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"
} |
Let $w = \frac{\sqrt{3}+i}{2}$ and $z=\frac{-1+i\sqrt{3}}{2}$ , where $i=\sqrt{-1}$ . Find the number of ordered pairs $(r, s)$ of positive integers not exceeding $100$ that satisfy the equation $i\cdot w^r=z^s$ . | <details><summary>Solution</summary>We want $i \cdot w^r = z^s$ . Re-writing $i, w,$ and $z$ in exponential form, we see that $e^{i \cdot \frac{\pi}{2}} \cdot e^{i \cdot \frac{\pi}{6} \cdot r} = e^{i \cdot \frac{2\pi}{3} \cdot s}$ . Simplifying, we have that $4s = r +3$ . From this, we have the pairs $(1, 1), (5, 2), (9, 3), \ldots, (97, 25)$ . BUT, and this is a big BUT, we must notice that any $r \pm 12$ (as you can probably see, this won't change anything in our list of pairs) and any $s \pm 3$ also work. (This is because, in exponential form, $\theta$ and $\theta + 2\pi$ result in the same complex number.) So, we see that our final group of pairs is: $$ (1, 1), (1, 4), (1, 7), \ldots, (1, 100) $$ $$ (5, 2), (5, 5), (5, 8), \ldots, (5, 98) $$ $$ (9, 3), (9, 6), (9, 9), \ldots, (9, 99) $$ $$ (13, 1), (13, 4), (13, 7), \ldots, (13, 100) $$ $$ . $$ $$ . $$ $$ . $$ $$ (97, 1), (97, 4), (97, 7), \ldots, (97, 100). $$ Which is a total of $34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34$ $= 8(34 + 33 + 33) + 34 = 800 + 34 = \boxed{834}$ .</details> | [
"Note that $i \\cdot w^{r}$ is periodic for $r \\bmod 12$ and $z^s$ is periodic for $s \\bmod 3$ . ",
"Note that $z = w^4$ and get $4s-r \\equiv 3 \\pmod{12}$ ",
"i got <details><summary>this</summary>$w=e^{\\pi i/6}, z=e^{2\\pi i/3},$ then the equation becomes $ie^{r\\pi i/6}=e^{2s\\pi i/3},$ then... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 57,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777219.json"
} |
A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ . | I think this would have been the intended solution:
[hide = Legit solution that never uses the word "vector"]
Define $m$ as the number of minutes they swam for.
Let their meeting point be $A$ . In an alternative reality, there would be no current. Then, had they swum facing the same direction that they had in the real universe, they would've met at a point west of $A$ . Precisely, since the water moves at $14$ meters per minute, this alternative reality meeting point would have been $14m$ meters to the left of $A$ .
So, our alternative reality is just a geometry problem now:
[asy]
unitsize(0.02cm);
draw((0,0)--(0,264)--(550,264)--(550,0)--cycle);
pair B = (198,264);
dot(B);
draw((0,0)--B,dashed);
draw((550,0)--B,dashed);
label(" $60m$ ", (0,0)--B, E);
label(" $80m$ ", (550,0)--B, W);
label(" $264$ ", (0,0)--(0,264), W);
label(" $\frac{D}{2} - 14m$ ", (0,264)--B, N);
label(" $\frac{D}{2} + 14m$ ", B--(550,264), N);
label(" $D$ ", (0,0)--(550,0), S);
[/asy]
(I skipped many steps here, but it should be somewhat self-explanatory where all these numbers came from)
Note that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides $60m$ , $80m$ and $D$ is a right triangle yet, so we cannot use that information.
By Pythagorean, we have
\begin{align*}
264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2}
264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}.
\end{align*}
Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that
\begin{align*}264^{2} + 36^{2} m^{2} &= 60m^{2}
264^{2} &= 96 \cdot 24 \cdot m^{2}
11^{2} &= 4 \cdot m^{2}
m &= \frac{11}{2}.
\end{align*}
So $D = 100m = \boxed{550}$ .
[/hide]
~~Please upvote this post, as it took me like forever to do the asymptote diagram.~~ | [
"when you need ap physics vectors knowledge on the AIME ",
"It's just a trapezoid",
"answer is 550?",
"when you put 275 because you forgot to times 2",
"<blockquote>when you need ap physics vectors knowledge on the AIME</blockquote>\n\nyes wth even is this problem :sadge:",
"Consider the perspective of th... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1036,
"boxed": false,
"end_of_proof": false,
"n_reply": 100,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777223.json"
} |
Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega$ . Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A$ , $\omega_B$ , and $\omega_C$ meet in six points $-$ two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC$ , and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC$ . The side length of the smaller equilateral triangle can be written as $\sqrt{a}-\sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a+b$ . | another solution but too many point names :(
Let $O_A$ be the center of the circle tangent to $AB$ , $AC$ , and $\omega$ (name this circle $\Gamma_A$ , and define $\Gamma_B$ , $\Gamma_C$ similarly). Let $r$ be its radius, and $E$ the point of tangency of $\Gamma_A$ to $AB$ . Let $D$ be the point diametrically opposite $A$ in $\omega$ , $X$ and $Y$ the points of intersection of $\Gamma_A$ with $BC$ where $BX < BY$ .
Then PoP gives $AE^2=AP\cdot AD=36(36-2r)$ , and $BE^2=BX\cdot BY$ . We’ll have to figure out $BX$ and $BY$ . Define another point— let $M$ be the midpoint of $BC$ . $XO_A=r$ and $O_AM=r-9$ , so $XM=\sqrt{r^2-(r-9)^2}=\sqrt{18r-81}$ . Now $BC=18\sqrt3$ , $BX=9\sqrt3-\sqrt{18r-81}$ , and $BY=9\sqrt3+\sqrt{18r-81}$ , hence $BE^2=324-18r$ . $BE+AE=AB=18\sqrt3$ , so solving $\sqrt{36(36-2r)}+\sqrt{324-18r}=18\sqrt3$ gives $r=12$ .
Let $K$ be the intersection of $\Gamma_B$ and $\Gamma_C$ that’s closer to $O$ . We want $OK$ . Let $O_K$ be the center of the circle tangent to $BA$ , $BC$ , and $\omega$ . It’s easy to see that $\angle O_KOK=120$ , so by the Law of Cosines,
\[OO_K^2+OK^2+OO_K\cdot OK=O_KK^2.\]
Note $OO_K=6$ and $O_KK=12$ , so we have $OK^2+6OK-108=0$ , hence $OK=\sqrt{117}-3$ . Thus the desired side length is $\sqrt3(\sqrt{117}-3)=\sqrt{351}-\sqrt{27}$ and the answer is $\boxed{378}$ . | [
"got $\\sqrt{351}-\\sqrt{27}$ for 378",
"Use $30-60-90$ triangles to find the smaller radii of $w_{a}, w_{b}, w_{c}$ , $r$ . We obtain $r=\\frac{36-r}{2} \\implies r = 12$ . Let the intersection that yields the smaller equilateral triangles of circles $w_{a}$ and $w_{b}$ be $I_{c}$ . Since the distanc... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1102,
"boxed": true,
"end_of_proof": false,
"n_reply": 64,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777225.json"
} |
Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ . | <details><summary>Solution</summary>The idea is to take the cross sections that pass through two of $A$ , $B$ , and $C$ , and their corresponding sphere centers, which I'm labelling $X$ , $Y$ , and $Z$ , respectively
Let $r$ be the radii of the congruent circles. let $d_X$ , $d_Y$ , and $d_Z$ be the positive distances from $X$ , $Y$ , and $Z$ , respectively, to the plane.
Note that $d_Y-d_X$ is the length of one of the legs of a right triangle with hypotenuse $XY$ and third side equal to $AB$ . Therefore $d_Y-d_X=\sqrt{(11+13)^2-560}=4$ .
As well, note that $d_X^2+r^2=121$ and $d_Y^2+r^2=169$ , so $d_Y+d_X=\frac{169-121}{4}=12$ , and therefore $d_Y=\frac{12+4}{2}=8$ , $d_X=\frac{12-4}{2}=4$ , and $r=\sqrt{121-16}=\sqrt{105}$ .
Therefore $d_Z=\sqrt{361-105}=16$ , so $AC^2=(11+19)^2-(16-4)^2=\boxed{756}$ .</details> | [
"Pretty misplaced imo - just take the 11-13 and 11-19 cross sections separately where the circles are congruent lines and just do 2D Pythag geo\n\nTook like 2 mins on the test",
"DeToasty3... legend\n\nanyways the numbers were very clean (4,8,16) as heights\n\n@below A,B, and C are projections, so they shouldn't ... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1050,
"boxed": true,
"end_of_proof": false,
"n_reply": 38,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777231.json"
} |
Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$ . A circle tangent to sides $\overline{DA}$ , $\overline{AB}$ , and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$ , as shown. Suppose that $AP = 3$ , $PQ = 9$ , and $QC = 16$ . Then the area of $ABCD$ can be expressed in the form $m\sqrt n$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
[asy]
defaultpen(linewidth(0.6)+fontsize(11));
size(8cm);
pair A,B,C,D,P,Q;
A=(0,0);
label(" $A$ ", A, SW);
B=(6,15);
label(" $B$ ", B, NW);
C=(30,15);
label(" $C$ ", C, NE);
D=(24,0);
label(" $D$ ", D, SE);
P=(5.2,2.6);
label(" $P$ ", (5.8,2.6), N);
Q=(18.3,9.1);
label(" $Q$ ", (18.1,9.7), W);
draw(A--B--C--D--cycle);
draw(C--A);
draw(Circle((10.95,7.45), 7.45));
dot(A^^B^^C^^D^^P^^Q);
[/asy] | <blockquote>Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$ . A circle tangent to sides $\overline{DA}$ , $\overline{AB}$ , and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$ , as shown. Suppose that $AP = 3$ , $PQ = 9$ , and $QC = 16$ . Then the area of $ABCD$ can be expressed in the form $m\sqrt n$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
[asy]
defaultpen(linewidth(0.6)+fontsize(11));
size(8cm);
pair A,B,C,D,P,Q;
A=(0,0);
label(" $A$ ", A, SW);
B=(6,15);
label(" $B$ ", B, NW);
C=(30,15);
label(" $C$ ", C, NE);
D=(24,0);
label(" $D$ ", D, SE);
P=(5.2,2.6);
label(" $P$ ", (5.8,2.6), N);
Q=(18.3,9.1);
label(" $Q$ ", (18.1,9.7), W);
draw(A--B--C--D--cycle);
draw(C--A);
draw(Circle((10.95,7.45), 7.45));
dot(A^^B^^C^^D^^P^^Q);
[/asy]</blockquote>
<details><summary>Long-winded solution</summary>[asy]
import olympiad;
size(250);
defaultpen(linewidth(0.7)+fontsize(10));
real h = 5, r = 3*sqrt(3), s = 26;
pair A = (0,0), B = (2,5), X = (B.x,0), C = (12,5), D = A+C-B;
real r = inradius(A,B,X);
path P = scale(B.y/(2*r))*incircle(A,B,X);
pair I = B.y/(2r) * incenter(A,B,X);
pair X = foot(I,A,D), Y = foot(I,B,C), Z = foot(I,A,B), R = foot(C,A,D);
draw(X--Y,gray(0.5));
draw(A--B--C--D--A);
draw(P);
pair[] inter = intersectionpoints(P,A--C);
draw(A--C);
label(" $A$ ",A,SW);
label(" $B$ ",B,NW);
label(" $C$ ",C,NE);
label(" $D$ ",D,SE);
label(" $P$ ",inter[1],dir(250));
label(" $Q$ ",inter[0],dir(330));
label(" $X$ ",X,S,gray(0.5));
label(" $Y$ ",Y,N,gray(0.5));
label(" $Z$ ",Z,NW,gray(0.5));
pair O=(X+Y)/2;
draw((O-(0.2,0))--(O+(0.2,0)),gray(0.5));
label(" $O$ ",O,dir(45),gray(0.5));
draw(A--O--Z^^O--B,RGB(13,83,76));
draw(rightanglemark(O,Z,A));
[/asy]
As in djmathman's solution, let $X,Y,Z$ be the touchpoints of $\omega$ on $\overline{AD},\overline{BC},\overline{AB}$ respectively. As mentioned above one may verify by PoP that $AX=6,CY=20$ .**Claim:** $\omega$ has radius $3\sqrt3$ .
*Proof:*By similar triangles $PX/QX=1/2$ and $PY/QY=5/4$ so we may let $PX=a,QX=2a$ and $PY=5b,QY=5b$ for some positive reals $a,b$ .
Because $\overline{AD}\parallel\overline{BC}$ , $\overline{XY}$ is a diameter so $\angle XZY=\angle XQY=90^\circ$ .
Pythagoras implies
\[a^2+(5b)^2=XY^2=(2a)^2+(4b)^2\]
whence $a=b\sqrt3$ .
We may derive a second equation in $a,b$ by noting that angles $PAX$ and $QCB$ are equal and hence have the same cosines; LoC implies
\[\frac{3^2+6^2-a^2}{2\cdot3\cdot6}=\frac{4^2+5^2-b^2}{2\cdot4\cdot5}.\]
Expressing everything in terms of $b$ gives $7b^2/120=9/40\Rightarrow b=3\sqrt3/\sqrt7$ . Hence $a=9/\sqrt7$ and the diameter evaluates as
\[XY=\sqrt{a^2+(5b)^2}=6\sqrt3.\qquad\qquad\square\]
Returning to the original problem, let $O$ be the center of $\omega$ . Observing that $\angle AOB=\angle AOZ+\angle ZOB=\angle XOZ/2+\angle ZOY=\pi/2$ , similar right triangles implies $BY=BZ=OZ^2/AZ=9/2.$ Now we may compute the area as $BC\cdot XY=(20+9/2)(6\sqrt3)=147\sqrt3\Rightarrow\boxed{150}$ .</details> | [
"Mine.\n\n<details><summary>Solution</summary>Let $X$ , $Y$ , and $Z$ denote the tangency points of $\\omega$ with $AD$ , $BC$ , and $AB$ , respectively. Furthermore, let $R$ be the foot of the perpendicular from $C$ to $AD$ .\n\n[asy]\n\timport olympiad;\nsize(250);\ndefaultpen(linewidth(0.7)+fontsi... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1220,
"boxed": true,
"end_of_proof": false,
"n_reply": 54,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777232.json"
} |
For positive integers $a$ , $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ , $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ . | First, we note that $a=1$ , otherwise we can not make 1 cent. We must also have $b-1$ of these, as otherwise, we can not make $b-1$ cents (any more are useless, because it is more optimal to take $b$ ).
Now, we need to make every value from $b$ to $c-1$ with only $a$ and $b$ as denominations. We know we have $b-1$ coins with worth $a$ , and assume we have $k$ with worth $b$ . Then, we know that
\[kb+b-1\geq c\implies k\geq\left\lceil\frac{c+1}b\right\rceil-1\]
where this is clearly sufficient, as the coins with denomination $a$ form every residue class. Similarly, we can already form every residue class modulo $c$ , so we just need enough coins of denomination $c$ . In particular, if we have $\ell$ coins of worth $c$ , then
\[\underbrace{\ell c}_{\text{coins with denomination $c$ }}+\underbrace{c-1}_{\text{all other coins}}\geq 1000\implies \ell\left\lceil\frac{1001}c\right\rceil-1\]
However, note that that we could have some extra coins, meaning our count could be off by one. Thus, we get that
\[f(a,b,c)=b+\left\lceil\frac{c+1}b\right\rceil+\left\lceil\frac{1001}c\right\rceil-3(-1)\]
so thus we want to find the solutions to
\[b+\left\lceil\frac{c+1}b\right\rceil+\left\lceil\frac{1001}c\right\rceil=100(+1)\]
Clearly, we need $\left\lceil\frac{1001}c\right\rceil<100$ , so $c>10$ . We shall find all solutions.
Now, we know that $2\leq b\leq c-1$ , and that $b+\frac{c+1}b$ has its maximum at the endpoints. Thus, we get that the maximum value of the function is either
\[c-1+\left\lceil\frac{c+1}{c-1}\right\rceil=c+1\qquad\text{or}\qquad 2+\left\lceil\frac{c+1}2\right\rceil\]
and the first is obviously bigger. Thus, we know that
\[c+1+\frac{1001}c+1> c+1+\left\lceil\frac{1001}c\right\rceil\geq 100\]
so thus
\[1001+2c+c^2>100c\]
Solving this, we get that
\[c<49-10\sqrt{14}\qquad c>49+10\sqrt{14}\]
Approximating, we get $c\leq 11$ or $c\geq 87$ . We can verify the ordered pair $(1,7,11)$ works. $(1,86,87)$ doesn't because we need only 10, not 11, stamps of denomination 87. We verify and $(1,87,88)$ and $(1,87,89)$ work, so the smallest values of $c$ sum to $11+88+89=\boxed{188}$ . | [
"<blockquote>First, we note that $a=1$ , otherwise we can not make 1 cent. We must also have $b-1$ of these, as otherwise, we can not make $b-1$ cents (any more are useless, because it is more optimal to take $b$ ).\n\n\nNow, we need to make every value from $b$ to $c-1$ with only $a$ and $b$ as denom... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1060,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782913.json"
} |
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ , $O_1O_2 = 15$ , $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon.
[asy]
import geometry;
size(10cm);
point O1=(0,0),O2=(15,0),B=9*dir(30);
circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0];
filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black);
draw(w1);
draw(w2);
draw(O1--O2,dashed);
draw(o);
dot(O1);
dot(O2);
dot(A);
dot(D);
dot(C);
dot(B);
label(" $\omega_1$ ",8*dir(110),SW);
label(" $\omega_2$ ",5*dir(70)+(15,0),SE);
label(" $O_1$ ",O1,W);
label(" $O_2$ ",O2,E);
label(" $B$ ",B,N+1/2*E);
label(" $A$ ",A,N+1/2*W);
label(" $C$ ",C,S+1/4*W);
label(" $D$ ",D,S+1/4*E);
label(" $15$ ",midpoint(O1--O2),N);
label(" $16$ ",midpoint(C--D),N);
label(" $2$ ",midpoint(A--B),S);
label(" $\Omega$ ",o.C+(o.r-1)*dir(270));
[/asy] | <details><summary>Solution</summary>I claim the answer is $\boxed{140}$ . First,**<span style="color:blue">Claim:</span>** Let $\ell$ be the radical axis of $\omega_1$ and $\omega_2$ . Then $BC$ , $AD$ and $\ell$ concur.
*Proof.* Immediate from Radical Axis theorem on $\omega_1$ , $\omega_2$ , and $\Omega$ . $\square$ **<span style="color:blue">Claim:</span>** The line $CD$ is parallel to the reflection of $AB$ across $O_1O_2$ .
*Proof.* I claim that $\ell$ is the internal angle bisector of $\angle(\overline{BC},\overline{AD})$ . (One can gues this claim from a good diagram/intuition).
Indeed, suppose $BC$ and $AD$ intersct at $T$ , and furthermore, $BC$ intersects $O_1O_2$ at $X$ and $AD$ intersects $O_1O_2$ at $Y$ . Then using the fact that $O_1$ and $O_2$ are the arc midpoints of $BC$ and $AD$ , we find \[ \angle TXY = \angle CXO_1 = 180 - \angle O_1CO_2 \] where we use the so-called Shooting Lemma which gives $\triangle O_1XC \sim \triangle O_1CO_2$ . Similarly, \[ \angle TYX = \angle DYO_2 = 180 - \angle O_2DO_1 \] from which it follows that $TX = TY$ .
