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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"
}
|
Let $w = \frac{\sqrt{3}+i}{2}$ and $z=\frac{-1+i\sqrt{3}}{2}$ , where $i=\sqrt{-1}$ . Find the number of ordered pairs $(r, s)$ of positive integers not exceeding $100$ that satisfy the equation $i\cdot w^r=z^s$ .
|
<details><summary>Solution</summary>We want $i \cdot w^r = z^s$ . Re-writing $i, w,$ and $z$ in exponential form, we see that $e^{i \cdot \frac{\pi}{2}} \cdot e^{i \cdot \frac{\pi}{6} \cdot r} = e^{i \cdot \frac{2\pi}{3} \cdot s}$ . Simplifying, we have that $4s = r +3$ . From this, we have the pairs $(1, 1), (5, 2), (9, 3), \ldots, (97, 25)$ . BUT, and this is a big BUT, we must notice that any $r \pm 12$ (as you can probably see, this won't change anything in our list of pairs) and any $s \pm 3$ also work. (This is because, in exponential form, $\theta$ and $\theta + 2\pi$ result in the same complex number.) So, we see that our final group of pairs is: $$ (1, 1), (1, 4), (1, 7), \ldots, (1, 100) $$ $$ (5, 2), (5, 5), (5, 8), \ldots, (5, 98) $$ $$ (9, 3), (9, 6), (9, 9), \ldots, (9, 99) $$ $$ (13, 1), (13, 4), (13, 7), \ldots, (13, 100) $$ $$ . $$ $$ . $$ $$ . $$ $$ (97, 1), (97, 4), (97, 7), \ldots, (97, 100). $$ Which is a total of $34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34 + 33 + 33 + 34$ $= 8(34 + 33 + 33) + 34 = 800 + 34 = \boxed{834}$ .</details>
|
[
"Note that $i \\cdot w^{r}$ is periodic for $r \\bmod 12$ and $z^s$ is periodic for $s \\bmod 3$ . ",
"Note that $z = w^4$ and get $4s-r \\equiv 3 \\pmod{12}$ ",
"i got <details><summary>this</summary>$w=e^{\\pi i/6}, z=e^{2\\pi i/3},$ then the equation becomes $ie^{r\\pi i/6}=e^{2s\\pi i/3},$ then $i=e^{\\pi i/2}$ so $e^{r\\pi i/6+\\pi i/2}=e^{2s\\pi i/3},$ so then\n\\begin{align*}\n\\frac{r\\pi i}{6}+\\frac{\\pi i}{2}&=\\frac{2s\\pi i}{3}\nr\\pi i+3\\pi i=4s\\pi i\nr+3=4s\n\\end{align*}\nthen, the lowest $(r, s)$ is $(1, 1)$ and the highest is $(97, 25)$ , so the answer is $25-1+1=\\boxed{025}.$</details>",
"<blockquote>i got <details><summary>025</summary>$w=e^{\\pi i/6}, z=e^{2\\pi i/3},$ then the equation becomes $ie^{r\\pi i/6}=e^{2s\\pi i/3},$ then $i=e^{\\pi i/2}$ so $e^{r\\pi i/6+\\pi i/2}=e^{2s\\pi i/3},$ so then\n\\begin{align*}\n\\frac{r\\pi i}{6}+\\frac{\\pi i}{2}&=\\frac{2s\\pi i}{3}\nr\\pi i+3\\pi i=4s\\pi i\nr+3=4s\n\\end{align*}\nthen, the lowest $(r, s)$ is $(1, 1)$ and the highest is $(97, 25)$ , so the answer is $25-1+1=\\boxed{025}.$</details></blockquote>\nW\nait no but r+3=4s (MOD 12) right?\n",
"there are multiple s for one r",
"<blockquote><blockquote>i got <details><summary>025</summary>$w=e^{\\pi i/6}, z=e^{2\\pi i/3},$ then the equation becomes $ie^{r\\pi i/6}=e^{2s\\pi i/3},$ then $i=e^{\\pi i/2}$ so $e^{r\\pi i/6+\\pi i/2}=e^{2s\\pi i/3},$ so then\n\\begin{align*}\n\\frac{r\\pi i}{6}+\\frac{\\pi i}{2}&=\\frac{2s\\pi i}{3}\nr\\pi i+3\\pi i=4s\\pi i\nr+3=4s\n\\end{align*}\nthen, the lowest $(r, s)$ is $(1, 1)$ and the highest is $(97, 25)$ , so the answer is $25-1+1=\\boxed{025}.$</details></blockquote>\nW\nait no but r+3=4s (MOD 12) right?</blockquote>\n\nyeah, shouldve noticed that",
"<details><summary>Sketch</summary>Rewrite in exponential form using $e^{i\\theta}=\\cos{\\theta}+i\\sin{\\theta}$ . Take cases on $s$ being $0, 1, 2$ mod $3$ . This gives $\\boxed{834}$ , which makes sense as its ~ $\\frac{10000}{12}$ .</details>",
"<blockquote>Note that $z = w^4$ and get $4s-r \\equiv 3 \\pmod{12}$ </blockquote>\n\nYe I got the same thing but how do I keep get $838$ xd",
"I Got 025 bruh",
"Forgot to include s=100 -> 825 :(",
"i got 025 for this",
"i sillied this problem bruh",
"Yo I almost sillied but then caught my mistake to get $\\boxed{834}$ ",
" $z^3=1$ and $w^3=i$ . So $w^4=z$ (I didn't use this in the test, so it took me more time). \n\nCase 1: $s\\equiv 0\\pmod3$ . \nThen $w^r=-i\\implies r\\equiv 9\\pmod{12}$ . This gives a total of $33\\cdot 8=264$ . \n\nCase 2: $s\\equiv 1\\pmod 3$ . \nThen $i\\cdot w^r=z$ , so $r\\equiv 1\\pmod{12}$ . This gives a total of $34\\cdot 9=306$ . \n\nCase 3: $s\\equiv 2\\pmod 3$ . \nThen $i\\cdot w^r=w^8\\implies r\\equiv 5\\pmod{12}$ . This gives a total of $33\\cdot 8=264$ . \n\nAnswer is $264+306+264=\\boxed{834}$ .",
"Did you guys get the answer for this, realize that $264$ was part of this answer, and wonder if the $264$ in the next question was confirming your answer for this question?",
"nah my brain was too dead to notice that <blockquote>Did you guys get the answer for this, realize that $264$ was part of this answer, and wonder if the $264$ in the next question was confirming your answer for this question?</blockquote>\n\n",
"I got $834$ ",
"025 gang",
"exponential form ftw",
"I got 025 then I realized that was wrong, then got 209 :(",
"bruh I got 833\nsad",
"you can do it using Polar coordinates I think, but my brain size was too limited to get the right answer that way.",
"rip I got 025",
"Write $i=e^{i\\frac{\\pi}{2}}$ , it turns to: $\\frac{\\pi}{6}(3+r)=\\frac{4n\\pi}{6}$ , so $3+r=4s+12k$ it follows a pattern that $s=1,r=1,13....97$ has 9 values; $s=2,r=5,17...89$ 8 values and $s=3,r=9,21,...93$ 8 values. \n\nSo the answer is $33*(9+8+8)+9=834$ ",
"<details><summary>Solution</summary>The idea is that w = cos(30) + isin(30) and z = cos(120) + isin(120). Since the magnitude. of both sides is 1, we just need the angle on both sides to be congruent mod 360. The left is 30r + 90 and the right is 120s. Notice that since 120*3 = 0, the right hand side repeats based on mod 3. Now note that 30(r + 12) works if r works, so the right is based on mod 12. From here its just counting as in other solutions in this thread to get 9*34+8*33+8*33 = 834</details>",
"Shoot, I explicitly left out the 100 case because I misread as \"less than 100\"",
"<blockquote>Shoot, I explicitly left out the 100 case because I misread as \"less than 100\"</blockquote>\n\nripp :(\n\nThis was my favorite problem on the test :cool:",
"<blockquote><blockquote>Shoot, I explicitly left out the 100 case because I misread as \"less than 100\"</blockquote>\n\nripp :(\n\nThis was my favorite problem on the test :cool:</blockquote>\n\nI disagree 14 and 15 were better",
"<blockquote><blockquote><blockquote>Shoot, I explicitly left out the 100 case because I misread as \"less than 100\"</blockquote>\n\nripp :(\n\nThis was my favorite problem on the test :cool:</blockquote>\n\nI disagree 14 and 15 were better</blockquote>\n\nI cannot say because I am not pr0 and couldn't solve those problems",
"<blockquote>\nThis was my favorite problem on the test :cool:</blockquote>\n - somebody who hasn't done enough cheap complex number exponent combo problems",
"<blockquote><blockquote>\nThis was my favorite problem on the test :cool:</blockquote>\n - somebody who hasn't done enough cheap complex number exponent combo problems</blockquote>\n\nYeah, it was my first complex number combo problem...so I was pretty fascinated by it...",
"I put on complex plane\nThen 90 degree rotation \nThen casework",
"I got 024 :(",
"I got 834 :( ",
"<blockquote>I got 834 :(</blockquote>\n\n:huh: :huuh:?",
"<blockquote>Note that $i \\cdot w^{r}$ is periodic for $r \\bmod 12$ and $z^s$ is periodic for $s \\bmod 3$ .</blockquote>\n\nYes if you spend a lot of time multiplying you get that.",
"<blockquote><blockquote>I got 834 :(</blockquote>\n\n:huh: :huuh:?</blockquote>\n\nIt was a joke ",
"1. Noted that z=iw,\nthe equation can changed to i*(z/i)^r=z^s,\nthen i*(z^r)*(i^(-r))=z^s,\nz^(r-s)=i^(r-1),\nRe(right)=0 or Im(right)=0, so Re(left)=0 or Im(left)=0.\n2. Noted that z^3=1,\nso Im(left)=0,\n(r-s)=0(mod 3) and (r-1)=0(mod 4).\n3. If r=1, s=1,4,7,…,100 (I missed this case in the exam)\n If r=5, s=2,5,8,…,98\n If r=9, s=3,6,9,…,99\n ……\n ……\n ……\n If r=97, s=1,4,7,…,100\nso the answer is: (34+33+33)*8+34=834.\n(I'm a student from China, so my English is not as well as yours, be sorry about that)",
"[tip=sol]note that w = cos 30 + i sin 30 and z = cos 120 + i sin 120 and then we get 90+30r (mod 360)≡ 120s (mod 360)[/hide]\nedit: notation error",
"Suppose we have a complex number $x=a+bi$ : then $xi=ai-b$ , which is clearly a $90^o$ rotation of $x$ counterclockwise. Going back to the problem since $w=cis(30)$ and $z=cis(120)$ using periodic $12$ and $3$ there are $3$ cases:**Case 1:** $iw=z$ Here there are $9 \\cdot 34$ cases\n*Case 2:* $iw^5 = z^2$ Here there are $8 \\cdot 33$ cases**Case 3:** $iw^9 = z^3$ Here there are again $8 \\cdot 33$ cases\n\nThe answer is $$ 33 \\cdot 16 + 8 \\cdot 34 = \\boxed{834} $$ ",
"If you want to now more accurately about my solution, you can read this: \n(如果你们想更准确地了解我的想法[解法], 你们可以看看下面的步骤):\n1. 注意到 z=iw,\n方程化为i·(z/i)^r=z^s,\n即i·z^r·i^(-r)=z^s,z^(r-s)=i^(r-1),\nRe(右)=0或Im(右)=0, Re(左)=0或Im(左)=0.\n2. 注意到 z^3=1,\n所以Im(左)=0,\n(r-s)=0(mod 3) 且 (r-1)=0(mod 4),\n3. 若r=1, s=1,4,7,…,100 (我考试的时候漏了这种情况)\n 若r=5, s=2,5,8,…,98\n 若r=9, s=3,6,9,…,99\n ……\n ……\n ……\n若r=97, s=1,4,7,…,100\n所以最终结果是: (34+33+33)*8+34=834.",
"@ above\n我的过程和你的完全一样",
"<blockquote>bruh I got 833\nsad</blockquote>\n\nDid you do $\\lfloor 10000/12 \\rfloor$ ?",
"Make it into $e$ . ",
"<blockquote><blockquote><blockquote>I got 834 :(</blockquote>\n\n:huh: :huuh:?</blockquote>\n\nIt was a joke</blockquote>\n\n[s]i laughed",
"<details><summary>Solution</summary>Note that $w=\\mathrm{cis}~30^\\circ$ and $z=\\mathrm{cis}~120^\\circ$ . Consider the equation in polar form. Since $\\vert i\\cdot w^r\\vert=\\vert z^s\\vert=1$ , all we have to do is check when $\\mathrm{arg}(i\\cdot w^r)=\\mathrm{arg}(z^s)$ , or equivalently, \\[90+30r\\equiv120s\\pmod{360}.\\]\n\nTo simplify, divide everything by $30$ including the modulus, to get \\[r\\equiv4s-3\\pmod{12}.\\]\n\nNow, do casework on $s\\mod3$ .\n\nIf $s\\in\\{3,6,\\dots,99\\}, r\\in\\{9,21,\\dots93\\}$ , for a total of $33\\cdot8=264$ possibilities.\n\nIf $s\\in\\{1,4,\\dots,100\\}, r\\in\\{1,13,\\dots,97\\}$ , for a total of $34\\cdot9=306$ possibilities.\n\nIf $s\\in\\{2,5,\\dots,98\\}, r\\in\\{5,17,\\dots89\\}$ , for a total of $33\\cdot8=264$ possibilities.\n\nIn total there are $\\boxed{834}$ ordered pairs.</details>",
"<blockquote><blockquote>Note that $i \\cdot w^{r}$ is periodic for $r \\bmod 12$ and $z^s$ is periodic for $s \\bmod 3$ .</blockquote>\n\nYes if you spend a lot of time multiplying you get that.</blockquote>\n\nIf you rewrite in exponential form you can note that they are roots of unity and thus are periodic every 3 or 12 terms",
" $\\alpha = \\beta \\implies e^{i \\alpha} = e^{i \\beta}$ , not $e^{i \\alpha} = e^{i \\beta} \\implies \\alpha = \\beta$ ",
"By expressing stuff as exponential and we have $r+3=4s$ so we have $25$ solutions.",
"why is this in contests and programs did it get moved?",
"<blockquote>why is this in contests and programs did it get moved?</blockquote>\n\naime problems and stuff like that are in C&P",
"Write both sides of equation in the form a(cos(x) + isin(x))\nSetting real and imaginary parts equal, and solving both equations by using shift by pi/2, and sum/difference to product, we get r - 4s = 9 mod 12.",
"Considering the angles one can obtain the relation $4s - r \\equiv 3 \\pmod{12}$ . Observe that $r = 4t + 1$ for some integer $t$ , so it reduces to $4s - 4t - 1 \\equiv 3 \\pmod{12} \\implies s - t \\equiv 1 \\pmod 3.$ Recall that $1 \\le r, s \\le 100$ , so $0 \\le t \\le 24$ . Now the rest is simple cases:**Case 1:** $s \\equiv 0 \\pmod 3, t \\equiv 2 \\pmod 3$ There are $33$ such $s$ , and $8$ such $t$ yielding $33 \\cdot 8 = 264$ .**Case 2:** $s \\equiv 1 \\pmod 3, t \\equiv 0 \\pmod 3$ There are $34$ such $s$ , and $9$ such $t$ yielding $34 \\cdot 9 = 306$ .**Case 3:** $s \\equiv 2 \\pmod 3, t \\equiv 1 \\pmod 3$ There are $33$ such $s$ , and $8$ such $t$ yielding $33 \\cdot 8 = 264$ .\n\nOur total is $264 + 306 + 264 = \\boxed{834}$ .\n",
"Note that $z = w^4$ so that we require,\n\\begin{align*}\ni \\cdot w^r = w^{4s}\n\\end{align*}\nwhich reduces neatly to $$ i = w^{4s - r} $$ Thus it is necessary that $$ 4s - r \\equiv 3 \\pmod{12} $$ from which we may answer extract to find $\\boxed{834}$ .",
"[Video Solution](https://youtu.be/tCpUJ7_j0U0?si=-Pn5Q2dAd0JBxwhy)",
"For some reason, i got 832, really confused"
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 57,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777219.json"
}
|
A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ .
|
I think this would have been the intended solution:
[hide = Legit solution that never uses the word "vector"]
Define $m$ as the number of minutes they swam for.
Let their meeting point be $A$ . In an alternative reality, there would be no current. Then, had they swum facing the same direction that they had in the real universe, they would've met at a point west of $A$ . Precisely, since the water moves at $14$ meters per minute, this alternative reality meeting point would have been $14m$ meters to the left of $A$ .
So, our alternative reality is just a geometry problem now:
[asy]
unitsize(0.02cm);
draw((0,0)--(0,264)--(550,264)--(550,0)--cycle);
pair B = (198,264);
dot(B);
draw((0,0)--B,dashed);
draw((550,0)--B,dashed);
label(" $60m$ ", (0,0)--B, E);
label(" $80m$ ", (550,0)--B, W);
label(" $264$ ", (0,0)--(0,264), W);
label(" $\frac{D}{2} - 14m$ ", (0,264)--B, N);
label(" $\frac{D}{2} + 14m$ ", B--(550,264), N);
label(" $D$ ", (0,0)--(550,0), S);
[/asy]
(I skipped many steps here, but it should be somewhat self-explanatory where all these numbers came from)
Note that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides $60m$ , $80m$ and $D$ is a right triangle yet, so we cannot use that information.
By Pythagorean, we have
\begin{align*}
264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2}
264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}.
\end{align*}
Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that
\begin{align*}264^{2} + 36^{2} m^{2} &= 60m^{2}
264^{2} &= 96 \cdot 24 \cdot m^{2}
11^{2} &= 4 \cdot m^{2}
m &= \frac{11}{2}.
\end{align*}
So $D = 100m = \boxed{550}$ .
[/hide]
~~Please upvote this post, as it took me like forever to do the asymptote diagram.~~
|
[
"when you need ap physics vectors knowledge on the AIME ",
"It's just a trapezoid",
"answer is 550?",
"when you put 275 because you forgot to times 2",
"<blockquote>when you need ap physics vectors knowledge on the AIME</blockquote>\n\nyes wth even is this problem :sadge:",
"Consider the perspective of the water. Then the point will end up at $(-14t, 0)$ .\n\nOr something",
"<blockquote>Consider the perspective of the water. Then the point will end up at $(-14t, 0)$ .\n\nOr something</blockquote>\n\nyea and then the distances are 60t and 80t\nthen you guess 3-4-5 which actually works and done",
"LETS GO I GUESSED IT RIGHT \n",
"I think I tried the equation $60 + 14 \\cos \\theta = 80 - 14 \\cos \\theta$ which yielded an non-integer answer of $220 \\sqrt{6}$ so I gave up and guessed $630$ (by finding different pythagorean triples) lol",
"God this was awful... Do not want to see this problem again!",
"I wasted 90 minutes on this",
":( Why physics",
"guess right triangle = gg",
"did anyone else spend like 25 minutes trying to comprehend what it was asking, not understanding, and guessing something random? ;-;",
"^^^^^^^^",
"There's no way I was the only one who didn't understand this right... if I interpreted this correctly, I would be guaranteed JMO.",
"Is this problem vectors? because i feel like it should be some clever geometric interpretation (the distances tripped me up)",
"<blockquote>Is this problem vectors? because i feel like it should be some clever geometric interpretation (the distances tripped me up)</blockquote>\n\nyea i did that",
"I got a non-integer answer as well and inputted 220. Why doesn't this work?",
"<blockquote>Is this problem vectors? because i feel like it should be some clever geometric interpretation (the distances tripped me up)</blockquote>\n\nit's a trapezoid\n\ntrippy",
"I got 770...",
"Roughly speaking, there is an east-pointing velocity component of $14$ for both people, and they also have equivalent north-pointing velocity components because they got across the river at the same time. We can use the Pythagorean Theorem to find the other component of their velocity to get something like\n\\[14 + \\sqrt{60^2 - v^2} = \\sqrt{80^2 - v^2} - 14\\]\nwhere $v$ is the north-pointing component. Also, there are two cases for Sherry's east-west component's direction but one case has no solution. \n\nSolving gives $v = 48$ . It's a bit hard to solve, but guessing 3-4-5 triangles and testing it did work for me... ( $48 = \\frac{3}{5} \\cdot 80 = \\frac{4}{5} \\cdot 60$ ) The combined horizontal velocity (including the river speed) can be calculated as $50$ by plugging $v$ back in. Hence, $t = \\frac{264}{48} = 5.5$ , so $D = 2 \\cdot 50 \\cdot 5.5 = \\boxed{550}$ .",
"<blockquote>when you need ap physics vectors knowledge on the AIME</blockquote>\n\nnah I did it without that\n\nwas painful tho :(",
"bruh physics on AIME smh\n\nanyways couldn't solved this and just skipped it",
"this is the only problem i can feel proud of myself for getting right lol. All the others were easy (1 and 2) and thats it. I literally got 1 and 2 and this nonsense problem right. Fumbled 6,7,8,9. Just break down vector components and set as equal. At least this gives me some hope for physics olympiad because im probably giving up math competitions for good after this disaster",
"<blockquote>There's no way I was the only one who didn't understand this right... if I interpreted this correctly, I would be guaranteed JMO.</blockquote>\n\nI spent a good 20 minutes wondering how they moved at different speeds yet traveled the same distance at the same time. I also questioned whether the \"relative to the water\" applied past the comma.",
"I got this wrong :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red:",
"did nobody else shift it, notice 3x and 4x, and guess 3-4-5 and it actually worked (no pythag needed)?",
"I created a variable for how many minutes they swam for, then shifted their meeting point back by a distance to compensate for the river flow, which turned it into a geometry problem.",
"I SPENT ONE HOUR ON THIS AND STILL GOT IT WRONG :wallbash_red: ",
"<blockquote>Roughly speaking, there is an east-pointing velocity component of $14$ for both people, and they also have equivalent north-pointing velocity components because they got across the river at the same time. We can use the Pythagorean Theorem to find the other component of their velocity to get something like\n\\[14 + \\sqrt{60^2 - v^2} = \\sqrt{80^2 - v^2} - 14\\]\nwhere $v$ is the north-pointing component. Also, there are two cases for Sherry's east-west component's direction but one case has no solution. \n\nSolving gives $v = 48$ . It's a bit hard to solve, but guessing 3-4-5 triangles and testing it did work for me... ( $48 = \\frac{3}{5} \\cdot 80 = \\frac{4}{5} \\cdot 60$ ) The combined horizontal velocity (including the river speed) can be calculated as $50$ by plugging $v$ back in. Hence, $t = \\frac{264}{48} = 5.5$ , so $D = 2 \\cdot 50 \\cdot 5.5 = \\boxed{550}$ .</blockquote>\ndid something like this and somehow solved this but missed 3 and 7\n\n",
"f=ma extensive prep kicked in\nbasically, you get a 60-80-28 triangle with a median to the 28\nlet x be the angle between the median and 28 side\nthen tan(x)=264/(D/2)",
"This was such a weird problem",
"<blockquote>This was such a weird problem</blockquote>\n\nthe whole test was weird",
"Absolutely trivial what was this problem \nSolved in under a minute and checked 2 times. \nMaybe if I didn’t check the second time I would have caught 1/84",
"I spent like 15 min just trying to understand how the problem worked :( but after than just split into x and y components and bash d=rt and pythag",
"i spent like 30 min on trying to use similar triangles on the trapezoid instead of pythag",
"My solution \nLet m be the number of\nMinutes. Note that 60m, 80m, 100m form a right triangle with altitude 264, to the hypotenuse\nM is 550 done\nLol idk what y’all r doing",
"yeah that's called guessing",
"Yeah that’s what I did\nI ain’t busting out no vector addition to find it was a trapezoid blah blah blah\n\nShould’ve spent more time on 1/288 smh smh rip jmo",
"<blockquote>did anyone else spend like 25 minutes trying to comprehend what it was asking, not understanding, and guessing something random? ;-;</blockquote>\n\n\n\nme",
"this was the worst possible aime imaginable; so many bashes and this stupid question",
"<blockquote>when you put 275 because you forgot to times 2</blockquote>\n\nomg I thought it's just me ._.",
"<blockquote>this was the worst possible aime imaginable; so many bashes and this stupid question</blockquote>\n\nnot really?",
"yay got this",
"Boat blocked from amc turns into river blocked XD\n[size=50]Luckily I got it tho.\n",
"<blockquote>A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ .</blockquote>\n\nThe one time where AP Physics knowledge actually helped me",
"Somebody at MAA must've really hated water",
"<blockquote>Boat blocked from amc turns into river blocked XD\n[size=50]Luckily I got it tho.</blockquote>\n\nevergreen",
"Anyone else use wrong configuration and get $100$ :(",
"tfw big D is too hard",
"Wait can someone explain what this problem means?\n",
"first thing i thought when i saw this problem was that $264^2=69696$ ",
"also idk why but i started using related rates at first and it just got no where probably because the wording was terrible",
"<blockquote>I think this would have been the intended solution:\n[hide = Legit solution that never uses the word \"vector\"]\nDefine $m$ as the number of minutes they swam for.\n\nLet their meeting point be $A$ . In an alternative reality, there would be no current. Then, had they swum facing the same direction that they had in the real universe, they would've met at a point west of $A$ . Precisely, since the water moves at $14$ meters per minute, this alternative reality meeting point would have been $14m$ meters to the left of $A$ .\n\nSo, our alternative reality is just a geometry problem now:\n[asy]\nunitsize(0.02cm);\ndraw((0,0)--(0,264)--(550,264)--(550,0)--cycle);\npair B = (198,264);\ndot(B);\ndraw((0,0)--B,dashed);\ndraw((550,0)--B,dashed);\n\nlabel(\" $60m$ \", (0,0)--B, E);\nlabel(\" $80m$ \", (550,0)--B, W);\nlabel(\" $264$ \", (0,0)--(0,264), W);\nlabel(\" $\\frac{D}{2} - 14m$ \", (0,264)--B, N);\nlabel(\" $\\frac{D}{2} + 14m$ \", B--(550,264), N);\nlabel(\" $D$ \", (0,0)--(550,0), S);\n[/asy]\n(I skipped many steps here, but it should be somewhat self-explanatory where all these numbers came from)\nNote that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides $60m$ , $80m$ and $D$ is a right triangle yet, so we cannot use that information.\n\nBy Pythagorean, we have\n\\begin{align*}\n264^{2} + \\left( \\frac{D}{2} - 14m \\right) ^{2} &= 3600m^{2} \n264^{2} + \\left( \\frac{D}{2} + 14m \\right) ^{2} &= 6400m^{2}.\n\\end{align*}\nSubtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that\n\\begin{align*}264^{2} + 36^{2} m^{2} &= 60m^{2} \n264^{2} &= 96 \\cdot 24 \\cdot m^{2} \n11^{2} &= 4 \\cdot m^{2} \nm &= \\frac{11}{2}.\n\\end{align*}\nSo $D = 100m = \\boxed{550}$ .\n[/hide]\n~~Please upvote this post, as it took me like forever to do the asymptote diagram.~~</blockquote>\n\nBrilliant solution!",
"I read an article about this test paper and it said that #5 is a pure physics contest problem.",
"F=MA and AIME probably mixed up their shortlists",
"<blockquote>F=MA and AIME probably mixed up their shortlists</blockquote>\n\nlol :P",
"<blockquote>F=MA and AIME probably mixed up their shortlists</blockquote>\n\ntbh this problem could have probably been on the f=ma and no one would have questioned it lmaoo",
"<blockquote>Wait can someone explain what this problem means?</blockquote>\n\n",
"<blockquote>Roughly speaking, there is an east-pointing velocity component of $14$ for both people, and they also have equivalent north-pointing velocity components because they got across the river at the same time. We can use the Pythagorean Theorem to find the other component of their velocity to get something like\n\\[14 + \\sqrt{60^2 - v^2} = \\sqrt{80^2 - v^2} - 14\\]\nwhere $v$ is the north-pointing component. Also, there are two cases for Sherry's east-west component's direction but one case has no solution. \n\nSolving gives $v = 48$ . It's a bit hard to solve, but guessing 3-4-5 triangles and testing it did work for me... ( $48 = \\frac{3}{5} \\cdot 80 = \\frac{4}{5} \\cdot 60$ ) The combined horizontal velocity (including the river speed) can be calculated as $50$ by plugging $v$ back in. Hence, $t = \\frac{264}{48} = 5.5$ , so $D = 2 \\cdot 50 \\cdot 5.5 = \\boxed{550}$ .</blockquote>\n\nSolving $$ v^2+(y-14)^2=60^2 $$ $$ v^2+(y+14)^2=80^2 $$ is the same idea but easier to solve.\n\n",
"make variables, for Sherry's x travel rate be x and y travel rate be y, for Melanie's x travel rate be a and y travel rate be b (sorry for confuse variables) Then we note that b is equal to y and x is equal to a - 28. Set another variable for the amount of minutes that the girls need to meet, maybe m, then use D, m, x, y, a, and b and use a system to solve for m, then subsequently D. no vectors! :)",
"Alternatively, Law of Cosines can be used.\n\nLet the magnitude of the resultant velocity for both people be $x$ . Let the (acute) angle between the velocity and the horizontal be $\\theta$ . Note that by Law of Cosines,\n\\begin{align*}\nx^2 + 196 - 28x\\cos\\theta &= 3600 \nx^2 + 196 + 28x\\cos\\theta &= 6400\n\\end{align*}\nSubtracting the two equations, we get $x\\cos\\theta = 50$ . Substituting into either equation gives $x = 2\\sqrt{1201}$ and $\\cos\\theta = \\frac{25}{\\sqrt{1201}}$ . This gives $\\tan\\theta = \\frac{24}{25}$ , so $D = 2\\cdot 264 \\cdot \\frac{25}{24} = \\boxed{550}$ ",
"<blockquote>when you put 275 because you forgot to times 2</blockquote>\n\nyes",
"Consider the reference frame of the river. Here the point that they both reach moves left constantly. Since Melanie swims 20 meters faster she must use this to catch up with how much it moved left. So they both swim 60 meters upward and Melanie also swims 20 meters to the left. They reach the point at 264/60 = 4.4 amperes, and Melanie swam an extra 20*4.4 = 88 meters, which must be the D.",
"bruh literally just put that they swim at 60 and 80 meters per minute....WHY WOULD YOU PUT \"RELATIVE TO WATER\" :( :( :( made me think there is something special going on like they are only swimming with the current or smth...ended up spending like 30 min just getting #5 correct -_-",
"<blockquote>bruh literally just put that they swim at 60 and 80 meters per minute....WHY WOULD YOU PUT \"RELATIVE TO WATER\" :( :( :( made me think there is something special going on like they are only swimming with the current or smth...ended up spending like 30 min just getting #5 correct -_-</blockquote>\n\nwhen the aime is actually hard!!!!!",
"i just assumed like 4 things and got it right lol\n\ni barely can explain how I got my answer",
"<blockquote>I think this would have been the intended solution:\n[hide = Legit solution that never uses the word \"vector\"]\nDefine $m$ as the number of minutes they swam for.\n\nLet their meeting point be $A$ . In an alternative reality, there would be no current. Then, had they swum facing the same direction that they had in the real universe, they would've met at a point west of $A$ . Precisely, since the water moves at $14$ meters per minute, this alternative reality meeting point would have been $14m$ meters to the left of $A$ .\n\nSo, our alternative reality is just a geometry problem now:\n[asy]\nunitsize(0.02cm);\ndraw((0,0)--(0,264)--(550,264)--(550,0)--cycle);\npair B = (198,264);\ndot(B);\ndraw((0,0)--B,dashed);\ndraw((550,0)--B,dashed);\n\nlabel(\" $60m$ \", (0,0)--B, E);\nlabel(\" $80m$ \", (550,0)--B, W);\nlabel(\" $264$ \", (0,0)--(0,264), W);\nlabel(\" $\\frac{D}{2} - 14m$ \", (0,264)--B, N);\nlabel(\" $\\frac{D}{2} + 14m$ \", B--(550,264), N);\nlabel(\" $D$ \", (0,0)--(550,0), S);\n[/asy]\n(I skipped many steps here, but it should be somewhat self-explanatory where all these numbers came from)\nNote that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides $60m$ , $80m$ and $D$ is a right triangle yet, so we cannot use that information.\n\nBy Pythagorean, we have\n\\begin{align*}\n264^{2} + \\left( \\frac{D}{2} - 14m \\right) ^{2} &= 3600m^{2} \n264^{2} + \\left( \\frac{D}{2} + 14m \\right) ^{2} &= 6400m^{2}.\n\\end{align*}\nSubtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that\n\\begin{align*}264^{2} + 36^{2} m^{2} &= 60m^{2} \n264^{2} &= 96 \\cdot 24 \\cdot m^{2} \n11^{2} &= 4 \\cdot m^{2} \nm &= \\frac{11}{2}.\n\\end{align*}\nSo $D = 100m = \\boxed{550}$ .\n[/hide]\n~~Please upvote this post, as it took me like forever to do the asymptote diagram.~~</blockquote>\n\nim pretty sure this is still vectors (just simplified)",
"I skipped this because I didn't understand the wording \n<span style=\"color:#fff\">No one asked but I don't care</span>",
"Looks like if I didn't procrastinate my precalc problem set I would've gotten this...there was literally a problem with a boat and relative speed to the water...this is why you shouldn't procrastinate!!!",
"if i didn't assume right triangle, this was too hard for me.",
"doesn't it have to be a right triangle??? cause the point on the opposite shore is on the perpendicular bisector of the segment between the two people",
"<blockquote>Alternatively, Law of Cosines can be used.\n\nLet the magnitude of the resultant velocity for both people be $x$ . Let the (acute) angle between the velocity and the horizontal be $\\theta$ . Note that by Law of Cosines,\n\\begin{align*}\nx^2 + 196 - 28x\\cos\\theta &= 3600 \nx^2 + 196 + 28x\\cos\\theta &= 6400\n\\end{align*}\nSubtracting the two equations, we get $x\\cos\\theta = 50$ . Substituting into either equation gives $x = 2\\sqrt{1201}$ and $\\cos\\theta = \\frac{25}{\\sqrt{1201}}$ . This gives $\\tan\\theta = \\frac{24}{25}$ , so $D = 2\\cdot 264 \\cdot \\frac{25}{24} = \\boxed{550}$ </blockquote>\n\nMy solution too. Nice physics problem :coolspeak:",
"Hopefully this can be helpful to someone who hasn't taken physics yet.\n\nMy solution that doesn't require any physics math:\n\nNormalize their movements based on the current (a.k.a pretend the current is nonexistent and change the swimming based on that).\nLet $x=\\dfrac{D}{2}.$ Let $\\theta_1$ be the counterclockwise angle Melanie's swimming path makes with the river, and similarly for Sherry.\nThe parametric equation for Melanie is $$ (x+80\\cos(\\theta_1)t+14t, 80\\sin(\\theta_1)t) $$ and the parametric equation for Sherry is $$ (-x+60\\cos(\\theta_2)t+14t, 60\\sin(\\theta_2)t). $$ Since these two have to intersect, set the y-values to be equal. We get $80\\sin(\\theta_1)t=60\\sin(\\theta_2)t.$ From here we guess that $\\theta_1=\\arcsin(\\tfrac{3}{5}), \\theta_2=\\arcsin(\\tfrac{4}{5}).$ We can check this is valid and from here it's trivial geometry.\n\nNote I phrased this very technically but you can follow all the same steps without any knowledge of parametric equations and such, although may basic trigonometry could help.\n\nNote: I am taking the AIME II, I mocked the AIME I for practice. I was able to get a 6/15.",
"May as well post my sol.\nNote that the speeds of Melanie and Sherry wrt river in the x,y direction must be the same. So we can say $v_S=v_x\\hat{i}+v_y\\hat{j}$ and $v_M=-(v_x+28)\\hat{i}+v_y\\hat{j}$ . So\n\\begin{align*}\nv_x^2+v_y^2&=60^2 \n(v_x+28)^2+v_y^2&=80^2.\n\\end{align*}\nSubtracting them gives $28(2v_x+28)=80^2-60^2$ so $v_x=36$ . Recognizing this as a 3-4-5 triple, $v_y=48$ . So we have\n\\begin{align*}\n\\frac{48}{36+14}=\\frac{264}{D/2}\\implies D=550.\n\\end{align*}\nRemark: Never had to solve a quadratic.",
"Why are people saying that you need physics (you could put it under the category of \"projectile motion\", but there isn't even gravity, so you don't need physics) for this? Sure, the intution you get from solving physics problems might help, but not that much. It's mostly math and logic.",
"vectors is needed\nand vectors is basic physics",
"<blockquote>Why are people saying that you need physics (you could put it under the category of \"projectile motion\", but there isn't even gravity, so you don't need physics) for this? Sure, the intution you get from solving physics problems might help, but not that much. It's mostly math and logic.</blockquote>\n\nTrue. But the entire concept of splitting motion into x and y components comes from physics",
"<blockquote>vectors is needed\nand vectors is basic physics</blockquote>\n\nno it isn't\ni solved it with basic pythag\n...\ntook me less time than 2",
"the only way you can get pythag is by putting it in its x & y forms, and that concept comes from physics",
"vectors are in the aops precalc book, right?",
"<blockquote>the only way you can get pythag is by putting it in its x & y forms, and that concept comes from physics</blockquote>\n\n^^^^^^^^^^^\n\nthe whole concept of shifting the finishing point over is vectors",
"me who knows physics:",
"me who doesn't know physics:",
"me who doesn't know physics but somehow still got it right:",
"solvable with prealgebra knowledge, completely misplaced should be a -1 >:( $ $ ",
"<blockquote>solvable with prealgebra knowledge, completely misplaced should be a -1 >:( $ $ </blockquote>\n\n#6, #7, #9, #13 were solvable with prealgebra knowledge so they must be easy",
"<blockquote><blockquote>solvable with prealgebra knowledge, completely misplaced should be a -1 >:( $ $ </blockquote>\n\n#6, #7, #9, #13 were solvable with prealgebra knowledge so they must be easy</blockquote>\n\nah yes\nevery problem was solvable with prealgebra knowledge\n",
"ah yes math is solvable with prealgebra knowledge, just derive everything from prealgebra",
"nah math is solvable with axioms",
"started attempting this problem 283 days ago and just finished it",
"nah math solvable with enough guessing and checking",
"Just a quick note…for people like me who forgot the physics they learned oop (just need to refresh my memory though)\n\nIn physics, if the object is traveling at an angle, it’s final/total speed depends on the horizontal and vertical velocity. $v = \\sqrt{v_x^2 + v_y^2}$ \n\n[asy]\npair A, B, C; \nA = (0,0);\nB= (1,0);\nC=(1,2);\n\ndraw(A--B--C--cycle);\ndraw((1,0)--(1,0.1)--(0.9, 0.1)--(0.9,0)--cycle);\n\nlabel(“ $v_x$ ”, (0.5,0), S);\nlabel(“ $v_y$ ”, (1,1), E);\nlabel(“ $v_f$ “, (0.5, 1), NW);\n[/asy]",
"What am I doing wrong here?\n<details><summary>bad Sol</summary>Let the point at which Sherry starts be A and the point at which Melanie starts be B, and the point which they are swimming to be C. Then we want to find $AB$ . First, notice that $$ r_S t_S=\\overline{AC}=\\overline{BC} = r_M t_M=r_M t_S $$ so $$ r_S=r_M $$ where $r_M, r_S, t_M, t_S$ are Melanie's total rate, Shelly's total rate, Melanie's time, and Shelly's time to get to point $C$ . Now, we find their total rates. To do this, we notice that the rates have a direction and that the rates that we are equating are the x components of their velocities. Thus, we use the cosine of the angle, which is equal to the adjacent over the hypotenuse. Remembering that we also must account for the rate at which the river is flowing, we find that $$ r_S = 60\\frac{D/2}{\\sqrt{(D/2)^2+264^2}}+14 = r_M = 80\\frac{D/2}{\\sqrt{(D/2)^2+264^2}}-14 $$ $$ \\implies 30D+28\\sqrt{(D/2)^2+264^2}=40D $$ $$ \\implies \\frac{D^2}{4}+264^2=\\frac{25D^2}{196} $$ Which is impossible.</details>",
"My solution when I mocked this:\n\nWhen I mocked this, I used pythag on $2$ triangles and got that it is right angle so hypotenuse is $100r$ , i did some more computation and using pyuthag again r=5.5 so we have $\\boxed{550}.$ ",
"If I didn't know physics vector intuition, I probably would have timesunk an insane amount of time into this. \nBasically the diagram is like <details><summary>this</summary>[img ]https://cdn.artofproblemsolving.com/attachments/6/4/e9dfff6e08657e53a346892d6860696734d306.png[/img]</details> with an isosceles triangle and then note the supplementary $\\theta$ angles. Also the sum of the vectors must be the same vertical distance because of the time condition so we have two LOC equations in $x$ and $\\theta$ . Solve it and bash to finish. $\\tan{\\theta}$ ends up being $\\frac{24}{25}$ and $d = \\boxed{550}$ ",
"this problem is actually really nice, especially since im also preparing for f=ma\n\nsol outline: the main idea is to dissolve the velocity vector of Sherry and Melanie into x and y components, w.r.t to the river. now just set up equations",
"I'm grinding physics to prepare for questions like this "
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1036,
"boxed": false,
"end_of_proof": false,
"n_reply": 100,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777223.json"
}
|
Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega$ . Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A$ , $\omega_B$ , and $\omega_C$ meet in six points $-$ two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC$ , and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC$ . The side length of the smaller equilateral triangle can be written as $\sqrt{a}-\sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a+b$ .
|
another solution but too many point names :(
Let $O_A$ be the center of the circle tangent to $AB$ , $AC$ , and $\omega$ (name this circle $\Gamma_A$ , and define $\Gamma_B$ , $\Gamma_C$ similarly). Let $r$ be its radius, and $E$ the point of tangency of $\Gamma_A$ to $AB$ . Let $D$ be the point diametrically opposite $A$ in $\omega$ , $X$ and $Y$ the points of intersection of $\Gamma_A$ with $BC$ where $BX < BY$ .
Then PoP gives $AE^2=AP\cdot AD=36(36-2r)$ , and $BE^2=BX\cdot BY$ . We’ll have to figure out $BX$ and $BY$ . Define another point— let $M$ be the midpoint of $BC$ . $XO_A=r$ and $O_AM=r-9$ , so $XM=\sqrt{r^2-(r-9)^2}=\sqrt{18r-81}$ . Now $BC=18\sqrt3$ , $BX=9\sqrt3-\sqrt{18r-81}$ , and $BY=9\sqrt3+\sqrt{18r-81}$ , hence $BE^2=324-18r$ . $BE+AE=AB=18\sqrt3$ , so solving $\sqrt{36(36-2r)}+\sqrt{324-18r}=18\sqrt3$ gives $r=12$ .
Let $K$ be the intersection of $\Gamma_B$ and $\Gamma_C$ that’s closer to $O$ . We want $OK$ . Let $O_K$ be the center of the circle tangent to $BA$ , $BC$ , and $\omega$ . It’s easy to see that $\angle O_KOK=120$ , so by the Law of Cosines,
\[OO_K^2+OK^2+OO_K\cdot OK=O_KK^2.\]
Note $OO_K=6$ and $O_KK=12$ , so we have $OK^2+6OK-108=0$ , hence $OK=\sqrt{117}-3$ . Thus the desired side length is $\sqrt3(\sqrt{117}-3)=\sqrt{351}-\sqrt{27}$ and the answer is $\boxed{378}$ .
|
[
"got $\\sqrt{351}-\\sqrt{27}$ for 378",
"Use $30-60-90$ triangles to find the smaller radii of $w_{a}, w_{b}, w_{c}$ , $r$ . We obtain $r=\\frac{36-r}{2} \\implies r = 12$ . Let the intersection that yields the smaller equilateral triangles of circles $w_{a}$ and $w_{b}$ be $I_{c}$ . Since the distance from the centers of $w_{a}, w_{b}, w_{c}$ to the center of the big circle is $6$ , the triangle $\\triangle w_{a}w_{b}w_{c}$ has side length $6\\sqrt{3}$ , since its circumradius is $6$ . Now consider triangle $\\triangle I_{c}w_{a}w_{b}$ . Clearly $\\overline{I_{c}w_{a}}=\\overline{I_{c}w_{b}}=12$ , and the base is $6\\sqrt{3}$ . Thus the height is $\\sqrt{12^2-(3\\sqrt{3})^2}=3\\sqrt{13}$ . Thus the distance from the center of the large circle to point $I_{c}$ is $3\\sqrt{13}-3$ . Thus the side length of the triangle would be $(3\\sqrt{13}-3)(\\sqrt{3})=3\\sqrt{39}-3\\sqrt{3} \\implies 9 \\cdot 39 + 9 \\cdot 3 \\implies \\boxed{378}$ .",
"i thought this was p hard for its pos? basically use homothety to incircle to get radius = 12, since O lies on raxis of two circles by symmetry and distance from O to each of the centers is 6 u can find circumradius and multiply by sqrt(3) ",
"A better title would be \"Mixtilinear Incircles on the AIME\". I don't like this geo very much, to say the least. As post #6 states, coordbash makes quick work of this.",
"circumcircle lies on the line segment connecting points of tangency and that simplifies a lot of bash",
"COORDBASHH",
"trivial by egmo",
"mixtillinear incircles :smiling_face_with_3_hearts:",
"Too long and too hard to draw. Didn't even try. ",
"For some reason, I immediately thought of [Lemma 25](https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-02/mr_2_2020_mixtilinear.pdf) after reading the question. Once I calmed down, however, I realized that symmetry and Pythag made this problem very approachable. Taking a homothety yields the radius of the mixtilinear incircles, and calculations follow easily from here on out.",
"Let $O$ be the center of $\\omega$ , $X$ be the intersection of $\\omega_B,\\omega_C$ further from $A,$ and $O_A$ be the centers of $\\omega_A.$ Also define $Y, Z, O_B, O_C$ similarly. It's well-known that the $A$ -mixtilinear inradius $R_A$ is $\\tfrac{r}{\\cos^2\\left(\\frac{\\angle A}{2}\\right)} = \\tfrac{9}{\\cos^2\\left(\\tfrac{\\pi}{6}\\right)} = 12$ (never thought I'd use this in contest), so in particular this means that $OO_B = 18 - R_B = 6 = OO_C.$ Since $\\measuredangle O_BOO_C = \\measuredangle BOC = \\measuredangle 120^\\circ,$ it follows by Law of Cosines (or just 30-60-90) on $\\triangle OO_BO_C$ that $O_BO_C = 6\\sqrt{3}.$ Then Pythag gives that the altitude of $O_BO_CX$ is $\\sqrt{117},$ so $OY = OX = \\text{dist}(X, YZ) - \\text{dist}(O, YZ) = \\sqrt{117} - 3$ and $YZ = \\tfrac{O_BO_C\\cdot OY}{OO_B} = \\tfrac{6\\sqrt{3}(\\sqrt{117} - 3)}{6}$ so the answer is $351 + 27 = \\boxed{378}.$ ",
"Diagram\n[asy]\nunitsize(0.3cm);\ndraw(circle((0,0),18));\npair A = (9 * sqrt(3), -9);\npair B = (-9 * sqrt(3), -9);\npair C = (0,18);\ndraw(A--B--C--cycle);\ndraw(circle((0,-6),12), gray);\ndraw(circle((3*sqrt(3),3),12), gray);\ndraw(circle((-3*sqrt(3),3),12), gray);\n\npair X = (0, 3-sqrt(117));\npair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 );\npair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 );\ndot(X);\ndot(Y);\ndot(Z);\n\ndraw(X--Y--Z--cycle, dashed);\n[/asy]",
"Coordbash galore after finding the radius of $\\omega_A$ ",
"<blockquote>Coordbash galore after finding the radius of $\\omega_A$ </blockquote>\n\nno, pythagorean theorem and done",
"key observation: the 6 tangency points trisect the sides\nthen u can focus only on the triangle formed by midpoints of arcs",
"Did anybody else find this one easier than P3?",
"<blockquote>Did anybody else find this one easier than P3?</blockquote>\n\nYes, definitely. Easy coord geo",
"Bashed:\n\nLet bottom left point as the origin, the radius of each circle is $36/3=12$ , note that three centers for circles are $(9\\sqrt{3},3),(12\\sqrt{3},12),(6\\sqrt{3},12)$ \nIt is not hard to find that one intersection point lies on $\\frac{\\sqrt{3}x}{3}$ , plug it into equation $(x-9\\sqrt{3})^2+(\\frac{\\sqrt{3}x}{3}-3)^2=12^2$ , getting that $x=\\frac{15\\sqrt{3}+3\\sqrt{39}}{2}$ , the length is $2*(\\frac{15\\sqrt{3}+3\\sqrt{39}-18\\sqrt{3}}{2})=3\\sqrt{39}-3\\sqrt{3}$ , leads to the answer $378$ ",
"Too many intersection points for me",
"how would i know what analogously means\ndidn't attempt it :(",
"I got the right answer, but I misread as \"The side length of the smaller equilateral triangle can be written as $\\sqrt{a}-b$ \" instead of \"The side length of the smaller equilateral triangle can be written as $\\sqrt{a}-\\sqrt{b}$ \"\nSo I left it blank.\nARGHHH!!!",
"<details><summary>Solution</summary>Consider just $\\omega_a$ and $\\omega_b$ . Let their centers be X and Y and let the center of $\\omega$ be O. Consider $\\omega_a$ . Draw the tangent radii onto AB and AC, which gives AX = 2r with similar triangles. Hence, 3r = 36 and r =12. This means that OX = OY = 6, and by symmetry the angle between them is 120, so XY = $6\\sqrt{3}$ . Now let the midpoint of XY be M and let the circles intersect at P and Q, with Q closer to O. Its not hard to see that P, O, and Q are collinear(all are equidistant from X and Y so they all lie on the radical axis of these circles. Alternatively you can see that since B is equidistant from X and Y, P and Q lie on the diameter including point B in $\\omega$ and notice that O lies on that diameter as well.). We have that OQ = MQ - OM. Now applying pythagorean on XMQ and XMO gives that OQ = $\\sqrt{117} - \\sqrt{9}$ . Now by symmetry that is the distance from the other closer points as well, and since we know the angle between each of these distances is 120, we have the side length being $\\sqrt{3}(OQ) = \\sqrt{351} - \\sqrt{27}$ for an answer of 378</details>",
"oh jeez I found this super hard",
"BRUHHH I COULD'VE SOLVED THIS WHY DIDN'T I TRY IT :cri:",
"find radius by 30-60-90 then trivial by letting center of circle by 0 then coord bashing for the distance between the intersection point and the center then using that and the fact that its an equilateral to solve.",
"<blockquote><blockquote>Coordbash galore after finding the radius of $\\omega_A$ </blockquote>\n\nno, pythagorean theorem and done</blockquote>\n\nI just used coordbash and it worked :)",
"rip forgot about the -3 and got 3sqrt39 and was confused :(\npretty easy problem\n",
"i got 042 because I left out the factor of 3 in $3(\\sqrt{39}-\\sqrt{3})$ :noo: ",
"<details><summary>prev</summary>Coordbash galore after finding the radius of $\\omega_A$ <details><summary>quote</summary></blockquote></details>\n\nno, pythagorean theorem and done <details><summary>quote</summary></blockquote></details>\n\nI just used coordbash and it worked :) <details><summary>quote</summary></blockquote></details></details>\n\ni used coord bash and also worked... but sillied a factor of 3 out of the answer :noo:",
"<blockquote>I got the right answer, but I misread as \"The side length of the smaller equilateral triangle can be written as $\\sqrt{a}-b$ \" instead of \"The side length of the smaller equilateral triangle can be written as $\\sqrt{a}-\\sqrt{b}$ \"\nSo I left it blank.\nARGHHH!!!</blockquote>\n\nDang it! You could have got another easy point!",
"redacted",
"Sketch (holy crap this took me an hour intest??)\n\nDrop a perpendicular from $O_A$ to $AB$ to get $r=12$ ; then the distance from $O_A$ to the center of the circumcircle is $6$ ; by symmetry this is true for $O_B,O_C$ . Thus $O_AO_B=6\\sqrt{3}$ , and then you can use Pythag to find where the intersection of the two circles is. Find the distance from said intersection to the center of the circumcircle again, multiply by $\\sqrt{3}$ , that's it.",
"This was a high-quality mid AIME geo problem.",
"<blockquote>Diagram\n[asy]\nunitsize(0.3cm);\ndraw(circle((0,0),18));\npair A = (9 * sqrt(3), -9);\npair B = (-9 * sqrt(3), -9);\npair C = (0,18);\ndraw(A--B--C--cycle);\ndraw(circle((0,-6),12), gray);\ndraw(circle((3*sqrt(3),3),12), gray);\ndraw(circle((-3*sqrt(3),3),12), gray);\n\npair X = (0, 3-sqrt(117));\npair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 );\npair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 );\ndot(X);\ndot(Y);\ndot(Z);\n\ndraw(X--Y--Z--cycle, dashed);\n[/asy]</blockquote>\n\nThank u so much my life has been saved",
"Don't be afraid to Sith!...oops synth...\n\ngoveganddomath supporting that coordbash :sunglas:",
"This was the hardest problem on the AIME. They should have given a diagram. It is impossible to draw this correctly. ",
"<blockquote>This was the hardest problem on the AIME.</blockquote>\n\nWith diagram: no\nWithout diagram: still no",
"<blockquote>This was the hardest problem on the AIME. They should have given a diagram. It is impossible to draw this correctly.</blockquote>\n\nIndeed this makes the problem hard, but def not the hardest",
"Compass-ruler construction makes it pretty easy IMO\n\n1. Draw a equilateral triangle and its circumcircle (shouldn't be hard!)\n2. Draw the intersection of the angle bisector of each angle and the circumcircle (these are the tangency points on the circumcircle by symmetry)\n3. Use your compass to find the center of the tangent circles and draw them (this can be done by rough estimation; it shouldn't be too bad because you can eyeball. Also if you want you can coordbash to find the right distances)",
"<blockquote>Compass-ruler construction makes it pretty easy IMO\n\n1. Draw a equilateral triangle and its circumcircle (shouldn't be hard!)\n2. Draw the intersection of the angle bisector of each angle and the circumcircle (these are the tangency points on the circumcircle by symmetry)\n3. Use your compass to find the center of the tangent circles and draw them (this can be done by rough estimation; it shouldn't be too bad because you can eyeball. Also if you want you can coordbash to find the right distances)</blockquote>\n\nYea, the easiest way to solve this is through a coord bash imo",
"<blockquote><blockquote>This was the hardest problem on the AIME.</blockquote>\n\nWith diagram: no\nWithout diagram: still no</blockquote>\n\nI attempted to draw this by hand, no compass no ruler all by hand! lol ",
"tbh a “very accurate” diagram for this problem was not necessary\n\nit’s killed by noting midpoints of arcs (mixtillinear incircle theorem) and trisection points which both are not necessary for “very accurate diagrams”",
"I mean, all problems involving just equilateral triangles and circles make for an easy coord bash",
"I drew it by hand no problem. It was straightforward to see that the ratio of the mixtilinear circle to the incircle was 4:3. Then you only needed to draw 2 of the mixtilinear circles to get 1 of the 3 necessary points. The other 2 points are found by 120 degree rotations about the center of the diagram.\n\nMaybe the harder part is drawing only one third of the diagram and then visualizing the other 2/3 analogously.",
"<blockquote>key observation: the 6 tangency points trisect the sides\nthen u can focus only on the triangle formed by midpoints of arcs</blockquote>\n\nhello, how do i solve the problem after knowing 6 tangency points trisect the sides? also how to prove 6 tangency points trisect the side? ",
"<blockquote><blockquote>key observation: the 6 tangency points trisect the sides\nthen u can focus only on the triangle formed by midpoints of arcs</blockquote>\n\nhello, how do i solve the problem after knowing 6 tangency points trisect the sides? also how to prove 6 tangency points trisect the side?</blockquote>\n\neh, I'd argue key observation is that you can express side length of an equilateral triangle with its circumradius",
"<blockquote><blockquote><blockquote>key observation: the 6 tangency points trisect the sides\nthen u can focus only on the triangle formed by midpoints of arcs</blockquote>\n\nhello, how do i solve the problem after knowing 6 tangency points trisect the sides? also how to prove 6 tangency points trisect the side?</blockquote>\n\neh, I'd argue key observation is that you can express side length of an equilateral triangle with its circumradius</blockquote>\n\nhello, i am still a little confused ? can you please explain how express side length of equilateral triangle with its circumradius helps? thank you.\n\noh no why is there a 6 post limit",
"Once you find the radius of the mixtillinear circles as 12 (using 30-60-90 triangles), you know that the circumradius that connects all the centers of the circles has length 6 (because its 18-12 to the center of the triangle). You can use pythagorean theorem after this to find the circumradius of the equilateral triangle from the intersection points. Though drawing a diagram isn't necessary in-test, I'd recommend trying to draw one here and seeing if you understand. (Check out post 14 for the diagram). My solution in post 3 should be reasonable to understand with the diagram. ",
"<blockquote>Once you find the radius of the mixtillinear circles as 12 (using 30-60-90 triangles), you know that the circumradius that connects all the centers of the circles has length 6 (because its 18-12 to the center of the triangle). You can use pythagorean theorem after this to find the circumradius of the equilateral triangle from the intersection points. Though drawing a diagram isn't necessary in-test, I'd recommend trying to draw one here and seeing if you understand. (Check out post 14 for the diagram). My solution in post 3 should be reasonable to understand with the diagram.</blockquote>\n\nI think they're asking how to solve it v4913's way given the 6 tangency points trisect the sides..",
"Help why did it take me so long to figure this out\n\n<details><summary>Outline of Solution</summary>1) By symmetry, all three small circles are of the same radii. Hence, focus on one circle and drop altitudes from the center to each of the sides. You should find that by **Vivani's Theorem,** $r+r+r-9 = 27$ , which means that $r=12$ . \n\n2) Make an equilateral triangle between the centers of the triangles (they will be inside the big equilateral triangle). Make a right triangle between the midpoint of the newly formed equilateral triangle, a desired intersection of two of the circles (say $X$ ), and a center. If we extend the longer leg of the right triangle to the side to intersect the midpoint at point $M$ , then let $XM=p$ and use the Pythagorean Theorem to find that $p=12-3\\sqrt{13}$ .\n\n3) The circumcenter of our desired equilateral triangle is thus $9-p = 3\\sqrt{13}-3$ and our side length is $(3\\sqrt{13}-3) \\cdot \\sqrt{3} = 3\\sqrt{39}-3\\sqrt{3} = \\sqrt{351}-\\sqrt{27}$ , and hence $m+n = \\boxed{378}$ .</details>",
"Let $XYZ$ be the smaller triangle ; $O,O_A,O_B,O_C$ be centers of $\\omega,\\omega_A,\\omega_B,\\omega_C$ , respectively. Observe $OO_B = 6$ and $O_BX = 12$ . Let $OX = k$ . Cosine rule in $\\triangle OO_BX$ gives \n[asy]\nsize(200);\npair A=dir(90),B=dir(210),C=dir(330),O=(0,0),OA=4/3*O-1/3*A,OB=4/3*O-1/3*B,OC=4/3*O-1/3*C;\npath wA=circle(OA,2/3),wB=circle(OB,2/3),wC=circle(OC,2/3);\npair X=IP(wB,O-- -A),Y=IP(wA,O-- -B),Z=IP(wA,O-- -C);\nfill(O--OB--X--O--cycle,cyan+white+white);\ndraw(unitcircle);\ndraw(wA^^wB^^wC,blue);\ndot(\" $A$ \",A,dir(A));\ndot(\" $B$ \",B,dir(B));\ndot(\" $C$ \",C,dir(C));\ndot(\" $O$ \",O,dir(180));\ndot(\" $O_A$ \",OA,dir(130));\ndot(\" $O_B$ \",OB,dir(-60));\ndot(\" $O_C$ \",OC,dir(40));\ndot(\" $X$ \",X,dir(5));\ndot(\" $Y$ \",Y,dir(115));\ndot(\" $Z$ \",Z,dir(240));\ndot(2*foot(X,OB,OC)-X);\ndot(2*foot(Y,OA,OC)-Y);\ndot(2*foot(Z,OA,OB)-Z);\ndraw(A--B--C--A-- -A^^B-- -B^^ C-- -C, red);\ndraw(X--OB,brown);\nlabel(\" $18$ \",1/2*(B+O)+1/15*dir(120),green);\nlabel(\" $6$ \",1/2*(O+OB)+1/18*dir(120),green);\nlabel(\" $12$ \",1/2*(X+OB)+1/12*dir(-60),green);\n[/asy]\n\\begin{align*}\nk^2 + 6^2 + 6k = 12^2 &\\implies k^2 + 6k = 108 \\implies (k+3)^2 = 117 \\implies k+3 = \\pm \\sqrt{117} \\implies k = \\sqrt{117} - 3 \n&\\therefore ~ \\text{side length of } \\triangle XYZ = k \\sqrt{3} = (\\sqrt{117} - 3) \\sqrt{3} = \\boxed{\\sqrt{351} - \\sqrt{27}}\n\\end{align*}",
"<blockquote>Let $XYZ$ be the smaller triangle ; $O,O_A,O_B,O_C$ be centers of $\\omega,\\omega_A,\\omega_B,\\omega_C$ , respectively. Observe $OO_B = 6$ and $O_BX = 12$ . Let $OX = k$ . Cosine rule in $\\triangle OO_BX$ gives \n[asy]\nsize(200);\npair A=dir(90),B=dir(210),C=dir(330),O=(0,0),OA=4/3*O-1/3*A,OB=4/3*O-1/3*B,OC=4/3*O-1/3*C;\npath wA=circle(OA,2/3),wB=circle(OB,2/3),wC=circle(OC,2/3);\npair X=IP(wB,O-- -A),Y=IP(wA,O-- -B),Z=IP(wA,O-- -C);\nfill(O--OB--X--O--cycle,cyan+white+white);\ndraw(unitcircle);\ndraw(wA^^wB^^wC,blue);\ndot(\" $A$ \",A,dir(A));\ndot(\" $B$ \",B,dir(B));\ndot(\" $C$ \",C,dir(C));\ndot(\" $O$ \",O,dir(180));\ndot(\" $O_A$ \",OA,dir(130));\ndot(\" $O_B$ \",OB,dir(-60));\ndot(\" $O_C$ \",OC,dir(40));\ndot(\" $X$ \",X,dir(5));\ndot(\" $Y$ \",Y,dir(115));\ndot(\" $Z$ \",Z,dir(240));\ndot(2*foot(X,OB,OC)-X);\ndot(2*foot(Y,OA,OC)-Y);\ndot(2*foot(Z,OA,OB)-Z);\ndraw(A--B--C--A-- -A^^B-- -B^^ C-- -C, red);\ndraw(X--OB,brown);\nlabel(\" $18$ \",1/2*(B+O)+1/15*dir(120),green);\nlabel(\" $6$ \",1/2*(O+OB)+1/18*dir(120),green);\nlabel(\" $12$ \",1/2*(X+OB)+1/12*dir(-60),green);\n[/asy]\n\\begin{align*}\nk^2 + 6^2 + 6k = 12^2 &\\implies k^2 + 6k = 108 \\implies (k+3)^2 = 117 \\implies k+3 = \\pm \\sqrt{117} \\implies k = \\sqrt{117} - 3 \n&\\therefore ~ \\text{side length of } \\triangle XYZ = k \\sqrt{3} = (\\sqrt{117} - 3) \\sqrt{3} = \\boxed{\\sqrt{351} - \\sqrt{27}}\n\\end{align*}</blockquote>\n\nDANG ",
"why was it so trivial \n\ndid it like @2above did ",
"This can be solved with a knowledge of basic angles dilation 30-60-90 triangles and Pythagoras the only hard part is drawing it",
"Coord bashing and homothety are overkill, just draw in some radii and use the definition of tangency",
"<blockquote>why was it so trivial \n\ndid it like @2above did</blockquote>\n\nSame! My solution just needed LoC like once and then I was done ",
"I just drew a bunch of lines and inferred where the 30-60-90 triangles were \ntrying to rigorously solve would be more difficult\n",
"[Video Solution](https://youtu.be/q6_LslAfFpI)",
"coordbash works fine for this problem but it took me ~30 minutes",
"<blockquote>took me so long it's embarrassing. don't even ask why i used loc to find the dist between the centers (beats me as well)</blockquote>\n\n:omighty: :omighty: :omighty: :omighty: :omighty: ",
"Yo I kinda guessed that the two tangent points trisect the side and this trivializes the problem a lot. How do you explicitly prove it tho?",
"In @#63, side $AB$ is split into two segments of length $12\\sqrt{3}$ and $6\\sqrt{3}$ by the first tangency point. By symmetry, the second tangency point splits it into segments of length $6\\sqrt{3}$ and $12\\sqrt{3}$ (in the other direction), and so the two tangency points together split the segment into lengths $6\\sqrt{3}$ , $6\\sqrt{3}$ , and $6\\sqrt{3}$ , the desired trisection.",
"coordbash 'cause synthetic's impossible\n[asy]\nsize(7cm);\npen greenfill,greendraw,turquoisedraw,lightbluedraw,bluedraw,purpledraw,lightpurpledraw,pinkdraw,greydraw,fadedpink,slateblue,royalblue;\ngreenfill = RGB(204,255,204);\ngreendraw = RGB(0,187,0);\nturquoisedraw = RGB(43, 198, 207);\nlightbluedraw = RGB(193, 223, 247);\nbluedraw = RGB(82, 79, 255);\npurpledraw = RGB(147, 65, 250);\nlightpurpledraw=RGB(231, 212, 255);\npinkdraw = RGB(255,17,255);\ngreydraw = RGB(196, 201, 204);\nfadedpink = RGB(176, 109, 169);\nslateblue = RGB(131, 156, 199);\nroyalblue = RGB(82, 116, 250);\n\npair A = (0, 18);\npair B = (-9*sqrt(3), -9);\npair C = (9*sqrt(3), -9);\npair D = (0, 3-sqrt(117));\npair E = (sqrt(3)*(sqrt(117)-3)/2, (sqrt(117)-3)/2);\npair F = (-sqrt(3)*(sqrt(117)-3)/2, (sqrt(117)-3)/2);\npair X = (0, 3+sqrt(117));\npair Y = (-sqrt(3)*(3+sqrt(117))/2,-(3+sqrt(117))/2);\npair Z = (sqrt(3)*(3+sqrt(117))/2,-(3+sqrt(117))/2);\ndraw(circumcircle(A, B, C),greydraw);\nfilldraw(A--B--C--cycle, 0.07*fadedpink+white,fadedpink);\ndraw(circle((0, -6), 12), slateblue);\ndraw(circle((3*sqrt(3), 3), 12), slateblue);\ndraw(circle((-3*sqrt(3), 3), 12), slateblue);\nfilldraw(D--E--F--cycle, 0.07*fadedpink+0.1*royalblue+white, royalblue);\ndot(\" $A$ \", A, dir(A));\ndot(\" $B$ \", B, dir(B));\ndot(\" $C$ \", C, dir(C));\ndot(\" \", D, dir(D));\ndot(\" \", E, dir(E));\ndot(\" \", F, dir(F));\n[/asy]**Setup:** Let $A=(0, 18)$ , $B=(-9\\sqrt3, -9)$ , $C=(9\\sqrt3, -9)$ . We'll find the equation for $\\omega_1$ first. $\\omega_1$ is tangent to $(ABC)$ at $(0, -18)$ , and let its center be $K=(0, -p)$ where $p>0$ . The tangency condition translates to \"the distance from $K$ to $(0, -18)$ is equal to the distance from $K$ to $AB$ (symmetrically $AC$ )\".\n\nThe distance from $K$ to $(0, -18)$ is $18-p$ (since clearly $p<18$ ). Note that the equation of line $AB$ is $\\sqrt3x-y+18=0$ , so the distance from $K$ to $AB$ is $\\tfrac{p+18}{2}$ via distance from point to line formula. Equating gives $p=6$ , so $K=(0, -6)$ . This means the radius of $\\omega_1$ is $12$ , and the distance from $K$ to the origin is $6$ .\n\nSo the centers of $\\omega_2$ and $\\omega_3$ are $(-3\\sqrt3, 3)$ and $(3\\sqrt3, 3)$ , respectively, hence their equations are\n\\begin{align*}\n(x+3\\sqrt3)^2 + (y-3)^2 &= 144\n(x-3\\sqrt3)^2 + (y-3)^2 &= 144\n\\end{align*}\nso the downmost vertex of the small triangle's coordinates are $(0, 3-\\sqrt{117})$ . The distance from here to the origin is $\\sqrt{117}-3$ , and so the side of the small triangle is $2\\cdot \\tfrac{\\sqrt3}{2}\\cdot (\\sqrt{117}-3)=\\sqrt{351}-\\sqrt{27}$ , and the answer is $351+27=\\boxed{378}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1102,
"boxed": true,
"end_of_proof": false,
"n_reply": 64,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777225.json"
}
|
Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .
|
<details><summary>Solution</summary>The idea is to take the cross sections that pass through two of $A$ , $B$ , and $C$ , and their corresponding sphere centers, which I'm labelling $X$ , $Y$ , and $Z$ , respectively
Let $r$ be the radii of the congruent circles. let $d_X$ , $d_Y$ , and $d_Z$ be the positive distances from $X$ , $Y$ , and $Z$ , respectively, to the plane.
Note that $d_Y-d_X$ is the length of one of the legs of a right triangle with hypotenuse $XY$ and third side equal to $AB$ . Therefore $d_Y-d_X=\sqrt{(11+13)^2-560}=4$ .
