problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
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 $\implies \angle AGC=180-\angle CFA=180-90=90. \square$ $\bullet$ $\angle AGC=90:$ $AGCD$ is cyclic. Let $\angle AGE=\angle DGE=\angle GDC=\alpha \implies CAD=\angle CGD=180-\alpha-\angle GCD=180-\alpha-(180-\angle GAD)=90-2\alpha \implies \angle GAC=\alpha \implies \angle DAF=\alpha \implies \angle CBA=\angle CAB=90-\alpha \implies CA=CB. \blacksquare$ | [
" $GM\\parallel BC, AB\\parallel BC$ , 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... | [
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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 possible number of chesslike squares on the board.
*Proposed by Nikola Velov* | [] | [
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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 Petrusevski* | 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 $$ \sqrt{a^2+Cac+c^2}+\sqrt{b^2+Cbd+d^2}<\sqrt{a^2+Cad+d^2}+\sqrt{b^2+Cbc+c^2} $$ Since $C\in(-2,2)$ we can write it as $C=-2\cos\theta$ for some $\theta\in(0,\pi)$ . Therefore we can interpret $f(x,y)$ as the distance $XY$ of the points $X$ and $Y$ on two halflines with an angle $\theta$ between them, with the same origin $O$ , so that $OX=x$ and $OY=y$ .
So if $O,A,B$ are on the half line $Or$ in this order and $O,C,D$ in this order on the half line $Os$ , it suffices to show $AC+BD<AD+BC$ . To do this simply pick $X=AD\cap BC$ , and by triangle inequality we have the strict inequality $$ AC+BD<AX+CX+BX+DX=AD+BC $$ as wanted. | [] | [
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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 $x\geq y$ . $\textbf{Case 1.}$ $y=1$ . $x^2+x+z+2=xz+1 \Rightarrow (x-1)z=x^2+x+1>0 \Rightarrow x>1$ . $z=g(x,1)=\frac{x^2+x+1}{x-1}=x+2+\frac{3}{x-1} \in \mathbb{Z}$ $\Rightarrow x=2, 4$ . $(x,y,z)=(2,1,7), (4,1,7)$ . $\textbf{Case 2.}$ $x=y\geq 2$ . $g(x,x)=\frac{2x^2+2x-1}{x^2-1}=2+\frac{2x+1}{x^2-1} \Rightarrow 2x+1 \geq x^2-1 \Rightarrow x=2$ .
But when $x=y=2$ , $g(x,x)=2+\frac{5}{3} \notin \mathbb{N*}$ , so no triplets satisfy (*) in this case.
Thus we have $x>y$ . $\textbf{Case 3.}$ $y=2$ . $g(x,2)=\frac{x^2+x+5}{2x-1}$ . $4g(x,2)=\frac{4x^2+4x+20}{2x-1}=2x+3+\frac{23}{2x-1}\in \mathbb{N*}$ $\Rightarrow (2x-1)|23 \Rightarrow x=12\Rightarrow (x,y,z)=(12,2,7)$ . $\textbf{Case 4.}$ $y\geq 3 \wedge x=y+1$ . $g(y+1,y)=\frac{2y^2+4y+1}{y^2+y-1}=2+\frac{2y+3}{y^2+y-1}\in \mathbb{N*}$ $\Rightarrow 2y+3\geq y^2+y-1$ which contradicts with $y\geq 3$ . $\textbf{Case 5.}$ $y\geq 3 \wedge x\geq y+2$ .
Suppose $(x,y,z)$ satisfies $x^2+y^2+x+y+z=xyz+1 (*)$ . $z=g(x,y)=\frac{x^2+y^2+x+y-1}{xy-1}\geq \frac{2xy+x+y-1}{xy-1} = 2+\frac{x+y+1}{xy-1} > 2 \Rightarrow z\geq 3$ .
Fix $y,z$ , and then $a^2-(yz-1)a+y^2+y+z-1=0$ is a quadratic, with one root $x$ and so the other is $\frac{y^2+y+z-1}{x}=yz-x-1$ . Because $y^2+y+z-1 > 0$ , $yz-x-1 \in \mathbb{N*}$ .
\begin{align*}
y'<y\quad \Leftrightarrow\quad &yz-1-x<y
\Leftrightarrow\quad &y\frac{x^2+y^2+x+y-1}{xy-1}<x+y+1
\Leftrightarrow\quad &x^2y+y^3+xy+y^2-y<x^2y+xy^2+xy-x-y-1
\Leftrightarrow\quad &y^3+y^2-y<xy^2-x-y-1
\Leftarrow\quad &y^3+y^2-y<(y+2)(y^2-1)-y-1
\Leftrightarrow\quad &y^2-y-3\geq 0
\Leftarrow\quad &y\geq 3.