So the foot from $T$ to $O_1O_2$ bisects $\angle XTY$ and $TY$ , or equivalently $\ell$ bisects $\angle XTY$ .
But it is well-known that in cyclic quadrilateral $ABCD$ , the internal angle bisector of $\angle(\overline{BC},\overline{AD})$ is perpendicular to the internal angle bisector of $\angle(\overline{AB},\overline{CD})$ ; this implies that $O_1O_2$ is parallel to the latter, which is equivalent to the fact that $CD$ is parallel to the reflection of $AB$ across $O_1O_2$ . $\square$ Thus, we consider the reflections $C'$ and $D'$ of $C$ and $D$ across the perpendicular bisector of $O_1O_2$ respectively. Then
- By the claim above, it follows $C'D' \parallel AB$ ,
- Thus we have $BD' = AC'$ ,
- We have $O_1B = O_2C'$ , so $15 = O_1O_2 = BC'$ ,
- We have $O_2A = O_1D'$ , so $15 = O_2O_1 = AD'$ ,
- Thus we have $\triangle O_1BD' \cong \triangle O_2C'A$
But then $ABC'D'$ is an isosceles trapezoid with bases of length $2$ and $16$ and diagonals of length $15$ . Then we can easily solve for the legs to be length $\sqrt{193}$ .
Moreover, $O_1B + O_1D' = 15$ , yet $\triangle O_1BD'$ has side lengths $O_1B$ , $O_1D'$ and $\sqrt{193}$ . Note also that \[ \theta \coloneqq \angle BO_1D' = 180 - \angle BC'D' \] so it follows that $\sin \theta = \frac 45$ and $\cos \theta = -\frac 35$ ; then \[ 193 = O_1B^2 + O_1D'^2 + 2\cdot O_1B\cdot O_1D' \cdot \frac 35 = (O_1B+O_1D')^2 - 2\cdot O_1B\cdot O_1D' \cdot \frac 25 \] which is from Law of Cosines. Then we can solve to find $O_1B\cdot O_1D' = 40$ ; then \[ [O_1BD'] = \frac 12 \cdot O_1B\cdot O_1D' \cdot \frac 45 = \frac 12 \cdot 40 \cdot \frac 45 = 16 \]
Then it follows that \[ [ABO_1CDO_2] = [O_1BAO_2C'D'] = 2\cdot [O_1BD'] + [BAC'D'] = 2\cdot 16 + \frac{2+16}{2}\cdot 12 = 32 + 108 = 140 \] as claimed.</details>
<details><summary>Motivation</summary>The first claim is olympiad-flavored; then from a good diagram/intuition, since $O_1$ and $O_2$ are arc midpoints, it seemed like O_1O_2 should be an angle bisector of $AB$ and $CD$ , which would be equivalent to $\ell$ being an angle bisector. But the concurrency and some angle chasing showed this to actually be true.
From that claim (and a little courage), I was inspired to reflect $CD$ so I could gain a parallelism and thus an isosceles trapezoid, as the total area is also preserved. The rest falls through.</details> | [
"Nice problem. Didn't even try synthetic because well, who cares about trying geometry.\n\nLet the radius of $\\omega_1$ be $r$ and that of $\\omega_2$ be $15-r$ . We recall the following important fact: in cyclic quadrilateral $ABCD$ with sides $a,b,c,$ and $d$ , with circumradius $R$ , area $A$ , an... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1252,
"boxed": true,
"end_of_proof": true,
"n_reply": 26,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782914.json"
} |
Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ .
[asy]
import geometry;
size(10cm);
draw((-2,0)--(13,0));
draw((0,4)--(10,4));
label(" $\ell_A$ ",(-2,0),W);
label(" $\ell_B$ ",(0,4),W);
point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2);
draw(B1--A1--B2);
draw(B1--A2--B2);
draw(B1--A3--B2);
label(" $A_1$ ",A1,S);
label(" $A_2$ ",A2,S);
label(" $A_3$ ",A3,S);
label(" $B_1$ ",B1,N);
label(" $B_2$ ",B2,N);
label("1",centroid(A1,B1,I1));
label("2",centroid(B1,I1,I3));
label("3",centroid(B1,B2,I3));
label("4",centroid(A1,A2,I1));
label("5",(A2+I1+I2+I3)/4);
label("6",centroid(B2,I2,I3));
label("7",centroid(A2,A3,I2));
label("8",centroid(A3,B2,I2));
dot(A1);
dot(A2);
dot(A3);
dot(B1);
dot(B2);
[/asy] | <details><summary>Click to expand</summary>When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points.
WLOG, let the point be on line $\ell_A$ . We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$ . Let there be $s$ points on lines $\ell_A$ and $\ell_B$ before $P$ was added. For each of the $s$ points on $\ell_B$ , we subtract the total number of lines formed, which is $s^2$ , not counting $P$ . Considering all possible points on $\ell_B$ , we get $(s^2-s)+(s^2-2s)\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$ . Simplifying, we get $s^3-s\sum_{i=1}^{s}{i}+s\Longrightarrow s(s^2-\sum_{i=1}^{s}{i}+1)$ . Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\ell_A$ .
Case 2: If a new point $P$ is added to the right of a line that has one less point than the other line.
Continuing on case one, let this point $P$ be on line $\ell_B$ . With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$ , we get $(s(s+1)-s)+(s(s+1)-2s)\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\ell_A$ . Simplying, we get $s(s+1)^2-s\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions.
For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$ . Notice that adding a final point on $\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$ , there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\cdots(s(s+1)-s(s+1))+s\Longrightarrow s^2(s+1)-(s+1)\sum_{i=1}^{s}{i}+s$ . Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=\boxed{244}$ .
.</details>
| [
"<details><summary>Sketch</summary>Let $f(m, n)$ denote the number of bounded regions. Then $f$ satisfies the recursion $f(m+1, n) = f(m, n) + m\\binom{n}{2} + n$ . Compute $f(5, 7) = 244$ .\n\nEdit: By using $f(1, 1) = 0$ , we can derive the explicit formula as shown in the below post.</details>",
"Person... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1102,
"boxed": false,
"end_of_proof": false,
"n_reply": 46,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782916.json"
} |
Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ , $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ . | <blockquote>
Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\lfloor \tfrac n4\rfloor$ , $\lfloor\tfrac n5\rfloor$ , and $\lfloor\tfrac n6\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ .
</blockquote>
this solution's relatively dumb oops
Let $\lfloor\tfrac n4\rfloor = a$ , $\lfloor\tfrac n5\rfloor = b$ , $\lfloor\tfrac n6\rfloor = c$ . Then $4a\le n \le 4a+3$ , $5b\le n \le 5b+4$ , $6c\le n \le 6c + 5$ ; and if we want $n$ to be uniquely determined, we require these bounds be "tight", in the sense that there's a lot of equality. This means
\[\max(4a, 5b, 6c)=\min(4a+3, 5b+4, 6c+5)\]
so.. casework.
- $n=4a$ : $n$ clearly cannot equal $4a+3$ or $6c+5$ , as both are odd. So $4a=5b+4$ , giving $n=4$ , $24$ , $44$ , $\dots$ , $584$ -- $30$ solutions.
- $n=5b$ : it must equal $4a+3$ or $6c+5$ .
[list]
- If $4a+3$ : $n$ is $15$ , $35$ , $55$ , $\dots$ , $595$ -- $30$ solutions.
- If $6c+5$ : $n$ is $5$ , $35$ , $65$ , $\dots$ , $575$ -- $20$ solutions.
[/list]
- $n=6c$ : then $5b+4$ , so $n$ is $4$ , $34$ , $64$ , $\dots$ , $574$ -- $20$ solutions.
If not for the overlapping between $n=4a$ and $n=6c$ , or between $5b=4a+3$ and $5b=6c+3$ , we'd have $100$ as our answer. Between $4a$ and $6c$ are $n=4$ , $64$ , $94$ , $\dots$ , $544$ -- $10$ values. Between $4a+3$ and $6c+3$ are $35$ , $95$ , $155$ , $\dots$ , $575$ , so $10$ solutions. Then the answer is $100-10-10=\boxed{080}$ . | [
"<details><summary>Solution</summary>Let $\\lfloor \\frac{n}{4} \\rfloor = a$ , and similarly for $b$ and $c$ . We need to find the number of integers such that $n$ increases the value of at least one of $a$ , $b$ , or $c$ (from $n-1$ ), and $n+1$ also increases the value of at least one of $a$ , $b$... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1148,
"boxed": false,
"end_of_proof": false,
"n_reply": 42,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782918.json"
} |
A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | [hide = MAA Should be Replaced as the "Misplaced Armadillo Association"!]
First notice that $DO_1$ is a straight line because $DXY$ is an isosceles triangle(or you can realize it by symmetry). That means, because $DO_1$ is a straight line, so angle $BDO_2$ = angle $ADO_1,$ triangle $ADO_1$ is similar to triangle $BDO_2$ . Also name $DO_2 = x$ . By our similar triangles, $\frac{BO_2}{AO_1} = \frac{1}{4} = \frac{x}{x+30}$ . Solving we get $x = 10 = DO_2$ . Pythagorean Theorem on triangle $DBO_2$ shows $BD = \sqrt{10^2 - 6^2} = 8$ . By similar triangles, $DA = 4 \cdot 8 = 32$ which means $AB = DA - DB = 32 - 8 = 24$ . Because $BE = CE = AE, AB = 2 \cdot BE = 24$ . $BE = 12,$ which means $CE = 12$ . $CD = DO_2$ (its value found earlier in this solution) + $CO_2$ ( $O_2$ 's radius) $= 10 + 6 = 16$ . The area of $DEF$ is $\frac{1}{2} \cdot CD \cdot EF = CD \cdot CE$ (because $CE$ is $\tfrac{1}{2}$ of $EF$ ) $= 16 \cdot 12 = 192$ .
[/hide] | [
"wow so misplaced",
"This looks so easy ... I didn't do it ofc but ...\n\nP7?\n\nMake up your mind, MAA. P10 on the AIME I and now this?",
"Bruh..........",
"lol what\ntwo 3-4-5's similar triangles 16*12=192",
"this would make problem 7 on an amc 10/12",
"Who else did coordbash",
"Overkilled by incircl... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782923.json"
} |
Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Let $T_b$ and $T_a$ denote the number of people at the concert before and after the bus's arrival, respectively. Since $\tfrac{11}{25}T_a$ is the number of adults after the bus's arrival, $T_a$ is a multiple of $25$ . But $T_a-T_b=50$ , the number of people on the bus, is also a multiple of $25$ , so $T_b$ is a multiple of $25$ . Furthermore, as $\tfrac{5}{12}T_b$ is the number of adults before the bus's arrival, $T_b$ is a multiple of $12$ as well. Therefore, $T_b$ is at least $\text{lcm}(12,25)=300$ , $T_a$ is at least $300+50=350$ , and the number of adults after the bus's arrival is at least $\tfrac{11}{25}T_a=\boxed{154}$ , as requested. | [
"<details><summary>AAAAAAAAAAAA</summary>$x = 0 \\pmod{12}$ $x + 50 = 0 \\pmod{25} \\to x = 0 \\pmod{25}$ $x = 0 \\pmod{300} \\to x = 300$ $\\text{adults}_0 = 125$ $\\text{adults} = \\boxed{154}$</details>",
"Technically we could have had an initial crowd of $0$ , followed by $22$ adults out of $50$ arri... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1034,
"boxed": false,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782926.json"
} |
Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | Denote by $P$ the midpoint of $\overline{CD}$ . The key claim is as follows.**<span style="color:#f00">Claim:</span>** $\triangle ABP\sim \triangle APD\sim\triangle PCD$ *Proof.* Suppose lines $AB$ and $CD$ meet at $E$ . Since two angle bisectors of $\triangle ADE$ intersect at $P$ , the third one also passes through $P$ . Hence, $\overrightarrow{EP}$ is the angle bisector of $\angle AED$ . Now, looking at $\triangle EBC$ , we see that $\overline{EP}$ is both an angle bisector and a median, so $\triangle EBC$ must be isosceles.
Now, let $\angle BAD = 2\theta$ and $\angle CAD = 2\phi$ . We compute $\angle AED = 180^\circ - 2\theta-2\phi$ and consequently $\angle EBC = \angle ECB = \theta + \phi$ . Hence, $\angle ABP = \angle PCD = 180^\circ - \theta - \phi$ . Also, $\angle BAP = \theta$ and $\angle CDP = \phi$ , so $\angle BPA = \phi$ and $\angle DPC = \theta$ . Hence, $\triangle ABP\sim\triangle PCD$ by AA similarity. Furthermore, $\angle PAD = \theta$ and $\angle PDA = \phi$ , so these two triangles are also similar to $\triangle APD$ by AA, as desired $\square$ Let $BP = CP = x$ . From similarity, we have $\frac{AB}{x} = \frac{x}{BC}\implies x = \sqrt{AB\cdot BC} = \sqrt{6}$ . Now, let $a = AP$ and $b = BP$ . Similarity gives $\frac{AB}{a} = \frac{a}{AD}$ and $\frac{CD}{b} = \frac{b}{AD}$ . Hence, $a = \sqrt{AB\cdot AD} = \sqrt{14}$ and $b = \sqrt{CD\cdot AD} = \sqrt{21}$ . Now, Heron's gives $[ABP] = \sqrt{5}$ . By similarity, we have $[ADP] = (\sqrt{14}/2)^2[ABP]$ and $[PCD] = (\sqrt{3/2})^2[ABP]$ . Thus, summing the three areas, we obtain
\[ [ABCD] = \left(1 + (\sqrt{14}/2)^2 + (\sqrt{3/2})^2\right)\cdot \sqrt{5} = \sqrt{180},\]
and squaring gives $\boxed{180}$ . | [
"Mine.\n\nHere is a very \"European\" styled solution. There exists a very \"American\" styled solution, too, that I came up with first; I'll let someone else post that method.\n\n<details><summary>Solution</summary>Let $E$ and $F$ denote the reflections of $B$ and $C$ across $AM$ and $DM$ , respectivel... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1180,
"boxed": true,
"end_of_proof": false,
"n_reply": 43,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782927.json"
} |
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | We proceed with casework based on Carl's first opponent.
If Carl's first opponent is Azar, which occurs with probability $\tfrac{1}{3}$ , Carl first beats Azar in the semifinals with probability $\tfrac{1}{3}$ , then beats either Jon or Sergey in the finals with probability $\tfrac{3}{4}$ , so the total probability here is $\tfrac{1}{3}\cdot\tfrac{1}{3}\cdot\tfrac{3}{4}=\tfrac{1}{12}$ .
If Carl's first opponent is Jon or Sergey, which occurs with probability $\tfrac{2}{3}$ , Carl first beats Jon or Sergey in the semifinals with probability $\tfrac{3}{4}$ . If his opponent in the finals is Azar with a $\tfrac{3}{4}$ chance, he wins in the finals with probability $\tfrac{1}{3}$ , but if his opponent in the finals is the other of Jon and Sergey with a $\tfrac{1}{4}$ chance, he wins with a probability of $\tfrac{3}{4}$ . Hence, the total the probability here is $$ \frac{2}{3}\cdot\frac{3}{4}\cdot\left(\frac{3}{4}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{3}{4}\right)=\frac{7}{32}. $$ In all, the probability that Carl wins the tournament is $\tfrac{1}{12}+\tfrac{7}{32}=\tfrac{29}{96}$ , which makes the requested answer $\boxed{125}$ . | [
"Harder than p12 tbh.\n\nTook me ten minutes to realize it doesn't matter whether Jon or Sergey wins :P ",
"Anyone else got $\\frac{25}{96}$ ?",
"I got 29/96",
"yeah i got that too\nidk why i said i got 25/96? \n\n<details><summary>AAAAAAAAAAAAAAAA</summary>1. azar vs carl in first round $\\to \\frac{1}{3}... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782928.json"
} |
A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | Let the pyramid be $ABCDE$ with $O$ the center of its circumsphere, where square $BCDE$ is the base and has center $P$ . Furthermore, let $R$ denote the circumradius of $ABCDE$ .
As $ABCDE$ has volume $54$ , while its base $BCDE$ has area $6^2=36$ , the corresponding height $AP$ has length $\tfrac{3\cdot54}{36}=\tfrac{9}{2}$ . As a result, $OP=AP-AO=\tfrac{9}{2}-R$ , and since $BP=3\sqrt{2}$ , by the Pythagorean theorem on $BOP$ we have that $$ BO=\sqrt{BP^2+OP^2}=\sqrt{(3\sqrt{2})^2+\left(\frac{9}{2}-R\right)^2}=\sqrt{R^2-9R+\frac{153}{4}}. $$ Setting this equal to $R$ and squaring, $R^2-9R+\tfrac{153}{4}=R^2$ , so $R=\tfrac{17}{4}$ . The requested answer is then $\boxed{021}$ . | [
"cross section trivializes",
"I got 21? 17/4?",
"I didn't take the test(87 on 10A and 10B) , but man does this look easy...\nCircumradius formula and cross-section trivializes :|",
"Also got 021",
"wait how\nI got $r^2=(3\\sqrt{2})^2+(r-3)^2$ so $r=9/2$ oh the height is 9/2\nim dumb",
"How would r be 9... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1042,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782929.json"
} |
There is a polynomial $P(x)$ with integer coefficients such that $$ P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)} $$ holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | <details><summary>Solution</summary>We will use the fact that $x^{kn} - 1 = (x^k - 1)(x^{k(n-1)} + x^{k(n-2)} + ... + x^k + 1)$ . We can express $(x^{2310}-1)$ in this manner 4 different ways, with $(k,n)$ being $(105,22)$ , $(70,33)$ , $(42,55)$ , and $(30,77)$ . After cancelation we have $$ P(x) = (x^{105 \cdot 21} + x^{105 \cdot 20} + ... + x^{105} + 1)(x^{70 \cdot 32} + ... + 1)(x^{42 \cdot 54} + ... + 1)(x^{30 \cdot 76} + ... + 1)(x^{2310}-1)^2. $$ Our problem boils down to finding the number of ways we can multiply one term from each grouping to yield $x^{2022}$ , or, equivalently, the number of ways to add the exponents of one $x$ term from each grouping to total $2022$ .
