id	question	solution	final_answer	context	image	modality	difficulty	is_multiple_answer	unit	answer_type	error	question_type	subfield	subject	language
0	"Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that $\operatorname{arc} \gamma_{2}$ lies between the $\operatorname{arcs} \gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be three rays starting at $B$, lying in the same half-plane, $h_{2}$ being between $h_{1}$ and $h_{3}$. For $i, j=1,2,3$, denote by $V_{i j}$ the point of intersection of $h_{i}$ and $\gamma_{j}$ (see the Figure below).

Denote by $\overparen{V_{i j} V_{k j}} \overparen{k_{k \ell} V_{i \ell}}$ the curved quadrilateral, whose sides are the segments $V_{i j} V_{i \ell}, V_{k j} V_{k \ell}$ and $\operatorname{arcs} V_{i j} V_{k j}$ and $V_{i \ell} V_{k \ell}$. We say that this quadrilateral is circumscribed if there exists a circle touching these two segments and two arcs.

Prove that if the curved quadrilaterals $\sqrt{V_{11} V_{21}} \sqrt{V_{22} V_{12}}, \sqrt{V_{12} V_{22}} \sqrt{V_{23} V_{13}}, \sqrt{V_{21} V_{31}} \sqrt{V_{32} V_{22}}$ are circumscribed, then the curved quadrilateral $\overparen{V_{22} V_{32}} \overparen{V_{33} V_{23}}$ is circumscribed, too.

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Fig. 1"	['Denote by $O_{i}$ and $R_{i}$ the center and the radius of $\\gamma_{i}$, respectively. Denote also by $H$ the half-plane defined by $A C$ which contains the whole configuration. For every point $P$ in the half-plane $H$, denote by $d(P)$ the distance between $P$ and line $A C$. Furthermore, for any $r>0$, denote by $\\Omega(P, r)$ the circle with center $P$ and radius $r$.\n\nLemma 1. For every $1 \\leq i<j \\leq 3$, consider those circles $\\Omega(P, r)$ in the half-plane $H$ which are tangent to $h_{i}$ and $h_{j}$.\n\n(a) The locus of the centers of these circles is the angle bisector $\\beta_{i j}$ between $h_{i}$ and $h_{j}$.\n\n(b) There is a constant $u_{i j}$ such that $r=u_{i j} \\cdot d(P)$ for all such circles.\n\nProof. Part (a) is obvious. To prove part (b), notice that the circles which are tangent to $h_{i}$ and $h_{j}$ are homothetic with the common homothety center $B$ (see Fig. 2). Then part (b) also becomes trivial.\n\nLemma 2. For every $1 \\leq i<j \\leq 3$, consider those circles $\\Omega(P, r)$ in the half-plane $H$ which are externally tangent to $\\gamma_{i}$ and internally tangent to $\\gamma_{j}$.\n\n(a) The locus of the centers of these circles is an ellipse arc $\\varepsilon_{i j}$ with end-points $A$ and $C$.\n\n(b) There is a constant $v_{i j}$ such that $r=v_{i j} \\cdot d(P)$ for all such circles.\n\nProof. (a) Notice that the circle $\\Omega(P, r)$ is externally tangent to $\\gamma_{i}$ and internally tangent to $\\gamma_{j}$ if and only if $O_{i} P=R_{i}+r$ and $O_{j}=R_{j}-r$. Therefore, for each such circle we have\n\n$$\nO_{i} P+O_{j} P=O_{i} A+O_{j} A=O_{i} C+O_{j} C=R_{i}+R_{j} .\n$$\n\nSuch points lie on an ellipse with foci $O_{i}$ and $O_{j}$; the diameter of this ellipse is $R_{i}+R_{j}$, and it passes through the points $A$ and $C$. Let $\\varepsilon_{i j}$ be that arc $A C$ of the ellipse which runs inside the half plane $H$ (see Fig. 3.)\n\nThis ellipse arc lies between the $\\operatorname{arcs} \\gamma_{i}$ and $\\gamma_{j}$. Therefore, if some point $P$ lies on $\\varepsilon_{i j}$, then $O_{i} P>R_{i}$ and $O_{j} P<R_{j}$. Now, we choose $r=O_{i} P-R_{i}=R_{j}-O_{j} P>0$; then the\n\n\n\n<img_3334>\n\nFig. 2\n\n<img_3279>\n\nFig. 3\n\ncircle $\\Omega(P, r)$ touches $\\gamma_{i}$ externally and touches $\\gamma_{j}$ internally, so $P$ belongs to the locus under investigation.\n\n(b) Let $\\vec{\\rho}=\\overrightarrow{A P}, \\vec{\\rho}_{i}=\\overrightarrow{A O_{i}}$, and $\\vec{\\rho}_{j}=\\overrightarrow{A O_{j}}$; let $d_{i j}=O_{i} O_{j}$, and let $\\vec{v}$ be a unit vector orthogonal to $A C$ and directed toward $H$. Then we have $\\left|\\vec{\\rho}_{i}\\right|=R_{i},\\left|\\vec{\\rho}_{j}\\right|=R_{j},\\left|\\overrightarrow{O_{i} P}\\right|=$ $\\left|\\vec{\\rho}-\\vec{\\rho}_{i}\\right|=R_{i}+r,\\left|\\overrightarrow{O_{j} P}\\right|=\\left|\\vec{\\rho}-\\vec{\\rho}_{j}\\right|=R_{j}-r$, hence\n\n$$\n\\begin{gathered}\n\\left(\\vec{\\rho}-\\vec{\\rho}_{i}\\right)^{2}-\\left(\\vec{\\rho}-\\vec{\\rho}_{j}\\right)^{2}=\\left(R_{i}+r\\right)^{2}-\\left(R_{j}-r\\right)^{2} \\\\\n\\left(\\vec{\\rho}_{i}^{2}-\\vec{\\rho}_{j}^{2}\\right)+2 \\vec{\\rho} \\cdot\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right)=\\left(R_{i}^{2}-R_{j}^{2}\\right)+2 r\\left(R_{i}+R_{j}\\right) \\\\\nd_{i j} \\cdot d(P)=d_{i j} \\vec{v} \\cdot \\vec{\\rho}=\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right) \\cdot \\vec{\\rho}=r\\left(R_{i}+R_{j}\\right)\n\\end{gathered}\n$$\n\nTherefore,\n\n$$\nr=\\frac{d_{i j}}{R_{i}+R_{j}} \\cdot d(P)\n$$\n\nand the value $v_{i j}=\\frac{d_{i j}}{R_{i}+R_{j}}$ does not depend on $P$.\n\nLemma 3. The curved quadrilateral $\\mathcal{Q}_{i j}=\\overparen{V_{i, j} V_{i+1}, j} \\overparen{V_{i+1, j+1} V_{i, j+1}}$ is circumscribed if and only if $u_{i, i+1}=v_{j, j+1}$.\n\nProof. First suppose that the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed and $\\Omega(P, r)$ is its inscribed circle. By Lemma 1 and Lemma 2 we have $r=u_{i, i+1} \\cdot d(P)$ and $r=v_{j, j+1} \\cdot d(P)$ as well. Hence, $u_{i, i+1}=v_{j, j+1}$.\n\nTo prove the opposite direction, suppose $u_{i, i+1}=v_{j, j+1}$. Let $P$ be the intersection of the angle bisector $\\beta_{i, i+1}$ and the ellipse arc $\\varepsilon_{j, j+1}$. Choose $r=u_{i, i+1} \\cdot d(P)=v_{j, j+1} \\cdot d(P)$. Then the circle $\\Omega(P, r)$ is tangent to the half lines $h_{i}$ and $h_{i+1}$ by Lemma 1 , and it is tangent to the $\\operatorname{arcs} \\gamma_{j}$ and $\\gamma_{j+1}$ by Lemma 2. Hence, the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed.\n\nBy Lemma 3, the statement of the problem can be reformulated to an obvious fact: If the equalities $u_{12}=v_{12}, u_{12}=v_{23}$, and $u_{23}=v_{12}$ hold, then $u_{23}=v_{23}$ holds as well.']			['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proof	Geometry	Math	English
1	"Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively.

<image_1>

S-tetrominoes

<image_2>

Z-tetrominoes

Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove than no matter how we tile $P$ using only $\mathrm{S}$ - and Z-tetrominoes, we always use an even number of Z-tetrominoes."	"['We may assume that polygon $P$ is the union of some squares of an infinite chessboard. Colour the squares of the chessboard with two colours as the figure below illustrates.\n\n<img_3847>\n\nObserve that no matter how we tile $P$, any S-tetromino covers an even number of black squares, whereas any Z-tetromino covers an odd number of them. As $P$ can be tiled exclusively by S-tetrominoes, it contains an even number of black squares. But if some S-tetrominoes and some Z-tetrominoes cover an even number of black squares, then the number of Z-tetrominoes must be even.'
 'Let us assign coordinates to the squares of the infinite chessboard in such a way that the squares of $P$ have nonnegative coordinates only, and that the first coordinate increases as one moves to the right, while the second coordinate increases as one moves upwards. Write the integer $3^{i} \\cdot(-3)^{j}$ into the square with coordinates $(i, j)$, as in the following figure:\n\n| $\\vdots$ |  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 81 | $\\vdots$ |  |  |  |  |\n| -27 | -81 | $\\vdots$ |  |  |  |\n| 9 | 27 | 81 | $\\cdots$ |  |  |\n| -3 | -9 | -27 | -81 | $\\ldots$ |  |\n| 1 | 3 | 9 | 27 | 81 | $\\ldots$ |\n\nThe sum of the numbers written into four squares that can be covered by an $S$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+3+3 \\cdot(-3)+3^{2} \\cdot(-3)\\right)=-32 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\n(for the first type of $S$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3 \\cdot(-3)+(-3)+(-3)^{2}\\right)=0\n$$\n\nand thus divisible by 32. For this reason, the sum of the numbers written into the squares of $P$, and thus also the sum of the numbers covered by $Z$-tetrominoes in the second covering, is likewise divisible by 32 . Now the sum of the entries of a $Z$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3^{2}+(-3)+3 \\cdot(-3)\\right)=0\n$$\n\n(for the first type of $Z$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+(-3)+3 \\cdot(-3)+3 \\cdot(-3)^{2}\\right)=16 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\ni.e., 16 times an odd number. Thus in order to obtain a total that is divisible by 32, an even number of the latter kind of $Z$-tetrominoes needs to be used. Rotating everything by $90^{\\circ}$, we find that the number of $Z$-tetrominoes of the first kind is even as well. So we have even proven slightly more than necessary.']"			['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proof	Combinatorics	Math	English
2	"An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once:

<image_1>

Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once."	"[""Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\\cdots+n$, it follows that they form a permutation of $1,2, \\ldots, n$.\n\n<img_3838>\n\nFigure 1\n\n<img_4049>\n\nFigure 2\n\nConsider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\\prime}$, has side length $\\ell \\geqslant\\lceil(n-2) / 2\\rceil$. Since $T^{\\prime}$ obeys the anti-Pascal rule, it contains $\\ell$ pairwise distinct positive integers $a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{\\ell}^{\\prime}$, where $a_{1}^{\\prime}$ is at the apex, and $a_{k}^{\\prime}$ and $b_{k}^{\\prime}=a_{1}^{\\prime}+a_{2}^{\\prime}+\\cdots+a_{k}^{\\prime}$ are the two neighbours below $b_{k-1}^{\\prime}$ for each $k=2,3 \\ldots, \\ell$. Since the $a_{k}$ all lie outside $T^{\\prime}$, and they form a permutation of $1,2, \\ldots, n$, the $a_{k}^{\\prime}$ are all greater than $n$. Consequently,\n\n$$\n\\begin{array}{r}\nb_{\\ell}^{\\prime} \\geqslant(n+1)+(n+2)+\\cdots+(n+\\ell)=\\frac{\\ell(2 n+\\ell+1)}{2} \\\\\n\\geqslant \\frac{1}{2} \\cdot \\frac{n-2}{2}\\left(2 n+\\frac{n-2}{2}+1\\right)=\\frac{5 n(n-2)}{8},\n\\end{array}\n$$\n\nwhich is greater than $1+2+\\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.\n\nSo it is not possible.""]"			['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proof	Combinatorics	Math	English
3	"Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$.

<image_1>"	['We are to prove that $\\angle E X Y=\\angle E F X$; alternatively, but equivalently, $\\angle A Y X+\\angle X A Y=\\angle B Y F+\\angle X B Y$.\n\nSince the quadrangle $A B C D$ is cyclic, the triangles $X A D$ and $X B C$ are similar, and since $A D_{1}$ and $B C_{1}$ are corresponding medians in these triangles, it follows that $\\angle X A Y=\\angle X A D_{1}=\\angle X B C_{1}=\\angle X B Y$.\n\nFinally, $\\angle A Y X=\\angle B Y F$, since $X$ and $M$ are corresponding points in the similar triangles $A B Y$ and $C_{1} D_{1} Y$ : indeed, $\\angle X A B=\\angle X D C=\\angle M C_{1} D_{1}$, and $\\angle X B A=\\angle X C D=\\angle M D_{1} C_{1}$.']			['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proof	Geometry	Math	English
4	"Let $H$ be the orthocenter and $G$ be the centroid of acute-angled triangle $\triangle A B C$ with $A B \neq A C$. The line $A G$ intersects the circumcircle of $\triangle A B C$ at $A$ and $P$. Let $P^{\prime}$ be the reflection of $P$ in the line $B C$. Prove that $\angle C A B=60^{\circ}$ if and only if $H G=G P^{\prime}$.


<image_1>"	"[""Let $\\omega$ be the circumcircle of $\\triangle A B C$. Reflecting $\\omega$ in line $B C$, we obtain circle $\\omega^{\\prime}$ which, obviously, contains points $H$ and $P^{\\prime}$. Let $M$ be the midpoint of $B C$. As triangle $\\triangle A B C$ is acute-angled, then $H$ and $O$ lie inside this triangle.\n\nLet us assume that $\\angle C A B=60^{\\circ}$. Since\n\n$$\n\\angle C O B=2 \\angle C A B=120^{\\circ}=180^{\\circ}-60^{\\circ}=180^{\\circ}-\\angle C A B=\\angle C H B,\n$$\n\nhence $O$ lies on $\\omega^{\\prime}$. Reflecting $O$ in line $B C$, we obtain point $O^{\\prime}$ which lies on $\\omega$ and this point is the center of $\\omega^{\\prime}$. Then $O O^{\\prime}=2 O M=2 R \\cos \\angle C A B=A H$, so $A H=O O^{\\prime}=H O^{\\prime}=A O=R$, where $R$ is the radius of $\\omega$ and, naturally, of $\\omega^{\\prime}$. Then quadrilateral $A H O^{\\prime} O$ is a rhombus, so $A$ and $O^{\\prime}$ are symmetric to each other with respect to $H O$. As $H, G$ and $O$ are collinear (Euler line), then $\\angle G A H=\\angle H O^{\\prime} G$. Diagonals of quadrilateral $G O P O^{\\prime}$ intersects at $M$. Since $\\angle B O M=60^{\\circ}$, so\n\n$$\nO M=M O^{\\prime}=\\operatorname{ctg} 60^{\\circ} \\cdot M B=\\frac{M B}{\\sqrt{3}}\n$$\n\nAs $3 M O \\cdot M O^{\\prime}=M B^{2}=M B \\cdot M C=M P \\cdot M A=3 M G \\cdot M P$, then $G O P O^{\\prime}$ is a cyclic. Since $B C$ is a perpendicular bisector of $O O^{\\prime}$, so the circumcircle of quadrilateral $G O P O^{\\prime}$ is symmetrical with respect to $B C$. Thus $P^{\\prime}$ also belongs to the circumcircle of $G O P O^{\\prime}$, hence $\\angle G O^{\\prime} P^{\\prime}=\\angle G P P^{\\prime}$. Note that $\\angle G P P^{\\prime}=\\angle G A H$ since $A H \\| P P^{\\prime}$. And as it was proved $\\angle G A H=\\angle H O^{\\prime} G$, then $\\angle H O^{\\prime} G=\\angle G O^{\\prime} P^{\\prime}$. Thus triangles $\\triangle H O^{\\prime} G$ and $\\triangle G O^{\\prime} P^{\\prime}$ are equal and hence $H G=G P^{\\prime}$.\n\nNow we will prove that if $H G=G P^{\\prime}$ then $\\angle C A B=60^{\\circ}$. Reflecting $A$ with respect to $M$, we get $A^{\\prime}$. Then, as it was said in the first part of solution, points $B, C, H$ and $P^{\\prime}$ belong to $\\omega^{\\prime}$. Also it is clear that $A^{\\prime}$ belongs to $\\omega^{\\prime}$. Note that $H C \\perp C A^{\\prime}$ since $A B \\| C A^{\\prime}$ and hence $H A^{\\prime}$ is a diameter of $\\omega^{\\prime}$. Obviously, the center $O^{\\prime}$ of circle $\\omega^{\\prime}$ is midpoint of $H A^{\\prime}$. From $H G=G P^{\\prime}$ it follows that $\\triangle H G O^{\\prime}$ is equal to $\\triangle P^{\\prime} G O^{\\prime}$. Therefore $H$ and $P^{\\prime}$ are symmetric with respect to $G O^{\\prime}$. Hence $G O^{\\prime} \\perp H P^{\\prime}$ and $G O^{\\prime} \\| A^{\\prime} P^{\\prime}$. Let $H G$ intersect $A^{\\prime} P^{\\prime}$ at $K$ and $K \\not \\equiv O$ since $A B \\neq A C$. We conclude that $H G=G K$, because line $G O^{\\prime}$ is midline of the triangle $\\triangle H K A^{\\prime}$. Note that $2 G O=H G$. since $H O$ is Euler line of triangle $A B C$. So $O$ is midpoint of segment $G K$. Because of $\\angle C M P=\\angle C M P^{\\prime}$, then $\\angle G M O=\\angle O M P^{\\prime}$. Line $O M$, that passes through $O^{\\prime}$, is an external angle bisector of $\\angle P^{\\prime} M A^{\\prime}$. Also we know that $P^{\\prime} O^{\\prime}=O^{\\prime} A^{\\prime}$, then $O^{\\prime}$ is the midpoint of arc $P^{\\prime} M A^{\\prime}$ of the circumcircle of triangle $\\triangle P^{\\prime} M A^{\\prime}$. It\n\n\n\n<img_3844>\n\nfollows that quadrilateral $P^{\\prime} M O^{\\prime} A^{\\prime}$ is cyclic, then $\\angle O^{\\prime} M A^{\\prime}=\\angle O^{\\prime} P^{\\prime} A^{\\prime}=\\angle O^{\\prime} A^{\\prime} P^{\\prime}$. Let $O M$ and $P^{\\prime} A^{\\prime}$ intersect at $T$. Triangles $\\triangle T O^{\\prime} A^{\\prime}$ and $\\triangle A^{\\prime} O^{\\prime} M$ are similar, hence $O^{\\prime} A^{\\prime} / O^{\\prime} M=O^{\\prime} T / O^{\\prime} A^{\\prime}$. In the other words, $O^{\\prime} M \\cdot O^{\\prime} T=O^{\\prime} A^{\\prime 2}$. Using Menelaus' theorem for triangle $\\triangle H K A^{\\prime}$ and line $T O^{\\prime}$, we obtain that\n\n$$\n\\frac{A^{\\prime} O^{\\prime}}{O^{\\prime} H} \\cdot \\frac{H O}{O K} \\cdot \\frac{K T}{T A^{\\prime}}=3 \\cdot \\frac{K T}{T A^{\\prime}}=1\n$$\n\nIt follows that $K T / T A^{\\prime}=1 / 3$ and $K A^{\\prime}=2 K T$. Using Menelaus' theorem for triangle $T O^{\\prime} A^{\\prime}$ and line $H K$ we get that\n\n$$\n1=\\frac{O^{\\prime} H}{H A^{\\prime}} \\cdot \\frac{A^{\\prime} K}{K T} \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{1}{2} \\cdot 2 \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{T O}{O O^{\\prime}}\n$$\n\nIt means that $T O=O O^{\\prime}$, so $O^{\\prime} A^{\\prime 2}=O^{\\prime} M \\cdot O^{\\prime} T=O O^{\\prime 2}$. Hence $O^{\\prime} A^{\\prime}=O O^{\\prime}$ and, consequently, $O \\in \\omega^{\\prime}$. Finally we conclude that $2 \\angle C A B=\\angle B O C=180^{\\circ}-\\angle C A B$, so $\\angle C A B=60^{\\circ}$.\n<img_4029>""
 'Let $O^{\\prime}$ and $G^{\\prime}$ denote the reflection of $O$ and $G$, respectively, with respect to the line $B C$. We then need to show $\\angle C A B=60^{\\circ}$ iff $G^{\\prime} H^{\\prime}=G^{\\prime} P$. Note that $\\triangle H^{\\prime} O P$ is isosceles and hence\n\n\n\n$G^{\\prime} H^{\\prime}=G^{\\prime} P$ is equivalent to $G^{\\prime}$ lying on the bisector $\\angle H^{\\prime} O P$. Let $\\angle H^{\\prime} A P=\\varepsilon$. By the assumption $A B \\neq A C$, we have $\\varepsilon \\neq 0$. Then $\\angle H^{\\prime} O P=2 \\angle H^{\\prime} A P=2 \\varepsilon$, hence $G^{\\prime} H^{\\prime}=G^{\\prime} P$ iff $\\angle G^{\\prime} O H^{\\prime}=\\varepsilon$. But $\\angle G O^{\\prime} H=\\angle G^{\\prime} O H^{\\prime}$. Let $D$ be the midpoint of $O O^{\\prime}$. It is known that $\\angle G D O=\\angle G A H=\\varepsilon$. Let $F$ be the midpoint of $H G$. Then $H G=F O$ (Euler line). Let $\\angle G O^{\\prime} H=\\delta$. We then have to show $\\delta=\\varepsilon$ iff $\\angle C A B=60^{\\circ}$. But by similarity $\\left(\\triangle G D O \\sim \\triangle F O^{\\prime} O\\right.$ ) we have $\\angle F O^{\\prime} O=\\varepsilon$. Consider the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$. By the sine law and since the segments $H G$ and $F O$ are of equal length we deduce that the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$ are symmetric with respect to the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$. Obviously, $O^{\\prime}$ is the common point of these two circles. Hence $O^{\\prime}$ must be fixed after the symmetry about the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$ so we have $\\varepsilon=\\delta$ iff $\\triangle H O O^{\\prime}$ is isosceles. But $H O^{\\prime}=H^{\\prime} O=R$, and so\n\n$$\n\\varepsilon=\\delta \\Longleftrightarrow O O^{\\prime}=R \\Longleftrightarrow O D=\\frac{R}{2} \\Longleftrightarrow \\cos \\angle C A B=\\frac{1}{2} \\Longleftrightarrow \\angle C A B=60^{\\circ} \\text {. }\n$$']"			['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proof	Geometry	Math	English
5	"In the diagram, two circles are tangent to each other at point $B$. A straight line is drawn through $B$ cutting the two circles at $A$ and $C$, as shown. Tangent lines are drawn to the circles at $A$ and $C$. Prove that these two tangent lines are parallel.

