies on the perpendicular bisector of segment $A E$, so $\\triangle A F E$ is right-angled and isosceles. Let $M$ be an intersection of $B C$ and $A F$. Note that $\\triangle A M B$ is isosceles since $B F$ is a bisector and altitude in this triangle. Thus $B F$ is a symmetry line of $\\triangle A M B$. Then $\\angle F D A=\\angle F E A=\\angle M E F=45^{\\circ}, A F=F E=F M$ and $\\angle D A E=\\angle E M C$. Let us show that $E C=C M$. Indeed,\n\n$$\n\\begin{aligned}\n\\angle C E M & =180^{\\circ}-(\\angle A E D+\\angle F E A+\\angle M E F)=90^{\\circ}-\\angle A E D= \\\\\n& =\\angle D A E=\\angle E M C .\n\\end{aligned}\n$$\n\nIt follows that $F M C E$ is a kite, since $E F=F M$ and $M C=C E$. Hence $\\angle E F C=\\angle C F M=\\angle E D F=45^{\\circ}$, so $F C$ is tangent to $\\omega$.' 'Let the tangent to $\\omega$ at $F$ intersect $C D$ at $C^{\\prime}$. Let $\\angle A B F=$ $\\angle F B C=\\beta$. It follows that $\\angle C^{\\prime} F E=45^{\\circ}$ since $C^{\\prime} F$ is tangent. We have\n\n$$\n\\frac{\\sin \\angle B D C}{\\sin \\angle C D F} \\cdot \\frac{\\sin \\angle D F C^{\\prime}}{\\sin \\angle C^{\\prime} F B} \\cdot \\frac{\\sin \\angle F B C}{\\sin \\angle C B D}=\\frac{\\sin 90^{\\circ}}{\\sin 45^{\\circ}} \\cdot \\frac{\\sin \\left(90^{\\circ}-\\beta\\right)}{\\sin 45^{\\circ}} \\cdot \\frac{\\sin \\beta}{\\sin 2 \\beta}=\\frac{2 \\sin \\beta \\cos \\beta}{\\sin 2 \\beta}=1 .\n$$\n\nSo by trig Ceva on triangle $\\triangle B D F$, lines $F C^{\\prime}, D C$ and $B C$ are concurrent (at $C$ ), so $C=C^{\\prime}$. Hence $C F$ is tangent to $\\omega$.']" [] Text-only Competition True Theorem proof Geometry Math English 414 "Let $n, m$ be integers greater than 1 , and let $a_{1}, a_{2}, \ldots, a_{m}$ be positive integers not greater than $n^{m}$. Prove that there exist positive integers $b_{1}, b_{2}, \ldots, b_{m}$ not greater than $n$, such that $$ \operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right)1$, let $a_{1}, \\ldots, a_{m}$ be positive integers with at least one $a_{i} \\leq n^{2^{m-1}}$. Then there are integers $b_{1}, \\ldots, b_{m}$, each equal to 1 or 2 , such that $\\operatorname{gcd}\\left(a_{1}+b_{1}, \\ldots, a_{m}+b_{m}\\right)1$ are all pairwise coprime, since for any two of them, there is some $i>1$ with $a_{i}+1$ appearing in one and $a_{i}+2$ in the other. Since each of these $2^{m-1}$ integers divides $a_{1}+1$, and each is $\\geq n$ with at most one equal to $n$, it follows that $a_{1}+1 \\geq n(n+1)^{2^{m-1}-1}$ so $a_{1} \\geq n^{2^{m-1}}$. The same is true for each $a_{i}, i=1, \\ldots, n$, a contradiction.']" [] Text-only Competition True Theorem proof Algebra Math English 415 "Determine whether there exists an infinite sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers which satisfies the equality $$ a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}} $$ for every positive integer $n$." "['The answer is no.\n\nSuppose that there exists a sequence $\\left(a_{n}\\right)$ of positive integers satisfying the given condition. We will show that this will lead to a contradiction.\n\nFor each $n \\geq 2$ define $b_{n}=a_{n+1}-a_{n}$. Then, by assumption, for $n \\geq 2$ we get $b_{n}=\\sqrt{a_{n}+a_{n-1}}$ so that we have\n\n$$\nb_{n+1}^{2}-b_{n}^{2}=\\left(a_{n+1}+a_{n}\\right)-\\left(a_{n}+a_{n-1}\\right)=\\left(a_{n+1}-a_{n}\\right)+\\left(a_{n}-a_{n-1}\\right)=b_{n}+b_{n-1} .\n$$\n\nSince each $a_{n}$ is a positive integer we see that $b_{n}$ is positive integer for $n \\geq 2$ and the sequence $\\left(b_{n}\\right)$ is strictly increasing for $n \\geq 3$. Thus $b_{n}+b_{n-1}=\\left(b_{n+1}-b_{n}\\right)\\left(b_{n+1}+b_{n}\\right) \\geq b_{n+1}+b_{n}$, whence $b_{n-1} \\geq b_{n+1}$ - a contradiction to increasing of the sequence $\\left(b_{i}\\right)$.\n\nThus we conclude that there exists no sequence $\\left(a_{n}\\right)$ of positive integers satisfying the given condition of the problem.' 'Suppose that such a sequence exists. We will calculate its members one by one and get a contradiction.\n\nFrom the equality $a_{3}=a_{2}+\\sqrt{a_{2}+a_{1}}$ it follows that $a_{3}>a_{2}$. Denote positive integers $\\sqrt{a_{3}+a_{2}}$ by $b$ and $a_{3}$ by $a$, then we have $\\sqrt{2 a}>b$. Since $a_{4}=a+b$ and $a_{5}=a+b+\\sqrt{2 a+b}$ are positive integers, then $\\sqrt{2 a+b}$ is positive integer.\n\nConsider $a_{6}=a+b+\\sqrt{2 a+b}+\\sqrt{2 a+2 b+\\sqrt{2 a+b}}$. Number $c=\\sqrt{2 a+2 b+\\sqrt{2 a+b}}$ must be positive integer, obviously it is greater than $\\sqrt{2 a+b}$. But\n\n$$\n(\\sqrt{2 a+b}+1)^{2}=2 a+b+2 \\sqrt{2 a+b}+1=2 a+2 b+\\sqrt{2 a+b}+(\\sqrt{2 a+b}-b)+1>c^{2}\n$$\n\nSo $\\sqrt{2 a+b}5$ members and satisfies\n\n$$\na_{n+2}=a_{n+1}+\\sqrt{a_{n+1}+a_{n}} \\tag{1}\n$$\n\nfor all $n=1, \\ldots, N-2$. Moreover, we will describe all such sequences with five members.\n\nSince every $a_{i}$ is a positive integer it follows from (1) that there exists such positive integer $k$ (obviously $k$ depends on $n$ ) that\n\n$$\na_{n+1}+a_{n}=k^{2} \\tag{2}\n$$\n\nFrom (1) we have $\\left(a_{n+2}-a_{n+1}\\right)^{2}=a_{n+1}+a_{n}$, consider this equality as a quadratic equation with respect to $a_{n+1}$ :\n\n$$\na_{n+1}^{2}-\\left(2 a_{n+2}+1\\right) a_{n+1}+a_{n+2}^{2}-a_{n}=0\n$$\n\nObviously its solutions are $\\left(a_{n+1}\\right)_{1,2}=\\frac{2 a_{n+2}+1 \\pm \\sqrt{D}}{2}$, where\n\n$$\nD=4\\left(a_{n}+a_{n+2}\\right)+1 \\tag{3}\n$$\n\nSince $a_{n+2}>a_{n+1}$ we have\n\n$$\na_{n+1}=\\frac{2 a_{n+2}+1-\\sqrt{D}}{2}\n$$\n\n\n\nFrom the last equality, using that $a_{n+1}$ and $a_{n+2}$ are positive integers, we conclude that $D$ is a square of some odd number i.e. $D=(2 m+1)^{2}$ for some positive integer $m \\in \\mathbb{N}$, substitute this into (3):\n\n$$\na_{n}+a_{n+2}=m(m+1) \\text {. } \\tag{4}\n$$\n\nNow adding $a_{n}$ to both sides of (1) and using (2) and (4) we get $m(m+1)=k^{2}+k$ whence $m=k$. So\n\n$$\n\\left\\{\\begin{array}{l}\na_{n}+a_{n+1}=k^{2} \\\\\na_{n}+a_{n+2}=k^{2}+k\n\\end{array}\\right. \\tag{5}\n$$\n\nfor some positive integer $k$ (recall that $k$ depends on $n$ ).\n\nWrite equations (5) for $n=2$ and $n=3$, then for some positive integers $k$ and $\\ell$ we get\n\n$$\n\\left\\{\\begin{array}{l}\na_{2}+a_{3}=k^{2}, \\\\\na_{2}+a_{4}=k^{2}+k, \\\\\na_{3}+a_{4}=\\ell^{2}, \\\\\na_{3}+a_{5}=\\ell^{2}+\\ell .\n\\end{array}\\right. \\tag{6}\n$$\n\nSolution of this linear system is\n\n$$\na_{2}=\\frac{2 k^{2}-\\ell^{2}+k}{2}, \\quad a_{3}=\\frac{\\ell^{2}-k}{2}, \\quad a_{4}=\\frac{\\ell^{2}+k}{2}, \\quad a_{5}=\\frac{\\ell^{2}+2 \\ell+k}{2} . \\tag{7}\n$$\n\nFrom $a_{2}\\ell^{2}$ and $2\\left(\\ell^{2}-k^{2}-k\\right)^{2}>2 k^{2}-\\ell^{2}+k$.' ""It is easy to see that $\\left(a_{n}\\right)$ is increasing for large enough $n$. Hence\n\n$$\na_{n+1}1$. Anastasia partitions the integers $1,2, \ldots, 2 m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to $n$. "['Define the following ordered partitions:\n\n$$\n\\begin{aligned}\n& P_{1}=(\\{1,2\\},\\{3,4\\}, \\ldots,\\{2 m-1,2 m\\}), \\\\\n& P_{2}=(\\{1, m+1\\},\\{2, m+2\\}, \\ldots,\\{m, 2 m\\}), \\\\\n& P_{3}=(\\{1,2 m\\},\\{2, m+1\\},\\{3, m+2\\}, \\ldots,\\{m, 2 m-1\\}) .\n\\end{aligned}\n$$\n\nFor each $P_{j}$ we will compute the possible values for the expression $s=a_{1}+\\ldots+a_{m}$, where $a_{i} \\in P_{j, i}$ are the chosen integers. Here, $P_{j, i}$ denotes the $i$-th coordinate of the ordered partition $P_{j}$.\n\nWe will denote by $\\sigma$ the number $\\sum_{i=1}^{m} i=\\left(m^{2}+m\\right) / 2$.\n\n- Consider the partition $P_{1}$ and a certain choice with corresponding sum $s$. We find that\n\n$$\nm^{2}=\\sum_{i=1}^{m}(2 i-1) \\leq s \\leq \\sum_{i=1}^{m} 2 i=m^{2}+m\n$$\n\nHence, if $nm^{2}+m$, this partition gives a positive answer.\n\n- Consider the partition $P_{2}$ and a certain choice with corresponding $s$. We find that\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\quad(\\bmod m)\n$$\n\nHence, if $m^{2} \\leq n \\leq m^{2}+m$ and $n \\not \\equiv \\sigma(\\bmod m)$, this partition solves the problem.\n\n- Consider the partition $P_{3}$ and a certain choice with corresponding $s$. We set\n\n$$\nd_{i}= \\begin{cases}0 & \\text { if } a_{i}=i \\\\ 1, & \\text { if } a_{i} \\neq i\\end{cases}\n$$\n\nWe also put $d=\\sum_{i=1}^{m} d_{i}$, and note that $0 \\leq d \\leq m$. Note also that if $a_{i} \\neq i$, then $a_{i} \\equiv i-1$ $(\\bmod m)$. Hence, for all $a_{i} \\in P_{3, i}$ it holds that\n\n$$\na_{i} \\equiv i-d_{i} \\quad(\\bmod m)\n$$\n\nHence,\n\n$$\ns \\equiv \\sum_{i=1}^{m} a_{i} \\equiv \\sum_{i=1}^{m}\\left(i-d_{i}\\right) \\equiv \\sigma-d \\quad(\\bmod m)\n$$\n\nwhich can only be congruent to $\\sigma$ modulo $m$ if all $d_{i}$ are equal, which forces $s=\\left(m^{2}+m\\right) / 2$ or $s=\\left(3 m^{2}+m\\right) / 2$. Since $m>1$, it holds that\n\n$$\n\\frac{m^{2}+m}{2}m^{2}+m$, this partition gives a positive answer.\n\n- Consider the partition $P_{2}$ and a certain choice with corresponding $s$. We find that\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\quad(\\bmod m)\n$$\n\nHence, if $m^{2} \\leq n \\leq m^{2}+m$ and $n \\not \\equiv \\sigma(\\bmod m)$, this partition solves the problem.\n\nGiven the analysis of $P_{1}$ and $P_{2}$ as above, we may conclude (noting that $\\sigma \\equiv m(m+1) / 2(\\bmod m))$ that if $m$ is odd then $m^{2}$ and $m^{2}+m$ are the only candidates for counterexamples $n$, while if $m$ is even then $m^{2}+\\frac{m}{2}$ is the only candidate.\n\nThere are now various ways to proceed as alternatives to the partition $P_{3}$.\n\nConsider the partition $(\\{1, m+2\\},\\{2, m+3\\}, \\ldots,\\{m-1,2 m\\},\\{m, m+1\\})$. We consider possible sums $\\bmod m+1$. For the first $m-1$ pairs, the elements of each pair are congruent $\\bmod m+1$, so the sum of one element of each pair is $(\\bmod m+1)$ congruent to $\\frac{1}{2} m(m+1)-m$, which is congruent to 1 if $m+1$ is odd and $1+\\frac{m+1}{2}$ if $m+1$ is even. Now the elements of the last pair are congruent to -1 and 0 , so any achievable value of $n$ is congruent to 0 or 1 if $m+1$ is odd, and to 0 or 1 plus $\\frac{m+1}{2}$ if $m+1$ is even. If $m$ is even then $m^{2}+\\frac{m}{2} \\equiv 1+\\frac{m}{2}$, which is not congruent to 0 or 1 . If $m$ is odd then $m^{2} \\equiv 1$ and $m^{2}+m \\equiv 0$, neither of which can equal 0 or 1 plus $\\frac{m+1}{2}$.' 'Define the following ordered partitions:\n\n$$\n\\begin{aligned}\n& P_{1}=(\\{1,2\\},\\{3,4\\}, \\ldots,\\{2 m-1,2 m\\}), \\\\\n& P_{2}=(\\{1, m+1\\},\\{2, m+2\\}, \\ldots,\\{m, 2 m\\}), \\\\\n& P_{3}=(\\{1,2 m\\},\\{2, m+1\\},\\{3, m+2\\}, \\ldots,\\{m, 2 m-1\\}) .\n\\end{aligned}\n$$\n\nFor each $P_{j}$ we will compute the possible values for the expression $s=a_{1}+\\ldots+a_{m}$, where $a_{i} \\in P_{j, i}$ are the chosen integers. Here, $P_{j, i}$ denotes the $i$-th coordinate of the ordered partition $P_{j}$.\n\nWe will denote by $\\sigma$ the number $\\sum_{i=1}^{m} i=\\left(m^{2}+m\\right) / 2$.\n\n- Consider the partition $P_{1}$ and a certain choice with corresponding sum $s$. We find that\n\n$$\nm^{2}=\\sum_{i=1}^{m}(2 i-1) \\leq s \\leq \\sum_{i=1}^{m} 2 i=m^{2}+m\n$$\n\nHence, if $nm^{2}+m$, this partition gives a positive answer.\n\n- Consider the partition $P_{2}$ and a certain choice with corresponding $s$. We find that\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\quad(\\bmod m)\n$$\n\nHence, if $m^{2} \\leq n \\leq m^{2}+m$ and $n \\not \\equiv \\sigma(\\bmod m)$, this partition solves the problem.\n\nGiven the analysis of $P_{1}$ and $P_{2}$ as above, we may conclude (noting that $\\sigma \\equiv m(m+1) / 2(\\bmod m))$ that if $m$ is odd then $m^{2}$ and $m^{2}+m$ are the only candidates for counterexamples $n$, while if $m$ is even then $m^{2}+\\frac{m}{2}$ is the only candidate.\n\nThere are now various ways to proceed as alternatives to the partition $P_{3}$.\nConsider the partition $(\\{1, m\\},\\{2, m+1\\}, \\ldots,\\{m-1,2 m-2\\},\\{2 m-1,2 m\\})$, this time considering sums of elements of pairs $\\bmod m-1$. If $m-1$ is odd, the sum is congruent to 1 or 2 ; if $m-1$ is even, to 1 or 2 plus $\\frac{m-1}{2}$. If $m$ is even then $m^{2}+\\frac{m}{2} \\equiv 1+\\frac{m}{2}$, and this can only be congruent to 1 or 2 when $m=2$. If $m$ is odd, $m^{2}$ and $m^{2}+m$ are congruent to 1 and 2 , and these can only be congruent to 1 or 2 plus $\\frac{m-1}{2}$ when $m=3$. Now the cases of $m=2$ and $m=3$ need considering separately (by finding explicit partitions excluding each $n$ ).' 'Define the following ordered partitions:\n\n$$\n\\begin{aligned}\n& P_{1}=(\\{1,2\\},\\{3,4\\}, \\ldots,\\{2 m-1,2 m\\})\n\\end{aligned}\n$$\n\nFor each $P_{j}$ we will compute the possible values for the expression $s=a_{1}+\\ldots+a_{m}$, where $a_{i} \\in P_{j, i}$ are the chosen integers. Here, $P_{j, i}$ denotes the $i$-th coordinate of the ordered partition $P_{j}$.\n\nWe will denote by $\\sigma$ the number $\\sum_{i=1}^{m} i=\\left(m^{2}+m\\right) / 2$.\n\n- Consider the partition $P_{1}$ and a certain choice with corresponding sum $s$. We find that\n\n$$\nm^{2}=\\sum_{i=1}^{m}(2 i-1) \\leq s \\leq \\sum_{i=1}^{m} 2 i=m^{2}+m\n$$\n\nHence, if $nm^{2}+m$, this partition gives a positive answer.\n\nThis solution does not use modulo arguments. Use only $P_{1}$ from above to conclude that $m^{2} \\leq n \\leq m^{2}+m$. Now consider the partition $(\\{1,2 m\\},\\{2,3\\},\\{4,5\\}, \\ldots,\\{2 m-$ $2,2 m-1\\})$. If 1 is chosen from the first pair, the sum is at most $m^{2}$; if $2 m$ is chosen, the sum is at least $m^{2}+m$. So either $n=m^{2}$ or $n=m^{2}+m$. Now consider the partition $(\\{1,2 m-$ $1\\},\\{2,2 m\\},\\{3,4\\},\\{5,6\\}, \\ldots,\\{2 m-3,2 m-2\\})$. Sums of one element from each of the last $m-2$ pairs are in the range from $(m-2) m=m^{2}-2 m$ to $(m-2)(m+1)=m^{2}-m-2$ inclusive. Sums of one element from each of the first two pairs are $3,2 m+1$ and $4 m-1$. In the first case we have $n \\leq m^{2}-m+1m^{2}+m$. So these three partitions together have eliminated all $n$.']" [] Text-only Competition True Theorem proof Combinatorics Math English 417 "There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$. Prove that $a_{1}=a_{2}=\cdots=a_{n}$." "['Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}c_{\\ell-1} \\geqslant 1$\n\nFrom $c_{\\ell-1} c_{\\ell} c_{\\ell+1}c_{\\ell} \\geqslant 1$, which is a contradiction to our choice of $\\ell$.\n\nCase 2: $c_{\\ell-1} \\geqslant c_{\\ell} \\geqslant 1$\n\nOnce again looking at the inequality (5) we can find that\n\n$$\nc_{\\ell-2} c_{\\ell-1} c_{\\ell} \\geqslant c_{\\ell-1}^{2} c_{\\ell} \\Longrightarrow c_{\\ell-2} \\geqslant c_{\\ell-1} \\tag{7}\n$$\n\nNote that we only needed $c_{\\ell-1} \\geqslant c_{\\ell} \\geqslant 1$ to show $c_{\\ell-2} \\geqslant c_{\\ell-1} \\geqslant 1$. So using induction we can easily show $c_{\\ell-s-1} \\geqslant c_{\\ell-s}$ for all $s$.\n\nSo\n\n$$\nc_{1} \\leqslant c_{2} \\leqslant \\cdots \\leqslant c_{n} \\leqslant c_{1} \\tag{8}\n$$\n\na contradiction to our innitial assumption.\n\nSo our innitial assumtion must have been wrong, which implies that all the $a_{i}$ must have been equal from the start.']" [] Text-only Competition True Theorem proof Algebra Math English 418 "We are given an acute triangle $A B C$. Let $D$ be the point on its circumcircle such that $A D$ is a diameter. Suppose that points $K$ and $L$ lie on segments $A B$ and $A C$, respectively, and that $D K$ and $D L$ are tangent to circle $A K L$. Show that line $K L$ passes through the orthocentre of $A B C$. The altitudes of a triangle meet at its orthocentre." "['\nLet $M$ be the midpoint of $K L$. We will prove that $M$ is the orthocentre of $A B C$. Since $D K$ and $D L$ are tangent to the same circle, $|D K|=|D L|$ and hence $D M \\perp K L$. The theorem of Thales in circle $A B C$ also gives $D B \\perp B A$ and $D C \\perp C A$. The right angles then give that quadrilaterals $B D M K$ and $D M L C$ are cyclic.\n\nIf $\\angle B A C=\\alpha$, then clearly $\\angle D K M=\\angle M L D=\\alpha$ by angle in the alternate segment of circle $A K L$, and so $\\angle M D K=\\angle L D M=\\frac{\\pi}{2}-\\alpha$, which thanks to cyclic quadrilaterals gives $\\angle M B K=\\angle L C M=\\frac{\\pi}{2}-\\alpha$. From this, we have $B M \\perp A C$ and $C M \\perp A B$, and so $M$ indeed is the orthocentre of $A B C$.' 'Let $A B C$ be a triangle with circumcircle $\\Gamma$. Let $X$ be a point in the plane. The Simson line (Wallace-Simson line) is defined via the following theorem. Drop perpendiculars from $X$ to each of the three side lines of $A B C$. The feet of these perpendiculars are collinear (on\n\n\n\nthe Simson line of $X$ ) if and only if $X$ lies of $\\Gamma$. The Simson line of $X$ in the circumcircle bisects the line segment $X H$ where $H$ is the orthocentre of triangle $A B C$. See Figure 2\n\n\n\nFigure 2: The Wallace-Simson configuration\n\nWhen $X$ is on $\\Gamma$, we can enlarge from $X$ with scale factor 2 (a homothety) to take the Simson line to the doubled Simson line which passes through the orthocentre $H$ and contains the reflections of $X$ in each of the three sides of $A B C$.