Note that the exponent from the first grouping will be in the form $105a$ where $a$ is a nonegative integer less than $22$ . Similar arguments are made for the next three groupings. The $(x^{2310}-1)^2$ grouping will never contribute to this sum, as its smallest term is $-2x^{2310}$ . Thus, we have the following equation over the nonegative integers: $$ 105a + 70b + 42c + 30d = 2022 $$ where $a<22$ , $b<33$ , $c<55$ , and $d<77$ . Examine this equation modulo $2$ , $3$ , $5$ , and $7$ to derive $a \equiv 0 \text{ (mod } 2 \text{)}$ , $b \equiv 0 \text{ (mod } 3 \text{)}$ , $c \equiv 1 \text{ (mod } 5 \text{)}$ , and $d \equiv 3 \text{ (mod } 7 \text{)}$ , respectively. Let $a=2w$ , $b=3x$ , $c=5y+1$ , and $d=7z+3$ . We now have $$ 105(2w) + 70(3x) + 42(5y+1) + 30(7z+3) = 2022 $$ $$ \Rightarrow w+x+y+z=9 $$ where each of $w$ , $x$ , $y$ , and $z$ is a nonegative integer less than 11. We can represent this as having $9$ stars and $3$ bars, with the bars splitting the stars into $4$ groups corresponding to the values of $w$ , $x$ , $y$ , and $z$ . Our answer is thus $\frac{12!}{(9!)(3!)} = \boxed{220}$ .</details> | [
"2022=30a+42b+72c+105d\n\nI wrote $\\frac{1}{1-x^n}$ as $1+x+x^2+...$ I think...",
"I got 220 for this one, can somebody please confirm? Also I think the last power was 30 @op?",
"<blockquote>I got 220 for this one, can somebody please confirm? Also I think the last power was 30 @op?</blockquote>\n\no oops"... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1102,
"boxed": true,
"end_of_proof": false,
"n_reply": 42,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782931.json"
} |
Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that $$ \frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1. $$ Find the least possible value of $a+b.$ | Another personal favorite. I LOVE this one. Kudos to the author.
<span style="color:#f00">Edit: This claim is wrong. Will figure out later.</span>
We can interpret the problem as following: an ellipse centered at the origin has semi-major axis $a$ and foci at $(\pm 4,0)$ . Another ellipse is centered at $(20,11)$ with semi-major axis $b$ and foci $(20,11\pm 1)$ . We claim that minimizing $a+b$ means that these ellipses are tangent - indeed, assume they are not. Then, we can reduce the major axis of the first extremely slightly, by some $\varepsilon>0$ , such that the circles still intersect, but $a+b$ decreased, a contradiction.[1\baselineskip]
We first have the following crucial claim about tangent ellipses. Assume that ellipses $\mathfrak E_1$ and $\mathfrak E_2$ , with foci $E_1,E_2$ and $F_1,F_2$ are tangent at $T$ .**Claim.** $T$ is either $E_1F_2\cap E_2F_1$ or $E_1F_1\cap E_2F_2$ .
[asy]
import geometry;
ellipse ellipse(point A, point B, point C) {
return ellipse(A, B, (abs(A-C)+abs(B-C))/2);
}
size(9cm);
point E1=(-4,0),E2=(4,0),F2=(10,12),F1=(10,5.6),T=(6,4);
draw(ellipse(E1,E2,T));
draw(ellipse(F1,F2,T));
draw(E1--F1);
draw(E2--F2);
label(" $E_1$ ",E1,S);
label(" $E_2$ ",E2,S);
label(" $F_1$ ",F1,E);
label(" $F_2$ ",F2,E);
label(" $T$ ",T,E);
dot(E1);
dot(E2);
dot(F1);
dot(F2);
dot(T);
[/asy]
We can fix $E_1,E_2,F_1,F_2$ , and we show that if $P=E_1F_1\cap E_2F_2$ (given that $P$ lies on both line segments; otherwise look at $E_1F_2\cap E_2F_1$ ), then the ellipses are tangent. Clearly, there is at most one ellipse with foci at $F_1$ and $F_2$ that is tangent to a given ellipse, so it suffices to prove that is the one.
We can assume there is another point $Q$ on both ellipses. This means that
\[E_1Q+E_2Q=E_1P+E_2P\]
\[F_1Q+F_2Q=F_1P+F_2P\]
or upon addition
\[E_1Q+E_2Q+F_1Q+F_2Q=E_1P+E_2P+E_1P+E_2Q\tag{1}\]
However, by the triangle inequality in $\triangle E_1F_1Q$ and $\triangle E_2F_2Q$ , we get
\[E_1Q+F_1Q\geq E_1F_1=E_1P+PF_1\]
\[E_2Q+F_2Q\geq E_2F_2=E_2P+PF_2\]
with both equalities impossible, as that means $Q=E_1F_1\cap E_2F_2=P$ , impossible. Thus, we know that
\[E_1Q+E_2Q+F_1Q+F_2Q>E_1P+E_2P+E_1P+E_2Q\]
a contradiction to (1).
Now, the rest of the problem is easy - we see the desired lines are the ones connecting $E_1=(4,0)$ with $F_1=(20,12)$ , and $E_2=(-4,0)$ with $F_2=(20,10)$ . We can compute this point to be $T=(14,7.5)$ . Thus, the semi-major axis, $a$ and $b$ , are just half of the sums of the distances from the foci. In particular,
\begin{align*}
a=\frac{E_1T+E_2T}2&=\frac{\sqrt{(4-14)^2+(0-7.5)^2}+\sqrt{(-4-14)^2+(0-7.5)^2}}{2}
&=\frac{12.5+19.5}{2}=16
\end{align*}
\begin{align*}
b=\frac{F_1T+F_2T}2&=\frac{\sqrt{(20-14)^2+(12-7.5)^2}+\sqrt{(20-14)^2+(10-7.5)^2}}{2}
&=\frac{7.5+6.5}{2}=7
\end{align*}
Thus, we get the minimum value of $a+b=\boxed{023}$ . | [
"CONICS oh yeah ;)\n\nUse the definition of foci and then the lemma that the point minimizing AP+BP+CP+DP is the intersection of the diagonals of quadrilateral ABCD.",
"My favorite problem on the test\n<details><summary>Solution</summary>Consider points $A_1 = (4, 0), A_2 = (-4, 0), B_1 = (20, 12), B_2 = (20, 10... | [
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"boxed": false,
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"n_reply": 31,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782937.json"
} |
Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | Let the three labels be $20\ge x > y > z\ge 1$ . Then, we have $x - y = p$ , $y - z = q$ , and $x - z = r$ for primes $p,q,r$ . Adding the first two equations gives $x - z = p + q$ . But $x - z = r$ , so we must have $p + q = r$ . For parity reasons, one of $p,q$ is $2$ and the other is odd.
Assume WLOG $p = 2$ (multiply by $2$ at the end since we can freely swap $p,q$ ). Then $2 + q = r$ . Noting that $q,r\in [1,20]$ , we see that $q\in \{3,5,11,17\}$ . The rest is just casework on the value of $q$ .
Case 1: $q = 3$ . Then we have $r = 5$ , so $(x,y,z)$ can be any triple of the form $(a + 5, a + 2, a)$ . Since $1\le a\le 15$ , there are $15$ such triples
Case 2: $q = 5$ . Then $(x,y,z)$ is $(a + 7, a + 2, a)$ , where $1\le a\le 13$ , giving $13$ triples
Case 3: $q = 11$ . Then $(x,y,z)$ is $(a + 13, a + 2, a)$ , where $1\le a\le 7$ , giving $7$ triples
Case 4: $q = 17$ . Then $(x,y,z)$ is $(a + 19, a + 2, a)$ where $1\le a\le 1$ , giving $1$ triple.
Hence, there are $15 + 13 + 7 + 1 = 36$ triples. Multiplying by $2$ gives $\boxed{72}$ . | [
"Mine. I like this one; it's a cute twist on a classic.\n\n<details><summary>Solution</summary>Suppose $i$ , $j$ , and $k$ are the labels of the three vertices of a triangle; without loss, let $i > j > k$ . Note that $(i-j) + (j-k) = (i-k)$ , so one of $i-j$ or $j-k$ must be $2$ , and furthermore the o... | [
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"2022 Contests",
"2022 AIME Problems"
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"answer_score": 1082,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782948.json"
} |
There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that $$ \log_{20x} (22x)=\log_{2x} (202x). $$ The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | <details><summary>Solution</summary>Let $\log (n)$ be the logarithm base $10$ . By change of base formula: $$ \frac{\log(22x)}{\log(20x)} = \frac{\log(202x)}{\log(2x)} $$ $$ \Rightarrow \log{22x} \cdot \log{2x} = \log{20x} \cdot \log{202x} $$ $$ \Rightarrow \log{22x} \cdot \log{2x} = (1 + \log{2x}) \cdot \log{202x} $$ $$ \Rightarrow \log{22x} \cdot \log{2x} = \log{202x} + \log{2x} \cdot \log{202x} $$ $$ \Rightarrow \log{2x} \cdot (\log{22x} - \log{202x}) = \log{202x} $$ $$ \Rightarrow \log{202x} = \log{2x} \cdot \log{(\frac{11}{101})} $$ Thus, we have $$ \log _{20x} (22x) = \frac{\log(22x)}{\log(20x)} = \frac{\log(202x)}{\log(2x)} = \frac{\log{2x} \cdot \log{(\frac{11}{101})}}{\log{2x}} = \log{(\frac{11}{101})} $$ giving us an answer of $11+101 = \boxed{112}$ .</details> | [
"~~which problem was this~~ probably #4",
"Notice if $\\frac{a}{b} = \\frac{c}{d}$ then $\\frac{a}{b}=\\frac{c}{d} = \\frac{a-c}{b-d}$ Apply the lemma, so $\\frac{a}{b} = \\frac{\\ln(\\frac{10}{101})}{\\ln(10)}$ And you are done. $112$ .",
"Friend got $11 + 101 = 121$ . rip\n<details><summary>AAAAAAAAAAAA... | [
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"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1034,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782952.json"
} |
Find the remainder when $$ \binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2} $$ is divided by $1000$ . | <blockquote><details><summary>Solution</summary>Consider a committee of $n$ -people as follows. $$ C = \{x_1, x_2, \cdots, x_n \} $$ Now, consider the set of all possible committees of pairs of these $n$ -peoples. $$ P = \{ \{x_1, x_2 \}, \{x_1, x_3 \}, \cdots, \{x_{n-1}, x_{n} \} \}. $$ Notice that when we choose two such pairs, the two sets of pairs we choose either have a person in common or no people in common. Therefore, $$ \dbinom{\binom{n}{2}}{2} = n \binom{n-1}{2} + \left(\frac{\binom{4}{2}}{2}\right)\dbinom{n}{4} = 3 \bigg( \dbinom{n}{3} + \dbinom{n}{4} \bigg) = 3 \bigg(\dbinom{n+1}{4} \bigg). $$ Therefore, the answer is $$ 3 \bigg( \dbinom{4}{4} + \dbinom{5}{4} + \cdots + \dbinom{41}{4} \bigg) = 3 \times \dbinom{42}{5} \equiv \boxed{004} \pmod{1000}. $$</details>
I am wondering if there was some direct bijection to go from $\dbinom{\binom{n}{2}}{2}$ to $3 \dbinom{n+1}{4}.$ </blockquote>
Well if you have a list of $n+1$ values, including integers $1$ thru $n$ and a "duplicate" value, then any choice of $4$ values from these $n+1$ yields $3$ possible arrangements, hence $3\binom{n+1}{4}$ . | [
"I got 004 here",
"3(42C5) = 2552004 by hockey stick. Note NC2C2 = 3(N+1C4)",
"<span style=\"font-size:50%\">i may or may not have used the last hour to bash this one out</span>",
"Note that the expression is equivalent to\n\\[ \\sum_{n=3}^{40} \\frac{(n-2)(n-1)n(n+1)}{8} = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1038,
"boxed": false,
"end_of_proof": false,
"n_reply": 55,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782974.json"
} |
Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | <blockquote>
Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
</blockquote>
what an intuitive problem
The conditions imply that the sum of the negative numbers is $-\tfrac12$ and those of the positive numbers is $\tfrac12$ . Therefore, at some term $x_i$ , the sum of everything before is $-\tfrac12$ and everything else (after or equal to $x_i$ ) is $\tfrac12$ . So we'd like to make $x_{16}$ as negative as possible and $x_{76}$ as positive as we can.
This means "pushing" the things in between ( $x_{17}$ , $x_{18}$ , $\dots$ , $x_{75}$ ) to be $0$ to maximize distance. Equality should hold, meaning $x_1=x_2=\dots=x_{16}$ and $x_{76}=x_{77}=\dots=x_{100}$ . So $x_{16}=-\tfrac12\cdot\tfrac1{16}=-\tfrac1{32}$ and $x_{76}=\tfrac12\cdot\tfrac1{25}=\tfrac{1}{50}$ , $x_{76}-x_{16}=\tfrac{41}{800}$ , obtaining $\boxed{841}$ as the desired answer. | [
"Got 841, solution set is $x_{1, \\ldots, 16} = -\\frac{1}{32}, x_{17, \\ldots 75} = 0, x_{76, \\ldots, 100} = \\frac{1}{50}$ giving $\\frac{41}{800}$ .",
"confirm 841",
"<details><summary>Kinda fakesolved it</summary>We want $x_{76}$ to be as large as posssible, and $x_{16}$ to be as small as possible. ... | [
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] | {
"answer_score": 1056,
"boxed": true,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782983.json"
} |
A polygon is called *convex* if all its internal angles are smaller than 180 $^{\circ}$ . Given a convex polygon, prove that one can find three distinct vertices $A$ , $P$ , and $Q$ , where $PQ$ is a side of the polygon, such that the perpendicular from $A$ to the line $PQ$ meets the segment $PQ$ (possible at $P$ of $Q$ ).
| [
"Thank you!",
"I drew an equilateral triangle APQ. Then, I labeled midpoint of PQ as M and drew AM. From there it's pretty easy to see/prove that AM is perpendicular to PQ.",
"<blockquote>I drew an equilateral triangle APQ. Then, I labeled midpoint of PQ as M and drew AM. From there it's pretty easy to see/prov... | [
"origin:aops",
"2022 Contests",
"2022 BAMO"
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793577.json"
} | |
The game of pool includes $15$ balls that fit within a triangular rack as shown:
[asy]
// thanks Ritwin for this diagram :D
unitsize(0.6cm);
pair pos(real i, real j) {
return i*dir(60) + (j,0);
}
for (int i = 0; i <= 4; ++i) {
for (int j = 0; j <= 4-i; ++j) {
draw(circle(pos(i,j), .5));
}
}
pair A = pos(0,0);
pair B = pos(0,4);
pair C = pos(4,0);
pair dd = dir(270) * .5;
pair ul = dir(150) * .5;
pair ur = dir( 30) * .5;
real S = 1.75;
draw(A+dd -- B+dd ^^ B+ur -- C+ur ^^ C+ul -- A+ul );
draw(A+dd*S -- B+dd*S ^^ B+ur*S -- C+ur*S ^^ C+ul*S -- A+ul*S);
draw(arc(A, A+ul*S, A+dd*S));
draw(arc(B, B+dd*S, B+ur*S));
draw(arc(C, C+ur*S, C+ul*S));
[/asy]
Seven of the balls are "striped" (not colored with a single color) and eight are "solid" (colored with a single color). Prove that no matter how the $15$ balls are arranged in the rack, there must always be a pair of striped balls adjacent to each other. | This one was cute :)
Divide it into the following $6$ triangular sections, each with $3$ balls:
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
pair x = (0,0), y = 3*dir(240), z = 3*dir(240) + 3*dir(0);
dot(x,red); dot(x+dir(240),red); dot(x+dir(300),red);
dot(y,red); dot(y+dir(240),red); dot(y+dir(300),red);
dot(z,red); dot(z+dir(240),red); dot(z+dir(300),red);
[/asy]
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
dot(2*dir(240),red); dot(3*dir(240)+dir(0),red); dot(2*dir(240)+dir(0),red);
[/asy]
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
dot(3*dir(240)+dir(0),red); dot(4*dir(240)+2*dir(0),red); dot(3*dir(240)+2*dir(0),red);
[/asy]
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
dot(2*dir(240)+dir(0),red); dot(2*dir(240)+2*dir(0),red); dot(3*dir(240)+2*dir(0),red);
[/asy]
Pigeonhole tells us there are two striped balls in the same section, so they are adjacent. | [
"Also BAMO-8 Problem C.\n\nNote that First Row: striped, Second Row: solid, solid, Third Row: striped, solid, striped, Fourth Row: solid, solid, solid, solid, Fifth Row: striped, solid, striped, solid, striped is the max that striped balls could be placed without any being adjacent to each other. If we place o... | [
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"2022 Contests",
"2022 BAMO"
] | {
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793698.json"
} |
Suppose that $p,p+d,p+2d,p+3d,p+4d$ , and $p+5d$ are six prime numbers, where $p$ and $d$ are positive integers. Show that $d$ must be divisible by $2,3,$ and $5$ . | For the sake of contradiction, let's consider if $2\nmid p$ , $3\nmid p$ , and $5\nmid p$ .
If $2\nmid p$ , then $p$ and $p+d$ are $0$ and $1$ $mod$ $2$ in some order. Regardless, this configuration gives that $3$ of the values given are even primes, contradicting the fact that there is only one even prime, so $2\mid p$ must be true.
If $3\nmid p$ , then $p, p+d, p+2d$ are $0, 1, 2$ $mod$ $3$ in some order. However, this configuration gives that $2$ of the values given are primes divisible by $3$ , contradicting the fact that the only prime divisible by $3$ is $3$ itself, so $3\mid p$ must be true.
If $5\nmid p$ , then at least one of the numbers given must be a multiple of $5$ , forcing it to be $5$ . Since $p$ and $d$ are positive integers, we do cases on $p = 2$ and $p = 3$ , since if p is more than $3$ , it is impossible for any number in the sequence to be equal to $5$ .
If $p = 2$ , then $d = 3$ or $d = 1$ gives a term equal to $5$ . We only have these cases because any larger values of $d$ make it impossible to achieve $5$ . However, both sequences contain nonprimes. Similarly, if $p = 3$ , then $d = 2$ or $d = 1$ gives a term equal to $5$ . However, both sequences still contain non-primes. Thus, $5\mid p$ is also forced to be true.
If $30\mid d$ , one choice of $p$ to make it work is $p = 7$ , which gives the sequence $7, 37, 67, 97, 127, 157$ . | [
"Hint: We can see that $d$ is divisibile by $2$ , else the $p+d$ is divisibile by $2$ , so it is not prime.\nAnalogous with $3$ : if $d$ it is not divisible by $3$ , we can have $d=3k+1$ , or $d=3k+2$ but in these case one of the 6 numbers it is divisible by 3, etc.",
"Bad problem,\nFirst note that ... | [
"origin:aops",
"2022 Contests",
"2022 BAMO"
] | {
"answer_score": 92,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793699.json"
} |
Ten birds land on a $10$ -meter-long wire, each at a random point chosen uniformly along the wire. (That is, if we pick out any $x$ -meter portion of the wire, there is an $\tfrac{x}{10}$ probability that a given bird will land there.) What is the probability that every bird sits more than one meter away from its closest neighbor? | The answer is $\boxed{\tfrac{1}{10^{10}}}$ .
We model the wire as the interval $[0,10]$ on the real line. We can view a point $x$ in the interval as a pair $(\lfloor x\rfloor, \{x\})$ (more precisely a bijection). The key insight is to see that floor of the positions of birds must all be distinct i.e, they are some permutation of $0,2,\ldots, 9$ and the probability is precisely $\tfrac{10!}{10^{10}}$ . We can then vary the fractional parts independently. But with an additional condition that the fractional parts must form an order statistic which has a chance of $1^{10}\tfrac{1}{10!}$ . Therefore the answer is $\tfrac{10!}{10^{10}}\cdot\tfrac{1}{10!}=\tfrac{1}{10^{10}}$ as desired. | [
"We claim the probability is $\\frac{1}{10^{10}}$ .\n\nLabel the birds from $1$ to $10$ , and divide the wire into $10$ disjoint segments, each with length $1$ meter. Each bird must occupy a different segment, or else two birds would be at most one meter away. There are $10!$ ways to assign the birds to t... | [
"origin:aops",
"2022 Contests",
"2022 BAMO"
] | {
"answer_score": 1016,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793705.json"
} |
Sofiya and Marquis are playing a game. Sofiya announces to Marquis that she's thinking of a polynomial of the form $f(x)=x^3+px+q$ with three integer roots that are not necessarily distinct. She also explains that all of the integer roots have absolute value less than (and not equal to) $N$ , where $N$ is some fixed number which she tells Marquis. As a "move" in this game, Marquis can ask Sofiya about any number $x$ and Sofiya will tell him whether $f(x)$ is positive negative, or zero. Marquis's goal is to figure out Sofiya's polynomial.
