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Let $ABC$ be an acute triangle with altitude $AD$ ( $D \in BC$ ). The line through $C$ parallel to $AB$ meets the perpendicular bisector of $AD$ at $G$ . Show that $AC = BC$ if and only if $\angle AGC = 90^{\circ}$ .
$\bullet$ $CA=CB:$ Let $E$ and $F$ be midpoints of $AD$ and $AB$ ,respectively. Since $GE||BC$ we get $F-E-G$ are collinear $\implies AF=FB=FD$ . $\angle GCA=\angle CAB=\angle CBA=\angle GFA \implies GCAF$ is cyclic $\implies \angle AGC=180-\angle CFA=180-90=90. \square$ $\bullet$ $\angle AGC=90:...
[ " $GM\\parallel BC, AB\\parallel BC$ , implies $AMCG$ is a parallelogram. $\\angle AGC=90^\\circ\\Leftrightarrow \\angle AMC=90^\\circ\\Leftrightarrow AC=BC$ since $M$ is midpoint of $AB$ .", "Let $M$ be the midpoint of $\\overline{AB}$ and note $BCGM$ is a parallelogram. Then, $MD=MB=GC$ so $CDM...
[ "origin:aops", "2022 Contests", "2022 3rd Memorial "Aleksandar Blazhevski-Cane"" ]
{ "answer_score": 130, "boxed": false, "end_of_proof": true, "n_reply": 3, "path": "Contest Collections/2022 Contests/2022 3rd Memorial "Aleksandar Blazhevski-Cane"/2759376.json" }
A $6 \times 6$ board is given such that each unit square is either red or green. It is known that there are no $4$ adjacent unit squares of the same color in a horizontal, vertical, or diagonal line. A $2 \times 2$ subsquare of the board is *chesslike* if it has one red and one green diagonal. Find the maximal po...
[]
[ "origin:aops", "2022 Contests", "2022 3rd Memorial "Aleksandar Blazhevski-Cane"" ]
{ "answer_score": 0, "boxed": false, "end_of_proof": false, "n_reply": 0, "path": "Contest Collections/2022 Contests/2022 3rd Memorial "Aleksandar Blazhevski-Cane"/2759377.json" }
Given an integer $n\geq2$ , let $x_1<x_2<\cdots<x_n$ and $y_1<y_2<\cdots<y_n$ be positive reals. Prove that for every value $C\in (-2,2)$ (by taking $y_{n+1}=y_1$ ) it holds that $\hspace{122px}\sum_{i=1}^{n}\sqrt{x_i^2+Cx_iy_i+y_i^2}<\sum_{i=1}^{n}\sqrt{x_i^2+Cx_iy_{i+1}+y_{i+1}^2}$ . *Proposed by Mirko Petru...
We can use a similar argument as in the proof of rearrangement inequality. Letting $f(x,y)=\sqrt{x^2+Cxy+y^2}$ , it suffices to show the case $n=2$ , which corresponds to a single transposition in the general case. Al we have to show that if $a<b$ and $c<d$ , then $$ \sqrt{a^2+Cac+c^2}+\sqrt{b^2+Cbd+d^2}<\sqrt{a...
[]
[ "origin:aops", "2022 Contests", "2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;" ]
{ "answer_score": 50, "boxed": false, "end_of_proof": false, "n_reply": 1, "path": "Contest Collections/2022 Contests/2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;/2759383.json" }
Find all triplets of positive integers $(x, y, z)$ such that $x^2 + y^2 + x + y + z = xyz + 1$ . *Proposed by Viktor Simjanoski*
<details><summary>Solution (using Vieta's Jumping Root Method)</summary>$\wedge$ means 'and'. $\mathbb{N*}$ means $\{n | n\in \mathbb{Z} \wedge n>0\}$ . $(*)$ stands for the equation $x^2+y^2+x+y+z=xyz+1$ . Define $g(x,y):=\frac{x^2+y^2+x+y-1}{xy-1}$ . WLOG assume $x\geq y$ . $\textbf{Case 1.}$ $y=1$ . $x^2+x+...
[ "<details><summary>Hint</summary>Vieta jumping. Solutions exists only for $z=7$ There are two series of solutions with first terms $1,2$ and $1,4$</details>", "Vieta jumping method and pell equation.", "what is your motivation to prove z=7 please?", "Does anybody have a complete solution?\n" ]
[ "origin:aops", "2022 Contests", "2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;" ]
{ "answer_score": 128, "boxed": false, "end_of_proof": false, "n_reply": 5, "path": "Contest Collections/2022 Contests/2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;/2759385.json" }
Find all positive integers $n$ such that the set $S=\{1,2,3, \dots 2n\}$ can be divided into $2$ disjoint subsets $S_1$ and $S_2$ , i.e. $S_1 \cap S_2 = \emptyset$ and $S_1 \cup S_2 = S$ , such that each one of them has $n$ elements, and the sum of the elements of $S_1$ is divisible by the sum of the el...
We claim the answer is all $n \not\equiv 5 \pmod 6$ . Let $\sum_{i \in S_1} i=A$ and $\sum_{i \in S_2} i=B$ . Then, $A+B=n(2n+1)$ and $A \mid B$ . Note that $A \geq 1+2+\ldots+n=\dfrac{n(n+1)}{2}$ and $B \leq 2n+(2n-1)+\ldots+(n+1)=\dfrac{n(3n+1)}{2}.$ Therefore, $B \leq \dfrac{n(3n+1)}{2} <\dfrac{3n(n+1)}{2} \...
