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Let $ABC$ be an acute triangle with altitude $AD$ ( $D \in BC$ ). The line through $C$ parallel to $AB$ meets the perpendicular bisector of $AD$ at $G$ . Show that $AC = BC$ if and only if $\angle AGC = 90^{\circ}$ .
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$\bullet$ $CA=CB:$ 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 $CDMG$ is a cyclic isosceles trapezoid. Suppose $\\angle AGC=90.$ Then, $ADCG$ and therefore $AMDCG$ is cyclic so $\\angle MAC=\\angle MGC=\\angle CBM$ and $AC=BC.$ Conversely, if $AC=BC,$ then $\\angle MAC=\\angle ABC=\\angle MGC$ so $AMDCG$ is cyclic and $\\angle AGC=90.$ $\\square$ "
] |
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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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"origin:aops",
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"2022 3rd Memorial "Aleksandar Blazhevski-Cane""
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"answer_score": 128,
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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 let $T=\\sum_{i\\in S_1} i$ and $R=\\sum_{j\\in S_2} j$ and assume $T\\le R$ . Then $T\\mid R \\implies T\\mid T+R \\implies T\\mid (2k+1)(4k+3)$ . Since $|S_1|=2k+1$ , we have $T\\geq (k+1)(2k+1)$ . $\\bullet$ **$k\\not\\equiv 2\\pmod{3}$**: In this case we can find $S_1$ such that $T=\\frac{(2k+1)(4k+3)}{3}$ , because $\\frac{(2k+1)(4k+3)}{2}>\\frac{(2k+1)(4k+3)}{3}>(k+1)(2k+1)$ . $\\bullet$ **$k\\equiv 2\\pmod{3}$**: Observe that $T\\geq (k+1)(2k+1)>\\frac{(2k+1)(4k+3)}{4}$ . Since $ T\\mid (2k+1)(4k+3)$ and $2$ and $3$ doesn't divide $(2k+1)(4k+3)$ we can't find $K$ . \nSo we are done!"
] |
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"origin:aops",
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"answer_score": 90,
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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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"origin:aops",
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"2022 3rd Memorial "Aleksandar Blazhevski-Cane""
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"answer_score": 30,
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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 .\nChoose for example $(2^n-5)^2$ where $n\\ge 5$ : this perfect square has $n+1$ digits $1$ and $n-1$ digits $0$ in its binary representation.\n\n"
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"answer_score": 1006,
"boxed": true,
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"path": "Contest Collections/2022 Contests/2022 3rd Memorial "Aleksandar Blazhevski-Cane"/2759392.json"
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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 ${n-1\\choose r-1}^2={n-1\\choose r-1}{n-1\\choose n-r}$ times, summed from $r=1$ to $r=n$ . By Vandermonde's this sum becomes ${2n-2\\choose n-1}$ , so summing over all values of $k$ , we get $S_n=n{2n-2\\choose n-1}$ . From here, a computation gives that the answer is $\\frac{2022\\cdot 2\\cdot 4041}{2021^2}$ , so taking mod 1000, $p+q=\\boxed{245}$ .",
"Uh I think I successfully engineered but panicked and forgot that Vandermonde's existed when I used it correctly for 2020 AIME I #7 :stretcher:",
"<blockquote>Uh I think I successfully engineered</blockquote>\n\nsame here but took me the first 30 min of the test",
"We first find $S_n$ for any $n$ .\n\nConsider the expected value of $|A\\cap B|$ when $|A|=|B|=k$ . Of the $k$ elements of $A$ , there is a $\\frac{k}{n}$ chance any one of them are in $B$ , so by Linearity of Expectation, this expected value is $\\frac{k^2}{n}$ . There are $\\binom{n}{k}^2$ pairs $(A,B)$ , so of all the pairs with $k$ elements each, the sum of $|A\\cap B|$ is $\\frac{k^2}{n}\\cdot \\binom{n}{k}^2=n\\binom{n-1}{k-1}^2$ . By Vandermonde's, summing this up for all $k$ gives $S_n=n\\binom{2n-2}{n-1}$ , as desired.\n\nNow math gives $245$ as the answer.",
"engineer induction + polynomial interpolation",
"spent thirty minutes bashing recursion and missed the two minute sol",
"Once you realize how disgusting this problem seems, attacking from another perspective, i.e. Global, follows naturally :-D. ",
"I used expected value then Vandermonde's",
"engineer's induction go brrrr",
"On the test, I thought in the wrong direction (or at least I couldn't find anything when thinking in that direction). Wasted over half an hour on this. \n\nWe'll find the number of times some integer $k$ from $1$ to $n$ is included in intersection.\n\nWe have that $k$ is included $\\sum_{i=1}^n \\binom{n-1}{i-1}^2=\\sum_{i=1}^n \\binom{n-1}{i-1}\\binom{n-1}{n-i}=\\binom{2n-2}{n-1}$ by Vandermonde's. \n\nSumming over all $k$ gives $S_n=n\\binom{2n-2}{n-1}$ . \n\nWe have $\\binom{4042}{2021}=\\frac{4042!}{2021!2021!}$ and $\\binom{4040}{2020}=\\frac{4040!}{2020!2020!}$ . Dividing gives $\\frac{4041\\cdot 4042}{2021^2}=\\frac{4041\\cdot 2}{2021}$ . \n\nSo it's totally $\\frac{2022\\cdot 4041\\cdot 2}{2021^2}$ , so the answer is $22\\cdot 41\\cdot 2+441=1804+441=2\\boxed{245}$ . ",
"Unusually low solve rate compared to 7 and 8, potentially due to placement after the fairly hard #11. But it’s basically easy double counting + vandermonde. Also, this was my first solve on the test before #1 weirdly.",
"Overall thought AIME was easy due to many misplaced last 5 problems. \n\nWe take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \\cdot \\frac{k}{n} \\cdot \\frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $\\frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $\\binom{n}{k}^2$ . Summing, we get $$ \\sum_{k=1}^{n} \\frac{k^2}{n} \\binom{n}{k}^2 $$ Notice that we can rewrite this as $$ \\sum_{k=1}^{n} \\frac{1}{n} \\left(\\frac{k \\cdot n!}{(k)!(n - k)!}\\right)^2 = \\sum_{k=1}^{n} \\frac{1}{n} n^2 \\left(\\frac{(n-1)!}{(k - 1)!(n - k)!}\\right)^2 = n \\sum_{k=1}^{n} \\binom{n - 1}{k - 1}^2 = n \\sum_{k=1}^{n} \\binom{n - 1}{k - 1}\\binom{n - 1}{n - k} $$ We can simplify this using Vandermonde's identity to get $n \\binom{2n - 2}{n - 1}$ . Evaluating this for $2022$ and $2021$ gives $$ \\frac{2022\\binom{4042}{2021}}{2021\\binom{4040}{2020}} = \\frac{2022 \\cdot 4042 \\cdot 4041}{2021^3} = \\frac{2022 \\cdot 2 \\cdot 4041}{2021^2} $$ Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1\\boxed{245}$ ",
"<blockquote>spent thirty minutes bashing recursion and missed the two minute sol</blockquote>\nbroooo same\n",
"<blockquote><blockquote>spent thirty minutes bashing recursion and missed the two minute sol</blockquote>\nbroooo same</blockquote>\n\nspent thirty minutes thinking in the wrong direction",
"spent 30 minutes engineer's inducting",
"<blockquote>spent 30 minutes engineer's inducting</blockquote>\n\nbro same \nproudest solve tho ",
"<blockquote>\nbro same \nproudest solve tho</blockquote>\n\"solve\"\nsame here though",
"i used engineers induction as a last resort and not my first idea oops",
"<blockquote><blockquote>\nbro same \nproudest solve tho</blockquote>\n\"solve\"\nsame here though</blockquote>\n\ni call everything that I get correct a solve",
"<blockquote><blockquote><blockquote>\nbro same \nproudest solve tho</blockquote>\n\"solve\"\nsame here though</blockquote>\n\ni call everything that I get correct a solve</blockquote>\n\neven if you guessed it",
"<blockquote><blockquote><blockquote>\n\"solve\"\nsame here though</blockquote>\n\ni call everything that I get correct a solve</blockquote>\n\neven if you guessed it</blockquote>\n\nno not guesses but if I legitimately solved it then it's a solve no matter what method",
"Consider each element $x\\in\\{1,2,...,n\\}$ individually. It's not hard to see that the number of ways to choose $A$ and $B$ such that $x\\in A\\cap B$ is $$ \\sum_{k=0}^{n-1}\\binom{n-1}{k}^2=\\binom{2(n-1)}{n-1} $$ By considering each possibility for the number of elements in $A$ and summing. So by double-counting $$ S_n=n\\binom{2n-2}{n-1} $$ And the rest is easy.",
"Bruh i would have solved this but i didn't notice |A|=|B| ",
"^\n\nsame\n\noops i guess we need to read more carefully",
"<blockquote>\nno not guesses but if I legitimately solved it then it's a solve no matter what method</blockquote>\nengineer's induction, anyone?",
"<details><summary>Solution</summary>Each element of $\\{1,2,\\dots,n\\}$ shows up in the sum $\\binom{n-1}{0}^2$ times in sets with cardinality $1$ , $\\binom{n-1}{1}^2$ times in sets with cardinality $2$ , up until $\\binom{n-1}{n-1}^2$ times in sets with cardinality $n$ . Therefore \\[S_n=n\\left(\\binom{n-1}{0}^2+\\binom{n-1}{1}^2+\\dots+\\binom{n-1}{n-1}^2\\right).\\]\n\nUsing Vandermonde's, \\[\\binom{n-1}{0}\\binom{n-1}{n-1}+\\binom{n-1}{1}\\binom{n-1}{n-2}+\\dots+\\binom{n-1}{n-1}\\binom{n-1}{0}=\\binom{2n-2}{n-1},\\]\n\nso \\[S_n=n\\binom{2n-2}{n-1}.\\]\n\nThe fraction is thus $\\frac{2022\\binom{4042}{2021}}{2021\\binom{4040}{2020}}$ which reduces down to a numerator that ends in $804$ and a denominator ending in $441$ , for an answer of $\\boxed{245}$ .</details>",
"kinda called this (idea is similar, but computation is different) \nMOST had <details><summary>this</summary>In Bringo Hall, there are two rooms; with $2021$ girls in one room, and $2021$ boys in the other. $M\\ge0$ girls are selected and $N\\ge0$ boys are selected in each room, and afterwards are led into the next room. Let $a$ be the number of ways for $M$ to equal $N$ . Find the largest three digit prime divisor of $a$ .</details> problem in their Gold Division",
"Remarkable that SADGIME P6 is also linearity of expectation + enumerative combo\n\nEdit: though the motivation here is less apparent\n\n<blockquote>Cherri walks from the bottom left square to the top right square of a $99 \\times 99$ square grid, taking any one of the possible routes with equal probability. She leaves a trail of integers tracing her path. First, she places $1$ in her starting square. On her nth step, she travels one unit right or one unit up to reach a new square and places $n + 1$ in this new square for all integers $1 \\leq n \\leq 196$ ; provided that she ends on the top right square. Find the expected value of the sum of the integers Cherri places in the middle row of the grid.\n</blockquote>",
"SADGIME P6 was more slightly engineer's induction-able though ",
"bruh such an easy problem, there were a lot of problems like this on intermediate C&P\ncouldn't get this during the test because I tried to bash 11 and 12 then added 1 instead of 3 on 13",
"idk though, for some reason it was statistically the second hardest problem on the AIME I",
"i think ppl just got scared by notation\n\nbecause this was like rlly similar to 2020 AIME I #7",
"<blockquote>bruh such an easy problem, there were a lot of problems like this on intermediate C&P\ncouldn't get this during the test because I tried to bash 11 and 12 then added 1 instead of 3 on 13</blockquote>\n\nikr but I just skipped it ;-;",
"This reminds me of a problem in 112 combo, but I forgot which problem exactly\n\nLet $k\\in \\{1, 2, \\dots, n\\}$ be an integer, and consider how many times $k$ contributes to $S_n$ . This happens when $k\\in A$ and $k\\in B$ . It remains to add some number of elements of $\\{1, 2, \\dots, n\\}\\backslash \\{k\\}$ to $A$ and an equal number of elements to $B$ . There are $$ \\binom{n-1}{0}^2 + \\binom{n-1}{1}^2 + \\dots + \\binom{n-1}{n-1}^2 = \\binom{2n-2}{n-1} $$ ways to do this. Therefore, each $k\\in \\{1, 2, \\dots, n \\}$ contributes $\\binom{2n-2}{n-1}$ to the total sum. Hence, $S_n = n\\binom{2n-2}{n-1}$ . The rest is computation:\n\\[ \\frac{S_{2022}}{S_{2021}} = \\frac{2022\\binom{4042}{2021}}{2021\\binom{4040}{2020}} = \\frac{2022}{2021}\\cdot \\frac{4042!}{2021!^2}\\cdot\\frac{2020!^2}{4040!} = \\frac{4044\\cdot 4041}{2021^2}.\\]\nWe get $p + q = 4044\\cdot 4041 + 2021^2\\equiv 44\\cdot 41 + 21^2\\equiv \\boxed{245}\\pmod{1000}$ .",
"[Video Solution](https://youtu.be/cXJmHV5BnfY)",
"Wow clean math work I wish :P",
"let's actually not engineer induct\n\nConsider all pairs of $m$ elements subsets. In each pair, the expected value of the intersection by LOE is $$ n\\cdot (m/n)^2=\\frac{m^2}{n}. $$ Since there are $\\binom{m}{n}^2$ such pairs, we have $$ S_n=\\sum_{m=0}^n \\binom{m}{n}^2 \\cdot \\frac{m^2}{n}=1/n \\sum_{m=0}^n \\binom{m}{n}^2 \\cdot m^2=1/n (n^3 C_{n-1})=n\\binom{2n-2}{n-1}. $$ Simply plugging in gives the answer of 245.",
"John post on your youtube channel pls :)))",
"First we find a general form for $S_n$ . As we typically do in these problems, count each individual element seperately based on how many times it appears in $S_n$ . In it clear that every element is symmetric, so just consider $1$ . It is counted only once for the one element susbets. It is counted $(n-1)^2$ times for the two element subsets because for $A$ you choose one more element out of the remainding $n-1$ and the same for $B$ (remember, ordered and not nessacerily distinct). Similarly for three elements it is counted $\\left( \\binom{n-1}{2} \\right)^2$ , and on. Summing over all elements, $$ S_n = n \\left( \\left( \\binom{n-1}{0} \\right)^2 + \\left( \\binom{n-1}{1} \\right)^2 + \\left( \\binom{n-1}{2} \\right)^2 + \\dots + \\left( \\binom{n-1}{n-1} \\right) ^2 \\right) = n \\cdot \\binom{2n-2}{n-1} $$ So $$ \\frac{S_{2022}}{S_{2021}} = \\frac{2022 \\cdot \\frac{4042!}{2021! \\cdot 2021!}}{2021 \\cdot \\frac{4040!}{2020! \\cdot 2020!}} = \\frac{2 \\cdot 2022 \\cdot 4041}{2021 \\cdot 2021} $$ And the answer is $$ 2 \\cdot 2022 \\cdot 4041 + 2021 \\cdot 2021 \\equiv 804+441 \\equiv \\boxed{245} \\pmod{1000} $$ ",
"We use indicator functions. We have $$ S_{n} = \\sum_{i=1}^{n} \\sum_{|A| = |B| = i} |A \\cap B| = \\sum_{i=1}^{n} \\sum_{|A|=|B| = i} \\sum_{j = 1}^{n} 1_{j \\in A}1_{j \\in B} = \\sum_{i=1}^{n} \\left(\\sum_{|A|=i}^{n}\\sum_{j=1}^{n} 1_{j \\in A}\\right)^{2}. $$ Now, $$ \\sum_{|A|=i}^{n}\\sum_{j=1}^{n} 1_{j \\in A} = \\sum_{j=1}^{n} \\sum_{j \\in A, |A| = i} 1 = \\sum_{j=1}^{n}{n-1 \\choose i-1} = n {n-1 \\choose i-1}, $$ so we get $$ S_{n} = \\sum_{i=1}^{n} n {n-1 \\choose i-1}^{2} = n {2n-2 \\choose n-1}. $$ When we divide these, we get $$ \\frac{S_{2022}}{S_{2021}} = \\frac{2022 \\cdot 8082}{2021^{2}}, $$ and since the numerator and denominator here are relatively prime, we get $2022 \\cdot 8082+2021^{2} \\equiv \\boxed{245} \\pmod 1000$ ",
"Begin by assuming that $|A|=|B|=k$ and $|A \\cap B|=a$ . \n\n\nThus after we have determined these variables, we begin by picking the common elements of $A$ and $B$ in $\\binom{n}{a}$ ways, and then selecting the remaining elements of A and B (in that order) in $\\binom{n-a}{k-a}\\binom{n-k}{k-a}$ ways. Thus we desire $\\sum_{k=1}^n\\sum_{a=1}^k a\\binom{n}{a} \\binom{n-a}{k-a} \\binom{n-k}{k-a}$ . \n\n\nBegin by noting $\\binom{n}{a}\\binom{n-a}{k-a} = \\frac{n!}{a! (k-a)! (n-k)!} = \\frac{n!k!}{a!(k-a)!(n-k)!k!} = \\binom{n}{k}\\binom{k}{a}$ . Since $\\binom{n}{k}$ is independent of $a$ , we can move it out of the inner summation and we are left with $\\sum_{a=1}^k a \\binom{k}{a} \\binom{n-k}{k-a}$ . This equals $\\sum_{a=1}^k k\\binom{k-1}{a-1} \\binom{n-k}{k-a}$ . From here, we have a direct application of Vandermonde's identity and this entire summation collapses into $\\binom{n-1}{k-1}$ .\n\n\nNow we are just left to deal with $\\sum_{k=1}^n k\\binom{n}{k} \\binom{n-1}{k-1} = n\\sum_{k=1}^n \\binom{n-1}{n-k}\\binom{n-1}{k-1} = n\\binom{2n-2}{n-1}$ from another direct application of Vandermonde's. This is actually our answer, and as a verification we do indeed have $S_2=2 \\binom{2}{1}=4$ .",
"This was a really fun question. First we want to do casework on $|A|$ , which can range from $0$ to $n$ . This variable will be represented as $k$ . There are $\\binom{n}{k}^2$ ways to choose $k$ elements for both $a$ and $b$ . For any given set of $k$ elements, there is a $\\frac{k}{n}$ probability that a given element is common to both $A$ and $B$ , so then by linearity of expectation the expected value of elements is $\\frac{k^2}{n}$ We end up getting the summation $\\sum_{k=0}^{n}\\frac{k^2}{n}\\binom{n}{k}^2,$ which we can reduce to $n\\binom{2n-2}{n-2}$ . Plug in our given values, and we end up getting $\\frac{8082\\cdot2022}{2021^2}$ , which ultimately yields a final value of $\\boxed{245}$ .",
"Use Vandermonde's and bash the double sum",
"I HATE this question.\n\nMy anti-combo brain can't think of anything but engineer's induction for this one",
"We count the number of times any element in the set of $n$ integers appears. We concern ourselves with the number $j$ . \n\nIn particular, for the set of subsets with $m$ elements, $j$ appears ${{{n-1} \\choose {m-1}} \\choose {2}}$ times. Of course, because the problem is really bad and includes order for the sake of getting a nice, round answer extraction, we have to multiply by $2$ . Moreover, we also have to include the fact that you can get numbers by putting subsets with themselves, so now we have to add ${{n-1} \\choose {m-1}}^2$ . \n\nIn the sum $n\\left(2\\sum_{m = 2}^{n-1} {{{n-1} \\choose {m-1}} \\choose {2}} + {{{n-1} \\choose {m-1}}^2} \\right)$ , a lot of stuff cancels and results in the form $n {{2n-2} \\choose {n-1}}$ . From here the answer extraction, which proved to be highly nontrivial to me, results in $245$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1048,
"boxed": false,
"end_of_proof": false,
"n_reply": 48,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777199.json"
}
|
Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a, b, c, $ or $d$ is nonzero. Let $N$ be the number of distinct numerators when numbers in $S$ are written as fractions in lowest terms. For example, 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 :/",
"i thought this was kinda misplaced? basically casework on if gcd(n, 9999) = 1, for n=27k w gcd(k, 1111) = 1, for n = 121k w gcd(k, 909) = 1, and n = 27 * 121k w gcd(k, 101) = 1 ",
"<blockquote>i thought this was kinda misplaced? basically casework on if gcd(n, 9999) = 1, for n=27k w gcd(k, 1111) = 1, for n = 121k w gcd(k, 909) = 1, and n = 27 * 121k w gcd(k, 101) = 1</blockquote>\n\nyeah could've honestly been a #10 or smth ",
"1992 AIME #5 is the same thing as this",
"Do casework on the denominator. If it is $9999$ , then there are $6000$ possible numerators (covering all numbers not divisible by $3, 11, 101$ ). If the denominator is $1111$ , the only new numerators are multiples of $3$ that are not divisible by $11, 101$ , so we can see there are $334$ of them. Similarly, for the denominator of $909$ we get $55$ numbers (numerators must be divisible by 11, but not 11 or 101). If the denominator is $101$ the only new numbers are divisible by $3, 11$ , so there are $3$ of them. It is easy to see any other denominators don't give new numbers, so the answer is $6000+334+55+3=6\\boxed{392}$ ",
"Got this after a 30 min bash, confirmed with python afterwards. Cannot believe I got out of this error free...",
"<blockquote>\n\nsadge moment when you forget to delete the three multiples of $303$ in the $1111$ set :/</blockquote>\n\nanyone else silly this problem this way?",
"i forgot to delete the multiples of 33 in the multiples of 11 set -> 416",
"<blockquote>i forgot to delete the multiples of 33 in the multiples of 11 set -> 416</blockquote>\n\nc'mon at least you didn't get 395 and missed the answer by 3",
"<blockquote><blockquote>\n\nsadge moment when you forget to delete the three multiples of $303$ in the $1111$ set :/</blockquote>\n\nanyone else silly this problem this way?</blockquote>\n\nMe",
"sillied this put 416\n\n",
"<blockquote>395 gang anyone?</blockquote>\n\nI GOT 393 AND OBOED",
"Welp not the only 449\n\ndid phi alright but afterwards forgot that problem was numerators and overcounted the following:\n\n3: denominators 11 and 101 are actually SUBSETS of 1111\n\n11: denominators 101 and 303 are subsets of 909\n\n33: well wasn't aware but didn't hurt cuz the only denominator was 101",
"Since everything relatively prime to 9999 and less than or equal to $9999$ works, suppose the numerator shares a prime factor with $3^2\\cdot 11\\cdot 101$ (\\phi(9999)=6000). \n\nIf it's divisible by $101$ , then it must be less than or equal to $99$ , contradiction. \n\nIf it's divisible by $3$ , then it must be less than or equal to $1111$ . \n\nIf it's divisibly by $11$ , then it must be less than or equal to $909$ . \n\nIf it's divisible by $33$ , then it must be less than or equal to $99$ . \n\nFirst we count all the multiples of $3$ less than or equal to $1111$ , and all the multiples of $11$ less than or equal to $909$ . The multiples of $33$ from $909$ to $1111$ are counted once and not allowed, so we need to subtract $6$ . The multiples of $33$ from $101$ to $909$ are counted twice and not allowed, so we need to subtract $24\\cdot 2=48$ . The multiples of $33$ from $1$ to $99$ are counted twice and are allowed, so we need to subtract $3$ . Now we also need to subtract $3$ for $303, 606, 909$ . \n\nAnswer is $370+82-6-48-3-3=\\boxed{392}$ ",