\n\n\n\nFigure 3: Three circles do the work\n\nLet $\\Gamma$ be the circle $A B C, \\Sigma$ be the circle $A K L$ with centre $O$, and $\\Omega$ be the circle on diameter $O D$ so $K$ and $L$ are on this circle by converse of Thales. Let $\\Omega$ and $\\Gamma$ meet at $D$ and $F$. By Thales in both circles, $\\angle A F D$ and $\\angle O F D$ are both right angles so $A O F$ is a line. Let $A F$ meet $\\Sigma$ again at $T$ so $A T$ (containing $O$ ) is a diameter of this circle and by Thales, $T L \\perp A C$.\n\nLet $G$ (on $\\Sigma$ ) be the reflection of $K$ in $A F$. Now $A T$ is the internal angle bisector of $\\angle G A K$ so, by an upmarket use of angles in the same segment (of $\\Sigma$ ), $T L$ is the internal angle bisector of $\\angle G L K$. Thus the line $G L$ is the reflection of the line $K L$ in $T L$, and so also the reflection of $K L$ in the line $A C$ (internal and external angle bisectors).\n\nOur next project is to show that $L G F$ are collinear. Well $\\angle F L K=\\angle F O K$ (angles in the same segment of $\\Omega$ ) and $\\angle G L K=\\angle G A K$ (angles in the same segment of $\\Sigma$ ) $=2 \\angle O A K$ $(A K G$ is isosceles with apex $A)=\\angle T O K$ (since $O A K$ is isosceles with apex $O$, and this is an external angle at $O$ ). The point $T$ lies in the interior of the line segment $F O$ so $\\angle T O K=\\angle F O K$. Therefore $\\angle F L K=\\angle G L K$ so $L G F$ is a line.\n\nNow from the second paragraph, $F$ is on the reflection of $K L$ in $A C$. By symmetry, $F$ is also on the reflection of $K L$ in $A B$. Therefore the reflections of $F$ in $A B$ and $A C$ are both on $K L$ which must therefore be the doubled Wallace-Simson line of $F$. Therefore the orthocentre of $A B C$ lies on $K L$.' 'Let $H$ be the orthocentre of triangle $A B C$ and $\\Sigma$ the circumcircle of $A K L$ with centre $O$. Let $\\Omega$ be the circle with diameter $O D$, which contains $K$ and $L$ by Thales, and let $\\Gamma$ be the circumcircle of $A B C$ containing $D$. Denote the second intersection of $\\Omega$ and $\\Gamma$ by $F$. Since $O D$ and $A D$ are diameters of $\\Omega$ and $\\Gamma$ we have $\\angle O F D=\\frac{\\pi}{2}=\\angle A F D$, so the points $A, O, F$ are collinear. Let $M$ and $N$ be the second intersections of $C H$ and $B H$ with $\\Gamma$, respectively. It is well-known that $M$ and $N$ are the reflections of $H$ in $A B$ and $A C$, respectively (because $\\angle N C A=\\angle N B A=\\angle A C M=\\angle A B M$ ). By collinearity of $A, O, F$ and the angles in $\\Gamma$ we have\n\n$$\n\\angle N F O=\\angle N F A=\\angle N B A=\\frac{\\pi}{2}-\\angle B A C=\\frac{\\pi}{2}-\\angle K A L .\n$$\n\nSince $D L$ is tangent to $\\Sigma$ we obtain\n\n$$\n\\angle N F O=\\frac{\\pi}{2}-\\angle K L D=\\angle L D O,\n$$\n\nwhere the last equality follows from the fact that $O D$ is bisector of $\\angle L D K$ since $L D$\n\n\n\nFigure 4: Diagram to Solution 3\n\n\n\nand $K D$ are tangent to $\\Sigma$. Furthermore, $\\angle L D O=\\angle L F O$ since these are angles in $\\Omega$. Hence, $\\angle N F O=\\angle L F O$, which implies that points $N, L, F$ are collinear. Similarly points $M, K, F$ are collinear. Since $N$ and $M$ are reflections of $H$ in $A C$ and $A B$ we have\n\n$$\n\\angle L H N=\\angle H N L=\\angle B N F=\\angle B M F=\\angle B M K=\\angle K H B .\n$$\n\nHence,\n\n$$\n\\angle L H K=\\angle L H N+\\angle N H K=\\angle K H B+\\angle N H K=\\pi\n$$\n\nand the points $L, H, K$ are collinear.' 'Let $M$ and $N$ be the reflections of the orthocentre in $A B$ and $A C$. Let $\\angle B A C=\\alpha$. Then $\\angle N D M=\\pi-\\angle M A N=\\pi-2 \\alpha$.\n\nLet $M K$ and $N L$ intersect at $F$. See Figure 3.\n\nClaim. $\\angle N F M=\\pi-2 \\alpha$, so $F$ lies on the circumcircle.\n\nProof. Since $K D$ and $L D$ are tangents to circle $A K L$, we have $|D K|=|D L|$ and $\\angle D K L=\\angle K L D=\\alpha$, so $\\angle L D K=\\pi-2 \\alpha$.\n\nBy definition of $M, N$ and $D, \\angle M N D=\\angle A N D-\\angle A N M=\\frac{\\pi}{2}-\\left(\\frac{\\pi}{2}-\\alpha\\right)=\\alpha$ and analogously $\\angle D M N=\\alpha$. Hence $|D M|=|D N|$.\n\nFrom $\\angle N D M=\\angle L D K=\\pi-2 \\alpha$ if follows that $\\angle L D N=\\angle K D M$. Since $|D K|=|D L|$ and $|D M|=|D N|$, triangles $M D K$ and $N D L$ are related by a rotation about $D$ through angle $\\pi-2 \\alpha$, and hence the angle between $M K$ and $N L$ is $\\pi-2 \\alpha$, which proved the claim.\n\nWe now finish as below:\n\n$$\n\\begin{gathered}\n\\angle M H K=\\angle K M H=\\angle F M C=\\angle F A C, \\\\\n\\angle L H N=\\angle H N L=\\angle B N F=\\angle B A F .\n\\end{gathered}\n$$\n\nAs $\\angle B A F+\\angle F A C=\\alpha$, we have $\\angle L H K=\\alpha+\\angle N H M=\\alpha+\\pi-\\alpha=\\pi$, so $H$ lies on $K L$.' 'Since $A D$ is a diameter, it is well known that $D B H C$ is a parallelogram (indeed, both $B D$ and $C H$ are perpendicular to $A B$, hence parallel, and similarly for $D C \\| B H)$. Let $B^{\\prime}, C^{\\prime}$ be the reflections of $D$ in lines $A K B$ and $A L C$, respectively; since $A B D$ and $A C D$ are right angles, these are also the factor-2 homotheties of $B$ and $C$ with respect to $D$, hence $H$ is the midpoint of $B^{\\prime} C^{\\prime}$. We will prove that $B^{\\prime} K C^{\\prime} L$ is a parallelogram: it will then follow that the midpoint of $B^{\\prime} C^{\\prime}$, which is $H$, is also the midpoint of $K L$, and in particular is on the line, as we wanted to show.\n\nWe will prove $B^{\\prime} K C^{\\prime} L$ is a parallelogram by showing that $B^{\\prime} K$ and $C^{\\prime} L$ are the same length and direction. Indeed, for lengths we have $K B^{\\prime}=K D=L D=L C^{\\prime}$, where the first and last equalities arise from the reflections defining $B^{\\prime}$ and $C^{\\prime}$, and the middle one\n\n\n\nis equality of tangents. For directions, let $\\alpha, \\beta, \\gamma$ denote the angles of triangle $A K L$. Immediate angle chasing in the circle $A K L$, and the properties of the reflections, yield\n\n$$\n\\begin{gathered}\n\\angle C^{\\prime} L C=\\angle C L D=\\angle A K L=\\beta \\\\\n\\angle B K B^{\\prime}=\\angle D K B=\\angle K L A=\\gamma \\\\\n\\angle L D K=2 \\alpha-\\pi\n\\end{gathered}\n$$\n\nand therefore in directed angles $(\\bmod 2 \\pi)$ we have\n\n$\\angle\\left(C^{\\prime} L, B^{\\prime} K\\right)=\\angle C^{\\prime} L C+\\angle C L D+\\angle L D K+\\angle D K B+\\angle B K B^{\\prime}=2 \\alpha+2 \\beta+2 \\gamma-\\pi=\\pi$\n\nand hence $C^{\\prime} L$ and $B^{\\prime} K$ are parallel and in opposite directions, i.e. $C^{\\prime} L$ and $K B^{\\prime}$ are in the same direction, as claimed.' 'There are a number of ""phantom point"" arguments which define $K^{\\prime}$ and $L^{\\prime}$ in terms of angles and then deduce that these points are actually $K$ and $L$.\n\nNote: In these solutions it is necessary to show that $K$ and $L$ are uniquely determined by the conditions of the problem. One example of doing this is the following:\n\nTo prove uniqueness of $K$ and $L$, let us consider that there exist two other points $K^{\\prime}$ and $L^{\\prime}$ that satisfy the same properties ( $K^{\\prime}$ on $A B$ and $L^{\\prime}$ on $A C$ such that $D K^{\\prime}$ and $D L^{\\prime}$ are tangent to the circle $\\left.A K^{\\prime} L^{\\prime}\\right)$.\n\nThen, we have that $D K=D L$ and $D K^{\\prime}=D L^{\\prime}$. We also have that $\\angle K D L=\\angle K^{\\prime} D L^{\\prime}=$ $\\pi-2 \\angle A$. Hence, we deduce $\\angle K D K^{\\prime}=\\angle K D L-\\angle K^{\\prime} D L=\\angle K^{\\prime} D L^{\\prime}-\\angle K^{\\prime} D L=\\angle L D L^{\\prime}$ Thus we have that $\\triangle K D K^{\\prime} \\equiv \\triangle L D L^{\\prime}$, so we deduce $\\angle D K A=\\angle D K K^{\\prime}=\\angle D L L^{\\prime}=$ $\\pi-\\angle A L D$. This implies that $A K D L$ is concyclic, which is clearly a contradiction since $\\angle K A L+\\angle K D L=\\pi-\\angle B A C$.' 'We will use the usual complex number notation, where we will use a capital letter (like $Z$ ) to denote the point associated to a complex number (like $z$ ). Consider $\\triangle A K L$ on the unit circle. So, we have $a \\cdot \\bar{a}=k \\cdot \\bar{k}=l \\cdot \\bar{l}=1$ As point $D$ is the intersection of the tangents to the unit circle at $K$ and $L$, we have that\n\n$$\nd=\\frac{2 k l}{k+l} \\text { and } \\bar{d}=\\frac{2}{k+l}\n$$\n\nDefining $B$ as the foot of the perpendicular from $D$ on the line $A K$, and $C$ as the foot of the perpendicular from $D$ on the line $A L$, we have the formulas:\n\n$$\nb=\\frac{1}{2}\\left(d+\\frac{(a-k) \\bar{d}+\\bar{a} k-a \\bar{k}}{\\bar{a}-\\bar{k}}\\right)\n$$\n\n\n\n$$\nc=\\frac{1}{2}\\left(d+\\frac{(a-l) \\bar{d}+\\bar{a} l-a \\bar{l}}{\\bar{a}-\\bar{l}}\\right)\n$$\n\nSimplyfing these formulas, we get:\n\n$$\n\\begin{gathered}\nb=\\frac{1}{2}\\left(d+\\frac{(a-k) \\frac{2}{k+l}+\\frac{k}{a}-\\frac{a}{k}}{\\frac{1}{a}-\\frac{1}{k}}\\right)=\\frac{1}{2}\\left(d+\\frac{\\frac{2(a-k)}{k+l}+\\frac{k^{2}-a^{2}}{a k}}{\\frac{k-a}{a k}}\\right) \\\\\nb=\\frac{1}{2}\\left(\\frac{2 k l}{k+l}-\\frac{2 a k}{k+l}+(a+k)\\right)=\\frac{k(l-a)}{k+l}+\\frac{1}{2}(k+a) \\\\\nc=\\frac{1}{2}\\left(d+\\frac{(a-l) \\frac{2}{k+l}+\\frac{l}{a}-\\frac{a}{l}}{\\frac{1}{a}-\\frac{1}{l}}\\right)=\\frac{1}{2}\\left(d+\\frac{\\frac{2(a-l)}{k+l}+\\frac{l^{2}-a^{2}}{a l}}{\\frac{l-a}{a l}}\\right) \\\\\nc=\\frac{1}{2}\\left(\\frac{2 k l}{k+l}-\\frac{2 a l}{k+l}+(a+l)\\right)=\\frac{l(k-a)}{k+l}+\\frac{1}{2}(l+a)\n\\end{gathered}\n$$\n\nLet $O$ be the the circumcenter of triangle $\\triangle A B C$. As $A D$ is the diameter of this circle, we have that:\n\n$$\no=\\frac{a+d}{2}\n$$\n\nDefining $H$ as the orthocentre of the $\\triangle A B C$, we get that:\n\n$$\n\\begin{gathered}\nh=a+b+c-2 \\cdot o=a+\\left(\\frac{k(l-a)}{k+l}+\\frac{1}{2}(k+a)\\right)+\\left(\\frac{l(k-a)}{k+l}+\\frac{1}{2}(l+a)\\right)-(a+d) \\\\\nh=a+\\frac{2 k l}{k+l}-\\frac{a(k+l)}{k+l}+\\frac{1}{2} k++\\frac{1}{2} l++a-\\left(a+\\frac{2 k l}{k+l}\\right) \\\\\nh=\\frac{1}{2}(k+l)\n\\end{gathered}\n$$\n\nHence, we conclude that $H$ is the midpoint of $K L$, so $H, K, L$ are collinear.' 'Let us employ the barycentric coordinates. Set $A(1,0,0), K(0,1,0), L(0,0,1)$.\n\nThe tangent at $K$ of $(A K L)$ is $a^{2} z+c^{2} x=0$, and the tangent of of $L$ at $(A K L)$ is $a^{2} y+b^{2} x=0$. Their intersection is\n\n$$\nD\\left(-a^{2}: b^{2}: c^{2}\\right)\n$$\n\nSince $B \\in A K$, we can let $B(1-t, t, 0)$. Solving for $\\overrightarrow{A B} \\cdot \\overrightarrow{B D}=0$ gives\n\n$$\nt=\\frac{3 b^{2}+c^{2}-a^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)} \\Longrightarrow B=\\left(\\frac{-a^{2}-b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, \\frac{-a^{2}+3 b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, 0\\right)\n$$\n\nLikewise, $C$ has the coordinate\n\n$$\nC=\\left(\\frac{-a^{2}+b^{2}-c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, 0, \\frac{-a^{2}+b^{2}+3 c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right) .\n$$\n\n\n\nThe altitude from $B$ for triangle $A B C$ is\n\n$$\n-b^{2}\\left(x-z-\\frac{-a^{2}-b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)+\\left(c^{2}-a^{2}\\right)\\left(y-\\frac{-a^{2}+3 b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)=0 .\n$$\n\nAlso the altitude from $C$ for triangle $A B C$ is\n\n$$\n-c^{2}\\left(x-y-\\frac{-a^{2}+b^{2}-c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)+\\left(a^{2}-b^{2}\\right)\\left(z-\\frac{-a^{2}+b^{2}+3 c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)=0 .\n$$\n\nThe intersection of these two altitudes, which is the orthocenter of triangle $A B C$, has the barycentric coordinate\n\n$$\nH=(0,1 / 2,1 / 2)\n$$\n\nwhich is the midpoint of the segment $K L$.']" [] Text-only Competition True Theorem proof Geometry Math English 419 "We are given a positive integer $s \geqslant 2$. For each positive integer $k$, we define its twist $k^{\prime}$ as follows: write $k$ as $a s+b$, where $a, b$ are non-negative integers and $bs^{2}$. Since the reminder when a power of $s$ is divided by $s^{2}-1$ is either 1 or $s$, there exists a positive integer $m$ such that $s^{m}-n$ is non-negative and divisible by $s^{2}-1$. By our assumption $m \\geqslant 3$. We also take the smallest such $m$, so that $n>s^{m-2}$. The quotient $\\frac{s^{m}-n}{s^{2}-1}$ is therefore smaller than $s^{m-2}$, so there exist $b_{1}, \\ldots, b_{m-2} \\in\\{0,1, \\ldots, s-1\\}$ such that $\\frac{s^{m}-n}{s^{2}-1}=\\sum_{i=1}^{m-2} b_{i} s^{i-1}$. It follows that\n\n$$\nn=s^{m}-\\sum_{i=1}^{m-2} b_{i}\\left(s^{i+1}-s^{i-1}\\right)\n$$\n\nWe now show that\n\n$$\nd_{j}=s^{m+1-j}-\\sum_{i=1}^{m-1-j} b_{i}\\left(s^{i+1}-s^{i-1}\\right) \\tag{10}\n$$\n\nfor $j=1,2, \\ldots, m-2$ by induction on $j$. For $j=1$ this follows from $d_{1}=n$. Assume now that (10) holds for some $j\n\nPart I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\nPart II: Consider the triangle $A S_{b} S_{c}$. Note that since $T$ is the intersection of the tangents at $S_{b}$ and $S_{c}$ to the circumcircle of $A S_{b} S_{c}$, it follows that $A T$ is the symmedian of $A$ in this triangle. Let $X$ denote the second intersection of the symmedian $A T$ with $\\Omega$. We wish to show that $X$ is also on $I N_{a}$.\n\nNote that $A N_{a}$ is the external angle bisector of angle $A$, and therefore it is parallel to $S_{b} S_{c}$. Let $M$ denote the midpoint of $S_{b} S_{c}$, and let $Y$ be the second intersection of $A M$ with $\\Omega$. Since in $A S_{b} S_{c}, A X T$ is the symmedian and $A M Y$ is the median, it follows that $X Y$ is also parallel to $S_{b} S_{c}$. Thus, reflecting in the perpendicular bisector of $S_{b} S_{c}$ sends the line $A M Y$ to line $N_{a} M X$.\n\nNext, consider the quadrilateral $A S_{b} I S_{c}$. From the trillium theorem we have $\\left|S_{b} A\\right|=\\left|S_{b} I\\right|$ and $\\left|S_{c} A\\right|=\\left|S_{c} I\\right|$, thus the quadrilateral is a kite, from which it follows that the reflection of the line $A M$ in $S_{b} S_{c}$ is the line $I M$. But previously we have seen that this is also the line $N_{a} M X$. Thus $M, I, N_{a}$ and $X$ are collinear, as we wanted to show.' '\n\nPart I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\nWe begin by showing that the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$ as above.\n\nPart II: We introduce the point $S_{a}$ with the obvious meaning. Observe that the incentre $I$ of $A B C$ is the orthocentre of $S_{a} S_{b} S_{c}$ either because this is well-known, or because of an angle argument that $A$ reflects in $S_{b} S_{c}$ to $I$ (and similar results by cyclic change of letters). Therefore $A S_{a}$ is perpendicular to $S_{b} S_{c}$.\n\n\n\nFigure 7: A reflections argument for Solution\n\nLet $M$ denote the midpoint of $S_{b} S_{c}$. Then $A$ is the reflection of $S_{a}$ in the diameter parallel to $S_{b} S_{c}$, so the reflection of $A$ in the diameter perpendicular to $S_{b} S_{c}$ is $N_{a}$, the antipode of $S_{a}$. Let the reflection of $X$ in $T M$ be $Y$, so $T Y$ passes through $N_{a}$ and is the reflection of $T X$ in $T M$.\n\nNow $S_{b} S_{c}$ is the polar line of $T$ with respect to $\\Omega$, so $A Y$ and $N_{a} X$ meet on this line, and by symmetry at its midpoint $M$. The line $N_{a} M X$ is therefore the reflection of the line $Y M A$ in $S_{b} S_{c}$, and so $N_{a} M X$ passes through $I$ (the reflection of $A$ in $S_{b} S_{c}$ ).\n\nThe triangle $A S_{c} S_{b}$ can be taken as generic, and from the argument above we can extract the fact that the symmedian point and the centroid are isogonal conjugates in that triangle.' 'Part I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\n\n\nFigure 8: Diagram to Solution\n\nPart II: By the trillelium theorem, $S_{c} S_{b}$ bisects $A I$, and since $N_{a} A \\| S_{b} S_{c}$, then $O T$ is a bisector of $A N_{a}$. This implies $\\left|M N_{a}\\right|=|M A|=|M I|$, since $M$ is the midpoint of $S_{c} S_{b}$ and lies also on $O T$. Hence, $M$ is the circumcentre of triangle $I A N_{a}$. But this triangle has a right angle at $A$ (since $A I$ and $A N_{a}$ are the inner and outer angle bisector at $A$ ), hence $M$ lies on $I N_{a}$.\n\nAgain, let $X$ be the second intersection of $T A$ and $\\Omega$. By the above, it suffices to prove that $X$ lies on the line $N_{a} M$. From the power of point $T$ with respect to $\\Omega$ we get $|T A| \\cdot|T X|=\\left|T S_{c}\\right|^{2}$. Since $M$ is the foot of the altitude of right triangle $T S_{c} O$, we obtain $\\left|T S_{c}\\right|^{2}=|T M| \\cdot|T O|$. Hence, $|T A| \\cdot|T X|=|T M| \\cdot|T O|$ so the points $O, M, A, X$ are concyclic. It follows that $\\angle M X A=\\angle M O A=\\frac{1}{2} \\angle N_{a} O A=\\angle N_{a} X A$. Hence, $X$ lies on the line $N_{a} M$.\n\nRemark. To show that $O M A X$ is cyclic, one can also invert the line $T A X$ in the circumcircle of the triangle $A B C$.