[ "The answer is all $n \\not \\equiv 5\\pmod{6}$ .**Constraction for $n=2k$** : $S_1=\\{1,2,...,k\\}\\cup \\{3k+1,3k+2,...,4k\\}$ and $S_2=S\\setminus S_1$ .\nFor $n\\equiv 1,3 \\pmod{6}$ I will not give a construction but I will show that it's possible to construct $S_1$ and $S_2$ .\nLet $n=2k+1$ and le...
[ "origin:aops", "2022 Contests", "2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;" ]
{ "answer_score": 90, "boxed": false, "end_of_proof": false, "n_reply": 2, "path": "Contest Collections/2022 Contests/2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;/2759387.json" }
Let $ABC$ be an acute triangle with incircle $\omega$ , incenter $I$ , and $A$ -excircle $\omega_{a}$ . Let $\omega$ and $\omega_{a}$ meet $BC$ at $X$ and $Y$ , respectively. Let $Z$ be the intersection point of $AY$ and $\omega$ which is closer to $A$ . The point $H$ is the foot of the altitude...
It's well known $XZ \perp BC$ . Let $AX$ and $HZ$ meet at $S$ , Note that $ZX || AH$ so $S$ lies on median of $AH$ in triangle $AYH$ so we must prove $IY$ is median of $AH$ . Note that $I$ is midpoint of $XZ$ and $AH || XZ$ so $IY$ is median of $AH$ . we're Done.
[ "From \"Diameter of Incircle\" Lemma we know that $X-I-Z$ are collinear. So in $\\triangle AHY$ $YI$ is median and $ZX||AH$ . So from Ceva's Theorem we get $AX-HZ-IY$ are concurrent.", " $XZ$ is diameter of $\\omega$ and $AH$ parallel $XZ$ .Remaning easy." ]
[ "origin:aops", "2022 Contests", "2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;" ]
{ "answer_score": 30, "boxed": false, "end_of_proof": false, "n_reply": 3, "path": "Contest Collections/2022 Contests/2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;/2759390.json" }
We say that a positive integer $n$ is *memorable* if it has a binary representation with strictly more $1$ 's than $0$ 's (for example $25$ is memorable because $25=(11001)_{2}$ has more $1$ 's than $0$ 's). Are there infinitely many memorable perfect squares? *Proposed by Nikola Velov*
$n^2=2^k \cdot a_k + ... + 2^1 \cdot a_1 + 2^0 a_0$ Next number $$ \boxed {(2^{k+2} + 1)n} $$
[ "<blockquote>We say that a positive integer $n$ is *memorable* if it has a binary representation with strictly more $1$ 's than $0$ 's (for example $25$ is memorable because $25=(11001)_{2}$ has more $1$ 's than $0$ 's). Are there infinitely many memorable perfect squares?</blockquote>\nYes, there are .\n...
[ "origin:aops", "2022 Contests", "2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;" ]
{ "answer_score": 1006, "boxed": true, "end_of_proof": false, "n_reply": 2, "path": "Contest Collections/2022 Contests/2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;/2759392.json" }
For any integer $n\geq1$ , we consider a set $P_{2n}$ of $2n$ points placed equidistantly on a circle. A *perfect matching* on this point set is comprised of $n$ (straight-line) segments whose endpoints constitute $P_{2n}$ . Let $\mathcal{M}_{n}$ denote the set of all non-crossing perfect matchings on $P_{2n...
[]
[ "origin:aops", "2022 Contests", "2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;" ]
{ "answer_score": 0, "boxed": false, "end_of_proof": false, "n_reply": 0, "path": "Contest Collections/2022 Contests/2022 3rd Memorial &quot;Aleksandar Blazhevski-Cane&quot;/2759397.json" }
![Image](https://data.artofproblemsolving.com/images/maa_logo.png) These problems are copyright $\copyright$ [Mathematical Association of America](http://maa.org).
[]
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 0, "boxed": false, "end_of_proof": false, "n_reply": 0, "path": "Contest Collections/2022 Contests/2022 AIME Problems/1100560.json" }
For any finite set $X$ , let $|X|$ denote the number of elements in $X.$ Define $$ S_n = \sum |A \cap B|, $$ where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\{1, 2, 3, …, n\}$ with $|A| = |B|.$ For example, $S_2 = 4$ because the sum is taken over the pai...
<blockquote>For any finite set $X$ , let $|X|$ denote the number of elements in $X.$ Define $$ S_n = \sum |A \cap B|, $$ where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\{1, 2, 3, …, n\}$ with $|A| = |B|.$ For example, $S_2 = 4$ because the sum is taken ...
[ "245, basically S_n = n(2n -2 choose n-1) from chairperson and vandermonde spam", "proudest solve lesgo", "Consider how many times any given number $k$ is counted in the intersection of $A, B$ , in the expression for $S_n$ . If $A, B$ each contain $r$ numbers, then it is counted ${n-1\\choose r-1}^2={...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1048, "boxed": false, "end_of_proof": false, "n_reply": 48, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777199.json" }
Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a, b, c, $ or $d$ is nonzero. Let $N$ be the number of distinct numerators when numbers in $S$ are written as fractions in lowest terms. For example, bot...
The factors of $9999$ are $1, 3, 9, 11, 33, 99, 101, 303, 909, 1111, 3333, $ and $9999$ . For any integer in the range $[1, 9998]$ , it can be a numerator if there exists a factor of $9999$ that is relatively prime to that integer (because that factor can be its denominator). We now break this big interval into...
[ "395 gang anyone?", "i got 449", "answer is 392 from 6392 confirmed with code", "<blockquote>answer is 392 from 6392 confirmed with code</blockquote>\n\nyeah same here I immediately wrote a Java code after the test \n\nsadge moment when you forget to delete the three multiples of $303$ in the $1111$ set :...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1188, "boxed": true, "end_of_proof": false, "n_reply": 48, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777200.json" }
Find the number of ordered pairs of integers $(a, b)$ such that the sequence $$ 3, 4, 5, a, b, 30, 40, 50 $$ is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Clearly just picking from the set $\{3, 4, 5, 30, 40, 50\}$ we cannot find an arithmetic progression. Case $1$ : The arithmetic progression contains only $a$ or only $b$ . Note that $6 \leq a \leq 28$ and $7 \leq b \leq 29$ . Clearly $a = 6$ fails from $3, 4, 5, a$ . Next $a/b, 30, 40, 50$ causes $a ...
[ "Note that $7\\le a<b\\le 29$ and $a\\ne 20, b\\ne 20$ . The only other restrictions are $(7,9)$ , $(12,21)$ , and $(16,28)$ . So the answer is $\\binom{23}{2}-9-13-3=\\boxed{228}$ . ", "I put 237 lmao\n\n(forgot to remove (20, 21), (20, 22), (20, 23), ..., (20, 29) ah stupid me)", "I got 236 oof", "Fo...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1074, "boxed": false, "end_of_proof": false, "n_reply": 61, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777203.json" }
Ellina has twelve blocks, two each of red $\left({\bf R}\right),$ blue $\left({\bf B}\right),$ yellow $\left({\bf Y}\right),$ green $\left({\bf G}\right),$ orange $\left({\bf O}\right),$ and purple $\left({\bf P}\right).$ Call an arrangement of blocks *even* if there is an even number of blocks between each...
Wow, this problem was actually so amazing and reminded me of why I enjoy comp math.**<span style="color:#f00">Claim:</span>** There exists a bijection between even arrangements with $n$ pairs of colored blocks and ways to order the evens and the odds (separately) from $1$ to $2n$ . *Proof.* Label the ordering of t...
[ "Note that the even positions (2,4,6,8,10,12) have one of each color and same thing with odd positions. There are totally $\\frac{12!}{64}$ permutations. So the fraction is \\[\\frac{6!^2\\cdot 64}{12!}=\\frac{6!\\cdot 64}{7\\cdot 8\\cdot 9\\cdot 10\\cdot 11\\cdot 12}=\\frac{720\\cdot 8}{7\\cdot 9\\cdot 10\\cdot ...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1122, "boxed": false, "end_of_proof": false, "n_reply": 53, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777204.json" }
Let $a, b, c, d, e, f, g, h, i$ be distinct integers from $1$ to $9$ . The minimum possible positive value of $$ \frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} $$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
Note that $\frac{2\cdot 3\cdot 6-1\cdot 5\cdot 7}{4\cdot 8\cdot 9}=\frac{1}{288}$ . We claim this is the minimum, which gives an answer of $\boxed{289}$ . Suppose there was something less. Then $abc-def=1$ . If $9$ was in $a,b,c,d,e,f$ , then we would need $ghi=6\cdot 7\cdot 8$ . Now $a,b,c,d,e,f$ is some pe...
[ "Note that $(6,2,3,7,5,1,4,8,9)$ gives $\\tfrac{1}{288}$ , for an answer of $1+288=\\boxed{289}$ . Otherwise, if $abc - def = 2$ , then the minimum possible value is $\\tfrac{2}{7 \\cdot 8 \\cdot 9} = \\tfrac{1}{252}$ . ", "i got 289? bsically let x = abc, y = def, then xy(x-y)/9! min which is x = 35 y = 36...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1052, "boxed": false, "end_of_proof": false, "n_reply": 31, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777205.json" }
Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients of $2$ and $-2$ , respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53)$ . Find ${P(0) + Q(0)}$ .
