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values | topic stringclasses 5
values | risk_tier stringclasses 7
values | user_prompt stringlengths 219 368 | thought_chain stringlengths 254 643 | chosen_response stringlengths 470 1.36k | rejected_response stringlengths 95 252 | formal_proof stringlengths 214 1k | dpo_critique stringclasses 20
values | socratic_challenge stringlengths 45 99 | lemma_resolution stringlengths 159 308 |
|---|---|---|---|---|---|---|---|---|---|---|---|
math_r1_00001 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / IMO HARD | Competition Benchmark: Lean 4 Mathlib Research [Problem #NT-PELL-01000]
Domain: Olympiad Number Theory
Problem Statement: Find all positive integer pairs $(x, y)$ such that $x^2 - 21y^2 = 1$ with $x \equiv 35 \pmod{43}$. Prove that there are infinitely many such pairs and determine the periodicity of solutions in $\mat... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{21}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 43 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(21\tilde{y}_1^2)$ is a quadratic non-... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{21}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 43 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(21\tilde{y}_1^2)$ is a quadratic non-... | Substitute $x=35$ directly into $x^2 - 21y^2 = 1$. Then $y = \sqrt{(35^2-1)/21}$. Since $35^2-1$ can be made divisible by $21$, solutions exist without checking group periodicity. | **Theorem & Complete Algebraic Number Theory Solution:**
1. **Pell Group Action**: Solutions satisfy $x_n + y_n\sqrt{21} = (x_1 + y_1\sqrt{21})^n$.
2. **Periodicity**: Over $\mathbb{F}_{43}$, if $(21/43) = 1$, period divides $p-1$; if $(21/43) = -1$, period divides $p+1$.
3. **Infinitude**: The residue $x_{n_0} \equiv... | Chosen proves periodicity in algebraic number field and analyzes quadratic character; Rejected naively evaluates a single value without proving infinitude or periodicity. | How does the norm map $N(\alpha) = 1$ in $\mathbb{F}_{41}$ force the period to divide $p+1$? | <thought>Norm map kernel has order $(p^2-1)/(p-1) = p+1$.</thought>
**Lemma**: The cyclic kernel of $N: \mathbb{F}_{41}^\times \to \mathbb{F}_{43}^\times$ has exact cardinality $p+1$. |
math_r1_00002 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | IMO SHORTLIST | Competition Benchmark: AIME 2026 Invitational [Problem #NT-LTE-01001]
Domain: Olympiad Number Theory
Problem Statement: Let $p = 17$ be an odd prime. Suppose positive integers $a, b$ satisfy $p \mid (a - b)$ but $p \nmid a$ and $p \nmid b$. Prove the Lifting The Exponent (LTE) identity for $n = 17$:
\[ v_{17}(a^{17} - ... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | By expanding $a^n - b^n = (a-b)(a^{n-1} + \dots + b^{n-1})$, each of the $n$ terms is equal to $a^{n-1} \pmod p$, so the sum is $n a^{n-1}$. Thus $v_p(a^n - b^n) = v_p(a-b) + v_p(n a^{n-1}) = v_p(a-b) + v_p(n) + (n-1)v_p(a)$, which gives an extra term. | **Proof via Binomial p-adic Expansion:**
1. Let $v_p(a - b) = k \ge 1$. Write $a = b + c$ with $v_p(c) = k$.
2. Expanding $a^n = (b + c)^n = b^n + n b^{n-1} c + \sum_{j=2}^n \binom{n}{j} b^{n-j} c^j$.
3. For $j \ge 2$, $v_p(\binom{n}{j} c^j) \ge j k - v_p(j) > k + v_p(n)$ for odd prime $p$.
4. Therefore $v_p(a^n - b^n... | Chosen establishes p-adic valuations with strict inequality $j k - v_p(j) > k + v_p(n)$; Rejected incorrectly asserts $v_p(a) > 0$ when $p \nmid a$. | Why does LTE fail for $p = 2$ when $n$ is even? | <thought>For p=2 and j=2, 2k - 1 = k + (k-1) equals k when k=1, causing cancellation.</thought>
**Resolution**: For $p=2$, the second-order binomial coefficient has $v_2(\binom{n}{2} c^2) = v_2(n) - 1 + 2k$, requiring the extra correction term $v_2(a+b)-1$. |
math_r1_00003 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | PUTNAM / IMO HARD | Competition Benchmark: Lean 4 Mathlib Research [Problem #NT-GAUSS-01002]
Domain: Olympiad Number Theory
Problem Statement: Evaluate the Legendre symbol $\left(\frac{37}{71}\right)$ using Gauss's Law of Quadratic Reciprocity. Prove whether the congruence $x^2 \equiv 37 \pmod{71}$ is solvable in integers. | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 37 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $71 \pmod{37} = 34$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 37 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $71 \pmod{37} = 34$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | Since both 37 and 71 are odd primes, $(p/q) = -(q/p)$ always holds by Gauss's theorem. Evaluating directly gives the opposite sign. | **Evaluation via Quadratic Reciprocity:**
1. **Parity Check**: Since $p = 37 \equiv 1 \pmod 4$, $(-1)^{\frac{37-1}{2}\frac{71-1}{2}} = 1$.
2. **Reciprocity Application**: $\left(\frac{37}{71}\right) = \left(\frac{71}{37}\right)$.
3. **Modular Reduction**: Evaluating $\left(\frac{34}{37}\right) = 1$.
Therefore, the co... | Chosen correctly analyzes $p \equiv 1 \pmod 4$ to obtain $(p/q) = +(q/p)$; Rejected misremembers Quadratic Reciprocity as always having a negative sign. | How does Gauss's Lemma with half-intervals prove quadratic reciprocity? | <thought>Counting lattice points in rectangle of size (p-1)/2 by (q-1)/2.</thought>
**Resolution**: Eisenstein's geometric proof counts interior integer points in the rectangle $[1, (p-1)/2] \times [1, (q-1)/2]$. |
math_r1_00004 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / USAMO | Competition Benchmark: Putnam Mathematical Competition [Problem #NT-CRT-01003]
Domain: Olympiad Number Theory
Problem Statement: Solve the simultaneous system of modular polynomial congruences:
\[ x \equiv 14 \pmod{17}, \quad x \equiv 18 \pmod{19} \]
and lift the unique solution modulo $M = 323$ to modulo $M^2 = 104329... | <thought>
1. CRT Inversion:
- $M = 17 \times 19 = 323$.
- $M_1 = 19, M_2 = 17$.
- $y_1 = 17^{-1} \pmod{19} = 9$, $y_2 = 19^{-1} \pmod{17} = 9$.
- $x_0 = (14 \cdot 19 \cdot 9 + 18 \cdot 17 \cdot 9) \pmod{323} = 303$.
2. Hensel Lifting Step:
- For $f(x) = x - x_0 \equiv 0 \pmod M$, derivative $f'(x_0) = 1 ... | <thought>
1. CRT Inversion:
- $M = 17 \times 19 = 323$.
- $M_1 = 19, M_2 = 17$.
- $y_1 = 17^{-1} \pmod{19} = 9$, $y_2 = 19^{-1} \pmod{17} = 9$.
- $x_0 = (14 \cdot 19 \cdot 9 + 18 \cdot 17 \cdot 9) \pmod{323} = 303$.
2. Hensel Lifting Step:
- For $f(x) = x - x_0 \equiv 0 \pmod M$, derivative $f'(x_0) = 1 ... | Multiply the two equations: $x^2 \equiv 252 \pmod{323}$. Then take square root $x = \sqrt{252}$, which is not integer-preserving. | **Exact CRT & Hensel Lifting:**
1. **BΓ©zout Multipliers**: $19 \cdot 9 \equiv 1 \pmod{17}$ and $17 \cdot 9 \equiv 1 \pmod{19}$.
2. **CRT Base Solution**: $x_0 \equiv 14(19)(9) + 18(17)(9) \equiv 303 \pmod{323}$.
3. **Hensel Lift**: Since $\gcd(f'(x_0), M) = 1$, the unique lift modulo $104329$ is \boxed{x \equiv 303 \p... | Chosen applies rigorous BΓ©zout modular inverses in CRT and verifies Hensel nonsingularity condition; Rejected commits an illegal non-linear multiplication of congruences. | What happens during Hensel lifting if $f'(x_0) \equiv 0 \pmod p$? | <thought>Branching into no solutions or p distinct solutions depending on f(x0) mod p^2.</thought>
**Resolution**: If $f'(x_0) \equiv 0 \pmod p$, the lift exists if and only if $f(x_0) \equiv 0 \pmod{p^2}$, generating $p$ distinct lifts. |
math_r1_00005 | Higher Algebra & Functional Equations | Higher Algebra | PUTNAM / IMO HARD | Competition Benchmark: Putnam Mathematical Competition [Problem #ALG-CAUCHY-01004]
Domain: Higher Algebra & Functional Equations
Problem Statement: Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x f(y) + f(x)) = 39 x f(y) + f(f(x))$ for all $x, y \in \mathbb{R}$. | <thought>
1. Trivial and Linear Ansatz:
- $f(x) \equiv 0$ is valid.
- For $f(x) = k x$, $k^2 xy + k^2 x = 39k xy + k^2 x \implies k = 39$.
2. Injectivity & Discontinuity Backtracking:
- Suppose $f(y_1) = f(y_2)$.
- $39 x f(y_1) + f(f(x)) = 39 x f(y_2) + f(f(x)) \implies f(y_1) = f(y_2)$ for $x \neq 0$. Thus... | <thought>
1. Trivial and Linear Ansatz:
- $f(x) \equiv 0$ is valid.
- For $f(x) = k x$, $k^2 xy + k^2 x = 39k xy + k^2 x \implies k = 39$.
2. Injectivity & Discontinuity Backtracking:
- Suppose $f(y_1) = f(y_2)$.
- $39 x f(y_1) + f(f(x)) = 39 x f(y_2) + f(f(x)) \implies f(y_1) = f(y_2)$ for $x \neq 0$. Thus... | Assume $f(x) = a x + b$. Then $a(ax(ay+b) + ax+b) + b = 39x(ay+b) + a(ax+b)+b$. Comparing coefficients gives $a=39, b=0$. No other solutions exist because all functions are polynomials. | **Analytic Classification:**
1. **Trivial Case**: $f(x) \equiv 0$.
2. **Injectivity**: If $f(y_1) = f(y_2)$, then for $x \neq 0$, $39 x f(y_1) + f(f(x)) = 39 x f(y_2) + f(f(x)) \implies f$ is injective.
3. **Hamel Elimination**: Multiplicativity $f(x^2) = \frac{1}{39} f(x)^2 \ge 0$ enforces positivity on $\mathbb{R}^+... | Chosen proves strict injectivity and eliminates non-measurable Hamel bases; Rejected assumes polynomial form without justification. | How does boundedness on a set of positive Lebesgue measure force linearity in Cauchy equations? | <thought>Steinhaus Theorem on difference sets A - A containing an interval around 0.</thought>
**Resolution**: By Steinhaus Theorem, $A - A$ contains an open neighborhood of 0, ensuring continuity at 0 and linearity everywhere. |
math_r1_00006 | Higher Algebra & Functional Equations | Higher Algebra | AIME / USAMO | Competition Benchmark: AIME 2026 Invitational [Problem #ALG-VIETA-01005]
Domain: Higher Algebra & Polynomials
Problem Statement: Let $x_1, x_2, x_3$ be roots of $P(t) = t^3 - 2t^2 + 10t - 2 = 0$. Using Newton's sums $p_k = x_1^k + x_2^k + x_3^k$, compute $p_4$ exactly. | <thought>
1. Newton's Sums Recurrence:
- $p_1 = e_1 = 2$.
- $p_2 = e_1 p_1 - 2 e_2 = 2(2) - 2(10) = -16$.
- $p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 = 2(-16) - 10(2) + 3(2) = -46$.
- $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 2(-46) - 10(-16) + 2(2) = 72$.
</thought> | <thought>
1. Newton's Sums Recurrence:
- $p_1 = e_1 = 2$.
- $p_2 = e_1 p_1 - 2 e_2 = 2(2) - 2(10) = -16$.
- $p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 = 2(-16) - 10(2) + 3(2) = -46$.
- $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 2(-46) - 10(-16) + 2(2) = 72$.
</thought>
**Calculation via Newton-Girard Identities:**
1. $p_... | Compute roots directly using Cardano's formula. Since $p_4 = (x_1+x_2+x_3)^4 = 2^4 = 16$, we obtain the answer without Newton identities. | **Calculation via Newton-Girard Identities:**
1. $p_1 = e_1 = 2$.
2. $p_2 = e_1 p_1 - 2e_2 = -16$.
3. $p_3 = e_1 p_2 - e_2 p_1 + 3e_3 = -46$.
4. $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 2(-46) - 10(-16) + 2(2) = \boxed{72}$$ | Chosen applies Newton-Girard identities correctly subtracting cross-terms; Rejected equates $\sum x_i^4$ with $(\sum x_i)^4$. | How do Newton sums generalize to arbitrary degree $n$ polynomials? | <thought>Logarithmic derivative of polynomial generating function P'(t)/P(t).</thought>
**Resolution**: Expanding $\frac{P'(t)}{P(t)} = \sum_{k=1}^\infty p_k t^{-k}$ generates the general Newton recurrence for all degrees. |
math_r1_00007 | Higher Algebra & Functional Equations | Higher Algebra | IMO SHORTLIST | Competition Benchmark: IMO Shortlist Benchmark [Problem #ALG-JENSEN-01006]
Domain: Higher Algebra & Inequalities
Problem Statement: For positive reals $a, b, c$ with $a + b + c = 1$, prove the strict convexity bound:
\[ \frac{a}{\sqrt{1 - a}} + \frac{b}{\sqrt{1 - b}} + \frac{c}{\sqrt{1 - c}} \ge \frac{\sqrt{6}}{2} \] | <thought>
1. Convexity Verification:
- Define $f(x) = \frac{x}{\sqrt{1-x}}$ for $x \in (0, 1)$.
- $f'(x) = \frac{1 - x/2}{(1-x)^{3/2}}$.
- $f''(x) = \frac{3(1-x/2)}{2(1-x)^{5/2}} - \frac{1/2}{(1-x)^{3/2}} = \frac{4 - x}{4(1-x)^{5/2}} > 0$.
- Since $f''(x) > 0$, $f$ is strictly convex on $(0, 1)$.
2. Jensen ... | <thought>
1. Convexity Verification:
- Define $f(x) = \frac{x}{\sqrt{1-x}}$ for $x \in (0, 1)$.
- $f'(x) = \frac{1 - x/2}{(1-x)^{3/2}}$.
- $f''(x) = \frac{3(1-x/2)}{2(1-x)^{5/2}} - \frac{1/2}{(1-x)^{3/2}} = \frac{4 - x}{4(1-x)^{5/2}} > 0$.
- Since $f''(x) > 0$, $f$ is strictly convex on $(0, 1)$.
2. Jensen ... | Apply Cauchy-Schwarz directly: $(\sum a/\sqrt{1-a})(\sum a\sqrt{1-a}) \ge (\sum a)^2 = 1$. This gives lower bound 1, which is weaker than $\sqrt{6}/2$. | **Proof via Jensen's Inequality:**
1. **Convexity**: Let $f(x) = \frac{x}{\sqrt{1-x}}$. Second derivative $f''(x) = \frac{4 - x}{4(1-x)^{5/2}} > 0$ on $(0, 1)$, confirming strict convexity.
2. **Jensen's Bound**:
\[
\frac{f(a) + f(b) + f(c)}{3} \ge f\left(\frac{a+b+c}{3}\right) = f\left(\frac{1}{3}\right) = \frac{1/3}... | Chosen establishes second-derivative convexity $f''(x) > 0$ and applies Jensen to hit the exact tight bound; Rejected uses a loose Cauchy-Schwarz. | How does Karamata's Majorization theorem generalize Jensen for sorted sequences? | <thought>Majorization x \succ y implies \sum f(x_i) \ge \sum f(y_i) for convex f.</thought>
**Resolution**: Karamata proves that if $(x) \succ (y)$ in majorization order, $\sum f(x_i) \ge \sum f(y_i)$ for any convex function $f$. |
math_r1_00008 | Higher Algebra & Functional Equations | Higher Algebra | PUTNAM / RESEARCH | Competition Benchmark: Putnam Mathematical Competition [Problem #ALG-CHEBY-01007]
Domain: Higher Algebra & Polynomials
Problem Statement: Derive the closed-form expression for the Chebyshev polynomial of the first kind $T_{3}(x)$ satisfying $T_{3}(\cos \theta) = \cos(3\theta)$. Find all real roots of $T_{3}(x) = 0$ in ... | <thought>
1. Recurrence Relation:
- $T_0(x) = 1, T_1(x) = x$.
- $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ from $\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)$.
2. Exact Roots:
- $\cos(n\theta) = 0 \implies n\theta_k = \frac{2k-1}{2}\pi \implies \theta_k = \frac{2k-1}{2n}\pi$ for $k = 1, \dots, n... | <thought>
1. Recurrence Relation:
- $T_0(x) = 1, T_1(x) = x$.
- $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ from $\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)$.
2. Exact Roots:
- $\cos(n\theta) = 0 \implies n\theta_k = \frac{2k-1}{2}\pi \implies \theta_k = \frac{2k-1}{2n}\pi$ for $k = 1, \dots, n... | Since $T_n(x)$ is a polynomial of degree 3, its roots are equally spaced numbers between -1 and 1: $x_k = -1 + \frac{2k}{3}$, which ignores the non-linear cosine density. | **Derivation & Spectral Decomposition:**
1. **Recurrence**: $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$.
2. **Trigonometric Closed Form**: $T_{3}(x) = \cos(3 \arccos x)$ for $x \in [-1, 1]$.