"this was like first 5 difficulty but with more computation",
"pls tell me someone else got 416",
"[pywindow]\ndef gcd(a,b):\n if a==0:\n return b\n elif b==0:\n return a\n elif a>b:\n return gcd(a%b,b)\n else:\n return gcd(a,b%a)\n\nnumerators = []\n\nfor x in range(1,10000):\n y = x/gcd(x,9999)\n if y not in numerators:\n numerators.append(y)\n\nprint(len(numerators))\n[/pywindow]",
"<blockquote>pls tell me someone else got 416</blockquote>\n\nI did as well ",
"i also did",
"<blockquote><blockquote>pls tell me someone else got 416</blockquote>\n\nI did as well</blockquote>\n\n<blockquote>i also did</blockquote>\n\nok good so I'm not the only one",
"my brain:\n\n*you don't need to do 9 because they obvious will be counted in the 3 case*\n\nthen:\n\n*that means you don't have to count 33 because you already counted 11 and 3!!!!*",
"LOL SAME",
"<blockquote>395 gang anyone?</blockquote>\n\nI got $505$ because i got $200$ instead of $55$ for the multiples-of- $11$ case... :wallbash: ",
"put 416 :(",
"<details><summary>nice</summary>[cpp]\n#include <vector>\n#include <algorithm>\n#include <iostream>\nusing namespace std;\nint main(){\n auto gcd = [](int a, int b) -> int {\n while(b) b ^= a ^= b ^= a %= b;\n return a;\n };\n vector<int> nums;\n for (int x = 1; x < 10000; ++x ){\n int y = x/gcd(x, 9999);\n auto val = std::find(nums.begin(), nums.end(), y);\n if (val == nums.end()) \n nums.push_back(y); //add y until end of loop\n }\n cout << nums.size();\n return 0;\n}[/cpp]\nOUTPUT:\n[img width= 50]https://media.discordapp.net/attachments/819417070185480202/941086026338816100/IMG_0736.png[/img]</details>\n\nproblem kinda easy to silly though f",
"I got 962 lol",
"<blockquote>1992 AIME #5 is the same thing as this</blockquote>\n\nIts pretty close but not the same thing.\n\n",
"449 :blush: I think this problem was definitely misplaced but pretty tricky too ",
"<blockquote><details><summary>nice</summary>[cpp]\n#include <vector>\n#include <algorithm>\n#include <iostream>\nusing namespace std;\n{\n if (a == 0)\n return b;\n if (b == 0)\n return a;\n if (a == b)\n return a;\n if (a > b)\n return gcd(a-b, b);\n return gcd(a, b-a); //just recursive Euclidean algorithm\n}\nint main(){\n vector<int> nums;\n for (int x = 1; x < 10000; ++x ){\n int y = x/gcd(x, 9999);\n auto val = std::find(nums.begin(), nums.end(), y);\n if (val == nums.end()) \n nums.push_back(y); //add y until end of loop\n }\n cout << nums.size();\n return 0;\n}[/cpp]\nOUTPUT:\n[img width= 50]https://media.discordapp.net/attachments/819417070185480202/941086026338816100/IMG_0736.png[/img]</details>\n\nproblem kinda easy to silly though f</blockquote>\n\n[java]\nimport java.util.*;\npublic class chungus {\n\n\tpublic static void main(String[] args) {\n\t\tArrayList <Integer> minecraft = new <Integer> ArrayList (); \n\t\tfor (int i = 1; i < 10000; i++) {\n\t\t\tint status = 1;\n\t\t\tfor (int j = 0; j < minecraft.size(); j++) {\n\t\t\t\tif (minecraft.get(j) == numerator(i, 9999)) {\n\t\t\t\t\tstatus = 0;\n\t\t\t\t}\n\t\t\t}\n\t\t\tif (status == 1) {\n\t\t\t\tminecraft.add(numerator(i, 9999));\n\t\t\t}\n\t\t}\n\t\tSystem.out.println(minecraft.size());\n\n\t}\n\n\tpublic static int numerator (int n1, int n2) {\n\t int temp1 = n1;\n\t int temp2 = n2; \n\n\t while (n1 != n2) {\n\t if(n1 > n2)\n\t n1 = n1 - n2;\n\t else\n\t n2 = n2 - n1;\n\t } \n\n\t int n3 = temp1 / n1 ;\n\t return n3;\n\t}\n}\n[/java]\ndoes the job too",
"just use millions of PIE and get 392(i felt this is a bit misplaced cuz found it easy than Q3 and Q8).\nthere's a total of 9998 numbers, firstly we subtract all numerators that contain a factor of 9999.\nthat is: 3332+908+98-302-32-8+2 = 3998\n9998-3998=6000 numerators after subtracting numbers contain some kind of factor.\nhowever we need to add numbers contain more than three \"3\"s, more than two \"11\"s, cuz these numerators also should be counted, e.g. 27/9999 = 3/1111:\nall number is a factor of 27: 370\nall number contain a factor of 27 and 11: 33\nall number contain a factor of 27 and 101: 3\nall number is a factor of 121: 82\nall number contain a factor of 3 and 121: 27\nall number is a factor of 27*121: 3\nUse PIE:\n370-33-3+82-27+3 = 392\n",
"I think I got like all numbers not multiples of $1331, 101, 2187$ work or something someone confirm this is completely wrong",
"<blockquote>Welp not the only 449\n</blockquote> yay 449 gang!",
"Am I the only one who tried totient function $\\implies 000$ ._.",
"<blockquote>Am I the only one who tried totient function $\\implies 000$ ._.</blockquote>\n\nyou had to consider other cases, that was literally just the $9999$ case ",
"<blockquote>I think I got like all numbers not multiples of $1331, 101, 2187$ work or something someone confirm this is completely wrong</blockquote> $132$ ",
"<blockquote><blockquote><details><summary>nice</summary>[cpp]\n#include <vector>\n#include <algorithm>\n#include <iostream>\nusing namespace std;\n{\n if (a == 0)\n return b;\n if (b == 0)\n return a;\n if (a == b)\n return a;\n if (a > b)\n return gcd(a-b, b);\n return gcd(a, b-a); //just recursive Euclidean algorithm\n}\nint main(){\n vector<int> nums;\n for (int x = 1; x < 10000; ++x ){\n int y = x/gcd(x, 9999);\n auto val = std::find(nums.begin(), nums.end(), y);\n if (val == nums.end()) \n nums.push_back(y); //add y until end of loop\n }\n cout << nums.size();\n return 0;\n}[/cpp]\nOUTPUT:\n[img width= 50]https://media.discordapp.net/attachments/819417070185480202/941086026338816100/IMG_0736.png[/img]</details>\n\nproblem kinda easy to silly though f</blockquote>\n\n[java]\nimport java.util.*;\npublic class chungus {\n\n\tpublic static void main(String[] args) {\n\t\tArrayList <Integer> minecraft = new <Integer> ArrayList (); \n\t\tfor (int i = 1; i < 10000; i++) {\n\t\t\tint status = 1;\n\t\t\tfor (int j = 0; j < minecraft.size(); j++) {\n\t\t\t\tif (minecraft.get(j) == numerator(i, 9999)) {\n\t\t\t\t\tstatus = 0;\n\t\t\t\t}\n\t\t\t}\n\t\t\tif (status == 1) {\n\t\t\t\tminecraft.add(numerator(i, 9999));\n\t\t\t}\n\t\t}\n\t\tSystem.out.println(minecraft.size());\n\n\t}\n\n\tpublic static int numerator (int n1, int n2) {\n\t int temp1 = n1;\n\t int temp2 = n2; \n\n\t while (n1 != n2) {\n\t if(n1 > n2)\n\t n1 = n1 - n2;\n\t else\n\t n2 = n2 - n1;\n\t } \n\n\t int n3 = temp1 / n1 ;\n\t return n3;\n\t}\n}\n[/java]\ndoes the job too</blockquote>\n\nwait f i forgot to copy the \"int gcd(int a, int b)\" part\nbut imagine java \nnice work nonetheless \n\n@fidgetboss Also, just a small note, your code works fine but line 5 uses an unchecked operation in that the initialization of the arraylist is not really done correctly, and this could, in practice, lead to some security issues, so i think the compiler is not really seeing that all the entries of the array list should be set as an integral value. I'd replace line 5 with\n[java]ArrayList <Integer> minecraft = new ArrayList <Integer> ();[/java] to clear the warning\njust a small thing i just noticed though, and something to keep in mind going forward lol",
"I only tried totient function",
"I got 392. I bashed too much with $\\phi$ and got the right answer.",
"<details><summary>Solution</summary>Basically, the question is asking how many distinct numerators their are among the simplified versions of $\\left \\{\\frac{1}{9999},\\frac{2}{9999},\\dots,\\frac{9999}{9999}\\right \\}$ .\n\nAll numerators relatively prime to $9999$ are not reduced, so they all work, and there are $\\phi(9999)=6000$ of them.\n\nNow we have to count the number of numerators that aren't relatively prime to $9999$ . Any numerator that would be divisible by $3$ after simplification must be divisible by $27$ before simplification, so any numerator divisible by $3$ would have to be at most $\\frac{9999}{9}=1111$ . There are $370-33-3=334$ multiples of $3$ not divisible by $11$ or $101$ .\n\nSimilarly, multiples of $11$ can be at max $\\frac{9999}{11}=909$ . There are $82-27-0=55$ multiples of $11$ not divisible by $3$ or $101$ There are no multiples of $101$ that can show up. However, there are multiples of both $3$ and $11$ , which can be at max $\\frac{9999}{99}=101$ , of which there are $3$ .\n\nIn total, there are $6\\boxed{392}$ possible numerators.</details>",
"449 sobad",
"First, every numbers relatively prime to $9999$ less than $9999$ works. This gives $\\varphi(9999)=9999\\cdot \\frac{100}{101}\\cdot \\frac{2}{3}\\cdot \\frac{10}{11}=6000$ . Next, we just use PIE. If a number is divisible by $27$ , then it is in the form $\\frac{3n}{1111}=\\frac{3n}{11\\cdot 101}$ . Therefore, we not want it to be divisible by $11$ and $101$ . This gives $370-33-3=334$ . Furthermore, we have the case with $11$ which is $82-27=55$ . \n\nWe can also have multiples of both $3^3$ and $11^2$ , which yields $3$ . Thus, the sum is $6000+334+55+3=6\\boxed{392}$ . ",
"[Video Solution](https://youtu.be/0FZyjuIOHnA)",
"<details><summary>solution sketch</summary>Note that any numerator relatively prime to $9999$ works automatically. Using totient function there are $6000$ such numerators. Other than that the numerator would be divisible by $3 \\cdot 9 = 27$ beforehand, any such numerator is at most $\\frac{9999}{\\frac{27}{3}} = 1111 = 11 \\cdot 101$ . So, the multiple of $3$ can't be divisible by $11$ or $101$ , there are $334$ such numbers. Do something similar for multiples of $11$ (the numbers 3, 11 come from 9999) which gives a total of $58$ . Adding gives $6392$ for an answer of $392$ .</details> ",
"Off by 3 in mock :what?: \nNote that $S = \\{ \\frac{x}{3^2 \\cdot 11 \\cdot 101} \\}, x \\in \\{1, 2, \\dots 9999 \\}$ Also $\\phi{(9999)} \\equiv 0 \\pmod{1000}$ so we can ignore those trivial cases. The only ones we didnt count were when the numerator was divisible by $3$ or $11$ . It cannot be divisible by $101$ because $101^2 > 9999$ . If the top is divisible by $3$ but not $11$ then $27 | x$ but $11$ and $101$ does not divide $x$ . If it did then we would have repeats. If $x = 9k$ then $k = 1, 2, \\dots 370$ but eliminating gives $334$ . The case where the top is divisble by $11$ but not $3$ gives $55$ similarly. Now the case where both are divisible means $3^3 \\cdot 11^2 | x$ which has $3$ cases. Sum to get $\\boxed{392}$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1188,
"boxed": true,
"end_of_proof": false,
"n_reply": 48,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777200.json"
}
|
Find the number of ordered pairs of integers $(a, b)$ such that the sequence $$ 3, 4, 5, a, b, 30, 40, 50 $$ is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
|
Clearly just picking from the set $\{3, 4, 5, 30, 40, 50\}$ we cannot find an arithmetic progression.
Case $1$ : The arithmetic progression contains only $a$ or only $b$ . Note that $6 \leq a \leq 28$ and $7 \leq b \leq 29$ .
Clearly $a = 6$ fails from $3, 4, 5, a$ . Next $a/b, 30, 40, 50$ causes $a = 20$ 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",
"Forgot $a \\neq 20,$ subtracted (6, 20) twice $\\rightarrow$ 236 :(",
"I PUT 227 BECAUSE I THOUGHT 20,35 WAS BAD AAAAAAH :(",
"I subtracted $(6,20)$ twice and got 227 :(",
"Put 229 since I forgot 3, 5, 7, 9 :( ",
"<blockquote>I PUT 227 BECAUSE I THOUGHT 20,35 WAS BAD AAAAAAH :(</blockquote>\n\nredfiretruck did that too bestie :love:",
"I thought this was extremely easy because I thought the sequences could only be consecutive.",
"got 227 :love: ",
"extremely sad story:\n\nwhen first solving, I forgot (7,9) doesn't work, and accidentally subtract (20,35) again so got the right answer\n\n15 minutes left\n\nnoticed the (20,35) error\n\nchanged to 229\n\nrip my life",
"NO I GOT 229 :(\nI FORGOT THAT (7,9) DOESNT WORK",
"i got 227 :(",
"<blockquote>I put 237 lmao\n\n(forgot to remove (20, 21), (20, 22), (20, 23), ..., (20, 29) ah stupid me)</blockquote>\n\nSAMEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE :CRI:",
"Notice that 6 and 20 cannot work. Also notice that a<b, so you simply have to choose two values a and b from the 22 remaining numbers between 7 and 29 (excluding 20). Then notice (16,28), (7,9) and (12,21) all don't work so subtract 3, yielding nCr(22,2)-3 = 231-3=**228**",
"We consider pairs of terms from the sequence $(X, Y)$ such that $X$ and $Y$ are the first and last terms of an arithmetic progression. It's clear that $(a, b)$ cannot form such a progression. \n\nIf one of $a, b$ is a part of the pair, then we have $10$ possible sequences to consider. Casework eliminates $$ a = 6; b = 20; a = 20; (a, b) = (7, 9). $$ \n\nNow, we only need to consider pairs containing neither $a$ nor $b$ . Taking modulo $3$ yields $$ (3, 30); (4, 40); (5, 50) $$ as the only possible pairs. Thus, we also eliminate $$ (a, b) = (12, 21); (16, 28). $$ A basic overcounting argument gives $$ \\binom{24}{2} - 23 - 3 - (14 - 1) - 9 = \\boxed{228} $$ as our answer.",
"thank god I got the correct answer",
"I GOT 229 AHHHHHHHHHHHHHHHHHHHH",
"Oops I forgot to subtract for complimentary counting\nI think I still got it wrong though because my answer was 49. I realized (20,35) failed but then forgot to remove it",
"<span style=\"color:#fff\">8charlimit</span>",
"<blockquote> $$ \\binom{24}{2} - 23 - 3 - (14 - 1) - 9 = \\boxed{228} $$ </blockquote>\n\nI got this exact same setup but, apparently, I decided that $13+9=21$ :wallbash_red: :wallbash: ",
"<blockquote>I PUT 227 BECAUSE I THOUGHT 20,35 WAS BAD AAAAAAH :(</blockquote>\n\nJoin the club",
"<blockquote>extremely sad story:\n\nwhen first solving, I forgot (7,9) doesn't work, and accidentally subtract (20,35) again so got the right answer\n\n15 minutes left\n\nnoticed the (20,35) error\n\nchanged to 229\n\nrip my life</blockquote>\n\nI did the same thing rip",
"<blockquote>I subtracted $(6,20)$ twice and got 227 :(</blockquote>\n\n",
"<blockquote>I PUT 227 BECAUSE I THOUGHT 20,35 WAS BAD AAAAAAH :(</blockquote>\n\nUH OH I MIGHT HAVE DONE THAT TOO",
"Lmao getting the wrong answer",
"I didn't read not necessarily consecutive and put 240\n\n<details><summary>Solution</summary>Firstly, consider sequences where only one of a or b are used. Only 3,4,5 and 30,40,50 have restrictions. If a > 6 then we don't have to worry about 3,4,5, and as long as a and b aren't 20 then we don't have to worry about 30,40,50. Now consider the sequences with both. If the two other terms are on one side of a and b, then with some checking the only additional restriction is that (7, 9). doesn't work.Now if they are on different sides of a and b, notice that one of 3,4 or 5 + 3(b - a) has to equal one of 30, 40,50 in this case. Using mod 3, the pairs are 3,30, 4,40, and 5,50 and only the former 2 yield new restrictions. From here we have 23c2 - 13 - 9 - 3 = 228 after counting the number of cases where a or b is 20.</details>",
"I skipped and came back and solved in last the 10 minutes.\nMiraculously, I didn't silly :)\n\nI just realized that I did silly, except that I sillied so many times that they cancelled out and I got the right answer :|",
"I got 364 because I thought it was three terms instead of four terms :maybe: ",
"I think I might have made 2 mistakes because I forgot the 3, 5, 7, 9 case but I still put 228\nedit: found my mistake, forgot about 7, 9 case and subtracted 20, 35 case instead\ni used up all of my luck for the year on this question",
"<blockquote>I think I might have made 2 mistakes because I forgot the 3, 5, 7, 9 case but I still put 228</blockquote>\n\nthe mistakes b canceling out",
"<blockquote>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 $\\binom{23}{2}-9-13-3=\\boxed{228}$ .</blockquote>\n\nI didn't consider (7, 9), so I got it wrong.",
"<blockquote>I PUT 227 BECAUSE I THOUGHT 20,35 WAS BAD AAAAAAH :(</blockquote>\n\nlol I was gonna put 227 because 10,20 but I was like, \"wait they'll never have two identical answers on one test...\"",
"<blockquote>extremely sad story:\n\nwhen first solving, I forgot (7,9) doesn't work, and accidentally subtract (20,35) again so got the right answer\n\n15 minutes left\n\nnoticed the (20,35) error\n\nchanged to 229\n\nrip my life</blockquote>\n\nWait this is literally exactly what happened to me",
"somehow got 226 sadge ",
"i forgot to exclude $a \\ne 20$ , but i excluded $(10,20,30,40)$ and $(5,20,35,50)$ by mistake, leading to $235$ (genius)",
"I DID EVERYTHING CORRECTLY BUT THEN ADDED BACK THE SUBSET OF (20, 20) TO GET 229\nI thought I was so smart for catching that",
"Both a and b cannot be 6 or 20, but they can be any integer from 7 to 19, as well as 21 to 29, giving a choice of 22 integers. $\\binom{22}{2}=231$ . \nNow notice that $(a,b)$ cannot be $(7,9),(12,21),(16,28)$ . Hence the answer is $231-3=228$ ",
"I did some stupid bash and got 188.",
"Anyone put 225? XD",
"I was second guessing myself for this problem, thinking \"how can this be only one more than the answer to problem 2?\".",
"<blockquote>I was second guessing myself for this problem, thinking \"how can this be only one more than the answer to problem 2?\".</blockquote>\n\nIf you took the 2018 AIME II and solved both problems 9 and 15, you would have thought “How can this [15] be only one more than the answer to number 9?”",
"<blockquote>I was second guessing myself for this problem, thinking \"how can this be only one more than the answer to problem 2?\".</blockquote>\n\nFrom memory, I could only recall that there are 2 tests that had two different questions have the same answer, I don't remember which tests though.",
"<blockquote>I put 237 lmao\n\n(forgot to remove (20, 21), (20, 22), (20, 23), ..., (20, 29) ah stupid me)</blockquote>\n\nI GOT 228 AT FIRST AND CHANGED TO 237 aewfsjd;aijsdf;",
"<blockquote><blockquote>I was second guessing myself for this problem, thinking \"how can this be only one more than the answer to problem 2?\".</blockquote>\n\nFrom memory, I could only recall that there are 2 tests that had two different questions have the same answer, I don't remember which tests though.</blockquote>\n\nOn the <details><summary>Some Past AIME</summary>AIME II 2021</details>, problems <details><summary>Two Numbers</summary>$11$ and $15$</details> had the same answer.\n",
"(20,35) more like bad :wallbash_red: ",
"remember, note that 20 and 6 aren't allowed for either a or b",
"It suffices to ensure that $a,b\\ne6$ , $a,b\\ne20$ , and: $$ (a,b)\\notin\\{(7,9),(12,21),(16,28)\\} $$ (where obviously $a<b$ and $a,b\\in[6,29]$ ). Then a simple counting argument finishes.",
"Is it just me, or is this way too easy for problem number 6?",
"easy but also easy to silly",
"Let $X = \\{3,4,5\\}$ and $Y =\\{30,40,50\\}$ and $Z =\\{a,b\\}$ Note that the only possible cases for four arithmetic sequences are $(x_1, x_2, x_3, z_1), (x_1, x_2, a,b), (a,b,y_1,y_2), (z_1, y_1, y_2,y_3).$ Define the set of possible tuples in the above as $S_1, S_2, S_3, S_4.$ A quick (20 min) PIE bash. Note that only intersections of at most 2 sets will be positive, so it isnt that bad.\n\nSubtract from $\\binom{24}{2}$ gives the answer",
"<blockquote>Is it just me, or is this way too easy for problem number 6?</blockquote>\n\nthe concept is easy for #6, but its hard to actually get the right answer bcz of all the cases",
"I got so lucky. \nWhat I did on the test:\nI thought that $a$ and $b$ had to be between 5 and 30. Clearly, they can't be $6$ or $20$ , so that leaves 22 numbers to choose from. In order to make a set that is an arithmetic sequence, a bit of mod bashing reveals that the first and last terms must be congruent to each other, and this determines what $a$ and $b$ are. Thus, the answer is $\\binom{22}{2}-3=228$ ",
"Oooof I’m bad I got $236$ Clearly by the condition $7 \\le a < b \\le 29$ since $a \\neq 6$ . There are $22+21+ \\cdots + 2+1 = \\frac{22(23)}{2}$ or $\\binom{23}{2}$ . Now, there are $2$ scenarios: either $a$ or $b$ form an arithmetic progression with $3$ other fixed terms, or $a$ and $b$ form an arithmetic progression with $2$ other fixed terms.**Case 1:** $a=20$ or $b=20$ . The former has $9$ cases, while the latter has $13$ cases.**Case 2A:** We pair with lower terms and get $(7,9)$ .**Case 2B:** We pair with $3$ and a higher term, for which $30$ works with an increment of $9$ . This yields $(a,b) = (12,21)$ . **Case 2C:** Repeat Case 2B with with $4$ and $5$ , the only valid solution of which comes with $4$ .\n\nThe answer is $$ \\binom{23}{2}-22-3=\\boxed{228} $$ ",
"<details><summary>Solution</summary>Note that neither $a$ nor $b$ can be equal to $6$ or $20$ . Other than that, the cases $(7,9)$ , $(12,21)$ , and $(16,28)$ all don't work. In total there are $\\binom{24-2}{2}-3=\\boxed{228}$ ordered pairs.</details>",
"I don't feel the need to post a solution, but I just wanted to remark that this problem is both less-scary than it may seem after reading it the first time, but also requires meticulous casework.",