As well, note that $d_X^2+r^2=121$ and $d_Y^2+r^2=169$ , so $d_Y+d_X=\frac{169-121}{4}=12$ , and therefore $d_Y=\frac{12+4}{2}=8$ , $d_X=\frac{12-4}{2}=4$ , and $r=\sqrt{121-16}=\sqrt{105}$ .
Therefore $d_Z=\sqrt{361-105}=16$ , so $AC^2=(11+19)^2-(16-4)^2=\boxed{756}$ .</details>
|
[
"Pretty misplaced imo - just take the 11-13 and 11-19 cross sections separately where the circles are congruent lines and just do 2D Pythag geo\n\nTook like 2 mins on the test",
"DeToasty3... legend\n\nanyways the numbers were very clean (4,8,16) as heights\n\n@below A,B, and C are projections, so they shouldn't be compared to the tangent portions of the spheres.",
"Shouldn't we have $AB > O_{11} O_{13}?$ @bove: Wait oops you're right :(",
"Let the distances from $A, B, C$ to the corresponding centers of the spheres be $a, b, c$ . Then the congruent circles condition becomes $11^2-a^2=13^2-b^2=19^2-c^2$ . The $AB^2=560$ condition becomes $(11+13)^2-(b-a)^2=560$ (this can be seen by taking a 2D cross section perpendicular to the plane), so we get $b-a=4$ (it is clearly positive as $b^2-a^2=13^2-11^2=48$ . This also gives $b+a=12$ , so $b=8, a=4$ . From here we can find $c=16$ , and get $AC^2=(11+19)^2-(c-a)^2=\\boxed{756}$ ",
"idts $ $ ",
"petition to change the name to \"Three big balls\".",
"<blockquote>petition to change the name to \"Three big balls\".</blockquote>\n\nNo.",
"Bruh I solved 90% of this and then gave up",
"Didn't read that they are congruent circles :wallbash_red: ",
"<blockquote>Bruh I solved 90% of this and then gave up</blockquote>\n\nreminds me of 2021 AMC 12A #24 lol",
"<blockquote>Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .</blockquote>\n<details><summary>Solution</summary>Let the common radius of the three circles be $r$ . Considering the distances from the centers to the plane, the given length condition translates as\n\\[(11+13)^2-(\\sqrt{13^2-r^2}-\\sqrt{11^2-r^2})=560\\text{, whence}\\]\n\\[13^2+11^2-2r^2-2\\sqrt{(13^2-r^2)(11^2-r^2)}=16.\\]\nRearranging, dividing by $2$ and squaring yields\n\\[(137-r^2)^2=(13^2-r^2)(11-r^2)\\]\nso $r^2=105$ .\n\nThe requested expression is then\n\\[(11+19)^2-(\\sqrt{19^2-r^2}-\\sqrt{11^2-r^2})^2=900-(16-4)^2=\\boxed{756}.\\]</details>",
"i wasted 15 minutes on this because i can't read and didn't see congruent",
"woops silly",
"I did the whole problem right but I did $30^2 - 12^2 = 18^2$ :wallbash_red: ",
"I guessed this completely randomly and got it right :o :cry:",
"<blockquote>I guessed this completely randomly and got it right :o :cry:</blockquote>\n\nOMG WUT PR0",
"Someone pls give diagram? Mine sucks... ",
"<blockquote>petition to change the name to \"Three big balls\".</blockquote>\n\nLMAO \ni need to unupvote a post and upvote this one",
"i normally despise geometry but this is literally my favorite question on the test\n\nthe 3d -> 2d strat works really well for this question; just draw the cross section containing $A$ , $B$ , and the centers of the $A$ and $B$ spheres. then use pythag and done",
"<blockquote>Pretty misplaced imo - just take the 11-13 and 11-19 cross sections separately where the circles are congruent lines and just do 2D Pythag geo\n\nTook like 2 mins on the test</blockquote>\n\nagree this question was really easy",
"<details><summary>Nice sol without too much algebra</summary>Take the cross section with $A$ , $B$ , and the centers of the spheres with radii $11$ and $13$ (let them be $A'$ and $B'$ , respectively). Let $AA' = h_1$ and $BB' = h_2$ . Also, let the radius be $r$ . Then, by shifting $A'B'$ $h_1$ units down such that $A'$ coincides with $A$ , we obtain $560 + (h_2 - h_1)^2 = 576$ and thus $h_2 - h_1 = 4$ .\n\nUsing the spheres, we also have $h_1^2 + r^2 = 121$ and $h_2^2 + r^2 = 169$ . Thus subtracting, we have $h_2^2 - h_1^2 = 48$ . Dividing by $h_2 - h_1$ , we get $h_2 + h_1 = 12$ , and thus $h_2 = 8$ , $h_2 = 4$ , and $r = \\sqrt{105}$ . \n\nDefine $C'$ and $h_3$ as the same as in the first paragraph.\n\nNow, let's use the cross-section with $A$ , $C$ , $A'$ , and $C'$ . We have $h_3 = \\sqrt{361 - 105} = 16$ , and thus $h_3 - h_1 = 12$ . Finally, shifting $A'C'$ $h_1$ units down, we get $30^2 - 12^2 = 756$ , which is our solution.</details>",
"my diagram for this question somehow had ovals in it. I don't know why, but I still got the correct answer",
"<blockquote><blockquote>petition to change the name to \"Three big balls\".</blockquote>\n\nNo.</blockquote>\n\nlol eagles ratioed you\n\nseems like a pretty fun question",
"not as hard as i thought. $O_A, O_B, O_C$ denote the respective centers of spheres and $a$ , $b$ , and $c$ denote the respective distances from center of circle to center of sphere. \n\nBy the equal radii condition, we have $121-a^2=169-b^2=361-c^2$ . \n\nBy taking perpendicular cross section, we get $(b-a)^2+560=576\\implies b-a=4$ . \n\nBy taking another perpendicular cross section, we get $(c-a)^2+AC^2=900$ . \n\nSo $121-a^2=169-(a+4)^2\\implies a=4$ . $c^2=256\\implies c=16$ . So $12^2+AC^2=900\\implies \\boxed{756}$ . ",
"Anyone thought the spheres lined up like me? : (",
"Not a tough 3D geometry problem. ",
"Done by cross section and two pythagorean theorems. ",
"Draw the radius of the spheres when taking a cross section of $AB$ . Since the cutted circle have same radius, then set up the equation gives $\\sqrt{169-r^2}-\\sqrt{121-r^2}=4$ Solve gives $r^2=105$ . Now the $AC$ section would simply be a pythagorean theorem $$ AC^2=(19+11)^2-(\\sqrt{361-r^2}-\\sqrt{121-r^2})^2=30^2-(16-4)^2=\\boxed{756} $$ ",
"Let $O_1, O_2, O_3$ be the centers of the three spheres, the radius of the three congruent circles be $r$ , and $P, Q$ be the projections of $O_1$ onto $O_2B, O_3C$ respectively. It's clear that $$ r^2 = 11^2 - O_1A^2 = 13^2 - O_2B^2 = 19^2 - O_3C^2 $$ and $$ 560 = AB^2 = O_1P^2 = (11+13)^2 - O_2P^2 $$ $$ = 576 - (O_2B - PB)^2 = 576 - (O_2B - O_1A)^2. $$ Solving the system yields $O_1A = 4$ and $O_2B = 8$ , so $O_3C = 16$ . To finish, we compute $$ AC^2 = O_1Q^2 = (11 + 19)^2 - O_3Q^2 = 900 - (O_3C - QC)^2 $$ $$ = 900 - (O_3C - O_1A)^2 = \\boxed{756.} $$ **Remark:** To help with visualization, it's worth noting that $$ O_1A < O_2B < O_3C $$ must hold.",
"Wow the geo on this test is really nice (p3 p10 p11 were all really fun)",
"I'm just disappointed that the title wasn't changed as per my suggestion :(",
"Hi, can someone please share an image of the two cross sections? Thank you!",
"<blockquote>Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .</blockquote>\n\nSmth nobody's seemed to notice the incircle of the triangle with vertices the centers of the spheres touches the sides at the tangency points of the spheres because of the $s-a, s-b, s-c$ trick. i wonder if i can find a unique solution based on this observation.",
"I wonder too...\n",
"last year i got a 2 on aime with a distribution of #2 and #10 correct",
"<blockquote><blockquote>Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .</blockquote>\n\nSmth nobody's seemed to notice the incircle of the triangle with vertices the centers of the spheres touches the sides at the tangency points of the spheres because of the $s-a, s-b, s-c$ trick. i wonder if i can find a unique solution based on this observation.</blockquote>\n\nGood observation, but the incenter of said triangle doesn't seem to be of much relevance here, especially since we're length-chasing rather than angle-chasing. I would also be interested in such a solution though.",
"Let the distance of the centers of the three spheres to the plane be $d_1, d_2, d_3$ . Then: $$ 11^2-d_1^2 = 13^2 - d_2^2 = 19^2 - d_3^2 $$ But the distance between two centers can be seen as the vertical leg of a trapezoid with two right angles by dropping altitudes, hence $$ (d_1-d_2)^2 + 560 = (11+13)^2 \\implies d_1=4, d_2=8 $$ accompanied by our first set of equations. Now $d_3=16$ and its trivial to use the same method to find the answer of $30^2 - 16^2 = \\boxed{756}$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1050,
"boxed": true,
"end_of_proof": false,
"n_reply": 38,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777231.json"
}
|
Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$ . A circle tangent to sides $\overline{DA}$ , $\overline{AB}$ , and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$ , as shown. Suppose that $AP = 3$ , $PQ = 9$ , and $QC = 16$ . Then the area of $ABCD$ can be expressed in the form $m\sqrt n$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
[asy]
defaultpen(linewidth(0.6)+fontsize(11));
size(8cm);
pair A,B,C,D,P,Q;
A=(0,0);
label(" $A$ ", A, SW);
B=(6,15);
label(" $B$ ", B, NW);
C=(30,15);
label(" $C$ ", C, NE);
D=(24,0);
label(" $D$ ", D, SE);
P=(5.2,2.6);
label(" $P$ ", (5.8,2.6), N);
Q=(18.3,9.1);
label(" $Q$ ", (18.1,9.7), W);
draw(A--B--C--D--cycle);
draw(C--A);
draw(Circle((10.95,7.45), 7.45));
dot(A^^B^^C^^D^^P^^Q);
[/asy]
|
<blockquote>Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$ . A circle tangent to sides $\overline{DA}$ , $\overline{AB}$ , and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$ , as shown. Suppose that $AP = 3$ , $PQ = 9$ , and $QC = 16$ . Then the area of $ABCD$ can be expressed in the form $m\sqrt n$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
[asy]
defaultpen(linewidth(0.6)+fontsize(11));
size(8cm);
pair A,B,C,D,P,Q;
A=(0,0);
label(" $A$ ", A, SW);
B=(6,15);
label(" $B$ ", B, NW);
C=(30,15);
label(" $C$ ", C, NE);
D=(24,0);
label(" $D$ ", D, SE);
P=(5.2,2.6);
label(" $P$ ", (5.8,2.6), N);
Q=(18.3,9.1);
label(" $Q$ ", (18.1,9.7), W);
draw(A--B--C--D--cycle);
draw(C--A);
draw(Circle((10.95,7.45), 7.45));
dot(A^^B^^C^^D^^P^^Q);
[/asy]</blockquote>
<details><summary>Long-winded solution</summary>[asy]
import olympiad;
size(250);
defaultpen(linewidth(0.7)+fontsize(10));
real h = 5, r = 3*sqrt(3), s = 26;
pair A = (0,0), B = (2,5), X = (B.x,0), C = (12,5), D = A+C-B;
real r = inradius(A,B,X);
path P = scale(B.y/(2*r))*incircle(A,B,X);
pair I = B.y/(2r) * incenter(A,B,X);
pair X = foot(I,A,D), Y = foot(I,B,C), Z = foot(I,A,B), R = foot(C,A,D);
draw(X--Y,gray(0.5));
draw(A--B--C--D--A);
draw(P);
pair[] inter = intersectionpoints(P,A--C);
draw(A--C);
label(" $A$ ",A,SW);
label(" $B$ ",B,NW);
label(" $C$ ",C,NE);
label(" $D$ ",D,SE);
label(" $P$ ",inter[1],dir(250));
label(" $Q$ ",inter[0],dir(330));
label(" $X$ ",X,S,gray(0.5));
label(" $Y$ ",Y,N,gray(0.5));
label(" $Z$ ",Z,NW,gray(0.5));
pair O=(X+Y)/2;
draw((O-(0.2,0))--(O+(0.2,0)),gray(0.5));
label(" $O$ ",O,dir(45),gray(0.5));
draw(A--O--Z^^O--B,RGB(13,83,76));
draw(rightanglemark(O,Z,A));
[/asy]
As in djmathman's solution, let $X,Y,Z$ be the touchpoints of $\omega$ on $\overline{AD},\overline{BC},\overline{AB}$ respectively. As mentioned above one may verify by PoP that $AX=6,CY=20$ .**Claim:** $\omega$ has radius $3\sqrt3$ .
*Proof:*By similar triangles $PX/QX=1/2$ and $PY/QY=5/4$ so we may let $PX=a,QX=2a$ and $PY=5b,QY=5b$ for some positive reals $a,b$ .
Because $\overline{AD}\parallel\overline{BC}$ , $\overline{XY}$ is a diameter so $\angle XZY=\angle XQY=90^\circ$ .
Pythagoras implies
\[a^2+(5b)^2=XY^2=(2a)^2+(4b)^2\]
whence $a=b\sqrt3$ .
We may derive a second equation in $a,b$ by noting that angles $PAX$ and $QCB$ are equal and hence have the same cosines; LoC implies
\[\frac{3^2+6^2-a^2}{2\cdot3\cdot6}=\frac{4^2+5^2-b^2}{2\cdot4\cdot5}.\]
Expressing everything in terms of $b$ gives $7b^2/120=9/40\Rightarrow b=3\sqrt3/\sqrt7$ . Hence $a=9/\sqrt7$ and the diameter evaluates as
\[XY=\sqrt{a^2+(5b)^2}=6\sqrt3.\qquad\qquad\square\]
Returning to the original problem, let $O$ be the center of $\omega$ . Observing that $\angle AOB=\angle AOZ+\angle ZOB=\angle XOZ/2+\angle ZOY=\pi/2$ , similar right triangles implies $BY=BZ=OZ^2/AZ=9/2.$ Now we may compute the area as $BC\cdot XY=(20+9/2)(6\sqrt3)=147\sqrt3\Rightarrow\boxed{150}$ .</details>
|
[
"Mine.\n\n<details><summary>Solution</summary>Let $X$ , $Y$ , and $Z$ denote the tangency points of $\\omega$ with $AD$ , $BC$ , and $AB$ , respectively. Furthermore, let $R$ be the foot of the perpendicular from $C$ to $AD$ .\n\n[asy]\n\timport olympiad;\nsize(250);\ndefaultpen(linewidth(0.7)+fontsize(10));\nreal h = 5, r = 3*sqrt(3), s = 26;\npair A = (0,0), B = (2,5), X = (B.x,0), C = (12,5), D = A+C-B;\nreal r = inradius(A,B,X);\npath P = scale(B.y/(2*r))*incircle(A,B,X);\npair I = B.y/(2r) * incenter(A,B,X);\npair X = foot(I,A,D), Y = foot(I,B,C), Z = foot(I,A,B), R = foot(C,A,D);\ndraw(X--Y^^C--R--D^^rightanglemark(C,R,D),gray(0.5));\ndraw(A--B--C--D--A);\ndraw(P);\npair[] inter = intersectionpoints(P,A--C);\ndraw(A--C);\nlabel(\" $A$ \",A,SW);\nlabel(\" $B$ \",B,NW);\nlabel(\" $C$ \",C,NE);\nlabel(\" $D$ \",D,SE);\nlabel(\" $P$ \",inter[1],dir(250));\nlabel(\" $Q$ \",inter[0],dir(330));\nlabel(\" $X$ \",X,S,gray(0.5));\nlabel(\" $Y$ \",Y,N,gray(0.5));\nlabel(\" $Z$ \",Z,NW,gray(0.5));\nlabel(\" $R$ \",R,S,gray(0.5));\n[/asy]\n\nBy Power of a Point, $AX = \\sqrt{AP\\cdot AQ} = 6$ and $CY = \\sqrt{CQ\\cdot CP} = 20$ . This means $AR = AX + YC = 26$ , so\n\\[\nXY = CR = \\sqrt{AC^2 - AR^2} = \\sqrt{28^2 - 26^2} = 6\\sqrt 3.\n\\]\nNow let $BY = BZ = t$ ; then $AB = 6 + t$ , and so another application of Pythagorean Theorem yields\n\\[\n6\\sqrt 3 = \\sqrt{(6+t)^2 - (6-t)^2} = 2\\sqrt{6t}.\n\\]\nThus $t = \\tfrac 92$ , so $BC = \\tfrac{49}2$ and the area of $ABCD$ is $\\tfrac{49}2\\cdot 6\\sqrt 3 = 147\\sqrt 3$ . The requested sum is $\\boxed{150}$ .</details>",
"Use two tangent, use power of a point to determine two of those variables, project A to BC and use pythaogreans for the last variable, and then base times height to finish.\n(oops didn't see @above)",
":( I sillied the pop computation for this.",
"Let the tangent of the circle on $\\overline{AB}$ be $H$ . By PoP, $\\overline{BH}=20$ . Let the point on $\\overline{CD}$ tangent to the circle be $G$ . $\\overline{GC}=6$ again by PoP. Extend out $\\overline{AB}$ to $I$ s.t $I$ is directly over $D$ . Then we have right triangle $\\triangle DIB$ . Since we have $\\overline{DB}=28$ and $\\overline{IB}=26$ , $\\overline{DI}=6\\sqrt{3}$ . Then we can use the rule of tangents to find $\\overline{BH}$ . Let $J$ be the point directly below $A$ Looking at the triangle $\\triangle DAJ$ , we have $(6+\\overline{BH})^2=(6\\sqrt{3})^2+(6-\\overline{BH})^2 \\implies BH=\\frac{9}{2}$ . The area of $ABCD$ is then $(20+\\frac{9}{2})(6\\sqrt{3})=147\\sqrt{3} \\implies \\boxed{150}$ .",
"pop and then just use loc on ABC (yk cos ABC = - cos BAD) and cos BAD = 6-x/6+x where x is len of tangent from B",
"When pythag managed to crack this one, we all knew this was DJ",
"so easy i couldn't solve it",
"Made this more complicated than it should've been...\n\n[asy]\nunitsize(0.4cm);\npair A, B, C, D, O, P, Q, W, X, Y, Z, R;\nA = (0,0);\nB = (3/2, 6sqrt(3));\nC = (26, 6sqrt(3));\nD = (49/2, 0);\nO = (6, 3sqrt(3));\nP = intersectionpoints(A--C, circle(O,3sqrt(3)))[0];\nQ = intersectionpoints(A--C, circle(O,3sqrt(3)))[1];\nW = (6, 0);\nX = (P+Q)/2;\nY = intersectionpoints(circle(O,3sqrt(3)), circle(B, 9/2))[1];\nZ = intersectionpoints(circle(O,3sqrt(3)), circle(B, 9/2))[0];\nR = (26, 0);\ndraw(A--B--C--D--cycle);\ndraw(C--R--D, dashed);\ndraw(circle(O,3sqrt(3)));\ndraw(A--C^^O--P^^O--Q^^O--W^^O--X^^O--Y^^O--Z^^O--A^^O--B^^O--C);\ndraw(rightanglemark(A, W, Z, 20)^^rightanglemark(O, X, Q, 20)^^rightanglemark(O, Y, B, 20)^^rightanglemark(O, Z, C, 20)^^rightanglemark(C, R, D, 20));\ndot(\" $A$ \", A, SW);\ndot(\" $B$ \", B, NW);\ndot(\" $C$ \", C, NE);\ndot(\" $D$ \", D, SW);\ndot(\" $P$ \", P, ESE);\ndot(\" $Q$ \", Q, SE);\ndot(\" $W$ \", W, S);\ndot(\" $X$ \", X, SE);\ndot(\" $Y$ \", Y, NW);\ndot(\" $Z$ \", Z, N);\ndot(\" $R$ \", R, SE);\ndot(O);\nlabel(\" $O$ \", O, dir(55));\n[/asy]\n\nLet the center of the circle be $O$ . Drop perpendiculars from $O$ to $\\overline{AD}$ , $\\overline{AC}, \\overline{AB}$ and $\\overline{BC}$ at $W, X, Y, Z$ respectively. Connect $\\overline{OP}, \\overline{OQ}, \\overline{OA}, \\overline{OB}$ , and $\\overline{OC}$ . Drop a perpendicular from $C$ to the extension of $AD$ with the foot of the perpendicular at $R$ . \n\nLet the radius of $\\odot O$ be $r$ . \n\nSince $OP = OQ$ , it is obvious that $PX = QX = 4.5$ . By the Pythagorean Theorem on $\\triangle OPX$ , $OX^2 = OP^2 - PX^2 = r^2 - 4.5^2$ . Also, $AX = AP + PX = 3 + 4.5 = 7.5$ . By the Pythagorean Theorem on $\\triangle OAX$ , $OA^2 = OX^2 + AX^2 = r^2 - 4.5^2 + 7.5^2 = r^2 + 36$ . By the Pythagorean Theorem on $\\triangle OYA$ , $AY^2 = OA^2 - OY^2 = r^2 + 36 - r^2 = 36$ . Therefore, $AY = \\sqrt{36} = 6$ . Applying a similar procedure on $\\triangle OQX$ , $\\triangle OCX$ , and $\\triangle OZC$ gives that $ZC = 20$ . \n\nIt is easy to prove that $\\triangle OBY \\cong \\triangle OBZ$ , so let $BZ = BY = y$ . It is also well known that $\\angle BOA = 90^\\circ$ if $\\overline{BC} \\parallel \\overline{AD}$ and $\\odot O$ is a circle tangent to $\\overline{BC}, \\overline{AB}$ , and $\\overline {AD}$ . The proof is left to the reader. Since $\\angle BOA = 90^\\circ$ , $\\angle YBO + \\angle YAO = 90^\\circ$ . Since $\\angle YBO + \\angle YOB = 90^\\circ$ , $\\angle YOB = \\angle YAO$ . Since $\\angle BYO = \\angle OYA = 90^\\circ$ , $\\triangle YOB \\sim \\triangle YAO$ . \n\nHence, $\\frac{BY}{YO} = \\frac{YO}{AY} \\implies \\frac{y}{r} = \\frac{r}{6} \\implies y = \\frac{r^2}{6}$ . Thus, $BZ = y = \\frac{r^2}{6}$ , so the base of the parallelogram is $BC = BZ + ZC = \\frac{r^2}{6} + 20$ . Now, we only need to find the height, which is $2r$ , and solve for $r$ . \n\nObserve that $ZCRW$ is a rectangle, so $WR = ZC = 20$ . Thus, $AR = AW + WR = 6 + 20 = 26$ . In right $\\triangle CAR$ , $AC = AP + PQ + QC = 3 + 9 + 16 = 28$ and $CR = ZW = 2r$ . By the Pythagorean Theorem on $\\triangle CAR$ , $AR^2 + CR^2 = AC^2 \\implies 26^2 + (2r)^2 = 28^2 \\implies r = 3\\sqrt{3}$ . Plugging this into the expression for the base gives $BC = \\frac{r^2}{6} + 20 = \\frac{(3\\sqrt{3})^2}{6} + 20 = \\frac{27}{6} + 20 = \\frac{49}{2}$ . \n\nFinally, the area of the parallelogram is equal to $BC \\cdot 2r = \\frac{49}{2} \\cdot 2 \\cdot 3\\sqrt{3} = 147\\sqrt{3}$ , so the desired answer is $147 + 3 = \\boxed{150}$ . \n\n",
"Well this one use pop and set variables. Then bash out $2AB^2+2BC^2=AC^2+BD^2$ . ",
"<blockquote>Well this one use pop and set variables. Then bash out $2AB^2+2BC^2=AC^2+BD^2$ .</blockquote>\n\nThis is also known as the parallelogram law",
"solved it like #2 did: only final five solve",
"<details><summary>solution</summary>Why did I not do this during the test?\n\nDefine $R$ , $S$ , and $T$ to be the points where the circle is tangent to sides $\\overline{DA}$ , $\\overline{AB}$ , and $\\overline{BC}$ , respectively. The Power of a Point Theorem tells us \\[AR=AS=\\sqrt{AP\\cdot AQ}=\\sqrt{3\\cdot 12}=6\\] and similarly \\[CT=\\sqrt{CQ\\cdot CP}=\\sqrt{16\\cdot 25}=20.\\] Now define $H$ to be the point on the extension of $\\overline{CB}$ past $B$ such that $\\overline{AH}\\perp\\overline{BC}$ . Then $AHTR$ is a rectangle, so $HT=AR=6$ , and $CH=CT+HT=26$ . Thus, by the Pythagorean Theorem on $triangle AHC$ , \\[AH^{2}=AC^{2}-HC^{2}=28^{2}-26^{2}=108\\] so $AH=6\\sqrt{3}$ . Define $BT=x=BS$ , so $BH=6-x$ and $AB=6+x$ . Then by the Pythagorean theorem on $\\triangle AHB$ , \\[108+(6-x)^{2}=(6+x)^{2}\\] which solves to $x=4.5$ . Therefore, \\[BC=CT+BT=20+4.5=24.5\\] and the area of $ABCD$ is simply \\[AH\\cdot BC=6\\sqrt{3}\\cdot 24.5=147\\sqrt{3}\\] from which we can extract an answer of $147+3=\\boxed{150}$ .</details>",
"Oops I constructed $AC \\cap XY = R$ and used a bunch of similar triangle relations, including $AXR \\sim CYR$ , instead of considering the right triangle with hypotenuse $AC$ ... what a waste of time.",
"I did it but it took me ~45 minutes and i kinda failed the rest of aime so... yeah",
"I wrote $2\\sqrt{6} t$ instead of $2\\sqrt{6t}$ and got it wrong :wacko: ",
"I probably would have solved this if I was given like 3 more minutes",
"i guessed 144sqrt(3) lol",
"I guessed 165 in the last few seconds. 15 off isn't that bad",
"<blockquote>Mine.\n\n<details><summary>Solution</summary>Let $X$ , $Y$ , and $Z$ denote the tangency points of $\\omega$ with $AD$ , $BC$ , and $AB$ , respectively. Furthermore, let $R$ be the foot of the perpendicular from $C$ to $AD$ .\n\n[asy]\n\timport olympiad;\nsize(250);\ndefaultpen(linewidth(0.7)+fontsize(10));\nreal h = 5, r = 3*sqrt(3), s = 26;\npair A = (0,0), B = (2,5), X = (B.x,0), C = (12,5), D = A+C-B;\nreal r = inradius(A,B,X);\npath P = scale(B.y/(2*r))*incircle(A,B,X);\npair I = B.y/(2r) * incenter(A,B,X);\npair X = foot(I,A,D), Y = foot(I,B,C), Z = foot(I,A,B), R = foot(C,A,D);\ndraw(X--Y^^C--R--D^^rightanglemark(C,R,D),gray(0.5));\ndraw(A--B--C--D--A);\ndraw(P);\npair[] inter = intersectionpoints(P,A--C);\ndraw(A--C);\nlabel(\" $A$ \",A,SW);\nlabel(\" $B$ \",B,NW);\nlabel(\" $C$ \",C,NE);\nlabel(\" $D$ \",D,SE);\nlabel(\" $P$ \",inter[1],dir(250));\nlabel(\" $Q$ \",inter[0],dir(330));\nlabel(\" $X$ \",X,S,gray(0.5));\nlabel(\" $Y$ \",Y,N,gray(0.5));\nlabel(\" $Z$ \",Z,NW,gray(0.5));\nlabel(\" $R$ \",R,S,gray(0.5));\n[/asy]\n\nBy Power of a Point, $AX = \\sqrt{AP\\cdot AQ} = 6$ and $CY = \\sqrt{CQ\\cdot CP} = 20$ . This means $AR = AX + YC = 26$ , so\n\\[\nXY = CR = \\sqrt{AC^2 - AR^2} = \\sqrt{28^2 - 26^2} = 6\\sqrt 3.\n\\]\nNow let $BY = BZ = t$ ; then $AB = 6 + t$ , and so another application of Pythagorean Theorem yields\n\\[\n6\\sqrt 3 = \\sqrt{(6+t)^2 - (6-t)^2} = 2\\sqrt{6t}.\n\\]\nThus $t = \\tfrac 92$ , so $BC = \\tfrac{49}2$ and the area of $ABCD$ is $\\tfrac{49}2\\cdot 6\\sqrt 3 = 147\\sqrt 3$ . The requested sum is $\\boxed{150}$ .</details></blockquote>\n\nThat's exactly how I solved it. Really cool problem!",
"i tried LoC on triangle ABC and got quartic in r, then proceeded to find the fast sol\noop",
"Same as #2\n\nLet the circle touch $\\overline{AB},\\overline{BC},$ and $\\overline{AD}$ at $X,Y,$ and $Z.$ Let $x=BX$ and $y=YZ.$ By PoP, $AX=6$ and $CY=20.$ Let $W$ and $V$ be the feet from $B$ and $C$ to $\\overline{AD}.$ By Pythagoras on $\\triangle BWA$ and $\\triangle ACV,$ $$ (6+x)^2=y^2+(6-x)^2 $$ and $$ 28^2=y^2+26^2. $$ Hence, $y=6\\sqrt{3}$ and $x=9/2$ and $[ABCD]=(x+20)y=147\\sqrt{3}$ or $\\boxed{150}.$ ",
"<blockquote>Mine.\n\n<details><summary>Solution</summary>Let $X$ , $Y$ , and $Z$ denote the tangency points of $\\omega$ with $AD$ , $BC$ , and $AB$ , respectively. Furthermore, let $R$ be the foot of the perpendicular from $C$ to $AD$ .\n\n[asy]\n\timport olympiad;\nsize(250);\ndefaultpen(linewidth(0.7)+fontsize(10));\nreal h = 5, r = 3*sqrt(3), s = 26;\npair A = (0,0), B = (2,5), X = (B.x,0), C = (12,5), D = A+C-B;\nreal r = inradius(A,B,X);\npath P = scale(B.y/(2*r))*incircle(A,B,X);\npair I = B.y/(2r) * incenter(A,B,X);\npair X = foot(I,A,D), Y = foot(I,B,C), Z = foot(I,A,B), R = foot(C,A,D);\ndraw(X--Y^^C--R--D^^rightanglemark(C,R,D),gray(0.5));\ndraw(A--B--C--D--A);\ndraw(P);\npair[] inter = intersectionpoints(P,A--C);\ndraw(A--C);\nlabel(\" $A$ \",A,SW);\nlabel(\" $B$ \",B,NW);\nlabel(\" $C$ \",C,NE);\nlabel(\" $D$ \",D,SE);\nlabel(\" $P$ \",inter[1],dir(250));\nlabel(\" $Q$ \",inter[0],dir(330));\nlabel(\" $X$ \",X,S,gray(0.5));\nlabel(\" $Y$ \",Y,N,gray(0.5));\nlabel(\" $Z$ \",Z,NW,gray(0.5));\nlabel(\" $R$ \",R,S,gray(0.5));\n[/asy]\n\nBy Power of a Point, $AX = \\sqrt{AP\\cdot AQ} = 6$ and $CY = \\sqrt{CQ\\cdot CP} = 20$ . This means $AR = AX + YC = 26$ , so\n\\[\nXY = CR = \\sqrt{AC^2 - AR^2} = \\sqrt{28^2 - 26^2} = 6\\sqrt 3.\n\\]\nNow let $BY = BZ = t$ ; then $AB = 6 + t$ , and so another application of Pythagorean Theorem yields\n\\[\n6\\sqrt 3 = \\sqrt{(6+t)^2 - (6-t)^2} = 2\\sqrt{6t}.\n\\]\nThus $t = \\tfrac 92$ , so $BC = \\tfrac{49}2$ and the area of $ABCD$ is $\\tfrac{49}2\\cdot 6\\sqrt 3 = 147\\sqrt 3$ . The requested sum is $\\boxed{150}$ .</details></blockquote>\n\nMy pre-discussion-typed solution was exactly the same as yours! Thanks dj, you always make really good problems :D",
"looks like mine is bashy(didn't take the test but as a prep for Aime ii)\n\nLet the circle tangent to $BC,AD,AB$ at $P,Q,M$ separately, denote that $\\angle{ABC}=\\angle{D}=\\alpha$ Using POP, it is very clear that $PC=20,AQ=AM=6$ , let $BM=BP=x,QD=14+x$ , using LOC in $\\triangle{ABP}$ , $x^2+(x+6)^2-2x(x+6)\\cos\\alpha=36+PQ^2$ , similarly, use LOC in $\\triangle{DQC}$ , getting that $(14+x)^2+(6+x)^2-2(6+x)(14+x)\\cos\\alpha=400+PQ^2$ . We use the second equation to minus the first equation, getting that $28x+196-(2x+12)*14*\\cos\\alpha=364$ , we can get $\\cos\\alpha=\\frac{2x-12}{2x+12}$ .\n\nNow applying LOC in $\\triangle{ADC}$ , getting $(6+x)^2+400-2(6+x)*20*\\frac{2x-12}{2x+12}=(3+9+16)^2$ , solving this equation to get $x=\\frac{9}{2}$ , then $\\cos\\alpha=-\\frac{1}{7}$ , $\\sin\\alpha=\\frac{4\\sqrt{3}}{7}$ , the area is $\\frac{21}{2}*\\frac{49}{2}*\\frac{4\\sqrt{3}}{7}=147\\sqrt{3}$ leads to $\\boxed{150}$ ",
"this was literally the only geo I did.",
"<blockquote>solved it like #2 did: only final five solve</blockquote>\n\nsame",
"Didn't try this because of its placement. Turns out it was easier than I thought. \n\nLet $X$ be the point of tangency of $AB$ and the circle, $Y$ be the point of tangency of $AD$ and the circle, $Z$ be the point of tangency of $BC$ and the circle, \n\nBy Power of a Point, $AY=\\sqrt{AP\\cdot AQ}=6$ , $AX=6$ , $CZ=\\sqrt{CQ\\cdot PQ}=20$ . \n\nLet $H$ be foot of altitude from $C$ to $AD$ . Then $AH=6+20=26$ . Since $AC=28$ , we have $CH=6\\sqrt{3}$ . \n\nLet $BX=BZ=x$ . Then $AB=CD=x+6$ and $BC=x+20$ . So \\[DH^2=CD^2-CH^2=x^2+12x+36-108=x^2+12x-72.\\]\n\nSo \\[\\sqrt{x^2+12x-72}=6-x\\implies x^2+12x-72=x^2-12x+36\\implies 24x=108\\implies x=\\frac{9}{2}\\]\n\nNow the area is $CH\\cdot BC=6\\sqrt{3}\\cdot \\frac{49}{2}=147\\sqrt{3}\\implies \\boxed{150}$ ",
"<blockquote>solved it like #2 did: only final five solve</blockquote>\n\nhm. i solved all final five *but* this one...",
"HOW DID I GET 88sqrt3\nI USED THE SAME METHOD",
"<blockquote>HOW DID I GET 88sqrt3\nI USED THE SAME METHOD</blockquote>\n\nI got 138rt3 for a moment...",
"I forgot secant tangent pop during the test(thought t^2 = ab instead of t^2 = a(a + b)).\n\n<details><summary>Solution</summary>Let the circle be tangent to AB at x, BC at y, and AD at z. We have AZ^2 = 3(3+9) -> AZ = AX = 6. Similarly, CY^2 = 16(16+9) -> CY = 20. Now let E be the point on line BC such tthat AEB = 90. We have EC = AZ + CY = 26, meaning that AE or the height of the parallelogram is $6\\sqrt{3}$ . Finally, let BX = BY = g. We have that (6 + g)^2 - (6 - g)^2 = 24g = 108, meaning that g = 9/2. Now the area is just BC* $6\\sqrt{3}$ = 49/2* $6\\sqrt{3} = 147\\sqrt{3}$ , giving 150.</details>",
"<details><summary>Similar Triangles Solution</summary>Let the tangency point on $BC$ be $S$ and let the tangency point on $AD$ be $Q$ . First notice $BC = 20, AQ = 6$ from PoP. Note that $PQ$ is an altitude. Let $SQ \\cup AC = R$ , so $\\triangle SRC \\sim \\triangle QRA$ . Let $QR = x$ , so $PR = 9-x$ and hence $\\frac{16+x}{12-x} = \\frac{10}{3} \\implies x = \\frac{72}{13} = QR, PR = \\frac{45}{13}$ . Then we can easily find $BQ = 6\\sqrt3$ . Next, let $BS = a$ . Now drop an altitude from $B$ to $AD$ , which allows us to make the equation $(6-a)^2 + 108 = (6+a)^2 \\implies a = \\frac{9}{2}$ . Therefore, the answer is $6\\sqrt3\\left(20+\\frac{9}{2}\\right) = 147\\sqrt3 \\implies \\boxed{150}$ .</details>",
"<details><summary>last minute coord bash</summary>By power of a point, tangent lengths from $A$ are 6 and tangent lengths from $C$ are 20. Let $O$ be the center of the circle. Then, as $AD || BC,$ the altitude from $O$ to $AD$ is the same as that from $O$ to $BC.$ Let $AD$ be the x axis, $A = (0,0)$ , $C = (26,2r).$ If $B = (a,2r)$ , then $D = (26-a,0)$ . The length of the tangent from $B$ is $BC-20=6-a.$ Then, $AB = 12-a = \\sqrt{a^2+4r^2}$ , so $r^2=36-6a$ . Also, $AC = 28 = \\sqrt{26^2+4r^2}$ , so $r^2=27.$ Solving, $a = \\frac{3}{2}.$ Then the area is $(6\\sqrt{3})(26-\\frac{3}{2}) = 147\\sqrt{3} \\implies \\boxed{150}$ .</details>",
"**Remark:** This problem secretly hides the Iran Lemma!\n\n\n(I'm using the diagram in from Post #2.) Let $M$ and $N$ denote the midpoints of $AB$ and $CD$ respectively, let $O$ be the center of the aforementioned circle, and suppose $T \\ne A$ is a point on line $AD$ such that $A$ isn't between $D$ and $T$ . Because $O$ is the midpoint of $XY$ , we know $M, N, O$ are collinear. \n\nIt's clear that $MN$ coincides with the $B$ -midline of $ABT$ . Now, since $AO$ bisects $\\angle BAT$ , the Iran Lemma yields $\\angle AOB = 90^{\\circ}$ .",
"I did everything right on this problem, didn’t overcomplicate it, and overall was very proud of myself for solving a #11 tho it was a really ez one... then I looked back at my paper after the test and realized I wrote down $\\sqrt{108} = 6\\sqrt{2}$ and my final answer was 149.\n\nThis one hurts, it really does, especially with my already extremely low score.",
"<blockquote>Well this one use pop and set variables. Then bash out $2AB^2+2BC^2=AC^2+BD^2$ .</blockquote>\n\nLol I just realized this formula existed.",
"Yay dj geo!\n\n<details><summary>Solution</summary>Let $M$ and $N$ be the tangency points of the circle on $AD$ and $BC$ respectively. Let $\\theta=\\angle CAD=\\angle BAC$ .\n\nNote that $AM=\\sqrt{AP\\cdot AQ}=6$ and $CN=\\sqrt{CQ\\cdot CP}=20$ .\n\nLet $X$ be the foot of the altitude from $C$ down to $AD$ . Note that $CX=\\sqrt{(3+9+16)^2-(6+20)^2}=6\\sqrt{3}$ . As well, $CX=MN$ , so the radius of the circle is $3\\sqrt{3}$ .\n\nTherefore, $\\tan\\theta=\\frac{3\\sqrt{3}}{6}=\\frac{\\sqrt{3}}{2}$ . This means $\\tan BAD=\\tan 2\\theta=\\frac{2\\tan\\theta}{1-\\tan^2\\theta}=\\frac{\\sqrt{3}}{\\frac{1}{4}}=4\\sqrt{3}$ .\n\nTherefore $BA=\\frac{\\sqrt{1^2+(4\\sqrt{3})^2}}{4\\sqrt{3}}\\cdot6\\sqrt{3}=\\frac{21}{2}$ . This means $BC=20+\\left(\\frac{21}{2}-6\\right)=\\frac{49}{2}$ . The area of $ABCD$ is thus $BC\\cdot MN=147\\sqrt{3}\\rightarrow\\boxed{150}$ .</details>",
"<details><summary>Solution</summary>Let the circle intersect $AB,BC,$ and $AD$ at $X,Y,$ and $Z,$ respectively. By Power of the Point, \n\\begin{align*}\nAX=AZ&=\\sqrt{AP \\cdot AQ}=\\sqrt{3 \\cdot 12}=6, \nCY&=\\sqrt{CQ \\cdot CP}=\\sqrt{16 \\cdot 25}=20.\n\\end{align*}\nLet $O_1$ be the point of the altitude from $C$ to $AD$ and $O_2$ be the point of the altitude from $A$ to $BC.$ Notice that, $$ AO_1=AZ+O_1Z=6+20=26 \\text{ and } CO_2=O_2Y+CY=26. $$ Since $AC=3+9+16=28,$ by Pythagorean Theorem, we get that $$ CO_1=\\sqrt{\\left(AC\\right)^2-\\left(AO_1\\right)^2}=\\sqrt{28^2-26^2}=6\\sqrt{3}. $$ Let $O_1D=O_2B=a,$ so $BY=BX=6-a$ and since $AZ=AX=6,$ we get $$ CD=BX+AX=6-a+6=12-a. $$ By Pythagorean Theorem, $$ CD=\\sqrt{\\left(CO_1\\right)^2+\\left(DO_1\\right)^2}=\\sqrt{108+a^2}. $$ Notice that $$ 12-a=\\sqrt{108-a^2} \\implies 144-24a+a^2=108-a^2 \\implies a=\\frac{3}{2}. $$ Thus, we get that $6-a=\\frac{9}{2},$ so $$ [ABCD]=BC \\cdot CO_1=\\left(\\frac{9}{2}+20\\right) \\cdot 6\\sqrt{3}=147\\sqrt{3}. $$ Thus, $m+n=147+3=\\boxed{150}.$</details>",
"<details><summary>Notations</summary>$O$ is center of circle ; $X,Y,Z$ are tangency points with $\\overline{AD},\\overline{BC},\\overline{AB}$ , respectively ; $T = \\overline{PQ} \\cap \\overline{XY}$ .</details> Note $AX = AX = 6$ and $CY = 20$ as $AP \\cdot AQ = 3 \\cdot 12 = 6^2$ and $CQ \\cdot CP = 16 \\cdot 25 = 20^2$ . Now $\\triangle ATX \\sim \\triangle CTY$ . Thus, $$ \\frac{AT}{CT} = \\frac{AX}{CY} = \\frac{6}{20} = \\frac{3}{10} $$ Using $AC = 28$ gives $$ AT = 28 \\cdot \\frac{3}{13} = \\frac{84}{13} \\implies PT = \\frac{84}{13} - 3 = \\frac{45}{13} ~~~,~~~ CT = 28 \\cdot \\frac{10}{13} = \\frac{280}{13} \\implies QT = \\frac{280}{13} - 16 = \\frac{72}{13} $$ Observe $\\overline{XY}$ is a diameter (as tangents at $X,Y$ are parallel). Let $TX = 3a,TY = 10a$ . Then PoP at $T$ gives \n[asy]\nsize(250);\npath w=circle((0,0),3*sqrt(3));\ndraw(w);\npair O=(0,0),A=sqrt(63)*dir(140),Z=IP(circle(A,6),arc(O,waypoint(w,0),waypoint(w,0.4))),X=2*foot(Z,A,O)-Z,Y=-X,B=2*Y*Z/(Y+Z),C=49/9*Y-40/9*B,D=A+C-B,P=IP(A--2/3*A+1/3*C,w),Q=2*foot(O,A,C)-P,T=extension(A,C,X,Y);\ndot(\" $O$ \",O,dir(60));\ndot(\" $A$ \",A,dir(A));\ndot(\" $Z$ \",Z,dir(Z));\ndot(\" $X$ \",X,dir(X));\ndot(\" $Y$ \",Y,dir(Y));\ndot(\" $Z$ \",Z,dir(Z));\ndot(\" $B$ \",B,dir(B));\ndot(\" $C$ \",C,dir(C));\ndot(\" $D$ \",D,dir(D));\ndot(\" $P$ \",P,dir(-10));\ndot(\" $Q$ \",Q,dir(-90));\ndot(\" $T$ \",T,dir(50));\ndraw(A--B--C--D--A--C);\ndraw(X--Y^^O--Z);\nlabel(\" $3$ \",1/2*(A+P)+1/2*dir(20),fuchsia);\nlabel(\" $9$ \",1/2*(P+Q)+dir(-20),fuchsia);\nlabel(\" $6$ \",1/2*(A+X)+1/2*dir(180),green);\nlabel(\" $\\frac{45}{13}$ \",1/2*(P+T)+3/4*dir(200),orange);\nlabel(\" $\\frac{72}{13}$ \",1/2*(T+Q)+dir(180),orange);\nlabel(\" $3a$ \",1/2*(X+T)+3/5*dir(-90),brown);\nlabel(\" $10a$ \",1/2*(T+Y)+2/3*dir(-90),brown);\nlabel(\" $16$ \",1/2*(Q+C)+dir(0),fuchsia);\nlabel(\" $3 \\sqrt{3}$ \",1/2*(O+Z)+3/2*dir(0),brown);\nlabel(\" $6$ \",1/2*(A+Z)+2/3*dir(90),green);\nlabel(\" $\\frac{9}{2}$ \",1/2*(B+Z)+dir(90),green);\nlabel(\" $\\frac{9}{2}$ \",1/2*(B+Y)+1/2*dir(0),green);\nlabel(\" $20$ \",1/2*(C+Y)+dir(0),green);\n[/asy]\n\\begin{align*}\n30a^2 = TX \\cdot TY &= TP \\cdot TQ = \\frac{45}{13} \\cdot \\frac{72}{13} = 40 \\left( \\frac{9}{13} \\right)^2 \\implies a= \\frac{2}{\\sqrt{3}} \\cdot \\frac{9}{13} = \\frac{6 \\sqrt{3}}{13} \n&\\therefore ~ OZ = \\text{radius} = \\frac{XY}{2} = \\frac{13a}{2} = \\frac{13}{2} \\cdot \\frac{6 \\sqrt{3}}{13} = 3 \\sqrt{3}\n\\end{align*}\nNow since $\\angle AOB = \\frac{1}{2} \\angle XOY = 90^\\circ$ , hence in $\\triangle AOB$ we obtain $$ ZA \\cdot ZB = ZO^2 = (3 \\sqrt{3})^2 = 27 \\implies ZB = 27 \\div 6 = \\frac{9}{2} $$ So now $BZ = BY = \\frac{9}{2}$ . Basically, we know each side of $\\triangle ABC$ : $$ AC = 28 ~,~ AB = 6 + \\frac{9}{2} = \\frac{21}{2} ~,~ CB = \\frac{9}{2} + 20 = \\frac{49}{2} $$ We can now use Heron's formula to find $[ABC]$ . First scale $\\triangle ABC$ by $\\frac{2}{7}$ to obtain $\\triangle A'B'C'$ with side lengths $$ A'C' = 8, A'B' = 3, C'B' = 7 $$ [asy]\nsize(200);\npath w=circle((0,0),3*sqrt(3));\npair O=(0,0),A=sqrt(63)*dir(140),Z=IP(circle(A,6),arc(O,waypoint(w,0),waypoint(w,0.4))),X=2*foot(Z,A,O)-Z,Y=-X,B=2*Y*Z/(Y+Z),C=49/9*Y-40/9*B,D=A+C-B,P=IP(A--2/3*A+1/3*C,w),Q=2*foot(O,A,C)-P,T=extension(A,C,X,Y);\ndot(\" $A'$ \",A,dir(A));\ndot(\" $B'$ \",B,dir(B));\ndot(\" $C'$ \",C,dir(C));\ndraw(A--B--C--A);\nlabel(\" $3$ \",1/2*(A+B)+dir(90),green);\nlabel(\" $7$ \",1/2*(B+C)+dir(0),green);\nlabel(\" $8$ \",1/2*(A+C)+dir(180),fuchsia);\n[/asy]\nObserve $s = \\frac{8 + 3 + 7}{2} = 9$ . Now Heron's formula just gives \n\\begin{align*}\n[A'B'C'] &= \\sqrt{(9)(9-8)(9-3)(9-7)} = \\sqrt{9 \\cdot 1 \\cdot 6 \\cdot 2} = 3 \\sqrt{12} = 6 \\sqrt{3} \n&\\implies [ABC] = \\left( \\frac{7}{2} \\right)^2 \\cdot [A'B'C'] = \\frac{49}{4} \\cdot 6 \\sqrt{3} = \\frac{147}{2} \\sqrt{3} \n& \\therefore ~ [ABCD] = 2 [ ABC] = \\boxed{147 \\sqrt{3}}\n\\end{align*}\n",
"<details><summary>Solution Sketch</summary>To summarize, use Power of a Point (PoP), then wishing that $YC$ could have a parallel and equal length below, drop the perpendicular to $R$ , then use the Pythagorean theorem to get $XY$ , then use the Pythagorean theorem again to compute $AD$ by computing $BY=AD$ with some algebra, then solve using base multiplied by height.</details>",
"<blockquote>**Remark:** This problem secretly hides the Iran Lemma!\n\n\n(I'm using the diagram in from Post #2.) Let $M$ and $N$ denote the midpoints of $AB$ and $CD$ respectively, let $O$ be the center of the aforementioned circle, and suppose $T \\ne A$ is a point on line $AD$ such that $A$ isn't between $D$ and $T$ . Because $O$ is the midpoint of $XY$ , we know $M, N, O$ are collinear. \n\nIt's clear that $MN$ coincides with the $B$ -midline of $ABT$ . Now, since $AO$ bisects $\\angle BAT$ , the Iran Lemma yields $\\angle AOB = 90^{\\circ}$ .</blockquote>\n\nThis problem also hides overkills like this. $\\angle AOB=180^\\circ-\\angle BAO-\\angle ABO=180^\\circ-\\tfrac{\\angle ABC}{2}-\\tfrac{\\angle DAB}{2}=180^\\circ-\\tfrac{180^\\circ}{2}=90^\\circ$ Keep things simple when you can.",
"Bow!\nCheck out the video: https://www.youtube.com/watch?v=5_Mb_RZtyus",
"Really nice problem!\n\nLet $\\omega$ be the circle in the diagram. Let $T = \\omega \\cap AB$ , $R = \\omega\\cap BC$ , and $S = \\omega\\cap AD$ . By the power of $A$ wrt $\\omega$ , we have $$ AS^2 = AT^2 = AP\\cdot AQ = 3\\cdot 12 = 36\\implies AT = AS= 6. $$ Similarly, by the power of $C$ wrt $\\omega$ , $CR^2 = CQ\\cdot CP = 400\\implies CR = 20$ . Now, let $BR = BT = x$ . Then, $BC = 20 + x$ , and since $AD =BC$ , we have $20 + x = AS + SD = 6 + SD\\implies SD = 14 + x$ . Let $F$ be the foot of the perpendicular from $C$ to line $AD$ . Note that $DF = CR - SD = 6 - x$ . Thus, $AF = AS + SD + DF = 6 + 14 + x + 6 - x = 26$ . Pythag on right $\\triangle ACF$ gives $$ CF = \\sqrt{AC^2 - AF^2} = \\sqrt{28^2 - 26^2} = 6\\sqrt{3}. $$ Now, notice that $CD = AB = 6 + x$ , and since $DF = 6 - x$ , Pythag on right $\\triangle CDF$ gives\n\\[ (6 + x)^2 = (6\\sqrt{3})^2 + (6 - x)^2\\implies x = 9/2.\\]\nThus, $AD = AS + SD = 20 + x = 49/2$ . Finally, $[ABCD] = AD\\cdot CF = \\frac{49}{2}\\cdot 6\\sqrt{3} = 147\\sqrt{3}\\to\\boxed{150}$ .",
"Use linearity of PoP on circle omega and the point circle A. Define f(X) = pow_A(X) - pow_omega(X) for a point X. Now we know that f is linear so f(A) + f(C) = f(B) + f(D). This gives $t$ , the length of the tangent from B to omega. Now use pythagorean thm (R+t)^2 = (R-t)^2 + (2R)^2 to get the value of R. Then the area is 2R*(20 + t)",
"Let the circle meets $AB,BC,AD$ at $M,P,N$ respectively. POP tells that $AM=AN=6,CP=20$ . Extend $AD$ to $H$ such $CH\\bot AD$ . \n\nLet $PN\\cap AC=Q$ , Similar triangle tells that $\\frac{QN}{CH}=\\frac{3}{13}$ As $\\triangle{AQN}\\sim \\triangle{CQP}, AQ=\\frac{84}{13}, QN=\\frac{18\\sqrt{3}}{13}$ , thus, the height of the parallelogram is $6\\sqrt{3}$ Now, we call $BM=BP=k, ND=14+k, DH=6-k, CD^2-DH^2=24k=h^2=108, k=\\frac{9}{2}$ , thus, the area is $\\frac{49}{2}\\cdot 6\\sqrt{3}=\\boxed{147\\sqrt{3}}$ leads to $150$ ",
"Let the points where the circle touches AB and CD be M and T.\n\nSo, $AT^2=AP\\cdot AQ$ and $CM^2=CQ\\cdot CP$ (By **Power of a Point**) $\\implies AT=6$ & $CM=20$ Now join M and T. It is obvious that the line MT will be perpendicular to the parallel sides. First use similarity and then pythagoras' theorem to the two right angled triangle formed and you will get the value of $MT$ .\n\nNow assume the length $AB=t$ By the property of tangents, $MB=BN$ and $NA=AT$ So, $AB=t-14$ and $BC=t$ Now, use heron's formula to find the area of the triangle $\\triangle ABC$ and then equating it with half of the area of the parallelogram, you'll get $t$ . Then, just find the area. It will be $147\\sqrt{3}$ . So, the answer is $\\boxed{150}$ ",
"Let $O$ be the center and $r$ be radius of the given circle.\nLet the circle touch $AB$ , $AD$ and $BC$ at $Y$ , $W$ and $Z$ , respectively.\nLet $BO$ meet $AD$ at $N$ .\nLet $BD$ and $AC$ at $M$ .\nSince $BO=ON$ and $BM=MD$ , $OM\\parallel AD$ .\nPower of $A$ gives $AW=\\sqrt{3\\cdot 12}=6$ .\nPower of $C$ gives $CZ=\\sqrt{16\\cdot 25 }=20$ Let $WM$ meet $BC$ at $R$ . $AW=CR=6$ and $RZ=20-6=14$ . So $OM=7$ .\nPower of $M$ gives $MO^2-r^2=MQ\\cdot MP = 2\\cdot 11 = 22 \\Longrightarrow r=3\\sqrt 3$ .\nSince $AO$ and $BO$ are angle bisectors of parallelogram, $BOA$ is a right triangle.\nApplying Euclides we get $BY = \\dfrac{27}{6} = \\dfrac 92$ .\nArea will be $\\left( 20 + \\dfrac 92 \\right )6\\sqrt 3 = 147\\sqrt 3$ ",
"<details><summary>Click to expand</summary>Denote the center of the circle point $O$ . Let points $M$ , $N$ , and $L$ be the tangent points of lines $BC$ , $AD$ , and $AB$ respectively to the circle. By Power of a Point, $16\\cdot{25}=({MC})^2 \\Longrightarrow MC=20$ . Similarly, $({AL})^2=3\\cdot{12} \\Longrightarrow AL=6$ . That means $AN$ is also $6$ since quadrilateral $LONA$ is symmetrical. Let $AC$ intersect $MN$ at point $K$ . Then, $\\bigtriangleup{KMC}$ is similar to $\\bigtriangleup{AKN}$ . Therefore, $\\frac{CK}{MC}=\\frac{AK}{AN}$ . Let the length of $PK=l$ , then $\\frac{25-l}{20}=\\frac{3+l}{6}$ . Solving we get $l=\\frac{45}{13}$ . Doing the Pythagorean theorem on triangles $KMC$ and $AKN$ for sides $MK$ and $KN$ respectively, we obtain the equation $\\sqrt{(\\frac{280}{13})^2-400} +\\sqrt{(\\frac{84}{13})^2-36}=MN=2r_1$ where $r_1$ denotes the radius of the circle. Solving, we get $MN=6\\sqrt{3}$ . Additionally, quadrilateral $OLBM$ is symmetrical so $OL=OM$ . Let $OL=OM=x$ and extend a perpendicular foot from $B$ to $AD$ and call it $R$ . Then, $\\bigtriangleup{ABR}$ is right with $AR=6-x$ , $AB=6+x$ , and $RB=2r_1=MN=6\\sqrt{3}$ . Taking the difference of squares, we get $108=24x \\Longrightarrow x=\\frac{9}{2}$ . The area of $ABCD$ is $MN\\cdot{BC}=(20+x)\\cdot{MN} \\Longrightarrow \\frac{49}{2}\\cdot{6\\sqrt{3}}=147\\sqrt{3}$ . Therefore, the answer is $147+3=\\boxed{150}$</details>",
"<details><summary>solution</summary>Let $X$ be the point of contact of the circle with $AB.$ Let $BX = x.$ Let $Y$ be the foot of the perpendicular from $B$ to $AD.$ Note that $$ \\sin(\\angle{ABY}) = -\\cos(\\angle(ABC)). $$ In other words, $$ \\frac{6-a}{6+a} = -\\frac{(a+6)^2+(a+20)^2-28^2}{2(a+b)(a+20)}. $$ Simplifying, we have $a = \\frac{9}{2}.$ So, the answer is $$ (20+a)\\sqrt{24a} = \\frac{49}{2}\\cdot\\sqrt{108} = \\boxed{147\\sqrt{3}}. $$</details>",
"Let the circle have radius $r$ and center $O$ and let it be tangent to $DA$ , $AB$ , and $BC$ at $E$ , $F$ , and $G$ , respectively. By PoP, $AE=AF=\\sqrt{3(3+9)}=6$ and $CG=\\sqrt{16(16+9)}=20$ . Since $\\angle AOB=90^\\circ$ , $BF=BG=\\frac{r^2}{6}$ . Notice that $(AE+GC)^2+GE^2=AC^2$ so $26^2+4r^2=28^2$ so $r=3\\sqrt3$ . Then, $[ABCD]=BC\\cdot EG=2r(20+\\frac{r^2}{6})=\\boxed{147\\sqrt3}$ .",
"Let the circle touch $\\overline{AD}$ at point $X$ , $\\overline{AB}$ at point $Y$ , and $\\overline{BC}$ at point $Z$ . Furthermore, drop an altitude from $B$ to $\\overline{AD}$ and denote the foot as point $E$ . Note that from Power of a Point we get $AX^2 = AP \\cdot AQ$ and $CZ^2 = CP \\cdot CQ$ . Calculating the exact values yields $AX = 6$ and $CZ = 20$ , so we know that $AE = 26$ . Then, Pythagorean Theorem on $\\triangle AEC$ yields $CE = 6\\sqrt{3}$ .\n\nLet $BY = BZ = k$ . We see that $AB = CD = 6+k$ and $BC = AD = 20+k$ . Since $DE = AE-AD$ , we get that $DE = 6-k$ . Hence, the Pythagorean Theorem on $\\triangle CDE$ yields\n\n\\[(6+k)^2 = (6\\sqrt{3})^2 + (6-k)^2,\\]\n\nupon which solving yields $k = \\tfrac{9}{2}$ and we can easily solve the area to be $\\boxed{147\\sqrt{3}}$ .",
"Suppose that the circle is tangent to $AB$ , $BC$ , and $DA$ at $P$ , $Q$ , and $R$ , respectively. By PoP on $A$ and $C$ , we obtain $AP=6$ and $CQ=20$ . Define $$ BQ=x\\implies BP=x. $$ Let the foot of the perpendicular from $A$ to $BC$ be $X$ . Notice that $$ XC=6+20=26\\implies AX = 6\\sqrt{3} $$ by the Pythagorean Theorem. Since $XB=6-x$ , we have $$ (6\\sqrt{3})^2+(6-x)^2=(6+x)^2\\implies 108=24x\\implies x = \\frac{9}{2}. $$ The area is $\\boxed{147\\sqrt{3}\\implies 150}$ .",
"good problem\n\nLet $X, Y, Z$ denote the tangency points of the circle with $BC, AD, AB,$ respectively.\nBy POP, $AY = 6$ while $CY = 20.$ Let $W$ be the antipode of $Z$ in the circle\nConstruct $C’D’$ such that $ABC’D’$ is a rhombus. \nObserve that $C’C = D’D = 20 - 6 = 14.$ Define $E$ the foot from $A$ to $BC.$ Suppose the radius of the circle was $r,$ and the center was $I.$ We have that $\\frac{IX}{AE} = \\frac12.$ By parallel lines, $EC’ = 2XC’,$ so $EC = EC’ + CC’ = 12 + 14 = 26.$ Since $AC = 28$ and $CE = 26,$ we have $AE = 6\\sqrt{3},$ implying that $r = 3\\sqrt3.$ Thus, $BX = \\frac92.$ Hence, the area is $6\\sqrt3(\\frac92 + 20) = \\boxed{147\\sqrt3}$ ",
" $\\mathfrak{The \\;Twenty-Fifth\\; Of\\; October,\\; 2025}$ ʕ•ᴥ•ʔ\n<details><summary>Solution - grinding AIME Geo</summary>$$ \\color{red} \\spadesuit\\color{red} \\boxed{\\textbf{Diagram.}}\\color{red} \\spadesuit $$ \n[asy]\nimport graph; size(10.35518cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-3.07052,xmax=7.28466,ymin=-2.85104,ymax=9.68698; \npen wwccff=rgb(0.4,0.8,1.), ffzzqq=rgb(1.,0.6,0.), ffwwzz=rgb(1.,0.4,0.6), wwffqq=rgb(0.4,1.,0.), zzwwff=rgb(0.6,0.4,1.); \npair X=(1.7,-0.5), Y=(3.34,0.44), Z=(0.09410821939479797,2.4895870324799385), B=(2.7857729024820777,-0.4936888936921343), C=(6.61797941816967,5.962308436745928), D=(2.14002599705641,5.936280120863568), A=(-1.6921805186311825,-0.5197172095744937), R=(3.31815929556369,7.92104409402812), O=(1.7170541096973988,1.4647935162399692), P=(-0.09998441618160706,0.7222150815653431), Q=(2.8012377732324336,2.9852036137325504); \nD(CR((1.6888392686500604,1.420110212142448),1.9201426480076889),linewidth(2.)+wwccff); D(A--D,linewidth(2.)+ffzzqq); D(D--C,linewidth(2.)+ffzzqq); D(C--B,linewidth(2.)+ffzzqq); D(B--A,linewidth(2.)+ffzzqq); D(D--R,linewidth(2.)+ffwwzz); D(R--C,linewidth(2.)+wwffqq); D(B--(-0.5140472201239029,1.465046763590057),linewidth(2.)+wwffqq); D(Z--Y,linewidth(2.)+wwffqq); D(A--C,linewidth(2.)+zzwwff); \nD(X); MP(\"X\",(1.4815,-1.2006),NE*lsf); D(Y); MP(\"Y\",(3.85068,0.07716),NE*lsf); D(Z); MP(\"Z\",(-0.83444,2.71254),NE*lsf); D(B,linewidth(4.pt)); MP(\"B\",(2.97222,-1.04088),NE*lsf); D(C,linewidth(4.pt)); MP(\"C\",(6.77888,5.8537),NE*lsf); D(D,linewidth(4.pt)); MP(\"D\",(1.58798,6.01342),NE*lsf); D(A,linewidth(4.pt)); MP(\"A\",(-2.45826,-0.5351),NE*lsf); D(R,linewidth(4.pt)); MP(\"R\",(3.42476,8.14302),NE*lsf); D((-0.5140472201239029,1.465046763590057),linewidth(4.pt)); MP(\"S\",(-1.10064,1.54126),NE*lsf); D(O,linewidth(4.pt)); MP(\"O\",(1.82756,1.67436),NE*lsf); D(P,linewidth(4.pt)); MP(\"P\",(-0.27542,0.05054),NE*lsf); D(Q,linewidth(4.pt)); MP(\"Q\",(2.65278,3.3248),NE*lsf); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n[/asy] $$ \\color{blue} \\clubsuit\\color{blue} \\boxed{\\textbf{Solution.}}\\color{blue} \\clubsuit $$ Let points be as denoted on the diagram, where $S$ is $B$ 's projection on $AD$ and $R$ is $C$ 's.\nNote that, by $\\textbf{Power of Point}$ , we have $$ AP \\cdot AQ = AX^2 \\implies AX=6; $$ similarly, we find that $CY=20$ .\nNow notice that, since $Y$ and $Z$ are antipodes in $\\odot XZY$ (to see why that is the case, it is enough to consider the center $O$ ), we have $AR=AZ+YC=26$ ; futhermore, since $AC=6+9+16=28$ , we are easily able to find that $$ CR=\\sqrt{28^2-26^2}=6\\sqrt{3}; $$ thus, all that remains is to find $BY$ to finish the problem.\nLet $BY=BX=x$ - then we have, due to $\\triangle ABS$ , that $$ (6+x)^2=(6-x)^2+(6 \\sqrt{3})^2 $$ or that $x= \\frac 92$ ; thus, all that remains is to compute $$ [ABCD]= 6 \\sqrt{3} \\left(20 +\\frac 92 \\right )=\\boxed{147 \\sqrt{3}}. $$</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1220,
"boxed": true,
"end_of_proof": false,
"n_reply": 54,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2777232.json"
}
|
For positive integers $a$ , $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ , $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ .
|
First, we note that $a=1$ , otherwise we can not make 1 cent. We must also have $b-1$ of these, as otherwise, we can not make $b-1$ cents (any more are useless, because it is more optimal to take $b$ ).
Now, we need to make every value from $b$ to $c-1$ with only $a$ and $b$ as denominations. We know we have $b-1$ coins with worth $a$ , and assume we have $k$ with worth $b$ . Then, we know that
\[kb+b-1\geq c\implies k\geq\left\lceil\frac{c+1}b\right\rceil-1\]
where this is clearly sufficient, as the coins with denomination $a$ form every residue class. Similarly, we can already form every residue class modulo $c$ , so we just need enough coins of denomination $c$ . In particular, if we have $\ell$ coins of worth $c$ , then
\[\underbrace{\ell c}_{\text{coins with denomination $c$ }}+\underbrace{c-1}_{\text{all other coins}}\geq 1000\implies \ell\left\lceil\frac{1001}c\right\rceil-1\]
However, note that that we could have some extra coins, meaning our count could be off by one. Thus, we get that
\[f(a,b,c)=b+\left\lceil\frac{c+1}b\right\rceil+\left\lceil\frac{1001}c\right\rceil-3(-1)\]
so thus we want to find the solutions to
\[b+\left\lceil\frac{c+1}b\right\rceil+\left\lceil\frac{1001}c\right\rceil=100(+1)\]
Clearly, we need $\left\lceil\frac{1001}c\right\rceil<100$ , so $c>10$ . We shall find all solutions.