\n\nSolution 4. \nPart I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\n\n\nFigure 9: Diagram to Solution\n\nPart II: We show that $A N_{a} \\| S_{b} S_{c}$. In particular, $\\angle N_{a} O T=\\angle T O A$. The conclusion of the problem trivially holds if $|A B|=|A C|$, therefore we assume without loss of generality that $|A C|>|A B|$. Let $S_{a}$ be the midpoint of the $\\operatorname{arc} B C$ which does not contain $A$. Then $N_{a} S_{a}$ is a diameter, so $\\angle S_{a} B N_{a}=\\frac{\\pi}{2}=\\angle O S_{c} T$. We also compute $\\angle B N_{a} S_{a}=\\angle B A S_{a}=\\frac{1}{2} \\angle B A C=\\frac{1}{2} \\angle S_{c} T S_{b}=\\angle S_{c} T O$. It follows that the triangles $T S_{c} O$ and $N_{a} B S_{a}$ are similar. In particular,\n\n$$\n\\frac{\\left|N_{a} B\\right|}{\\left|T S_{c}\\right|}=\\frac{\\left|S_{a} B\\right|}{\\left|O S_{c}\\right|} \\tag{12}\n$$\n\nNext we compute\n\n$$\n\\angle I S_{a} B=\\angle N_{a} S_{a} B-\\angle N_{a} S_{a} I=\\angle T O S_{c}-\\frac{1}{2} \\angle N_{a} O A=\\angle T O S_{c}-\\angle T O A=\\angle A O S_{c} \\tag{13}\n$$\n\n\n\nand\n\n$$\n\\angle I B N_{a}=\\angle C B N_{a}-\\angle C B I=\\frac{1}{2}\\left(\\pi-\\angle B N_{a} C\\right)-\\frac{1}{2} \\angle C B A=\\frac{1}{2} \\angle A C B=\\angle A C S_{c}=\\angle A S_{c} T \\tag{14}\n$$,\n\nhence\n\n$$\n\\angle S_{a} B I=\\frac{\\pi}{2}-\\angle I B N_{a}=\\frac{\\pi}{2}-\\angle A S_{c} T=\\angle O S_{c} A .\n$$\n\nTogether with (13) it follows that the triangles $I B S_{a}$ and $A S_{c} O$ are similar, so $\\frac{\\left|S_{a} B\\right|}{\\left|O S_{c}\\right|}=$ $\\frac{|I B|}{\\left|A S_{c}\\right|}$, and (12) implies $\\frac{\\left|N_{a} B\\right|}{\\left|T S_{c}\\right|}=\\frac{|I B|}{\\left|A S_{c}\\right|}$. Consequently, by (14) the triangles $T S_{c} A$ and $N_{a} B I$ are similar and therefore $\\angle S_{c} T A=\\angle B N_{a} I$. Now let $Q$ be the second intersection of $N_{a} I$ with $\\Omega$. Then $\\angle B N_{a} I=\\angle B N_{a} Q=\\angle B A Q$, so $\\angle S_{c} T A=\\angle B A Q$. Since $A B$ is parallel to $T S_{c}$, we get $A Q \\| T A$, i.e. $A, T, Q$ are collinear.\n\nRemark. After proving similarity of triangles $T S_{c} O$ and $N_{a} B S_{a}$ one can use spiral symmetry to show similarity of triangles $T S_{c} A$ and $N_{a} B I$.' 'Part I: First we show that $A$ lies on the radical axis between $\\omega_{b}$ and $\\omega_{c}$.\n\nLet $T$ be the radical center of the circumcircle, $\\omega_{b}$ and $\\omega_{c}$; then $T S_{b}$ and $T S_{c}$ are common tangents of the circles, as shown in Figure 5a. Moreover, let $P_{b}=A B \\cap S_{b} S_{c}$ and $P_{c}=$ $A C \\cap S_{b} S_{c}$. The triangle $T S_{c} S_{b}$ is isosceles: $A B \\| T S_{c}$ and $A C \\| T S_{b}$ so\n\n$$\n\\angle A P_{b} P_{c}=\\angle T S_{c} S_{b}=\\angle S_{c} S_{b} T=\\angle P_{b} P_{c} A .\n$$\n\nFrom these angles we can see that $\\omega_{b}$ passes through $P_{b}, \\omega_{c}$ passes through $P_{c}$, and finally $A P_{b}$ and $A P_{c}$ are equal tangents to $\\omega_{b}$ and $\\omega_{c}$, so $A$ lies on the radical axis.\n\n\n\nFigure 5a\n\n\n\nFigure $5 b$\n\n\n\nPart II. Let the radical axis $T A$ meet the circumcircle again at $X$, let $S_{a}$ be the midpoint of the $\\operatorname{arc} B C$ opposite to $A$, and let $X I$ meet the circumcirlce again at $N$. (See Figure 2.) For solving the problem, we have prove that $N_{a}=N$.\n\nThe triples of points $A, I, S_{a} ; B, I, S_{b}$ and $C, I, S_{c}$ are collinear because they lie on the angle bisectors of the triangle $A B C$.\n\nNotice that the quadrilateral $A S_{c} X S_{b}$ is harmonic, because the tangents at $S_{b}$ and $S_{c}$, and the line $A X$ are concurrent at $T$. This quadrilateral can be projected (or inverted) to the quadrilateral $S_{a} C N B$ through $I$. So, $S_{a} C N B$ also is a harmonic quadrilateral. Due to $S_{a} B=S_{a} C$, this implies $N B=N C$, so $N=N_{a}$. Done.' ""Part I: First let's show that this is equivalent to proving that $T A$ and $N_{a} I$ intersect in $\\Omega$.\n\nLemma: Let's recall that if we have two circles $\\omega_{1}$ and $\\omega_{2}$ which are internally tangent at point $X$ and if we have a line $A B$ tangent to $\\omega_{2}$ at $Y$. Let $M$ be the midpoint of the arc $A B$ not containing $Z$. We have that $Z, Y, M$ are collinear.\n\n\n\n\n\nLet $P_{b}=A B \\cap \\omega_{b}$ and $P_{c}=A C \\cap \\omega_{c}$. We can notice by the lemma that $S_{b}, P_{b}$ and $S_{c}$ are collinear, and similarly $S_{c}, P_{c}$ and $S_{b}$ are also collinear. Therefore $S_{c}, P_{b}, P_{c}$, and $S_{b}$ are collinear, and since $\\angle A P_{b} P_{c}=\\frac{\\angle A B C}{2}+\\frac{\\angle A C B}{2}=\\angle A P_{c} P_{b}$ then $A P_{b}=A P_{c}$ so $A$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $T$ be the intersection of the tangent lines of $\\Omega$ through $S_{c}$ and $S_{b}$. Since $T S_{c}=T S_{b}$ then $A T$ is the radical axis between $\\omega_{b}$ and $\\omega_{c}$.\n\nPart II: $T A$ and $N_{a} I$ intersect in $\\Omega$.\n\n\n\nLet $\\omega_{a}$ the $A$-mixtilinear incircle (that is, the circle internally tangent to $\\Omega$, and tangent to both $A B$ and $A C$ ), and let $X=\\Omega \\cap \\omega_{a}$. It is known that $N_{a}, I, X$ are collinear.\n\nLet $M_{c}$ and $M_{b}$ be the tangent points of $\\omega_{A}$ to $A B$ and $A C$ respectively, then by the lemma $X, M_{c}, S_{c}$ are collinear and $X, M_{b}, S_{b}$ are collinear. We can see that $S_{c} T S_{b}$ and $M_{c} A M_{b}$ are homothetic with respect to $X$; therefore $T$ and $A$ are homothetic with respect to $X$, impying that $T, A, X$ are collinear.\n\n\n\nFigure 10: Diagram to Solution""]" [] Text-only Competition True Theorem proof Geometry Math English 421 "The Sieve of Sundaram uses the following infinite table of positive integers: | 4 | 7 | 10 | 13 | $\cdots$ | | :---: | :---: | :---: | :---: | :---: | | 7 | 12 | 17 | 22 | $\cdots$ | | 10 | 17 | 24 | 31 | $\cdots$ | | 13 | 22 | 31 | 40 | $\cdots$ | | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | | The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown. Prove that if $N$ is an entry in the table, then $2 N+1$ is composite." ['First, we prove that the number in the $R$ th row and $C$ th column equals $2RC+R+C$.\nProof. First, we determine the first entry in the $R$ th row.\n\nSince the first column is an arithmetic sequence with common difference 3 , then the $R$ th entry in the first column (that is, the first entry in the $R$ th row) is $4+(R-1)(3)$ or $4+3 R-3=3 R+1$.\n\nSecond, we determine the common difference in the $R$ th row by determining the second entry in the $R$ th row.\n\nSince the second column is an arithmetic sequence with common difference 5 , then the $R$ th entry in the second column (that is, the second entry in the $R$ th row) is $7+(R-1)(5)$ or $7+5 R-5=5 R+2$.\n\nTherefore, the common difference in the $R$ th row must be $(5 R+2)-(3 R+1)=2 R+1$. Thus, the $C$ th entry in the $R$ th row (that is, the number in the $R$ th row and the $C$ th column) is\n\n$$\n3 R+1+(C-1)(2 R+1)=3 R+1+2 R C+C-2 R-1=2 R C+R+C\n$$\n\nSuppose that $N$ is an entry in the table, say in the $R$ th row and $C$ th column.\n\nFrom the proof above, then $N=2 R C+R+C$ and so $2 N+1=4 R C+2 R+2 C+1$.\n\nNow $4 R C+2 R+2 C+1=2 R(2 C+1)+2 C+1=(2 R+1)(2 C+1)$.\n\nSince $R$ and $C$ are integers with $R \\geq 1$ and $C \\geq 1$, then $2 R+1$ and $2 C+1$ are each integers that are at least 3 .\n\nTherefore, $2 N+1=(2 R+1)(2 C+1)$ must be composite, since it is the product of two integers that are each greater than 1.'] [] Text-only Competition True Theorem proof Algebra Math English 422 "Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$. Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$. Suppose that $A=\{f(1), f(2), f(3), \ldots\}$ and $B=\{g(1), g(2), g(3), \ldots\}$; that is, $A$ is the range of $f$ and $B$ is the range of $g$. Prove that every positive integer $m$ is an element of exactly one of $A$ or $B$." "['We want to show that each positive integer $m$ is in the range of $f$ or the range of $g$, but not both.\n\nTo do this, we first try to better understand the ""complicated"" term of each of the functions - that is, the term involving the greatest integer function.\n\nIn particular, we start with a positive integer $k \\geq 1$ and try to determine the positive integers $n$ that give $\\left\\lfloor\\frac{1+\\sqrt{8 n-7}}{2}\\right\\rfloor=k$.\n\nBy definition of the greatest integer function, the equation $\\left\\lfloor\\frac{1+\\sqrt{8 n-7}}{2}\\right\\rfloor=k$ is equivalent to the inequality $k \\leq \\frac{1+\\sqrt{8 n-7}}{2}\n\nWe know that $C$ has coordinates $(p, 0)$ and $D$ has coordinates $(0, p)$.\n\nThus, the slope of line segment $C D$ is -1 .\n\nSince the points $A$ and $B$ both lie on the line $y=x$, then the slope of line segment $A B$ is 1 .\n\nTherefore, $A B$ is perpendicular to $C D$, so $C A D B$ is a kite, and so its area is equal to $\\frac{1}{2}(A B)(C D)$.\n\n(We could derive this by breaking quadrilateral $C A D B$ into $\\triangle C A B$ and $\\triangle D A B$.)\n\nSince $C$ has coordinates $(p, 0)$ and $D$ has coordinates $(0, p)$, then $C D=\\sqrt{p^{2}+(-p)^{2}}=$ $\\sqrt{2 p^{2}}$.\n\n(We do not know if $p$ is positive, so this is not necessarily equal to $\\sqrt{2} p$.)\n\nWe know that $A$ has coordinates $(a, a)$ and $B$ has coordinates $(b, b)$, so\n\n$$\n\\begin{aligned}\nA B & =\\sqrt{(a-b)^{2}+(a-b)^{2}} \\\\\n& =\\sqrt{2 a^{2}-4 a b+2 b^{2}} \\\\\n& =\\sqrt{2\\left(a^{2}+b^{2}\\right)-4 a b} \\\\\n& =\\sqrt{2 r^{2}-4\\left(\\frac{1}{2}\\left(p^{2}-r^{2}\\right)\\right)} \\\\\n& =\\sqrt{4 r^{2}-2 p^{2}}\n\\end{aligned}\n$$\n\nTherefore, the area of quadrilateral $C A D B$ is\n\n$$\n\\frac{1}{2}(A B)(C D)=\\frac{1}{2} \\sqrt{4 r^{2}-2 p^{2}} \\sqrt{2 p^{2}}=\\sqrt{2 r^{2} p^{2}-p^{4}}\n$$\n\n\n\nTo maximize this area, we must maximize $2 r^{2} p^{2}-p^{4}=2 r^{2}\\left(p^{2}\\right)-\\left(p^{2}\\right)^{2}$.\n\nSince $r$ is fixed, we can consider this as a quadratic polynomial in $p^{2}$. Since the coefficient of $\\left(p^{2}\\right)^{2}$ is negative, then this is a parabola opening downwards, so we find its maximum value by finding its vertex.\n\nThe vertex of $2 r^{2}\\left(p^{2}\\right)-\\left(p^{2}\\right)^{2}$ is at $p^{2}=-\\frac{2 r^{2}}{2(-1)}=r^{2}$.\n\nSo the maximum area of the quadrilateral occurs when $p$ is chosen so that $p^{2}=r^{2}$.\n\nSince $p^{2}=r^{2}$, then $(a+b)^{2}=p^{2}=r^{2}$ so $a^{2}+2 a b+b^{2}=r^{2}$.\n\nSince $a^{2}+b^{2}=r^{2}$, then $2 a b=0$ so either $a=0$ or $b=0$, and so either $A$ has coordinates $(0,0)$ or $B$ has coordinates $(0,0)$, ie. either $A$ is the origin or $B$ is the origin.'] [] Text-only Competition True Theorem proof Geometry Math English 425 Prove that the integer $\frac{(1 !)(2 !)(3 !) \cdots(398 !)(399 !)(400 !)}{200 !}$ is a perfect square. (In this fraction, the numerator is the product of the factorials of the integers from 1 to 400 , inclusive.) ['Let $N=\\frac{(1 !)(2 !)(3 !) \\cdots(398 !)(399 !)(400 !)}{200 !}$.\n\nFor each integer $k$ from 1 to 200 , inclusive, we rewrite $(2 k)$ ! as $2 k \\cdot(2 k-1)$ !.\n\nTherefore, $(2 k-1) !(2 k) !=(2 k-1) ! \\cdot 2 k \\cdot(2 k-1) !=2 k((2 k-1) !)^{2}$.\n\n(In particular, $(1 !)(2 !)=2(1 !)^{2},(3 !)(4 !)=4(3 !)^{2}$, and so on.)\n\nThus,\n\n$$\nN=\\frac{2(1 !)^{2} \\cdot 4(3 !)^{2} \\cdots \\cdot 398(397 !)^{2} \\cdot 400(399 !)^{2}}{200 !}\n$$\n\nRe-arranging the numerator of the expression, we obtain\n\n$$\nN=\\frac{(1 !)^{2}(3 !)^{2} \\cdots(397 !)^{2}(399 !)^{2} \\cdot(2 \\cdot 4 \\cdot \\cdots 398 \\cdot 400)}{200 !}\n$$\n\nWe can now re-write $2 \\cdot 4 \\cdot \\cdots 398 \\cdot 400$ as $(2 \\cdot 1) \\cdot(2 \\cdot 2) \\cdots(2 \\cdot 199) \\cdot(2 \\cdot 200)$.\n\nSince there are 200 sets of parentheses, we obtain\n\n$$\nN=\\frac{(1 !)^{2}(3 !)^{2} \\cdots(397 !)^{2}(399 !)^{2} \\cdot 2^{200} \\cdot(1 \\cdot 2 \\cdots \\cdots 199 \\cdot 200)}{200 !}\n$$\n\nSince $1 \\cdot 2 \\cdot \\cdots \\cdot 199 \\cdot 200=200$ !, we can conclude that\n\n$$\nN=2^{200}(1 !)^{2}(3 !)^{2} \\cdots(397 !)^{2}(399 !)^{2}\n$$\n\nTherefore,\n\n$$\n\\sqrt{N}=2^{100}(1 !)(3 !) \\cdots(397 !)(399 !)\n$$\n\nwhich is a product of integers and thus an integer itself.\n\nSince $\\sqrt{N}$ is an integer, $N$ is a perfect square, as required.'] [] Text-only Competition True Theorem proof Number Theory Math English 426 Prove that, for all integers $K$ and $L$, there is at least one pair of integers $(a, b)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$. ['Suppose that $K$ and $L$ are integers.\n\nThen\n\n$$\n\\begin{aligned}\n& (K+L)^{2}+(K-L)^{2}-(K+L)(K-L) \\\\\n& \\quad=\\left(K^{2}+2 K L+L^{2}\\right)+\\left(K^{2}-2 K L+L^{2}\\right)-\\left(K^{2}-L^{2}\\right) \\\\\n& \\quad=K^{2}+3 L^{2}\n\\end{aligned}\n$$\n\nTherefore, the integers $a=K+L$ and $b=K-L$ satisfy the equation $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$, and so for all integers $K$ and $L$, there is at least one pair of integers $(a, b)$ that satisfy the equation.\n\nHow could we come up with this? One way to do this would be trying some small values\n\n\n\nof $K$ and $L$, calculating $K^{2}+3 L^{2}$ and using this to make a guess, which can then be proven algebraically as above. In particular, here are some values:\n\n| $K$ | $L$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $a$ | $b$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 1 | 1 | 4 | 2 | 0 |\n| 2 | 1 | 7 | 3 | 1 |\n| 3 | 1 | 12 | 4 | 2 |\n| 1 | 2 | 13 | 3 | -1 |\n| 2 | 2 | 16 | 4 | 0 |\n| 3 | 2 | 21 | 5 | 1 |\n\nThe columns for $a$ and $b$ might lead us to guess that $a=K+L$ and $b=K-L$, which we proved above does in fact work.'] [] Text-only Competition True Theorem proof Number Theory Math English 427 Prove that, for all integers $a$ and $b$, there is at least one pair of integers $(K, L)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$. ['Suppose that $a$ and $b$ are integers.\n\nIf $a$ is even, then $\\frac{a}{2}$ is an integer and so\n\n$$\n\\left(\\frac{a}{2}-b\\right)^{2}+3\\left(\\frac{a}{2}\\right)^{2}=\\frac{a^{2}}{4}-2 \\cdot \\frac{a}{2} \\cdot b+b^{2}+\\frac{3 a^{2}}{4}=a^{2}+b^{2}-a b\n$$\n\nThus, if $K=\\frac{a}{2}-b$ and $L=\\frac{a}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nIf $b$ is even, then $\\frac{b}{2}$ is an integer and so a similar algebraic argument shows that\n\n$$\n\\left(\\frac{b}{2}-a\\right)^{2}+3\\left(\\frac{b}{2}\\right)^{2}=a^{2}+b^{2}-a b\n$$\n\nand so if $K=\\frac{b}{2}-a$ and $L=\\frac{b}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nIf $a$ and $b$ are both odd, then $a+b$ and $a-b$ are both even, which means that $\\frac{a+b}{2}$ and $\\frac{a-b}{2}$ are both integers, and so\n\n$\\left(\\frac{a+b}{2}\\right)^{2}+3\\left(\\frac{a-b}{2}\\right)^{2}=\\frac{a^{2}+2 a b+b^{2}}{4}+\\frac{3 a^{2}-6 a b+3 b^{2}}{4}=\\frac{4 a^{2}+4 b^{2}-4 a b}{4}=a^{2}+b^{2}-a b$\n\nThus, if $K=\\frac{a+b}{2}$ and $L=\\frac{a-b}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nTherefore, in all cases, for all integers $a$ and $b$, there is at least one pair of integers $(K, L)$ with $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nNow, trying some small cases might help us make a guess of possible expressions for $K$ and $L$ in terms of $a$ and $b$ :\n\n| $a$ | $b$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $K$ | $L$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 1 | 1 | 1 | 1 | 0 |\n| 2 | 1 | 3 | 0 | 1 |\n| 3 | 1 | 7 | 2 | 1 |\n| 4 | 1 | 13 | 1 | 2 |\n| 1 | 2 | 3 | 0 | 1 |\n| 2 | 2 | 4 | 1 | 1 |\n| 3 | 2 | 7 | 2 | 1 |\n| 4 | 2 | 12 | 3 | 1 |\n| 5 | 3 | 19 | 4 | 1 |\n\n\n\nWhile there might not initially seem to be useful patterns here, re-arranging the rows and adding some duplicates might help show a pattern:\n\n| $a$ | $b$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $K$ | $L$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 2 | 1 | 3 | 0 | 1 |\n| 4 | 1 | 13 | 1 | 2 |\n| 2 | 2 | 4 | 1 | 1 |\n| 4 | 2 | 12 | 3 | 1 |\n| 1 | 2 | 3 | 0 | 1 |\n| 3 | 2 | 7 | 2 | 1 |\n| 4 | 2 | 12 | 3 | 1 |\n| 1 | 1 | 1 | 1 | 0 |\n| 3 | 1 | 7 | 2 | 1 |\n| 5 | 3 | 19 | 4 | 1 |'] [] Text-only Competition True Theorem proof Number Theory Math English 428 "For each positive integer $n \geq 1$, let $C_{n}$ be the set containing the $n$ smallest positive integers; that is, $C_{n}=\{1,2, \ldots, n-1, n\}$. For example, $C_{4}=\{1,2,3,4\}$. We call a set, $F$, of subsets of $C_{n}$ a Furoni family of $C_{n}$ if no element of $F$ is a subset of another element of $F$. Suppose that $n$ is a positive integer and that $F$ is a Furoni family of $C_{n}$. For each non-negative integer $k$, define $a_{k}$ to be the number of elements of $F$ that contain exactly $k$ integers. Prove that $$ \frac{a_{0}}{\left(\begin{array}{l} n \\ 0 \end{array}\right)}+\frac{a_{1}}{\left(\begin{array}{l} n \\ 1 \end{array}\right)}+\frac{a_{2}}{\left(\begin{array}{l} n \\ 2 \end{array}\right)}+\cdots+\frac{a_{n-1}}{\left(\begin{array}{c} n \\ n-1 \end{array}\right)}+\frac{a_{n}}{\left(\begin{array}{l} n \\ n \end{array}\right)} \leq 1 $$ (The sum on the left side includes $n+1$ terms.) (Note: If $n$ is a positive integer and $k$ is an integer with $0 \leq k \leq n$, then $\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}$ is the number of subsets of $C_{n}$ that contain exactly $k$ integers, where $0 !