If $N=3\cdot 2^k$ for some positive integer $k$ , prove that there is a strategy which allows Marquis to identify the polynomial after making at most $2k+1$ "moves". | Neat problem!
<span style="font-size:50%"> ~~Though my writeup was way too long :**(~~ </span>
<details><summary>Solution</summary>The key idea is to use binary search.
Note that the coefficient of $x^2$ in $f$ is $0$ , so if the roots are $r$ , $s$ , and $t$ , we know that $r+s+t=0$ . Thus we only need to find two roots to reconstruct the polynomial (as we know it's monic).
[list=1]
[*] Query $f(0)$ .
- If $f(0) = 0$ then we have found a root. Binary search on the positive root in step 2; however if we think $1$ is the positive root we must query it again, as it is possible that $f(x) = x^3$ has only roots $0$ .
- If $f(0) > 0$ , two roots are positive while one is negative. This case is symmetric to the next.
- If $f(0) < 0$ , two roots are negative while one is positive. We will binary search on the interval $[1, 3\cdot2^k-1]$ for this root.
[/list]
- If we binary search naively, we will use $2k+2$ queries. However we can optimize it by shifting the first query. Start by querying $f(2^{k+1})$ . [list]
- If $f(2^{k+1}) = 0$ , we have found a positive root. Skip to step 3
- If $f(2^{k+1}) > 0$ , the positive root is in the interval $[1, 2^{k+1}-1]$ (length $2^{k+1}-1$ ). Binary search on this interval for the root in $k$ steps.
- If $f(2^{k+1}) < 0$ , the positive root is in the interval $[2^{k+1}+1, 3\cdot2^k-1]$ (length $2^k-1$ ). Binary search on this interval for the root in $k-1$ steps.
[/list]
- So now we have a positive root, call it $r$ . We know that $s+t = -r$ so (at least) one of $s$ or $t$ is in the interval $\big[\!-\!\tfrac r2,0\big]$ . Binary search on this interval for one root, with at most $\lceil \log_2r \rceil - 2$ queries.
Now we have two roots, and can find the third root from $t = -r-s$ , so we know the polynomial. Carefully counting the queries, we see that in all cases we use at most $2k+1$ queries (we offset the second query---for $f(2^{k+1})$ ---to balance the number of queries out more). $\square$ (man)
*Note: The binary search we do is equivalent to looking for the $0$ in a hidden list $[a,a,a,\ldots,a,0,b,\ldots,b,b,b]$ (where $a\neq b$ are known and nonzero, also note that the $0$ could be on either end) where we can query the value at any index. We can prove by induction that it takes at most $k$ queries, where the length of the list is $2^{k+1}-1$ . The base case is that it takes one query for $k=1$ (so the list has $3$ elements).*
I like this problem because I've never seen binary search in a math problem before!</details>
There were several tricky things in this problem that you need to be careful about. Hopefully I caught all of them. | [
"Sad, I could only prove at most $2k+2$ moves ... I felt like I was almost there but I didn't think of splitting $N$ into powers of $2$ ."
] | [
"origin:aops",
"2022 Contests",
"2022 BAMO"
] | {
"answer_score": 188,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793708.json"
} |
You are bargaining with a salesperson for the price of an item. Your first offer is $a$ dollars and theirs is $b$ dollars. After you raise your offer by a certain percentage and they lower their offer by the same percentage, you arrive at an agreed price. What is that price, in terms of $a$ and $b$ ? | $\frac{p}{a}-1=1-\frac{p}{b}$ so $p=\boxed{\frac{2ab}{a+b}}$ | [
"<details><summary>Sketch of what I did.</summary>NOTE: $b>a$ and $x<1$ . Let $x$ be the percentage. then $a+ax = b(1-x)$ (this is the final price). Alg. bash to get $x=\\frac{b-a}{a+b}$ . Plug into price equation to get answer as $a+(\\frac{ab-a^2}{a+b})$ . Use some values of $a$ and $b$ to check.</de... | [
"origin:aops",
"2022 Contests",
"2022 BAMO"
] | {
"answer_score": 1004,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793751.json"
} |
If I have 100 cards with all the numbers 1 through 100 on them, how should I put them in order to create the largest possible number? | The no. of digits in such a number is $192$ . The largest 192-digit number is $99....999$ where $9$ occurs $192$ times. This gives us an idea that we need numbers having $9$ in their leftmost place first following with $8$ and so on.
So the number should be
99 9 98 97 96 95 94 93 92 91 90 89 88 8 87 ... 22 2 21 20 19 18 17 16 15 14 13 12 11 1 10 100
Notice the placement of the single digit numbers. | [
"<details><summary>Solution</summary>99 9 98 97 96 95 94 93 92 91 90 89 88 8 ... 20 19 18 17 16 15 14 13 12 11 1 10 100.</details>",
"@above, your solution is what I got, but what happened to Card #10? (I placed it right before Card #100)\n\noh and what happened to card #15 and card #13?",
"please correct me if... | [
"origin:aops",
"2022 Contests",
"2022 BAMO"
] | {
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793759.json"
} |
The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$ . The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$ . Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$ , prove that $ \angle BKC \equal{} \angle CDB$ . | Cute and easy!! :D
Denote $\boxed{MA=x,MB=w,MC=y,CD=z}$ . Then by the given condition, we have $xy+xz=w(MD) \implies \boxed{MD=\frac{x(y+z)}{w}}$ .
Define $O=BD \cap KC$ then $CO$ is the angle bisector of $\angle{MCD}$ . By internal angle bisector theorem, we have $\frac{MO}{OD}=\frac{MC}{CD}=\frac{y}{z}$ and $MO+OD=MD=\frac{x(y+z)}{w}$ . Thus, $\boxed{MO=\frac{xy}{w}}$ and $\boxed{OD=\frac{xz}{w}}$ . Now note that $MO.MB=xy=MA.MC$ implying $BAOC$ is cyclic. Thus, $\angle{KBD}=\angle{ABO}=\angle{ACO}=\angle{ACK}=\angle{KCD} \implies \boxed{\angle{KBD}=\angle{KCD}}$ . Thus $BKDC$ is also cyclic implying $\angle{BKC}=\angle{CDB}$ as desired. $\blacksquare$ ( $\mathcal{QED}$ ) | [
"Here is what I have so far:\r\nI will first prove the converse. We will assume KBCD is cyclic. Let KC and BD intersect at P. Then ABCP is also cyclic. By power of a point, $ MA \\cdot MC \\equal{} MP \\cdot MB$ . So we need to show $ MA \\cdot CD \\equal{} MB(MD \\minus{} MP) \\equal{} MB \\cdot PD$ .\r\nOr $ \... | [
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
] | {
"answer_score": 1132,
"boxed": true,
"end_of_proof": true,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/228384.json"
} |
For a given value $t$ , we consider number sequences $a_1, a_2, a_3,...$ such that $a_{n+1} =\frac{a_n + t}{a_n + 1}$ for all $n \ge 1$ .
(a) Suppose that $t = 2$ . Determine all starting values $a_1 > 0$ such that $\frac43 \le a_n \le \frac32$ holds for all $n \ge 2$ .
(b) Suppose that $t = -3$ . Investigate whether $a_{2020} = a_1$ for all starting values $a_1$ different from $-1$ and $1$ . | **(a)** $a_{n+1} =\frac{a_n +2}{a_n + 1}$ $\frac43 \le \frac{a_1 +2}{a_1 + 1} \le \frac32$ $4a_1+4 \le 3a_1+6 \implies a_1\le 2$ and $2a_1+4 \le 3a_1+3 \implies 1\le a_1$ , thus $2\geq a_1\geq 1$ .
Suppose $\frac43 \le a_{n}\le \frac32$ , then we want to show that for all $n\geq 1$ we have $\frac43 \le 1+\frac{1}{a_n + 1} \leq \frac{3}{2} \Longleftrightarrow 1 \leq a_n\leq 2$ , which is true. Thus, $\boxed{a_1=[1,2]}$ .**(b)** $a_{n+1} =\frac{a_n -3}{a_n + 1}$ We want to show that $a_4=a_1$ , then $a_4=\frac{\frac{a_2 -3}{a_2 + 1}-3}{\frac{a_2 -3}{a_2 + 1} + 1}=\frac{\frac{a_2 -3-3(a_2+1)}{a_2 + 1}}{\frac{a_2 -3+(a_2+1)}{a_2 + 1}}=\frac{a_2 -3-3(a_2+1)}{a_2 -3+(a_2+1)}=\frac{-2a_2-6}{2a_2 -2}=-\frac{a_2+3}{a_2 -1}=-\frac{\frac{a_1 -3}{a_1 + 1}+3}{\frac{a_1 -3}{a_1 + 1}-1}=-\frac{\frac{a_1 -3+3(a_1+1)}{a_1 + 1}}{\frac{a_1 -3-1(a_1+1)}{a_1 + 1}}=-\frac{a_1 -3+3(a_1+1)}{a_1 -3-1(a_1+1)}=-\frac{4a_1}{-4}=a_1$ .
Notice that we cannot have $a_2=-1\Longleftrightarrow a_1=1$ or $a_1=-1$ .
Hence, we have showed that $a_4=a_1$ , hence $a_{3k+1}=a_1$ and we are done. | [] | [
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Let $a_1, a_2, \cdots, a_{2n}$ be $2n$ elements of $\{1, 2, 3, \cdots, 2n-1\}$ ( $n>3$ ) with the sum $a_1+a_2+\cdots+a_{2n}=4n$ . Prove that exist some numbers $a_i$ with the sum is $2n$ . | It is well-known that among any $n$ integers we can find a non-empty subset with sum divisible by $n$ .
Apply this to $a_1,\dots,a_n$ to find that (w.l.o.g.) $a_1+\dots+a_k$ is divisible by $n$ for some $1 \le k \le n$ .
If $a_1+\dots+a_k=2n$ , we are done.
If $a_1+\dots+a_k=3n$ , then we must have $k=n$ and the remaining $n$ numbers $a_{n+1},\dots,a_{2n}$ all must be equal to $1$ .
Now just take any subsum of $a_1+\dots+a_n$ which is in $[n,2n]$ (this surely exists) and add as many of the ones as necessary.
Finally, assume that $a_1+\dots+a_k=n$ . But now we can just apply the same reasoning to the numbers $a_{n+1},\dots,a_{2n}$ and find that the only remaining case is that $a_{m}+\dots+a_{2n}=n$ for some $m \ge n+1$ .
But then these two sums together give the desired $2n$ . Done. | [
"thank you :)))",
"wlog $1<=a_1<=.....<=a_{2n}<=2n-1$ Consider the numbers $a_1,a_1+a_2,....,a_1+...+a_{2n-1}<4n$ and the sets $(1,2n+1),(2,2n+2),...,(2n-1,4n-1)$ if one of the numbers equal to $2n$ we are done.\n\nOtherwise we have $2n-1$ numbers and $2n-1$ sets.\nIf two numbers go to the some set we ar... | [
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Let $a, b, c$ be positive real numbers such that: $$ ab - c = 3 $$ $$ abc = 18 $$ Calculate the numerical value of $\frac{ab}{c}$ | <details><summary>Just to fill the storage</summary>$ab - c = 3$ $\implies$ $ab=3+c$ $abc = 18$ $\implies$ $ab=\frac{18}{c}$ $ab=ab$ $3+c=\frac{18}{c}$ $c^2+3c-18=0$ $(c+6)(c-3)=0$ $\implies$ $c+6=0$ $\wedge$ $c-3=0$ $c=-6$ $\wedge$ $c=3$ Since $c\in$ $R^+$ we get that $c=3$ $ab=\frac{18}{c}$ $ab=6$ $\frac{ab}{c}=2$</details> | [
"<details><summary>answer</summary>$2$</details> i just plugged in values lol but there was some algebra ",
"Since ab+(-c)=3 and ab(-c)=-18,\nab and -c are two solutions to c^2-3x-18=0\nSince a, b, c positive, ab=6 and -c=-3.\nHence (ab)/c=2.",
"solved also [here](https://artofproblemsolving.com/community/c6h27... | [
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Let $\vartriangle ABC$ be a triangle in which $\angle ACB = 40^o$ and $\angle BAC = 60^o$ . Let $D$ be a point inside the segment $BC$ such that $CD =\frac{AB}{2}$ and let $M$ be the midpoint of the segment $AC$ . How much is the angle $\angle CMD$ in degrees? | $\angle CMD=40^o$ .
let midpoint $AB$ is $E$ and $F$ on $BC$ such that $MEBF$ parallelogram ( $F$ between $B$ and $D$ ).
easy to check triangle $AME$ similar to triangle $ABC$ , therefore $\angle BEM=100^o$ and $\angle MFD=80^o$ .
by sine rule we got $\dfrac {sin(80)}{sin(\angle CDM)}=\dfrac {sin(40)}{sin(140-\angle CDM)}, \angle CDM=100^o, \angle CMD=40^o$ .
CMIIW | [
"A good drawing may drastically shorten the solution; a non trigo one at [https://stanfulger.blogspot.com/2022/02/aops-brasil-girls-math-tournament-2021.html](https://stanfulger.blogspot.com/2022/02/aops-brasil-girls-math-tournament-2021.html)\n\nBest regards,\nsunken rock",
"solved also [here](https://artofprobl... | [
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A natural number is called *chaotigal* if it and its successor both have the sum of their digits divisible by $2021$ . How many digits are in the smallest chaotigal number? | [
"Let S(n) denote digit sum of natural number n.\nLet N be a smallest chaotigal number.\n\nSay that k nines appear consecutively at the right end of N. Adding 1 to this number will cause k nines turn to k zeros, increase one of the digits by 1 and rest will not change. Therefore we can see that S(N)-S(N+1)=9k-1.\n\n... | [
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Mariana plays with an $8\times 8$ board with all its squares blank. She says that two houses are *neighbors* if they have a common side or vertex, that is, two houses can be neighbors vertically, horizontally or diagonally. The game consists of filling the $64$ squares on the board, one after the other, each with a number according to the following rule: she always chooses a house blank and fill it with an integer equal to the number of neighboring houses that are still in White. Once this is done, the house is no longer considered blank.
Show that the value of the sum of all $64$ numbers written on the board at the end of the game does not depend in the order of filling. Also, calculate the value of this sum.
Note: A house is not neighbor to itself.
<details><summary>original wording</summary>Mariana brinca com um tabuleiro 8 x 8 com todas as suas casas em branco. Ela diz que duas
casas s˜ao vizinhas se elas possu´ırem um lado ou um v´ertice em comum, ou seja, duas casas podem ser vizinhas
verticalmente, horizontalmente ou diagonalmente. A brincadeira consiste em preencher as 64 casas do tabuleiro,
uma ap´os a outra, cada uma com um n´umero de acordo com a seguinte regra: ela escolhe sempre uma casa
em branco e a preenche com o n´umero inteiro igual `a quantidade de casas vizinhas desta que ainda estejam em
branco. Feito isso, a casa n˜ao ´e mais considerada em branco.
Demonstre que o valor da soma de todos os 64 n´umeros escritos no tabuleiro ao final da brincadeira n˜ao depende
da ordem do preenchimento. Al´em disso, calcule o valor dessa soma.
Observa¸c˜ao: Uma casa n˜ao ´e vizinha a si mesma</details> | Consider a pair of neighboring houses. Since no two squares are filled simultaneously, for every pair of neighboring houses one house is filled after the other, so the house that is filled first will definitely include the second one in its count, but the second one will not count the first house. Therefore, the total contribution of the pair to the sum of all numbers is exactly one. Since this is true for all pairs of houses, the total sum is just the number of pairs of neighboring houses.
To compute this number, we can count sum of the number of neighbors each house has and divide it by 2, since every pair is included twice: every house in the center of the board has 8 neighbors, every 'edge' house has 5, and the 'corner' houses each have 3: $\frac{36*8+24*5+4*3}{2}=210$ . | [
"solved also [here](https://artofproblemsolving.com/community/c6h2731190p23794410)"
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Find all pairs $(a,b)$ of positive integers, such that for **every** $n$ positive integer, the equality $a^n+b^n=c_n^{n+1}$ is true, for some $c_n$ positive integer. | We claim that the only solution is the pair $(2,2)$ which clearly works. Now we show that this is the only pair that works.**<span style="color:#00f">Claim:</span> We cannot have an odd prime $p$ and an odd positive integer $n$ so that $a$ and $b$ are not divisible by $p$ and $p \mid a^n + b^n.$**
<u>*proof*</u>
Assume for contradiciton that the *Claim* is false. Then let $p$ and $n$ as above. Since $n$ is odd and $p$ does not divide $a$ and $b,$ we must have by LTE that for all odd positive integer $k:$ $$ v_p (a^{nk} + b^{nk}) = v_p(a^n+b^n) +v_p(k) $$ On the other hand, by problems condition $$ nk+1 \mid v_p(a^{nk} + b^{nk}) $$ Conbining the two relation we get an absurd since we can make $nk+1$ arbitraly large while $v_p(a^n+b^n)+v_p(k)$ is nonzero and can be small (for instance, take $k$ sufficiently large with $p\nmid k$ ) $.\square$ [rule]
Now we claim that if $a\ne b$ then there exist $p$ and $n$ as in the *Claim;* it will force $a=b.$ **<span style="color:#00f">Lemma:</span> If $a$ and $b$ are distinct positive integers, then exists an odd integer $n$ and an odd prime $p$ so that $p\nmid a,b$ and $p\mid a^n+b^n$**
<u>*proof:*</u>
First write $d=gcd(a,b)$ and then let $x={a}{/d}$ and $y={b}{/d}.$ Take $q$ a sufficiently large odd prime, more precisely, take $q$ so that $q>a+b$ and $q>r,$ for all $r\mid d.$ Then take $n=q$ and $p$ as an odd prime dividing $$ \frac{x^q + y^q}{x+y} $$ For seeing that such $p$ exists, first note that, since $a\ne b$ by assumption, we cannot have $x=y=1$ and hence ${(x^q+y^q)}{/(x+y)}>1;$ now to show that $p$ is odd, note that since $gcd(x,y)=1,$ we must have that either $x,y$ are both odds or have different parities. In the later case, we would have that ${(x^q+y^q)}{/(x+y)}$ is odd and we are done; in the former case, just note that $$ \frac{x^q+y^q}{x+y} = x^{q-1}-x^{q-2}y+\dots-xy^{q-2}+y^{q-1} $$ is odd since it is the sum of an odd number of odd numbers. So indeed $p$ exists and is odd.