Now, we know that $2\leq b\leq c-1$ , and that $b+\frac{c+1}b$ has its maximum at the endpoints. Thus, we get that the maximum value of the function is either
\[c-1+\left\lceil\frac{c+1}{c-1}\right\rceil=c+1\qquad\text{or}\qquad 2+\left\lceil\frac{c+1}2\right\rceil\]
and the first is obviously bigger. Thus, we know that
\[c+1+\frac{1001}c+1> c+1+\left\lceil\frac{1001}c\right\rceil\geq 100\]
so thus
\[1001+2c+c^2>100c\]
Solving this, we get that
\[c<49-10\sqrt{14}\qquad c>49+10\sqrt{14}\]
Approximating, we get $c\leq 11$ or $c\geq 87$ . We can verify the ordered pair $(1,7,11)$ works. $(1,86,87)$ doesn't because we need only 10, not 11, stamps of denomination 87. We verify and $(1,87,88)$ and $(1,87,89)$ work, so the smallest values of $c$ sum to $11+88+89=\boxed{188}$ .
|
[
"<blockquote>First, we note that $a=1$ , otherwise we can not make 1 cent. We must also have $b-1$ of these, as otherwise, we can not make $b-1$ cents (any more are useless, because it is more optimal to take $b$ ).\n\n\nNow, we need to make every value from $b$ to $c-1$ with only $a$ and $b$ as denominations. We know we have $b-1$ coins with worth $a$ , and assume we have $k$ with worth $b$ . Then, we know that\n\\[kb+b-1\\geq c\\implies k\\geq\\left\\lceil\\frac{c+1}b\\right\\rceil-1\\]\nwhere this is clearly sufficient, as the coins with denomination $a$ form every residue class. Similarly, we can already form every residue class modulo $c$ , so we just need enough coins of denomination $c$ . In particular, if we have $\\ell$ coins of worth $c$ , then\n\\[\\underbrace{\\ell c}_{\\text{coins with denomination $c$ }}+\\underbrace{c-1}_{\\text{all other coins}}\\geq 1000\\implies \\ell\\left\\lceil\\frac{1001}c\\right\\rceil-1\\]\nThus, we get that\n\\[f(a,b,c)=b+\\left\\lceil\\frac{c+1}b\\right\\rceil+\\left\\lceil\\frac{1001}c\\right\\rceil-3\\]\nso thus we want to find the solutions to\n\\[b+\\left\\lceil\\frac{c+1}b\\right\\rceil+\\left\\lceil\\frac{1001}c\\right\\rceil=100\\]\nClearly, we need $\\left\\lceil\\frac{1001}c\\right\\rceil<100$ , so $c>10$ . We shall find all solutions.\n\n\nNow, we know that $2\\leq b\\leq c-1$ , and that $b+\\frac{c+1}b$ has its maximum at the endpoints. Thus, we get that the maximum value of the function is either\n\\[c-1+\\left\\lceil\\frac{c+1}{c-1}\\right\\rceil=c+1\\qquad\\text{or}\\qquad 2+\\left\\lceil\\frac{c+1}2\\right\\rceil\\]\nand the first is obviously bigger. Thus, we know that\n\\[c+1+\\frac{1001}c+1> c+1+\\left\\lceil\\frac{1001}c\\right\\rceil\\geq 100\\]\nso thus\n\\[1001+2c+c^2>100c\\]\nSolving this, we get that\n\\[c<49-10\\sqrt{14}\\qquad c>49+10\\sqrt{14}\\]\nApproximating, we get $c\\leq 11$ or $c\\geq 87$ . We can verify the ordered pairs $(1,7,11)$ , $(1,86,87)$ , and $(1,86,88)$ all work, so the smallest values of $c$ sum to $11+87+88=\\boxed{186}$ .</blockquote>\n\nIsn't $f(1,86,87)$ equal to 96?\n85 stamps of 1, 1 stamp of 86, and 10 stamps of 87 seem to suffice.",
"I agree with above. I got 11+88+89",
"f(1, 86, 87) was 97, not 96 afaik.",
"Fixed the answer - I was wondering why the answer I put didn't match.",
"2019 AIME II Problem 14 on steroids",
"Can you just casework mod a using chicken McNugget? I’m trying to find a cleaner way rather than just bashing, but can’t find one.",
"<blockquote>Can you just casework mod a using chicken McNugget? I’m trying to find a cleaner way rather than just bashing, but can’t find one.</blockquote>\n\nchicken mcnugget?",
"<blockquote><blockquote>Can you just casework mod a using chicken McNugget? I’m trying to find a cleaner way rather than just bashing, but can’t find one.</blockquote>\n\nchicken mcnugget?</blockquote>\n\nThey were referring to the [url = https://artofproblemsolving.com/wiki/index.php/Chicken_McNugget_Theorem]Chicken McNugget Theorem[/url].",
"the equation $b+\\left\\lceil \\frac{c}{b}\\right\\rceil + \\left\\lceil \\frac{1001-b\\lceil \\frac{c}{b}\\rceil}{c} \\right\\rceil = 99$ is equivalent i think\n\nedit: maybe not oops",
"wait actually $(91, 182)$ satisfies $b+\\left\\lceil\\frac{c+1}b\\right\\rceil+\\left\\lceil\\frac{1001}c\\right\\rceil=100$ but you can have numbers of coins $(90, 1, 5)$ changing to $b+\\left\\lceil\\frac{c}b\\right\\rceil+\\left\\lceil\\frac{1001}c\\right\\rceil=100$ doesnt seem to work either because $(1, 91, 186)$ you can do $(90, 2, 4)$ ",
"oops triple post\ni think the error is $kb+b-1\\geq c$ it should be $kb+b-1\\geq c-1$ since you only need to make up to c-1 (you can make c by just using one of the c coins)",
"2019 aime ii flashbacks",
"[Video Solution](https://youtu.be/jptMMVCuj34)",
"It's probably easier to think of and work with the equation $b-1+\\left\\lfloor \\frac{c-1}{b} \\right\\rfloor+\\left \\lfloor\\frac{1000}{c} \\right\\rfloor=97$ instead. Kind of a joyless problem either way."
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1060,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782913.json"
}
|
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ , $O_1O_2 = 15$ , $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon.
[asy]
import geometry;
size(10cm);
point O1=(0,0),O2=(15,0),B=9*dir(30);
circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0];
filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black);
draw(w1);
draw(w2);
draw(O1--O2,dashed);
draw(o);
dot(O1);
dot(O2);
dot(A);
dot(D);
dot(C);
dot(B);
label(" $\omega_1$ ",8*dir(110),SW);
label(" $\omega_2$ ",5*dir(70)+(15,0),SE);
label(" $O_1$ ",O1,W);
label(" $O_2$ ",O2,E);
label(" $B$ ",B,N+1/2*E);
label(" $A$ ",A,N+1/2*W);
label(" $C$ ",C,S+1/4*W);
label(" $D$ ",D,S+1/4*E);
label(" $15$ ",midpoint(O1--O2),N);
label(" $16$ ",midpoint(C--D),N);
label(" $2$ ",midpoint(A--B),S);
label(" $\Omega$ ",o.C+(o.r-1)*dir(270));
[/asy]
|
<details><summary>Solution</summary>I claim the answer is $\boxed{140}$ . First,**<span style="color:blue">Claim:</span>** Let $\ell$ be the radical axis of $\omega_1$ and $\omega_2$ . Then $BC$ , $AD$ and $\ell$ concur.
*Proof.* Immediate from Radical Axis theorem on $\omega_1$ , $\omega_2$ , and $\Omega$ . $\square$ **<span style="color:blue">Claim:</span>** The line $CD$ is parallel to the reflection of $AB$ across $O_1O_2$ .
*Proof.* I claim that $\ell$ is the internal angle bisector of $\angle(\overline{BC},\overline{AD})$ . (One can gues this claim from a good diagram/intuition).
Indeed, suppose $BC$ and $AD$ intersct at $T$ , and furthermore, $BC$ intersects $O_1O_2$ at $X$ and $AD$ intersects $O_1O_2$ at $Y$ . Then using the fact that $O_1$ and $O_2$ are the arc midpoints of $BC$ and $AD$ , we find \[ \angle TXY = \angle CXO_1 = 180 - \angle O_1CO_2 \] where we use the so-called Shooting Lemma which gives $\triangle O_1XC \sim \triangle O_1CO_2$ . Similarly, \[ \angle TYX = \angle DYO_2 = 180 - \angle O_2DO_1 \] from which it follows that $TX = TY$ .
So the foot from $T$ to $O_1O_2$ bisects $\angle XTY$ and $TY$ , or equivalently $\ell$ bisects $\angle XTY$ .
But it is well-known that in cyclic quadrilateral $ABCD$ , the internal angle bisector of $\angle(\overline{BC},\overline{AD})$ is perpendicular to the internal angle bisector of $\angle(\overline{AB},\overline{CD})$ ; this implies that $O_1O_2$ is parallel to the latter, which is equivalent to the fact that $CD$ is parallel to the reflection of $AB$ across $O_1O_2$ . $\square$ Thus, we consider the reflections $C'$ and $D'$ of $C$ and $D$ across the perpendicular bisector of $O_1O_2$ respectively. Then
- By the claim above, it follows $C'D' \parallel AB$ ,
- Thus we have $BD' = AC'$ ,
- We have $O_1B = O_2C'$ , so $15 = O_1O_2 = BC'$ ,
- We have $O_2A = O_1D'$ , so $15 = O_2O_1 = AD'$ ,
- Thus we have $\triangle O_1BD' \cong \triangle O_2C'A$
But then $ABC'D'$ is an isosceles trapezoid with bases of length $2$ and $16$ and diagonals of length $15$ . Then we can easily solve for the legs to be length $\sqrt{193}$ .
Moreover, $O_1B + O_1D' = 15$ , yet $\triangle O_1BD'$ has side lengths $O_1B$ , $O_1D'$ and $\sqrt{193}$ . Note also that \[ \theta \coloneqq \angle BO_1D' = 180 - \angle BC'D' \] so it follows that $\sin \theta = \frac 45$ and $\cos \theta = -\frac 35$ ; then \[ 193 = O_1B^2 + O_1D'^2 + 2\cdot O_1B\cdot O_1D' \cdot \frac 35 = (O_1B+O_1D')^2 - 2\cdot O_1B\cdot O_1D' \cdot \frac 25 \] which is from Law of Cosines. Then we can solve to find $O_1B\cdot O_1D' = 40$ ; then \[ [O_1BD'] = \frac 12 \cdot O_1B\cdot O_1D' \cdot \frac 45 = \frac 12 \cdot 40 \cdot \frac 45 = 16 \]
Then it follows that \[ [ABO_1CDO_2] = [O_1BAO_2C'D'] = 2\cdot [O_1BD'] + [BAC'D'] = 2\cdot 16 + \frac{2+16}{2}\cdot 12 = 32 + 108 = 140 \] as claimed.</details>
<details><summary>Motivation</summary>The first claim is olympiad-flavored; then from a good diagram/intuition, since $O_1$ and $O_2$ are arc midpoints, it seemed like O_1O_2 should be an angle bisector of $AB$ and $CD$ , which would be equivalent to $\ell$ being an angle bisector. But the concurrency and some angle chasing showed this to actually be true.
From that claim (and a little courage), I was inspired to reflect $CD$ so I could gain a parallelism and thus an isosceles trapezoid, as the total area is also preserved. The rest falls through.</details>
|
[
"Nice problem. Didn't even try synthetic because well, who cares about trying geometry.\n\nLet the radius of $\\omega_1$ be $r$ and that of $\\omega_2$ be $15-r$ . We recall the following important fact: in cyclic quadrilateral $ABCD$ with sides $a,b,c,$ and $d$ , with circumradius $R$ , area $A$ , and semi-perimeter $s$ , then\n\\[R^2=\\frac 1{16}\\cdot\\frac{(ab+cd)(ac+bd)(ad+bc)}{A^2}\\qquad A^2=(s-a)(s-b)(s-c)(s-d)\\]**Claim.** The above formulas are true.[/b]**Proof.** We first show that the value of $A^2$ is indeed what we expect.\n[asy]\n import geometry;\n size(5cm);\n pair O=(0,0),A=dir(20),B=dir(100),C=dir(180),D=dir(260);\n draw(circle(O,1));\n draw(A--B--C--D--cycle);\n draw(A--C);\n label(\" $A$ \",A,A);\n label(\" $B$ \",B,B);\n label(\" $C$ \",C,C);\n label(\" $D$ \",D,D);\n label(\" $a$ \",A--B,S);\n label(\" $b$ \",C--B,SE);\n label(\" $c$ \",C--D,NE);\n label(\" $d$ \",A--D,NW);\n[/asy]\nWe know the area is\n\\[A=[ABCD]=[ABC]+[ACD]=\\frac 12ab\\sin\\angle B+\\frac 12cd\\sin\\angle D=\\frac 12\\sin\\angle B(ab+cd)\\]\nso\n\\[A^2=\\frac 14(ab+cd)^2(1-\\cos^2\\angle B)\\]\nHowever, by the Law of Cosines, we know\n\\[a^2+b^2-2ab\\cos\\angle B=AC^2=c^2+d^2-2cd\\cos\\angle D=c^2+d^2+2cd\\cos\\angle B\\]\nThus, we get that\n\\[2(ab+cd)\\cos\\angle B=a^2+b^2-c^2-d^2\\implies\\cos\\angle B=\\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\\tag{1}\\]\nThus, we get\n\\begin{align*}\nA^2&=\\frac 14(ab+cd)^2(1-\\cos^2\\angle B)\n&=\\frac 1{16}\\left((2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2\\right)\n&=\\frac 1{16}(a^2+b^2+2ab-c^2-d^2+2cd)(c^2+d^2+2cd-a^2-b^2+2ab)\n&=\\frac 1{16}((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\n&=\\frac 1{16}(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)\n&=(s-a)(s-b)(s-c)(s-d)\n\\end{align*}\nas desired. Now, we show the formula for $R$ .\n\nLet $AC=p$ and $BD=q$ . Then, by Ptomely's Theorem, $ac+bd=pq$ , and\n\\[R=\\frac{abp}{4[ABC]}=\\frac{cdp}{4[ACD]}\\]\nThus, we get that\n\\[[ABC]=\\frac{abp}{4R}\\qquad[ACD]=\\frac{cdp}{4R}\\]\nThus, this means that\n\\[A=[ABC]+[ACD]=\\frac{abp}{4R}+\\frac{cdp}{4R}=\\frac{(ab+cd)p}{4R}\\]\nThus, we need to find $p$ . Looking at (1), we know\n\\begin{align*}\np^2&=a^2+b^2-2ab\\cos B\n&=a^2+b^2-2ab\\cdot\\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\n&=\\frac{(a^2+b^2)ab-(a^2+b^2)ab+(a^2+b^2)cd-(c^2+d^2)ab}{ab+cd}\n&=\\frac{a^2cd+b^2cd+abc^2+abd^2}{ab+cd}\n&=\\frac{(ac+bd)(ad+bc)}{ab+cd}\n\\end{align*}\nThus, substituting, we get\n\\[R^2=\\frac 1{16}\\cdot\\frac{(ab+cd)p^2}{A^2}=\\frac 1{16}\\cdot\\frac{(ab+cd)(ac+bd)(ad+bc)}{A^2}\\]\nas desired.\n\nNow, this means that the cyclic quadrilaterals with sides $2,15,15-r,r$ and $16,15,r,15-r$ have the same circumradius. We can calculate the semiperimeters as $16$ and $23$ , so\n\\[16R^2=\\frac{(2\\cdot 15+r(15-r))(2\\cdot r+15\\cdot(15-r))(15\\cdot r+2\\cdot (15-r)}{(16-r)(16-(15-r))(16-15)(16-2)}\\]\n\\[16R^2=\\frac{(16\\cdot 15+r(15-r))(16\\cdot r+15\\cdot(15-r))(15\\cdot r+16\\cdot (15-r)}{(23-r)(23-(15-r))(23-15)(23-16)}\\]\nWe note this will give us a sextic, which would be pretty hard to solve. Instead, we let $x=r(15-r)$ , which means that\n\\[16R^2=\\frac{(30+x)(169x+6750)}{14(16+x)}=\\frac{(240+x)(x+54000)}{56(184+x)}\\]\nThis rearranges to\n\\[4(x+30)(169x+6750)(x+184)=(x+240)(x+54000)(x+16)\\]\nExpanding would be a pain, so we look to cleverly bound and use the rational root theorem. We know that $(x+30)(x+184)\\approx(x+16)(x+240)$ , so we need $4(169x+6750)\\approx x+54000$ . In particular, this means $x\\approx 40$ , so we try $x=40$ , which works.\n\nNow, we need to find the area. We know that it is\n\\begin{align*}\n[ABO_1CDO_2]=&[ABO_1O_2]+[CDO_1O_2]\n=&\\sqrt{(16-r)(16-(15-r))(16-15)(16-2)}\n&+\\sqrt{(23-r)(23-(15-r))(23-15)(23-16)}\n=&\\sqrt{14(x+16)}+\\sqrt{56(x+30)}\n=&28+112=\\boxed{140}\n\\end{align*}",
"Credit to not me: Rearrange the sides of the cyclic hexagon so it's $x, 15-x, 16, x, 15-x, 2$ around the hexagon instead. Now solve and use Brahmagupta's twice to extract answer. The point is, it's now an isosceles trapezoid without changing the circumradius.\n\nMy solve during test: Use Brahmagupta's to calculate area, then do a second time using Ptolemy ratios. Equate to obtain a cubic, then guess integer roots for the value of the product of the radii of the two circles. RRT finishes. You can also do a diophantine assuming the area of the top and bottom quadrilateral are integers, and finish that way too after assuming the product of the radii is an integer.",
"Isn't $A^2 = (s-a)(s-b)(s-c)(s-d)$ well-known as Brahmagupta's",
"Mine. In short, I'm sorry, but like, not really.\n\n<details><summary>Solution</summary>First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $A'B'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $B'A'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2B'$ and $A'O_1C$ are congruent, so $B'D = A'C$ and quadrilateral $B'A'CD$ is an isosceles trapezoid.\n[asy]\n\timport olympiad;\nsize(180);\ndefaultpen(linewidth(0.7));\npair Bp = dir(105), Ap = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175);\ndraw(unitcircle^^Bp--Ap--O1--C--D--O2--cycle);\nlabel(\" $B'$ \",Bp,dir(origin--Bp));\nlabel(\" $A'$ \",Ap,dir(origin--Ap));\nlabel(\" $O_1$ \",O1,dir(origin--O1));\nlabel(\" $C$ \",C,dir(origin--C));\nlabel(\" $D$ \",D,dir(origin--D));\nlabel(\" $O_2$ \",O2,dir(origin--O2));\ndraw(O2--O1,linetype(\"4 4\"));\ndraw(Bp--D^^Ap--C,linetype(\"2 2\"));\n[/asy]\nNext, remark that $A'O_1 = DO_2$ , so quadrilateral $A'O_1DO_2$ is also an isosceles trapezoid; in turn, $A'D = O_1O_2 = 15$ , and similarly $B'C = 15$ . Thus, Ptolemy's Theorem on $B'A'CD$ yields $B'D\\cdot A'C + 2\\cdot 16 = 15^2$ , whence $B'D = A'C = \\sqrt{193}$ . Let $\\alpha = \\angle B'A'D$ . The Law of Cosines on triangle $B'A'D$ yields\n\\[\n\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,\n\\]\nand hence $\\sin\\alpha = \\tfrac 45$ . Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle B'A'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $B'A'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$ .\n\nNow let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$ ; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$ . Furthermore, angles $B'O_2D$ and $B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$ , which implies the measure of angle $B'O_2D$ is $180^\\circ - \\alpha$ . Therefore, the Law of Cosines applied to triangle $B'O_2D$ yields\n\\begin{align*}\n193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\n&= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2.\n\\end{align*}\nThus $r_1r_2 = 40$ , and so the area of triangle $B'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$ .\n\nAll in all, the area of the hexagon is $108 + 2\\cdot 16 = \\boxed{140}$ .</details>",
"\"Mine. In short, I'm sorry, but like, not really\"\n~David Altizio, 2022",
"you should upvote posts 5 and 6",
"you should also upvote post 7",
"can pascal's be used? i just loc bashed with cyclic quads",
"after an hour and a half of work...\n\nLet $X$ be the intersection of $AO_1$ and $BO_2$ and let $Y$ be the intersection of $DO_1$ and $CO_2$ . Then we get \n\\[\\triangle BXA\\sim \\triangle O_1XO_2\\]\n\\[\\triangle CYD\\sim \\triangle O_1YO_2\\]\n\\[\\triangle O_1XO_2\\cong \\triangle O_1YO_2.\\] \nSet $O_1X=15a$ and $O_2X=15b$ , and let the radius of $\\omega_1$ be $r$ . Ptolemy's Theorem on $O_1O_2AB$ and $O_1O_2DC$ gives \n\\[240a^2+481ab+240b^2=240+r(15-r)\\]\n\\[30a^2+229ab+30b^2=30+r(15-r),\\]\nfrom which we obtain \n\\[5a^2+6ab+5b^2=5.\\]\nSince $\\triangle BXO_1\\sim \\triangle AXO_2$ we get $\\frac{b}{a}=\\frac{15-r}{r}$ . Then we get $193a^2=r^2$ and $b=\\frac{15a}{r}-a=\\frac{15\\sqrt{193}}{193}-a.$ Plugging into $5a^2+6ab+5b^2=5$ gives us \n\\[a^2-\\frac{15\\sqrt{193}}{193}a+\\frac{40}{193}=0\\]\nand solving gives $a=\\frac{15+\\sqrt{65}}{2}\\cdot \\frac{\\sqrt{193}}{193}$ (the other solution to the quadratic simply swaps the circles $\\omega_1$ and $\\omega_2$ ) so that $r=\\frac{15+\\sqrt{65}}{2}$ . Finally Brahmagupta on $O_1O_2AB$ and $O_1O_2DC$ gives $$ [O_1O_2AB]+[O_1O_2DC]=28+112=\\boxed{140}, $$ and we are done.\n\nactually tried trigbash at the start but got nowhere, then realized what the equal side lengths in cyclic quad meant\n\nprobably a faster way ig cuz i used a ton of b ash still",
"<blockquote>Nice problem. Didn't even try synthetic because well, who cares about trying geometry.\n\nLet the radius of $\\omega_1$ be $r$ and that of $\\omega_2$ be $15-r$ . We recall the following important fact: in cyclic quadrilateral $ABCD$ with sides $a,b,c,$ and $d$ , with circumradius $R$ , area $A$ , and semi-perimeter $s$ , then\n\\[R^2=\\frac 1{16}\\cdot\\frac{(ab+cd)(ac+bd)(ad+bc)}{A^2}\\qquad A^2=(s-a)(s-b)(s-c)(s-d)\\]**Claim.** The above formulas are true.[/b]**Proof.** We first show that the value of $A^2$ is indeed what we expect.\n[asy]\n import geometry;\n size(5cm);\n pair O=(0,0),A=dir(20),B=dir(100),C=dir(180),D=dir(260);\n draw(circle(O,1));\n draw(A--B--C--D--cycle);\n draw(A--C);\n label(\" $A$ \",A,A);\n label(\" $B$ \",B,B);\n label(\" $C$ \",C,C);\n label(\" $D$ \",D,D);\n label(\" $a$ \",A--B,S);\n label(\" $b$ \",C--B,SE);\n label(\" $c$ \",C--D,NE);\n label(\" $d$ \",A--D,NW);\n[/asy]\nWe know the area is\n\\[A=[ABCD]=[ABC]+[ACD]=\\frac 12ab\\sin\\angle B+\\frac 12cd\\sin\\angle D=\\frac 12\\sin\\angle B(ab+cd)\\]\nso\n\\[A^2=\\frac 14(ab+cd)^2(1-\\cos^2\\angle B)\\]\nHowever, by the Law of Cosines, we know\n\\[a^2+b^2-2ab\\cos\\angle B=AC^2=c^2+d^2-2cd\\cos\\angle D=c^2+d^2+2cd\\cos\\angle B\\]\nThus, we get that\n\\[2(ab+cd)\\cos\\angle B=a^2+b^2-c^2-d^2\\implies\\cos\\angle B=\\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\\tag{1}\\]\nThus, we get\n\\begin{align*}\nA^2&=\\frac 14(ab+cd)^2(1-\\cos^2\\angle B)\n&=\\frac 1{16}\\left((2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2\\right)\n&=\\frac 1{16}(a^2+b^2+2ab-c^2-d^2+2cd)(c^2+d^2+2cd-a^2-b^2+2ab)\n&=\\frac 1{16}((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\n&=\\frac 1{16}(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)\n&=(s-a)(s-b)(s-c)(s-d)\n\\end{align*}\nas desired. Now, we show the formula for $R$ .\n\nLet $AC=p$ and $BD=q$ . Then, by Ptomely's Theorem, $ac+bd=pq$ , and\n\\[R=\\frac{abp}{4[ABC]}=\\frac{cdp}{4[ACD]}\\]\nThus, we get that\n\\[[ABC]=\\frac{abp}{4R}\\qquad[ACD]=\\frac{cdp}{4R}\\]\nThus, this means that\n\\[A=[ABC]+[ACD]=\\frac{abp}{4R}+\\frac{cdp}{4R}=\\frac{(ab+cd)p}{4R}\\]\nThus, we need to find $p$ . Looking at (1), we know\n\\begin{align*}\np^2&=a^2+b^2-2ab\\cos B\n&=a^2+b^2-2ab\\cdot\\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\n&=\\frac{(a^2+b^2)ab-(a^2+b^2)ab+(a^2+b^2)cd-(c^2+d^2)ab}{ab+cd}\n&=\\frac{a^2cd+b^2cd+abc^2+abd^2}{ab+cd}\n&=\\frac{(ac+bd)(ad+bc)}{ab+cd}\n\\end{align*}\nThus, substituting, we get\n\\[R^2=\\frac 1{16}\\cdot\\frac{(ab+cd)p^2}{A^2}=\\frac 1{16}\\cdot\\frac{(ab+cd)(ac+bd)(ad+bc)}{A^2}\\]\nas desired.\n\nNow, this means that the cyclic quadrilaterals with sides $2,15,15-r,r$ and $16,15,r,15-r$ have the same circumradius. We can calculate the semiperimeters as $16$ and $23$ , so\n\\[16R^2=\\frac{(2\\cdot 15+r(15-r))(2\\cdot r+15\\cdot(15-r))(15\\cdot r+2\\cdot (15-r)}{(16-r)(16-(15-r))(16-15)(16-2)}\\]\n\\[16R^2=\\frac{(16\\cdot 15+r(15-r))(16\\cdot r+15\\cdot(15-r))(15\\cdot r+16\\cdot (15-r)}{(23-r)(23-(15-r))(23-15)(23-16)}\\]\nWe note this will give us a sextic, which would be pretty hard to solve. Instead, we let $x=r(15-r)$ , which means that\n\\[16R^2=\\frac{(30+x)(169x+6750)}{14(16+x)}=\\frac{(240+x)(x+54000)}{56(184+x)}\\]\nThis rearranges to\n\\[4(x+30)(169x+6750)(x+184)=(x+240)(x+54000)(x+16)\\]\nExpanding would be a pain, so we look to cleverly bound and use the rational root theorem. We know that $(x+30)(x+184)\\approx(x+16)(x+240)$ , so we need $4(169x+6750)\\approx x+54000$ . In particular, this means $x\\approx 40$ , so we try $x=40$ , which works.\n\nNow, we need to find the area. We know that it is\n\\begin{align*}\n[ABO_1CDO_2]=&[ABO_1O_2]+[CDO_1O_2]\n=&\\sqrt{(16-r)(16-(15-r))(16-15)(16-2)}\n&+\\sqrt{(23-r)(23-(15-r))(23-15)(23-16)}\n=&\\sqrt{14(x+16)}+\\sqrt{56(x+30)}\n=&28+112=\\boxed{140}\n\\end{align*}</blockquote>\n\nthis solution must have taken a long time orz",
"One of my friends was very close to getting the correct answer; he put 147 instead of 140.",
"<blockquote>can pascal's be used? i just loc bashed with cyclic quads</blockquote>\nIdk how that would help though? There are no parallel lines\n",
"Let $\\omega_i$ have radius $r_i$ . As mentioned already, it's enough to figure out what $r_1$ and $r_2$ are, since the answer follows from Brahmagupta on $ABO_1O_2$ and $CDO_2O_1$ . We already know $r_1 + r_2 = 15$ , so we just need to squeeze out one more piece of information. \n\nWe do this by calculating $AO_1\\cdot BO_2 \\cdot CO_2 \\cdot DO_1$ in two ways. On one hand, by Ptolemy, $AO_1\\cdot BO_2 = 30 + r_1r_2$ and $CO_2\\cdot DO_1 = 240 + r_1r_2$ . On the other hand, $O_1A \\cdot O_1D$ is the power of $O_1$ with respect to $\\omega_2$ (say, by $\\sqrt{bc}$ and Fact 5 on $\\triangle AO_1D$ ), which is $r_1(15 + r_2)$ . Similarly, $O_2B \\cdot O_2C = r_2(15 + r_1)$ , so we obtain \\[(30 + r_1r_2)(240 + r_1r_2) = r_1r_2(15 + r_1)(15 + r_2).\\] Using $r_1 + r_2 = 15$ , this comes out cleanly to $r_1r_2 = 40$ . From here, answer extraction with Brahmagupta is pretty straightforward.",
":O op solution",
"Found a nice way to cheese the question by using the information that the answer to an AIME problem must be an integer:\n[https://www.youtube.com/watch?v=NHNB9IWR62I](https://www.youtube.com/watch?v=NHNB9IWR62I)",
"there are people who are able to solve this problem, and then there's me who looks at this problem and is like \"that's a pentagon!\"",
"redacted",
"<blockquote>there are people who are able to solve this problem, and then there's me who looks at this problem and is like \"that's a pentagon!\"</blockquote>\n\nthat was me when i first saw the problem",
"It took me a while to realize that this solution is basically the same as the one in post #2, but the idea feels exceedingly natural. The key lemma is the following:**<span style=\"color:#9a00ff\">Lemma (Parameshvara's formula).</span>** The circumradius of a cyclic quadrilateral with side lengths $a, b, c, d$ is given by $$ R = \\frac 14\\sqrt{\\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}. $$ The idea is to compute the area in two ways: via Brahmagupta's formula and via the formula $AC \\cdot BD \\cdot \\sin \\theta$ , where $\\theta$ is the angle between the diagonals and use strong Ptolemy. The proof itself is emitted for time purposes (indeed, I had memorized the formula before but just stated it here in case it was not well-known).\n\nNow we proceed with the lemma. Let the radii of $\\omega_1$ and $\\omega_2$ be $r_1, r_2$ respectively. By the previous lemma, we have $$ \\frac{(2 \\cdot 15 + r_1r_2)(2r_1+15r_2)(2r_2+15r_1)}{(16-2)(16-15)(16-r_1)(16-r_2)} = (4R)^2 = \\frac{(15 \\cdot 16+r_1r_2)(15r_1+16r_2)(15r_2+16r_1)}{(23-15)(23-16)(23-r_1)(23-r_2)}. $$ Simplifying this, $$ \\frac{(30+r_1r_2)(30r_1^2+30r_2^2+229r_1r_2)}{(256-16r_1-16r_2+r_1r_2) \\cdot 14 \\cdot 1} = \\frac{(240+r_1r_2)(240r_1^2+240r_2^2+481r_1r_2)}{8 \\cdot 7 \\cdot (529-23r_1-23r_2+r_1r_2)}. $$ Now let $x = r_1r_2$ , and observe that $r_1+r_2 = 15$ . We obtain\n\\begin{align*}\n\\frac{(30+x)(30 \\cdot 15^2 + 169x)}{14 \\cdot (256-16 \\cdot 15 + x)} &= \\frac{(240+x)(240 \\cdot 15^2+x)}{56 \\cdot (529-23 \\cdot 15 + x)} \n4(x+30)(169x+6750)(x+184) &= (x+16)(x+240)(x+54000).\n\\end{align*}\nThere are a couple of ways to look through this monstrous cubic to find the integer solution. For example, one can reckon that equality holds in one or more of the sub-expressions, the most reasonable of which is $$ (x+30)(x+184) = (x+16)(x+240) \\iff x = 40. $$ Indeed, this works as $$ 4(169 \\cdot 40 + 6750) = 4(6760+6750) = 54040 = 40+54000. $$ Then, by Brahmagupta the total area is equal to $$ A = \\sqrt{14 \\cdot (40+16)} + \\sqrt{56 \\cdot (184+40)} = (56+14) \\cdot 2 = \\boxed{140}. $$ ",
"Okay bud.. This problem is very good. Actually I will be sort of \"cheating\" in this solution, which means I wont be completely solving the problem with full proof. You know what that means? Haha yes I will make use of the very pleasant fact that all answers in the AIME are integers :-D \n\nWe use the brahmagupta's formula, which states that the area of a cyclic quadrilateral with side-lengths \\(a,b,c,d\\) is \\[\\sqrt{(s-a)(s-b)(s-c)(s-d)}\\] where \\(s\\) is the semi-perimeter.\n\nNow, Let the radii of the two circles be \\(x\\) and \\(y\\) respectively. I will calculate \\([ABO_1O_2]\\) and \\([CDO_1O_2]\\) separately and then sum them up.\n\nNow using the brahmagupta formula on the cyclic quadrilateral \\(ABO_1O_2\\), we first find that the semi-perimeter is \\(16\\). So basically, our area is: \\[\\sqrt{(16-x)(16-y)(14)(1)}=\\sqrt{14(16+xy)}\\] using the fact that \\(x+y=16\\). Similarly, the area of the cyclic quadrilateral \\(CDO_1O_2\\) is \\[2\\sqrt{14(184+xy)}\\] So that means we want to find \\[2\\sqrt{14(184+xy)}+\\sqrt{14(16+xy)}\\] Okayy now here comes the magic. From common intuition, we may expect that both the terms in the radicals are perfect squares! That practically means that \\(xy+184=14k^2\\) and \\(xy+16=14l^2\\). Subtracting gives \\(k^2-l^2=12\\) and so \\(k=4\\) and \\(l=2\\). That implies \\(xy=40\\)! So, we have that \\[[ABO_2CDO_1]=[ABO_1O_2]+[CDO_1O_2]= 2\\sqrt{14(184+xy)}+\\sqrt{14(16+xy)}=2\\times 14\\times 4+14\\times2=140\\] and we are donneee!!",
"Alright, this is a great solution(even though it isn't rigorous)\nIt reduces the difficulty by a LOT.",
"[Video Solution](https://youtu.be/TeAWS5H5bcc)",
"<blockquote>after an hour and a half of work...\n\nLet $X$ be the intersection of $AO_1$ and $BO_2$ and let $Y$ be the intersection of $DO_1$ and $CO_2$ . Then we get \n\\[\\triangle BXA\\sim \\triangle O_1XO_2\\]\n\\[\\triangle CYD\\sim \\triangle O_1YO_2\\]\n\\[\\triangle O_1XO_2\\cong \\triangle O_1YO_2.\\] \nSet $O_1X=15a$ and $O_2X=15b$ , and let the radius of $\\omega_1$ be $r$ . Ptolemy's Theorem on $O_1O_2AB$ and $O_1O_2DC$ gives \n\\[240a^2+481ab+240b^2=240+r(15-r)\\]\n\\[30a^2+229ab+30b^2=30+r(15-r),\\]\nfrom which we obtain \n\\[5a^2+6ab+5b^2=5.\\]\nSince $\\triangle BXO_1\\sim \\triangle AXO_2$ we get $\\frac{b}{a}=\\frac{15-r}{r}$ . Then we get $193a^2=r^2$ and $b=\\frac{15a}{r}-a=\\frac{15\\sqrt{193}}{193}-a.$ Plugging into $5a^2+6ab+5b^2=5$ gives us \n\\[a^2-\\frac{15\\sqrt{193}}{193}a+\\frac{40}{193}=0\\]\nand solving gives $a=\\frac{15+\\sqrt{65}}{2}\\cdot \\frac{\\sqrt{193}}{193}$ (the other solution to the quadratic simply swaps the circles $\\omega_1$ and $\\omega_2$ ) so that $r=\\frac{15+\\sqrt{65}}{2}$ . Finally Brahmagupta on $O_1O_2AB$ and $O_1O_2DC$ gives $$ [O_1O_2AB]+[O_1O_2DC]=28+112=\\boxed{140}, $$ and we are done.\n\nactually tried trigbash at the start but got nowhere, then realized what the equal side lengths in cyclic quad meant\n\nprobably a faster way ig cuz i used a ton of b ash still</blockquote>\n\nHow do you get $193a^2 = r^2$ ?",
"[quote name=\"asdf334\" url=\"/community/p24450442\"]\nafter an hour and a half of work...\n\nLet $X$ be the intersection of $AO_1$ and $BO_2$ and let $Y$ be the intersection of $DO_1$ and $CO_2$ . Then we get \n\\[\\triangle BXA\\sim \\triangle O_1XO_2\\]\n\\[\\triangle CYD\\sim \\triangle O_1YO_2\\]\n\\[\\triangle O_1XO_2\\cong \\triangle O_1YO_2.\\] \nSet $O_1X=15a$ and $O_2X=15b$ , and let the radius of $\\omega_1$ be $r$ . Ptolemy's Theorem on $O_1O_2AB$ and $O_1O_2DC$ gives \n\\[240a^2+481ab+240b^2=240+r(15-r)\\]\n\\[30a^2+229ab+30b^2=30+r(15-r),\\]\nfrom which we obtain \n\\[5a^2+6ab+5b^2=5.\\]\nSince $\\triangle BXO_1\\sim \\triangle AXO_2$ we get $\\frac{b}{a}=\\frac{15-r}{r}$ . Then we get $193a^2=r^2$ and $b=\\frac{15a}{r}-a=\\frac{15\\sqrt{193}}{193}-a.$ Plugging into $5a^2+6ab+5b^2=5$ gives us \n\\[a^2-\\frac{15\\sqrt{193}}{193}a+\\frac{40}{193}=0\\]\nand solving gives $a=\\frac{15+\\sqrt{65}}{2}\\cdot \\frac{\\sqrt{193}}{193}$ (the other solution to the quadratic simply swaps the circles $\\omega_1$ and $\\omega_2$ ) so that $r=\\frac{15+\\sqrt{65}}{2}$ . Finally Brahmagupta on $O_1O_2AB$ and $O_1O_2DC$ gives $$ [O_1O_2AB]+[O_1O_2DC]=28+112=\\boxed{140}, $$ and we are done.\n\nactually tried trigbash at the start but got nowhere, then realized what the equal side lengths in cyclic quad meant\n\nprobably a faster way ig cuz i used a ton of b ash still\n</blockquote>\ncan someone explain why $\\triangle O_1XO_2\\cong\\triangle O_1YO_2$ ? it seems intuitive but i want a proper proof.",
"Hardest problem to ever appear on AIME (?)\n\n-----\n\nNotice that as $O_2A=O_2D$ , $O_1O_2$ bisects $\\angle AO_1D$ . Similarly, $O_1O_2$ bisects $\\angle BO_2C$ .\n\nLet $Q=O_1D\\cap CO_2$ and $R=O_1A\\cap BO_2$ . Then by ASA Congruence $\\triangle O_1RO_2\\cong\\triangle O_1QO_2$ , so $\\angle ARO_2=\\angle DQO_2$ .\n\nAs $\\triangle O_1QO_2\\cong\\triangle CQD$ , $\\frac{QO_2}{QD}=\\frac{QO_1}{QC}=\\frac{O_1O_2}{CD}=\\frac{15}{16}$ . Similarly, $\\frac{RO_2}{RA}=\\frac{RO_1}{RB}=\\frac{O_1O_2}{AB}=\\frac{15}{2}$ .\n\nFrom this we obtain $AR:RO_2:O_2A:DQ:QO_2:O_2D=2:15:x:16:15:x$ , as $O_2A=O_2D$ and $O_2R=O_2Q$ . If we let $A'$ be the reflection of $A$ across $O_1O_2$ , then $AO_2=A'O_2$ , and as $A$ is on $O_1R$ , $A'$ is on $O_1Q$ (as $Q$ is the reflection of $R$ across $O_1O_2$ ) and $A'Q=AR$ . Thus, $\\triangle O_2QA'\\cong\\triangle O_2RA$ this ratio can be rewritten as $A'Q:QO_2:O_2A':DQ:QO_2:O_2D=2:15:x:16:15:x$ . Looking at $\\triangle AO_2D$ alone ( $M$ is the projection of $O_2$ onto $A'D$ ):\n\n[asy]\ndraw((2,0)--(0,0)--(9,12)--(16,0)--(2,0)--(9,12)--(9,0));\nlabel(\" $Q$ \",(0,0),SW);label(\" $A'$ \",(2,0),S);label(\" $D$ \",(16,0),SE);label(\" $O_2$ \",(9,12),N);label(\" $M$ \",(9,0),S);\n[/asy]\n\nLet $r_1$ and $r_2$ be the radii of $\\omega_1$ and $\\omega_2$ respectively. We have $QA':QD:QO_2=2:16:15$ and $A'O_2=DO_2$ , so $QA':A'M:M'D:QO_2=2:7:7:15$ , so $QM:M'D:QO_2:DO_2=9:7:15:12:\\sqrt{193}$ and $\\sin\\angle O_2QD=\\frac{4}{5}$ and $\\cos\\angle O_2QD=\\frac{3}{5}$ . We know $\\angle ARO_2=\\angle DQO_2$ , so $\\sin\\angle ARO_2=\\frac{4}{5}$ . Thus \\begin{align*}[ABO_1CDO_2]&=[ABO_1O_2]+[DCO_1O_2]&=\\frac12BO_2\\cdot AO_1\\sin\\angle ARO_2+\\frac12CO_2\\cdot DO_1\\sin\\angle DQO_2&=\\frac25(AO_1\\cdot BO_2+CO_2\\cdot DO_1)&=\\frac25(AB\\cdot O_1O_2+BO_1\\cdot AO_2+CD\\cdot O_1O_2+CO_1\\cdot DO_2)\\text{ by Ptolemy's}&=\\frac25(270+2r_1r_2).\\end{align*}\n\nWe have $O_2Q:r_2=15:\\sqrt{193}$ . As $\\triangle QO_2D\\sim\\triangle QO_1C$ , $O_1Q:r_1=15:\\sqrt{193}$ . As $\\cos\\angle DQO_2=\\frac{3}{5}$ , $\\cos\\angle O_1QO_2=-\\frac{3}{5}$ . Then by Law of Cosines, $15^2=\\left(r_1\\cdot\\frac{15}{\\sqrt{193}}\\right)^2+\\left(r_2\\cdot\\frac{15}{\\sqrt{193}}\\right)^2+\\frac{6}{5}\\left(r_1\\cdot\\frac{15}{\\sqrt{193}}\\right)\\left(r_2\\cdot\\frac{15}{\\sqrt{193}}\\right)$ and $193=r_1^2+r_2^2+2r_1r_2-\\frac{4}{5}r_1r_2$ . As $r_1+r_2=15$ , $193=225-\\frac{4}{5}r_1r_2$ and $r_1r_2=40$ . Thus $[ABO_1CDO_2]=\\frac{2}{5}(270+2\\cdot40)=\\boxed{140}$ . $\\Box$ **Remark.** If $X$ is the point of tangency between $\\omega_1$ and $\\omega_2$ , $X$ is the incenter of triangles $ADO_1$ and $BCO_2$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1252,
"boxed": true,
"end_of_proof": true,
"n_reply": 26,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782914.json"
}
|
Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ .
[asy]
import geometry;
size(10cm);
draw((-2,0)--(13,0));
draw((0,4)--(10,4));
label(" $\ell_A$ ",(-2,0),W);
label(" $\ell_B$ ",(0,4),W);
point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2);
draw(B1--A1--B2);
draw(B1--A2--B2);
draw(B1--A3--B2);
label(" $A_1$ ",A1,S);
label(" $A_2$ ",A2,S);
label(" $A_3$ ",A3,S);
label(" $B_1$ ",B1,N);
label(" $B_2$ ",B2,N);
label("1",centroid(A1,B1,I1));
label("2",centroid(B1,I1,I3));
label("3",centroid(B1,B2,I3));
label("4",centroid(A1,A2,I1));
label("5",(A2+I1+I2+I3)/4);
label("6",centroid(B2,I2,I3));
label("7",centroid(A2,A3,I2));
label("8",centroid(A3,B2,I2));
dot(A1);
dot(A2);
dot(A3);
dot(B1);
dot(B2);
[/asy]
|
<details><summary>Click to expand</summary>When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points.
WLOG, let the point be on line $\ell_A$ . We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$ . Let there be $s$ points on lines $\ell_A$ and $\ell_B$ before $P$ was added. For each of the $s$ points on $\ell_B$ , we subtract the total number of lines formed, which is $s^2$ , not counting $P$ . Considering all possible points on $\ell_B$ , we get $(s^2-s)+(s^2-2s)\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$ . Simplifying, we get $s^3-s\sum_{i=1}^{s}{i}+s\Longrightarrow s(s^2-\sum_{i=1}^{s}{i}+1)$ . Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\ell_A$ .
Case 2: If a new point $P$ is added to the right of a line that has one less point than the other line.
Continuing on case one, let this point $P$ be on line $\ell_B$ . With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$ , we get $(s(s+1)-s)+(s(s+1)-2s)\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\ell_A$ . Simplying, we get $s(s+1)^2-s\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions.
For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$ . Notice that adding a final point on $\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$ , there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\cdots(s(s+1)-s(s+1))+s\Longrightarrow s^2(s+1)-(s+1)\sum_{i=1}^{s}{i}+s$ . Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=\boxed{244}$ .
.</details>
|
[
"<details><summary>Sketch</summary>Let $f(m, n)$ denote the number of bounded regions. Then $f$ satisfies the recursion $f(m+1, n) = f(m, n) + m\\binom{n}{2} + n$ . Compute $f(5, 7) = 244$ .\n\nEdit: By using $f(1, 1) = 0$ , we can derive the explicit formula as shown in the below post.</details>",
"Personally my favorite problem.\n\nLet $\\mathfrak N(m,n)$ be the number of regions as dictated in the problem, and let $\\mathfrak I(m,n)$ be the number of intersection points of lines strictly between $\\ell_A$ and $\\ell_B$ . We claim\n\\[\\mathfrak N(m,n)=\\mathfrak I(m,n)+mn-1\\]\nWe can do this by doubly induction on $n$ and then $m$ (note that we can assume $m\\geq n$ because it is symmetric). The base case is $n=1$ , in which case we have $\\mathfrak I(m,1)=0$ and $\\mathfrak N(m,1)=m-1$ , as the picture is just $m-1$ triangles, none of which overlap.\n\nAssume we want to show it for some pair $(m,n)$ , and already know that it works for $(m-1,n)$ . Without loss of generality, assume that we add the last point between $A_1$ and $A_2$ . Then, we can take a look at each triangle $A_1A_2B_k$ for $1\\leq k\\leq n$ . In particular, this triangle is divided into $\\mathfrak I+1$ regions, assuming there are $\\mathfrak I$ lines inside of the triangle. Thus, if we add this over all triangles, we get $m+\\sum\\mathfrak I$ , or $m$ more than the total number of new intersection points. This is effectively what we wanted to show, so we are done with the induction.\n\nTo count $\\mathfrak I(m,n)$ is simple - $A_wB_x$ and $A_yB_z$ intersect if and only if $(w-y)(x-z)<0$ . Thus, for any $1\\leq w<y\\leq m$ and $1\\leq z<x\\leq n$ , we get 1 intersection point. Thus,\n\\[\\mathfrak I(m,n)=\\binom m2\\binom n2\\implies\\mathfrak N(m,n)=\\binom m2\\binom n2+mn-1\\]\nThus, $\\mathfrak N(7,5)=\\boxed{244}$ .",
"O I used Euler characteristic to get it and wrote $\\binom 52=15$ oops\n\nI couldnt get recursion to work because I am bad at geometry to count number of intersection pts and new regions XD",
"35 lines to get\n0 1 1 1 1 1 1\n7 6 5 4 3 2 1\n13 11 9 7 5 3 1\n19 16 13 10 7 4 1\n25 21 17 13 9 5 1 \n\ndone",
"I just put 5 points on top and put 1,2 and 3 points on the bottom and got 4,19, and 44 so just assumed it was 5*n^2-1. Got pretty lucky it worked out",
"I used V-E+F=2 and removed the lines extending past the whole points and got 244",
"I drew the entire thing out and got 241. :(",
"<details><summary>scuffed solution</summary>I started with $(m,n)=(1,5)$ which has $4$ regions, and increasing $m$ one at a time. I counted the regions created by each new line by looking at how many of the old lines they crossed. This gave me $4+15+25+35+45+55+65=\\boxed{244}$ . \n\nTo check my answer I then started with $(m,n)=(7,1)$ and increased $n$ , giving me $6+28+49+70+91=\\boxed{244}$</details>",
"Label all of the interior intersection points with dots. Then we have a planar graph with all of the dotted points (including the $A_i$ s and the $B_i$ s). So we can use $V+F-E = 2$ to solve for $F$ . \n\nFirst, we have $V = [\\text{number of interior intersection points}] + 5 + 7$ . To count the number of interior intersection points, we can do the following:\n\nPick 2 points from $\\ell_B$ and 2 from $\\ell_A$ . Then, notice exactly one point on the interior is defined by both of these lines, and each interior point corresponds to exactly one set of 4 points. So there are $\\binom{7}{2}\\binom{5}{2} = 210$ interior points. \nThus $V = 222$ . \n\nNow for $E$ , we will calculate the number of edges going out from each point, then add those together and divide by 2 because every edge is counted twice. \n\nEach interior point has $4$ edges going out from it, for a total of $210\\cdot 4 = 840$ . $B_1$ and $B_5$ each have $8$ lines going out from them, and $B_2, B_3,$ and $B_4$ all have $9$ . $A_1$ and $A_7$ both have $6$ , and $A_2, A_3, A_4, A_5,$ and $A_6$ all have $7$ . \n\nThe total is $840 + 2\\cdot 8 + 3\\cdot 9 + 2\\cdot 6 + 5 \\cdot 7 = 930$ . So $E = \\frac{930}{2} = 465$ . \n\n\nThus $F = E+2-V = 245$ . This includes the infinite face, so we have to subtract that off. This gives $\\boxed{244}$ .",
"I drew the entire thing out, miscounting a total of <details><summary>hmm</summary>15</details> regions, attaining an answer of <details><summary>hmm</summary>$244 - 15 = 229$</details>. How sad.",
"<blockquote>Ez clapz</blockquote>\n\npls go see a doctor",
"<blockquote>Ez clapz</blockquote>\nIs grid paper allowed? :ewpu: ",
"<blockquote><blockquote>Ez clapz</blockquote>\nIs grid paper allowed? :ewpu:</blockquote>\n\nno :yaw:",
"<details><summary>Motivation</summary>I think what motivated my solution was this old MATHCOUNTS Minis video (#42) I watched probably around 5 years ago which stated that any time an intersection occurs it should create another region. \n\nI didn't want to draw everything out so I tried to find a pattern by looking at smaller values and I would eventually make a formula for any $m$ and $n$ .</details>\n<details><summary>Solution</summary>The plane starts out as 1 region. There are $mn$ lines. \n\nEach line creates another region as it \"cuts\" across the plane (unless you're talking $A_1B_1$ and $A_mB_n$ as they don't do anything). So we have $mn - 1$ regions.\n\nHowever, when two lines intersect they form another intersection as well (and max 2 lines intersect at one point). After trying $(m, n) = (3, 2), (5, 2), (5, 3), (4, 4)$ I found a few interesting discoveries (you can try them yourself by drawing the lines for each of the cases): $B_1A_k$ do not create any new intersections. (assuming we draw the lines by incrementing $n$ then $m$ ) $B_2A_k$ creates $m-1$ intersections for $k = 1$ , then $m-2$ for $k = 2$ , then $m-3$ , ..., then $1$ then $0$ for $k = m$ . $B_3A_k$ creates $2(m-1)$ intersections for $k = 1$ , then $2(m-2)$ for $k = 2$ , then $2(m-3)$ for $k = 3$ , .... $B_4A_k$ creates $3(m-1)$ intersections for $k = 1$ , then $3(m-2)$ for $k = 2$ , then $3(m-3)$ for $k = 3$ , ...\n\nSo $B_nA_k$ creates $(n-1)\\frac{m(m-1)}{2}$ intersections. Summing everything up we get $\\frac{m(m-1)(n)(n-1)}{4}$ intersections, so our final formula looks like: $$ R = (mn-2) + 1 + (\\frac{m(m-1)(n)(n-1)}{4}) $$ I wasn't able to prove it, but it worked for the 4 cases above, so I plugged in $m = 7, n = 5$ and got $\\boxed{244}$ .</details>",
"<details><summary>Solution</summary>This is trivial with $V-E+F=2$ . In order for this to be true, none of the lines can intersect, and thus every intersection point must be a vertex.. We see that $V = m+n+{m\\choose 2}{n\\choose 2}$ , and $E = mn+2{m\\choose 2}{n\\choose 2}+(m-1)+(n-1)$ , thus $F = {m\\choose 2}{n\\choose 2}+mn$ . However, this counts the external face, which we do not count, thus we see that for arbitrary m and n, we get ${m\\choose 2}{n\\choose 2}+mn-1$ . \n Thus, $m = 7$ , $n = 5$ gives $244$</details>",
"<blockquote><blockquote><blockquote>Ez clapz</blockquote>\nIs grid paper allowed? :ewpu:</blockquote>\n\nno :yaw:</blockquote>\n\noh :noo: ",
"but rulers are",
"IMO a natural way to count the edges is to see that each segment $\\overline{A_iB_j}$ contains $(i-1)(5-j) + (j-1)(7-i) = -2ij + 6i + 8j - 12$ vertices on that segment (not including endpoints $A_i$ , $B_j$ , so there are $-2ij+6i+8j-11$ edges on segment $\\overline{A_iB_j}$ . The number of edges between $\\ell_A$ and $\\ell_B$ is\n\n\\[ \\sum_{i=1}^7 \\sum_{j=1}^5 (-2ij+6i+8j-11). \\]\nRewrite $-2ij+6i+8j-11$ as $-2(i-4)(j-3)+13$ . The above sum equals\n\\[13 \\times 35 - 2\\sum_{i=1}^7 \\sum_{j=1}^5 (i-4)(j-3) = 455 - 2\\sum_{i=1}^7 (i-4) \\sum_{j=1}^5 (j-3)\\]\nThis double sum appears to magically vanish! (since $\\sum_{i=1}^7 (i-4) = (-3)+(-2)+\\ldots + 3 = 0$ ). So the number of edges $|E|$ is simply $455 + 10 = 465$ (the $+10$ accounts for the edges on $\\ell_A$ and $\\ell_B$ ).",
"<blockquote>I drew the entire thing out, miscounting a total of <details><summary>hmm</summary>15</details> regions, attaining an answer of <details><summary>hmm</summary>$244 - 15 = 229$</details>. How sad.</blockquote>\n\nsame :( but i miscounted by 2 so i ended up getting 242 as my answer",
"Name the points $A_1, A_2, \\dots, A_m$ in that order on $\\ell_{A}$ and the points $B_1, B_2, \\dots, B_n$ in that order on $\\ell_{B}$ . Suppose we draw the lines connecting $A_1$ through $A_m$ to $B_1$ through $B_{n-1}$ . \n\nNow, consider what happens when we draw $A_1B_n$ : it intersects each line with endpoints in $A_2, A_3, \\dots, A_m$ and $B_1, B_2, \\dots, B_{n-1}$ , as well as $\\ell_{B}$ , for a total of $(m-1)(n-1)+1$ new regions.\n\nNext, consider $A_2B_n$ . It intersection each line with endpoints in $A_3, A_4, \\dots, A_m$ and $B_1, B_2, \\dots, B_{n-1}$ as well as $\\ell_{B}$ , for a total of $(m-2)(n-1)+1$ new regions.\n\nIf we continue this logic, the number of new regions is \n\\[\\sum_{i=0}^{m-1}(i(n-1)+1)=m+(n-1)\\left(\\frac{m^2-m}{2}\\right).\\]\nThen our total will be \n\\begin{align*}\n\\sum_{i=1}^{n}\\left(m+(n-1)\\left(\\frac{m^2-m}{2}\\right)\\right)-1&=mn+\\frac{m^2-m}{2}\\left(\\sum_{i=1}^{n}(n-1)\\right)-1\n&=\\left(\\frac{m^2-m}{2}\\right)\\left(\\frac{n^2-n}{2}\\right)+mn-1\n&=\\binom{m}{2}\\binom{n}{2}+mn-1\n&=\\binom{7}{2}\\binom{5}{2}+35-1\n&=\\boxed{244},\n\\end{align*}\nand we are done. \n\n(The $-1$ comes from the line $A_1B_1$ , which does not add any new regions.)",
"[quote name=\"Bole\" url=\"/community/p24453228\"]\nEz clapz\n</blockquote>\n\nwas this what you did in test? must have taken a long time...",
"<blockquote>[quote name=\"Bole\" url=\"/community/p24453228\"]\nEz clapz\n</blockquote>\n\nwas this what you did in test? must have taken a long time...</blockquote>\n\nDrawing took ~15 minutes and accurately counting took ~20 minutes",
"<blockquote>Legend :heart_eyes:</blockquote>\n\ni might be dumb but the computer generated diagram looks pretty easy to count-",
"<blockquote><blockquote>Legend :heart_eyes:</blockquote>\n\ni might be dumb but the computer generated diagram looks pretty easy to count-</blockquote>\n\nIt isn't too hard, but the way they chose the points creates way too many points where more than 2 lines intersect ",
"Seeing this made me glad that I took the AIME I.\n\ni managed to solve this in like 25 minutes lel i feel like i've gotten significantly dumber since the AIME I.",
"Pretty standard recursion. IMO there was a huge jump in difficulty from 1-9 to 10-15 (except 11 and 13).",
"This reminds me of the Intro Geometry? message board? problem on counting intersections in some figure. It was also a planar graph :)",
"<blockquote><blockquote><blockquote>Legend :heart_eyes:</blockquote>\n\ni might be dumb but the computer generated diagram looks pretty easy to count-</blockquote>\n\nIt isn't too hard, but the way they chose the points creates way too many points where more than 2 lines intersect</blockquote>\n\nik, but you can just add regions for each point depending on the number of lines pass thru it",
"this is application of graph theory with E=V+F-2",
"<blockquote>This felt a little bit too hard for a #9 (probably should’ve been switched with #10 or #11), but to make it easier for people to understand I present a visual solution in the form of a light-hearted comic</blockquote>\n\nThat makes it so much easier to understand the solution",
"Redacted",
"Treat the given configuration as a graph. We first compute the number of vertices and edges, then the number of faces by Euler's characteristic.\n\nFirst, we find $V$ , the number of vertices. Observe that each intersection point $P$ in the interior of $A_1A_7B_5B_1$ corresponds to $2$ points $A_i,A_j$ on $\\overline{A_1A_7}$ and $2$ points $B_k,B_l$ on $\\overline{B_1B_5}$ , where $i<j$ and $k<l$ . This is because $A_i,A_j,B_k,B_l$ can be obtained from $P$ by looking at the two segments $\\overline{A_iB_l}$ and $\\overline{A_jB_k}$ that were drawn to intersect at $P$ , and $P$ can be obtained from $A_i,A_j,B_k,B_l$ by taking the intersection of the diagonals of $A_iA_jB_lB_k$ , so we have a bijection. Hence, the number of interior intersections is $\\tbinom{7}{2}\\tbinom{5}{2}=21\\cdot10=210$ , and adding the $A_i$ and $B_j$ themselves yields $V=210+7+5=222$ total vertices.\n\nThen, we find $E$ , the number of edges. There are $6+4=10$ edges along $\\overline{A_1A_7}$ and $\\overline{B_1B_5}$ , and $7\\cdot5=35$ segments of the form $\\overline{A_iB_j}$ . Also, each of the $210$ interior vertices contributes $2$ additional edges, creating $1$ new edge on each of the $2$ segments it lies on. Therefore, the number of total edges is $E=10+35+2\\cdot210=465$ , so by Euler's characteristic, $$ F=E-V+2=465-222+2=245. $$ Subtracting $1$ for the face outside $A_1A_7B_5B_1$ yields $\\boxed{244}$ , as requested.",
"Some related problems:\n\n\n- Into how many distinct regions can $n$ lines divide the plane into?\n- Draw $n$ points on a circle, and connect every pair of points with a line. Into how many regions can the lines divide the circle?\n\nIt's interesting to note that the latter problem, in particular, was considered especially difficult a few decades back. To quote from Paul Zeitz,\n<blockquote>\nI had first seen it in high school (the 5 circles problem is part of 'folklore'), although I had never worked out the correct formula. twenty years later, during the IMO training camp, a guest lecturer by the name of Bjorn Poonen discussed this problem, among others. Bjorn derived the formula for the numver of regions, using a tricky argument that employed fairly esoteric ideas. Yet the formula looked very simple, containing only binomial coefficients. I asked him, 'Surely something with such a combinatorial formula should have a simple combinatorial derivation. Is there one that you know of?\"' He replied that he had never found one.\n\nWhen Bjorn Poonen says that he could not do something, everyone pays attention, for he is one of the most talented young problemists in the country. He is oonly the second person in history to have placed in the top 5 on the Putnam Exam during each of his four years in college, and he composes very original, beautiful problems for a number of math contests and journals. So if someone as gifted and accomplished as Bjorn cannot do it, it must be hard.\n\nNevertheless, by the end of his lecture, one of the IMO team members had come up with a very elementary derivation, a combinatorial argument which used only the most basic ideas of the subject. we were all impressed with his brilliance...\n</blockquote>\nThat being said, however, the story goes that while conducting a problem-solving seminar, Paul Zeitz split the students into groups and gave them the problem. A few days later, his \"weakest\" group (consisting of a psychology major, an environmental studies major, and a guy who had, at the time, recently switched from majoring in history to math) was able to solve the problem in a manner similar to the IMO team member.\n\nSo yes, very motivating. Something something problem-solving can be taught something something. Big insight. 最强大脑.\n",
"[Video Solution](https://youtu.be/OWNHkKlEo2A)",
"This problem reminds me of a mathcounts target round where you looked a triangular numbers, and after I remembered that you could use recursion it was extremely straightforward.",
"Btw, I got $219$ on test since I thought $25\\cdot 6=125$ ",
"Reminds me a lot of Moser's circle problem. Wonder if there's a similar solution.",
"<blockquote>Reminds me a lot of Moser's circle problem. Wonder if there's a similar solution.</blockquote>\n\nwhat I was thinking! its also exactly what winmac posted \n\nsketch: We can use the fact that $V-E+R=2.$ there is a $1-1$ correspondence between points on the interior, and 2 intersecting lines. There is also a $1-1$ correspondence between those and quadrilaterals, so there are $210$ quadrilaterals and thus $210$ interior points. Then, we can count the number of edges using the fact that each interior point has $4$ edges coming out of it, and count the ones on the edges, then divide by $2$ for overcounting. Then, we simply plug those in to find $R,$ which gives $244$ (maybe)",
"I used Euler's Formula ( $V-E+F=1$ )\n<details><summary>Solution</summary>We will first generalize to $m$ and $n$ points on the lines, and then simply plug in $m=5,$ and $n=7.$ First, we compute the number $V.$ Obviously, there are the original $m+n$ vertices. Also, there are ${m \\choose 2}{n \\choose 2}$ ways to choose $4$ points which will make exactly one intersection. Thus, overall, there are $$ m+n+{m \\choose 2}{n \\choose 2} = m+n+\\frac{mn(m-1)(n-1)}{4} $$ vertices.\nFor the edges, obviously the lines are split into $m+n-2$ segments by the points. Now, to compute how many segments each segment connecting points on opposite lines are split into, take the endpoints of this segment to be the $i$ th and $j$ th points on the lines with $m$ and $n$ points, respectively. Then, we will compute the number of segments the segment between them is split into by computing the number of intersections it has. Any segment which intersects the segment between the $i$ th and $j$ th points on the lines must have its endpoints on opposite sides of the $i$ th and $j$ th points. On one side, we get $(i-1)(n-j)$ lines which intersect this segment, and $(m-i)(j-1)$ on the other side. So, we get $(i-1)(n-j) + (m-i)(j-1)+1$ separate segments. Summing over all $i,j,$ we get $$ n+m-2+\\sum_{i=1}^{m} \\sum_{j=1}^{n} ni+mj-2ij - n-m+1 = n+m-2\\left( \\sum_{i=1}^{m} \\sum_{j=1}^{n} (n-1)i+(m-1)j\\right) - 2 \\left(\\sum_{i=1}^{m} i \\right)\\left(\\sum_{i=1}^{m} j \\right)+mn(-n-m+1)=n+m-2\\left(\\frac{n^{2}(n-1)(n+1)}{2}+\\frac{m^{2}(m-1)(m+1)}{2}\\right) + \\left(\\frac{mn(n+1)(m+1)}{2}\\right)+mn(-n-m+1) = n+m-2 +mn+\\frac{m(m-1)n(n-1)}{2}. $$ Thus, by Euler's formula, we get $$ F = \\frac{n(n-1)m(m-1)}{4}+mn-1, $$ giving $$ F= 10 \\cdot 21 + 35-1 = 244, $$ our answer.</details>",
"We seek a recursion. Let $f(m, n)$ be the answer to the problem for general $m, n$ . We seek to find a general form for $f(m+1, n)$ in terms of $f(m, n)$ i.e. the number of regions we add by adding an extra point. It's not hard to see that this is exactly the number of intersections when drawing the $n$ new lines (including the intersections on the parallel lines). Casework on the number of intersections with each of the $n$ points gives us \n\\begin{align*}\nf(m+1, n) = f(m, n) + \\underbrace{1+(m+1)+(2m+1)+\\dots}_{\\text{n terms}} \n= f(m,n) + n + m(1+2+\\dots + n-1) \n= f(m, n) + n \\left( 1+\\frac{m(n-1)}{2} \\right)\n\\end{align*}\nTedious computation from $f(3, 2) = 8$ gives the answer of $\\boxed{244}$ ",
"Fix $7$ points at the bottom and put $m$ points at the top, as $m$ varies from $0$ to $5$ . As we add the $m+1$ th point at the top, we draw $7$ lines, which form $6m+1$ , $5m+1$ , $\\dots$ , $1$ new region(s) respectively, for a total of $21m + 7$ .\nHowever, the first line does not create a new region, so we must subtract 1.\n\nHence \\[-1 + \\sum_{m=0}^{4}(21m+7) = \\boxed{244}.\\]\n\n",
"<blockquote>I used V-E+F=2 and removed the lines extending past the whole points and got 244</blockquote>\n\nThat is what I also used, I just randomly saw a FTW problem back then and then I just remembered that formula and used it....",
"We can use recursion to tackle this problem. Assume that $f(m, n)$ denotes the number of regions formed when we have $m$ points on $\\ell_A$ and $n$ points on $\\ell_B$ . \n\nConsider adding one more point to $\\ell_B$ at the very end. It is easy to see that we have an additional $m$ sections at the end and an additional $n \\cdot \\binom{m}{2}$ internal intersections. Hence our recursion becomes,\n\\begin{align*}\nf(m, n + 1) = f(m, n) + m + n \\cdot \\binom{m}{2}\n\\end{align*}\nBeginning from $f(2, 2) = 4$ we have:\n\n\n- $f(2, 3) = f(2, 2) + 2 + 2 \\cdot \\binom{2}{2} = 8$\n- $f(3, 3) = f(3, 2) + 3 + 2 \\cdot \\binom{3}{2} = 17$\n- $f(3, 4) = f(3, 3) + 3 + 3 \\cdot \\binom{3}{2} = 29$\n- $f(4, 4) = f(4, 3) + 4 + 3 \\cdot \\binom{4}{2} = 51$\n- $f(4, 5) = f(4, 4) + 4 + 4 \\cdot \\binom{4}{2} = 79$\n- $f(5, 5) = f(5, 4) + 5 + 4 \\cdot \\binom{5}{2} = 124$\n- $f(6, 5) = f(5, 5) + 5 + 5 \\cdot \\binom{5}{2} = 179$\n- $f(5, 7) = f(5, 6) + 5 + 6 \\cdot \\binom{5}{2} = 244$\n\nfor a final answer of $\\boxed{244}$ . ",
"When I mocked this, I did a bit of casework and pattern-finding\nwhen i let m=1, i change the values of n from 1 onwards to get\nthe pattern 0,1,2,3,4...\n\nwhen i let m=2, i change the values of n from 1 onwards to get\nthe pattern 1,4,8,13...\n\nwhen i let m=3, i change the values of n from 1 onwards to get\nthe pattern 2,8,17,29...\n\nyou can kind of observe a pattern between the differences of the terms\nand when u extend this to m=7, n=5, you get the result of 244",
"<blockquote>When I mocked this, I did a bit of casework and pattern-finding\nwhen i let m=1, i change the values of n from 1 onwards to get\nthe pattern 0,1,2,3,4...\n\nwhen i let m=2, i change the values of n from 1 onwards to get\nthe pattern 1,4,8,13...\n\nwhen i let m=3, i change the values of n from 1 onwards to get\nthe pattern 2,8,17,29...\n\nyou can kind of observe a pattern between the differences of the terms\nand when u extend this to m=7, n=5, you get the result of 244</blockquote>\n\nI did something similar. I found some random pattern and hoped it would extend nicely, and it did."
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1102,
"boxed": false,
"end_of_proof": false,
"n_reply": 46,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782916.json"
}
|
Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ , $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ .
|
<blockquote>
Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\lfloor \tfrac n4\rfloor$ , $\lfloor\tfrac n5\rfloor$ , and $\lfloor\tfrac n6\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ .
</blockquote>
this solution's relatively dumb oops
Let $\lfloor\tfrac n4\rfloor = a$ , $\lfloor\tfrac n5\rfloor = b$ , $\lfloor\tfrac n6\rfloor = c$ . Then $4a\le n \le 4a+3$ , $5b\le n \le 5b+4$ , $6c\le n \le 6c + 5$ ; and if we want $n$ to be uniquely determined, we require these bounds be "tight", in the sense that there's a lot of equality. This means
\[\max(4a, 5b, 6c)=\min(4a+3, 5b+4, 6c+5)\]
so.. casework.
- $n=4a$ : $n$ clearly cannot equal $4a+3$ or $6c+5$ , as both are odd. So $4a=5b+4$ , giving $n=4$ , $24$ , $44$ , $\dots$ , $584$ -- $30$ solutions.
- $n=5b$ : it must equal $4a+3$ or $6c+5$ .
[list]
- If $4a+3$ : $n$ is $15$ , $35$ , $55$ , $\dots$ , $595$ -- $30$ solutions.
- If $6c+5$ : $n$ is $5$ , $35$ , $65$ , $\dots$ , $575$ -- $20$ solutions.
[/list]
- $n=6c$ : then $5b+4$ , so $n$ is $4$ , $34$ , $64$ , $\dots$ , $574$ -- $20$ solutions.