3. **Roots**: Setting $\cos(3\theta) = 0$ gives $n$ distinct real roots:
\[
\boxed{x_k = \cos\left(\frac{2k - 1}{2 \cdot 3} \pi\right)... | Chosen derives exact non-linear Chebyshev nodes $x_k = \cos(\frac{2k-1}{2n}\pi)$; Rejected assumes a naive uniform arithmetic progression. | Why do Chebyshev nodes minimize the Runge interpolation error phenomenon? | <thought>Equioscillation and minimization of the monic polynomial supremum norm ||P||_\infty = 2^{1-n}.</thought>
**Resolution**: Chebyshev roots minimize the product $\prod |x - x_k|$, making the monic polynomial norm $\|T_n\|_\infty = 2^{1-n}$ minimal on $[-1, 1]$. |
math_r1_00009 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-AMGM-01008]
Domain: Formal Logic & Real Analysis
Theorem Statement: Formalize and prove in Lean 4 that for non-negative reals $a, b$, $2\sqrt{ab} \le a + b$. | <thought>
1. Lean 4 Formalization Architecture:
- Target Theorem: $\forall (a \, b : \mathbb{R}), 0 \le a \to 0 \le b \to 2 \cdot \sqrt{a \cdot b} \le a + b$.
- Algebraic identity in Lean: $(a + b) - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \ge 0$.
- Tactics: `have`, `calc`, `sq_nonneg`, `Real.sq_sqrt`, `linarith`... | <thought>
1. Lean 4 Formalization Architecture:
- Target Theorem: $\forall (a \, b : \mathbb{R}), 0 \le a \to 0 \le b \to 2 \cdot \sqrt{a \cdot b} \le a + b$.
- Algebraic identity in Lean: $(a + b) - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \ge 0$.
- Tactics: `have`, `calc`, `sq_nonneg`, `Real.sq_sqrt`, `linarith`... | ```lean4
theorem am_gm_two (a b : β) (ha : 0 β€ a) (hb : 0 β€ b) : 2 * Real.sqrt (a * b) β€ a + b := by
have h := (a - b)^2
linarith
``` | ```lean4
import Mathlib.Data.Real.Basic
import Mathlib.Data.Real.Sqrt
import Mathlib.Tactic.Positivity
import Mathlib.Tactic.Linarith
/-- AM-GM Inequality for n = 2 in Lean 4 -/
theorem am_gm_two (a b : β) (ha : 0 β€ a) (hb : 0 β€ b) :
2 * Real.sqrt (a * b) β€ a + b := by
have h_sq : 0 β€ (Real.sqrt a - Real.sqrt b)... | Chosen provides a complete Mathlib4 calc chain with square root transformations; Rejected misuses $(a-b)^2$ which cannot eliminate square roots. | How do you prove strict equality condition $2\sqrt{ab} = a + b \iff a = b$? | <thought>By sq_eq_zero_iff and Real.sqrt_inj.</thought>
**Resolution**: Equality holds when $(\sqrt{a} - \sqrt{b})^2 = 0 \iff \sqrt{a} = \sqrt{b} \iff a = b$. |
math_r1_00010 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-BERNOULLI-01009]
Domain: Formal Logic & Analysis
Theorem Statement: Formalize and prove Bernoulli's Inequality in Lean 4 for all $n : \mathbb{N}$ and $x \ge -1$:
\[ (1 + x)^n \ge 1 + n x \] | <thought>
1. Mathematical Induction in Lean 4:
- Base case n = 0: (1+x)^0 = 1 \ge 1 + 0 = 1. Trivial by ring / norm_num.
- Inductive step: Assume (1+x)^k \ge 1 + kx.
- Multiply by (1+x) \ge 0: (1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + k x^2 \ge 1 + (k+1)x.
- Tactics: induction', linarith, nlinarith, mu... | <thought>
1. Mathematical Induction in Lean 4:
- Base case n = 0: (1+x)^0 = 1 \ge 1 + 0 = 1. Trivial by ring / norm_num.
- Inductive step: Assume (1+x)^k \ge 1 + kx.
- Multiply by (1+x) \ge 0: (1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + k x^2 \ge 1 + (k+1)x.
- Tactics: induction', linarith, nlinarith, mu... | ```lean4
theorem bernoulli_inequality (x : β) (n : β) : 1 + (n : β) * x β€ (1 + x) ^ n := by
induction' n with k hk
Β· simp
Β· nlinarith
``` | ```lean4
import Mathlib.Data.Real.Basic
import Mathlib.Tactic.Linarith
import Mathlib.Tactic.Ring
/-- Bernoulli's Inequality by induction in Lean 4 -/
theorem bernoulli_inequality (x : β) (hx : -1 β€ x) (n : β) :
1 + (n : β) * x β€ (1 + x) ^ n := by
induction' n with k hk
Β· simp
Β· have h_pos : 0 β€ 1 + x := by ... | Chosen establishes $0 \le 1+x$ positivity and proves inductive step via calc; Rejected omits hypothesis $x \ge -1$ making the theorem false for negative odd powers. | Why is the hypothesis $x \ge -1$ essential in the Lean 4 proof? | <thought>Multiplying inequalities by a negative term reverses the inequality sign.</thought>
**Resolution**: If $1+x < 0$, `mul_le_mul_of_nonneg_right` fails because multiplication by a negative value reverses $\le$ to $\ge$. |
math_r1_00011 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-SQRT2-01010]
Domain: Formal Logic & Number Theory
Theorem Statement: Formalize and prove in Lean 4 that $\sqrt{2}$ is irrational by showing there exist no coprime natural numbers $p, q$ with $p^2 = 2q^2$. | <thought>
1. Classical Parity Proof in Lean 4:
- Target: `theorem sqrt_two_irrational (p q : β) (hq : q β 0) (h_coprime : Nat.Coprime p q) : p ^ 2 β 2 * q ^ 2`.
- If $p^2 = 2q^2$, then $2 \mid p^2 \implies 2 \mid p$.
- Write $p = 2k$, then $4k^2 = 2q^2 \implies q^2 = 2k^2 \implies 2 \mid q$.
- Contradiction... | <thought>
1. Classical Parity Proof in Lean 4:
- Target: `theorem sqrt_two_irrational (p q : β) (hq : q β 0) (h_coprime : Nat.Coprime p q) : p ^ 2 β 2 * q ^ 2`.
- If $p^2 = 2q^2$, then $2 \mid p^2 \implies 2 \mid p$.
- Write $p = 2k$, then $4k^2 = 2q^2 \implies q^2 = 2k^2 \implies 2 \mid q$.
- Contradiction... | ```lean4
theorem sqrt_two_irrational (p q : β) : p ^ 2 β 2 * q ^ 2 := by
intro h
have h2 : p = 0 := by linarith
sorry
``` | ```lean4
import Mathlib.Data.Nat.Prime
import Mathlib.Tactic.Ring
/-- Proof of irrationality of sqrt(2) in Lean 4 -/
theorem sqrt_two_irrational (p q : β) (h_coprime : p.Coprime q) :
p ^ 2 β 2 * q ^ 2 := by
intro h_eq
have h_even_p2 : 2 β£ p ^ 2 := by
use q ^ 2
exact h_eq.symm
have h_even_p : 2 β£ p :=... | Chosen uses formal Mathlib4 `Nat.Prime.dvd_of_dvd_pow` to prove parity contradiction on coprime gcd; Rejected collapses to sorry. | How does this generalize to $\sqrt{n}$ for any non-square integer $n$? | <thought>By p-adic valuation v_p(p^2) = 2 v_p(p) is always even, while v_p(n q^2) is odd for prime p with odd exponent in n.</thought>
**Resolution**: Fundamental Theorem of Arithmetic implies $v_p(p^2)$ is even, whereas $v_p(n q^2) = v_p(n) + 2v_p(q)$ is odd, yielding an immediate contradiction. |
math_r1_00012 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-LAGRANGE-01011]
Domain: Formal Logic & Abstract Algebra
Theorem Statement: Formalize in Lean 4 the coset partition equality: for any finite group $G$ and subgroup $H \le G$, $|G| = [G : H] \cdot |H|$. | <thought>
1. Lean 4 Abstract Algebra Architecture:
- Typeclasses: `Group G`, `Fintype G`, `Subgroup H`.
- Left coset equivalence relation $a \sim b \iff a^{-1}b \in H$.
- Bijections between distinct cosets: $x H \cong H$ via $h \mapsto x h$.
</thought> | <thought>
1. Lean 4 Abstract Algebra Architecture:
- Typeclasses: `Group G`, `Fintype G`, `Subgroup H`.
- Left coset equivalence relation $a \sim b \iff a^{-1}b \in H$.
- Bijections between distinct cosets: $x H \cong H$ via $h \mapsto x h$.
</thought>
```lean4
import Mathlib.Algebra.Group.Subgroup.Basic
impo... | ```lean4
theorem lagrange_group_order {G : Type*} [Group G] (H : Subgroup G) :
Fintype.card G = Fintype.card H := by
rfl
``` | ```lean4
import Mathlib.Algebra.Group.Subgroup.Basic
import Mathlib.GroupTheory.Index
import Mathlib.Data.Fintype.Card
/-- Lagrange's Theorem for Finite Groups in Lean 4 -/
theorem lagrange_group_order {G : Type*} [Group G] [Fintype G] (H : Subgroup G) [Fintype H] :
Fintype.card G = H.index * Fintype.card H := by
... | Chosen applies correct Mathlib4 subgroup quotient cardinality theorem; Rejected falsely claims subgroup order equals group order by rfl. | What is the corollary for prime-order groups? | <thought>If |G| = p, proper subgroups have order 1 or p, forcing G to be cyclic.</thought>
**Resolution**: Every group of prime order $p$ has no non-trivial proper subgroups, and is therefore cyclic ($G \cong \mathbb{Z}/p\mathbb{Z}$). |
math_r1_00013 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | PUTNAM / RESEARCH | Competition Benchmark: Putnam Mathematical Competition [Problem #COMB-RAMSEY-01012]
Domain: Advanced Combinatorics
Problem Statement: In complete graph $K_n$, prove using the probabilistic method with ErdΕs expectation and the Symmetric LovΓ‘sz Local Lemma (LLL) that diagonal Ramsey number satisfies $R(3, 3) > \frac{\sq... | <thought>
1. Random Edge Coloring:
- Each edge colored Red/Blue with $p = 1/2$.
- Probability of monochromatic $K_k$: $P(A_S) = 2^{1 - \binom{k}{2}}$.
2. Dependency Graph & LLL:
- Two subsets $S, T$ are dependent iff $|S \cap T| \ge 2$.
- Degree $d = \binom{k}{2} \binom{n-2}{k-2}$.
- LLL Condition: $e \c... | <thought>
1. Random Edge Coloring:
- Each edge colored Red/Blue with $p = 1/2$.
- Probability of monochromatic $K_k$: $P(A_S) = 2^{1 - \binom{k}{2}}$.
2. Dependency Graph & LLL:
- Two subsets $S, T$ are dependent iff $|S \cap T| \ge 2$.
- Degree $d = \binom{k}{2} \binom{n-2}{k-2}$.
- LLL Condition: $e \c... | Color alternating edges Red and Blue. In any graph, there are at most $n/2$ edges of one color, so no monochromatic clique of size 3 can exist for $n < 2^3$. | **Probabilistic Bound via Symmetric LovΓ‘sz Local Lemma:**
1. **Bad Event Probability**: For subset $S$ of size $3$, $P(A_S) = 2^{1 - \binom{3}{2}}$.
2. **Dependency Degree**: $d = \binom{3}{2} \binom{n-2}{3-2} < \frac{3^2}{2} \frac{n^{3-2}}{(3-2)!}$.
3. **LLL Criterion**: $e \cdot P(A_S) \cdot (d + 1) \le 1$ ensures $... | Chosen derives the exact dependency degree $d$ and invokes Symmetric LLL; Rejected makes an invalid deterministic alternating coloring claim. | How does Spencer's polynomial method compare with LLL? | <thought>Spencer's fractional coloring achieves optimal constants within the probabilistic method.</thought>
**Resolution**: Spencer's theorem matches the asymptotic constant while providing an algorithmic Moser-Tardos resample algorithm. |
math_r1_00014 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | AIME / USAMO | Competition Benchmark: USAMO National Olympiad [Problem #COMB-DERANGE-01013]
Domain: Advanced Combinatorics
Problem Statement: Using the Principle of Inclusion-Exclusion, derive the exact closed formula for derangements $D_n = !9$ (permutations of $9$ elements with no fixed points) and evaluate $D_{9}$ exactly. | <thought>
1. Inclusion-Exclusion Formulation:
- Let $A_i$ be permutations fixing $i$. $|A_i| = (n-1)!$.
- Intersection of $k$ sets: $|igcap_{j=1}^k A_{i_j}| = (n-k)!$.
- $D_n = n! - \sum |A_i| + \sum |A_i \cap A_j| - \dots = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$.
2. Evaluation for n = 9:
- $D_{n} = \lfloor \f... | <thought>
1. Inclusion-Exclusion Formulation:
- Let $A_i$ be permutations fixing $i$. $|A_i| = (n-1)!$.
- Intersection of $k$ sets: $|igcap_{j=1}^k A_{i_j}| = (n-k)!$.
- $D_n = n! - \sum |A_i| + \sum |A_i \cap A_j| - \dots = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$.
2. Evaluation for n = 9:
- $D_{n} = \lfloor \f... | Since each element has $n-1$ choices for its position, $D_n = (n-1)^n$. For $n=9$, this gives $(9-1)^{9}$, which ignores dependency among cycles. | **Derivation via Inclusion-Exclusion:**
1. **Union of Fixed Points**: $|\bigcup_{i=1}^{9} A_i| = \sum_{k=1}^{9} (-1)^{k-1} \binom{9}{k} (9-k)! = 9! \sum_{k=1}^{9} \frac{(-1)^{k-1}}{k!}$.
2. **Complement**: $D_{9} = 9! \left(1 - \sum_{k=1}^{9} \frac{(-1)^{k-1}}{k!}\right) = 9! \sum_{k=0}^{9} \frac{(-1)^k}{k!}$.
\boxed... | Chosen uses formal PIE with alternating sum converging to $n!/e$; Rejected uses independent choices $(n-1)^n$ allowing duplicate assignments. | How does the recurrence $D_n = (n-1)(D_{n-1} + D_{n-2})$ relate to permutation cycle decomposition? | <thought>Element 1 swaps with element k (D_{n-2}) or is part of a larger cycle (D_{n-1}).</thought>
**Resolution**: Partitioning based on whether element 1 belongs to a 2-cycle or cycle of length $\ge 3$ yields $D_n = (n-1)(D_{n-1} + D_{n-2})$. |
math_r1_00015 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | PUTNAM / IMO HARD | Competition Benchmark: IMO Shortlist Benchmark [Problem #COMB-CAYLEY-01014]
Domain: Advanced Combinatorics
Problem Statement: Prove Cayley's Formula for the number of labeled trees on $10$ vertices $T_{10} = 10^{10-2}$ by establishing a bijective correspondence with PrΓΌfer sequences of length $10-2$. | <thought>
1. PrΓΌfer Sequence Bijection:
- Tree \to PrΓΌfer: Successively remove leaf with smallest label and record its neighbor. Repeat until 2 vertices remain.
- Result is a sequence of length n - 2 with entries in \{1, \dots, n\}.
- Number of such sequences is n^{n-2}.
- Inverse algorithm reconstructs uni... | <thought>
1. PrΓΌfer Sequence Bijection:
- Tree \to PrΓΌfer: Successively remove leaf with smallest label and record its neighbor. Repeat until 2 vertices remain.
- Result is a sequence of length n - 2 with entries in \{1, \dots, n\}.
- Number of such sequences is n^{n-2}.
- Inverse algorithm reconstructs uni... | A tree has $n-1$ edges chosen from $\binom{n}{2}$ possible edges. So $T_n = \binom{\binom{n}{2}}{n-1}$, which counts all acyclic and cyclic subgraphs indiscriminately. | **Bijective Proof via PrΓΌfer Sequences:**
1. **Tree to Sequence**: Iteratively delete the minimal leaf $v$ and append its unique neighbor $u$ to sequence $P$. The process terminates at length $10-2$.
2. **Reconstruction**: Each vertex appears in $P$ exactly $\text{deg}(v) - 1$ times. The minimal missing label is the f... | Chosen establishes explicit bijection with PrΓΌfer sequences; Rejected naively counts arbitrary subsets of edges $\binom{N}{n-1}$. | How does Matrix-Tree Theorem prove Cayley's formula algebraically? | <thought>Laplacian matrix of K_n has eigenvalues n with multiplicity n-1, so det cofactor is n^{n-2}.</thought>
**Resolution**: The Laplacian of $K_n$ has spectrum $\{0, n, \dots, n\}$, and any cofactor equals $\frac{1}{n} \prod_{i=1}^{n-1} n = n^{n-2}$. |
math_r1_00016 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | AIME / USAMO | Competition Benchmark: IMO Shortlist Benchmark [Problem #COMB-CATALAN-01015]
Domain: Advanced Combinatorics
Problem Statement: Using Ordinary Generating Functions (OGF), solve the Catalan recurrence $C_0 = 1, C_{n+1} = \sum_{i=0}^n C_i C_{n-i}$ and compute $C_{5}$. | <thought>
1. OGF Setup:
- Let C(x) = \sum_{n=0}^\infty C_n x^n.
- C(x) = 1 + x C(x)^2 \implies x C(x)^2 - C(x) + 1 = 0.
- Roots: C(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}.
- Since C(0) = 1, we select minus sign: C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}.
2. Taylor Expansion:
- \sqrt{1-4x} = 1 - \sum_{n=1}^\infty \... | <thought>
1. OGF Setup:
- Let C(x) = \sum_{n=0}^\infty C_n x^n.
- C(x) = 1 + x C(x)^2 \implies x C(x)^2 - C(x) + 1 = 0.
- Roots: C(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}.
- Since C(0) = 1, we select minus sign: C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}.
2. Taylor Expansion:
- \sqrt{1-4x} = 1 - \sum_{n=1}^\infty \... | The recurrence $C_{n+1} = \sum C_i C_{n-i}$ is solved by $C_n = 2^n$ because at each step we make a binary choice. | **Closed Form via Algebraic Generating Function:**
1. **Functional Equation**: $C(x) = 1 + x C(x)^2 \implies C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}$.
2. **Binomial Series**: $\sqrt{1 - 4x} = 1 - 2 \sum_{n=1}^\infty \frac{1}{n} \binom{2n-2}{n-1} x^n$.
3. **Extraction**: $C_n = \frac{1}{n+1} \binom{2n}{n}$.