<image_1>"	"['Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nDesignate two points $W$ and $X$ on either side of $A$ on one tangent line, and two points $Y$ and $Z$ on either side of $C$ on the other tangent line.\n\n<img_3553>\n\nLet $\\angle X A B=\\theta$.\n\nSince $W X$ is tangent to the circle with centre $O_{1}$ at $A$, then $O_{1} A$ is perpendicular to $W X$, so $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince $O_{1} A=O_{1} B$ because both are radii, then $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} B A=$ $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince the two circles are tangent at $B$, then the line segment joining $O_{1}$ and $O_{2}$ passes through $B$, ie. $O_{1} B O_{2}$ is a straight line segment.\n\nThus, $\\angle O_{2} B C=\\angle O_{1} B A=90^{\\circ}-\\theta$, by opposite angles.\n\nSince $O_{2} B=O_{2} C$, then similarly to above, $\\angle O_{2} C B=\\angle O_{2} B C=90^{\\circ}-\\theta$.\n\nSince $Y Z$ is tangent to the circle with centre $O_{2}$ at $C$, then $O_{2} C$ is perpendicular to $Y Z$. Thus, $\\angle Y C B=90^{\\circ}-\\angle O_{2} C B=\\theta$.\n\nSince $\\angle X A B=\\angle Y C B$, then $W X$ is parallel to $Y Z$, by alternate angles, as required.'
 'Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nSince $A O_{1}$ and $B O_{1}$ are radii of the same circle, $A O_{1}=B O_{1}$ so $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} A B=\\angle O_{1} B A$.\n\n<img_3245>\n\nSince $\\mathrm{BO}_{2}$ and $\\mathrm{CO}_{2}$ are radii of the same circle, $B O_{2}=C_{2}$ so $\\triangle B O_{2} C$ is isosceles, so $\\angle O_{2} B C=\\angle O_{2} C B$.\n\nSince the two circles are tangent at $B$, then $O_{1} B O_{2}$ is a line segment (ie. the line segment joining $O_{1}$ and $O_{2}$ passes through the point of tangency of the two circles).\n\nSince $O_{1} B O_{2}$ is straight, then $\\angle O_{1} B A=\\angle O_{2} B C$, by opposite angles.\n\nThus, $\\angle O_{1} A B=\\angle O_{1} B A=\\angle O_{2} B C=\\angle O_{2} C B$.\n\nThis tells us that $\\triangle A O_{1} B$ is similar to $\\triangle B O_{2} C$, so $\\angle A O_{1} B=\\angle B O_{2} C$ or $\\angle A O_{1} O_{2}=$ $\\angle C O_{2} O_{1}$.\n\nTherefore, $A O_{1}$ is parallel to $C_{2}$, by alternate angles.\n\nBut $A$ and $C$ are points of tangency, $A O_{1}$ is perpendicular to the tangent line at $A$ and $\\mathrm{CO}_{2}$ is perpendicular to the tangent line at $C$.\n\nSince $A O_{1}$ and $C O_{2}$ are parallel, then the two tangent lines must be parallel.']"			['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64bW/HH2WJgdyW08nf1RAin86nsf2ctH3eZq2vajeSk5ZoVSLd9d24/rXtdFAHA6d8G/AumlGXQ0uJF/juZXkz9VJ2/pXXadoWk6Qu3TdLs7IelvAsf8AIVoUUAIPpS0UUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQB//Z']	Multimodal	Competition	True				Theorem proof	Geometry	Math	English
6	"A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below:

<image_1>
Prove that there is no positive integer $n$ such that $f(n)=2005$ and there are infinitely many positive integers $n$ such that $f(n)=f(2005)$."	['First, we calculate $f(n)$ for $n$ from 1 to 32 , to get a feeling for what happens. We obtain $1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11,1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11$. This will help us to establish some patterns.\n\nNext, we establish two recursive formulas for $f(n)$.\n\nFirst, from our pattern, it looks like $f(2 m)=f(2 m-1)$.\n\nWhy is this true in general?\n\nConsider a row of $2 m$ lockers.\n\nOn the first pass through, Josephine shuts all of the even numbered lockers, leaving open lockers $1,3, \\ldots, 2 m-1$.\n\nThese are exactly the same open lockers as if she had started with $2 m-1$ lockers in total. Thus, as she starts her second pass from right to left, the process will be the same now whether she started with $2 m$ lockers or $2 m-1$ lockers.\n\nTherefore, $f(2 m)=f(2 m-1)$.\n\nThis tells us that we need only focus on the values of $f(n)$ where $n$ is odd.\n\nSecondly, we show that $f(2 m-1)=2 m+1-2 f(m)$.\n\n(It is helpful to connect $n=2 m-1$ to a smaller case.)\n\nWhy is this formula true?\n\nStarting with $2 m-1$ lockers, the lockers left open after the first pass are $1,3, \\ldots, 2 m-1$, ie. $m$ lockers in total.\n\nSuppose $f(m)=p$. As Josephine begins her second pass, which is from right to left, we can think of this as being like the first pass through a row of $m$ lockers.\n\nThus, the last open locker will be the $p$ th locker, counting from the right hand end, from the list $1,3, \\ldots, 2 m-1$.\n\nThe first locker from the right is $2 m-1=2 m+1-2(1)$, the second is $2 m-3=2 m+1-2(2)$, and so on, so the $p$ th locker is $2 m+1-2 p$.\n\nTherefore, the final open locker is $2 m+1-2 p$, ie. $f(2 m-1)=2 m+1-2 p=2 m+1-2 f(m)$.\n\nUsing these two formulae repeatedly,\n\n$$\n\\begin{aligned}\nf(4 k+1) & =f(2(2 k+1)-1) \\\\\n& =2(2 k+1)+1-2 f(2 k+1) \\\\\n& =4 k+3-2 f(2(k+1)-1) \\\\\n& =4 k+3-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+3-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-3\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\nf(4 k+3) & =f(2(2 k+2)-1) \\\\\n& =2(2 k+2)+1-2 f(2 k+2) \\\\\n& =4 k+5-2 f(2 k+1) \\\\\n& =4 k+5-2 f(2(k+1)-1) \\\\\n& =4 k+5-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+5-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-1\n\\end{aligned}\n$$\n\nFrom our initial list of values of $f(n)$, it appears as if $f(n)$ cannot leave a remainder of 5 or 7 when divided by 8 . So we use these recursive relations once more to try to establish this:\n\n$$\n\\begin{aligned}\nf(8 l+1) & =4 f(2 l+1)-3 \\quad(\\text { since } 8 l+1=4(2 l)+1) \\\\\n& =4(2 l+3-2 f(l+1))-3 \\\\\n& =8 l+9-8 f(l+1) \\\\\n& =8(l-f(l+1))+9 \\\\\nf(8 l+3) & =4 f(2 l+1)-1 \\quad(\\text { since } 8 l+3=4(2 l)+3) \\\\\n& =4(2 l+3-2 f(l+1))-1 \\\\\n& =8 l+11-8 f(l+1) \\\\\n& =8(l-f(l+1))+11\n\\end{aligned}\n$$\n\nSimilarly, $f(8 l+5)=8 l+9-8 f(l+1)$ and $f(8 l+7)=8 l+11-8 f(l+1)$.\n\nTherefore, since any odd positive integer $n$ can be written as $8 l+1,8 l+3,8 l+5$ or $8 l+7$, then for any odd positive integer $n, f(n)$ is either 9 more or 11 more than a multiple of 8 . Therefore, for any odd positive integer $n, f(n)$ cannot be 2005, since 2005 is not 9 more or 11 more than a multiple of 8 .\n\nThus, for every positive integer $n, f(n) \\neq 2005$, since we only need to consider odd values of $n$.\n\nNext, we show that there are infinitely many positive integers $n$ such that $f(n)=f(2005)$. We do this by looking at the pattern we initially created and conjecturing that\n\n$$\nf(2005)=f\\left(2005+2^{2 a}\\right)\n$$\n\nif $2^{2 a}>2005$. (We might guess this by looking at the connection between $f(1)$ and $f(3)$ with $f(5)$ and $f(7)$ and then $f(1)$ through $f(15)$ with $f(17)$ through $f(31)$. In fact, it appears to be true that $f\\left(m+2^{2 a}\\right)=f(m)$ if $2^{2 a}>m$.)\n\n\n\nUsing our formulae from above,\n\n$$\n\\begin{aligned}\n& f\\left(2005+2^{2 a}\\right)=4 f\\left(502+2^{2 a-2}\\right)-3 \\quad\\left(2005+2^{2 a}=4\\left(501+2^{2 a-2}\\right)+1\\right) \\\\\n& =4 f\\left(501+2^{2 a-2}\\right)-3 \\\\\n& =4\\left(4 f\\left(126+2^{2 a-4}\\right)-3\\right)-3 \\quad\\left(501+2^{2 a-2}=4\\left(125+2^{2 a-4}\\right)+1\\right) \\\\\n& =16 f\\left(126+2^{2 a-4}\\right)-15 \\\\\n& =16 f\\left(125+2^{2 a-4}\\right)-15 \\\\\n& =16\\left(4 f\\left(32+2^{2 a-6}\\right)-3\\right)-15 \\\\\n& =64 f\\left(32+2^{2 a-6}\\right)-63 \\\\\n& =64 f\\left(31+2^{2 a-6}\\right)-63 \\\\\n& =64\\left(4 f\\left(8+2^{2 a-8}\\right)-1\\right)-63 \\\\\n& =256 f\\left(8+2^{2 a-8}\\right)-127 \\\\\n& =256 f\\left(7+2^{2 a-8}\\right)-127 \\\\\n& =256\\left(4 f\\left(2+2^{2 a-10}\\right)-1\\right)-127 \\quad\\left(7+2^{2 a-8}=4\\left(1+2^{2 a-10}\\right)+3\\right) \\\\\n& =1024 f\\left(2+2^{2 a-10}\\right)-383 \\\\\n& =1024 f\\left(1+2^{2 a-10}\\right)-383 \\\\\n& \\left(125+2^{2 a-4}=4\\left(31+2^{2 a-6}\\right)+1\\right) \\\\\n& \\left(31+2^{2 a-6}=4\\left(7+2^{2 a-8}\\right)+3\\right)\n\\end{aligned}\n$$\n\n(Notice that we could have removed the powers of 2 from inside the functions and used this same approach to show that $f(2005)=1024 f(1)-383=641$.)\n\nBut, $f\\left(2^{2 b}+1\\right)=1$ for every positive integer $b$.\n\nWhy is this true? We can prove this quickly by induction.\n\nFor $b=1$, we know $f(5)=1$.\n\nAssume that the result is true for $b=B-1$, for some positive integer $B \\geq 2$.\n\nThen $f\\left(2^{2 B}+1\\right)=f\\left(4\\left(2^{2 B-2}\\right)+1\\right)=4 f\\left(2^{2 B-2}+1\\right)-3=4(1)-3=1$ by our induction hypothesis.\n\nTherefore, if $a \\geq 6$, then $f\\left(1+2^{2 a-10}\\right)=f\\left(1+2^{2(a-5)}\\right)=1$ so\n\n$$\nf\\left(2005+2^{2 a}\\right)=1024(1)-383=641=f(2005)\n$$\n\nso there are infinitely many integers $n$ for which $f(n)=f(2005)$.']			['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proof	Combinatorics	Math	English
7	"In the diagram, $C$ lies on $B D$. Also, $\triangle A B C$ and $\triangle E C D$ are equilateral triangles. If $M$ is the midpoint of $B E$ and $N$ is the midpoint of $A D$, prove that $\triangle M N C$ is equilateral.

<image_1>"	"['Consider $\\triangle B C E$ and $\\triangle A C D$.\n\n<img_3765>\n\nSince $\\triangle A B C$ is equilateral, then $B C=A C$.\n\nSince $\\triangle E C D$ is equilateral, then $C E=C D$.\n\nSince $B C D$ is a straight line and $\\angle E C D=60^{\\circ}$, then $\\angle B C E=180^{\\circ}-\\angle E C D=120^{\\circ}$.\n\nSince $B C D$ is a straight line and $\\angle B C A=60^{\\circ}$, then $\\angle A C D=180^{\\circ}-\\angle B C A=120^{\\circ}$.\n\nTherefore, $\\triangle B C E$ is congruent to $\\triangle A C D$ (""side-angle-side"").\n\nSince $\\triangle B C E$ and $\\triangle A C D$ are congruent and $C M$ and $C N$ are line segments drawn from the corresponding vertex ( $C$ in both triangles) to the midpoint of the opposite side, then $C M=C N$.\n\nSince $\\angle E C D=60^{\\circ}$, then $\\triangle A C D$ can be obtained by rotating $\\triangle B C E$ through an angle of $60^{\\circ}$ clockwise about $C$.\n\nThis means that after this $60^{\\circ}$ rotation, $C M$ coincides with $C N$.\n\nIn other words, $\\angle M C N=60^{\\circ}$.\n\nBut since $C M=C N$ and $\\angle M C N=60^{\\circ}$, then\n\n$$\n\\angle C M N=\\angle C N M=\\frac{1}{2}\\left(180^{\\circ}-\\angle M C N\\right)=60^{\\circ}\n$$\n\nTherefore, $\\triangle M N C$ is equilateral, as required.'
 'We prove that $\\triangle M N C$ is equilateral by introducing a coordinate system.\n\nSuppose that $C$ is at the origin $(0,0)$ with $B C D$ along the $x$-axis, with $B$ having coordinates $(-4 b, 0)$ and $D$ having coordinates $(4 d, 0)$ for some real numbers $b, d>0$.\n\nDrop a perpendicular from $E$ to $P$ on $C D$.\n\n<img_3660>\n\nSince $\\triangle E C D$ is equilateral, then $P$ is the midpoint of $C D$.\n\nSince $C$ has coordinates $(0,0)$ and $D$ has coordinates $(4 d, 0)$, then the coordinates of $P$ are $(2 d, 0)$.\n\nSince $\\triangle E C D$ is equilateral, then $\\angle E C D=60^{\\circ}$ and so $\\triangle E P C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle and so $E P=\\sqrt{3} C P=2 \\sqrt{3} d$.\n\nTherefore, the coordinates of $E$ are $(2 d, 2 \\sqrt{3} d)$.\n\nIn a similar way, we can show that the coordinates of $A$ are $(-2 b, 2 \\sqrt{3} b)$.\n\nNow $M$ is the midpoint of $B(-4 b, 0)$ and $E(2 d, 2 \\sqrt{3} d)$, so the coordinates of $M$ are $\\left(\\frac{1}{2}(-4 b+2 d), \\frac{1}{2}(0+2 \\sqrt{3} d)\\right)$ or $(-2 b+d, \\sqrt{3} d)$.\n\nAlso, $N$ is the midpoint of $A(-2 b, 2 \\sqrt{3} b)$ and $D(4 d, 0)$, so the coordinates of $N$ are $\\left(\\frac{1}{2}(-2 b+4 d), \\frac{1}{2}(2 \\sqrt{3} b+0)\\right)$ or $(-b+2 d, \\sqrt{3} b)$.\n\nTo show that $\\triangle M N C$ is equilateral, we show that $C M=C N=M N$ or equivalently that $C M^{2}=C N^{2}=M N^{2}$ :\n\n$$\n\\begin{aligned}\nC M^{2} & =(-2 b+d-0)^{2}+(\\sqrt{3} d-0)^{2} \\\\\n& =(-2 b+d)^{2}+(\\sqrt{3} d)^{2} \\\\\n& =4 b^{2}-4 b d+d^{2}+3 d^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nC N^{2} & =(-b+2 d-0)^{2}+(\\sqrt{3} b-0)^{2} \\\\\n& =(-b+2 d)^{2}+(\\sqrt{3} b)^{2} \\\\\n& =b^{2}-4 b d+4 d^{2}+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nM N^{2} & =((-2 b+d)-(-b+2 d))^{2}+(\\sqrt{3} d-\\sqrt{3} b)^{2} \\\\\n& =(-b-d)^{2}+3(d-b)^{2} \\\\\n& =b^{2}+2 b d+d^{2}+3 d^{2}-6 b d+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2}\n\\end{aligned}\n$$\n\nTherefore, $C M^{2}=C N^{2}=M N^{2}$ and so $\\triangle M N C$ is equilateral, as required.\n\n\n\n#']"			['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']	Multimodal	Competition	True				Theorem proof	Geometry	Math	English
8	"In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.

<image_1>
Prove that $P Q R S$ is a rectangle."	['In a parallelogram opposite angles are equal.\n\nSince $D F$ and $B E$ bisect the two angles, let $\\angle A D F=\\angle C D F=\\angle A B E=\\angle C B E$\n\n$$\n=x \\text { (in degrees) }\n$$\n\nAlso $\\angle C D F=\\angle A F D=x$ (alternate angles)\n\nLet $\\angle D A M=\\angle B A M=\\angle D C N=\\angle B C N=y$ (in degrees)\n\n\n\nFor any parallelogram, any two consecutive angles add to $180^{\\circ}, \\therefore 2 x+2 y=180$ or, $x+y=90$.\n\n<img_3560>\n\nTherefore in $\\triangle P A F, \\angle A P F=90^{\\circ}$.\n\n<img_3544>\n\nUsing similar reasoning and properties of parallel lines we get right angles at $Q, R$ and $S$.\n\nThus $P Q R S$ is a rectangle.']			['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']	Multimodal	Competition	True				Theorem proof	Geometry	Math	English
9	"In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.

<image_1>
Prove that $P R=a-b$."	['Since $A M$ is a bisector of $\\angle D A B$, let $\\angle D A M=\\angle B A M=y$.\n\nAlso, $\\angle D M A=y$ (alternate angles)\n\nThis implies that $\\triangle A D M$ is isosceles.\n\nUsing the same reasoning in $\\triangle C B N$, we see that it is also isosceles and so the diagram may now be labelled as:\n\n<img_3709>\n\n$A N=a-b$\n\nThus $\\triangle A D M$ and $\\triangle C B N$ are identical isosceles triangles.\n\nAlso, $A M \\| N C$ (corresponding angles)\n\nor, $A P \\| N R$.\n\nBy using properties of isosceles triangles (or congruency), $A P=N R$ implying that $A P R N$ is a parallelogram.\n\nThus $A N=P R$ and since $A N=a-b, P R=a-b$ (as required)']			['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']	Multimodal	Competition	True				Theorem proof	Geometry	Math	English
10	"An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$.

<image_1>"	['Let $\\angle R Q C=\\theta$ and from $S$ draw a line perpendicular to the base at $P$.\n\nThen $\\angle T Q B=180-(90+\\theta)=90-\\theta$.\n\nLet $s$ be the length of the side of the square.\n\nFrom $R$ draw a line perpendicular to $B C$ at $D$ and then through $S$ draw a line parallel to $B C$. From $R$ draw a line perpendicular to this line at E.\n\n<img_3178>\n\nFrom $\\triangle R Q D, R D=s \\sin \\theta$.\n\nSince $\\angle Q R D=90-\\theta$ then $\\angle S R E=\\theta$.\n\nFrom $\\triangle S E R, E R=s \\cos \\theta$.\n\nThe perpendicular distance from $S$ to $B C$ is $R D+E R=s \\sin \\theta+s \\cos \\theta$ which we must now show is a constant.\n\nWe can now take each of the lengths $D C, D Q, P F, F B$ and express them in terms of $s$.\n\nFrom $\\triangle R D C$ which is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, $\\frac{D C}{R D}=\\frac{1}{\\sqrt{3}}$.\n\nSince $R D=s \\sin \\theta$ (from above)\n\n$$\nD C=\\frac{1}{\\sqrt{3}}(s \\sin \\theta)=\\frac{\\sqrt{3}}{3} s \\sin \\theta\n$$\n\n\n\nFrom $\\Delta R D Q, \\frac{Q D}{R Q}=\\cos \\theta, Q D=s \\cos \\theta$.\n\nFrom $\\triangle T F Q, \\sin \\theta=\\frac{F Q}{s}$ and $\\cos \\theta=\\frac{T F}{s}$. or, $F Q=s \\sin \\theta$ and $T F=s \\sin \\theta$.\n\nFrom $\\triangle T F B, \\frac{B F}{T F}=\\frac{1}{\\sqrt{3}}, B F=\\frac{1}{\\sqrt{3}} T F=\\frac{1}{\\sqrt{3}} s \\cos \\theta=\\frac{\\sqrt{3}}{3} s \\cos \\theta$.\n\nSince $D C+Q D+F Q+B F=2, \\frac{\\sqrt{3}}{3} s \\sin \\theta+s \\cos \\theta+s \\sin \\theta+\\frac{\\sqrt{3}}{3} s \\cos \\theta=2$.\n\n$$\n\\begin{aligned}\n\\frac{\\sqrt{3}}{3}(s \\cos \\theta+s \\sin \\theta)+(s \\cos \\theta+s \\sin \\theta) & =2 \\\\\ns \\cos \\theta+s \\sin \\theta & =\\frac{2}{\\left(\\frac{\\sqrt{3}}{3}+1\\right)}\n\\end{aligned}\n$$\n\nThus $s \\cos \\theta+s \\sin \\theta$ is a constant and the path traced out by $S$ is a straight line parallel to $B 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proof	Geometry	Math	English
11	"In the diagram, line segment $F C G$ passes through vertex $C$ of square $A B C D$, with $F$ lying on $A B$ extended and $G$ lying on $A D$ extended. Prove that $\frac{1}{A B}=\frac{1}{A F}+\frac{1}{A G}$.

<image_1>"	"['Without loss of generality, suppose that square $A B C D$ has side length 1 .\n\nSuppose next that $B F=a$ and $\\angle C F B=\\theta$.\n\nSince $\\triangle C B F$ is right-angled at $B$, then $\\angle B C F=90^{\\circ}-\\theta$.\n\nSince $G C F$ is a straight line, then $\\angle G C D=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nTherefore, $\\triangle G D C$ is similar to $\\triangle C B F$, since $\\triangle G D C$ is right-angled at $D$.\n\nThus, $\\frac{G D}{D C}=\\frac{B C}{B F}$ or $\\frac{G D}{1}=\\frac{1}{a}$ or $G D=\\frac{1}{a}$.\n\nSo $A F=A B+B F=1+a$ and $A G=A D+D G=1+\\frac{1}{a}=\\frac{a+1}{a}$.\n\nThus, $\\frac{1}{A F}+\\frac{1}{A G}=\\frac{1}{1+a}+\\frac{a}{a+1}=\\frac{a+1}{a+1}=1=\\frac{1}{A B}$, as required.'
 'We attach a set of coordinate axes to the diagram, with $A$ at the origin, $A G$ lying along the positive $y$-axis and $A F$ lying along the positive $x$-axis.\n\nWithout loss of generality, suppose that square $A B C D$ has side length 1 , so that $C$ has coordinates $(1,1)$. (We can make this assumption without loss of generality, because if the square had a different side length, then each of the lengths in the problem would be scaled by the same factor.)\n\n\n\nSuppose that the line through $G$ and $F$ has slope $m$.\n\nSince this line passes through $(1,1)$, its equation is $y-1=m(x-1)$ or $y=m x+(1-m)$. The $y$-intercept of this line is $1-m$, so $G$ has coordinates $(0,1-m)$.\n\nThe $x$-intercept of this line is $\\frac{m-1}{m}$, so $F$ has coordinates $\\left(\\frac{m-1}{m}, 0\\right)$. (Note that $m \\neq 0$ as the line cannot be horizontal.)\n\nTherefore,\n\n$$\n\\frac{1}{A F}+\\frac{1}{A G}=\\frac{m}{m-1}+\\frac{1}{1-m}=\\frac{m}{m-1}+\\frac{-1}{m-1}=\\frac{m-1}{m-1}=1=\\frac{1}{A B}\n$$\n\nas required.'
 'Join $A$ to $C$.\n\nWe know that the sum of the areas of $\\triangle G C A$ and $\\triangle F C A$ equals the area of $\\triangle G A F$.\n\nThe area of $\\triangle G C A$ (thinking of $A G$ as the base) is $\\frac{1}{2}(A G)(D C)$, since $D C$ is perpendicular to $A G$.\n\nSimilarly, the area of $\\triangle F C A$ is $\\frac{1}{2}(A F)(C B)$.\n\nAlso, the area of $\\triangle G A F$ is $\\frac{1}{2}(A G)(A F)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A G)(D C)+\\frac{1}{2}(A F)(C B) & =\\frac{1}{2}(A G)(A F) \\\\\n\\frac{(A G)(D C)}{(A G)(A F)(A B)}+\\frac{(A F)(C B)}{(A G)(A F)(A B)} & =\\frac{(A G)(A F)}{(A G)(A F)(A B)} \\\\\n\\frac{1}{A F}+\\frac{1}{A G} & =\\frac{1}{A B}\n\\end{aligned}\n$$\n\nas required, since $A B=D C=C B$.']"			['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proof	Geometry	Math	English
12	"A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$.

<image_1>"	['Since both the circle with its centre on the $y$-axis and the graph of $y=|x|$ are symmetric about the $y$-axis, then for each point of intersection between these two graphs, there should be a corresponding point of intersection symmetrically located across the $y$-axis. Thus, since there are exactly three points of intersection, then one of these points must be on the $y$-axis, ie. has $x$-coordinate 0 . Since this point is on the graph of $y=|x|$, then this point must be $(0,0)$.\n\nSince the circle has centre on the $y$-axis (say, has coordinates $(0, b)$ ), then its radius is equal to $b$ (and $b$ must be positive for there to be three points of intersection). So the circle has equation $x^{2}+(y-b)^{2}=b^{2}$. Where are the other two points of intersection? We consider the points with $x$ positive and use symmetry to get the other point of intersection.\n\n\n\nWhen $x \\geq 0$, then $y=|x|$ has equation $y=x$. Substituting into the equation of the circle,\n\n$$\n\\begin{aligned}\nx^{2}+(x-b)^{2} & =b^{2} \\\\\n2 x^{2}-2 b x & =0 \\\\\n2 x(x-b) & =0\n\\end{aligned}\n$$\n\nTherefore, the points of intersection are $(0,0)$ and $(b, b)$ on the positive side of the $y$-axis, and so at the point $(-b, b)$ on the negative side of the $y$ axis.\n\nThus the points $O, A$ and $B$ are the points $(0,0)$, $(b, b)$ and $(-b, b)$.\n\n<img_3620>\n\nSince the radius of the circle is $b$, then the area of the circle is $\\pi b^{2}$.\n\nTriangle $O A B$ has a base from $(-b, b)$ to $(b, b)$ of length $2 b$, and a height from the line $y=b$ to the point $(0,0)$ of length $b$, and so an area of $\\frac{1}{2} b(2 b)=b^{2}$.\n\nTherefore, the ratio of the area of the triangle to the area of the circle is $b^{2}: \\pi b^{2}=1: \\pi$.']			['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proof	Geometry	Math	English
13	"In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$.