<blockquote>We have $P(x)=2x^2+ax+b$ . So $512+16a+b=54\implies 16a+b=-458$ . Also, $800+20a+b=53$ , so $20a+b=-747$ . Thus, $4a=-289$ . So $-1156+b=-458\implies b=698$ . Also, $Q(x)=-2x^2+cx+d$ . So $-512+16c+d=54\implies 16c+d=566$ . Also, $-800+20c+d=53\implies 20c+d=853$ . So $4c=287$ . So $1148+d=566\i...
[ "Mine.\n\n<details><summary>Solution</summary>The polynomial $R(x) := P(x) + Q(x)$ is linear, and we may compute $R(16) = 108$ and $R(20) = 106$ . Hence $R(0) = \\boxed{116}$ .</details>", "<details><summary>other sol</summary>The line that passes through both points is $y=-\\frac{1}{4}x+58$ .\n Hence, ...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1032, "boxed": true, "end_of_proof": false, "n_reply": 37, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777211.json" }
Find the three-digit positive integer $\underline{a} \ \underline{b} \ \underline{c}$ whose representation in base nine is $\underline{b} \ \underline{c} \ \underline{a}_{\hspace{.02in}\text{nine}}$ , where $a$ , $b$ , and $c$ are (not necessarily distinct) digits.
We claim $\boxed{227}$ works. Proof: $227=2\cdot 81+7\cdot 9+2$ . $\blacksquare$ .
[ " $227$ works I think. What I did was set up the equation $99a=71b+8c$ and casework on $a$ .", "<blockquote> $227$ works I think.</blockquote>\n\nGot that too", "<blockquote> $227$ works I think. What I did was set up the equation $99a=71b+8c$ and casework on $a$ .</blockquote>\n\ntaking mod 9 works be...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1106, "boxed": false, "end_of_proof": false, "n_reply": 41, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777212.json" }
Let $x$ , $y$ , and $z$ be positive real numbers satisfying the system of equations \begin{align*} \sqrt{2x - xy} + \sqrt{2y - xy} & = 1 \sqrt{2y - yz} + \hspace{0.1em} \sqrt{2z - yz} & = \sqrt{2} \sqrt{2z - zx\vphantom{y}} + \sqrt{2x - zx\vphantom{y}} & = \sqrt{3}. \end{align*}Then $\big[ (1-x)(1-y)(1-z) \big] ^2...
Favorite problem on the test. Extremely clean. (Solution close to that in post #2) First, we note that we can let a triangle exist with side lengths $\sqrt{2x}$ , $\sqrt{2z}$ , and opposite altitude $\sqrt{xz}$ . This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal...
[ "Magical solution communicated to me by a girl in my school who doesn't even do competition math and got this during the test. Let $x=2\\sin^2\\alpha, y=2\\sin^2\\beta, z=2\\sin^2\\theta$ . The given conditions rewrite themselves as:\n\\begin{align*}\n2\\sin(\\alpha+\\beta)&=1 \n2\\sin(\\beta+\\theta)&=\\sqrt{2} \...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1158, "boxed": true, "end_of_proof": false, "n_reply": 43, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777215.json" }
In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ .
<blockquote><blockquote>Diagram: [asy] unitsize(0.016cm); pair A = (-300,324.4); pair B = (300, 324.4); pair C = (375, 0); pair D = (-375, 0); draw(A--B--C--D--cycle); pair W = (-42,0); pair X = (42, 0); pair Y = (-33,324.4); pair Z = (33,324.4); pair P = (-171, 162.2); pair Q = (171, 162.2); dot(P); dot(Q); draw(A--...
[ "Solution (related to the title):\n\nTranslate points $B, Q,$ and $C$ by $PQ$ units to the left, as shown. Let $PQ = x$ .\n[asy]\nsize(250);\nlabel((0,0), \"D\", SW);\nlabel((1.166666666, 6), \"A\", NW);\nlabel((4,2.6), \"P, Q'\", S);\nlabel((8,3), \"Q\", S);\nlabel((12,0), \"C\", SE);\nlabel((10.833333, 6),...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1178, "boxed": true, "end_of_proof": false, "n_reply": 74, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777216.json" }
Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the splitting line of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are po...
<blockquote>Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the splitting line of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and ...
[ "Consider the splitting line through $M$ . Extend $D$ on ray $BC$ such that $CD=CA$ . Then the splitting line bisects segment $BD$ , so in particular it is the midline of triangle $ABD$ and thus it is parallel to $AD$ . But since triangle $ACD$ is isosceles, we can easily see $AD$ is parallel to the a...
[ "origin:aops", "2022 Contests", "2022 AIME Problems" ]
{ "answer_score": 1198, "boxed": true, "end_of_proof": false, "n_reply": 39, "path": "Contest Collections/2022 Contests/2022 AIME Problems/2777218.json" }
End of preview. Expand in Data Studio