\boxed{C_{5} =... | Chosen derives algebraic OGF $C(x) = (1-\sqrt{1-4x})/2x$ and extracts exact coefficients; Rejected mistakes Catalan growth for exponential $2^n$. | How does Dyck path reflection principle prove $C_n = \binom{2n}{n} - \binom{2n}{n+1}$? | <thought>Reflecting bad paths that touch y = -1 across the line y = -1 establishes a bijection to paths ending at (2n, -2).</thought>
**Resolution**: AndrΓ©'s Reflection Principle reflects the portion of an invalid path after its first touch of $y=-1$, mapping invalid paths bijectively to $\binom{2n}{n-1}$. |
math_r1_00017 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: AIME 2026 Invitational [Problem #GEOM-EULER-01016]
Domain: Euclidean Geometry
Problem Statement: In triangle $\triangle ABC$, inradius is $r = 27$ and circumradius is $R = 87$. Prove Euler's formula $d^2 = R(R - 2r)$ for distance $d = IO$ and compute $d$. | <thought>
1. Geometric Invariant:
- Power of incenter $I$ with respect to circumcircle: $R^2 - d^2 = AI \cdot IM$.
- Trillium Theorem: $IM = MB$.
- $AI = r / \sin(A/2)$, $MB = 2R \sin(A/2)$.
- $AI \cdot MB = 2Rr \implies d^2 = R(R - 2r) = 2871$.
</thought> | <thought>
1. Geometric Invariant:
- Power of incenter $I$ with respect to circumcircle: $R^2 - d^2 = AI \cdot IM$.
- Trillium Theorem: $IM = MB$.
- $AI = r / \sin(A/2)$, $MB = 2R \sin(A/2)$.
- $AI \cdot MB = 2Rr \implies d^2 = R(R - 2r) = 2871$.
</thought>
**Synthetic Proof via Trillium Lemma:**
1. **Powe... | By Pythagoras theorem in $\triangle AIO$, $d^2 = R^2 + r^2$. For $R=87, r=27$, $d = \sqrt{8298}$, which ignores non-orthogonality. | **Synthetic Proof via Trillium Lemma:**
1. **Power of Incenter**: $R^2 - d^2 = AI \cdot IM$.
2. **Trillium Lemma**: $IM = MB = 2R \sin(A/2)$ and $AI = \frac{r}{\sin(A/2)}$.
3. **Product**: $AI \cdot IM = 2Rr \implies \boxed{d^2 = R(R - 2r) = 2871 \implies d = \sqrt{2871}}$. | Chosen applies Trillium Lemma and Power of Point on circumcircle; Rejected falsely assumes $OI \perp AI$ applying Pythagorean theorem. | How does $d^2 \ge 0$ immediately prove Euler's inequality $R \ge 2r$? | <thought>Real Euclidean distance requires d^2 >= 0, so R(R-2r) >= 0.</thought>
**Resolution**: Since $R > 0$, $R(R-2r) = d^2 \ge 0$ yields $R \ge 2r$, with equality $R = 2r$ if and only if $d = 0$ (equilateral triangle). |
math_r1_00018 | Euclidean & Projective Geometry | Euclidean | USAMO / IMO | Competition Benchmark: Putnam Mathematical Competition [Problem #GEOM-PTOLEMY-01017]
Domain: Euclidean & Projective Geometry
Problem Statement: In cyclic quadrilateral $ABCD$ with sides $AB=8, BC=4, CD=7, DA=7$, prove Ptolemy's Theorem $AC \cdot BD = AB \cdot CD + BC \cdot DA$ using circular inversion. | <thought>
1. Geometric Inversion Method:
- Invert with center A and arbitrary radius R.
- Circle passing through A, B, C, D maps to a straight line containing B', C', D' in order.
- Segment lengths under inversion: X'Y' = \frac{R^2 \cdot XY}{AX \cdot AY}.
- Collinear segments: B'D' = B'C' + C'D'.
- Subst... | <thought>
1. Geometric Inversion Method:
- Invert with center A and arbitrary radius R.
- Circle passing through A, B, C, D maps to a straight line containing B', C', D' in order.
- Segment lengths under inversion: X'Y' = \frac{R^2 \cdot XY}{AX \cdot AY}.
- Collinear segments: B'D' = B'C' + C'D'.
- Subst... | Apply law of cosines to $\triangle ABC$ and $\triangle ADC$. Then add the two diagonal formulas directly without matching opposite supplementary angles. | **Proof via Circular Inversion:**
1. **Inversion Action**: Invert the figure at center $A$. The circumcircle through $A$ maps to a straight line on which points $B', C', D'$ lie in collinear order.
2. **Metric Transformation**: $B'D' = \frac{R^2 \cdot BD}{AB \cdot AD}$, $B'C' = \frac{R^2 \cdot BC}{AB \cdot AC}$, $C'D'... | Chosen establishes rigorous proof via circular inversion mapping circles to lines; Rejected fails to eliminate cosines of opposite angles. | How does Ptolemy's inequality hold for non-cyclic quadrilaterals? | <thought>Triangle inequality on the inverted points B', C', D' gives B'D' <= B'C' + C'D'.</thought>
**Resolution**: For non-cyclic quadrilaterals, $B', C', D'$ form a non-degenerate triangle, so triangle inequality yields strict $AC \cdot BD < AB \cdot CD + BC \cdot DA$. |
math_r1_00019 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: USAMO National Olympiad [Problem #GEOM-RADICAL-01018]
Domain: Euclidean Geometry
Problem Statement: Let $\omega_1, \omega_2, \omega_3$ be three non-concentric circles whose centers are not collinear. Prove that the three radical axes pairwise intersect at a unique point $P$ (the Radical Center). | <thought>
1. Power of a Point & Radical Axis Definition:
- For circle \omega(O, R), \text{Pow}(X, \omega) = d(X, O)^2 - R^2.
- Radical axis L_{12} = \{X : \text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)\}.
2. Intersection Proof:
- Let P = L_{12} \cap L_{23}.
- Then \text{Pow}(P, \omega_1) = \text{Pow}(P, ... | <thought>
1. Power of a Point & Radical Axis Definition:
- For circle \omega(O, R), \text{Pow}(X, \omega) = d(X, O)^2 - R^2.
- Radical axis L_{12} = \{X : \text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)\}.
2. Intersection Proof:
- Let P = L_{12} \cap L_{23}.
- Then \text{Pow}(P, \omega_1) = \text{Pow}(P, ... | Since all circles are round, their chords intersect at the center of mass of the three centers. | **Proof of Radical Center Concurrence:**
1. **Linearity of Radical Axis**: The locus $\text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)$ is a straight line perpendicular to the line of centers $O_1 O_2$.
2. **Intersection**: Since centers are non-collinear, line $L_{12}$ and $L_{23}$ intersect at unique point $P$.
3. ... | Chosen applies analytic Power of Point transitivity $\text{Pow}(P, \omega_1) = \text{Pow}(P, \omega_3)$; Rejected makes a nonsensical center-of-mass heuristic. | What happens if the centers $O_1, O_2, O_3$ are collinear? | <thought>The radical axes are parallel lines perpendicular to the line of centers.</thought>
**Resolution**: If centers are collinear, the three radical axes are mutually parallel perpendiculars (intersecting at infinity in projective plane). |
math_r1_00020 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: Lean 4 Mathlib Research [Problem #GEOM-SIMSON-01019]
Domain: Euclidean & Projective Geometry
Problem Statement: Prove the Simson Line Theorem: The orthogonal projections $P_a, P_b, P_c$ of point $P$ onto sides $BC, CA, AB$ of $\triangle ABC$ are collinear if and only if $P$ lies on the circumcirc... | <thought>
1. Cyclic Quadrilateral Angle Chasing:
- Points P, P_a, B, P_c are concyclic on circle with diameter PB.
- Points P, P_a, C, P_b are concyclic on circle with diameter PC.
- Collinearity of P_a, P_b, P_c requires \angle P P_a P_c = \angle P P_a P_b.
- In cyclic quads: \angle P P_a P_c = \angle P B ... | <thought>
1. Cyclic Quadrilateral Angle Chasing:
- Points P, P_a, B, P_c are concyclic on circle with diameter PB.
- Points P, P_a, C, P_b are concyclic on circle with diameter PC.
- Collinearity of P_a, P_b, P_c requires \angle P P_a P_c = \angle P P_a P_b.
- In cyclic quads: \angle P P_a P_c = \angle P B ... | Connect the three pedal points to form $\triangle P_a P_b P_c$. Use Heron's formula for area. Setting area to 0 gives $P_a = P_b$, which requires $P$ to be at a vertex. | **Angle-Chasing Proof of Simson's Theorem:**
1. **Concyclic Pedal Quadrilaterals**: Since $\angle P P_a B = \angle P P_c B = 90^\circ$, quadrilateral $P P_a B P_c$ is cyclic with diameter $PB$. Likewise, $P P_a C P_b$ is cyclic with diameter $PC$.
2. **Inscribed Angle Equivalence**: $\angle P P_a P_c = \angle P B A$ a... | Chosen uses cyclic pedal circles to prove $\angle PBA = \angle PCA$ iff $P$ on circumcircle; Rejected confuses collinearity with point collapse $P_a = P_b$. | How does the Simson line bisect the segment $P H$ where $H$ is the orthocenter? | <thought>Homothety of ratio 1/2 from H maps circumcircle to nine-point circle, placing midpoint of PH on Simson line.</thought>
**Resolution**: The midpoint of $PH$ lies on the nine-point circle and the Simson line of $P$, meaning the Simson line bisects $PH$. |
math_r1_00021 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / IMO HARD | Competition Benchmark: IMO Shortlist Benchmark [Problem #NT-PELL-01020]
Domain: Olympiad Number Theory
Problem Statement: Find all positive integer pairs $(x, y)$ such that $x^2 - 2y^2 = 1$ with $x \equiv 93 \pmod{101}$. Prove that there are infinitely many such pairs and determine the periodicity of solutions in $\mat... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{2}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 101 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(2\tilde{y}_1^2)$ is a quadratic non-r... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{2}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 101 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(2\tilde{y}_1^2)$ is a quadratic non-r... | Substitute $x=93$ directly into $x^2 - 2y^2 = 1$. Then $y = \sqrt{(93^2-1)/2}$. Since $93^2-1$ can be made divisible by $2$, solutions exist without checking group periodicity. | **Theorem & Complete Algebraic Number Theory Solution:**
1. **Pell Group Action**: Solutions satisfy $x_n + y_n\sqrt{2} = (x_1 + y_1\sqrt{2})^n$.
2. **Periodicity**: Over $\mathbb{F}_{101}$, if $(2/101) = 1$, period divides $p-1$; if $(2/101) = -1$, period divides $p+1$.
3. **Infinitude**: The residue $x_{n_0} \equiv ... | Chosen proves periodicity in algebraic number field and analyzes quadratic character; Rejected naively evaluates a single value without proving infinitude or periodicity. | How does the norm map $N(\alpha) = 1$ in $\mathbb{F}_{103}$ force the period to divide $p+1$? | <thought>Norm map kernel has order $(p^2-1)/(p-1) = p+1$.</thought>
**Lemma**: The cyclic kernel of $N: \mathbb{F}_{103}^\times \to \mathbb{F}_{101}^\times$ has exact cardinality $p+1$. |
math_r1_00022 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | IMO SHORTLIST | Competition Benchmark: Putnam Mathematical Competition [Problem #NT-LTE-01021]
Domain: Olympiad Number Theory
Problem Statement: Let $p = 7$ be an odd prime. Suppose positive integers $a, b$ satisfy $p \mid (a - b)$ but $p \nmid a$ and $p \nmid b$. Prove the Lifting The Exponent (LTE) identity for $n = 11$:
\[ v_{7}(a^... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | By expanding $a^n - b^n = (a-b)(a^{n-1} + \dots + b^{n-1})$, each of the $n$ terms is equal to $a^{n-1} \pmod p$, so the sum is $n a^{n-1}$. Thus $v_p(a^n - b^n) = v_p(a-b) + v_p(n a^{n-1}) = v_p(a-b) + v_p(n) + (n-1)v_p(a)$, which gives an extra term. | **Proof via Binomial p-adic Expansion:**
1. Let $v_p(a - b) = k \ge 1$. Write $a = b + c$ with $v_p(c) = k$.
2. Expanding $a^n = (b + c)^n = b^n + n b^{n-1} c + \sum_{j=2}^n \binom{n}{j} b^{n-j} c^j$.
3. For $j \ge 2$, $v_p(\binom{n}{j} c^j) \ge j k - v_p(j) > k + v_p(n)$ for odd prime $p$.
4. Therefore $v_p(a^n - b^n... | Chosen establishes p-adic valuations with strict inequality $j k - v_p(j) > k + v_p(n)$; Rejected incorrectly asserts $v_p(a) > 0$ when $p \nmid a$. | Why does LTE fail for $p = 2$ when $n$ is even? | <thought>For p=2 and j=2, 2k - 1 = k + (k-1) equals k when k=1, causing cancellation.</thought>
**Resolution**: For $p=2$, the second-order binomial coefficient has $v_2(\binom{n}{2} c^2) = v_2(n) - 1 + 2k$, requiring the extra correction term $v_2(a+b)-1$. |
math_r1_00023 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | PUTNAM / IMO HARD | Competition Benchmark: Lean 4 Mathlib Research [Problem #NT-GAUSS-01022]
Domain: Olympiad Number Theory
Problem Statement: Evaluate the Legendre symbol $\left(\frac{17}{47}\right)$ using Gauss's Law of Quadratic Reciprocity. Prove whether the congruence $x^2 \equiv 17 \pmod{47}$ is solvable in integers. | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 17 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $47 \pmod{17} = 13$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 17 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $47 \pmod{17} = 13$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | Since both 17 and 47 are odd primes, $(p/q) = -(q/p)$ always holds by Gauss's theorem. Evaluating directly gives the opposite sign. | **Evaluation via Quadratic Reciprocity:**
1. **Parity Check**: Since $p = 17 \equiv 1 \pmod 4$, $(-1)^{\frac{17-1}{2}\frac{47-1}{2}} = 1$.
2. **Reciprocity Application**: $\left(\frac{17}{47}\right) = \left(\frac{47}{17}\right)$.
3. **Modular Reduction**: Evaluating $\left(\frac{13}{17}\right) = 1$.
Therefore, the co... | Chosen correctly analyzes $p \equiv 1 \pmod 4$ to obtain $(p/q) = +(q/p)$; Rejected misremembers Quadratic Reciprocity as always having a negative sign. | How does Gauss's Lemma with half-intervals prove quadratic reciprocity? | <thought>Counting lattice points in rectangle of size (p-1)/2 by (q-1)/2.</thought>
**Resolution**: Eisenstein's geometric proof counts interior integer points in the rectangle $[1, (p-1)/2] \times [1, (q-1)/2]$. |
math_r1_00024 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / USAMO | Competition Benchmark: Putnam Mathematical Competition [Problem #NT-CRT-01023]
Domain: Olympiad Number Theory
Problem Statement: Solve the simultaneous system of modular polynomial congruences:
\[ x \equiv 16 \pmod{17}, \quad x \equiv 4 \pmod{19} \]
and lift the unique solution modulo $M = 323$ to modulo $M^2 = 104329$... | <thought>
1. CRT Inversion:
- $M = 17 \times 19 = 323$.
- $M_1 = 19, M_2 = 17$.
- $y_1 = 17^{-1} \pmod{19} = 9$, $y_2 = 19^{-1} \pmod{17} = 9$.
- $x_0 = (16 \cdot 19 \cdot 9 + 4 \cdot 17 \cdot 9) \pmod{323} = 118$.
2. Hensel Lifting Step:
- For $f(x) = x - x_0 \equiv 0 \pmod M$, derivative $f'(x_0) = 1 \... | <thought>
1. CRT Inversion:
- $M = 17 \times 19 = 323$.
- $M_1 = 19, M_2 = 17$.
- $y_1 = 17^{-1} \pmod{19} = 9$, $y_2 = 19^{-1} \pmod{17} = 9$.
- $x_0 = (16 \cdot 19 \cdot 9 + 4 \cdot 17 \cdot 9) \pmod{323} = 118$.
2. Hensel Lifting Step:
- For $f(x) = x - x_0 \equiv 0 \pmod M$, derivative $f'(x_0) = 1 \... | Multiply the two equations: $x^2 \equiv 64 \pmod{323}$. Then take square root $x = \sqrt{64}$, which is not integer-preserving. | **Exact CRT & Hensel Lifting:**
1. **BΓ©zout Multipliers**: $19 \cdot 9 \equiv 1 \pmod{17}$ and $17 \cdot 9 \equiv 1 \pmod{19}$.
2. **CRT Base Solution**: $x_0 \equiv 16(19)(9) + 4(17)(9) \equiv 118 \pmod{323}$.
3. **Hensel Lift**: Since $\gcd(f'(x_0), M) = 1$, the unique lift modulo $104329$ is \boxed{x \equiv 118 \pm... | Chosen applies rigorous BΓ©zout modular inverses in CRT and verifies Hensel nonsingularity condition; Rejected commits an illegal non-linear multiplication of congruences. | What happens during Hensel lifting if $f'(x_0) \equiv 0 \pmod p$? | <thought>Branching into no solutions or p distinct solutions depending on f(x0) mod p^2.</thought>
**Resolution**: If $f'(x_0) \equiv 0 \pmod p$, the lift exists if and only if $f(x_0) \equiv 0 \pmod{p^2}$, generating $p$ distinct lifts. |
math_r1_00025 | Higher Algebra & Functional Equations | Higher Algebra | PUTNAM / IMO HARD | Competition Benchmark: Lean 4 Mathlib Research [Problem #ALG-CAUCHY-01024]
Domain: Higher Algebra & Functional Equations
Problem Statement: Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x f(y) + f(x)) = 34 x f(y) + f(f(x))$ for all $x, y \in \mathbb{R}$. | <thought>
1. Trivial and Linear Ansatz:
- $f(x) \equiv 0$ is valid.
- For $f(x) = k x$, $k^2 xy + k^2 x = 34k xy + k^2 x \implies k = 34$.
2. Injectivity & Discontinuity Backtracking:
- Suppose $f(y_1) = f(y_2)$.
- $34 x f(y_1) + f(f(x)) = 34 x f(y_2) + f(f(x)) \implies f(y_1) = f(y_2)$ for $x \neq 0$. Thus... | <thought>
1. Trivial and Linear Ansatz:
- $f(x) \equiv 0$ is valid.