If not for the overlapping between $n=4a$ and $n=6c$ , or between $5b=4a+3$ and $5b=6c+3$ , we'd have $100$ as our answer. Between $4a$ and $6c$ are $n=4$ , $64$ , $94$ , $\dots$ , $544$ -- $10$ values. Between $4a+3$ and $6c+3$ are $35$ , $95$ , $155$ , $\dots$ , $575$ , so $10$ solutions. Then the answer is $100-10-10=\boxed{080}$ .
|
[
"<details><summary>Solution</summary>Let $\\lfloor \\frac{n}{4} \\rfloor = a$ , and similarly for $b$ and $c$ . We need to find the number of integers such that $n$ increases the value of at least one of $a$ , $b$ , or $c$ (from $n-1$ ), and $n+1$ also increases the value of at least one of $a$ , $b$ , or $c$ (from $n$ ). This is equivalent to finding all $n$ such that both $n$ and $n+1$ is divisible by at least one of $4$ , $5$ , and $6$ . Also note one of $n$ or $n+1$ must be odd, meaning it must be an odd multiple of $5$ .\n\nWe examine all possible $n$ less than $60$ and find $8$ solutions (being $4$ , $5$ , $15$ , $24$ , $35$ , $44$ , $54$ , and $55$ ). Since $\\text{lcm} (4, 5, 6) = 60$ , there will be $8$ solutions for every multiple of $60$ . Thus, our answer is $8 \\cdot \\frac{600}{60} = \\boxed{080}$ .</details>",
"<details><summary>basically</summary>$n$ has to be $-1$ mod (one of 4, 5, 6) and $0$ mod (4, 5, 6) for it to be determined uniquely-- can solve from here</details>",
" $n$ satisfies this condition as long as at least one of these values differ from those of $n+1$ and $n-1$ . In particular, we want that\n\\[\\left\\{\\left\\lfloor \\frac n4\\right\\rfloor,\\left\\lfloor\\frac n5\\right\\rfloor,\\left\\lfloor\\frac n6\\right\\rfloor\\right\\}\\neq\\left\\{\\left\\lfloor \\frac {n+1}4\\right\\rfloor,\\left\\lfloor\\frac {n+1}5\\right\\rfloor,\\left\\lfloor\\frac {n+1}6\\right\\rfloor\\right\\}\\]\nand similarly with $n-1$ instead of $n+1$ . This means that one of $4,5,6$ divide $n$ , and one of them divide $n+1$ , but clearly the same number can not divide both. Furthermore, $5$ divides at least one of them, as $\\gcd(n,n-1)=1$ . As\n\\[\\left\\lfloor \\frac {n+60}4\\right\\rfloor=15+\\left\\lfloor \\frac n4\\right\\rfloor\\qquad\\left\\lfloor \\frac {n+60}5\\right\\rfloor=12+\\left\\lfloor \\frac n5\\right\\rfloor\\qquad \\left\\lfloor \\frac {n+60}6\\right\\rfloor=10+\\left\\lfloor \\frac n6\\right\\rfloor\\]\nit suffices to work modulo 60 and then multiply by $10$ . Thus, it suffices to count the solutions to this. We shall assume that $5$ divides $n$ , and multiply by $2$ to count for the other case. If $4\\mid n+1$ , we get there are a total of\n\\[\\frac{60}{5\\times 4}=3\\]\nsolutions by the Chinese Remainder Theorem. If $6\\mid n+1$ , we get there are a total of\n\\[\\frac{60}{5\\times 6}=2\\]\nsolutions. Now, if both 4 and 6 divided $n+1$ (clearly one can not divide $n$ and the other $n+1$ , as $n$ and $n+1$ are relatively prime), there are a total of\n\\[\\frac{60}{5\\times \\text{lcm}(4,6)}=1\\]\nsolution. In particular, this gives $3+2-1=4$ solutions if $5\\mid n$ , and by symmetry, 4 if $5\\mid n+1$ . This implies that there are 8 solutions to this every 60 numbers.\n\nThus, we get that there are $8\\times 10=\\boxed{080}$ such numbers.",
"Mine. Inspired by me getting confused as to how weapon upgrades worked in *Hollow Knight*.\n\n<details><summary>Solution 1 (Michael Tang)</summary>Say an integer $n$ is *winnable* if it is possible to win the game starting with $n$ . Suppose $n$ is a winnable integer. Then the ordered triples $(\\lfloor\\tfrac n4\\rfloor,\\lfloor\\tfrac n5\\rfloor,\\lfloor\\tfrac n6\\rfloor)$ and $(\\lfloor\\tfrac{n-1}4\\rfloor,\\lfloor\\tfrac{n-1}5\\rfloor,\\lfloor\\tfrac{n-1}6\\rfloor)$ are not identical, and so they must differ in at least one coordinate; this means $n$ is a multiple of $4$ , $5$ , or $6$ . Analogously, $n+1$ is a multiple of $4$ , $5$ , or $6$ . Since it is impossible for both $n$ and $n+1$ to be even, one of these multiples must be $5$ . These conditions are both necessary and sufficient.\n\nAssume first that $n$ is a multiple of $5$ , so that $n\\equiv 0\\pmod 5$ . Then $n+1$ is a multiple of $4$ or $6$ (or both); in either case, $n$ is odd. There are $2$ possible remainders for $n$ upon division by $4$ and $3$ possible remainders for $n$ upon division by $6$ ; of the $2\\cdot 3 = 6$ total possibilities, four of them satisfy the requested condition. By Chinese Remainder Theorem on the coprime moduli $3$ , $4$ , and $5$ , there are $4$ values of $n$ in this case.\n\nThe case where $n+1$ is a multiple of $5$ is completely analogous and leads to another $4$ values of $n$ . This yields $8$ total winnable integers $n$ between $1$ and $60$ , and so $8\\cdot 10 = \\boxed{80}$ are winnable in the set $\\{1,2,\\ldots, 600\\}$ .</details>\n<details><summary>Solution 2 (Own)</summary>This second solution is essentially equivalent but phrased in a more visual way. I, personally, found this visual argument easier to wrap my head around, but it does have the downside of being longer and not as rigorous.\n\nFix a positive integer $n$ ; assume for now that $n$ is between $1$ and $60=\\operatorname{lcm}(4,5,6)$ . The set of integers $m$ such that $\\lfloor \\tfrac m4\\rfloor = \\lfloor\\tfrac n4\\rfloor$ is an interval of four consecutive integers; the smallest of these integers is divisible by $4$ . Similarly, the set of integers $m$ such that $\\lfloor \\tfrac m5\\rfloor = \\lfloor\\tfrac n5\\rfloor$ is an interval of five consecutive integers, the smallest of which is divisible by $5$ ; and the set of integers $m$ such that $\\lfloor \\tfrac m6\\rfloor = \\lfloor\\tfrac n6\\rfloor$ is an interval of six consecutive integers, the smallest of which is divisible by $6$ . Under this reformulation of the problem, $n$ is winnable if and only if these three intervals intersect in exactly one point.\n\nThere are two key points to observe about this configuration. First, these three intervals are guaranteed to have nontrivial intersection, since $n$ lies in all three intervals. Second the intervals of lengths $4$ and $6$ must intersect to an interval of even length, since the leftmost numbers in both intervals have the same parity. From these two observations, all relative positions of the three intervals can be determined. Indeed, fixing the position of the interval of length $6$ , there are four locations for the interval of length $4$ ; then, by the even length condition, there are two ways to place the interval of length $5$ so that all three intervals intersect at a single point. Thus, there are eight ways to position the intervals relative to each other to obtain the desired condition; these eight configurations are displayed below.\n\n[asy]\nunitsize(1cm);\ndefaultpen(linewidth(0.7));\nreal s = 0.1, unit=0.4, vert = -1.7;\nvoid drawX(pair P)\n{\ndraw((P.x-s,P.y-s)--(P.x+s,P.y+s)^^(P.x+s,P.y-s)--(P.x-s,P.y+s));\n}\nvoid drawBoldX(pair P)\n{\ndraw((P.x-s,P.y-s)--(P.x+s,P.y+s)^^(P.x+s,P.y-s)--(P.x-s,P.y+s),linewidth(1.7));\n}\nvoid stack(int a, int b, int c, pair Q)\n{\nfor(int i=0;i<=5;i=i+1)\ndrawX(Q+(i*unit,0));\n\nfor(int i=0;i<=3;i=i+1)\ndrawX(Q+((i+a)*unit,unit));\n\nfor(int i=0;i<=4;i=i+1)\ndrawX(Q+((i+b)*unit,-unit));\n\ndrawBoldX(Q+(c*unit,unit));\ndrawBoldX(Q+(c*unit,0));\ndrawBoldX(Q+(c*unit,-unit));\n}\nstack(-2,-4,0,(0,0));\nstack(-2,1,1,(4.8,0));\nstack(0,-4,0,(0,vert));\nstack(0,3,3,(4.8-2*unit,vert));\nstack(2,-2,2,(-2*unit,2*vert));\nstack(2,5,5,(4.8-4*unit,2*vert));\nstack(4,0,4,(-4*unit,3*vert));\nstack(4,5,5,(4.8-4*unit,3*vert));\n[/asy]\n\nFinally, observe that by the Chinese Remainder Theorem, each configuration above can be achieved by a unique integer $n$ . This means that $8$ values of $n$ are winnable in the set $\\{1,2,\\ldots, 60\\}$ , and so $8\\cdot 10 = \\boxed{80}$ are winnable in the set $\\{1,2,\\ldots, 600\\}$ .</details>",
"Funny solution:\n\nNote that there are several symmetries involved, notably $\\pm$ doesn't matter to some degree.\n\nSo just do the first $30 = \\frac{1}{2} lcm(4, 5, 6)$ numbers to find $4$ solutions, then take $\\frac{4}{30} \\cdot 600 = 80$ .",
"(0, 4, 0), (0, 4, 2), (0, 4, 4), (1, 0, 5), (2, 4, 0), (3, 0, 1), (3, 0, 3), (3, 0, 5) i think",
"Bashed from 0~59 and then multiplied by 10 to get 080",
"NOOO I PUT 010 :C",
"I put 081, because though 600 is indistinguishable from 601 given the three pieces of information, only 600 is within the bound $n\\le600$ .\n\nThis was a terribly worded question.",
"I thought this one was pretty obviously 81? imo both should be accepted, this wording issue could have been easily avoided too (by swapping 600 with 599 or 601, for example)",
"I may accidentay omit the \"l\" key sometimes. \n\nCall these values $a$ , $b$ , and $c$ , respectivey. \n\nWe have $4a\\le n\\le 4a+3$ , $5b\\le n\\le 5b+3$ , $6c\\le n\\le 6c+3$ . \n\nFor $n-1$ to not be in the range for at least one of these, we must have $n$ is either a multipe of $4$ , $5$ , or $6$ . For $n+1$ to not be in the range for at least one of these, we have $n+1$ is a multipe of $4,5$ or $6$ . All such $n$ work. $lcm(4,5,6)=60$ , so we work only on $n\\le 60$ and multiply by $10$ . The vaues of $n$ that work are $4,24,44, 5,15,35,55, 54$ , so $8$ total ways. Answer is $8\\cdot 10=\\boxed{080}$ . ",
"@2above and @3above: As a person who got both interpretations during the test, I think the answer is $\\boxed{080}$ . I think this is because its wording is as follows:\n\n<blockquote>Find the number of positive integers $n \\le 600$ whose value can be uniquely determined when the values of $\\left\\lfloor \\frac n4\\right\\rfloor$ , $\\left\\lfloor\\frac n5\\right\\rfloor$ , and $\\left\\lfloor\\frac n6\\right\\rfloor$ are given, where $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to the real number $x$ .</blockquote>\n\nAnd not\n<blockquote>Among the first 600 positive integers, find the number of positive integers whose value can be uniquely determined when the values of $\\left \\lfloor \\frac n4 \\right \\rfloor$ , $ \\left \\lfloor \\frac n5 \\right \\rfloor$ , and $\\left \\lfloor \\frac n6 \\right \\rfloor$ are given, where $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to the real number $x$ .</blockquote>\n\nThere is a subtle difference between them, the first says that you are given the values, and you need to find how many of them can be uniquely determined given the three alone (and not the bound $n \\leq 600$ ) and $n=600$ is not uniquely determined given those three values.\n\nWhereas the second problem implies that you only look at the first 600 positive integers, hence $n=600$ is actually uniquely determined since we're only considering the first 600 positive integers (so actually $n=600$ is actually unique among them).\n\nI do agree there is some ambiguity in the initial problem, so yeah I understand your concern.\n\nEdit @4below: Ah yes it's never stated that $n$ is a positive integer so the definition of \"uniquely determined\" isn't consistent. Nice document! Although it's a technical issue, I also hope 80 and 81 are both accepted as valid answers.",
"I also put $81,$ I assumed that it meant within the bounds of $[0,600].$ ",
"News flash: $600$ does not work.\n\nHere is my very very long solution lol\n\nSuppose $(x,y,z)=(\\lfloor \\tfrac{n}{4} \\rfloor, \\lfloor \\tfrac{n}{5} \\rfloor, \\lfloor \\tfrac{n}{6} \\rfloor)$ . We desire all integers $n$ such that the ordered pairs for $n-1$ and $n+1$ are distinct from $n$ : this happens when $\\mod a$ is $0$ and $\\mod b$ is 1 for any $a,b \\in \\{4,5,6 \\}$ and $a \\neq b$ . Clearly only the pairings $(4,5)$ and $(5,6)$ work, yielding the solution equations $$ 15+20(x-1) $$ $$ 4+20(x-1) $$ $$ 24+30(x-1) $$ $$ 5+30(x-1) $$ where $x$ is a positive integer. Clearly there is overlap between two pairs of equations, so some computation yields the final answer to be $$ (30+20-10)+(20+30-10) \\implies \\boxed{080} $$ ",
"You also have to consider 600 thos since although it wouldn't be a solution if the range was bigger than 600, because it's the max, letting n = 600 gives us a solution\nhence, answer must be 81",
"<blockquote>You also have to consider 600 thos since although it wouldn't be a solution if the range was bigger than 600, because it's the max, letting n = 600 gives us a solution\nhence, answer must be 81</blockquote>\n\nI agree with this reasoning. I emailed the MAA with a dispute about this problem and attached the following argument for both 80 and 81 to be accepted.\n\nUPDATE: Shortly after my email, someone at MAA replied, acknowledging my dispute and saying that MAA will get back with a decision. As of 2/21/22, I have not heard from MAA again about this.\n\nUPDATE: 80 and 81 are now both acceptable answers. See post #34 by djmathman.\n\n**Attachments:**\n\n[AIME_II_2022_Problem_Dispute_Names_Obscured.pdf](https://cdn.artofproblemsolving.com/attachments/0/4/670dfc2c061a877f109cd8b12ead103fbbc68f.pdf)",
"I agree, since it set a restriction on the value of $n.$ It doesn't really make sense how $600$ isn't accepted because there is a value OUTSIDE the range specified, that causes it to be false.",
"rip sillied and got 070",
"<blockquote>You also have to consider 600 thos since although it wouldn't be a solution if the range was bigger than 600, because it's the max, letting n = 600 gives us a solution\nhence, answer must be 81</blockquote>\n\nThis reasoning is still not correct since it does not say that $n \\le 600$ would be given as information along with the other values. It just states a restriction for your final answer.",
"it doesn't say n is a positive integer would be information either, so why assume that?\n\n~~the real answer is 1~~",
"~~this is why you spend more time reading books and getting better at reading comprehension~~\n\nbtw the condition was there because otherwise you would have infinite solutions (i know its amazing omg!)",
"<blockquote>it doesn't say n is a positive integer would be information either, so why assume that?\n\n~~the real answer is 1~~</blockquote>\n\nThis is actually a very good argument for why it should be 81. Still, I put 80 since it seems like the answer they would want.",
"This is such a bad and ambiguous question. I put 81 because 600 is unique within the range.",
"<blockquote>I agree, since it set a restriction on the value of $n.$ It doesn't really make sense how $600$ isn't accepted because there is a value OUTSIDE the range specified, that causes it to be false.</blockquote>\n\n~~Ah yes 5 and 5.1 yield the same result so 5 is not accepted.~~",
"<blockquote><blockquote>You also have to consider 600 thos since although it wouldn't be a solution if the range was bigger than 600, because it's the max, letting n = 600 gives us a solution\nhence, answer must be 81</blockquote>\n\nI agree with this reasoning. I emailed the MAA with a dispute about this problem and attached the following argument for both 80 and 81 to be accepted.</blockquote>\n\nDid they reply??? The aops aime section said that the answer is 80. Hope they accepted both tho",
"<details><summary>Casework Solution: Organize Using Modular Arithmetic</summary>Before we begin, define $\\left\\lfloor \\frac n4\\right\\rfloor = a, \\ \\left\\lfloor\\frac n5\\right\\rfloor = b, \\ $ and $\\left\\lfloor\\frac n6\\right\\rfloor = c$ Suppose that $\\lfloor \\frac{n}{4} \\rfloor$ and $\\lfloor \\frac{n}{6} \\rfloor$ uniquely determined $n.$ Then, since $$ 4a \\le n \\le 4a+3, \\ \\ 6c \\le n \\le 6c+5, $$ this either implies that $4a + 3 = 6c = n$ or $6c+5 = 4a = n$ since $n$ is uniquely determined, both of which are impossible due to the fact that $4$ and $6$ are both even numbers. Therefore, we proceed as follows.\n\n\nSuppose that $\\left\\lfloor \\frac{n}{4} \\right\\rfloor$ and $\\left\\lfloor \\frac{n}{5} \\right\\rfloor$ uniquely determine $n.$ Then notice that we have $$ 4a \\le n \\le 4a+3, \\ \\ 5b \\le n \\le 5b+4, $$ and since $n$ is uniquely determined by these two conditions, we have that either $n = 4a+3 = 5b$ , or that $n = 4a = 5b+4.$ Taking the first case, we get that $a \\equiv 3 \\pmod{5}$ which gives $\\boxed{n \\equiv 15 \\pmod{20}}$ for $40$ solutions. The second case gives that $b \\equiv 0 \\pmod{4}, \\ \\ \\implies \\ \\ \\boxed{n \\equiv 4 \\pmod{20}}$ for a total of another $40$ solutions.\n\nSuppose that $\\left\\lfloor \\frac{n}{5} \\right\\rfloor$ and $\\left\\lfloor \\frac{n}{6} \\right\\rfloor$ uniquely determine $n.$ Then notice that we have $$ 5b \\le n \\le 5b+4, \\ \\ 6c \\le n \\le 6c+5, $$ and since $n$ is uniquely determined by these two conditions, we have that either $n = 5b+4 = 6c$ , or that $n = 5b = 6c+5.$ Taking the first case, we get that $b \\equiv 4 \\pmod{6}$ which gives $\\boxed{n \\equiv 24 \\pmod{30}}$ for another $20$ solutions. The second case gives that $c \\equiv 0 \\pmod{5}, \\ \\ \\implies \\ \\ \\boxed{n \\equiv 5 \\pmod{30}}$ for a total of another $20$ solutions.\n\nTherefore, our total number of solutions that we have counted so far is $$ 40+40+20+20 = 100. $$ Now, we need to check for overcounting in our four solutions sets $$ \\{15 \\pmod{20} \\} $$ $$ \\{ 4 \\pmod{20} \\} $$ $$ \\{ 24 \\pmod{30} \\} $$ $$ \\{5 \\pmod{30} \\} $$ Isolating these solution classes by modular 2, notice that $$ \\{15 \\pmod{20}, \\ \\ 5 \\pmod{30} \\} \\ \\ \\implies \\ \\ \\{35 \\pmod{60} \\} $$ which means we overcounted $10$ solutions, and that $$ \\{ 4 \\pmod{20}, \\ \\ 24 \\pmod{30} \\} \\ \\ \\implies \\ \\ \\{24 \\pmod{60} \\} $$ for a total of another overcounted $10$ solutions. Therefore, the answer is $$ 100 - 10 - 10 = \\boxed{080}. $$ Suprised that noone saw this. I would argue though that $\\boxed{081}$ is the correct answer though. Since you know that $n \\le 600,$ when $\\left\\lfloor \\frac n4\\right\\rfloor = 150, \\ \\left\\lfloor\\frac n5\\right\\rfloor = 120, \\ $ and $\\left\\lfloor\\frac n6\\right\\rfloor = 100$ you can uniquely determine that $n = 600.$</details>",
"<blockquote><blockquote>You also have to consider 600 thos since although it wouldn't be a solution if the range was bigger than 600, because it's the max, letting n = 600 gives us a solution\nhence, answer must be 81</blockquote>\n\nI agree with this reasoning. I emailed the MAA with a dispute about this problem and attached the following argument for both 80 and 81 to be accepted.</blockquote>\n\nI also emailed them with a PDF I wrote similarly explaining why 081 should be the answer. I argued that if they were to accept only one answer, 081 would be correct, but it would also be reasonable to accept both 080 and 081 due to ambiguity in the problem. Anyone else who put 081 should consider writing to them (pretty sure the proper email for this is amcinfo@maa.org) with your dispute as well because it should make our case stronger if they know that many people have the same argument. The main idea behind it is that if it is assumed that \"n is a positive integer\" is given info, then it is only logical that \"n<=600\" must also be given, resulting in an answer of 80 + 1 = 081. The problem doesn't make it explicitly clear whether or not \"n<=600\" is given info or not, so this makes the most sense. Anyways, anyone else with this problem should email them and can use this as their general idea.",
"<blockquote>This is such a bad and ambiguous question. I put 81 because 600 is unique within the range.</blockquote>\n\nI agree; read above",
"<blockquote>I also put $81,$ I assumed that it meant within the bounds of $[0,600].$ </blockquote>\n\ncheck above",
"When I did this problem I assumed that the problem solver looking for $n$ given the values of $\\left\\lfloor \\frac n4\\right\\rfloor$ , $\\left\\lfloor\\frac n5\\right\\rfloor$ , and $\\left\\lfloor\\frac n6\\right\\rfloor$ doesn't know that $n$ must be less than or equal to 600. The key to the dispute is whether or not the problem solver knows this restriction. \n\nimo both 080 and 081 should be considered correct.",
"<blockquote><blockquote>You also have to consider 600 thos since although it wouldn't be a solution if the range was bigger than 600, because it's the max, letting n = 600 gives us a solution\nhence, answer must be 81</blockquote>\n\nI agree with this reasoning. I emailed the MAA with a dispute about this problem and attached the following argument for both 80 and 81 to be accepted.\n\nUPDATE: Shortly after my email, someone at MAA replied, acknowledging my dispute and saying that MAA will get back with a decision. As of 2/21/22, I have not heard from MAA again about this.</blockquote>\nAny update yet?\n\n",
"@above\nNo further update yet.",
"According to the [MAA website](\"https://www.maa.org/math-competitions\"), 080 and 081 are now both considered acceptable answers to this problem.\n\nApologies for the issues with this problem; this was not my original wording (mine involved winning a \"popular new computer game\"), but said original wording was terrible enough that further changes were necessary.",
"That's alright, glad to hear that both answers are now accepted :)",
"By experimenting, we can see that $n$ has to be a multiple of $4$ , $5$ , or $6$ in order for it to be uniquely determined by $\\lfloor\\tfrac{n}{4}\\rfloor$ , $\\lfloor\\tfrac{n}{5}\\rfloor$ , and $\\lfloor\\tfrac{n}{6}\\rfloor$ , since otherwise these values are the same as those for the largest multiple of $4$ , $5$ , or $6$ below it. Furthermore, if $n+1$ isn't a multiple of $4$ , $5$ , or $6$ , these values are the same as those for $n+1$ , so the given conditions hold iff $n$ and $n+1$ are both multiples of $4$ , $5$ , or $6$ . Listing out the multiples of $4$ , $5$ , and $6$ from $1$ to $60$ , we obtain $8$ such values, and we have the same number of such values in each of the intervals $[61,120]$ , $[121,180]$ and so on since $60$ is a multiple of all of $4$ , $5$ , and $6$ . The requested answer is then $8\\cdot\\tfrac{600}{60}=\\boxed{80}$ .",
"[Video Solution](https://youtu.be/mW9YQPNZqQg)",
"This problem you just have to list the multiples of $4, 5, 6$ below $\\gcd(4, 5, 6)$ and find the number of consecutive pairs. There are $8$ of the, so multiply by $10$ gives $\\boxed{80}$ . ",
"[quote name=\"naman12\" url=\"/community/p24447128\"]\nFind the number of positive integers $n \\le 600$ whose value can be uniquely determined when the values of $\\left\\lfloor \\frac n4\\right\\rfloor$ , $\\left\\lfloor\\frac n5\\right\\rfloor$ , and $\\left\\lfloor\\frac n6\\right\\rfloor$ are given, where $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to the real number $x$ .\n</blockquote>\nextremely long solution oops\n\nWe do a little experimenting. Note that out of the first 20 integers, only 4, 5, 15 work. Why? Because 4 is a multiple of 4 and one less than a multiple of 5, 5 is a multiple of 5 but one less than a multiple of 6, and 15 is a multiple of 5 but one less than a multiple of 4. To see why this, and only this works, observe that if we let $\\left\\lfloor\\frac{n}{4}\\right\\rfloor=a$ , $\\left\\lfloor\\frac{n}{5}\\right\\rfloor=b$ , and $\\left\\lfloor\\frac{n}{6}\\right\\rfloor=c$ , then $4a\\leq n\\leq 4a+3$ , $5b\\leq n\\leq 5b+4$ , and $6c\\leq n\\leq 6c+5$ . So if $n$ is a multiple of 5 and 1 less than a multiple of 4, then $n=4a+3=5b$ , so the bounds force $n$ to be $n$ , if you get what I mean.\n\nAlright, time to start counting. Note that $n$ cannot be a multiple of 4 and 1 less than a multiple of 6 since then $n$ would be odd, which is a contradiction. We have 4 cases:\n\n<span style=\"color:#00f\">**Case 1:**</span> $n$ is a multiple of 4 and one less than a multiple of 5. The $n$ that work are\n\\begin{align*}\n4&=5\\cdot 1-1\n24&=5\\cdot5-1\n44&=5\\cdot9-1\n&\\vdots\n585&=5\\cdot 117-1.\n\\end{align*}\nThere are 30 of these.\n\n<span style=\"color:#00f\">**Case 2:**</span> $n$ is a multiple of 6 and one less than a multiple of 5. The $n$ that work are\n\\begin{align*}\n24&=5\\cdot5-1\n54&=5\\cdot11-1\n84&=5\\cdot17-1\n&\\vdots\n594&=5\\cdot119-1.\n\\end{align*}\nHowever, the one's that start with even numbers are overcounted, so we're actually looking at $54$ , $114$ , $\\dots$ , $594$ . There are 10 of these.\n\n<span style=\"color:#00f\">**Case 3:**</span> $n$ is a multiple of 5 and one less than a multiple of 4. The $n$ that work are\n\\begin{align*}\n15&=4\\cdot4-1\n35&=4\\cdot9-1\n55&=4\\cdot14-1\n&\\dots\n595&=4\\cdot 149-1.\n\\end{align*}\nThere are 30 of these.\n\n<span style=\"color:#00f\">**Case 4:**</span> $n$ is a multiple of 5 and one less than a multiple of 6. The $n$ that work are\n\\begin{align*}\n5&=6\\cdot1-1\n35&=6\\cdot6-1\n65&=6\\cdot11-1\n&\\dots\n575&=6\\cdot 96-1.\n\\end{align*}\nHowever, the ones that start with an odd number are overcounted, so we're actually looking at $5$ , $65$ , $\\dots$ , $545$ . There are 10 of these.\n\nThe answer is then $30+10+30+10=\\boxed{080}$ .",
"[quote name=\"naman12\" url=\"/community/p24447128\"]\nFind the number of positive integers $n \\le 600$ whose value can be uniquely determined when the values of $\\left\\lfloor \\frac n4\\right\\rfloor$ , $\\left\\lfloor\\frac n5\\right\\rfloor$ , and $\\left\\lfloor\\frac n6\\right\\rfloor$ are given, where $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to the real number $x$ .\n</blockquote>\nextremely long solution oops\n\nWe do a little experimenting. Note that out of the first 20 integers, only 4, 5, 15 work. Why? Because 4 is a multiple of 4 and one less than a multiple of 5, 5 is a multiple of 5 but one less than a multiple of 6, and 15 is a multiple of 5 but one less than a multiple of 4. To see why this, and only this works, observe that if we let $\\left\\lfloor\\frac{n}{4}\\right\\rfloor=a$ , $\\left\\lfloor\\frac{n}{5}\\right\\rfloor=b$ , and $\\left\\lfloor\\frac{n}{6}\\right\\rfloor=c$ , then $4a\\leq n\\leq 4a+3$ , $5b\\leq n\\leq 5b+4$ , and $6c\\leq n\\leq 6c+5$ . So if $n$ is a multiple of 5 and 1 less than a multiple of 4, then $n=4a+3=5b$ , so the bounds force $n$ to be $n$ , if you get what I mean.\n\nAlright, time to start counting. Note that $n$ cannot be a multiple of 4 and 1 less than a multiple of 6 since then $n$ would be odd, which is a contradiction. We have 4 cases:\n\n<span style=\"color:#00f\">**Case 1:**</span> $n$ is a multiple of 4 and one less than a multiple of 5. The $n$ that work are\n\\begin{align*}\n4&=5\\cdot 1-1\n24&=5\\cdot5-1\n44&=5\\cdot9-1\n&\\vdots\n585&=5\\cdot 117-1.\n\\end{align*}\nThere are 30 of these.\n\n<span style=\"color:#00f\">**Case 2:**</span> $n$ is a multiple of 6 and one less than a multiple of 5. The $n$ that work are\n\\begin{align*}\n24&=5\\cdot5-1\n54&=5\\cdot11-1\n84&=5\\cdot17-1\n&\\vdots\n594&=5\\cdot119-1.\n\\end{align*}\nHowever, the one's that start with even numbers are overcounted, so we're actually looking at $54$ , $114$ , $\\dots$ , $594$ . There are 10 of these.\n\n<span style=\"color:#00f\">**Case 3:**</span> $n$ is a multiple of 5 and one less than a multiple of 4. The $n$ that work are\n\\begin{align*}\n15&=4\\cdot4-1\n35&=4\\cdot9-1\n55&=4\\cdot14-1\n&\\dots\n595&=4\\cdot 149-1.\n\\end{align*}\nThere are 30 of these.\n\n<span style=\"color:#00f\">**Case 4:**</span> $n$ is a multiple of 5 and one less than a multiple of 6. The $n$ that work are\n\\begin{align*}\n5&=6\\cdot1-1\n35&=6\\cdot6-1\n65&=6\\cdot11-1\n&\\dots\n575&=6\\cdot 96-1.\n\\end{align*}\nHowever, the ones that start with an odd number are overcounted, so we're actually looking at $5$ , $65$ , $\\dots$ , $545$ . There are 10 of these.\n\nThe answer is then $30+10+30+10=\\boxed{080}$ .",
"pretty sure this repeats every $\\text{lcm}(4,5,6)=60$ values\nso we can just focus on that part and multiply by 10",
"<details><summary>Solution</summary>The key thing to notice in this problem is that $n$ is uniquely determined if and only if $k \\lfloor \\frac{n}{k} \\rfloor$ for $k=4,5,6$ contains consecutive values. Checking for $n \\le 60,$ we find $\\frac{600}{60} \\cdot 8 = 80$ to be our answer</details>"
] |
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"2022 Contests",
"2022 AIME Problems"
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{
"answer_score": 1148,
"boxed": false,
"end_of_proof": false,
"n_reply": 42,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782918.json"
}
|
A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.
|
[hide = MAA Should be Replaced as the "Misplaced Armadillo Association"!]
First notice that $DO_1$ is a straight line because $DXY$ is an isosceles triangle(or you can realize it by symmetry). That means, because $DO_1$ is a straight line, so angle $BDO_2$ = angle $ADO_1,$ triangle $ADO_1$ is similar to triangle $BDO_2$ . Also name $DO_2 = x$ . By our similar triangles, $\frac{BO_2}{AO_1} = \frac{1}{4} = \frac{x}{x+30}$ . Solving we get $x = 10 = DO_2$ . Pythagorean Theorem on triangle $DBO_2$ shows $BD = \sqrt{10^2 - 6^2} = 8$ . By similar triangles, $DA = 4 \cdot 8 = 32$ which means $AB = DA - DB = 32 - 8 = 24$ . Because $BE = CE = AE, AB = 2 \cdot BE = 24$ . $BE = 12,$ which means $CE = 12$ . $CD = DO_2$ (its value found earlier in this solution) + $CO_2$ ( $O_2$ 's radius) $= 10 + 6 = 16$ . The area of $DEF$ is $\frac{1}{2} \cdot CD \cdot EF = CD \cdot CE$ (because $CE$ is $\tfrac{1}{2}$ of $EF$ ) $= 16 \cdot 12 = 192$ .
[/hide]
|
[
"wow so misplaced",
"This looks so easy ... I didn't do it ofc but ...\n\nP7?\n\nMake up your mind, MAA. P10 on the AIME I and now this?",
"Bruh..........",
"lol what\ntwo 3-4-5's similar triangles 16*12=192",
"this would make problem 7 on an amc 10/12",
"Who else did coordbash",
"Overkilled by incircle and excircle.",
"[quote name=\"Hayabusa1\" url=\"/community/p24467164\"]\nOverkilled by incircle and excircle.\n</blockquote>\n\nha yeah. me thinking fact 5 was needed for the first 3 minutes of looking at this problem :eyes:",
"Two minutes is enough to kill this problem",
"I just wanted to remark that the internal tangent line bisects both line segments with one external tangent line on both circle, because the internal tangent line is the radical axis of both circles. This makes computing the final answer much quicker. I don't feel the need to post a solution, as Post #7 has a solution already in their diagram.",
"Most misplaced problem on this AIME. Two similar triangles killed this problem :)",
"this seems like an amc 10 problem 13"
] |
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"2022 Contests",
"2022 AIME Problems"
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{
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782923.json"
}
|
Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.
|
Let $T_b$ and $T_a$ denote the number of people at the concert before and after the bus's arrival, respectively. Since $\tfrac{11}{25}T_a$ is the number of adults after the bus's arrival, $T_a$ is a multiple of $25$ . But $T_a-T_b=50$ , the number of people on the bus, is also a multiple of $25$ , so $T_b$ is a multiple of $25$ . Furthermore, as $\tfrac{5}{12}T_b$ is the number of adults before the bus's arrival, $T_b$ is a multiple of $12$ as well. Therefore, $T_b$ is at least $\text{lcm}(12,25)=300$ , $T_a$ is at least $300+50=350$ , and the number of adults after the bus's arrival is at least $\tfrac{11}{25}T_a=\boxed{154}$ , as requested.
|
[
"<details><summary>AAAAAAAAAAAA</summary>$x = 0 \\pmod{12}$ $x + 50 = 0 \\pmod{25} \\to x = 0 \\pmod{25}$ $x = 0 \\pmod{300} \\to x = 300$ $\\text{adults}_0 = 125$ $\\text{adults} = \\boxed{154}$</details>",
"Technically we could have had an initial crowd of $0$ , followed by $22$ adults out of $50$ arrive, but the problem does say \" $50$ MORE people arrived.\"",
"125x, 175x\ny, 50-y\n\nsolve to get y = 7x+22 we want to minimize 125x+y=132x+22 so 154",
"I put 350 I am such an idiot\n\nHC9's_brain.exe has stopped responding",
"<details><summary>My Solution</summary>Let the number of people in the crowd be $N$ , and the number of new adults from the bus be $a$ . Then,\n\\[ \\frac{5}{12}N + a = \\frac{11}{25}(N+50) = \\frac{11}{25}N + 22 \\iff a = \\frac{7}{300}N + 22. \\] Clearly $N \\geq 300$ , so take $N=300$ to obtain $\\frac{5}{12}N=125$ . Hence, the number of adults is $125+7+22 = \\boxed{154}$ .</details>",
"The total number of people at the end must be $25x$ for some positive integer $x$ , and the number of people before must be $25(x-2)$ , so the initial number of people is a multiple of $25$ . Clearly $$ \\text{lcm} (12,25) = 300 $$ Therefore $x=14$ works, or $$ 14 \\cdot 11 = \\boxed{154} $$ ",
"Somehow the USAMO qualifier from our school got $500$ ._.",
"I attempted this and got 286 but realized my mistake\n\n<details><summary>my sol</summary>We can make the fraction 5y/12y, in which y can be any number\nWe can then add 50-x to the numerator and 50 in the denominator\n\nWe get that\n(5y + 50 -x)/(12y + 50)\n\n\n\nequals 11/25, we can cross multiply and put the variables on one side:\n\n25x + 7y = 700\n\nWe want y to be as small as possible, and x and y are integers, so we can have y = 25, so x = 21.\nWe can substitute that in the numerator\n5 * 25 + 50 - 21 = 125 + 50 - 21 = 154</details>",
"<blockquote>I put 350 I am such an idiot\n\nHC9's_brain.exe has stopped responding</blockquote>\n\nExit or Wait for the program to respond",
"<blockquote><blockquote>I put 350 I am such an idiot\n\nHC9's_brain.exe has stopped responding</blockquote>\n\nExit or Wait for the program to respond</blockquote>\n\nexit :)",
"<blockquote><blockquote><blockquote>I put 350 I am such an idiot\n\nHC9's_brain.exe has stopped responding</blockquote>\n\nExit or Wait for the program to respond</blockquote>\n\nexit :)</blockquote>\n\nYou do not have permission to access this program. Only the user of this device may choose to exit HC9's_brain.exe.",
"<blockquote>Technically we could have had an initial crowd of $0$ </blockquote>\n\nThat's not a crowd of people.",
"<blockquote><blockquote>Technically we could have had an initial crowd of $0$ </blockquote>\n\nThat's not a crowd of people.</blockquote>\n\nTake a crowd of people. Remove one person, and you still have a crowd of people.\n\nThree's a crowd. Remove one person and get two people, which is a crowd by definition. Repeat twice to get zero people, which is a crowd by induction. $\\blacksquare$ ",
"Let there be $5x$ adults and $7x$ kids. \n\nSo after $50$ (which is a multiple of 25) more people arrived, the population must be a multiple of $25$ . So $25\\mid 12x\\implies 25\\mid x$ . \n\nFrom \" $50$ **more** people\", we know that $x>0$ . Let $x=25$ . Then there are $12x+50=350$ people after the bus arrived. So the answer is $\\frac{11}{25}\\cdot 350=11\\cdot 14=\\boxed{154}$ >",
"<details><summary>Solution</summary>Let $x$ equal the number of people at first and $x+50$ represent the number of people after the bus arrives. Since $x\\equiv 0 \\mod 12$ and $x+50\\equiv x\\equiv 0 \\mod 25$ , we just want to find the least number that is both a multiple of $12$ and $25$ , which is just $12\\cdot 25=300$ . Therefore, the number of people at the concert after the bus arrives is $300+50=350$ . Finally, the amount of adults at this time is just $300\\cdot \\frac{11}{25}=14\\cdot 11=\\boxed{154}.$</details>",
"Imagine sillying this and putting 350 :(((\n",
"<details><summary>sol</summary>Let $5a$ be the number of adults at the party, and $x$ be the number of adults that come on the bus.\n\nWe have the equation $\\frac{5a + x}{12a + 50}=\\frac{11}{25}$ . That simplifies to $25x = 7a + 550$ .\n\nClearly $7a$ has to be a multiple of $25$ so $a = 25$ and $x= 7 + 22 = 29$ . Our answer is $5a + x = 5\\cdot 25 + 29 = 154$ .</details>\n\nBut of course I am a clown and put $5a=125$ instead of $5a + x=154$ as the answer.",
"<blockquote><details><summary>sol</summary>Let $5a$ be the number of adults at the party, and $x$ be the number of adults that come on the bus.\n\nWe have the equation $\\frac{5a + x}{12a + 50}=\\frac{11}{25}$ . That simplifies to $25x = 7a + 550$ .\n\nClearly $7a$ has to be a multiple of $25$ so $a = 25$ and $x= 7 + 22 = 29$ . Our answer is $5a + x = 5\\cdot 25 + 29 = 154$ .</details>\n\nBut of course I am a clown and put $5a=125$ instead of $5a + x=154$ as the answer.</blockquote>\nthats exactly what I did when mocking it, but I got 286 on the first try then got 154",
"Only problem I could solve on this AIME while trying out the problems for fun :/\n\n<details><summary>Sol</summary>Let their be a people at the concert initially, then 5/12 of those people are adults, so a is a multiple of 12. \nBy a similar logic, (a+50) is a multiple of 25, so a is a multiple of 25.\nThis means the least value of a is lcm(12, 25)=300.\nThen there would be (11/25)(a+50)=154 adults after the bus arrived.</details>",
"[quote name=\"HumanCalculator9\" url=\"/community/p24447836\"]\nI put 350 I am such an idiot\n\nHC9's_brain.exe has stopped responding\n</blockquote>\n\nI did that too! Luckily I caught my mistake while double checking. But then I changed 8 from 080 to 070 so :shrug:",
"<blockquote>[quote name=\"HumanCalculator9\" url=\"/community/p24447836\"]\nI put 350 I am such an idiot\n\nHC9's_brain.exe has stopped responding\n</blockquote>\n\nI did that too! Luckily I caught my mistake while double checking. But then I changed 8 from 080 to 070 so :shrug:</blockquote>\n\nInvalid response. Please select Exit or Wait for the program to respond.",
"Let there be $x$ total people at the party at the beginning. We know that $x \\equiv 0 \\pmod {12}$ and $x \\equiv -50 \\equiv 0 \\pmod {25}$ . $lcm(12, 25) = 300$ , so $\\frac{350\\cdot11}{25}=154$ .",
"I like the concert theme. Very classical and sophisticated :gleam: ",
"so we hafe 5/12N=125 so 125+7+22=154",
"<blockquote><blockquote><blockquote>Technically we could have had an initial crowd of $0$ </blockquote>\n\nThat's not a crowd of people.</blockquote>\n\nTake a crowd of people. Remove one person, and you still have a crowd of people.\n\nThree's a crowd. Remove one person and get two people, which is a crowd by definition. Repeat twice to get zero people, which is a crowd by induction. $\\blacksquare$ </blockquote>\n\nWait that's actually so true\nAt what point does a crown neq a crowd?\n\nSus\n\nIs there a def, like <50% of the original or smth..? if the original is considered an abundant crowd and percentage diminishes with lack of abundance",
"As the no. Of adults after and before the arrival of bus is an integer \nThen let X be the no. Of crowd \nNow 5x/12 is an integer when X=12 \nBut \n11(x+50)/25 is also an integer and 12 and 25 do not share any common factor \nX=12.25\nHence before the arrival of bus adults are \n5(12.25)/12 =125 \nAnd after the arrival \nIs 11(12.25+50)/25 =154 !!\n"
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"2022 Contests",
"2022 AIME Problems"
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{
"answer_score": 1034,
"boxed": false,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782926.json"
}
|
Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$
|
Denote by $P$ the midpoint of $\overline{CD}$ . The key claim is as follows.**<span style="color:#f00">Claim:</span>** $\triangle ABP\sim \triangle APD\sim\triangle PCD$ *Proof.* Suppose lines $AB$ and $CD$ meet at $E$ . Since two angle bisectors of $\triangle ADE$ intersect at $P$ , the third one also passes through $P$ . Hence, $\overrightarrow{EP}$ is the angle bisector of $\angle AED$ . Now, looking at $\triangle EBC$ , we see that $\overline{EP}$ is both an angle bisector and a median, so $\triangle EBC$ must be isosceles.
Now, let $\angle BAD = 2\theta$ and $\angle CAD = 2\phi$ . We compute $\angle AED = 180^\circ - 2\theta-2\phi$ and consequently $\angle EBC = \angle ECB = \theta + \phi$ . Hence, $\angle ABP = \angle PCD = 180^\circ - \theta - \phi$ . Also, $\angle BAP = \theta$ and $\angle CDP = \phi$ , so $\angle BPA = \phi$ and $\angle DPC = \theta$ . Hence, $\triangle ABP\sim\triangle PCD$ by AA similarity. Furthermore, $\angle PAD = \theta$ and $\angle PDA = \phi$ , so these two triangles are also similar to $\triangle APD$ by AA, as desired $\square$ Let $BP = CP = x$ . From similarity, we have $\frac{AB}{x} = \frac{x}{BC}\implies x = \sqrt{AB\cdot BC} = \sqrt{6}$ . Now, let $a = AP$ and $b = BP$ . Similarity gives $\frac{AB}{a} = \frac{a}{AD}$ and $\frac{CD}{b} = \frac{b}{AD}$ . Hence, $a = \sqrt{AB\cdot AD} = \sqrt{14}$ and $b = \sqrt{CD\cdot AD} = \sqrt{21}$ . Now, Heron's gives $[ABP] = \sqrt{5}$ . By similarity, we have $[ADP] = (\sqrt{14}/2)^2[ABP]$ and $[PCD] = (\sqrt{3/2})^2[ABP]$ . Thus, summing the three areas, we obtain
\[ [ABCD] = \left(1 + (\sqrt{14}/2)^2 + (\sqrt{3/2})^2\right)\cdot \sqrt{5} = \sqrt{180},\]
and squaring gives $\boxed{180}$ .
|
[
"Mine.\n\nHere is a very \"European\" styled solution. There exists a very \"American\" styled solution, too, that I came up with first; I'll let someone else post that method.\n\n<details><summary>Solution</summary>Let $E$ and $F$ denote the reflections of $B$ and $C$ across $AM$ and $DM$ , respectively, where $M$ is the midpoint of $\\overline{BC}$ . Note that, since $AM$ and $DM$ are angle bisectors, $E$ and $F$ lie on segment $\\overline{AD}$ . Then $AE = 2$ and $DF = 3$ , so $EF = 2$ . Furthermore, $ME = MF$ by the definition of reflection, so \n\\[\n\\angle ABM = \\angle AEM = \\angle MFD = \\angle MCD\n\\]\nand, if $N$ is the foot of the altitude from $M$ to $\\overline{AD}$ , $EN = NF = 1$ .\n\nNow let $\\angle BAM = \\angle MAD =: \\alpha$ , $\\angle CDM = \\angle MDA =: \\beta$ , and $\\angle ABC = \\angle BCD =: \\gamma$ . Since the sum of the measures of the angles of a quadrilateral is $360^\\circ$ , $\\alpha + \\beta + \\gamma = 180^\\circ$ . This implies\n\\[\n\\angle BMA = 180^\\circ - \\gamma - \\alpha = \\beta,\n\\]\nand likewise $\\angle CMD = \\alpha$ . Thus $\\triangle ABM\\sim\\triangle MCD$ , so $BM = MC = \\sqrt{AB\\cdot CD} = \\sqrt 6$ .\n\nFinally, Pythagorean Theorem on $\\triangle MEN$ yields $MN = \\sqrt 5$ , and so\n\\begin{align*}\n\\operatorname{Area}(ABCD) &= 2\\operatorname{Area}(AEM) + \\operatorname{Area}(MEF) + 2\\operatorname{Area}(DFM)\n &= 2(\\tfrac12\\cdot 2\\cdot\\sqrt 5) + \\tfrac12\\cdot 2\\cdot\\sqrt 5 + 2(\\tfrac12\\cdot 3\\cdot\\sqrt 5) = 6\\sqrt 5.\n\\end{align*}\nThe requested answer is $(6\\sqrt 5)^2 = \\boxed{180}$ .</details>",
"The 3 triangles that the two angle bisectors split ABCD into are all similar ",
"We use bary. Extend $AB$ and $CD$ to meet at some point $P$ . Let $P=(1,0,0),D=(0,1,0),A=(0,0,1)$ , with $a=AD=7$ , $b=AP$ , $c=DP$ . Then $B=(2:0:b-2),C=(3:c-3:0)$ . Let the midpoint of $BC$ be $M$ so that $M=(a:b:c)$ .\n\nStandardizing coordinate sums, we have $B=(2c(a+b+c):0:c(b-2)(a+b+c)),C=(3b(a+b+c):b(c-3)(a+b+c):0),M=(abc:b^2c:bc^2)$ .\nThen $$ b(c-3)(a+b+c)=2b^2c\\implies (c-3)(a+b+c)=2bc $$ and $$ c(b-2)(a+b+c)=2bc^2\\implies (b-2)(a+b+c)=2bc\\implies c-3=b-2\\implies c=b+1 $$ We also have $$ (a+b+c)(2c+3b)=2abc=a(b-2)(a+b+c)\\implies 5b+2=7b-14\\implies b=8\\implies c=9 $$ Now LoC gives $\\cos\\angle APD=\\frac{2}{3}\\implies \\sin\\angle APD=\\frac{\\sqrt{5}}{3}$ , so $[ABCD]=[APD]-[BPC]=\\frac{1}{2}\\cdot\\frac{\\sqrt{5}}{3}(8\\cdot 9-6\\cdot 6)=6\\sqrt{5}$ , so the answer is $\\boxed{180}$ .",
"yoooooo my sketchy solution was correct lets go",
"does this remind anyone else of aaime p15//",
":love: \n\nLet $M$ be the midpoint of $BC.$ Extend $AB,CD$ to meet at $F.$ So $M$ is the incenter of $DAF.$ The fact that $M$ is the midpoint of $BC$ gives $FM \\perp BC$ and $BCF$ is isosceles. Then, angle chase gives $\\triangle ABM \\sim \\triangle MCD \\implies BM = \\sqrt{6}.$ \n\nIf the feet of $M$ to $AF$ and $DF$ are $X,Y$ respectively, $2+BX+3+CY = AX+DY = DA = 7,$ but also $BX=CY,$ so they are both $1.$ So by similar triangle relationships $BF = FC = 6.$ \n\nSo $MF = \\sqrt{6^2 -(\\sqrt{6})^2} = \\sqrt{30}$ and the area of $BCF$ can be computed to be $6\\sqrt{5}.$ The area of $DAF$ is then $6\\sqrt{5} \\cdot \\frac{8}{6} \\cdot \\frac{9}{6} = 12\\sqrt{5}.$ Subtracting yields $6\\sqrt{5} \\implies \\boxed{180}.$ ",
"Extended AB and CD, got area of the triangle formed, and forgot the problem was asking for area of the quadrilateral. Put 720 :(",
"<blockquote>:love: \n\nLet $M$ be the midpoint of $BC.$ Extend $AB,CD$ to meet at $F.$ So $M$ is the incenter of $DAF.$ The fact that $M$ is the midpoint of $BC$ gives $FM \\perp BC$ and $BCF$ is isosceles. Then, angle chase gives $\\triangle ABM \\sim \\triangle MCD \\implies BM = \\sqrt{6}.$ \n\nIf the feet of $M$ to $AF$ and $DF$ are $X,Y$ respectively, $2+BX+3+CY = AX+DY = DA = 7,$ but also $BX=CY,$ so they are both $1.$ So by similar triangle relationships $BF = FC = 6.$ \n\nSo $MF = \\sqrt{6^2 -(\\sqrt{6})^2} = \\sqrt{30}$ and the area of $BCF$ can be computed to be $6\\sqrt{5}.$ The area of $DAF$ is then $6\\sqrt{5} \\cdot \\frac{8}{6} \\cdot \\frac{9}{6} = 12\\sqrt{5}.$ Subtracting yields $6\\sqrt{5} \\implies \\boxed{180}.$ </blockquote>\n\nhow do you know that $M$ must be incenter of $DAF,$ what if $M$ is outside of $DAF.$ ",
"The problem says $\\angle DAB$ and $\\angle ADC$ are both acute, that's how you know $M$ is inside $\\triangle ADF$ .",
"heron's kills i think\n\nyou get BC = 2sqrt5",
"Using @above's diagram:\n\nIt's easy to see that $GF=EH=1$ . Now set $AF=AE=x$ ; we obtain $r=\\sqrt{x}$ and $A=\\sqrt{(x+7)(x)(4)(3)}$ and since $s=x+7$ we have $x=5$ . Then $r=\\sqrt{5}$ , $s=12$ , so $[ABC]=12\\sqrt{5}$ while $[AGH]=IF\\cdot AG=6\\sqrt{5}$ and the answer is $\\left(12\\sqrt{5}-6\\sqrt{5}\\right)^2=\\boxed{180}$ .",
"Another bary solution with a different finish.\n\nLet $AB\\cap DC=P$ , let $M$ be the midpoint of $BC$ , and employ barycentrics on $\\triangle PAD$ with $P=(1,0,0),A=(0,1,0),D=(0,0,1)$ . Let $PB=b-2$ and $PC=c-3$ and $AD=a=7$ . Then we can crank out $B=(2:b-2:0)$ and $C=(3:0:c-3)$ . Normalizing, we have $B=\\left(\\frac2b,\\frac{b-2}b,0\\right)$ and $C=\\left(\\frac3c,0,\\frac{c-3}c\\right)$ . Since $M$ is the midpoint of $BC$ , it has coordinates $$ M=\\frac{B+C}2=\\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right). $$ However, $M$ is also the incenter, so it has coordinates $$ M=(a:c:b)=\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right). $$ Hence, we have $M=\\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right)=\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right)$ and we can compare components to get the vector-component equation $$ \\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right)-\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right)=(3 b^2 - 9 b c + 21 b + 2 c^2 + 14 c,c(b^2 - b c + 5 b - 2 c - 14),b(-b c - 3 b + c^2 + 4 c - 21))=(0,0,0). $$ where we used unhomogenized coordinates. We can solved this to get $(b,c)=(8,9)$ . So $B=(2:6:0)$ and $C=(3:0:6)$ and we can normalize these to $B=(1/4,3/4,0)$ and $C=(1/3,0,2/3)$ .\n\nNote that the semiperimiter of $\\triangle ABC$ is $\\frac{7+8+9}2=12$ . By Heron's, we have $[ABC]=\\sqrt{12(12-7)(12-8)(12-9)}=12\\sqrt{5}$ . To finish, note that $[ABCD]=[BAD]+[DCB]$ . We have $$ [BAD]=\\begin{vmatrix}1/4&3/4&0 0&1&0 0&0&1\\end{vmatrix}\\cdot[ABC]=\\frac14\\cdot 12\\sqrt{5}=3\\sqrt5 $$ and $$ [DCB]=\\begin{vmatrix}0&0&1 1/3&0&2/3 1/4&3/4&0\\end{vmatrix}\\cdot[ABC]=\\frac14\\cdot 12\\sqrt{5}=3\\sqrt5 $$ so $[ABCD]=6\\sqrt5$ , so the answer is $\\boxed{180}$ .\n",
"This is a very nice problem! Unfortunately, I didn't see the clever similarity in djmathman's solution.\nHere's a synthetic solution with a minor bash.\n<details><summary>Solution</summary>Let $M$ be the midpoint of $\\overline{BC}$ . Reflect $B$ and $C$ across $\\overline{AM}$ and $\\overline{DM}$ to obtain $B'$ and $C'$ , respectively. Since $M$ lies on the angle bisectors, $B'$ and $C'$ lie on $\\overline{AD}$ . \n\nLet $N=AB\\cap CD$ . Note that $\\triangle MB'C'$ is isosceles, and $$ \\angle MB'C'=180^{\\circ}-\\angle MB'A=180^{\\circ}-\\angle MBA=\\angle NBM, $$ so $\\triangle MB'C'\\sim\\triangle NBC$ .\n\nSince $AB'=2$ and $C'D=3$ , $B'C'=2$ . Let $BM=B'M=CM=C'M=x$ . Then by similar triangles, $NC=x^2$ . From law of cosines on $\\triangle MB'C'$ and $\\triangle NAD$ , and letting $y=x^2$ , we have:\n\\begin{align*}\n\\frac{(y+2)^2+(y+3)^2-7^2}{2(y+2)(y+3)} &= \\frac{2y-2^2}{2y} \n\\frac{2y^2+10y-36}{2(y^2+5y+6)} &= \\frac{y-2}{y} \n\\frac{y^2+5y-18}{y^2+5y+6} &= \\frac{y-2}{y} \ny^3+5y^2-18y &= y^3+3y^2-4y-12 \n2y^2-14y+12 &= 0 \n(y-1)(y-6) &= 0 \\end{align*}\nThis means $y=1,6$ , or $x=1,\\sqrt{6}$ . $x=1$ fails the triangle inequality, so $x=\\sqrt{6}$ .\n\nTo finish, we have the distance from $M$ to $\\overline{AD}$ is $\\sqrt5$ , so $[ABCD]=\\sqrt{5}+2\\sqrt{5}+3\\sqrt{5}=6\\sqrt5\\implies\\boxed{180}$ .</details>",
"Let's mix everyone's solutions (including mine) shall we:\n\nLet the extensions of $AB$ and $DC$ meet at $F$ . It is clear that $M$ is the incenter of $\\triangle AFD$ . This means $\\angle CFM = \\angle BFM$ which implies that $\\triangle BFC$ is isosceles and that $MF$ is the perpendicular bisector of the triangle.\n\nNext, let $\\angle ABM = \\theta$ and $BF=CF=x$ . Here $$ \\angle DCM = 180 - \\angle MCF = 180 - \\angle CBF = \\theta $$ which means $\\triangle ABM \\sim \\triangle MCD$ , giving $BM=CM=\\sqrt{6}$ Stewart's Theorem on cevian $MB$ suffices to show that $AM=10+\\frac{24}{x}$ , so similar triangles again on $\\triangle AMD$ and $\\triangle ABM$ gives $$ 10+\\frac{24}{x}=14 \\implies x=6 $$ Heron's formula on $AFD$ gives $$ \\sqrt{12(3)(4)(5)} = 12 \\sqrt{5} $$ $$ [ABCD]=[AFD]-[BFC] \\implies 12 \\sqrt{5} - [BFC] $$ By Pythagorean theorem $FM=\\sqrt{30}$ , so $[BFC]=6 \\sqrt{5}$ . The answer is $$ (12 \\sqrt{5} - 6 \\sqrt{5})^2 = (6 \\sqrt{5})^2 = \\boxed{180} $$ ",
"Had to get a couple hints from **i3435**.\n\nLet $E=AB\\cap CD$ and let $M$ be the midpoint of $BC$ . Notice that $M$ is the incenter of $\\triangle{ADE}$ . Then drawing in the angle bisector of $E$ we find that $EB=EC$ from the angle-bisector theorem. This implies $\\triangle{EBM}\\cong \\triangle{ECM}$ from SSS.\n\nLet $\\angle{EAM}=\\angle{MAD}=x$ and let $\\angle{EDM}=\\angle{MDA}=y$ . Then notice that $\\angle{AMD}=180-x-y$ and that $\\angle{AED}=180-2x-2y$ . The latter implies $\\angle{EBM}=x+y$ so $\\angle{ABM}=180-x-y$ . This means that $\\triangle{ABM}$ and $\\triangle{AMD}$ are similar, and doing a similar angle chase reveals that these two triangles are also similar to $\\triangle{MCD}$ . \n\nThen we can figure out $BM=MC=\\sqrt{6}$ . Using this we find $AM=\\sqrt{14}$ and $DM=\\sqrt{21}$ . Using Heron's on $\\triangle{ABM}$ gives us $[ABM]=\\sqrt{5}$ . Then from similar triangles, $[MCD]=\\frac{3\\sqrt{5}}{2}$ and $[AMD]=\\frac{7\\sqrt{5}}{2}$ . Thus $[ABCD]=6\\sqrt{5}\\implies 180$ .",
"first attempt: t r i g b a s h\n\nLet $M$ be the midpoint of $BC$ . As $M$ lies on the bisector of $\\angle DAB$ , $M$ is equidistant from $AB$ and $AD$ , and as it also lies on the bisector of $\\angle ADC$ , it is also equidistant from $CD$ and $AD$ . As a result, there exists a circle with radius $r$ centered at $M$ that is tangent to $AB$ , $CD$ , and $AD$ . Let $P$ , $Q$ , and $R$ be the respective points of tangency, which are equivalently the projections of $M$ onto those sides. \n\nSince $BM=CM=\\tfrac{BC}{2}$ , $MP=MQ=r$ , and $\\angle BPM=\\angle CQM=90^\\circ$ , triangles $BMP$ and $CMQ$ are congruent by HL. Set $BP=CQ=k$ . We also have that triangles $AMP$ and $ARM$ are congruent by HL, as are $DQM$ and $DRM$ , so $AR=AP=2+k$ and $DR=DQ=3+k$ . Then, $$ (2+k)+(3+k)=AR+DR=AD=7\\implies k=1. $$ It follows that $AR=3$ and $DR=4$ .\n\nNow, the sum of $\\angle AMR=\\arctan\\tfrac{3}{r}$ , $\\angle DMR=\\arctan\\tfrac{4}{r}$ , $\\angle AMB=\\angle AMP-\\angle BMP=\\arctan\\tfrac{3}{r}-\\arctan\\tfrac{1}{r}$ , and $\\angle DMC=\\angle DMQ-\\angle CMQ=\\arctan\\tfrac{4}{r}-\\tfrac{1}{r}$ are $180^\\circ$ by looking along $BC$ , so $$ \\arctan\\frac{3}{r}+\\arctan\\frac{4}{r}+\\left(\\arctan\\frac{3}{r}-\\arctan\\frac{1}{r}\\right)+\\left(\\arctan{4}{r}-\\arctan\\frac{1}{r}\\right)=2\\arctan\\frac{3}{r}+2\\arctan\\frac{4}{r}-2\\arctan\\frac{1}{r}=180^\\circ. $$ By the angle addition formulas, $$ \\arctan\\frac{3}{r}+\\arctan\\frac{4}{r}-\\arctan\\frac{1}{r}=\\arctan\\frac{7r}{1-12r^2}-\\arctan\\frac{1}{r}=\\arctan\\frac{6r^2+12}{r^3-5r}, $$ and doubling gives $$ 2\\arctan\\frac{6r^2+12}{r^3-5r}=\\frac{2\\cdot\\frac{6r^2+12}{r^3-5r}}{1-\\frac{(6r^2+12)^2}{(r^3-5r)^2}}=\\frac{2(r^3-5r)(6r^2+12)}{(r^3-5r)^2-(6r^2+12)^2}. $$ Setting this equal to $180^\\circ$ and taking $\\tan$ of both sides, $$ \\frac{2(r^3-5r)(6r^2+12)}{(r^3-5r)^2-(6r^2+12)^2}=0\\implies2(r^3-5r)(6r^2+12)=0, $$ whose only positive real solution is $r=\\sqrt{5}$ . Finally, $$ [ABCD]=[ABM]+[CDM]+[ADM]=\\frac{2r}{2}+\\frac{3r}{2}+\\frac{7r}{2}=6r=6\\sqrt{5}, $$ and squaring yields $\\boxed{180}$ .",
"<blockquote>There exists a very \"American\" styled solution, too, that I came up with first; I'll let someone else post that method.</blockquote>\nOops, it's been a while; I don't think anyone's posted the \"American\" styled solution I referred to. For what it's worth, here was the original formulation of the problem, back when it was going to appear on [Problem Stash 2](\"https://davidaltizio.web.illinois.edu/stash-contest2p.pdf\") (rest in peace).**Original Problem (February 13, 2020).** Let $ABC$ be an acute triangle with circumcircle $\\Omega$ . Let $\\omega$ be the $A$ -*mixtillinear incircle* of $\\triangle ABC$ -- that is, the unique circle internally tangent to $\\Omega$ that is also tangent to $\\overline{AB}$ and $\\overline{AC}$ at $X$ and $Y$ , respectively. Suppose $XB = 2$ , $BC = 7$ , and $CY = 3$ . Compute the square of the area of $\\triangle ABC$ .\n\nThis reformulation doesn't kill the problem, surprisingly, but it does offer alternate motivation for some of the steps.",
"<details><summary>a lemma</summary>Let $\\triangle ABC$ be a triangle. Note that if $X$ and $Y$ are the $A$ -mixtilinar touch-points on $\\overline {AB}$ and $\\overline{AC}$ respectively, then $AX = AY = \\tfrac{bc}{s}$ : this can be seen from the fact that a $\\sqrt{bc}$ inversion swaps the excircle with the mixtilinar incircle.</details>\n\nDefine $P\\equiv\\overline{AB}\\cap\\overline{CD}$ , and note that $B$ and $C$ are the $P$ -mixtilinear touch-points of $\\triangle PAD$ . If we let $x\\equiv PA$ , then we have, by the above lemma, $$ x = \\frac{(x+2)(x+3)}{\\frac{1}{2}\\big((x + 2) + 7 + (x + 3)\\big)} \\implies x = 6. $$ Now, the area of $[PAD] = 12\\sqrt{5}$ by Heron's formula, so $$ [ABCD] = [PAD] - [PBC] = 12\\sqrt{5} - 12\\sqrt{5}\\cdot\\tfrac{6}{8}\\cdot\\tfrac{6}{9} = 6\\sqrt{5}, $$ and therefore our desired answer is $(6\\sqrt{5})^2 = \\boxed{180}$ .",
"once you figure out that $\\triangle ABM \\sim \\triangle MCD \\sim \\triangle AMD$ it is not too hard to finish with Heron's (if you are used to this type of radical Heron's and even if you aren't it's not too difficult).",
"I found this solution while mocking:\n\nLet $I$ be the midpoint of $BC$ . Extend $AB$ and $CD$ to meet at $P$ . note that $I$ is the incenter of $\\triangle APD$ . Furthermore, $PI \\perp BC$ .\nLet $PC = PB = x$ . Let $K = [APD]$ . We have\n\\[x = \\frac{PI}{\\sin \\frac P2} = \\frac{r}{\\sin \\frac P2 \\cos \\frac P2} = \\frac{2r}{\\sin P} = \\frac{\\frac{2K}{s}}{\\frac{2K}{(x+3)(x+2)}} = \\frac{(x+3)(x+2)}{x+6}\\]\nSolving, we get $x=6$ . We use Heron's to get $K = 12\\sqrt{5} \\implies [PBC] = 6 \\sqrt{5} \\implies [ABCD] = 6\\sqrt{5}$ .",
"Let $M$ be the midpoint of $BC$ , and extend $AB$ and $DC$ . Let the intersection point be $P$ . $M$ is the incenter of $\\triangle{PAD}$ . Thus, $\\angle{MPB}\\cong \\angle{MPC}$ . Now, angle chasing on $\\triangle{ABM}, \\triangle{DCB}$ shows that they are similar triangles, which yields $BM=MC=\\sqrt{6}$ . Construct $MP$ and since $BM=MC$ and $\\angle{MPB}\\cong \\angle{MPC}$ , $MP\\perp BC$ . Now, we can construct the law of cosines equations and set them equal: $$ \\frac{(x+2)^2+(x+3)^2-7^2}{2(x+2)(x+3)}=\\frac{2x^2-24}{2x^2} $$ $$ x=6 $$ From that, we can find the cosine value of $\\angle{APD}$ via law of cosines, and we need the sine value to compute the area. Trig gives $\\sin \\angle{APD}=\\frac{\\sqrt{5}}{3}$ . Now finish with $\\frac{1}{2}\\cdot \\frac{\\sqrt{5}}{3}\\left(8\\cdot 9-6^2\\right)=6\\sqrt{5}$ and we need the square so $\\boxed{180}$ .",
"What does the title mean? \"MAA likes misplacing problems\"\n\nIs it too easy or too hard for AIME 11?",
"Easy $ $ ",
"No i think it is fine for a number 11. Extending and noticing the perpendicularity takes some thinking and then expressing <DCB is terms of alpha and beta also requires some thought. So yea i would say well suited for 11. \n\nNice problem tho. My sol was basically the same as the others. Notice similar triangles $ABM$ , $AMD$ and $DMC$ where $M$ is the mid point of BC. Then doing sim triangles we get $AM = \\sqrt{14}$ and $MD = \\sqrt{21}$ . Then notice if the height is dropped for $\\triangle{AMD}$ from $M$ it slipts the base into 3,4. So area is 7/2 sqrt(5). And rest is just similar areas and add to get 6sqrt5. Hence 180.",