AoPS: Art of Problem Solving Competition Mathematics

Dataset Description

This dataset is a collection of 80,661 competition mathematics problems and solutions obtained from the Art of Problem Solving (AoPS) community wiki and forums. It covers a wide range of mathematical contests and olympiads, including problems from events such as AIME, BAMO, IMO, and various national and memorial competitions.

The dataset was curated by AI-MO (Project Numina), an initiative focused on building AI systems capable of mathematical reasoning at the olympiad level.

Dataset Structure

Fields

Column Type Description
problem string The mathematical problem statement, typically formatted in LaTeX.
solution string A solution or proof for the problem. May be empty for some entries.
candidates list[string] Alternative or candidate solutions contributed by the community.
tags list[string] Metadata tags indicating the origin, contest name, and year (e.g., "origin:aops", "2022 AIME Problems").
metadata dict Additional metadata about the problem (see below).

Metadata Fields

Field Type Description
answer_score int64 Community score or rating of the answer.
boxed bool Whether the answer contains a boxed final result (e.g., \boxed{42}).
end_of_proof bool Whether the solution includes a complete proof ending.
n_reply int64 Number of community replies or comments on the problem thread.
path string Source path in the AoPS collection (e.g., Contest Collections/2022 Contests/...).

Splits

Split Examples
train 80,661

Example

{
    "problem": "Let $ABC$ be an acute triangle with altitude $AD$ ($D \\in BC$). The line through $C$ parallel to $AB$ meets the perpendicular bisector of $AD$ at $G$. Show that $AC = BC$ if and only if $\\angle AGC = 90°$.",
    "solution": "...",
    "candidates": ["..."],
    "tags": ["origin:aops", "2022 Contests", "2022 3rd Memorial \"Aleksandar Blazhevski-Cane\""],
    "metadata": {
        "answer_score": 130,
        "boxed": false,
        "end_of_proof": true,
        "n_reply": 3,
        "path": "Contest Collections/2022 Contests/2022 3rd Memorial .../2759376.json"
    }
}

Topic Coverage

Problems span a broad range of competition mathematics topics, including:

  • Geometry -- triangle properties, cyclic quadrilaterals, angle chasing
  • Number Theory -- divisibility, modular arithmetic, Diophantine equations
  • Algebra -- inequalities, polynomials, functional equations
  • Combinatorics -- counting, graph theory, board coloring problems

Usage

from datasets import load_dataset

dataset = load_dataset("AI-MO/aops")

# Access a problem
print(dataset["train"][0]["problem"])
print(dataset["train"][0]["solution"])

Intended Use

  • Training and evaluating mathematical reasoning models
  • Benchmarking LLMs on competition-level mathematics
  • Studying solution quality and problem difficulty distributions
  • Building retrieval-augmented generation (RAG) systems for math tutoring

Source

All problems and solutions originate from the Art of Problem Solving community.

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