"<blockquote>I put 237 lmao\n\n(forgot to remove (20, 21), (20, 22), (20, 23), ..., (20, 29) ah stupid me)</blockquote>\n\nkind of late but i got 228 then changed it to 237 :wallbash: ",
"I solved it by just finding the range of possible values of $a, b$ and removing the pesky cases; that gave 23c2 - 25 = 228",
"here is a meticulous casework that i did. this problem can be pretty trippy, just have to be really careful.\n\nFirst, we know the range for a and b that must be in is $5<a<b<30$ . Now, we look for all the values of $a$ and $b$ where four terms form an arithmetic progression, and b/c we don't want this to happen, we subtract that from the total number of ways--complementary counting. \n\n(1) 3 of the four terms in the arithmetic progression comes from the first group of numbers {3,4,5}\nThe next choice in the arithmetic progression must be from a or b, not the largest group, {30,40,50}, as it's too large here.\nThen, we have that $a\\neq6$ and that $b\\neq 6$ . So, either one of a or b cannot equal 6 at all--we can completely remove this number from the range that a and be can be in. \n\n(2) 2 of the four terms comes from the first group of numbers {3,4,5}\nIt can't be {3,4} b/c a can't be 5; it can't be {4,5} because we already know that $a,b\\neq6$ . Then, {3,5} gives that (a,b)=<u>(7,9)</u> doesn't work. **1 to subtract out here.**\n\n(3) 1 of the four terms comes from the first group of numbers {3,4,5} and then we must have 1 number, not two (b/c that doesn't work after quick checking), from {30,40,50}\nit goes smallNum, a, b, bigNum. So, bigNum - smallNum must be a multiple of 3 because they are all integers so the common difference is an integer as well. Then we subtract out when it's 3, <u>(12,21)</u>, 30; 4, <u>(16,28)</u>, 5, (20,35), 50 but 35 already exceeds the range so that won't work. **So only 2 possibilities to subtract here**. \n\n(4) Now, no term from the first group and 2 terms from the second group\nSo we have a, b, then two terms from {30,40,50}. That means that (a,b)≠(10,20) and that's the only thing that doesn't work here\n\n(5) Finally, one of a and b, and all three terms of 30, 40, and 50\nSo that means, like case 1, a and b can't be another number: $~~a,b\\neq 20$ . That encompasses all the possibilities when either one or both of a and b equals twenty, so we just subtract this big case out and don't even have to account for the one above. \n\nFinal ans. 24-2=22 numbers to choose from now accounting for that we can't have 6 or 20. So, $\\binom{22}{2} - 3 = \\boxed{228}$ .\n\nSorry if this is kindof messy and wordy and long but it made sense to me",
"So the bounds are from 7 to 29, if we only incorporate one new term. The only differences that can't work are d=2 (7,9); d=9 (12, 21), and d=12 (16, 28), and we can't have 20 as any term. If we just have small terms and both new terms, we can only use d=2. If we just use small terms, and both new terms, we can't include 20 anywhere. To incorporate the larger and smaller terms, we need their difference to be divisible by 3, for two cases--d=9 and d=12. So that's three pairs we can't use, 23 choose 2, and terms with 20 leave 9 and 13 cases. So it's 253-9-13-3=228."
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
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{
"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 11\\cdot 12}=\\frac{60\\cdot 8}{7\\cdot 9\\cdot 10\\cdot 11}=\\frac{16}{231}\\implies \\boxed{247}\\]",
"somehow got 5/231 -> 236 rip ",
"Note that one block of each of the colors must go in an odd numbered spot, and one must go in an even numbered spot, to ensure each pair has an even number of blocks between them. Thus, we have $6$ different colored balls to put in the $6$ odd spots (i.e. 1st, 3rd, 5th, etc.), so there are $6!$ ways. Similarly there are $6!$ ways to put the blocks in the even spots. The total number of ways to arrange the blocks is $\\frac{12!}{2^6}$ , so a computation gives us an answer of $\\frac{16}{231}$ , so $p+q=\\boxed{247}$ .",
"First, the total number of arrangements is equal to $\\frac{12!}{2! \\cdot 2! \\cdot 2! \\cdot 2! \\cdot 2! \\cdot 2!} = \\frac{12!}{64}$ . We now seek to count the number of *even*arrangements.\n\nLet the positions of the blocks be numbered $1, 2, 3, \\dots, 12$ , from left to right. In the example given by the problem, the two red blocks would be in positions $1$ and $8$ , the two blue blocks would be in positions $2$ and $3$ , and so on.\n\nThe key observation is to notice that as long as the two blocks of every color are in positions with opposite parity numberings, then there will be an even number of blocks between them. This is because the number of blocks between two positions is equal to the difference between the two numberings minus one, so if the difference between the two numberings is odd, then the number of blocks between the two positions is even. \n\nHence, there must be one even-numbered block and one odd-numbered block for each of the six colors.\n\nWith this in mind, the problem can be thought of as distributing the numbers $1, 3, 5, 7, 9, 11$ to the six colors, and then separately distributing the numbers $2, 4, 6, 8, 10, 12$ to the six colors. This ensures that each color has an even and odd block. \n\nThere are $6! \\cdot 6!$ ways to distribute the numbers, so the final result is\n\\[\\frac{6! \\cdot 6!}{\\frac{12!}{64}} = \\frac{16}{231} \\implies \\boxed{247}\\]\n\nAlternatively, the first red block can have any position, so the probability is $1$ . The second red block must be in an opposite parity position, so the probability is $\\frac{6}{11}$ . Continuing for the other colors gives $1 \\cdot \\frac{6}{11} \\cdot 1 \\cdot \\frac{5}{9} \\cdot 1 \\cdot \\frac{4}{7} \\cdot 1 \\cdot \\frac{3}{5} \\cdot 1 \\cdot \\frac{2}{3} \\cdot 1 \\cdot \\frac{1}{1} = \\frac{16}{231} \\implies \\boxed{247}$ . ",
"Notice that one block of each color will be in an odd position and the other block of each color will be in an even position. So, it's just (6!*6!)/(12!/64)\nwhich is 16/231 = 247 :)",
"can you change wording to [latex]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 $m+n$ .\n[/latex]",
"<blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nGot this because I thought that the positions of the same colors had to be same parities. Realized this was wrong, bc they have to be different parities. ",
"uhhh i forgot to italicize, can you change wording (again) to\n[latex]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 $m+n$ .\n[/latex]",
"<blockquote><blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nGot this because I thought that the positions of the same colors had to be same parities. Realized this was wrong, bc they have to be different parities.</blockquote>\ni think i got that they are different parities but when i tried to count it i assumed they had the same parities :stretcher: ",
"From a geo main, this is easier than #3. Ridiculous.",
"did anyone else think that this was Catalan numbers ;-; (analogous to the Catalan parenthesis argument)\napparently there can be an odd number of a certain color between two of the same colors.\n\nthree people in my testing room got the same wrong answer due to this misinterpretation",
"We can use recursion and a think-about-it argument. Since there are six letters, call the answer to this question $W_6$ . WLOG, let the leftmost block be $R$ . If the other $R$ splits the other ten letters into two groups of letters, just **take the leftmost letter-group and reverse it, and then the other ten letters just form a 5-letter version of this problem**. The probability that the other $R$ goes into a \"legal\" spot is $\\frac{6}{11}$ so $W_6 = \\frac{6}{11} W_5$ .\n\nWe repeat this argument, and using $W_1 = 1$ we get $W_6 = \\frac{6}{11} \\cdot \\frac{5}{9} \\cdot \\frac{4}{7} \\cdot \\frac{3}{5} \\cdot \\frac{2}{3} \\cdot 1 = \\frac{16}{231}$ so our answer is $\\boxed{247}$ .",
"upvoted for the title",
"One must go odd and one must go even. This gives $$ \\frac{6!^2\\cdot 64}{12!}=\\frac{720\\cdot 64}{12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7}=\\frac{64}{12\\cdot 11\\cdot 7}=\\frac{16}{231} $$ and thus yields $\\boxed{247}$ . ",
"not me labeling my blanks as $$ \\bigcirc \\underline{~} \\bigcirc \\underline{~} \\bigcirc \\underline{~} \\bigcirc \\underline{~} \\bigcirc \\underline{~} \\bigcirc \\underline{~} $$ and still not noticing what even means",
"Easiest combo on the test.",
"<blockquote><blockquote><blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nGot this because I thought that the positions of the same colors had to be same parities. Realized this was wrong, bc they have to be different parities.</blockquote>\ni think i got that they are different parities but when i tried to count it i assumed they had the same parities :stretcher:</blockquote>\n\nYeah, I caught my mistake at the end. ",
"<blockquote>did anyone else think that this was Catalan numbers ;-; (analogous to the Catalan parenthesis argument)\napparently there can be an odd number of a certain color between two of the same colors.\n\nthree people in my testing room got the same wrong answer due to this misinterpretation</blockquote>\n\nThis is exactly what I did... only one I actually solved that was wrong rip :(",
"<blockquote>did anyone else think that this was Catalan numbers ;-; (analogous to the Catalan parenthesis argument)\napparently there can be an odd number of a certain color between two of the same colors.\n\nthree people in my testing room got the same wrong answer due to this misinterpretation</blockquote>\n\nYES, I DID THIS. oops",
"Guess who thought there were $12!$ ways total to arrange the blocks",
"Me, trying to do 6-way PIE and complementary counting and spending 30 minutes with extremely bashy numbers and eventually giving up",
"<blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nAlmost put the same, but I found the problem after testing my logic on only 2 colours.",
"<blockquote><blockquote><blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nGot this because I thought that the positions of the same colors had to be same parities. Realized this was wrong, bc they have to be different parities.</blockquote>\ni think i got that they are different parities but when i tried to count it i assumed they had the same parities :stretcher:</blockquote>\n\nBruh I did the exact same thing bruhbruhrurburbuhrbr\nNow I'm surely not making jmo sadge",
"<blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nOh ouch. Made an incorrect bijection, got the same.",
"<blockquote><blockquote>somehow got 5/231 -> 236 rip</blockquote>\n\nGot this because I thought that the positions of the same colors had to be same parities. Realized this was wrong, bc they have to be different parities.</blockquote>\n\nSame :(",
"I seriously messed up at the last step in the denominator(I somehow divided 12! by 2^5 instead of 2^6 I have no idea how I even messed that up)\n\n<details><summary>Solution</summary>The key observation is that if the blocks have an even difference then they have indices of opposite parity(n and n + 2k+1 have 2k blocks between them). Now its just counting. There are 6 choices for the odd red and 6 choices for the even red. Then 5 choices for the odd blue and 5 choices for the even blue and so on, yielding. the numerator to be 6!^2. Since there are 12!/2^6 ways to arrange the blocks, our answer is 64/12C6 = 16/231.</details>",
"Oops I'm very late to posting this because AoPS was blocked for me. This was one of the most misplaced(easy) problems on the test, but it was possibly my favorite as well.\n\n<details><summary>Imagine trying to use recursion for 10 straight minutes</summary>Note that there are $\\dfrac{12!}{2^6}$ ways to arrange the blocks without any restrictions. The problem condition implies that blocks of the same color must be placed in positions of opposite parity. Therefore, each of the $6$ positions of each parity correspond to an arrangement of the $6$ differently colored blocks, in any order. The answer is thus $$ \\dfrac{6!\\cdot 6!}{\\dfrac{12!}{2^6}}=\\dfrac{16}{231}\\implies \\boxed{247}. $$</details>",
"Was it just me, or is this problem JMO 1/4 difficulty (the main observation felt similar in depth to JMO 2020/1)",
"<blockquote>Was it just me, or is this problem JMO 1/4 difficulty</blockquote>\n\n??? I thought it was literally AMC 10 #20 difficulty and I thought i did something wrong so i spent like 10 minutes trying to check it ",
"i was being clown",
"<blockquote>Was it just me, or is this problem JMO 1/4 difficulty</blockquote>\n\ndidn't you spend four and a half hours on 2021 jmo 1",
"<blockquote>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 11\\cdot 12}=\\frac{60\\cdot 8}{7\\cdot 9\\cdot 10\\cdot 11}=\\frac{16}{231}\\implies \\boxed{247}\\]</blockquote>\n\nI did this except argued with two sets $A,B$ where $A$ is the set of indices of the leftmost letter in each pair of blocks with the same color.",
"lmao I think I did $\\frac{6^2 \\cdot 5^2 \\cdot 4^2 \\cdot 3^2 \\cdot 2^2 \\cdot 1^2}{\\tfrac{12!}{2^6}}.$ Basically, I first counted the number of ways the red blocks can be placed, which is $6^2 = 36.$ Then I counted the number of ways the green blocks can be placed, which I observed was $5^2 = 25$ regardless of the arrangement of the original red blocks. This then repeats for all the pairs of blocks, so yeah. ",
"most misplaced math problem in the history of maa please fix ty",
"my solution: notice pattern, with 2 blocks answer is $\\frac{2}{3}$ , then 4 blocks is $\\frac{2}{5}$ . In the process of calculating, notice something cool that leads to a recursion. Let the leftmost block be $A$ WLOG. Then, for any of the $n$ possible spaces out of the $2n-1$ remaining spaces, the number of ways to arrange the rest of the blocks is always the same as if there are $2n-2$ blocks. Therefore, we get the recursion $a_n=\\frac{n}{2n-1}a_{n-1}$ and we can find the answer. (i didnt prove this but idc)",
"<blockquote>most misplaced math problem in the history of maa please fix ty</blockquote>\n\nmeh i thought #7 was more misplaced but ok",
"wow ok ig im the only engineer here.... :huh: \n\nlike i did it with 1, 2, 3, and 4 colors of blocks and guessed its $\\frac{(n!)^2 \\times 2^n}{(2n)!}$ for n colors",
"<blockquote>Was it just me, or is this problem JMO 1/4 difficulty (the main observation felt similar in depth to JMO 2020/1)</blockquote>\n\nits just you",
"<details><summary>solution</summary>If her arrangement is even, the odd places (1, 3, 5, 7, 9, 11) must contain all six block colors, as must the even places (2, 4, 6, 8, 10, 12). There are $\\displaystyle\\frac{12!}{(2!)^{6}}$ total arrangements and $(6!)^{2}$ work, so the probability that her arrangement is even is \\[(2!)^{6}\\cdot\\displaystyle\\frac{(6!)^{2}}{12!}.\\] Some computation gives that this equals $\\tfrac{16}{231}$ , and the answer is $\\boxed{247}$ .</details>",
"<blockquote>wow ok ig im the only engineer here.... :huh: \n\nlike i did it with 1, 2, 3, and 4 colors of blocks and guessed its $\\frac{(n!)^2 \\times 2^n}{(2n)!}$ for n colors</blockquote>\n\nNote that in order for an arrangement to be even, one block of a color has to be in an odd-numbered spot and the other block has to be in an even-numbered spot. The problem would be harder if it were the number of odd arrangements instead... \nThus, for 19 colors, there are like, 19!19! ways to arrange them according to the definition @above.",
"<details><summary>Solution</summary>Note that for each color, one block has to be in an even-indexed position and the other has to be in an odd-indexed position. Therefore, the $6$ even indexed positions must have one block of each color, and the same for the odds. All arrangements of this form work, for a total of $6!^2$ possibilities out of $\\frac{12!}{2^6}$ total possibilities, which reduces to $\\frac{16}{231}\\rightarrow \\boxed{247}$ .</details>",
"<blockquote>Was it just me, or is this problem JMO 1/4 difficulty (the main observation felt similar in depth to JMO 2020/1)</blockquote>\n\ni mean the odd/even thing wasn't jmo-level",
"<blockquote>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 11\\cdot 12}=\\frac{60\\cdot 8}{7\\cdot 9\\cdot 10\\cdot 11}=\\frac{16}{231}\\implies \\boxed{247}\\]</blockquote>\n\nwait how did they get that the both the even and odd positions have one of each color? how does this guarantee that her arrangement will be even?",
"<blockquote><blockquote>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 11\\cdot 12}=\\frac{60\\cdot 8}{7\\cdot 9\\cdot 10\\cdot 11}=\\frac{16}{231}\\implies \\boxed{247}\\]</blockquote>\n\nwait how did they get that the both the even and odd positions have one of each color? how does this guarantee that her arrangement will be even?</blockquote>\n\nthe possible slots, in terms of parity, go OEOEOEOEOEOE. The 'distance' between identical letters is even if and only if one is in an even slot and the other is in an odd slot. If both are in even slots or both are in odd slots, then the distance will be odd. ",
"trivial combo yay except i can only do p1 combo in the real test so ",
"goal: not skip problems",
"Total permutations = $12!/2^6$ For having even number of blocks in between two blocks of the same colour one must have one of the block placed in the even position and the other block in the old position \nTo place them in odd position i.e. 1st,3rd,5th....11th ,we have 6 positions and 6 blocks of different colours\nTherefore arrangements of these blocks in 6 positions will be $6!$ \nSimilarly for even positions , the blocks can be arranged in $6!$ ways \nSo probability= $6!.6!/(12!/2^6)$ = 16/231=m/n\nSo, m+n = 247",
"<blockquote>\nEllina 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 $m+n$ .\n\n</blockquote>\n<details><summary>solution</summary>Notice that if we index the positions of the blocks as\n\\[\\text{\\bf 1 2 3 4 5 6 7 8 9 10 11 12}\\]\nthen an odd-indexed block cannot be the same color as another odd-indexed block (as there are an odd number of blocks between them), and similarly an even-indexed block can't be the same as another even-indexed block. Therefore, the six colors are permuted throughout the odd-indexed blocks ( $6!$ ways), and they are also permuted through the even-indexed blocks ( $6!$ ways here too), and it is easy to see that an arrangement created in this manner is always even.\n\nIf you were to put the blocks randomly, there would be $\\tbinom{12}{2}\\tbinom{10}{2}\\tbinom{8}{2}\\tbinom{6}{2}\\tbinom{4}{2}$ ways to do so. You can see that this actually is $12! / 2^6$ , so number crunching a bit gives that the desired probability is $\\tfrac{16}{231}$ and answer extraction is $16+231=\\boxed{247}$ .</details>",
"Took me forever to realize this problem can be solved by constructive counting (because aime #9 is usually hard lol)\n\nsolution while mocking: Consider a line of $12$ empty spaces, numbered from $1$ to $12$ . Observe that blocks of the same color must go on spaces with different parities. There are $6$ odd and $6$ even spaces. WLOG, we select spaces for the red blocks, then the blue blocks, and so on. For the red block, we have $6^2$ choices. The other cases are similar, so in total there are $(6!)^2$ successful arrangements.\n\nIn total, there are $\\frac{12!}{2^6}$ arrangements. We compute: $$ \\frac{(6!)^2}{\\frac{12!}{2^6}}=\\frac{16}{231} $$ ",
"<details><summary>Storage</summary>We can assign $6$ different colours in the first place, wolog say <span style=\"color:#f00\">red</span>. Now we have $6$ more places to place the other <span style=\"color:#f00\">red</span>. At the second position we have $5$ different colour choices, wolog say <span style=\"color:#00f\">blue</span>. Now we have $5$ more places to place the other <span style=\"color:#00f\">blue</span>. Continue in this manner to get $(6!)^2$ ways. The total number of arrangements $= \\dfrac{12!}{2!2!2!2!2!2!}$ . So $\\dfrac{m}{n} = \\dfrac{16}{231}$ . So $m+n = \\boxed{247}$ .</details>",
"<details><summary>Misplaced frfr</summary>Note that there will be an even distance between each color in one is on an odd position(from the left), and another in on an even position. Therefore we get $$ \\frac{6!\\cdot6!}{\\frac{12!}{64}} = \\frac{16}{231}. $$ Hence $$ \\boxed{247.} $$</details>",
"<details><summary>storage</summary>Number the positions 1 through 12. No two blocks of the same color can be in the same parity position. Therefore all odd positions have one block of each color and so do all evens. This gives (6!)^2 arrangements. The total number of arrangements is 12!/2^6. Hence the probability is 16/231.</details>"
] |
[
"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 i believe",
" $\\frac{7}{4} < 2$ ",
"WHY DID I PUT 085",