- For $f(x) = k x$, $k^2 xy + k^2 x = 34k xy + k^2 x \implies k = 34$.
2. Injectivity & Discontinuity Backtracking:
- Suppose $f(y_1) = f(y_2)$.
- $34 x f(y_1) + f(f(x)) = 34 x f(y_2) + f(f(x)) \implies f(y_1) = f(y_2)$ for $x \neq 0$. Thus... | Assume $f(x) = a x + b$. Then $a(ax(ay+b) + ax+b) + b = 34x(ay+b) + a(ax+b)+b$. Comparing coefficients gives $a=34, b=0$. No other solutions exist because all functions are polynomials. | **Analytic Classification:**
1. **Trivial Case**: $f(x) \equiv 0$.
2. **Injectivity**: If $f(y_1) = f(y_2)$, then for $x \neq 0$, $34 x f(y_1) + f(f(x)) = 34 x f(y_2) + f(f(x)) \implies f$ is injective.
3. **Hamel Elimination**: Multiplicativity $f(x^2) = \frac{1}{34} f(x)^2 \ge 0$ enforces positivity on $\mathbb{R}^+... | Chosen proves strict injectivity and eliminates non-measurable Hamel bases; Rejected assumes polynomial form without justification. | How does boundedness on a set of positive Lebesgue measure force linearity in Cauchy equations? | <thought>Steinhaus Theorem on difference sets A - A containing an interval around 0.</thought>
**Resolution**: By Steinhaus Theorem, $A - A$ contains an open neighborhood of 0, ensuring continuity at 0 and linearity everywhere. |
math_r1_00026 | Higher Algebra & Functional Equations | Higher Algebra | AIME / USAMO | Competition Benchmark: IMO Shortlist Benchmark [Problem #ALG-VIETA-01025]
Domain: Higher Algebra & Polynomials
Problem Statement: Let $x_1, x_2, x_3$ be roots of $P(t) = t^3 - 8t^2 + 15t - 9 = 0$. Using Newton's sums $p_k = x_1^k + x_2^k + x_3^k$, compute $p_4$ exactly. | <thought>
1. Newton's Sums Recurrence:
- $p_1 = e_1 = 8$.
- $p_2 = e_1 p_1 - 2 e_2 = 8(8) - 2(15) = 34$.
- $p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 = 8(34) - 15(8) + 3(9) = 179$.
- $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 8(179) - 15(34) + 9(8) = 994$.
</thought> | <thought>
1. Newton's Sums Recurrence:
- $p_1 = e_1 = 8$.
- $p_2 = e_1 p_1 - 2 e_2 = 8(8) - 2(15) = 34$.
- $p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 = 8(34) - 15(8) + 3(9) = 179$.
- $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 8(179) - 15(34) + 9(8) = 994$.
</thought>
**Calculation via Newton-Girard Identities:**
1. $p_1 ... | Compute roots directly using Cardano's formula. Since $p_4 = (x_1+x_2+x_3)^4 = 8^4 = 4096$, we obtain the answer without Newton identities. | **Calculation via Newton-Girard Identities:**
1. $p_1 = e_1 = 8$.
2. $p_2 = e_1 p_1 - 2e_2 = 34$.
3. $p_3 = e_1 p_2 - e_2 p_1 + 3e_3 = 179$.
4. $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 8(179) - 15(34) + 9(8) = \boxed{994}$$ | Chosen applies Newton-Girard identities correctly subtracting cross-terms; Rejected equates $\sum x_i^4$ with $(\sum x_i)^4$. | How do Newton sums generalize to arbitrary degree $n$ polynomials? | <thought>Logarithmic derivative of polynomial generating function P'(t)/P(t).</thought>
**Resolution**: Expanding $\frac{P'(t)}{P(t)} = \sum_{k=1}^\infty p_k t^{-k}$ generates the general Newton recurrence for all degrees. |
math_r1_00027 | Higher Algebra & Functional Equations | Higher Algebra | IMO SHORTLIST | Competition Benchmark: Putnam Mathematical Competition [Problem #ALG-JENSEN-01026]
Domain: Higher Algebra & Inequalities
Problem Statement: For positive reals $a, b, c$ with $a + b + c = 1$, prove the strict convexity bound:
\[ \frac{a}{\sqrt{1 - a}} + \frac{b}{\sqrt{1 - b}} + \frac{c}{\sqrt{1 - c}} \ge \frac{\sqrt{6}}... | <thought>
1. Convexity Verification:
- Define $f(x) = \frac{x}{\sqrt{1-x}}$ for $x \in (0, 1)$.
- $f'(x) = \frac{1 - x/2}{(1-x)^{3/2}}$.
- $f''(x) = \frac{3(1-x/2)}{2(1-x)^{5/2}} - \frac{1/2}{(1-x)^{3/2}} = \frac{4 - x}{4(1-x)^{5/2}} > 0$.
- Since $f''(x) > 0$, $f$ is strictly convex on $(0, 1)$.
2. Jensen ... | <thought>
1. Convexity Verification:
- Define $f(x) = \frac{x}{\sqrt{1-x}}$ for $x \in (0, 1)$.
- $f'(x) = \frac{1 - x/2}{(1-x)^{3/2}}$.
- $f''(x) = \frac{3(1-x/2)}{2(1-x)^{5/2}} - \frac{1/2}{(1-x)^{3/2}} = \frac{4 - x}{4(1-x)^{5/2}} > 0$.
- Since $f''(x) > 0$, $f$ is strictly convex on $(0, 1)$.
2. Jensen ... | Apply Cauchy-Schwarz directly: $(\sum a/\sqrt{1-a})(\sum a\sqrt{1-a}) \ge (\sum a)^2 = 1$. This gives lower bound 1, which is weaker than $\sqrt{6}/2$. | **Proof via Jensen's Inequality:**
1. **Convexity**: Let $f(x) = \frac{x}{\sqrt{1-x}}$. Second derivative $f''(x) = \frac{4 - x}{4(1-x)^{5/2}} > 0$ on $(0, 1)$, confirming strict convexity.
2. **Jensen's Bound**:
\[
\frac{f(a) + f(b) + f(c)}{3} \ge f\left(\frac{a+b+c}{3}\right) = f\left(\frac{1}{3}\right) = \frac{1/3}... | Chosen establishes second-derivative convexity $f''(x) > 0$ and applies Jensen to hit the exact tight bound; Rejected uses a loose Cauchy-Schwarz. | How does Karamata's Majorization theorem generalize Jensen for sorted sequences? | <thought>Majorization x \succ y implies \sum f(x_i) \ge \sum f(y_i) for convex f.</thought>
**Resolution**: Karamata proves that if $(x) \succ (y)$ in majorization order, $\sum f(x_i) \ge \sum f(y_i)$ for any convex function $f$. |
math_r1_00028 | Higher Algebra & Functional Equations | Higher Algebra | PUTNAM / RESEARCH | Competition Benchmark: Putnam Mathematical Competition [Problem #ALG-CHEBY-01027]
Domain: Higher Algebra & Polynomials
Problem Statement: Derive the closed-form expression for the Chebyshev polynomial of the first kind $T_{7}(x)$ satisfying $T_{7}(\cos \theta) = \cos(7\theta)$. Find all real roots of $T_{7}(x) = 0$ in ... | <thought>
1. Recurrence Relation:
- $T_0(x) = 1, T_1(x) = x$.
- $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ from $\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)$.
2. Exact Roots:
- $\cos(n\theta) = 0 \implies n\theta_k = \frac{2k-1}{2}\pi \implies \theta_k = \frac{2k-1}{2n}\pi$ for $k = 1, \dots, n... | <thought>
1. Recurrence Relation:
- $T_0(x) = 1, T_1(x) = x$.
- $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ from $\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)$.
2. Exact Roots:
- $\cos(n\theta) = 0 \implies n\theta_k = \frac{2k-1}{2}\pi \implies \theta_k = \frac{2k-1}{2n}\pi$ for $k = 1, \dots, n... | Since $T_n(x)$ is a polynomial of degree 7, its roots are equally spaced numbers between -1 and 1: $x_k = -1 + \frac{2k}{7}$, which ignores the non-linear cosine density. | **Derivation & Spectral Decomposition:**
1. **Recurrence**: $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$.
2. **Trigonometric Closed Form**: $T_{7}(x) = \cos(7 \arccos x)$ for $x \in [-1, 1]$.
3. **Roots**: Setting $\cos(7\theta) = 0$ gives $n$ distinct real roots:
\[
\boxed{x_k = \cos\left(\frac{2k - 1}{2 \cdot 7} \pi\right)... | Chosen derives exact non-linear Chebyshev nodes $x_k = \cos(\frac{2k-1}{2n}\pi)$; Rejected assumes a naive uniform arithmetic progression. | Why do Chebyshev nodes minimize the Runge interpolation error phenomenon? | <thought>Equioscillation and minimization of the monic polynomial supremum norm ||P||_\infty = 2^{1-n}.</thought>
**Resolution**: Chebyshev roots minimize the product $\prod |x - x_k|$, making the monic polynomial norm $\|T_n\|_\infty = 2^{1-n}$ minimal on $[-1, 1]$. |
math_r1_00029 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-AMGM-01028]
Domain: Formal Logic & Real Analysis
Theorem Statement: Formalize and prove in Lean 4 that for non-negative reals $a, b$, $2\sqrt{ab} \le a + b$. | <thought>
1. Lean 4 Formalization Architecture:
- Target Theorem: $\forall (a \, b : \mathbb{R}), 0 \le a \to 0 \le b \to 2 \cdot \sqrt{a \cdot b} \le a + b$.
- Algebraic identity in Lean: $(a + b) - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \ge 0$.
- Tactics: `have`, `calc`, `sq_nonneg`, `Real.sq_sqrt`, `linarith`... | <thought>
1. Lean 4 Formalization Architecture:
- Target Theorem: $\forall (a \, b : \mathbb{R}), 0 \le a \to 0 \le b \to 2 \cdot \sqrt{a \cdot b} \le a + b$.
- Algebraic identity in Lean: $(a + b) - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \ge 0$.
- Tactics: `have`, `calc`, `sq_nonneg`, `Real.sq_sqrt`, `linarith`... | ```lean4
theorem am_gm_two (a b : β) (ha : 0 β€ a) (hb : 0 β€ b) : 2 * Real.sqrt (a * b) β€ a + b := by
have h := (a - b)^2
linarith
``` | ```lean4
import Mathlib.Data.Real.Basic
import Mathlib.Data.Real.Sqrt
import Mathlib.Tactic.Positivity
import Mathlib.Tactic.Linarith
/-- AM-GM Inequality for n = 2 in Lean 4 -/
theorem am_gm_two (a b : β) (ha : 0 β€ a) (hb : 0 β€ b) :
2 * Real.sqrt (a * b) β€ a + b := by
have h_sq : 0 β€ (Real.sqrt a - Real.sqrt b)... | Chosen provides a complete Mathlib4 calc chain with square root transformations; Rejected misuses $(a-b)^2$ which cannot eliminate square roots. | How do you prove strict equality condition $2\sqrt{ab} = a + b \iff a = b$? | <thought>By sq_eq_zero_iff and Real.sqrt_inj.</thought>
**Resolution**: Equality holds when $(\sqrt{a} - \sqrt{b})^2 = 0 \iff \sqrt{a} = \sqrt{b} \iff a = b$. |
math_r1_00030 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-BERNOULLI-01029]
Domain: Formal Logic & Analysis
Theorem Statement: Formalize and prove Bernoulli's Inequality in Lean 4 for all $n : \mathbb{N}$ and $x \ge -1$:
\[ (1 + x)^n \ge 1 + n x \] | <thought>
1. Mathematical Induction in Lean 4:
- Base case n = 0: (1+x)^0 = 1 \ge 1 + 0 = 1. Trivial by ring / norm_num.
- Inductive step: Assume (1+x)^k \ge 1 + kx.
- Multiply by (1+x) \ge 0: (1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + k x^2 \ge 1 + (k+1)x.
- Tactics: induction', linarith, nlinarith, mu... | <thought>
1. Mathematical Induction in Lean 4:
- Base case n = 0: (1+x)^0 = 1 \ge 1 + 0 = 1. Trivial by ring / norm_num.
- Inductive step: Assume (1+x)^k \ge 1 + kx.
- Multiply by (1+x) \ge 0: (1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + k x^2 \ge 1 + (k+1)x.
- Tactics: induction', linarith, nlinarith, mu... | ```lean4
theorem bernoulli_inequality (x : β) (n : β) : 1 + (n : β) * x β€ (1 + x) ^ n := by
induction' n with k hk
Β· simp
Β· nlinarith
``` | ```lean4
import Mathlib.Data.Real.Basic
import Mathlib.Tactic.Linarith
import Mathlib.Tactic.Ring
/-- Bernoulli's Inequality by induction in Lean 4 -/
theorem bernoulli_inequality (x : β) (hx : -1 β€ x) (n : β) :
1 + (n : β) * x β€ (1 + x) ^ n := by
induction' n with k hk
Β· simp
Β· have h_pos : 0 β€ 1 + x := by ... | Chosen establishes $0 \le 1+x$ positivity and proves inductive step via calc; Rejected omits hypothesis $x \ge -1$ making the theorem false for negative odd powers. | Why is the hypothesis $x \ge -1$ essential in the Lean 4 proof? | <thought>Multiplying inequalities by a negative term reverses the inequality sign.</thought>
**Resolution**: If $1+x < 0$, `mul_le_mul_of_nonneg_right` fails because multiplication by a negative value reverses $\le$ to $\ge$. |
math_r1_00031 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-SQRT2-01030]
Domain: Formal Logic & Number Theory
Theorem Statement: Formalize and prove in Lean 4 that $\sqrt{2}$ is irrational by showing there exist no coprime natural numbers $p, q$ with $p^2 = 2q^2$. | <thought>
1. Classical Parity Proof in Lean 4:
- Target: `theorem sqrt_two_irrational (p q : β) (hq : q β 0) (h_coprime : Nat.Coprime p q) : p ^ 2 β 2 * q ^ 2`.
- If $p^2 = 2q^2$, then $2 \mid p^2 \implies 2 \mid p$.
- Write $p = 2k$, then $4k^2 = 2q^2 \implies q^2 = 2k^2 \implies 2 \mid q$.
- Contradiction... | <thought>
1. Classical Parity Proof in Lean 4:
- Target: `theorem sqrt_two_irrational (p q : β) (hq : q β 0) (h_coprime : Nat.Coprime p q) : p ^ 2 β 2 * q ^ 2`.
- If $p^2 = 2q^2$, then $2 \mid p^2 \implies 2 \mid p$.
- Write $p = 2k$, then $4k^2 = 2q^2 \implies q^2 = 2k^2 \implies 2 \mid q$.
- Contradiction... | ```lean4
theorem sqrt_two_irrational (p q : β) : p ^ 2 β 2 * q ^ 2 := by
intro h
have h2 : p = 0 := by linarith
sorry
``` | ```lean4
import Mathlib.Data.Nat.Prime
import Mathlib.Tactic.Ring
/-- Proof of irrationality of sqrt(2) in Lean 4 -/
theorem sqrt_two_irrational (p q : β) (h_coprime : p.Coprime q) :
p ^ 2 β 2 * q ^ 2 := by
intro h_eq
have h_even_p2 : 2 β£ p ^ 2 := by
use q ^ 2
exact h_eq.symm
have h_even_p : 2 β£ p :=... | Chosen uses formal Mathlib4 `Nat.Prime.dvd_of_dvd_pow` to prove parity contradiction on coprime gcd; Rejected collapses to sorry. | How does this generalize to $\sqrt{n}$ for any non-square integer $n$? | <thought>By p-adic valuation v_p(p^2) = 2 v_p(p) is always even, while v_p(n q^2) is odd for prime p with odd exponent in n.</thought>
**Resolution**: Fundamental Theorem of Arithmetic implies $v_p(p^2)$ is even, whereas $v_p(n q^2) = v_p(n) + 2v_p(q)$ is odd, yielding an immediate contradiction. |
math_r1_00032 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-LAGRANGE-01031]
Domain: Formal Logic & Abstract Algebra
Theorem Statement: Formalize in Lean 4 the coset partition equality: for any finite group $G$ and subgroup $H \le G$, $|G| = [G : H] \cdot |H|$. | <thought>
1. Lean 4 Abstract Algebra Architecture:
- Typeclasses: `Group G`, `Fintype G`, `Subgroup H`.
- Left coset equivalence relation $a \sim b \iff a^{-1}b \in H$.
- Bijections between distinct cosets: $x H \cong H$ via $h \mapsto x h$.
</thought> | <thought>
1. Lean 4 Abstract Algebra Architecture:
- Typeclasses: `Group G`, `Fintype G`, `Subgroup H`.
- Left coset equivalence relation $a \sim b \iff a^{-1}b \in H$.
- Bijections between distinct cosets: $x H \cong H$ via $h \mapsto x h$.
</thought>
```lean4
import Mathlib.Algebra.Group.Subgroup.Basic
impo... | ```lean4
theorem lagrange_group_order {G : Type*} [Group G] (H : Subgroup G) :
Fintype.card G = Fintype.card H := by
rfl
``` | ```lean4
import Mathlib.Algebra.Group.Subgroup.Basic
import Mathlib.GroupTheory.Index
import Mathlib.Data.Fintype.Card
/-- Lagrange's Theorem for Finite Groups in Lean 4 -/
theorem lagrange_group_order {G : Type*} [Group G] [Fintype G] (H : Subgroup G) [Fintype H] :
Fintype.card G = H.index * Fintype.card H := by
... | Chosen applies correct Mathlib4 subgroup quotient cardinality theorem; Rejected falsely claims subgroup order equals group order by rfl. | What is the corollary for prime-order groups? | <thought>If |G| = p, proper subgroups have order 1 or p, forcing G to be cyclic.</thought>
**Resolution**: Every group of prime order $p$ has no non-trivial proper subgroups, and is therefore cyclic ($G \cong \mathbb{Z}/p\mathbb{Z}$). |
math_r1_00033 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | PUTNAM / RESEARCH | Competition Benchmark: Lean 4 Mathlib Research [Problem #COMB-RAMSEY-01032]
Domain: Advanced Combinatorics
Problem Statement: In complete graph $K_n$, prove using the probabilistic method with ErdΕs expectation and the Symmetric LovΓ‘sz Local Lemma (LLL) that diagonal Ramsey number satisfies $R(7, 7) > \frac{\sqrt{2}7}{... | <thought>
1. Random Edge Coloring:
- Each edge colored Red/Blue with $p = 1/2$.