Now we have to show that $p$ satisfies the other condition, i.e., $p\nmid a,b.$ First note that since $x$ and $y$ are coprime we have $p\nmid x,y$ and so $p\mid a,b \iff p \mid d.$ Now note that since $q$ is odd, we have that $$ p\mid x^q+y^q \Rightarrow {(-xy^{-1})}^q \equiv 1 \pmod p $$ where $y^{-1}$ denotes the multiplicative inverse of $y$ modulo $p.$ Then let $k$ denote the order of $-xy^{-1}$ modulo $p.$ By the above congruence, we have that $k \mid q$ and since $q$ is prime, it follows that $k=q$ or $k=1.$ If the former case happens, then we have that $p-1\ge k =q$ which implies that $p\nmid d$ by the choice of $q$ (we have selected $q$ so that it is greater than all divisors of $d$ ) hence $p\nmid a,b$ and we are done. In the later case, we have that $p\mid x+y;$ hence by LTE we have $$ v_p\left(\frac{x^q+y^q}{x+y}\right)=v_p(q) $$ since $p$ divides the above number by definition, we must have that $v_p(q)\ge 1,$ i.e. $p=q.$ Then $q \mid x+y\mid a+b$ which is an absurd since we have choosen $q>a+b.$ So the Lemma is proven. $\square.$ [rule]
From the *Claim* and the *Lemma,* it follows that we must have $a=b$ and the condition becomes $2a^n$ is an $n+1$ power for all $n.$ We claim that $a$ must be $2.$ Indeed, let $l=v_2(a),$ then problems condition becomes $n+1 \mid 1 + nl$ for all positive integer $n.$ This is equivalen to $n+1 \mid l-1$ for all positive integer $n,$ which can only happen if $l-1 = 0$ (since $l$ is fixed an $n$ can be arbitraly large) i.e. $l=1.$
Now let $p$ be any odd prime, then the condition implies $n+1 \mid nv_p(a)$ for all positive integer $n;$ since $n+1$ and $n$ are coprime, this is equivalent to $n+1 \mid v_p(a)$ for all positive integer $n,$ which similarly implies $v_p(a)=0.$ Then we have that $v_2(a)=1$ and $v_p(a)=0,$ for all odd prime $p,$ i.e. $a=2.$ Clearly $(2,2)$ satisfies problems condition and so we are done. | [
"If wlog $a\\geq b$ , assume we're in the case $a>b$ , and consider all the $n>3$ . By Zsigmondy's theorem, there exists a prime $p_n$ dividing $a^n+b^n$ , which doesn't divide the previous number. Since $p_n|c_n^{n+1}$ , it follows that $p_n^{n+1}|c_n^{n+1}=a^n+b^n$ .\nThis implies $p_n<p_n^{1+1/n}\\leq ... | [
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Let $P=(x^4-40x^2+144)(x^3-16x)$ . $a)$ Factor $P$ as a product of irreducible polynomials. $b)$ We write down the values of $P(10)$ and $P(91)$ . What is the greatest common divisor of the two numbers? | [
"a)P(x)=x(x+6)(x-6)(x+4)(x-4)(x+2)(x-2) b)(P(1),P(91))=3^2*5*7=245"
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Let $\triangle ABC$ have $AB = 1$ cm, $BC = 2$ cm and $AC = \sqrt{3}$ cm. Points $D$ , $E$ and $F$ lie on segments $AB$ , $AC$ and $BC$ respectively are such that $AE = BD$ and $BF = AD$ . The angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ADE$ for the second time at $M$ and the angle bisector of $\angle ABC$ intersects the circumcircle of $\triangle BDF$ at $N$ . Determine the length of $MN$ . | [] | [
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Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality. | Let $a,b,c$ be positive real numbers. Prove that $$ \left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+9 $$ [https://artofproblemsolving.com/community/c6h486634p5449967](https://artofproblemsolving.com/community/c6h486634p5449967)
<details><summary>Nice.</summary><blockquote><blockquote>Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.</blockquote>
The inequality $a)$ is false for $a = 0^+, b = c = 1$ . As for b), using the inequality $x^2 + y^2 + z^2 \geq xy + yz + zx$ we have
\begin{align*}
\sum \left( \frac{a + b}{c} \right)^2 &\geq \sum \frac{(a + b)(b + c)}{ca} = \sum \frac{ab + ca + b^2 + bc}{ca}
&= \sum \frac{b}{c} + 3 + \sum \frac{b^2}{ca} + \sum \frac{b}{a}
&\geq \sum \frac{a}{b} + 3 + 3 + 3
&= \sum \frac{a}{b} + 9
\end{align*}</blockquote></details> | [
"Let $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9 $$ <details><summary>Nice.</summary><blockquote>Given the inequalities: $a)$ $\\left(\\frac{2a}{b+c}\\right)^2+\\le... | [
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Let $p = (a_{1}, a_{2}, \ldots , a_{12})$ be a permutation of $1, 2, \ldots, 12$ .
We will denote \[S_{p} = |a_{1}-a_{2}|+|a_{2}-a_{3}|+\ldots+|a_{11}-a_{12}|\]We'll call $p$ $\textit{optimistic}$ if $a_{i} > \min(a_{i-1}, a_{i+1})$ $\forall i = 2, \ldots, 11$ . $a)$ What is the maximum possible value of $S_{p}$ . How many permutations $p$ achieve this maximum? $\newline$ $b)$ What is the number of $\textit{optimistic}$ permtations $p$ ? $c)$ What is the maximum possible value of $S_{p}$ for an $\textit{optimistic}$ $p$ ? How many $\textit{optimistic}$ permutations $p$ achieve this maximum? | In any permutation $p$ , call $a_i$ a $peak$ if $a_{i}>\max (a_{i-1},a_{i+1})$ , and a $bottom$ if $a_{i}<\min (a_{i-1},a_{i+1}),$ (to make the definition valid for the endpoints, we let $a_{0}=a_{2}$ and $a_{13}=a_{11}).$ Now we can easily see that for any permutation $p,$ the value of $S_{p}$ only depends on peaks and bottoms
Let $p'=(13-a_{1},13-a_{2},...,13-a_{12}),$ then $S_{p}=S_{p'}$ , and so for the sake of evaluation (for the sake of counting, we simply multiply by 2), we may assume wlog that $a_{1}$ is a bottom. If $a_{12}$ was bottom as well, then for some $1\le k\le 5$ we have (aside of $a_{1},a_{12},$ the $p_{i}$ 's are the peaks and the $b_{j}$ 's are the bottoms of $p$ ):
\[S_{p}=-a_{1}+2(p_{1}-b_{1}+p_{2}-b_{2}+...-b_{k-1}+p_{k})-a_{12}=2(p_{1}+p_{2}+...+p_{k})-2(b_{1}+...+b_{k-1})-a_{1}-a_{12}\]
\[\le 2(12+11+...+(13-k))-2(1+...+(k-1))-k-(k+1)=2k(12-k)-1\le 69\]
And if $a_{12}$ was peak, then for some $0\le k\le 5$ we have
\[S_{p}=-a_{1}+2(p_{1}-b_{1}+p_{2}-b_{2}+...+p_{k}-b_{k})+a_{12}=2(p_{1}+p_{2}+...+p_{k})+a_{12}-2(b_{1}+...+b_{k})-a_{1}\]
\[\le 2(12+11+...+(13-k))+(12-k)-2(1+...+k)-(k+1)=2k(11-k)+11\le 71\]
So $S_{p}\le 71,$ and the equality case should then satisfies the following conditions:
1) $a_{1}=6$ and $a_{12}=7$ 2) $(a_{2},a_{4},a_{6},a_{8},a_{10})$ is a permutation of (8,9,10,11,12)
3) $(a_{3},a_{5},a_{7},a_{9},a_{11})$ is a permutation of (1,2,3,4,5)
This counts a total of $5!\cdot 5!$ equality cases,
so by returning generality we see that the maximum value of $S_{p}$ is 71, and it is achievable through $2\cdot 5!\cdot 5!=28800$ permutations. This concludes part (a).
Now let's consider parts b),c). Notice that the $optimistic$ condition is simply saying that there should be no bottoms, except possibly for the endpoints, so we can only have one peak (if there were 2 peaks, a bottom should appear in between, which violates the optimistic condition, and a peak exists because '12' is always peak) and this peak is $a_{i}=12$ .
But this would imply that $a_{1}<a_{2}<...<a_{i}=12>a_{i+1}>...>a_{12}\;\;^{(*)}$ , and so $$ S_{p}=2a_{i}-a_{1}-a_{12}=24-a_{1}-a_{12}\le 21 $$ This is achievable when $\{a_{1},a_{12}\} =\{1,2\}$ and the rest can be arbitrarywhile preserving (*), this can be easily counted, we shall get a total of $2^{10}=1024$ permutations satisfying $S_{p}=21$ and $p$ optimistic. This concludes part (c)
For part (b), we just ignore the condition $\{a_{1},a_{12}\} =\{1,2\}$ , we'll get a total of $2^{11}=2048$ optimistic permutations | [] | [
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Let $f(x)$ be a quadratic function with integer coefficients. If we know that $f(0)$ , $f(3)$ and $f(4)$ are all different and elements of the set $\{2, 20, 202, 2022\}$ , determine all possible values of $f(1)$ . | $f(x) = ax^2+bx+c$ $f(0) = c $ $f(3) = 9a+3b+c $ $f(4) = 16a+4b+c$ Case $f(0)=2$ $f(3) = 9a+3b+2 \equiv 20 \mod 3 $ ( 2 cannot be used since f(0)=2) $9a+3b=18,3a+b=6$ $f(4) = 16a+4b+2 \equiv (202,2022) \mod 4$ [rule]
subCase (f(4)=202) $16a+4b=200$ $4a+b = 50$ $a+b+c = 44+6-44*3+2 = 8-44*2 = -80$ subCase (f(4) 2022) $16a+4b=2020$ $4a+b = 505$ $3a+b=6$ $a+b+c=-990$ [rule]
case $f(0)=20$ $f(4) = 16a+4b+20 \equiv 0 \mod 4$ forcing f(4) = 20 contradiction
case $f(0) = 202$ $f(3) = 9a+3b+202 \equiv 1 \mod 3$ forcing $f(3) = 202$ contradiction
case $f(0) = 2022$ $f(3) = 9a+3b+2022 \equiv 0 \mod 3 $ forcing $f(3) = 2022$ contradiction | [
"f(1)=-80 or f(1)=-990 \n"
] | [
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Let $\triangle ABC$ have median $CM$ ( $M\in AB$ ) and circumcenter $O$ . The circumcircle of $\triangle AMO$ bisects $CM$ . Determine the least possible perimeter of $\triangle ABC$ if it has integer side lengths. | <details><summary>Solution</summary>$\underline{\textbf{Claim:}}$ $AC=CM$ $\textit{Proof 1:}$ Let the midpoint of $CM$ be point $N$ . Since $\angle OMA=90^{\circ}$ and $N\in (AMO)$ , this means that $\angle ONA=90^{\circ}$ . If $\overline{AN}\cap (ABC)=X$ , since $ON\perp AN$ , this implies that $N$ is the midpoint of $AX$ . Now $NA=NX$ and $NC=NM$ , so $ACXM$ is a parallelogram, so
\[\angle CAM=\angle CAX+\angle XAM=\angle CBX+\angle CXA=\angle CBX+\angle CBA=\angle ABX=\angle AMN=\angle AMC\Longrightarrow CA=CM\] $\textit{Proof 2:}$ Let $S$ be the midpoint of $AC$ . Now $AMONS$ is cyclic, but $SN$ is a midsegment in $\triangle AMC$ , so $SN\parallel AM$ and $AMNS$ is cyclic, therefore $AMNS$ is an isosceles trapezoid. Now $AS=NM\Longrightarrow CA=CM$ .
Now, if $a=BC$ , $b=AC=CM$ , $c=AB$ we can use the median formula ( $m_{c}=\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$ ) to get that:
\[b^2=CM^2=\frac{1}{4}(2a^2+2b^2-c^2)\Longrightarrow c^2=2(a-b)(a+b)\]
We want $a,b,c$ to be positive integers with minimal sum (the perimeter of $\triangle ABC$ ). If $\gcd(a,b)=d>1$ , then $\left(\frac{a}{d},\frac{b}{d},\frac{c}{d}\right)$ are the sidelengths of a triangle which satisfies the conditions in the problem with smaller perimeter. Thus we can assume that $a,b,c$ are two by two coprime. Now we have two cases: $\textbf{Case 1:}$ $a-b=2s^2$ , $a+b=4t^2$ for positive integers $s,t$ .
Then $a=s^2+2t^2$ , $b=2t^2-s^2$ , $c=4st$ , so the perimeter is $4t^2+4st$ . If $s=1,t=1$ we get $(a,b,c)=(3,1,4)$ , impossible. If $s=1, t=2$ , then $(a,b,c)=(9,7,8)$ , so the least possible perimeter in this case is $24$ . $\textbf{Case 2:}$ $a-b=4s^2$ , $a+b=2t^2$ for positive integers $s,t$ .
Then $a=2s^2+t^2$ , $b=t^2-2s^2$ , $c=4st$ , so the perimeter is $2t^2+4st$ . If $s=1,t=2$ we get $(a,b,c)=(6,2,8)$ , impossible. If $s=1, t=3$ , then $(a,b,c)=(11,7,12)$ , so the least possible perimeter in this case is $30$ .
In both cases, it should be briefly explained why those are the minimal perimeter cases though it's pretty obvious. Finally, the answer is $\boxed{24}$ achieved when $(a,b,c)=(9,7,8)$ .</details>
<details><summary>Note</summary>Although it may be intuitive to think that $\triangle ABC$ is acute, this is not necessarily the case. We can have $AC<BC$ , $\angle C>90^{\circ}$ and $(AMO)$ bisecting $CM$ . This, however, isn't a problem for the solution above.</details> | [] | [
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Find all primes $p$ , such that there exist positive integers $x$ , $y$ which satisfy $$ \begin{cases}
p + 49 = 2x^2
p^2 + 49 = 2y^2
\end{cases} $$ | Wow, this can be solved exactly like USAMO 2022/4, held two days ago. So, here is a <details><summary>sketch</summary>Note that $p>y$ , otherwise $y \leq 7$ and case check. Now, subtracting the two equations, we obtain $p(p-1)=2(y-x)(y+x)$ . Obviously $p=2$ doesn't work, and $p$ can't divide $y-x$ . So $p|y+x$ and if $x+y$ is not equal to $p$ , then $x+y \geq 2p > p+y$ , so $x>p$ and $p+49>2p^2$ , case check. So $x+y=p, y-x=\frac {p-1}{2}$ , calculate $x, y$ in terms of $p$ and plug in the system, done.</details> | [
"Way older than USAMO 2022. Very similar idea in [Bundeswettbewerb Mathematik 1997.\n](https://artofproblemsolving.com/community/c1058102_1997_bundeswettbewerb_mathematik)",
"Even removing the requirement that $p$ is prime, this is an elliptic curve and it is possible to determine all integral points on it.\n\n... | [
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14 students attend the IMO training camp. Every student has at least $k$ favourite numbers. The organisers want to give each student a shirt with one of the student's favourite numbers on the back. Determine the least $k$ , such that this is always possible if: $a)$ The students can be arranged in a circle such that every two students sitting next to one another have different numbers. $b)$ $7$ of the students are boys, the rest are girls, and there isn't a boy and a girl with the same number. | This nice problem was proposed by Miroslav Marinov. Part a) is just for warm up. Obviously, $k=1$ is not enough and it's possible for $k=2$ . The worst case scenario is when all the students have the same favorite numbers, say $1$ and $2$ . In this case we arrange the numbers as $1-2-1-\dots 2$ . **Part b).** $k=4$ . I follow in general the author's solution. We omit the counterexample that it is not possible for $k=3$ and let's prove that it's always possible when $k=4$ . Consider a bipartite graph $G(S, N)$ where $S=\{1,2,\dots,14\}$ are the students and $N=\{n_1,n_2,\dots,n_k\}$ are all the numbers the students like. A vertex (student) $s\in S$ is connected with $n_i\in N$ if $n_i$ is a favorite number of $s$ . The given condition says that every $s$ is connected with at least $4$ vertices in $N$ , i.e. $d(s)\ge 4, \forall s\in S$ . Let us partition the vertices $S$ into two groups $S_0$ (boys) and $S_1$ (girls). The problem actually asks to prove the following claim. $\textbf{Claim.}$ The vertices in $N$ can be partitioned into two disjoint sets $N_0$ and $N_1$ , such that each vertex in $S_0$ is connected with a vertex in $N_0$ and each vertex in $S_1$ is connected with a vertex in $N_1$ . $\textit{Proof.}$ Each partition of $N$ into two sets can be interpret as assigning to each $n_i\in N$ either a value $0$ if $n_i\in N_0$ or $1$ in case $n_i\in N_1$ . Clearly the family $\mathcal{A}$ of all assignments (partitions) consists of $2^k$ elements.
For each $s\in S$ let us denote by $B_s\subset \mathcal{A}$ the set of "bad" assignments for the student $s$ . That is, in case $s\in S_0$ , $B_s$ consists of all assignments for which to all neighbors of $s$ is assigned $1$ , and in case $s\in S_1,$ $B_s$ are those assignments for which $s$ is connected only with $0$ 's. It means $$ |B_s|= \frac{2^k}{2^d} $$ where $d=d(s)$ . Since $d(s)\ge 4$ we get $|B_s|\le 2^{k-4}$ . Let $B:=\bigcup_{s\in S}B_s$ be all bad assignments. It yields $$ |B|=\left|\bigcup_{s\in S}B_s\right|\le \sum_{s\in S}|B_s|\le 14\cdot 2^{k-4}<2^k. $$ Therefore $|B|<|\mathcal{A}|$ which means there exists an assignment $A\in \mathcal{A}\setminus B$ . Obviously, $A$ comply with all the requirements of the Claim. $\textbf{Comment}.$ Actually, it's a disguised probabilistic approach (or vise versa :) )Note that in the above proof it's not essential the students are $7$ boys and $7$ girls. The size of $S_0$ and $S_1$ can be whatever we want, providing $|S_0\cup S_1|=14$ . Moreover, the same holds even if the number of students is $16$ . But, in this case a further argument is needed. Namely, for any $i,j\in S$ that are of the same sex, that is either $i,j\in S_0$ or $i,j\in S_1$ , it holds $B_i\cap B_j\neq \emptyset$ . | [
"Seems closely related to All-Russian 2005 9.8 & 11.8.\n@below Ok, actually you're right.",
"@above I don't think so. It is correct that a) requires a greedy algorithm (though much simpler than the ARO problem) and for b) I doubt that anything of this sort would work."
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If $x, y, z \in \mathbb{R}$ are solutions to the system of equations $$ \begin{cases}
x - y + z - 1 = 0
xy + 2z^2 - 6z + 1 = 0
\end{cases} $$ what is the greatest value of $(x - 1)^2 + (y + 1)^2$ ? | $y-x=z-1$ $xy=-2z^2+6z-1$ $(y-x)^2+4xy\geq 0\Rightarrow \frac{1}{7}\leq z\leq 3$ $(x-1)^2+(y+1)^2=(y-x)^2+2xy+2(y-x)+2=z^2-2z+1-4z^2+12z-2+2z-2+2=-3z^2+12z-1=-3(z-2)^2+11$ its maximum value is $11$ when $z=2$ . | [
"bump........anyone? anyidea?"