"<blockquote>WHY DID I PUT 085</blockquote>\n\nSAME :(",
"Answer is $\\frac{6\\cdot3\\cdot2-7\\cdot5\\cdot1}{9\\cdot8\\cdot4}\\Rightarrow\\boxed{289}$ .",
"just list out those nums close to each other like (20, 21), (44, 45), (35, 36) and realized it's (35, 36)\n\ni didnt do it rigorously",
"**Smart Solution:** First, we find all $9$ -tuples such that $a \\cdot b \\cdot c - d \\cdot e \\cdot f = 1$ by testing pairs of consecutive positive integers. Once we see $$ \\frac{2 \\cdot 3 \\cdot 6 - 1 \\cdot 5 \\cdot 7}{4 \\cdot 8 \\cdot 9} = \\frac{1}{288} $$ is the best possible $9$ -tuple for this condition, it becomes clear that this is the minimum, as $$ \\frac{1}{288} = \\frac{2}{576} < \\frac{2}{7 \\cdot 8 \\cdot 9} = \\frac{2}{504}. $$ Unfortunately, I spent $45$ minutes on this question during the test, as I tried to bound the denominator (instead of the numerator) :(.",
"I PUT 217 CRIIIIII",
"Had 1/216 until the very end before I realized a difference of 1 was possible :oops_sign: ",
"Here's a way to rigorously solve the $abc - def = 1$ case. (This is not my problem, but rather the official solution.)\n\nLet $X = abc$ . Then\n\\[\n\\frac{abc - def}{ghi} = \\frac{1}{ghi} = \\frac{abcdef}{9!} = \\frac{X(X-1)}{9!}.\n\\]\nMinimizing this fraction then boils down to minimizing $X$ .\n\nObserve that $X(X-1) = abcdef \\geq 6!$ , and therefore $X\\geq 28$ . Either $X$ or $X-1$ is odd and is, therefore, a product of three distinct odd factors from $\\{1,2,\\ldots, 9\\}$ .\n\nNow it's not *too* bad to check that $X = 28$ fails, and the next smallest possible value of $X$ , namely $X = 36$ , works. More specifically, $36 = 6\\cdot 2\\cdot 3$ and $35 = 1\\cdot 5\\cdot 7$ . Hence $\\{g,h,i\\} = \\{4,8,9\\}$ , and the minimum value is $\\tfrac{1}{4\\cdot 8\\cdot 9} = \\tfrac{1}{288}$ .",
"Numerator must be 1, I used the fact that $2 \\cdot 3 = 6$ , and $n^2-(n-1)(n+1)=1$ , set $n=6$ , plugged in the rest",
"This problem is just trying to get the numerator to be $1$ , feels misplaced?",
"Do a small bash to get the best possible case where the numerator is $1$ is $\\frac{2\\cdot 3\\cdot 6 - 1\\cdot 5\\cdot 7}{4\\cdot 8\\cdot 9}=\\frac{1}{288}$ .\n\nThis is best because it is better than the best-case $\\frac{2}{7\\cdot 8\\cdot 9}$ with numerator $2$ .",
"I thought this was hugely misplaced. This should have switched with #3 imo",
"@Above I agree\n\nI got 289 also",
"The key idea of this problem was to minimize the numerator, not maximize the denominator",
"1/90 whoops",
"just check the 10 products and the numbers within one of those",
"289 \n(6*3*2-7*5*1)/(9*8*4)",
"<blockquote>The key idea of this problem was to minimize the numerator, not maximize the denominator</blockquote>\n\nyou still have to maximize the denominator, it's just that minimizing the numerator leads to a pretty good denominator already",
"how do you do this without guess and check (i got it wrong)",
"<details><summary>Solution</summary>Ideally, $abc-def=1$ . Guessing and checking without using $9$ or $8$ in the numerator gives $\\frac{6\\cdot3\\cdot2-7\\cdot5\\cdot1}{9\\cdot8\\cdot4}$ . Since $abc-def=1$ , then one of $\\{a,b,c\\}$ or $\\{d,e,f\\}$ must contain only odd factors, and trying the $1,3,5$ case means one of them must be $7$ or $9$ . As well, it's impossible to get $\\frac{1}{9\\cdot8\\cdot6}$ by checking all of the cases, leaving $\\frac{6\\cdot3\\cdot2-7\\cdot5\\cdot1}{9\\cdot8\\cdot4}=\\frac{1}{288}$ as the minimum positive fraction achievable, for an answer of $\\boxed{289}$ .</details>",
"Am I the only one who feels this was harder than 2021 AIME I #7?",
"i mean abc-def=1 means that you only need to test (1, 3, 5, 7, 9) triples cuz parity stuff",
"This is my problem; the solution in the packet is my original solution. I originally thought the answer was 1/216 and only when I worked out a solution rigorously did I find 1/288. ",
"<blockquote>This is my problem; the solution in the packet is my original solution. I originally thought the answer was 1/216 and only when I worked out a solution rigorously did I find 1/288.</blockquote>\n\nWhat was your motivation for looking at numerator cases first then optimizing the denominator? ",
"I guess you just have to see that halving the numerator from 2 to 1 is as \"effective\" as doubling the denominator, which is somewhat difficult. Also, to find the lowest possible value for the numerator, use parity: since 1 is odd, you can only use odd numbers for def.",
"<blockquote>WHY DID I PUT 085</blockquote>\n\nOh, the first thing i did in my mock was take $9,8,7$ in the denominator, but then i tested other values and turns out that it is $\\boxed{289}$ ",
"why is this so misplaced, literally requires 4th grade math knowledge + common sense. this should've been p1 or p2"
] |
[
"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, the polynomials are\n $$ P(x)=2(x-16)(x-20)-\\frac{1}{4}x+58 $$ $$ Q(x)=-2(x-16)(x-20)-\\frac{1}{4}x+58. $$ Plugging in $x=0$ and adding, one would see that the product of the binomials cancel each other out, and the $-\\frac{1}{4}x$ becomes $0$ . What is left is $2\\cdot 58=\\boxed{116}$ .</details>",
"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 $-1156+b=-458\\implies b=698$ .\n\nAlso, $Q(x)=-2x^2+cx+d$ . So $-512+16c+d=54\\implies 16c+d=566$ . Also, $-800+20c+d=53\\implies 20c+d=853$ . So $4c=287$ . So $1148+d=566\\implies d=-582$ .\n\nAnswer is $698-582=\\boxed{116}$ .",
"Congratulations to **djmathman** for finally cracking the first 5 with a brilliant #1! :)",
"I had the same solution as megarnie.",
"how the heck did I get 124\n\nEDIT: lol I thought 747 - 458 = 291",
"I got 066 smh \nI got 698 - 632",
"<details><summary>Alt Sol + Motivation</summary>So, we first set $P(x)=2x^2+ax+b$ and $Q(x)=-2x^2+cx+d$ . Intuitively, you might want to add these two equations for $x=16, 20$ because the quadratic terms cancel. In addition, we are looking for $P(0)+Q(0)$ , which is just adding the two functions. So, after you add the two equations you get $108=16(a+c)+b+d$ and $106=20(a+c)+b+d$ . This means that $a+c=-\\frac{1}{2}$ . Plugging this sum into a previous equation gives us that $b+d=116$ . The final sum is just this value, so our final answer is going to be $\\boxed{116}$ .</details>",
"116 no silly :)",
"The idea is that the average of the quadratics is the line containing the two given points, making the problem trivial.\n\nI was only able to come up with this instantly after bashing on Year 4 CIME II P2, when the idea was revealed.",
"Oh yea I did that problem too but somehow i was stupid enough to decide to bash it out :) \n\nluckily only took 10 min and got it right",
"I checked my answer in the final 5 minutes on this problem cuz I knew I wouldn't get any other answer in 5 minutes",
"058 gang\n\nAlso this is basically just [2021 CIME II/2](https://artofproblemsolving.com/community/c1677139h2477781p20782061)",
"<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></blockquote>\n\nWow, the intended solution was actually clean.",
"<blockquote>Congratulations to **djmathman** for finally cracking the first 5 with a brilliant #1! :)</blockquote>\n\nYup this was definitely in the top three best problems!",
"I solved for $P(0)$ and $Q(0)$ separately, and added them up. \nAhh took me so much time, and I didn't get the correct answer until the last 10 minute",
"\\(P(x)=2x^2+ax+b,Q(x)=-2x^2+cx+d\\)\n\\(T(x)=P(x)+Q(x)=(a+c)x+b+d\\) \nRequired is \\(b+d\\)\n\\(16(a-c)+b+d=108\\)\n\\(20(a-c)+b+d=106\\)\n\\((a-c)=-1/2\\)\n\\(b+d=116\\)",
"either you can bash out P(0) or just find out they're linear ",
"<blockquote>I solved for $P(0)$ and $Q(0)$ separately, and added them up. \nAhh took me so much time, and I didn't get the correct answer until the last 10 minute</blockquote>\n\nThat's what I did, but it was pretty quick for me. ",
"CIME #2 go BRRRRRRR\nALmost at the same level as Joe BRRRRRRRow",
"Suppose $P(x) = 2x^2+ax+b$ and $Q(x)=-2x^2+cx+d$ . We plug in both points and compute $P(16)-P(20)$ to find $a=-\\frac{289}{4}$ , for which substituting back in gets $b=698$ . We now compute $Q(16)-Q(20)$ to get $c=\\frac{287}{4}$ , or $d=-582$ . Therefore $$ 698-582=\\boxed{116} $$ By far the most straightforward solution but definitely not the most elegant",
"<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></blockquote>\n\nRe this sol: I did the exact same thing but forgot to add the y values (incorrectly getting $R(16)=54$ , etc). Thus I was off by a factor of $2$ What a way to miss USAMO. (If I got this problem right, I'd likely have made, whereas now I'm almost guaranteed not to. Oh well.)",
"<blockquote><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></blockquote>\n\nRe this sol: I did the exact same thing but forgot to add the y values (incorrectly getting $R(16)=54$ , etc). Thus I was off by a factor of $2$ What a way to miss USAMO. (If I got this problem right, I'd likely have made, whereas now I'm almost guaranteed not to. Oh well.)</blockquote> $\\text{slope}=-1/4$ rip.",
"did i\ndid i subtract wrong",
"<details><summary>Solution</summary>Let $P(x)=2x^2+ax+b$ and $Q(x)=-2x^2+cx+d$ . Then the two points both polynomials pass through give the equations:\n\\begin{align*}\n16a+b&=54-2\\cdot16^2\n16c+d&=54+2\\cdot16^2\n20a+b&=53-2\\cdot20^2\n20c+d&=53+2\\cdot20^2\n\\end{align*}\n\nAdding the first two equations and the third and fourth equations gives $16(a+b)+(c+d)=108$ and $20(a+b)+(c+d)=106$ . Therefore, $a+b=-\\frac{1}{2}$ and $c+d=P(0)+Q(0)=\\boxed{116}$ .</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></blockquote>\n\ni also did that :o",
"<details><summary>Division Algorithm</summary>We are given: $P(16)=Q(16)=54$ , $P(20)=Q(20)=53$ .\nFrom division algorithm, we can write: $P(x)=2(x-16)(x-20)+58-\\frac{x}{4}$ $Q(x)=-2(x-16)(x-20)+58-\\frac{x}{4}$ . \nSo, $P(x)+Q(x)=116-\\frac{x}{2}$ .\nHence, $P(0)+Q(0)=\\boxed{116}$</details>",
"Tons of equations bashing to get 698-582=116 ",
"Best AIME #1 I've ever seen.\n\nLet $P'(x)=P(x)-54=2(x-16)(x-a)$ and $Q'(x)=Q(x)-54=-2(x-16)(x-b)$ since $16$ are roots by plugging it in, then plug in $20$ for x to obtain that $-1=P(20)-54=P'(x)=2(20-16)(20-a)$ to get $a=\\tfrac{161}{8},$ similarly get $b=\\tfrac{159}{8}.$ Hence, $P'(0)+Q'(0)=P(0)-54+Q(0)-54=2 \\cdot 16 \\cdot \\tfrac{161}{8} - 2 \\cdot 16 \\cdot \\tfrac{159}{8}=8,$ so $P(0)+Q(0)=\\boxed{116}.$ ",
"Along with 2022 AIME II P1, both P1's last year were very cool",
"We let $g(x)=p(x)+q(x)$ , we have $g(16)=108$ and $g(20)=106$ so we add $2\\cdot4=8$ , so $108+8=\\boxed{116}.$ ",
"Let $P(x)=2x^2+b_1x+c_1$ and $Q(x)=-2x^2+b_2x+c_2.$ Define the function $R(x)=P(x)+Q(x).$ Note that $R(x)$ will be linear since the quadratic terms will cancel each other out. Evaluating $R$ at $16$ and $20,$ respectively, we have $R(16)=108$ and $R(20)=106.$ Subtracting the latter from the former, we have $16(b_1+b_2)-20(b_1+b_2)=2 \\implies -4(b_1+b_2)=2 \\implies b_1+b_2=-\\frac{1}{2}.$ Plugging this back into $R(16),$ we have $(-\\frac{1}{2})16+(c_1+c_2)=108 \\implies -8+(c_1+c_2)=108 \\implies c_1+c_2=116,$ hence $R(0)=P(0)+Q(0)=\\boxed{116}.$ ",
"Note B'B=x and C'C=x and also PQ=x\nThen from theorem;\n500-x+650-x=666\nx=242=PQ.",
"Note that $P(x) + Q(x) = -\\frac{1}{4}x + 58$ and hence $P(0) + Q(0) =2 \\cdot 58$ which is just $\\boxed{116}$ .",
"Let\n\\[\nF(x) = Q(x) + P(x)\n\\]\n\nSince\n\\[\n[x^2]P(x) = 2 \\quad \\text{and} \\quad [x^2]Q(x) = -2,\n\\]\nwe have\n\\[\n[x^2]F(x) = [x^2]P(x) + [x^2]Q(x) = 2 + (-2) = 0.\n\\]\nTherefore,\n\\[\n\\deg(F(x)) = 1,\n\\]\nso \\( F(x) \\) is a linear function.\n\nWe are given that\n\\[\n(16,\\ 108),\\quad (20,\\ 106) \\in \\{(x, F(x)) \\mid x \\in \\mathbb{R}\\}\n\\]\n(since the graph is a straight line).\n\nThen,\n\\[\n\\frac{108 - 106}{16 - 20} = \\frac{F(x) - 106}{x - 20}\n\\]\n\nSo,\n\\[\nF(x) = -\\frac{x}{2} + 116\n\\]",
" $\\mathfrak{The \\;Twenty-Ninth\\; Of\\; October,\\; 2025}$ ʕ•ᴥ•ʔ\n<details><summary>Solution - grinding problems</summary>$$ \\color{red} \\spadesuit\\color{red} \\boxed{\\textbf{Wordless, two line solution.}}\\color{red} \\spadesuit $$ $$ \\exists a,b \\textbf{ s.t. }R(x)=P(x)+Q(x)=ax+b \\: \\forall x; $$ $$ \\implies R(4 \\cdot 5)=106, R(4 \\cdot 4)=108, R(4 \\cdot 3)=110, \\dots, R(4 \\cdot 0)=\\boxed{116}. $$</details>"
] |
[
"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 better\n",
"I got 227 as well",
"very ez, could've been switched with #1, got $\\boxed{227}$ ",
"easier than #1",
"<blockquote>easier than #1</blockquote>\n\nyea should've been switched with it imo",
"<blockquote>easier than #1</blockquote>\n\nno it was harder\n",
"After solving #2 you feel like it's easier, since it's so easy to confirm if you're correct, and the problem statement is so simple.\n#1 is a little more complicated, but anybody (except me rip) can bash out the system of equations and get the correct answer, without spotting the intended solution.",
"<details><summary>Solution</summary>We have that $\\overline{abc}_{10} = \\overline{bca}_{9} \\Rightarrow 100a + 10b + c = 81b + 9c + a \\Rightarrow 99a = 71b + 8c$ . Let's assume that $a=1$ . Then, $71b + 8c = 99$ . Since $8$ is not a factor of $99$ , $b \\ne 0$ and must be $1$ . But, even then, there are no digits that could satisfy the equation. Ok, now let's assume that $a = 2$ . So, we have that $71b + 8c = 198$ . Again, since $8$ is not a factor of $198$ , $b \\ne 0$ . If we assume $b=1$ , then we need to find a digit $c$ such that $8c = 127$ . But, using the same logic that $8$ is not a factor of $127$ , we see that this is impossible. If we assume $b=2$ , we see that $8c = 56 \\Rightarrow c = 7$ . We can stop here as we have found the answer to the problem as $\\boxed{227}$ . (There is no reason to continue this casework as we have already found *the* three-digit positive integer which fulfills the condition of this problem.)</details>",
"<blockquote><blockquote>easier than #1</blockquote>\n\nno it was harder</blockquote>\n\nhow?",
"<blockquote><blockquote><blockquote>easier than #1</blockquote>\n\nno it was harder</blockquote>\n\nhow?</blockquote>\nbecause 1 is just easy system of equations",
"@above yeah but it's more bashy and takes more time (and more sillyable too ig?)",
"Easiest problem on the test by far",
"change latex i guess\n[latex]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.[/latex]",
"Change problem statement to\n\n<blockquote>Find the three-digit positive integer $\\underline{a}~\\underline{b}~\\underline{c}$ whose representation in base nine is $\\underline{b}~\\underline{c}~\\underline{a}_9,$ where $a$ , $b$ , and $c$ are (not necessarily distinct) digits.</blockquote>\n\n",
"<blockquote>Change problem statement to\n\n<blockquote>Find the three-digit positive integer $\\underline{a}~\\underline{b}~\\underline{c}$ whose representation in base nine is $\\underline{b}~\\underline{c}~\\underline{a}_9,$ where $a$ , $b$ , and $c$ are (not necessarily distinct) digits.</blockquote></blockquote>\n\nwhy did you change nine to 9? i think we want to stay with official wording",
"<details><summary>Solution</summary>We have $100a + 10b + c = 81b + 9c + a$ , meaning $99a = 71b + 8c$ or $99a - 71b = 8c$ . Taking mod 8 we have $3a \\equiv 7b$ . If a. = 1, then b = 5, but that doesn't. have a value for c. Then if a = 2, b = 2, which gives c = 7. Hence, the answer is 227.</details>",
"<blockquote>Change problem statement to\n\n<blockquote>Find the three-digit positive integer $\\underline{a}~\\underline{b}~\\underline{c}$ whose representation in base nine is $\\underline{b}~\\underline{c}~\\underline{a}_9,$ where $a$ , $b$ , and $c$ are (not necessarily distinct) digits.</blockquote></blockquote>\n\n/downvote\nThe packet said $\\text{nine}$ for clarity.",
"<blockquote><blockquote>Change problem statement to\n\n<blockquote>Find the three-digit positive integer $\\underline{a}~\\underline{b}~\\underline{c}$ whose representation in base nine is $\\underline{b}~\\underline{c}~\\underline{a}_9,$ where $a$ , $b$ , and $c$ are (not necessarily distinct) digits.</blockquote></blockquote>\n\n/downvote\nThe packet said $\\text{nine}$ for clarity.</blockquote>\n\nI can confirm. ",
"This is like AMC 12 problem 10 level",
"\\(99a=71b+8c\\)\n\\(a\\) belongs to \\((0,8]\\)\n\\(b\\) belongs to \\([0,8]\\)\n\\(c\\) belongs to \\([0,8]\\)\nNow note \n\\(b+c \\equiv 0 mod 9\\)\nwhich gives us a few cases \\((1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)\\)\n\\(3a+b \\equiv 0 mod 8\\)\nwhich gives us a few cases \\((1,5),(2,2),(3,7),(4,4),(5,1),(6,6),(7,3)\\) \nnow taking intersection of both cases and checking we get \\(227\\) works",
"I just took mod 9 and mod 11 and then went through a few cases. \nmod 8 seems a bit faster but this was still pretty quick",
"notice that 99 - 71 = 28 so doubling that gives a multiple of 8, which is 56, so 227",
"note that the answer to this is one less than the answer to problem 6, so the answer is $228-1=\\boxed{227}$ . ",
"This is your standard base problem. We have the restriction $(a,b,c) < 9$ and $$ 100a+10b+c=81b+9c+a \\implies 8c=99a-71b $$ Taking $\\mod 8$ we have $$ 3a + b \\equiv 0 \\mod 8 $$ Clearly $(a,b)=(2,2)$ works, and plugging in gives $c=7$ , which satisfies our condition. The answer is $\\boxed{227}$ ",
"yea use the base equation and bash on the var $a$ ",
"<details><summary>how I did it</summary>You get the equation $100a+10b+c=81b+9c+a \\implies 99a=71b+8c$ . Take mod $99$ to get $-28b+8c\\equiv 0\\pmod{99}$ and you instantly get $b=2, c=7$ . Substitute back to get $a=2$ , so the answer is $\\boxed{227}$ .</details>",
"<details><summary>Solution</summary>Expanding,\n\\[100a+10b+c=81b+9c+a\\leftrightarrow99a=71b+8c.\\]\nNote that $a\\equiv5b\\pmod8$ . Testing, if $a=1$ , $b=5$ , but then there is no value of $c$ that works. If $a=2$ , $b=2$ as well, and then $c=7$ works, for an answer of $\\boxed{227}$ .</details>",
"<details><summary>remark</summary>This problem kind of easy ngl. [\\hide]</details>",
"Get the equation $99a=71b+8c$ , take it modulo $9$ to get $0 \\equiv b+c \\pmod{9}$ and then modulo $7$ to get $a \\equiv b+c \\pmod{7}$ . So $b+c=9$ since $0$ obviously can't work and if $b+c = 18$ then $a$ can't be a positive integer. Next $a \\equiv 2 \\pmod{7}$ and since if $a$ were $9$ then even if you maximized $b$ and $c$ to $9$ the initial equation would fail, $a=2$ . \nSo we have $a=2$ and $b+c=9$ . Next take the initial equation $\\pmod{4}$ . Then $3a \\equiv 3b \\pmod{4}$ . So $a, b$ must have the same parity. Then just use casework for $b=\\{0, 2, 4, 6, 8\\}$ to get $b=2$ as the one that works so $c=7$ so the answer is $\\boxed{227}$ .",
"<blockquote><details><summary>remark</summary>This problem kind of easy ngl. [\\hide]</blockquote>\n\nhow insightful of you</details>",
"When you think the second is $CBA$ instead of $BCA$ :stink: ",
"<blockquote>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.</blockquote>\n\nWe have $100a+10b + c = 81b + 9c + a$ . Taking mod 9, we have $a +b + c = a$ mod 9. Therefore, $b+c$ has a remainder of $0$ when divided by $9$ . Simplifying our equation, we have $99a=71b + 8c$ . We can now do casework.",
"The equation relating the two numbers is $100a+10b+c=81b+9c+a$ which simplifies to $99a=71b+8c.$ Taking mod $7$ yields $a \\equiv b+c \\pmod{7}.$ Since the RHS can range from $0$ to $16$ and the LHS ranges from $0$ to $7,$ $b+c=a + 7k$ for $k=-1,0,1,2.$ The $-1$ and $2$ cases are unlikely, so check the $k=0,1$ cases first. Rearrange the equation as $$ 99a-8(b+c)=63b, $$ then the $k=0$ cases yields $91a=63b$ or $13a=9b,$ which is clearly impossible. The $k=1$ case yields $91a=63b-56$ or $9b+8=13a.$ Testing, $a=2$ works, yielding $b=2$ and $c=7.$ Our answer is $\\boxed{227}.$ ",
"<blockquote>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.</blockquote>\n\nJohn's solution. Mine as well.\n<details><summary>Sol</summary>Say we have\n\\[81b + 9c + a = 100a + 10b + c \\]\nAccording to the problem statement. This gives us $71b + 8c - 99a = 0,$ a diophantine equation (how fun!) Taking mod 9, we note that $71 \\equiv -1 \\pmod{9}, 8 \\equiv -1 \\pmod{9}.$ Note that in this case we need $b + c \\equiv 0 \\pmod{9}.$ Proof: $(b+c) \\cdot (-1) \\equiv 0 \\pmod{9},$ what we want. Thus, $b + c$ must equal nine, as $0$ and $18$ (last one by inspection) clearly don't work.\nBy more inspection, note that $b = 2, c = 7$ gives us with $142 + 56 = 198 = 2 \\cdot 99,$ thus resulting in $227$ as our answer.</details>",
"We have $99a=71b+8c$ , using casework, $\\boxed{227}$ is the answer.",
"abc(10)------->bca(9)\n9⁰ * a + 9¹ * c + 9² * b = abc\na + 9c + 81b = 100a + 10b + c\n8c + 71b = 99a \na= 2 b=2 c=7\n227",
"As by all the previous equations \n99a=71b+8c \nNoting down those 11 cases of b+c=9 .\nNow if we take again mod 11 on both sides we get \n5b+3c is a multiple of 11 \nWhich satisfies only when b=2 c=7 and a=2\n227",
"Taking $\\pmod 8$ , $\\pmod 9$ and then bounding $a<7$ finishes the problem."