=1$ and, if $m$ is a positive integer, $m$ ! represents the product of the integers from 1 to $m$, inclusive.)" "['Suppose that $n$ is a positive integer and $F$ is a Furoni family of $C_{n}$ that contains $a_{k}$ elements that contain exactly $k$ integers each, for each integer $k$ from 0 to $n$, inclusive.\n\nConsider each element $E$ of $F$.\n\nEach $E$ is a subset of $C_{n}$. Suppose that a particular choice for $E$ contains exactly $k$ elements.\n\nWe use $E$ to generate $k !(n-k)$ ! permutations $\\sigma$ of the integers in $C_{n}=\\{1,2,3, \\ldots, n\\}$ by starting with a permutation $\\alpha$ of the elements of $E$ and appending a permutation $\\beta$ of the elements in $C_{n}$ not in $E$.\n\nSince there are $k$ elements in $E$, there are $k$ ! possible permutations $\\alpha$.\n\nSince there are $n-k$ elements in $C_{n}$ that are not in $E$, there are $(n-k)$ ! possible permutations $\\beta$.\n\nEach possible $\\alpha$ can have each possible $\\beta$ appended to it, so there are $k !(n-k)$ ! possible permutations $\\sigma=\\alpha \\mid \\beta$. (The notation "" $\\alpha \\mid \\beta$ "" means the permutation of $C_{n}$ formed by writing out the permutation $\\alpha$ (of the elements of $E$ ) followed by writing out the permutation $\\beta$ (of the elements of $C_{n}$ not in $E$ ).)\n\nEach of these $k !(n-k)$ ! permutations generated by $E$ is indeed different, since if two permutations $\\sigma=\\alpha \\mid \\beta$ and $\\sigma^{\\prime}=\\alpha^{\\prime} \\mid \\beta^{\\prime}$ are equal, then since $\\alpha$ and $\\alpha^{\\prime}$ are both permutations of the elements of $E$, then they have the same length and so $\\alpha\\left|\\beta=\\alpha^{\\prime}\\right| \\beta^{\\prime}$ means $\\alpha=\\alpha^{\\prime}$.\n\n\n\nThis then means that $\\beta=\\beta^{\\prime}$ and so the permutations started out the same.\n\nWe repeat this process for each of the elements $E$ of $F$.\n\nSince, for each $k$, there are $a_{k}$ subsets of size $k$ in $F$, then the total number of permutations that this generates is\n\n$$\na_{0} 0 !(n-0) !+a_{1} 1 !(n-1) !+\\cdots+a_{n-1}(n-1) !(n-(n-1)) 1 !+a_{n} n !(n-n) !\n$$\n\nIf each of these permutations is different, then this total is at most $n$ !, since this is the total number of permutations of the elements of $C_{n}$.\n\nIs it possible that two elements $E$ and $G$ of $F$ generate identical permutations of the elements of $C_{n}$ in this way?\n\nSuppose that two permutations $\\sigma=\\alpha \\mid \\beta$ (generated by $E$ ) and $\\sigma^{\\prime}=\\alpha^{\\prime} \\mid \\beta^{\\prime}$ (generated by $G)$ are identical.\n\nSuppose that $E$ contains $k$ elements and $G$ contains $k^{\\prime}$ elements.\n\nEither $k \\leq k^{\\prime}$ or $k^{\\prime} \\leq k$ (or both, if they are equal).\n\nWithout loss of generality, suppose that $k \\leq k^{\\prime}$.\n\nThen the length of $\\alpha$ (which is $k$ ) is less than or equal to the length of $\\alpha^{\\prime}$ (which is $k^{\\prime}$ ). But $\\alpha\\left|\\beta=\\alpha^{\\prime}\\right| \\beta^{\\prime}$, so this means that the first $k$ entries in $\\alpha^{\\prime}$ are equal to the first $k$ entries in $\\alpha$.\n\nBut the entries in $\\alpha$ are the elements of $E$ and the entries of $\\alpha^{\\prime}$ are the elements of $G$, so this means that $E$ is a subset of $G$, which cannot be the case. This is a contradiction. Therefore, each of the permutations generated by each of the subsets of $C_{n}$ contained in $F$ is unique.\n\nTherefore,\n\n$$\na_{0} 0 !(n-0) !+a_{1} 1 !(n-1) !+\\cdots+a_{n-1}(n-1) !(n-(n-1)) 1 !+a_{n} n !(n-n) ! \\leq n !\n$$\n\nDividing both sides by $n$ !, we obtain successively\n\n$$\n\\begin{array}{r}\na_{0} 0 !(n-0) !+a_{1} 1 !(n-1) !+\\cdots+a_{n-1}(n-1) !(n-(n-1)) 1 !+a_{n} n !(n-n) ! \\leq n ! \\\\\na_{0} \\frac{0 !(n-0) !}{n !}+a_{1} \\frac{1 !(n-1) !}{n !}+\\cdots+a_{n-1} \\frac{(n-1) !(n-(n-1)) 1 !}{n !}+a_{n} \\frac{n !(n-n) !}{n !} \\leq 1 \\\\\na_{0} \\frac{1}{\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)}+a_{1} \\frac{1}{\\left(\\begin{array}{l}\nn \\\\\n1\n\\end{array}\\right)}+\\cdots+a_{n-1} \\frac{1}{\\left(\\begin{array}{c}\nn \\\\\nn-1\n\\end{array}\\right)}+a_{n} \\frac{1}{\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)} \\leq 1 \\\\\n\\frac{a_{0}}{\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)}+\\frac{a_{1}}{\\left(\\begin{array}{l}\nn \\\\\n1\n\\end{array}\\right)}+\\cdots+\\frac{a_{n-1}}{\\left(\\begin{array}{c}\nn \\\\\nn-1\n\\end{array}\\right)}+\\frac{a_{n}}{\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)} \\leq 1\n\\end{array}\n$$\n\nas required.' 'Suppose that $n$ is a positive integer and that $F$ is a randomly chosen Furoni family of $C_{n}$. Consider $L=\\{\\{\\},\\{1\\},\\{1,2\\},\\{1,2,3\\},\\{1,2,3, \\ldots, n\\}\\}$.\n\nThe probability that the intersection of $L$ and $F$ is non-empty is at most 1 .\n\nNote that since each element of $L$ is a subset of all of those to its right in the listing of $L$, then at most one of the elements of $L$ can be in $F$.\n\nIf $k$ is an integer with $k \\geq 0$, the probability that $\\{1,2,3, \\ldots, k\\}$ is an element of $F$ is $\\frac{a_{k}}{\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)}$, where $a_{k}$ is the number of elements in $F$ that contain exactly $k$ integers:\n\nThere are $\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)$ subsets of $C_{n}$ that contain exactly $k$ integer.\n\nThe probability that any particular one of these subsets is $\\{1,2,3, \\ldots, k\\}$ equals\n\n$\\frac{1}{\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)}$.\n\nSince $a_{k}$ of these subsets are in $F$, then the probability that one of these $a_{k}$ subsets is $\\{1,2,3, \\ldots, k\\}$ equals $\\frac{a_{k}}{\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)}$.\n\n(Note that we use the convention that if $k=0$, then $\\{1,2,3, \\ldots, k\\}=\\{\\}$.)\n\nThe probability that any of the elements of $L$ is in $F$ is the sum of the probability of each element being in $F$, since at most one of the elements in $L$ is in $F$.\n\nTherefore,\n\n$$\n\\frac{a_{0}}{\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)}+\\frac{a_{1}}{\\left(\\begin{array}{l}\nn \\\\\n1\n\\end{array}\\right)}+\\cdots+\\frac{a_{n-1}}{\\left(\\begin{array}{c}\nn \\\\\nn-1\n\\end{array}\\right)}+\\frac{a_{n}}{\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)} \\leq 1\n$$\n\nas required.']" [] Text-only Competition True Theorem proof Combinatorics Math English 429 "A Skolem sequence of order $n$ is a sequence $\left(s_{1}, s_{2}, \ldots, s_{2 n}\right)$ of $2 n$ integers satisfying the conditions: i) for every $k$ in $\{1,2,3, \ldots, n\}$, there exist exactly two elements $s_{i}$ and $s_{j}$ with $s_{i}=s_{j}=k$, and ii) if $s_{i}=s_{j}=k$ with $i\\frac{1}{2} \\quad(\\text { since } 8 b c>0)\n\\end{aligned}\n$$\n\nSince $2 a\\frac{1}{2}$ implies $A<60^{\\circ}$, as required.' 'We use the notation $A=\\angle B A C, B=\\angle A B C$ and $C=\\angle A C B$.\n\nWe need to show that $A<\\frac{1}{2}(B+C)$. Since the sum of the angles in $\\triangle A B C$ is $180^{\\circ}$, then $B+C=180^{\\circ}-A$, and so this inequality is equivalent to $A<\\frac{1}{2}\\left(180^{\\circ}-A\\right)$ which is equivalent to $\\frac{3}{2} A<90^{\\circ}$ or $A<60^{\\circ}$.\n\nSo we need to show that $A<60^{\\circ}$.\n\nWe know that $a<\\frac{1}{2}(b+c)$ which implies $2 a0 \\text { for } 0^{\\circ}0$. Next, we use the trigonometric formula $\\sin B+\\sin C=2 \\sin \\left(\\frac{B+C}{2}\\right) \\cos \\left(\\frac{B-C}{2}\\right)$.\n\nSince $\\cos \\theta \\leq 1$ for any $\\theta$, then $\\sin B+\\sin C \\leq 2 \\sin \\left(\\frac{B+C}{2}\\right) \\cdot 1=2 \\sin \\left(\\frac{B+C}{2}\\right)$.\n\n\n\nTherefore,\n\n$$\n\\begin{aligned}\n2 \\sin A & <\\sin B+\\sin C \\leq 2 \\sin \\left(\\frac{B+C}{2}\\right) \\\\\n2 \\sin A & <2 \\sin \\left(\\frac{B+C}{2}\\right) \\\\\n2 \\sin A & <2 \\sin \\left(\\frac{180^{\\circ}-A}{2}\\right) \\\\\n4 \\sin \\left(\\frac{1}{2} A\\right) \\cos \\left(\\frac{1}{2} A\\right) & <2 \\sin \\left(90^{\\circ}-\\frac{1}{2} A\\right) \\\\\n2 \\sin \\left(\\frac{1}{2} A\\right) \\cos \\left(\\frac{1}{2} A\\right) & <\\cos \\left(\\frac{1}{2} A\\right)\n\\end{aligned}\n$$\n\nSince $0^{\\circ}0$, so $\\sin \\left(\\frac{1}{2} A\\right)<\\frac{1}{2}$.\n\nSince $2 a1$. ( $t$ can be written in this form in at least one way, so we take one of these possibilities.)\n\nIn this case, consider $f(t+1)-f(t)$.\n\nWe can write this as\n\n$$\n\\begin{aligned}\nf(t+1)-f(t)= & \\left\\lfloor\\frac{t+1}{1^{2}+1}\\right\\rfloor+\\left\\lfloor\\frac{2(t+1)}{2^{2}+1}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{(t-1)(t+1)}{(t-1)^{2}+1}\\right\\rfloor+\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor \\\\\n& -\\left\\lfloor\\frac{t}{1^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{2 t}{2^{2}+1}\\right\\rfloor-\\cdots-\\left\\lfloor\\frac{(t-1) t}{(t-1)^{2}+1}\\right\\rfloor\n\\end{aligned}\n$$\n\nWe re-write this as\n\n$$\n\\begin{gathered}\nf(t+1)-f(t)=\\left(\\left\\lfloor\\frac{t+1}{1^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{t}{1^{2}+1}\\right\\rfloor\\right)+\\left(\\left\\lfloor\\frac{2(t+1)}{2^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{2 t}{2^{2}+1}\\right\\rfloor\\right)+\\cdots \\\\\n+\\left(\\left\\lfloor\\frac{(t-1)(t+1)}{(t-1)^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{(t-1) t}{(t-1)^{2}+1}\\right\\rfloor\\right)+\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor\n\\end{gathered}\n$$\n\nIn the $t-1$ sets of parentheses, we have terms of the form $\\left\\lfloor\\frac{k(t+1)}{k^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{k t}{k^{2}+1}\\right\\rfloor$ for each integer $k$ from 1 to $t-1$.\n\nWe know that $\\frac{k(t+1)}{k^{2}+1}>\\frac{k t}{k^{2}+1}$ because both $k$ and $t$ are positive, the denominators are\n\n\n\nequal and $k(t+1)>k t$.\n\nThus, $\\left.\\left\\lfloor\\frac{k(t+1)}{k^{2}+1}\\right\\rfloor \\geq \\frac{k t}{k^{2}+1}\\right\\rfloor$. (The greatest integer less than or equal to the first fraction must be at least as large as the greatest integer less than or equal to the second fraction.) This means that the $t-1$ differences in parentheses, each of which is an integer, is at least 0 .\n\nTo show that $f(t+1)-f(t) \\neq 2$, we show that there are at least 2 places where the difference is at least 1, and that the final term is at least 1 . This will tell us that $f(t+1)-f(t) \\geq 3$ and so $f(t+1)-f(t) \\neq 2$, which will tell us that $t$ cannot be composite, and so $t$ must be prime, as required.\n\nConsider $\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor$.\n\nSince $t(t+1)=t^{2}+t \\geq t^{2}+1$, then $\\frac{t(t+1)}{t^{2}+1} \\geq 1$, which means that $\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor \\geq 1$.\n\nConsider $\\left\\lfloor\\frac{t+1}{1^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{t}{1^{2}+1}\\right\\rfloor=\\left\\lfloor\\frac{t+1}{2}\\right\\rfloor-\\left\\lfloor\\frac{t}{2}\\right\\rfloor$.\n\nSince $t$ is odd, then we write $t=2 u+1$ for some positive integer $u$, which gives\n\n$$\n\\left\\lfloor\\frac{t+1}{2}\\right\\rfloor-\\left\\lfloor\\frac{t}{2}\\right\\rfloor=\\left\\lfloor\\frac{2 u+2}{2}\\right\\rfloor-\\left\\lfloor\\frac{2 u+1}{2}\\right\\rfloor=\\lfloor u+1\\rfloor-\\left\\lfloor u+\\frac{1}{2}\\right\\rfloor=(u+1)-u=1\n$$\n\nRecall that $t=r s$ with $r \\geq s>1$.\n\nConsider the term $\\left\\lfloor\\frac{r(t+1)}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r t}{r^{2}+1}\\right\\rfloor$.\n\nWe have\n\n$$\n\\left\\lfloor\\frac{r(t+1)}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r t}{r^{2}+1}\\right\\rfloor=\\left\\lfloor\\frac{r(r s+1)}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r \\cdot r s}{r^{2}+1}\\right\\rfloor=\\left\\lfloor\\frac{r^{2} s+r}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r^{2} s}{r^{2}+1}\\right\\rfloor\n$$\n\nWe note that $\\frac{r^{2} s+r}{r^{2}+1} \\geq \\frac{r^{2} s+s}{r^{2}+1}=s$ and $\\frac{r^{2} s}{r^{2}+1}<\\frac{r^{2} s+s}{r^{2}+1}=s$.\n\nThus, $\\left\\lfloor\\frac{r^{2} s+r}{r^{2}+1}\\right\\rfloor \\geq s$.\n\nAlso, $\\left\\lfloor\\frac{r^{2} s}{r^{2}+1}\\right\\rfloor\\frac{k t}{k^{2}+1}$.\n\n\n\nThis pair of inequalities is equivalent to the pair of inequalities $t+1 \\geq N \\cdot \\frac{k^{2}+1}{k}>t$ which is in turn equivalent to $t+1 \\geq N k+\\frac{N}{k}>t$.\n\nThe following three pairs $(N, k)$ of integers satisfy this equation:\n\n- $k=1$ and $N=\\frac{t+1}{2}$ (noting that $t$ is odd), which give $N k+\\frac{N}{k}=t+1$;\n- $k=r$ and $N=s$, which give $N k+\\frac{N}{k}=r s+\\frac{s}{r}$ (noting that $\\left.\\frac{s}{r}<1\\right)$\n- $k=t$ and $N=1$, which give $N k+\\frac{N}{k}=t+\\frac{1}{t}$.\n\nThis shows that $f(t+1)-f(t) \\geq 3$ when $t$ is odd and composite, as required.'] [] Text-only Competition True Theorem proof Number Theory Math English 438 "Consider a set $S$ that contains $m \geq 4$ elements, each of which is a positive integer and no two of which are equal. We call $S$ boring if it contains four distinct integers $a, b, c, d$ such that $a+b=c+d$. We call $S$ exciting if it is not boring. For example, $\{2,4,6,8,10\}$ is boring since $4+8=2+10$. Also, $\{1,5,10,25,50\}$ is exciting. Prove that, if $S$ is an exciting set of $m \geq 4$ positive integers, then $S$ contains an integer greater than or equal to $\frac{m^{2}-m}{4}$." ['Suppose that $S$ is an exciting set that contains exactly $m$ elements.\n\nThere are $\\left(\\begin{array}{c}m \\\\ 2\\end{array}\\right)=\\frac{m(m-1)}{2}$ pairs of elements of $S$.\n\nSince $S$ is exciting, the sums of these pairs of elements are all distinct positive integers.\n\nThis means that the largest of these sums is greater than or equal to $\\frac{m(m-1)}{2}$.\n\nWhen two numbers add to $\\frac{m(m-1)}{2}$ or greater, then at least one of them must be at least as large as $\\frac{1}{2} \\cdot \\frac{m(m-1)}{2}=\\frac{m^{2}-m}{4}$.\n\nTherefore, there is an element of $S$ that is greater than or equal to $\\frac{m^{2}-m}{4}$.'] [] Text-only Competition True Theorem proof Number Theory Math English 439 Cyclic quadrilateral $A B C D$ has $A B=A D=1, C D=\cos \angle A B C$, and $\cos \angle B A D=-\frac{1}{3}$. Prove that $B C$ is a diameter of the circumscribed circle. "['Consider $\\triangle B A D$. Since we know the lengths of sides $B A$ and $A D$ and the cosine of the angle between them, we can calculate the length of $B D$ using the cosine law:\n\n$$\n\\begin{aligned}\nB D & =\\sqrt{B A^{2}+A D^{2}-2(B A)(A D) \\cos \\angle B A D} \\\\\n& =\\sqrt{2-2\\left(-\\frac{1}{3}\\right)} \\\\\n& =\\sqrt{\\frac{8}{3}}\n\\end{aligned}\n$$\n\n\n\nNext, let $x=\\cos \\angle A B C$. Note that $D C=x$.\n\n\n\nSince $A B C D$ is a cyclic quadrilateral, then $\\angle A D C=180^{\\circ}-\\angle A B C$, and so $\\cos \\angle A D C=-\\cos \\angle A B C=-x$.\n\nSimilarly, $\\cos \\angle B C D=-\\cos \\angle B A D=\\frac{1}{3}$ (since $A B C D$ is a cyclic quadrilateral).\n\nSo we can now use the cosine law simultaneously in $\\triangle A D C$ and $\\triangle A B C$ (since side $A C$ is common) in order to try to solve for $B C$ :\n\n$$\n\\begin{aligned}\n1^{2}+x^{2}-2(1)(x) \\cos \\angle A D C & =1^{2}+B C^{2}-2(1)(B C) \\cos \\angle A B C \\\\\n1^{2}+x^{2}-2(1)(x)(-x) & =1^{2}+B C^{2}-2(1)(B C)(x) \\\\\n0 & =B C^{2}-2(B C) x-3 x^{2} \\\\\n0 & =(B C-3 x)(B C+x)\n\\end{aligned}\n$$\n\nSince $x$ is already a side length, then $x$ must be positive (ie. $\\angle A B C$ is acute), so $B C=3 x$.\n\nSince $\\cos \\angle B C D=\\frac{1}{3}$ and sides $D C$ and $B C$ are in the ratio $1: 3$, then $\\triangle B C D$ must indeed be right-angled at $D$. (We could prove this by using the cosine law to calculate $B D^{2}=8 x^{2}$ and then noticing that $D C^{2}+B D^{2}=B C^{2}$.)\n\nSince $\\triangle B C D$ is right-angled at $D$, then $B C$ is a diameter of the circle.' 'Let $x=\\cos \\angle A B C=C D$, and let $B C=y$.\n\nSince the opposite angles in a cyclic quadrilateral are supplementary, their cosines are negatives of each other. Thus, $\\cos \\angle A D C=-x$ and $\\cos \\angle B C D=\\frac{1}{3}$.\n\n\n\nNext, we use the cosine law four times: twice to calculate $A C^{2}$ in the two triangles $A B C$ and $A D C$, and then twice to calculate $B D^{2}$ in the triangles $A D B$ and $C D B$ to obtain:\n\n$$\n\\begin{aligned}\n1^{2}+x^{2}-2(1)(x) \\cos \\angle A D C & =1^{2}+y^{2}-2(1)(y) \\cos \\angle A B C \\\\\n1+x^{2}-2 x(-x) & =1+y^{2}-2 y(x) \\\\\n0 & =y^{2}-2 x y-3 x^{2} \\\\\n0 & =(y-3 x)(y+x)\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\n1^{2}+1^{2}-2(1)(1) \\cos \\angle B A D & =x^{2}+y^{2}-2 x y \\cos \\angle B C D \\\\\n2-2\\left(-\\frac{1}{3}\\right) & =x^{2}+y^{2}-2 x y\\left(\\frac{1}{3}\\right) \\\\\n\\frac{8}{3} & =x^{2}+y^{2}-\\frac{2}{3} x y\n\\end{aligned}\n$$\n\nFrom the first equation, since $x$ is already a side length and so is positive, we must have that $y=3 x$.\n\nSubstituting into the second equation, we obtain\n\n\n\n$$\n\\begin{aligned}\n& \\frac{8}{3}=x^{2}+(3 x)^{2}-\\frac{2}{3} x(3 x) \\\\\n& \\frac{8}{3}=8 x^{2} \\\\\n& x=\\frac{1}{\\sqrt{3}}\n\\end{aligned}\n$$\n\nsince $x$ must be positive. Thus, since $y=3 x$, then $y=\\sqrt{3}$.