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Let $\triangle ABC$ have incenter $I$ . The line $CI$ intersects the circumcircle of $\triangle ABC$ for the second time at $L$ , and $CI=2IL$ . Points $M$ and $N$ lie on the segment $AB$ , such that $\angle AIM =\angle BIN = 90^{\circ}$ . Prove that $AB=2MN$ . | Interesting Problem.
Let $IH$ be perpendicular to $AB$ . First Lets have some angle chasing stuff. $\frac{\angle A}{2} = \angle HAI = \angle HIM,\frac{\angle B}{2} = \angle HBI = \angle HIN \implies \angle AIN = \frac{\angle C}{2} = \angle BIM$ .
Claim : $ANI$ and $AIC$ are similar.
Proof : $\angle NAI = \angle IAC$ and $AIN = \angle ACI$ .
Let $I'$ be reflection of $I$ across $N$ and $I_c$ be reflection of $C$ across $I$ . Note that $CI=2IL$ so $IL = LI_c$ . Now we have $L$ is center of $AIBI_c$ .
Claim : $I'$ lies on $AIBI_c$ .
Proof : we had $ANI$ and $AIC$ are similar and $I'N = NI , I_cI = IC$ so $AI'I$ and $AI_cC$ are similar so $\angle AI'I = \angle AI_cI$ so $I'$ lies on $AIBI_c$ .
Claim : $AN = NH$ .
Proof : we had $\angle HIN = \angle HBI$ so $NH.NB = NI^2 = NI.NI' = NA.NB \implies NH = NA$ .
with same approach we have $MH = MB$ so $MN = \frac{AB}{2}$ .
we're Done. | [] | [
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A permutation $\sigma$ of the numbers $1,2,\ldots , 10$ is called $\textit{bad}$ if there exist integers $i, j, k$ which satisfy
\[1 \leq i < j < k \leq 10 \quad \text{ and }\quad \sigma(j) < \sigma(k) < \sigma(i)\]
and $\textit{good}$ otherwise. Find the number of $\textit{good}$ permutations. | These permutations are called $312$ avoiding permutations or stack-realizable permutations, because they can be sorted by the following recursively defined procedure $S$ . Let $w$ be the permutation(string) we want to sort in increasing order as $123\dots n$ . Let $w=u1v$ where $u,v$ are strings. Then $S(w)=1S(u)S(v)$ . The number of those permutations with $n$ elements is the Catalan number $\frac{1}{n+1}\binom{2n}{n}$ . Indeed, if $w=u1v$ is a permutation like that then any number in $u$ is less than any number in $v$ (because $w$ is $312$ avoiding permutation). It means that all those permutations can be constructed recursively by putting $1$ into some position, say $j$ -th and then constructing all $312$ -free permutations with the numbers $2,3,\dots,j$ on the left side and with the numbers $j+1,j+2,n$ on the right side. It means, denoting by $g(n)$ the number of all $312$ free permutations of $1,2,\dots,n$ , it holds $$ g(n)=\sum_{j=0}^{n-1} g(j)g(n-1-j) $$ and, as it is well known, this recurrence formula characterizes the Catalan numbers. | [] | [
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Find the smallest odd prime $p$ , such that there exist coprime positive integers $k$ and $\ell$ which satisfy
\[4k-3\ell=12\quad \text{ and }\quad \ell^2+\ell k +k^2\equiv 3\text{ }(\text{mod }p)\] | No need to use overcomplicated ideas. (Hopefully I have not made a mistake in these rushed calculations.) Clearly $k=3x$ and then $\ell = 4x-4$ , so the congruence becomes $37x^2 - 44x + 13 \equiv 0 \pmod p$ . For $p=3$ we get $(x-1)^2 \equiv 0 \pmod 3$ and $k$ and $\ell$ are both divisible by $3$ . For $p=5$ we get $2x^2 - 4x + 3 \equiv 0 \pmod 5$ , impossible by direct check. For $p=7$ we get $2x^2 - 2x + 6 \equiv 0 \pmod 7$ , again impossible by direct check. For $p=11$ we get $4x^2 + 2 \equiv 0 \pmod {11}$ , with $x\equiv 4 \pmod {11}$ as a solution. To ensure that $k$ and $\ell$ are coprime just pick $x=59$ (hmm even $x=15$ works actually). | [
"//wrong//",
"<details><summary>Soln sketch (alternative)</summary>Proceed as below to get $37x^2-44x+13 \\equiv 0 (mod p)$ and multiply it by $37$ and transform it into the form $m^2 \\equiv 3 (mod p)$ then use quadratic residues and Quadratic Reciprocity. Answer is $p=11$ ; construction can be done with ... | [
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Solve the equation
\[(x+1)\log^2_{3}x+4x\log_{3}x-16=0\] | [
" $ x= \\frac {1} {81} $ "
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A circle through the vertices $A$ and $B$ of $\triangle ABC$ intersects segments $AC$ and $BC$ at points $P$ and $Q$ respectively. If $AQ=AC$ , $\angle BAQ=\angle CBP$ and $AB=(\sqrt{3}+1)PQ$ , find the measures of the angles of $\triangle ABC$ . | [] | [
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In every cell of a table with $n$ rows and $m$ columns is written one of the letters $a$ , $b$ , $c$ . Every two rows of the table have the same letter in at most $k\geq 0$ positions and every two columns coincide at most $k$ positions. Find $m$ , $n$ , $k$ if
\[\frac{2mn+6k}{3(m+n)}\geq k+1\] | <details><summary>Solution sketch</summary>The idea is to double count the triplets (column, column, row), where the two columns coincide in the position which represents the row. If the number of triplets is $T$ , then the idea is to prove $\frac{m(m-1)}{2}k \geq T \geq \frac {\frac{m^2}{3}-m}{2}n=\frac{(m^2-3m)n}{6}$ where the first one is due to the problem statement, the second one due to the following: if there are $n_i$ cells of each color in a fixed row, then the number of pairs of columns which coincide in position of this row is $\sum \frac{n_i(n_i-1)}{2}$ and you can find lower bound of this by CS. The rest of the problem is to see equality holds.</details> | [] | [
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Let $n \geq 2$ be a positive integer. The set $M$ consists of $2n^2-3n+2$ positive rational numbers. Prove that there exists a subset $A$ of $M$ with $n$ elements with the following property: $\forall$ $2 \leq k \leq n$ the sum of any $k$ (not necessarily distinct) numbers from $A$ is not in $A$ . | I hope this works because it seems extremely beautiful. <details><summary>Solution</summary>If a set $S$ satisfies the property which we want for $A$ (but is not necessarily a subset of $M$ ), we will call it $\textit{sum-free}$ . First, notice that if $B$ is $\textit{sum-free}$ , then $B'=\{cx|x\in S\}$ is also $\textit{sum-free}$ for any $c>0$ . Thus if we multiply all numbers in $M$ by the lcm of the denominators we will get a new set $M^{*}$ consisting only of positive integers (since we multiply a positive rational number by a $c>0$ to get an integer, this means that the new set $M$ consists of positive integers).
Let $p$ be a prime of the form $(2n-1)q-n+1$ . Since $\gcd(n-1,2n-1)=1$ , by Dirichlet's prime number theorem we know that infinitely many such prime numbers exist, so we can choose a big enough $q$ (for example, $q>\max_{x\in M^{*}} x$ will suffice). Now, notice that the set:
\[S=\{q,q+1, q+2\ldots ,2q-1\}\] is $\textit{sum-free}$ . Indeed, if we look at the extremal cases, we notice that if $\sum$ denotes the sum of $k$ integers from the set $S$ , then:
\[2q-1<q+q\leq \sum\leq n(2q-1)<p+q\]
Now the key observation is that the sets $S_{t}=\{tx (\text{mod } p)|x\in S, 1\leq t\leq p-1\}$ are sum-free since we noted above that if $B$ is $\textit{sum-free}$ , then also is $B'=\{cx|x\in B\}$ and the result follows from the fact that $S$ is $\textit{sum-free}$ . Notice that $S\equiv S_{1}$ and now let's construct a $q\times (p-1)$ table (with $q$ columns and $p-1$ rows). In the $i$ -th row we write the numbers from $S_{i}$ , so that every cell contains a single number between $1$ and $q-1$ (we picked $q>\max_{x\in M^{*}} x$ earlier). Since $p$ is prime, that means that each column contains every residue modulo $p$ . That is, if we have the numbers $xt (\text{mod }p)$ and $xs(\text{mod }p)$ in rows $t$ and $s$ respectively, then $q\leq x\leq 2q-1$ , so $p\nmid x$ and $|t-s|\leq |(p-1)-1|<p$ , so $xs-xt\not\equiv 0(\text{mod }p)$ , thus every residue from $1$ to $p-1$ appears in every column. Thus, if for every number in $M^{*}$ we write a $1$ in the cells of the table which are congruent to it modulo $p$ (here since $p$ is huge, the numbers from $M^{*}$ are themselves residues mod $p$ , but nevermind), we would have $|M^{*}|\cdot (q-1)$ ones since every number appears in every column once. However $|M^{*}|\cdot q=(2n^2-3n+2)q$ . On the other hand, notice that the sets $S_{1},S_{2},\ldots ,S_{n-1}$ don't contain numbers from $M^{*}$ because $\min_{x\in S_{t}} x=tx (\text{mod }p)>q>\max_{x\in M^{*}} x$ and $\max_{x\in S_{t}} x=(2q-1)t \leq (2q-1)(n-1)<p<p+\min_{x\in M^{*}} x$ . Thus, by the PHP we have a row with at least
\[\Bigg\lceil \frac{(2n^2-3n+2)q}{(p-1)-(n-1)}\Bigg\rceil=\Bigg\lceil \frac{(2n^2-3n+2)q}{(2n-1)(q-1)}\Bigg\rceil\quad 1\text{'s}\]
Note that the numbers in this set form a $\textit{sum-free}$ subset $A$ of $M^{*}$ since they are a subset of some $\textit{sum-free}$ $S_{i}$ . Now we only have to prove that:
\[\frac{(2n^2-3n+2)q}{(2n-1)(q-1)}>n-1\]
\[\iff 2n^2q-3nq+2q>2n^2q-2n^2-nq+n-2nq+2n+q-1\]
\[\iff q>-2n^2+3n-1\]
which is obviously true even for small $q$ , so we're done since
\[\frac{(2n^2-3n+2)q}{(2n-1)(q-1)}>n-1\Longrightarrow \Bigg\lceil \frac{(2n^2-3n+2)q}{(2n-1)(q-1)}\Bigg\rceil\geq n\]</details>
| [
"It is indeed a very nice problem! The best problem of this year's tournament, as I see the problems you've posted. Could you write its author, please? And is this the official solution?",
"<blockquote>It is indeed a very nice problem! The best problem of this year's tournament, as I see the problems you've poste... | [
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$ABCD$ is circumscribed in a circle $k$ , such that $[ACB]=s$ , $[ACD]=t$ , $s<t$ . Determine the smallest value of $\frac{4s^2+t^2}{5st}$ and when this minimum is achieved. | Since $4s^2+t^2 \ge 4st$ , the ratio is at least $4/5$ . For an example, it suffices to consider $|AB|=|BC|=1$ , $|CD|=|DA|=\sqrt{2}$ , $\angle ADC = \angle ABC = \pi/2$ . (Check that $|BD|=2\sqrt{2}/\sqrt{3}$ by Ptolemy and everything works out). | [] | [
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Let $ABCDV$ be a regular quadrangular pyramid with $V$ as the apex. The plane $\lambda$ intersects the $VA$ , $VB$ , $VC$ and $VD$ at $M$ , $N$ , $P$ , $Q$ respectively. Find $VQ : QD$ , if $VM : MA = 2 : 1$ , $VN : NB = 1 : 1$ and $VP : PC = 1 : 2$ . | <details><summary>Solution</summary>Let $AB$ be of length $6$ . The $VM=4,$ $VN=3,$ $VP=2.$ Let $O$ be the intersection of $MP$ and $QN.$ Then $O$ lies on the altitude of the pyramid, which also bisects the right angles $\angle MVP$ and $\angle QVN.$ Let points $Q'$ and $N'$ lie on segments $VA$ and $VC$ , respectively, such that $VQ'=VQ$ and $VN'=VN$ . Then $Q'N'$ and $MP$ also intersect at $H.$ We have now simplified a 3D problem into a 2D problem and can be graphed like so: [asy]defaultpen(fontsize(10pt));
pair V,Q,M,A,H,P,N,C;
V=(0,0);Q=(0,12/5);M=(0,4);A=(0,6);H=(4/3,4/3);P=(2,0);N=(3,0);C=(6,0);dot(" $V(0,0)$ ",V,SW);dot(" $A(0,6)$ ",A,W);dot(" $M(4,0)$ ",M,W);dot(" $Q'$ ",Q,W);draw(A--V);
dot(" $P(2,0)$ ",P,S);dot(" $N'(3,0)$ ",N,S);dot(" $C(6,0)$ ",C,S);draw(C--V--H);draw(Q--N);draw(M--P);dot(" $O$ ",H,NE);[/asy]
Now we can do some light coordinate bashing to find that $O=\left(\tfrac43,\tfrac43\right),$ and then from there $Q'=\left(0,\tfrac{12}5\right).$ So $VQ:QD=VQ':Q'A=\frac{12}5:6-\frac{12}5=\fbox{2:3}.$</details>
I haven't done math in years; hope this solution is ok | [] | [
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} |
Let $P,Q\in\mathbb{R}[x]$ , such that $Q$ is a $2021$ -degree polynomial and let $a_{1}, a_{2}, \ldots , a_{2022}, b_{1}, b_{2}, \ldots , b_{2022}$ be real numbers such that $a_{1}a_{2}\ldots a_{2022}\neq 0$ . If for all real $x$ \[P(a_{1}Q(x) + b_{1}) + \ldots + P(a_{2021}Q(x) + b_{2021}) = P(a_{2022}Q(x) + b_{2022})\]
prove that $P(x)$ has a real root. | I think I got that all $a_i$ are equal, otherwise we can get that there exists $i$ , such that $a_i.Q(x)+b_i=a_{2022}.Q(x)+b_{2022}$ has a root since $Q$ has odd degree, and then we get that $P(x)$ can't have the same sign for all $x$ , so it has a root. The harder case seems to be when all $a_i$ are equal, and no $b_i$ is equal to $b_{2022}$ . How to proceed then?
@below Oh yes, indeed, you're right, comparing leading coefficients and taking large x seems to suffice, thanks. | [
"Similar question in old russian olympiad :))",
"@above when all $a_i$ are equal the left hand side will have a much larger leading coefficient and so the equality will fail for sufficiently large $x$ , I think?"
] | [
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Let $m$ and $n$ be positive integers and $p$ be a prime number. Find the greatest positive integer $s$ (as a function of $m,n$ and $p$ ) such that from a random set of $mnp$ positive integers we can choose $snp$ numbers, such that they can be partitioned into $s$ sets of $np$ numbers, such that the sum of the numbers in every group gives the same remainder when divided by $p$ . | ha,ha, it's correct. Knowing Erdos-Ginzburg-Ziv theorem makes it easy. It says that in any set of $2p-1$ integers (p may not be a prime) there are $p$ of them with sum $0\pmod{p}$ . In the official solution, the proposer proves EGZ thm in case $p$ is prime (because it's easier) in a separate lemma. After that, it's trivial. | [
"Is the answer m-1? Just spam EGZ and we can find mn-1 disjoint sets with size p and their sum is divisible by p, and our set is the union of (m-1)n such sets.",
"Please tell me if it is incorrect, since this is too easy."
] | [
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I think we are allowed to discuss since its after 24 hours
How do you do this
Prove that $d(1)+d(3)+..+d(2n-1)\leq d(2)+d(4)+...d(2n)$ which $d(x)$ is the divisor function | Notice that the inequality holds for $n=1,2,3$ , for now on assume $n>3$ . We will prove the following equivalent statement \[ \sum_{k=1}^n d(2k) > \sum_{k=2}^n d(2k-1) \] For any positive integer $\ell$ let $f_\ell(x)$ denote the number of elements in the set $\{2,4,6 \ldots, 2x \}$ with exactly $\ell$ divisors and $g_\ell(x)$ denote the number of elements in $\{ 3,5,7, \ldots, 2x-1 \}$ with exactly $\ell$ divisors.**Claim:** For all $\ell > 2$ we have $f_\ell(x) \geq g_\ell(x)$ Let $a_m$ is the $m$ 'th smallest odd number with $\ell$ divisors. We can change one of the primes dividing $a_m$ into a $2$ creating a new even number $b_\ell$ with $\ell$ divisors. So, we have a new sequence of evens $b_1<b_2< \cdots$ all with $\ell$ divisors and obeys $b_i < a_i$ for all $i$ . Thus, the $m$ 'th smallest even number with $\ell$ divisors is strictly less that the $m$ 'th smallest odd with $\ell$ divisors. This implies the claim. $\square$ Notice that \[ \sum_{k=1}^n d(2k) > \sum_{k=2}^n d(2k-1) \Longleftrightarrow \sum_{k=2}^n kf_k(n) > \sum_{k=1}^n kg_k(n)\] We see $g_2(n) > f_2(n)$ since this just counts primes. (Handwaving) The LHS is putting more weight on bigger numbers, so it is bigger. The RHS is only putting more weight on $2$ than the LHS, which is the smallest number.
| [
"<details><summary>Solution</summary>Count number of pairs $(x,y)$ such that $x|y$ . Each odd $x$ gives at most 1 to LHS-RHS but x=2 gives n to RHS-LHS, GG</details>",
"not sure if discussion is allowed yet, but if it is,\n\n<details><summary>idea</summary>double count each side by how many times each number... | [
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} |
If $ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0$ , find the value of $b\sqrt{a^2+b}+a\sqrt{b^2+a}$ | The first equation becomes $(ab+\sqrt{a^2+b}\sqrt{a+b^2})^2=(-\sqrt{ab+1})^2$ . This rearranges to $a^3+2a^2b^2+b^3+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$ . Now let $x=b\sqrt{a^2+b}+a\sqrt{b^2+a}$ ; then we get $x^2=a^3+2a^2b^3+b^2+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$ , so $x=\pm 1$ . Now suppose $x=-1$ . Note that $ab\leq 0$ , since $\sqrt{a^2+b}\sqrt{a+b^2}+\sqrt{ab+1}\geq 0$ . Then one of the variables is $\leq 0$ , one is $\geq 0$ . WLOG let $a\leq 0, b\geq 0$ . Then by $x=-1$ we get $(a\sqrt{b^2+a})^2=(-1-b\sqrt{a^2+b})^2$ , which rearranges to $a^3-b^3=1+2b\sqrt{a^2+b}$ . But since $a^3-b^3\leq 0$ and $\geq 0$ , this is impossible, so $\boxed{x=1}$ .
Side note, is a construction needed for this problem? The phrasing made it seem like you didn't, and I couldn't actually find one in contest (I think the exact phrasing is something like "Suppose real numbers $a, b$ satisfy...find, with prove, the value of...") | [
"Nice I got the answer $1$ as well\nBut I kinda guessed.\n\nI cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds\n\nAssume that $b>0,a<0$ and then I got $b^3=\\frac{\\sqrt{5}-1}{2}$ , the desired value is $\\sqrt{b^6+b^3}=1$ ",
"I somehow provided a wrong construc... | [
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"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799959.json"
} |
Vishal starts with $n$ copies of the number $1$ written on the board. Every minute, he takes two numbers $a, b$ and replaces them with either $a+b$ or $\min(a^2, b^2)$ . After $n-1$ there is $1$ number on the board. Let the maximal possible value of this number be $f(n)$ . Prove $2^{n/3}<f(n)\leq 3^{n/3}$ . | **The lower bound:****Claim:** let $n$ be a natural number greater than 3, then: $f(n)> 2^{\frac{n}{3}+\frac{1}{2}}$ .