] |
[
"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} \n2\\sin(\\theta+\\alpha)&=\\sqrt{3}.\n\\end{align*}\nAssuming $\\alpha,\\beta,\\theta\\in [0,\\pi/2]$ , we have $\\alpha=\\pi/8, \\beta=\\pi/24, \\theta=5\\pi/24$ . To finish, we may compute:\n\\[ [(1-x)(1-y)(1-z)]^2=(\\cos^2(\\pi/4)\\cos^2(\\pi/12)\\cos^2(5\\pi/12))^2=1/32,\\]\nwhere the first step comes from double angle.",
"Was this one 033",
"Let $a=1-x, b=1-y, c=1-z$ . Then squaring the original equations and simplifying gives $\\sqrt{(1-a^2)(1-b^2)}-ab=-\\frac{1}{2}$ and analogous equations. From here we can trig sub $a=\\sin p, b=\\sin q, c=\\sin r$ for acute angles $p, q, r$ . Then our equations become $\\cos(p+q)=-\\frac{1}{2}, \\cos(q+r)=0, \\cos(p+r)=\\frac{1}{2}$ . We can then find $p=45^\\circ, q=75^\\circ, r=15^\\circ$ . From here we can find that $a=\\frac{1}{\\sqrt{2}}, b=\\frac{\\sqrt{2}+\\sqrt{6}}{4}, c=\\frac{\\sqrt{6}-\\sqrt{2}}{4}$ . We want to find $(abc)^2$ , and a computation gives us an answer of $\\frac{1}{32}$ , to get $\\boxed{33}$ .",
"three triangles, each triangle is made up of a pair of similar triangles\n\nyou get a+b=30, b+c=45, c+a=60 and you want (cos 2a cos 2b cos 2c)^2",
"you can do this without trig sub also, let a=1-x, b=1-y, c=1-z. Then you get a quadratic for a, and bc = (1-a^2)/2",
"Let $x=2\\sin^2\\theta, y=2\\sin^2\\beta, z=2\\sin^2\\alpha$ . Plug it back in this gives $$ \\sin(\\theta+\\beta)=\\frac{1}{2} $$ $$ \\sin(\\beta+\\alpha)=\\frac{\\sqrt{2}}{2} $$ $$ \\sin(\\alpha+\\theta)=\\frac{\\sqrt{3}}{2} $$ As a result, let the arcsines be $\\frac{\\pi}{6}$ , $\\frac{\\pi}{4}$ , and $\\frac{\\pi}{3}$ . As a result, we have $$ \\theta+\\beta=\\frac{\\pi}{6} $$ $$ \\alpha+\\beta=\\frac{\\pi}{4} $$ $$ \\alpha+\\theta=\\frac{\\pi}{3} $$ Add all the equation gives $\\alpha+\\beta+\\theta=\\frac{1}{2}\\cdot \\frac{3\\pi}{4}=\\frac{3\\pi}{8}$ . \n\n\nSubtract from each equation gives $\\alpha=\\frac{5\\pi}{24}$ , $\\beta=\\frac{\\pi}{24}$ and $\\theta=\\frac{\\pi}{8}$ . What we want to find is $$ \\prod_{cyc} (1-x)^2 $$ $1-x=1-(2\\sin^2\\theta)=1-(2-2\\cos^2\\theta)=1-2+2\\cos^2\\theta=2\\cos^2\\theta-1=\\cos(2\\theta)$ . Therefore, this is essentially $$ \\cos^2(2\\cdot \\frac{5\\pi}{24})\\cos^2(2\\cdot \\frac{5\\pi}{12})\\cos^2(2\\cdot \\frac{\\pi}{8})=\\cos^2(\\frac{\\pi}{12}\\cdot \\cos^2(\\frac{5\\pi}{12})\\cdot \\cos^2(\\frac{\\pi}{4})=\\frac{1}{32^2}\\cdot (\\sqrt{6}-\\sqrt{2})^2(\\sqrt{6}+\\sqrt{2})^2\\cdot 2=\\frac{1}{32} $$ This yield the answer $\\boxed{033}$ ",
"Nice problem! (and first #15 solved in contest :))\n\nThe key is to divide both sides of all equations by $2$ and use the substitutions $\\sin{a} = \\sqrt{\\tfrac{x}{2}},$ $\\sin{b} = \\sqrt{\\tfrac{y}{2}}, \\sin{c} = \\sqrt{\\tfrac{z}{2}}$ for first quadrant angles $a,b,c,$ so that the conditions turn into sin addition formulas:\\begin{align*}\\sin{(a+b)} &= \\frac{1}{2} \\sin{(b+c)} &= \\frac{\\sqrt{2}}{2} \\sin{(c+a)} &= \\frac{\\sqrt{3}}{2}.\\end{align*}\nHere, we can quickly find working angles $(a,b,c) = (\\tfrac{45}{2}, \\tfrac{15}{2}, \\tfrac{75}{2}).$ Then,\\begin{align*}[(1-x)(1-y)(1-z)]^2 &= [(1-2\\sin^2{a})(1-2\\sin^2{b})(1-2\\sin^2{c})]^2 &= [\\cos{2a}\\cos{2b}\\cos{2c}]^2 &= [\\cos{15}\\cos{45}\\cos{75}]^2,\\end{align*}from which this can be computed to $\\tfrac{1}{32},$ for an answer of $\\boxed{033}$ $\\blacksquare$ ",
"(It's always trigonometry. Don't trust square roots)",
" $a=1-x$ , $b=1-y$ , $c=1-z$ . $\\sqrt{-(a-1)(b+1)}+\\sqrt{-(a+1)(b-1)}=1$ . $\\sqrt{-(b-1)(c+1)}+\\sqrt{-(b+1)(c-1)}=\\sqrt{2}$ . $\\sqrt{-(a-1)(c+1)}+\\sqrt{-(a+1)(c-1)}=\\sqrt{3}$ . \n\nLet $a=\\sin \\alpha, b=\\sin \\beta, c=\\sin \\theta$ . \n\nNow we square the equations. \n\nThe first is $2\\sqrt{(1-a^2)(1-b^2)}-2ab=-1\\implies \\sqrt{(1-a^2)(1-b^2)}-ab=-\\frac{1}{2}$ . \n\nThe second is $\\sqrt{(1-b^2)(1-c^2)}-bc=0$ . \n\nThe third is $\\sqrt{(1-a^2)(1-c^2)}-ac=\\frac{1}{2}$ . \n\nSo $\\cos(\\alpha+\\beta)=-\\frac{1}{2}$ , $\\cos(\\beta+\\theta)=0$ , $\\cos(\\alpha+\\theta)=\\frac{1}{2}$ . \n\nSo $\\alpha+\\beta=120, \\beta+\\theta=90, \\alpha+\\theta=60$ . Thus, $\\alpha=45^{\\circ}, \\beta=75^{\\circ}, \\theta=15^{\\circ}$ . \n\nSo $c=\\frac{\\sqrt{6}-\\sqrt{2}}{4}$ (from memory). This implies $b=\\frac{\\sqrt{6}+\\sqrt{2}}{4}$ , and $a=\\frac{\\sqrt{2}}{2}$ . $abc=\\frac{\\sqrt{2}}{8}$ , so $a^2b^2c^2=\\frac{2}{64}=\\frac{1}{32}\\implies \\boxed{033}$ . ",
"what is the motivation for trig substitution?",
"<blockquote>what is the motivation for trig substitution?</blockquote>\n\nI see none-- i subsituted $a,b,c=1-x,1-y,1-z$ .",
"<blockquote>what is the motivation for trig substitution?</blockquote>\n\nfor me it was $b^2+c^2=1$ ",
"<blockquote>Let $x$ , $y$ , and $z$ be positive real numbers satisfying the system of equations\n\\begin{align*}\n\\sqrt{2x - xy} + \\sqrt{2y - xy} & = 1\n\\sqrt{2y - yz} + \\hspace{0.1em} \\sqrt{2z - yz} & = \\sqrt{2}\n\\sqrt{2z - zx\\vphantom{y}} + \\sqrt{2x - zx\\vphantom{y}} & = \\sqrt{3}.\n\\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 $m+n$ .</blockquote>\n\nThe requested expression is very \"sus\" indeed.\n<details><summary>Subsitution solution</summary>Let $a,b,c=1-x,1-y,1-z$ respectively. Rewrite the equations as\n\\[\\sqrt{(1-a)(1+b)}+\\sqrt{(1+a)(1-b)}=1\\]\nand cyclic variants.\n\nSquare and rearrange, giving $2\\sqrt{(1-a^2)(1-b^2)}=2ab-1$ ; then dividing by 2 and squaring again yields $a^2b^2-a^2-b^2+1=a^2b^2-ab+1/4$ , whence $a^2-ab+b^2=3/4$ .\n\nCombining it with its cyclic variants yields\n\\begin{align*}\na^2-ab+b^2&=3/4;\nb^2+c^2&=1;\nc^2+a^2+ca&=3/4.\n\\end{align*}\nSubtract the first and third equations, then factor, yielding \n\\[(c+b)(c-b+a)=0.\\]\nThere are two cases from here:**Case 1: b + c = 0.** $b=-c$ implies $(b,c)=(\\pm1/\\sqrt2,\\mp1/\\sqrt2)$ . Then we get $a^2-a/\\sqrt2-1/4=0$ , which does not yield any rational values of $a^2b^2c^2=a^2$ . Probably extraneous.**Case 2: b = a + c.** System reduces to\n\\begin{align*}\na^2+2c(a+c)=1;\na^2+c(a+c)=3/4;\n\\end{align*}\nHence $a^2=1/2$ and $c(a+c)=1/4$ ; WLOG $a=1/\\sqrt2$ ; then $c=(-1\\pm\\sqrt3)/2\\sqrt2$ and $b=a+c=(1+\\pm\\sqrt3)/2\\sqrt2$ .\nWe get $abc=1/4\\sqrt2\\Rightarrow1/32\\Rightarrow\\boxed{033}.$ **Remark:**Almost forgot a factor of $1/2$ in $abc$ which would've been a silly.</details>",
"<blockquote><blockquote>what is the motivation for trig substitution?</blockquote>\n\nI see none-- i subsituted $a,b,c=1-x,1-y,1-z$ .</blockquote>\n\nhow would you think of that though",
"yay got this one\n@above its something you're trying to find the product of lol",
"The expression you're trying to find all are 1-[variable], so it's pretty natural to solve for what you're trying to find. Probably much simpler than solving the original thing and then subtracting it from 1 three times",
"Is there a pure-algebraic solution to this?",
"<blockquote>yay got this one\n@above its something you're trying to find the product of lol</blockquote>\n\nwhat do you mean?",
"<blockquote><blockquote>yay got this one\n@above its something you're trying to find the product of lol</blockquote>\n\nwhat do you mean?</blockquote>\n\nif we let a=1-x, b=1-y, c=1-z, then we can just find (abc)^2 with trig substitutions",
"<blockquote>Is there a pure-algebraic solution to this?</blockquote>\n\n<blockquote><blockquote>Let $x$ , $y$ , and $z$ be positive real numbers satisfying the system of equations\n\\begin{align*}\n\\sqrt{2x - xy} + \\sqrt{2y - xy} & = 1\n\\sqrt{2y - yz} + \\hspace{0.1em} \\sqrt{2z - yz} & = \\sqrt{2}\n\\sqrt{2z - zx\\vphantom{y}} + \\sqrt{2x - zx\\vphantom{y}} & = \\sqrt{3}.\n\\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 $m+n$ .</blockquote>\n\nThe requested expression is very \"sus\" indeed.\n<details><summary>Subsitution solution</summary>Let $a,b,c=1-x,1-y,1-z$ respectively. Rewrite the equations as\n\\[\\sqrt{(1-a)(1+b)}+\\sqrt{(1+a)(1-b)}=1\\]\nand cyclic variants.\n\nSquare and rearrange, giving $2\\sqrt{(1-a^2)(1-b^2)}=2ab-1$ ; then dividing by 2 and squaring again yields $a^2b^2-a^2-b^2+1=a^2b^2-ab+1/4$ , whence $a^2-ab+b^2=3/4$ .\n\nCombining it with its cyclic variants yields\n\\begin{align*}\na^2-ab+b^2&=3/4;\nb^2+c^2&=1;\nc^2+a^2+ca&=3/4.\n\\end{align*}\nSubtract the first and third equations, then factor, yielding \n\\[(c+b)(c-b+a)=0.\\]\nThere are two cases from here:**Case 1: b + c = 0.** $b=-c$ implies $(b,c)=(\\pm1/\\sqrt2,\\mp1/\\sqrt2)$ . Then we get $a^2-a/\\sqrt2-1/4=0$ , which does not yield any rational values of $a^2b^2c^2=a^2$ . Probably extraneous.**Case 2: b = a + c.** System reduces to\n\\begin{align*}\na^2+2c(a+c)=1;\na^2+c(a+c)=3/4;\n\\end{align*}\nHence $a^2=1/2$ and $c(a+c)=1/4$ ; WLOG $a=1/\\sqrt2$ ; then $c=(-1\\pm\\sqrt3)/2\\sqrt2$ and $b=a+c=(1+\\pm\\sqrt3)/2\\sqrt2$ .\nWe get $abc=1/4\\sqrt2\\Rightarrow1/32\\Rightarrow\\boxed{033}.$ **Remark:**Almost forgot a factor of $1/2$ in $abc$ which would've been a silly.</details></blockquote>\n\n",
"<blockquote><blockquote><blockquote>yay got this one\n@above its something you're trying to find the product of lol</blockquote>\n\nwhat do you mean?</blockquote>\n\nif we let a=1-x, b=1-y, c=1-z, then we can just find (abc)^2 with trig substitutions</blockquote>\n\nsubstitutions are simply a way of making apparent observations that would otherwise be very obscure...",
"How I did it in test. Why trig when you can use real geometry? (I also find this more natural than trig sub.)\n\nSubstitute $a=1-x,$ $b=1-y, c=1-z.$ \n\nSo $\\sqrt{-(a-1)(b+1)}+\\sqrt{-(a+1)(b-1)}=1,$ $\\sqrt{-(b-1)(c+1)}+\\sqrt{-(b+1)(c-1)}=\\sqrt{2},$ $\\sqrt{-(a-1)(c+1)}+\\sqrt{-(a+1)(c-1)}=\\sqrt{3}.$ Squaring, we get $2-2ab + 2\\sqrt{(1-a^2)(1-b^2)}=1,$ $2-2bc + 2\\sqrt{(1-b^2)(1-c^2)}=2, 2-2bc +2\\sqrt{(1-b^2)(1-c^2)}=3.$ Isolating the radical and squaring again yields $a^2-ab+b^2 = \\frac{3}{4}, b^2+c^2=1, c^2+ca+a^2 = \\frac{3}{4}.$ \n\nThe fun part: take the cyclic quadrilateral $WXYZ$ where $\\angle XYZ = \\angle ZWX= 90^\\circ,$ $XZ = 1,$ $\\angle WXY = 60^\\circ, \\angle YZW= 120^\\circ,$ $ZW=WX.$ Easy computation yields $ZW=WX = \\frac{1}{\\sqrt{2}}, XY = \\frac{\\sqrt{6}+\\sqrt{2}}{4}, YZ = \\frac{\\sqrt{6}-\\sqrt{2}}{4},$ and $WY = \\frac{\\sqrt{3}}{2}.$ It's easy to see w/ LoC that setting $ZW=WX=a,XY=b,YZ=c$ works. The rest is computation. $\\blacksquare$ ",
"Hello, LoC is not *real* geometry.",
"unpopular opinion\neasiest problem on the test",
"<blockquote>unpopular opinion\neasiest problem on the test</blockquote>\n\nfor its placement yes maybe",
"<blockquote><blockquote>unpopular opinion\neasiest problem on the test</blockquote>\n\nfor its placement yes maybe</blockquote>\n\nno\neasier than 2\n>:( $ $ ",
"<blockquote><blockquote>unpopular opinion\neasiest problem on the test</blockquote>\n\nfor its placement yes maybe</blockquote>\n\nnah 13 is easier for its placement",
"<blockquote>what is the motivation for trig substitution?</blockquote>\n\nThere was also the fact that the RHS of the equations are all 2 times common trig values. I believe this is already stated but doing the factorization for the LHS leads to an expression format extremely similar to that of the sin addition formula.",
"gUeSs AnD cHeCk?!?!?!?",
"No need to use Trig\nwhen reaching here \n\\begin{align*}\na^2-ab+b^2&=3/4; & (1) \nb^2+c^2&=1; & (2)\nc^2+a^2+ca&=3/4.& (3) \n\\end{align*}\nBased on (1) & (3): $-ab+b^2=c^2+ca$ , i.e. $(b+c)(b-c-a)=0$ .\nTwo situations $b=-c$ or $b=c+a$ .\nCheck $b=c+a$ first. Based on (2), $(a+c)^2+c^2=1$ , Combine this with (3) we get $a^2=1/2$ .\n\nThen from $b=c+a$ , we have $(b-c)^2=a^2=1/2$ , then $bc$ = $1/4$ . Thus $a^2b^2c^2=1/2*(1/4)^2=1/32.$ So the answer is $1+32=33$ . (the other situation $b=-c$ not working.)\n",
"Let $1-x=a;1-y=b;1-z=c$ , rewrite those equations $\\sqrt{(1-a)(1+b)}+\\sqrt{(1+a)(1-b)}=1$ ; $\\sqrt{(1-b)(1+c)}+\\sqrt{(1+b)(1-c)}=\\sqrt{2}$ $\\sqrt{(1-a)(1+c)}+\\sqrt{(1-c)(1+a)}=\\sqrt{3}$ square both sides, get three equations: $2ab-1=2\\sqrt{(1-a^2)(1-b^2)}$ $2bc=2\\sqrt{(1-b^2)(1-c^2)}$ $2ac+1=2\\sqrt{(1-c^2)(1-a^2)}$ Getting that $a^2+b^2-ab=\\frac{3}{4}$ $b^2+c^2=1$ $a^2+c^2+ac=\\frac{3}{4}$ Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$ , $a=b-c$ Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\\frac{3}{4}$ , $bc=\\frac{1}{4}$ Since $a^2=b^2+c^2-2bc=\\frac{1}{2}$ , the final answer is $\\frac{1}{4}*\\frac{1}{4}*\\frac{1}{2}=\\frac{1}{32}$ the final answer is 33\n",
"<details><summary>Intuition</summary>It is quite strange that the values $x, y, z$ are in the range $(0, 2)$ , and there are a bunch of $1-a$ and $2-a$ expressions. To make what we want to find simpler, we substitute $x = 1-a, y=1-b, z=1-c$ , where now $a, b, c \\in (-1, 1)$ . This range reminds us of the standard trigonometric functions, yet we shouldn't do something like the square of sine or cosine because it doesn't match up with addition. This leaves only the half-angle formulas, in which the answer comes out.</details>",
"Let $x=2\\sin^2a,y=2\\sin^2b,z=2\\sin^2c$ with $a,b,c\\in[0,2\\pi)$ . (The $2$ is to psuedo-homogenize (make the coordinates all $1$ ) the equations while the trig sub comes from factorizing $\\sqrt{x-xy}$ into $\\sqrt{x(1-y)}$ and also noticing that the RHS's become $\\frac12,\\frac{\\sqrt2}2,\\frac{\\sqrt3}2$ . I did try both $x=2\\cos^2a$ and $x=2\\sin^2a$ . It is a leap of faith to assume that $0\\le x,y,z\\le2$ , but if we find one solution with this condition the problem is solved.)\n\nThen we have: $$ \\sin(a+b)=\\frac12,\\sin(b+c)=\\frac{\\sqrt2}2,\\sin(c+a)=\\frac{\\sqrt3}2, $$ so $a+b=\\frac\\pi6$ , $b+c=\\frac\\pi4$ , $c+a=\\frac\\pi3$ . Then easily solving this system gives $a=\\frac{5\\pi}{24}$ , $b=\\frac\\pi{24}$ , $c=\\frac\\pi8$ , so by spamming double angle and bashing a bit, the answer is $\\frac1{32}\\Rightarrow\\boxed{033}$ .\n",
"<blockquote>gUeSs AnD cHeCk?!?!?!?</blockquote>\n\nhow bruh ",
"We note that the solution calls for $(1-x)^2(1-y)^2(1-z)^2$ , which is difficult to work with so we substitute, $1-x = a$ , $1-y=b$ , $1-z = c$ . Now, our problem is simply\n\n\\[\\sqrt{(1-a)(1+b)}+\\sqrt{(1-b)(1+a)} = 1\\]\n\\[\\sqrt{(1-b)(1+c)}+\\sqrt{(1-c)(1+b)} = \\sqrt{2}\\]\n\\[\\sqrt{(1-a)(1+c)}+\\sqrt{(1-c)(1+a)} = \\sqrt{3}\\]\n\nWe square both sides to get\n\n\\[\\sqrt{(1-a^2)(1-b^2)} = ab-\\frac{1}{2}\\]\n\\[\\sqrt{(1-b^2)(1-c^2)} = bc\\]\n\\[\\sqrt{(1-a^2)(1-c^2)} = ac+\\frac{1}{2}\\]\n\nOnce again, we square both sides to get\n\\[\\frac{3}{4} = a^2-ab+b^2\\]\n\\[1=b^2+c^2\\]\n\\[\\frac{3}{4} = a^2+ac+c^2\\]\n\nWe can subtract the first equation from the third equation to get\n\n\\[0 = b^2-c^2-ab-ac = (b+c)(b-c-a)\\]\n\nNote that we can't have $c = -b$ because then, one of the equations that's supposed to be a square root is negative, so we have that $b = a+c$ . We substitute this into the second equation to get the system of equations\n\n\\[1=a^2+2ac+2c^2\\]\n\\[\\frac{3}{4} = a^2+ac+c^2\\]\n\nThis means that $a^2 = \\frac{1}{2}$ . We solve for $c$ to get\n\n\\[1=\\frac{1}{2}+\\sqrt{2}c+2c^2\\]\n\\[c = \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\]\n\nThus, $b = \\frac{\\sqrt{6}+\\sqrt{2}}{4}$ . Our answer is $a^2b^2c^2 = \\frac{1}{2}\\times(\\frac{4}{16})^2 = \\frac{1}{32} \\implies \\boxed{033}$ .\n",
"An actually solvable number 15",
"<blockquote>what is the motivation for trig substitution?</blockquote>\n\nI have the same question as this - does anyone have concrete motivation? (I know it's the day before AIME; I'm asking in case something like this appears on the test)",