\end{align*}
\begin{align*}
x'-y'<x-y\quad\Leftrightarrow\quad &y-(yz-x-1)<x-y
\Leftrightarrow\quad &yz>2y+1
\Leftarrow\quad &z\geq 3 \wedge y\geq 2
\end{align*}
Define $f(x,y,z):=(y,yz-x-1,z)$ .
Then if $(x,y,z)$ satisfies (*), so does $f(x,y,z)$ .
Suppose $(x,y,z)$ satisfies (*), and then we can replace it with $f(x,y,z)$ finite times until it does not satisfy the condition $x-2\geq y\geq 3$ , because each time $y$ and $x-y$ strictly decreases. At last we will certainly have $(x,y,z)=(2,1,7),(4,1,7)$ or $(12,2,7)$ . Surprisingly $f(12,2,7)=(2,1,7)$ . Thus for all $(x,y,z)$ satisfying (*), we can replace it with $f(x,y,z)$ finite times to become (2,1,7) or (4,1,7). (Thus $z=7$ .)
If $(x,y,z)\in \mathbb{N*}^3$ , then $f^{-1}(x,y,z)=(xz-y-1,x,z)\in \mathbb{N*}^3$ . Let $(x_0,y_0)=(2,1)$ or $(4,1)$ .
Let $(x_{n+1},y_{n+1})=(7x_n-y_n-1,x_n)$ . Then the two sequences of triplets $(x_n,y_n,7)$ (with different $(x_0,y_0)$ ) contains "all" the triplets satisfying (*). Be careful! $(y_n,x_n,7)$ satisfies (*) as well, because WLOG at the beginning.
We have $y_{n+1}=x_n$ and so $y_{n+2}=7y_{n+1}-y_n-1$ . The only thing left to do is to solve two sequences.</details> | [
"<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"
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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 elements in $S_2$ .
*Proposed by Viktor Simjanoski* | 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, $B \leq \dfrac{n(3n+1)}{2} <\dfrac{3n(n+1)}{2} \leq 3A$ Since $A \mid B$ , this implies that $B \in \{A,2A \}$ . We distinguish two cases.**Case 1:** $B=A$ . Then, $A=\dfrac{n(2n+1)}{2},$ and so $n$ must be even. For all even $n$ , we may take $S_1=\{1,2n \} \cup \{2,2n-1 \} \cup \ldots \cup (\dfrac{n}{2},(2n+1)-\dfrac{n}{2})$ . It is straightforward to check that $|S_1|=n$ and $A=n(2n+1)$ .**Case 2:** $B=2A$ . Then, $A=\dfrac{n(2n+1)}{3}$ , and so $3 \mid n(2n+1)$ , i.e. $n \not\equiv 2 \pmod 3$ . Consider the collection $\mathcal{F}$ of all sets $X \subseteq \{1,2,\ldots, 2n \}$ such that $|X|=n$ . Note that the minimum sum of the elements of a set belonging in $\mathcal{F}$ is $m=1+2+\ldots+n=\dfrac{n(2n+1)}{2}$ , and the maximum is $M=2n+(2n-1)+\ldots+(n+1)=\dfrac{n(3n+1)}{2}$ . Note that $m<A<M$ .
We claim that all intermediate sums in $[m,M]$ can be achieved by a set in collection $\mathcal{F}$ . Indeed, assume all sums in $[m,t]$ have be achieved for some $t \geq m$ . If $t=M$ , we are done. If not, we want to find a set that achieves $t+1$ . Let $T=\{x_1,\ldots,x_n \}$ be a set such that its elements sum to $t$ .
If there are two elements of $T$ that are not consecutive, we may increment the smallest one of them by one and finish. Moreover, if $x_n \neq 2n$ , we may increment $x_n$ by one and finish. If neither of these happens, set $T$ must necessarily be $\{n+1,n+2,\ldots,2n \}$ , which is a contradiction as we assumed $t \neq M$ .
To sum up, the working $n$ are the evens and the $n \not\equiv 2 \pmod 3$ , that is all $n \not\equiv 5 \pmod 6$ . | [
"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... | [
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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 from $A$ . Show that $HZ$ , $IY$ and $AX$ are concurrent.
*Proposed by Nikola Velov* | 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."