*Proof:* notice that $f(2x) \geq {f(x)}^2$ and $f(2x+1) \geq {f(x)}^2+1$ . Also we can check manually the <u>original</u> lower bound inequality for $\{1,2\}$ and the <u>strong</u> lower bound inequality for $\{3,4,5\}$ . Then we proceed by strong induction; write for $n \geq 3$ : $\bullet f(2n) \geq {f(n)}^2 > 2^{\frac{2n}{3}+1} \geq 2^{\frac{2n}{3}+\frac{1}{2}}$ $\bullet f(2n+1) \geq {f(n)}^2 > 2^{\frac{2n}{3}+1} > 2^{\frac{2n}{3}+\frac{1}{2}+\frac{1}{3}} = 2^{\frac{2n+1}{3}+\frac{1}{2}}$ And the claim is proved. $\square$ **The upper bound:**
We proceed by strong induction. Notice that the last pair on the board can be written as $f(m),f(n-m)$ since we need to maximize $f(n)$ .
We encounter two cases: $\bullet f(n)=min\{{f(m)}^2,{f(n-m)}^2\} \leq {f(\left \lfloor \frac{n}{2} \right \rfloor)}^2 \leq 3^{\frac{2}{3}\left \lfloor \frac{n}{2} \right \rfloor} \leq 3^\frac{n}{3}$ $ \bullet f(n)=f(m)+f(n-m)$ : check manually the cases where $n \in \{1,2,3,4,5\}$ , and assume WLOG $m \leq \left \lfloor \frac{n}{2} \right \rfloor$ :
<span style="color:#ff9a00">First case</span> $m=1$ : Notice that when $n \geq 6$ the following holds: $1+3^{\frac{n}{3}} \leq 3^{\frac{1+n}{3}} \Leftrightarrow 1 \leq (\sqrt[3]{3}-1)3^{\frac{n}{3}}$ <span style="color:#ff9a00">Second case</span> $m\neq 1$ : Notice that $f(n) = f(m)+f(n-m) \leq f(m)f(n-m) \leq 3^{\frac{n}{3}}$ So we are done. $\blacksquare$ | [
"optimal strategy:\n\nfor 1,2,3,4,5: just add them all\n\nfor n>5:\n\nfor odd numbers n, take a 1 away, construct the best possible construction for the rest of the 1's, then add the 1 back in (wait does this work)\n\nfor even numbers n, split it in half, take the best construction for each half, then min(squares)"... | [
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"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799970.json"
} |
Call a set of $n$ lines *good* if no $3$ lines are concurrent. These $n$ lines divide the Euclidean plane into regions (possible unbounded). A *coloring* is an assignment of two colors to each region, one from the set $\{A_1, A_2\}$ and the other from $\{B_1, B_2, B_3\}$ , such that no two adjacent regions (adjacent meaning sharing an edge) have the same $A_i$ color or the same $B_i$ color, and there is a region colored $A_i, B_j$ for any combination of $A_i, B_j$ .
A number $n$ is *colourable* if there is a coloring for any set of $n$ good lines. Find all colourable $n$ . | <blockquote>The answer should be $n\ge 5$ I proved that the graph of regions is bipartite and if two regions have distance $\ge 5$ then we are done.
Then I failed to prove two regions have distance $\ge 5$ .</blockquote>
If $n\le 4$ , setting $n$ parallel lines divide the plane into $n+1$ regions, so it fails.
It remains to show $n\ge 5$ works. Consider a graph where each vertex corresponds to a region and there is an edge between two vertices if and only if their corresponding regions intersect at a line. We can represent each region with a binary number with $n$ digits, where for $j=1,\cdots,n$ , a 1 on bit $j$ indicates it is on one fixed side of line $j$ , and 0 on bit $j$ indicates it is on the other side.
Claim 1: This graph is bipartite.
Proof: It suffices to show that all cycles of the graph have even length. Note two vertices are adjacent if and only if their binary numbers differ by exactly one digit. Therefore, each line must be crossed an even number of times for me to go back to the region I started with.
Claim 2: This graph has 2 vertices with distance $5$ (in fact, $n$ ).
Proof: Note $n\ge 5$ . If a line is not parallel to any of the other lines, it must intersect with all regions, so we can induct down.
Otherwise, each line is parallel to another line, so we can casework on $n=4$ and $n=5$ .
Claim 3: A coloring exists if the distance between two vertices is at least 5.
Proof: Suppose $dist(u,v)\ge 5$ . Let $S_j=\{ w | dist(u,w)=j\}$ for $j\in \mathbb{N}$ . Since the graph is bipartite, the induced subgraph of $S_j$ is an anticlique. By the definition of distance, there are no edges between $S_i$ and $S_j$ if $|i-j|\ge 2$ .
We color all vertices in $S_j$ with color $A_{(j \bmod\; 2)}, B_{(j\bmod\; 3)}$ , which works for reasons explained above. | [
"I'm probably misunderstanding the problem, but why are the $B_i$ colors needed? We can do a 2-coloring of the regions as in [EGMO 2017/3](https://artofproblemsolving.com/community/c6h1424941p8024557) and so all $n$ work?",
"<blockquote>I'm probably misunderstanding the problem, but why are the $B_i$ colors... | [
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A pentagon is inscribed in a circle, such that the pentagon has an incircle. All $10$ sets of $3$ vertices from the pentagon are chosen, and the incenters of each of the $10$ resulting triangles are drawn in. Prove these $10$ incenters lie on $2$ concentric circles.
Note: I spent nearly no time on this, so if anyone took CMO and I misremembered just let me know. | Let $I_A,I_B,I_C,I_D,I_E$ be the incenters of triangles $EAB,ABC,BCD,CDE,DEA,$ respectively. Let $J_A,J_B,J_C,J_D,J_E$ be the incenters of triangles $ACD,BDE,CAE,DAB,EBC.$ **<span style="color:red">Claim:</span>** $AI_AJ_DI_BB$ is cyclic.
*Proof.* Notice \begin{align*}\angle AI_AB&=90+\tfrac{1}{2}\angle AEB&=90+\tfrac{1}{2}\angle ACB=\angle AI_BB&=90+\tfrac{1}{2}\angle ADB=\angle AJ_DB.\end{align*} $\blacksquare$ Similarly, $BI_BJ_EI_CC,$ etc, are cyclic.**<span style="color:red">Claim:</span>** $AJ_CJ_AC$ is cyclic.
*Proof.* We know $$ \angle AJ_CC=90+\tfrac{1}{2}\angle DEA=90+\tfrac{1}{2}\angle CDA=\angle AJ_AC. $$ $\blacksquare$ Similarly, $AJ_DJ_AD,$ etc, are cyclic.**<span style="color:red">Claim:</span>** $I_AI_BI_CI_DI_E$ is cyclic.
*Proof.* We see $$ II_A\cdot IA=II_B\cdot IB=II_C\cdot IC=II_D\cdot ID=II_E\cdot IE $$ so there is an inversion at $I$ that maps $A$ to $I_A$ and so on. Since $ABCDE$ is cyclic, $I_AI_BI_CI_DI_E$ is also cyclic. $\blacksquare$ **<span style="color:red">Claim:</span>** A circle concentric with $(I_AI_BI_CI_DI_E)$ circumscribes $J_AJ_BJ_CJ_DJ_E.$ *Proof.* By symmetry, it suffices to show the perpendicular bisector of $\overline{I_BI_C}$ is congruent to the perpendicular bisector of $\overline{J_DJ_A}.$ Note $$ \angle I_BI_CJ_A=360-(180-\angle I_BBC)-(180-\angle CDJ_A)=\tfrac{1}{2}\angle ABC+\tfrac{1}{2}\angle ADC=90. $$ Similarly, all other angles of $I_BI_CJ_AJ_D$ are right, so it is a rectangle. $\blacksquare$ $\square$ | [
"I think the statement is correct",
"qwerty_ytrewq solved it in 30 minutes, but sadly I don't understand his solution\n\nDiscord links: [ page 1 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390087827496/image0.jpg)\n[ page 2 ](https://cdn.discordapp.com/attachments/927412307817005097/95202139... | [
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} |
Let $\triangle{ABC}$ has circumcircle $\Gamma$ , drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$ , similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$ . Prove that if $AB=DE, \angle{ACB}=60^{\circ}$ (sorry it is from my memory I can't remember the exact problem, but it means the same) | Let $AD \cap BC = X$ and $BE \cap CA = Y$ . The Orthocenter Reflection Lemma yields $$ XY = \frac{DE}{2} = \frac{AB}{2}. $$ Now, since $CXY \overset{-}{\sim} CAB$ , we know $$ | \cos ACB | = \frac{CX}{CA} = \frac{XY}{AB} = \frac{1}{2} $$ so $\angle ACB = 60^{\circ}$ or $\angle ACB = 120^{\circ}$ must hold. $\blacksquare$ **Remarks:** I'm pretty sure $\angle ACB = 120^{\circ}$ is possible too.
Also, what is the CJMO? Personally, I only know of one [CJMO](https://artofproblemsolving.com/community/c3012738), and this problem isn't from there. | [
":( lucky imagine getting a doable geo\n\nour geo was so hard\n\nanyway, AB=DE implies AE || BD which implies $\\angle EAD =\\angle EBD$ which by orthocenter reflection stuff blah blah implies BHD and AHE both equilateral so we're done",
"This is my solution provided during the exam\n<details><summary>solution<... | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Canadian Junior Mathematical Olympiad/2799974.json"
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You have an infinite stack of T-shaped tetrominoes (composed of four squares of side length 1), and an n × n board. You are allowed to place some tetrominoes on the board, possibly rotated, as long as no two tetrominoes overlap and no tetrominoes extend off the board. For which values of n can you cover the entire board?
| The answer is $\boxed{n \text{ divisible by } 4}$ . The construction is obvious by copypasting the $4 \times 4$ construction across the board. **<span style="color:#f00">Claim</span>** $4 \mid n$ .
*Proof.* It is easy to see, from say area, that $2 \mid n$ . To show that we require $4 \mid n$ consider a checkerboard coloring. Each tetrominoe covers exactly $3$ squares of one of the colors, and $1$ of the other. Thus we require an even number of tetronimoes. This implies that $8 \mid n^2$ . However it is easy to see then that we must have $4 \mid n$ as claimed. $\square$ | [
" $4\\mid n$ only.\n\nClearly $n$ is even. If $4\\nmid n$ , then the number of T's is odd, and a checkerboard coloring gives contradiction.\n\nFor $4\\mid n$ we have \n\nTTTS\nATSS\nAABS\nABBB",
"<blockquote> $4\\mid n$ only.\n\nClearly $n$ is even. If $4\\nmid n$ , then the number of T's is odd, and a ... | [
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} |
Positive integers $a$ , $b$ , $c$ are given. It is known that $\frac{c}{b}=\frac{b}{a}$ , and the number $b^2-a-c+1$ is a prime. Prove that $a$ and $c$ are double of a squares of positive integers. | We have that $b^2=ac$ so by substituting it here we get $$ ac-a-c+1=(a-1)(c-1)=p $$ so one of them is equal $2$ so another one must be $2k^2$ which finishes the proof $\blacksquare$ | [
"We have that $(a-1)(c-1)$ is prime, so one of $a$ and $c$ equals $2$ . Since $ac$ is a perfect square, it follows that the other of $a$ and $c$ is also double a perfect square.",
"This is quite simple. Simply note that $b^2=ac$ . Thus, \n\\[ac-a-c+1=(a-1)(c-1)\\]\nis a prime. This requires one of ... | [
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} |
In parallelogram $ABCD$ , points $E$ and $F$ on segments $AD$ and $CD$ are such that $\angle BCE=\angle BAF$ . Points $K$ and $L$ on segments $AD$ and $CD$ are such that $AK=ED$ and $CL=FD$ . Prove that $\angle BKD=\angle BLD$ . | $\angle BAF = \angle BCE \implies \angle EAF = \angle ECF \implies AEFC$ is cyclic. $\angle DAF = \angle ECD , \angle ADF = \angle CDE \implies ADF$ and $CDE$ are similar. $\angle BAK = \angle BCL$ and $\frac{BA}{BC} = \frac{CD}{AD} = \frac{ED}{DF} = \frac{AK}{CL}$ so $BKA$ and $BLC$ are similar so $\angle BKD = \angle BLD$ . | [] | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800865.json"
} |
Pete wrote down $21$ pairwise distinct positive integers, each not greater than $1,000,000$ . For every pair $(a, b)$ of numbers written down by Pete, Nick wrote the number $$ F(a;b)=a+b -\gcd(a;b) $$
on his piece of paper. Prove that one of Nick’s numbers differs from all of Pete’s numbers. | Let WLOG $a_{21}$ be the largest number. By plugging $(a_{21}, a_i)$ for $i=1,2,...,20$ and supposing otherwise we obtain $lcm(a_1, a_2,..., a_{20})/a_{21}$ , so $2^{20} \leq a_{21}$ , size contradiction.
@below, ok you're right. We have to prove that $a_k/a_{k+1}$ (otherwise $F(a_k, a_{k+1})$ is between $a_k$ and $a_{k+1}$ ) | [
"<blockquote>Let WLOG $a_{21}$ be the largest number. By plugging $(a_{21}, a_i)$ for $i=1,2,...,20$ and supposing otherwise we obtain $lcm(a_1, a_2,..., a_{20})/a_{21}$ , so $2^{20} \\leq a_{21}$ , size contradiction.</blockquote> $\\mathrm{lcm}$ of 20 different numbers can be less than $2^{20}$ . for ex... | [
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"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800868.json"
} |
Do there exist 2021 points with integer coordinates on the plane such that the pairwise distances between them are pairwise distinct consecutive integers? | [
"Compare with [this problem](\"https://artofproblemsolving.com/community/c6h2800893_pairwise_distances_are_consecutive_numbers\")"
] | [
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} | |
Let $S$ be the set of all $5^6$ positive integers, whose decimal representation consists of exactly 6 odd digits. Find the number of solutions $(x,y,z)$ of the equation $x+y=10z$ , where $x\in S$ , $y\in S$ , $z\in S$ . | <blockquote><blockquote>Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits
We have $x_6+y_6=10$ $x_5+y_5 \geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \geq 10$ $x_3+y_3 \geq 10$ $x_2+y_2 \geq 10$ $x_1+y_1 \geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \geq 10$ has $15$ solutions
So total we have $5*15^5$ solutions.</blockquote>
Hi,good solution but i think that x1+y1 can take any value,what about you?</blockquote>
Hi, if $x_1+y_1$ less than 10, x+y can't be 7 digit number but 10z is 7 digit number | [
"Who can send solution of this problem",
"<blockquote>Who can send solution of this problem</blockquote>\n\nİ dont have any idea :play_ball: ",
"Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits\n\nWe have $x_6+y_6=10$ $x_5+y_5 \\geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \\geq 10$ $x_... | [
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16 NHL teams in the first playoff round divided in pairs and to play series until 4 wins (thus the series could finish with score 4-0, 4-1, 4-2, or 4-3). After that 8 winners of the series play the second playoff round divided into 4 pairs to play series until 4 wins, and so on. After all the final round is over, it happens that $k$ teams have non-negative balance of wins (for example, the team that won in the first round with a score of 4-2 and lost in the second with a score of 4-3 fits the condition: it has $4+3=7$ wins and $2+4=6$ losses). Find the least possible $k$ . | $k=2$
In the first round everyone gets $4-3$ points, in the second round everyone gets $4-2$ points and in the third round everyone gets $4-0$ points, and in the last round the goals difference of 2 players is not negative | [] | [
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Point $P$ is chosen on the leg $CB$ of right triangle $ABC$ ( $\angle ACB = 90^\circ$ ). The line $AP$ intersects the circumcircle of $ABC$ at point $Q$ . Let $L$ be the midpoint of $PB$ . Prove that $QL$ is tangent to a fixed circle independent of the choice of point $P$ . | Let $M$ be midpoint of $AB$ . we claim that our circle is a circle with center $M$ which touches $BC$ . Note that we need to prove $M$ lies on external angle bisector of $\angle PLQ$ . from Thales Theorem we have $ML || AQ$ so $\angle MLP = \angle APB = \angle LPQ$ so we need to prove $PL = QL$ . Note that $PL = LB$ so we need to prove $\angle PQB = \angle 90$ which is true. | [
"We claim the desired circle is the circle whose center is the midpoint $O$ of $AB$ and that touches $BC$ .\n\nWe use complex numbers with $a=-1,b=1$ . Then by intersection formula, $p=\\frac{2cq+q-c}{c+q},l=\\frac 12(b+p)=\\frac{q(c+1)}{c+q}$ . Now the reflection of $O$ over $QL$ is given by $\\frac{\\... | [
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Paul can write polynomial $(x+1)^n$ , expand and simplify it, and after that change every coefficient by its reciprocal. For example if $n=3$ Paul gets $(x+1)^3=x^3+3x^2+3x+1$ and then $x^3+\frac13x^2+\frac13x+1$ . Prove that Paul can choose $n$ for which the sum of Paul’s polynomial coefficients is less than $2.022$ . | As stated in the title of the problem we need to show that we can choose $n$ , such that: $$ \sum_{i=0}^{n}\frac{1}{{n \choose i}} < 2.022 $$ Notice that for each $i$ , $1 < i < n$ , we have that: $$ \frac{1}{{n \choose 2}} \geq \frac{1}{{n \choose i}} $$ thus we have that: $$ LHS = 2+\sum_{i=1}^{n-1}\frac{1}{{n \choose i}} \leq 2+(n-2)\frac{2}{n(n-1)} < 2.022 $$ which implies that we must show that, there exists an $n$ such that: $$ \frac{n-2}{n(n-1)} < 0.011 $$ but notice that we have that: $$ \lim_{n \rightarrow \infty} \frac{n-2}{n(n-1)} = 0 $$ Which means that as $n$ gets bigger the fraction gets smaller and smaller, thus there exists an $n$ so that the sum is less than $2.022$ | [
"Related to USA TST 2000/4 https://artofproblemsolving.com/community/c6h61959p371496. This result can be obtained by using the recursion in post #10 in that thread (we have to prove that the sum of the reciprocals of the coefficients has lim=2).",
"We have to prove that \n\\[\\mathop {\\lim }\\limits_{n \\to \\in... | [
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Given a rectangular table with 2 rows and 100 columns. Dima fills the cells of the first row with numbers 1, 2 or 3. Prove that Alex can fill the cells of the second row with numbers 1, 2, 3 in such a way that the numbers in each column are different and the sum of the numbers in the second row equals 200. | <blockquote>Given a rectangular table with 2 rows and 100 columns. Dima fills the cells of the first row with numbers 1, 2 or 3. Prove that Alex can fill the cells of the second row with numbers 1, 2, 3 in such a way that the numbers in each column are different and the sum of the numbers in the second row equals 200.</blockquote> $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $x$ be the number of $1's$ , $y$ the number of $3's$ and $z$ the number of $2's$ $\Rightarrow x+y+z=100...(I)$ Let $S$ be the sum of the numbers in the second row
Let $2_i$ the $i$ -th $2's$ Let " $a\to b$ " the transformation that Alex makes, that is, if Dima places $a$ above, Alex places $b$ below $\color{red}\boxed{\textbf{If the number of 2's is even}}$ $\color{red}\rule{24cm}{0.3pt}$ Numbers of $2's=z=2k$ Alex's strategy is: $$ 2_{2i-1}\to 1, \forall 1\le i\le k $$ $$ 2_{2i}\to 3, \forall 1\le i\le k $$ $$ 1\to 2 $$ $$ 3\to 2 $$ $$ \Rightarrow S=(1)(k)+(3)(k)+(2)(x)+(2)(y) $$ $$ \Rightarrow S=4k+2x+2y $$ $$ \Rightarrow S=2x+2y+2z $$ By $(I):$ $$ \Rightarrow S=200_\blacksquare $$ $\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If the number of 2's is odd}}$ $\color{red}\rule{24cm}{0.3pt}$ Numbers of $2's=z=2k+1$ If there is at least one $1$ Alex's strategy is: $$ 2_{2i-1}\to 1, \forall 1\le i\le k $$ $$ 2_{2i}\to 3, \forall 1\le i\le k $$ $$ 2_{2k+1}\to 1 $$ $$ 1\to 2, \text{ except the last one that will become 3} $$ $$ 3\to 2 $$ $$ \Rightarrow S=(1)(k)+(3)(k)+1+(2)(x-1)+3+(2)(y) $$ $$ \Rightarrow S=4k+2+2x+2y $$ $$ \Rightarrow S=2x+2y+2z $$ By $(I):$ $$ \Rightarrow S=200_\blacksquare $$ If there is at least one $3$ Alex's strategy is: $$ 2_{2i-1}\to 1, \forall 1\le i\le k $$ $$ 2_{2i}\to 3, \forall 1\le i\le k $$ $$ 2_{2k+1}\to 3 $$ $$ 1\to 2 $$ $$ 3\to 2, \text{ except the last one that will become 1} $$ $$ \Rightarrow S=(1)(k)+(3)(k)+3+(2)(x)+1+(2)(y-1) $$ $$ \Rightarrow S=4k+2+2x+2y $$ $$ \Rightarrow S=2x+2y+2z $$ By $(I):$ $$ \Rightarrow S=200_\blacksquare $$ $\color{red}\rule{24cm}{0.3pt}$ $$ \Rightarrow \boxed{\textbf{Alex can achieve his goal}}_\blacksquare $$ $\color{blue}\rule{24cm}{0.3pt}$ | [] | [
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Let $\omega$ is tangent to the sides of an acute angle with vertex $A$ at points $B$ and $C$ . Let $D$ be an arbitrary point onn the major arc $BC$ of the circle $\omega$ . Points $E$ and $F$ are chosen inside the angle $DAC$ so that quadrilaterals $ABDF$ and $ACED$ are inscribed and the points $A,E,F$ lie on the same straight line. Prove that lines $BE$ and $CF$ intersectat $\omega$ . | Equivalent, but without the use of spiral similarity basic trickery (I know it is easier to figure out with it, but just for the sake of inexperienced people).