<image_1>"	"['Since $M$ is the midpoint of a diameter of the circle, $M$ is the centre of the circle.\n\nJoin $P$ to $M$. Since $Q P$ is tangent to the circle, $P M$ is perpendicular to $Q P$.\n\nSince $P M$ and $B M$ are both radii of the circle, then $P M=M B$.\n\n<img_3432>\n\nTherefore, $\\triangle Q P M$ and $\\triangle Q B M$ are congruent (Hypotenuse - Side).\n\nThus, let $\\angle M Q B=\\angle M Q P=\\theta$. So $\\angle Q M B=\\angle Q M P=90^{\\circ}-\\theta$\n\nThen $\\angle P M C=180^{\\circ}-\\angle P M Q-\\angle B M Q=180^{\\circ}-\\left(90^{\\circ}-\\theta\\right)-\\left(90^{\\circ}-\\theta\\right)=2 \\theta$.\n\nBut $\\triangle P M C$ is isosceles with $P M=M C$ since $P M$ and $M C$ are both radii.\n\nTherefore, $\\angle C P M=\\frac{1}{2}\\left(180^{\\circ}-\\angle P M C\\right)=90^{\\circ}-\\theta$.\n\nBut then $\\angle C P M=\\angle P M Q$, and since $P M$ is a transversal between $A C$ and $Q M$, then $Q M$ is parallel to $A C$ because of equal alternating angles.'
 'Join $M$ to $P$ and $B$ to $P$.\n\nSince $Q P$ and $Q B$ are tangents to the circle coming from the same point, they have the same length. Since $Q M$ joins the point of intersection of the tangents to the centre of the circle, then by symmetry, $\\angle P Q M=\\angle B Q M$ and $\\angle P M Q=\\angle B M Q$. So let $\\angle P Q M=\\angle B Q M=x$ and $\\angle P M Q=\\angle B M Q=y$.\n\n<img_3687>\n\nLooking at $\\triangle Q M B$, we see that $x+y=90^{\\circ}$, since $\\triangle Q M B$ is right-angled.\n\nNow if we consider the chord $P B$, we see that its central angle is $2 y$, so any angle that it subtends on the circle (eg. $\\angle P C B$ ) is equal to $y$.\n\nThus, $\\angle A C B=\\angle Q M B$, so $Q M$ is parallel to $A C$.'
 'Join $P B$.\n\nSince $Q P$ is tangent to the circle, then by the Tangent-Chord Theorem, $\\angle Q P B=\\angle P C B=x$ (ie. the inscribed angle of a chord is equal to the angle between the tangent and chord.\n\n<img_3658>\n\nSince $B C$ is a diameter of the circle, then $\\angle C P B=90^{\\circ}$ and so $\\angle A P B=90^{\\circ}$, whence\n\n$\\angle A P Q=90^{\\circ}-\\angle Q P B=90^{\\circ}-x$.\n\nLooking at $\\triangle A B C$, we see that $\\angle P A Q=90^{\\circ}-x$, so $\\angle P A Q=\\angle A P Q$, and so $A Q=Q P$.\n\nBut $Q P$ and $Q B$ are both tangents to the circle $(Q B$ is tangent since it is perpendicular to a radius), so $Q P=Q B$.\n\nBut then $A Q=Q B$ and $B M=M C$, so $Q$ is the midpoint of $A B$ and $M$ is the midpoint of $B C$. Thus we can conclude that $Q M$ is parallel to $A C$.\n\n(To justify this last statement, we can show very easily that $\\triangle Q B M$ is similar to $\\triangle A B C$, and so show that $\\angle C A B=\\angle M Q B$.)']"			['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proof	Geometry	Math	English
14	"A large square $A B C D$ is drawn, with a second smaller square $P Q R S$ completely inside it so that the squares do not touch. Line segments $A P, B Q, C R$, and $D S$ are drawn, dividing the region between the squares into four nonoverlapping convex quadrilaterals, as shown. If the sides of $P Q R S$ are not parallel to the sides of $A B C D$, prove that the sum of the areas of quadrilaterals $A P S D$ and $B C R Q$ equals the sum of the areas of quadrilaterals $A B Q P$ and $C D S R$. (Note: A convex quadrilateral is a quadrilateral in which the measure of each of the four interior angles is less than $180^{\circ}$.)

<image_1>"	"['We begin by ""boxing in"" square $P Q R S$ by drawing horizontal and vertical lines through its vertices to form rectangle $W X Y Z$, as shown. (Because the four quadrilaterals $A B Q P$, $B C R Q, C D S R$, and $D A P S$ are convex, there will not be any configurations that look substantially different from this the diagram below.) We also label the various areas.\n\n<img_3366>\n\nSince $W X$ is parallel to $A B$, then quadrilateral $A B X W$ is a trapezoid. Similarly, quadrilaterals $B C Y X, C D Z Y$, and $D A W Z$ are trapezoids.\n\nWe use the notation $|A B Q P|$ to denote the area of quadrilateral $A B Q P$, and similar notation for other areas.\n\nSuppose that the side length of square $A B C D$ is $x$ and the side length of square $P Q R S$ is $y$.\n\nAlso, we let $\\angle W P Q=\\theta$.\n\nSince each of $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ is right-angled and each of the four angles of square $P Q R S$ is $90^{\\circ}$, then $\\angle W P Q=\\angle X Q R=\\angle Y R S=\\angle Z S P=\\theta$. This is because, for example,\n\n$\\angle X Q R=180^{\\circ}-\\angle P Q R-\\angle W Q P=90^{\\circ}-\\left(180^{\\circ}-\\angle P W Q-\\angle W P Q\\right)=90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$\n\nThis fact, together with the fact that $P Q=Q R=R S=S P=y$, allows us to conclude that the four triangles $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ are congruent.\n\nIn particular, this tells us\n\n* the four areas labelled $e, f, g$ and $h$ are equal (that is, $e=f=g=h$ ),\n* $P Z=Q W=R X=S Y=y \\sin \\theta$, and\n\n$* W P=X Q=Y R=Z S=y \\cos \\theta$.\n\n\n\nCombining these last two facts tells us that $W Z=X W=Y X=Z Y$, since, for example, $W Z=W P+P Z=Z S+S Y=Z Y$. In other words, $W X Y Z$ is a square, with side length $z$, say.\n\nNext, we show that $(a+r)+(c+n)$ is equal to $(b+m)+(d+s)$.\n\nNote that the sum of these two quantities is the total area between square $A B C D$ and square $W X Y Z$, so equals $x^{2}-z^{2}$.\n\nThus, to show that the quantities are equal, it is enough to show that $(a+r)+(c+n)$ equals $\\frac{1}{2}\\left(x^{2}-z^{2}\\right)$.\n\nLet the height of trapezoid $A B X W$ be $k$ and the height of trapezoid $Z Y C D$ be $l$.\n\nThen $|A B X W|=a+r=\\frac{1}{2} k(A B+W X)=\\frac{1}{2} k(x+z)$.\n\nAlso, $|Z Y C D|=c+n=\\frac{1}{2} l(D C+Z Y)=\\frac{1}{2} l(x+z)$.\n\nSince $A B, W X, Z Y$, and $D C$ are parallel, then the sum of the heights of trapezoid $A B X W$, square $W X Y Z$, and trapezoid $Z Y C D$ equals the height of square $A B C D$, so $k+z+l=x$, or $k+l=x-z$.\n\nTherefore,\n\n$$\n(a+r)+(c+n)=\\frac{1}{2} k(x+z)+\\frac{1}{2} l(x+z)=\\frac{1}{2}(x+z)(k+l)=\\frac{1}{2}(x+z)(x-z)=\\frac{1}{2}\\left(x^{2}-z^{2}\\right)\n$$\n\nas required.\n\nTherefore, $(a+r)+(c+n)=(b+m)+(d+s)$. We label this equation $(*)$.\n\nNext, we show that $r+n=m+s$.\n\nNote that $r=|\\triangle Q X B|$. This triangle can be viewed as having base $Q X$ and height equal to the height of trapezoid $A B X W$, or $k$.\n\nThus, $r=\\frac{1}{2}(y \\cos \\theta) k$.\n\nNote that $n=|\\triangle S Z D|$. This triangle can be viewed as having base $S Z$ and height equal to the height of trapezoid $Z Y C D$, or $l$.\n\nThus, $n=\\frac{1}{2}(y \\cos \\theta) l$.\n\nCombining these facts, we obtain\n\n$$\nn+r=\\frac{1}{2}(y \\cos \\theta) k+\\frac{1}{2}(y \\cos \\theta) l=\\frac{1}{2} y \\cos \\theta(k+l)=\\frac{1}{2} y \\cos \\theta(x-z)\n$$\n\nWe note that this sum depends only on the side lengths of the squares and the angle of rotation of the inner square, so is independent of the position of the inner square within the outer square.\n\nThis means that we can repeat this analysis to obtain the same expression for $m+s$.\n\nTherefore, $n+r=m+s$. We label this equation $(* *)$.\n\nWe subtract $(*)-(* *)$ to obtain $a+c=b+d$.\n\nFinally, we can combine all of this information:\n\n$$\n\\begin{aligned}\n& (|A B Q P|+|C D S R|)-(|B C R Q|+|A P S D|) \\\\\n& \\quad=(a+e+s+c+g+m)-(b+f+r+d+h+n) \\\\\n& \\quad=((a+c)-(b+d))+((m+s)-(n+r))+((e+g)-(f+h)) \\\\\n& \\quad=0+0+0\n\\end{aligned}\n$$\n\nsince $a+c=b+d$ and $n+r=m+s$ and $e=f=g=h$.\n\nTherefore, $|A B Q P|+|C D S R|=|B C R Q|+|A P S D|$, as required.']"			['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proof	Geometry	Math	English
15	"In triangle $A B C, \angle A B C=90^{\circ}$. Rectangle $D E F G$ is inscribed in $\triangle A B C$, as shown. Squares $J K G H$ and $M L F N$ are inscribed in $\triangle A G D$ and $\triangle C F E$, respectively. If the side length of $J H G K$ is $v$, the side length of $M L F N$ is $w$, and $D G=u$, prove that $u=v+w$.

<image_1>"	['Let $\\angle B A C=\\theta$. Then by parallel lines, $\\angle D J H=\\angle B D E=\\theta$.\n\nThus, $\\angle B E D=90^{\\circ}-\\theta$ and so\n\n$\\angle N E M=\\theta$ since\n\n$\\angle D E F=90^{\\circ}$.\n\nSince $D G=u$ and $H G=v$,\n\nthen $D H=u-v$.\n\n<img_3789>\n\nSimilarly, $E N=u-w$.\n\nLooking at $\\triangle D H J$ and $\\triangle M N E$, we see that $\\tan \\theta=\\frac{u-v}{v}$ and $\\tan \\theta=\\frac{w}{u-w}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{u-v}{v} & =\\frac{w}{u-w} \\\\\n(u-v)(u-w) & =v w \\\\\nu^{2}-u v-u w+v w & =v w \\\\\nu(u-v-w) & =0\n\\end{aligned}\n$$\n\nand since $u \\neq 0$, we must have $u-v-w=0$ or $u=v+w$.\n\n[Note: If $u=0$, then the height of rectangle $D E F G$ is 0 , ie. $D$ coincides with point $A$ and $E$ coincides with point $C$, which says that we must also have $v=w=0$, ie. the squares have no place to go!]']			['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proof	Geometry	Math	English
16	"In the diagram, quadrilateral $A B C D$ has points $M$ and $N$ on $A B$ and $D C$, respectively, with $\frac{A M}{A B}=\frac{N C}{D C}$. Line segments $A N$ and $D M$ intersect at $P$, while $B N$ and $C M$ intersect at $Q$. Prove that the area of quadrilateral $P M Q N$ equals the sum of the areas of $\triangle A P D$ and $\triangle B Q C$.

<image_1>"	"['We use the notation $|P M Q N|$ to represent the area of quadrilateral $|P M Q N|,|\\triangle A P D|$ to represent the area of $\\triangle A P D$, and so on.\n\nWe want to show that $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$.\n\nThis is equivalent to showing\n\n$$\n|P M Q N|+|\\triangle D P N|+|\\triangle C Q N|=|\\triangle A P D|+|\\triangle D P N|+|\\triangle B Q C|+|\\triangle C Q N|\n$$\n\nwhich is equivalent to showing\n\n$$\n|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|\n$$\n\nsince combining quadrilateral $P M Q N$ with $\\triangle D P N$ and $\\triangle C Q N$ gives $\\triangle D M C$, combining $\\triangle A P D$ with $\\triangle D P N$ gives $\\triangle D A N$, and combining $\\triangle B Q C$ with $\\triangle C Q N$ gives $\\triangle C B N$. Suppose that $D C$ has length $x$ and $D N$ has length $t x$ for some $t$ with $0<t<1$.\n\nThen $N C=D C-D N=x-t x=(1-t) x$.\n\nSuppose also that the height of $A$ above $D C$ is $a$, the height of $B$ above $D C$ is $b$ and the height of $M$ above $D C$ is $m$.\n\n<img_3814>\n\nFigure 1\n\nThen $|\\triangle D A N|=\\frac{1}{2}(t x)(a)$ and $|\\triangle C B N|=\\frac{1}{2}((1-t) x) b$ so\n\n$$\n|\\triangle D A N|+|\\triangle C B N|=\\frac{1}{2}(t x a+(1-t) x b)=\\frac{1}{2} x(t a+(1-t) b)\n$$\n\nAlso, $|\\triangle D M C|=\\frac{1}{2} x m$.\n\nIn order to prove that $|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|$, we need to show that $\\frac{1}{2} x m$ equals\n\n\n\n$\\frac{1}{2} x(t a+(1-t) b)$ which is equivalent to showing that $m$ is equal to $t a+(1-t) b$.\n\nIn Figure 2, we draw a horizontal line from $A$ to $B G$, meeting $M F$ at $R$ and $B G$ at $S$.\n\nSince $M F$ and $B G$ are vertical and $A R S$ is horizontal, then these line segments are perpendicular.\n\nSince $A E=a, M F=m$ and $B G=b$, then $M R=m-a$ and $B S=b-a$.\n\n<img_3348>\n\nFigure 2\n\nNow $\\triangle A R M$ is similar to $\\triangle A S B$, since each is right-angled and they share a common angle at $A$.\n\nTherefore, $\\frac{M R}{B S}=\\frac{A M}{A B}=\\frac{N C}{D C}$.\n\nSince $M R=m-a$ and $B S=b-a$, then $\\frac{M R}{B S}=\\frac{m-a}{b-a}$.\n\nSince $\\frac{A M}{A B}=\\frac{N C}{D C}$, then $\\frac{M R}{B S}=\\frac{(1-t) x}{x}=1-t$.\n\nComparing these two expressions, we obtain $\\frac{m-a}{b-a}=(1-t)$ or $m-a=(b-a)(1-t)$ or $m=a+b(1-t)+(t-1) a=t a+(1-t) b$, as required.\n\nThis concludes the proof, and so $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$, as required.'
 'Let $A M=x$ and $M B=y$. Then $A B=x+y$ and so $\\frac{A M}{A B}=\\frac{x}{x+y}$.\n\nLet $N C=n x$ for some real number $n$.\n\nSince $\\frac{N C}{D C}=\\frac{A M}{A B}$, then $\\frac{n x}{D C}=\\frac{x}{x+y}$ and so $D C=n(x+y)$.\n\nThis tells us that $D N=D C-N C=n(x+y)-n x=n y$.\n\nJoin $M$ to $N$ and label the areas as shown in the diagram:\n\n<img_3372>\n\nWe repeatedly use the fact that triangles with a common height have areas in proportion to the lengths of their bases.\n\nFor example, $\\triangle M D N$ and $\\triangle M N C$ have a common height from line segment to $D C$ to $M$ and so the ratio of their areas equals the ratio of the lengths of their bases.\n\nIn other words, $\\frac{w+r}{u+v}=\\frac{n x}{n y}=\\frac{x}{y}$. Thus, $w+r=\\frac{x}{y}(u+v)$.\n\n\n\nAlso, the ratio of the area of $\\triangle N A M$ to the area of $\\triangle N M B$ equals the ratio of $A M$ to $M B$.\n\nThis gives $\\frac{k+v}{s+w}=\\frac{x}{y}$ or $k+v=\\frac{x}{y}(s+w)$.\n\nNext, we join $A$ to $C$ and relabel the areas divided by this new line segment as shown:\n\n<img_3488>\n\n(The unlabelled triangle adjacent to the one labelled $k_{1}$ has area $k_{2}$ and the unlabelled triangle adjacent to the one labelled $r_{2}$ has area $r_{1}$.)\n\nConsider $\\triangle A N C$ and $\\triangle A D N$.\n\nAs above, the ratio of their areas equals the ratio of their bases.\n\nThus, $\\frac{k_{2}+v_{2}+w_{2}+r_{2}}{z+u}=\\frac{n x}{n y}=\\frac{x}{y}$, and so $k_{2}+v_{2}+w_{2}+r_{2}=\\frac{x}{y}(z+u)$.\n\nConsider $\\triangle C A M$ and $\\triangle C M B$.\n\nAs above, the ratio of their areas equals the ratio of their bases.\n\nThus, $\\frac{k_{1}+v_{1}+w_{1}+r_{1}}{s+t}=\\frac{x}{y}$, and so $k_{1}+v_{1}+w_{1}+r_{1}=\\frac{x}{y}(s+t)$.\n\nAdding $k_{2}+v_{2}+w_{2}+r_{2}=\\frac{x}{y}(z+u)$ and $k_{1}+v_{1}+w_{1}+r_{1}=\\frac{x}{y}(s+t)$ gives\n\n$$\n\\left(k_{1}+k_{2}\\right)+\\left(v_{1}+v_{2}\\right)+\\left(w_{1}+w_{2}\\right)+\\left(r_{1}+r_{2}\\right)=\\frac{x}{y}(s+t+z+u)\n$$\n\nor\n\n$$\nk+v+w+r=\\frac{x}{y}(s+t+z+u)\n$$\n\nSince $k+v=\\frac{x}{y}(s+w)$ and $w+r=\\frac{x}{y}(u+v)$, then\n\n$$\n\\frac{x}{y}(s+w)+\\frac{x}{y}(u+v)=\\frac{x}{y}(s+t+z+u)\n$$\n\nwhich gives\n\n$$\ns+w+u+v=s+t+z+u\n$$\n\nor\n\n$$\nw+v=t+z\n$$\n\nBut $w+v$ is the area of quadrilateral $P M Q N, z$ is the area of $\\triangle A P D$ and $t$ is the area of $\\triangle B Q C$. In other words, the area of quadrilateral $P M Q N$ equals the sum of the areas of $\\triangle A P D$ and $\\triangle P Q C$, as required.']"			['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proof	Geometry	Math	English
17	"In the diagram, $A B$ and $B C$ are chords of the circle with $A B<B C$. If $D$ is the point on the circle such that $A D$ is perpendicular to $B C$ and $E$ is the point on the circle such that $D E$ is parallel to $B C$, carefully prove, explaining all steps, that $\angle E A C+\angle A B C=90^{\circ}$.

<image_1>"	"['Join $A$ to $E$ and $C$, and $B$ to $E$.\n\n<img_3770>\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nNow $\\angle E A C=\\angle E B C$ since both are subtended by $E C$.\n\nTherefore, $\\angle E A C+\\angle A B C=\\angle E B C+\\angle A B C=\\angle E B A$ which is indeed equal to $90^{\\circ}$ as required, since $A E$ is a diameter.'
 'Join $A$ to $E$ and $C$.\n\n<img_3932>\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nThus, $\\angle E C A=90^{\\circ}$.\n\nNow $\\angle A B C=\\angle A E C$ since both are subtended by $A C$.\n\nNow $\\angle E A C+\\angle A B C=\\angle E A C+\\angle A E C=180^{\\circ}-\\angle E C A$ using the sum of the angles in $\\triangle A E C$.\n\nBut $\\angle E C A=90^{\\circ}$, so $\\angle E A C+\\angle A E C=90^{\\circ}$.'
 'Join $A$ to $E$ and $C$, and $C$ to $D$.\n\n<img_3930>\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nNow $\\angle A B C=\\angle A D C$ since both are subtended by $A C$.\n\nAlso $\\angle E A C=\\angle E D C$ since both are subtended by $E C$.\n\nSo $\\angle E A C+\\angle A B C=\\angle E D C+\\angle A D C=\\angle A D E=90^{\\circ}$.']"			['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proof	Geometry	Math	English
18	"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

<image_1>

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer."	['In an $(m, n)$-sawtooth sequence, the sum of the terms is $n\\left(m^{2}-1\\right)+1$.\n\nIn each tooth, there are $(m-1)+(m-1)=2 m-2$ terms (from 2 to $m$, inclusive, and from $m-1$ to 1 , inclusive).\n\nThis means that there are $n(2 m-2)+1$ terms in the sequence.\n\nThus, the average of the terms in the sequence is $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}$.\n\nWe need to prove that this is not an integer for all pairs of positive integers $(m, n)$ with $m \\geq 2$.\n\nSuppose that $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$ for some integer $k$. We will show, by contradiction, that this is not possible.\n\nSince $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$, then\n\n$$\n\\begin{aligned}\n\\frac{m^{2} n-n+1}{2 m n-2 n+1} & =k \\\\\nm^{2} n-n+1 & =2 m n k-2 n k+k \\\\\nm^{2} n-2 m n k+(2 n k-n-k+1) & =0\n\\end{aligned}\n$$\n\nWe treat this as a quadratic equation in $m$.\n\nSince $m$ is an integer, then this equation has integer roots, and so its discriminant must be a perfect square.\n\nThe discriminant of this quadratic equation is\n\n$$\n\\begin{aligned}\n\\Delta & =(-2 n k)^{2}-4 n(2 n k-n-k+1) \\\\\n& =4 n^{2} k^{2}-8 n^{2} k+4 n^{2}+4 n k-4 n \\\\\n& =4 n^{2}\\left(k^{2}-2 k+1\\right)+4 n(k-1) \\\\\n& =4 n^{2}(k-1)^{2}+4 n(k-1) \\\\\n& =(2 n(k-1))^{2}+2(2 n(k-1))+1-1 \\\\\n& =(2 n(k-1)+1)^{2}-1\n\\end{aligned}\n$$\n\nWe note that $(2 n(k-1)+1)^{2}$ is a perfect square and $\\Delta$ is supposed to be a perfect square. But these perfect squares differ by 1 , and the only two perfect squares that differ by 1 are\n\n\n\n1 and 0.\n\n(To justify this last fact, we could look at the equation $a^{2}-b^{2}=1$ where $a$ and $b$ are non-negative integers, and factor this to obtain $(a+b)(a-b)=1$ which would give $a+b=a-b=1$ from which we get $a=1$ and $b=0$.)\n\nSince $(2 n(k-1)+1)^{2}=1$ and $2 n(k-1)+1$ is non-negative, then $2 n(k-1)+1=1$ and so $2 n(k-1)=0$.\n\nSince $n$ is positive, then $k-1=0$ or $k=1$.\n\nTherefore, the only possible way in which the average is an integer is if the average is 1.\n\nIn this case, we get\n\n$$\n\\begin{aligned}\nm^{2} n-2 m n+(2 n-n-1+1) & =0 \\\\\nm^{2} n-2 m n+n & =0 \\\\\nn\\left(m^{2}-2 m+1\\right) & =0 \\\\\nn(m-1)^{2} & =0\n\\end{aligned}\n$$\n\nSince $n$ and $m$ are positive integers with $m \\geq 2$, then $n(m-1)^{2} \\neq 0$, which is a contradiction.\n\nTherefore, the average of the terms in an $(m, n)$-sawtooth sequence cannot be an integer.']			['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KKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKAP//Z']	Multimodal	Competition	True				Theorem proof	Number Theory	Math	English
19	"In the diagram, $A B C D$ is a square. Points $E$ and $F$ are chosen on $A C$ so that $\angle E D F=45^{\circ}$. If $A E=x, E F=y$, and $F C=z$, prove that $y^{2}=x^{2}+z^{2}$.