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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... | [
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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 $P_{2n}$ . A perfect matching $M\in \mathcal{M}_{n}$ is said to be *centrally symmetric*, if it is invariant under point reflection at the circle center. Determine, as a function of $n$ , the number of centrally symmetric perfect matchings within $\mathcal{M}_{n}$ .
*Proposed by Mirko Petrusevski* | [] | [
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 These problems are copyright $\copyright$ [Mathematical Association of America](http://maa.org). | [] | [
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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 pairs of subsets $$ (A, B) \in \{ (\emptyset, \emptyset), (\{1\}, \{1\}), (\{1\}, \{2\}), (\{2\}, \{1\}), (\{2\}, \{2\}), (\{1, 2\}, \{1, 2\})\}, $$ giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4.$ Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by $1000.$ | <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 over the pairs of subsets $$ (A, B) \in \{ (\emptyset, \emptyset), (\{1\}, \{1\}), (\{1\}, \{2\}), (\{2\}, \{1\}), (\{2\}, \{2\}), (\{1, 2\}, \{1, 2\})\}, $$ giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4.$ Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by $1000.$ </blockquote>
<details><summary>Non-rigorous solution</summary>Engineer's induct; after evaluating $n=2,3,4,5$ one may observe that $n\mid S_n$ ; then it is apparent that the sequence $\{S_n/n\}$ is $2,6,20,70$ , so probably $S_n/n=\dbinom{2(n-1)}{n-1}$ .
Finally we turn our attention to the grueling task of answer extraction:
\[\frac{S_{2022}}{S_{2021}}=\frac{2022}{2021}\frac{\dbinom{4042}{2021}}{\dbinom{4040}{2020}}=\frac{2022}{2021}\frac{4042\cdot4041}{2021^2}\]
\[=\frac{2\cdot2022\cdot4041}{2021^2}.\]
The requested sum is
\[2\cdot2022\cdot4041+2021^2\equiv2\cdot22\cdot41+21^2\equiv804+441\equiv\boxed{245}\pmod{1000}.\]</details> | [
"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={... | [
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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, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$ | 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 five smaller intervals:
-----------
<u>Interval 1</u>: $[1,100]$ All of these numbers are relatively prime to $101$ , so all of them work. We get **100** from this interval.
-----
<u>Interval 2</u>: $[101, 908]$ All numbers in this interval work except multiples of $101$ and $33$ (as they are the numbers that cannot be put over $909$ and $1111$ ). Since there are no multiples of $101 \cdot 33$ in this interval, we can just take the size of this interval minus multiples of $101$ and $33$ . The multiples of $33$ are $33 \cdot 4 = 132$ through $33 \cdot 27 = 891$ for a total of $27-4+1=24$ multiples of $33$ , and the multiples of $101$ are $101 \cdot 1 = 101$ to $101 \cdot 8 = 808$ for a total of $8$ multiples of $101$ . Since the size of the interval is $908-101+1=808$ , we get $808-24-8 = $ **776** possibilities from this interval.
-----------
<u>Interval 3</u>: $[909,1110]$ All numbers in this interval work except multiples of $11$ and multiples of $101$ (as they are the ones that cannot be put over $1111$ ). We do the same thing as in the previous case.The multiples of $11$ are $11 \cdot 83 = 913$ to $11 \cdot 100 = 1100$ for a total of $100-83+1=18$ multiples of $11$ . The multiples of $101$ are $101 \cdot 9 = 909$ and $101 \cdot 10 = 1010$ for a total of $2$ multiples of $101$ . Since there are $1110-909+1=202$ numbers in this interval, we get a total of $202 - 18 - 2 = $ **182** possibilities from this case.
---------
<u>Interval 4</u>: $[1111,3332]$ We repeat the same process as in the previous two intervals. Multiples of $3$ , $11$ , and $101$ cannot be put on the numerator of $3333$ or $9999$ . We use PIE to remove these multiples.
The multiples of $3$ in this interval are $3 \cdot 371 = 1113$ to $3 \cdot 1110 = 3330$ for a total of $1110-371+1=740$ multiples of $3$ .
The multiples of $11$ in this interval are $11 \cdot 101 = 1111$ to $11 \cdot 302 = 3322$ for a total of $302-101+1=202$ multiples of $11$ .
The multiples of $101$ in this interval are $101 \cdot 11 = 1111$ to $101 \cdot 32 = 3232$ for a total of $32-11+1=22$ multiples of $101$ .
The multiples of $33$ in this interval are $33 \cdot 34 = 1122$ to $33 \cdot 100 = 3300$ for a total of $100-34+1=67$ multiples of $33$ .
The multiples of $303$ in this interval are $303 \cdot 4 = 1212$ to $303 \cdot 10 = 3030$ for a total of $10-4+1=7$ multiples of $303$ .