"square roots\n\nalso just intuition from problems\nits useful to just incorporate it into your toolbox",
"<blockquote><blockquote>what is the motivation for trig substitution?</blockquote>\n\nI have the same question as this - does anyone have concrete motivation? (I know it's the day before AIME; I'm asking in case something like this appears on the test)</blockquote>\n\nThere are many examples of trigsubbing for expressions such as $x(1-y).$ However, $x(2-y)$ is a bit more complicated. It should still use the same idea, though, and that's how you come up with the $\\sqrt{2}$ factor.",
"[Video Solution](https://youtu.be/aa_VY4e4OOM?si=oJ2H67mBGO7YtOuh)",
"Putting $1-x=a , 1-y=b\\; \\&\\; 1-z=c$ So, the system of equations reduces to: $(1-a)(1+b)+(1+a)(1-b)=1\n(1-c)(1+a)+(1+c)(1-a)=\\sqrt{2}\n(1-b)(1+c)+(1+b)(1-c)=\\sqrt{3}$ Now, just squaring and after manipulating a bit, I think you would get $(abc)^2=\\frac{1}{32}$ .\nSo, $m+n=\\boxed{33}$ ",
"<details><summary>solution</summary>We use a trig substitution; $x=2 \\cos^2 a$ , $y=2 \\cos^2 b$ , $z=2 \\cos^2 c$ . The top equation can be rewritten as\n\t\\begin{align*}\n\t\t2\\cos a \\sin b + 2\\cos b \\sin a &= 1 \n\t\t2 \\sin(a+b) &= 1 \n\t\t\\sin(a+b) &= 1/2\n\t\\end{align*}\n\tSimilarly, we can write\n\t\\begin{align*}\n\t\t\\sin(b+c) &= \\sqrt2/2 \n\t\t\\sin(c+a) &= \\sqrt3/2\n\t\\end{align*}\n\tAll of these values are common on the unit circle, and we can easy find that $(a,b,c)=(45/2,15/2,75/2)$ . $abc$ is then equal to\n\t\\begin{align*}\n\t\t\\prod_{\\text{cyc}} (1-x) &= \\prod_{\\text{cyc}} (1-2 \\cos^2a) \n\t\t&= \\prod_{\\text{cyc}} (-\\cos 2a) \n\t\t&= - \\cos 45 \\cos 15 \\cos 75 \n\t\t&= \\sqrt2/8\n\t\\end{align*}\nIt is then obvious that the answer is $\\boxed{33}$</details>"
] |
[
"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), \"B\", NE);\nlabel((6.8333333, 6), \"B'\", N);\nlabel((8,0), \"C'\", S);\ndraw((0,0)--(12,0)--(10.8333333,6)--(1.16666666,6)--(0,0));\ndot((6.83333333,6));\ndot((8,0));\ndraw((4,3)--(8,3), dashed);\ndraw((1.1666666,6)--(4,3), dashed);\ndraw((0,0)--(4,3), dashed);\ndot((8,3));\ndraw((6.8333333, 6)--(4,3)--(8,0)--cycle, dashed);\ndot((4,3));\nlabel((6,3), \"x\", N);\nlabel((8.83333333, 6), \"x\", N);\nlabel((10, 0), \"x\", S);\nlabel((4,0), \"650-x\", S);\nlabel((4,6), \"500-x\", N);\nlabel((0.83333333, 3), \"333\", NW);\ndraw((10.833333, 6)--(8,3)--(12,0), dashed);\nlabel((11.1666666, 3), \"333\", NE);\n[/asy]\n\nNotice that after this translation, $B'Q'$ still bisects $AB'C'$ , and $C'Q'$ still bisects $AC'B'$ . Therefore, the intersection of the angle bisectors in quadrilateral $AB'C'D$ exists and it is point $P$ . So, $AB'C'D$ is tangential. By the Pitot Theorem, we have $AB' + DC' = AD + B'C'$ , so $(500-x) + (650-x) = 333+333$ , meaning $x = \\boxed{242}$ .",
"575-333=242 essentially we know PQ lies on midline and then if M and N are midpoints of AD and BC than MP+NQ=333 so qed.",
"So apparently nobody else at my testing center did this? Drop altitudes from P to AB, AD, CD, and from Q to AB, BC, CD. Area=area immediately gives a linear equation in PQ.",
"you can trig this also",
"Let $E$ on $\\overline{AB}$ and $F$ on $\\overline{CD}$ be collinear with $P$ , s.t $\\overline{EF}$ is perpendicular to $\\overline{AB}$ and $\\overline{CD}$ . Then $\\triangle DPF~ \\triangle DAP~ \\triangle APE$ . By a simple length-chase, $\\overline{EP}=\\overline{PF}$ , so $\\overline{PQ}$ is contained in the median of trapezoid $ABCD$ . Extend $\\overline{PQ}$ to hit $\\overline{AD}$ at $S$ and $\\overline{BC}$ at $T$ . Then $SP=QT=\\frac{333}{2}$ since it is a median to the hypotenuse of right triangle $\\triangle ADP$ . Thus $PQ=\\frac{AB+CD}{2}-\\frac{AD+BC}{2}=\\frac{500+650}{2}-2\\cdot \\frac{333}{2} = \\boxed{242}$ .",
"<blockquote>you can trig this also</blockquote>\nTrig was clean.\n\n<details><summary>Click to expand</summary>Let $\\angle C = x$ so that $\\angle B = 180^\\circ - x.$ It follows that $\\angle QBC = \\tfrac{180^\\circ - x}{2}$ and $\\angle QCB = \\tfrac{x}{2},$ hence $\\angle Q = 90^\\circ.$ If $B'$ is the projection of $B$ onto $\\overline{CD},$ $B'C = \\tfrac{650 - 500}{2} = 75,$ hence $\\cos x = \\tfrac{75}{333} = \\tfrac{25}{111}$ and $\\cos \\tfrac{x}{2} = \\sqrt{\\tfrac{1 + \\tfrac{25}{111}}{2}} = \\sqrt{\\tfrac{68}{111}}.$ It follows that $CQ = 3\\sqrt{111 \\cdot 68},$ but $CQ^2 = BC \\cdot CQ'$ by similar triangles, where $Q'$ is the projection of $Q$ onto $\\overline{CD}.$ Mass cancellation reveals that $CQ' = 3 \\cdot 68 = 204,$ hence $PQ = 650 - 2 \\cdot 204 = \\boxed{242},$ the requested answer.</details>\n\nHastily written in like two minutes, but you get the point.",
"<details><summary>In contest Solution</summary>Sit at this problem and do nothing (only draw the diagram and don't start solving) for at least 20 minutes total.</details>",
"My trig approach:\n<details><summary>Click to expand</summary>Project P to CD and name it E. We then know that $PQ = 650-2DE$ , and $ DE = AD \\cdot \\cos^2(\\angle \\frac{D}{2})$ . We also know that $\\cos(D) = \\frac{75}{333}$ , so double angle finishes it off.</details>",
"I spent like 30 minutes attempting this and trying to use similar triangles but didn't work lol\n",
"IT TOOK ME 30 minutes to get it :imdumb:\n\n@above ikr i just cord-bashed",
"**30 Second Solution:**\n\nLet $K$ be the projection of $A$ onto $CD$ and $AP$ meet $CD$ at $X$ . Angle chasing yields $DA = DX$ , and symmetry implies $$ DK = \\frac{650 - 500}{2} = 75. $$ Because $P$ is equidistant from $AB$ and $CD$ , symmetry and properties of midlines gives $$ PQ = 500 - KX = 500 - (333 - 75) = \\boxed{242.} $$ ",
"This literally took me like 20-30 minutes for some reason, more than $7$ and $9$ combined... either misplaced or I'm just that bad at geo ",
"trig bash o",
"YESS I SPENT 1 HOUR ON THIS AND GOT IT RIGHT :D",
"I got 250 for some reason",
"<details><summary>Sketch</summary>Extend $AP$ to $CD$ , call the intersection $S$ . $ADS$ is an isosceles, so $CS$ is $333$ , now we coor-trig bash (not that long of a bash) to find that though $PQ=242$</details>",
"Favorite problem on the test. Took me 25 minutes though... Definitely felt this was harder than P7, OTIS consensus seems to be that this problem was massively misplaced.",
"hm i thought this was p easy?",
"<blockquote>Favorite problem on the test. Took me 25 minutes though... Definitely felt this was harder than P7, OTIS consensus seems to be that this problem was massively misplaced.</blockquote>\n\nI think there's a previous AIME problem with the idea that $P$ and $Q$ lie on the midline. (I can't find it.) Still, I wasn't able to observe this immediately.\n\nHowever, after this, the problem falls apart.",
"<blockquote>hm i thought this was p easy?</blockquote>\n\nWhat?? This was harder than problems 6, 7 and 9.",
"Similar triangles kills it... solved in 5-10min",
"Diagram:\n[asy]\nunitsize(0.016cm);\npair A = (-250,324.4);\npair B = (250, 324.4);\npair C = (325, 0);\npair D = (-325, 0);\ndraw(A--B--C--D--cycle);\npair W = (8,0);\npair X = (-8, 0);\npair Y = (-83,324.4);\npair Z = (83,324.4);\n\npair P = (-121, 162.2);\npair Q = (121, 162.2);\ndot(P);\ndot(Q);\n\ndraw(A--W, dashed);\ndraw(B--X, dashed);\ndraw(C--Y, dashed);\ndraw(D--Z, dashed);\nlabel(\" $A$ \", A, N);\nlabel(\" $B$ \", B, N);\nlabel(\" $Y$ \", Y, N);\nlabel(\" $Z$ \", Z, N);\nlabel(\" $C$ \", C, S);\nlabel(\" $D$ \", D, S);\nlabel(\" $W$ \", W, SE);\nlabel(\" $X$ \", X, SW);\nlabel(\" $P$ \", P, N);\nlabel(\" $Q$ \", Q, N);\n[/asy]\n\nExtend 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.\n\nClaim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.\n\nProof: 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.\n\nExtend 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, $PQ = RS - RP - QS = \\boxed{242}$ .",
"This took me one hour. I just coordbashed",
"<blockquote>Similar triangles kills it... solved in 5-10min</blockquote>\n\nThat's EXACTLY what I did...notice that we have one triangle with height $2x-y$ and base $500$ , the other with base $PQ$ and height $x-y$ , and another one with base $16$ and height $y$ .\n\nSo, $\\frac{2x-y}{500} = \\frac{y}{16} \\Rightarrow x = \\frac{516}{32}y$ . So...our triangle with base $PQ$ has height $\\frac{516}{32}y - y = \\frac{484}{32}y$ . Now, we see that $PQ = \\frac{484}{32} \\cdot 16 = \\boxed{242}$ .",
"<blockquote>Diagram:\n[asy]\nunitsize(0.016cm);\npair A = (-300,324.4);\npair B = (300, 324.4);\npair C = (375, 0);\npair D = (-375, 0);\ndraw(A--B--C--D--cycle);\npair W = (-42,0);\npair X = (42, 0);\npair Y = (-33,324.4);\npair Z = (33,324.4);\n\npair P = (-171, 162.2);\npair Q = (171, 162.2);\ndot(P);\ndot(Q);\n\ndraw(A--W, dashed);\ndraw(B--X, dashed);\ndraw(C--Y, dashed);\ndraw(D--Z, dashed);\nlabel(\" $A$ \", A, N);\nlabel(\" $B$ \", B, N);\nlabel(\" $Y$ \", Y, N);\nlabel(\" $Z$ \", Z, N);\nlabel(\" $C$ \", C, S);\nlabel(\" $D$ \", D, S);\nlabel(\" $W$ \", W, S);\nlabel(\" $X$ \", X, S);\nlabel(\" $P$ \", P, N);\nlabel(\" $Q$ \", Q, N);\n[/asy]\n\nExtend 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.\n\nClaim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.\n\nProof: 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.\n\nExtend 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>\n\nummmmmm answer is $242$ i thought...",
"500+650 = 2(675) o",
"<blockquote>This took me one hour. I just coordbashed</blockquote>\n\nI spent like 1 hour doing some congruent triangles nonsense and somehow got $PQ = 333$ , but then thought for a long time thinking \"this can't be right...\" and then realized the right way to do it via similar $\\triangle$ s.",
"<blockquote><blockquote>hm i thought this was p easy?</blockquote>\n\nWhat?? This was harder than problems 6, 7 and 9.</blockquote>\n\nyea I agree\n\n~~though I sillied #6 and #7 :(~~",
"in test solution was similar triangles\n\nright triangles works too though",
"<details><summary>Angle Bisector Theorem sol</summary>Drop perpendiculars from points $A$ and $B$ to $CD$ and label the points $A'$ and $B'$ . Label the intersection of $AA'$ and $PD$ $P'$ . Drop a perpendicular from $P$ and $Q$ , and call them $M$ and $N$ , respectively. Let the height of the trapezoid be $h$ .\n\nBy the Angle Bisector theorem, $\\frac{A'P'}{AP'} = \\frac{75}{333}$ . Thus, $A'P' = \\frac{75h}{408}$ . Also, $MP = \\frac{h}{2}$ . This is because we can draw a circle that is tangent to $AB$ , $CD$ , and $AD$ . The center of this circle is the intersection of the angle bisectors, $P$ , and thus $P$ is equidistant from the two bases. Now, $DA'P'$ and $DMP$ are similar, so the ratio between $DA'$ and $DM$ is $\\frac{75}{204}$ , the ratio between $A'P'$ and $MP'$ . Thus, $DM = 204$ .\n\nWe can do a similar process on the other side of the trapezoid. Then, observe that $MNQP$ is a rectangle, and thus $PQ = MN$ , so $PQ = MN = 650 - DM - CN = \\boxed{242.}$</details>",
"<blockquote>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>\n\nNice problem.\n<details><summary>Put everything onto CD</summary>Because $\\overline{PQ}\\parallel\\overline{AB}\\parallel\\overline{CD}$ by symmetry, it is equivalent to finding the distance between the projections of $P$ and $Q$ on $\\overline{CD}$ , which is along said segment.\n\nLet $B'$ be the reflection of $B$ in $\\overline{CQ}$ , so that $B'\\in\\overline{CD}$ . By symmetry about $\\overline{CQ}$ , $B'\\in\\overline{CD}$ ; because $Q$ is the midpoint of $\\overline{BB'}$ , its projection onto $\\overline{CD}$ is the midpoint of the segment formed by the projections of $B,B'$ onto $\\overline{CD}$ . Letting $X$ be the projection of $B$ onto $\\overline{CD}$ , we have $CX=(CD-AB)/2=75$ and $XB'=258$ . Finally we let $Y$ be the projection of $Q$ onto $\\overline{CD}$ , whence $XY=XB'/2=129$ ; the final answer is then $500-2(129)=\\boxed{242}.$</details>",
"Let $PQ=x$ and notice $\\angle APD=90.$ Let $Q'$ be the foot from $Q$ to $\\overline{AB}.$ Similarly, let $P'$ be the foot from $P$ to $\\overline{AB}.$ Then, $\\triangle QQ'D\\sim\\triangle CQB$ so $$ \\frac{\\frac{500-x}{2}}{QB}=\\frac{Q'B}{QB}=\\frac{QB}{BC}=\\frac{QB}{333}. $$ Hence, $QB^2=\\frac{(500-x)333}{2}.$ Similarly, $QC^2=\\frac{(650-x)333}{2}.$ Therefore, $$ \\frac{(500-x)333}{2}+\\frac{(650-x)333}{2}=333^2 $$ so $x=\\boxed{242}.$ ",
" $P,Q$ lies on the median of the trapezoid( easy to prove because of parallel line and angle bisectors)\n\nExtend $PQ$ to meet at two sides at $M,N$ , $PM=QN=\\frac{1}{2}*333$ , answer is $(650+500)/2-333=242$ ",
"The extend-AP-and-BQ-and-notice-that-APD-is-90 solution I used reminded me of [2011 AIME I](https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_4)",
"I got something with 325/83 to solve did anyone else do this?",
"how did i get 245 :noo:",
"<details><summary>Solution</summary>Extend the bisector of angle A till it meats CD at a point X. Notice that D = 180 - A/2, so the final angle is A/2, so DX = 333. Now notice that angle APD is 90 so DP is the altitude onto AX in triangle ADX, meaning ADP and DPX are congruent, so P is halfway down the trapezoid. So, the horizontal distance between P and AD is 333/2. Since the other side is the same, we have (500+650)/2 - 333 = 242</details>",
"wait bruh i misread this and took 30-45 minutes bashing it because i thought p and q were the intersection of the bisector of <a, <b, and <c, <d\nnow i'm mad at myself\n\nanyone else do this?",
"<blockquote>how did i get 245 :noo:</blockquote>\n\nyou thought you were solving this but you were actually solving problem 12",
"Using this diagram, DW=333, and CX=333, so XW = |650-333-333| = 16, AB = 500, PQ is in the middle heightwise, so PQ = (500-16)/2 = 242...\nDid anyone else do it this way or just me?\n<blockquote>Diagram:\n[asy]\nunitsize(0.016cm);\npair A = (-250,324.4);\npair B = (250, 324.4);\npair C = (325, 0);\npair D = (-325, 0);\ndraw(A--B--C--D--cycle);\npair W = (8,0);\npair X = (-8, 0);\npair Y = (-83,324.4);\npair Z = (83,324.4);\n\npair P = (-121, 162.2);\npair Q = (121, 162.2);\ndot(P);\ndot(Q);\n\ndraw(A--W, dashed);\ndraw(B--X, dashed);\ndraw(C--Y, dashed);\ndraw(D--Z, dashed);\nlabel(\" $A$ \", A, N);\nlabel(\" $B$ \", B, N);\nlabel(\" $Y$ \", Y, N);\nlabel(\" $Z$ \", Z, N);\nlabel(\" $C$ \", C, S);\nlabel(\" $D$ \", D, S);\nlabel(\" $W$ \", W, SE);\nlabel(\" $X$ \", X, SW);\nlabel(\" $P$ \", P, N);\nlabel(\" $Q$ \", Q, N);\n[/asy]\n\nExtend 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.\n\nClaim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.\n\nProof: 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.\n\nExtend 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, $PQ = RS - RP - QS = \\boxed{242}$ .</blockquote>\n\n",
"Did anyone make an inscribed circle? Fairly easy to solve",
"<blockquote>Did anyone make an inscribed circle? Fairly easy to solve</blockquote>\n\nyes (helps to consider it as two infinitely long (:smirk:) triangles",
"<details><summary>trig</summary>Let $\\theta = \\angle PDA$ , and $X$ be the foot from $A$ to $CD$ . Then $\\angle D = 2\\theta$ , so $\\angle XAD = 90-2\\theta.$ Since $\\angle PAD = 90-\\theta$ , $\\angle PAX = \\theta.$ Now, let $x$ be the distance from $P$ to $AX.$ Note that $\\angle APD = 90.$ From trig ratios, $\\cos(2\\theta) = \\frac{75}{333}$ , $\\sin (\\theta) = \\frac{AP}{333} = \\frac{x}{AP}.$ Then $x = 333\\sin^2 (\\theta).$ From double angle, $x = 129.$ Note that by symmetry, the distances from $P$ and $Q$ to $CD$ are equal. Thus the answer is $500-2 \\cdot 129 = 242$ .</details>",