<image_1>"	"['Rotate $\\triangle D F C$ through an angle of $90^{\\circ}$ counterclockwise about $D$, so that $D C$ now lies along $D A$ and $F^{\\prime}$ is outside the square, as shown.\n\nJoin $F^{\\prime}$ to $E$.\n\n<img_3177>\n\nSince $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nSince $\\angle E A D=45^{\\circ}$ and $\\angle F^{\\prime} A D=\\angle F C D=45^{\\circ}$, then $\\angle F^{\\prime} A E=45^{\\circ}+45^{\\circ}=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle F^{\\prime} A E$, we have $F^{\\prime} E^{2}=F^{\\prime} A^{2}+A E^{2}$.\n\nBut $F^{\\prime} A=F C=z$ and $A E=x$, so $F^{\\prime} E^{2}=z^{2}+x^{2}$.\n\nTo show that $y^{2}=x^{2}+z^{2}$, it is sufficient to show that $F^{\\prime} E=y$.\n\nConsider $\\triangle F^{\\prime} D E$ and $\\triangle F D E$.\n\nNote that $F^{\\prime} D=F D$ and $\\angle F^{\\prime} D A=\\angle F D C$ by construction and $E D=E D$.\n\nAlso, $\\angle F^{\\prime} D E=\\angle F^{\\prime} D A+\\angle E D A=\\angle F D C+\\angle E D A=90^{\\circ}-\\angle E D F=45^{\\circ}$, which tells us that $\\angle F^{\\prime} D E=\\angle F D E=45^{\\circ}$.\n\nTherefore, $\\triangle F^{\\prime} D E$ is congruent to $\\triangle F D E$ (side-angle-side), and so $F^{\\prime} E=F E=y$.\n\nThis gives us the desired conclusion that $y^{2}=x^{2}+z^{2}$.'
 'Since $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nLet $\\angle A D E=\\theta$.\n\nSince the angles in a triangle have a sum of $180^{\\circ}$, then\n\n$$\n\\angle A E D=180^{\\circ}-\\angle E A D-\\angle A D E=180^{\\circ}-45^{\\circ}-\\theta=135^{\\circ}-\\theta\n$$\n\nSince $A E F$ is a straight angle, then $\\angle D E F=180^{\\circ}-\\angle A E D=180^{\\circ}-\\left(135^{\\circ}-\\theta\\right)=45^{\\circ}+\\theta$. Continuing in this way, we find that $\\angle E F D=90^{\\circ}-\\theta, \\angle D F C=90^{\\circ}+\\theta$, and $\\angle F D C=45^{\\circ}-\\theta$.\n\n<img_3414>\n\nUsing the sine law in $\\triangle A E D$, we see that $\\frac{A E}{\\sin \\angle A D E}=\\frac{E D}{\\sin \\angle E A D}$ or $\\frac{x}{\\sin \\theta}=\\frac{E D}{\\sin 45^{\\circ}}$.\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{E D}{\\sin \\angle E F D}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{E D}{\\sin \\left(90^{\\circ}-\\theta\\right)}$.\n\n\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{F D}{\\sin \\angle D E F}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{F D}{\\sin \\left(45^{\\circ}+\\theta\\right)}$.\n\nUsing the sine law in $\\triangle D F C$, we get $\\frac{F C}{\\sin \\angle F D C}=\\frac{F D}{\\sin \\angle D C F}$ or $\\frac{z}{\\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{F D}{\\sin 45^{\\circ}}$.\n\nDividing the first of these equations by the second, we obtain $\\frac{x \\sin 45^{\\circ}}{y \\sin \\theta}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{x}{y}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right) \\sin \\theta}{\\sin ^{2} 45^{\\circ}}$.\n\nDividing the fourth of these equations by the third, we obtain $\\frac{z \\sin 45^{\\circ}}{y \\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{z}{y}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)}{\\sin ^{2} 45^{\\circ}}$.\n\nSince $\\sin \\left(90^{\\circ}-\\alpha\\right)=\\cos \\alpha$ for every angle $\\alpha$, then $\\sin \\left(90^{\\circ}-\\theta\\right)=\\cos \\theta$.\n\nAlso, $\\sin \\left(45^{\\circ}+\\theta\\right)=\\sin \\left(90^{\\circ}-\\left(45^{\\circ}-\\theta\\right)\\right)=\\cos \\left(45^{\\circ}-\\theta\\right)$.\n\nUsing this and the fact that $\\frac{1}{\\sin ^{2} 45^{\\circ}}=\\frac{1}{(1 / \\sqrt{2})^{2}}=2$, the expressions for $\\frac{x}{y}$ and $\\frac{z}{y}$ become\n\n$$\n\\frac{x}{y}=2 \\cos \\theta \\sin \\theta=\\sin 2 \\theta\n$$\n\nand\n\n$$\n\\frac{z}{y}=2 \\cos \\left(45^{\\circ}-\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)=\\sin \\left(2\\left(45^{\\circ}-\\theta\\right)\\right)=\\sin \\left(90^{\\circ}-2 \\theta\\right)=\\cos 2 \\theta\n$$\n\nFinally, this tells us that\n\n$$\n\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=\\left(\\frac{x}{y}\\right)^{2}+\\left(\\frac{z}{y}\\right)^{2}=\\sin ^{2} 2 \\theta+\\cos ^{2} 2 \\theta=1\n$$\n\nSince $\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=1$, then $x^{2}+z^{2}=y^{2}$, as required.\n\n\n\n\n#']"			['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proof	Geometry	Math	English
20	"In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$.

<image_1>"	['Join $O$ to $A, B$ and $C$.\n\n<img_3419>\n\nSince $A B$ is tangent to the circle at $A$, then $\\angle O A B=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle O A B$, we get $O A^{2}+A B^{2}=O B^{2}$ or $r^{2}+p^{2}=O B^{2}$.\n\nIn $\\triangle O D C$, we have $O D=D C=q$ and $O C=r$.\n\nBy the cosine law,\n\n$$\n\\begin{aligned}\nO C^{2} & =O D^{2}+D C^{2}-2(O D)(D C) \\cos (\\angle O D C) \\\\\nr^{2} & =q^{2}+q^{2}-2 q^{2} \\cos (\\angle O D C) \\\\\n\\cos (\\angle O D C) & =\\frac{2 q^{2}-r^{2}}{2 q^{2}}\n\\end{aligned}\n$$\n\nIn $\\triangle O D B$, we have $\\angle O D B=\\angle O D C$.\n\nThus, using the cosine law again,\n\n$$\n\\begin{aligned}\nO B^{2} & =O D^{2}+D B^{2}-2(O D)(D B) \\cos (\\angle O D B) \\\\\n& =q^{2}+(2 q)^{2}-2(q)(2 q)\\left(\\frac{2 q^{2}-r^{2}}{2 q^{2}}\\right) \\\\\n& =q^{2}+4 q^{2}-2\\left(2 q^{2}-r^{2}\\right) \\\\\n& =q^{2}+2 r^{2}\n\\end{aligned}\n$$\n\nSo $O B^{2}=r^{2}+p^{2}=q^{2}+2 r^{2}$, which gives $p^{2}=q^{2}+r^{2}$, as required.']			['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proof	Geometry	Math	English
21	"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below.
<image_1>

Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$.
Prove that $f(n, k)=f(n-1, k)+f(n-2, k-1)$ for all integers $n \geq 3$ and $k \geq 2$."	"[""An allowable string $p_{1} p_{2} \\cdots p_{n-1} p_{n}$ has $\\left(p_{1}, p_{n}\\right)=(1,0),(0,1)$, or $(0,0)$.\n\nDefine $g(n, k, 1,0)$ to be the number of allowable strings of length $n$, containing $k 1$ 's, and with $\\left(p_{1}, p_{n}\\right)=(1,0)$.\n\nWe define $g(n, k, 0,1)$ and $g(n, k, 0,0)$ in a similar manner.\n\nNote that $f(n, k)=g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0)$.\n\nConsider the strings counted by $g(n, k, 0,1)$.\n\nSince $p_{n}=1$, then $p_{n-1}=0$. Since $p_{1}=0$, then $p_{2}$ can equal 0 or 1 .\n\nWe remove the first and last digits of these strings.\n\nWe obtain strings $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ that is strings of length $n-2$ containing $k-11$ 's.\n\nSince $p_{n-1}=0$, then the first and last digits of these strings are not both 1 . Also, since the original strings did not contain two consecutive 1's, then these new strings does not either.\n\nTherefore, $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ are allowable strings of length $n-2$ containing $k-1$ 1's, with $p_{n-1}=0$ and $p_{2}=1$ or $p_{2}=0$.\n\nThe number of such strings with $p_{2}=1$ and $p_{n-1}=0$ is $g(n-2, k-1,1,0)$ and the number of such strings with $p_{2}=0$ and $p_{n-1}=0$ is $g(n-2, k-1,0,0)$.\n\nThus, $g(n, k, 0,1)=g(n-2, k-1,1,0)+g(n-2, k-1,0,0)$.\n\nConsider the strings counted by $g(n, k, 0,0)$.\n\nSince $p_{1}=0$ and $p_{n}=0$, then we can remove $p_{n}$ to obtain strings $p_{1} p_{2} \\cdots p_{n-1}$ of length $n-1$ containing $k 1$ 's. These strings are allowable since $p_{1}=0$ and the original strings were allowable.\n\nNote that we have $p_{1}=0$ and $p_{n-1}$ is either 0 or 1 .\n\nSo the strings $p_{1} p_{2} \\cdots p_{n-1}$ are allowable strings of length $n-1$ containing $k$ 1's, starting with 0 , and ending with 0 or 1 .\n\nThe number of such strings with $p_{1}=0$ and $p_{n-1}=0$ is $g(n-1, k, 0,0)$ and the number of such strings with $p_{1}=0$ and $p_{n-1}=1$ is $g(n-1, k, 0,1)$.\n\nThus, $g(n, k, 0,0)=g(n-1, k, 0,0)+g(n-1, k, 0,1)$.\n\n\n\nConsider the strings counted by $g(n, k, 1,0)$.\n\nHere, $p_{1}=1$ and $p_{n}=0$. Thus, $p_{n-1}$ can equal 0 or 1 . We consider these two sets separately.\n\nIf $p_{n-1}=0$, then the string $p_{1} p_{2} \\cdots p_{n-1}$ is an allowable string of length $n-1$, containing $k 1$ 's, beginning with 1 and ending with 0 .\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=0$ is equal to $g(n-1, k, 1,0)$.\n\nIf $p_{n-1}=1$, then the string $p_{2} p_{3} \\cdots p_{n-1}$ is of length $n-2$, begins with 0 and ends with 1 . Also, it contains $k-1$ 1's (having removed the original leading 1) and is allowable since the original string was.\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=1$ is equal to $g(n-2, k-1,0,1)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n, k)= & g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0) \\\\\n= & (g(n-1, k, 1,0)+g(n-2, k-1,0,1)) \\\\\n& \\quad+(g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n& \\quad+(g(n-1, k, 0,0)+g(n-1, k, 0,1)) \\\\\n= & (g(n-1, k, 1,0)+g(n-1, k, 0,1)+g(n-1, k, 0,0)) \\\\\n& \\quad+(g(n-2, k-1,0,1)+g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n= & f(n-1, k)+f(n-2, k-1)\n\\end{aligned}\n$$\n\nas required.""
 ""We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nTo prove the desired fact, we will use the fact that $\\left(\\begin{array}{c}m \\\\ r\\end{array}\\right)=\\left(\\begin{array}{c}m-1 \\\\ r\\end{array}\\right)+\\left(\\begin{array}{c}m-1 \\\\ r-1\\end{array}\\right)$, which we prove at the end.\n\nNow\n\n$$\n\\begin{aligned}\nf & n-1, k)+f(n-2, k-1) \\\\\n& =\\left(\\begin{array}{c}\n(n-1)-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-1)-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1) \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1)-1 \\\\\n(k-1)-1\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\quad \\text { (using the identity above) } \\\\\n& =f(n, k)\n\\end{aligned}\n$$\n\nas required.\n\nTo prove the identity, we expand the terms on the right side:\n\n$$\n\\begin{aligned}\n\\left(\\begin{array}{c}\nm-1 \\\\\nr\n\\end{array}\\right)+\\left(\\begin{array}{c}\nm-1 \\\\\nr-1\n\\end{array}\\right) & =\\frac{(m-1) !}{r !(m-r-1) !}+\\frac{(m-1) !}{(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r-1) !(m-r)}+\\frac{r(m-1) !}{r(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r) !}+\\frac{r(m-1) !}{r !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r+r)}{r !(m-r) !} \\\\\n& =\\frac{(m-1) ! m}{r !(m-r) !} \\\\\n& =\\frac{m !}{r !(m-r) !} \\\\\n& =\\left(\\begin{array}{c}\nm \\\\\nr\n\\end{array}\\right)\n\\end{aligned}\n$$\n\nas required.""]"			['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proof	Algebra	Math	English
22	"In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant.)

<image_1>"	['Since the lengths of $A D, A B$ and $B C$ form a geometric sequence, we suppose that these lengths are $a$, ar and $a r^{2}$, respectively, for some real numbers $a>0$ and $r>0$.\n\nSince the angles at $A$ and $B$ are both right angles, we assign coordinates to the diagram, putting $B$ at the origin ( 0,0$), C$ on the positive $x$-axis at $\\left(a r^{2}, 0\\right), A$ on the positive $y$-axis at $(0, a r)$, and $D$ at $(a, a r)$.\n\n<img_3911>\n\nTherefore, the slope of the line segment joining $B(0,0)$ and $D(a, a r)$ is $\\frac{a r-0}{a-0}=r$.\n\nAlso, the slope of the line segment joining $A(0, a r)$ and $C\\left(a r^{2}, 0\\right)$ is $\\frac{a r-0}{0-a r^{2}}=-\\frac{1}{r}$.\n\nSince the product of the slopes of these two line segments is -1 , then the segments are perpendicular, as required.']			['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CiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKAOU+IXg+Lxv4TuNJLiK4DCW2lYcJKoOM+xBKn61xng/xvqXhDSbfw9420XUrWWyUQwX8Ns00MsY4Xlc8gADjOR6V65t5pcUAcmnii517Nv4c0+8+fAN/fWrwQxL/eCvhpG9ABj1Irp7W3+zWsUHmSy+WgXzJWy7YHUnuTUoXHelHAoAKKKKAP/9k=']	Multimodal	Competition	True				Theorem proof	Geometry	Math	English
23	"In the graph, the parabola $y=x^{2}$ has been translated to the position shown.

Prove that $d e=f$.

<image_1>"	['Since the given graph is congruent to $y=x^{2}$ and has $x$-intercepts $-d$ and $e$, its general form is $y=(x+d)(x-e)$.\n\nTo find the $y$-intercept, let $x=0$. Therefore $y$-intercept $=-d e$.\n\nWe are given that the $y$-intercept is $-f$.\n\nTherefore $-f=-d e$ or $f=d e$.']			['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proof	Geometry	Math	English
24	"In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$.

<image_1>"	"['Establish a coordinate system with $K(0,0), D(2 a, 0)$ on the $x$-axes. Let $W$ be $(2 b, 2 c)$ and $A$ be $(2 d, 2 e)$.\n\nThus $M$ is $(b, c)$ and $N$ is $(a+d, e)$.\n\n$K D$ has slope 0 and slope $W A=\\frac{e-c}{d-b}$.\n\nSince $M N=\\frac{1}{2}(A W+D K)$\n\n$$\n\\begin{aligned}\n& \\sqrt{(a+d-b)^{2}+(e-c)^{2}} \\\\\n= & \\frac{1}{2}\\left(2 a+\\sqrt{(2 d-2 b)^{2}+(2 e-2 c)^{2}}\\right) \\\\\n= & \\frac{1}{2}\\left(2 a+2 \\sqrt{(d-b)^{2}+(e-c)^{2}}\\right)\n\\end{aligned}\n$$\n\n<img_3380>\n\nSquaring both sides gives,\n\n$$\n\\begin{aligned}\n& (a+d-b)^{2}+(e-c)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}+(e-c)^{2} \\\\\n& a^{2}+2 a(d-b)+(d-b)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}\n\\end{aligned}\n$$\n\nSimplifying and dividing by $2 a$ we have, $d-b=\\sqrt{(d-b)^{2}+(e-c)^{2}}$.\n\nSquaring, $(d-b)^{2}=(d-b)^{2}+(e-c)^{2}$.\n\nTherefore $(e-c)^{2}=0$ or $e=c$.\n\nSince $e=c$ then slope of $W A$ is 0 and $K D \\| A W$.'
 'Join $A$ to $K$ and call $P$ the mid-point of $A K$.\n\nJoin $M$ to $P, N$ to $P$ and $M$ to $N$.\n\nIn $\\triangle K A W, P$ and $M$ are the mid-points of $K A$ and $K W$.\n\nTherefore, $M P=\\frac{1}{2} W A$.\n\nSimilarly in $\\triangle K A D, P N=\\frac{1}{2} K D$.\n\nTherefore $M P+P N=M N$.\n\n<img_3772>\n\nAs a result $M, P$ and $N$ cannot form the vertices of a triangle but must form a straight line.\n\nSo if $M P N$ is a straight line with $M P \\| W A$ and $P N \\| K D$ then $W A \\| K D$ as required.'
 'We are given that $\\overrightarrow{A N}=\\overrightarrow{N D}$ and $\\overrightarrow{W M}=\\overrightarrow{M K}$.\n\nUsing vectors,\n\n(1) $\\overrightarrow{M N}=\\overrightarrow{M W}+\\overrightarrow{W A}+\\overrightarrow{A N}$ (from quad. $M W A N$ )\n\n(2) $\\overrightarrow{M N}=\\overrightarrow{M K}+\\overrightarrow{K D}+\\overrightarrow{D N} \\quad$ (from quad. $K M N D$ )\n\nIt is also possible to write, $\\overrightarrow{M N}=-\\overrightarrow{M W}+\\overrightarrow{K D}-\\overrightarrow{A N}$,\n\n(3) (This comes from taking statement (2) and making appropriate substitutions.)\n\n\n\nIf we add (1) and (3) we find, $2 \\overrightarrow{M N}=\\overrightarrow{W A}+\\overrightarrow{K D}$.\n\nBut it is given that $2|\\overrightarrow{M N}|=|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nFrom these two previous statements, $\\overrightarrow{M N}$ must be parallel to $\\overrightarrow{W A}$ and $\\overrightarrow{K D}$ otherwise $2|\\overrightarrow{M N}|<|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nTherefore, $W A \\| K D$.']"			['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proof	Geometry	Math	English
25	"Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal.

<image_1>"	"[""The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, k(2 n-k+1), \\ldots, p(2 n-p+1), \\ldots, n(n+1)$.\n\nIn essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'\n\n$$\n\\begin{aligned}\nk(2 n-k+1) & =p(2 n-p+1) \\\\\n2 n k-k^{2}+k=2 n p-p^{2}+p & \\text { (supposing them to be equal) } \\\\\np^{2}-k^{2}+2 n k-2 n p+k-p & =0 \\\\\n(p-k)(p+k)+2 n(k-p)+(k-p) & =0 \\\\\n(p-k)[(p+k)-2 n-1] & =0 \\\\\n(p-k)(p+k-2 n-1) & =0\n\\end{aligned}\n$$\n\nSince $p$ and $k$ are both less than or equal to $n$, it follows $p+k-2 n-1 \\neq 0$. Therefore $p=k$ and they represent the same pair. Thus the required is proven.""
 'The products are $1(2 n+1-1), 2(2 n+1-2), 3(2 n+1-3), \\ldots, n(2 n+1-n)$.\n\nConsider the function, $y=x(2 n+1-x)=-x^{2}+(2 n+1) x=f(x)$.\n\n\n\nThe graph of this function is a parabola, opening down, with its vertex at $x=n+\\frac{1}{2}$.\n\nThe products are the $y$-coordinates of the points on the parabola corresponding to $x=1,2,3, \\ldots, n$. Since all the points are to the left of the vertex, no two have the same $y$ coordinate.\n\nThus the products are distinct.\n\n<img_3942>'
 'The sum of these numbers is $\\frac{2 n(2 n+1)}{2}$ or $n(2 n+1)$.\n\nTheir average is $\\frac{n(2 n+1)}{2 n}=n+\\frac{1}{2}$.\n\nThe $2 n$ numbers can be rewritten as,\n\n$$\nn+\\frac{1}{2}-\\left(\\frac{2 n-1}{2}\\right), \\cdots, n+\\frac{1}{2}-\\frac{3}{2}, n+\\frac{1}{2}-\\frac{1}{2}, n+\\frac{1}{2}+\\frac{1}{2}, n+\\frac{1}{2}+\\frac{3}{2}, \\cdots, n+\\frac{1}{2}+\\left(\\frac{2 n-1}{2}\\right) .\n$$\n\nThe product pairs, starting from the middle and working outward are\n\n$$\n\\begin{gathered}\nP_{1}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{1}{4} \\\\\nP_{2}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{9}{4} \\\\\n\\vdots \\\\\nP_{n}=\\left(n+\\frac{1}{2}\\right)^{2}-\\left(\\frac{2 n-1}{2}\\right)^{2}\n\\end{gathered}\n$$\n\nEach of the numbers $\\left(\\frac{2 k-1}{2}\\right)^{2}$ is distinct for $k=1,2,3, \\ldots, n$ and hence no terms of $P_{k}$ are equal.'
 'The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, n[2 n-(n-1)]$.\n\nThis sequence has exactly $n$ terms.\n\nWhen the $k$ th term is subtracted from the $(k+1)$ th term the difference is $(k+1)[2 n-k]-k[2 n-(k-1)]=2(n-k)$. Since $n>k$, this is a positive difference.\n\nTherefore each term is greater than the term before, so no two terms are equal.']"			['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proof	Number Theory	Math	English
26	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Consider the hideout map $M$ below.

<image_2>

Show that one Cop can always catch the Robber."	['Have the Cop stay at $A$ for 2 days. If the Robber is not at $A$ the first day, he must be at one of $B_{1}-B_{6}$, and because the Robber must move along an edge every night, he will be forced to go to $A$ on day 2 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proof	Combinatorics	Math	English
27	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The map shown below is $\mathcal{C}_{6}$, the cyclic graph with six hideouts. Show that three Cops are sufficient to catch the Robber on $\mathcal{C}_{6}$, so that $C\left(\mathcal{C}_{6}\right) \leq 3$.