"Once you prove (or guess! It's AIME!) that the three triangles are similar, then similarity ratios give you that $\\overline{BC} = \\sqrt 6 $ and Heron's formula (3 times, or use similar areas) does the rest.\nThe √2:√3:√7 side length ratios make the alternate forms of Heron's formula very nice to use.\nI added these to [https://artofproblemsolving.com/wiki/index.php?title=Area#Other_formulas_equivalent_to_Heron.27s](https://artofproblemsolving.com/wiki/index.php?title=Area#Other_formulas_equivalent_to_Heron.27s) today.",
"How did @djmathman get the \"original formulation\" 2 years before the AIME 2022? Did you create that problem and then contribute it to the AIME? ",
"<blockquote>How did @djmathman get the \"original formulation\" 2 years before the AIME 2022? Did you create that problem and then contribute it to the AIME?</blockquote>\nYes; the original formulation was the first version of the problem I came up with. This final version came after removing the mixtillinear incircle and instead focusing the statement on quadrilateral $XBCY$ .",
"Disagree with the title. I found this way harder than #11 on AIME I at least. I think this could switch with #13, considering some #13 geos have also been easier imo (2019 AIME I, 2020 AIME I come to mind).",
"The hard part is first: noticing the condition that the midpoint of $BC$ is the incenter of triangle. \n\nSecond: Notice the similarity and find $BM=CM=\\sqrt{6}$ . The rest is computational. ",
"<blockquote>Another bary solution with a different finish.\n\nLet $AB\\cap DC=P$ , let $M$ be the midpoint of $BC$ , and employ barycentrics on $\\triangle PAD$ with $P=(1,0,0),A=(0,1,0),D=(0,0,1)$ . Let $PB=b-2$ and $PC=c-3$ and $AD=a=7$ . Then we can crank out $B=(2:b-2:0)$ and $C=(3:0:c-3)$ . Normalizing, we have $B=\\left(\\frac2b,\\frac{b-2}b,0\\right)$ and $C=\\left(\\frac3c,0,\\frac{c-3}c\\right)$ . Since $M$ is the midpoint of $BC$ , it has coordinates $$ M=\\frac{B+C}2=\\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right). $$ However, $M$ is also the incenter, so it has coordinates $$ M=(a:c:b)=\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right). $$ Hence, we have $M=\\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right)=\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right)$ and we can compare components to get the vector-component equation $$ \\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right)-\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right)=(3 b^2 - 9 b c + 21 b + 2 c^2 + 14 c,c(b^2 - b c + 5 b - 2 c - 14),b(-b c - 3 b + c^2 + 4 c - 21))=(0,0,0). $$ where we used unhomogenized coordinates. We can solved this to get $(b,c)=(8,9)$ . So $B=(2:6:0)$ and $C=(3:0:6)$ and we can normalize these to $B=(1/4,3/4,0)$ and $C=(1/3,0,2/3)$ .\n\nNote that the semiperimiter of $\\triangle ABC$ is $\\frac{7+8+9}2=12$ . By Heron's, we have $[ABC]=\\sqrt{12(12-7)(12-8)(12-9)}=12\\sqrt{5}$ . To finish, note that $[ABCD]=[BAD]+[DCB]$ . We have $$ [BAD]=\\begin{vmatrix}1/4&3/4&0 0&1&0 0&0&1\\end{vmatrix}\\cdot[ABC]=\\frac14\\cdot 12\\sqrt{5}=3\\sqrt5 $$ and $$ [DCB]=\\begin{vmatrix}0&0&1 1/3&0&2/3 1/4&3/4&0\\end{vmatrix}\\cdot[ABC]=\\frac14\\cdot 12\\sqrt{5}=3\\sqrt5 $$ so $[ABCD]=6\\sqrt5$ , so the answer is $\\boxed{180}$ .</blockquote>\n\nhow long did it take you to type that",
"I think the \"European\" solution in #2 and #23 is the simplest.\n\nIt's reflecting across a bisector (twice) and using the midpoint to build an isosceles triangle that helps in the angle-chasing to find similar triangles. Then Heron or Pythagoras to get the basic Area, and some AIME bookkeeping computations to get the answer.\nThe hardest part IMHO is organizing all the details to find which angles to look at. In particular, you have two overlapping diagrams: the original quadrilateral with 2 angle bisectors (3 triangles); and the original quadrilateral with the isosceles triangle made by reflecting sides across the angle bisectors. \n\n It feels like a \"big\" version of an AMC 10 #20-25 problem. \n\nNo circle needed, no external P triangle needed.",
"This bary bash is so clean I have to write it up here for storage.\n\n<details><summary>Bary Bashing AIME Geo</summary>Let $E = AB \\cap CD$ . Note that the midpoint $I$ of $BC$ is the incenter of $ADE$ . Additionally, since $I$ lies on the angle bisector of $\\angle BEC$ , we know $BI = IC \\iff BC \\perp IE \\iff BE = CE$ . Let $BE = CE = x$ .\n\nWe apply barycentric coordinates on $ADE$ .\n\\[B = (x:2:0); I = (x+3:7:x+2); C = (0:3:x)\\]\n\nClearly, the equation for line $BC$ is as follows: $(a:b:c)$ lies on $BC$ if and only if $2a - xb + 3c = 0$ . Plugging $I$ into this, since $B, I, C$ are collinear, we may solve to get $x = 6$ . Applying Heron's gives $[ADE] = 12 \\sqrt5$ . Since $[BCE] = [ADE] \\cdot \\frac{2}{2 + 6} \\cdot \\frac{3}{3 + 6} = [ADE] \\cdot \\frac12 = 6 \\sqrt5$ , the area of $[ABCD] = 12 \\sqrt 5 - 6 \\sqrt 5 = 6 \\sqrt5$ .</details>",
"Let $X=AB\\cap CD$ , and let $I$ be the midpoint of $BC$ . Then it is also the incenter of triangle $XAD$ . Let $a=XB=XC$ .\n\nThus $B,C$ are the tangency points of the $X$ -mixtilinear incircle. The $\\sqrt{bc}$ inverse of this circle is the $X$ -excircle. The lengths of the tangents from $X$ to these circles are $a$ and $a+6$ (half of perimeter), respectively. Thus $a(a+6)=(a+2)(a+3)$ , so $a=6$ .\n\nFinally, by Heron we get $[XAD]=12\\sqrt5$ and $\\frac{6\\cdot 6}{8\\cdot 9}=\\frac12$ , so the answer is $\\left(12\\sqrt5\\left(1-\\frac12\\right)\\right)^2=\\boxed{180}$ .",
"Favorite problem on this high quality test.\n\nLet $AB \\cap DC = E$ , and the midpoint of $BC = I$ . Then $I$ is the incenter of $\\triangle EAD$ , from which is follows that $\\angle BEI = \\angle CEI \\implies EB = EC$ by Angle Bisector. Then, let $AE = x+3$ and $DE = x+4$ . Applying the Law of Cosines on $\\triangle EAD$ wrt to $\\angle A = \\theta_1$ and $\\angle D = \\theta_2$ gives us $$ \\cos{(\\theta_1)} = \\frac{21-x}{7(x+3)}, \\cos{(\\theta_2)} = \\frac{x+28}{7(x+4)} $$ Drop the altitude from $B$ and $C$ to $AD$ , let them be $F$ and $G$ respectively. Let the tangency point of the incircle with $AD$ be $X$ . The semiperimeter of the triangle is $x+7$ from which it follows from the $s-a$ formulas that $AX=3, DX=4$ . Then, $AF = 2 \\cos{(\\theta_1)} = \\frac{2(21-x)}{7(x+3)}$ and $AG = 7 - 3 \\cos{(\\theta_2)} = 7 - \\frac{3(x+28)}{7(x+4)}$ . Now, because $I$ is the midpoint similar triangles gives us that $AX = \\frac{AF+AG}{2}$ so we solve the equation $$ \\frac{ \\frac{2(21-x)}{7(x+3)} + 7 - \\frac{3(x+28)}{7(x+4)}}{2}=3 \\implies x=5, 0 $$ Now we have a $7-8-9$ triangle which by Herons is $12\\sqrt{5}$ , We have an isosceles triangle cut out from it inscribed inside which by ratios has area $12\\sqrt{5} \\cdot \\frac68 \\frac69= 6\\sqrt{5}$ . The desired answer is $(12\\sqrt{5}-6\\sqrt{5})^2 = \\boxed{180}$ ",
"We extend $AB$ and $CD$ to a point $E.$ Then, if $I$ is the midpoint of $BC,$ we have that $I$ is the incenter of $\\triangle AED.$ Let $EB=EC = x.$ Now, drawing a height from $I$ to $ED$ and $AE,$ and calling them $G$ and $H,$ we get that $EG = EH = x-1,$ so $EI = x\\sqrt{x-1}.$ Let $F = AI \\cap AD.$ By Stewart's on $AF,$ and angle bisector theorem on $AIF,$ we get $$ AI^{2} = \\frac{(2x-2)(x+2)(x+3)}{(2x+12)} = x(x-1), $$ so $x=6.$ Thus, by simple computations, the area of the quadrilateral is $6\\sqrt{5}.$ ",
"<details><summary>Solution</summary>Let the foot altitude from $M$ to $\\overline{AB}$ be denoted as $E$ and the foot of the altitude from $M$ to $\\overline{CD}$ as $F$ . Finally, let the foot of the altitude from $M$ to $\\overline{AD}$ be $G$ , and the point at which $\\overline{AB}$ and $\\overline{CD}$ meet $H$ . First, we know that \\[\\triangle EBM \\cong \\triangle FCM\\] since they are both right triangles, their hypotenuses are equal ( $BM = MC$ ), and $EM=FM$ since $M$ is the incenter of $\\triangle ADH$ . Then, we know that \\[7 = AG+DG = AE+DF = 5+EB+FC = 5+2EB\\] implying that \\[EB = FC = 1.\\] Next, we wish to show that \\[\\triangle ABM \\sim \\triangle MCD.\\] Since $\\angle EBM = \\angle FCM$ due to the congruent triangles, we know that \\[\\angle ABM = \\angle MCD. \\tag{1}\\] Next, let $\\angle ABC = \\angle BCD = \\alpha$ , $\\angle BAC = \\beta$ , and $\\angle ADC = \\gamma$ . Then, we know that: \\[2\\alpha+\\beta+\\gamma = 360\\] and \\[\\angle CMD = 180-\\alpha - \\frac{\\gamma}{2} = \\frac{\\alpha}{2} = \\angle BAM.\\] Thus, we have indeed proved $(1)$ . With this, we know that \\[\\frac{AB}{BM}=\\frac{MC}{CD}\\] implying that \\[BM = CM = \\sqrt{6}.\\] By Pythagorean Theorem, we can say that \\[EM = FM = GM = \\sqrt{5}.\\] Finally, we can say that\n\\begin{align*}\n[ABCD] &= [AEM]-[BEM]+[DFM]-[CFM]+[AMD] \n&= \\dfrac{AE\\cdot EM}{2}-\\dfrac{BE\\cdot EM}{2}+\\dfrac{DF\\cdot FM}{2}-\\dfrac{CF\\cdot FM}{2}+\\dfrac{AD\\cdot MG}{2} \n&= \\dfrac{3\\sqrt{5}}{2}-\\dfrac{\\sqrt{5}}{2}+\\dfrac{4\\sqrt{5}}{2}-\\dfrac{\\sqrt{5}}{2}+\\dfrac{7\\sqrt{5}}{2} \n&= \\boxed{6\\sqrt{5}}\n\\end{align*}\nas required. $\\square$</details>",
" $\\mathfrak{The \\;Tenth\\; Of\\; August,\\; 2025}$ ʕ•ᴥ•ʔ\nWill hopefully typeset a diagram tomorrow if I do not forget.\n<details><summary>Solution - grinding Computational Geometry</summary>$$ \\color{red} \\spadesuit\\color{red} \\boxed{\\textbf{Solution.}}\\color{red} \\spadesuit $$ Let $M=mid. BC$ ; $ F \\in CD, MF \\perp CD; E \\in AB, ME \\perp AB$ - since $DM$ and $AM$ are both angle bisectors, we have (where $HM=$ altitude in $\\triangle AMD$ ): $$ ME=MH=MF, $$ which implies $\\triangle BEM \\cong \\triangle CFM$ and then $\\triangle AEM \\cong \\triangle AHM, \\triangle MFD \\cong \\triangle MHD$ from angle bisectors and cyclic $HMFD, AEMH$ . From here, we obtain that $$ CF=BE=\\frac{7-3-2}{2}=1. $$ We also have $\\triangle ABM \\sim \\triangle CDM \\implies BM=CM=\\sqrt{6}$ , from where $ME=\\sqrt{5}$ by Pythagoras and thus, $\\boxed{[ABCD]=6\\sqrt{5}}$ .\n[asy]\n\nimport graph; size(6.5cm); \nreal labelscalefactor = 0.5; /* changes label-to-point distance */\npen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ \npen dotstyle = black; /* point style */ \nreal xmin = -4.3, xmax = 28.32, ymin = -13.16, ymax = 6.3; /* image dimensions */\npen ffzztt = rgb(1.,0.6,0.2); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzwwff = rgb(0.6,0.4,1.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen qqccqq = rgb(0.,0.8,0.); pen ttffqq = rgb(0.2,1.,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); \n\ndraw(arc((12.36,-6.46),0.6,-2.602562202499808,38.82755729189866)--(12.36,-6.46)--cycle, linewidth(2.) + qqwuqq); \ndraw(arc((12.36,-6.46),0.6,38.82755729189866,72.43498531270703)--(12.36,-6.46)--cycle, linewidth(2.) + qqwuqq); \ndraw(arc((15.,1.88),0.6,-107.56501468729299,-78.339070745046)--(15.,1.88)--cycle, linewidth(2.) + qqwuqq); \ndraw(arc((15.,1.88),0.6,-78.339070745046,-45.)--(15.,1.88)--cycle, linewidth(2.) + qqwuqq); \n /* draw figures */\ndraw((12.36,-6.46)--(15.,1.88), linewidth(2.) + ffzztt); \ndraw((15.,1.88)--(17.2,-0.32), linewidth(2.) + xdxdff); \ndraw((12.36,-6.46)--(15.,-6.58), linewidth(2.) + zzwwff); \ndraw((15.,-6.58)--(17.2,-0.32), linewidth(2.) + ffdxqq); \ndraw((15.,1.88)--(16.1,-3.45), linewidth(2.) + qqccqq); \ndraw((12.36,-6.46)--(16.1,-3.45), linewidth(2.) + ttffqq); \ndraw((12.36,-6.46)--(15.955752577319588,-6.623443298969073), linewidth(2.) + xdxdff); \ndraw((15.955752577319588,-6.623443298969073)--(16.1,-3.45), linewidth(2.) + red); \ndraw((16.1,-3.45)--(18.215,-1.335), linewidth(2.) + red); \ndraw((18.215,-1.335)--(17.2,-0.32), linewidth(2.) + xdxdff); \ndraw((16.1,-3.45)--(13.566651926424237,-2.648076868796161), linewidth(2.) + red); \n /* dots and labels */\ndot((12.36,-6.46),dotstyle); \nlabel(\" $A$ \", (11.68,-6.74), NE * labelscalefactor); \ndot((15.,1.88),dotstyle); \nlabel(\" $B$ \", (15.08,2.08), NE * labelscalefactor); \ndot((17.2,-0.32),dotstyle); \nlabel(\" $C$ \", (17.28,-0.12), NE * labelscalefactor); \ndot((15.,-6.58),dotstyle); \nlabel(\" $D$ \", (14.82,-7.1), NE * labelscalefactor); \ndot((16.1,-3.45),linewidth(4.pt) + dotstyle); \nlabel(\" $E$ \", (16.28,-4.06), NE * labelscalefactor); \ndot((15.955752577319588,-6.623443298969073),linewidth(4.pt) + dotstyle); \nlabel(\" $F$ \", (15.88,-7.2), NE * labelscalefactor); \ndot((18.215,-1.335),linewidth(4.pt) + dotstyle); \nlabel(\" $G$ \", (18.3,-1.18), NE * labelscalefactor); \ndot((13.566651926424237,-2.648076868796161),linewidth(4.pt) + dotstyle); \nlabel(\" $H$ \", (13.64,-2.48), NE * labelscalefactor); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n /* end of picture */\n[/asy]</details>",
"<details><summary>Solution</summary>We get $\\angle MCD=\\angle ABM$ . After some angle chasing we get $MCD,ABM,AMD$ are similar. \nSo $BM=CM=\\sqrt 6$ , $AM=\\sqrt {14}$ , $MD=\\sqrt{21}$ . $\\cos AMD=-\\frac{1}{\\sqrt{6}}$ , so $\\sin AMD=\\frac{\\sqrt 5}{\\sqrt 6}$ $[ABCD]=\\frac{12}{7}AMD=\\frac{6}{7} \\sqrt {14} \\sqrt{21}\\frac{\\sqrt 5}{\\sqrt 6}=6 \\sqrt 5$</details>",
"<details><summary>Solution using trig</summary>Let $\\angle CDM = \\theta = \\angle MDA, \\angle DAM = \\angle MAB = \\alpha$ By sine rule:- $\\frac{\\sin{\\theta}}{MC} = \\frac{\\sin{\\angle DCM}}{DM}$ , $\\frac{\\sin{\\alpha}}{MB} = \\frac{\\sin{\\angle MBA}}{MA}$ , $\\frac{\\sin{\\theta}}{\\sin{\\alpha}} = \\frac{AM}{DM}$ Which gives $\\sin{\\angle DCM} = \\sin{\\angle ABM} \\implies \\angle DCM = \\angle ABM$ because trapezoid can't hold (and if this case works it works since this is AIME jk jk)\nThen $MCD, ABM, AMD$ are similar then after that $MB = MC = \\sqrt{6}$ by similarity ratios, after this it's just computation using herons formula to get $[ABCD] = 6\\sqrt{5}$</details>",
"Let the intersection of the bisectors be $M$ , extend lines $AB$ and $CD$ , note that they intersect at $E$ and that $\\triangle BEC$ is isosceles. Additionally, note $$ \\triangle ABM \\sim \\triangle AMD\\sim \\triangle MCD $$ through basic angle chasing. We find that $$ BM=MC=\\sqrt{6}, \\text{ and } AM=\\sqrt{14}, \\ MD=\\sqrt{21}. $$ Using herons on the similar triangles gives us $\\frac{12}{7}\\cdot \\frac{7\\sqrt{5}}{2}=6\\sqrt{5}. \\ \\ \\Box$ ",
"We abuse trigonometry. Let the angle bisectors meet at \\(M\\), and let \n\\[\\angle BAM=\\angle MAD=\\alpha\\]\n\\[\\angle MDC=\\angle ADM=\\beta\\]\n\\[\\angle MBA=\\gamma\\]\nSince all angles of quadrilateral sum to \\(2\\pi\\), we have:\n\\[\\angle DCB=2\\pi-2\\alpha-2\\beta-\\gamma\\]\nThen by sine law on \\(\\Delta DMA\\):\n\\[\\frac{AM}{\\sin\\angle ADM}=\\frac{DM}{\\sin\\angle MAD}\\]\n\\[\\frac{AM}{\\sin\\beta}=\\frac{DM}{\\sin\\alpha}\\]\n\\[\\frac{AM}{DM}=\\frac{\\sin\\beta}{\\sin\\alpha}\\dots\\dots (1)\\]\nUsing sine law on \\(\\Delta MBA\\), \\(\\Delta DCM\\):\n\\[\\frac{BM}{\\sin\\angle BAM}=\\frac{AM}{\\sin\\angle MBA}\\]\n\\[\\frac{BM}{\\sin\\alpha}=\\frac{AM}{\\sin\\gamma}\\dots\\dots (2)\\]\n\\[\\frac{CM}{\\sin\\angle MDC}=\\frac{DM}{\\sin\\angle DCM}\\]\n\\[\\frac{CM}{\\sin\\beta}=\\frac{DM}{\\sin(2\\pi-2\\alpha-2\\beta-\\gamma)}\\dots\\dots (3)\\]\nMultiplying \\((2)\\) by the reciprocal of \\((3)\\) and using \\(BM=CM\\), and (1):\\begin{align*}\n&\\frac{BM\\sin\\beta}{CM\\sin\\alpha}=\\frac{AM\\sin(2\\pi-2\\alpha-2\\beta-\\gamma)}{DM\\sin\\gamma}\n&1=-\\frac{\\sin(2\\alpha+2\\beta+\\gamma)}{\\sin\\gamma}\n&\\sin\\gamma=-\\sin(2\\alpha+2\\beta+\\gamma)\n\\end{align*}\nNow we have either:\n\\[\\gamma=2\\alpha+2\\beta+\\gamma-\\pi\\]\nwhich simplifies to\n\\[\\alpha+\\beta=\\frac{1}{2}\\pi\\]\nThis is not possible as \\(\\angle DAB=2\\alpha\\) and \\(\\angle ADC=2\\beta\\) have to be acute.\nOr\n\\[\\gamma=2\\pi-2\\alpha-2\\beta-\\gamma\\]\n\\[\\alpha+\\beta+\\gamma=\\pi\\]\nThis implies:\n\\[\\angle DMA=\\pi-\\angle MAD-\\angle ADM= \\gamma\\]\n\\[\\angle DCB=\\angle DCM=2\\pi-2\\alpha-2\\beta-\\gamma=\\gamma\\]\nSo we have:\n\\[\\angle MBA= \\angle DMA= \\angle DCM=\\gamma\\]\nAnd \n\\[\\angle BAM= \\angle MAD= \\alpha\\]\nThis gives \\(\\Delta MAD \\sim \\Delta BAM\\).\nAlso:\n\\[\\angle MDC= \\angle ADM= \\beta\\]\nThis gives \\(\\Delta MAD \\sim \\Delta CMD\\).\nAltogether: \n\\[\\Delta BAM \\sim \\Delta MAD \\sim \\Delta CMD\\]\nUsing similar ratios:\n\\[\\frac{BM}{BA}=\\frac{CD}{CM}\\]\n\\[BM\\cdot CM=6\\]\n\\[BM=CM=\\sqrt{6}\\]\nAnd:\n\\[\\frac{AM}{BA}=\\frac{AD}{AM}\\]\n\\[AM=\\sqrt{14}\\]\n\\[\\frac{DM}{DC}=\\frac{AD}{DM}\\]\n\\[DM=\\sqrt{21}\\]\nNow we see that all the triangles are similar to \\(\\sqrt{2}-\\sqrt{3}-\\sqrt{7}\\) triangle. We could use Heron, but for the sake of continuing trig, we use cosine law on this triangle:\n\\[7=2+3-2\\sqrt{6}\\cos\\theta\\]\n\\[\\cos\\theta=-\\frac{1}{\\sqrt{6}}\\]\n\\[\\sin\\theta=\\frac{\\sqrt{5}}{\\sqrt{6}}\\]\nwhere \\(\\theta\\) is the angle between \\(\\sqrt{2},\\sqrt{3}\\) sides. Now we find the area of this triangle using:\n\\[\\frac{1}{2}\\sqrt{2}\\sqrt{3}\\sin\\theta=\\frac{\\sqrt{5}}{2}\\]\n\\(\\Delta BAM\\) has side lengths which are \\(\\sqrt{2}\\) times longer so its area is:\n\\[\\frac{\\sqrt{5}}{2}\\cdot(\\sqrt{2})^2=\\sqrt{5}\\]\n\\(\\Delta MAD\\) has side lengths which are \\(\\sqrt{7}\\) times longer so its area is:\n\\[\\frac{\\sqrt{5}}{2}\\cdot(\\sqrt{7})^2=\\frac{7}{2}\\sqrt{5}\\]\n\\(\\Delta CMD\\) has side lengths which are \\(\\sqrt{3}\\) times longer so its area is:\n\\[\\frac{\\sqrt{5}}{2}\\cdot(\\sqrt{3})^2=\\frac{3}{2}\\sqrt{5}\\]\nSumming these areas to find the area of \\(ABCD\\) to be \\({6\\sqrt{5}}\\). Then the desired answer is \\(\\boxed{180}\\). This took way too long to type out.",
"<blockquote><blockquote>Another bary solution with a different finish.\n\nLet $AB\\cap DC=P$ , let $M$ be the midpoint of $BC$ , and employ barycentrics on $\\triangle PAD$ with $P=(1,0,0),A=(0,1,0),D=(0,0,1)$ . Let $PB=b-2$ and $PC=c-3$ and $AD=a=7$ . Then we can crank out $B=(2:b-2:0)$ and $C=(3:0:c-3)$ . Normalizing, we have $B=\\left(\\frac2b,\\frac{b-2}b,0\\right)$ and $C=\\left(\\frac3c,0,\\frac{c-3}c\\right)$ . Since $M$ is the midpoint of $BC$ , it has coordinates $$ M=\\frac{B+C}2=\\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right). $$ However, $M$ is also the incenter, so it has coordinates $$ M=(a:c:b)=\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right). $$ Hence, we have $M=\\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right)=\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right)$ and we can compare components to get the vector-component equation $$ \\left(\\frac{3b+2c}{2bc},\\frac{bc-2c}{2bc},\\frac{bc-3b}{2bc} \\right)-\\left(\\frac{7}{7+b+c},\\frac{c}{7+b+c},\\frac{b}{7+b+c} \\right)=(3 b^2 - 9 b c + 21 b + 2 c^2 + 14 c,c(b^2 - b c + 5 b - 2 c - 14),b(-b c - 3 b + c^2 + 4 c - 21))=(0,0,0). $$ where we used unhomogenized coordinates. We can solved this to get $(b,c)=(8,9)$ . So $B=(2:6:0)$ and $C=(3:0:6)$ and we can normalize these to $B=(1/4,3/4,0)$ and $C=(1/3,0,2/3)$ .\n\nNote that the semiperimiter of $\\triangle ABC$ is $\\frac{7+8+9}2=12$ . By Heron's, we have $[ABC]=\\sqrt{12(12-7)(12-8)(12-9)}=12\\sqrt{5}$ . To finish, note that $[ABCD]=[BAD]+[DCB]$ . We have $$ [BAD]=\\begin{vmatrix}1/4&3/4&0 0&1&0 0&0&1\\end{vmatrix}\\cdot[ABC]=\\frac14\\cdot 12\\sqrt{5}=3\\sqrt5 $$ and $$ [DCB]=\\begin{vmatrix}0&0&1 1/3&0&2/3 1/4&3/4&0\\end{vmatrix}\\cdot[ABC]=\\frac14\\cdot 12\\sqrt{5}=3\\sqrt5 $$ so $[ABCD]=6\\sqrt5$ , so the answer is $\\boxed{180}$ .</blockquote>\n\nhow long did it take you to type that</blockquote>\n\nMy solution probably took longer to type out. :("
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1180,
"boxed": true,
"end_of_proof": false,
"n_reply": 43,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782927.json"
}
|
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
|
We proceed with casework based on Carl's first opponent.
If Carl's first opponent is Azar, which occurs with probability $\tfrac{1}{3}$ , Carl first beats Azar in the semifinals with probability $\tfrac{1}{3}$ , then beats either Jon or Sergey in the finals with probability $\tfrac{3}{4}$ , so the total probability here is $\tfrac{1}{3}\cdot\tfrac{1}{3}\cdot\tfrac{3}{4}=\tfrac{1}{12}$ .
If Carl's first opponent is Jon or Sergey, which occurs with probability $\tfrac{2}{3}$ , Carl first beats Jon or Sergey in the semifinals with probability $\tfrac{3}{4}$ . If his opponent in the finals is Azar with a $\tfrac{3}{4}$ chance, he wins in the finals with probability $\tfrac{1}{3}$ , but if his opponent in the finals is the other of Jon and Sergey with a $\tfrac{1}{4}$ chance, he wins with a probability of $\tfrac{3}{4}$ . Hence, the total the probability here is $$ \frac{2}{3}\cdot\frac{3}{4}\cdot\left(\frac{3}{4}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{3}{4}\right)=\frac{7}{32}. $$ In all, the probability that Carl wins the tournament is $\tfrac{1}{12}+\tfrac{7}{32}=\tfrac{29}{96}$ , which makes the requested answer $\boxed{125}$ .
|
[
"Harder than p12 tbh.\n\nTook me ten minutes to realize it doesn't matter whether Jon or Sergey wins :P ",
"Anyone else got $\\frac{25}{96}$ ?",
"I got 29/96",
"yeah i got that too\nidk why i said i got 25/96? \n\n<details><summary>AAAAAAAAAAAAAAAA</summary>1. azar vs carl in first round $\\to \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{3}{4} = \\frac{1}{12}$ 2. azar vs carl in finals $\\to \\frac{2}{3} \\cdot \\frac{3}{4} \\cdot \\frac{3}{4} \\cdot \\frac{1}{3} = \\frac{1}{8}$ 3. carl vs someone else in finals $\\to \\frac{2}{3} \\cdot \\frac{3}{4} \\cdot \\frac{1}{4} \\cdot \\frac{3}{4} = \\frac{3}{32}$ $\\frac{1}{12} + \\frac{1}{8} + \\frac{3}{32} = \\boxed{\\frac{29}{96}}$</details>",
"also getting 29/96",
"Oops did not know “random assignment” mattered and got $\\frac{29}{32}$ at first\n\nBasically for each case we have a $\\frac{1}{3}$ chance of actually having the case. The final probability is $$ \\frac{1}{3} \\left( \\frac{1}{4} + 2 \\cdot \\frac{3}{4} \\cdot \\left( \\frac{3}{4} \\cdot \\frac{1}{3} + \\frac{1}{4} \\cdot \\frac{3}{4} \\right) \\right) $$ $$ \\implies \\frac{29}{96} \\implies \\boxed{125} $$ ",
"Case 1: Semifinal matchup is $A\\to C$ and $S\\to J$ . \nThen it's just $\\frac{1}{3}\\cdot \\frac{1}{3}\\cdot 1\\cdot \\frac{3}{4}=\\frac{1}{12}$ . \n\nCase 2: Semifinal matchup is $A\\to S$ and $C\\to J$ or $A\\to J$ and $C\\to S$ . \nThen it's \\[\\frac{2}{3}\\cdot \\frac{3}{4} \\cdot \\frac{1}{4}\\cdot \\frac{3}{4}+\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4} \\cdot \\frac{1}{3}=\\frac{3}{32}+\\frac{1}{8}=\\frac{7}{32}\\]\n\nSo the answer is $\\frac{7}{32}+\\frac{1}{12}=\\frac{21}{97}+\\frac{8}{96}=\\frac{29}{96}\\implies \\boxed{125}$ .",
"<details><summary>Solution</summary>We proceed with casework: $$ \\frac{1}{3}\\cdot \\frac{1}{3}\\cdot \\frac{3}{4}+\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4}\\cdot \\frac{1}{3}+\\frac{2}{3}\\cdot \\frac{3}{4}+\\frac{1}{4}+\\frac{3}{4}=\\frac{1}{12}+\\frac{1}{8}+\\frac{7}{32}=\\frac{29}{96} $$ Therefore, the final answer is $29+96=\\boxed{125}.$</details>",
"<blockquote>\n\nCase 2: Semifinal matchup is $A\\to S$ and $C\\to J$ or $A\\to J$ and $C\\to S$ . \nThen it's \\[\\frac{2}{3}\\cdot \\frac{3}{4} \\cdot \\frac{1}{4}\\cdot \\frac{3}{4}+\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4} \\cdot \\frac{2}{3}=\\frac{3}{32}+\\frac{1}{8}=\\frac{7}{32}\\]\n\n</blockquote>\n\nShould be \\[\\frac{2}{3}\\cdot \\frac{3}{4} \\cdot \\frac{1}{4}\\cdot \\frac{3}{4}+\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4} \\cdot \\frac{1}{3}=\\frac{3}{32}+\\frac{1}{8}=\\frac{7}{32}\\] ( $\\frac{1}{3}$ instead of $\\frac{2}{3}$ )",
"Mocked it and got 29/96 -> 125\n\n@3above I think you typoed. it isnt 21/97 its 21/96",
"[quote name=\"Geometry285\" url=\"/community/p24448531\"]\nOops did not know “random assignment” mattered and got $\\frac{29}{32}$ at first\n\nBasically for each case we have a $\\frac{1}{3}$ chance of actually having the case. The final probability is $$ \\frac{1}{3} \\left( \\frac{1}{4} + 2 \\cdot \\frac{3}{4} \\cdot \\left( \\frac{3}{4} \\cdot \\frac{1}{3} + \\frac{1}{4} \\cdot \\frac{3}{4} \\right) \\right) $$ $$ \\implies \\frac{29}{96} \\implies \\boxed{125} $$ </blockquote>\n\nsame, and I was so confused as to how C had a higher chance of winning than A.",
"How did I get this wrong :(",
"<blockquote><blockquote>\n\nCase 2: Semifinal matchup is $A\\to S$ and $C\\to J$ or $A\\to J$ and $C\\to S$ . \nThen it's \\[\\frac{2}{3}\\cdot \\frac{3}{4} \\cdot \\frac{1}{4}\\cdot \\frac{3}{4}+\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4} \\cdot \\frac{2}{3}=\\frac{3}{32}+\\frac{1}{8}=\\frac{7}{32}\\]\n\n</blockquote>\n\nShould be \\[\\frac{2}{3}\\cdot \\frac{3}{4} \\cdot \\frac{1}{4}\\cdot \\frac{3}{4}+\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4} \\cdot \\frac{1}{3}=\\frac{3}{32}+\\frac{1}{8}=\\frac{7}{32}\\] ( $\\frac{1}{3}$ instead of $\\frac{2}{3}$ )</blockquote>\n\nEdited",
"If Carl, plays against, Azar, first:\nThe, probability, of this happening, is $\\frac{1}{3}$ , now, Carl, has a $\\frac{1}{3}$ , chance, of winning against, Azar. Then, Carl, beats, either of the finalsts, with probability, $\\frac{3}{4}$ , hence, we have, $\\frac{1}{3}\\cdot \\frac{1}{3}\\cdot \\frac{3}{4}=\\frac{1}{12}.$ If Carl, plays against, either, Jon or Sergey, first:\nThe probability, of this happening, is $\\frac{2}{3}$ , now, Carl, wins with probability, $\\frac{3}{4}$ , now if Azar, is in the finals, with a $\\frac{3}{4},$ chance, Carl, wins with $\\frac{1}{3},$ chance. Hence, $\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{3}{4}\\cdot \\frac{1}{3}=\\frac{18}{144}=\\frac{1}{8}.$ Now, if his opponent, in the finals, is Jon or Sergey, this happens with $\\frac{1}{4}$ , chance, then, Carl, wins with probability, $\\frac{3}{4}$ , hence, $\\frac{2}{3}\\cdot \\frac{3}{4}\\cdot \\frac{1}{4}\\cdot \\frac{3}{4}=\\frac{18}{192}=\\frac{3}{32}$ , hence, $\\frac{1}{8}+\\frac{3}{32}=\\frac{7}{32}.$ Hence, $\\frac{1}{12}+\\frac{7}{32}=\\frac{29}{96} \\implies 29+96=\\boxed{125}.$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
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{
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782928.json"
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|
A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
|
Let the pyramid be $ABCDE$ with $O$ the center of its circumsphere, where square $BCDE$ is the base and has center $P$ . Furthermore, let $R$ denote the circumradius of $ABCDE$ .
As $ABCDE$ has volume $54$ , while its base $BCDE$ has area $6^2=36$ , the corresponding height $AP$ has length $\tfrac{3\cdot54}{36}=\tfrac{9}{2}$ . As a result, $OP=AP-AO=\tfrac{9}{2}-R$ , and since $BP=3\sqrt{2}$ , by the Pythagorean theorem on $BOP$ we have that $$ BO=\sqrt{BP^2+OP^2}=\sqrt{(3\sqrt{2})^2+\left(\frac{9}{2}-R\right)^2}=\sqrt{R^2-9R+\frac{153}{4}}. $$ Setting this equal to $R$ and squaring, $R^2-9R+\tfrac{153}{4}=R^2$ , so $R=\tfrac{17}{4}$ . The requested answer is then $\boxed{021}$ .
|
[
"cross section trivializes",
"I got 21? 17/4?",
"I didn't take the test(87 on 10A and 10B) , but man does this look easy...\nCircumradius formula and cross-section trivializes :|",
"Also got 021",
"wait how\nI got $r^2=(3\\sqrt{2})^2+(r-3)^2$ so $r=9/2$ oh the height is 9/2\nim dumb",
"How would r be 9/2? The height is 9/2..",
"am also getting 17/4",
"umm I got $\\fbox{017}$ from 13/4",
"I also got $\\frac{13}{4}$ at first, but I quickly realized my mistake (phew!)\n\n<details><summary>My Solution</summary>Let $P$ be a point on $ABCD$ so that the pyramid with apex $T$ satisfies $TP \\perp ABCD$ . Now the length of $TP$ is the height of the pyramid, which is\n\\[ \\frac{1}{3} \\times 6^2 \\times h = 54 \\iff h = \\frac{9}{2}. \\]\nLet $R$ be the radius of the sphere. Notice the centre of the sphere lies on $TP$ , since $P$ is equidistant to $A,B,C,D$ and all points on the \"perpendicular bisector\" of the centre of the base is equidistant.\n\nThen, $\\left ( \\frac{9}{2} - R \\right)^2 + 3^2 + 3^2 = R^2 \\iff R = \\frac{17}{4}$ , so the answer is $\\boxed{021}$ .</details>\n\nI agree this might not be very difficult for the AIME, perhaps because last week's AIME was too hard for the first 5 problems, maybe they overcompensated a bit here?",
"Clearly the “circumcenter” must be at the center of the pyramid, or in other words must lie on the height of the pyramid. If the distance to the apex is $x$ , then after finding the height $h=\\frac{9}{2}$ we have $$ x^2=(\\tfrac{9}{2}-x)^2+18 \\implies 9x=18+\\frac{81}{4} \\implies x=2+\\frac{9}{4} = \\frac{17}{4} $$ $$ 17+4=\\boxed{021} $$ ",
"this was like 10x easier than AIME1 #3 geo",
"first 5 so easy compared to the other test",
"Did this feel more like an AMC style problem for anyone else?",
"Why wasn't this on the AIME I?\n\nNote that the height must be $\\frac{54\\cdot 3}{36}=\\frac{9}{2}$ . \n\nLet the vertices of the square base be $(0,0,0), (0,6,0), (6,0,0), (6,6,0)$ . So the apex is $\\left(3,3,\\frac{9}{2}\\right)$ . Now we note that the center of the sphere is on $(3,3,x)$ for some $x<\\frac{9}{2}$ . \n\nThe distance from $(3,3,x)$ to $(0,0,0)$ is $\\sqrt{x^2+18}$ and the distance from $x$ to the apex is $\\frac{9}{2}-x$ . \n\nThus, \\[\\left(x-\\frac{9}{2}\\right)^2=x^2+18\\implies \\frac{81}{4}-9x=18\\implies 9x=\\frac{9}{4}\\implies x=\\frac{1}{4}.\\]\n\nSo it's just $\\frac{9}{2}-\\frac{1}{4}=\\frac{17}{4}\\implies \\fbox{021}$ . ",
"<blockquote>Did this feel more like an AMC style problem for anyone else?</blockquote>\n\nFax bro this one was a 20 sec solve ",
"I just drew three point in a straight line as P(Apex point ) ,O(center of sphere) and H the center of square \nLet THE SQUARE BE ABCD AS FOLLOWS NOW THE DISTANCE OM=√R^2-18 \nWHERE R IS THE RADIUS \nCH IS JUST HALF OF THE DIAGONAL =3√2 \nNOW , OM+PO=9/2 \n√R^2-18+R=9/2 \n√R^2-18=9/2-R \nSQUARING BOTH SIDES AND SIMPLIFYING\n9R=81/4+18 \nWHICH FURTHER GIVES R=17/4 \nHENCE (17+4)=21 OUR ANSWER\n"
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1042,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782929.json"
}
|
There is a polynomial $P(x)$ with integer coefficients such that $$ P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)} $$ holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$
|
<details><summary>Solution</summary>We will use the fact that $x^{kn} - 1 = (x^k - 1)(x^{k(n-1)} + x^{k(n-2)} + ... + x^k + 1)$ . We can express $(x^{2310}-1)$ in this manner 4 different ways, with $(k,n)$ being $(105,22)$ , $(70,33)$ , $(42,55)$ , and $(30,77)$ . After cancelation we have $$ P(x) = (x^{105 \cdot 21} + x^{105 \cdot 20} + ... + x^{105} + 1)(x^{70 \cdot 32} + ... + 1)(x^{42 \cdot 54} + ... + 1)(x^{30 \cdot 76} + ... + 1)(x^{2310}-1)^2. $$ Our problem boils down to finding the number of ways we can multiply one term from each grouping to yield $x^{2022}$ , or, equivalently, the number of ways to add the exponents of one $x$ term from each grouping to total $2022$ .
Note that the exponent from the first grouping will be in the form $105a$ where $a$ is a nonegative integer less than $22$ . Similar arguments are made for the next three groupings. The $(x^{2310}-1)^2$ grouping will never contribute to this sum, as its smallest term is $-2x^{2310}$ . Thus, we have the following equation over the nonegative integers: $$ 105a + 70b + 42c + 30d = 2022 $$ where $a<22$ , $b<33$ , $c<55$ , and $d<77$ . Examine this equation modulo $2$ , $3$ , $5$ , and $7$ to derive $a \equiv 0 \text{ (mod } 2 \text{)}$ , $b \equiv 0 \text{ (mod } 3 \text{)}$ , $c \equiv 1 \text{ (mod } 5 \text{)}$ , and $d \equiv 3 \text{ (mod } 7 \text{)}$ , respectively. Let $a=2w$ , $b=3x$ , $c=5y+1$ , and $d=7z+3$ . We now have $$ 105(2w) + 70(3x) + 42(5y+1) + 30(7z+3) = 2022 $$ $$ \Rightarrow w+x+y+z=9 $$ where each of $w$ , $x$ , $y$ , and $z$ is a nonegative integer less than 11. We can represent this as having $9$ stars and $3$ bars, with the bars splitting the stars into $4$ groups corresponding to the values of $w$ , $x$ , $y$ , and $z$ . Our answer is thus $\frac{12!}{(9!)(3!)} = \boxed{220}$ .</details>
|
[
"2022=30a+42b+72c+105d\n\nI wrote $\\frac{1}{1-x^n}$ as $1+x+x^2+...$ I think...",
"I got 220 for this one, can somebody please confirm? Also I think the last power was 30 @op?",
"<blockquote>I got 220 for this one, can somebody please confirm? Also I think the last power was 30 @op?</blockquote>\n\no oops",
"<details><summary>Sketch</summary>Reduces to $(x^{2310}-1)^2 (1+x^{105}+\\cdots)(1+x^{70}+\\cdots)(1+x^{42}+\\cdots)(1+x^{30}+\\cdots)$ . Note that the remaining two factors of $x^{2310}-1$ do not matter anymore. A subsequent casework bash gives $220$ as the answer.</details>",
"<blockquote>2022=30a+42b+72c+105d\n\nI wrote $\\frac{1}{1-x^n}$ as $1+x^n+x^{2n}...$ I think...</blockquote>\n\nI got 12C3 as well",
"<blockquote>2022=30a+42b+72c+105d\n\nI wrote $\\frac{1}{1-x^n}$ as $1+x+x^2+...$ I think...</blockquote>\n\nI did that but forgot to consider the case where d=0, so I said 165 instead of 220 :( :( :(",
"30a + 42b + 70c + 105d = 2022\ndo some mods stuff \na = 7w + 3\nb = 5x + 1\nc = 3y\nd = 2z\n\nthen w + x + y + z = 9 -> 12c3 = 220",
"So it's \\[(1+x^{105}+x^{210}+\\ldots+x^{2205})(x^{70}+x^{140}+\\ldots+x^{2240})(x^{42}+x^{84}+\\ldots+x^{2268})(x^{30}+x^{60}+\\ldots+x^{2280})(x^{4620}-2x^{2310}+1).\\]\n\nThus, $105a+70b+42c+30d=2022$ . \n\nBy mod $2$ , we have $a=2p$ . \n\nBy mod $3$ , we have $b=3q$ . \n\nBy mod $5$ , we have $c=5r+1$ . \n\nBy mod $7$ , we have $d=7s+3$ . \n\nSo $210p+210q+210r+210s+42+90=2022\\implies 210(p+q+r+s)=1890\\implies p+q+r+s=9$ . $\\binom{12}{3}=\\boxed{220}$ ways by Stars and Bars.",
"the answer is the chapter number in kaguya sama love is war manga where kaguya and miyuki ",
"the post is missing a period xd",
"I did some weird mod stuff to 2022=30a+42b+72c+105d, but was wondering what the general way to solve aw+bx+cy+dz=k was, and possibly with more terms.",
"<blockquote>I did some weird mod stuff to 2022=30a+42b+72c+105d, but was wondering what the general way to solve aw+bx+cy+dz=k was, and possibly with more terms.</blockquote>\n\nI believe the way you do it is through generating functions\nand yes, I did the sketch sol too\nHere's what I did:\n<details><summary>Sol</summary>We see that $30, 42, 70, 105$ are all factors of $210$ . $30 = \\frac{210}{7}$ , $42 = \\frac{210}{5}$ , $70 = \\frac{210}{3}$ , $105 = \\frac{210}{2}$ . Thus, we see that any multiple of 210 is 0 mod 2, 3, 5, and 7. However, 2022 is 0 mod 2, 0 mod 3, 2 mod 5, and 6 mod 7. So, we must include a 42 and 3 30s into this sum, making the remaining part 1890. This is $9\\cdot 210$ , and using stars and bars gives ${12\\choose 3} = 220$</details>",
"This was my problem, hope you all enjoyed it. It was originally just the equation $2022 = 30a+42b+70c+105d$ , but then I had the idea to wrap it inside a generating function.",
"Nice problem, but too easy for a 13. I think #9 is much harder than this, but maybe that's just me. Since $0<x<1$ , the formula for an infinite geometric series holds so we may write\n\\[ P(x)=(x^{2310}-1)^6(1+x^{105}+x^{210}+\\dots)(1+x^{70}+x^{140}+\\dots)(1+x^{42}+x^{84}+\\dots)(1+x^{30}+x^{60}+\\dots). \\] $(x^{2310}-1)^6$ is now irrelevant since all of its terms are too big, so effectively, we can write\n\\[ P(x)=(1+x^{105}+x^{210}+\\dots)(1+x^{70}+x^{140}+\\dots)(1+x^{42}+x^{84}+\\dots)(1+x^{30}+x^{60}+\\dots).\\]\nIt's not hard to see this boils down to the number of sols in nonnegative integers to the equation $105a+70b+42c+30d=2022$ . Rewriting this as $(3\\cdot 5\\cdot 7)a+(2\\cdot 5\\cdot 7)b+(2\\cdot 3\\cdot 7)c+(2\\cdot 3\\cdot 5)d=2022$ , we find $a\\equiv 0\\pmod{2}, b\\equiv 0\\pmod{3}, c\\equiv 1\\pmod{5}, d\\equiv 3\\pmod{7}$ , so write $a=2a_1, b=3b_1, c=5c_1+1, d=7d_1+3$ for nonnegative integers $a_1, b_1, c_1, d_1$ . Finally, we have\n\\[ (3\\cdot 5\\cdot 7)(2\\cdot a_1)+(2\\cdot 5\\cdot 7)(3\\cdot b_1)+(2\\cdot 3\\cdot 7)(5c_1+1)+(2\\cdot 3\\cdot 5)(7d_1+3)=2022\\implies 210(a_1+b_1+c_1+d_1)=1890\\implies a_1+b_1+c_1+d_1=9,\\]\nafter simplification. By stars and bars, this is $\\tbinom{12}{3}=220$ .",
"same solution as everyone else but oh well\n\nThe problem reduces to finding all ordered quadruplets $(a, b, c, d)$ of nonnegative integers satisfying $105a+70b+42c+30d=2022$ . Then we must have $a\\equiv 0\\pmod 2, b\\equiv 0\\pmod 3, c\\equiv 1\\pmod 5, d\\equiv 3\\pmod 7$ so that $a=2w,b=3x,c=5y+1,d=7z+3\\implies w+x+y+z=9$ . Stars and bars gives $\\binom{12}{3}=\\boxed{220}$ , and we are done.",
"[quote name=\"v_Enhance\" url=\"/community/p24456812\"]\nThis was my problem, hope you all enjoyed it. It was originally just the equation $2022 = 30a+42b+72c+105d$ , but then I had the idea to wrap it inside a generating function.\n</blockquote>\n\nWhat was the proper way to solve that equation. When I did i remember doing like 5 modulo analysis's and then recursing back up and somehow getting the right answer. but I feel like there is a proper \"slick\" way to do this.",
"i feel like that is the slick way",
" $$ P(x)=\\left(1+x^{105}+x^{105\\cdot 2}+...+x^{105\\cdot 22}\\right)\\left(1+x^{70}+x^{70\\cdot 2}+...+x^{70\\cdot 32}\\right)\\left(1+x^{42}+x^{42\\cdot 2}+...+x^{42\\cdot 54}\\right)\\left(1+x^{30}+x^{30\\cdot 2}+...+x^{30\\cdot 76}\\right)(x^{2310}-1)^2 $$ This generalizes $105a+70b+42c+30d=2022$ . Since $2022\\equiv 6\\pmod{7}$ , therefore, $d\\equiv 3\\pmod{7}$ . Substitute it in, we get $105a+70b+42c+30(7z+3)=105a+70b+42c+210z=1932=276\\cdot 7$ . This simplifies to $15a+10b+6c+30z=276$ . $276\\equiv 0\\pmod{3}$ , therefore, $3\\mid b$ . Let $b=3x$ , then we have $5a+10x+2c+10z=92$ . By looking at the parity, then $2\\mid a$ . Thus, let $a=2y$ , then we have $10(x+y+z)+2c=92$ . This is basically asking how many ordered pair of nonnegative integers $(x,y,z)$ gives sum of less than or equal to $9$ . Thus, the answer is ${11\\choose 2}+...+{2\\choose 2}={12\\choose 3}=\\boxed{220}$ .",
"<blockquote>So it's \\[(1+x^{105}+x^{210}+\\ldots+x^{2205})(x^{70}+x^{140}+\\ldots+x^{2240})(x^{42}+x^{84}+\\ldots+x^{2268})(x^{30}+x^{60}+\\ldots+x^{2280})(x^{4620}-2x^{2310}+1).\\]\n\nThus, $105a+70b+42c+30d=2022$ . \n\nBy mod $2$ , we have $a=2p$ . \n\nBy mod $3$ , we have $b=3q$ . \n\nBy mod $5$ , we have $c=5r+1$ . \n\nBy mod $7$ , we have $d=7s+3$ . \n\nSo $210p+210q+210r+210s+42+90=2022\\implies 210(p+q+r+s)=1890\\implies p+q+r+s=9$ . $\\binom{12}{3}=\\boxed{220}$ ways by Stars and Bars.</blockquote>\n\nI used this solution. The motivation for me, in using this solution, was that for any three out of four of the exponents, there was a GCD (greatest common divisor) that could be used to reduce the $2022$ exponent to make the problem simpler in terms of casework and computation, and then wishful thinking led me to realize that modular arithmetic forced specific residues for this common factor of the other three exponents. Then solve this like a Diophantine equation, then get a Stars and Bars/Sticks and Stones equation that can be solved to get the answer. This was my favorite problem on the AIME II.",
"Why didn't I take the II, this problem was literally trivial",
"Overcomplicated engineer sol here we go. I got that $w+x+y+z=9$ but then I didn't think that all of those solutions would satisfy the original. :clown:\n\nJust like everybody else, notice that we are trying to find the number of solutions to the equation $105a+70b+42c+30d=2022$ . Notice that $2\\mid a$ and $0\\le a\\le 18$ . So we do some casework bash. If $a=18$ we find one solution, namely $(18, 0, 1, 3)$ . Now notice that when we decrease $a$ by 2, we're decreasing the entire sum by 210. So in order to make up for this, either we increase $b$ by 3, increase $c$ by 5 or increase $d$ by 7. Any other combination of the three numbers wouldn't work. So we continue with our casework. If $a=16$ we find 3 solution\n\n \\begin{tabular}{|c|c|}\n \\hline\n $a$ & Number of solutions given $a$ [0.5ex]\n \\hline\n 18 & (18, 0, 1, 3) \n 16 & (16, 3, 1, 3), (16, 0, 6, 3), (16, 0, 1, 10) \n 14 & (14, 6, 1, 3), (14, 3, 6, 3), (14, 3, 1, 10), (14, 0, 11, 3), (14, 0, 6, 10), (14, 0, 1, 17) \n \\hline\n \\end{tabular}\nDoing the same for $a=12$ gives us 10 solutions. Now do you see what I'm seeing? 1, 3, 6, 10. Do these numbers look familiar? Of course! It is just $\\binom{n}{2}$ starting from $n=2$ ! So we know 18 corresponds with $\\binom{2}{2}$ and 16 corresponds with $\\binom{3}{2}$ so then 0 corresponds with $\\binom{11}{2}$ . Then we have\n\\[ \\binom{2}{2}+\\binom{3}{2}+\\dots+\\binom{11}{2}=\\binom{12}{3} \\] from Hockey Stick.",
"Using $\\tfrac{x^{mn}-1}{x^n-1}=x^{(m-1)n}+x^{(m-2)n}+\\dots+x^n+1$ , we can simplify $P(x)$ as follows: $$ P(x)=(x^{21\\cdot105}+x^{20\\cdot105}+\\dots+x^{105}+1)(x^{32\\cdot70}+x^{31\\cdot70}+\\dots+x^{70}+1)(x^{54\\cdot42}+x^{53\\cdot42}+\\dots+x^{53}+1)(x^{76\\cdot30}+x^{75\\cdot30}+\\dots+x^{30}+1)(x^{2310}-1)^2. $$ We seek the number of ways to choose a term from each of the $6$ factors above such that the product of all $6$ terms is $x^{2022}$ . A term from the first factor is of the form $x^{105a}$ , where $a$ is nonnegative; similarly, terms from the second, third, and fourth factors are of the form $x^{70b}$ , $x^{42c}$ , and $x^{30d}$ , respectively. The terms chosen from the last two factors are forced to be $-1$ , since $x^{2310}$ causes the degree of the term to already be too high. Hence, we seek the number of solutions in nonnegative integers to the equation $$ 105a+70b+42c+30d=2022. $$ Motivated by the fact that $22\\cdot105=33\\cdot70=55\\cdot42=77\\cdot30=2310$ from above, we see that each of $105$ , $70$ , $42$ , and $30$ aren't divisible by a prime that is divisible by the other three. These $4$ such primes are in fact $2$ , $3$ , $5$ , and $7$ , and taking the equation modulo each of these primes yields that $a=2a'$ , $b=3b'$ , $c=5c'+1$ , and $d=7d'+2$ for some nonnegative $a'$ , $b'$ , $c'$ , and $d'$ . Substituting, $$ 105(2a')+70(3b')+42(5c'+1)+30(7d'+3)=210a'+210b'+210c'+210d'+132=2022\\iff a'+b'+c'+d'=9. $$ By Stars and Bars, the number of solutions $(a,b,c,d)$ in nonnegative integers to the above equation is $\\tbinom{9+4-1}{4-1}=\\boxed{220}$ , as requested.",
"<details><summary>solution</summary>Note that $$ P(x)=(x^{2310}-1)^7\\cdot (x^{2280}+x^{2250}\\cdots+1)\\cdot (x^{2268}+x^{2226}+\\cdots+1)\\cdot (x^{2240}+x^{2170}+\\cdots+1)\\cdot (x^{2205}+x^{2100}+\\cdots+1). $$ The absolute value of the coefficient of $x^{2022}$ in $P(x)$ is therefore the same as the absolute value of the coefficient of $x^{2022}$ in $$ (x^{2280}+x^{2250}\\cdots+1)\\cdot (x^{2268}+x^{2226}+\\cdots+1)\\cdot (x^{2240}+x^{2170}+\\cdots+1)\\cdot (x^{2205}+x^{2100}+\\cdots+1). $$ We have now reduced the problem to finding the number of solutions to $$ 30a+42b+70c+105d=2022, $$ where $a,b,c,$ and $d$ are nonnegative integers. Taking mod $2,$ we find that $d\\equiv 0\\pmod {2}.$ Taking mod $3,$ we find that $c\\equiv 0\\pmod {3}.$ Taking mod $5,$ we find that $b\\equiv 1\\pmod {5}.$ Taking mod $7,$ we find that $a\\equiv 3\\pmod {7}.$ \n\nLet $a=7a'+3,$ $b=5b'+1,$ $c=3c',$ and $d=2d'.$ Note that $a',b',c',$ and $d'$ are all nonnegative integers. Then, the equation $$ 30a+42b+70c+105d=2022 $$ reduces to $$ a'+b'+c'+d'=9, $$ which has $220$ solutions by Stars and Bars.</details>",
"[Video Solution](https://youtu.be/v2SFCqWOBjs)",
"wait what this is actually so clean\n\n@below I think this problem is mostly just NT, the algebra part is non-trivial (especially if you are well-versed with Generating Functions)**Claim**: $\\frac{1}{x^n-1}=(1+x^n+ \\cdots )$ Proof is *self-explanatory*\n\nFrom this we get that the numerator's power is far too large for it to contribute anything to the sum. Therefore $105a+70b+42c+30d=2022$ where $a$ , $b$ , $c$ , and $d$ are the number of times its corresponding power contributes to the sum. We are motivated by mods, so we try to isolate conditions for each variable based on taking them modulo the greatest common divisor of the other three elements combined. This is shown below: $$ \\gcd(70,42,30)=2 \\implies a \\equiv 0 \\mod 2 $$ $$ \\gcd(105,42,30)=3 \\implies b \\equiv 0 \\mod 3 $$ $$ \\gcd(105,70,30)=5 \\implies c \\equiv 1 \\mod 5 $$ $$ \\gcd(105,70,42)=7 \\implies d \\equiv 3 \\mod 7 $$ Incorporating these conditions into a new equation gives $210(a'+b'+c'+d')=2022-(90+42) \\implies a'+b'+c'+d'=9$ From here there are no restrictions so by stars and bars the answer is $\\binom{12}{3} \\implies \\boxed{220}$ ",
"I agree that the number theory part of this is very clean, but tf is the flavortext of this problem",
"Since $2310 > 2022$ , we just want the number of solutions to $30a+42b+70c+105d=2022$ over nonnegative integers $a,b,c,d$ . Taking $\\pmod5$ gives $b=5e+1$ so $6a+42e+14c+21d=396$ . Taking $\\pmod7$ gives $a=7f+3$ so $6f+6e+2c+3d=54$ . Taking $\\pmod 3$ gives $c=3g$ so $2f+2e+2g+d=18$ . Parity gives $d=2h$ so $f+e+g+h=9$ . Therefore, our answer is $\\binom{12}{3}=220$ .",
"<blockquote>\nThere is a polynomial $P(x)$ with integer coefficients such that $$ P(x)=\\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)} $$ holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ .\n</blockquote>\nWrite $P(x)$ as\n\\[(x^{2310}-1)^2\\left(\\sum_{i=0}^{21}x^{105i}\\right)\n\\left(\\sum_{i=0}^{32}x^{70i}\\right)\n\\left(\\sum_{i=0}^{54}x^{42i}\\right)\n\\left(\\sum_{i=0}^{76}x^{30i}\\right)\\]\nand since $2310>2022$ , we have to find the number of nonnegative integer solutions to the linear diophantine equation\n\\[105 a + 70 b + 42 c + 30 d = 2022.\\]\nPrime factorize each of the constants\n\\begin{align}3\\cdot 5\\cdot 7 \\cdot a + 2\\cdot 5 \\cdot 7\\cdot b + 2\\cdot 3 \\cdot 7\\cdot c+ 2\\cdot 3\\cdot 5 \\cdot d = 2022\n\\end{align}\nand take mods:\n\\begin{align*}\n\\pmod 2&\\implies a\\equiv 0\\pmod 2 \n\\pmod 3&\\implies b\\equiv 0\\pmod 3 \n\\pmod 5&\\implies c\\equiv 1\\pmod 5 \n\\pmod 7&\\implies d\\equiv 3\\pmod 7\n\\end{align*}\nwhich motivates the substitutions $a = 2a'$ , $b = 3b'$ , $c = 5c' + 1$ , $d = 7d' + 3$ . Implementing these into $(1)$ , simplifying things, and dividing by $2\\cdot 3\\cdot 5\\cdot 7 = 210$ gives\n\\[a'+b'+c'+d'=9\\]\nwhich has $\\tbinom{9+4-1}{4-1}=\\boxed{220}$ solutions by stars and bars.",
"They really included $0<x<1$ for no reason",
"[quote name=\"Sagnik123Biswas\" url=\"/community/p28738420\"]\nThey really included $0<x<1$ for no reason\n</blockquote>\n\nThere was actually a reason, but not a good one: it was to make the statement shorter. Issue is that when $x = \\pm1$ the right-hand side is nominally undefined, so I thought it was shorter to just write $0 < x < 1$ rather than \"all real numbers $x$ other than $\\pm 1$ \", since they're obviously equivalent.",
"We convert the terms in the denominator by expressing them as infinite geometric series - essentially a generating functions argument. Since we want the coefficient of $x^{2022}$ , the term $\\left(x^{2310} - 1\\right)^{11}$ is irrelevant other than the constant term of $-1$ , which can be ignored since we seek the absolute value.\n\nThe problem reduces to finding the number of solutions to \\[30a + 42b + 70c + 105d = 2022.\\] We use mod 2, 3, 5, and 7 to get \\[e + f + g + h = 9 \\implies \\binom {12}{3} = \\boxed{220}.\\]",
"<details><summary>Solution</summary>Using the fact that $$ \\frac{a^{n}-1}{a-1} = a^{n-1}+...+1, $$ our expression is equivalent to $$ (x^{2310}-1)^{2}(x^{2205}+...+1)(x^{2240}+...+1)(x^{2268}+...+1)(x^{2280}+...+1). $$ Then, notice that since $2022<2310,$ the term of $(x^{2310}-1)^{2}$ will contribute a $(-1)^{2}=1$ factor to the coefficient of $x^{2022}$ , and hence it can be neglected. \nNow, note that $2022$ is smaller than all the largest terms in each of the terms. Also, we can see that every term in the first expression, with a nonzero coefficient has a degree that is a multiple of $105$ , and similarly for the others. Thus, let $105x$ be the degree of the term used to form $2022$ , and define $70y$ , $42z$ , and $30w$ similarly. To find the coefficient of $x^{2022}$ , we must find the number of positive integer solutions to $$ 30w+42z+70y+105x=2022. $$ Taking modulo $7,$ we have that $x \\equiv 3(\\textrm{mod }7),$ and so we let $x = 7x_{1}+3.$ Similarly, $y =5y_{1}+1$ , $z_{1} = 3z_{1}$ , and $w = 2w_{1}.$ Now, substituting this into our original diophantine equation, we conclude that $$ w_{1}+x_{1}+y_{1}+z_{1} = 9, $$ and then by Stars and Bars, there are ${12 \\choose 3} = \\boxed{220}$ solutions, and we are done.<details><summary>Click to expand</summary></details></details>",
"So misplaced.\n\nThe extra two factors of $x^{2310}-1$ on the top die because $2022 < 2310$ so we want to count the number of solutions to $105a+70b+42c+30d=2022$ . It's not hard to see that $c \\equiv 1 \\pmod{5}$ so if $c = 5c' + 1$ then $21a+14b+42c' + 6d=396$ . Then $b \\equiv = \\pmod{3}$ so if $b=3b'$ then $7a+14b'+14c'+2d=132$ . Then $d \\equiv 3 \\pmod{7}$ so if $d = 7d' + 3$ then $a+2b'+2c'+2d'=18$ . Finally $a$ is even so if $a=2a'$ then $a'+b'+c'+d'=9$ from which stars and bars gives us $\\binom{12}{3} = \\boxed{220}$ ",
"<details><summary>Cute</summary>$p(x) = (x^{2310} - 1)^6(1 + x^{105} + x^{210} + ......)(1 + x^{70} + x^{140} + ......)(1 + x^{42} + x^{84} + .....)(1 + x^{30} + x^{60} + .....)$ since $0 < x < 1$ Notice that there's no term containing $x^{2022}$ in the expansion of $(x^{2310} - 1)^6$ Now, $p(x) = (1 + x^{105} + x^{210} + ......)(1 + x^{70} + x^{140} + ......)(1 + x^{42} + x^{84} + .....)(1 + x^{30} + x^{60} + .....)$ \n\nSo, the problem reduces to finding the non negative integral solutions of the equation $105a + 70b + 42c + 30d = 2022$ $105a + 70b + 42c + 30d = 2022 ----{1}$ By mod $2$ , a = 0(mod $2$ ) $a = 2x$ By mod $3$ , $b = 0$ (mod $3$ ) $b = 3y$ By mod $5$ , $c = 1$ (mod $5$ ) $c = 5z + 1$ By mod $7$ , $d = 3$ (mod $7$ ) $d = 7w + 3$ On plugging the values of $a,b,c$ and $d$ in equation ${1}$ , we get $210(x + y + z + w) = 1890$ $x + y + z + w = 9$ Stars and Bars gives $\\binom{12}{3} = \\boxed{220}$ :D</details>",
"My solution is probably similar/identical to others but i just thought i did this problem so many times i might aswell posrt my sol:\n\nNotice that the numerator does not contribute to the coefficient of $x^{2022}$ . I have done this problem dozens of times now, and I do it same every time. The denominator can be re-written as $\\frac{1}{(1-x^{30})(1-x^{42})(1-x^{70})(1-x^{105})}$ . From generating functions experience or infnite geometric series, this equals $(1+x^{30}+x^{60}+\\dots+)(1+x^{42}+x^{84}+\\dots+)(1+x^{70}+x^{140}+\\dots+)(1+x^{105}+x^{210}+\\dots+)$ . Now let the choice of exponent in each expresion be $30a, 42b, 70c, 210d$ where $a,b,c,d$ are all non-negative integers. We want the number of solutions to $30a+42b+70c+105d=2022$ . Upon taking modulo $7, 5, 3, 2$ respectively reveals we have the following substitutions: $a=7a'+3, b=5b'+1, c=3c', d=2d'$ . Substituting all of these in and simplifying reveals we are searching for solutions to $a'+b'+c'+d'=9$ (each of which has a unique $(a, b, c, d)$ quadruple) which by stars and bars has $\\binom{12}{3}=220$ solutions.\n",
"Should've been switched with #9 ",
"We can write that $P(x) = (x^{2310}-1)^2(1+x^{30}+\\dots)(1+x^{42}+\\dots)(1+x^{70}+\\dots)(1+x^{105}+\\dots)$ , where the dots are due to the fact that we only need the coefficient of $x^{2022}$ . In addition, we can also drop the first term for the same reason, so we only need to consider \\[P(x) = (1+x^{30}+\\dots)(1+x^{42}+\\dots)(1+x^{70}+\\dots)(1+x^{105}+\\dots).\\] This is equivalent to finding the number of quadruplets of non-negative integers $a$ , $b$ , $c$ , $d$ such that \\[30a+42b+70c+105d = 2022\\]. Due to modulos, we can find that\n\\begin{align*}\na &\\equiv 3 \\pmod{7} = 7a_1+3 \nb &\\equiv 1 \\pmod{5} = 5b_1+1 \nc &\\equiv 0 \\pmod{3} = 3c_1 \nd &\\equiv 0 \\pmod{2} = 2d_1. \n\\end{align*}\nSubstitution gives that $a_1+b_1+c_1+d_1 = 9$ for non-negative integers $a_1$ , $b_1$ , $c_1$ , $d_1$ . By stars and bars, we have $\\binom{12}{3}$ or $\\boxed{220}$ solutions.",
"Too easy, only intimidating cus it's a gen func?\nWe note that $2310 = 2*3*5*7*11$ meaning all the numbers in the denominator are factors of it. We use the well-known identity and represent: $P(x) = (1 + x^{105} + ... + x^{2205})(1 + x^{70} + ... + x^{2240})(1+ x^{42} + ... + x^{2268})(1 + x^{30} + ... + x^{2280})(x^{2310} - 1)^2$ We can get rid of the $(x^{2310} - 1)^2$ since $2310 > 2022$ . Since $2205,2240,2268,2280 > 2022$ , we don't have to put any \"bounds\" on our new goal which is to find the number of solutions to the equation: $$ 105a + 70b + 42c + 30d = 2022 $$ We note that $c = 5e+1$ for some $e$ , and $a=2f$ for some $f$ . We can reduce the equation to now: $$ 210f + 70b + 210e + 30d = 1980 \\Rightarrow 21(f+e) + 7b + 3d = 198 $$ We can also reduce this by noting that $d=7g+3$ , so we get: $$ 21(f+e)+7b+21g = 189 \\Rightarrow 3(e+f+g) + b = 27 $$ Now we need to find all solutions to this much simpler solution, as if we let $s=e+f+g$ , we have $3s + b = 27$ . We know that $0 \\le s \\le 9$ and $s$ uniquely determines the $b$ , so we can use stars and bars and hockey stick to get the number of solutions is: $\\binom{2}{2} + \\binom{3}{2} + ... + \\binom{11}{2} = \\binom{12}{3} = \\boxed{220}$ ",
" $\\mathfrak{The \\;Ninth\\; Of\\; October,\\; 2025}$ ʕ•ᴥ•ʔ\n<details><summary>Solution - grinding Modular Arithmetic</summary>$$ \\color{red} \\spadesuit\\color{red} \\boxed{\\textbf{Reducing to geometric series.}}\\color{red} \\spadesuit $$ Notice that the nominator does not contribute to $x^{2022}$ by virtue of $2310$ being a greater number than $2022$ ; thus, we can consider it as $1$ and expand into a $\\textbf{sum of an infinite geometric sequence}$ to obtain $$ (1+x^{30}+x^{60}+x^{90}+\\dots)\\cdot (1+x^{42}+x^{84}+x^{126}+\\dots)\\cdot (1+x^{70}+x^{140}+x^{210}+\\dots)\\cdot (1+x^{105}+x^{210}+x^{315}+\\dots) $$ Thus, we have a generating function. $$ \\color{blue} \\clubsuit\\color{blue} \\boxed{\\textbf{Generating functions!!!}}\\color{blue} \\clubsuit $$ Notice that finding the coefficient in front of $x^{2022}$ is equivalent to solving $$ 30a+42b+70c+210d=2022 $$ in integers. Since $30 \\cdot 7 = 42 \\cdot 5 = 70 \\cdot 3 = 210$ , we can easily see that $$ \\begin{aligned}\n\\bmod{7} \\implies & a=7x+3 \n\\bmod{5} \\implies & b=5y+1 \n\\bmod{3} \\implies & c=3z \n & d=t \n\\end{aligned} $$ for some $x,y,z,t \\in \\mathbb{N}$ . $$ \\color{magenta} \\bigstar\\color{magenta} \\boxed{\\textbf{Stars \\& bars.}}\\color{magenta} \\bigstar $$ \nNotice that the above is equivalent to finding the number of non-negative solutions to $$ x+y+z+t=9, $$ which by $\\textbf{Stars \\& Bars}$ is $\\binom{12}{3}=\\boxed{220}$ .</details>",
"<details><summary>solution</summary>The equation is equivalent to\n\t\\[ (x^{2310}-1)^2 \\cdot \\frac{x^{2310}-1}{x^{30}-1} \\cdot \\frac{x^{2310}-1}{x^{42}-1} \\cdot \\frac{x^{2310}-1}{x^{70}-1} \\cdot \\frac{x^{2310}-1}{x^{105}-1}\\]\n\tWe can safely ignore the $(x^{2310}-1)^2$ term, since it will not affect the coefficient of $x^{2022}$ anyway. Now, notice that when $n \\mid 2310$ ,\n\t\\[\\frac{x^{2310}-1}{x^{n}-1} = \\sum_{k=1}^{2310/n} x^{n(k-1)}\\]\n\tFrom here, it is easy to see that the problem is equivalent to finding the number of nonnegative solutions $(a,b,c,d)$ to the diophantine equation\n\t\\[30a+42b+70c+105d=2022\\]\n\tSome simplification tells us that this has the same number of solutions to\n\t\\[w+x+y+z=9\\]\n\tso the answer is $\\binom{12}{3}=220$ .</details>",
" $P(x) = (1 + x^{105} + x^{210} + \\dots)(1 + x^{70} + x^{140} + \\dots)(1 + x^{42} + x^{84} + \\dots)(1+x^{30} + x^{60} + \\dots)(x^{2310} - 1)^2$ We can ignore the last factor, so this is equivalent to $105a + 70b + 42c + 30d = 2022$ $a \\equiv 0 \\pmod 2$ so let $a = 2p$ $b \\equiv 0 \\pmod 3$ so let $b = 3q$ $c \\equiv 1 \\pmod 5$ so let $c = 5r + 1$ $d \\equiv 3 \\pmod 7$ so let $d = 7r + 3$ $105(2p) + 70(3q) + 42(5r + 1) + 30(7r + 3) = 2022 \\implies 210(p+q+r+s) = 1890 \\implies p+q+r+s=9$ Stars & bars gives ${12 \\choose 3} = \\boxed{220}$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1102,
"boxed": true,
"end_of_proof": false,
"n_reply": 42,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782931.json"
}
|
Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that $$ \frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1. $$ Find the least possible value of $a+b.$
|
Another personal favorite. I LOVE this one. Kudos to the author.