"<details><summary>83/325 solution</summary>[asy]\nunitsize(0.016cm);\npair A = (-250,324.4);\npair B = (250, 324.4);\npair C = (325, 0);\npair D = (-325, 0);\ndraw(A--B--C--D--cycle);\npair W = (8,0);\npair X = (-8, 0);\npair Y = (-83,324.4);\npair Z = (83,324.4);\n\npair P = (-121, 162.2);\npair Q = (121, 162.2);\ndot(P);\ndot(Q);\n\ndraw(A--W, dashed);\ndraw(B--X, dashed);\ndraw(C--Y, dashed);\ndraw(D--Z, dashed);\nlabel(\" $A$ \", A, N);\nlabel(\" $B$ \", B, N);\nlabel(\" $Y$ \", Y, N);\nlabel(\" $Z$ \", Z, N);\nlabel(\" $C$ \", C, S);\nlabel(\" $D$ \", D, S);\nlabel(\" $W$ \", W, SE);\nlabel(\" $X$ \", X, SW);\nlabel(\" $P$ \", P, N);\nlabel(\" $Q$ \", Q, N);\n[/asy]\n\nDenote $YQ \\cup PZ = X$ . Let $DP = a, PX = b$ . Then by similarity, we have $a = b + \\frac{166}{500}(a+b) \\implies 121a = 204b$ . We also have $\\frac{b}{a+b} = \\frac{1}{\\frac{a}{b}+1} = \\frac{1}{\\frac{204}{121}+1}=\\frac{121}{325}$ . Therefore, the answer is $650 \\cdot \\frac{121}{325} = \\boxed{242}$ .</details>",
"I spent like 30 min on this and ended up coordinate bashing with the slope-tangent formula. The nice way is to note that the points on the angle bisector of $\\angle A$ are equidistant from $AB$ and $AD$ , and the points on the angle bisector of $\\angle D$ are equidistant from $AD$ and $CD$ , which implies that $P$ is on the midline, likewise with $Q$ . Then let $AP$ hit $CD$ at $X$ and $BQ$ hit $CD$ at $Y$ . Additionally, $$ \\angle ADX=180-\\angle DAB=180-2\\angle DAX\\implies \\angle ADX=\\angle DXA\\implies DA=DX $$ Thus, since the midline has length $575$ , the answer is $$ 575-2\\left(\\frac{333}{2}\\right)=242 $$ ",
"Let $\\angle{ADC}=2x$ . Angle chasing yields $\\angle{APD}=90$ . Construct the line perpendicular to $PQ$ that intersects it at $P$ . Call its intersections with $AB$ and $CD$ points $E$ and $F$ respectively. From here, $\\angle{ADP}=\\angle{APE}=\\angle{PDF}=x$ . Let $PQ=2n$ . Then, $AE=250-n$ and $DF=325-n$ . From $\\triangle{DFP}$ , $\\cos{x}=\\frac{325-n}{DP}$ . From $\\triangle{ADP}$ , $\\cos{x}=\\frac{DP}{333}$ . Therefore $DP^2=333(325-n)$ . From $\\triangle{APE}$ , $\\sin{x}=\\frac{250-n}{AP}$ . From $\\triangle{ADP}$ , $\\sin{x}=\\frac{AP}{333}$ . Therefore $AP^2=333(250-n)$ . By the Pythagorean Theorem, $DP^2+AP^2=AD^2=333^2$ . Simplifying gives $2n=\\boxed{242}.$ ",
"i found that APD is 90 degrees and the used similar triangles \nwhoops didn't see @above's solution, just realized that i did the same thing as them",
"Who else forgot to do the final step and got 204?",
"<blockquote>Using this diagram, DW=333, and CX=333, so XW = |650-333-333| = 16, AB = 500, PQ is in the middle heightwise, so PQ = (500-16)/2 = 242...\nDid anyone else do it this way or just me?\n<blockquote>Diagram:\n[asy]\nunitsize(0.016cm);\npair A = (-250,324.4);\npair B = (250, 324.4);\npair C = (325, 0);\npair D = (-325, 0);\ndraw(A--B--C--D--cycle);\npair W = (8,0);\npair X = (-8, 0);\npair Y = (-83,324.4);\npair Z = (83,324.4);\n\npair P = (-121, 162.2);\npair Q = (121, 162.2);\ndot(P);\ndot(Q);\n\ndraw(A--W, dashed);\ndraw(B--X, dashed);\ndraw(C--Y, dashed);\ndraw(D--Z, dashed);\nlabel(\" $A$ \", A, N);\nlabel(\" $B$ \", B, N);\nlabel(\" $Y$ \", Y, N);\nlabel(\" $Z$ \", Z, N);\nlabel(\" $C$ \", C, S);\nlabel(\" $D$ \", D, S);\nlabel(\" $W$ \", W, SE);\nlabel(\" $X$ \", X, SW);\nlabel(\" $P$ \", P, N);\nlabel(\" $Q$ \", Q, N);\n[/asy]\n\nExtend 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.\n\nClaim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.\n\nProof: 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.\n\nExtend 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, $PQ = RS - RP - QS = \\boxed{242}$ .</blockquote></blockquote>\n\nyeah i did it this way. better than any other solution",
"<blockquote>you can trig this also</blockquote>\n\nmore bashy approach but still fine",
"If I haven't done 2011 AIME I P4 I would have taken an entire hour for this single problem",
"<blockquote><blockquote>you can trig this also</blockquote>\n\nmore bashy approach but still fine</blockquote>\n\nCheck post 49 I think it’s pretty clean",
"TRIG FTW\n\nSuppose $\\angle ADP = \\angle BQC = \\theta$ . It is pretty easy to find that $\\angle APD = \\angle PQC = 90^o$ . Therefore if $R$ and $S$ are the foot of the perpendicular from $P$ and $Q$ to $DC$ respectively, then $DR=CS=333 \\cos^2 (\\theta)$ , which means $$ 650-2DR=PQ $$ $$ \\implies \\text{By double angle } \\cos^2 (\\theta) = \\frac{204}{333} $$ $$ 650-2(204)=650-408=\\boxed{242} $$ This is actually the cleanest solution in the thread since $cos^2(\\theta)$ is pretty easy to find",
"completely different solution I made a rhombus XCBY with X on DC and Y on AB and then drew diagonals, dropped altitudes, extremely quick solve",
"drew and measured a diagram but only got 236 rip",
"Yall wacky.\n\n\t\t\tLet $M$ be the midpoint of $AD$ and $N$ be the midpoint of $BC$ . Note that $\\angle APD$ is a right angle, since $\\angle ADP+\\angle DAP=\\frac{1}{2}(\\angle BAD+\\angle ADC)=90^{\\circ}$ . Thus $MP=\\frac{1}{2}AD=\\frac{333}{2}$ . Also note that $MP$ is parallel to $CD$ as $\\angle MPD=\\angle MDP=\\angle MDC$ . Since $MN$ is also parallel to $CD$ , we know that $P$ lies on $MN$ ; similarly, $Q$ lies on $MN$ . Then \n\t\t\t\\[MN=MP+PQ+QN=\\frac{AD}{2}+PQ+\\frac{BC}{2}=333+PQ.\\]\n\t\t\tSince $MN=\\frac{AB+CD}{2}=\\frac{500+650}{2}=575$ , our answer is $PQ=242$ .",
"sorry for double solution, but note that P is the center of a rhombus with A and D with its vertices, and we can use coord to get the coord of P, then double the x coord to get $\\colorbox{yellow}{242}$ .",
"tfw F is the \"incenter\" and I is the point of tangency",
"<blockquote>**30 Second Solution:**\n\nLet $K$ be the projection of $A$ onto $CD$ and $AP$ meet $CD$ at $X$ . Angle chasing yields $DA = DX$ , and symmetry implies $$ DK = \\frac{650 - 500}{2} = 75. $$ Because $P$ is equidistant from $AB$ and $CD$ , symmetry and properties of midlines gives $$ PQ = 500 - KX = 500 - (333 - 75) = \\boxed{242.} $$ </blockquote>\n\nSorry, but can someone explain how DA=DX is found? I can't see the angle-chasing.",
"<blockquote><blockquote>**30 Second Solution:**\n\nLet $K$ be the projection of $A$ onto $CD$ and $AP$ meet $CD$ at $X$ . Angle chasing yields $DA = DX$ , and symmetry implies $$ DK = \\frac{650 - 500}{2} = 75. $$ Because $P$ is equidistant from $AB$ and $CD$ , symmetry and properties of midlines gives $$ PQ = 500 - KX = 500 - (333 - 75) = \\boxed{242.} $$ </blockquote>\n\nSorry, but can someone explain how DA=DX is found? I can't see the angle-chasing.</blockquote>\n\nsince $\\angle XAB = \\angle DAX $ (angle bisector) and $\\angle XAB = \\angle AXD $ (alternate interior) then $\\angle DAX = \\angle AXD$ . \nSo triangle DAX is isoceles with DA=DX",
"Slick Title",
"I noticed that PQ is in the middle of AB and CD, using a incircle thingy. From there, I dropped a perpendicular segment from P and Q to AD and BC, and also to the midpoints of AD and BC. Now we have two triangles that are similar to the triangle formed by A, D, and the point dropped from A to CD. Hope that made sense. ",
"<details><summary>Solution</summary>Sorry for no diagram. I should learn Asy eventually.\n\nLet $M$ and $N$ be the feet of the altitudes from $P$ and $Q$ to $CD$ , respectively. Since $MNPQ$ is a rectangle, finding $PQ$ is equivalent to finding $MN$ , which is $650-2DM$ by symmetry. Let $\\theta=\\angle PDC=\\angle PDA$ .\n\nClaim: $\\angle APD=90^\\circ$ . Proof: $\\angle APD=180^\\circ-\\angle PAD-\\angle PDA=180^\\circ-\\frac{1}{2}\\angle BAD-\\frac{1}{2}\\angle CDA=180^\\circ-\\frac{1}{2}(180^\\circ)=90^\\circ$ .\n\nTherefore $DM=DP\\cos\\theta=DA\\cos^2\\theta$ . Since the trapezoid is isosceles, $\\cos 2\\theta=\\frac{\\frac{650-500}{2}}{333}=2\\cos^2\\theta-1$ , so $DM=333\\cdot\\frac{204}{333}=204$ and $PQ=MN=650-408=\\boxed{242}$ .</details>",
"Simplest solution so far:\n\n<details><summary>solution</summary>Extend lines $AP$ and $BQ$ to meet $CD$ at $P^{\\prime}$ and $Q^{\\prime}$ , respectively. Then $\\triangle ADP^{\\prime}$ and $BCQ^{\\prime}$ are isosceles, as for both triangles one of the bisectors is also a height. Therefore, $P$ and $Q$ are the midpoints of $AP^{\\prime}$ and $BQ^{\\prime}$ , respectively. Since $DP^{\\prime}=CQ^{\\prime}=333$ , we have that the (directed) length $P^{\\prime}Q^{\\prime}$ is equal to $650-(333+333)=-16$ . Since $PQ$ is a midline, $PQ=\\displaystyle\\frac{500+(-16)}{2}=\\boxed{242}$ .</details>",
"I missed AIME by 1.5 :wallbash_red: :wallbash: \n\nBut actually this problem is easy...\n\n<details><summary>Solution</summary>We can first note that $\\angle PAD = \\frac{180 - \\angle ADC}{2} = 90^{\\circ} - \\frac{\\angle ADC}{2} = 90^{\\circ} - \\angle ADP$ . (adenosine diphosphate?? :P)\n\nTherefore $\\angle APD = 90^{\\circ}$ and $\\triangle APD$ is right. By similar reasoning, $\\triangle BQC$ is also right. \n\nExtend $PQ$ to meet $AD$ and $BC$ at $E$ and $F$ respectively. Notice that $EF$ is parallel to $AB$ and $DC$ . Hence, $\\angle AEF = \\angle ADC = \\angle EDP + \\angle EPD$ by the Exterior Angle Theorem. Thus $\\angle EPD = \\angle EDP$ and $E,F$ are the midpoints of $AD,BC$ respectively! Also, $AE=ED=EP$ and $BF=FC=FQ$ because of right angles. Hence $EF = PQ+2 \\cdot \\frac{333}{2} = PQ+333$ . \n\nNow $EF$ is the median of the trapezoid, so $EF = \\frac{650+500}{2} = 575$ . Our answer is $$ PQ = 575-333 = \\boxed{242} $$</details>",
"Similarity: $333\\cdot2=666$ which is a bit bigger than $650$ . Move the right side of the picture left $16$ units, the upper base becomes $500-16=484$ so the midsegment is $484/2=\\boxed{242}$ .",
"<blockquote>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), \"B\", NE);\nlabel((6.8333333, 6), \"B'\", N);\nlabel((8,0), \"C'\", S);\ndraw((0,0)--(12,0)--(10.8333333,6)--(1.16666666,6)--(0,0));\ndot((6.83333333,6));\ndot((8,0));\ndraw((4,3)--(8,3), dashed);\ndraw((1.1666666,6)--(4,3), dashed);\ndraw((0,0)--(4,3), dashed);\ndot((8,3));\ndraw((6.8333333, 6)--(4,3)--(8,0)--cycle, dashed);\ndot((4,3));\nlabel((6,3), \"x\", N);\nlabel((8.83333333, 6), \"x\", N);\nlabel((10, 0), \"x\", S);\nlabel((4,0), \"650-x\", S);\nlabel((4,6), \"500-x\", N);\nlabel((0.83333333, 3), \"333\", NW);\ndraw((10.833333, 6)--(8,3)--(12,0), dashed);\nlabel((11.1666666, 3), \"333\", NE);\n[/asy]\n\nNotice that after this translation, $B'Q'$ still bisects $AB'C'$ , and $C'Q'$ still bisects $AC'B'$ . Therefore, the intersection of the angle bisectors in quadrilateral $AB'C'D$ exists and it is point $P$ . So, $AB'C'D$ is tangential. By the Pitot Theorem, we have $AB' + DC' = AD + B'C'$ , so $(500-x) + (650-x) = 333+333$ , meaning $x = \\boxed{242}$ .</blockquote>\n\nwow. brilliant solution. For some reason wasn't able to solve it when I mocked the test, i got 6/15.",
"<blockquote>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), \"B\", NE);\nlabel((6.8333333, 6), \"B'\", N);\nlabel((8,0), \"C'\", S);\ndraw((0,0)--(12,0)--(10.8333333,6)--(1.16666666,6)--(0,0));\ndot((6.83333333,6));\ndot((8,0));\ndraw((4,3)--(8,3), dashed);\ndraw((1.1666666,6)--(4,3), dashed);\ndraw((0,0)--(4,3), dashed);\ndot((8,3));\ndraw((6.8333333, 6)--(4,3)--(8,0)--cycle, dashed);\ndot((4,3));\nlabel((6,3), \"x\", N);\nlabel((8.83333333, 6), \"x\", N);\nlabel((10, 0), \"x\", S);\nlabel((4,0), \"650-x\", S);\nlabel((4,6), \"500-x\", N);\nlabel((0.83333333, 3), \"333\", NW);\ndraw((10.833333, 6)--(8,3)--(12,0), dashed);\nlabel((11.1666666, 3), \"333\", NE);\n[/asy]\n\nNotice that after this translation, $B'Q'$ still bisects $AB'C'$ , and $C'Q'$ still bisects $AC'B'$ . Therefore, the intersection of the angle bisectors in quadrilateral $AB'C'D$ exists and it is point $P$ . So, $AB'C'D$ is tangential. By the Pitot Theorem, we have $AB' + DC' = AD + B'C'$ , so $(500-x) + (650-x) = 333+333$ , meaning $x = \\boxed{242}$ .</blockquote>\n\nTo expand on the Pitot theorem, the isosceles trapezoid $AB'C'D$ has an incenter, in that there is a common point of concurrence for all of the angle bisectors of its internal angles. Therefore, a circle can be inscribed in $AB'C'D$ because when the altitudes from the incenter (as the center of the inscribed circle) are dropped to the edges of $AB'C'D$ , you have congruent triangles via SAS congruency. This is a specialized case of the Pitot Theorem of course, since $AB'C'D$ is an isosceles trapezoid, but this is a rigorous proof for why $AB'C'D$ can have a circle inscribed inside of it and because it can have a circle inscribed inside of it you can solve the problem by continuing on in the quoted above solution.\n",
"<blockquote>**30 Second Solution:**\n\nLet $K$ be the projection of $A$ onto $CD$ and $AP$ meet $CD$ at $X$ . Angle chasing yields $DA = DX$ , and symmetry implies $$ DK = \\frac{650 - 500}{2} = 75. $$ Because $P$ is equidistant from $AB$ and $CD$ , symmetry and properties of midlines gives $$ PQ = 500 - KX = 500 - (333 - 75) = \\boxed{242.} $$ </blockquote>\n\nCan anyone elaborate how $PQ$ is equal to $500-KX$ ",
"<blockquote>This took me one hour. I just coordbashed</blockquote>\n\norz\nthats probably the easiest way",
"<blockquote><blockquote>This took me one hour. I just coordbashed</blockquote>\n\norz\nthats probably the easiest way</blockquote>\n\nLOL i failed :(",
"In my mock, I took 4 minutes for this problem, I think it was relatively easy.\n\nDraw height $BP$ , we have $PC=\\frac{650-500}{2}=75$ , now let $\\angle C=x$ , now $\\angle B=180-x$ and this implies that $\\cos x=\\frac{75}{333}=\\frac{25}{111}$ , now using half angle formulas, $\\cos \\frac{x}{2}=\\sqrt{\\frac{68}{111}}$ so $QC=3\\sqrt{68\\cdot111}$ note that $CQ^2=BC\\cdot RC$ if $R$ is the height $QR$ , so $RC=204$ , so $PQ=650-2\\cdot204=\\boxed{242}.$ ",
"Extend line $PQ$ such that is intersects $AD$ at $R$ and $BC$ at $S$ . Since $AB||CD$ , we have $\\angle RPA = \\angle PAB$ and $\\angle DPR = \\angle PDC$ . Now, we know that $AR = DR = RP = \\frac{333}{2}$ because of isosceles triangles, so $RS = 575$ . This also applies to the other side of the trapezoid by symmetry. Hence, $$ PQ = 575 - RP - QS = 575 - AR - BS = 575 - (AR + BS) = 575 - 333 = \\boxed{242}. $$ "
] |
[
"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 angle bisector of $C$ , so the splitting line is also parallel to this bisector, and similar for the splitting line through $N$ . Some simple angle chasing reveals the condition is now equivalent to $\\angle A=120^\\circ$ , so law of cosines gives $b^2+bc+c^2=219^2$ .This becomes $(2b+c)^2+3c^2=438^2$ , and from here we can follow the proof of finding all Pythagorean triples to reveal that the general solution of $3x^2+y^2=z^2$ is $x=2kmn$ , $z=k(m^2+3n^2)$ , and $y=k(3n^2-m^2)$ where $m, n$ are integers, and $2k$ is an integer. From here, it is easy to see that $k=\\frac{1}{2}$ , $n=17$ , and $m=1$ . Then from here we can get $x, y=189, 51$ so the perimeter is $\\boxed{459}$ .\n\nNote: anyone else find the NT harder than the geo? ",
"@above I found the NT part harder, too. Had to bash nearly a hundred cases for it.",
"Same, I only did the geo and couldn't do the NT",
"Could've finished NT using mods and substitution!",
"Can you italicize splitting line? (I forgot to :P)",
"i found the NT answer extraction (unless they intended for this to be the main part of the problem) utterly ridiculous; the problem would have been quite nice otherwise",