<image_2>"	['The Cops should stay at $\\left\\{A_{1}, A_{3}, A_{5}\\right\\}$ for 2 days. If the Robber evades capture the first day, he must have been at an even-numbered hideout. Because he must move, he will be at an odd-numbered hideout the second day. Equivalently, the Cops could stay at $\\left\\{A_{2}, A_{4}, A_{6}\\right\\}$ for 2 days.']			['/9j/2wCEAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDIBCQkJDAsMGA0NGDIhHCEyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMv/AABEIAVABvwMBIgACEQEDEQH/xAGiAAABBQEBAQEBAQAAAAAAAAAAAQIDBAUGBwgJCgsQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2drh4uPk5ebn6Onq8fLz9PX29/j5+gEAAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoLEQACAQIEBAMEBwUEBAABAncAAQIDEQQFITEGEkFRB2FxEyIygQgUQpGhscEJIzNS8BVictEKFiQ04SXxFxgZGiYnKCkqNTY3ODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqCg4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2dri4+Tl5ufo6ery8/T19vf4+fr/2gAMAwEAAhEDEQA/APf6KKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKQ0ALketFcB8SfHg8MJYaTp7q2t6nMkcI4PkozbTIR+YHqc9dpFd6p60AOooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKADNZevazDoeltdvG88pYR29vGfnnlY4VF+p79hkngGr9xNFbQPNPKkUUalndyAqgckknoK4e10aPx7NH4hvpr23sUz/ZEdrcvA6xkENOxUg7pAeB2QjuxoA4Txpo0sHjzwDDfSrPqt/qZu9QkXpkPFtRc9EUAqPpnqa93HrXh2q6RBZ/tDeE9Ohur64SO3e4Y3l5JcFW2ytgF2JH3BwPY969xHpQAtFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQBka/4a0zxNaC01WKea2/iiS5liV84+8EYbunfOOam0nRbPRNLj02wWaO0jXaiPcSSlBjAAZ2LADsAcDtWjRQByEnw18Ly6wusPZXjakhyt2dSufMHUcN5mQME8V1cMYhiWNd21VABZix4Hcnk/U81JRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUZxQAUhpayte8Q6V4b0832rX8NpAM4MjcuQM4UDlj7AE+1AGdrnj/wr4dv0stV1u2t7pmCmLl2Qn+8FB2jnqcVv2V5a6haR3Vlcw3NvIMpLC4dGHsRwa+GdbvLjUNbvb+5dnlupmnMhTZu3HcGC9gc8Dng19N/ACC+h+GwN3kQyXcjWgJziPCg49t4c/iT3oA9SooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAoozjrSHrjPNAC1FPNFDA00sqRxICzOzbQoHcnoB9a43xZ8S9H8NXK6dbpLqutyHEWm2Q3yFvRsZ2/z9AeawI/BHifx5NHd+Pb37JpobdHoVhIQvsZXycnj178FelAFjUvibd63fSaL8PdN/ti8XCy6jKNtpBnuTxu4Deg443dKl0f4Z20F03iLxtqJ17V1Uu0lyP9GtgOSEQ8YHPJGO4VTXeaXpdjo9hHZadZw2ltGPlihQKo98Dv79TXE/FzU518OWvhzTiP7S8Q3C2MXP3YyR5jH/AGcEA+zUAY/w60a08Z3+seOtY06C5F/d7NNS6hV/JhiO1WUEHaeAD7oT3r1lRgYwAOwFUtF0u20PRbLS7QYgtIViTjkgDGT7nqav0AFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUVBdXVvZ28lzczxwwRqWeSRwqqB1JJ4FeaXvxH1bxPeyaV8OtN+3Mh2TatdKVtYPp/ePOf5Bh0AO58R+KdF8K2H2zWNQitYyDtUnLyeyqOSenQV5+b/AMb/ABLUDSo5fC/ht/8Al8lH+mXC/wCwB9wHjkH6MeRWz4d+F9nY6ida8R3UniDXWIJubsZji74jQ5AA7enbHSu/Xvnv7UAc14V8CaD4Ot/L0q0/fsMTXc2Glm9dzY/HAAFdIoxTqKAEbpj+VeW+HR/wmfxe1bxGT5mm6Cv9nWBxwZj/AK1x24ywz3DKe1dP8R/Ev/CK+Cb6/ifF46/Z7QDk+c4wuB3IGWx6A1L8PvDI8J+C9P0tlAuQnm3R67pn+Zue+M7QfRRQB0wz3paKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiuf8U+MdD8I2YuNXvo4WIJjhX5pZP91ByfTPT1IoA3zziuE8UfE3S9GvP7J0u3l1zXWO1bCy+faeh3sMhcenJ9h1rB/4rr4m9PN8KeGX/wDA25T9NoI+nX+MV3XhfwZofg+y+zaPYpESB5kzfNJIfVmxnueBwM8AUAcXb/D/AMQeMriPUPiJqG6BW3xaHZOVgjPYuwOWI56HP+1jivS7CxtdNso7Oytora2iXEcMSBVUewHFWBS0AFFFFABSNnHFLWb4g1i28P6BfatdnEFpC0rDOC2Bwo9ycAe5oA8/1lf+Ey+Mem6MmW0zw1GL274+VrlseWh7EgYYf8DFeorXn/wj0e4tvC82u6io/tTX521C4YLjCsfkUe2PmA7bq9BFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFNbkUAOqtf31rp1nJeXtzFbW0Q3SSyuFVR6kmuH8SfFCysdROieHLSXxBrzfKLW0OY4z6yOOAB3x+JXrWdY/DjVfFF5Hq3xE1D7cyt5kOkWrFLWE++D8x7foSwoAbP4/1/xncy6b8PNOBgRvLm1y+UrBGe+xSPmPTqPquOa2fDPww03Rrw6vq00mua85Dvf3nzbT/sKchcdjyR2wOK7S0tYLK3S2tYI4LeJQsccShVUegA4FT0AIOvTtS0UUAFFFFABRRRQAjfzry/4nSSeJfEGgeALZmC30ovNRKcbbaM5xn3IOP9pV9a9OlkSGNpZGCogLMzHAAAySa8x+F8T+JNZ17x9dRsP7SmNrp6sPuW0Zxx6ZIAPuh9aAPTokSKJY40CIoAVQMAD0Ap9IvfrS0AFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFJketAC0ZqlqmqWOj2Ml7qN3Da20Yy8szhVH59z6V5tJ438TePJXs/AdgbPTt22TXr5CEx38pCDk+/PuF60Adj4t8c6D4Ntll1W8xMwzFaRfPNKewC5/U4Hv0rizp3jf4ljOpvL4W8NSdbOI/6ZcL33kj5AeeCPqG610fhX4a6R4cuDqVyZdW1yQ7pdRvjvk3eq5J2/UZPOMmu0UdTzz60AY3hzwto/hSwFno9hFbRnl2Ay8h9Wbqe/wBO2BW0KWigAooooAKKKKACiiigAoopD/k0Aee/F7V7iDwzB4e05lOp+IJxYQqT0RseYx9sEKT2312eh6Tb6FolnpVopEFpCsSZ6nA6n3PU+5rzzw//AMVp8YNV185fTPDqnT7Lj5WnOfMYfT5h9GU16kowKAFooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAozjrSEgdSOK5bxd4+0LwhGsd7cNNfycQ2FsN88hPT5ewPqce3NAHU5B54rzrXvilD/aP9ieDrJvEOtHIxAf8AR4P9p3zjGe2QODkjjOYdA8Z/EnD+I5X8OeHX5XS7V/8ASJl9JWPQe2PT5cjNeh6F4d0nw3p62OkafFaW46hByx9Wbqx9yaAOH034ZXmuX0esfELUP7XvFOYdOiJW0ts9gvG48DPrjndwa9KgijgiWKKNY4kAVEVQFUegA6U8ZpaACiiigAooooAKKKKACiiigAooooAK5T4i+Jx4T8E6hqUbf6UV8i1UHlpn4XH0+9j0U11Rry7Xj/wmXxh0rQUJbTfDqDULwfwtOQPLQ9iQMEexb0oA6j4deGf+ET8E6fpsi4uinnXTdzM/Lc98cLn0UV1VNXvTqACiiigAooooAKKK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proof	Combinatorics	Math	English
28	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that for all maps $M, C(M)<h(M)$."	['Let $n=h(M)$. The following strategy will always catch a Robber within two days using $n-1$ Cops, which proves that $C(M) \\leq n-1$. Choose any subset $\\mathcal{S}$ of $n-1$ hideouts and position $n-1$ Cops at the hideouts of $\\mathcal{S}$ for 2 days. If the Robber is not caught on the first day, he must have been at the hideout not in $\\mathcal{S}$, and therefore must move to a hideout in $\\mathcal{S}$ on the following day.']			['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proof	Combinatorics	Math	English
29	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that $C(M) \leq 3$ for the map below.

<image_2>"	['The following strategy guarantees capture using three Cops for four consecutive days, so $C(M) \\leq 3$. Position three Cops at $\\{B, E, H\\}$ for two days, which will catch any Robber who starts out at $B, C, D, E, F, G$, or $H$, because a Robber at $C$ or $D$ would have to move to either $B$ or $E$, and a Robber at $F$ or $G$ would have to move to either $E$ or $H$. So if the Robber is not yet caught, he must have been at $I, J, K, L$, or $A$ for those two days. In this case, an analogous argument shows that placing Cops at $B, H, K$ for two consecutive days will guarantee a capture.']			['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']	Multimodal	Competition	True				Theorem proof	Combinatorics	Math	English
30	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
A path on $n$ hideouts is a map with $n$ hideouts, connected in one long string. (More formally, a map is a path if and only if two hideouts are adjacent to exactly one hideout each and all other hideouts are adjacent to exactly two hideouts each.) It is denoted by $\mathcal{P}_{n}$. The maps $\mathcal{P}_{3}$ through $\mathcal{P}_{6}$ are shown below.

<image_3>

Show that $D\left(\mathcal{P}_{n}, 1\right) \leq 2 n$ for $n \geq 3$."	"[""The following argument shows that $C\\left(\\mathcal{P}_{n}\\right)=1$, and that capture occurs in at most $2 n-4$ days. It helps to draw the hideout map as in the following diagram, so that odd-numbered hideouts are all on one level and even-numbered hideouts are all on another level; the case where $n$ is odd is shown below. (A similar argument applies where $n$ is even.)\n\n<img_3525>\n\n\n\nEach night, the Robber has the choice of moving diagonally left or diagonally right on this map, but he is required to move from top to bottom or vice-versa.\n\nFor the first $n-2$ days, the Cop should search hideouts $A_{2}, A_{3}, \\ldots, A_{n-1}$, in that order. For the next $n-2$ days, the Cop should search hideouts $A_{n-1}, A_{n-2}, \\ldots, A_{2}$, in that order.\n\nThe Cop always moves from top to bottom, or vice-versa, except that he searches $A_{n-1}$ two days in a row. If the Cop is lucky, the Robber chose an even-numbered hideout on the first day. In this case, the Cop and the Robber are always on the same level (top or bottom) for the first half ( $n-2$ days) of the search. The Cop is searching from left to right, and the Robber started out to the right of the Cop (or at $A_{2}$, the first hideout searched), so eventually the Robber is caught (at $A_{n-1}$, if not earlier).\n\nIf the Robber chose an odd-numbered hideout on the first day, then the Cop and Robber will be on different levels for the first half of the search, but on the same level for the second half. For the second half of the search, the Robber is to the left of the Cop (or possibly at $A_{n-1}$, which would happen if the Robber moved unwisely or if he spent day $n-2$ at hideout $A_{n}$ ). But now the Cop is searching from right to left, so again, the Cop will eventually catch the Robber.\n\nThe same strategy can be justified without using the zig-zag diagram above. Suppose that the Robber is at hideout $A_{R}$ on a given day, and the Cop searches hideout $A_{C}$. Whenever the Cop moves to the next hideout or the preceding hideout, $C$ changes by \\pm 1 , and the Robber's constraint forces $R$ to change by \\pm 1 . Thus if the Cop uses the strategy above, on each of the first $n-2$ days, either the difference $R-C$ stays the same or it decreases by 2 . If on the first day $R-C$ is even, either $R-C$ is 0 (the Robber was at $A_{2}$ and is caught immediately) or is positive (because $C=2$ and $R \\geq 4$ ). Because the difference $R-C$ is even and decreases (if at all) by 2 each day, it cannot go from positive to negative without being zero. If on the first day $R-C$ is odd, then the Robber avoids capture through day $n-2$ (because $R-C$ is still odd), but then on day $n-1, R$ changes (by \\pm 1 ) while $C$ does not. So $R-C$ is now even (and is either 0 or negative), and henceforth either remains the same or increases by 2 each day, so again, $R-C$ must be zero at some point between day $n-1$ and day $2 n-4$, inclusive.""]"			['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rue				Theorem proof	Combinatorics	Math	English
31	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.

Let $M$ be a map with $n \geq 3$ hideouts. Prove that $2 \leq W(M) \leq n$, and that these bounds cannot be improved. In other words, prove that for each $n \geq 3$, there exist maps $M_{1}$ and $M_{2}$ such that $W\left(M_{1}\right)=2$ and $W\left(M_{2}\right)=n$."	['A single Cop can only search one hideout in a day, so as long as $M$ has two or more hideouts, there is no strategy that guarantees that a lone Cop captures the Robber the first day. Then either more than one Cop will have to search on the first day, or a lone Cop will have to search for at least 2 days; in either of these cases, $W(M) \\geq 2$. On the other hand, $n$ Cops can always guarantee a capture simply by searching all $n$ hideouts on the first day, so $W(M) \\leq n$.\n\nTo show that the lower bound cannot be improved, consider the star on $n$ hideouts; that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout. (The map in 1a is $\\mathcal{S}_{7}$.) Then $W\\left(\\mathcal{S}_{n}\\right)=2$, because a single Cop can search the central hideout for 2 days; if the Robber is at one of the outer hideouts on the first day, he must go to the central hideout on the second day. For the upper bound, the complete map $\\mathcal{K}_{n}$ is an example of a map with $W(M)=n$. As argued above, the minimum number of Cops needed to guarantee a catch is $n-1$, and using $n-1$ Cops requires two days, for a total of $2 n-2$ Cop workdays. So the optimal strategy for $\\mathcal{K}_{n}$ (see 2c) is to use $n$ Cops for a single day.']			['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proof	Combinatorics	Math	English
32	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.

Prove that if $M$ is bipartite, then $C(M) \leq n / 2$."	['Suppose $M$ is bipartite, and let $\\mathcal{A}$ and $\\mathcal{B}$ be the sets of hideouts referenced in the definition. Because $\\mathcal{A}$ and $\\mathcal{B}$ are disjoint, either $|\\mathcal{A}| \\leq n / 2$ or $|\\mathcal{B}| \\leq n / 2$ or both. Without loss of generality, suppose that $|\\mathcal{A}| \\leq n / 2$. Then position Cops at each hideout in $\\mathcal{A}$ for two days. If the Robber was initially on a hideout in $\\mathcal{B}$, he must move the following day, and because no hideout in $\\mathcal{B}$ is connected to any other hideout in $\\mathcal{B}$, his new hideout must be a hideout in $\\mathcal{A}$.']			['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proof	Combinatorics	Math	English
33	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.

Prove that $C(M) \leq n / 2$ for any map $M$ with the property that, for all hideouts $H_{1}$ and $H_{2}$, either all paths from $H_{1}$ to $H_{2}$ contain an odd number of edges, or all paths from $H_{1}$ to $H_{2}$ contain an even number of edges."	['The given condition actually implies that the graph is bipartite. Let $A_{1}$ be a hideout in $M$, and let $\\mathcal{A}$ be the set of all hideouts $V$ such that all paths from $A_{1}$ to $V$ have an even number of edges, as well as $A_{1}$ itself; let $\\mathcal{B}$ be the set of all other hideouts in $M$. Notice that there are no edges from $A_{1}$ to any other element $A_{i}$ in $\\mathcal{A}$, because if there were, that edge would create a path from $A_{1}$ to $A_{i}$ with an odd number of edges (namely 1). So $A_{1}$ has edges only to hideouts in $\\mathcal{B}$. In fact, there can be no edge from any hideout $A_{i}$ in $\\mathcal{A}$ to any other hideout $A_{j}$ in $\\mathcal{A}$, because if there were, there would be a path with an odd number of edges from $A_{1}$ to $A_{j}$ via $A_{i}$. Similarly, there can be no edge from any hideout $B_{i}$ in $\\mathcal{B}$ to any other hideout $B_{j}$ in $\\mathcal{B}$, because if there were such an edge, there would be a path from $A_{1}$ to $B_{j}$ with an even number of edges via $B_{i}$. Thus hideouts in $\\mathcal{B}$ are connected only to hideouts in $\\mathcal{A}$ and vice versa; hence $M$ is bipartite.']			['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proof	Combinatorics	Math	English
34	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.

A map $M$ is called $k$-perfect if its hideout set $H$ can be partitioned into equal-sized subsets $\mathcal{A}_{1}, \mathcal{A}_{2}, \ldots, \mathcal{A}_{k}$ such that for any $j$, the hideouts of $\mathcal{A}_{j}$ are only adjacent to hideouts in $\mathcal{A}_{j+1}$ or $\mathcal{A}_{j-1}$. (The indices are taken modulo $k$ : the hideouts of $\mathcal{A}_{1}$ may be adjacent to the hideouts of $\mathcal{A}_{k}$.) Show that if $M$ is $k$-perfect, then $C(M) \leq \frac{2 n}{k}$."	"['Because a Robber in $\\mathcal{A}_{i}$ can only move to a hideout in $\\mathcal{A}_{i-1}$ or $\\mathcal{A}_{i+1}$, this map is essentially the same as the cyclic map $\\mathcal{C}_{k}$. So the Cops should apply a similar strategy. First, position $n / k$ Cops at the hideouts of set $\\mathcal{A}_{1}$ and $n / k$ Cops at the hideouts of set $\\mathcal{A}_{2}$. On day 2 , leave the first $n / k$ Cops at $\\mathcal{A}_{1}$, but move the second group of Cops to $\\mathcal{A}_{3}$. Continue until the second set of Cops is at $\\mathcal{A}_{k}$; then ""wait"" one turn (search $\\mathcal{A}_{k}$ again), and then search backward $\\mathcal{A}_{k-1}$, $\\mathcal{A}_{k-2}$, etc.']"			['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proof	Combinatorics	Math	English
35	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.

Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.

Find an example of a map $M$ with 2012 hideouts such that $C(M)=17$ and $W(M)=34$, or prove that no such map exists."	['There are many examples. Perhaps the simplest to describe is the complete bipartite map on the hideouts $\\mathcal{A}=\\left\\{A_{1}, A_{2}, \\ldots, A_{17}\\right\\}$ and $\\mathcal{B}=\\left\\{B_{1}, B_{2}, \\ldots, B_{1995}\\right\\}$, which is often denoted $M=\\mathcal{K}_{17,1995}$. That is, let $M$ be the map with hideouts $\\mathcal{A} \\cup \\mathcal{B}$ such that $A_{i}$ is adjacent to $B_{j}$ for all $i$ and $j$, and that $A_{i}$ is not adjacent to $A_{j}$, nor is $B_{i}$ adjacent to $B_{j}$, for any $i$ and $j$.\n\nIf 17 Cops search the 17 hideouts in $\\mathcal{A}$ for two consecutive days, then they are guaranteed to catch the Robber. This shows that $C(M) \\leq 17$ and that $W(M) \\leq 34$. If fewer than 17 Cops search on a given day, then the Robber has at least one safe hideout: he has exactly one choice if the Cops search 16 of the hideouts $A_{i}$ and the Robber was hiding at some $B_{j}$ the previous day, 1995 choices if the Robber was hiding at some $A_{i}$ the previous day. Thus $C(M)=17$.\n\nOn the other hand, fewer than 34 Cop workdays cannot guarantee catching the Robber. Unless the Cops search every $A_{i}$ (or every $B_{j}$ ) on a given day, they gain no information about where the Robber is (unless he is unlucky enough to be caught).\n\nThe complete bipartite map is not the simplest in terms of the number of edges. Try to find a map with 2012 hideouts and as few edges as possible that has a Cop number of 17 and a workday number of 34 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proof	Combinatorics	Math	English
36	"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<image_1>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.