\n\nLooking then at $\\triangle B D C$, we have side lengths $B C=\\sqrt{3}, C D=\\frac{1}{\\sqrt{3}}$ and $B D=\\sqrt{\\frac{8}{3}}$. (The last is from the left side of the second cosine law equation.) Thus, $B C^{2}=C D^{2}+B D^{2}$, and so $\\triangle B D C$ is right-angled at $D$, whence $B C$ is a diameter of the circle.']" [] Text-only Competition True Theorem proof Geometry Math English 440 "A positive integer $n$ is called ""savage"" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that i) the sum of the elements in each of $A, B$, and $C$ is the same, ii) $A$ contains only odd numbers, iii) $B$ contains only even numbers, and iv) C contains every multiple of 3 (and possibly other numbers). Show that 8 is a savage integer." ['To show that 8 is a savage integer, we must partition the set $\\{1,2,3,4,5,6,7,8\\}$ according to the given criteria.\n\nSince the sum of the integers from 1 to 8 is 36 , then the sum of the elements in each of the sets $A, B$, and $C$ must be 12 .\n\n$C$ must contain both 3 and 6 .\n\n$A$ can contain only the numbers $1,5,7$, and may not contain all of these.\n\n$B$ can contain only the numbers $2,4,8$, and may not contain all of these.\n\nSo if we let $C=\\{1,2,3,6\\}, A=\\{5,7\\}$ and $B=\\{4,8\\}$, then these sets have the desired properties.\n\nTherefore, 8 is a savage integer.'] [] Text-only Competition True Theorem proof Algebra Math English 441 "A positive integer $n$ is called ""savage"" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that i) the sum of the elements in each of $A, B$, and $C$ is the same, ii) $A$ contains only odd numbers, iii) $B$ contains only even numbers, and iv) C contains every multiple of 3 (and possibly other numbers). Prove that if $n$ is an even savage integer, then $\frac{n+4}{12}$ is an integer." "[""We use the strategy of putting all of the multiples of 3 between 1 and $n$ in the set $C$, all of the remaining even numbers in the set $B$, and all of the remaining numbers in the set $A$. The sums of these sets will not likely all be equal, but we then try to adjust the sums to by moving elements out of $A$ and $B$ into $C$ to try to make these sums equal. (Notice that we can't move elements either into $A$ or $B$, or out of $C$.) We will use the notation $|C|$ to denote the sum of the elements of $C$.\n\nSince we are considering the case of $n$ even and we want to examine multiples of 3 less than or equal to $n$, it makes sense to consider $n$ as having one of the three forms $6 k$, $6 k+2$ or $6 k+4$. (These forms allow us to quickly tell what the greatest multiple of 3 less than $n$ is.)\n\nCase 1: $n=6 k$\n\nIn this case, $C$ contains at least the integers $3,6,9, \\ldots, 6 k$, and so the sum of $C$ is greater than one-third of the sum of the integers from 1 to $n$, since if we divide the integers from 1 to $n=6 k$ into groups of 3 consecutive integers starting with 1,2, 3 , then the set $C$ will always contain the largest of the 3 .\n\n\n\nCase 2: $n=6 k+4$\n\nHere, the sum of the integers from 1 to $n=6 k+4$ is $\\frac{1}{2}(6 k+4)(6 k+5)=18 k^{2}+27 k+10=3\\left(6 k^{2}+9 k+3\\right)+1$, which is never divisible by 3 . Therefore, $n$ cannot be savage in this case because the integers from 1 to $n$ cannot be partitioned into 3 sets with equal sums.\n\nCase 3: $n=6 k+2$\n\nHere, the sum of the integers from 1 to $n=6 k+2$ is\n\n$\\frac{1}{2}(6 k+2)(6 k+3)=18 k^{2}+15 k+3$, so the sum of the elements of each of the sets $A, B$ and $C$ should be $6 k^{2}+5 k+1$, so that the sums are equal.\n\nIn this case $C$, contains at least the integers $3,6,9, \\ldots, 6 k$, and so $|C| \\geq 3+6+9+\\cdots 6 k=3(1+2+3+\\cdots+2 k)=3\\left(\\frac{1}{2}(2 k)(2 k+1)\\right)=6 k^{2}+3 k$\n\nThe set $A$ contains at most the integers $1,3,5,7, \\ldots, 6 k+1$, but does not contain the odd multiples of 3 less than $n$, ie. the integers $3,9,15, \\ldots, 6 k-3$. Therefore, $|A| \\leq(1+3+5+\\cdots+6 k+1)-(3+9+\\cdots+6 k-3)$\n\n$=\\frac{1}{2}(3 k+1)[1+6 k+1]-\\frac{1}{2}(k)[3+6 k-3]$\n\n$=(3 k+1)(3 k+1)-k(3 k)$\n\n$=6 k^{2}+6 k+1$\n\n(To compute the sum of each of these arithmetic sequences, we use the fact that the sum of an arithmetic sequence is equal to half of the number of terms times the sum of the first and last terms.)\n\nThe set $B$ contains at most the integers $2,4,6,8, \\ldots, 6 k+2$, but does not contain the even multiples of 3 less than $n$, ie. the integers $6,12, \\ldots, 6 k$. Therefore, $|B| \\leq(2+4+6+\\cdots+6 k+2)-(6+12+\\cdots+6 k)$\n\n$=\\frac{1}{2}(3 k+1)[2+6 k+2]-\\frac{1}{2}(k)[6+6 k]$\n\n$=(3 k+1)(3 k+2)-k(3 k+3)$\n\n$=6 k^{2}+6 k+2$\n\nThus, the set $C$ is $2 k+1$ short of the desired sum, while the set $A$ has a sum that is $k$ too big and the set $B$ has a sum that is $k+1$ too big.\n\nSo in order to correct this, we would like to move elements from $A$ adding to $k$, and elements from $B$ which add to $k+1$ all to set $C$.\n\n\n\nSince we are assuming that $n$ is savage, then this is possible, which means that $k+1$ must be even since every element in $B$ is even, so the sum of any number of elements of $B$ is even.\n\nTherefore, $k$ is odd, and so $k=2 l+1$ for some integer $l$, and so\n\n$n=6(2 l+1)+2=12 l+8$, ie. $\\frac{n+4}{12}$ is an integer.\n\nHaving examined all cases, we see that if $n$ is an even savage integer, then $\\frac{n+4}{12}$ is an integer.""]" [] Text-only Competition True Theorem proof Algebra Math English 442 Prove that it is not possible to create a sequence of 4 numbers $a, b, c, d$, such that the sum of any two consecutive terms is positive, and the sum of any three consecutive terms is negative. "['Assume such a sequence $a, b, c, d$ exists. (We proceed by contradiction.)\n\nSince the sum of any two consecutive terms is positive, $a+b>0, b+c>0$, and\n\n$c+d>0$. Adding these three inequalities, $(a+b)+(b+c)+(c+d)>0$ or\n\n$a+2 b+2 c+d>0$.\n\nWe are going to show that this statement contradicts the facts that are known about the sequence. We are told that the sum of any three consecutive terms is negative, ie.\n\n$a+b+c<0$ and $b+c+d<0$. Adding these two inequalities, $(a+b+c)+(b+c+d)<0$ or $a+2 b+2 c+d<0$.\n\nThis is a contradiction, since the two conditions $a+2 b+2 c+d>0$ and $a+2 b+2 c+d<0$ cannot occur simultaneously.\n\nTherefore, our original assumption is false, and so no such sequence exists.' 'Assume such a sequence $a, b, c, d$ exists. (We proceed by contradiction.)\n\nWe consider two cases.\n\nCase 1: $a \\leq 0$\n\nIn this case, $b>0$ since $a+b>0$\n\nThen, since $a+b+c<0$, we must have that $c<0$.\n\nBut $c+d>0$, so $d>0$.\n\nThis means that we have $b>0$ and $c+d>0$, ie. $b+c+d>0$.\n\nBut from the conditions on the sequence, $b+c+d<0$, a contradiction.\n\nTherefore, no such sequence exists with $a \\leq 0$.\n\nCase 2: $a>0$\n\nIn this case, it is not immediately clear whether $b$ has to be positive or negative.\n\nHowever, we do know that $a+b>0$ and $a+b+c<0$, so it must be true that $c<0$.\n\nThen since $b+c>0$ and $c+d>0$, we must have both $b>0$ and $d>0$. But then $b+c+d=b+(c+d)>0$ since $c+d>0$ and $b>0$.\n\nThis is again a contradiction.\n\nTherefore, no such sequence exists with $a>0$.']" [] Text-only Competition True Theorem proof Algebra Math English 443 "A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer. Show that every Pythagorean triangle is Heronian." ['Consider a Pythagorean triangle with integer side lengths $a, b, c$ satisfying $a^{2}+b^{2}=c^{2}$. To show that this triangle is Heronian, we must show that it has an integer area. Now we know that the area is equal to $\\frac{1}{2} a b$, so we must show that either $a$ or $b$ is an even integer.\n\n\n\nSuppose that both $a$ and $b$ are odd. (We proceed by contradiction.)\n\nIn this case, let $a=2 k+1$ and $b=2 l+1$. Then both $a^{2}$ and $b^{2}$ are odd, and so $c^{2}$ is even since $a^{2}+b^{2}=c^{2}$. Therefore, $c$ itself must be even, so let $c=2 m$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n(2 k+1)^{2}+(2 l+1)^{2} & =(2 m)^{2} \\\\\n4 k^{2}+4 k+1+4 l^{2}+4 l+1 & =4 m^{2} \\\\\n4\\left(k^{2}+k+l^{2}+l\\right)+2 & =4\\left(m^{2}\\right)\n\\end{aligned}\n$$\n\nBut the right side is a multiple of 4 , and the left side is not a multiple of 4 . This is a contradiction.\n\nTherefore, one of $a$ or $b$ must be even, and so the area of the triangle is an integer. Thus, any Pythagorean triangle is Heronian.'] [] Text-only Competition True Theorem proof Geometry Math English 444 "A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer. Show that every odd integer greater than 1 is a side length of some Pythagorean triangle." ['We examine the first few smallest Pythagorean triples:\n\n$$\n345 \\quad\\left(3^{2}=4+5\\right)\n$$\n\n\n\n$$\n\\begin{array}{llll}\n5 & 12 & 13 & \\left(5^{2}=12+13\\right) \\\\\n6 & 8 & 10 & \\text { (Does not fit pattern) } \\\\\n7 & 24 & 25 & \\left(7^{2}=24+25\\right)\n\\end{array}\n$$\n\nIt appears from the first few examples that perhaps we can form a Pythagorean triple by using any odd number greater than 1 as its shortest leg.\n\nNext, we notice from the pattern that the sum of the second leg and the hypotenuse is the square of the shortest leg, and that these two side lengths differ by 1 .\n\nWill this pattern always hold?\n\nLet $a=2 k+1$ with $k \\geq 1$. (This formula will generate all odd integers greater than or equal to 3.) Can we always find $b$ so that $c=b+1$ and $a^{2}+b^{2}=c^{2}$ ?\n\nConsider the equation\n\n$$\n\\begin{aligned}\n(2 k+1)^{2}+b^{2} & =(b+1)^{2} \\\\\n4 k^{2}+4 k+1+b^{2} & =b^{2}+2 b+1 \\\\\n4 k^{2}+4 k & =2 b \\\\\nb & =2 k^{2}+2 k\n\\end{aligned}\n$$\n\nSo we can always find a $b$ to make the equation true. Therefore, since $a$ can be any odd integer greater than or equal to 3 , then we can make any odd number the shortest leg of a Pythagorean triangle, namely the Pythagorean triangle $a=2 k+1, b=2 k^{2}+2 k$, $c=2 k^{2}+2 k+1$. (Check that $a^{2}+b^{2}=c^{2}$ does indeed hold here!)'] [] Text-only Competition True Theorem proof Geometry Math English 445 An auditorium has a rectangular array of chairs. There are exactly 14 boys seated in each row and exactly 10 girls seated in each column. If exactly 3 chairs are empty, prove that there are at least 567 chairs in the auditorium. ['Suppose that the auditorium with these properties has $r$ rows and $c$ columns of chairs.\n\nThen there are $r c$ chairs in total.\n\nEach chair is empty, is occupied by a boy, or is occupied by a girl.\n\nSince there are 14 boys in each row, then there are $14 r$ chairs occupied by boys.\n\nSince there are 10 girls in each column, then there are 10c chairs occupied by girls.\n\nSince there are exactly 3 empty chairs, then the total number of chairs can also be written as $14 r+10 c+3$.\n\nTherefore, $r c=14 r+10 c+3$.\n\nWe proceed to find all pairs of positive integers $r$ and $c$ that satisfy this equation. We note that since there are 14 boys in each row, then there must be at least 14 columns (that is, $c \\geq 14$ ) and since there are 10 girls in each column, then there must be at least 10 rows (that is, $r \\geq 10$ ).\n\nManipulating the equation,\n\n$$\n\\begin{aligned}\nr c & =14 r+10 c+3 \\\\\nr c-14 r & =10 c+3 \\\\\nr(c-14) & =10 c+3 \\\\\nr & =\\frac{10 c+3}{c-14} \\\\\nr & =\\frac{10 c-140+143}{c-14} \\\\\nr & =\\frac{10 c-140}{c-14}+\\frac{143}{c-14} \\\\\nr & =10+\\frac{143}{c-14}\n\\end{aligned}\n$$\n\n\n\nSince $r$ is an integer, then $10+\\frac{143}{c-14}$ is an integer, so $\\frac{143}{c-14}$ must be an integer.\n\nTherefore, $c-14$ is a divisor of 143 . Since $c \\geq 14$, then $c-14 \\geq 0$, so $c-14$ is a positive divisor of 143 .\n\nSince $143=11 \\times 13$, then its positive divisors are 1,11,13, 143 .\n\nWe make a table of the possible values of $c-14$ along with the resulting values of $c, r$ (calculated using $r=10+\\frac{143}{c-14}$ ) and $r c$ :\n\n| $c-14$ | $c$ | $r$ | $r c$ |\n| :---: | :---: | :---: | :---: |\n| 1 | 15 | 153 | 2295 |\n| 11 | 25 | 23 | 575 |\n| 13 | 27 | 21 | 567 |\n| 143 | 157 | 11 | 1727 |\n\nTherefore, the four possible values for $r c$ are 567,575,1727,2295. That is, the smallest possible number of chairs in the auditorium is 567 .'] [] Text-only Competition True Theorem proof Combinatorics Math English 446 "Suppose that $a, b, c$ are three consecutive terms in an arithmetic sequence. Prove that $a^{2}-b c, b^{2}-a c$, and $c^{2}-a b$ are also three consecutive terms in an arithmetic sequence. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms.)" "['Since $a, b$ and $c$ are consecutive terms in an arithmetic sequence, then $b=a+d$ and $c=a+2 d$ for some number $d$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n& a^{2}-b c=a^{2}-(a+d)(a+2 d)=a^{2}-a^{2}-3 a d-2 d^{2}=-3 a d-2 d^{2} \\\\\n& b^{2}-a c=(a+d)^{2}-a(a+2 d)=a^{2}+2 a d+d^{2}-a^{2}-2 a d=d^{2} \\\\\n& c^{2}-a b=(a+2 d)^{2}-a(a+d)=a^{2}+4 a d+4 d^{2}-a^{2}-a d=3 a d+4 d^{2}\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=d^{2}-\\left(-3 a d-2 d^{2}\\right)=3 d^{2}+3 a d\n$$\n\nand\n\n$$\n\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)=\\left(3 a d+4 d^{2}\\right)-d^{2}=3 d^{2}+3 a d\n$$\n\nTherefore, $\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)$, so the sequence $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ is arithmetic.' 'Since $a, b$ and $c$ are consecutive terms in an arithmetic sequence, then $a=b-d$ and $c=b+d$ for some number $d$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n& a^{2}-b c=(b-d)^{2}-b(b+d)=b^{2}-2 b d+d^{2}-b^{2}-b d=-3 b d+d^{2} \\\\\n& b^{2}-a c=b^{2}-(b-d)(b+d)=b^{2}-b^{2}+d^{2}=d^{2} \\\\\n& c^{2}-a b=(b+d)^{2}-(b-d) b=b^{2}+2 b d+d^{2}-b^{2}+b d=3 b d+d^{2}\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=d^{2}-\\left(-3 b d+d^{2}\\right)=3 b d\n$$\n\nand\n\n$$\n\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)=\\left(3 b d+d^{2}\\right)-d^{2}=3 b d\n$$\n\nTherefore, $\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)$, so the sequence $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ is arithmetic.' 'To show that $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ form an arithmetic sequence, we can show that $\\left(c^{2}-a b\\right)+\\left(a^{2}-b c\\right)=2\\left(b^{2}-a c\\right)$.\n\nSince $a, b$ and $c$ form an arithmetic sequence, then $a+c=2 b$.\n\nNow\n\n$$\n\\begin{aligned}\n\\left(c^{2}-a b\\right)+\\left(a^{2}-b c\\right) & =c^{2}+a^{2}-b(a+c) \\\\\n& =c^{2}+a^{2}+2 a c-b(a+c)-2 a c \\\\\n& =(c+a)^{2}-b(a+c)-2 a c \\\\\n& =(c+a)(a+c-b)-2 a c \\\\\n& =2 b(2 b-b)-2 a c \\\\\n& =2 b^{2}-2 a c \\\\\n& =2\\left(b^{2}-a c\\right)\n\\end{aligned}\n$$\n\nas required.']" [] Text-only Competition True Theorem proof Algebra Math English 447 "Consider the following statement: There is a triangle that is not equilateral whose side lengths form a geometric sequence, and the measures of whose angles form an arithmetic sequence. Show that this statement is true by finding such a triangle or prove that it is false by demonstrating that there cannot be such a triangle." ['We prove that there cannot be such a triangle.\n\nWe prove this by contradiction. That is, we suppose that there is such a triangle and prove that there is then a logical contradiction.\n\nSuppose that $\\triangle A B C$ is not equilateral, has side lengths that form a geometric sequence, and angles whose measures form an arithmetic sequence.\n\nSuppose that $\\triangle A B C$ has side lengths $B C=a, A C=a r$, and $A B=a r^{2}$, for some real numbers $a>0$ and $r>1$. (These lengths form a geometric sequence, and we can assume that this sequence is increasing, and that the sides are labelled in this particular order.) Since $B C\n\nWe could proceed using the cosine law:\n\n$$\n\\begin{aligned}\nA C^{2} & =B C^{2}+A B^{2}-2 \\cdot B C \\cdot A B \\cdot \\cos (\\angle A B C) \\\\\n(a r)^{2} & =a^{2}+\\left(a r^{2}\\right)^{2}-2 a\\left(a r^{2}\\right) \\cos \\left(60^{\\circ}\\right) \\\\\na^{2} r^{2} & =a^{2}+a^{2} r^{4}-2 a^{2} r^{2} \\cdot \\frac{1}{2} \\\\\na^{2} r^{2} & =a^{2}+a^{2} r^{4}-a^{2} r^{2} \\\\\n0 & =a^{2} r^{4}-2 a^{2} r^{2}+a^{2} \\\\\n0 & =a^{2}\\left(r^{4}-2 r^{2}+1\\right) \\\\\n0 & =a^{2}\\left(r^{2}-1\\right)^{2}\n\\end{aligned}\n$$\n\nThis tells us that $a=0$ (which is impossible) or $r^{2}=1$ (and thus $r= \\pm 1$, which is impossible).\n\nTherefore, we have reached a logical contradiction and so such a triangle cannot exist.