The multiples of $1111$ in this interval are $1111$ and $2222$ for a total of $2$ multiples of $1111$ .
Because there are no multiples of $3333$ in this interval, our total number of failure numbers is $740+202+22-67-7-2=888$ so our total number of succeeding numbers is $2222-888=$ **1334** possibilities.
-----------
<u>Interval 5</u>: $[3333, 9998]$ Because $3333$ shares the same prime factors as $9999$ , we can just take the totient function of $9999$ , multiplied by two thirds. $\frac{2}{3} \phi ( 9999) = \frac{2}{3} \cdot 9999 \cdot \frac{2}{3} \cdot \frac{10}{11} \cdot \frac{100}{101} = $ **4000** possibilities from this case.
----------
So our final answer is $100+776+182+1334+4000 = 6392$ -> $\boxed{392}$ .
-----
Unfortunately for me, I said that there are $6$ multiples of $303$ in the range $[1111,3332]$ instead of $7$ and ended up with an answer of $393$ . | [
"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 :... | [
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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 $a = 20$ and $b = 20$ to fail. Now we can check that there are no other arithmetic sequences only containing $a$ , or $b$ that fail.
Case $2$ : The arithmetic progression contains both $a$ and $b$ .
Then we have $(a, b) = (7, 9)$ fails from considering $(3, 5, a, b)$ . We also have $(a, b) = (10, 20)$ fails by taking $(a, b, 30, 40)$ but we have already counted this because $b \neq 20$ . Next assume we have an arithmetic sequence of the form $\{x, a, b, y\}$ . Then clearly $3 \mid y - x$ . Checking yields the possible triples $\{3, a, b, 30\}$ , $\{4, a, b, 40\}$ and $\{5, a, b, 50\}$ . These yield the bad combinations $(a, b) = (12, 21)$ , $(a, b) = (16, 28)$ and $(a, b) = (20, 35)$ which we do not need to care about due to the bounds on $(a, b)$ .
Now consider choosing $(a, b)$ from $\{7, 8, \dots, 19, 21, \dots, 29\}$ . We can do this in $\binom{22}{2}$ ways. However the pairs $(7, 9)$ , $(12, 21)$ and $(16, 28)$ are all bad. Our final count is then $231 - 4 = \boxed{228}$ . | [
"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 pair of blocks of the same color. For example, the arrangement $$ {\text {\bf R B B Y G G Y R O P P O}} $$ is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 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 the $2n$ numbers as $1, 2, 3, \dots 2n-1, 2n$ . We require the numbering of each color to be of different parities, so we can count each parity separately. $\square$ Plug in $n=6$ to get $6!^2$ . Since there are $\frac{12!}{2^6}$ total ways to order the blocks, our answer is $\boxed{\frac{16}{231}}$ upon simplification. $\blacksquare$ **Remark.** This problem :love: :love: | [
"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 permutation of $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 9=1080$ . No two factors of $1080$ have difference $1$ , contradiction.
So $9$ is in the denominator.
Case 1: $a,b,c$ all odd.
Then if $a,b,c=1,3,5$ , then $def=14$ , contradiction.
If $a,b,c=1,3,7$ , then $def=20$ , contradiction.
If $a,b,c=1,5,7$ , then $def=34$ , contradiction.
If $a,b,c=3,5,7$ , then $def=104$ , contradiction.
Case 2: $d,e,f$ all odd.
Then $def\in \{16,22,36,106\}$ . All except $36$ don't work. So $abc=36$ and $d,e,f,=1,5,7$ . So $a,b,c=2,3,6$ , which is what our answer was. | [
"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 $1148+d=566\implies d=-582$ .
Answer is $698-582=\boxed{116}$ .</blockquote>
yea I did it the same way but
OMG that's so smart!:
<blockquote>Mine.
<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></blockquote> | [
"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 $\big[ (1-x)(1-y)(1-z) \big] ^2$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 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 to the sum on the right - let this be $l$ for symmetry purposes. So, we note that if the angle opposite the side with length $\sqrt{2x}$ has a value of $\sin(\theta)$ , then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$ so $x=2\sin^2(\theta)$ and the triangle side with length $\sqrt{2x}$ is equal to $2\sin(\theta)$ .