Find an example of a map $M$ with 4 or more hideouts such that $W(M)=3$, or prove that no such map exists."	"['We denote a map as star, that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout.\n\nNo such map exists. Let $M$ be a map with at least four hideouts. The proof below shows that either $W(M)>3$ or else $M$ is a star, in which case $W(M)=2$.\n\nFirst, suppose that there are four distinct hideouts $A_{1}, A_{2}, B_{1}$, and $B_{2}$ such that $A_{1}$ and $A_{2}$ are adjacent, as are $B_{1}$ and $B_{2}$. If the Cops make fewer than four searches, then they cannot search $\\left\\{A_{1}, A_{2}\\right\\}$ twice and also search $\\left\\{B_{1}, B_{2}\\right\\}$ twice. Without loss of generality, assume the Cops search $\\left\\{A_{1}, A_{2}\\right\\}$ at most once. Then the Robber can evade capture by moving from $A_{1}$ to $A_{2}$ and back again, provided that he is lucky enough to start off at the right one. Thus\n\n\n\n$W(M)>3$ in this case. For the second case, assume that whenever $A_{1}$ is adjacent to $A_{2}$ and $B_{1}$ is adjacent to $B_{2}$, the four hideouts are not distinct. Start with two adjacent hideouts, and call them $A_{1}$ and $A_{2}$. Consider any hideout, say $B$, that is not one of these two. Then $B$ must be adjacent to some hideout, say $C$. By assumption, $A_{1}, A_{2}, B$, and $C$ are not distinct, so $C=A_{1}$ or $C=A_{2}$. That is, every hideout $B$ is adjacent to $A_{1}$ or to $A_{2}$.\n\nIf there are hideouts $B_{1}$ and $B_{2}$, distinct from $A_{1}$ and $A_{2}$ and from each other, such that $B_{1}$ is adjacent to $A_{1}$ and $B_{2}$ is adjacent to $A_{2}$, then it clearly violates the assumption of this case. That is, either every hideout $B$, distinct from $A_{1}$ and $A_{2}$, is adjacent to $A_{1}$, or every such hideout is adjacent to $A_{2}$. Without loss of generality, assume the former.\n\nBecause every hideout is adjacent to $A_{1}$ (except for $A_{1}$ itself), the foregoing proves that the map $M$ is a star. It remains to show that there are no ""extra"" edges in the map. For the sake of contradiction, suppose that $B_{1}$ and $B_{2}$ are adjacent hideouts, both distinct from $A_{1}$. Because the map has at least four hideouts, choose one distinct from these three and call it $A_{2}$. Then these four hideouts violate the assumption of this case.']"			['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proof	Combinatorics	Math	English
37	"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.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<image_1>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<image_2>
Prove that the following signature is possible.
$(123,132,213)$,"	['The signature is possible, because it is the 3 -signature of 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proof	Combinatorics	Math	English
38	"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.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<image_1>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<image_2>
Prove that the following signature is impossible:
$(321,312,213)$."	['The signature is impossible. Let a 5 -label be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. The second window of (ii) implies $a_{3}<a_{4}$, whereas the third window implies $a_{3}>a_{4}$, a contradiction.']			['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proof	Combinatorics	Math	English
39	"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.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<image_1>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<image_2>
Show that $(312,231,312,132)$ is not a unique 3 -signature."	['The $p$-signature is not unique because it equals both $S_{3}[625143]$ and $S_{3}[635142]$.']			['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proof	Combinatorics	Math	English
40	"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.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<image_1>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<image_2>
Show that $(231,213,123,132)$ is a unique 3 -signature."	['Let $L=a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We have $a_{4}<a_{6}<a_{5}$ (from window \\#4), $a_{3}<a_{1}<a_{2}$ (from window \\#1), and $a_{2}<a_{4}$ (from window \\#2). Linking these inequalities, we get\n\n$$\na_{3}<a_{1}<a_{2}<a_{4}<a_{6}<a_{5} \\quad \\Rightarrow \\quad L=231465\n$$\n\nso $S_{3}[L]$ is unique.']			['/9j/2wCEAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDIBCQkJDAsMGA0NGDIhHCEyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMv/AABEIAOwBugMBIgACEQEDEQH/xAGiAAABBQEBAQEBAQAAAAAAAAAAAQIDBAUGBwgJCgsQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2drh4uPk5ebn6Onq8fLz9PX29/j5+gEAAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoLEQACAQIEBAMEBwUEBAABAncAAQIDEQQFITEGEkFRB2FxEyIygQgUQpGhscEJIzNS8BVictEKFiQ04SXxFxgZGiYnKCkqNTY3ODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqCg4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2dri4+Tl5ufo6ery8/T19vf4+fr/2gAMAwEAAhEDEQA/APf6KKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiikYkYwKAFoqnaanZX09zBa3cE01q/lzpG4YxNjowHTv+R9DVygAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKQkjoKqXWqWVjPbQXV3BBLdP5cCSOFaRsdFB6n/EetAFyikUk5zS0AFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRTWbbigAZtuK5jxd4pXR4odOsrdr7XL7KWdlG2GY9C7EfdQckn2PTBId4r8VroawWNlbm+1y8ytnZIeT6u3og7nj8OSG+E/CZ0c3Gp6pOL7X70A3d4RjA7Rp/dQYGPXA9BgA860bwXrPwtvk8TSyLq1ps8u/ittweJCFJkXJ+fa24c9ueMnb7Hp+oWmqWMN7Y3Ec9tOm+OVDww/z1HUVZYDvyD1rzu/0+8+Hl/NrGj28k/hyd/M1HTYutqehlhHp0yv8h90A9FBzS1U0/ULTVLGG9sbiOe2nXfHKh4Yf56jqKtA5oAWiiigAooooAKKKKACiiigAooooAKQnHpQTj0qpqWp2mkafPf6hOkFrAu6SRzwB/Uk8AetACanqVnpOnT3+oTrb20C7pJGOMD/E9AOpryDV/BOt/E+8bxN5i6RAU2afbXG4u8YBKu2D8m5sdO3POBu6fTdMvPH2owa7rtu8GhwN5mmaXL1l44mmHfOeF6fUcn0MDPc0Acz4R8UprMc2n3lu1jrljhLyykYkg8AOpJyyHjB9x1yCemVt2a5jxZ4TOsGDU9LnFjr9lk2l4BnI7xv6oecjtk+py7wp4rXXFnsb23Njrlnhbyyc8j0dfVD2PP48EgHT0U1W3Zp1ABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFNZtuKABm24rmvFfitdDWCxsrc32uXmVs7JDyfV29EHc8fhyQeK/Fa6GsFjZW5vtcvMrZ2SHk+rt6IO54/DkhvhPwodGafU9UuPt2v3uDd3ZGAPSNB2QdvX8gAA8J+FDozT6nqlx9u1+9wbu7IwB6RoOyDt6/kB1AUDOM80BQM4zzS0ABGaaQB16Hr706gjNAHnN/p958PL+bWNHt5J/Dk7+ZqOmxdbU9DLCPTplf5D7vd6fqFpqljDe2NxHPbTrvjlQ8MP89R1FWSAOvQ9fevO7/T7z4eX82saPbyT+HJ38zUdNi62p6GWEenTK/yH3QD0UHNLVTT9QtNUsYb2xuI57add8cqHhh/nqOoq0DmgBaKKKACiiigAooooAKQnHpQTj0qpqWp2mkafPf6hOkFrAu6SRzwB/Uk8AetABqWp2mkafPf6hOkFrAu6SRzwB/Uk8AetcNpumXnj7UYNd123eDQ4G8zTNLl6y8cTTDvnPC9PqOSabpl54+1GDXddt3g0OBvM0zS5esvHE0w75zwvT6jk+hj60AA+tOxRiigBCoOM54rl/FnhQ6y0Gp6XcfYdfssm0uwMg+sbjuh7+n5g9TSFQcZzxQBzPhTxWuuLPY3tubHXLPC3lk55Ho6+qHsefx4J6VW3ZrmPFnhQ6y0Gp6XcfYdfssm0uwMg+sbjuh7+n5gu8KeK11xZ7G9tzY65Z4W8snPI9HX1Q9jz+PBIB09FNVt2adQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUU1m24oAGbbiua8V+K10NYLGytzfa5eZWzskPJ9Xb0Qdzx+HJB4r8VroawWNlbm+1y8ytnZIeT6u3og7nj8OSG+E/Ch0Zp9T1S4+3a/e4N3dkYA9I0HZB29fyAADwn4UOjNPqeqXH27X73Bu7sjAHpGg7IO3r+QHUBQM4zzQFAzjPNLQAUUUUAFFFFAARmmkAdeh6+9OoIzQB5zf6fefDy/m1jR7eSfw5O/majpsXW1PQywj06ZX+Q+73en6haapYw3tjcRz206745UPDD/PUdRVkgDr0PX3rzu/0+8+Hl/NrGj28k/hyd/M1HTYutqehlhHp0yv8AIfdAPRQc0tVNP1C01SxhvbG4jntp13xyoeGH+eo6irQOaAFooooAKQnHpQTj0qpqWp2mkafPf6hOkFrAu6SRzwB/Uk8AetABqWp2mkafPf6hOkFrAu6SRzwB/Uk8AetcNpumXnj7UYNd123eDQ4G8zTNLl6y8cTTDvnPC9PqOSabpl54+1GDXddt3g0OBvM0zS5esvHE0w75zwvT6jk+hj60AA+tOxRiigAooooAKKKKAEKg4zniuX8WeFDrLQanpdx9h1+yybS7AyD6xuO6Hv6fmD1NIVBxnPFAHM+FPFa64s9je25sdcs8LeWTnkejr6oex5/HgnpVbdmuY8WeFDrLQanpdx9h1+yybS7AyD6xuO6Hv6fmC7wp4rXXFnsb23Njrlnhbyyc8j0dfVD2PP48EgHT0U1W3Zp1ABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUU1m24oAGbbiua8V+K10NYLGytzfa5eZWzskPJ9Xb0Qdzx+HJB4r8VroawWNlbm+1y8ytnZIeT6u3og7nj8OSG+E/Ch0Zp9T1S4+3a/e4N3dkYA9I0HZB29fyAADwn4UOjNPqeqXH27X73Bu7sjAHpGg7IO3r+QHUBQM4zzQFAzjPNLQAUUUUAFFFFABRRRQAUUUUABGaaQB16Hr706gjNAHnN/p958PL+bWNHt5J/Dk7+ZqOmxdbU9DLCPTplf5D7vd6fqFpqljDe2NxHPbTrvjlQ8MP89R1FWSAOvQ9fevO7/T7z4eX82saPbyT+HJ38zUdNi62p6GWEenTK/yH3QD0UHNBOPSqun6haapYw3tjOlxbTrvjlQ8MP8APUdsUmpanZ6Rp8+oX86QWsC7pJHPAH9Sew75oANS1O00jT57/UJ0gtYF3SSOeAP6kngD1rhtN0y88fajBruu27waHA3maZpcvWXjiaYd854Xp9RyTTdMvPH2owa7rtu8GhwN5mmaXL1l44mmHfOeF6fUcn0MfWgAH1p2KMUUAFFFFABRRRQAUUUUAFFFFACFQcZzxXL+LPCh1loNT0u4+w6/ZZNpdgZB9Y3HdD39PzB6mkKg4znigDmfCnitdcWexvbc2OuWeFvLJzyPR19UPY8/jwT0qtuzXMeLPCh1loNT0u4+w6/ZZNpdgZB9Y3HdD39PzBd4U8Vrriz2N7bmx1yzwt5ZOeR6Ovqh7Hn8eCQDp6Karbs06gAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiims23FACsSMV5z46+K9p4V1Q6RbWRu7/y8szuUjiZgNgPGT1BOOg9+nQeK/Fi6GkFjZW5vtcvMrZ2SHk9i7eiDuePw5Io+H/h9Z291FrWv7dU8RGTz5LuQnbG+MBUXptUdOPcYwMAF3wn4TOjNPqeqT/btfvcNdXZHA9I09EHT3x24A6cKBnGeaAMUtABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFAARmmkAdeh6+9OoIzQB5b4gkk+FF02s6Yiy6DfzbbjSt+3yZipw8JxgA7eR/8AWwvhtpPindJ4g1VFTRrKYraaVu3K0oUZklOMN97gdO2OufRtR0yy1fTp7C/t0uLWddskbjhv8D3BHINLp+nWmlWENjYwLBawrsjjToo/qe5J5JNAFgfWnYoxRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAhUHGc8VzHizwodYMGp6XcCx1+yybS7A4b1jk4OUPQ9cZ6HkHqKQjNAHAeA/ibb+Lr+XSprJrTUIot52tvjkwcMVPYZ6Z7V3xkUHGao2mh6XYaleajaWMMN3eEG4lRcFyPX/PPU1oUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRTWbbigAZtuK5rxX4rXQ1gsbK3N9rl5lbOyQ8n1dvRB3PH4ckHivxWuhrBY2Vub7XLzK2dkh5Pq7eiDuePw5Ib4T8KHRmn1PVLj7dr97g3d2RgD0jQdkHb1/IAAPCfhQ6M0+p6pcfbtfvcG7uyMAekaDsg7ev5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proof	Combinatorics	Math	English
41	"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.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<image_1>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<image_2>
Prove that $S_{5}[495138627]$ is unique."	['Let $L=a_{1}, \\ldots, a_{9}$ and suppose $S_{5}[L]=S_{5}[495138627]=\\left(\\omega_{1}, \\ldots, \\omega_{5}\\right)$. Then we get the following inequalities:\n\n| $a_{4}<a_{8}$ | $\\left[\\right.$ from $\\left.\\omega_{4}\\right]$ | $a_{8}<a_{5}$ | $\\left[\\right.$ from $\\left.\\omega_{4}\\right]$ |\n| :--- | :--- | :--- | :--- |\n| $a_{5}<a_{1}$ | $\\left[\\right.$ from $\\left.\\omega_{1}\\right]$ | $a_{1}<a_{3}$ | $\\left[\\right.$ from $\\left.\\omega_{1}\\right]$ |\n| $a_{3}<a_{7}$ | $\\left[\\right.$ from $\\left.\\omega_{3}\\right]$ | $a_{7}<a_{9}$ | $\\left[\\right.$ from $\\left.\\omega_{5}\\right]$ |\n| $a_{9}<a_{6}$ | $\\left[\\right.$ from $\\left.\\omega_{5}\\right]$ | $a_{6}<a_{2}$ | [from $\\left.\\omega_{2}\\right]$ |\n\nCombining, we get $a_{4}<a_{8}<a_{5}<a_{1}<a_{3}<a_{7}<a_{9}<a_{6}<a_{2}$, which forces $a_{4}=1, a_{8}=2, \\ldots, a_{2}=9$. So the label $L$ is forced and $S_{5}[495138627]$ is therefore unique.']			['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proof	Combinatorics	Math	English
42	"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.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<image_1>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<image_2>
Show that for each $k \geq 2$, the number of unique $2^{k-1}$-signatures on the set of $2^{k}$-labels is at least $2^{2^{k}-3}$."	"[""Let $s_{k}$ denote the number of such unique signatures. We proceed by induction with base case $k=2$. From $8(\\mathrm{c})$, a 2-signature for a label $L$ is unique if and only if consecutive numbers in $L$ appear together in some window. Because $k=2$, the consecutive numbers must be adjacent\n\n\nin the label. The $2^{2}$-labels 1234 and 4321 satisfy this condition, ${ }^{1}$ so their $2^{1}$-signatures are unique. Thus we have shown that $s_{2} \\geq 2=2^{2^{2}-3}$, and the base case is established.\n\nNow suppose $s_{k} \\geq 2^{2^{k}-3}$ for some $k \\geq 2$. Let $L_{k}$ be a $2^{k}$-label with a unique $2^{k-1}$-signature. Write $L_{k}=\\left(a_{1}, a_{2}, \\ldots, a_{2^{k}}\\right)$. We will expand $L_{k}$ to form a $2^{k+1}$ label by replacing each $a_{i}$ above with the numbers $2 a_{i}-1$ and $2 a_{i}$ (in some order). This process produces a valid $2^{k+1}$-label, because the numbers produced are all the integers from 1 to $2^{k+1}$. Furthermore, different $L_{k}$ 's will produce different labels: if the starting labels differ at place $i$, then the new labels will differ at places $2 i-1$ and $2 i$. Therefore, each starting label produces $2^{2^{k}}$ distinct $2^{k+1}$ labels through this process. Summarizing, each valid $2^{k}$-label can be expanded to produce $2^{2^{k}}$ distinct $2^{k+1}$-labels, none of which could be obtained by expanding any other $2^{k}$-label.\n\nIt remains to be shown that the new label has a unique $2^{k-1}$-signature. Because $L_{k}$ has a unique $2^{k-1}$-signature, for all $i \\leq 2^{k}-1$, both $i$ and $i+1$ appeared in some $2^{k-1}$-window. Therefore, there were fewer than $2^{k-1}-1$ numbers between $i$ and $i+1$. When the label is expanded, $2 i$ and $2 i-1$ are adjacent, $2 i+1$ and $2 i+2$ are adjacent, and $2 i$ and $2 i+1$ are fewer than $2 \\cdot\\left(2^{k-1}-1\\right)+2=2^{k}$ places apart. Thus, every pair of adjacent integers is within some $2^{k}$-window.\n\nSince each pair of consecutive integers in our new $2^{k+1}$-label coexists in some $2^{k}$-window for every possible such expansion of $L_{k}$, that means all $2^{2^{k}}$ ways of expanding $L_{k}$ to a $2^{k+1}$-label result in labels with unique $2^{k}$-signatures. We then get\n\n$$\n\\begin{aligned}\ns_{k+1} & \\geq 2^{2^{k}} \\cdot s_{k} \\\\\n& \\geq 2^{2^{k}} \\cdot 2^{2^{k}-3} \\\\\n& =2^{2^{k+1}-3}\n\\end{aligned}\n$$\n\nwhich completes the induction.""]"			['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proof	Combinatorics	Math	English
43	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Without using Descartes' Circle Formula, Show that the circles marked off and sold on day 1 are centered at $\left(0, \pm \frac{2}{3}\right)$ with radius $\frac{1}{3}$."	['By symmetry, $P_{1}, P_{2}$, the two plots sold on day 1 , are centered on the $y$-axis, say at $(0, \\pm y)$ with $y>0$. Let these plots have radius $r$. Because $P_{1}$ is tangent to $U, y+r=1$. Because $P_{1}$ is tangent to $C$, the distance from $(0,1-r)$ to $\\left(\\frac{1}{2}, 0\\right)$ is $r+\\frac{1}{2}$. Therefore\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{2}\\right)^{2}+(1-r)^{2} & =\\left(r+\\frac{1}{2}\\right)^{2} \\\\\n1-2 r & =r \\\\\nr & =\\frac{1}{3} \\\\\ny=1-r & =\\frac{2}{3} .\n\\end{aligned}\n$$\n\nThus the plots are centered at $\\left(0, \\pm \\frac{2}{3}\\right)$ and have radius $\\frac{1}{3}$.']			['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proof	Geometry	Math	English
44	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Given three mutually tangent circles with curvatures $a, b, c>0$, suppose that $(a, b, c, 0)$ does not satisfy Descartes' Circle Formula. Show that there are two distinct values of $r$ such that there is a circle of radius $r$ tangent to the given circles."	"[""Apply Descartes' Circle Formula to yield\n\n$$\n(a+b+c+x)^{2}=2 \\cdot\\left(a^{2}+b^{2}+c^{2}+x^{2}\\right),\n$$\n\na quadratic equation in $x$. Expanding and rewriting in standard form yields the equation\n\n$$\nx^{2}-p x+q=0\n$$\n\nwhere $p=2(a+b+c)$ and $q=2\\left(a^{2}+b^{2}+c^{2}\\right)-(a+b+c)^{2}$.\n\nThe discriminant of this quadratic is\n\n$$\n\\begin{aligned}\np^{2}-4 q & =8(a+b+c)^{2}-8\\left(a^{2}+b^{2}+c^{2}\\right) \\\\\n& =16(a b+a c+b c)\n\\end{aligned}\n$$\n\nThis last expression is positive because it is given that $a, b, c>0$. Therefore the quadratic has two distinct real roots, say $d_{1}$ and $d_{2}$. These usually correspond to two distinct radii, $r_{1}=\\frac{1}{\\left|d_{1}\\right|}$ and $r_{2}=\\frac{1}{\\left|d_{2}\\right|}$.\n\nOne possible exception case to consider is if $r_{2}=r_{1}$, which would occur if $d_{2}=-d_{1}$. This case can be ruled out because $d_{1}+d_{2}=p=2(a+b+c)$, which must be positive if $a, b, c>0$. (Notice too that this inequality rules out the possibility that both circles have negative curvature, so that there cannot be two distinct circles to which the given circles are internally tangent.)\n\nWhen both roots $d_{1}$ and $d_{2}$ are positive, the three given circles are externally tangent to both fourth circles. When one is positive and one is negative, the three given circles are internally tangent to one circle and externally tangent to the other.\n\n\n\nWhile the foregoing answers the question posed, it is interesting to examine the result from a geometric perspective: why are there normally two possible fourth circles? Consider the case when one of $a, b$, and $c$ is negative (i.e., two circles are internally tangent to a third). Let $A$ and $B$ be circles internally tangent to $C$. Then $A$ and $B$ partition the remaining area of $C$ into two c-triangles, each of which has an incircle, providing the two solutions.\n\nIf, as in the given problem, $a, b, c>0$, then all three circles $A, B$, and $C$, are mutually externally tangent. In this case, the given circles bound a c-triangle, which has an incircle, corresponding to one of the two roots. The complementary arcs of the given circles bound an infinite region, and this region normally contains a second circle tangent to the given circles. To demonstrate this fact geometrically, consider shrink-wrapping the circles: the shrink-wrap is the border of the smallest convex region containing all three circles. (This region is called the convex hull of the circles). There are two cases to address. If only two circles are touched by the shrink-wrap, then one circle is wedged between two larger ones and completely enclosed by their common tangents. In such a case, a circle can be drawn so that it is tangent to all three circles as shown in the diagram below (shrink-wrap in bold; locations of fourth circle marked at $D_{1}$ and $D_{2}$ ).\n\n<img_3391>\n\nOn the other hand, if the shrink-wrap touches all three circles, then it can be expanded to make a circle tangent to and containing $A, B$, and $C$, as shown below.\n\n\n\n<img_3298>""]"			['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proof	Geometry	Math	English
45	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation."	"['In this case, the fourth ""circle"" is actually a line tangent to all three circles, as shown in the diagram below.\n\n<img_3396>']"			['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proof	Geometry	Math	English
46	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$.