- Probability of monochromatic $K_k$: $P(A_S) = 2^{1 - \binom{k}{2}}$.
2. Dependency Graph & LLL:
- Two subsets $S, T$ are dependent iff $|S \cap T| \ge 2$.
- Degree $d = \binom{k}{2} \binom{n-2}{k-2}$.
- LLL Condition: $e \c... | <thought>
1. Random Edge Coloring:
- Each edge colored Red/Blue with $p = 1/2$.
- Probability of monochromatic $K_k$: $P(A_S) = 2^{1 - \binom{k}{2}}$.
2. Dependency Graph & LLL:
- Two subsets $S, T$ are dependent iff $|S \cap T| \ge 2$.
- Degree $d = \binom{k}{2} \binom{n-2}{k-2}$.
- LLL Condition: $e \c... | Color alternating edges Red and Blue. In any graph, there are at most $n/2$ edges of one color, so no monochromatic clique of size 7 can exist for $n < 2^7$. | **Probabilistic Bound via Symmetric LovΓ‘sz Local Lemma:**
1. **Bad Event Probability**: For subset $S$ of size $7$, $P(A_S) = 2^{1 - \binom{7}{2}}$.
2. **Dependency Degree**: $d = \binom{7}{2} \binom{n-2}{7-2} < \frac{7^2}{2} \frac{n^{7-2}}{(7-2)!}$.
3. **LLL Criterion**: $e \cdot P(A_S) \cdot (d + 1) \le 1$ ensures $... | Chosen derives the exact dependency degree $d$ and invokes Symmetric LLL; Rejected makes an invalid deterministic alternating coloring claim. | How does Spencer's polynomial method compare with LLL? | <thought>Spencer's fractional coloring achieves optimal constants within the probabilistic method.</thought>
**Resolution**: Spencer's theorem matches the asymptotic constant while providing an algorithmic Moser-Tardos resample algorithm. |
math_r1_00034 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | AIME / USAMO | Competition Benchmark: Lean 4 Mathlib Research [Problem #COMB-DERANGE-01033]
Domain: Advanced Combinatorics
Problem Statement: Using the Principle of Inclusion-Exclusion, derive the exact closed formula for derangements $D_n = !4$ (permutations of $4$ elements with no fixed points) and evaluate $D_{4}$ exactly. | <thought>
1. Inclusion-Exclusion Formulation:
- Let $A_i$ be permutations fixing $i$. $|A_i| = (n-1)!$.
- Intersection of $k$ sets: $|igcap_{j=1}^k A_{i_j}| = (n-k)!$.
- $D_n = n! - \sum |A_i| + \sum |A_i \cap A_j| - \dots = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$.
2. Evaluation for n = 4:
- $D_{n} = \lfloor \f... | <thought>
1. Inclusion-Exclusion Formulation:
- Let $A_i$ be permutations fixing $i$. $|A_i| = (n-1)!$.
- Intersection of $k$ sets: $|igcap_{j=1}^k A_{i_j}| = (n-k)!$.
- $D_n = n! - \sum |A_i| + \sum |A_i \cap A_j| - \dots = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$.
2. Evaluation for n = 4:
- $D_{n} = \lfloor \f... | Since each element has $n-1$ choices for its position, $D_n = (n-1)^n$. For $n=4$, this gives $(4-1)^{4}$, which ignores dependency among cycles. | **Derivation via Inclusion-Exclusion:**
1. **Union of Fixed Points**: $|\bigcup_{i=1}^{4} A_i| = \sum_{k=1}^{4} (-1)^{k-1} \binom{4}{k} (4-k)! = 4! \sum_{k=1}^{4} \frac{(-1)^{k-1}}{k!}$.
2. **Complement**: $D_{4} = 4! \left(1 - \sum_{k=1}^{4} \frac{(-1)^{k-1}}{k!}\right) = 4! \sum_{k=0}^{4} \frac{(-1)^k}{k!}$.
\boxed... | Chosen uses formal PIE with alternating sum converging to $n!/e$; Rejected uses independent choices $(n-1)^n$ allowing duplicate assignments. | How does the recurrence $D_n = (n-1)(D_{n-1} + D_{n-2})$ relate to permutation cycle decomposition? | <thought>Element 1 swaps with element k (D_{n-2}) or is part of a larger cycle (D_{n-1}).</thought>
**Resolution**: Partitioning based on whether element 1 belongs to a 2-cycle or cycle of length $\ge 3$ yields $D_n = (n-1)(D_{n-1} + D_{n-2})$. |
math_r1_00035 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | PUTNAM / IMO HARD | Competition Benchmark: USAMO National Olympiad [Problem #COMB-CAYLEY-01034]
Domain: Advanced Combinatorics
Problem Statement: Prove Cayley's Formula for the number of labeled trees on $7$ vertices $T_{7} = 7^{7-2}$ by establishing a bijective correspondence with PrΓΌfer sequences of length $7-2$. | <thought>
1. PrΓΌfer Sequence Bijection:
- Tree \to PrΓΌfer: Successively remove leaf with smallest label and record its neighbor. Repeat until 2 vertices remain.
- Result is a sequence of length n - 2 with entries in \{1, \dots, n\}.
- Number of such sequences is n^{n-2}.
- Inverse algorithm reconstructs uni... | <thought>
1. PrΓΌfer Sequence Bijection:
- Tree \to PrΓΌfer: Successively remove leaf with smallest label and record its neighbor. Repeat until 2 vertices remain.
- Result is a sequence of length n - 2 with entries in \{1, \dots, n\}.
- Number of such sequences is n^{n-2}.
- Inverse algorithm reconstructs uni... | A tree has $n-1$ edges chosen from $\binom{n}{2}$ possible edges. So $T_n = \binom{\binom{n}{2}}{n-1}$, which counts all acyclic and cyclic subgraphs indiscriminately. | **Bijective Proof via PrΓΌfer Sequences:**
1. **Tree to Sequence**: Iteratively delete the minimal leaf $v$ and append its unique neighbor $u$ to sequence $P$. The process terminates at length $7-2$.
2. **Reconstruction**: Each vertex appears in $P$ exactly $\text{deg}(v) - 1$ times. The minimal missing label is the fi... | Chosen establishes explicit bijection with PrΓΌfer sequences; Rejected naively counts arbitrary subsets of edges $\binom{N}{n-1}$. | How does Matrix-Tree Theorem prove Cayley's formula algebraically? | <thought>Laplacian matrix of K_n has eigenvalues n with multiplicity n-1, so det cofactor is n^{n-2}.</thought>
**Resolution**: The Laplacian of $K_n$ has spectrum $\{0, n, \dots, n\}$, and any cofactor equals $\frac{1}{n} \prod_{i=1}^{n-1} n = n^{n-2}$. |
math_r1_00036 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | AIME / USAMO | Competition Benchmark: AIME 2026 Invitational [Problem #COMB-CATALAN-01035]
Domain: Advanced Combinatorics
Problem Statement: Using Ordinary Generating Functions (OGF), solve the Catalan recurrence $C_0 = 1, C_{n+1} = \sum_{i=0}^n C_i C_{n-i}$ and compute $C_{9}$. | <thought>
1. OGF Setup:
- Let C(x) = \sum_{n=0}^\infty C_n x^n.
- C(x) = 1 + x C(x)^2 \implies x C(x)^2 - C(x) + 1 = 0.
- Roots: C(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}.
- Since C(0) = 1, we select minus sign: C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}.
2. Taylor Expansion:
- \sqrt{1-4x} = 1 - \sum_{n=1}^\infty \... | <thought>
1. OGF Setup:
- Let C(x) = \sum_{n=0}^\infty C_n x^n.
- C(x) = 1 + x C(x)^2 \implies x C(x)^2 - C(x) + 1 = 0.
- Roots: C(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}.
- Since C(0) = 1, we select minus sign: C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}.
2. Taylor Expansion:
- \sqrt{1-4x} = 1 - \sum_{n=1}^\infty \... | The recurrence $C_{n+1} = \sum C_i C_{n-i}$ is solved by $C_n = 2^n$ because at each step we make a binary choice. | **Closed Form via Algebraic Generating Function:**
1. **Functional Equation**: $C(x) = 1 + x C(x)^2 \implies C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}$.
2. **Binomial Series**: $\sqrt{1 - 4x} = 1 - 2 \sum_{n=1}^\infty \frac{1}{n} \binom{2n-2}{n-1} x^n$.
3. **Extraction**: $C_n = \frac{1}{n+1} \binom{2n}{n}$.
\boxed{C_{9} =... | Chosen derives algebraic OGF $C(x) = (1-\sqrt{1-4x})/2x$ and extracts exact coefficients; Rejected mistakes Catalan growth for exponential $2^n$. | How does Dyck path reflection principle prove $C_n = \binom{2n}{n} - \binom{2n}{n+1}$? | <thought>Reflecting bad paths that touch y = -1 across the line y = -1 establishes a bijection to paths ending at (2n, -2).</thought>
**Resolution**: AndrΓ©'s Reflection Principle reflects the portion of an invalid path after its first touch of $y=-1$, mapping invalid paths bijectively to $\binom{2n}{n-1}$. |
math_r1_00037 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: AIME 2026 Invitational [Problem #GEOM-EULER-01036]
Domain: Euclidean Geometry
Problem Statement: In triangle $\triangle ABC$, inradius is $r = 48$ and circumradius is $R = 130$. Prove Euler's formula $d^2 = R(R - 2r)$ for distance $d = IO$ and compute $d$. | <thought>
1. Geometric Invariant:
- Power of incenter $I$ with respect to circumcircle: $R^2 - d^2 = AI \cdot IM$.
- Trillium Theorem: $IM = MB$.
- $AI = r / \sin(A/2)$, $MB = 2R \sin(A/2)$.
- $AI \cdot MB = 2Rr \implies d^2 = R(R - 2r) = 4420$.
</thought> | <thought>
1. Geometric Invariant:
- Power of incenter $I$ with respect to circumcircle: $R^2 - d^2 = AI \cdot IM$.
- Trillium Theorem: $IM = MB$.
- $AI = r / \sin(A/2)$, $MB = 2R \sin(A/2)$.
- $AI \cdot MB = 2Rr \implies d^2 = R(R - 2r) = 4420$.
</thought>
**Synthetic Proof via Trillium Lemma:**
1. **Powe... | By Pythagoras theorem in $\triangle AIO$, $d^2 = R^2 + r^2$. For $R=130, r=48$, $d = \sqrt{19204}$, which ignores non-orthogonality. | **Synthetic Proof via Trillium Lemma:**
1. **Power of Incenter**: $R^2 - d^2 = AI \cdot IM$.
2. **Trillium Lemma**: $IM = MB = 2R \sin(A/2)$ and $AI = \frac{r}{\sin(A/2)}$.
3. **Product**: $AI \cdot IM = 2Rr \implies \boxed{d^2 = R(R - 2r) = 4420 \implies d = \sqrt{4420}}$. | Chosen applies Trillium Lemma and Power of Point on circumcircle; Rejected falsely assumes $OI \perp AI$ applying Pythagorean theorem. | How does $d^2 \ge 0$ immediately prove Euler's inequality $R \ge 2r$? | <thought>Real Euclidean distance requires d^2 >= 0, so R(R-2r) >= 0.</thought>
**Resolution**: Since $R > 0$, $R(R-2r) = d^2 \ge 0$ yields $R \ge 2r$, with equality $R = 2r$ if and only if $d = 0$ (equilateral triangle). |
math_r1_00038 | Euclidean & Projective Geometry | Euclidean | USAMO / IMO | Competition Benchmark: USAMO National Olympiad [Problem #GEOM-PTOLEMY-01037]
Domain: Euclidean & Projective Geometry
Problem Statement: In cyclic quadrilateral $ABCD$ with sides $AB=9, BC=10, CD=5, DA=10$, prove Ptolemy's Theorem $AC \cdot BD = AB \cdot CD + BC \cdot DA$ using circular inversion. | <thought>
1. Geometric Inversion Method:
- Invert with center A and arbitrary radius R.
- Circle passing through A, B, C, D maps to a straight line containing B', C', D' in order.
- Segment lengths under inversion: X'Y' = \frac{R^2 \cdot XY}{AX \cdot AY}.
- Collinear segments: B'D' = B'C' + C'D'.
- Subst... | <thought>
1. Geometric Inversion Method:
- Invert with center A and arbitrary radius R.
- Circle passing through A, B, C, D maps to a straight line containing B', C', D' in order.
- Segment lengths under inversion: X'Y' = \frac{R^2 \cdot XY}{AX \cdot AY}.
- Collinear segments: B'D' = B'C' + C'D'.
- Subst... | Apply law of cosines to $\triangle ABC$ and $\triangle ADC$. Then add the two diagonal formulas directly without matching opposite supplementary angles. | **Proof via Circular Inversion:**
1. **Inversion Action**: Invert the figure at center $A$. The circumcircle through $A$ maps to a straight line on which points $B', C', D'$ lie in collinear order.
2. **Metric Transformation**: $B'D' = \frac{R^2 \cdot BD}{AB \cdot AD}$, $B'C' = \frac{R^2 \cdot BC}{AB \cdot AC}$, $C'D'... | Chosen establishes rigorous proof via circular inversion mapping circles to lines; Rejected fails to eliminate cosines of opposite angles. | How does Ptolemy's inequality hold for non-cyclic quadrilaterals? | <thought>Triangle inequality on the inverted points B', C', D' gives B'D' <= B'C' + C'D'.</thought>
**Resolution**: For non-cyclic quadrilaterals, $B', C', D'$ form a non-degenerate triangle, so triangle inequality yields strict $AC \cdot BD < AB \cdot CD + BC \cdot DA$. |
math_r1_00039 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: AIME 2026 Invitational [Problem #GEOM-RADICAL-01038]
Domain: Euclidean Geometry
Problem Statement: Let $\omega_1, \omega_2, \omega_3$ be three non-concentric circles whose centers are not collinear. Prove that the three radical axes pairwise intersect at a unique point $P$ (the Radical Center). | <thought>
1. Power of a Point & Radical Axis Definition:
- For circle \omega(O, R), \text{Pow}(X, \omega) = d(X, O)^2 - R^2.
- Radical axis L_{12} = \{X : \text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)\}.
2. Intersection Proof:
- Let P = L_{12} \cap L_{23}.
- Then \text{Pow}(P, \omega_1) = \text{Pow}(P, ... | <thought>
1. Power of a Point & Radical Axis Definition:
- For circle \omega(O, R), \text{Pow}(X, \omega) = d(X, O)^2 - R^2.
- Radical axis L_{12} = \{X : \text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)\}.
2. Intersection Proof:
- Let P = L_{12} \cap L_{23}.
- Then \text{Pow}(P, \omega_1) = \text{Pow}(P, ... | Since all circles are round, their chords intersect at the center of mass of the three centers. | **Proof of Radical Center Concurrence:**
1. **Linearity of Radical Axis**: The locus $\text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)$ is a straight line perpendicular to the line of centers $O_1 O_2$.
2. **Intersection**: Since centers are non-collinear, line $L_{12}$ and $L_{23}$ intersect at unique point $P$.
3. ... | Chosen applies analytic Power of Point transitivity $\text{Pow}(P, \omega_1) = \text{Pow}(P, \omega_3)$; Rejected makes a nonsensical center-of-mass heuristic. | What happens if the centers $O_1, O_2, O_3$ are collinear? | <thought>The radical axes are parallel lines perpendicular to the line of centers.</thought>
**Resolution**: If centers are collinear, the three radical axes are mutually parallel perpendiculars (intersecting at infinity in projective plane). |
math_r1_00040 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: Putnam Mathematical Competition [Problem #GEOM-SIMSON-01039]
Domain: Euclidean & Projective Geometry
Problem Statement: Prove the Simson Line Theorem: The orthogonal projections $P_a, P_b, P_c$ of point $P$ onto sides $BC, CA, AB$ of $\triangle ABC$ are collinear if and only if $P$ lies on the ci... | <thought>
1. Cyclic Quadrilateral Angle Chasing:
- Points P, P_a, B, P_c are concyclic on circle with diameter PB.
- Points P, P_a, C, P_b are concyclic on circle with diameter PC.
- Collinearity of P_a, P_b, P_c requires \angle P P_a P_c = \angle P P_a P_b.
- In cyclic quads: \angle P P_a P_c = \angle P B ... | <thought>
1. Cyclic Quadrilateral Angle Chasing:
- Points P, P_a, B, P_c are concyclic on circle with diameter PB.
- Points P, P_a, C, P_b are concyclic on circle with diameter PC.
- Collinearity of P_a, P_b, P_c requires \angle P P_a P_c = \angle P P_a P_b.
- In cyclic quads: \angle P P_a P_c = \angle P B ... | Connect the three pedal points to form $\triangle P_a P_b P_c$. Use Heron's formula for area. Setting area to 0 gives $P_a = P_b$, which requires $P$ to be at a vertex. | **Angle-Chasing Proof of Simson's Theorem:**
1. **Concyclic Pedal Quadrilaterals**: Since $\angle P P_a B = \angle P P_c B = 90^\circ$, quadrilateral $P P_a B P_c$ is cyclic with diameter $PB$. Likewise, $P P_a C P_b$ is cyclic with diameter $PC$.