We have $\angle DFE = 180^{\circ} - \angle AFD = \angle ABD = \angle BCD$ from the cyclic $ABDF$ and the tangency around $\omega$ ,
similarly $\angle AED = \angle ACD = \angle DBC$ . Hence $\triangle DBC \sim \triangle DEF$ . Hence $\frac{BD}{DC} = \frac{DE}{DF}$ , i.e. $\frac{BD}{DE} = \frac{DC}{DF}$ and also $\angle BDC = \angle EDF$ , i.e. $\angle BDE = \angle CDF$ . Hence $\triangle BDE \sim \triangle CDF$ , so $\angle DBE = \angle DCF$ , thus after extending $BE$ and $CF$ to intersect $\omega$ , the respective arcs would be equal and so the intersection points would coincide, done. | [
"Again easy P4 (Like lsast year's P4). But anyway, it's better than last year's P4.\nInvert around circle $(A,AB)$ and assume $X'$ is image of $X$ under this inversion. We get $D'=AD\\cap \\omega, E'=AE\\cap CD', F'=AF\\cap BD'$ and let $T=BE\\cap CF$ .\nSince $AE\\cdot AE'=AC^2$ we get $(CEE')$ tangen... | [
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Prove that infinitely many positive integers can be represented as $(a-1)/b + (b-1)/c + (c-1)/a$ , where $a$ , $b$ and $c$ are pairwise distinct positive integers greater than 1. | $a=(2k+1)×k$ $b=(2k+1)×k-1$ $c=2k+1$ $(a-1)/b + (b-1)/c + (c-1)/a=k+1$ $k$ is a positive integer greater than 1 | [] | [
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Do there exist 100 points on the plane such that the pairwise distances between them are pairwise distinct consecutive integer numbers larger than 2022? | [
"Compare with [this problem](\"https://artofproblemsolving.com/community/c6h2800872_pairwise_distances_of_lattice_points_are_consecutive_numbers\")"
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Let $ABC$ be an acute triangle. Let $P$ be a point on the circle $(ABC)$ , and $Q$ be a point on the segment $AC$ such that $AP\perp BC$ and $BQ\perp AC$ . Lot $O$ be the circumcenter of triangle $APQ$ . Find the angle $OBC$ . | Let $H$ be orthocenter of $ABC$ , $S$ be foot of $A$ on $BC$ and $QB$ meet $APQ$ at $K$ . we have $\angle APK = \angle 90 = \angle ASB \implies BS || KP$ . It's well known that $P$ is reflection of $H$ across $BC$ so $HB = BK$ so $B$ lies on perpendicular bisector of $KP$ so $OB \perp KP || BC \implies OB \perp BC \implies \angle OBC = \angle 90$ . | [
"Let $H$ be orthocenter of $\\triangle ABC$ and $D$ be foot of $A \\text{-altitude}$ in $\\triangle ABC$ . Define $E=BQ\\cap (APQ)$ , where $E\\ne Q$ . $\\angle EQA=90 \\implies O\\in EA$ . $\\angle APE=\\angle AQE=90\\implies BD||EP$ and since $DH=DP$ we get $BH=BE$ . So $OB$ is $E\\text{-midlin... | [
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There are $n > 2022$ cities in the country. Some pairs of cities are connected with straight two-ways airlines. Call the set of the cities {\it unlucky}, if it is impossible to color the airlines between them in two colors without monochromatic triangle (i.e. three cities $A$ , $B$ , $C$ with the airlines $AB$ , $AC$ and $BC$ of the same color).
The set containing all the cities is unlucky. Is there always an unlucky set containing exactly 2022 cities? | The answer is negative. We will construct a graph with 2023 vertices such that the whole graph is unlucky but every 2020 subgraph is not so. Note that if we add some empty vertices to the graph a counterexample for all $n$ is built.
<span style="color:#f00">**Observation 1.** </span> If we color $K_5$ properly every vertex has 2 edges of each color; otherwise 3 edges have same color and the triangle forming by their uncommon vertices cannot be colored properly.
<span style="color:#f00">**Observation 2.** </span> $K_5$ is colored quite uniquely. <details><summary>Proof</summary>Take a vertex $v_1$ and let it connect with color blue to $v_2, v_3$ and color red to $v_4, v_5$ . $v_4v_5$ must be blue and $v_2v_3$ must be red. WLOG assume that $v_2v_4$ is red, then $v_3v_4$ must be blue, so $v_3v_5$ must be red and $v_2v_5$ must be blue.</details>
<span style="color:#f00">**Observation 3.** </span> If we connect one vertex to four vertices of a properly colored $K_5$ , the edges of that vertex must be colored exactly like the missing vertex of $K_5$ . <details><summary>Proof</summary>This is because every one of four vertices due to observation 1 misses an specific color to be complete and hence the edges of any vertex connecting to these four has a unique way of coloring.</details>
It is now enough to make a cycle of $K_5$ s. Put 2023 vertices on a circle and connect each vertex to all 4 nearest points in each direction. Assume for the sake of contradiction that the said graph could be properly colored. Then consider 5 consecutive points on the circle. They form a $K_5$ . Now the next 5 points on the circle form a $K_5$ that only differs in one vertex and due to observation 3 must be colored with the same pattern. We can conclude that the edges that connect two consecutive points is colored in a repeating pattern of length 5. But 2023 is not divisible by 5 and hence such pattern cannot exist. The contradiction means the said graph with $2023$ vertices cannot be properly colored. Meanwhile every subgraph of size 2022 can be colored properly. We only need to color a $K_5$ and just as described color other $K_5$ s one by one in each direction.
<details><summary>Motivation</summary>Sticking to $K_5$ came to mind after noting that $R(3,3) = 6$ and so every $K_6$ cannot be properly colored, this suggest that a weak $K_6$ could be bound to some restriction. After noting that $K_5$ is restricted, working with a cycle to abuse the restriction is quite natural.</details> | [
"bump this gem"
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Let $a$ and $b$ be two positive real numbers, and $AB$ a segment of length $a$ on a plane. Let $C,D$ be two variable points on the plane such that $ABCD$ is a non-degenerate convex quadrilateral with $BC=CD=b$ and $DA=a$ . It is easy to see that there is a circle tangent to all four sides of the quadrilateral $ABCD$ .
Find the precise locus of the point $I$ . | @2 above since $ABCD$ is not degenrate, not all points on this circle work. My solution is same with at 2 above, but I find the locus as arc of this circle.
Let circle with center $B$ and radius $b$ be $\Gamma$ and let Homothety with center $A$ and radius $\frac{a}{a+b}$ maps $\Gamma$ to $\omega$ . Then it's obvious that $B$ lies on $\omega$ . Let $AX_1$ and $AX_2$ tangent to $\omega$ at $X_1,X_2$ , rescpectively. Then locus of $I$ is $\text{arc}$ $X_1BX_2$ of $\omega $ , except points $X_1,X_2$ and $B$ . | [
"Is a,b fixed?",
"<blockquote>Is a,b fixed?</blockquote>\n\nYes.",
"Since $ABCD$ is a deltoid, $I$ lies on its axis of symmetry $AC$ . Therefore, by bisector theorem on $BCA$ , we have $AB:BC=AI:IC$ , which implies $AI=\\frac{a}{a+b}AC$ . Since $C$ lies on a fixed circle of radius $b$ and center $B... | [
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Find the largest real number $\lambda$ with the following property: for any positive real numbers $p,q,r,s$ there exists a complex number $z=a+bi$ ( $a,b\in \mathbb{R})$ such that $$ |b|\ge \lambda |a| \quad \text{and} \quad (pz^3+2qz^2+2rz+s) \cdot (qz^3+2pz^2+2sz+r) =0. $$ | Take $p=q=r=s=1$ ,we got $\lambda \leq \sqrt3$ To prove $\lambda = \sqrt 3$ ,we suppose the roots of the two equations are $x_1,x_2,x_3;y_1,y_2,y_3$ we have(Viete) $\sum x_i \sum {y_i}=\sum\frac{1}{x_i}\sum\frac{1}{y_i}=4$ but one can easily prove that $\sum x_i \sum \frac{1}{x_i},\sum y_i \sum \frac{1}{y_i}>4$ when $|b|<\sqrt3 |a|$ <details><summary>reason</summary>When $x_i$ are reals this is just Cauchy (notice $x_i<0$ ),otherwise suppose $x_1$ is real we have $(x_2+x_3)(\frac{1}{x_2}+\frac{1}{x_3})=\frac{4x^2}{x^2+y^2}>1(x_{2,3}=x\pm yi)$ ,use Cauchy again.
(Notice $x<0$ ,otherwise $2x+x_1<0,x^2+y^2+2xx_1>0(Viete) \rightarrow 3x^2 \leq y^2$</details> $\square$ | [
"Not a difficult problem! The answer is sqrt(3).\nTo prove it we can simply assume the contradictory. Using the fact that these two equality have a solution in R, assume them out as x1, x2 to get another two quadratic ones, like x^2+A1x+B1. Then by the assumption, A1^2>B1. Use p/q and q/p,r/s and s/r to establish t... | [
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Find all positive integers $a$ such that there exists a set $X$ of $6$ integers satisfying the following conditions: for every $k=1,2,\ldots ,36$ there exist $x,y\in X$ such that $ax+y-k$ is divisible by $37$ . | Clearly $37 \nmid a$ . We will take the elements of $X$ as residues $\pmod{37}$ . Clearly all of them must be nonzero and distinct.
So if $\omega$ is any primitive $37$ th root of unity, then it is necessary and sufficient for $$ \left(\sum_{n\in X} \omega^{an}\right)\left(\sum_{n\in X} \omega^{n}\right)=\left(\sum_{n\in X} \omega^{a^2n}\right)\left(\sum_{n\in X} \omega^{an}\right) =\sum_{n=1}^{36} \omega^n = -1. $$ Consider the polynomial $\sum_{n \in X} x^{a^2n}-\sum_{n\in X} x^{n}$ . For each term $x^m$ in the polynomial, replace it with $x^{m\text{'s remainder mod 37}}$ . Then this polynomial has degree less than $37,$ but is divisible by $x^{37}-1.$ So it's the zero polynomial, and $a^2X \equiv X \pmod{37}$ . Thus, $ord_{37}(a^2) \mid 6.$ If $a^3 \equiv -1,$ $$ \left(\sum_{n\in X} \omega^{an}\right)\left(\sum_{n\in X} \omega^{n}\right)=\left(\sum_{n\in X} \omega^{-\frac{n}{a^2}}\right)\left(\sum_{n\in X} \omega^{n}\right)=\left(\sum_{n\in X} \omega^{-n}\right)\left(\sum_{n\in X} \omega^{n}\right) $$ which is the magnitude of $\sum_{n\in X} \omega^{n}$ squared, which cannot be $-1.$ If $a^3 \equiv 1,$ then $aX \equiv X \pmod{37}$ . So $$ \left(\sum_{n\in X} \omega^{n}\right)^2=\left(\sum_{n\in X} \omega^{an}\right)\left(\sum_{n\in X} \omega^{n}\right)=\sum_{n=1}^{36} \omega^n $$
for any primitive $37$ th root of unity $\omega$ . Take the polynomial $\left(\sum_{n\in X} x^{n}\right)^2-\sum_{n=1}^{36} x^n,$ and for each term $x^m$ in the polynomial, replace it with $x^{m\text{'s remainder mod 37}}$ . Then the result has degree less than $37$ and is divisible by $\frac{x^{37}-1}{x-1}$ and $x,$ so it is a zero polynomial, but it's not hard to see that this is impossible.
So $ord_{37}(a^2) = 2$ or $6.$ In the former case, $\boxed{a \equiv 6 \pmod{37}}$ works if $X = \{16,17,18,19,20,21 \}$ . Note that in this case, $$ \left(\sum_{n\in X} \omega^{6n}\right)\left(\sum_{n\in X} \omega^{n}\right) = \omega^{16} \cdot \frac{\omega^{6}-1}{\omega-1} \cdot \omega^{16 \cdot 6} \cdot \frac{\omega^{36}-1}{\omega^{6}-1} = \omega \cdot \frac{\omega^{36}-1}{\omega-1}=-1 $$ for any primitive $37$ th root of unity $\omega$ . $\boxed{a \equiv 31 \pmod{37}}$ works as well by an identical argument.
In the latter case, $X=\{x,a^2x,a^4x,\cdots,a^{10}x\}$ for some nonzero residue $x$ . We can assume WLOG $x=1,$ then this forces $X=\{1,27,26,36,10,11\}, aX=\{8,31,23,29,6,14\}.$ Note $26+6\equiv 1+31 \pmod{37}$ which is contradiction. $\square$ | [
"<details><summary>Solution</summary>Throughout the solution, a and elements of X are in $\\mathbb{Z}_{37}$ .\n\nThe answer is $a\\equiv \\pm 6$ only. \n\nConstruction: $X=\\{16,17,18,19,20,21\\}$ Proof of optimality: Let $\\omega$ be any primitive $37$ th root of unity. Then $\\left(\\sum_{t\\in X} \\omega... | [
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A conference is attended by $n (n\ge 3)$ scientists. Each scientist has some friends in this conference (friendship is mutual and no one is a friend of him/herself). Suppose that no matter how we partition the scientists into two nonempty groups, there always exist two scientists in the same group who are friends, and there always exist two scientists in different groups who are friends.
A proposal is introduced on the first day of the conference. Each of the scientists' opinion on the proposal can be expressed as a non-negative integer. Everyday from the second day onwards, each scientists' opinion is changed to the integer part of the average of his/her friends' opinions from the previous day.
Prove that after a period of time, all scientists have the same opinion on the proposal. | <details><summary>Wrong solution</summary>Solved with **bora_olmez**
The problem says the graph is connected and non-bipartite. Call the given operation *averaging* and denote it by $f$ . I claim that for a graph $G$ if $f^k(G) = G$ then all the vertices have the same number. Note that the sum of opinions cannot increase since the sum is now the floor of some sums. So since the sum has to eventually return to the same thing, the sum stays constant and for this, every vertex has all its neighbours as the same number. So since between any two vertices, we may find a path with an even number of edges between them (because its not bipartite), they must have the same number, so every vertex has the same number, as claimed. $\square$ .
So suppose eventually the graph is unchanged, then because there are only finitely many possibilities, it must cycle back to the same thing. But from the previous paragraph, this implies that averaging leaves this graph unchanged, as desired. $\blacksquare$</details> | [
"Consider the graph of friendships is connected. The condition of having two scientists in different groups is equivalent to saying the graph is connected (partition connected components). The condition of having two scientists in the same group is equivalent to saying the graph is not 2-colorable/bipartite (by def... | [
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On a blank piece of paper, two points with distance $1$ is given. Prove that one can use (only) straightedge and compass to construct on this paper a straight line, and two points on it whose distance is $\sqrt{2021}$ such that, in the process of constructing it, the total number of circles or straight lines drawn is at most $10.$
Remark: Explicit steps of the construction should be given. Label the circles and straight lines in the order that they appear. Partial credit may be awarded depending on the total number of circles/lines. | Pretty easy problem
Construction brings to mind the methods to multiply a segment. To multiply a segment directly by a factor $a$ , we must construct $a-1$ circles. Note that we can 'copy the segment' by marking points a unit away from each other.
The length $\sqrt{2021}$ immediately gives us the idea of using the Pythagorean theorem. The most obvious way to do this is note $2021 = 45^2 - 2^2$ . Copy the unit segment $45$ times and $2$ times, such that we now have marked a distance $A=45$ with endpoints $A_1, A_2$ and $B=2$ .
Now construct the perpendicular bisector of $A$ and set it to the center of a circle with diameter $45$ . At this point we have constructed $2$ lines and $4$ circles. Now draw a circle with radius $2$ at $A_1$ and let its intersection with the larger circle be $X$ . Then $A_2X$ will have our desired length of $\sqrt{2021}$ and we are done! $\blacksquare$ | [
"Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?",
"<blockquote>Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?</blockquote>\n\nIf so, then $5$ lines/circles suffice. On the line $l_1$ with the two points $A_0$ and $A_1$ ,copy the length ... | [
"origin:aops",
"2022 Contests",
"2022 China National Olympiad"
] | {
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"path": "Contest Collections/2022 Contests/2022 China National Olympiad/2742832.json"
} |
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