Prove that $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$."	['Note that\n\n$$\n\\begin{aligned}\n\\frac{1}{\\rho} & =\\frac{\\phi^{2}-\\phi}{\\phi+\\sqrt{\\phi}} \\\\\n& =\\phi-\\sqrt{\\phi}\n\\end{aligned}\n$$\n\n\n\nTherefore\n\n$$\n\\begin{aligned}\n\\left(\\rho-\\frac{1}{\\rho}\\right)^{2} & =(2 \\sqrt{\\phi})^{2}=4 \\phi \\\\\n& =2\\left(\\rho+\\frac{1}{\\rho}\\right) .\n\\end{aligned}\n$$\n\nMultiplying both sides of the equation by $\\rho^{2}$ gives $\\left(\\rho^{2}-1\\right)^{2}=2\\left(\\rho^{3}+\\rho\\right)$. Expand and isolate $\\rho^{4}$ to obtain $\\rho^{4}=2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1$.\n\nAlternate Proof: Because $\\phi^{2}=\\phi+1$, any power of $\\rho$ can be expressed as an integer plus integer multiples of $\\sqrt{\\phi}, \\phi$, and $\\phi \\sqrt{\\phi}$. In particular,\n\n$$\n\\begin{aligned}\n\\rho^{2} & =\\phi^{2}+2 \\phi \\sqrt{\\phi}+\\phi \\\\\n& =2 \\phi \\sqrt{\\phi}+2 \\phi+1, \\\\\n\\rho^{3} & =4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4, \\text { and } \\\\\n\\rho^{4} & =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 .\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1 & =2(4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4)+2(2 \\phi \\sqrt{\\phi}+2 \\phi+1)+2 \\rho-1 \\\\\n& =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 \\\\\n& =\\rho^{4}\n\\end{aligned}\n$$']			['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Vz/jTw2vinwne6ZnbcMnmWsgODHMvKMD25Az7ZpngTxC/ibwlZ6hcLsvQDDeRkYKTodrjHbkZA7BhQB0tFFFABRRRQAUUUUAFIeATS0HpQByHibxufDuvaLpX9k3M39qXkdqtyXVY1LFQcckkjd6Ae9daOteb/FL/AJD/AIB/7D8P8xXpVABRRRQAUUUUAFFFFAHM/EDRf+Eg8BazpyKWle2aSEA4PmJ86f8AjyirnhLWR4h8JaTq2QWurVJJMdA+MMPwYEfhWyfunjPtXB/CoCx0fWNAC7E0fV7m1hXP/LIt5iH8d5/KgDvaKKKACkPApaRiFUknAHJNAHh3w7tX8R/GnxTrlxseDS7qZIGA53ufKU59PLiI/wCBV7lXlvwHtHHgi51abBm1S/luGb2Hy4/NW/OvUqAEPSvO4Vlt/j40H2u6lgk8PNP5U0pZEczqp2r0HCjpXoteeyEf8NBxDPP/AAjB4/7eaAOm8U3M0WgyW1s5S7vmWzgYdUeQ7S4/3VJf/gNZuqeBdP12VIdUkuDpltEkVnYwXDxRoFH322kFm7DJIAAx1NXJ/wDiY+NLaAcw6Vb/AGlv+u0u6NPxCLL/AN9iovEniqy0qaPSl1OxtNSuVLK11KqJAnQyMCRn2Xqx9AGZQDE+Fy6jaDxDo91ezXtjpepPa2VxO259oAJQnvtyPzPoAPQq53wtd+H0g/sjQtRgvfsy+ZM8MwmJZySXkZeN7Nub1PPFdFQAUUUUAFFFFABRRRQAUUUUAIelcB4c/wCJD8V/EehgBbXVYU1i2VRgK/8Aq5vxLbWr0A9K4Lxp/wAS7x34I1rfsQ3kunSc43CaM7Af+BJQB3tFNGc4p1ABRRRQAUUUUAFIehpaKAPMfiYbm41/wf8AZdO1K5Fjq8dzctbWM0ixxgrkkqpB+gJr0iCQTRJIofDLkb0KH8VPI+h5qaigAooooAKKKKACiiigBD0rhPDyGw+L3i+1LYS9trO9jTHAwrRsfzAruz0rhJ4/s/x0s7jkC78Pyw+xKTq354NAHeUUUUAB6VzvjPxPpvhTw3c3+oXXkh0aOAAEmSQqSqrjvx16DvgV0J6V5b8bfBWoeLPDtpdaYwa40xpJWgZ9qyRlcnAPBYbRj6tQBY+B+r6XefDqx02yulkurDcLqLBDRl5HYZyOc54Iz068GvS6+dvhD8LrzUdHk1y513VdKgvFUW39kXnkySKGYMZPlPGQMD6+1ek/8Ks/6nzxx/4OP/sKAPQDWW3h3RDqP9pHR9P+3A7vtP2ZPNz67sZ/WuT/AOFWf9T544/8HH/2FH/CrP8AqfPHH/g4/wDsKAO4jtYIZ5po4lWWdg0rAcuQoUZ/AD8qzL3wn4c1K7ku7/QNKu7mQ/PNPZxu7Y4GWKknAAH4VzX/AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAdfpeh6VowkXS9MsrFZDmRbWBYgxHTO0DPWtA9K4D/AIVZ/wBT544/8HH/ANhR/wAKtx/zPnjj/wAHH/2FAG14u8caH4KtYp9Zumjafd5EUaF3lx1AHTjI5JA5q34Y8U6R4u0z+0dGuvtEAcxvlSrI2M7SCBg4I9vQmvHfiP8ABjWruKzudD1XVdcljVlli1W+EkijjHlkgDHXIz6Ve+H3wa1PT9HlfV9f1vR7ydwTb6RqAjUKBgbyFIZs7uhwBjBNAHt9Fef/APCrP+p88cf+Dj/7Cj/hVn/U+eOP/Bx/9hQB6BRXn/8Awqz/AKnzxx/4OP8A7Cj/AIVZ/wBT544/8HH/ANhQB6BRXn//AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAd+3CnnFeWfGPxLo2lQaJb3V0q6hbana6jHAEJby0chm46cbvrg1pn4W4/5nzxwf8AuMf/AGFeUfFf4W6ppl5a6pZ6ne6tbTGK1MmpXYkuBIzEKNxCjbyMehznigD6G0TW9P8AEOlwanpVytzZzg7JACMkEg5B5BBHQ1pV494M+Dl5pnh2GHUfFPiHTryRmklt9J1DyoVJOBxtOWwBk/gOma6H/hVn/U+eOP8Awcf/AGFAHoFFef8A/CrP+p88cf8Ag4/+wo/4VZ/1Pnjj/wAHH/2FAHoFFef/APCrP+p88cf+Dj/7Cj/hVn/U+eOP/Bx/9hQB6BRXn/8Awqz/AKnzxx/4OP8A7Cj/AIVZ/wBT544/8HH/ANhQB6BRXn//AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAegUV5/wD8Ks/6nzxx/wCDj/7Cj/hVn/U+eOP/AAcf/YUAegUV5/8A8Ks/6nzxx/4OP/sKP+FWf9T544/8HH/2FAHoFFef/wDCrP8AqfPHH/g4/wDsKP8AhVn/AFPnjj/wcf8A2FAHft9015B4p8f+GdO+LmhvNqeBp0N1bX7ojMsbOF2KcDnkHOM4OM4xW+fhbgZ/4TzxwfY6v/8AYV4j4p+EXiG18fR6Tayi8GqPLNa3NxMN7qvLGU/3xkZOPmzkd8AH1TFMssSSRsXjdQysvIIPQ0/efRvyrN8P6Y2ieHNM0rzjL9jtY7fzNv3tqhc/TitHLf3v/HDQBJUVzCtxaywuAVkQowPoRipaQ9KAPOvgbdrdfCrTIw2WtpJon9j5jMB+TCvRq8b+Ct+lj4j8Z+Fiyg2upSzxLnkjeY2/Lan517JQAUUUUAFFFFABRRRQAUUUjfdPOKAKmqalb6Tp0t7clvLjAwqLuZ2JwqqO7EkAD1Irjfhbr2peJrLXdV1IlHbVJIIoA+5YY0RAFH4k5Pc5Pelm1a+1LVxq6aFeX2jWQL2LxzQqsj8hpiHcEgDIXjGCWGcjFL4GRMPhpa3D8tdXM8xOep3lf/ZTQB6VRRRQAUUUUAB6VwXxIVru98HaZG3zz69BM68fNHEGdv6V3p6V5/dFdc+NlhAvzQ+H9NkndgeFmnwgU++wE0Ad8cYrz34v3t1ZeE7caZdXdvq13ew2tm1vcPGS7NkghSN2QCOQeteh9q8u8VW//CWfGDQ9BE8sdvo9pJqU7wttYSMQqc9mB2n6NQBF8QHvvh9oGnatpGtajNfC6jga2u7p7hLzIO4bXJIPHVcEV6lCxaNCy7GKjK+ntXkF/bL4Z+M+g217LPrkOpxnyH1FzLLYuM/NH/CBwM8Z6817Ep56/hQA6iiigAooooAKKKKACiiigAooooAKKKKAA9K4bU8XHxn8Pxct9l0q6nx027mRM+9dwehrhLJFvPjlqdypyLDQobVh6NJK0n8hQB3lFFFABQaKQnAoA8E1GVvB3xp1TXApSyiuoDdnput7mMKzn2SVB+de9j2rzPxnpVtN8SNJS9jLWPiDTbjR7hs8KR+9jP8AvZ3AfStP4bavdNYXPhnV3H9taEwtpuf9dFj91KPUFcDPXjnrQB3VFFFABRRRQAUUUUAB6Vz3ivRdV1/SnsLDWI9NhlG2dvshld17qDvXbkcH2PauhooAx9T06+vPD0mnWd7b2c0sBhM/2YsqgqQSibxjHbJOPeqHgXwvdeD/AA5b6JJqMN7b227ymW1MLjcxY7vnYHlu2K6eigAooooAKKKD0oAhvLmGysp7q5lWKCGNpJJGOAiqMkn6AZriPhdbTXenal4rvIil34gujdKGHzJbr8sKn2C8j/eqH4g3EviDUbDwHp7sr6hifUpUPMFmp5+hcjaPXkd67y2ghtLeK3t41ihiQIkaDCqoGAB6AAUASTBzC4iZVkI+VmXIB7ZGRn8xXB6d4J8Q6Z4l1jXYvEmnSXWqlPN83SXIjVBhVTE4OAMdc9BXf0UAcfongSOz8St4m1fU59X1ryzFFNJGI44EPGI0HTgkZyTz7knsKKKACiiigAooooAKKKKACiiigAooooAKKKKAEPSuC+H4W/8AFPjjWwD+/wBVFkCe4t0CZHtkmuw1nUY9H0S/1OYZjtLd52HqFUnH6Vzfws0+Ww+HWkm5Yvc3aNezOerNKxkyffDD8qAOyooooAKDRRQBw3xVs55PBp1W0TdeaJcxanCM4/1bZb/xwtVXxZY3Nwum+PvCqm41C2twzQIP+P8As2AYx8fxAHcvXnPXiu/uIIrq2lt50WSGVCjo3RlIwQfwrhfhdcS2elaj4UvHZrzw/dtagv8AeeBiXhc+xXp7LQB1Ph/XrDxLo9tq2mTebbXC5GeCh7qR2YHOR+vStWvNtZ0rU/A2t3Pijw3bvdaVdMZNX0pG6t3nhHZxzuHGf1XttE1vTvEWmwanpV0lxaSjKsv6gjqCOhBoA06KKKACiiigAooooAKKKKACiig9KAEPSsDxd4ptPCehPf3MbTzORFa2sfL3MzfdjUe/48Z61L4m8T6Z4U0p7/VJiqk7IokGZJ3PREXqT/Lqa5nwx4d1PW9bTxl4uiEd4EK6bph5WwiPOW9ZD3Pbp6BQDR8CeGrvSLa61bW3WbxDqzie+kHSMY+WJf8AZQcfXPtXYU0dRTqACiiigAooooAKKKKACiiigAooooAKKKKACiiigAoPSig9KAOB+K00t14fsfDdszC416+isgUPKRZDSP8AQKuD7NXdQxJDGkUahY0UKqjooHAArgrPPiX4xXV5jdYeGrX7LCexuphmQg+yYUj1Ir0GgAooooAKKKKAEb7przvxQT4T+IWj+KF+TTtSUaVqTDOEYnMEhA44OVJPQcd69FrL8SaFa+JfDl/o14P3F3EYy2MlD1Vh7ggEe4oA0TwCAM8dK4PV/BWoaTq0/iDwPcRWd/Md15p0xP2W+PqR/A/+0MZJ5xkk3Phzr11quhyadqxI1zR5DZXyk5LMvCye4ZRnPcg12ZoA4/w78QNN1i9OlahDLo+ux8Sade/KxPrG3Rx3BHbnGK68VjeIvC2ieKbL7LrOnRXSL9wtwyH/AGWHIrlR4f8AHHhVf+Kf1uPXbBOlhrJxMq/7M68k9AA3AFAHolFefL8UI9MBTxX4d1jQXX787QG4tgfQSxg5/Kt2w8feEtTCfZf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proof	Geometry	Math	English
47	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$.
Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have common ratio $\rho$."	"[""If the radii are in geometric progression, then so are their reciprocals (i.e., curvatures). Without loss of generality, let $(a, b, c, d)=\\left(a, a r, a r^{2}, a r^{3}\\right)$ for $r>1$. By Descartes' Circle Formula,\n\n$$\n\\left(a+a r+a r^{2}+a r^{3}\\right)^{2}=2\\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\\right)\n$$\n\nCancel $a^{2}$ from both sides of the equation to obtain\n\n$$\n\\left(1+r+r^{2}+r^{3}\\right)^{2}=2\\left(1+r^{2}+r^{4}+r^{6}\\right) .\n$$\n\nBecause $1+r+r^{2}+r^{3}=(1+r)\\left(1+r^{2}\\right)$ and $1+r^{2}+r^{4}+r^{6}=\\left(1+r^{2}\\right)\\left(1+r^{4}\\right)$, the equation can be rewritten as follows:\n\n$$\n\\begin{aligned}\n(1+r)^{2}\\left(1+r^{2}\\right)^{2} & =2\\left(1+r^{2}\\right)\\left(1+r^{4}\\right) \\\\\n(1+r)^{2}\\left(1+r^{2}\\right) & =2\\left(1+r^{4}\\right) \\\\\nr^{4}-2 r^{3}-2 r^{2}-2 r+1 & =0 .\n\\end{aligned}\n$$\n\nUsing the identity from 4a, $r=\\rho$ is one solution; because the polynomial is palindromic, another real solution is $r=\\rho^{-1}=\\phi-\\sqrt{\\phi}$, but this value is less than 1 . The product of the corresponding linear factors is $r^{2}-2 \\phi r+\\phi^{2}-\\phi=r^{2}-2 \\phi r+1$. Division verifies that the other quadratic factor of the polynomial is $r^{2}+(2 \\phi-2) r+1$, which has no real roots because $(\\sqrt{5}-1)^{2}<1$.""]"			['/9j/2wCEAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDIBCQkJDAsMGA0NGDIhHCEyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMv/AABEIATABxwMBIgACEQEDEQH/xAGiAAABBQEBAQEBAQAAAAAAAAAAAQIDBAUGBwgJCgsQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2drh4uPk5ebn6Onq8fLz9PX29/j5+gEAAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoLEQACAQIEBAMEBwUEBAABAncAAQIDEQQFITEGEkFRB2FxEyIygQgUQpGhscEJIzNS8BVictEKFiQ04SXxFxgZGiYnKCkqNTY3ODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqCg4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2dri4+Tl5ufo6ery8/T19vf4+fr/2gAMAwEAAhEDEQA/APf6KKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKQ8CgBaKztU13StFhEuqanZ2SHobiZY930yefwrlH+LXh+dnTRbbVtekQ4ZdMsJJNp/3iAP1oA7yiuD/wCEr8b3rgaf4BeGJhkT6jqMcWPrGoZqao+K1wTvfwnZxkcBVnlcdOvIHrQB31FcN/YnxFlO5/GWmQHpti0gMPzZ6aPDvxDDs/8AwntqQQMIdFjwv/j2aAO7orhv7I+I8GfL8V6Rc9x5+llB9PlemLP8U7VyZrXwpfxjosEs8Ln/AL6BAoA7yiuB/wCE38VWAZtZ+HupLGvRtNuYrwsPZQQamtfix4UluEtr27n0m6brDqds9uR9SRtH50AdxRVazv7TUbdbiyuoLmBxlZYZFdT+IOKsA84PWgBaKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooPtTXcKpYnAHc9qAFJwKhuLqG0t3uLmWOGGMbpJJGCqo9STgAVxGofEV7+9fSvBennXr9Tte6DbLO3Pq8vRj0O1evOD2qO0+HEuszx33jnVZNbuVYOlkmYrOE/wCygwWxyMt1BORQBJcfE6G+uHs/COk3niS6Q7Glth5dqjejTN8o45GM5qM+HvHfiMf8T3xNHotq3Wz0OPEhXtmd+Q3rtGK7y2tbezt47e2gjghjXakcShVQegA4FS4HpQBx+l/DDwjp0pnfSY7+7bBe51BjcyO3qS5IB+gFdekaRoqIiqqjCqowB9KdRQAhAPUUYFLRQAYowKKKAEwD2pcUUUAJgVDd2drewNBd20M8LcNHKgZT+BqeigDiLz4VeGnuGu9KjutDvSAPtGk3DQED/dHyY/Cqn2b4jeGsNa3lj4qshn9zcj7JdY7AOMofq3NehYpNoNAHF6X8TdEurwadqy3Og6nwPsuqR+Vu7fK/3WGeBzk+ldmDk1R1fRdM12xaz1WxgvIG/gmQNj3BPQ+45riW8G+IfCDGfwPqhmsxnOiapI0kWPSKQ/Mh9Acgk8mgD0aiuQ8P/ECw1a/Gj6lbTaNroGW0684ZveNuki8HGOeDxXWq2SeelADqKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigApCcUprkfFnjVNAlt9L0+1bU/EF3/x7afCfmA/56Of4EGDyf5AkAGl4k8VaV4V043uqXIRSdsUKDdLM/ZUXqxP5euK5BND8R/ET994nMui+HnwYtGgciacf9PD9h/sDHXkArzreGfBD2+pf8JD4mul1bxC4+WUr+5tAf4IVPQDpu6n25z2mBnOOaAKmm6ZY6RYx2WnWkNraxjCxQqFUfh3PvVvFLRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFJtU54HPX3paKAMbxH4W0bxTp/2PV7GO4Qf6tzkSRH1Rxyp+n45FcZ/aPiT4bYGqyXHiHwwM4v1G67sk/6agf6xMfxDkYOccCvTKawBGMDB4oAq6ZqdnrFhDf6fdxXVpMu6OWI5Df8A1+xH8quV5zqfhTVPCOoT694GRSkjeZfaExxDcju0f/POTH4Hj0weo8L+LNO8WaZ9ssHZWjbZcW8o2yW8g6o47Ec+1AG9RSDOaWgAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACkPShjhcj+dct408Wnw9ZQ2lhb/bdd1BvJ0+yx99/7zeiLnJPHHcdQAVvGHjC50y6tdC0KCK+8RX5Hk27Z2QJ3llxyEH69ulWvB3g6Hw5FNd3c51DXL0iS+1GUZeVuOF/uoMYCjHQU3wX4PHhyCe7v5/t2u3zebfX7Dl2/ur6IOgHH07DqwMUAJgZzS0UUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFADSOK4bxZ4QvF1UeK/CZjt/EMI/fQk4i1CMY/dyf7RAwG9hz0I7sjPWkI4oA57wj4tsvFmltc2yvBdQN5V3ZyjEltIOqsODjg4Pf8CB0I61wXjHw5f2Wqr4z8LR51i3QC7tBwuoQDqh4++APlPtjngV03hnxDYeKNEg1XTZN0EowVbho2HVGHZgf84IJANiiiigAooooAKKKKACiiigAooooAKKKKACiiigAoopCaAFopOc0ZP+RQAtFIKWgAooooAKKKKACiiigApDxS0jHAzQBl+INdsvDeh3Wr6lJstbZN7cZZj2Ue5JAFcz4G0C9uL2bxl4kixrmoJiC3Y5FhbdVjX/AGjnLH1OMA5znon/AAsTx80p+fwz4cnwg/hvL0dT7qn6npkE49JXr+Hf/P1oAUAA5xz0zS0UUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUANZeOBz6mvNvEEUvw68SyeLLFCdB1GRU1m2UE+S5OFuVA9zhh3z6nI9KIBGCM1Dd21vdWk1tcxJJbzIySxuPldSDkH2IzQA6CZLiJJonWSJ1DI6kEMDyCD3BqWvOfBdxceEvEVx4Cv5JHtwpudEuJTky2+TuiJ7sh/TsBivRQc0ALRRRQAUUUUAFFFFABRRRQAUh4FDZ4x61zHijxxpvhlorMrLfavcD/RtNtF3zSe5H8K8H5j6HGcYoA6YtjJrkNU+JvhzT7s2FpNPrGpD/lz0qI3Mnv8Ad+UY4zkislfCXiPxkfO8aag1jp78roemybVx1xNKDlz6gHHAI9K7bSND0rQbQWmlafb2cA6pDGFz7n1PuaAOQGs/EjWudN8O6XosOeJNWuTLIy+oSL7p9iaePC3jq7lL33xAMKMP9TY6XEgX6OxLV3uBRigDhh8PdQb5pfHvigyHqUuI0X8tlNbwR4igjIsPiDrEbD7puoIbgfkVH867ujFAHBfY/idpYHkavoOtoB832y1e2kb6eWSufrTB8SLzSBjxf4V1PR0HLXkAF5bKPVnj5XPpg16BikIB6igDP0nXdM12z+16VqFvewdC8Dhtp9CByD7HmtAEnqK4vWfhro95eHU9HeXQNYwdt5p3yBj1+dB8rgnBI4JwOaz4fGOt+EJ47Lx3bxm0ZgkOvWaEwtnp5yAZjY+uNuenAJoA9FoqK3niuYUmgkSSKRQyOhDAg8ggjgipaACiiigBDXHfELXrzTtHt9K0dv8AieazMLOyA6x5+/L7BFyc9jiuxbpXnnhHb4u8cat4vkUNZWedM0kkcFVOZZl/3mOAR2yD3oA63w1oFl4Z8P2ejWC/6PaxhNx6u3UsfcnJ/GtfFGMUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABQQD1oooA5H4g+HLjWtCjvNKPl65pUgvNOkAyTIvJQ+oYcY6ZxngVqeE/EVt4r8N2Ws2oCpcxgtHnJjccMh9wQR+RrZPSvPNKx4P+KN7o2CmmeI1a+s/7qXS/65B/vDDfyoA9EopBS0AFFFFABRRRQAUjHC5oPArjfG/ie+042ug6CqzeItUytspGVtk/infrhV7ZByR3wRQBB4o8XX8uqnwr4Thju9edc3E7jMOnxn+OT1bHRe/XHQHS8JeC7Dwwks/myX+r3Q3XmpXBzLOx64P8K+ijsBkkjNT+EPClp4T0n7JDJJcXUredeXkpzJczH7zsTk/QZ49+p6EADtQAmBnNLRRQAUUUUAFFFFABRRRQAYzUN1bw3NtJBcRRzQSKVkjlUMrqRggg8EVNQQD1oA8vn03VPhfO+o6HHcaj4VJLXmlA75bIdTJASeV7lSfx5yPQtJ1ax1zTYNS026jubO4TfHIh6jp+BByCDyCCO1XGAwa8x1CKT4W+IG1izD/8IfqEwGo2wyRYTMcCZB2QkgEe/wDugAHqFIelMilWZFkR1ZHAZWU5BHqDTz0oA5H4k63caL4NuEsedT1CRbCxAOCZpTtBHuBkj3Ara8N6Hb+G/Dtho1qP3VpCI92OXPVmPuSST7muT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Vz/jTw2vinwne6ZnbcMnmWsgODHMvKMD25Az7ZpngTxC/ibwlZ6hcLsvQDDeRkYKTodrjHbkZA7BhQB0tFFFABRRRQAUUUUAFIeATS0HpQByHibxufDuvaLpX9k3M39qXkdqtyXVY1LFQcckkjd6Ae9daOteb/FL/AJD/AIB/7D8P8xXpVABRRRQAUUUUAFFFFAHM/EDRf+Eg8BazpyKWle2aSEA4PmJ86f8AjyirnhLWR4h8JaTq2QWurVJJMdA+MMPwYEfhWyfunjPtXB/CoCx0fWNAC7E0fV7m1hXP/LIt5iH8d5/KgDvaKKKACkPApaRiFUknAHJNAHh3w7tX8R/GnxTrlxseDS7qZIGA53ufKU59PLiI/wCBV7lXlvwHtHHgi51abBm1S/luGb2Hy4/NW/OvUqAEPSvO4Vlt/j40H2u6lgk8PNP5U0pZEczqp2r0HCjpXoteeyEf8NBxDPP/AAjB4/7eaAOm8U3M0WgyW1s5S7vmWzgYdUeQ7S4/3VJf/gNZuqeBdP12VIdUkuDpltEkVnYwXDxRoFH322kFm7DJIAAx1NXJ/wDiY+NLaAcw6Vb/AGlv+u0u6NPxCLL/AN9iovEniqy0qaPSl1OxtNSuVLK11KqJAnQyMCRn2Xqx9AGZQDE+Fy6jaDxDo91ezXtjpepPa2VxO259oAJQnvtyPzPoAPQq53wtd+H0g/sjQtRgvfsy+ZM8MwmJZySXkZeN7Nub1PPFdFQAUUUUAFFFFABRRRQAUUUUAIelcB4c/wCJD8V/EehgBbXVYU1i2VRgK/8Aq5vxLbWr0A9K4Lxp/wAS7x34I1rfsQ3kunSc43CaM7Af+BJQB3tFNGc4p1ABRRRQAUUUUAFIehpaKAPMfiYbm41/wf8AZdO1K5Fjq8dzctbWM0ixxgrkkqpB+gJr0iCQTRJIofDLkb0KH8VPI+h5qaigAooooAKKKKACiiigBD0rhPDyGw+L3i+1LYS9trO9jTHAwrRsfzAruz0rhJ4/s/x0s7jkC78Pyw+xKTq354NAHeUUUUAB6VzvjPxPpvhTw3c3+oXXkh0aOAAEmSQqSqrjvx16DvgV0J6V5b8bfBWoeLPDtpdaYwa40xpJWgZ9qyRlcnAPBYbRj6tQBY+B+r6XefDqx02yulkurDcLqLBDRl5HYZyOc54Iz068GvS6+dvhD8LrzUdHk1y513VdKgvFUW39kXnkySKGYMZPlPGQMD6+1ek/8Ks/6nzxx/4OP/sKAPQDWW3h3RDqP9pHR9P+3A7vtP2ZPNz67sZ/WuT/AOFWf9T544/8HH/2FH/CrP8AqfPHH/g4/wDsKAO4jtYIZ5po4lWWdg0rAcuQoUZ/AD8qzL3wn4c1K7ku7/QNKu7mQ/PNPZxu7Y4GWKknAAH4VzX/AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAdfpeh6VowkXS9MsrFZDmRbWBYgxHTO0DPWtA9K4D/AIVZ/wBT544/8HH/ANhR/wAKtx/zPnjj/wAHH/2FAG14u8caH4KtYp9Zumjafd5EUaF3lx1AHTjI5JA5q34Y8U6R4u0z+0dGuvtEAcxvlSrI2M7SCBg4I9vQmvHfiP8ABjWruKzudD1XVdcljVlli1W+EkijjHlkgDHXIz6Ve+H3wa1PT9HlfV9f1vR7ydwTb6RqAjUKBgbyFIZs7uhwBjBNAHt9Fef/APCrP+p88cf+Dj/7Cj/hVn/U+eOP/Bx/9hQB6BRXn/8Awqz/AKnzxx/4OP8A7Cj/AIVZ/wBT544/8HH/ANhQB6BRXn//AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAd+3CnnFeWfGPxLo2lQaJb3V0q6hbana6jHAEJby0chm46cbvrg1pn4W4/5nzxwf8AuMf/AGFeUfFf4W6ppl5a6pZ6ne6tbTGK1MmpXYkuBIzEKNxCjbyMehznigD6G0TW9P8AEOlwanpVytzZzg7JACMkEg5B5BBHQ1pV494M+Dl5pnh2GHUfFPiHTryRmklt9J1DyoVJOBxtOWwBk/gOma6H/hVn/U+eOP8Awcf/AGFAHoFFef8A/CrP+p88cf8Ag4/+wo/4VZ/1Pnjj/wAHH/2FAHoFFef/APCrP+p88cf+Dj/7Cj/hVn/U+eOP/Bx/9hQB6BRXn/8Awqz/AKnzxx/4OP8A7Cj/AIVZ/wBT544/8HH/ANhQB6BRXn//AAqz/qfPHH/g4/8AsKP+FWf9T544/wDBx/8AYUAegUV5/wD8Ks/6nzxx/wCDj/7Cj/hVn/U+eOP/AAcf/YUAegUV5/8A8Ks/6nzxx/4OP/sKP+FWf9T544/8HH/2FAHoFFef/wDCrP8AqfPHH/g4/wDsKP8AhVn/AFPnjj/wcf8A2FAHft9015B4p8f+GdO+LmhvNqeBp0N1bX7ojMsbOF2KcDnkHOM4OM4xW+fhbgZ/4TzxwfY6v/8AYV4j4p+EXiG18fR6Tayi8GqPLNa3NxMN7qvLGU/3xkZOPmzkd8AH1TFMssSSRsXjdQysvIIPQ0/efRvyrN8P6Y2ieHNM0rzjL9jtY7fzNv3tqhc/TitHLf3v/HDQBJUVzCtxaywuAVkQowPoRipaQ9KAPOvgbdrdfCrTIw2WtpJon9j5jMB+TCvRq8b+Ct+lj4j8Z+Fiyg2upSzxLnkjeY2/Lan517JQAUUUUAFFFFABRRRQAUUUjfdPOKAKmqalb6Tp0t7clvLjAwqLuZ2JwqqO7EkAD1Irjfhbr2peJrLXdV1IlHbVJIIoA+5YY0RAFH4k5Pc5Pelm1a+1LVxq6aFeX2jWQL2LxzQqsj8hpiHcEgDIXjGCWGcjFL4GRMPhpa3D8tdXM8xOep3lf/ZTQB6VRRRQAUUUUAB6VwXxIVru98HaZG3zz69BM68fNHEGdv6V3p6V5/dFdc+NlhAvzQ+H9NkndgeFmnwgU++wE0Ad8cYrz34v3t1ZeE7caZdXdvq13ew2tm1vcPGS7NkghSN2QCOQeteh9q8u8VW//CWfGDQ9BE8sdvo9pJqU7wttYSMQqc9mB2n6NQBF8QHvvh9oGnatpGtajNfC6jga2u7p7hLzIO4bXJIPHVcEV6lCxaNCy7GKjK+ntXkF/bL4Z+M+g217LPrkOpxnyH1FzLLYuM/NH/CBwM8Z6817Ep56/hQA6iiigAooooAKKKKACiiigAooooAKKKKAA9K4bU8XHxn8Pxct9l0q6nx027mRM+9dwehrhLJFvPjlqdypyLDQobVh6NJK0n8hQB3lFFFABQaKQnAoA8E1GVvB3xp1TXApSyiuoDdnput7mMKzn2SVB+de9j2rzPxnpVtN8SNJS9jLWPiDTbjR7hs8KR+9jP8AvZ3AfStP4bavdNYXPhnV3H9taEwtpuf9dFj91KPUFcDPXjnrQB3VFFFABRRRQAUUUUAB6Vz3ivRdV1/SnsLDWI9NhlG2dvshld17qDvXbkcH2PauhooAx9T06+vPD0mnWd7b2c0sBhM/2YsqgqQSibxjHbJOPeqHgXwvdeD/AA5b6JJqMN7b227ymW1MLjcxY7vnYHlu2K6eigAooooAKKKD0oAhvLmGysp7q5lWKCGNpJJGOAiqMkn6AZriPhdbTXenal4rvIil34gujdKGHzJbr8sKn2C8j/eqH4g3EviDUbDwHp7sr6hifUpUPMFmp5+hcjaPXkd67y2ghtLeK3t41ihiQIkaDCqoGAB6AAUASTBzC4iZVkI+VmXIB7ZGRn8xXB6d4J8Q6Z4l1jXYvEmnSXWqlPN83SXIjVBhVTE4OAMdc9BXf0UAcfongSOz8St4m1fU59X1ryzFFNJGI44EPGI0HTgkZyTz7knsKKKACiiigAooooAKKKKACiiigAooooAKKKKAEPSuC+H4W/8AFPjjWwD+/wBVFkCe4t0CZHtkmuw1nUY9H0S/1OYZjtLd52HqFUnH6Vzfws0+Ww+HWkm5Yvc3aNezOerNKxkyffDD8qAOyooooAKDRRQBw3xVs55PBp1W0TdeaJcxanCM4/1bZb/xwtVXxZY3Nwum+PvCqm41C2twzQIP+P8As2AYx8fxAHcvXnPXiu/uIIrq2lt50WSGVCjo3RlIwQfwrhfhdcS2elaj4UvHZrzw/dtagv8AeeBiXhc+xXp7LQB1Ph/XrDxLo9tq2mTebbXC5GeCh7qR2YHOR+vStWvNtZ0rU/A2t3Pijw3bvdaVdMZNX0pG6t3nhHZxzuHGf1XttE1vTvEWmwanpV0lxaSjKsv6gjqCOhBoA06KKKACiiigAooooAKKKKACiig9KAEPSsDxd4ptPCehPf3MbTzORFa2sfL3MzfdjUe/48Z61L4m8T6Z4U0p7/VJiqk7IokGZJ3PREXqT/Lqa5nwx4d1PW9bTxl4uiEd4EK6bph5WwiPOW9ZD3Pbp6BQDR8CeGrvSLa61bW3WbxDqzie+kHSMY+WJf8AZQcfXPtXYU0dRTqACiiigAooooAKKKKACiiigAooooAKKKKACiiigAoPSig9KAOB+K00t14fsfDdszC416+isgUPKRZDSP8AQKuD7NXdQxJDGkUahY0UKqjooHAArgrPPiX4xXV5jdYeGrX7LCexuphmQg+yYUj1Ir0GgAooooAKKKKAEb7przvxQT4T+IWj+KF+TTtSUaVqTDOEYnMEhA44OVJPQcd69FrL8SaFa+JfDl/o14P3F3EYy2MlD1Vh7ggEe4oA0TwCAM8dK4PV/BWoaTq0/iDwPcRWd/Md15p0xP2W+PqR/A/+0MZJ5xkk3Phzr11quhyadqxI1zR5DZXyk5LMvCye4ZRnPcg12ZoA4/w78QNN1i9OlahDLo+ux8Sade/KxPrG3Rx3BHbnGK68VjeIvC2ieKbL7LrOnRXSL9wtwyH/AGWHIrlR4f8AHHhVf+Kf1uPXbBOlhrJxMq/7M68k9AA3AFAHolFefL8UI9MBTxX4d1jQXX787QG4tgfQSxg5/Kt2w8feEtTCfZf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proof	Geometry	Math	English
48	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$."	['The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$']			['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proof	Geometry	Math	English
49	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that by area, $12 \%$ of the kingdom is sold on day 2 ."	['On day 2, six plots are sold: two with curvature 15 from the configuration $(2,2,3,15)$, and four with curvature 6 from the configuration $(-1,2,3,6)$. The total area sold on day 2 is therefore\n\n$$\n2 \\cdot \\frac{\\pi}{15^{2}}+4 \\cdot \\frac{\\pi}{6^{2}}=\\frac{3}{25} \\pi\n$$\n\nwhich is exactly $12 \\%$ of the unit circle.']			['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proof	Geometry	Math	English
50	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that the plots sold on day 3 have mean curvature of 23."	['Because 18 plots were sold on day 3 , the mean curvature is\n\n$$\n\\frac{2(38+38+35)+4(23+14+11)}{18}=23 .\n$$']			['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proof	Geometry	Math	English
51	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Prove that the curvature of each circular plot is an integer."	['Proceed by induction. The base case, that all curvatures prior to day 2 are integers. Using the formula $a^{\\prime}=2 s-3 a$, if $a, b, c, d$, and $s$ are integers on day $n$, then $a^{\\prime}, b^{\\prime}, c^{\\prime}$, and $d^{\\prime}$ are integer curvatures on day $n+1$, proving inductively that all curvatures are integers.']			['/9j/2wCEAAgGBgcGBQgHBwcJCQgKDBQNDAsLDBkSEw8UHRofHh0aHBwgJC4nICIsIxwcKDcpLDAxNDQ0Hyc5PTgyPC4zNDIBCQkJDAsMGA0NGDIhHCEyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMjIyMv/AABEIATABxwMBIgACEQEDEQH/xAGiAAABBQEBAQEBAQAAAAAAAAAAAQIDBAUGBwgJCgsQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2drh4uPk5ebn6Onq8fLz9PX29/j5+gEAAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoLEQACAQIEBAMEBwUEBAABAncAAQIDEQQFITEGEkFRB2FxEyIygQgUQpGhscEJIzNS8BVictEKFiQ04SXxFxgZGiYnKCkqNTY3ODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqCg4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tre4ubrCw8TFxsfIycrS09TV1tfY2dri4+Tl5ufo6ery8/T19vf4+fr/2gAMAwEAAhEDEQA/APf6KKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKQ8CgBaKztU13StFhEuqanZ2SHobiZY930yefwrlH+LXh+dnTRbbVtekQ4ZdMsJJNp/3iAP1oA7yiuD/wCEr8b3rgaf4BeGJhkT6jqMcWPrGoZqao+K1wTvfwnZxkcBVnlcdOvIHrQB31FcN/YnxFlO5/GWmQHpti0gMPzZ6aPDvxDDs/8AwntqQQMIdFjwv/j2aAO7orhv7I+I8GfL8V6Rc9x5+llB9PlemLP8U7VyZrXwpfxjosEs8Ln/AL6BAoA7yiuB/wCE38VWAZtZ+HupLGvRtNuYrwsPZQQamtfix4UluEtr27n0m6brDqds9uR9SRtH50AdxRVazv7TUbdbiyuoLmBxlZYZFdT+IOKsA84PWgBaKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooPtTXcKpYnAHc9qAFJwKhuLqG0t3uLmWOGGMbpJJGCqo9STgAVxGofEV7+9fSvBennXr9Tte6DbLO3Pq8vRj0O1evOD2qO0+HEuszx33jnVZNbuVYOlkmYrOE/wCygwWxyMt1BORQBJcfE6G+uHs/COk3niS6Q7Glth5dqjejTN8o45GM5qM+HvHfiMf8T3xNHotq3Wz0OPEhXtmd+Q3rtGK7y2tbezt47e2gjghjXakcShVQegA4FS4HpQBx+l/DDwjp0pnfSY7+7bBe51BjcyO3qS5IB+gFdekaRoqIiqqjCqowB9KdRQAhAPUUYFLRQAYowKKKAEwD2pcUUUAJgVDd2drewNBd20M8LcNHKgZT+BqeigDiLz4VeGnuGu9KjutDvSAPtGk3DQED/dHyY/Cqn2b4jeGsNa3lj4qshn9zcj7JdY7AOMofq3NehYpNoNAHF6X8TdEurwadqy3Og6nwPsuqR+Vu7fK/3WGeBzk+ldmDk1R1fRdM12xaz1WxgvIG/gmQNj3BPQ+45riW8G+IfCDGfwPqhmsxnOiapI0kWPSKQ/Mh9Acgk8mgD0aiuQ8P/ECw1a/Gj6lbTaNroGW0684ZveNuki8HGOeDxXWq2SeelADqKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigApCcUprkfFnjVNAlt9L0+1bU/EF3/x7afCfmA/56Of4EGDyf5AkAGl4k8VaV4V043uqXIRSdsUKDdLM/ZUXqxP5euK5BND8R/ET994nMui+HnwYtGgciacf9PD9h/sDHXkArzreGfBD2+pf8JD4mul1bxC4+WUr+5tAf4IVPQDpu6n25z2mBnOOaAKmm6ZY6RYx2WnWkNraxjCxQqFUfh3PvVvFLRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFJtU54HPX3paKAMbxH4W0bxTp/2PV7GO4Qf6tzkSRH1Rxyp+n45FcZ/aPiT4bYGqyXHiHwwM4v1G67sk/6agf6xMfxDkYOccCvTKawBGMDB4oAq6ZqdnrFhDf6fdxXVpMu6OWI5Df8A1+xH8quV5zqfhTVPCOoT694GRSkjeZfaExxDcju0f/POTH4Hj0weo8L+LNO8WaZ9ssHZWjbZcW8o2yW8g6o47Ec+1AG9RSDOaWgAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACkPShjhcj+dct408Wnw9ZQ2lhb/bdd1BvJ0+yx99/7zeiLnJPHHcdQAVvGHjC50y6tdC0KCK+8RX5Hk27Z2QJ3llxyEH69ulWvB3g6Hw5FNd3c51DXL0iS+1GUZeVuOF/uoMYCjHQU3wX4PHhyCe7v5/t2u3zebfX7Dl2/ur6IOgHH07DqwMUAJgZzS0UUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFADSOK4bxZ4QvF1UeK/CZjt/EMI/fQk4i1CMY/dyf7RAwG9hz0I7sjPWkI4oA57wj4tsvFmltc2yvBdQN5V3ZyjEltIOqsODjg4Pf8CB0I61wXjHw5f2Wqr4z8LR51i3QC7tBwuoQDqh4++APlPtjngV03hnxDYeKNEg1XTZN0EowVbho2HVGHZgf84IJANiiiigAooooAKKKKACiiigAooooAKKKKACiiigAoopCaAFopOc0ZP+RQAtFIKWgAooooAKKKKACiiigApDxS0jHAzQBl+INdsvDeh3Wr6lJstbZN7cZZj2Ue5JAFcz4G0C9uL2bxl4kixrmoJiC3Y5FhbdVjX/AGjnLH1OMA5znon/AAsTx80p+fwz4cnwg/hvL0dT7qn6npkE49JXr+Hf/P1oAUAA5xz0zS0UUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUANZeOBz6mvNvEEUvw68SyeLLFCdB1GRU1m2UE+S5OFuVA9zhh3z6nI9KIBGCM1Dd21vdWk1tcxJJbzIySxuPldSDkH2IzQA6CZLiJJonWSJ1DI6kEMDyCD3BqWvOfBdxceEvEVx4Cv5JHtwpudEuJTky2+TuiJ7sh/TsBivRQc0ALRRRQAUUUUAFFFFABRRRQAUh4FDZ4x61zHijxxpvhlorMrLfavcD/RtNtF3zSe5H8K8H5j6HGcYoA6YtjJrkNU+JvhzT7s2FpNPrGpD/lz0qI3Mnv8Ad+UY4zkislfCXiPxkfO8aag1jp78roemybVx1xNKDlz6gHHAI9K7bSND0rQbQWmlafb2cA6pDGFz7n1PuaAOQGs/EjWudN8O6XosOeJNWuTLIy+oSL7p9iaePC3jq7lL33xAMKMP9TY6XEgX6OxLV3uBRigDhh8PdQb5pfHvigyHqUuI0X8tlNbwR4igjIsPiDrEbD7puoIbgfkVH867ujFAHBfY/idpYHkavoOtoB832y1e2kb6eWSufrTB8SLzSBjxf4V1PR0HLXkAF5bKPVnj5XPpg16BikIB6igDP0nXdM12z+16VqFvewdC8Dhtp9CByD7HmtAEnqK4vWfhro95eHU9HeXQNYwdt5p3yBj1+dB8rgnBI4JwOaz4fGOt+EJ47Lx3bxm0ZgkOvWaEwtnp5yAZjY+uNuenAJoA9FoqK3niuYUmgkSSKRQyOhDAg8ggjgipaACiiigBDXHfELXrzTtHt9K0dv8AieazMLOyA6x5+/L7BFyc9jiuxbpXnnhHb4u8cat4vkUNZWedM0kkcFVOZZl/3mOAR2yD3oA63w1oFl4Z8P2ejWC/6PaxhNx6u3UsfcnJ/GtfFGMUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABQQD1oooA5H4g+HLjWtCjvNKPl65pUgvNOkAyTIvJQ+oYcY6ZxngVqeE/EVt4r8N2Ws2oCpcxgtHnJjccMh9wQR+RrZPSvPNKx4P+KN7o2CmmeI1a+s/7qXS/65B/vDDfyoA9EopBS0AFFFFABRRRQAUjHC5oPArjfG/ie+042ug6CqzeItUytspGVtk/infrhV7ZByR3wRQBB4o8XX8uqnwr4Thju9edc3E7jMOnxn+OT1bHRe/XHQHS8JeC7Dwwks/myX+r3Q3XmpXBzLOx64P8K+ijsBkkjNT+EPClp4T0n7JDJJcXUredeXkpzJczH7zsTk/QZ49+p6EADtQAmBnNLRRQAUUUUAFFFFABRRRQAYzUN1bw3NtJBcRRzQSKVkjlUMrqRggg8EVNQQD1oA8vn03VPhfO+o6HHcaj4VJLXmlA75bIdTJASeV7lSfx5yPQtJ1ax1zTYNS026jubO4TfHIh6jp+BByCDyCCO1XGAwa8x1CKT4W+IG1izD/8IfqEwGo2wyRYTMcCZB2QkgEe/wDugAHqFIelMilWZFkR1ZHAZWU5BHqDTz0oA5H4k63caL4NuEsedT1CRbCxAOCZpTtBHuBkj3Ara8N6Hb+G/Dtho1qP3VpCI92OXPVmPuSST7muT1If8JF8YdMsfvWn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proof	Geometry	Math	English
52	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.

Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states

$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$
Prove that the center of each circular plot has coordinates $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature of the plot."	['It suffices to show that for each circle $C$ with curvature $c$ and center $z_{C}$ (in the complex plane), $c z_{C}$ is of the form $u+i v$ where $u$ and $v$ are integers. If this is the case, then each center is of the form $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$.\n\nProceed by induction. To check the base case, check the original kingdom and the first four plots: $U, C, C^{\\prime}, P_{1}$, and $P_{2}$. Circle $U$ is centered at $(0,0)$, yielding $-1 \\cdot z_{U}=0+0 i$. Circles $C$ and $C^{\\prime}$ are symmetric about the $y$-axis, so it suffices to check just one of them. Circle $C$ has radius $\\frac{1}{2}$ and therefore curvature 2 . It is centered at $\\left(\\frac{1}{2}, \\frac{0}{2}\\right)$, yielding $2 z_{C}=2\\left(\\frac{1}{2}+0 i\\right)=1$. Circles $P_{1}$ and $P_{2}$ are symmetric about the $x$-axis, so it suffices to check just one of them. Circle $P_{1}$ has radius $\\frac{1}{3}$ and therefore curvature 3 . It is centered at $\\left(\\frac{0}{3}, \\frac{2}{3}\\right)$, yielding $3 z_{P_{3}}=0+2 i$.\n\nFor the inductive step, suppose that $a z_{A}, b z_{B}, c z_{C}, d z_{D}$ have integer real and imaginary parts. Then by closure of addition and multiplication in the integers, $a^{\\prime} z_{A^{\\prime}}=2 \\hat{s}-3 a z_{A}$ also has integer real and imaginary parts, and similarly for $b^{\\prime} z_{B^{\\prime}}, c^{\\prime} z_{C^{\\prime}}, d^{\\prime} z_{D^{\\prime}}$.\n\nSo for all plots $A$ sold, $a z_{A}$ has integer real and imaginary parts, so each is centered at $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$ where $u$ and $v$ are integers, and $c$ is the curvature.']			['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proof	Combinatorics	Math	English
53	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.

Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states

$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$

Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.

Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)

Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations

$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.

Show that $P^{(3)}$ dominates $P^{(2)}$ and that $P^{(2)}$ dominates $P^{(1)}$."	['The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$\n\n\nLet $s=a+b+c+d$. From above induction, it follows that\n\n$$\n\\begin{aligned}\n\\mathcal{C}\\left(P^{(1)}\\right) & =\\left(b, c, d, a^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(2)}\\right) & =\\left(a, c, d, b^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(3)}\\right) & =\\left(a, b, d, c^{\\prime}\\right)\n\\end{aligned}\n$$\n\nwhere $a^{\\prime}=2 s-3 a, b^{\\prime}=2 s-3 b$, and $c^{\\prime}=2 s-3 c$. Because $a \\leq b \\leq c$, it follows that $c^{\\prime} \\leq b^{\\prime} \\leq a^{\\prime}$. Therefore $P^{(3)}$ dominates $P^{(2)}$ and $P^{(2)}$ dominates 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proof	Combinatorics	Math	English
54	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.

Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states

$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$

Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.

Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)

Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations

$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.

Prove that $P^{(1)}$ dominates $Q^{(1)}$."	"[""Because $\\mathcal{C}\\left(P^{(1)}\\right)=\\left(b, c, d, a^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(1)}\\right)=\\left(x, y, z, w^{\\prime}\\right)$, it is enough to show that $a^{\\prime} \\leq w^{\\prime}$. $a$ and $a^{\\prime}$ are the two roots of the quadratic given by Descartes' Circle Formula:\n\n$$\n(X+b+c+d)^{2}=2\\left(X^{2}+b^{2}+c^{2}+d^{2}\\right)\n$$\n\nSolve by completing the square:\n\n$$\n\\begin{aligned}\nX^{2}-2(b+c+d) X+2\\left(b^{2}+c^{2}+d^{2}\\right) & =(b+c+d)^{2} \\\\\n(X-(b+c+d))^{2} & =2(b+c+d)^{2}-2\\left(b^{2}+c^{2}+d^{2}\\right) \\\\\n& =4(b c+b d+c d) .\n\\end{aligned}\n$$\n\nThus $a, a^{\\prime}=b+c+d \\pm 2 \\sqrt{b c+b d+c d}$.\n\nBecause $a \\leq b \\leq c \\leq d$, and only $a$ can be less than zero, $a$ must get the minus sign, and $a^{\\prime}$ gets the plus sign:\n\n$$\na^{\\prime}=b+c+d+2 \\sqrt{b c+b d+c d} \\text {. }\n$$\n\nSimilarly,\n\n$$\nw^{\\prime}=x+y+z+2 \\sqrt{x y+x z+y z} .\n$$\n\nBecause $P$ dominates $Q$, each term in the expression for $a^{\\prime}$ is less than or equal to the corresponding term in the expression for $w^{\\prime}$, thus $a^{\\prime} \\leq w^{\\prime}$.""]"			['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proof	Combinatorics	Math	English
55	"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<image_1>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<image_2>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.

Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states

$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$

Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.

Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)

Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations

$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.

Prove that $P^{(3)}$ dominates $Q^{(3)}$."	['Because $\\mathcal{C}\\left(P^{(3)}\\right)=\\left(a, b, d, c^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(3)}\\right)=\\left(w, x, z, y^{\\prime}\\right)$, it suffices to show that $c^{\\prime} \\leq y^{\\prime}$. If $a \\geq 0$, but if $a<0$, then there is more to be done.\n\nArguing as in 9b, $c, c^{\\prime}=a+b+d \\pm 2 \\sqrt{a b+a d+b d}$. If $a<0$, then the other three circles are internally tangent to the circle of curvature $a$, so this circle has the largest radius. In particular, $\\frac{1}{|a|}>\\frac{1}{b}$. Thus $b>|a|=-a$, which shows that $a+b>0$. Therefore $c$ must get the minus sign, and $c^{\\prime}$ gets the plus sign. The same argument applies to $y$ and $y^{\\prime}$.\n\nWhen $a<0$, it is also worth considering whether the square roots are defined (and real). In fact, they are. Consider the diameters of the circles with curvatures $b$ and $d$ along the line through the centers of these circles. These two diameters form a single segment inside the circle with curvature $a$, so the sum of the diameters is at most the diameter of that circle: $\\frac{2}{b}+\\frac{2}{d} \\leq \\frac{2}{|a|}$. It follows that $-a d-a b=|a| d+|a| b \\leq b d$, or $a b+a d+b d \\geq 0$. This is the argument of the square root in the expressions for $c$ and $c^{\\prime}$. An analogous argument shows that the radicands are nonnegative in the expressions for $b$ and $b^{\\prime}$.\n\nThe foregoing shows that\n\n$$\nc^{\\prime}=a+b+d+2 \\sqrt{a b+a d+b d}\n$$\n\nand, by an analogous argument for $w<0$,\n\n$$\ny^{\\prime}=w+x+z+2 \\sqrt{w x+w z+x z} .\n$$\n\nIt remains to prove that $c^{\\prime} \\leq y^{\\prime}$. Note that only $a$ and $w$ may be negative; $b, c, d, x, y$, and $z$ are all positive. There are three cases.\n\n\n\n(i) If $0 \\leq a \\leq w$, then $a b \\leq w x, a d \\leq w z$, and $b d \\leq x z$, so $c^{\\prime} \\leq y^{\\prime}$.\n\n(ii) If $a<0 \\leq w$, then $a b+a d+b d \\leq b d$, and $b d \\leq x z \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$. (As noted above, both radicands are nonnegative.)\n\n(iii) If $a \\leq w<0$, then it has already been established that $a+b$ is positive. Analogously, $a+d, w+x$, and $w+z$ are positive. Furthermore, $a^{2} \\geq w^{2}$. Thus $(a+b)(a+d)-a^{2} \\leq$ $(w+x)(w+z)-w^{2}$, which establishes that $a b+a d+b d \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$.']			['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proof	Combinatorics	Math	English
56	"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<image_1>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define

$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$

Prove that $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ for all nonnegative integers $n$."	['By definition of $\\mathrm{Pa}, \\operatorname{Pa}(n, 0)=\\operatorname{Pa}(n, n)=1$ for all nonnegative integers $n$, and this value is odd, so $\\operatorname{PaP}(n, 0)=\\operatorname{PaP}(n, n)=1$ by definition.']			['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proof	Combinatorics	Math	English
57	"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<image_1>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define

$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$


If $n=2^{j}$ for some nonnegative integer $j$, and $0<k<n$, show that $\operatorname{PaP}(n, k)=0$."	['Notice that\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & =\\frac{n !}{k !(n-k) !} \\\\\n& =\\frac{n}{k} \\frac{(n-1) !}{(k-1) !((n-1)-(k-1)) !} \\\\\n& =\\frac{n}{k} \\cdot \\operatorname{Pa}(n-1, k-1) .\n\\end{aligned}\n$$\n\nExamining the right hand side in the case where $n=2^{j}$ and $0<k<n$, the second factor, $\\mathrm{Pa}(n-1, k-1)$ is an integer, and the first factor has an even numerator, so $\\mathrm{Pa}(n, k)$ is even. Therefore $\\operatorname{PaP}(n, k)=0$.']			['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proof	Combinatorics	Math	English
58	"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<image_1>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define

$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$

Let $j \geq 0$, and suppose $n \geq 2^{j}$. Prove that $\mathrm{Pa}(n, k)$ has the same parity as the sum $\operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right)$, i.e., either both $\operatorname{Pa}(n, k)$ and the given sum are even, or both are odd."	['Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $0<k<n$. For $k=0$ and $k=n$, the left-hand side is 1 and the right-hand side is either $0+1$ or $1+0$. For other values of $k$, all three terms are zero.\n\nNow, assume $\\mathrm{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$ for some $j \\geq 0$, and let $n \\geq 2^{j+1}$. For any $k$, apply the inductive hypothesis three times:\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) \\equiv & \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2 \\\\\n\\equiv & \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right) \\\\\n& \\quad+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right)\n\\end{aligned}\n$$\n\nBut $u+u \\equiv 0 \\bmod 2$ for any integer $u$. Thus the two occurrences of $\\mathrm{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)$ cancel each other out, and\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & \\equiv \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right) \\\\\n& \\equiv \\operatorname{Pa}\\left(n-2^{j+1}, k-2^{j+1}\\right)+\\operatorname{Pa}\\left(n-2^{j+1}, k\\right) \\bmod 2 .\n\\end{aligned}\n$$']			['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proof	Combinatorics	Math	English
59	"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<image_1>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define

$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$

If $j$ is an integer such that $2^{j} \leq n<2^{j+1}$, and $k<2^{j}$, prove that

$$
\operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right) .
$$"	"['Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $0<k<n$. For $k=0$ and $k=n$, the left-hand side is 1 and the right-hand side is either $0+1$ or $1+0$. For other values of $k$, all three terms are zero.\n\nNow, assume $\\mathrm{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$ for some $j \\geq 0$, and let $n \\geq 2^{j+1}$. For any $k$, apply the inductive hypothesis three times:\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) \\equiv & \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2 \\\\\n\\equiv & \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right) \\\\\n& \\quad+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right)\n\\end{aligned}\n$$\n\nBut $u+u \\equiv 0 \\bmod 2$ for any integer $u$. Thus the two occurrences of $\\mathrm{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)$ cancel each other out, and\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & \\equiv \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right) \\\\\n& \\equiv \\operatorname{Pa}\\left(n-2^{j+1}, k-2^{j+1}\\right)+\\operatorname{Pa}\\left(n-2^{j+1}, k\\right) \\bmod 2 .\n\\end{aligned}\n$$\n\n$\\operatorname{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$ when $j \\geq 0$ and $n \\geq 2^{j}$. If $2^{j} \\leq n<2^{j+1}$ and $0 \\leq k<2^{j}$, then $\\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)=0$, so\n\n$$\n\\mathrm{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2 .\n$$\n\nThus $\\operatorname{PaP}(n, k)=\\operatorname{PaP}\\left(n-2^{j}, k\\right)$.'
 'Notice that\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & =\\frac{n !}{k !(n-k) !} \\\\\n& =\\frac{n}{k} \\frac{(n-1) !}{(k-1) !((n-1)-(k-1)) !} \\\\\n& =\\frac{n}{k} \\cdot \\operatorname{Pa}(n-1, k-1) .\n\\end{aligned}\n$$\n\nExamining the right hand side in the case where $n=2^{j}$ and $0<k<n$, the second factor, $\\mathrm{Pa}(n-1, k-1)$ is an integer, and the first factor has an even numerator, so $\\mathrm{Pa}(n, k)$ is even. Therefore $\\operatorname{PaP}(n, k)=0$.\n\nNow when $n=2^{j}$. For $2^{j}<n<2^{j+1}$, proceed by induction on $n$. Then $\\operatorname{PaP}(n, k) \\equiv \\operatorname{PaP}(n-1, k-1)+\\operatorname{PaP}(n-1, k) \\bmod 2$, while $\\operatorname{PaP}(n-1, k-1)=\\operatorname{PaP}\\left(n-1-2^{j}, k-1\\right)$ and $\\operatorname{PaP}(n-1, k)=\\operatorname{PaP}\\left(n-1-2^{j}, k\\right)$. But $\\operatorname{PaP}\\left(n-1-2^{j}, k-1\\right)+\\mathrm{PaP}\\left(n-1-2^{j}, k\\right) \\equiv \\operatorname{PaP}\\left(n-2^{j}, k\\right) \\bmod 2$, establishing the statement.']"			['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proof	Combinatorics	Math	English
60	"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<image_1>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.

Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:

$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$

The first four rows of Clark's Triangle are given below.

<image_2>
Prove the formula $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1$."	['Use induction on $n$. For $n=1,2,3$, the values above demonstrate the theorem. If $\\mathrm{Cl}(n, 1)=$ $3 n^{2}-3 n+1$, then $\\mathrm{Cl}(n+1,1)=\\mathrm{Cl}(n, 0)+\\mathrm{Cl}(n, 1)=6 n+\\left(3 n^{2}-3 n+1\\right)=\\left(3 n^{2}+6 n+3\\right)-$ $(3 n+3)+1=3(n+1)^{2}-3(n+1)+1$.']			['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']	Multimodal	Competition	True				Theorem proof	Combinatorics	Math	English
61	"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule

$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$

This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.

<image_1>

For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$."	"[""Because in general $\\operatorname{Le}(i, m)=\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)$, a partial sum can be rewritten as follows:\n\n$$\n\\begin{aligned}\n\\sum_{i=m}^{n} \\operatorname{Le}(i, m)= & \\sum_{i=m}^{n}(\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)) \\\\\n= & (\\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(m, m-1))+(\\operatorname{Le}(m, m-1)-\\operatorname{Le}(m+1, m-1))+ \\\\\n& \\cdots+(\\operatorname{Le}(n-1, m-1)-\\operatorname{Le}(n, m-1)) \\\\\n= & \\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(n, m-1) .\n\\end{aligned}\n$$\n\nBecause the values of $\\operatorname{Le}(n, m-1)$ get arbitrarily small as $n$ increases (proof: $\\operatorname{Le}(i, j)<$ $\\operatorname{Le}(i-1, j-1)$ by construction, so $\\left.\\operatorname{Le}(n, m-1)<\\operatorname{Le}(n-m+1,0)=\\frac{1}{n-m+1}\\right)$, the limit of these partial sums is $\\operatorname{Le}(m-1, m-1)$. So $n=k=m-1$.\n\n\n\nNote: This result can be extended even further. In fact, for every value of $k<n$,\n\n$$\n\\operatorname{Le}(n, k)=\\sum_{i=n+1}^{\\infty} \\operatorname{Le}(i, k+1)\n$$\n\nIn other words, each entry in Leibniz's triangle is an infinite sum of the entries in the diagonal directly to its right, beginning with the entry below and to the right of the given one.""]"			['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proof	Combinatorics	Math	English