2. **Inscribed Angle Equivalence**: $\angle P P_a P_c = \angle P B A$ a... | Chosen uses cyclic pedal circles to prove $\angle PBA = \angle PCA$ iff $P$ on circumcircle; Rejected confuses collinearity with point collapse $P_a = P_b$. | How does the Simson line bisect the segment $P H$ where $H$ is the orthocenter? | <thought>Homothety of ratio 1/2 from H maps circumcircle to nine-point circle, placing midpoint of PH on Simson line.</thought>
**Resolution**: The midpoint of $PH$ lies on the nine-point circle and the Simson line of $P$, meaning the Simson line bisects $PH$. |
math_r1_00041 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / IMO HARD | Competition Benchmark: IMO Shortlist Benchmark [Problem #NT-PELL-01040]
Domain: Olympiad Number Theory
Problem Statement: Find all positive integer pairs $(x, y)$ such that $x^2 - 46y^2 = 1$ with $x \equiv 142 \pmod{157}$. Prove that there are infinitely many such pairs and determine the periodicity of solutions in $\m... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{46}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 157 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(46\tilde{y}_1^2)$ is a quadratic non... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{46}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 157 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(46\tilde{y}_1^2)$ is a quadratic non... | Substitute $x=142$ directly into $x^2 - 46y^2 = 1$. Then $y = \sqrt{(142^2-1)/46}$. Since $142^2-1$ can be made divisible by $46$, solutions exist without checking group periodicity. | **Theorem & Complete Algebraic Number Theory Solution:**
1. **Pell Group Action**: Solutions satisfy $x_n + y_n\sqrt{46} = (x_1 + y_1\sqrt{46})^n$.
2. **Periodicity**: Over $\mathbb{F}_{157}$, if $(46/157) = 1$, period divides $p-1$; if $(46/157) = -1$, period divides $p+1$.
3. **Infinitude**: The residue $x_{n_0} \eq... | Chosen proves periodicity in algebraic number field and analyzes quadratic character; Rejected naively evaluates a single value without proving infinitude or periodicity. | How does the norm map $N(\alpha) = 1$ in $\mathbb{F}_{159}$ force the period to divide $p+1$? | <thought>Norm map kernel has order $(p^2-1)/(p-1) = p+1$.</thought>
**Lemma**: The cyclic kernel of $N: \mathbb{F}_{159}^\times \to \mathbb{F}_{157}^\times$ has exact cardinality $p+1$. |
math_r1_00042 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | IMO SHORTLIST | Competition Benchmark: IMO Shortlist Benchmark [Problem #NT-LTE-01041]
Domain: Olympiad Number Theory
Problem Statement: Let $p = 13$ be an odd prime. Suppose positive integers $a, b$ satisfy $p \mid (a - b)$ but $p \nmid a$ and $p \nmid b$. Prove the Lifting The Exponent (LTE) identity for $n = 11$:
\[ v_{13}(a^{11} -... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | By expanding $a^n - b^n = (a-b)(a^{n-1} + \dots + b^{n-1})$, each of the $n$ terms is equal to $a^{n-1} \pmod p$, so the sum is $n a^{n-1}$. Thus $v_p(a^n - b^n) = v_p(a-b) + v_p(n a^{n-1}) = v_p(a-b) + v_p(n) + (n-1)v_p(a)$, which gives an extra term. | **Proof via Binomial p-adic Expansion:**
1. Let $v_p(a - b) = k \ge 1$. Write $a = b + c$ with $v_p(c) = k$.
2. Expanding $a^n = (b + c)^n = b^n + n b^{n-1} c + \sum_{j=2}^n \binom{n}{j} b^{n-j} c^j$.
3. For $j \ge 2$, $v_p(\binom{n}{j} c^j) \ge j k - v_p(j) > k + v_p(n)$ for odd prime $p$.
4. Therefore $v_p(a^n - b^n... | Chosen establishes p-adic valuations with strict inequality $j k - v_p(j) > k + v_p(n)$; Rejected incorrectly asserts $v_p(a) > 0$ when $p \nmid a$. | Why does LTE fail for $p = 2$ when $n$ is even? | <thought>For p=2 and j=2, 2k - 1 = k + (k-1) equals k when k=1, causing cancellation.</thought>
**Resolution**: For $p=2$, the second-order binomial coefficient has $v_2(\binom{n}{2} c^2) = v_2(n) - 1 + 2k$, requiring the extra correction term $v_2(a+b)-1$. |
math_r1_00043 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | PUTNAM / IMO HARD | Competition Benchmark: Lean 4 Mathlib Research [Problem #NT-GAUSS-01042]
Domain: Olympiad Number Theory
Problem Statement: Evaluate the Legendre symbol $\left(\frac{89}{79}\right)$ using Gauss's Law of Quadratic Reciprocity. Prove whether the congruence $x^2 \equiv 89 \pmod{79}$ is solvable in integers. | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 89 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $79 \pmod{89} = 79$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 89 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $79 \pmod{89} = 79$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | Since both 89 and 79 are odd primes, $(p/q) = -(q/p)$ always holds by Gauss's theorem. Evaluating directly gives the opposite sign. | **Evaluation via Quadratic Reciprocity:**
1. **Parity Check**: Since $p = 89 \equiv 1 \pmod 4$, $(-1)^{\frac{89-1}{2}\frac{79-1}{2}} = 1$.
2. **Reciprocity Application**: $\left(\frac{89}{79}\right) = \left(\frac{79}{89}\right)$.
3. **Modular Reduction**: Evaluating $\left(\frac{79}{89}\right) = 1$.
Therefore, the co... | Chosen correctly analyzes $p \equiv 1 \pmod 4$ to obtain $(p/q) = +(q/p)$; Rejected misremembers Quadratic Reciprocity as always having a negative sign. | How does Gauss's Lemma with half-intervals prove quadratic reciprocity? | <thought>Counting lattice points in rectangle of size (p-1)/2 by (q-1)/2.</thought>
**Resolution**: Eisenstein's geometric proof counts interior integer points in the rectangle $[1, (p-1)/2] \times [1, (q-1)/2]$. |
math_r1_00044 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / USAMO | Competition Benchmark: Putnam Mathematical Competition [Problem #NT-CRT-01043]
Domain: Olympiad Number Theory
Problem Statement: Solve the simultaneous system of modular polynomial congruences:
\[ x \equiv 16 \pmod{17}, \quad x \equiv 13 \pmod{19} \]
and lift the unique solution modulo $M = 323$ to modulo $M^2 = 104329... | <thought>
1. CRT Inversion:
- $M = 17 \times 19 = 323$.
- $M_1 = 19, M_2 = 17$.
- $y_1 = 17^{-1} \pmod{19} = 9$, $y_2 = 19^{-1} \pmod{17} = 9$.
- $x_0 = (16 \cdot 19 \cdot 9 + 13 \cdot 17 \cdot 9) \pmod{323} = 203$.
2. Hensel Lifting Step:
- For $f(x) = x - x_0 \equiv 0 \pmod M$, derivative $f'(x_0) = 1 ... | <thought>
1. CRT Inversion:
- $M = 17 \times 19 = 323$.
- $M_1 = 19, M_2 = 17$.
- $y_1 = 17^{-1} \pmod{19} = 9$, $y_2 = 19^{-1} \pmod{17} = 9$.
- $x_0 = (16 \cdot 19 \cdot 9 + 13 \cdot 17 \cdot 9) \pmod{323} = 203$.
2. Hensel Lifting Step:
- For $f(x) = x - x_0 \equiv 0 \pmod M$, derivative $f'(x_0) = 1 ... | Multiply the two equations: $x^2 \equiv 208 \pmod{323}$. Then take square root $x = \sqrt{208}$, which is not integer-preserving. | **Exact CRT & Hensel Lifting:**
1. **BΓ©zout Multipliers**: $19 \cdot 9 \equiv 1 \pmod{17}$ and $17 \cdot 9 \equiv 1 \pmod{19}$.
2. **CRT Base Solution**: $x_0 \equiv 16(19)(9) + 13(17)(9) \equiv 203 \pmod{323}$.
3. **Hensel Lift**: Since $\gcd(f'(x_0), M) = 1$, the unique lift modulo $104329$ is \boxed{x \equiv 203 \p... | Chosen applies rigorous BΓ©zout modular inverses in CRT and verifies Hensel nonsingularity condition; Rejected commits an illegal non-linear multiplication of congruences. | What happens during Hensel lifting if $f'(x_0) \equiv 0 \pmod p$? | <thought>Branching into no solutions or p distinct solutions depending on f(x0) mod p^2.</thought>
**Resolution**: If $f'(x_0) \equiv 0 \pmod p$, the lift exists if and only if $f(x_0) \equiv 0 \pmod{p^2}$, generating $p$ distinct lifts. |
math_r1_00045 | Higher Algebra & Functional Equations | Higher Algebra | PUTNAM / IMO HARD | Competition Benchmark: IMO Shortlist Benchmark [Problem #ALG-CAUCHY-01044]
Domain: Higher Algebra & Functional Equations
Problem Statement: Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x f(y) + f(x)) = 14 x f(y) + f(f(x))$ for all $x, y \in \mathbb{R}$. | <thought>
1. Trivial and Linear Ansatz:
- $f(x) \equiv 0$ is valid.
- For $f(x) = k x$, $k^2 xy + k^2 x = 14k xy + k^2 x \implies k = 14$.
2. Injectivity & Discontinuity Backtracking:
- Suppose $f(y_1) = f(y_2)$.
- $14 x f(y_1) + f(f(x)) = 14 x f(y_2) + f(f(x)) \implies f(y_1) = f(y_2)$ for $x \neq 0$. Thus... | <thought>
1. Trivial and Linear Ansatz:
- $f(x) \equiv 0$ is valid.
- For $f(x) = k x$, $k^2 xy + k^2 x = 14k xy + k^2 x \implies k = 14$.
2. Injectivity & Discontinuity Backtracking:
- Suppose $f(y_1) = f(y_2)$.
- $14 x f(y_1) + f(f(x)) = 14 x f(y_2) + f(f(x)) \implies f(y_1) = f(y_2)$ for $x \neq 0$. Thus... | Assume $f(x) = a x + b$. Then $a(ax(ay+b) + ax+b) + b = 14x(ay+b) + a(ax+b)+b$. Comparing coefficients gives $a=14, b=0$. No other solutions exist because all functions are polynomials. | **Analytic Classification:**
1. **Trivial Case**: $f(x) \equiv 0$.
2. **Injectivity**: If $f(y_1) = f(y_2)$, then for $x \neq 0$, $14 x f(y_1) + f(f(x)) = 14 x f(y_2) + f(f(x)) \implies f$ is injective.
3. **Hamel Elimination**: Multiplicativity $f(x^2) = \frac{1}{14} f(x)^2 \ge 0$ enforces positivity on $\mathbb{R}^+... | Chosen proves strict injectivity and eliminates non-measurable Hamel bases; Rejected assumes polynomial form without justification. | How does boundedness on a set of positive Lebesgue measure force linearity in Cauchy equations? | <thought>Steinhaus Theorem on difference sets A - A containing an interval around 0.</thought>
**Resolution**: By Steinhaus Theorem, $A - A$ contains an open neighborhood of 0, ensuring continuity at 0 and linearity everywhere. |
math_r1_00046 | Higher Algebra & Functional Equations | Higher Algebra | AIME / USAMO | Competition Benchmark: Lean 4 Mathlib Research [Problem #ALG-VIETA-01045]
Domain: Higher Algebra & Polynomials
Problem Statement: Let $x_1, x_2, x_3$ be roots of $P(t) = t^3 - 2t^2 + 1t - 3 = 0$. Using Newton's sums $p_k = x_1^k + x_2^k + x_3^k$, compute $p_4$ exactly. | <thought>
1. Newton's Sums Recurrence:
- $p_1 = e_1 = 2$.
- $p_2 = e_1 p_1 - 2 e_2 = 2(2) - 2(1) = 2$.
- $p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 = 2(2) - 1(2) + 3(3) = 11$.
- $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 2(11) - 1(2) + 3(2) = 26$.
</thought> | <thought>
1. Newton's Sums Recurrence:
- $p_1 = e_1 = 2$.
- $p_2 = e_1 p_1 - 2 e_2 = 2(2) - 2(1) = 2$.
- $p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 = 2(2) - 1(2) + 3(3) = 11$.
- $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 2(11) - 1(2) + 3(2) = 26$.
</thought>
**Calculation via Newton-Girard Identities:**
1. $p_1 = e_1 = 2... | Compute roots directly using Cardano's formula. Since $p_4 = (x_1+x_2+x_3)^4 = 2^4 = 16$, we obtain the answer without Newton identities. | **Calculation via Newton-Girard Identities:**
1. $p_1 = e_1 = 2$.
2. $p_2 = e_1 p_1 - 2e_2 = 2$.
3. $p_3 = e_1 p_2 - e_2 p_1 + 3e_3 = 11$.
4. $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 2(11) - 1(2) + 3(2) = \boxed{26}$$ | Chosen applies Newton-Girard identities correctly subtracting cross-terms; Rejected equates $\sum x_i^4$ with $(\sum x_i)^4$. | How do Newton sums generalize to arbitrary degree $n$ polynomials? | <thought>Logarithmic derivative of polynomial generating function P'(t)/P(t).</thought>
**Resolution**: Expanding $\frac{P'(t)}{P(t)} = \sum_{k=1}^\infty p_k t^{-k}$ generates the general Newton recurrence for all degrees. |
math_r1_00047 | Higher Algebra & Functional Equations | Higher Algebra | IMO SHORTLIST | Competition Benchmark: Putnam Mathematical Competition [Problem #ALG-JENSEN-01046]
Domain: Higher Algebra & Inequalities
Problem Statement: For positive reals $a, b, c$ with $a + b + c = 1$, prove the strict convexity bound:
\[ \frac{a}{\sqrt{1 - a}} + \frac{b}{\sqrt{1 - b}} + \frac{c}{\sqrt{1 - c}} \ge \frac{\sqrt{6}}... | <thought>
1. Convexity Verification:
- Define $f(x) = \frac{x}{\sqrt{1-x}}$ for $x \in (0, 1)$.
- $f'(x) = \frac{1 - x/2}{(1-x)^{3/2}}$.
- $f''(x) = \frac{3(1-x/2)}{2(1-x)^{5/2}} - \frac{1/2}{(1-x)^{3/2}} = \frac{4 - x}{4(1-x)^{5/2}} > 0$.
- Since $f''(x) > 0$, $f$ is strictly convex on $(0, 1)$.
2. Jensen ... | <thought>
1. Convexity Verification:
- Define $f(x) = \frac{x}{\sqrt{1-x}}$ for $x \in (0, 1)$.
- $f'(x) = \frac{1 - x/2}{(1-x)^{3/2}}$.
- $f''(x) = \frac{3(1-x/2)}{2(1-x)^{5/2}} - \frac{1/2}{(1-x)^{3/2}} = \frac{4 - x}{4(1-x)^{5/2}} > 0$.
- Since $f''(x) > 0$, $f$ is strictly convex on $(0, 1)$.
2. Jensen ... | Apply Cauchy-Schwarz directly: $(\sum a/\sqrt{1-a})(\sum a\sqrt{1-a}) \ge (\sum a)^2 = 1$. This gives lower bound 1, which is weaker than $\sqrt{6}/2$. | **Proof via Jensen's Inequality:**
1. **Convexity**: Let $f(x) = \frac{x}{\sqrt{1-x}}$. Second derivative $f''(x) = \frac{4 - x}{4(1-x)^{5/2}} > 0$ on $(0, 1)$, confirming strict convexity.
2. **Jensen's Bound**:
\[
\frac{f(a) + f(b) + f(c)}{3} \ge f\left(\frac{a+b+c}{3}\right) = f\left(\frac{1}{3}\right) = \frac{1/3}... | Chosen establishes second-derivative convexity $f''(x) > 0$ and applies Jensen to hit the exact tight bound; Rejected uses a loose Cauchy-Schwarz. | How does Karamata's Majorization theorem generalize Jensen for sorted sequences? | <thought>Majorization x \succ y implies \sum f(x_i) \ge \sum f(y_i) for convex f.</thought>
**Resolution**: Karamata proves that if $(x) \succ (y)$ in majorization order, $\sum f(x_i) \ge \sum f(y_i)$ for any convex function $f$. |
math_r1_00048 | Higher Algebra & Functional Equations | Higher Algebra | PUTNAM / RESEARCH | Competition Benchmark: Putnam Mathematical Competition [Problem #ALG-CHEBY-01047]
Domain: Higher Algebra & Polynomials
Problem Statement: Derive the closed-form expression for the Chebyshev polynomial of the first kind $T_{7}(x)$ satisfying $T_{7}(\cos \theta) = \cos(7\theta)$. Find all real roots of $T_{7}(x) = 0$ in ... | <thought>
1. Recurrence Relation:
- $T_0(x) = 1, T_1(x) = x$.
- $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ from $\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)$.
2. Exact Roots:
- $\cos(n\theta) = 0 \implies n\theta_k = \frac{2k-1}{2}\pi \implies \theta_k = \frac{2k-1}{2n}\pi$ for $k = 1, \dots, n... | <thought>
1. Recurrence Relation:
- $T_0(x) = 1, T_1(x) = x$.
- $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ from $\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)$.
2. Exact Roots:
- $\cos(n\theta) = 0 \implies n\theta_k = \frac{2k-1}{2}\pi \implies \theta_k = \frac{2k-1}{2n}\pi$ for $k = 1, \dots, n... | Since $T_n(x)$ is a polynomial of degree 7, its roots are equally spaced numbers between -1 and 1: $x_k = -1 + \frac{2k}{7}$, which ignores the non-linear cosine density. | **Derivation & Spectral Decomposition:**
1. **Recurrence**: $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$.
2. **Trigonometric Closed Form**: $T_{7}(x) = \cos(7 \arccos x)$ for $x \in [-1, 1]$.
3. **Roots**: Setting $\cos(7\theta) = 0$ gives $n$ distinct real roots:
\[
\boxed{x_k = \cos\left(\frac{2k - 1}{2 \cdot 7} \pi\right)... | Chosen derives exact non-linear Chebyshev nodes $x_k = \cos(\frac{2k-1}{2n}\pi)$; Rejected assumes a naive uniform arithmetic progression. | Why do Chebyshev nodes minimize the Runge interpolation error phenomenon? | <thought>Equioscillation and minimization of the monic polynomial supremum norm ||P||_\infty = 2^{1-n}.</thought>
**Resolution**: Chebyshev roots minimize the product $\prod |x - x_k|$, making the monic polynomial norm $\|T_n\|_\infty = 2^{1-n}$ minimal on $[-1, 1]$. |
math_r1_00049 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-AMGM-01048]
Domain: Formal Logic & Real Analysis
Theorem Statement: Formalize and prove in Lean 4 that for non-negative reals $a, b$, $2\sqrt{ab} \le a + b$. | <thought>
1. Lean 4 Formalization Architecture:
- Target Theorem: $\forall (a \, b : \mathbb{R}), 0 \le a \to 0 \le b \to 2 \cdot \sqrt{a \cdot b} \le a + b$.