We can symmetrically apply this to the two other triangles, and since by law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R \to R=1$ is the circumradius of that triangle. Hence. we calculate that with $l=1, \sqrt{2}$ , and $\sqrt{3}$ , the angles from the third side with respect to the circumcenter are $120^{\circ}, 90^{\circ}$ , and $60^{\circ}$ . This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$ , $x=2\sin^2(\beta)$ , and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$ , $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$ , and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$ . Solving, we get $\alpha=\frac{135^{\circ}}{2}$ , $\beta=\frac{105^{\circ}}{2}$ , and $\gamma=\frac{165^{\circ}}{2}$ .
We notice that $$ [(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2 $$ $$ =\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare $$ | [
"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--W, dashed);
draw(B--X, dashed);
draw(C--Y, dashed);
draw(D--Z, dashed);
label(" $A$ ", A, N);
label(" $B$ ", B, N);
label(" $Y$ ", Y, N);
label(" $Z$ ", Z, N);
label(" $C$ ", C, S);
label(" $D$ ", D, S);
label(" $W$ ", W, S);
label(" $X$ ", X, S);
label(" $P$ ", P, N);
label(" $Q$ ", Q, N);
[/asy]
Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$ , respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$ , respectively.
Claim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$ , $\angle ADP + \angle PAD = 90^{\circ}$ . Therefore, triangles $APD$ , $APZ$ , $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$ , so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.
Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$ , respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$ . Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$ , respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 675$ . Finally, $PQ = RS - RP - QS = \boxed{342}$ .</blockquote>
ummmmmm answer is $242$ i thought...</blockquote>
lmao i made a mistake originally bc i remembered the problem wrong
i thought that i saved the texer but ig not.
<details><summary>fixed now</summary>Diagram:
[asy]
unitsize(0.016cm);
pair A = (-250,324.4);
pair B = (250, 324.4);
pair C = (325, 0);
pair D = (-325, 0);
draw(A--B--C--D--cycle);
pair W = (8,0);
pair X = (-8, 0);
pair Y = (-83,324.4);
pair Z = (83,324.4);
pair P = (-121, 162.2);
pair Q = (121, 162.2);
dot(P);
dot(Q);
draw(A--W, dashed);
draw(B--X, dashed);
draw(C--Y, dashed);
draw(D--Z, dashed);
label(" $A$ ", A, N);
label(" $B$ ", B, N);
label(" $Y$ ", Y, N);
label(" $Z$ ", Z, N);
label(" $C$ ", C, S);
label(" $D$ ", D, S);
label(" $W$ ", W, SE);
label(" $X$ ", X, SW);
label(" $P$ ", P, N);
label(" $Q$ ", Q, N);
[/asy]
Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$ , respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$ , respectively.
Claim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$ , $\angle ADP + \angle PAD = 90^{\circ}$ . Therefore, triangles $APD$ , $APZ$ , $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$ , so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.
Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$ , respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$ . Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$ , respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 575$ . Finally, $PQ = RS - RP - QS = \boxed{242}$ .</details> | [
"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 positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ , respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^{\circ}$ . Find the perimeter of $\triangle ABC$ . | <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 $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ , respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^{\circ}$ . Find the perimeter of $\triangle ABC$ .</blockquote>
<details><summary>Non-rigorous solution</summary>From drawing a diagram it seems that $a$ must be longest, after which the solution should be rigorous. Call the sides $a=BC=219,b=AC,c=AB$ .**Claim:** $\angle A=120^\circ.$ *Proof:* Let $X,Y$ be points on line $BC$ with $X,B,C,Y$ in that order with $CX=a+c,BY=a+b$ . It is easily verifiable that the line through the midpoints of $\overline{CX}$ and $\overline{AC}$ is the splitting line of $N$ wrt $\triangle ABC$ . By midlines the said splitting lines of $N,M$ are parallel to $\overline{AX},\overline{AY}$ respectively. The given condition implies that we want $\angle XAY=150^\circ.$ The length conditions imply $BX=c=AB,CY=b=AC$ , so isosceles triangles imply $\angle BAX=\angle BXA=\angle B/2$ , and similarly $\angle CAY=\angle CYA=\angle C/2$ .
Finally $150^\circ=\angle XAY=\angle A+\angle B/2+\angle C/2=(A+180^\circ)/2$ , from which we derive the claimed statement. $\qquad\square$
The problem reduces to solving the Diophantine equation $b^2+bc+c^2=219^2$ in positive integers $b,c$ . A lengthy enumeration gives us $(51,189)$ , yielding a perimeter of $219+51+189=\boxed{459}$ .**Remark:** The last Diophantine equation is more easily solved by replacing $219$ with $219/3=73$ , then multiplying those solutions by $3$ . It is then tractable to discover $b=63$ works in $b^2+bc+c^2=73$ .</details> | [
"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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