\n\nAlternatively, we could proceed using the sine law, noting that\n\n$$\n\\begin{aligned}\n& \\angle B A C=\\theta=(\\theta+\\delta)-\\delta=60^{\\circ}-\\delta \\\\\n& \\angle A C B=\\theta+2 \\delta=(\\theta+\\delta)+\\delta=60^{\\circ}+\\delta\n\\end{aligned}\n$$\n\nBy the sine law,\n\n$$\n\\frac{B C}{\\sin (\\angle B A C)}=\\frac{A C}{\\sin (\\angle A B C)}=\\frac{A B}{\\sin (\\angle A C B)}\n$$\n\nfrom which we obtain\n\n$$\n\\frac{a}{\\sin \\left(60^{\\circ}-\\delta\\right)}=\\frac{a r}{\\sin \\left(60^{\\circ}\\right)}=\\frac{a r^{2}}{\\sin \\left(60^{\\circ}+\\delta\\right)}\n$$\n\n\n\nSince $a \\neq 0$, from the first two parts,\n\n$$\nr=\\frac{a r}{a}=\\frac{\\sin 60^{\\circ}}{\\sin \\left(60^{\\circ}-\\delta\\right)}\n$$\n\nSince $a r \\neq 0$, from the second two parts,\n\n$$\nr=\\frac{a r^{2}}{a r}=\\frac{\\sin \\left(60^{\\circ}+\\delta\\right)}{\\sin 60^{\\circ}}\n$$\n\nEquating expressions for $r$, we obtain successively\n\n$$\n\\begin{aligned}\n\\frac{\\sin 60^{\\circ}}{\\sin \\left(60^{\\circ}-\\delta\\right)} & =\\frac{\\sin \\left(60^{\\circ}+\\delta\\right)}{\\sin 60^{\\circ}} \\\\\n\\sin ^{2} 60^{\\circ} & =\\sin \\left(60^{\\circ}-\\delta\\right) \\sin \\left(60^{\\circ}+\\delta\\right) \\\\\n\\left(\\frac{\\sqrt{3}}{2}\\right)^{2} & =\\left(\\sin 60^{\\circ} \\cos \\delta-\\cos 60^{\\circ} \\sin \\delta\\right)\\left(\\sin 60^{\\circ} \\cos \\delta+\\cos 60^{\\circ} \\sin \\delta\\right) \\\\\n\\frac{3}{4} & =\\left(\\frac{\\sqrt{3}}{2} \\cos \\delta-\\frac{1}{2} \\sin \\delta\\right)\\left(\\frac{\\sqrt{3}}{2} \\cos \\delta+\\frac{1}{2} \\sin \\delta\\right) \\\\\n\\frac{3}{4} & =\\frac{3}{4} \\cos ^{2} \\delta-\\frac{1}{4} \\sin ^{2} \\delta \\\\\n\\frac{3}{4} & =\\frac{3}{4} \\cos ^{2} \\delta+\\frac{3}{4} \\sin ^{2} \\delta-\\sin ^{2} \\delta \\\\\n\\frac{3}{4} & =\\frac{3}{4}\\left(\\cos ^{2} \\delta+\\sin ^{2} \\delta\\right)-\\sin ^{2} \\delta \\\\\n\\frac{3}{4} & =\\frac{3}{4}-\\sin ^{2} \\delta \\\\\n\\sin ^{2} \\delta & =0\n\\end{aligned}\n$$\n\nwhich means that $\\delta=0^{\\circ}$. (Any other angle $\\delta$ with $\\sin \\delta=0$ would not produce angles in a triangle.)\n\nTherefore, all three angles in the triangle are $60^{\\circ}$, which means that the triangle is equilateral, which it cannot be.\n\nTherefore, we have reached a logical contradiction and so such a triangle cannot exist.\n\n#'] [] Text-only Competition True Theorem proof Geometry Math English 448 "Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$. Determine, with justification, a polynomial $f(n)$ for which - $P(n) \geq \frac{f(n)}{n^{2}}$ for all positive integers $n$ with $n \geq 2$, and - $P(n)=\frac{f(n)}{n^{2}}$ for infinitely many positive integers $n$ with $n \geq 2$. (A polynomial $f(x)$ is an algebraic expression of the form $f(x)=a_{m} x^{m}+a_{m-1} x^{m-1}+\cdots+a_{1} x+a_{0}$ for some integer $m \geq 0$ and for some real numbers $a_{m}, a_{m-1}, \ldots, a_{1}, a_{0}$.)" ['Let $n$ be a positive integer with $n \\geq 2$.\n\nConsider $f(n)=2 n-1$. This is a polynomial in $n$.\n\nWe demonstrate that $P(n) \\geq \\frac{2 n-1}{n^{2}}$ for all positive integers $n$ with $n \\geq 2$ and that $P(n)=\\frac{2 n-1}{n^{2}}$ for infinitely many positive integers $n$ with $n \\geq 2$.\n\nSuppose that there are two bags, each containing $n$ balls labelled from 1 to $n$.\n\nSince there are $n$ balls in each bag, there are $n^{2}$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen.\n\nThe pairs\n\n$$\n(a, b)=(1, n),(2, n), \\ldots,(n-1, n),(n, n),(n, n-1), \\ldots,(n, 2),(n, 1)\n$$\n\neach have the property that $a b$ is divisible by $n$.\n\nThere are $(n-1)+1+(n-1)=2 n-1$ of these pairs.\n\nTherefore, at least $2 n-1$ of the pairs of balls that can be chosen have labels whose product is divisible by $n$.\n\nSince there are $n^{2}$ pairs that can be chosen and the number of pairs of balls with the desired property is at least $2 n-1$, then $P(n) \\geq \\frac{2 n-1}{n^{2}}$.\n\nThis proves the first part of what we needed to prove.\n\n\n\nNext, suppose that $n=p$, a prime number.\n\nFor $a b$ to be divisible by $p$, then either $a$ is divisible by $p$ or $b$ is divisible by $p$ (or both). (This property is not true when $p$ is not a prime number; for example, $2 \\cdot 2$ is divisible by 4 even though neither factor is divisible by 4.)\n\nSince $1 \\leq a \\leq p$ and $1 \\leq b \\leq p$, then if either $a$ is divisible by $p$ or $b$ is divisible by $p$ (or both), we must have either $a=p$ or $b=p$, or both.\n\nIn other words, $a b$ is divisible by $p$ exactly when $(a, b)$ is in the list\n\n$$\n(1, p),(2, p), \\ldots,(p-1, p),(p, p),(p, p-1), \\ldots,(p, 2),(p, 1)\n$$\n\nThere are $2 p-1$ pairs in this list and these are the only pairs for which $a b$ is divisible by $p$.\n\nTherefore, $P(n)=\\frac{2 n-1}{n^{2}}$ when $n$ is a prime number.\n\nSince there are infinitely many prime numbers, then $P(n)=\\frac{2 n-1}{n^{2}}$ for infinitely many positive integers $n$ with $n \\geq 2$.\n\nThus, $f(n)=2 n-1$ is a polynomial with the desired properties.'] [] Text-only Competition True Theorem proof Combinatorics Math English 449 "Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$. Prove there exists a positive integer $m$ for which $P(m)>\frac{2016}{m}$." ['Let $N=2^{k}$, where $k$ is a positive integer with $k \\geq 2$.\n\nSuppose that there are two bags, each containing $N$ balls labelled from 1 to $N$.\n\nSince there are $N$ balls in each bag, there are $N^{2}$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen.\n\nLet $j$ be a positive integer with $1 \\leq j \\leq k-1$.\n\nConsider pairs of the form $(a, b)=\\left(2^{j} x, 2^{k-j} y\\right)$ where $x$ and $y$ are odd positive integers that keep $a$ and $b$ in the desired range.\n\nNote that, in each case, $a b=\\left(2^{j} x\\right)\\left(2^{k-j} y\\right)=2^{k} x y$ which is divisible by $N=2^{k}$.\n\nSince $1 \\leq a \\leq 2^{k}$, then $1 \\leq 2^{j} x \\leq 2^{k}$ and so $x \\leq 2^{k-j}$.\n\nSince half of the positive integers from 1 to $2^{k-j}$ are odd, then there are $\\frac{1}{2} 2^{k-j}=2^{k-j-1}$ choices for $x$.\n\nSimilarly, there are $2^{k-(k-j)-1}=2^{j-1}$ choices for $y$.\n\nNote that each choice of $x$ and $y$ gives a unique pair $(a, b)$.\n\nFor any fixed $j$, there are $2^{k-j-1}$ choices for $x$ and $2^{j-1}$ choices for $y$.\n\nThus, there are $2^{k-j-1} \\cdot 2^{j-1}=2^{k-2}$ choices of this form for the pair $(a, b)$.\n\nSo for a fixed $j$ with $1 \\leq j \\leq k-1$, this method gives $2^{k-2}$ pairs $(a, b)$ for which $a b$ is divisible by $N$.\n\nSince there are $k-1$ different values for $j$, then there are at least $(k-1) 2^{k-2}$ pairs $(a, b)$ for which $a b$ is divisible by $N$. (Note that two pairs that come from different values of $j$ will in fact be different since the number of factors of 2 in their values of $a$ will be different.) Since there are $N^{2}$ choices for $(a, b)$, then\n\n$$\nP(N) \\geq \\frac{(k-1) 2^{k-2}}{N^{2}}=\\frac{(k-1) 2^{k} 2^{-2}}{N^{2}}=\\frac{k-1}{4} \\cdot \\frac{1}{N}\n$$\n\nWhen $\\frac{k-1}{4}>2016$, we will have $P(N)>2016 \\cdot \\frac{1}{N}$.\n\nThe inequality $\\frac{k-1}{4}>2016$ is equivalent to $k-1>8064$ or $k>8065$.\n\nWe want to show that there exists a positive integer $m$ with $P(m)>\\frac{2016}{m}$.\n\nSet $m=2^{8066}$.\n\nBy the work above, $P(m) \\geq \\frac{8065}{4} \\cdot \\frac{1}{m}>\\frac{2016}{m}$, as required.'] [] Text-only Competition True Theorem proof Combinatorics Math English 450 "For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$. For example, the value of $f(1,2)$ is 3 . If $a$ and $b$ are positive integers and $f(a, b)$ is an integer, prove that $f(a, b)$ must be a multiple of 3 ." ['Suppose that $a$ and $b$ are positive integers for which $f(a, b)$ is an integer.\n\nAssume that $k=f(a, b)$ is not a multiple of 3 .\n\nWe will show that there must be a contradiction, which will lead to the conclusion that $k$ must be a multiple of 3 .\n\nBy definition, $k=f(a, b)=\\frac{a}{b}+\\frac{b}{a}+\\frac{1}{a b}$.\n\nMultiplying by $a b$, we obtain $k a b=a^{2}+b^{2}+1$, which we re-write as $a^{2}-(k b) a+\\left(b^{2}+1\\right)=0$. We treat this as a quadratic equation in $a$ with coefficients in terms of the variables $b$ and $k$. Solving for $a$ in terms of $b$ and $k$ using the quadratic formula, we obtain\n\n$$\na=\\frac{k b \\pm \\sqrt{(-k b)^{2}-4(1)\\left(b^{2}+1\\right)}}{2}=\\frac{k b \\pm \\sqrt{k^{2} b^{2}-4 b^{2}-4}}{2}\n$$\n\nSince $a$ is an integer, then the discriminant $k^{2} b^{2}-4 b^{2}-4$ must be a perfect square. Re-writing the discriminant, we obtain\n\n$$\nk^{2} b^{2}-4 b^{2}-4=b^{2}\\left(k^{2}-4\\right)-4=b^{2}(k-2)(k+2)-4\n$$\n\nSince $k$ is not a multiple of 3 , then it is either 1 more than a multiple of 3 or it is 2 more than a multiple of 3.\n\nIf $k$ is 1 more than a multiple of 3 , then $k+2$ is a multiple of 3 .\n\nIf $k$ is 2 more than a multiple of 3 , then $k-2$ is a multiple of 3 .\n\nIn either case, $(k-2)(k+2)$ is a multiple of 3 , say $(k-2)(k+2)=3 m$ for some integer $m$. This means that the discriminant can be re-written again as\n\n$$\nb^{2}(3 m)-4=3\\left(b^{2} m-2\\right)+2\n$$\n\nIn other words, the discriminant is itself 2 more than a multiple of 3.\n\nHowever, every perfect square is either a multiple of 3 or one more than a multiple of 3 :\n\nSuppose that $r$ is an integer and consider $r^{2}$.\n\nThe integer $r$ can be written as one of $3 q, 3 q+1,3 q+2$, for some integer $q$.\n\nThese three cases give\n\n$$\n\\begin{aligned}\n(3 q)^{2} & =9 q^{2}=3\\left(3 q^{2}\\right) \\\\\n(3 q+1)^{2} & =9 q^{2}+6 q+1=3\\left(3 q^{2}+2 q\\right)+1 \\\\\n(3 q+2)^{2} & =9 q^{2}+12 q+4=3\\left(3 q^{2}+4 q+1\\right)+1\n\\end{aligned}\n$$\n\nand so $r^{2}$ is either a multiple of 3 or 1 more than a multiple of 3.\n\n\n\nWe have determined that the discriminant is a perfect square and is 2 more than a multiple of 3. This is a contradiction.\n\nThis means that our initial assumption must be incorrect, and so $k=f(a, b)$ must be a multiple of 3.'] [] Text-only Competition True Theorem proof Number Theory Math English 451 "Carlos has 14 coins, numbered 1 to 14. Each coin has exactly one face called ""heads"". When flipped, coins 1,2,3,..,13,14 land heads with probabilities $h_{1}, h_{2}, h_{3}, \ldots, h_{13}, h_{14}$, respectively. When Carlos flips each of the 14 coins exactly once, the probability that an even number of coins land heads is exactly $\frac{1}{2}$. Must there be a $k$ between 1 and 14, inclusive, for which $h_{k}=\frac{1}{2}$ ? Prove your answer." ['Suppose that, after flipping the first 13 coins, the probability that there is an even number of heads is $p$.\n\nThen the probability that there is an odd number of heads is $1-p$.\n\nWhen the 14th coin is flipped, the probability of heads is $h_{14}$ and the probability of not heads is $1-h_{14}$.\n\nAfter the 14th coin is flipped, there can be an even number of heads if the first 13 include an even number of heads and the 14th is not heads, or if the first 13 include an odd number of heads and the 14th is heads.\n\nThe probability of this is $p\\left(1-h_{14}\\right)+(1-p) h_{14}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\np\\left(1-h_{14}\\right)+(1-p) h_{14} & =\\frac{1}{2} \\\\\n2 p-2 p h_{14}+2 h_{14}-2 p h_{14} & =1 \\\\\n0 & =4 p h_{14}-2 p-2 h_{14}+1 \\\\\n0 & =2 p\\left(2 h_{14}-1\\right)-(2 h 14-1) \\\\\n0 & =(2 p-1)\\left(2 h_{14}-1\\right)\n\\end{aligned}\n$$\n\nTherefore, either $h_{14}=\\frac{1}{2}$ or $p=\\frac{1}{2}$.\n\nIf $h_{14}=\\frac{1}{2}$, we have proven the result.\n\nIf $h_{14} \\neq \\frac{1}{2}$, then $p=\\frac{1}{2}$.\n\nThis would mean that the probability of getting an even number of heads when the first 13 coins are flipped is $\\frac{1}{2}$.\n\nWe could repeat the argument above to conclude that either $h_{13}=\\frac{1}{2}$ or the probability of obtaining an even number of heads when the first 12 coins are flipped is $\\frac{1}{2}$.\n\nContinuing in this way, either one of $h_{14}, h_{13}, \\ldots, h_{3}, h_{2}$ will equal $\\frac{1}{2}$, or the probability of obtaining an even number of heads when 1 coin is flipped is $\\frac{1}{2}$.\n\nThis last statement is equivalent to saying that the probability of obtaining a head with the first coin is $\\frac{1}{2}$ (that is, $h_{1}=\\frac{1}{2}$ ).\n\nTherefore, at least one of $h_{1}, h_{2}, \\ldots, h_{13}, h_{14}$ must equal $\\frac{1}{2}$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 452 Serge and Lis each have a machine that prints a digit from 1 to 6. Serge's machine prints the digits $1,2,3,4,5,6$ with probability $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}$, respectively. Lis's machine prints the digits $1,2,3,4,5,6$ with probability $q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}$, respectively. Each of the machines prints one digit. Let $S(i)$ be the probability that the sum of the two digits printed is $i$. If $S(2)=S(12)=\frac{1}{2} S(7)$ and $S(7)>0$, prove that $p_{1}=p_{6}$ and $q_{1}=q_{6}$. ['For the sum of the two digits printed to be 2, each digit must equal 1.\n\nThus, $S(2)=p_{1} q_{1}$.\n\nFor the sum of the two digits printed to be 12, each digit must equal 6 .\n\nThus, $S(12)=p_{6} q_{6}$.\n\nFor the sum of the two digits printed to be 7 , the digits must be 1 and 6 , or 2 and 5 , or 3 and 4 , or 4 and 3 , or 5 and 2 , or 6 and 1 .\n\nThus, $S(7)=p_{1} q_{6}+p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}+p_{6} q_{1}$.\n\nSince $S(2)=S(12)$, then $p_{1} q_{1}=p_{6} q_{6}$.\n\nSince $S(2)>0$ and $S(12)>0$, then $p_{1}, q_{1}, p_{6}, q_{6}>0$.\n\nIf $p_{1}=p_{6}$, then we can divide both sides of $p_{1} q_{1}=p_{6} q_{6}$ by $p_{1}=p_{6}$ to obtain $q_{1}=q_{6}$.\n\nIf $q_{1}=q_{6}$, then we can divide both sides of $p_{1} q_{1}=p_{6} q_{6}$ by $q_{1}=q_{6}$ to obtain $p_{1}=p_{6}$.\n\nTherefore, if we can show that either $p_{1}=p_{6}$ or $q_{1}=q_{6}$, our result will be true.\n\nSuppose that $p_{1} \\neq p_{6}$ and $q_{1} \\neq q_{6}$.\n\nSince $S(2)=\\frac{1}{2} S(7)$ and $S(12)=\\frac{1}{2} S(7)$, then\n\n$$\n\\begin{aligned}\nS(7)-\\frac{1}{2} S(7)-\\frac{1}{2} S(7) & =0 \\\\\nS(7)-S(2)-S(12) & =0 \\\\\np_{1} q_{6}+p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}+p_{6} q_{1}-p_{1} q_{1}-p_{6} q_{6} & =0 \\\\\np_{1} q_{6}+p_{6} q_{1}-p_{1} q_{1}-p_{6} q_{6}+\\left(p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}\\right) & =0 \\\\\n\\left(p_{1}-p_{6}\\right)\\left(q_{6}-q_{1}\\right)+\\left(p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}\\right) & =0 \\\\\np_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2} & =-\\left(p_{1}-p_{6}\\right)\\left(q_{6}-q_{1}\\right) \\\\\np_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2} & =\\left(p_{1}-p_{6}\\right)\\left(q_{1}-q_{6}\\right)\n\\end{aligned}\n$$\n\nSince $p_{2}, p_{3}, p_{4}, p_{5}, q_{2}, q_{3}, q_{4}, q_{5} \\geq 0$, then $p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2} \\geq 0$.\n\nFrom this, $\\left(p_{1}-p_{6}\\right)\\left(q_{1}-q_{6}\\right) \\geq 0$.\n\nSince $p_{1} \\neq p_{6}$, then either $p_{1}>p_{6}$ or $p_{1}p_{6}$, then $p_{1}-p_{6}>0$ and so $\\left(p_{1}-p_{6}\\right)\\left(q_{1}-q_{6}\\right) \\geq 0$ tells us that $q_{1}-q_{6}>0$ which means $q_{1}>q_{6}$.\n\nBut we know that $p_{1} q_{1}=p_{6} q_{6}$ and $p_{1}, q_{1}, p_{6}, q_{6}>0$ so we cannot have $p_{1}>p_{6}$ and $q_{1}>q_{6}$. If $p_{1}0$ so we cannot have $p_{1}p_{6}$ or $p_{1}3 m$, which is a contradiction.\n\nTherefore our assumption that not all of the entries are equal must be false, which means that all of the entries are equal, which proves statement $(\\mathrm{P})$.\n\nProof of statement (Q)\n\nSuppose that all of the entries are positive real numbers.\n\nAssume that not all of the entries in the array are equal.\n\nAs in (b), define $S_{k}=A_{k}+B_{k}$ for each integer $k$.\n\nAlso, define $D_{k}=A_{k}-B_{k}$ for each integer $k$.\n\nStep 1: Prove that the numbers $S_{k}$ form an arithmetic sequence\n\nFrom (b), $S_{k+1}=2 S_{k}-S_{k-1}$.\n\nRe-arranging, we see $S_{k+1}-S_{k}=S_{k}-S_{k-1}$ for each integer $k$, which means that the differences between consecutive pairs of terms are equal.\n\nSince this is true for all integers $k$, then the difference between each pair of consecutive\n\n\n\nterms through the whole sequence is constant, which means that the sequence is an arithmetic sequence.\n\nStep 2: Prove that $S_{k}$ is constant\n\nSuppose that $S_{0}=c$. Since $A_{0}>0$ and $B_{0}>0$, then $S_{0}=c>0$.\n\nSince the terms $S_{k}$ form an arithmetic sequence, then the sequence is either constant, increasing or decreasing.\n\nIf the sequence of terms $S_{k}$ is increasing, then the common difference $d=S_{1}-S_{0}$ is positive.\n\nNote that $S_{-1}=c-d, S_{-2}=c-2 d$, and so on.\n\nSince $c$ and $d$ are constant, then if we move far enough back along the sequence, eventually $S_{t}$ will be negative for some integer $t$. This is a contradiction since $A_{t}>0$ and $B_{t}>0$ and $S_{t}=A_{t}+B_{t}$.\n\nThus, the sequence cannot be increasing.\n\nIf the sequence of terms $S_{k}$ is decreasing, then the common difference $d=S_{1}-S_{0}$ is negative.\n\nNote that $S_{1}=c+d, S_{2}=c+2 d$, and so on.\n\nSince $c$ and $d$ are constant, then if we move far along the sequence, eventually $S_{t}$ will be negative for some integer $t$. This is also a contradiction since $A_{t}>0$ and $B_{t}>0$ and $S_{t}=A_{t}+B_{t}$.\n\nThus, the sequence cannot be decreasing.\n\nTherefore, since all of the entries are positive and the sequence $S_{k}$ is arithmetic, then $S_{k}$ is constant, say $S_{k}=c>0$ for all integers $k$.\n\nStep 3: Determine range of possible values for $D_{k}$\n\nWe note that $S_{k}=A_{k}+B_{k}=c$ for all integers $k$ and $A_{k}>0$ and $B_{k}>0$.\n\nSince $A_{k}>0$, then $B_{k}=S_{k}-A_{k}=c-A_{k}0-c=-c$.\n\nIn other words, $-c0$ or $D_{0}<0$.