"<blockquote>i found the NT answer extraction (unless they intended for this to be the main part of the problem) utterly ridiculous; the problem would have been quite nice otherwise</blockquote>\n\nPersonally I do think the NT is the main part, or at least meant to be a significant part. I can't decide if I like or hate the idea of mixing two completely unrelated subjects in one problem.",
" $3(5-4\\sqrt{-3})^2(1-\\sqrt{-3})=3(-23-40\\sqrt{-3})(1-\\sqrt{-3})=3(-143-17\\sqrt{-3})$ since norm in $\\mathbb Z[\\sqrt{-3}]$ is multiplicative, hence the title of my thread for this problem",
"Let $N$ be the midpoint of arc $ABC$ and consider the Simson line corresponding to $N$ ......\n\nsadly failed the NT",
"Well known that splitting lines pass through the Spieker center, the incenter of the medial triangle. Then trivially $\\angle A=120$ .\n\nThen $b^2+bc+c^2=219^2$ . During contest, I guessed that side lengths were multiples of $3$ so scaled down to $73$ and then used quadratic formula to find a working pair.\n@below typo",
"it was 219...",
" $x^2+xy+y^2=219^2$ thus $438^2-3y^2=a^2$ and $(438+a)(438-a)=3y^2$ substitude $a=3m$ and $y=3n$ to get $(146+m)(146-m)=3n^2$ so we want two numbers summing to 292 to multiply to three times a perfect square\n\nso if the numbers are $c,d$ then $\\left(\\frac{c}{3},d\\right)$ starts at $(97, 1)$ and the first number decreases by 1 as the second increases by 3\n\ntest until you get $81,49$ and done\n\n\nedit: you get $146+m=243,146-m=49,n=63$ so that $a=291$ and $y=189$ which gives $x=\\frac{291-189}{2}=51$ so $(51,189,219)$ edit edit: wait if you start at the other end it gives $(1, 289)$ as the first one which literally removes any need to bash LOL",
"I literally got A = 120 and somehow couldn't finish whatttt",
"why was this the only hard NT problem :(",
"I was thinking of Menelaus' theorem during the test but couldn't get it to work.",
"The extraction is casework-free if you know a little bit of algebraic number theory. However, this is definitely not the intended approach, as the knowledge used is very advanced. The following is a minor modification of what I did in the exam.\n\nSo, we need to solve $a^2+ab+b^2 = 3^2\\cdot 73^2$ . A quick $(\\bmod 9)$ check gives that $3\\mid a$ and $3\\mid b$ . Thus, it's equivalent to solve $x^2+xy+y^2 = 73^2$ .\n\nLet $\\omega$ be one root of $\\omega^2+\\omega+1=0$ . Then, recall that $\\mathbb Z[\\omega]$ is the ring of integers of $\\mathbb Q[\\sqrt{-3}]$ and is a unique factorization domain. Notice that $N(x-y\\omega) = (x-y\\omega)(x-y\\omega^2) = x^2-xy+y^2$ . Therefore, it suffices to find an element of $\\mathbb Z[\\omega]$ with the norm $73^2$ .\n\nTo do so, we factor $73$ in $\\mathbb Z[\\omega]$ . Since it's $1\\pmod 3$ , it must split. A quick inspection gives $73 = (8-\\omega)(8-\\omega^2)$ . Thus, $N(8-\\omega) = 73$ , so\n\\begin{align*}\n73^2 &= N((8-\\omega)^2) \n&= N(64 - 16\\omega + \\omega^2) \n&= N(64 - 16\\omega + (-1-\\omega)) \n&= N(63 - 17\\omega),\n\\end{align*}\ngiving the solution $x=63$ and $y=17$ , yielding $a=189$ and $b=51$ . Since $8-\\omega$ and $8-\\omega^2$ are primes in $\\mathbb Z[\\omega]$ , the solution must divide $73^2$ . One can then easily check that this is the unique solution.**Remark:** You can also factor $3^2$ as well. However, things will become weird since $3$ ramifies in $\\mathbb Z[\\omega]$ .**Remark 2:** The fact that the splitting line through midpoint is parallel to an angle bisector is actually level 8.5 in the game Euclidea. I recognized this instantly in the test.",
"<blockquote>Well known that splitting lines pass through the Spieker center, the incenter of the medial triangle.</blockquote>\n\nin what universe...",
"Solution after we get that $a^2+ab+b^2=219^2=3^2\\cdot 73^2$ (because I didn't get the geo part). \n\nWe can check that $3\\mid a$ and $3\\mid b$ , so let $x=\\frac{a}{3}$ and $y=\\frac{b}{3}$ . So $x^2+xy+y^2=73^2$ .\n\nSo we must find $x$ and $y$ so that $(x-y\\omega)(x-y\\omega^2)=73^2$ , where $\\omega$ is a primitive $3$ rd root of unity. \n\nFor $73$ instead of $73^2$ , we have $(8-\\omega)(8-\\omega^2)=73$ , so we just square now. $(8-\\omega)^2=64-16\\omega+\\omega^2=63-17\\omega$ . \n\nThus, $x=63$ and $y=17$ , so the answer is $189+51+219=\\boxed{459}$ . ",
"<blockquote>The extraction is casework-free if you know a little bit of algebraic number theory. However, this is definitely not the intended approach, as the knowledge used is very advanced. The following is a minor modification of what I did in the exam.\n\nSo, we need to solve $a^2+ab+b^2 = 3^2\\cdot 73^2$ . A quick $(\\bmod 9)$ check gives that $3\\mid a$ and $3\\mid b$ . Thus, it's equivalent to solve $x^2+xy+y^2 = 73^2$ .\n\nLet $\\omega$ be one root of $\\omega^2+\\omega+1=0$ . Then, recall that $\\mathbb Z[\\omega]$ is the ring of integers of $\\mathbb Q[\\sqrt{-3}]$ and is a unique factorization domain. Notice that $N(x-y\\omega) = (x-y\\omega)(x-y\\omega^2) = x^2-xy+y^2$ . Therefore, it suffices to find an element of $\\mathbb Z[\\omega]$ with the norm $73^2$ .\n\nTo do so, we factor $73$ in $\\mathbb Z[\\omega]$ . Since it's $1\\pmod 3$ , it must split. A quick inspection gives $73 = (8-\\omega)(8-\\omega^2)$ . Thus, $N(8-\\omega) = 73$ , so\n\\begin{align*}\n73^2 &= N((8-\\omega)^2) \n&= N(64 - 16\\omega + \\omega^2) \n&= N(64 - 16\\omega + (-1-\\omega)) \n&= N(63 - 17\\omega),\n\\end{align*}\ngiving the solution $x=63$ and $y=17$ , yielding $a=189$ and $b=51$ . Since $8-\\omega$ and $8-\\omega^2$ are primes in $\\mathbb Z[\\omega]$ , the solution must divide $73^2$ . One can then easily check that this is the unique solution.**Remark:** You can also factor $3^2$ as well. However, things will become weird since $3$ ramifies in $\\mathbb Z[\\omega]$ .**Remark 2:** The fact that the splitting line through midpoint is parallel to an angle bisector is actually level 8.5 in the game Euclidea. I recognized this instantly in the test.</blockquote>\n\nnoo i solved up to chapter 7 in euclidea :(",
"oof this is one of the 3 problems I missed, but I still cant really visualize it, can anyone draw a diagram or something??\nthanks!",
"<blockquote>oof this is one of the 3 problems I missed, but I still cant really visualize it, can anyone draw a diagram or something??\nthanks!</blockquote>\n\nOK, diagram not to scale though:\n[asy]\nsize(10cm);\ndraw((6,0)--(0,8)--(-15,0));\ndraw((-32,0)--(18,0));\ndraw((3,4)--(-13,0),RGB(13,83,76));\ndraw((1.5,0)--((-7.5,4)),RGB(13,83,76));\ndraw((18,0)--(0,8)--(-32,0),linewidth(1)+RGB(13,83,76));\nlabel(\" $A$ \",(0,8),N);\nlabel(\" $B$ \",(-15,0),S);\nlabel(\" $C$ \",(6,0),S);\nlabel(\" $M$ \",(-7.5,4),S);\nlabel(\" $N$ \",(3,4),S);\nlabel(\" $X$ \",(-32,0),S);\nlabel(\" $Y$ \",(18,0),S);\n[/asy]",
"<blockquote><blockquote>oof this is one of the 3 problems I missed, but I still cant really visualize it, can anyone draw a diagram or something??\nthanks!</blockquote>\n\nOK, diagram not to scale though:\n[asy]\nsize(10cm);\ndraw((6,0)--(0,8)--(-15,0));\ndraw((-32,0)--(18,0));\ndraw((3,4)--(-13,0),RGB(13,83,76));\ndraw((1.5,0)--((-7.5,4)),RGB(13,83,76));\ndraw((18,0)--(0,8)--(-32,0),linewidth(1)+RGB(13,83,76));\nlabel(\" $A$ \",(0,8),N);\nlabel(\" $B$ \",(-15,0),S);\nlabel(\" $C$ \",(6,0),S);\nlabel(\" $M$ \",(-7.5,4),S);\nlabel(\" $N$ \",(3,4),S);\nlabel(\" $X$ \",(-32,0),S);\nlabel(\" $Y$ \",(18,0),S);\n[/asy]</blockquote>\n\nits kinda funny XY doesn't look straight",
"<blockquote><blockquote>oof this is one of the 3 problems I missed, but I still cant really visualize it, can anyone draw a diagram or something??\nthanks!</blockquote>\n\nOK, diagram not to scale though:\n[asy]\nsize(10cm);\ndraw((6,0)--(0,8)--(-15,0));\ndraw((-32,0)--(18,0));\ndraw((3,4)--(-13,0),RGB(13,83,76));\ndraw((1.5,0)--((-7.5,4)),RGB(13,83,76));\ndraw((18,0)--(0,8)--(-32,0),linewidth(1)+RGB(13,83,76));\nlabel(\" $A$ \",(0,8),N);\nlabel(\" $B$ \",(-15,0),S);\nlabel(\" $C$ \",(6,0),S);\nlabel(\" $M$ \",(-7.5,4),S);\nlabel(\" $N$ \",(3,4),S);\nlabel(\" $X$ \",(-32,0),S);\nlabel(\" $Y$ \",(18,0),S);\n[/asy]</blockquote>\n\nThanks!",
"Req title change to \"Beauty and the Bash\" bc geo :love: :love: :love:",
"<blockquote>Req title change to \"Beauty and the Bash\" bc geo :love: :love: :love:</blockquote>\n\nBeauty and the Bash\n\nFeaturing homotheties and calculations with four-digit squares!\n\nin theaters at your local testing center... a few days ago?",
"<blockquote>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 angle bisector of $C$ , so the splitting line is also parallel to this bisector, and similar for the splitting line through $N$ . Some simple angle chasing reveals the condition is now equivalent to $\\angle A=120^\\circ$ , so law of cosines gives $b^2+bc+c^2=219^2$ .This becomes $(2b+c)^2+3c^2=438^2$ , and from here we can follow the proof of finding all Pythagorean triples to reveal that the general solution of $3x^2+y^2=z^2$ is $x=2kmn$ , $z=k(m^2+3n^2)$ , and $y=k(3n^2-m^2)$ where $m, n$ are integers, and $2k$ is an integer. From here, it is easy to see that $k=\\frac{1}{2}$ , $n=17$ , and $m=1$ . Then from here we can get $x, y=189, 51$ so the perimeter is $\\boxed{459}$ .\n\nNote: anyone else find the NT harder than the geo?</blockquote>\n\nI think you meant $m = 3$ , and $b = 189, c = 51$ not $x$ and $y$ .",
"I think the geo part was harder than the NT part",
"Nobody has posted the solution that is the method that I solved the problem with during the test, so I'll post mine (for the geo part, the NT part is just a big bash).\n\nLet $PM$ and $QN$ be the splitting lines. Reflect $B$ across $Q$ to be $B'$ and $C$ across $P$ to be $C'$ . Take $S_B$ and $S_C$ , which are spiral similarity centers on the other side of $BC$ as $A$ such that $\\triangle S_BB'C \\sim \\triangle S_BBA$ and $\\triangle S_CC'B \\sim \\triangle S_CCA$ . This gets that because $\\angle S_BCB = \\angle S_BCB' = \\angle S_BAB$ and $\\angle S_CBC = \\angle S_CBC' = \\angle S_CAC$ , then $S_B$ and $S_C$ are on $\\triangle ABC$ 's circumcircle. Now, we know that $\\triangle S_BBB' \\sim \\triangle S_BAC$ and $\\triangle S_CCC' \\sim \\triangle S_CAB$ so because $BA=B'C$ and $CA=C'B$ , then $S_BB=SBB'$ and $S_CC=S_CC'$ and $S_BQ \\perp BC$ and $S_CP \\perp BC$ . \n\nWe also notice that because $Q$ and $N$ correspond on $\\triangle S_BBB'$ and $\\triangle S_BAC$ , and because $P$ and $M$ correspond on $\\triangle S_CCC' $ and $\\triangle S_CAB$ , then the angle formed by $NQ$ and $BA$ is equal to the angle formed by $B'C$ and $NQ$ which is equal to $\\angle BS_BQ = \\angle QS_BB'$ . Thus, $\\angle CBA=2\\angle CQN$ . Similarly, $\\angle BCA = 2\\angle QPM$ and so $\\angle CBA + \\angle BCA = 2\\angle PQN + 2\\angle QPM = 60^{\\circ}$ and $\\angle A = 120^{\\circ}$ .",
"<blockquote>Nobody has posted the solution that is the method that I solved the problem with during the test, so I'll post mine (for the geo part, the NT part is just a big bash).\n</blockquote>\n\nThere was a much better solution than a \"big bash\".",
"prolly the hardest aime problem in recent years, 2022 aime ii #14/15 and 2021 aime i #13/15 are not far behind.",
"Another approach for the nt part, find the bound of $m+n$ We skip all those steps mentioned above, only $m^2+n^2+mn=73^2$ We know that $(m+n)^2=73^2+mn, m+n> 73$ , moreover, according to AM-GM, we can attain $(m+n)^2\\leq 73^2+(\\frac{m+n}{2})^2, \\frac{3(m+n)^2}{4}\\leq 73^2, m+n\\leq 84$ \n\n we can try $\\pmod 5$ , it tells that $m^2+n^2+mn\\equiv 4\\pmod 5$ , we can inspect $m\\equiv 2\\pmod 5, n\\equiv {-2}\\pmod 5$ , which means that $m+n$ must be a multiple of $5$ , only two $m+n$ satisfy this requirement are $75,80$ When $m+n=80$ , $x^2+(80-x)^2+x(80-x)=73^2, x_1=17, x_2=63$ , thus the final answer is $219+3(17+63)=\\boxed{459}$ ",
"[Video Solution](https://youtu.be/T6zq1e1RZdg)",
"Problems like these are why we can't have good things in this world",
"Well instead of remembering random Pythagorean triples, if you're impatient like me and ready to dive in, here's an outline of a nice executable in-contest approach (takes ~15-30 min with no mistakes)\n\nDoing some simple geo analysis with angle chase gives $\\angle A = 120$ (easy part), so if $\\overline{AB}=A$ and $\\overline{AC}=B$ , then $A^2+B^2+AB=219^2 \\implies (A+B)^2-AB=219^2$ . Therefore if $x=A+B$ , then $AB=x^2-219^2$ . Let $A$ and $B$ be the roots of $P(t)$ . Then trivially $$ P(t)=t^2-xt+(x^2-219^2) $$ Applying quadratic formula gives $$ t=\\frac{x \\pm \\sqrt{438^2-3x^2}}{2} $$ Clearly $438^2-3x^2 = y^2$ for some integer $y>0$ and $x \\ge 220$ . Therefore using the discriminant $x < \\frac{438}{\\sqrt{3}} \\approx 260$ (a rough approximation suffices)\n\nTaking mod 3 gives $y \\equiv 0 \\mod 3$ , so it is also $0 \\mod 9$ . Then $x \\equiv 0 \\mod 9$ as well. Then resubstituting $x=3x'$ and $y=3y'$ gives $$ 146^2-3(x')^2=(y')^2 $$ where $74 \\le x' \\le 86$ . This is a sufficiently small list to bash out squares, so doing such gives $x'=80$ and $y'=46$ . Thus the perimeter is simply $$ p=219+\\frac{240 \\pm 46}{2} = 219+240=\\boxed{459} $$ ",
"Reflection to find angle $A.$ Then, you use roots of unity. Definitely the hardest on the test (I think)",
"Elementary/minimal bashing approach for the NT part :oops: \n\nWe want $b^2 + bc + c^2 = 219^2$ , modulo inspection yields $3 \\mid b, c$ so substituting $b = 3p, c = 3q$ yields the relation $p^2 + pq + q^2 = 73^2$ . It is easy to show that $p$ and $q$ are both odd, so we use the clever substitution $p = m - n, q = m + n$ to get\n\\[(m - n)^2 + (m - n)(m + n) + (m + n)^2 = 3m^2 + n^2 = 73^2 \\implies (73 - n)(73 + n) = 3m^2. \\]\nBut $\\text{gcd}(73 - n, 73 + n) = \\text{gcd}(73 - n, 146)$ which can only equal $1$ or $2$ . For the former, this implies \n\\[ \\{73 - n, 73 + n \\} = \\{3j^2, k^2 \\} \\implies 3j^2 + k^2 = 146\\] \nfor some integers $j, k$ which by inspection fails. If the latter, note that the $\\nu_2$ of one of $73 - n ,73 + n$ is exactly $1$ , implying one of them can be expressed in the form $2j^2$ or $6j^2$ for odd $j$ . From here, one can easily obtain the solution \n\\[ (73 - n, 73 + n) = (50, 96) \\implies (m, n) = (40, 23) \\implies (p, q) = (17, 63) \\implies (b, c) = (51, 189). \\]\nThe requested answer is $51 + 189 + 219 = \\boxed{459}$ .",
"probably top 3 of my favorite aime problems ",
"solution i had while mocking this test (requires no mods/advanced knowledge):\n\nI'll omit the geometry part, since previous solutions are pretty much the same as mine. After finding that $\\angle A = 120^{\\circ}$ , and letting the two sides other than $219$ be $m$ and $n$ , apply LoC to obtain $219^2=m^2+n^2+mn$ . Rearrange this as $mn=(m+n)^2-219^2$ . Now, make the substitution $a=m+n$ , which also implies $mn=a^2-219^2$ . Solving for $m$ with the quadratic formula reveals that the discriminant is $-3a^2+328^2$ . Let this equal a perfect square, $b^2$ . Rearrange and factor to get $3a^2=(438-b)(438+b)$ Observe that $3|438$ , so make the substitution $b=3k$ , yielding $$ a^2=3(146+k)(146-k). $$ We would like for one of the factors $146+k$ or $146-k$ to be a perfect square. We first let $146-k=12^2$ , however that does not yield a solution. $11^2$ also doesn't work. Now, observe that when $146-k=10^2$ , we have $k=46$ , and $146+k$ becomes $192$ , which is equivalent to $3\\cdot64$ . Thus, $k=46$ satisfies the equation, which makes $a=3\\cdot8\\cdot10=240$ . We extract $a+219=\\boxed{459}$ ."
] |
[
"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"
}
|
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