<span style="color:#f00">Edit: This claim is wrong. Will figure out later.</span>
We can interpret the problem as following: an ellipse centered at the origin has semi-major axis $a$ and foci at $(\pm 4,0)$ . Another ellipse is centered at $(20,11)$ with semi-major axis $b$ and foci $(20,11\pm 1)$ . We claim that minimizing $a+b$ means that these ellipses are tangent - indeed, assume they are not. Then, we can reduce the major axis of the first extremely slightly, by some $\varepsilon>0$ , such that the circles still intersect, but $a+b$ decreased, a contradiction.[1\baselineskip]
We first have the following crucial claim about tangent ellipses. Assume that ellipses $\mathfrak E_1$ and $\mathfrak E_2$ , with foci $E_1,E_2$ and $F_1,F_2$ are tangent at $T$ .**Claim.** $T$ is either $E_1F_2\cap E_2F_1$ or $E_1F_1\cap E_2F_2$ .
[asy]
import geometry;
ellipse ellipse(point A, point B, point C) {
return ellipse(A, B, (abs(A-C)+abs(B-C))/2);
}
size(9cm);
point E1=(-4,0),E2=(4,0),F2=(10,12),F1=(10,5.6),T=(6,4);
draw(ellipse(E1,E2,T));
draw(ellipse(F1,F2,T));
draw(E1--F1);
draw(E2--F2);
label(" $E_1$ ",E1,S);
label(" $E_2$ ",E2,S);
label(" $F_1$ ",F1,E);
label(" $F_2$ ",F2,E);
label(" $T$ ",T,E);
dot(E1);
dot(E2);
dot(F1);
dot(F2);
dot(T);
[/asy]
We can fix $E_1,E_2,F_1,F_2$ , and we show that if $P=E_1F_1\cap E_2F_2$ (given that $P$ lies on both line segments; otherwise look at $E_1F_2\cap E_2F_1$ ), then the ellipses are tangent. Clearly, there is at most one ellipse with foci at $F_1$ and $F_2$ that is tangent to a given ellipse, so it suffices to prove that is the one.
We can assume there is another point $Q$ on both ellipses. This means that
\[E_1Q+E_2Q=E_1P+E_2P\]
\[F_1Q+F_2Q=F_1P+F_2P\]
or upon addition
\[E_1Q+E_2Q+F_1Q+F_2Q=E_1P+E_2P+E_1P+E_2Q\tag{1}\]
However, by the triangle inequality in $\triangle E_1F_1Q$ and $\triangle E_2F_2Q$ , we get
\[E_1Q+F_1Q\geq E_1F_1=E_1P+PF_1\]
\[E_2Q+F_2Q\geq E_2F_2=E_2P+PF_2\]
with both equalities impossible, as that means $Q=E_1F_1\cap E_2F_2=P$ , impossible. Thus, we know that
\[E_1Q+E_2Q+F_1Q+F_2Q>E_1P+E_2P+E_1P+E_2Q\]
a contradiction to (1).
Now, the rest of the problem is easy - we see the desired lines are the ones connecting $E_1=(4,0)$ with $F_1=(20,12)$ , and $E_2=(-4,0)$ with $F_2=(20,10)$ . We can compute this point to be $T=(14,7.5)$ . Thus, the semi-major axis, $a$ and $b$ , are just half of the sums of the distances from the foci. In particular,
\begin{align*}
a=\frac{E_1T+E_2T}2&=\frac{\sqrt{(4-14)^2+(0-7.5)^2}+\sqrt{(-4-14)^2+(0-7.5)^2}}{2}
&=\frac{12.5+19.5}{2}=16
\end{align*}
\begin{align*}
b=\frac{F_1T+F_2T}2&=\frac{\sqrt{(20-14)^2+(12-7.5)^2}+\sqrt{(20-14)^2+(10-7.5)^2}}{2}
&=\frac{7.5+6.5}{2}=7
\end{align*}
Thus, we get the minimum value of $a+b=\boxed{023}$ .
|
[
"CONICS oh yeah ;)\n\nUse the definition of foci and then the lemma that the point minimizing AP+BP+CP+DP is the intersection of the diagonals of quadrilateral ABCD.",
"My favorite problem on the test\n<details><summary>Solution</summary>Consider points $A_1 = (4, 0), A_2 = (-4, 0), B_1 = (20, 12), B_2 = (20, 10)$ . Then, consider the ellipse with foci $A_1, A_2$ and semimajor axis $a$ . Also consider the ellipse with foci $B_1, B_2$ and semimajor axis $b$ . The given conditions are equivalent to these two ellipses intersecting, say at a point $X$ . Then, $XA_1 + XA_2 + XB_1 + XB_2 = 2(a+b)$ , which is minimized by Triangle Inequality when $X$ is the intersection of $A_1B_1$ and $A_2B_2$ . Thus, the minimum possible value is $\\frac{A_1B_1 + A_2B_2}{2} = \\boxed{023}$ .</details>",
"My only problem with this problem: $\\sqrt{20^2+11^2} \\approx 22.82 \\approx \\boxed{23}$ LOL just assume ellipses are circle!",
"Lmao after proving a+b>=21 I random guessed 25",
"Even I random guessed 025!! off just just by a 2 :(",
"I guessed 025 as well!! what",
"Tangent ellipses = dj problem?",
"<blockquote>Tangent ellipses = dj problem?</blockquote>\nProbably he would've said it was his problem",
"<blockquote>Tangent ellipses = dj problem?</blockquote>\n\nI was told this was not a djmathman problem but the proposer is just as creative with his iconic geometry problems.",
"The title of this thread disgusts me.\n\n@2below not what I meant >:(",
"<blockquote>kudos to the author</blockquote>\n\nthanks :yup:",
"<blockquote>The title of this thread disgusts me.</blockquote>\n\nI know, conics disgust me too. :surf:",
"I just saw the centers were sqrt(521) away from each other, so I thought the sum had to be less than 22.5 instead of greater than, I guessed 22 :( :(",
"ngl i dont like conics but this one is op\n\nbasically you are given the foci so if you let an intersection point be (m, n) then to minimize a+b you want to minimize the sum of the four distances which happens when that intersection point is at the intersection of line segments\n\nthen a+b is half the sum of the four line segments or 0.5(26 + 20) = 23",
"blah full solution\n\nThe first ellipse has foci $E_1=(4, 0)$ and $E_2=(-4, 0)$ and the second ellipse has foci $F_1=(20, 12)$ and $F_2=(20, 10)$ . Now consider the point $T=(x, y)$ . We have that $a+b=\\frac{E_1T+E_2T+F_1T+F_2T}{2}$ and since $E_1T+F_1T\\geq E_1F_1$ and $E_2T+F_2T\\geq E_2F_2$ , we have that $a+b\\geq \\frac{E_1F_1+E_2F_2}{2}$ , where equality occurs when $T$ is the intersection of $E_1F_1$ and $E_2F_2$ . Then $E_1F_1=20,E_2F_2=26$ , and the answer is $\\frac{26+20}{2}=\\boxed{023}$ .",
"People at my school agreed to guess 23 for any questions we didn’t know and here we are :)",
"why are there zeroes in front of every answer",
"<blockquote>why are there zeroes in front of every answer</blockquote>\n\nbecause that's the answer you're supposed to put on the AIME. the zeroes are not required though",
"because there are three bubbles on your answer sheet so you bubble a zero for smaller answers",
"<blockquote>People at my school agreed to guess 23 for any questions we didn’t know and here we are :)</blockquote>\n\norz i guess 029",
"ahh i see thanks @2above and @3above. see, i'e never taken the aime so i didnt know how it worked. i didn't even know that it happened on paper",
"Never thought I'd enjoy a conics problem.\n\nThe foci of the first ellipse are $(\\pm 4,0)$ and the foci of the second ellipse are $(20,10),(20,12)$ . By the definition of an ellipse, we have:\n\\begin{align*}\n\\sqrt{(x-4)^2+y^2}+\\sqrt{(x+4)^2+y^2}&=2a \n\\sqrt{(x-20)^2+(y-12)^2}+\\sqrt{(x-20)^2+(y-10)^2}&=2b.\n\\end{align*}\nSo\n\\begin{align*}\na+b&=\\frac{\\left(\\sqrt{(x+4)^2+y^2}+\\sqrt{(x-20)^2+(y-10)^2}\\right)+\\left(\\sqrt{(x-4)^2+y^2}+\\sqrt{(x-20)^2+(y-12)^2}\\right)}{2} \n&\\geq\\frac{\\text{dist}((-4,0),(20,10)+\\text{dist}((4,0),(20,12))}{2}=\\frac{20+26}{2}=23.\n\\end{align*}\nEquality can occur, just draw a picture or solve for the intersection of the lines if you aren't convinced yet.",
":love: The lines between foci intersect at (14,15/2) and so that is the minimum => 16+7 = 23 ",
"I was very slow 15 minute solve\nAIME II problems this year are so easy\nfind the foci and write out the alternative form of the equation for an ellipse\nthen 2a+2b is just the sum of the distances from (x,y) to (-4,0), (4,0), (20,10), (20,12)",
"This problem is so satisfying, managed to finish all the bash and then whoosh, an integer answer! :P \n\nWe reference [1985 AIME #11]( https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_11) to find that if $E_1$ and $E_2$ are the foci of the first ellipse and $F_1$ and $F_2$ are the foci of the second ellipse, then the lines $E_1F_2$ and $E_2F_1$ intersect at the tangency point. This can be proven intuitively since AM-GM equality gives that $a \\approx b$ minimizes any sum in general. Therefore we want them to be as small as possible while not too large, giving the assertion.\n\nWe now find the foci of the first ellipse to be $(\\pm 4,0)$ and $(20,11 \\pm 1)$ which gives the tangency point $T = \\left(14, \\frac{15}{2} \\right)$ by basic ellipse properties and coordinates. We then plug this point back into each equation to solve for $a$ and $b$ respectively using the substitution $a^2=m$ and $b^2=n$ . Miraculously, solving each quadratic gives $m = 256$ and $n=49$ , or $a=16$ and $b=7$ , which means $$ a+b=16+7=\\boxed{023} $$ ",
"[Video Solution](https://youtu.be/4qiu7GGUGIg)",
"Construct the two conics: $\\beta_1. \\beta_2$ . $\\beta_1$ is centered at the origin with a major axis of $2a$ and minor axis $2\\sqrt{a^2-16}$ , while $\\beta_2$ is centered at $(20, 11)$ with a major axis of $2b$ and minor axis of $2\\sqrt{b^2-1}$ . Now, let $F_1, F_2$ denote the foci of $\\beta_1$ and $F_3, F_4$ denote the foci of $\\beta_2$ , respectively. Let $D$ be the intersection of $\\beta_1$ and $\\beta_2$ . Spamming the definition of an ellipse gives that $a=\\frac{1}{2}(DF_1+DF_2), b=\\frac{1}{2}(DF_3+DF_4)$ . This problem is boiled down to finding the minimum value of $DF_1+DF_2+DF_3+DF_4$ .\n\nBy triangle inequality, the minimum occurs when $F_1, F_3$ and $D$ are collinear, and $F_2, F_4, D$ are collinear. The rest is just simple computation: $$ \\min(a+b)=\\frac{1}{2}\\left( \\sqrt{(20-4)^2+(12)^2}+\\sqrt{(20+4)^2+(10)^2}\\right)=\\boxed{023} $$ ",
"woah since when were conic problems this nice\n\nThe equation $$ \\frac{x^2}{a^2} + \\frac{y^2}{a^2-16} =1 $$ is that of an ellipse centered at $(0, 0)$ with foci at $(4, 0)$ and $(-4, 0)$ . Similarly, the equation $$ \\frac{(x-20)^2}{b^2-1}+\\frac{(y-11)^2}{b^2}=1 $$ is that of an ellipse centered at $(20, 11)$ with foci at $(20, 10)$ and $(20, 12)$ . Note that from the definition of an ellipse $2a$ is the sum the distances from any point on the ellipse to the two foci ( $2c$ is defined similarly). We know that $(x, y)$ lies on the two ellipses. Thus, we want to minimize the sum of the distances from $(x, y)$ to the two pairs of foci, which have coordinates $(4, 0)$ , $(-4, 0)$ , $(20, 10)$ , $(10, 12)$ . In other words, we want to minimize the sum of the distances from $(x, y)$ to those four points.\n\nThis occurs when $(x, y)$ is the intersection of the segments connecting $(4, 0)$ and $(20, 12)$ , $(-4, 0)$ and $(20, 10)$ (provable using triangle inequality). We can easily solve for $(x, y) = (14, 15/2)$ , so $2a+2b = 46 \\implies a+b=23$ (I skipped over some distance formula computation).",
"Least contrived conic problem.\nDefinition of conics gives us that $2a+2b$ is the sum of $P = (x, y)$ from the four foci. The four foci are literally just $A = (4, 0), B = (-4, 0), C = (20, 10), D = (20, 12)$ . But minimizing $PA+PB+PC+PD$ is minimizing $PA+PD$ and $PB+PC$ and in fact these minima can happen at the same time at $P = BC \\cap AD$ . Basic system of equations extracts $\\boxed{23}$ ",
"One of the coolest problems on AIME ever:\n\nWhen graphing it, the foci for the first ellipse are $(-4,0)$ and $(4,0)$ , and the foci for the second ellipse are $(20,10)$ and $(20,12)$ . For a pair $(x,y)$ to exist, the graphs must intersect. Let them intersect at $P(r,s)$ . Then $\\sqrt{(r-20)^2+(s-10)^2}+\\sqrt{(r-20)^2+(s-12)^2}=2b$ and $\\sqrt{(r-4)^2+s^2}+\\sqrt{(r+4)^2+s^2}=2a$ . $2(a+b)=\\sqrt{(r-20)^2+(s-10)^2}+\\sqrt{(r-20)^2+(s-12)^2}+\\sqrt{(r-4)^2+s^2}+\\sqrt{(r+4)^2+s^2}$ . To minimize this expression, we must have circles with centers at the 4 foci points intersect at the point $P(r,s)$ . We must minimize the sum of all the radii of these four circles. Consider the circles $\\omega_1$ and $\\omega_2$ at centers $(-4,0)$ and $(20,10)$ respectively. Clearly $r_1+r_2\\geq{\\sqrt{24^2+10^2}=26}$ . Now consider circles $\\omega_3$ and $\\omega_4$ at centers $(4,0)$ and $(20,12)$ respectively. Similarly, $r_3+r_4\\geq{\\sqrt{12^2+16^2}=20}$ . Both of these inequalities are equal when the two circles are tangent, and the pairs of circles are tangent at the same point, giving $r_1+r_2+r_3+r_4=\\geq{20+26} \\Longrightarrow 2(a+b)\\geq{46} \\Longrightarrow (a+b)\\geq{23}$ ."
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"2022 Contests",
"2022 AIME Problems"
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"answer_score": 1078,
"boxed": false,
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"n_reply": 31,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782937.json"
}
|
Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.
|
Let the three labels be $20\ge x > y > z\ge 1$ . Then, we have $x - y = p$ , $y - z = q$ , and $x - z = r$ for primes $p,q,r$ . Adding the first two equations gives $x - z = p + q$ . But $x - z = r$ , so we must have $p + q = r$ . For parity reasons, one of $p,q$ is $2$ and the other is odd.
Assume WLOG $p = 2$ (multiply by $2$ at the end since we can freely swap $p,q$ ). Then $2 + q = r$ . Noting that $q,r\in [1,20]$ , we see that $q\in \{3,5,11,17\}$ . The rest is just casework on the value of $q$ .
Case 1: $q = 3$ . Then we have $r = 5$ , so $(x,y,z)$ can be any triple of the form $(a + 5, a + 2, a)$ . Since $1\le a\le 15$ , there are $15$ such triples
Case 2: $q = 5$ . Then $(x,y,z)$ is $(a + 7, a + 2, a)$ , where $1\le a\le 13$ , giving $13$ triples
Case 3: $q = 11$ . Then $(x,y,z)$ is $(a + 13, a + 2, a)$ , where $1\le a\le 7$ , giving $7$ triples
Case 4: $q = 17$ . Then $(x,y,z)$ is $(a + 19, a + 2, a)$ where $1\le a\le 1$ , giving $1$ triple.
Hence, there are $15 + 13 + 7 + 1 = 36$ triples. Multiplying by $2$ gives $\boxed{72}$ .
|
[
"Mine. I like this one; it's a cute twist on a classic.\n\n<details><summary>Solution</summary>Suppose $i$ , $j$ , and $k$ are the labels of the three vertices of a triangle; without loss, let $i > j > k$ . Note that $(i-j) + (j-k) = (i-k)$ , so one of $i-j$ or $j-k$ must be $2$ , and furthermore the other two primes must be twin primes. A quick count reveals that\n\\[(i-j,j-k,i-k)\\in\\{(2,3,5),(2,5,7),(2,11,13),(2,17,19)\\}\\] and permutations where the first two coordinates are swapped. In particular, for any pairs of vertices $(a,a+d)$ , where $d\\in\\{5,7,13,19\\}$ , there are exactly two locations for the middle vertex which yield a triangle. Finally, observe that there are $20-d$ pairs of vertices $(a,a+d)$ for every $d$ between $1$ and $19$ . Summing over all $d$ gives an answer of $2(15+13+7+1) = \\boxed{72}$ .</details>\n<details><summary>Remark</summary>This was originally submitted to CMIMC. Here's a summary of how I came up with this problem that I wrote way back in late 2019:\n<blockquote>\n\n\n- I wanted to come up with a CS problem...\n- ...so I thought about BFS (instead of DFS, which was the subject of 2018 cs10)...\n- ...which led into complete graphs with a few edges removed, for some reason...\n- ...which, after BFS ended, segued into finding triangles in the graph...\n- ...which led to me realizing the problem was too easy with that graph...\n- ...which led to the graph $K$ in the problem statement.\n\n</blockquote></details>",
"<blockquote>Mine. </blockquote> thats like 4 problems on this test wow",
"djmathman orz",
"2 3 15\n2 5 13\n2 11 7\n2 17 1\n\n2(15 + 13 + 7 + 1) = 72",
"<blockquote>2 3 15\n2 5 13\n2 11 7\n2 17 1\n\n2(15 + 13 + 7 + 1) = 72</blockquote>\n\nNot sure, but did anyone get $\\boxed{086}$ ?\n\nI did the same thing as this, but also got an extra 14 triangles for 2, 11, 13. I just want to know what my mistake is.",
"@2above's 2 11 7 is a typo, it's 2 11 13.",
"<blockquote>@2above's 2 11 7 is a typo, it's 2 11 13.</blockquote>\n\nThanks! I found my mistake.",
"Call the labels of the triangle $x,y,z$ with $x>y>z$ . Then we get $(x-y, y-z, x-z)$ are all primes. Basically if none of them are $2$ , then they are of pairwise different parity, contradiction. So at least one of them must be $2$ . \n\nNow the difference between the other two elements is equal to $2$ . So our three primes can be a permutation of \\[\\{2,3,5\\}, \\{2,5,7\\}, \\{2,11,13\\}, \\{2,17,19\\}.\\]\n\nCase 1: $x-y=2$ . \nThen we get $x-2-z, x-z$ are primes which differ by $2$ . Now for $x<5$ , this can't be possible. If $5< x\\le 7$ , then we have $z$ so that $x-z=5$ . If $7<x\\le 13$ , then we can have $z$ so that $x-z=5$ or $7$ . If $13<x\\le 19$ , then we can have $z$ so that $x-z=5, 7,$ or $13$ . If $x=20$ , then we can have $z$ so that $x-z=5, 7, 13,$ or $19$ . This gives $2\\cdot 1+6\\cdot 2+6\\cdot 3+1\\cdot 4=36$ ways. \n\nCase 2: $x-z=2$ . \nThis implies $x-y=1$ , a contradiction. \n\nCase 3: $y-z=2$ . \nThen $x-z-2$ and $x-z$ are primes. Exact same as case 1. \n\n\nSo the answer is $36+0+36=\\boxed{072}$ . ",
"I got 36 :wallbash:",
"I didn't solve this one when I took the actual test, but the key is the work backwards because there are so many triangles that regular casework won't work (at least without taking a ton of time), so instead think about the condition itself and realize the condition that @djmathman came up with, the \"differ by\" means you can assume the numbers are strictly increasing so then (i - j) + (j - k) = (i - k). All three are primes, so some basic casework yields at least one is even but the only even prime is 2. From here use @djmathman's solution.",
"Of the three labels $a<b<c$ at the vertices of such a triangle, let $p_1=b-a$ be the distance between the smaller two, $p_2=c-b$ that between the larger two, and $p_3=c-a$ that between the smallest and largest. Then, since $(b-a)+(c-b)=(c-a)$ , we have that $p_1+p_2=p_3$ , where we wish $p_1$ , $p_2$ , and $p_3$ to be primes. As it is impossible for all of $p_1$ , $p_2$ , and $p_3$ to be odd, as an odd number plus an odd number is an even number, either $p_1=2$ and $p_2,p_3$ are twin primes or $p_2=2$ and $p_1,p_3$ are twin primes.\n\nWrite $(p_1,p_2,p_3)$ as $(2,p,p+2)$ or $(p,2,p+2)$ . It can be seen that $p$ can take on any of $3$ , $5$ , $11$ , or $17$ . Given a fixed $p$ , there are $18-p$ values for the largest label $c$ , as it ranges from $p+3$ to $20$ . Then, $b$ can either be $c-2$ or $c-p$ , and $a$ is fixed accordingly, so we have $$ 2[(18-3)+(18-5)+(18-11)+(18-17)]=2[15+13+7+1]=\\boxed{72} $$ total triangles, as requested.",
"<blockquote>\nTwenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.\n</blockquote>\ninitially overcomplicated this by actually drawing in chords for $p=2$ , $3$ :blobsob:\n\nLet $(a, b, c)$ be a triple of points, with indices in increasing clockwise order. Then $(b-a)+(c-b)=c-a$ , and let $b-a=p_1$ , $c-b=p_2$ , $c-a=p_3$ , so $p_1+p_2=p_3$ . Clearly $p_1$ and $p_2$ can't be of the same parity, so one of them has to be $2$ and the other a greater prime. Hence $(p_1, p_2, p_3)$ is of the forms $(2, p, p+2)$ , $(p, 2, p+2)$ for some prime $p$ .\n\nThis implies $p$ can be $3$ , $5$ , $11$ , or $17$ . For each of these cases, $p+2$ can vary from $20$ to $p+3$ ( $18-p$ options) and the \"first\" vertex (i.e. $a$ ) varies accordingly; but $c-b$ and $b-a$ could both be $2$ (or $p$ ), so that means $2(18-p)$ for each $p$ . Summing,\n\\[2\\cdot 15+2\\cdot 13 + 2\\cdot 7 + 2\\cdot 1=\\boxed{072}\\]\nas desired.",
"Label the vertices from $1$ to $20$ . Then say we take $a, b, c$ with $a<b<c$ , so we need $b-a, c-a, c-b$ to all be prime. If we add all of these, we get $2c - 2a$ which must be even, so at least one of them is equal to $2$ . \n\nCase 1: $b-a = c-a = c-b = 2$ . This is not possible.\n\nCase 2: $b - a = c - a = 2$ . Then we have $b = c$ , so this is not possible.\n\nCase 3: $b-a = c-b = 2$ . Then we have $b = a+2, c = a+4$ , which is also not possible.\n\nCase 4: $c-a = c-b = 2$ . Then we have $a = b$ , which is not possible.\n\nThus only one of $b-a, c-a, c-b$ can be equal to two.\n\nCase 5: $b - a = 2$ .\n\nThen we have $b = a+2$ . We can then have $c = a+5, a+7, a+13, a+19$ . If $a = 1$ there are $4$ ways, from $a = 2$ to $a = 7$ there are $3$ ways, from $a = 8$ to $a = 13$ there are $2$ ways, and from $a = 14$ to $a = 15$ there is one way. This gives $1 \\cdot 4 + 6 \\cdot 3 + 6 \\cdot 2 + 2 \\cdot 1 = 36$ ways.\n\nCase 6: $c - a = 2$ Then we have $c = a+2$ . This cannot happen since $b = a+1$ is impossible.\n\nCase 7: $c-b = 2$ .\n\nThis is symmetric to Case 5, so we get another $36$ ways.\n\nThe total is $36+36 = 72$ .\n\nNice problem!",
"Essentially, if $a$ is the lowest number in the triangle we seek primes $p, q$ such that $p+q$ is prime and $a+p+q \\leq 20$ . By parity, the primes can only be $(2, 3), (2, 5), (2, 11), (2, 17)$ . Note that we can permutate these primes in any order so we multiply our answer by two. The sums of the primes in these cases are $5, 7, 13, 19$ . Clearly only $19$ can be viable when $a=1$ and similarly only $13$ is viable when $13+a \\leq 20 \\implies a \\leq 7$ . Continue this to get $$ 2((20-19)+(20-13)+(20-7)+(20-5)) = \\boxed{72} $$ "
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"2022 Contests",
"2022 AIME Problems"
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"answer_score": 1082,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782948.json"
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|
There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that $$ \log_{20x} (22x)=\log_{2x} (202x). $$ The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
|
<details><summary>Solution</summary>Let $\log (n)$ be the logarithm base $10$ . By change of base formula: $$ \frac{\log(22x)}{\log(20x)} = \frac{\log(202x)}{\log(2x)} $$ $$ \Rightarrow \log{22x} \cdot \log{2x} = \log{20x} \cdot \log{202x} $$ $$ \Rightarrow \log{22x} \cdot \log{2x} = (1 + \log{2x}) \cdot \log{202x} $$ $$ \Rightarrow \log{22x} \cdot \log{2x} = \log{202x} + \log{2x} \cdot \log{202x} $$ $$ \Rightarrow \log{2x} \cdot (\log{22x} - \log{202x}) = \log{202x} $$ $$ \Rightarrow \log{202x} = \log{2x} \cdot \log{(\frac{11}{101})} $$ Thus, we have $$ \log _{20x} (22x) = \frac{\log(22x)}{\log(20x)} = \frac{\log(202x)}{\log(2x)} = \frac{\log{2x} \cdot \log{(\frac{11}{101})}}{\log{2x}} = \log{(\frac{11}{101})} $$ giving us an answer of $11+101 = \boxed{112}$ .</details>
|
[
"~~which problem was this~~ probably #4",
"Notice if $\\frac{a}{b} = \\frac{c}{d}$ then $\\frac{a}{b}=\\frac{c}{d} = \\frac{a-c}{b-d}$ Apply the lemma, so $\\frac{a}{b} = \\frac{\\ln(\\frac{10}{101})}{\\ln(10)}$ And you are done. $112$ .",
"Friend got $11 + 101 = 121$ . rip\n<details><summary>AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA</summary>$1 + \\log_{20x}(\\frac{22}{20}) = 1 + \\log_{2x}(101)$ $(20x)^n = \\frac{22}{20}$ $(2x)^n = 101$ $10^n = \\frac{11}{1010}$ $n = \\log_{10}(\\frac{11}{1010}) = \\log_{10}(\\frac{11}{101}) - 1$ $n + 1 = \\log_{10}(\\boxed{\\frac{11}{101}})$</details>",
"Why no logs on AIME I :sob:",
"set equal to x and divide",
"Let $log_{20x}(22x)=k$ . Then $(20x)^k=22x$ and $(2x)^k=202x\\implies (20x)^k=202x\\cdot 10^k\\implies 10^k=\\frac{11}{101}\\implies\\boxed{112}$ .",
"<blockquote>Let $log_{20x}(22x)=k$ . Then $(20x)^k=22x$ and $(2x)^k=202x\\implies (20x)^k=202x\\cdot 10^k\\implies 10^k=\\frac{11}{101}\\implies\\boxed{112}$ .</blockquote>\n\nBasically the same way I did it. I think this is the easiest problem on the test.",
"<blockquote><blockquote>Let $log_{20x}(22x)=k$ . Then $(20x)^k=22x$ and $(2x)^k=202x\\implies (20x)^k=202x\\cdot 10^k\\implies 10^k=\\frac{11}{101}\\implies\\boxed{112}$ .</blockquote>\n\nBasically the same way I did it. I think this is the easiest problem on the test.</blockquote>\n\n",
" $20^k\\cdot x^k=22x$ and $2^k\\cdot x^k=202x$ . So $10^k=\\frac{22}{202}=\\frac{11}{101}\\implies \\boxed{112}$ . ",
" $(20x)^y=22x$ and $(2x)^y=202x$ Divide gives $10^y=\\frac{11}{101}$ which $y=\\log_{10} \\frac{11}{101}$ gives an answer of $\\boxed{112}$ .",
"\n<details><summary>@#5</summary>[**juliankuang**](aops.com/community/user/467940) · Feb 17, 2022, 1:38 PM <span style=\"font-size:150%\">[···](aops.com/community/p24447460)</span><span style=\"color:transparent\">helo</span>\nFriend got $11 + 101 = 121$ . rip\n<details><summary>AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA</summary>$1 + \\log_{20x}(\\frac{22}{20}) = 1 + \\log_{2x}(101)$ $(20x)^n = \\frac{22}{20}$ $(2x)^n = 101$ $10^n = \\frac{11}{1010}$ $n = \\log_{10}(\\frac{11}{1010}) = \\log_{10}(\\frac{11}{101}) - 1$ $n + 1 = \\log_{10}(\\boxed{\\frac{11}{101}})$ -----------\n<span style=\"color:#5b7083\">[aops]x[/aops] 5[color=transparent]hellloolo</span> [aops]Y[/aops] 0 <span style=\"color:transparent\">hellloolo</span></details></details>\nit took me a solid 20 seconds to realise that $11+101\\neq121$ ",
"Let $k=\\log_{20x}(22x)=\\log_{2x}(202x)$ ; we seek $k$ . From $k=\\log_{20x}(22x)$ , $(20x)^k=22x\\implies20^kx^{k-1}=22$ ; and from $k=\\log_{2x}(202x)$ , $(2x)^k=202x\\implies2^kx^{k-1}=202$ . Dividing the two, $$ \\frac{20^kx^{k-1}}{2^kx^{k-1}}=\\frac{22}{202}\\implies10^k=\\frac{11}{101}, $$ from which $k=\\log_{10}(\\tfrac{11}{101})$ . The requested answer is then $\\boxed{112}$ .",
"Let $m = \\log_{20x}(22x)=\\log_{2x}(202x)$ . We know that $(20x)^m = 22x$ and $(2x)^m=202x$ . Dividing the first equation by the second, we get $10^m = \\frac{11}{101}$ , so $m=\\log_{10}(\\frac{11}{101})$ . Hence, $m+n=112$ ",
"aight chat so basically $\\frac{\\ln22+\\ln x}{\\ln20+\\ln x}=\\frac{\\ln202+\\ln x}{\\ln2+\\ln x}$ now do the subtracc thingy so $\\frac{\\ln22+\\ln x}{\\ln20+\\ln x}=\\frac{\\ln\\frac{11}{101}}{\\ln 10}$ ans is $11+101=\\boxed{112}$ "
] |
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"2022 Contests",
"2022 AIME Problems"
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"answer_score": 1034,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782952.json"
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|
Find the remainder when $$ \binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2} $$ is divided by $1000$ .
|
<blockquote><details><summary>Solution</summary>Consider a committee of $n$ -people as follows. $$ C = \{x_1, x_2, \cdots, x_n \} $$ Now, consider the set of all possible committees of pairs of these $n$ -peoples. $$ P = \{ \{x_1, x_2 \}, \{x_1, x_3 \}, \cdots, \{x_{n-1}, x_{n} \} \}. $$ Notice that when we choose two such pairs, the two sets of pairs we choose either have a person in common or no people in common. Therefore, $$ \dbinom{\binom{n}{2}}{2} = n \binom{n-1}{2} + \left(\frac{\binom{4}{2}}{2}\right)\dbinom{n}{4} = 3 \bigg( \dbinom{n}{3} + \dbinom{n}{4} \bigg) = 3 \bigg(\dbinom{n+1}{4} \bigg). $$ Therefore, the answer is $$ 3 \bigg( \dbinom{4}{4} + \dbinom{5}{4} + \cdots + \dbinom{41}{4} \bigg) = 3 \times \dbinom{42}{5} \equiv \boxed{004} \pmod{1000}. $$</details>
I am wondering if there was some direct bijection to go from $\dbinom{\binom{n}{2}}{2}$ to $3 \dbinom{n+1}{4}.$ </blockquote>
Well if you have a list of $n+1$ values, including integers $1$ thru $n$ and a "duplicate" value, then any choice of $4$ values from these $n+1$ yields $3$ possible arrangements, hence $3\binom{n+1}{4}$ .
|
[
"I got 004 here",
"3(42C5) = 2552004 by hockey stick. Note NC2C2 = 3(N+1C4)",
"<span style=\"font-size:50%\">i may or may not have used the last hour to bash this one out</span>",
"Note that the expression is equivalent to\n\\[ \\sum_{n=3}^{40} \\frac{(n-2)(n-1)n(n+1)}{8} = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 + 2 \\cdot 3 \\cdot 4 \\cdot 5 + \\cdots + 38 \\cdot 39 \\cdot 40 \\cdot 41}{8} \\] which is equal to $\\frac{\\frac{38 \\times 39 \\times 40 \\times 41 \\times 42}{5}}{8} = 38 \\times 39 \\times 41 \\times 42 = 2552 \\boxed{004}$ , as desired.",
"**Combinatorial Argument**\nSuppose we have a list of $n$ positive integers $1$ thru $n$ . Now consider a list of the $\\binom{n}{2}$ unordered pairs of two elements of our original list. Now we choose two pairs from this list.\nWe have two cases: Either all four numbers are distinct, giving $3\\binom{n}{4}$ , or one value is repeated, which gives $3\\binom{n}{3}$ . This sum yields $\\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4}$ .",
"[quote name=\"kootrapali\" url=\"/community/p24447807\"]**Combinatorial Argument**\nSuppose we have a list of $n$ positive integers $1$ thru $n$ . Now consider a list of the $\\binom{n}{2}$ unordered pairs of two elements of our original list. Now we choose two pairs from this list.\nWe have two cases: Either all four numbers are distinct, giving $3\\binom{n}{4}$ , or one value is repeated, which gives $3\\binom{n}{3}$ . This sum yields $\\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4}$ .\n</blockquote>\n\nthis was my solution. we simply choose 2 distinct pairs. we can have (a,b),(a,c) or (a,b),(c,d). Given any triplet, we get 3 solutions there. given any quadruplet, we also get 3 solutions. Then its just hockey stick",
"<blockquote>Note that the expression is equivalent to\n\\[ \\sum_{n=3}^{40} \\frac{(n-2)(n-1)n(n+1)}{8} = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 + 2 \\cdot 3 \\cdot 4 \\cdot 5 + \\cdots + 38 \\cdot 39 \\cdot 40 \\cdot 41}{8} \\] which is equal to $\\frac{\\frac{38 \\times 39 \\times 40 \\times 41 \\times 42}{5}}{8} = 38 \\times 39 \\times 41 \\times 42 = 2552 \\boxed{004}$ , as desired.</blockquote>\n\nDang i didn't think of this clever way\n\nI bashed hockey stick twice lol (when mocking)\n\nMy final closed form was $\\frac{1}{2} \\left [ \\dbinom{41}{3} \\left ( \\dbinom{40}{2} - 1 \\right ) - \\left ( 39 \\dbinom{41}{4} - \\dbinom{41}{5} \\right) \\right ].$ at least it's still the correct answer",
"We have $\\binom{\\binom{x}{2}}{2}=\\frac{\\dbinom{x}{2}\\cdot \\left(\\dbinom{x}{2}-1\\right)}{2}$ . We have \\[\\binom{x}{2}\\left(\\binom{x}{2}-1\\right)=\\binom{x}{2}^2-\\binom{x}{2}=\\frac{x^2(x-1)^2}{4}-\\frac{2x(x-1)}{4}=\\frac{x(x-1)(x+1)(x-2)}{4}\\]\n\nSo $\\binom{\\binom{x}{2}}{2}=\\frac{(x-2)(x-1)x(x+1)}{8}$ . \n\nNow we will compute $1\\cdot 2\\cdot 3\\cdot 4+2\\cdot 3\\cdot 4\\cdot5+\\ldots+38\\cdot 39\\cdot 40\\cdot 41$ . \n\nIt's equivalent to $4!\\left(\\binom{4}{4}+\\binom{5}{4}+\\binom{6}{4}+\\ldots+\\binom{41}{4}\\right)=24\\left(\\binom{42}{5}\\right)$ . \n\nDividing by $8$ gives us \\[3\\cdot \\binom{42}{5}=\\frac{3\\cdot 42!}{5!37!}=\\frac{3\\cdot 38\\cdot 39\\cdot 40\\cdot 41\\cdot 42}{5!}=38\\cdot 39\\cdot 41\\cdot 42\\equiv 482\\cdot 722\\equiv 0-18\\cdot 722\\equiv -(12960+36)=-12996\\equiv \\boxed{004}\\pmod{1000}.\\]",
"<details><summary>my scuffed solution</summary>I got this by writing out $\\binom{\\binom{3}{2}}{2},\\binom{\\binom{4}{2}}{2},\\binom{\\binom{5}{2}}{2},\\dots$ and noticing that they were all divisible by three. I divided them by three and got $\\binom{4}{4},\\binom{5}{4},\\binom{6}{4},\\dots$ I computed this pattern for the first 8 terms and knew it had to be true - this gives us $3\\left(\\binom{4}{4}+\\dots+\\binom{41}{4}\\right)=3\\binom{42}{5}=2552\\boxed{004}$ .\n\nI spent 5 minutes looking for a combinatorial argument to justify $\\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4}$ and gave up. My dumb self did eventually find the algebraic derivation after the test.</details>",
"it's pretty easy to get $\\frac{n^4-2n^3-n^2+2n}{8}$ then by grouping the top factors into $n^2(n^2-1)-2n(n^2-1)=n(n-2)(n+1)(n-1)$ from which you get $3\\binom{n+1}{4}$ ",
" $\\binom{n}{2}=\\frac{n(n-1)}{2} \\Rightarrow \\binom{\\binom{n}{2}}{2}=\\frac{\\frac{n(n-1)}{2}\\cdot(\\frac{n(n-1)}{2}-1)}{2}=\\frac{(n-2)(n-1)(n)(n+1)}{8}$ So we want $\\sum_{n=3}^{40} \\frac{(n-2)(n-1)(n)(n+1)}{8}$ .\nI cannot do sums, so we will proceed with induction.\nClaim: $\\sum_{n=3}^{x} \\frac{(n-2)(n-1)(n)(n+1)}{8}=\\frac{(x-2)(x-1)(x)(x+1)(x+2)}{40}$ .\nLet $\\Sigma(x)$ denote that sum.\nBase case: $\\Sigma(3)=\\frac{1\\cdot2\\cdot3\\cdot4}{8}=3=\\frac{1\\cdot2\\cdot3\\cdot4\\cdot5}{8\\cdot5}$ . The fives cancel.\nInduction: $\\Sigma(k+1)=\\frac{(k-1)(k)(k+1)(k+2)(k+3)}{40}$ by our claim. $\\Sigma(k+1)=\\Sigma(k)+\\frac{(k-1)(k)(k+1)(k+2)}{8}$ $=\\frac{(k-1)(k)(k+1)(k+2)}{8} + \\sum_{n=3}^{k} \\frac{(n-2)(n-1)(n)(n+1)}{8}$ $=\\frac{(k-1)(k)(k+1)(k+2)}{8}+\\frac{(k-2)(k-1)(k)(k+1)(k+2)}{40}$ $=\\frac{5[(k-1)(k)(k+1)(k+2)]+(k-2)(k-1)(k)(k+1)(k+2)}{40}$ $=\\frac{(k-1)(k)(k+1)(k+2)(5+k-2)}{40}$ $=\\frac{(k-1)(k)(k+1)(k+2)(k+3)}{40}$ $=\\Sigma(k+1)$ .\nAnd we're done. $\\sum_{n=3}^{40} \\frac{(n-2)(n-1)(n)(n+1)}{8}=\\frac{(40-2)(40-1)(40)(40+1)(40+2)}{40}=(1600-1)(1600-4)= 2560000-5(1600)+4 \\equiv\\boxed{004} \\pmod{1000}.$ ",
"My friend just added it out :/",
"Note that \n\\begin{align*}\n\\binom{\\tbinom{n}{2}}{2}&=\\binom{\\tfrac{n^2-n}{2}}{2}\n&=\\frac{(n^2-n)(n^2-n-2)}{8}\n&=\\frac{(n+1)n(n-1)(n-2)}{8}\n&=3\\binom{n+1}{4}\n\\end{align*}\nso we get \n\\begin{align*}\n\\binom{\\tbinom{3}{2}}{2} + \\binom{\\tbinom{4}{2}}{2} + \\dots + \\binom{\\tbinom{40}{2}}{2}&=3\\left(\\binom{4}{4}+\\binom{5}{4}+\\dots+\\binom{41}{4}\\right)\n&=3\\binom{42}{5}\n&=42\\cdot 41\\cdot 39\\cdot 38\n&=(40^2-4)(40^2-1)\n&=40^4-5(40^2)+4\n&\\equiv \\boxed{004}\\pmod{1000},\n\\end{align*}\nand we are done.",
"<blockquote>Note that \n\\begin{align*}\n\\binom{\\tbinom{n}{2}}{2}&=\\binom{\\tfrac{n^2-n}{2}}{2}\n&=\\frac{(n^2-n)(n^2-n-2)}{8}\n&=\\frac{n^4-2n^3-n^2+2n}{8}\n&=\\frac{(n+1)n(n-1)(n-2)}{8}\n&=3\\binom{n+1}{4}\n\\end{align*}\nso we get \n\\begin{align*}\n\\binom{\\tbinom{3}{2}}{2} + \\binom{\\tbinom{4}{2}}{2} + \\dots + \\binom{\\tbinom{40}{2}}{2}&=3\\left(\\binom{4}{4}+\\binom{5}{4}+\\dots+\\binom{41}{4}\\right)\n&=3\\binom{42}{5}\n&=42\\cdot 41\\cdot 39\\cdot 38\n&=(40^2-4)(40^2-1)\n&\\equiv \\boxed{004}\\pmod{1000},\n\\end{align*}\nand we are done.</blockquote>\n\nnice factorization at the end",
"This is the best problem on the test",
"<blockquote><span style=\"font-size:50%\">i may or may not have used the last hour to bash this one out</span></blockquote>\n\nme too me too",
"3rd hardest problem on the test, IMO. Spent an hour and couldn't get this. Turned it into finding $\\binom{3}{2}^2+\\binom{4}{2}^2+...+\\binom{40}{2}^2$ , and tried recursion + hockey stick, to no avail. :stink:",
"<blockquote>3rd hardest problem on the test, IMO. Spent an hour and couldn't get this. Turned it into finding $\\binom{3}{2}^2+\\binom{4}{2}^2+...+\\binom{40}{2}^2$ , and tried recursion + hockey stick, to no avail. :stink:</blockquote>\n\nHow is this the third hardest problem?",
"@above, at least for me, 15 > 14 > 10 > 12 > 13. 10 obviously has an easier solution but the motivation was a lot harder.",
" $$ {{n\\choose 2}\\choose 2}=\\frac{\\frac{n(n-1)((n)(n-1)-2)}{4}}{2!} $$ $$ =\\frac{1}{8}\\cdot n(n-1)(n^2-n-2)=\\frac{1}{8}\\cdot (n+1)(n)(n-1)(n-2)=3{n+1\\choose 4} $$ By hockey sticks, the answer is $$ 3\\cdot \\sum_{i=4}^{41} {i\\choose 4}=3\\cdot {42\\choose 5}=42\\cdot 41\\cdot 39\\cdot 38 $$ This is basically equivalent to $$ (40^2-1)(40^2-4)=40^4-5\\cdot 40^2+4=0-0+4=\\boxed{4}\\pmod{1000} $$ ",
"<details><summary>Solution</summary>Consider a committee of $n$ -people as follows. $$ C = \\{x_1, x_2, \\cdots, x_n \\} $$ Now, consider the set of all possible committees of pairs of these $n$ -peoples. $$ P = \\{ \\{x_1, x_2 \\}, \\{x_1, x_3 \\}, \\cdots, \\{x_{n-1}, x_{n} \\} \\}. $$ Notice that when we choose two such pairs, the two sets of pairs we choose either have a person in common or no people in common. Therefore, $$ \\dbinom{\\binom{n}{2}}{2} = n \\binom{n-1}{2} + \\left(\\frac{\\binom{4}{2}}{2}\\right)\\dbinom{n}{4} = 3 \\bigg( \\dbinom{n}{3} + \\dbinom{n}{4} \\bigg) = 3 \\bigg(\\dbinom{n+1}{4} \\bigg). $$ Therefore, the answer is $$ 3 \\bigg( \\dbinom{4}{4} + \\dbinom{5}{4} + \\cdots + \\dbinom{41}{4} \\bigg) = 3 \\times \\dbinom{42}{5} \\equiv \\boxed{004} \\pmod{1000}. $$</details>\n\nI am wondering if there was some direct bijection to go from $\\dbinom{\\binom{n}{2}}{2}$ to $3 \\dbinom{n+1}{4}.$ ",
"I just realized that proving $\\binom{\\binom{n}{2}}{2} = 3\\binom{n+1}{4}$ is *literally* an exercise in chapter 12 of intermediate c&p lol",
"<details><summary>@#15</summary><blockquote>Note that \n\\begin{align*}\n\\binom{\\tbinom{n}{2}}{2}&=\\binom{\\tfrac{n^2-n}{2}}{2}\n&=\\frac{(n^2-n)(n^2-n-2)}{8}\n&=\\frac{n^4-2n^3-n^2+2n}{8}\n&=\\frac{(n+1)n(n-1)(n-2)}{8}\n&=3\\binom{n+1}{4}\n\\end{align*}\nso we get \n\\begin{align*}\n\\binom{\\tbinom{3}{2}}{2} + \\binom{\\tbinom{4}{2}}{2} + \\dots + \\binom{\\tbinom{40}{2}}{2}&=3\\left(\\binom{4}{4}+\\binom{5}{4}+\\dots+\\binom{41}{4}\\right)\n&=3\\binom{42}{5}\n&=42\\cdot 41\\cdot 39\\cdot 38\n&=(40^2-4)(40^2-1)\n&=40^4-5(40^2)+4\n&\\equiv \\boxed{004}\\pmod{1000},\n\\end{align*}\nand we are done.</blockquote></details>\n\nWow, this is a great solution! :****D",
"Pretty easy for #10.",
"<blockquote>I just realized that proving $\\binom{\\binom{n}{2}}{2} = 3\\binom{n+1}{4}$ is *literally* an exercise in chapter 12 of intermediate c&p lol</blockquote>\n\nyeah the problem is much easier if someone has done that exercise before (speaking from personal experience)",
"[See this](https://artofproblemsolving.com/community/c6h2656258p23000979)!",
"The identity $$ \\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4} $$ shows up in AoPS inter counting and probability as one of the exercise problems",
"<blockquote>The identity $$ \\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4} $$ shows up in AoPS inter counting and probability as one of the exercise problems</blockquote>\n\nExercise 12.3.1, specifically",
"This is a clear case of where knowing the Hockey-Stick identity will solve the problem once you manage to rearrange the sum a bit. \nHere is a <details><summary>proof</summary>To prove $\\sum_{k=i}^{j}\\dbinom{k}{n} = \\dbinom{j+1}{n+1}$ where $i \\geq n$ , one proof is to use double-counting. Since $j \\geq i$ and $i \\geq n$ then $j - n \\geq 0$ . So consider a grid with $j-n$ columns and $n+1$ rows. Let this be, for the sake of the example, a neighborhood grid with a house in each square of the grid and I want to get from the bottom-left corner to the top-right corner. Anyway, consider the first time I get off the bottom row and move one up. There are a total of $\\dbinom{j+1}{n+1}$ ways to go through this grid, but also for each scenario of the first time I leave the bottom row there are from $\\dbinom{j}{n}$ to $\\dbinom{n}{n}$ ways starting from the left-most position one above from the bottom row to the right-most position one above from the bottom row. The sum has the same value read backwards and forwards, so therefore the Hockey-Stick identity is proven.</details> of the Hockey-Stick identity. Anyway use a little algebra to get \n<blockquote>The identity $$ \\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4} $$ shows up in AoPS inter counting and probability as one of the exercise problems.</blockquote>\nSo then the answer is $3 \\cdot \\dbinom{42}{5} \\pmod{1000} = \\boxed{004}$ .\n\n",
"<blockquote>@above, at least for me, 15 > 14 > 10 > 12 > 13. 10 obviously has an easier solution but the motivation was a lot harder.</blockquote>\n\nWhere did 11 go xd<blockquote>I just realized that proving $\\binom{\\binom{n}{2}}{2} = 3\\binom{n+1}{4}$ is *literally* an exercise in chapter 12 of intermediate c&p lol</blockquote>\n\nme who did ICP up to chapter 9: \n\n\n\ntried bashing it using (n^4-2n^3-n^2+n)/8 but doing that on a school bus isn't the best way to get a problem right ",
"10 minute solve because I was trying to find an identity for $\\dbinom{\\tbinom n2}{2}$ this was such an easy problem",
" $$ \\binom{\\binom{n}{2}}{2} = 3\\binom{n+1}{4} $$ By deduction.",
"We can start with the above solution, then we can use the formula for the summation of $x^4$ , and same for the other terms, then get the answer.\n",
"<details><summary>Counting argument for (n choose 2) choose 2 =2(n+1 choose 4)</summary>In a $n$ -gon, the number, of diagonals+edges is $\\binom{n}{2}$ , and the number of ways to choose a pair of them is $\\binom{\\binom{n}{2}}{2}$ . Or, we can add another imaginary vertex onto $n$ , $A$ , st. that we choose $4$ of these $n+1$ vertices, and if $A$ is not among the 4, we choose two lines that don't intersect at a vertex (3 ways). If $A$ is selected, we choose any pair of distinct diagonals from the other $3$ vertices. Thus $\\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4}$</details>",
"<blockquote>We can start with the above solution, then we can use the formula for the summation of $x^4$ , and same for the other terms, then get the answer.</blockquote>\n\nYep... Just some basic combo bash and Hockey Stick should suffice for this one. Still, it's a really cool problem!",
"Observe first that $$ \\binom{\\binom{k}{2}}{2}=\\frac{\\binom{k}{2}\\left(\\binom{k}{2}-1\\right)}{2}=\\frac{\\frac{k(k-1)}{2}\\left(\\frac{k(k-1)}{2}-1\\right)}{2}, $$ and since $\\tfrac{k(k-1)}{2}-1=\\tfrac{k^2-k-2}{2}=\\tfrac{(k+1)(k-2)}{2}$ , we have $\\tfrac{(k+1)(k)(k-1)(k-2)}{8}=3\\tbinom{k+1}{4}$ . As a result, $$ \\sum_{k=3}^{40}\\binom{\\binom{k}{2}}{2}=\\sum_{k=3}^{40}3\\binom{k+1}{4}=3\\sum_{k=4}^{41}\\binom{k}{4}, $$ which by the Hockey Stick Identity equals $$ 3\\binom{42}{5}=3\\cdot\\frac{42\\cdot41\\cdot40\\cdot39\\cdot38}{5\\cdot4\\cdot3\\cdot2\\cdot1}=42\\cdot41\\cdot39\\cdot38. $$ To finish, note that $42\\cdot38=40^2-2^2\\equiv-4\\pmod{1000}$ and $41\\cdot39=40^2-1^2\\equiv-1\\pmod{1000}$ , so $$ 42\\cdot41\\cdot39\\cdot38\\equiv(-4)\\cdot(-1)\\equiv\\boxed{004}\\pmod{1000}, $$ as requested.",
" $\\binom{\\binom{n}{2}}{2} = \\frac{(\\frac{n(n-1)}{2})(\\frac{n(n-1)}{2}-1)}{2}=\\frac{n^4}{8}-\\frac{n^3}{4}-\\frac{n^2}{8}+\\frac{n}{4}$ the answer is the sum of that from 2 to 40\nso it is $(\\frac{40(41)(81)(4919)}{240} - 1) - (\\frac{40(40)(41)(41)}{16}-1) - (\\frac{40(41)(81)}{48}-1) + (\\frac{40(41)}{8}-1)=2552\\boxed{004}$ ",
"Notice that in general, we have $$ \\binom{\\binom{k}{2}}{2} = \\binom{\\frac{k(k-1)}{2}}{2} = \\frac{\\frac{k^2-k}{2} \\cdot \\frac{k^2-k-2}{2}}{2} = \\frac{(k^2-k)(k^2-k-2)}{8}. $$ **Finish 1:** Expand to get $$ \\frac{k^4}{8} - \\frac{k^3}{4} - \\frac{k^2}{8} + \\frac{k}{4}. $$ Use summation formulas and modular arithmetic to obtain the result of $\\boxed{004}$ .**Finish 2:** Rewrite as $\\frac{(k-2)(k-1)(k)(k+1)}{8}$ , and notice that this is equivalent to $3\\binom{k+1}{4}$ . Apply the Hockey Stick Identity to obtain a result of $3\\binom{42}{5}$ . Extract $\\boxed{004}$ .",
"Motivation for @above's Finish 2 for me is that we started with a combo problem, so it would be really nice to tie this back into combo. Also that $(k+1)(k)(k-1)(k-2)$ looks hella sus. So, we try and use something with $\\binom {k+1}{4}$ since that would get us the numerator. Noticing the extra $3$ in the denominator, we multiply both sides by 3.",
"How do you solve this via generating functions?",
"I must agree.",
"<details><summary>Key observation</summary>${{n\\choose2}\\choose2}=\\frac{\\frac{n(n-1)}{2}\\left(\\frac{n(n-1)}{2}-1\\right)}{2}=\\frac{n(n-1)(n^2-n-2)}{8}=\\frac{n(n-1)(n+1)(n-2)}{8}=3{{n+1}\\choose4}$</details>\n<details><summary>Hockey Stick Finish</summary>$\\Rightarrow\\sum_{n=3}^{40}{{n\\choose2}\\choose2}=3\\sum_{n=3}^{40}{{n+1}\\choose4}=3\\left({4\\choose4}+{5\\choose4}+\\cdots+{41\\choose4}\\right)=3{{41+1}\\choose{4+1}}=3{42\\choose5}=2552\\boxed{004}$ .</details>\n",
"I found it funny on how I somehow solved this problem, but anyways here is my solution:\n\n<details><summary>Not a realistic solution but whatever</summary>Bashing out the first 5 chooses WITHOUT DIVIDING BY 2, we get the remainder to be 756. Bashing out the next 5 gives us a remainder of 256.\nWe see that for every 5 chooses, the remainder alternates between 756 and 256. This continues until we get to 38C2C2 + 39C2C2 + 40C2C2 + 41C2C2 + 42C2C2, which is expected to have a remainder of 256. Bashing out 41C2C2 and 42C2C2 gives us a remainder of 460 and 580 respectively. Adding these all up, dividing by 1000, and dividing by 2 we get: $((756 + 256)*4 - (460 + 580))/2 = (048 - 040)/2 = 008/2 = \\boxed{004}$</details>",
"Hopefully this year's AIME is littered with easy problems like this :D \n\nNote that $\\binom{\\binom{n}{2}}{2} = \\binom{\\frac{n(n - 1)}{2}}{2} = \\frac{\\left(\\frac{n(n - 1)}{2} \\right) \\left(\\frac{n(n - 1)}{2} - 1 \\right)}{2} = \\frac{n(n - 1)(n + 1)(n - 2)}{8} = 3 \\binom{n + 1}{4}$ .\n\nSuppose the requested sum is $S$ . Then $$ S = 3\\left[\\binom{4}{4} + \\binom{5}{4} + \\cdots + \\binom{41}{4} \\right] $$ $$ = 3\\binom{42}{5} \\equiv 4 \\pmod{1000}. $$ The final answer is $\\boxed{4}$ .",
"<blockquote>Hopefully this year's AIME is littered with easy problems like this :D \n\nNote that $\\binom{\\binom{n}{2}}{2} = \\binom{\\frac{n(n - 1)}{2}}{2} = \\frac{\\left(\\frac{n(n - 1)}{2} \\right) \\left(\\frac{n(n - 1)}{2} - 1 \\right)}{2} = \\frac{n(n - 1)(n + 1)(n - 2)}{8} = 3 \\binom{n + 1}{4}$ .\n\nSuppose the requested sum is $S$ . Then $$ S = 3\\left[\\binom{4}{4} + \\binom{5}{4} + \\cdots + \\binom{41}{4} \\right] $$ $$ = 3\\binom{42}{5} \\equiv 4 \\pmod{1000}. $$ The final answer is $\\boxed{4}$ .</blockquote>\n\nVery ez but good problem",
"<blockquote><blockquote>Hopefully this year's AIME is littered with easy problems like this :D \n\nNote that $\\binom{\\binom{n}{2}}{2} = \\binom{\\frac{n(n - 1)}{2}}{2} = \\frac{\\left(\\frac{n(n - 1)}{2} \\right) \\left(\\frac{n(n - 1)}{2} - 1 \\right)}{2} = \\frac{n(n - 1)(n + 1)(n - 2)}{8} = 3 \\binom{n + 1}{4}$ .\n\nSuppose the requested sum is $S$ . Then $$ S = 3\\left[\\binom{4}{4} + \\binom{5}{4} + \\cdots + \\binom{41}{4} \\right] $$ $$ = 3\\binom{42}{5} \\equiv 4 \\pmod{1000}. $$ The final answer is $\\boxed{4}$ .</blockquote>\n\nVery ez but good problem</blockquote>\n\nEz",
"ummm this is just 3(42C5) a problem in the counting and probability book",
"Bro i wish all AIME problems were like this :(",
"<details><summary>Solution</summary>$\\textbf{Claim}:$ For any positive integer $n \\ge 3,$ we have that $$ 3{n+1 \\choose 4} = {{n \\choose 2} \\choose 2}. $$ Proof: Notice that $$ {{n \\choose 2} \\choose 2} = {\\frac{n(n-1)}{2} \\choose 2} = \\frac{n(n-1)}{2} \\cdot \\frac{n^{2}-n-2}{4} = \\frac{(n+1)n(n-1)(n-2)}{8}, $$ which is the expression on the LHS. \nNow, by Hockey-Stick, we get $$ 3 {42 \\choose 5} = 3 \\frac{42 \\cdot 41 \\cdot 39 \\cdot 38}{3} = 42 \\cdot 41 \\cdot 39 \\cdot 38 \\equiv 596 \\cdot 599 \\equiv 4, $$ our answer.</details>",
"this ended up being the easiest problem for me because I came across the identity while randomly reading the intermediate c&p book. 2 min solve",
"<blockquote>this ended up being the easiest problem for me because I came across the identity while randomly reading the intermediate c&p book. 2 min solve</blockquote>\n\nSame! You could just apply the identity on pg. 260 of the textbook.",
"<blockquote><blockquote>this ended up being the easiest problem for me because I came across the identity while randomly reading the intermediate c&p book. 2 min solve</blockquote>\n\nSame! You could just apply the identity on pg. 260 of the textbook.</blockquote>\n\nmy weird \"hobby\" of reading random pages of aops textbooks finally became useful",
"Fun problem.\n\nLet the sum be $S$ . Write it as \n\\begin{align*}\nS = \\sum_{k=3}^{40} \\binom{\\binom{k}{2}}{2} \n= \\sum_{k=3}^{40} \\binom{\\frac{k(k-1)}{2}}{2} \n= \\sum_{k=3}^{40} \\frac{\\frac{k(k-1)}{2} \\cdot \\frac{(k-2)(k+1)}{2}}{2} \n= \\frac18 \\sum_{k=3}^{40} (k-2)(k-1)(k)(k+1) \n= \\frac18 \\sum_{k=3}^{40} 24 \\binom{k+1}{4} \n= \\frac18 \\cdot 24 \\cdot \\binom{42}{5} \n\\equiv 004 \\pmod{1000} \n\\end{align*} \nwhere we used the Hockey Stick Identity in the second to last line.",
"notice that $\n\\sum_{n=3}^{40} \\binom{\\binom{n}{2}}{2}\n= 3 \\left( \\sum_{n=3}^{40} \\binom{n}{3} + \\sum_{n=3}^{40} \\binom{n}{4} \\right)\n= 3 \\left( \\binom{41}{4} + \\binom{41}{5} \\right)\n$ by hockey stick\n\nalso notice that $ \\binom{41}{4}+\\binom{41}{5}=\\binom{42}{5}$ by pascal\n\n then you can eaisly evaluate this and find the $ \\mod 1000$ of the sum"
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1038,
"boxed": false,
"end_of_proof": false,
"n_reply": 55,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782974.json"
}
|
Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
|
<blockquote>
Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
</blockquote>
what an intuitive problem
The conditions imply that the sum of the negative numbers is $-\tfrac12$ and those of the positive numbers is $\tfrac12$ . Therefore, at some term $x_i$ , the sum of everything before is $-\tfrac12$ and everything else (after or equal to $x_i$ ) is $\tfrac12$ . So we'd like to make $x_{16}$ as negative as possible and $x_{76}$ as positive as we can.
This means "pushing" the things in between ( $x_{17}$ , $x_{18}$ , $\dots$ , $x_{75}$ ) to be $0$ to maximize distance. Equality should hold, meaning $x_1=x_2=\dots=x_{16}$ and $x_{76}=x_{77}=\dots=x_{100}$ . So $x_{16}=-\tfrac12\cdot\tfrac1{16}=-\tfrac1{32}$ and $x_{76}=\tfrac12\cdot\tfrac1{25}=\tfrac{1}{50}$ , $x_{76}-x_{16}=\tfrac{41}{800}$ , obtaining $\boxed{841}$ as the desired answer.
|
[
"Got 841, solution set is $x_{1, \\ldots, 16} = -\\frac{1}{32}, x_{17, \\ldots 75} = 0, x_{76, \\ldots, 100} = \\frac{1}{50}$ giving $\\frac{41}{800}$ .",
"confirm 841",
"<details><summary>Kinda fakesolved it</summary>We want $x_{76}$ to be as large as posssible, and $x_{16}$ to be as small as possible. Since we want to minimize the absolute values as well, from $x_{76} = x_{77} = \\cdots = x_{100}$ (so $x_{76}$ is as large as possible) and $x_1 = x_2 = \\cdots = x_{16}$ (so $x_{16}$ is as small as possible).\n\nLet's say $\\vert x_{17} \\vert + \\vert x_{18} \\vert + \\cdots + \\vert x_{75} \\vert = m$ and $x_{17} + \\cdots + x_{75} = n$ , then since $x_{16} < 0$ and $x_{76} > 0$ we have\n\\begin{align*}\n 25x_{76} - 16x_{16} &= 1 - m \n 25x_{76} + 16x_{16} &= -n\n\\end{align*}\nso since $m \\geq n$ , adding we have $50x_{76} = 1 -m-n \\leq 1 \\iff x_{76} \\leq \\frac{1}{50}$ (equality is reached when $m=n=0$ , which is attained by $x_{17} = x_{18} = \\cdots = x_{75} = 0$ ). Similarly, $x_{16} \\geq - \\frac{1}{32}$ , hence\n\\[ \\frac{1}{50} - \\left ( -\\frac{1}{32} \\right ) = \\frac{41}{800} \\] which means $m+n = \\boxed{841}$ .</details>",
"<details><summary>sketchy solution I used during the test</summary>We see that the sum of all positive numbers is $\\frac{1}{2}$ and the sum of all the negative numbers is $-\\frac{1}{2}$ . We must have at least 16 negative numbers and 25 positive numbers. So the minimum possible value of the maximum negative number is $-\\frac{1}{32}$ , and the maximum possible value of the minimum positive number is $\\frac{1}{50}$ . Subtracting gives $\\frac{41}{800}$</details>",
"Can confirm. Not too hard once you realize that for some index $i$ , the numbers before $i$ sum to $-1/2$ and the numbers from $i$ onwards sum to $1/2$ .",
"Rigorous sol: you can replace $x_1,\\cdots, x_{16}$ with their average since they should be all negative. Similarly make the last 25 numbers equal. Now use triangular inequality to prove $x_{16}\\ge \\frac{-1}{32}$ and $x_{76}\\le \\frac{1}{50}$ ",
"WLOG $x_{1,2,\\ldots,16}$ equal to $x$ and $x_{76,77,\\ldots,100}=y$ with $x<0$ and $y>0$ . We wish to maximize $y-x$ . \n\nWe have $25y-16x+|x_{17}|+|x_{18}|+\\ldots+|x_{75}|= 1$ and $25y+16x+x_{17}+x_{18}+\\ldots+x_{75}=0$ . \n\nSubtracting gives $-32x+(|x_{17}|-x_{17}|)+(|x_{18}|-x_{18})+\\ldots+(|x_{75}|-x_{75}|)=1\\implies -32x\\le 1\\implies x\\ge -\\frac{1}{32}$ . \n\nAdding gives $50y\\le 1\\implies y\\le \\frac{1}{50}$ . \n\nBoth equalities hold with $x_{17}=x_{18}=\\cdots=x_{75}=0$ . \n\nSo it's $\\frac{1}{50}+\\frac{1}{32}=\\frac{82}{1600}=\\frac{41}{800}\\implies \\boxed{841}$ . \n",
"One of the easier problems on the AIME II, imo, though I couldn't get to solve while mocking\n\n<details><summary>Sol</summary>The key to the problem is realizing the conditions in the problem imply that the sum of all the negatives is $-\\tfrac 12$ and the sum of all the positives is $\\tfrac 12$ . To maximize $x_{76}-x_{16}$ , we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible. Note that $x_{76}+x_{77}+\\cdots+x_{100}=\\tfrac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order, and similarly $x_{1}+x_{2}+\\cdots+x_{16}=-\\tfrac 12$ is less than or equal to $16x_{16}$ . So we now know that $\\tfrac 1{50}$ is the best we can do for $x_{76}$ , and $-\\tfrac 1{32}$ is the least we can do for $x_{16}$ .\n\nFinally, the maximum value of $x_{76}-x_{16}=\\tfrac 1{50}+\\tfrac 1{32}=\\tfrac{41}{800}$ so $\\boxed{841}$ is our answer.</details>",
"I got 841 in like 2 minutes and spent 20 minutes doubting whether my answer is correct. Proving that the answer is $\\frac{41}{800}$ is not that easy. ",
"it's just simple ineq i think?",
"<blockquote>Rigorous sol: you can replace $x_1,\\cdots, x_{16}$ with their average since they should be all negative. Similarly make the last 25 numbers equal. Now use triangular inequality to prove $x_{16}\\ge \\frac{-1}{32}$ and $x_{76}\\le \\frac{1}{50}$ </blockquote>\n\nAnother optimization one can do to be sure that $x_{17} = \\cdots = x_{75} = 0$ is the best case is: If $x_{17} < 0$ , then change $x_{17} \\to \\min(0,x_{18})$ and decrease $0 \\ge a = x_1 = \\cdots = x_{16}$ appropriately. Hence $x_{17} \\ge 0$ . Similarly, if $x_{75} > 0$ , then change $x_{75} \\to \\max(0,x_{74})$ and increase $0 \\le b = x_{76} = \\cdots = x_{100}$ appropriately. Hence $x_{75} \\le 0$ . This forces $x_{17} = \\cdots = x_{75} = 0$ , as desired. ",
"Since $x_1\\leq x_2\\leq\\dots\\leq x_{16}$ and $x_{76}\\leq x_{77}\\leq\\dots\\leq x_{100}$ , $x_{16}$ and $x_{76}$ are minimized and maximized respectively when $x_1=x_2=\\dots=x_{16}$ and $x_{76}=x_{77}=\\dots=x_{100}$ . Now, it is optimal for $x_{16}$ and $x_{76}$ to be positive and negative respectively, so we wish to maximize their absolute values. Hence, as the sum of all the absolute values is upper bounded, we set those of all \"irrelevant\" values $x_{17},x_{18},\\dots,x_{75}$ to $0$ , and divide $-\\tfrac{1}{2}$ and $\\tfrac{1}{2}$ equally to $x_1,x_2,\\dots,x_{16}$ and $x_{76},x_{77},\\dots,x_{100}$ , respectively. It follows that $x_{16}=-\\tfrac{1}{32}$ and $x_{76}=\\tfrac{1}{50}$ , which implies that $x_{76}-x_{16}$ is at most $\\tfrac{1}{50}+\\tfrac{1}{32}=\\tfrac{41}{800}$ . The requested answer is then $\\boxed{841}$ .",
"We clearly seek the cases where $x_1=x_2= \\dots = x_{16}$ , otherwise we can compensate for the smaller numbers to increase the given difference. Similarly $x_{76}=x_{78} = \\dots = x_{100}$ . Note that there must be positive and negative numbers and that $\\sum |\\text{negative numbers}| = \\frac12$ and $\\sum \\text{positive numbers} = \\frac12$ by our condition. Hence, to maximize $x_{76}$ and minimize $x_{16}$ we must let all in between be $0$ . Then, we have $\\frac12$ to split between the $25$ positive numbers giving $x_{76} = \\frac{1}{50}$ and similarly for negatives giving $x_{16} = \\frac{1}{32}$ . Subtracting gives $\\frac{41}{800} \\implies \\boxed{841}$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 AIME Problems"
] |
{
"answer_score": 1056,
"boxed": true,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2022 Contests/2022 AIME Problems/2782983.json"
}
|
A polygon is called *convex* if all its internal angles are smaller than 180 $^{\circ}$ . Given a convex polygon, prove that one can find three distinct vertices $A$ , $P$ , and $Q$ , where $PQ$ is a side of the polygon, such that the perpendicular from $A$ to the line $PQ$ meets the segment $PQ$ (possible at $P$ of $Q$ ).
|
[
"Thank you!",
"I drew an equilateral triangle APQ. Then, I labeled midpoint of PQ as M and drew AM. From there it's pretty easy to see/prove that AM is perpendicular to PQ.",
"<blockquote>I drew an equilateral triangle APQ. Then, I labeled midpoint of PQ as M and drew AM. From there it's pretty easy to see/prove that AM is perpendicular to PQ.</blockquote>\n\nThis is just an outline of my proof. If I used this as the basis of my proof, would it be a coherent and correct proof?",
"<details><summary>#6:</summary>[quote name=\"asimov\" url=\"/community/p24585029\"]\n<blockquote>I drew an equilateral triangle APQ. Then, I labeled midpoint of PQ as M and drew AM. From there it's pretty easy to see/prove that AM is perpendicular to PQ.</blockquote>\n\nThis is just an outline of my proof. If I used this as the basis of my proof, would it be a coherent and correct proof?\n</blockquote></details> You need to prove that the problem statement is true for all polygons, but you only proved it for equilateral triangles, so no",
"But it said \"prove that one **can**\" find the condition, so an example would not suffice?\n\nI might be very mistaken as this is my first proof-based competition",
"<details><summary>#8:</summary>[quote name=\"asimov\" url=\"/community/p24585126\"]\nBut it said \"prove that one **can**\" find the condition, so an example would not suffice?\n\nI might be very mistaken as this is my first proof-based competition\n</blockquote></details> In this case we would interpret it as \"given any convex polygon, prove that one can always find the condition\""
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793577.json"
}
|
|
The game of pool includes $15$ balls that fit within a triangular rack as shown:
[asy]
// thanks Ritwin for this diagram :D
unitsize(0.6cm);
pair pos(real i, real j) {
return i*dir(60) + (j,0);
}
for (int i = 0; i <= 4; ++i) {
for (int j = 0; j <= 4-i; ++j) {
draw(circle(pos(i,j), .5));
}
}
pair A = pos(0,0);
pair B = pos(0,4);
pair C = pos(4,0);
pair dd = dir(270) * .5;
pair ul = dir(150) * .5;
pair ur = dir( 30) * .5;
real S = 1.75;
draw(A+dd -- B+dd ^^ B+ur -- C+ur ^^ C+ul -- A+ul );
draw(A+dd*S -- B+dd*S ^^ B+ur*S -- C+ur*S ^^ C+ul*S -- A+ul*S);
draw(arc(A, A+ul*S, A+dd*S));
draw(arc(B, B+dd*S, B+ur*S));
draw(arc(C, C+ur*S, C+ul*S));
[/asy]
Seven of the balls are "striped" (not colored with a single color) and eight are "solid" (colored with a single color). Prove that no matter how the $15$ balls are arranged in the rack, there must always be a pair of striped balls adjacent to each other.
|
This one was cute :)
Divide it into the following $6$ triangular sections, each with $3$ balls:
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
pair x = (0,0), y = 3*dir(240), z = 3*dir(240) + 3*dir(0);
dot(x,red); dot(x+dir(240),red); dot(x+dir(300),red);
dot(y,red); dot(y+dir(240),red); dot(y+dir(300),red);
dot(z,red); dot(z+dir(240),red); dot(z+dir(300),red);
[/asy]
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
dot(2*dir(240),red); dot(3*dir(240)+dir(0),red); dot(2*dir(240)+dir(0),red);
[/asy]
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
dot(3*dir(240)+dir(0),red); dot(4*dir(240)+2*dir(0),red); dot(3*dir(240)+2*dir(0),red);
[/asy]
[asy]
size(75);
for (int i = 0; i < 5; ++i) {
for (int j = 0; j <= i; ++j) {
dot(i * dir(240) + j * dir(0));
}
}
dot(2*dir(240)+dir(0),red); dot(2*dir(240)+2*dir(0),red); dot(3*dir(240)+2*dir(0),red);
[/asy]
Pigeonhole tells us there are two striped balls in the same section, so they are adjacent.
|
[
"Also BAMO-8 Problem C.\n\nNote that First Row: striped, Second Row: solid, solid, Third Row: striped, solid, striped, Fourth Row: solid, solid, solid, solid, Fifth Row: striped, solid, striped, solid, striped is the max that striped balls could be placed without any being adjacent to each other. If we place one more, it will touch another striped ball.",
"<blockquote>Also BAMO-8 Problem C.\n\nNote that First Row: striped, Second Row: solid, solid, Third Row: striped, solid, striped, Fourth Row: solid, solid, solid, solid, Fifth Row: striped, solid, striped, solid, striped is the max that striped balls could be placed without any being adjacent to each other. If we place one more, it will touch another striped ball.</blockquote>\n\nThis is not a correct prove!",
"Hi, that wasn't my proof. i was just stating that that is the maximum striped balls can be filled without touching one another.",
"<blockquote>This one was cute :)\n</blockquote>\nI agree :D\n\nMy solution was a bit different; each section here can only include two striped balls at maximum.\n[asy]\nsize(75);\n\npen r=hsv(0,0.8,1);\npen g=hsv(120,0.8,0.8);\npen b=Cyan;\n\npair x = (1,0)+4*dir(240);\nfor (int d = 0; d <= 180; d += 60) {\n dot(x+dir(d),r);\n}\ndot(x,r);\n\npair y = (3,0)+3*dir(240);\nfor (int d = 120; d <= 300; d += 60) {\n dot(y+dir(d),g);\n}\ndot(y,g);\n\npair z = (0,0)+dir(240);\nfor (int d = -120; d <= 60; d += 60) {\n dot(z+dir(d),b);\n}\ndot(z,b);\n[/asy]",
"Does solving the problem with casework + pigeonhole give docks? "
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
] |
{
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793698.json"
}
|
Suppose that $p,p+d,p+2d,p+3d,p+4d$ , and $p+5d$ are six prime numbers, where $p$ and $d$ are positive integers. Show that $d$ must be divisible by $2,3,$ and $5$ .
|
For the sake of contradiction, let's consider if $2\nmid p$ , $3\nmid p$ , and $5\nmid p$ .