\n\nWe may assume that $D_{0}>0$. (If $D_{0}<0$, then we can switch the bottom and top rows of the array so that $D_{0}$ becomes positive.)\n\nSuppose that $D_{1} \\geq D_{0}>0$.\n\nThen $D_{2}=4 D_{1}-D_{0} \\geq 4 D_{1}-D_{1}=3 D_{1}$. Since $D_{1}>0$, this also means that $D_{2}>D_{1}>0$.\n\nSimilarly, $D_{3}=4 D_{2}-D_{1} \\geq 4 D_{2}-D_{2}=3 D_{2}>D_{2}>0$. Since $D_{2} \\geq 3 D_{1}$, then $D_{3} \\geq 9 D_{1}$. Continuing in this way, we see that $D_{4} \\geq 27 D_{1}$ and $D_{5} \\geq 81 D_{1}$ and so on, with $D_{k} \\geq 3^{k-1} D_{1}$ for each positive integer $k \\geq 2$. Since the value of $D_{1}$ is a fixed positive real number and $D_{k}4 D_{0}-D_{0}=3 D_{0}>D_{0}>0$.\n\nExtending this using a similar method, we see that $D_{-j}>3^{j} D_{0}$ for all positive integers $j$ which will lead to the same contradiction as above.\n\nTherefore, a contradiction is obtained in all cases and so it cannot be the case that $D_{k} \\neq 0$ for some integer $k$.\n\nSince $D_{k}=0$ and $S_{k}=c$ for all integers $k$, then $A_{k}=B_{k}=\\frac{1}{2} c$ for all integers $k$, which means that all entries in the array are equal.'] [] Text-only Competition True Theorem proof Algebra Math English 456 The equation $y=x^{2}+2 a x+a$ represents a parabola for all real values of $a$. Prove that each of these parabolas pass through a common point and determine the coordinates of this point. "['Since $y=x^{2}+2 a x+a$ for all $a, a \\in R$, it must be true for $a=0$ and $a=1$.\n\nFor $a=0, y=x^{2}$; for $a=1, y=x^{2}+2 x+1$.\n\nBy comparison, (or substitution)\n\n$$\n\\begin{aligned}\n& x^{2}=x^{2}+2 x+1 \\\\\n& \\therefore x=\\frac{-1}{2} \\\\\n& \\Rightarrow y=\\frac{1}{4}\n\\end{aligned}\n$$\n\nWe must verify that $x=\\frac{-1}{2}, y=\\frac{1}{4}$ satisfies the original.\n\nVerification: $y=x^{2}+2 a x+a=\\left(\\frac{-1}{2}\\right)^{2}+2 a\\left(\\frac{-1}{2}\\right)+a=\\frac{1}{4}-a+a=\\frac{1}{4}$\n\n$$\n\\therefore\\left(\\frac{-1}{2}, \\frac{1}{4}\\right) \\text { is a point on } y=x^{2}+2 a x+a, a \\in R \\text {. }\n$$' 'If $y=x^{2}+2 a x+a$ represents a parabola for all real values of $a$ then it is true for all $a$ and $b$ where $a \\neq b$.\n\nSo, $y=x^{2}+2 a x+a$ and $y=x^{2}+2 b x+b$ (by substitution of $a$ and $b$ into $y=x^{2}+2 a x+a$ ) Since we are looking for common point, $x^{2}+2 a x+a=x^{2}+2 b x+b$\n\n$$\n\\begin{aligned}\n& 2 a x-2 b x+a-b=0 \\\\\n& a(2 x+1)-b(2 x+1)=0 \\\\\n& (a-b)(2 x+1)=0\n\\end{aligned}\n$$\n\nSince $a \\neq b, 2 x+1=0 \\Rightarrow x=\\frac{-1}{2}$ and $y=\\frac{1}{4}$.']" [] Text-only Competition True Theorem proof Geometry Math English 457 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. Let $T$ be a full town and let $D \in \operatorname{dish}(T)$. Let $T^{\prime}$ be the town consisting of all residents of $T$ who do not know how to make $D$. Prove that $T^{\prime}$ is full." ['The town $T^{\\prime}$ consists of all residents of $T$ who do not know how to make $D$. Because $T$ is full, every pair of dishes $\\left\\{d_{i}, d_{j}\\right\\}$ in $\\operatorname{dish}(T)$ can be made by some resident $r_{i j}$ in $T$. If $d_{i} \\neq D$ and $d_{j} \\neq D$, then $r_{i j} \\in T^{\\prime}$. So every pair of dishes in $\\operatorname{dish}(T) \\backslash\\{D\\}$ can be made by some resident of $T^{\\prime}$. Hence $T^{\\prime}$ is full.'] [] Text-only Competition True Theorem proof Combinatorics Math English 458 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. Show that $\operatorname{gr}($ ARMLton $)=2$." ['Paul and Arnold cannot be in the same group, because they both make pie, and Arnold and Kelly cannot be in the same group, because they both make salad. Hence there must be at least two groups. But Paul and Kelly make none of the same dishes, so they can be in the same group. Thus a valid group assignment is\n\n$$\n\\begin{aligned}\n\\text { Paul } & \\mapsto 1 \\\\\n\\text { Kelly } & \\mapsto 1 \\\\\n\\text { Arnold } & \\mapsto 2\n\\end{aligned}\n$$\n\nHence $\\operatorname{gr}($ ARMLton $)=2$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 459 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. Show that gr(ARMLville $)=3$." ['Sally and Ross both make calzones, Ross and David both make pancakes, and Sally and David both make steak. So no two of these people can be in the same group, and $\\operatorname{gr}($ ARMLville $)=3$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 460 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. Show that $\operatorname{gr}\left(\mathcal{F}_{4}\right)=3$." ['Let the dishes be $d_{1}, d_{2}, d_{3}, d_{4}$ and let resident $r_{i j}$ make dishes $d_{i}$ and $d_{j}$, where $i\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town with at least two residents that has a single resident cycle that contains every resident. Prove that $T$ is homogeneous if and only if $\operatorname{pop}(T)$ is even." ['Note that for every $D \\in \\operatorname{dish}(T), \\operatorname{chef}_{T}(D) \\leq 2$, because otherwise, $r_{1}, r_{2}, \\ldots, r_{n}$ could not be a resident cycle. Without loss of generality, assume the cycle is $r_{1}, r_{2}, \\ldots, r_{n}$. If $n$ is even, assign resident $r_{i}$ to group 1 if $i$ is odd, and to group 2 if $i$ is even. This is a valid group assignment, because the only pairs of residents who cook the same dish are $\\left(r_{i}, r_{i+1}\\right)$ for $i=1,2, \\ldots, n-1$ and $\\left(r_{n}, r_{1}\\right)$. In each case, the residents are assigned to different groups. This proves $\\operatorname{gr}(T)=2$, so $T$ is homogeneous.\n\nOn the other hand, if $n$ is odd, suppose for the sake of contradiction that there are only two groups. Then either $r_{1}$ and $r_{n}$ are in the same group, or for some $i, r_{i}$ and $r_{i+1}$ are in the\n\n\n\nsame group. In either case, two residents in the same group share a dish, contradicting the requirement that no members of a group have a common dish. Hence $\\operatorname{gr}(T) \\geq 3$ when $n$ is odd, making $T$ heterogeneous.'] [] Text-only Competition True Theorem proof Combinatorics Math English 467 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=2$. Prove that there are finitely many resident cycles $C_{1}, C_{2}, \ldots, C_{j}$ in $T$ so that each resident belongs to exactly one of the $C_{i}$." ['First note that the condition $\\left|\\operatorname{chef}_{T}(D)\\right|=2$ for all $D$ implies that $\\operatorname{pop}(T)=|\\operatorname{dish}(T)|$, using the equation from problem 5 . So for the town in question, the population of the town equals the number of dishes in the town. Because no two chefs cook the same pair of dishes, it is impossible for such a town to have exactly two residents, and because each dish is cooked by exactly two chefs, it is impossible for such a town to have only one resident.\n\nThe claim is true for towns of three residents satisfying the conditions: such towns must have one resident who cooks dishes $d_{1}$ and $d_{2}$, one resident who cooks dishes $d_{2}$ and $d_{3}$, and one resident who cooks dishes $d_{3}$ and $d_{1}$, and those three residents form a cycle. So proceed by (modified) strong induction: assume that for some $n>3$ and for all positive integers $k$ such that $3 \\leq k0$, so by the inductive hypothesis, the residents of $T^{\\prime}$ can be divided into disjoint resident cycles.\n\nThus the statement is proved by strong induction.'] [] Text-only Competition True Theorem proof Combinatorics Math English 468 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=2$. Prove that if $\operatorname{pop}(T)$ is odd, then $T$ is heterogeneous." ['In order for $T$ to be homogeneous, it must be possible to partition the residents into exactly two dining groups. First apply 10a to divide the town into finitely many resident cycles $C_{i}$, and assume towards a contradiction that such a group assignment $f: T \\rightarrow$ $\\{1,2\\}$ exists. If $\\operatorname{pop}(T)$ is odd, then at least one of the cycles $C_{i}$ must contain an odd number of residents; without loss of generality, suppose this cycle to be $C_{1}$, with residents $r_{1}, r_{2}, \\ldots, r_{2 k+1}$. (By the restrictions noted in part a, $k \\geq 1$.) Now because $r_{i}$ and $r_{i+1}$ cook a dish in common, $f\\left(r_{i}\\right) \\neq f\\left(r_{i+1}\\right)$ for all $i$. Thus if $f\\left(r_{1}\\right)=1$, it follows that $f\\left(r_{2}\\right)=2$, and that $f\\left(r_{3}\\right)=1$, etc. So $f\\left(r_{i}\\right)=f\\left(r_{1}\\right)$ if $i$ is odd and $f\\left(r_{i}\\right)=f\\left(r_{2}\\right)$ if $i$ is\n\n\n\neven; in particular, $f\\left(r_{2 k+1}\\right)=f(1)$. But that equation would imply that $r_{1}$ and $r_{2 k+1}$ cook no dishes in common, which is impossible if they are the first and last residents in a resident cycle. So no such group assignment can exist, and $\\operatorname{gr}(T) \\geq 3$. Hence $T$ is heterogeneous.'] [] Text-only Competition True Theorem proof Combinatorics Math English 469 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=3$. Either find such a town $T$ for which $|\operatorname{dish}(T)|$ is odd, or show that no such town exists." "['First, we Prove that\n$$\n\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|=2 \\operatorname{pop}(T) .\n$$\n\nBecause each chef knows how to prepare exactly two dishes, and no two chefs know how to prepare the same two dishes, each chef is counted exactly twice in the sum $\\Sigma\\left|\\operatorname{chef}_{T}(D)\\right|$. More formally, consider the set of ""resident-dish pairs"":\n\n$$\nS=\\{(r, D) \\in T \\times \\operatorname{dish}(T) \\mid r \\text { makes } D\\}\n$$\n\nCount $|S|$ in two different ways. First, every dish $D$ is made by $\\left|\\operatorname{chef}_{T}(D)\\right|$ residents of $T$, so\n\n$$\n|S|=\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|\n$$\n\nSecond, each resident knows how to make exactly two dishes, so\n\n$$\n|S|=\\sum_{r \\in T} 2=2 \\operatorname{pop}(T)\n$$\n\n\nTherefore $\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|$ is even. But if $\\left|\\operatorname{chef}_{T}(D)\\right|=3$ for all $D \\in \\operatorname{dish}(T)$, then the sum is simply $3|\\operatorname{dish}(T)|$, so $|\\operatorname{dish}(T)|$ must be even.']" [] Text-only Competition True Theorem proof Combinatorics Math English 470 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=3$. Prove that if $T$ contains a resident cycle such that for every dish $D \in \operatorname{dish}(T)$, there exists a chef in the cycle that can prepare $D$, then $\operatorname{gr}(T)=3$." ['First we prove that for any town $T$ and any $D \\in \\operatorname{dish}(T), \\operatorname{gr}(T) \\geq\\left|\\operatorname{chef}_{T}(D)\\right|$.\n\nLet $D \\in \\operatorname{dish}(T)$. Suppose that $f$ is a valid group assignment on $T$. Then for $r, s \\in \\operatorname{chef}_{T}(D)$, if $r \\neq s$, it follows that $f(r) \\neq f(s)$. Hence there must be at least $\\left|\\operatorname{chef}_{T}(D)\\right|$ distinct groups in the range of $f$, i.e., $\\operatorname{gr}(T) \\geq\\left|\\operatorname{chef}_{T}(D)\\right|$.\n\nBy the conclusion above, it must be the case that $\\operatorname{gr}(T) \\geq 3$. Let $C=\\left\\{r_{1}, r_{2}, \\ldots, r_{n}\\right\\}$ denote a resident cycle such that for every dish $D \\in \\operatorname{dish}(T)$, there exists a chef in $C$ that can prepare $D$. Each resident is a chef for two dishes, and every dish can be made by two residents in $C$ (although by three in $T$ ). Thus the number of residents in the resident cycle $C$ is equal to $|\\operatorname{dish}(T)|$, which was proved to be even in the previous part.\n\nDefine a group assignment by setting\n\n$$\nf(r)= \\begin{cases}1 & \\text { if } r \\notin C \\\\ 2 & \\text { if } r=r_{i} \\text { and } i \\text { is even } \\\\ 3 & \\text { if } r=r_{i} \\text { and } i \\text { is odd. }\\end{cases}\n$$\n\nFor any $D \\in \\operatorname{dish}(T)$, there are exactly three $D$-chefs, and exactly two of them belong to the resident cycle $C$. Hence exactly one of the $D$-chefs $r$ will have $f(r)=1$. The remaining two $D$-chefs will be $r_{i}$ and $r_{i+1}$ for some $i$, or $r_{1}$ and $r_{n}$. In either case, the group assignment $f$ will assign one of them to 2 and the other to 3 . Thus any two residents who make a common dish will be assigned different groups by $f$, so $f$ is a valid group assignment, proving that $\\operatorname{gr}(T)=3$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 471 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $k$ be a positive integer, and let $T$ be a town in which $\left|\operatorname{chef}_{T}(D)\right|=k$ for every dish $D \in \operatorname{dish}(T)$. Suppose further that $|\operatorname{dish}(T)|$ is odd. Show that $k$ is even." "['First, we Prove that\n$$\n\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|=2 \\operatorname{pop}(T) .\n$$\n\nBecause each chef knows how to prepare exactly two dishes, and no two chefs know how to prepare the same two dishes, each chef is counted exactly twice in the sum $\\Sigma\\left|\\operatorname{chef}_{T}(D)\\right|$. More formally, consider the set of ""resident-dish pairs"":\n\n$$\nS=\\{(r, D) \\in T \\times \\operatorname{dish}(T) \\mid r \\text { makes } D\\}\n$$\n\nCount $|S|$ in two different ways. First, every dish $D$ is made by $\\left|\\operatorname{chef}_{T}(D)\\right|$ residents of $T$, so\n\n$$\n|S|=\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|\n$$\n\nSecond, each resident knows how to make exactly two dishes, so\n\n$$\n|S|=\\sum_{r \\in T} 2=2 \\operatorname{pop}(T)\n$$\n\nFrom above prove,\n\n$$\n2 \\operatorname{pop}(T)=\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|\n$$\n\nBecause $\\left|\\operatorname{chef}_{T}(D)\\right|=k$ for all $D \\in \\operatorname{dish}(T)$, the sum is $k \\cdot \\operatorname{dish}(T)$. Thus $2 \\operatorname{pop} T=$ $k \\cdot \\operatorname{dish}(T)$, and so $k \\cdot \\operatorname{dish}(T)$ must be even. By assumption, $|\\operatorname{dish}(T)|$ is odd, so $k$ must be even.']" [] Text-only Competition True Theorem proof Combinatorics Math English 472 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $k$ be a positive integer, and let $T$ be a town in which $\left|\operatorname{chef}_{T}(D)\right|=k$ for every dish $D \in \operatorname{dish}(T)$. Suppose further that $|\operatorname{dish}(T)|$ is odd. Prove the following: for every group in $T$, there is some $\operatorname{dish} D \in \operatorname{dish}(T)$ such that no one in the group is a $D$-chef." ['Suppose for the sake of contradiction that there is some $n$ for which the group $R=\\{r \\in$ $T \\mid f(r)=n\\}$ has a $D$-chef for every dish $D$. Because $f$ is a group assignment and $f$ assigns every resident of $R$ to group $n$, no two residents of $R$ make the same dish. Thus for every $D \\in \\operatorname{dish}(T)$, exactly one resident of $R$ is a $D$-chef; and each $D$-chef cooks exactly one other dish, which itself is not cooked by anyone else in $R$. Thus the dishes come in pairs: for each dish $D$, there is another dish $D^{\\prime}$ cooked by the $D$-chef in $R$ and no one else in $R$. However, if the dishes can be paired off, there must be an even number of dishes, contradicting the assumption that $|\\operatorname{dish}(T)|$ is odd. Thus for every $n$, the set $\\{r \\in T \\mid f(r)=n\\}$ must be missing a $D$-chef for some dish $D$.'] [] Text-only Competition True Theorem proof Combinatorics Math English 473 "In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. For each odd positive integer $d \geq 3$, prove that $\operatorname{gr}\left(\mathcal{F}_{d}\right)=d$." ['Fix $D \\in \\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)$. Then for every other dish $D^{\\prime} \\in \\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)$, there is exactly one chef who makes both $D$ and $D^{\\prime}$, hence $\\left|\\operatorname{chef}_{\\mathcal{F}_{d}}(D)\\right|=d-1$, which is even because $d$ is odd. Thus for each $D \\in \\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)$, $\\left|\\operatorname{chef}_{\\mathcal{F}_{d}}(D)\\right|$ is even. Because $\\left|\\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)\\right|=d$ is odd and $\\left|\\operatorname{chef}_{\\mathcal{F}_{d}}(D)\\right|=d-1$ for every dish in $\\mathcal{F}_{d}$, problem 12c applies, hence $\\operatorname{gr}\\left(\\mathcal{F}_{d}\\right)>d-1$.