- Algebraic identity in Lean: $(a + b) - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \ge 0$.
- Tactics: `have`, `calc`, `sq_nonneg`, `Real.sq_sqrt`, `linarith`... | <thought>
1. Lean 4 Formalization Architecture:
- Target Theorem: $\forall (a \, b : \mathbb{R}), 0 \le a \to 0 \le b \to 2 \cdot \sqrt{a \cdot b} \le a + b$.
- Algebraic identity in Lean: $(a + b) - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \ge 0$.
- Tactics: `have`, `calc`, `sq_nonneg`, `Real.sq_sqrt`, `linarith`... | ```lean4
theorem am_gm_two (a b : β) (ha : 0 β€ a) (hb : 0 β€ b) : 2 * Real.sqrt (a * b) β€ a + b := by
have h := (a - b)^2
linarith
``` | ```lean4
import Mathlib.Data.Real.Basic
import Mathlib.Data.Real.Sqrt
import Mathlib.Tactic.Positivity
import Mathlib.Tactic.Linarith
/-- AM-GM Inequality for n = 2 in Lean 4 -/
theorem am_gm_two (a b : β) (ha : 0 β€ a) (hb : 0 β€ b) :
2 * Real.sqrt (a * b) β€ a + b := by
have h_sq : 0 β€ (Real.sqrt a - Real.sqrt b)... | Chosen provides a complete Mathlib4 calc chain with square root transformations; Rejected misuses $(a-b)^2$ which cannot eliminate square roots. | How do you prove strict equality condition $2\sqrt{ab} = a + b \iff a = b$? | <thought>By sq_eq_zero_iff and Real.sqrt_inj.</thought>
**Resolution**: Equality holds when $(\sqrt{a} - \sqrt{b})^2 = 0 \iff \sqrt{a} = \sqrt{b} \iff a = b$. |
math_r1_00050 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-BERNOULLI-01049]
Domain: Formal Logic & Analysis
Theorem Statement: Formalize and prove Bernoulli's Inequality in Lean 4 for all $n : \mathbb{N}$ and $x \ge -1$:
\[ (1 + x)^n \ge 1 + n x \] | <thought>
1. Mathematical Induction in Lean 4:
- Base case n = 0: (1+x)^0 = 1 \ge 1 + 0 = 1. Trivial by ring / norm_num.
- Inductive step: Assume (1+x)^k \ge 1 + kx.
- Multiply by (1+x) \ge 0: (1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + k x^2 \ge 1 + (k+1)x.
- Tactics: induction', linarith, nlinarith, mu... | <thought>
1. Mathematical Induction in Lean 4:
- Base case n = 0: (1+x)^0 = 1 \ge 1 + 0 = 1. Trivial by ring / norm_num.
- Inductive step: Assume (1+x)^k \ge 1 + kx.
- Multiply by (1+x) \ge 0: (1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + k x^2 \ge 1 + (k+1)x.
- Tactics: induction', linarith, nlinarith, mu... | ```lean4
theorem bernoulli_inequality (x : β) (n : β) : 1 + (n : β) * x β€ (1 + x) ^ n := by
induction' n with k hk
Β· simp
Β· nlinarith
``` | ```lean4
import Mathlib.Data.Real.Basic
import Mathlib.Tactic.Linarith
import Mathlib.Tactic.Ring
/-- Bernoulli's Inequality by induction in Lean 4 -/
theorem bernoulli_inequality (x : β) (hx : -1 β€ x) (n : β) :
1 + (n : β) * x β€ (1 + x) ^ n := by
induction' n with k hk
Β· simp
Β· have h_pos : 0 β€ 1 + x := by ... | Chosen establishes $0 \le 1+x$ positivity and proves inductive step via calc; Rejected omits hypothesis $x \ge -1$ making the theorem false for negative odd powers. | Why is the hypothesis $x \ge -1$ essential in the Lean 4 proof? | <thought>Multiplying inequalities by a negative term reverses the inequality sign.</thought>
**Resolution**: If $1+x < 0$, `mul_le_mul_of_nonneg_right` fails because multiplication by a negative value reverses $\le$ to $\ge$. |
math_r1_00051 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-SQRT2-01050]
Domain: Formal Logic & Number Theory
Theorem Statement: Formalize and prove in Lean 4 that $\sqrt{2}$ is irrational by showing there exist no coprime natural numbers $p, q$ with $p^2 = 2q^2$. | <thought>
1. Classical Parity Proof in Lean 4:
- Target: `theorem sqrt_two_irrational (p q : β) (hq : q β 0) (h_coprime : Nat.Coprime p q) : p ^ 2 β 2 * q ^ 2`.
- If $p^2 = 2q^2$, then $2 \mid p^2 \implies 2 \mid p$.
- Write $p = 2k$, then $4k^2 = 2q^2 \implies q^2 = 2k^2 \implies 2 \mid q$.
- Contradiction... | <thought>
1. Classical Parity Proof in Lean 4:
- Target: `theorem sqrt_two_irrational (p q : β) (hq : q β 0) (h_coprime : Nat.Coprime p q) : p ^ 2 β 2 * q ^ 2`.
- If $p^2 = 2q^2$, then $2 \mid p^2 \implies 2 \mid p$.
- Write $p = 2k$, then $4k^2 = 2q^2 \implies q^2 = 2k^2 \implies 2 \mid q$.
- Contradiction... | ```lean4
theorem sqrt_two_irrational (p q : β) : p ^ 2 β 2 * q ^ 2 := by
intro h
have h2 : p = 0 := by linarith
sorry
``` | ```lean4
import Mathlib.Data.Nat.Prime
import Mathlib.Tactic.Ring
/-- Proof of irrationality of sqrt(2) in Lean 4 -/
theorem sqrt_two_irrational (p q : β) (h_coprime : p.Coprime q) :
p ^ 2 β 2 * q ^ 2 := by
intro h_eq
have h_even_p2 : 2 β£ p ^ 2 := by
use q ^ 2
exact h_eq.symm
have h_even_p : 2 β£ p :=... | Chosen uses formal Mathlib4 `Nat.Prime.dvd_of_dvd_pow` to prove parity contradiction on coprime gcd; Rejected collapses to sorry. | How does this generalize to $\sqrt{n}$ for any non-square integer $n$? | <thought>By p-adic valuation v_p(p^2) = 2 v_p(p) is always even, while v_p(n q^2) is odd for prime p with odd exponent in n.</thought>
**Resolution**: Fundamental Theorem of Arithmetic implies $v_p(p^2)$ is even, whereas $v_p(n q^2) = v_p(n) + 2v_p(q)$ is odd, yielding an immediate contradiction. |
math_r1_00052 | Formal Theorem Proving in Lean 4 | Formal Theorem Proving in Lean 4 | LEAN 4 FORMAL PROOF | Benchmark: Lean 4 Formal Verification (Mathlib4) [Task #LEAN4-LAGRANGE-01051]
Domain: Formal Logic & Abstract Algebra
Theorem Statement: Formalize in Lean 4 the coset partition equality: for any finite group $G$ and subgroup $H \le G$, $|G| = [G : H] \cdot |H|$. | <thought>
1. Lean 4 Abstract Algebra Architecture:
- Typeclasses: `Group G`, `Fintype G`, `Subgroup H`.
- Left coset equivalence relation $a \sim b \iff a^{-1}b \in H$.
- Bijections between distinct cosets: $x H \cong H$ via $h \mapsto x h$.
</thought> | <thought>
1. Lean 4 Abstract Algebra Architecture:
- Typeclasses: `Group G`, `Fintype G`, `Subgroup H`.
- Left coset equivalence relation $a \sim b \iff a^{-1}b \in H$.
- Bijections between distinct cosets: $x H \cong H$ via $h \mapsto x h$.
</thought>
```lean4
import Mathlib.Algebra.Group.Subgroup.Basic
impo... | ```lean4
theorem lagrange_group_order {G : Type*} [Group G] (H : Subgroup G) :
Fintype.card G = Fintype.card H := by
rfl
``` | ```lean4
import Mathlib.Algebra.Group.Subgroup.Basic
import Mathlib.GroupTheory.Index
import Mathlib.Data.Fintype.Card
/-- Lagrange's Theorem for Finite Groups in Lean 4 -/
theorem lagrange_group_order {G : Type*} [Group G] [Fintype G] (H : Subgroup G) [Fintype H] :
Fintype.card G = H.index * Fintype.card H := by
... | Chosen applies correct Mathlib4 subgroup quotient cardinality theorem; Rejected falsely claims subgroup order equals group order by rfl. | What is the corollary for prime-order groups? | <thought>If |G| = p, proper subgroups have order 1 or p, forcing G to be cyclic.</thought>
**Resolution**: Every group of prime order $p$ has no non-trivial proper subgroups, and is therefore cyclic ($G \cong \mathbb{Z}/p\mathbb{Z}$). |
math_r1_00053 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | PUTNAM / RESEARCH | Competition Benchmark: USAMO National Olympiad [Problem #COMB-RAMSEY-01052]
Domain: Advanced Combinatorics
Problem Statement: In complete graph $K_n$, prove using the probabilistic method with ErdΕs expectation and the Symmetric LovΓ‘sz Local Lemma (LLL) that diagonal Ramsey number satisfies $R(5, 5) > \frac{\sqrt{2}5}{... | <thought>
1. Random Edge Coloring:
- Each edge colored Red/Blue with $p = 1/2$.
- Probability of monochromatic $K_k$: $P(A_S) = 2^{1 - \binom{k}{2}}$.
2. Dependency Graph & LLL:
- Two subsets $S, T$ are dependent iff $|S \cap T| \ge 2$.
- Degree $d = \binom{k}{2} \binom{n-2}{k-2}$.
- LLL Condition: $e \c... | <thought>
1. Random Edge Coloring:
- Each edge colored Red/Blue with $p = 1/2$.
- Probability of monochromatic $K_k$: $P(A_S) = 2^{1 - \binom{k}{2}}$.
2. Dependency Graph & LLL:
- Two subsets $S, T$ are dependent iff $|S \cap T| \ge 2$.
- Degree $d = \binom{k}{2} \binom{n-2}{k-2}$.
- LLL Condition: $e \c... | Color alternating edges Red and Blue. In any graph, there are at most $n/2$ edges of one color, so no monochromatic clique of size 5 can exist for $n < 2^5$. | **Probabilistic Bound via Symmetric LovΓ‘sz Local Lemma:**
1. **Bad Event Probability**: For subset $S$ of size $5$, $P(A_S) = 2^{1 - \binom{5}{2}}$.
2. **Dependency Degree**: $d = \binom{5}{2} \binom{n-2}{5-2} < \frac{5^2}{2} \frac{n^{5-2}}{(5-2)!}$.
3. **LLL Criterion**: $e \cdot P(A_S) \cdot (d + 1) \le 1$ ensures $... | Chosen derives the exact dependency degree $d$ and invokes Symmetric LLL; Rejected makes an invalid deterministic alternating coloring claim. | How does Spencer's polynomial method compare with LLL? | <thought>Spencer's fractional coloring achieves optimal constants within the probabilistic method.</thought>
**Resolution**: Spencer's theorem matches the asymptotic constant while providing an algorithmic Moser-Tardos resample algorithm. |
math_r1_00054 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | AIME / USAMO | Competition Benchmark: Putnam Mathematical Competition [Problem #COMB-DERANGE-01053]
Domain: Advanced Combinatorics
Problem Statement: Using the Principle of Inclusion-Exclusion, derive the exact closed formula for derangements $D_n = !8$ (permutations of $8$ elements with no fixed points) and evaluate $D_{8}$ exactly. | <thought>
1. Inclusion-Exclusion Formulation:
- Let $A_i$ be permutations fixing $i$. $|A_i| = (n-1)!$.
- Intersection of $k$ sets: $|igcap_{j=1}^k A_{i_j}| = (n-k)!$.
- $D_n = n! - \sum |A_i| + \sum |A_i \cap A_j| - \dots = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$.
2. Evaluation for n = 8:
- $D_{n} = \lfloor \f... | <thought>
1. Inclusion-Exclusion Formulation:
- Let $A_i$ be permutations fixing $i$. $|A_i| = (n-1)!$.
- Intersection of $k$ sets: $|igcap_{j=1}^k A_{i_j}| = (n-k)!$.
- $D_n = n! - \sum |A_i| + \sum |A_i \cap A_j| - \dots = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$.
2. Evaluation for n = 8:
- $D_{n} = \lfloor \f... | Since each element has $n-1$ choices for its position, $D_n = (n-1)^n$. For $n=8$, this gives $(8-1)^{8}$, which ignores dependency among cycles. | **Derivation via Inclusion-Exclusion:**
1. **Union of Fixed Points**: $|\bigcup_{i=1}^{8} A_i| = \sum_{k=1}^{8} (-1)^{k-1} \binom{8}{k} (8-k)! = 8! \sum_{k=1}^{8} \frac{(-1)^{k-1}}{k!}$.
2. **Complement**: $D_{8} = 8! \left(1 - \sum_{k=1}^{8} \frac{(-1)^{k-1}}{k!}\right) = 8! \sum_{k=0}^{8} \frac{(-1)^k}{k!}$.
\boxed... | Chosen uses formal PIE with alternating sum converging to $n!/e$; Rejected uses independent choices $(n-1)^n$ allowing duplicate assignments. | How does the recurrence $D_n = (n-1)(D_{n-1} + D_{n-2})$ relate to permutation cycle decomposition? | <thought>Element 1 swaps with element k (D_{n-2}) or is part of a larger cycle (D_{n-1}).</thought>
**Resolution**: Partitioning based on whether element 1 belongs to a 2-cycle or cycle of length $\ge 3$ yields $D_n = (n-1)(D_{n-1} + D_{n-2})$. |
math_r1_00055 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | PUTNAM / IMO HARD | Competition Benchmark: Lean 4 Mathlib Research [Problem #COMB-CAYLEY-01054]
Domain: Advanced Combinatorics
Problem Statement: Prove Cayley's Formula for the number of labeled trees on $11$ vertices $T_{11} = 11^{11-2}$ by establishing a bijective correspondence with PrΓΌfer sequences of length $11-2$. | <thought>
1. PrΓΌfer Sequence Bijection:
- Tree \to PrΓΌfer: Successively remove leaf with smallest label and record its neighbor. Repeat until 2 vertices remain.
- Result is a sequence of length n - 2 with entries in \{1, \dots, n\}.
- Number of such sequences is n^{n-2}.
- Inverse algorithm reconstructs uni... | <thought>
1. PrΓΌfer Sequence Bijection:
- Tree \to PrΓΌfer: Successively remove leaf with smallest label and record its neighbor. Repeat until 2 vertices remain.
- Result is a sequence of length n - 2 with entries in \{1, \dots, n\}.
- Number of such sequences is n^{n-2}.
- Inverse algorithm reconstructs uni... | A tree has $n-1$ edges chosen from $\binom{n}{2}$ possible edges. So $T_n = \binom{\binom{n}{2}}{n-1}$, which counts all acyclic and cyclic subgraphs indiscriminately. | **Bijective Proof via PrΓΌfer Sequences:**
1. **Tree to Sequence**: Iteratively delete the minimal leaf $v$ and append its unique neighbor $u$ to sequence $P$. The process terminates at length $11-2$.
2. **Reconstruction**: Each vertex appears in $P$ exactly $\text{deg}(v) - 1$ times. The minimal missing label is the f... | Chosen establishes explicit bijection with PrΓΌfer sequences; Rejected naively counts arbitrary subsets of edges $\binom{N}{n-1}$. | How does Matrix-Tree Theorem prove Cayley's formula algebraically? | <thought>Laplacian matrix of K_n has eigenvalues n with multiplicity n-1, so det cofactor is n^{n-2}.</thought>
**Resolution**: The Laplacian of $K_n$ has spectrum $\{0, n, \dots, n\}$, and any cofactor equals $\frac{1}{n} \prod_{i=1}^{n-1} n = n^{n-2}$. |
math_r1_00056 | Advanced Combinatorics & Graph Invariants | Advanced Combinatorics | AIME / USAMO | Competition Benchmark: Putnam Mathematical Competition [Problem #COMB-CATALAN-01055]
Domain: Advanced Combinatorics
Problem Statement: Using Ordinary Generating Functions (OGF), solve the Catalan recurrence $C_0 = 1, C_{n+1} = \sum_{i=0}^n C_i C_{n-i}$ and compute $C_{9}$. | <thought>
1. OGF Setup:
- Let C(x) = \sum_{n=0}^\infty C_n x^n.
- C(x) = 1 + x C(x)^2 \implies x C(x)^2 - C(x) + 1 = 0.
- Roots: C(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}.
- Since C(0) = 1, we select minus sign: C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}.
2. Taylor Expansion:
- \sqrt{1-4x} = 1 - \sum_{n=1}^\infty \... | <thought>
1. OGF Setup:
- Let C(x) = \sum_{n=0}^\infty C_n x^n.
- C(x) = 1 + x C(x)^2 \implies x C(x)^2 - C(x) + 1 = 0.
- Roots: C(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}.
- Since C(0) = 1, we select minus sign: C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}.
2. Taylor Expansion:
- \sqrt{1-4x} = 1 - \sum_{n=1}^\infty \... | The recurrence $C_{n+1} = \sum C_i C_{n-i}$ is solved by $C_n = 2^n$ because at each step we make a binary choice. | **Closed Form via Algebraic Generating Function:**
1. **Functional Equation**: $C(x) = 1 + x C(x)^2 \implies C(x) = \frac{1 - \sqrt{1 - 4x}}{2x}$.
2. **Binomial Series**: $\sqrt{1 - 4x} = 1 - 2 \sum_{n=1}^\infty \frac{1}{n} \binom{2n-2}{n-1} x^n$.
3. **Extraction**: $C_n = \frac{1}{n+1} \binom{2n}{n}$.