If $2\nmid p$ , then $p$ and $p+d$ are $0$ and $1$ $mod$ $2$ in some order. Regardless, this configuration gives that $3$ of the values given are even primes, contradicting the fact that there is only one even prime, so $2\mid p$ must be true.
If $3\nmid p$ , then $p, p+d, p+2d$ are $0, 1, 2$ $mod$ $3$ in some order. However, this configuration gives that $2$ of the values given are primes divisible by $3$ , contradicting the fact that the only prime divisible by $3$ is $3$ itself, so $3\mid p$ must be true.
If $5\nmid p$ , then at least one of the numbers given must be a multiple of $5$ , forcing it to be $5$ . Since $p$ and $d$ are positive integers, we do cases on $p = 2$ and $p = 3$ , since if p is more than $3$ , it is impossible for any number in the sequence to be equal to $5$ .
If $p = 2$ , then $d = 3$ or $d = 1$ gives a term equal to $5$ . We only have these cases because any larger values of $d$ make it impossible to achieve $5$ . However, both sequences contain nonprimes. Similarly, if $p = 3$ , then $d = 2$ or $d = 1$ gives a term equal to $5$ . However, both sequences still contain non-primes. Thus, $5\mid p$ is also forced to be true.
If $30\mid d$ , one choice of $p$ to make it work is $p = 7$ , which gives the sequence $7, 37, 67, 97, 127, 157$ .
|
[
"Hint: We can see that $d$ is divisibile by $2$ , else the $p+d$ is divisibile by $2$ , so it is not prime.\nAnalogous with $3$ : if $d$ it is not divisible by $3$ , we can have $d=3k+1$ , or $d=3k+2$ but in these case one of the 6 numbers it is divisible by 3, etc.",
"Bad problem,\nFirst note that $p\\not =2,3,5$ ,so $d$ is even.\nSimilarly assume $3 \\not | d $ .Then consider the residue class and bash every case\n[Ii.e. $(p \\equiv 1 \\pmod 3,d \\equiv 1\\pmod 3),......(p \\equiv 1 \\pmod 5,d \\equiv 1 \\pmod 5)$ and so on]",
"If $p=2$ then $p+2d$ is not a prime, so $p$ is an odd integer. Hence, all six primes are odd, and then $d=(p+d) -p$ is even.\n\nIf $p=3$ then $p+3d$ is not a prime, so $p>3$ , then all such primes give $1,2 \\pmod 3$ . Participate our numbers into three groups $(p,p+3d), (p+d,p+4d), (p+2d, p+5d)$ . Suppose that $d$ is not divisible by $3$ , then $p, p+d, p+2d$ gives different remainders modulo $3$ and non of them gives the same remainder. But by PigeonHole Principle we have that there at least two of them must have the same residue class modulo $3$ , a contradiction. Hence, $3\\mid d$ For $p>5$ (of course, $p$ cannot be $5$ ), we assume by a contradiction that $d$ is not divisible by $5$ , then we have $12$ easy cases, in each of which we can find the number among our six that is divisible by $5$ .",
"If $2$ doesn't divide $d$ , then if $p$ is odd, $p+d$ is even and thus composite, and if $p=2$ , $p+2d$ is even and thus composite. So $2\\mid d$ .\n\nConsider $p+3d,p+4d,p+5d$ . If $3$ doesn't divide $d$ , then each of these numbers are different $\\pmod{3}$ , and therefore one of them must be divisible by $3$ . None of these can equal $3$ , by a simple size argument, and so we have a contradiction. Thus $3 \\mid d$ . We can use a similar argument for $5 \\mid d$ .",
"This was also BAMO-8 Problem D\n\nI only got until proving $2 | d$ , didn't know how to continue after that.\n\nAlso, note that $p=7$ and $d=30$ is a possible configuration.",
"<details><summary>Solution</summary>Suppose that $q\\nmid d$ for some $q\\in\\{2,3,5\\}$ . Then, $p$ , $p+d$ , $p+2d$ , $p+3d$ , $p+4d$ , and $p+5d$ contain all of the numbers between $0$ and $q-1$ mod $q$ , so there exists a multiple of $q$ . Therefore, exactly one of the numbers must be a multiple of $q$ . However, if $q\\in\\{2,3\\}$ , then at least two of the numbers are multiples of $q$ . Therefore, $6\\mid d$ , which means that we must have $p=q=5$ . This is a contradiction since $p+5d$ is also a multiple of $5$ . Therefore, $30\\mid d$ .</details>",
"Subtract $p+5d$ by $p+4d$ gives $d\\mid 2$ because primes greater than $2$ are all odd. \n\nNow to show its divisible by $3$ , FTSOC let $d\\equiv 1\\pmod{3}$ , then $p\\equiv 1\\pmod{3}$ . In this case, $p+2d\\equiv 0\\pmod{3}$ which is a contradiction. \n\nThe other case is when $d\\equiv 2\\pmod{3}$ and $p\\equiv 2\\pmod{3}$ , then $p+2d$ would still be divisible by $3$ . Thus, this proves the claim that $d\\mid 3$ . \n\nTo do $\\mod 5$ , turn the equations to mod 5 gives $p, p+d, p+2d, p-2d, p-d\\pmod{5}$ . FTSOC if $d\\nmid 5$ , then $d=5-k\\pmod{5}$ such that $k$ is a positive integer less than $5$ but greater than $0$ . $p\\neq 0, k\\pmod{5}$ in this case. Its clear that when $d$ is not divisible by $5$ then we can see a contradiction since $p\\neq 0\\pmod{5}$ and $d\\pmod{5}$ , then one of the five would always be divisible by $5$ . Thus, $d\\mid 5$ . \n\n*q.e.d*",
" $mod 2,3,5$ ",
"<blockquote>Therefore, $6\\mid d$ , which means that we must have $p=q=5$ .</blockquote>\n\nWhy is this true?",
"@samrocksnature , since we got that 6 | d, then since according to our claim one of the numbers should not divide two of the 6 numbers (p , p+d ...) so 5 does not divide 2 of them, but since 5 does not divide d, d, 2d, 3d, 4d, 5d , all have different residues (mod 5), so lets suppose p = 5, in this case we are done, otherwise p>5, in this case p+xd will be congruent to 0 in (mod 5) for some x (from 1 to 5) but since p>5 , p+xd > 5, but since 5 | p+xd, p+xd is not prime. And hence we are done.",
"<blockquote>@samrocksnature , since we got that 6 | d, then since according to our claim one of the numbers should not divide two of the 6 numbers (p , p+d ...) so 5 does not divide 2 of them, but since 5 does not divide d, d, 2d, 3d, 4d, 5d , all have different residues (mod 5), so lets suppose p = 5, in this case we are done, otherwise p>5, in this case p+xd will be congruent to 0 in (mod 5) for some x (from 1 to 5) but since p>5 , p+xd > 5, but since 5 | p+xd, p+xd is not prime. And hence we are done.</blockquote>\nYep and by the way, the smallest solution is p = 107 and d = 30"
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
] |
{
"answer_score": 92,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793699.json"
}
|
Ten birds land on a $10$ -meter-long wire, each at a random point chosen uniformly along the wire. (That is, if we pick out any $x$ -meter portion of the wire, there is an $\tfrac{x}{10}$ probability that a given bird will land there.) What is the probability that every bird sits more than one meter away from its closest neighbor?
|
The answer is $\boxed{\tfrac{1}{10^{10}}}$ .
We model the wire as the interval $[0,10]$ on the real line. We can view a point $x$ in the interval as a pair $(\lfloor x\rfloor, \{x\})$ (more precisely a bijection). The key insight is to see that floor of the positions of birds must all be distinct i.e, they are some permutation of $0,2,\ldots, 9$ and the probability is precisely $\tfrac{10!}{10^{10}}$ . We can then vary the fractional parts independently. But with an additional condition that the fractional parts must form an order statistic which has a chance of $1^{10}\tfrac{1}{10!}$ . Therefore the answer is $\tfrac{10!}{10^{10}}\cdot\tfrac{1}{10!}=\tfrac{1}{10^{10}}$ as desired.
|
[
"We claim the probability is $\\frac{1}{10^{10}}$ .\n\nLabel the birds from $1$ to $10$ , and divide the wire into $10$ disjoint segments, each with length $1$ meter. Each bird must occupy a different segment, or else two birds would be at most one meter away. There are $10!$ ways to assign the birds to the segments, and each assignment has $\\frac{1}{10^{10}}$ probability of occuring, so the probability each bird lies on a different segment is $\\frac{10!}{10^{10}}$ .\n\nLabel the segments from left to right as $1$ to $10$ . For segment $i$ , let $x_i$ be the distance from the left edge of the segment to the position of the bird on the segment. Each $x_i$ lies in the range $[0,1]$ , chosen uniformly at random. The probability two $x_i$ are equal to each other is $0$ . This is because if $9$ of the $x_i$ 's are fixed, the tenth has infinitely many possible values, and finitely many equal to one of the other $x_i$ . Therefore, we may assume all $x_i$ are distinct.\n\nEvery pair of neighboring birds is more than $1$ meter away from each other iff $x_1<x_2<x_3<\\ldots <x_{10}$ . But since every permutation of the $x_i$ is equally likely to be in sorted order, $P(x_1<x_2<x_3<\\ldots <x_{10})=\\frac{1}{10!}$ .\n\nTherefore, the desired probability is $\\frac{10!}{10^{10}}\\cdot\\frac{1}{10!}=\\frac{1}{10^{10}}$ .",
"The probability of placing the birds on a $10$ -meter wire such that they are all more than a meter away from their neighbor is in one to one correspondence to the probability that they all land on the first meter of the $10$ meter wire, since from there we can insert $1$ meter between every pair of birds to get a working $10$ -meter configuration, and from any $10$ -meter configuration we can remove a meter between every pair of birds to get a configuration of $10$ birds on the first meter.\n\nThus the probability we are looking for is the same as the probability that each of the birds land on a specific $\\frac{1}{10}$ of the strip. Since we have $10$ birds, that probability is $\\frac{1}{10^{10}}$ .",
"<blockquote>The probability of placing the birds on a $10$ -meter wire such that they are all more than a meter away from their neighbor is in one to one correspondence to the probability that they all land on the first meter of the $10$ meter wire, since from there we can insert $1$ meter between every pair of birds to get a working $10$ -meter configuration, and from any $10$ -meter configuration we can remove a meter between every pair of birds to get a configuration of $10$ birds on the first meter.\n\nThus the probability we are looking for is the same as the probability that each of the birds land on a specific $\\frac{1}{10}$ of the strip. Since we have $10$ birds, that probability is $\\frac{1}{10^{10}}$ .</blockquote>\n\nThe idea is obviously intuitive, but bijections don't work in measure/probability theory when everything has cardinality $\\mathfrak c$ . :D\n\nThe proper way to do this is rayfish's method. In measure theory, we WLOG and assume $x_1<\\dots<x_{10}$ , then the space of events that you want is the interior of a $10$ -dimensional shape (in fact, a $10$ -simplex) with side length $1$ and translated by $(0,1,2,3,4,5,6,7,8,9)$ , out of the whole space which is the same shape with side length $10$ . Obviously the ratio of hypervolumes is $1$ to $10^{10}$ .",
"I think you can get bijections working if you use the fact that it preserves the Jacobian and integrals.\n\nRecharacterize the probability space on the birds at $b_1, b_2, \\dots, b_{10}$ when sorted to be choosing gaps $g_0 = b_1 - \\frac{1}{2}, g_2 = b_2 - b_1, \\dots, g_{10} = b_{10} - b_9, g_{11} = 10\\frac{1}{2} - b_{10}$ where we have that $g_1 + \\dots + g_{10} \\le 10$ .\nIt can be shown that this doesn't change the probability of each event (formally, use the Jacobian).\nThe answer is then equal to \\[ \\frac{\\int_{g_1=0}^{10} \\int_{g_2 + \\dots + g_{10} \\le 10 - g_1, g_2 \\ge 1} 1}{\\int_{g_1=0}^{10} \\int_{g_2 + \\dots + g_{10} \\le 10 - g_1} 1} = \\frac{\\int_{g_1=0}^{1} \\frac{(1-g_1)^{9}}{9!}}{\\int_{g_1=0}^{10} \\frac{(10-g_1)^{9}}{9!}} = \\frac{\\int_{g_1=0}^{1} \\frac{g_1^{9}}{9!}}{\\int_{g_1=0}^{10} \\frac{g_1^{9}}{9!}} = \\frac{1}{10^{10}} \\] which finishes.\nwhere $g_i$ are the gaps.\n"
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
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{
"answer_score": 1016,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793705.json"
}
|
Sofiya and Marquis are playing a game. Sofiya announces to Marquis that she's thinking of a polynomial of the form $f(x)=x^3+px+q$ with three integer roots that are not necessarily distinct. She also explains that all of the integer roots have absolute value less than (and not equal to) $N$ , where $N$ is some fixed number which she tells Marquis. As a "move" in this game, Marquis can ask Sofiya about any number $x$ and Sofiya will tell him whether $f(x)$ is positive negative, or zero. Marquis's goal is to figure out Sofiya's polynomial.
If $N=3\cdot 2^k$ for some positive integer $k$ , prove that there is a strategy which allows Marquis to identify the polynomial after making at most $2k+1$ "moves".
|
Neat problem!
<span style="font-size:50%"> ~~Though my writeup was way too long :**(~~ </span>
<details><summary>Solution</summary>The key idea is to use binary search.
Note that the coefficient of $x^2$ in $f$ is $0$ , so if the roots are $r$ , $s$ , and $t$ , we know that $r+s+t=0$ . Thus we only need to find two roots to reconstruct the polynomial (as we know it's monic).
[list=1]
[*] Query $f(0)$ .
- If $f(0) = 0$ then we have found a root. Binary search on the positive root in step 2; however if we think $1$ is the positive root we must query it again, as it is possible that $f(x) = x^3$ has only roots $0$ .
- If $f(0) > 0$ , two roots are positive while one is negative. This case is symmetric to the next.
- If $f(0) < 0$ , two roots are negative while one is positive. We will binary search on the interval $[1, 3\cdot2^k-1]$ for this root.
[/list]
- If we binary search naively, we will use $2k+2$ queries. However we can optimize it by shifting the first query. Start by querying $f(2^{k+1})$ . [list]
- If $f(2^{k+1}) = 0$ , we have found a positive root. Skip to step 3
- If $f(2^{k+1}) > 0$ , the positive root is in the interval $[1, 2^{k+1}-1]$ (length $2^{k+1}-1$ ). Binary search on this interval for the root in $k$ steps.
- If $f(2^{k+1}) < 0$ , the positive root is in the interval $[2^{k+1}+1, 3\cdot2^k-1]$ (length $2^k-1$ ). Binary search on this interval for the root in $k-1$ steps.
[/list]
- So now we have a positive root, call it $r$ . We know that $s+t = -r$ so (at least) one of $s$ or $t$ is in the interval $\big[\!-\!\tfrac r2,0\big]$ . Binary search on this interval for one root, with at most $\lceil \log_2r \rceil - 2$ queries.
Now we have two roots, and can find the third root from $t = -r-s$ , so we know the polynomial. Carefully counting the queries, we see that in all cases we use at most $2k+1$ queries (we offset the second query---for $f(2^{k+1})$ ---to balance the number of queries out more). $\square$ (man)
*Note: The binary search we do is equivalent to looking for the $0$ in a hidden list $[a,a,a,\ldots,a,0,b,\ldots,b,b,b]$ (where $a\neq b$ are known and nonzero, also note that the $0$ could be on either end) where we can query the value at any index. We can prove by induction that it takes at most $k$ queries, where the length of the list is $2^{k+1}-1$ . The base case is that it takes one query for $k=1$ (so the list has $3$ elements).*
I like this problem because I've never seen binary search in a math problem before!</details>
There were several tricky things in this problem that you need to be careful about. Hopefully I caught all of them.
|
[
"Sad, I could only prove at most $2k+2$ moves ... I felt like I was almost there but I didn't think of splitting $N$ into powers of $2$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
] |
{
"answer_score": 188,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793708.json"
}
|
You are bargaining with a salesperson for the price of an item. Your first offer is $a$ dollars and theirs is $b$ dollars. After you raise your offer by a certain percentage and they lower their offer by the same percentage, you arrive at an agreed price. What is that price, in terms of $a$ and $b$ ?
|
$\frac{p}{a}-1=1-\frac{p}{b}$ so $p=\boxed{\frac{2ab}{a+b}}$
|
[
"<details><summary>Sketch of what I did.</summary>NOTE: $b>a$ and $x<1$ . Let $x$ be the percentage. then $a+ax = b(1-x)$ (this is the final price). Alg. bash to get $x=\\frac{b-a}{a+b}$ . Plug into price equation to get answer as $a+(\\frac{ab-a^2}{a+b})$ . Use some values of $a$ and $b$ to check.</details>",
" $$ a+\\left(\\frac{ab-a^2}{a+b}\\right)=\\frac{a^2+ab}{a+b}+\\frac{ab-a^2}{a+b}=\\frac{2ab}{a+b}=\\frac{2}{\\frac{1}{a}+\\frac{1}{b}} $$ so its just harmonic mean of a and b",
"<details><summary>my answer</summary>Let the agreed price be \\( x \\), and the percentage (as a decimal) be \\( p \\). Then:\n\\[\na(1 + p) = b(1 - p)\n\\]\nSolving for \\( p \\):\n\\begin{align*}\na(1 + p) &= b(1 - p) \na + ap &= b - bp \nap + bp &= b - a \np(a + b) &= b - a \np &= \\frac{b - a}{a + b}\n\\end{align*}\n\nNow substitute \\( p \\) into \\( x = a(1 + p) \\):\n\\[\nx = a\\left(1 + \\frac{b - a}{a + b}\\right) = a\\left(\\frac{2b}{a + b}\\right) = \\frac{2ab}{a + b}\n\\]\n\n\\[\n\\boxed{x = \\frac{2ab}{a + b}}\n\\]</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
] |
{
"answer_score": 1004,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793751.json"
}
|
If I have 100 cards with all the numbers 1 through 100 on them, how should I put them in order to create the largest possible number?
|
The no. of digits in such a number is $192$ . The largest 192-digit number is $99....999$ where $9$ occurs $192$ times. This gives us an idea that we need numbers having $9$ in their leftmost place first following with $8$ and so on.
So the number should be
99 9 98 97 96 95 94 93 92 91 90 89 88 8 87 ... 22 2 21 20 19 18 17 16 15 14 13 12 11 1 10 100
Notice the placement of the single digit numbers.
|
[
"<details><summary>Solution</summary>99 9 98 97 96 95 94 93 92 91 90 89 88 8 ... 20 19 18 17 16 15 14 13 12 11 1 10 100.</details>",
"@above, your solution is what I got, but what happened to Card #10? (I placed it right before Card #100)\n\noh and what happened to card #15 and card #13?",
"please correct me if I am wrong"
] |
[
"origin:aops",
"2022 Contests",
"2022 BAMO"
] |
{
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 BAMO/2793759.json"
}
|
The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$ . The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$ . Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$ , prove that $ \angle BKC \equal{} \angle CDB$ .
|
Cute and easy!! :D
Denote $\boxed{MA=x,MB=w,MC=y,CD=z}$ . Then by the given condition, we have $xy+xz=w(MD) \implies \boxed{MD=\frac{x(y+z)}{w}}$ .
Define $O=BD \cap KC$ then $CO$ is the angle bisector of $\angle{MCD}$ . By internal angle bisector theorem, we have $\frac{MO}{OD}=\frac{MC}{CD}=\frac{y}{z}$ and $MO+OD=MD=\frac{x(y+z)}{w}$ . Thus, $\boxed{MO=\frac{xy}{w}}$ and $\boxed{OD=\frac{xz}{w}}$ . Now note that $MO.MB=xy=MA.MC$ implying $BAOC$ is cyclic. Thus, $\angle{KBD}=\angle{ABO}=\angle{ACO}=\angle{ACK}=\angle{KCD} \implies \boxed{\angle{KBD}=\angle{KCD}}$ . Thus $BKDC$ is also cyclic implying $\angle{BKC}=\angle{CDB}$ as desired. $\blacksquare$ ( $\mathcal{QED}$ )
|
[
"Here is what I have so far:\r\nI will first prove the converse. We will assume KBCD is cyclic. Let KC and BD intersect at P. Then ABCP is also cyclic. By power of a point, $ MA \\cdot MC \\equal{} MP \\cdot MB$ . So we need to show $ MA \\cdot CD \\equal{} MB(MD \\minus{} MP) \\equal{} MB \\cdot PD$ .\r\nOr $ \\frac {MA}{MB} \\equal{} \\frac {PD}{CD} \\implies \\frac {MP}{PC} \\equal{} \\frac {PK}{KB}$ . But by Power of a point, $ PC \\cdot PK \\equal{} PD \\cdot PB$ . So we need to show $ \\frac {KB}{PB} \\equal{} \\frac {PD}{PM} \\implies \\frac {CD}{CP} \\equal{} \\frac {CD}{CM}$ (angle bisector theorem). But then that would mean CP=CM, where's my mistake?",
"<blockquote>\nOr $ \\frac {MA}{MB} \\equal{} \\frac {PD}{CD} \\implies \\frac {MP}{PC} \\equal{} \\frac {PK}{KB}$ . </blockquote>\r\n\r\nHere, I think.\r\n\r\nAnd if you prove the converse, you have to show it implies the original problem. :)",
"Note: $ MH.CD\\equal{}HD.MC$ (1)\r\n\r $ MH\\plus{}HD\\equal{}MD \\longrightarrow MC.MH\\plus{}MC.HD\\equal{}MC.MD \\longrightarrow$ according to(1), $ MC.MH\\plus{}CD.MH\\equal{}MC.MD$ \r\n\r\nmultiple the equation at $ \\frac{MA}{MH}$ ,then, $ MA.MC\\plus{}MA.CD\\equal{}\\frac{MA.MC.MD}{MH}$ (2)\r\n\r\naccording to assumption we have: $ \\frac{MA.MC.MD}{MH}\\equal{}MB.MD \\longrightarrow MA.MC\\equal{}MB.MH$ \r\n\r $ ABCH$ is cyclic,then $ \\angle ACH\\equal{}\\angle ABH\\equal{}\\angle HCD \\longrightarrow KBCD$ is cyclic and $ \\angle BKC\\equal{}\\angle CDB$ ",
"<blockquote>(1) draw a circle with center $ C$ and radius $ CD$ (2) let the intersection of that circle and $ AC$ be $ E$ (3) $ AM\\cdot MC$ + $ AC\\cdot CD$ = $ AM\\cdot ME$ = $ BM\\cdot MD$ $ \\Longrightarrow$ $ A,B,E,D$ are concyclic\n(4) $ \\angle ABD$ = $ \\angle KCD$ $ \\Longrightarrow$ $ K,B,C,D$ are concyclic\n(5) $ \\Longrightarrow\\angle BKC$ = $ \\angle CDB$ </blockquote>\r\n\r\nHow did you get (3) $ AM\\cdot MC$ + $ AC\\cdot CD$ = $ AM\\cdot ME$ .\r\n\r\nAM*MC+AC*CD=AM*ME=AM(MC+CE)=AM*MC+AM*CE=AM*MC+AM*CD ----> AC=AM which I think is impossible.",
"My solution(77ant`s idea):\r\n\r\nDraw a circle with center C and radius CD. Let the intersection of that circle and AC be E(E lies on the extension of AC).\r\nMA*MC+MA*CD=MA(MC+CD)=MA(MC+CE)=MA*ME\r\nMA*ME=MB*MD so A,B,E,D are concyclic.\r\n<CDE=<CED=<ACD/2=x\r\n<AED=<ABD=<KCD=x, so K,B,C,D are concyclic.\r\n\r\n---> <BKC=<CDB. :)",
"my mistake in typing :blush: I have corrected it right now.",
"Very interesting question\n\nNote that showing <BKC = <CDB is the same as showing KBCD cyclic.\n\nExtend AC and let a point D' be on AC with CD = CD'. Now the metric condition given becomes MA(MC+CD') = MB(MD) so MA(MD') = MB(MD). This implies that quadrilateral ABD'D is cyclic.\n\nNow let <CDD' = <CD'D = <AD'D = <ABD (by cyclic quads) = <1. Since <DCA is an exterior angle, it is the sum of <CDD' and <CD'D, so <DCA = 2<1.\nSince CK is the angle bisector of <DCA, <DCK = <KCA = <1. Now let X be the intersection of CK and BD. Since <XCA = <XBA, quadrilateral BAXC is cyclic as well.\n\nNow this question is just a simple angle chase. Let <CAB = <CXB = <2. Then <CKA = <CAB - <KCA (by exterior angles) = <2-<1. \n<CDM = <CXM - <DCX (by exterior angles again) = <2-<1. Thus, <CKA = <CDM, so quadrilateral KBCD is cyclic.\n\nFinally, this implies that <BKC = <CDB. QED\n\n(Sorry for no diagram or LaTeX, I don't know how to do either :()",
"<details><summary>solution</summary>Let $D'$ be the point on ray $AC$ such that $CD'=CD$ and let $A'$ be the point on ray $DC$ such that $AC=A'C$ . The condition tells us by PoP that $ABD'D$ is cyclic, and by PoP, $A'$ is also on the circle. Thus, $\\angle ACD = \\angle A'CD' = 2\\cdot \\angle AD'D$ . So $$ \\angle ABD = \\angle AD'D = \\dfrac{1}{2}\\angle ACD = \\angle KCD $$ so $BCDK$ is cyclic. Thus, $\\angle BKC = \\angle CDB$ as desired.</details>",
"Solution:\n\nLet $BD$ meet $KC$ at $N$ . $ MA \\cdot MC \\plus{}MA \\cdot CD \\equal{} MB \\cdot MD$ implies that $\\frac{MB}{MA}$ = $\\frac{MC+CD}{MD}$ .\nApplying angle bisector theorem, we get $\\frac{MC}{MN}$ = $\\frac{CD}{DN}$ ,\n\nor, $\\frac{MC\\cdot ND}{MN\\cdot MD}$ = $\\frac{CD}{MD}$ ,\n\nor, $\\frac{MC}{MN}$ = $\\frac{MC+CD}{MD}$ .\n\nThus, $\\frac{MC}{MN}$ = $\\frac{MB}{MA}$ ,\n\nor, $MC\\cdot MA$ = $MB\\cdot MN$ .\nHence, by the converse of angle bisector theorem, $ABCN$ is cyclic.\nThis implies that $\\angle ABD$ = $\\angle ACN$ ,\nor, $\\angle KBD$ = $\\angle KCD$ ,\nor, $KBCD$ is cyclic.\nSo, $\\angle BKC$ = $\\angle CDB$ .",
"Nice one. \nLet $\\omega$ be the circumcircle of $BCD$ . Let $E=AC\\cap \\omega$ and $K'$ be the point on $\\omega$ , that lies on angle bisector of $\\angle ACD$ . We ultimately want to show that $A,B,K'$ are collinear.\nWe have that $$ MA \\cdot MC +MA \\cdot CD = MB \\cdot MD, \\quad (1) $$ but by Power of a Point we have that $$ MB\\cdot MD=MC\\cdot ME. $$ Thus, $(1)$ takes the following form: $$ MA \\cdot MC +MA \\cdot CD =MC\\cdot ME \\Longleftrightarrow MA\\cdot CD=AE\\cdot MC\\Longleftrightarrow \n \\frac {MA}{AE} = \\frac {MC}{CD} \\quad (2) $$ We also have that $\\triangle MCD\\sim \\triangle MBE$ , since $\\angle ECD=\\angle EBD$ and $\\angle EMB=\\angle DMC$ , thus $$ \\frac {MC}{CD} = \\frac {MB}{BE} \\quad (3) $$ By $(2)$ and $(3)$ , we conclude that $\\frac {MA}{AE} = \\frac {MB}{BE}$ , which means that $AB$ is the angle bisector of $\\angle DBE$ . Also, since $K'$ is the defined as intersection of $\\omega$ and angle bisector of $\\angle ACD$ , we have that $\\triangle K'DE$ is isosceles, from here we have that $K'B$ is the angle bisector of $\\angle DBE$ . We conclude that $A,B,K'$ are collinear.",
"since it came back in front page due to the last solution, \nlet's say the origin of the problem: **2000 Belarusian MO B 10.3**\n\nalso posted and solved [here](https://artofproblemsolving.com/community/c6h296380p1605623), [here ](https://artofproblemsolving.com/community/c6h213009p1175536) and [here](https://artofproblemsolving.com/community/c6h1920219p13167817)",
"Let $D'$ be on $AC$ such that $C$ lies between $M,D'$ and $CD' = CD$ . $MA.MC + MA.CD = MB.MD \\implies MA.MD' = MB.MD \\implies ABD'D$ is cyclic. $\\angle KBD = \\angle ABD = \\angle AD'D = \\frac{\\angle ACD}{2} = \\angle KCD \\implies KBCD$ is cyclic so $\\angle BKC = \\angle BDC$ .",
"Let $N= BD\\cap KC$ . By the angle bisector theorem, we know that $\\frac{NM}{ND}= \\frac{MC}{CD}$ .(1)\n\nAs $MA(MC+CD)= MB\\cdot MD$ , by (1), we get $MA(MC+\\frac{MC\\cdot ND}{NM})= MB\\cdot MD$ $\\implies MA\\cdot MC(1+\\frac{ND}{NM})= MB\\cdot MD$ $\\implies MA\\cdot MC\\cdot\\frac{(ND+NM)}{NM}= MB\\cdot MD\\implies MA\\cdot MC\\cdot\\frac{MD}{NM}= MB\\cdot MD$ $\\implies MA\\cdot MC= NM\\cdot MB$ .\nSo, $ABCN$ is ciclic $\\implies \\angle DCN=\\angle NCM= \\angle NBA\\implies KBCD$ is ciclic, therefore $\\angle BKC=\\angle BDC$ , QED",
"Let the angle bisector of $\\angle{ACD}$ intersect $BD$ at $X$ .\n\nBy given condition, we know that $MA=\\frac{MB \\cdot MD}{MC+CD}$ . By angle bisector theorem, we get that $XM=\\frac{MC \\cdot MD}{MC+CD}$ .\n\nThen $XM \\cdot MB=\\frac{MB \\cdot MC \\cdot MD}{MC+CD}=AM \\cdot MC$ , implying that $ABCX$ is cyclic by converse of PoP.\n\nTherefore $\\angle{DCX}=\\angle{MCX}=\\angle{ABD}$ , implying that $BCDK$ is cyclic, and $\\angle{BKC}=\\angle{BDC}$ $\\blacksquare$ ",
"Extend $AC$ to meet $(ABD)$ at $D'$ . Then $MD = MD'$ , and from here some angle chasing shows us that $BKDC$ is cyclic as desired."
] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
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{
"answer_score": 1132,
"boxed": true,
"end_of_proof": true,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/228384.json"
}
|
For a given value $t$ , we consider number sequences $a_1, a_2, a_3,...$ such that $a_{n+1} =\frac{a_n + t}{a_n + 1}$ for all $n \ge 1$ .
(a) Suppose that $t = 2$ . Determine all starting values $a_1 > 0$ such that $\frac43 \le a_n \le \frac32$ holds for all $n \ge 2$ .
(b) Suppose that $t = -3$ . Investigate whether $a_{2020} = a_1$ for all starting values $a_1$ different from $-1$ and $1$ .
|
**(a)** $a_{n+1} =\frac{a_n +2}{a_n + 1}$ $\frac43 \le \frac{a_1 +2}{a_1 + 1} \le \frac32$ $4a_1+4 \le 3a_1+6 \implies a_1\le 2$ and $2a_1+4 \le 3a_1+3 \implies 1\le a_1$ , thus $2\geq a_1\geq 1$ .
Suppose $\frac43 \le a_{n}\le \frac32$ , then we want to show that for all $n\geq 1$ we have $\frac43 \le 1+\frac{1}{a_n + 1} \leq \frac{3}{2} \Longleftrightarrow 1 \leq a_n\leq 2$ , which is true. Thus, $\boxed{a_1=[1,2]}$ .**(b)** $a_{n+1} =\frac{a_n -3}{a_n + 1}$ We want to show that $a_4=a_1$ , then $a_4=\frac{\frac{a_2 -3}{a_2 + 1}-3}{\frac{a_2 -3}{a_2 + 1} + 1}=\frac{\frac{a_2 -3-3(a_2+1)}{a_2 + 1}}{\frac{a_2 -3+(a_2+1)}{a_2 + 1}}=\frac{a_2 -3-3(a_2+1)}{a_2 -3+(a_2+1)}=\frac{-2a_2-6}{2a_2 -2}=-\frac{a_2+3}{a_2 -1}=-\frac{\frac{a_1 -3}{a_1 + 1}+3}{\frac{a_1 -3}{a_1 + 1}-1}=-\frac{\frac{a_1 -3+3(a_1+1)}{a_1 + 1}}{\frac{a_1 -3-1(a_1+1)}{a_1 + 1}}=-\frac{a_1 -3+3(a_1+1)}{a_1 -3-1(a_1+1)}=-\frac{4a_1}{-4}=a_1$ .
Notice that we cannot have $a_2=-1\Longleftrightarrow a_1=1$ or $a_1=-1$ .
Hence, we have showed that $a_4=a_1$ , hence $a_{3k+1}=a_1$ and we are done.
|
[] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
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"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2352936.json"
}
|
Let $a_1, a_2, \cdots, a_{2n}$ be $2n$ elements of $\{1, 2, 3, \cdots, 2n-1\}$ ( $n>3$ ) with the sum $a_1+a_2+\cdots+a_{2n}=4n$ . Prove that exist some numbers $a_i$ with the sum is $2n$ .
|
It is well-known that among any $n$ integers we can find a non-empty subset with sum divisible by $n$ .
Apply this to $a_1,\dots,a_n$ to find that (w.l.o.g.) $a_1+\dots+a_k$ is divisible by $n$ for some $1 \le k \le n$ .
If $a_1+\dots+a_k=2n$ , we are done.
If $a_1+\dots+a_k=3n$ , then we must have $k=n$ and the remaining $n$ numbers $a_{n+1},\dots,a_{2n}$ all must be equal to $1$ .
Now just take any subsum of $a_1+\dots+a_n$ which is in $[n,2n]$ (this surely exists) and add as many of the ones as necessary.
Finally, assume that $a_1+\dots+a_k=n$ . But now we can just apply the same reasoning to the numbers $a_{n+1},\dots,a_{2n}$ and find that the only remaining case is that $a_{m}+\dots+a_{2n}=n$ for some $m \ge n+1$ .
But then these two sums together give the desired $2n$ . Done.
|
[
"thank you :)))",
"wlog $1<=a_1<=.....<=a_{2n}<=2n-1$ Consider the numbers $a_1,a_1+a_2,....,a_1+...+a_{2n-1}<4n$ and the sets $(1,2n+1),(2,2n+2),...,(2n-1,4n-1)$ if one of the numbers equal to $2n$ we are done.\n\nOtherwise we have $2n-1$ numbers and $2n-1$ sets.\nIf two numbers go to the some set we are done.\nIf $a_{2n}=2$ we are done.\nOtherwise $a_{2n}>=3$ which means that $a_1+...+a_{2n-1}<4n-2$ so we must have take the numberw $2n-1,2n-2$ whicj means that we have $2n-1$ ones contradiction.",
"<blockquote>wlog $1\\le a_1\\le \\cdots \\le a_{2n}\\le2n-1$ Consider the numbers $a_1,a_1+a_2,\\cdots,a_1+\\cdots+a_{2n-1}<4n$ and the sets $(1,2n+1),(2,2n+2),\\cdots,(2n-1,4n-1)$ if one of the numbers equal to $2n$ we are done.\n\nOtherwise we have $2n-1$ numbers and $2n-1$ sets.\nIf two numbers go to the some set we are done.\nIf $a_{2n}=2$ we are done.\nOtherwise $a_{2n}\\ge3$ which means that $a_1+\\cdots+a_{2n-1}<4n-2$ so we must have take the numberw $2n-1,2n-2$ whicj means that we have $2n-1$ ones contradiction.</blockquote>\n\n"
] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
] |
{
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2682466.json"
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|
Let $a, b, c$ be positive real numbers such that: $$ ab - c = 3 $$ $$ abc = 18 $$ Calculate the numerical value of $\frac{ab}{c}$
|
<details><summary>Just to fill the storage</summary>$ab - c = 3$ $\implies$ $ab=3+c$ $abc = 18$ $\implies$ $ab=\frac{18}{c}$ $ab=ab$ $3+c=\frac{18}{c}$ $c^2+3c-18=0$ $(c+6)(c-3)=0$ $\implies$ $c+6=0$ $\wedge$ $c-3=0$ $c=-6$ $\wedge$ $c=3$ Since $c\in$ $R^+$ we get that $c=3$ $ab=\frac{18}{c}$ $ab=6$ $\frac{ab}{c}=2$</details>
|
[
"<details><summary>answer</summary>$2$</details> i just plugged in values lol but there was some algebra ",
"Since ab+(-c)=3 and ab(-c)=-18,\nab and -c are two solutions to c^2-3x-18=0\nSince a, b, c positive, ab=6 and -c=-3.\nHence (ab)/c=2.",
"solved also [here](https://artofproblemsolving.com/community/c6h2731173p23794255)",
"This was easy. $ab = 3+c$ , so $c(3+c) = 18$ , and $c = 3$ . Since $c = 3, ab = 6$ , and $\\frac {ab}{c} = 2$ ."
] |
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"origin:aops",
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"2022 Contests"
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2780694.json"
}
|
Let $\vartriangle ABC$ be a triangle in which $\angle ACB = 40^o$ and $\angle BAC = 60^o$ . Let $D$ be a point inside the segment $BC$ such that $CD =\frac{AB}{2}$ and let $M$ be the midpoint of the segment $AC$ . How much is the angle $\angle CMD$ in degrees?
|
$\angle CMD=40^o$ .
let midpoint $AB$ is $E$ and $F$ on $BC$ such that $MEBF$ parallelogram ( $F$ between $B$ and $D$ ).
easy to check triangle $AME$ similar to triangle $ABC$ , therefore $\angle BEM=100^o$ and $\angle MFD=80^o$ .
by sine rule we got $\dfrac {sin(80)}{sin(\angle CDM)}=\dfrac {sin(40)}{sin(140-\angle CDM)}, \angle CDM=100^o, \angle CMD=40^o$ .
CMIIW
|
[
"A good drawing may drastically shorten the solution; a non trigo one at [https://stanfulger.blogspot.com/2022/02/aops-brasil-girls-math-tournament-2021.html](https://stanfulger.blogspot.com/2022/02/aops-brasil-girls-math-tournament-2021.html)\n\nBest regards,\nsunken rock",
"solved also [here](https://artofproblemsolving.com/community/c6h2731173p23794255)"
] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
] |
{
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"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2780698.json"
}
|
A natural number is called *chaotigal* if it and its successor both have the sum of their digits divisible by $2021$ . How many digits are in the smallest chaotigal number?
|
[
"Let S(n) denote digit sum of natural number n.\nLet N be a smallest chaotigal number.\n\nSay that k nines appear consecutively at the right end of N. Adding 1 to this number will cause k nines turn to k zeros, increase one of the digits by 1 and rest will not change. Therefore we can see that S(N)-S(N+1)=9k-1.\n\nBoth S(N) and S(N+1) are multiple of 2021 so 9k-1 also should be a multiple of 2021. This implies that k is at least 1572. k nines will be congruent to 1 (modulo 2021) so excluding those k nines, sum of digits must be congruent to 2020 modulo 2021. In other words, write N=m*10^k+999...9 then S(m) will be at least 2020. To minimize m we should make its rightmost digt 8 (since it can't be 9), put many nines as possible in the right. Thus minimum of m is 599...998 where there are 223 nines.\n\nHence we conclude that N=599...99899...99, where there are 223 nines in the first group of nines and 1572 nines in the second one. We finally get that minimum number of digits of N is 1798.",
"solved also [here](https://artofproblemsolving.com/community/c6h2731190p23794410)"
] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
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"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2780701.json"
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|
|
Mariana plays with an $8\times 8$ board with all its squares blank. She says that two houses are *neighbors* if they have a common side or vertex, that is, two houses can be neighbors vertically, horizontally or diagonally. The game consists of filling the $64$ squares on the board, one after the other, each with a number according to the following rule: she always chooses a house blank and fill it with an integer equal to the number of neighboring houses that are still in White. Once this is done, the house is no longer considered blank.
Show that the value of the sum of all $64$ numbers written on the board at the end of the game does not depend in the order of filling. Also, calculate the value of this sum.
Note: A house is not neighbor to itself.
<details><summary>original wording</summary>Mariana brinca com um tabuleiro 8 x 8 com todas as suas casas em branco. Ela diz que duas
casas s˜ao vizinhas se elas possu´ırem um lado ou um v´ertice em comum, ou seja, duas casas podem ser vizinhas
verticalmente, horizontalmente ou diagonalmente. A brincadeira consiste em preencher as 64 casas do tabuleiro,
uma ap´os a outra, cada uma com um n´umero de acordo com a seguinte regra: ela escolhe sempre uma casa
em branco e a preenche com o n´umero inteiro igual `a quantidade de casas vizinhas desta que ainda estejam em
branco. Feito isso, a casa n˜ao ´e mais considerada em branco.
Demonstre que o valor da soma de todos os 64 n´umeros escritos no tabuleiro ao final da brincadeira n˜ao depende
da ordem do preenchimento. Al´em disso, calcule o valor dessa soma.
Observa¸c˜ao: Uma casa n˜ao ´e vizinha a si mesma</details>
|
Consider a pair of neighboring houses. Since no two squares are filled simultaneously, for every pair of neighboring houses one house is filled after the other, so the house that is filled first will definitely include the second one in its count, but the second one will not count the first house. Therefore, the total contribution of the pair to the sum of all numbers is exactly one. Since this is true for all pairs of houses, the total sum is just the number of pairs of neighboring houses.
To compute this number, we can count sum of the number of neighbors each house has and divide it by 2, since every pair is included twice: every house in the center of the board has 8 neighbors, every 'edge' house has 5, and the 'corner' houses each have 3: $\frac{36*8+24*5+4*3}{2}=210$ .
|
[
"solved also [here](https://artofproblemsolving.com/community/c6h2731190p23794410)"
] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
] |
{
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"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2780707.json"
}
|
Find all pairs $(a,b)$ of positive integers, such that for **every** $n$ positive integer, the equality $a^n+b^n=c_n^{n+1}$ is true, for some $c_n$ positive integer.
|
We claim that the only solution is the pair $(2,2)$ which clearly works. Now we show that this is the only pair that works.**<span style="color:#00f">Claim:</span> We cannot have an odd prime $p$ and an odd positive integer $n$ so that $a$ and $b$ are not divisible by $p$ and $p \mid a^n + b^n.$**
<u>*proof*</u>
Assume for contradiciton that the *Claim* is false. Then let $p$ and $n$ as above. Since $n$ is odd and $p$ does not divide $a$ and $b,$ we must have by LTE that for all odd positive integer $k:$ $$ v_p (a^{nk} + b^{nk}) = v_p(a^n+b^n) +v_p(k) $$ On the other hand, by problems condition $$ nk+1 \mid v_p(a^{nk} + b^{nk}) $$ Conbining the two relation we get an absurd since we can make $nk+1$ arbitraly large while $v_p(a^n+b^n)+v_p(k)$ is nonzero and can be small (for instance, take $k$ sufficiently large with $p\nmid k$ ) $.\square$ [rule]
Now we claim that if $a\ne b$ then there exist $p$ and $n$ as in the *Claim;* it will force $a=b.$ **<span style="color:#00f">Lemma:</span> If $a$ and $b$ are distinct positive integers, then exists an odd integer $n$ and an odd prime $p$ so that $p\nmid a,b$ and $p\mid a^n+b^n$**
<u>*proof:*</u>
First write $d=gcd(a,b)$ and then let $x={a}{/d}$ and $y={b}{/d}.$ Take $q$ a sufficiently large odd prime, more precisely, take $q$ so that $q>a+b$ and $q>r,$ for all $r\mid d.$ Then take $n=q$ and $p$ as an odd prime dividing $$ \frac{x^q + y^q}{x+y} $$ For seeing that such $p$ exists, first note that, since $a\ne b$ by assumption, we cannot have $x=y=1$ and hence ${(x^q+y^q)}{/(x+y)}>1;$ now to show that $p$ is odd, note that since $gcd(x,y)=1,$ we must have that either $x,y$ are both odds or have different parities. In the later case, we would have that ${(x^q+y^q)}{/(x+y)}$ is odd and we are done; in the former case, just note that $$ \frac{x^q+y^q}{x+y} = x^{q-1}-x^{q-2}y+\dots-xy^{q-2}+y^{q-1} $$ is odd since it is the sum of an odd number of odd numbers. So indeed $p$ exists and is odd.
Now we have to show that $p$ satisfies the other condition, i.e., $p\nmid a,b.$ First note that since $x$ and $y$ are coprime we have $p\nmid x,y$ and so $p\mid a,b \iff p \mid d.$ Now note that since $q$ is odd, we have that $$ p\mid x^q+y^q \Rightarrow {(-xy^{-1})}^q \equiv 1 \pmod p $$ where $y^{-1}$ denotes the multiplicative inverse of $y$ modulo $p.$ Then let $k$ denote the order of $-xy^{-1}$ modulo $p.$ By the above congruence, we have that $k \mid q$ and since $q$ is prime, it follows that $k=q$ or $k=1.$ If the former case happens, then we have that $p-1\ge k =q$ which implies that $p\nmid d$ by the choice of $q$ (we have selected $q$ so that it is greater than all divisors of $d$ ) hence $p\nmid a,b$ and we are done. In the later case, we have that $p\mid x+y;$ hence by LTE we have $$ v_p\left(\frac{x^q+y^q}{x+y}\right)=v_p(q) $$ since $p$ divides the above number by definition, we must have that $v_p(q)\ge 1,$ i.e. $p=q.$ Then $q \mid x+y\mid a+b$ which is an absurd since we have choosen $q>a+b.$ So the Lemma is proven. $\square.$ [rule]
From the *Claim* and the *Lemma,* it follows that we must have $a=b$ and the condition becomes $2a^n$ is an $n+1$ power for all $n.$ We claim that $a$ must be $2.$ Indeed, let $l=v_2(a),$ then problems condition becomes $n+1 \mid 1 + nl$ for all positive integer $n.$ This is equivalen to $n+1 \mid l-1$ for all positive integer $n,$ which can only happen if $l-1 = 0$ (since $l$ is fixed an $n$ can be arbitraly large) i.e. $l=1.$
Now let $p$ be any odd prime, then the condition implies $n+1 \mid nv_p(a)$ for all positive integer $n;$ since $n+1$ and $n$ are coprime, this is equivalent to $n+1 \mid v_p(a)$ for all positive integer $n,$ which similarly implies $v_p(a)=0.$ Then we have that $v_2(a)=1$ and $v_p(a)=0,$ for all odd prime $p,$ i.e. $a=2.$ Clearly $(2,2)$ satisfies problems condition and so we are done.
|
[
"If wlog $a\\geq b$ , assume we're in the case $a>b$ , and consider all the $n>3$ . By Zsigmondy's theorem, there exists a prime $p_n$ dividing $a^n+b^n$ , which doesn't divide the previous number. Since $p_n|c_n^{n+1}$ , it follows that $p_n^{n+1}|c_n^{n+1}=a^n+b^n$ .\nThis implies $p_n<p_n^{1+1/n}\\leq (c_n^{n+1})^{1/n}=\\sqrt[n]{a^n+b^n}\\leq \\sqrt[n]{a^n\\cdot 2^n}=2a$ .\nHowever this implies that the sequence $p_n$ is bounded, in contradiction with the fact that the $p_n$ are primitive factors of $a^n+b^n$ , meaning that there is an infinite quantity of distinct $p_n$ s.\nSo we may assume $a=b$ , which gives $2a^n=c_n^{n+1}$ . Since $a=1$ doesn't work, let $a=2^{e_0}p_1^{e_1}\\cdots p_m^{e_m}$ , with $2<p_1<...<p_m$ and $e_i>0$ (except $e_0\\geq 0$ ). Taking $n+1>\\max\\{e_1,...,e_m\\}$ , this would imply $n+1|ne_i\\implies n+1|e_i\\implies n+1\\leq e_i$ which is a contradiction.\nSo we only have to look at $n_0$ , which gives us $n+1|1+ne_0\\implies n+1|n(e_0-1)\\implies n+1|e_0-1$ . Since this is true for all $n$ , this implies $e_0=1$ , i.e. $a=b=2$ .",
"Let $p$ be a prime such that $p|a+b$ then:\nSuppose that $p$ doesn't devise $a,b$ if we select $n$ such that $p$ doesn't devise $n$ we have: $U_p(a^n+b^n)=U_p(a+b)$ obviously a contradiction.\n\nSo if $p|a+b$ then $p|a$ , $p|b$ .\n\nLet $U_p(a)=x$ and $U_p(b)=y$ with $x>y$ then $U_p(a^n+b^n)=y*n$ with is a contradiction.(because $n+1|yn$ ).\n\nSo if $p|a+b$ then $p|a$ , $p|b$ and $U_p(a)=U_p(b)$ .\n\nLet $a=p^xa_1$ and $b=p^xb_1$ for some odd prime $p|a+b$ then: $p^{xn}(a_1^n+b_1^n)=c_n^{n+1}$ which means that: $n+1|xn+U_p(a_1^n+b_1^n)$ Let $k$ be the smallest number from which $p|a_1^k+b_1^k$ (there's such a $k$ otherwise $n+1|nx$ from every $n$ a contradiction).\nLet $n=kl$ then $kl+1|klx+Up(a_1^k+b_1^k)+U_p(l)$ .\nNow if we take $l$ such that $U_p(l)=0$ and $l$ go to infinity we have $x=U_p(a_1^k+b_1^k)$ Now if we take $l$ such that $U_p(l)=1$ and $l$ go to infinity we have $x=U_p(a_1^k+b_1^k)+1$ Contradiction.\n\nSo $a+b=2^f$ and $U_2(a)=U_2(b)$ and there is no obb prime divesi $a,b$ which gives $a=b=2^{f-1}$ and now easy that $a=b=2$ ",
"<blockquote>\nLet $k$ be the smallest number from which $p|a_1^k+b_1^k$ (there's such a $k$ otherwise $n+1|nx$ from every $n$ a contradiction).\nLet $n=kl$ then $kl+1|klx+Up(a_1^k+b_1^k)+U_p(l)$ .\n</blockquote>\nWhy $n=kl$ ?\n",
"Here's a somewhat different solution.\nNotice that $(a, b) = (2, 2)$ works. We'll prove that it is the only solution.\nIf $a=b$ , $2a^n=c_{n}^{n+1}$ . Let $p$ be an odd prime divisor of $a$ . Then $n+1 \\mid v_{p}(a)n \\Leftrightarrow n+1 \\mid v_{p}(a)$ which is fixed. Thus, $a=2^s$ and $n+1 \\mid ns+1 \\Leftrightarrow n+1 \\mid s-1$ for any $n \\Rightarrow s=1$ .\n\nNow assume $a>b$ . Let $(a, b)=d, a=xd, b=yd$ where $(x,y)=1$ . By Zsigmondy there exists an odd prime $p$ that divides $x^k+y^k$ for an odd $k$ . Taking $n=kp^e$ for some $e$ , and looking at the exponent of $p$ on the condition, $p^ek+1 \\mid p^ekv_{p}(d)+v_{p}(x^{kp^e}+y^{kp^e})$ which by LTE becomes $p^ek+1 \\mid e-v_{p}(d)+v_{p}(x^k+y^k)$ , a contradiction.",
"Cant believe no one mentioned the similarity to 2011 A2:\n\nAll such $c$ are less than $a+b$ for obvious reasons. That means there exists $X$ such that for infinitely many $n,$ $a^n+b^n=X^{n+1}.$ If either $a$ or $b$ are greater than $X$ we have $(a/X)^n >x$ for $n$ sufficiently large. Same holds for $b$ .\nNow, $X^{n+1}=a^n+b^n<=2X^n$ so $X$ is either $1$ or $2$ . $X$ cannot be $1$ because $a^n+b^n>=2$ . This gives $X=2$ and $a=b=2$ is the only possibility.",
"It is even older than 2011 A2. It is from Russia 2005: https://artofproblemsolving.com/community/q2h32168p200066",
"The answer is $a=b=2$ . If $a=b$ we get the solution described above. WLOG $a>b$ .**Claim 1**: $c_n$ is bounded**Pf:** $c_n^{n+1}=a^n+b^n<2a^n<a^{n+1}\\implies c_n<a$ However notice that $$ c_n^{2n+2}=(a^n+b^n)^2<2(a^{2n}+b^{2n})=2c_{2n}^{2n+1}\\leq c_{2n}^{2n+2} $$ so $c_n<c_{2n}$ . This contradicts with claim 1. $\\square$ "
] |
[
"origin:aops",
"2022 Brazil EGMO TST",
"2022 Contests"
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{
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"boxed": false,
"end_of_proof": true,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Brazil EGMO TST/2792230.json"
}
|
Let $P=(x^4-40x^2+144)(x^3-16x)$ . $a)$ Factor $P$ as a product of irreducible polynomials. $b)$ We write down the values of $P(10)$ and $P(91)$ . What is the greatest common divisor of the two numbers?
|
[
"a)P(x)=x(x+6)(x-6)(x+4)(x-4)(x+2)(x-2) b)(P(1),P(91))=3^2*5*7=245"
] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810680.json"
}
|
|
Let $\triangle ABC$ have $AB = 1$ cm, $BC = 2$ cm and $AC = \sqrt{3}$ cm. Points $D$ , $E$ and $F$ lie on segments $AB$ , $AC$ and $BC$ respectively are such that $AE = BD$ and $BF = AD$ . The angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ADE$ for the second time at $M$ and the angle bisector of $\angle ABC$ intersects the circumcircle of $\triangle BDF$ at $N$ . Determine the length of $MN$ .
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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{
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810682.json"
}
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|
Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.
|
Let $a,b,c$ be positive real numbers. Prove that $$ \left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+9 $$ [https://artofproblemsolving.com/community/c6h486634p5449967](https://artofproblemsolving.com/community/c6h486634p5449967)
<details><summary>Nice.</summary><blockquote><blockquote>Given the inequalities: $a)$ $\left(\frac{2a}{b+c}\right)^2+\left(\frac{2b}{a+c}\right)^2+\left(\frac{2c}{a+b}\right)^2\geq \frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ $b)$ $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.</blockquote>
The inequality $a)$ is false for $a = 0^+, b = c = 1$ . As for b), using the inequality $x^2 + y^2 + z^2 \geq xy + yz + zx$ we have
\begin{align*}
\sum \left( \frac{a + b}{c} \right)^2 &\geq \sum \frac{(a + b)(b + c)}{ca} = \sum \frac{ab + ca + b^2 + bc}{ca}
&= \sum \frac{b}{c} + 3 + \sum \frac{b^2}{ca} + \sum \frac{b}{a}
&\geq \sum \frac{a}{b} + 3 + 3 + 3
&= \sum \frac{a}{b} + 9
\end{align*}</blockquote></details>
|
[
"Let $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9 $$ <details><summary>Nice.</summary><blockquote>Given the inequalities: $a)$ $\\left(\\frac{2a}{b+c}\\right)^2+\\left(\\frac{2b}{a+c}\\right)^2+\\left(\\frac{2c}{a+b}\\right)^2\\geq \\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}$ $b)$ $\\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.</blockquote></details>",
"Let $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq 4\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\right) $$ ",
"<blockquote>Given the inequalities: $a)$ $\\left(\\frac{2a}{b+c}\\right)^2+\\left(\\frac{2b}{a+c}\\right)^2+\\left(\\frac{2c}{a+b}\\right)^2\\geq \\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}$ $b)$ $\\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.</blockquote>\n\nThe inequality $a)$ is false for $a = 0^+, b = c = 1$ . As for b), using the inequality $x^2 + y^2 + z^2 \\geq xy + yz + zx$ we have\n\\begin{align*}\n\\sum \\left( \\frac{a + b}{c} \\right)^2 &\\geq \\sum \\frac{(a + b)(b + c)}{ca} = \\sum \\frac{ab + ca + b^2 + bc}{ca} \n&= \\sum \\frac{b}{c} + 3 + \\sum \\frac{b^2}{ca} + \\sum \\frac{b}{a} \n&\\geq \\sum \\frac{a}{b} + 3 + 3 + 3 \n&= \\sum \\frac{a}{b} + 9\n\\end{align*}",
"The second is true and nice.",
"<details><summary>Solution sketch</summary>It is equivalent to $\\sum \\frac {a^2}{b^2}+\\sum \\frac {2ab} {c^2} \\geq \\sum \\frac {a}{b} +9$ (the first two sums are symmetrical, the third is cyclic). Now sum the bounds $\\frac {a^2}{b^2} \\geq 2\\frac {a}{b}-1$ cyclically (not symmetrically!) and apply AM-GM for all 12 terms you obtain at the end and we're done.</details>",
"<blockquote><blockquote>Given the inequalities: $a)$ $\\left(\\frac{2a}{b+c}\\right)^2+\\left(\\frac{2b}{a+c}\\right)^2+\\left(\\frac{2c}{a+b}\\right)^2\\geq \\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}$ $b)$ $\\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.</blockquote>\n\nThe inequality $a)$ is false for $a = 0^+, b = c = 1$ . As for b), using the inequality $x^2 + y^2 + z^2 \\geq xy + yz + zx$ we have\n\\begin{align*}\n\\sum \\left( \\frac{a + b}{c} \\right)^2 &\\geq \\sum \\frac{(a + b)(b + c)}{ca} = \\sum \\frac{ab + ca + b^2 + bc}{ca} \n&= \\sum \\frac{b}{c} + 3 + \\sum \\frac{b^2}{ca} + \\sum \\frac{b}{a} \n&\\geq \\sum \\frac{a}{b} + 3 + 3 + 3 \n&= \\sum \\frac{a}{b} + 9\n\\end{align*}</blockquote>\n\nBravo",
"<details><summary>TYT:</summary><blockquote><blockquote>Given the inequalities: $a)$ $\\left(\\frac{2a}{b+c}\\right)^2+\\left(\\frac{2b}{a+c}\\right)^2+\\left(\\frac{2c}{a+b}\\right)^2\\geq \\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}$ $b)$ $\\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9$ For each of them either prove that it holds for all positive real numbers $a$ , $b$ , $c$ or present a counterexample $(a,b,c)$ which doesn't satisfy the inequality.</blockquote>\n\nThe inequality $a)$ is false for $a = 0^+, b = c = 1$ . As for b), using the inequality $x^2 + y^2 + z^2 \\geq xy + yz + zx$ we have\n\\begin{align*}\n\\sum \\left( \\frac{a + b}{c} \\right)^2 &\\geq \\sum \\frac{(a + b)(b + c)}{ca} = \\sum \\frac{ab + ca + b^2 + bc}{ca} \n&= \\sum \\frac{b}{c} + 3 + \\sum \\frac{b^2}{ca} + \\sum \\frac{b}{a} \n&\\geq \\sum \\frac{a}{b} + 3 + 3 + 3 \n&= \\sum \\frac{a}{b} + 9\n\\end{align*}</blockquote></details>\nLet $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b} +6 $$ ",
"\\[\\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2=\\frac{1}{2}\\sum\\limits_{cyc} \\left(\\frac{a^2}{c^2}+\\frac{c^2}{b^2}\\right)+\\left[\\sum\\limits_{cyc} \\frac{b^2}{c^2}+2\\sum\\limits_{cyc}\\frac{ab}{c^2}\\right]\\overset{AMGM}{\\geq} \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+9\\]",
"<blockquote>\nLet $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b} +6 $$ </blockquote>\n\n\\[\\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2=\\frac{1}{2}\\sum\\limits_{cyc} \\left(\\frac{a^2}{c^2}+\\frac{c^2}{b^2}\\right)+\\frac{1}{2}\\sum\\limits_{cyc} \\left(\\frac{b^2}{c^2}+\\frac{c^2}{a^2}\\right)+2\\sum\\limits_{cyc}\\frac{ab}{c^2}\\overset{AMGM}{\\geq} \\sum\\limits_{cyc} \\frac{a}{b}+\\sum\\limits_{cyc} \\frac{b}{a}+6\\]",
"<blockquote>Let $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq 4\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\right) $$ </blockquote>\nSolution of Zhangyanzong: **By AM-GM,** $$ \\left(\\frac{a+b}{c}\\right)^2+\\left(\\frac{b+c}{a}\\right)^2+\\left(\\frac{c+a}{b}\\right)^2\\geq 4\\left(\\frac{a+b}{c}+\\frac{b+c}{a}+\\frac{c+a}{b}\\right)-12 $$ $$ =4\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\right)+4\\left(\\frac{b}{a}+\\frac{c}{b}+\\frac{a}{c}-3\\right)\\geq4\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\right) $$ ",
"Let $a,b,c$ be positive real numbers. Prove that $$ \\left(\\frac{b+c}{a}\\right)^2+\\frac{2(a+b)(c+a)}{bc}\\geq \\frac{4b}{a}+8\\sqrt{\\frac{a}{b}} $$ "
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|
Let $p = (a_{1}, a_{2}, \ldots , a_{12})$ be a permutation of $1, 2, \ldots, 12$ .
We will denote \[S_{p} = |a_{1}-a_{2}|+|a_{2}-a_{3}|+\ldots+|a_{11}-a_{12}|\]We'll call $p$ $\textit{optimistic}$ if $a_{i} > \min(a_{i-1}, a_{i+1})$ $\forall i = 2, \ldots, 11$ . $a)$ What is the maximum possible value of $S_{p}$ . How many permutations $p$ achieve this maximum? $\newline$ $b)$ What is the number of $\textit{optimistic}$ permtations $p$ ? $c)$ What is the maximum possible value of $S_{p}$ for an $\textit{optimistic}$ $p$ ? How many $\textit{optimistic}$ permutations $p$ achieve this maximum?
|
In any permutation $p$ , call $a_i$ a $peak$ if $a_{i}>\max (a_{i-1},a_{i+1})$ , and a $bottom$ if $a_{i}<\min (a_{i-1},a_{i+1}),$ (to make the definition valid for the endpoints, we let $a_{0}=a_{2}$ and $a_{13}=a_{11}).$ Now we can easily see that for any permutation $p,$ the value of $S_{p}$ only depends on peaks and bottoms
Let $p'=(13-a_{1},13-a_{2},...,13-a_{12}),$ then $S_{p}=S_{p'}$ , and so for the sake of evaluation (for the sake of counting, we simply multiply by 2), we may assume wlog that $a_{1}$ is a bottom. If $a_{12}$ was bottom as well, then for some $1\le k\le 5$ we have (aside of $a_{1},a_{12},$ the $p_{i}$ 's are the peaks and the $b_{j}$ 's are the bottoms of $p$ ):
\[S_{p}=-a_{1}+2(p_{1}-b_{1}+p_{2}-b_{2}+...-b_{k-1}+p_{k})-a_{12}=2(p_{1}+p_{2}+...+p_{k})-2(b_{1}+...+b_{k-1})-a_{1}-a_{12}\]
\[\le 2(12+11+...+(13-k))-2(1+...+(k-1))-k-(k+1)=2k(12-k)-1\le 69\]
And if $a_{12}$ was peak, then for some $0\le k\le 5$ we have
\[S_{p}=-a_{1}+2(p_{1}-b_{1}+p_{2}-b_{2}+...+p_{k}-b_{k})+a_{12}=2(p_{1}+p_{2}+...+p_{k})+a_{12}-2(b_{1}+...+b_{k})-a_{1}\]
\[\le 2(12+11+...+(13-k))+(12-k)-2(1+...+k)-(k+1)=2k(11-k)+11\le 71\]
So $S_{p}\le 71,$ and the equality case should then satisfies the following conditions:
1) $a_{1}=6$ and $a_{12}=7$ 2) $(a_{2},a_{4},a_{6},a_{8},a_{10})$ is a permutation of (8,9,10,11,12)
3) $(a_{3},a_{5},a_{7},a_{9},a_{11})$ is a permutation of (1,2,3,4,5)
This counts a total of $5!\cdot 5!$ equality cases,
so by returning generality we see that the maximum value of $S_{p}$ is 71, and it is achievable through $2\cdot 5!\cdot 5!=28800$ permutations. This concludes part (a).