\n\nLabel the dishes $D_{1}, D_{2}, \\ldots, D_{d}$, and label the residents $r_{i, j}$ for $1 \\leq id(b))$. Then $a-b=3^{\\alpha}\\left(a_{0}-3^{\\beta-\\alpha} b_{0}\\right)$. In this second factor, notice that the second term, $3^{\\beta-\\alpha} b_{0}$ is divisible by 3 but the first term $a_{0}$ is not, so their difference is not divisible by 3 . Thus $d(a, b)=3^{-\\alpha}=d(a)$. Therefore $d(a, b)=d(a)$ when $d(a)>d(b)$, and similarly $d(a, b)=d(b)$ when $d(b)>d(a)$. Hence $d(a, b) \\leq \\max \\{d(a), d(b)\\}$.'] [] Text-only Competition True Theorem proof Geometry Math English 477 "In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses. Prove that, for all $a, b$, and $c, d(a, c) \leq \max \{d(a, b), d(b, c)\}$." ['Note that $d(a, c)=d(a-c)=d((a-b)-(c-b))=d(a-b, c-b)$. Then $d(a, c) \\leq$ $\\max \\{d(a, b), d(c, b)\\}$ by part 5 a, and the fact that $d(c, b)=d(b, c)$.'] [] Text-only Competition True Theorem proof Geometry Math English 478 "In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses. Prove that, for all $a, b$, and $c, d(a, c) \leq d(a, b)+d(b, c)$." ['Because $d(x, y)>0$ for all $x \\neq y, \\max \\{d(a, b), d(b, c)\\}0$, so $d(a, c)1 / 9$ and $d(x)<1 / 9$. In the first case, $9 \\nmid x$ (either $x$ is not divisible by 3 , or $x$ is divisible by 3 and not 9 ). These houses do not move into $\\mathcal{N}(9): x \\not \\equiv 0 \\bmod 9 \\Rightarrow x+18 \\not \\equiv 0 \\bmod 9$. On the other hand, if $d(x)<1 / 9$, then $27 \\mid x$, i.e., $x=27 k$. Then $x+18=27 k+18$ which is divisible by 9 but not by 27 , so $d(x+18)=9$. So in fact every house $x$ of ARMLopolis with $d(x)<1 / 9$ will move into $\\mathcal{N}(9)$ !'] [] Text-only Competition True Theorem proof Geometry Math English 484 "In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. One day, Paul (house 23) and Sally (house 32), longtime residents of $\mathcal{N}(1)$, are discussing the "" 2 side"" of the neighborhood, that is, the set of houses $\{n \mid n=3 k+2, k \in \mathbb{Z}\}$. Paul says ""I feel like I'm at the center of the "" 2 side"": when I looked out my window, I realized that the ""2 side"" consists of exactly those houses whose distance from me is at most $1 / 3 . ""$ Prove that Paul is correct." ['If $n=3 k+2$, then $d(23, n)=d(3 k-21)$, and because $3 k-21=3(k-7), d(3 k-21) \\leq$ $1 / 3$. On the other hand, if $d(23, n) \\leq 1 / 3$, then $3 \\mid(n-23)$, so $n-23=3 k$, and $n=3 k+23=3(k+7)+2$.'] [] Text-only Competition True Theorem proof Geometry Math English 485 "In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. Sally (house 32). Prove that $\{n \mid n=3 k+2, k \in \mathbb{Z}\}$ consists exactly of those houses whose distance from Sally's house is at most $1 / 3$." ['If $n=3 k+2$, then $d(32, n)=d(3 k-30) \\leq 1 / 3$. Similarly, if $d(32, n) \\leq 1 / 3$, then $3 \\mid n-32 \\Rightarrow n=3 k+32=3(k+10)+2$.'] [] Text-only Competition True Theorem proof Geometry Math English 486 "In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. For any $x$, let $\mathcal{D}_{r}(x)=\{y \mid d(x, y) \leq r\}$, that is, $\mathcal{D}_{r}(x)$ is the disk of radius $r$. Prove that if $d(x, z)=r$, then $\mathcal{D}_{r}(x)=\mathcal{D}_{r}(z) . \quad[2 \mathrm{pts}]$" ['Show that $\\mathcal{D}_{r}(x) \\subseteq \\mathcal{D}_{r}(z)$ and conversely. Suppose that $y \\in \\mathcal{D}_{r}(x)$. Then $d(x, y) \\leq r$. Because $d(x, z)=r$, by $5 \\mathrm{~b}, d(y, z) \\leq \\max \\{d(x, y), d(x, z)\\}=r$. So $y \\in \\mathcal{D}_{r}(z)$. Thus $\\mathcal{D}_{r}(x) \\subseteq \\mathcal{D}_{r}(z)$. Similarly, $\\mathcal{D}_{r}(z) \\subseteq \\mathcal{D}_{r}(x)$, so the two sets are equal.'] [] Text-only Competition True Theorem proof Geometry Math English 487 "In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$. A longtime resident of IMOpia moves to ARMLopolis and hopes to keep his same address. When asked, he says ""My old address was $e$, that is, the sum $\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\ldots$ I'd really like to keep that address, because the addresses of my friends here are all partial sums of this series: Alice's house is $\frac{1}{0 !}$, Bob's house is $\frac{1}{0 !}+\frac{1}{1 !}$, Carol's house is $\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}$, and so on. Just let me know what I have to do in order to be near my friends!"" After some head-scratching, the ARMLopolitan planning council announces that this request cannot be satisfied: there is no number, rational or otherwise, that corresponds to the (infinite) sum, or that is arbitrarily close to the houses in this sequence. Prove that the council is correct (and not just bureaucratic)." "[""The houses in this sequence actually get arbitrarily far apart from each other, and so there's no single house which they approach. If $H_{n}=\\frac{1}{0 !}+\\frac{1}{1 !}+\\cdots+\\frac{1}{n !}$, then $d\\left(H_{n}, H_{n-1}\\right)=$ $d(1 / n !)=3^{k}$, where $k=\\lfloor n / 3\\rfloor+\\lfloor n / 9\\rfloor+\\lfloor n / 27\\rfloor \\ldots$. In particular, if $n !=3^{k} \\cdot n_{0}$, where $k>0$ and $3 \\nmid n_{0}$, then $d\\left(H_{n}, H_{n-1}\\right)=3^{k}$, which can be made arbitrarily large as $n$ increases.""]" [] Text-only Competition True Theorem proof Algebra Math English 488 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that any two consecutive odd positive integers are relatively prime to each other." ['Two consecutive odd integers differ by 2 . Thus any common divisor of the two integers must also divide 2 . However, the only prime divisor of 2 is 2 , and neither of the consecutive odd integers is a multiple of 2 . Therefore any two consecutive odd integers are relatively prime to each other, as desired.'] [] Text-only Competition True Theorem proof Algebra Math English 489 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that every sequence of three consecutive positive integers contains a cromulent element." ['Consider the sequence $a-1, a, a+1$. Note that $a$ is relatively prime to $a-1$ and $a+1$ because any common divisor must divide their difference, and the differences are both 1 . Thus $a$ is a cromulent element of the sequence.'] [] Text-only Competition True Theorem proof Algebra Math English 490 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that every sequence of four consecutive positive integers contains a cromulent element." ['First we prove that any two consecutive odd positive integers are relatively prime to each other.\n\nProof: Two consecutive odd integers differ by 2 . Thus any common divisor of the two integers must also divide 2 . However, the only prime divisor of 2 is 2 , and neither of the consecutive odd integers is a multiple of 2 . Therefore any two consecutive odd integers are relatively prime to each other, as desired.\n\nConsider a sequence of four consecutive integers. Notice that two of these integers will be odd, and by the proof above, these are relatively prime to each other. At most one of these two odd integers can be a multiple of 3 . Let $a$ be an odd integer in the sequence that is not a multiple of 3 . Then $a$ is cromulent by the following reasoning. Note that the difference between $a$ and any other element of the sequence is at most 3. Thus if $a$ shared a common factor greater than 1 with some other element of the sequence, this factor would have to be 2 or 3 . But because $a$ is odd and not a multiple of 3 , it follows that $a$ is cromulent.'] [] Text-only Competition True Theorem proof Algebra Math English 491 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that every sequence of five consecutive positive integers contains a cromulent element." ['Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at least one other number in the sequence. There are several cases to consider, namely whether the sequence starts with an even number or an odd number.\n\nIf the sequence starts with an even number, then the second and fourth numbers are both odd, and at least one of them is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence because it is neither a multiple of 2 nor 3 and hence is at least 5 away from the nearest integer with a common factor. Thus the sequence contains a cromulent element.\n\nIf the sequence starts with an odd number, then again, it contains an odd number that is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence, thus the sequence contains a cromulent element. In fact, it contains two such numbers if the first or last number is a multiple of 3 , and if the middle number is a multiple of 3 , then all three odd elements are cromulent.'] [] Text-only Competition True Theorem proof Algebra Math English 492 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=8$;" ['In any sequence of eight consecutive integers, four of them will be even and hence not cromulent. The primes 3,5 , and 7 are the only numbers that could possibly divide more than one number in the sequence, with 5 and 7 each dividing at most one of the four odd numbers. Note that for 7 to divide two numbers in the sequence, they must be the first and last numbers because only those differ by 7 . The prime 3 can divide two odd numbers in the sequence, but only if they differ by 6 , which means they must be either the 1st and 7th or the $2 \\mathrm{nd}$ and 8th numbers in the list. Therefore one of the two odd multiples of 3 would also be the only candidate to be an odd multiple of 7 when there are two multiples of 7 , and so at least one of the four odd numbers must be cromulent.'] [] Text-only Competition True Theorem proof Algebra Math English 493 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=9$." ['First we prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=8$;\n\nProof: In any sequence of eight consecutive integers, four of them will be even and hence not cromulent. The primes 3,5 , and 7 are the only numbers that could possibly divide more than one number in the sequence, with 5 and 7 each dividing at most one of the four odd numbers. Note that for 7 to divide two numbers in the sequence, they must be the first and last numbers because only those differ by 7 . The prime 3 can divide two odd numbers in the sequence, but only if they differ by 6 , which means they must be either the 1st and 7th or the $2 \\mathrm{nd}$ and 8th numbers in the list. Therefore one of the two odd multiples of 3 would also be the only candidate to be an odd multiple of 7 when there are two multiples of 7 , and so at least one of the four odd numbers must be cromulent.\n\nNow we turn to prove that given assumption. In a sequence of nine consecutive integers, again the only primes to check are $2,3,5$, and 7 . If there are five even numbers in the sequence, then at least one of the odd numbers is cromulent by similar reasoning to the proof above. The difference is that in this case the multiples of 7 can be either the 1st and 8th or the 2nd and 9th numbers, while the odd multiples of 3 must be the 2 nd and 8th numbers (because the 1st number is even) and hence overlap with the multiples of 7 again. If instead there are four even numbers in the sequence, then the 1st, 3rd, 5th, 7th, and 9th numbers are odd. Now 7 divides at most one of these odd numbers, 5 divides at most one, and 3 divides at most two, leaving at least one odd cromulent element.'] [] Text-only Competition True Theorem proof Algebra Math English 494 "In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. The goal of this problem is to prove the following claim: Claim: For $k \geq 300$, there is a sequence of $k$ consecutive integers with no cromulent elements. Let $\pi(n)$ denote the number of primes less than or equal to $n$. You may use the following lemma without proof in the problems that follow. Lemma: For $x \geq 75, \pi(2 x)-\pi(x) \geq 2\left\lfloor\log _{2}(x)\right\rfloor+2$. One argument for proving the above claim begins as follows. Let $m=\left\lfloor\frac{k}{4}\right\rfloor \geq 75$, let $p_{1}, p_{2}, \ldots, p_{r}$ denote the primes in the set $\{1,2, \ldots, m\}$ (note that $p_{1}=2$ ), and let $p_{r+1}, p_{r+2}, \ldots, p_{s}$ denote the primes in the set $\{m+1, m+2, \ldots, 2 m\}$. Prove that if each of the $k$ consecutive integers in the sequence is divisible by at least one of the primes $p_{1}, p_{2}, \ldots, p_{s}$, then the sequence has no cromulent elements." ['Let the sequence of $k$ integers be $a_{1}, a_{2}, \\ldots, a_{k}$, where $a_{n}=a_{1}+(n-1)$ for $1 \\leq n \\leq k$. For any $a_{j} \\in\\left\\{a_{1}, a_{2}, \\ldots, a_{2 m}\\right\\}$, suppose that $a_{j}$ is divisible by some $p \\in\\left\\{p_{1}, p_{2}, \\ldots, p_{s}\\right\\}$. Then $a_{j+p}=a_{j}+p$, so $a_{j+p}$ is also divisible by $p$. Hence neither $a_{j}$ nor $a_{j+p}$ is cromulent because both numbers are divisible by $p$ (also note that $j+p<4 m \\leq k$, so $a_{j+p}$ is one of the $k$ integers in the sequence). Thus none of $a_{1}, a_{2}, \\ldots, a_{2 m}$ are cromulent. Similarly, for any $a_{j^{\\prime}} \\in\\left\\{a_{2 m+1}, a_{2 m+2}, \\ldots, a_{k}\\right\\}$, suppose that $a_{j^{\\prime}}$ is divisible by some $p^{\\prime} \\in\\left\\{p_{1}, p_{2}, \\ldots, p_{s}\\right\\}$. Then $a_{j^{\\prime}-p^{\\prime}}=a_{j^{\\prime}}-p^{\\prime}$ is also divisible by $p^{\\prime}$ (also note that $11$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. The goal of this problem is to prove the following claim: Claim: For $k \geq 300$, there is a sequence of $k$ consecutive integers with no cromulent elements. Let $\pi(n)$ denote the number of primes less than or equal to $n$. You may use the following lemma without proof in the problems that follow. Lemma: For $x \geq 75, \pi(2 x)-\pi(x) \geq 2\left\lfloor\log _{2}(x)\right\rfloor+2$. One argument for proving the above claim begins as follows. Let $m=\left\lfloor\frac{k}{4}\right\rfloor \geq 75$, let $p_{1}, p_{2}, \ldots, p_{r}$ denote the primes in the set $\{1,2, \ldots, m\}$ (note that $p_{1}=2$ ), and let $p_{r+1}, p_{r+2}, \ldots, p_{s}$ denote the primes in the set $\{m+1, m+2, \ldots, 2 m\}$. Let $x$ be a solution to the system of congruences $x \equiv 1(\bmod 2)$ and $x \equiv 0\left(\bmod p_{2} p_{3} \cdots p_{r}\right)$. Then the integers $$ x-2 m, x-2 m+2, \ldots, x-2, x, x+2, \ldots, x+2 m-2, x+2 m $$ form a sequence of $2 m+1$ consecutive odd integers of the form $x \pm 2 y$, where $y$ varies from 0 to $m$. Prove that every number in the above sequence, except those in which $y$ is a power of 2 , is divisible by one of the primes $p_{2}, \ldots, p_{r}$." ['Note that $x \\equiv 1(\\bmod 2)$ implies that $x$ is odd. Because $x \\equiv 0\\left(\\bmod p_{2} p_{3} \\cdots p_{r}\\right)$, write $x=q p_{2} p_{3} \\cdots p_{r}$ for some odd integer $q$. Then $x$ is divisible by all of the primes $p_{2}, \\ldots, p_{r}$. The desired result will first be established for numbers of the form $x+2 y$, with $1 \\leq y \\leq m$. Because $x$ is divisible by each of $p_{2}, \\ldots, p_{r}$, it suffices to prove that $y$ is divisible by one of $p_{2}, \\ldots, p_{r}$ unless $y$ is a power of 2 . Note that because $p_{2}, \\ldots, p_{r}$ are all odd, if $y$ is a power of 2 (including $2^{0}=1$ ), then $2 y$ cannot be divisible by any of $p_{2}, \\ldots, p_{r}$. On the other hand, if $y$ has an odd factor greater than 1, by the Fundamental Theorem of Arithmetic, $y$ must be divisible by at least one of $p_{2}, \\ldots, p_{r}$ (note that $p_{2}=3$ and $p_{r} \\leq m$ ). Finally, because the desired result holds for $x+2 y$, with $1 \\leq y \\leq m$, it also holds for $x-2 y$ because $x-2 y=(x+2 y)-4 y$, thus completing the proof.'] [] Text-only Competition True Theorem proof Algebra Math English 496 "An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. Explain why the label 1234 has a unique 3-signature." ['We will prove this by contradiction. Suppose for some other 4-label $L$ we have $S_{3}[L]=$ $S_{3}[1234]=(123,123)$. Write out $L$ as $a_{1}, a_{2}, a_{3}, a_{4}$. From the first window of $S_{3}[L]$, we have $a_{1}a_{3}$. This is impossible, so the 3 -signature $(123,321)$ is impossible.'] [] Text-only Competition True Theorem proof Combinatorics Math English 498 "Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Show that if $n+k$ is even and $\frac{n}{2}\\frac{n}{2}$, some switches must be flipped twice in the first two minutes, and so $E(n, k) \\neq 2$.\n\nHere is a strategy by which Elizabeth can escape in three minutes. Note that because $n+k$ is even, it follows that $3 k-n=(k+n)+2(k-n)$ is also even. Then let $3 k-n=2 b$. Note that $b>0$ (because $k>\\frac{n}{2}$ ), and $b2$, there exists an ordered pair $(n, k)$ for which $I(n, k)>x$." "['First, we prove that if $n$ is even and $k$ is odd, then $E(n, k)=E(n,n-k)$.\n\n\nBecause $n$ is even, and because each switch must be flipped an odd number of times in order to escape, the total number of flips is even. Because $k$ must be odd, $E(n, k)$ must be even. To show this, consider the case where $E(n, k)$ is odd. If $E(n, k)$ is odd, then an odd number of flips happen an odd number of times, resulting in an odd number of total flips. This is a contradiction because $n$ is even.\n\nCall a switch ""non-flipped"" in any given minute if it is not among the switches flipped in that minute. Because $E(n, k)$ (i.e., the total number of minutes) is even, and each switch is flipped an odd number of times, each switch must also be non-flipped an odd number of times. Therefore any sequence of flips that solves the "" $(n, k)$ puzzle"" can be made into a sequence of flips that solves the "" $(n, n-k)$ "" puzzle by interchanging flips and non-flips. These sequences last for the same number of minutes, and therefore $E(n, k)=E(n, n-k)$.\n\n\nLet $n=2 x$ and $k=2 x-1$ for some positive integer $x$. Then $n$ is even and $k$ is odd, so the above prove applies, and $E(n, k)=E(n, n-k)=E(n, 1)=2 x$. Therefore $I(n, k)=2 x-\\frac{2 x}{2 x-1}$. Because $x>2$, it follows that $\\frac{2 x}{2 x-1}<2x$, as desired.']" [] Text-only Competition True Theorem proof Combinatorics Math English