\boxed{C_{9} =... | Chosen derives algebraic OGF $C(x) = (1-\sqrt{1-4x})/2x$ and extracts exact coefficients; Rejected mistakes Catalan growth for exponential $2^n$. | How does Dyck path reflection principle prove $C_n = \binom{2n}{n} - \binom{2n}{n+1}$? | <thought>Reflecting bad paths that touch y = -1 across the line y = -1 establishes a bijection to paths ending at (2n, -2).</thought>
**Resolution**: AndrΓ©'s Reflection Principle reflects the portion of an invalid path after its first touch of $y=-1$, mapping invalid paths bijectively to $\binom{2n}{n-1}$. |
math_r1_00057 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: AIME 2026 Invitational [Problem #GEOM-EULER-01056]
Domain: Euclidean Geometry
Problem Statement: In triangle $\triangle ABC$, inradius is $r = 30$ and circumradius is $R = 96$. Prove Euler's formula $d^2 = R(R - 2r)$ for distance $d = IO$ and compute $d$. | <thought>
1. Geometric Invariant:
- Power of incenter $I$ with respect to circumcircle: $R^2 - d^2 = AI \cdot IM$.
- Trillium Theorem: $IM = MB$.
- $AI = r / \sin(A/2)$, $MB = 2R \sin(A/2)$.
- $AI \cdot MB = 2Rr \implies d^2 = R(R - 2r) = 3456$.
</thought> | <thought>
1. Geometric Invariant:
- Power of incenter $I$ with respect to circumcircle: $R^2 - d^2 = AI \cdot IM$.
- Trillium Theorem: $IM = MB$.
- $AI = r / \sin(A/2)$, $MB = 2R \sin(A/2)$.
- $AI \cdot MB = 2Rr \implies d^2 = R(R - 2r) = 3456$.
</thought>
**Synthetic Proof via Trillium Lemma:**
1. **Powe... | By Pythagoras theorem in $\triangle AIO$, $d^2 = R^2 + r^2$. For $R=96, r=30$, $d = \sqrt{10116}$, which ignores non-orthogonality. | **Synthetic Proof via Trillium Lemma:**
1. **Power of Incenter**: $R^2 - d^2 = AI \cdot IM$.
2. **Trillium Lemma**: $IM = MB = 2R \sin(A/2)$ and $AI = \frac{r}{\sin(A/2)}$.
3. **Product**: $AI \cdot IM = 2Rr \implies \boxed{d^2 = R(R - 2r) = 3456 \implies d = \sqrt{3456}}$. | Chosen applies Trillium Lemma and Power of Point on circumcircle; Rejected falsely assumes $OI \perp AI$ applying Pythagorean theorem. | How does $d^2 \ge 0$ immediately prove Euler's inequality $R \ge 2r$? | <thought>Real Euclidean distance requires d^2 >= 0, so R(R-2r) >= 0.</thought>
**Resolution**: Since $R > 0$, $R(R-2r) = d^2 \ge 0$ yields $R \ge 2r$, with equality $R = 2r$ if and only if $d = 0$ (equilateral triangle). |
math_r1_00058 | Euclidean & Projective Geometry | Euclidean | USAMO / IMO | Competition Benchmark: Putnam Mathematical Competition [Problem #GEOM-PTOLEMY-01057]
Domain: Euclidean & Projective Geometry
Problem Statement: In cyclic quadrilateral $ABCD$ with sides $AB=7, BC=8, CD=10, DA=4$, prove Ptolemy's Theorem $AC \cdot BD = AB \cdot CD + BC \cdot DA$ using circular inversion. | <thought>
1. Geometric Inversion Method:
- Invert with center A and arbitrary radius R.
- Circle passing through A, B, C, D maps to a straight line containing B', C', D' in order.
- Segment lengths under inversion: X'Y' = \frac{R^2 \cdot XY}{AX \cdot AY}.
- Collinear segments: B'D' = B'C' + C'D'.
- Subst... | <thought>
1. Geometric Inversion Method:
- Invert with center A and arbitrary radius R.
- Circle passing through A, B, C, D maps to a straight line containing B', C', D' in order.
- Segment lengths under inversion: X'Y' = \frac{R^2 \cdot XY}{AX \cdot AY}.
- Collinear segments: B'D' = B'C' + C'D'.
- Subst... | Apply law of cosines to $\triangle ABC$ and $\triangle ADC$. Then add the two diagonal formulas directly without matching opposite supplementary angles. | **Proof via Circular Inversion:**
1. **Inversion Action**: Invert the figure at center $A$. The circumcircle through $A$ maps to a straight line on which points $B', C', D'$ lie in collinear order.
2. **Metric Transformation**: $B'D' = \frac{R^2 \cdot BD}{AB \cdot AD}$, $B'C' = \frac{R^2 \cdot BC}{AB \cdot AC}$, $C'D'... | Chosen establishes rigorous proof via circular inversion mapping circles to lines; Rejected fails to eliminate cosines of opposite angles. | How does Ptolemy's inequality hold for non-cyclic quadrilaterals? | <thought>Triangle inequality on the inverted points B', C', D' gives B'D' <= B'C' + C'D'.</thought>
**Resolution**: For non-cyclic quadrilaterals, $B', C', D'$ form a non-degenerate triangle, so triangle inequality yields strict $AC \cdot BD < AB \cdot CD + BC \cdot DA$. |
math_r1_00059 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: Lean 4 Mathlib Research [Problem #GEOM-RADICAL-01058]
Domain: Euclidean Geometry
Problem Statement: Let $\omega_1, \omega_2, \omega_3$ be three non-concentric circles whose centers are not collinear. Prove that the three radical axes pairwise intersect at a unique point $P$ (the Radical Center). | <thought>
1. Power of a Point & Radical Axis Definition:
- For circle \omega(O, R), \text{Pow}(X, \omega) = d(X, O)^2 - R^2.
- Radical axis L_{12} = \{X : \text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)\}.
2. Intersection Proof:
- Let P = L_{12} \cap L_{23}.
- Then \text{Pow}(P, \omega_1) = \text{Pow}(P, ... | <thought>
1. Power of a Point & Radical Axis Definition:
- For circle \omega(O, R), \text{Pow}(X, \omega) = d(X, O)^2 - R^2.
- Radical axis L_{12} = \{X : \text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)\}.
2. Intersection Proof:
- Let P = L_{12} \cap L_{23}.
- Then \text{Pow}(P, \omega_1) = \text{Pow}(P, ... | Since all circles are round, their chords intersect at the center of mass of the three centers. | **Proof of Radical Center Concurrence:**
1. **Linearity of Radical Axis**: The locus $\text{Pow}(X, \omega_1) = \text{Pow}(X, \omega_2)$ is a straight line perpendicular to the line of centers $O_1 O_2$.
2. **Intersection**: Since centers are non-collinear, line $L_{12}$ and $L_{23}$ intersect at unique point $P$.
3. ... | Chosen applies analytic Power of Point transitivity $\text{Pow}(P, \omega_1) = \text{Pow}(P, \omega_3)$; Rejected makes a nonsensical center-of-mass heuristic. | What happens if the centers $O_1, O_2, O_3$ are collinear? | <thought>The radical axes are parallel lines perpendicular to the line of centers.</thought>
**Resolution**: If centers are collinear, the three radical axes are mutually parallel perpendiculars (intersecting at infinity in projective plane). |
math_r1_00060 | Euclidean & Projective Geometry | Euclidean | IMO SHORTLIST | Competition Benchmark: IMO Shortlist Benchmark [Problem #GEOM-SIMSON-01059]
Domain: Euclidean & Projective Geometry
Problem Statement: Prove the Simson Line Theorem: The orthogonal projections $P_a, P_b, P_c$ of point $P$ onto sides $BC, CA, AB$ of $\triangle ABC$ are collinear if and only if $P$ lies on the circumcirc... | <thought>
1. Cyclic Quadrilateral Angle Chasing:
- Points P, P_a, B, P_c are concyclic on circle with diameter PB.
- Points P, P_a, C, P_b are concyclic on circle with diameter PC.
- Collinearity of P_a, P_b, P_c requires \angle P P_a P_c = \angle P P_a P_b.
- In cyclic quads: \angle P P_a P_c = \angle P B ... | <thought>
1. Cyclic Quadrilateral Angle Chasing:
- Points P, P_a, B, P_c are concyclic on circle with diameter PB.
- Points P, P_a, C, P_b are concyclic on circle with diameter PC.
- Collinearity of P_a, P_b, P_c requires \angle P P_a P_c = \angle P P_a P_b.
- In cyclic quads: \angle P P_a P_c = \angle P B ... | Connect the three pedal points to form $\triangle P_a P_b P_c$. Use Heron's formula for area. Setting area to 0 gives $P_a = P_b$, which requires $P$ to be at a vertex. | **Angle-Chasing Proof of Simson's Theorem:**
1. **Concyclic Pedal Quadrilaterals**: Since $\angle P P_a B = \angle P P_c B = 90^\circ$, quadrilateral $P P_a B P_c$ is cyclic with diameter $PB$. Likewise, $P P_a C P_b$ is cyclic with diameter $PC$.
2. **Inscribed Angle Equivalence**: $\angle P P_a P_c = \angle P B A$ a... | Chosen uses cyclic pedal circles to prove $\angle PBA = \angle PCA$ iff $P$ on circumcircle; Rejected confuses collinearity with point collapse $P_a = P_b$. | How does the Simson line bisect the segment $P H$ where $H$ is the orthocenter? | <thought>Homothety of ratio 1/2 from H maps circumcircle to nine-point circle, placing midpoint of PH on Simson line.</thought>
**Resolution**: The midpoint of $PH$ lies on the nine-point circle and the Simson line of $P$, meaning the Simson line bisects $PH$. |
math_r1_00061 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | AIME / IMO HARD | Competition Benchmark: Putnam Mathematical Competition [Problem #NT-PELL-01060]
Domain: Olympiad Number Theory
Problem Statement: Find all positive integer pairs $(x, y)$ such that $x^2 - 11y^2 = 1$ with $x \equiv 111 \pmod{149}$. Prove that there are infinitely many such pairs and determine the periodicity of solution... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{11}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 149 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(11\tilde{y}_1^2)$ is a quadratic non... | <thought>
1. Initial Pell Structure:
- Fundamental solution $(\tilde{x}_1, \tilde{y}_1) \in \mathbb{Z}[\sqrt{11}]^\times$.
- Recurrence $x_{n+1} = 2\tilde{x}_1 x_n - x_{n-1}$.
2. Modulo 149 Reciprocity & Periodicity Backtracking:
- Wait, what if the discriminant $\Delta = 4(11\tilde{y}_1^2)$ is a quadratic non... | Substitute $x=111$ directly into $x^2 - 11y^2 = 1$. Then $y = \sqrt{(111^2-1)/11}$. Since $111^2-1$ can be made divisible by $11$, solutions exist without checking group periodicity. | **Theorem & Complete Algebraic Number Theory Solution:**
1. **Pell Group Action**: Solutions satisfy $x_n + y_n\sqrt{11} = (x_1 + y_1\sqrt{11})^n$.
2. **Periodicity**: Over $\mathbb{F}_{149}$, if $(11/149) = 1$, period divides $p-1$; if $(11/149) = -1$, period divides $p+1$.
3. **Infinitude**: The residue $x_{n_0} \eq... | Chosen proves periodicity in algebraic number field and analyzes quadratic character; Rejected naively evaluates a single value without proving infinitude or periodicity. | How does the norm map $N(\alpha) = 1$ in $\mathbb{F}_{151}$ force the period to divide $p+1$? | <thought>Norm map kernel has order $(p^2-1)/(p-1) = p+1$.</thought>
**Lemma**: The cyclic kernel of $N: \mathbb{F}_{151}^\times \to \mathbb{F}_{149}^\times$ has exact cardinality $p+1$. |
math_r1_00062 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | IMO SHORTLIST | Competition Benchmark: IMO Shortlist Benchmark [Problem #NT-LTE-01061]
Domain: Olympiad Number Theory
Problem Statement: Let $p = 3$ be an odd prime. Suppose positive integers $a, b$ satisfy $p \mid (a - b)$ but $p \nmid a$ and $p \nmid b$. Prove the Lifting The Exponent (LTE) identity for $n = 15$:
\[ v_{3}(a^{15} - b... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | <thought>
1. LTE Lemma Exploration:
- Factorization: $a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$.
- Let $S = \sum_{k=0}^{n-1} a^{n-1-k} b^k$. We must prove $v_p(S) = v_p(n)$.
2. Backtracking & Binomial Expansion:
- Write $a = b + p^k m$ where $k = v_p(a-b) \ge 1$.
- $a^j = (b + p^k m)^j \equiv b^j ... | By expanding $a^n - b^n = (a-b)(a^{n-1} + \dots + b^{n-1})$, each of the $n$ terms is equal to $a^{n-1} \pmod p$, so the sum is $n a^{n-1}$. Thus $v_p(a^n - b^n) = v_p(a-b) + v_p(n a^{n-1}) = v_p(a-b) + v_p(n) + (n-1)v_p(a)$, which gives an extra term. | **Proof via Binomial p-adic Expansion:**
1. Let $v_p(a - b) = k \ge 1$. Write $a = b + c$ with $v_p(c) = k$.
2. Expanding $a^n = (b + c)^n = b^n + n b^{n-1} c + \sum_{j=2}^n \binom{n}{j} b^{n-j} c^j$.
3. For $j \ge 2$, $v_p(\binom{n}{j} c^j) \ge j k - v_p(j) > k + v_p(n)$ for odd prime $p$.
4. Therefore $v_p(a^n - b^n... | Chosen establishes p-adic valuations with strict inequality $j k - v_p(j) > k + v_p(n)$; Rejected incorrectly asserts $v_p(a) > 0$ when $p \nmid a$. | Why does LTE fail for $p = 2$ when $n$ is even? | <thought>For p=2 and j=2, 2k - 1 = k + (k-1) equals k when k=1, causing cancellation.</thought>
**Resolution**: For $p=2$, the second-order binomial coefficient has $v_2(\binom{n}{2} c^2) = v_2(n) - 1 + 2k$, requiring the extra correction term $v_2(a+b)-1$. |
math_r1_00063 | Olympiad Number Theory & Diophantine Equations | Olympiad Number Theory | PUTNAM / IMO HARD | Competition Benchmark: AIME 2026 Invitational [Problem #NT-GAUSS-01062]
Domain: Olympiad Number Theory
Problem Statement: Evaluate the Legendre symbol $\left(\frac{73}{43}\right)$ using Gauss's Law of Quadratic Reciprocity. Prove whether the congruence $x^2 \equiv 73 \pmod{43}$ is solvable in integers. | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 73 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $43 \pmod{73} = 43$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | <thought>
1. Quadratic Reciprocity Law:
- $(p/q)(q/p) = (-1)^{(p-1)/2 \cdot (q-1)/2}$.
- Here $p = 73 \equiv 1 \pmod 4$, so $(p-1)/2$ is even.
- Therefore $(p/q) = (q/p)$.
2. Reduction Modulo p:
- Compute $q \pmod p$: $43 \pmod{73} = 43$.
- Decompose into prime factors and apply Euler's criterion.
</thou... | Since both 73 and 43 are odd primes, $(p/q) = -(q/p)$ always holds by Gauss's theorem. Evaluating directly gives the opposite sign. | **Evaluation via Quadratic Reciprocity:**
1. **Parity Check**: Since $p = 73 \equiv 1 \pmod 4$, $(-1)^{\frac{73-1}{2}\frac{43-1}{2}} = 1$.
2. **Reciprocity Application**: $\left(\frac{73}{43}\right) = \left(\frac{43}{73}\right)$.
3. **Modular Reduction**: Evaluating $\left(\frac{43}{73}\right) = -1$.
Therefore, the c... | Chosen correctly analyzes $p \equiv 1 \pmod 4$ to obtain $(p/q) = +(q/p)$; Rejected misremembers Quadratic Reciprocity as always having a negative sign. | How does Gauss's Lemma with half-intervals prove quadratic reciprocity? | <thought>Counting lattice points in rectangle of size (p-1)/2 by (q-1)/2.</thought>
**Resolution**: Eisenstein's geometric proof counts interior integer points in the rectangle $[1, (p-1)/2] \times [1, (q-1)/2]$. |
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π Enterprise DeepSeek-R1 Autonomous Mathematical & Logic CoT SFT/DPO Dataset (2026)
High-precision multi-turn instruction tuning and preference optimization dataset with step-by-step hypothesis exploration, error discovery, and dynamic backtracking Chain-of-Thought (<thought>) reasoning trees for fine-tuning LLMs (DeepSeek-R1-Distill-Qwen, Qwen-2.5-Math, Llama-3.3, Mistral) into World-Class Olympiad Mathematicians and Formal Verification Agents.
π Dataset Architecture & 20 Mathematical Cores
- 20 Independent Theorem Cores: Olympiad Number Theory (Pell's Equations, LTE Lemma, Quadratic Reciprocity, Chinese Remainder & Hensel), Higher Algebra (Cauchy functional equations, Newton-Girard sums, Jensen convexity, Chebyshev recurrence), Formal Theorem Proving in Lean 4 (AM-GM, Bernoulli induction, Sqrt(2) irrationality, Lagrange group cosets), Advanced Combinatorics (Ramsey & LLL, Derangements & PIE, Cayley tree PrΓΌfer bijection, Catalan OGF/EGF), and Euclidean/Projective Geometry (Euler distance & Trillium, Ptolemy inversion, Radical axis coaxal systems, Simson line).
- Dynamic DeepSeek-R1 Backtracking: Explores hypotheses -> detects contradictions ("Wait, let me rethink...") -> derives rigorous proof.
- Hardware & VRAM Profiling: Pre-computed token quantiles ($P_{50} = 375$, $P_{99} = 497$) and GPU VRAM memory footprint matrix across model sizes.
- DPO Preference Pairs: 2,500 curated Chosen vs Subtle Mathematical Traps with automated formal logic critiques.
- Pre-packaged Tooling: 1-Click Unsloth Colab Notebook (
1_CLICK_UNSLOTH_MATH_R1_FINE_TUNING_COLAB.ipynb), LLaMA-Factory YAML, and Axolotl YML configs.
π Quick Start with Hugging Face datasets
from datasets import load_dataset
# Load from Parquet
dataset = load_dataset("parquet", data_files="DEEPSEEK_R1_MATH_LOGIC_COT_SFT_DPO_2026_100_SAMPLE.parquet", split="train")
print(f"Loaded {len(dataset)} sample mathematical training instances!")
print("Sample Problem:", dataset[0]["user_prompt"])
print("Sample Proof:", dataset[0]["formal_proof"])
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