Now let's consider parts b),c). Notice that the $optimistic$ condition is simply saying that there should be no bottoms, except possibly for the endpoints, so we can only have one peak (if there were 2 peaks, a bottom should appear in between, which violates the optimistic condition, and a peak exists because '12' is always peak) and this peak is $a_{i}=12$ .
But this would imply that $a_{1}<a_{2}<...<a_{i}=12>a_{i+1}>...>a_{12}\;\;^{(*)}$ , and so $$ S_{p}=2a_{i}-a_{1}-a_{12}=24-a_{1}-a_{12}\le 21 $$ This is achievable when $\{a_{1},a_{12}\} =\{1,2\}$ and the rest can be arbitrarywhile preserving (*), this can be easily counted, we shall get a total of $2^{10}=1024$ permutations satisfying $S_{p}=21$ and $p$ optimistic. This concludes part (c)
For part (b), we just ignore the condition $\{a_{1},a_{12}\} =\{1,2\}$ , we'll get a total of $2^{11}=2048$ optimistic permutations
|
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Let $f(x)$ be a quadratic function with integer coefficients. If we know that $f(0)$ , $f(3)$ and $f(4)$ are all different and elements of the set $\{2, 20, 202, 2022\}$ , determine all possible values of $f(1)$ .
|
$f(x) = ax^2+bx+c$ $f(0) = c $ $f(3) = 9a+3b+c $ $f(4) = 16a+4b+c$ Case $f(0)=2$ $f(3) = 9a+3b+2 \equiv 20 \mod 3 $ ( 2 cannot be used since f(0)=2) $9a+3b=18,3a+b=6$ $f(4) = 16a+4b+2 \equiv (202,2022) \mod 4$ [rule]
subCase (f(4)=202) $16a+4b=200$ $4a+b = 50$ $a+b+c = 44+6-44*3+2 = 8-44*2 = -80$ subCase (f(4) 2022) $16a+4b=2020$ $4a+b = 505$ $3a+b=6$ $a+b+c=-990$ [rule]
case $f(0)=20$ $f(4) = 16a+4b+20 \equiv 0 \mod 4$ forcing f(4) = 20 contradiction
case $f(0) = 202$ $f(3) = 9a+3b+202 \equiv 1 \mod 3$ forcing $f(3) = 202$ contradiction
case $f(0) = 2022$ $f(3) = 9a+3b+2022 \equiv 0 \mod 3 $ forcing $f(3) = 2022$ contradiction
|
[
"f(1)=-80 or f(1)=-990 \n"
] |
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|
Let $\triangle ABC$ have median $CM$ ( $M\in AB$ ) and circumcenter $O$ . The circumcircle of $\triangle AMO$ bisects $CM$ . Determine the least possible perimeter of $\triangle ABC$ if it has integer side lengths.
|
<details><summary>Solution</summary>$\underline{\textbf{Claim:}}$ $AC=CM$ $\textit{Proof 1:}$ Let the midpoint of $CM$ be point $N$ . Since $\angle OMA=90^{\circ}$ and $N\in (AMO)$ , this means that $\angle ONA=90^{\circ}$ . If $\overline{AN}\cap (ABC)=X$ , since $ON\perp AN$ , this implies that $N$ is the midpoint of $AX$ . Now $NA=NX$ and $NC=NM$ , so $ACXM$ is a parallelogram, so
\[\angle CAM=\angle CAX+\angle XAM=\angle CBX+\angle CXA=\angle CBX+\angle CBA=\angle ABX=\angle AMN=\angle AMC\Longrightarrow CA=CM\] $\textit{Proof 2:}$ Let $S$ be the midpoint of $AC$ . Now $AMONS$ is cyclic, but $SN$ is a midsegment in $\triangle AMC$ , so $SN\parallel AM$ and $AMNS$ is cyclic, therefore $AMNS$ is an isosceles trapezoid. Now $AS=NM\Longrightarrow CA=CM$ .
Now, if $a=BC$ , $b=AC=CM$ , $c=AB$ we can use the median formula ( $m_{c}=\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$ ) to get that:
\[b^2=CM^2=\frac{1}{4}(2a^2+2b^2-c^2)\Longrightarrow c^2=2(a-b)(a+b)\]
We want $a,b,c$ to be positive integers with minimal sum (the perimeter of $\triangle ABC$ ). If $\gcd(a,b)=d>1$ , then $\left(\frac{a}{d},\frac{b}{d},\frac{c}{d}\right)$ are the sidelengths of a triangle which satisfies the conditions in the problem with smaller perimeter. Thus we can assume that $a,b,c$ are two by two coprime. Now we have two cases: $\textbf{Case 1:}$ $a-b=2s^2$ , $a+b=4t^2$ for positive integers $s,t$ .
Then $a=s^2+2t^2$ , $b=2t^2-s^2$ , $c=4st$ , so the perimeter is $4t^2+4st$ . If $s=1,t=1$ we get $(a,b,c)=(3,1,4)$ , impossible. If $s=1, t=2$ , then $(a,b,c)=(9,7,8)$ , so the least possible perimeter in this case is $24$ . $\textbf{Case 2:}$ $a-b=4s^2$ , $a+b=2t^2$ for positive integers $s,t$ .
Then $a=2s^2+t^2$ , $b=t^2-2s^2$ , $c=4st$ , so the perimeter is $2t^2+4st$ . If $s=1,t=2$ we get $(a,b,c)=(6,2,8)$ , impossible. If $s=1, t=3$ , then $(a,b,c)=(11,7,12)$ , so the least possible perimeter in this case is $30$ .
In both cases, it should be briefly explained why those are the minimal perimeter cases though it's pretty obvious. Finally, the answer is $\boxed{24}$ achieved when $(a,b,c)=(9,7,8)$ .</details>
<details><summary>Note</summary>Although it may be intuitive to think that $\triangle ABC$ is acute, this is not necessarily the case. We can have $AC<BC$ , $\angle C>90^{\circ}$ and $(AMO)$ bisecting $CM$ . This, however, isn't a problem for the solution above.</details>
|
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"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810687.json"
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|
Find all primes $p$ , such that there exist positive integers $x$ , $y$ which satisfy $$ \begin{cases}
p + 49 = 2x^2
p^2 + 49 = 2y^2
\end{cases} $$
|
Wow, this can be solved exactly like USAMO 2022/4, held two days ago. So, here is a <details><summary>sketch</summary>Note that $p>y$ , otherwise $y \leq 7$ and case check. Now, subtracting the two equations, we obtain $p(p-1)=2(y-x)(y+x)$ . Obviously $p=2$ doesn't work, and $p$ can't divide $y-x$ . So $p|y+x$ and if $x+y$ is not equal to $p$ , then $x+y \geq 2p > p+y$ , so $x>p$ and $p+49>2p^2$ , case check. So $x+y=p, y-x=\frac {p-1}{2}$ , calculate $x, y$ in terms of $p$ and plug in the system, done.</details>
|
[
"Way older than USAMO 2022. Very similar idea in [Bundeswettbewerb Mathematik 1997.\n](https://artofproblemsolving.com/community/c1058102_1997_bundeswettbewerb_mathematik)",
"Even removing the requirement that $p$ is prime, this is an elliptic curve and it is possible to determine all integral points on it.\n\nMore precisely, we can rewrite it as $(2x^2 - 49)^2 = 2y^2 - 49$ . Multiplying by $2x^2$ gives us $$ Y^2 = X^3 - 98 X^2 + 2450 X $$ where $X = 2x^2$ and $Y = 2xy$ .\n\nA computer algebra system can be used to compute the Mordell-Weil rank, which is equal to $2$ , and to determine all the integral points (up to sign and ignoring the infinity point): $$ (X, Y) = (0, 0), (25, 125), (32, 104), (49, 49), (50, 50), (72, 204), (98, 490), (9800, 965300). $$ Translating back into $(x, y, p)$ , we find the following solutions: $$ (x, y, p) = (0, 35, -49), (4, 13, -17), (5, 5, 1), (6, 17, 23), (7, 35, 49), (70, 6895, 9751). $$ Among these, only $(6, 17, 23)$ gives prime value of $p$ ."
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|
14 students attend the IMO training camp. Every student has at least $k$ favourite numbers. The organisers want to give each student a shirt with one of the student's favourite numbers on the back. Determine the least $k$ , such that this is always possible if: $a)$ The students can be arranged in a circle such that every two students sitting next to one another have different numbers. $b)$ $7$ of the students are boys, the rest are girls, and there isn't a boy and a girl with the same number.
|
This nice problem was proposed by Miroslav Marinov. Part a) is just for warm up. Obviously, $k=1$ is not enough and it's possible for $k=2$ . The worst case scenario is when all the students have the same favorite numbers, say $1$ and $2$ . In this case we arrange the numbers as $1-2-1-\dots 2$ . **Part b).** $k=4$ . I follow in general the author's solution. We omit the counterexample that it is not possible for $k=3$ and let's prove that it's always possible when $k=4$ . Consider a bipartite graph $G(S, N)$ where $S=\{1,2,\dots,14\}$ are the students and $N=\{n_1,n_2,\dots,n_k\}$ are all the numbers the students like. A vertex (student) $s\in S$ is connected with $n_i\in N$ if $n_i$ is a favorite number of $s$ . The given condition says that every $s$ is connected with at least $4$ vertices in $N$ , i.e. $d(s)\ge 4, \forall s\in S$ . Let us partition the vertices $S$ into two groups $S_0$ (boys) and $S_1$ (girls). The problem actually asks to prove the following claim. $\textbf{Claim.}$ The vertices in $N$ can be partitioned into two disjoint sets $N_0$ and $N_1$ , such that each vertex in $S_0$ is connected with a vertex in $N_0$ and each vertex in $S_1$ is connected with a vertex in $N_1$ . $\textit{Proof.}$ Each partition of $N$ into two sets can be interpret as assigning to each $n_i\in N$ either a value $0$ if $n_i\in N_0$ or $1$ in case $n_i\in N_1$ . Clearly the family $\mathcal{A}$ of all assignments (partitions) consists of $2^k$ elements.
For each $s\in S$ let us denote by $B_s\subset \mathcal{A}$ the set of "bad" assignments for the student $s$ . That is, in case $s\in S_0$ , $B_s$ consists of all assignments for which to all neighbors of $s$ is assigned $1$ , and in case $s\in S_1,$ $B_s$ are those assignments for which $s$ is connected only with $0$ 's. It means $$ |B_s|= \frac{2^k}{2^d} $$ where $d=d(s)$ . Since $d(s)\ge 4$ we get $|B_s|\le 2^{k-4}$ . Let $B:=\bigcup_{s\in S}B_s$ be all bad assignments. It yields $$ |B|=\left|\bigcup_{s\in S}B_s\right|\le \sum_{s\in S}|B_s|\le 14\cdot 2^{k-4}<2^k. $$ Therefore $|B|<|\mathcal{A}|$ which means there exists an assignment $A\in \mathcal{A}\setminus B$ . Obviously, $A$ comply with all the requirements of the Claim. $\textbf{Comment}.$ Actually, it's a disguised probabilistic approach (or vise versa :) )Note that in the above proof it's not essential the students are $7$ boys and $7$ girls. The size of $S_0$ and $S_1$ can be whatever we want, providing $|S_0\cup S_1|=14$ . Moreover, the same holds even if the number of students is $16$ . But, in this case a further argument is needed. Namely, for any $i,j\in S$ that are of the same sex, that is either $i,j\in S_0$ or $i,j\in S_1$ , it holds $B_i\cap B_j\neq \emptyset$ .
|
[
"Seems closely related to All-Russian 2005 9.8 & 11.8.\n@below Ok, actually you're right.",
"@above I don't think so. It is correct that a) requires a greedy algorithm (though much simpler than the ARO problem) and for b) I doubt that anything of this sort would work."
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|
If $x, y, z \in \mathbb{R}$ are solutions to the system of equations $$ \begin{cases}
x - y + z - 1 = 0
xy + 2z^2 - 6z + 1 = 0
\end{cases} $$ what is the greatest value of $(x - 1)^2 + (y + 1)^2$ ?
|
$y-x=z-1$ $xy=-2z^2+6z-1$ $(y-x)^2+4xy\geq 0\Rightarrow \frac{1}{7}\leq z\leq 3$ $(x-1)^2+(y+1)^2=(y-x)^2+2xy+2(y-x)+2=z^2-2z+1-4z^2+12z-2+2z-2+2=-3z^2+12z-1=-3(z-2)^2+11$ its maximum value is $11$ when $z=2$ .
|
[
"bump........anyone? anyidea?"
] |
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"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810692.json"
}
|
Let $\triangle ABC$ have incenter $I$ . The line $CI$ intersects the circumcircle of $\triangle ABC$ for the second time at $L$ , and $CI=2IL$ . Points $M$ and $N$ lie on the segment $AB$ , such that $\angle AIM =\angle BIN = 90^{\circ}$ . Prove that $AB=2MN$ .
|
Interesting Problem.
Let $IH$ be perpendicular to $AB$ . First Lets have some angle chasing stuff. $\frac{\angle A}{2} = \angle HAI = \angle HIM,\frac{\angle B}{2} = \angle HBI = \angle HIN \implies \angle AIN = \frac{\angle C}{2} = \angle BIM$ .
Claim : $ANI$ and $AIC$ are similar.
Proof : $\angle NAI = \angle IAC$ and $AIN = \angle ACI$ .
Let $I'$ be reflection of $I$ across $N$ and $I_c$ be reflection of $C$ across $I$ . Note that $CI=2IL$ so $IL = LI_c$ . Now we have $L$ is center of $AIBI_c$ .
Claim : $I'$ lies on $AIBI_c$ .
Proof : we had $ANI$ and $AIC$ are similar and $I'N = NI , I_cI = IC$ so $AI'I$ and $AI_cC$ are similar so $\angle AI'I = \angle AI_cI$ so $I'$ lies on $AIBI_c$ .
Claim : $AN = NH$ .
Proof : we had $\angle HIN = \angle HBI$ so $NH.NB = NI^2 = NI.NI' = NA.NB \implies NH = NA$ .
with same approach we have $MH = MB$ so $MN = \frac{AB}{2}$ .
we're Done.
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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"answer_score": 64,
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"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810693.json"
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|
A permutation $\sigma$ of the numbers $1,2,\ldots , 10$ is called $\textit{bad}$ if there exist integers $i, j, k$ which satisfy
\[1 \leq i < j < k \leq 10 \quad \text{ and }\quad \sigma(j) < \sigma(k) < \sigma(i)\]
and $\textit{good}$ otherwise. Find the number of $\textit{good}$ permutations.
|
These permutations are called $312$ avoiding permutations or stack-realizable permutations, because they can be sorted by the following recursively defined procedure $S$ . Let $w$ be the permutation(string) we want to sort in increasing order as $123\dots n$ . Let $w=u1v$ where $u,v$ are strings. Then $S(w)=1S(u)S(v)$ . The number of those permutations with $n$ elements is the Catalan number $\frac{1}{n+1}\binom{2n}{n}$ . Indeed, if $w=u1v$ is a permutation like that then any number in $u$ is less than any number in $v$ (because $w$ is $312$ avoiding permutation). It means that all those permutations can be constructed recursively by putting $1$ into some position, say $j$ -th and then constructing all $312$ -free permutations with the numbers $2,3,\dots,j$ on the left side and with the numbers $j+1,j+2,n$ on the right side. It means, denoting by $g(n)$ the number of all $312$ free permutations of $1,2,\dots,n$ , it holds $$ g(n)=\sum_{j=0}^{n-1} g(j)g(n-1-j) $$ and, as it is well known, this recurrence formula characterizes the Catalan numbers.
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810694.json"
}
|
Find the smallest odd prime $p$ , such that there exist coprime positive integers $k$ and $\ell$ which satisfy
\[4k-3\ell=12\quad \text{ and }\quad \ell^2+\ell k +k^2\equiv 3\text{ }(\text{mod }p)\]
|
No need to use overcomplicated ideas. (Hopefully I have not made a mistake in these rushed calculations.) Clearly $k=3x$ and then $\ell = 4x-4$ , so the congruence becomes $37x^2 - 44x + 13 \equiv 0 \pmod p$ . For $p=3$ we get $(x-1)^2 \equiv 0 \pmod 3$ and $k$ and $\ell$ are both divisible by $3$ . For $p=5$ we get $2x^2 - 4x + 3 \equiv 0 \pmod 5$ , impossible by direct check. For $p=7$ we get $2x^2 - 2x + 6 \equiv 0 \pmod 7$ , again impossible by direct check. For $p=11$ we get $4x^2 + 2 \equiv 0 \pmod {11}$ , with $x\equiv 4 \pmod {11}$ as a solution. To ensure that $k$ and $\ell$ are coprime just pick $x=59$ (hmm even $x=15$ works actually).
|
[
"//wrong//",
"<details><summary>Soln sketch (alternative)</summary>Proceed as below to get $37x^2-44x+13 \\equiv 0 (mod p)$ and multiply it by $37$ and transform it into the form $m^2 \\equiv 3 (mod p)$ then use quadratic residues and Quadratic Reciprocity. Answer is $p=11$ ; construction can be done with CRT as well, if you're lazy to find concrete example.</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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{
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810695.json"
}
|
Solve the equation
\[(x+1)\log^2_{3}x+4x\log_{3}x-16=0\]
|
[
" $ x= \\frac {1} {81} $ "
] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810696.json"
}
|
|
A circle through the vertices $A$ and $B$ of $\triangle ABC$ intersects segments $AC$ and $BC$ at points $P$ and $Q$ respectively. If $AQ=AC$ , $\angle BAQ=\angle CBP$ and $AB=(\sqrt{3}+1)PQ$ , find the measures of the angles of $\triangle ABC$ .
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810697.json"
}
|
|
In every cell of a table with $n$ rows and $m$ columns is written one of the letters $a$ , $b$ , $c$ . Every two rows of the table have the same letter in at most $k\geq 0$ positions and every two columns coincide at most $k$ positions. Find $m$ , $n$ , $k$ if
\[\frac{2mn+6k}{3(m+n)}\geq k+1\]
|
<details><summary>Solution sketch</summary>The idea is to double count the triplets (column, column, row), where the two columns coincide in the position which represents the row. If the number of triplets is $T$ , then the idea is to prove $\frac{m(m-1)}{2}k \geq T \geq \frac {\frac{m^2}{3}-m}{2}n=\frac{(m^2-3m)n}{6}$ where the first one is due to the problem statement, the second one due to the following: if there are $n_i$ cells of each color in a fixed row, then the number of pairs of columns which coincide in position of this row is $\sum \frac{n_i(n_i-1)}{2}$ and you can find lower bound of this by CS. The rest of the problem is to see equality holds.</details>
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
] |
{
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810699.json"
}
|
Let $n \geq 2$ be a positive integer. The set $M$ consists of $2n^2-3n+2$ positive rational numbers. Prove that there exists a subset $A$ of $M$ with $n$ elements with the following property: $\forall$ $2 \leq k \leq n$ the sum of any $k$ (not necessarily distinct) numbers from $A$ is not in $A$ .
|
I hope this works because it seems extremely beautiful. <details><summary>Solution</summary>If a set $S$ satisfies the property which we want for $A$ (but is not necessarily a subset of $M$ ), we will call it $\textit{sum-free}$ . First, notice that if $B$ is $\textit{sum-free}$ , then $B'=\{cx|x\in S\}$ is also $\textit{sum-free}$ for any $c>0$ . Thus if we multiply all numbers in $M$ by the lcm of the denominators we will get a new set $M^{*}$ consisting only of positive integers (since we multiply a positive rational number by a $c>0$ to get an integer, this means that the new set $M$ consists of positive integers).
Let $p$ be a prime of the form $(2n-1)q-n+1$ . Since $\gcd(n-1,2n-1)=1$ , by Dirichlet's prime number theorem we know that infinitely many such prime numbers exist, so we can choose a big enough $q$ (for example, $q>\max_{x\in M^{*}} x$ will suffice). Now, notice that the set:
\[S=\{q,q+1, q+2\ldots ,2q-1\}\] is $\textit{sum-free}$ . Indeed, if we look at the extremal cases, we notice that if $\sum$ denotes the sum of $k$ integers from the set $S$ , then:
\[2q-1<q+q\leq \sum\leq n(2q-1)<p+q\]
Now the key observation is that the sets $S_{t}=\{tx (\text{mod } p)|x\in S, 1\leq t\leq p-1\}$ are sum-free since we noted above that if $B$ is $\textit{sum-free}$ , then also is $B'=\{cx|x\in B\}$ and the result follows from the fact that $S$ is $\textit{sum-free}$ . Notice that $S\equiv S_{1}$ and now let's construct a $q\times (p-1)$ table (with $q$ columns and $p-1$ rows). In the $i$ -th row we write the numbers from $S_{i}$ , so that every cell contains a single number between $1$ and $q-1$ (we picked $q>\max_{x\in M^{*}} x$ earlier). Since $p$ is prime, that means that each column contains every residue modulo $p$ . That is, if we have the numbers $xt (\text{mod }p)$ and $xs(\text{mod }p)$ in rows $t$ and $s$ respectively, then $q\leq x\leq 2q-1$ , so $p\nmid x$ and $|t-s|\leq |(p-1)-1|<p$ , so $xs-xt\not\equiv 0(\text{mod }p)$ , thus every residue from $1$ to $p-1$ appears in every column. Thus, if for every number in $M^{*}$ we write a $1$ in the cells of the table which are congruent to it modulo $p$ (here since $p$ is huge, the numbers from $M^{*}$ are themselves residues mod $p$ , but nevermind), we would have $|M^{*}|\cdot (q-1)$ ones since every number appears in every column once. However $|M^{*}|\cdot q=(2n^2-3n+2)q$ . On the other hand, notice that the sets $S_{1},S_{2},\ldots ,S_{n-1}$ don't contain numbers from $M^{*}$ because $\min_{x\in S_{t}} x=tx (\text{mod }p)>q>\max_{x\in M^{*}} x$ and $\max_{x\in S_{t}} x=(2q-1)t \leq (2q-1)(n-1)<p<p+\min_{x\in M^{*}} x$ . Thus, by the PHP we have a row with at least
\[\Bigg\lceil \frac{(2n^2-3n+2)q}{(p-1)-(n-1)}\Bigg\rceil=\Bigg\lceil \frac{(2n^2-3n+2)q}{(2n-1)(q-1)}\Bigg\rceil\quad 1\text{'s}\]
Note that the numbers in this set form a $\textit{sum-free}$ subset $A$ of $M^{*}$ since they are a subset of some $\textit{sum-free}$ $S_{i}$ . Now we only have to prove that:
\[\frac{(2n^2-3n+2)q}{(2n-1)(q-1)}>n-1\]
\[\iff 2n^2q-3nq+2q>2n^2q-2n^2-nq+n-2nq+2n+q-1\]
\[\iff q>-2n^2+3n-1\]
which is obviously true even for small $q$ , so we're done since
\[\frac{(2n^2-3n+2)q}{(2n-1)(q-1)}>n-1\Longrightarrow \Bigg\lceil \frac{(2n^2-3n+2)q}{(2n-1)(q-1)}\Bigg\rceil\geq n\]</details>
|
[
"It is indeed a very nice problem! The best problem of this year's tournament, as I see the problems you've posted. Could you write its author, please? And is this the official solution?",
"<blockquote>It is indeed a very nice problem! The best problem of this year's tournament, as I see the problems you've posted. Could you write its author, please? And is this the official solution?</blockquote>\n\nI completely agree, this is certainly a beautiful problem. I don't have the solutions or the names of the authors of each problem. I only have the problem statements and #2 is my solution (which I hope is correct) and not an official one. I may post the official solutions to the problems when they come out.",
"The author of this problem is Aleksandar Ivanov. The main idea is to make the given numbers integers and work modulo $ p$ where $ p$ is a large (prime) number. If we manage to choose a large \"sum-free\" subset $ M'$ modulo $ p$ the same set $ M'$ will be sum-free. The plan follows as a whole the author's idea of the official solution.\n\t $\\textbf{First observation}$ : If $ M'$ is a \"sum-free\" set of integers modulo $ p$ the same holds for $ s\\cdot M':=\\{s\\cdot m : m\\in M\\}$ , where $ s$ is a natural number, and vise versa. \n\t $\\textbf{Second observation}$ : The set $ I:= \\{i : i\\in [0..p-1], \\frac{p}{2n}\\le i\\le \\frac{p}{n}\\}$ is sum-free modulo $ p$ . Indeed, summing up two or more (but at most $ n$ ) elements of $ I$ will result in a sum that is in the interval $ \\left(\\frac{p}{n},p-1\\right]$ .\n\t $\\textbf{The key idea is}$ to find an integer $ s\\in [1.. p-1]$ such that $ M_s:=s\\cdot M$ has large number of elements in $ I$ . \n\t\n\t**Solution**. We can assume all numbers in $ M$ are positive integers, otherwise they can be made such by multiplication by an appropriate integer. Let us take a prime number $ p$ , greater than all the elements in $ M$ . Further we consider $ M$ as a set of residues modulo $ p$ . Denote by $ I$ the interval $ \\displaystyle \\left[\\frac{p}{2n-1}, \\frac{2p}{2n-1}\\right].$ Note that the set $ M'$ of all integers inside $ I$ have the property that any sum of no more than $ n$ of them is not an element of $ M'$ modulo $ p$ . For each $ a\\in M$ let $ 1_{a, I}$ be the random variable that takes value $ 1$ if for a randomly taken integer $ s\\in[1,p-1]$ we have $ s\\cdot a\\in I \\pmod{p}$ , and value $ 0$ otherwise. It holds\n\t $$ \\displaystyle \\mathbb{E}[1_{a,I}]> \\frac{ \\frac{2p}{2n-1}- \\frac{p}{2n-1}-1 }{p-1} $$ \twhich yields\n\t $$ \\displaystyle \\mathbb{E}[1_{a,I}]>\\frac{1}{2n-1}-\\frac{1}{p} $$ \tLet $ X$ be a random variable which equals $ \\#\\{a\\in M : s\\cdot a\\in I \\pmod{p}\\}$ where $ s$ is a randomly taken integer in $ [1,p-1]$ . Clearly, $X=\\sum_{a\\in M} 1_{a,I}.$ Using the linearity of expectation, we get\n\t $$ \\displaystyle \\mathbb{E}[X]=\\sum_{a\\in M}\\mathbb{E}[1_{a,I}] > m\\cdot \\left(\\frac{1}{2n-1}-\\frac{1}{p}\\right) $$ \tSuppose we have already checked that\n\t $$ \\displaystyle m\\cdot \\left(\\frac{1}{2n-1}-\\frac{1}{p}\\right)> n-1 \\qquad(1) $$ \tIt means $ \\displaystyle \\mathbb{E}[X]>n-1,$ therefore $X$ takes $ n$ as a value. Thus, there exists $ s\\in [1..p-1]$ such that the set $ M'_s:=\\{a\\in M : s\\cdot a\\in I\\}$ satisfies $ |M'_s|\\ge n$ . Hence, the elements of $ M'_s$ satisfy the requirement of the problem. \n\tIt remains to check that $ (1)$ holds. Since the prime $ p$ can be chosen as large as we want, it is enough to check\n\t $$ \\displaystyle m\\cdot \\left(\\frac{1}{2n-1}\\right)> n-1 $$ \twhich is a matter of simple calculation.\n\t\n\t**Comment**. The restriction of numbers being positive is redundant. The same proof can be used, just omitting the word \"positive\".\n\tThe official solution is like in $\\#$ 2 (without the glitches and typos). Btw, no one managed to solve it at the competition - all the contestants scored only 1 point. The claim also holds for any real numbers. More details and comments can be found [in my blog.](https://dgrozev.wordpress.com/2022/04/01/bulgarian-2022-spring-math-competition-part-1/)",
"Here is the proof of the general case, when all the numbers are real. The plan is to multiply all the numbers by some real number to make them all irrationals and then work modulo $ 1$ . That is, we consider $ a\\equiv b\\pmod 1$ if $ a-b$ is integer. That said, we could think of $ M$ as a set of irrational numbers in $ [0,1)$ and points $ 0$ and $ 1$ being \"glued\". Let $ I$ be the interval $ \\displaystyle \\left(\\frac{1}{2n-1}, \\frac{2}{2n-1}\\right).$ Note that it is impossible any sum of $ k$ terms in $ I$ ( $ 2\\le k\\le n$ ) to be also in $ I$ modulo $ 1$ . Thus, if we multiply all elements of $ M$ by some natural $ s\\in\\mathbb{N}$ modulo $ 1$ and if it happens (by chance) that at least $ n$ of the numbers $ s\\cdot M:=\\{sx : x\\in M \\}$ are in $ I$ , we are done, because the corresponding terms of $ M$ would be \"special sum-free\" modulo $ 1$ .\n\tSo, there exists $ a\\in \\mathbb{R}$ such that all the numbers $ ax, x\\in M$ are irrational. This is intuitive clear, one can see the proof [in my blog](https://dgrozev.wordpress.com/2022/04/03/bulgarian-2022-spring-math-competition-part-2/).\n\t\n\tFurther, we can assume $ M$ consists of irrational numbers. Let now $ N$ is sufficiently large positive integer, which will be determined later. Fix $ x\\in M$ and look at the numbers $ \\{kx\\pmod 1 : k=1,2,\\dots,N\\}$ . It is known that the infinite sequence $ \\big( ka\\pmod{1}\\big)_{k=1}^{\\infty}$ is uniformly distributed in $ [0,1]$ ([Equidistribution theorem](https://en.wikipedia.org/wiki/Equidistribution_theorem)). Hence, \n\t $$ \\displaystyle \\frac{\\#\\{k : kx\\in I\\pmod 1, 1\\le k\\le N\\}}{N}=|I| +o(1)\\qquad (1) $$ \twhere $ o(1)$ denotes a value that tends to $ 0$ when $ N\\to\\infty$ . As in the original problem, let $ 1_{x,I}$ be a random variable defined as follows. We take randomly $ k\\in [1..N]$ and if $ kx\\in I\\pmod 1$ then $ 1_{x,I}$ is $ 1$ otherwise it is $ 0$ . Usually, it's called indicator variable, because it indicates when $ kx\\in I\\pmod 1$ . By $ (1)$ we get\n\t $$ \\displaystyle \\mathbb{E}[1_{x,I}]=\\frac{1}{2n-1}+o(1). $$ \tLet $ X$ be a random variable equal to the number of terms of $ k\\cdot M=\\{kx:x\\in M\\}$ that are inside $ I$ where $ k$ is a randomly taken integer in $ [1..N]$ . Clearly\n\t $$ \\displaystyle X=\\sum_{x\\in M}1_{x,I}. $$ \tBy the linearity of expectation we obtain\n\t\\begin{align*}\n\t\\displaystyle \\mathbb{E}[X]&=\\sum_{x\\in M}\\mathbb{E}[1_{x,I}]\n\t\\displaystyle \\mathbb{E}[X]&= m\\cdot \\left(\\frac{1}{2n-1}\\right)+m\\cdot o(1).\n\t\\end{align*}\n\twhere $ m=|M|=2n^2-3n+2$ . It is a matter of calculation to check \n\t $$ \\displaystyle m\\left(\\frac{1}{2n-1} \\right)>n-1 $$ \tand since $ m\\cdot o(1)\\to 0$ as $ N\\to \\infty$ , it yields\n\t $$ \\mathbb{E}[X]>n-1. $$ \tIt implies that $ X$ takes a value $ n$ or greater, providing $ N$ is sufficiently large. That is, there exists $ k\\in [1..N]$ such that $ \\#\\{x\\in M : kx\\in I \\pmod 1\\}\\ge n$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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{
"answer_score": 134,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810700.json"
}
|
$ABCD$ is circumscribed in a circle $k$ , such that $[ACB]=s$ , $[ACD]=t$ , $s<t$ . Determine the smallest value of $\frac{4s^2+t^2}{5st}$ and when this minimum is achieved.
|
Since $4s^2+t^2 \ge 4st$ , the ratio is at least $4/5$ . For an example, it suffices to consider $|AB|=|BC|=1$ , $|CD|=|DA|=\sqrt{2}$ , $\angle ADC = \angle ABC = \pi/2$ . (Check that $|BD|=2\sqrt{2}/\sqrt{3}$ by Ptolemy and everything works out).
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
] |
{
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810701.json"
}
|
Let $ABCDV$ be a regular quadrangular pyramid with $V$ as the apex. The plane $\lambda$ intersects the $VA$ , $VB$ , $VC$ and $VD$ at $M$ , $N$ , $P$ , $Q$ respectively. Find $VQ : QD$ , if $VM : MA = 2 : 1$ , $VN : NB = 1 : 1$ and $VP : PC = 1 : 2$ .
|
<details><summary>Solution</summary>Let $AB$ be of length $6$ . The $VM=4,$ $VN=3,$ $VP=2.$ Let $O$ be the intersection of $MP$ and $QN.$ Then $O$ lies on the altitude of the pyramid, which also bisects the right angles $\angle MVP$ and $\angle QVN.$ Let points $Q'$ and $N'$ lie on segments $VA$ and $VC$ , respectively, such that $VQ'=VQ$ and $VN'=VN$ . Then $Q'N'$ and $MP$ also intersect at $H.$ We have now simplified a 3D problem into a 2D problem and can be graphed like so: [asy]defaultpen(fontsize(10pt));
pair V,Q,M,A,H,P,N,C;
V=(0,0);Q=(0,12/5);M=(0,4);A=(0,6);H=(4/3,4/3);P=(2,0);N=(3,0);C=(6,0);dot(" $V(0,0)$ ",V,SW);dot(" $A(0,6)$ ",A,W);dot(" $M(4,0)$ ",M,W);dot(" $Q'$ ",Q,W);draw(A--V);
dot(" $P(2,0)$ ",P,S);dot(" $N'(3,0)$ ",N,S);dot(" $C(6,0)$ ",C,S);draw(C--V--H);draw(Q--N);draw(M--P);dot(" $O$ ",H,NE);[/asy]
Now we can do some light coordinate bashing to find that $O=\left(\tfrac43,\tfrac43\right),$ and then from there $Q'=\left(0,\tfrac{12}5\right).$ So $VQ:QD=VQ':Q'A=\frac{12}5:6-\frac{12}5=\fbox{2:3}.$</details>
I haven't done math in years; hope this solution is ok
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
] |
{
"answer_score": 62,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810702.json"
}
|
Let $P,Q\in\mathbb{R}[x]$ , such that $Q$ is a $2021$ -degree polynomial and let $a_{1}, a_{2}, \ldots , a_{2022}, b_{1}, b_{2}, \ldots , b_{2022}$ be real numbers such that $a_{1}a_{2}\ldots a_{2022}\neq 0$ . If for all real $x$ \[P(a_{1}Q(x) + b_{1}) + \ldots + P(a_{2021}Q(x) + b_{2021}) = P(a_{2022}Q(x) + b_{2022})\]
prove that $P(x)$ has a real root.
|
I think I got that all $a_i$ are equal, otherwise we can get that there exists $i$ , such that $a_i.Q(x)+b_i=a_{2022}.Q(x)+b_{2022}$ has a root since $Q$ has odd degree, and then we get that $P(x)$ can't have the same sign for all $x$ , so it has a root. The harder case seems to be when all $a_i$ are equal, and no $b_i$ is equal to $b_{2022}$ . How to proceed then?
@below Oh yes, indeed, you're right, comparing leading coefficients and taking large x seems to suffice, thanks.
|
[
"Similar question in old russian olympiad :))",
"@above when all $a_i$ are equal the left hand side will have a much larger leading coefficient and so the equality will fail for sufficiently large $x$ , I think?"
] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
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{
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810703.json"
}
|
Let $m$ and $n$ be positive integers and $p$ be a prime number. Find the greatest positive integer $s$ (as a function of $m,n$ and $p$ ) such that from a random set of $mnp$ positive integers we can choose $snp$ numbers, such that they can be partitioned into $s$ sets of $np$ numbers, such that the sum of the numbers in every group gives the same remainder when divided by $p$ .
|
ha,ha, it's correct. Knowing Erdos-Ginzburg-Ziv theorem makes it easy. It says that in any set of $2p-1$ integers (p may not be a prime) there are $p$ of them with sum $0\pmod{p}$ . In the official solution, the proposer proves EGZ thm in case $p$ is prime (because it's easier) in a separate lemma. After that, it's trivial.
|
[
"Is the answer m-1? Just spam EGZ and we can find mn-1 disjoint sets with size p and their sum is divisible by p, and our set is the union of (m-1)n such sets.",
"Please tell me if it is incorrect, since this is too easy."
] |
[
"origin:aops",
"2022 Contests",
"2022 Bulgarian Spring Math Competition"
] |
{
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Bulgarian Spring Math Competition/2810704.json"
}
|
I think we are allowed to discuss since its after 24 hours
How do you do this
Prove that $d(1)+d(3)+..+d(2n-1)\leq d(2)+d(4)+...d(2n)$ which $d(x)$ is the divisor function
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Notice that the inequality holds for $n=1,2,3$ , for now on assume $n>3$ . We will prove the following equivalent statement \[ \sum_{k=1}^n d(2k) > \sum_{k=2}^n d(2k-1) \] For any positive integer $\ell$ let $f_\ell(x)$ denote the number of elements in the set $\{2,4,6 \ldots, 2x \}$ with exactly $\ell$ divisors and $g_\ell(x)$ denote the number of elements in $\{ 3,5,7, \ldots, 2x-1 \}$ with exactly $\ell$ divisors.**Claim:** For all $\ell > 2$ we have $f_\ell(x) \geq g_\ell(x)$ Let $a_m$ is the $m$ 'th smallest odd number with $\ell$ divisors. We can change one of the primes dividing $a_m$ into a $2$ creating a new even number $b_\ell$ with $\ell$ divisors. So, we have a new sequence of evens $b_1<b_2< \cdots$ all with $\ell$ divisors and obeys $b_i < a_i$ for all $i$ . Thus, the $m$ 'th smallest even number with $\ell$ divisors is strictly less that the $m$ 'th smallest odd with $\ell$ divisors. This implies the claim. $\square$ Notice that \[ \sum_{k=1}^n d(2k) > \sum_{k=2}^n d(2k-1) \Longleftrightarrow \sum_{k=2}^n kf_k(n) > \sum_{k=1}^n kg_k(n)\] We see $g_2(n) > f_2(n)$ since this just counts primes. (Handwaving) The LHS is putting more weight on bigger numbers, so it is bigger. The RHS is only putting more weight on $2$ than the LHS, which is the smallest number.
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[
"<details><summary>Solution</summary>Count number of pairs $(x,y)$ such that $x|y$ . Each odd $x$ gives at most 1 to LHS-RHS but x=2 gives n to RHS-LHS, GG</details>",
"not sure if discussion is allowed yet, but if it is,\n\n<details><summary>idea</summary>double count each side by how many times each number from $1 \\to 2n$ divides one of the numbers on each side</details>",
"<blockquote>not sure if discussion is allowed yet, but if it is,\n\n<details><summary>idea</summary>double count each side by how many times each number from $1 \\to 2n$ divides one of the numbers on each side</details></blockquote>\n\nI guess you can cuz it is after 24 hours\n\neveryone finished around 3pm est yesterday",
"<blockquote><details><summary>Solution</summary>Count number of pairs $(x,y)$ such that $x|y$ . Each odd $x$ gives at most 1 to LHS-RHS but x=2 gives n to RHS-LHS, GG</details></blockquote>\n\nThis is exactly what I did",
"I spent way too long trying to find a pairing argument.\n\n<details><summary>Solution</summary>Denote the LHS of the inequality with $A$ and the RHS with $B$ . Then $A$ counts the number of ordered pairs $(x,y)$ of positive integers with $x\\le y\\le 2n$ such that $x | y$ and $y$ is odd and $B$ counts the number of pairs such that $x | y$ and $y$ is even. If we count these quantities by fixing $x$ , we can see that each odd $x$ contributes at least $-1$ to $B-A$ and each even $x$ contributes exactly $\\left\\lfloor \\frac{2n}{x}\\right\\rfloor$ to $b-a$ . Hence, we obtain $B - A\\ge -n + \\sum_{i=1}^n \\left\\lfloor \\frac{2n}{2i}\\right\\rfloor > -n + \\sum_{i=1}^n \\left(\\frac{n}{i} - 1\\right) = n(H_n-2)$ which is much stronger than the desired inequality. It is easy to manually verify the $n = 1$ , $n = 2$ , and $n = 3$ cases where $H_n < 2$ .</details>\n\n<details><summary>Remark</summary>This solution can be modified slightly to prove that $n(H_n-2)\\le B-A\\le nH_n$ .</details>",
"<blockquote>I spent way too long trying to find a pairing argument.\n\n<details><summary>Solution</summary>Denote the LHS of the inequality with $A$ and the RHS with $B$ . Then $A$ counts the number of ordered pairs $(x,y)$ of positive integers with $x\\le y\\le 2n$ such that $x | y$ and $y$ is odd and $B$ counts the number of pairs such that $x | y$ and $y$ is even. If we count these quantities by fixing $x$ , we can see that each odd $x$ contributes at least $-1$ to $B-A$ and each even $x$ contributes exactly $\\left\\lfloor \\frac{2n}{x}\\right\\rfloor$ to $b-a$ . Hence, we obtain $B - A\\ge -n + \\sum_{i=1}^n \\left\\lfloor \\frac{2n}{2i}\\right\\rfloor > -n + \\sum_{i=1}^n \\left(\\frac{n}{i} - 1\\right) = n(H_n-2)$ which is much stronger than the desired inequality. It is easy to manually verify the $n = 1$ , $n = 2$ , and $n = 3$ cases where $H_n < 2$ .</details>\n\n<details><summary>Remark</summary>This solution can be modified slightly to prove that $n(H_n-2)\\le B-A\\le nH_n$ .</details></blockquote>\n\nA pairings argument should exist and work, where we match each term in the left hand side with each term in the right hand side such that <= is preserved. Although this is messy.",
"Let $d(k)$ denote the number of positive integer divisors of $k$ . For example, $d(6) = 4$ since $6$ has $4$ positive divisors, namely, $1, 2, 3$ , and $6$ . Prove that for all positive integers $n$ , $$ d(1) + d(3) + d(5) +...+ d(2n - 1)\\le d(2) + d(4) + d(6) + ... + d(2n). $$ ",
"Here you can find the official solutions https://cms.math.ca/competitions/cmo/",
"See that $$ \\sum_{1\\leq k\\leq 2n}(-1)^kd(k)=\\sum_{1\\leq 2j\\leq n}\\left\\lfloor\\frac{2n}{2j}\\right\\rfloor-\\sum_{1\\leq 2l+1\\leq 2n}\\left(\\left\\lfloor\\frac{2n}{2l+1}\\right\\rfloor\\text{ }(mod.2)\\right), $$ and since $\\sum_{1\\leq 2l+1\\leq 2n}\\left(\\left\\lfloor\\frac{2n}{2l+1}\\right\\rfloor\\text{ }(mod.2)\\right)\\leq n-1$ and $\\sum_{1\\leq 2j\\leq n}\\left\\lfloor\\frac{2n}{2j}\\right\\rfloor\\geq n$ , we get the desired result. $\\blacksquare$ ",
"Here is my solution: [https://calimath.org/pdf/CMO2022-2.pdf](https://calimath.org/pdf/CMO2022-2.pdf)\nAnd I uploaded the solution with motivation to: [https://youtu.be/W5S-mL5JT7I](https://youtu.be/W5S-mL5JT7I)",
"In fact, a much stronger inequality holds: $$ d(1) + d(3) + d(5) +...+ d(2n - 1)\\le d(1) + d(2) + d(3) + ... + d(n) $$ It is easy to see that, for $1\\le k\\le n$ , the number of multiples of $2k-1$ in $1,3,5,...,2n-1$ is $\\lfloor\\frac{n+k-1}{2k-1}\\rfloor$ , and the number of multiples of $k$ in $1,2,...,n$ is $\\lfloor\\frac{n}{k}\\rfloor$ .\nBy double counting, we have $d(1) + d(3) + d(5) +...+ d(2n - 1)=\\sum_{k=1}^n \\lfloor\\frac{n+k-1}{2k-1}\\rfloor$ and $d(1) + d(2) + d(3) + ... + d(n)=\\sum_{k=1}^n \\lfloor\\frac{n}{k}\\rfloor$ .\nBecause $\\frac{n}{k}-\\frac{n+k-1}{2k-1}=\\frac{(n-k)(k-1)}{k(2k-1)}\\ge 0$ , we have $\\lfloor\\frac{n+k-1}{2k-1}\\rfloor\\le\\lfloor\\frac{n}{k}\\rfloor$ for $1\\le k\\le n$ . Therefore $\\sum_{k=1}^n \\lfloor\\frac{n+k-1}{2k-1}\\rfloor\\le\\sum_{k=1}^n \\lfloor\\frac{n}{k}\\rfloor$ , so $d(1) + d(3) + d(5) +...+ d(2n - 1)\\le d(1) + d(2) + d(3) + ... + d(n)$ .\nIn my opinion, this form is more interesting than the original statement, because not only is it much stronger, but there are several equality cases: equality holds for $n=1,2,3,5$ .",
"Here is a solution with a pairing argument.\n\nWe will prove a much stronger inequality $$ d(1) + d(3) + d(5) + \\ldots + d(2n - 1)\\le d(1) + d(2) + d(3) + \\ldots + d(n). $$ Let $A\\subset\\mathbb N^2$ be the subset of pairs of positive integers $(x,y)$ such that $x,y$ are odd and $xy\\le 2n-1$ . Let $B\\subset\\mathbb N^2$ be the subset of pairs of positive integers $(x,y)$ such that $xy\\le n$ . We have $LHS=|A|$ and $RHS=|B|$ , so it is equivalent to prove that $|A|\\le |B|$ .\n\nDefine a function $f\\colon A\\to\\mathbb N^2$ by $f(x,y)=\\left(\\frac{x+1}{2},\\frac{y+1}{2}\\right)$ . The function $f$ is well-define since $x,y$ are both odd for every $(x,y)\\in A$ . Moreover, the function $f$ is injective. Hence, it suffices to prove that $\\text{Im}(f)\\subset B$ .\n\nLet us fix some $(p,q)\\in A$ . Since $p,q$ are both positive integers, it follows that:\n\\begin{align*}\n& (p-1)(q-1)\\ge 0\n&\\implies (p+1)(q+1)\\le 2(pq+1)\n&\\implies\\frac{p+1}{2}\\cdot\\frac{q+1}{2}\\le\\frac{pq+1}{2}\\le\\frac{2n-1+1}{2}=n\n&\\implies f(p,q)\\in B.\n\\end{align*}\n\nThis proves that $\\text{Im}(f)\\subset B$ .",
"Consider all the odd numbers up to $2n-1$ . Consider some odd number $o$ that the left side counts (with multiplicity). We claim that the number of times $o$ is counted by the LHS is at most $1$ greater than the number of times it is counted by the LHS. This is clear because all multiples of $o$ switch between even and odd, so done. Clearly no even numbers are counted by the LHS, but at least $n$ even numbers are counted by the RHS, so done.",
"Like a outline, almost a solution, but without numbers\n\n**Attachments:**\n\n[Canada 2022 NMO.pdf](https://cdn.artofproblemsolving.com/attachments/7/3/c54b31fe87dac00ad05e47ef35119f50f1542a.pdf)"
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"answer_score": 158,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799792.json"
}
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If $ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0$ , find the value of $b\sqrt{a^2+b}+a\sqrt{b^2+a}$
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The first equation becomes $(ab+\sqrt{a^2+b}\sqrt{a+b^2})^2=(-\sqrt{ab+1})^2$ . This rearranges to $a^3+2a^2b^2+b^3+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$ . Now let $x=b\sqrt{a^2+b}+a\sqrt{b^2+a}$ ; then we get $x^2=a^3+2a^2b^3+b^2+2ab\sqrt{a^2+b}\sqrt{a+b^2}=1$ , so $x=\pm 1$ . Now suppose $x=-1$ . Note that $ab\leq 0$ , since $\sqrt{a^2+b}\sqrt{a+b^2}+\sqrt{ab+1}\geq 0$ . Then one of the variables is $\leq 0$ , one is $\geq 0$ . WLOG let $a\leq 0, b\geq 0$ . Then by $x=-1$ we get $(a\sqrt{b^2+a})^2=(-1-b\sqrt{a^2+b})^2$ , which rearranges to $a^3-b^3=1+2b\sqrt{a^2+b}$ . But since $a^3-b^3\leq 0$ and $\geq 0$ , this is impossible, so $\boxed{x=1}$ .
Side note, is a construction needed for this problem? The phrasing made it seem like you didn't, and I couldn't actually find one in contest (I think the exact phrasing is something like "Suppose real numbers $a, b$ satisfy...find, with prove, the value of...")
|
[
"Nice I got the answer $1$ as well\nBut I kinda guessed.\n\nI cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds\n\nAssume that $b>0,a<0$ and then I got $b^3=\\frac{\\sqrt{5}-1}{2}$ , the desired value is $\\sqrt{b^6+b^3}=1$ ",
"I somehow provided a wrong construction for both $x=1$ and also somehow thought $x=-1$ worked because I found a wrong construction.",
"<blockquote>Nice I got the answer $1$ as well\nBut I kinda guessed.\n\nI cannot prove rigorously but when $a^2+b=0$ or $a+b^2=0$ is the only way the equation holds\n\nAssume that $b>0,a<0$ and then I got $b^3=\\frac{\\sqrt{5}-1}{2}$ , the desired value is $\\sqrt{b^6+b^3}=1$ </blockquote>\n\nThis seems false....... and wildly unnecessary",
"Assume that real numbers $a$ and $b$ satisfy $$ ab+\\sqrt{ab+1}+\\sqrt{a^2+b}\\sqrt{a+b^2}=0. $$ Find, with proof, the value of $$ b\\sqrt{a^2+b}+a\\sqrt{b^2+a}. $$ ",
"<blockquote><blockquote>Let $a,b$ be real numbers satisfy $ab+\\sqrt{ab+1}+\\sqrt{(a^2+b)(a+b^2)}=0.$ Prove that $$ b\\sqrt{a^2+b}+a\\sqrt{b^2+a}=1. $$ </blockquote></blockquote>\n\nWhere did you find this Chinese solution? Or you wrote this solution?",
"Let $a,b\\geq 0$ and $ab+\\sqrt{3ab+1}+\\sqrt{(a+b^2)(b+a^2)}=5.$ Prove that $$ 2\\sqrt 2\\leq a\\sqrt{a+b^2}+b\\sqrt{b+a^2}\\leq 4 $$ ",
"Let $a,b\\geq 0$ and $ab+\\sqrt{3ab+1}+\\sqrt{(a^2+b)(b^2+a)}=5.$ Prove that $$ 2\\sqrt 2\\leq a\\sqrt{a+b^2}+b\\sqrt{b+a^2}\\leq 4 $$ ",
"Notice that \n\\begin{align*}\n(b\\sqrt{a^2+b}+a\\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\\sqrt{ab+1})\n&=a^3+b^3-2ab\\sqrt{ab+1}\n&=a^3+b^3-(ab+\\sqrt{ab+1})^2+a^2b^2+ab+1\n&=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\n&=1\n\\end{align*}\nNow obviously $ab<0$ , suppose $a>0>b$ , then $a^3>b^3$ which implies $$ b^2(a^2+b)<a^2(b^2+a) $$ Hence $|b\\sqrt{a^2+b}|<|a\\sqrt{b^2+a}|$ and so $$ b\\sqrt{a^2+b}+a\\sqrt{b^2+a}>0 $$ which implies that the value is $1$ .",
"<blockquote>Notice that \n\\begin{align*}\n(b\\sqrt{a^2+b}+a\\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\\sqrt{ab+1})\n&=a^3+b^3-2ab\\sqrt{ab+1}\n&=a^3+b^3-(ab+\\sqrt{ab+1})^2+a^2b^2+ab+1\n&=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\n&=1\n\\end{align*}</blockquote>\n\nA large part of the problem is actually proving whether the value is $1$ or $-1$ ",
"<blockquote><blockquote>Notice that \n\\begin{align*}\n(b\\sqrt{a^2+b}+a\\sqrt{b^2+a})^2&=b^2(a^2+b)+a^2(b^2+a)+2ab(-ab-\\sqrt{ab+1})\n&=a^3+b^3-2ab\\sqrt{ab+1}\n&=a^3+b^3-(ab+\\sqrt{ab+1})^2+a^2b^2+ab+1\n&=a^3+b^3-(a^2+b)(b^2+a)+a^2b^2+ab+1\n&=1\n\\end{align*}</blockquote>\n\n\nA large part of the problem is actually proving whether the value is $1$ or $-1$ </blockquote>\nIndeed! Thanks very much for pointing out, just edited. :D \n",
"Does this work???\n\nRewrite the given condition as $$ ab+\\sqrt{a^2+b} \\cdot \\sqrt{b^2+a} = -\\sqrt{ab+1} $$ $$ a^3 + a b + 2 a^2 b^2 + b^3 + 2ab\\sqrt{a^2 + b}\\sqrt{a + b^2}=ab+1 $$ $$ a^3+ 2 a^2 b^2 + b^3 + 2ab\\sqrt{a^2 + b}\\sqrt{a + b^2}=1 $$ $$ (a\\sqrt{b^2+a}+b\\sqrt{a^2+b})^2=1 $$ $$ a\\sqrt{b^2+a}+b\\sqrt{a^2+b}= \\pm 1 $$ We prove that it is equal to $1,$ or equivalently that it is positive. Since $a,b$ have opposite signs, assume $b$ is negative and let $b=-c$ for some positive real $c.$ $$ a\\sqrt{b^2+a}+b\\sqrt{a^2+b}=a\\sqrt{c^2+a}-c\\sqrt{a^2-c}. $$ Note that we must have $a^2 \\geq c$ or $a \\geq \\sqrt{c}.$ $$ a\\sqrt{c^2+a}-c\\sqrt{a^2-c} \\geq \\sqrt{c^3+c\\sqrt{c}}\\geq 0. $$ <details><summary>Remarks</summary>The first few steps were motivated by me attempting to use Cauchy Schwarz then stumbling upon the factorization through wishful thinking</details>",
"Any other approach to solve this problem?\n:')",
"<details><summary>cute for algebra</summary>we have $(ab+\\sqrt{ab+1})^2=a^3+a^2b^2+a+b^3\\implies 2ab\\cdot \\sqrt{ab+1}=a^3+b^3-1 \\qquad \\qquad (\\spadesuit)$ now consider $X=b\\sqrt{a^2+b}+a\\sqrt{b^2+a}\\implies X^2=a^3+b^3-2ab\\sqrt{ab+1}$ from $(\\spadesuit)$ we get $X^2=1\\implies X=1$ or $-1$ now if $X=-1$ then $b\\sqrt{a^2+b}=-a\\sqrt{b^2+a}-1$ on squaring we get $b^3=a^3+1+2a\\sqrt{b^2+a} \\qquad \\qquad (\\star)$ clearly we can't have $a$ and $b$ with same sign due to equation given in the problem also we can't have $a$ and $b$ of opposite signs from $(\\star)$ hence we have $X=1$ which upon checking works\nhence we have $\\boxed{ b\\sqrt{a^2+b}+a\\sqrt{b^2+a}=1}$</details>",
"From the first equation,\n\\[ a^2b^2 + ab + 1 + 2ab\\sqrt{ab + 1} = (a^2 + b)(b^2 + a) = a^2b^2 + b^3 + a^3 + ab,\\]\n\\[ 1 + 2ab\\sqrt{ab + 1} = a^3 +b^3.\\]\nLet $k$ be the desired quantity. Then,\n\\[ k^2 = 2a^2b^2 + a^3 + b^3 + 2ab\\sqrt{a^2 + b}\\sqrt{b^2 + a},\\]\n\\[ k^2 = 2a^2b^2 + 2ab\\sqrt{ab + 1} + 1 + 2ab(-ab - \\sqrt{ab + 1}) = 1.\\]\nHence, $k = \\pm 1$ . We claim $k = -1$ is impossible, so assume for contradiction $k = -1$ . Note that $ab \\le 0$ , so $a\\le 0\\le b$ or $b\\le 0\\le a$ ; WLOG let it be the second one. We have $b\\sqrt{a^2 + b} + a\\sqrt{b^2 + a} = -1$ , and subtracting $a\\sqrt{b^2 + a}$ and squaring gives\n\\[ b^2(a^2 + b) = 1 + a^2(b^2 + a) + 2a\\sqrt{b^2 + a}\\]\n\\[ \\implies b^3 = 1 + a^3 + 2a\\sqrt{b^2 + a}\\implies b^3 - a^3 = 1 + 2a\\sqrt{b^2 +a}.\\]\nBut this is impossible because $b^3 - a^3\\le 0$ and $1 + 2a\\sqrt{b^2 +a}\\ge 1$ . Hence, $k = +1$ .\n\n<details><summary>Comment</summary>I figured out the squaring idea relatively quickly. However, realizing that ab <= 0 is actually useful information took a long time oops (even though this was literally the first thing I noticed when I looked at the problem).</details>\n",
"<blockquote>\nWe prove that it is equal to $1,$ or equivalently that it is positive. Since $a,b$ have opposite signs, assume $b$ is negative and let $b=-c$ for some positive real $c.$ $$ a\\sqrt{b^2+a}+b\\sqrt{a^2+b}=a\\sqrt{c^2+a}-c\\sqrt{a^2-c}. $$ Note that we must have $a^2 \\geq c$ or $a \\geq \\sqrt{c}.$ $$ a\\sqrt{c^2+a}-c\\sqrt{a^2-c} \\geq \\sqrt{c^3+c\\sqrt{c}}\\geq 0. $$ </blockquote>\n\nI may be misunderstanding some part, but the first part of the last inequality isn't necessarily true, right?. Consider a=3 and c=4, where $3\\sqrt{19}-4\\sqrt{5} \\approx 4.13 < \\sqrt{72} \\approx 8.49$ . \n\n",
"I'm not sure if this is correct (especially the plus-minus thing, I'm bad at this type of algebra :( ) , but I think no one write in this approach so here is mine. (I was actually surprised that the answer is constant!)\n\n\nLet $a\\sqrt{a+b^2}+b\\sqrt{a^2+b}=X$ . Consider two equations:\\begin{align*}\n\\bullet (\\sqrt{a^2+b}+a)(\\sqrt{a+b^2}+b)&=\\sqrt{a^2+b}\\cdot \\sqrt{a+b^2}+ab+X=X-\\sqrt{ab+1} \n\\bullet (\\sqrt{a^2+b}-a)(\\sqrt{a+b^2}-b)&=\\sqrt{a^2+b}\\cdot \\sqrt{a+b^2}+ab-X=-X-\\sqrt{ab+1}\n\\end{align*}\n\nMultiply these two: \\begin{align*}\nab\n&=(\\sqrt{ab+1}+X)(\\sqrt{ab+1}-X) \n&= ab+1-X^2\n\\end{align*}\nTherefore, $X=1$ or $X=-1$ .\n\nFrom now, let's prove that $X\\ge 0$ . WLOG $b$ is negative and let $c=-b$ . Then $a, c$ are positive and\\begin{align*}\nX &\\ge 0 \n\\Longleftrightarrow a\\sqrt{a^2+c} &\\ge c\\sqrt{a^2-c} \n\\Longleftrightarrow a^3 &\\ge -c^3\n\\end{align*}\nwhich is obviously true.",
"Get that $(ab+\\sqrt{a^2+b}\\sqrt{b^2+a}) ^ 2 = a^2b^2 + 2ab\\sqrt{a^2+b}\\sqrt{b^2+a}+(a^2+b)(b^2+a) = ab+1$ so that $2a^2b^2 + 2ab\\sqrt{a^2+b}\\sqrt{b^2+a} + a^3+b^3 = 1 = (b\\sqrt{a^2+b}+a\\sqrt{b^2+a})^2$ so the result is either $1$ or $-1$ . Suppose it's $-1$ . Then either $a$ is negative, $b$ is negative, or both are negative. First, both cannot be negative because then the first equation can never be 0. WLOG $a$ is negative, and $b$ is positive (cannot be 0 for same reason). Then $ab \\le 1$ , and $1 + b\\sqrt{a^2+b} = a\\sqrt{b^2-a}$ , but $1+b\\sqrt{a^2+b} > 1+ab > ab > a\\sqrt{b^2-a}$ , contradiction.",
"<blockquote>If $ab+\\sqrt{ab+1}+\\sqrt{a^2+b}\\sqrt{a+b^2}=0$ , find the value of $b\\sqrt{a^2+b}+a\\sqrt{b^2+a}$ </blockquote>\n\nLet $x=b\\sqrt{a^2+b}+a\\sqrt{a+b^2}\\Rightarrow x^2=a^3+2a^2b^2+b^3+2ab\\sqrt{(a^2+b)(a+b^2)}$ $\\Leftrightarrow x^2= a^2b^2+2ab\\sqrt{(a^2+b)(a+b^2)}+a^3+a^2b^2+b^3+ab-ab=a^2b^2+2ab\\sqrt{(a^2+b)(a+b^2)}+(a^2+b)(a+b^2)-ab$ $\\Leftrightarrow x^2= (ab+\\sqrt{(a^2+b)(b+a^2)})^2-ab$ Because $ab+\\sqrt{ab+1}+\\sqrt{a^2+b}\\sqrt{a+b^2}=0\\Rightarrow x^2= (ab+\\sqrt{(a^2+b)(b+a^2)})^2-ab=ab+1-ab=1$ $ab+\\sqrt{ab+1}+\\sqrt{a^2+b}\\sqrt{a+b^2}=0\\Rightarrow ab\\le 0$ . WLOG, let $a< 0, b> 0$ If $x<0\\Rightarrow x=-1\\Leftrightarrow b\\sqrt{a^2+b}=-a\\sqrt{b^2+a}-1$ Squaring both sides and rearranging, we've got: $b^3-a^3=2a\\sqrt{b^2+a}+1$ . This is false because $LHS<0, RHS>0$ Therefore, $x=1$ "
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"origin:aops",
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"answer_score": 1036,
"boxed": false,
"end_of_proof": false,
"n_reply": 19,
"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799959.json"
}
|
Vishal starts with $n$ copies of the number $1$ written on the board. Every minute, he takes two numbers $a, b$ and replaces them with either $a+b$ or $\min(a^2, b^2)$ . After $n-1$ there is $1$ number on the board. Let the maximal possible value of this number be $f(n)$ . Prove $2^{n/3}<f(n)\leq 3^{n/3}$ .
|
**The lower bound:****Claim:** let $n$ be a natural number greater than 3, then: $f(n)> 2^{\frac{n}{3}+\frac{1}{2}}$ .
*Proof:* notice that $f(2x) \geq {f(x)}^2$ and $f(2x+1) \geq {f(x)}^2+1$ . Also we can check manually the <u>original</u> lower bound inequality for $\{1,2\}$ and the <u>strong</u> lower bound inequality for $\{3,4,5\}$ . Then we proceed by strong induction; write for $n \geq 3$ : $\bullet f(2n) \geq {f(n)}^2 > 2^{\frac{2n}{3}+1} \geq 2^{\frac{2n}{3}+\frac{1}{2}}$ $\bullet f(2n+1) \geq {f(n)}^2 > 2^{\frac{2n}{3}+1} > 2^{\frac{2n}{3}+\frac{1}{2}+\frac{1}{3}} = 2^{\frac{2n+1}{3}+\frac{1}{2}}$ And the claim is proved. $\square$ **The upper bound:**
We proceed by strong induction. Notice that the last pair on the board can be written as $f(m),f(n-m)$ since we need to maximize $f(n)$ .
We encounter two cases: $\bullet f(n)=min\{{f(m)}^2,{f(n-m)}^2\} \leq {f(\left \lfloor \frac{n}{2} \right \rfloor)}^2 \leq 3^{\frac{2}{3}\left \lfloor \frac{n}{2} \right \rfloor} \leq 3^\frac{n}{3}$ $ \bullet f(n)=f(m)+f(n-m)$ : check manually the cases where $n \in \{1,2,3,4,5\}$ , and assume WLOG $m \leq \left \lfloor \frac{n}{2} \right \rfloor$ :
<span style="color:#ff9a00">First case</span> $m=1$ : Notice that when $n \geq 6$ the following holds: $1+3^{\frac{n}{3}} \leq 3^{\frac{1+n}{3}} \Leftrightarrow 1 \leq (\sqrt[3]{3}-1)3^{\frac{n}{3}}$ <span style="color:#ff9a00">Second case</span> $m\neq 1$ : Notice that $f(n) = f(m)+f(n-m) \leq f(m)f(n-m) \leq 3^{\frac{n}{3}}$ So we are done. $\blacksquare$
|
[
"optimal strategy:\n\nfor 1,2,3,4,5: just add them all\n\nfor n>5:\n\nfor odd numbers n, take a 1 away, construct the best possible construction for the rest of the 1's, then add the 1 back in (wait does this work)\n\nfor even numbers n, split it in half, take the best construction for each half, then min(squares)",
"For the upper bound, the general idea is to consider $S=\\log_3 x_1+\\log_3 x_2+\\dots+\\log_3 x_k$ where $x_1, \\dots, x_k$ are the numbers on the board. Unless Vishal makes the move $(b, 1)\\rightarrow b+1$ , the value of this sum is (nonstrictly) decreasing. Let $x$ be the number of times he makes the move $(1, 1)\\rightarrow 2$ , and let $y$ be the number of times he goes $(2, 1)\\rightarrow 3$ . Some bounding (namely $2x+y\\leq n$ and $x\\geq y$ shows us that $S$ increases by at most $\\frac{n}{3}$ during this process, giving the desired upper bound.\n\nFor the lower bound, I did some sketchy induction by generalizing the lower bound to $f(n)\\geq 2^{\\lceil (n+1)/2\\rceil\\cdot 2/3}$ , which felt suspiciously like a fakesolve...but I basically said that if $n=2m$ , $f(n)\\geq f(m)^2\\geq 2^{\\lceil (m+1)/2\\rceil\\cdot 4/3}\\geq 2^{\\lceil (2m+1)/2\\rceil\\cdot 2/3}$ . Similarly if $n=2m+1$ , $f(n)\\geq f(m)^2\\geq 2^{\\lceil (m+1)/2\\rceil\\cdot 4/3}\\geq 2^{\\lceil (2m+2)/2\\rceil\\cdot 2/3}$ . I'm not sure why this strikes me as a fakesolve, but it just feels wrong :shrug:\n",
"I did an very suspicious bash on the fact that if the final number came from a,b then a,b has to be maximal constructions\n\nidk this was a wacky problem\n\nupper bound wasn't too bad\n\nlower bound for evens was easy\n\nlower bound for odds was murder",
"<details><summary>What are you guys doing? This is a VERY NICE problem</summary>The key idea is to consider how the process ends, and use divide and conquer. I will prove $2^{\\frac{n+1}{3}} \\le f(n)\\le 3^{\\frac n3}$ For $2^{\\frac{n+1}{3}}\\le f(n)$ : we can check this holds for $n=2,3$ , and for $n$ odd we use the first $\\frac{n-1}{2}$ numbers to form $f(\\frac{n-1}{2})$ and other $\\frac{n+1}{2}$ numbers to form $f(\\frac{n+1}{2})$ . Then we get $\\min\\{f(\\frac{n-1}{2}), f(\\frac{n+1}{2})\\}^2\\ge 2^{2\\frac{\\frac{n-1}{2}+1}{3}}=2^{2\\frac{n+1}{6}}=2^{\\frac{n+1}{3}}$ . $n$ even is similar.\n\nFor $f(n)\\le 3^{\\frac n3}$ . Consider how the process ends. $t$ numbers contribute to one number while $n-t$ numbers contribute to the other. Clearly, $f(t)f(n-t)\\le 3^{\\frac n3}$ by i.h. Check $f(t)+f(n-t)\\le 3^{\\frac n3}$ which is boring (of course I finished all the details in the test)</details>",
"oh are zee\n\ni did something kinda similar but i completely just threw the $n$ odd part.",
"Does this work?\n\n<details><summary>Upper bound</summary>Let $a_1$ , $a_2$ , $a_3$ , $\\dots$ denote the number of moves where $1$ is added to $1$ , $2$ , $3$ , $\\dots$ , respectively. Let $k$ be the maximal index such that $a_k > 0$ . We can immediately make the following observations:\n\n1) $2a_1 + a_2 + \\dots + a_k\\le n$ for all valid move sequences.\n\n2) For any valid move sequence, $a_2\\le a_1$ .\n\n3) If $x$ and $y$ are nonnegative integers such that $2x + y\\le n$ and $y\\le x$ , then there exists a valid move sequence with $a_1 = x$ , $a_2 = y$ , and $a_i = 0$ for all $i\\ge 3$ .\n\nClaim: the maximum value of $A = \\prod_{i=1}^{k} \\left(\\frac{i+1}{i}\\right)^{a_i}$ occurs when $a_i = 0$ for all $i\\ge 3$ .\n\nProof: consider any valid move sequence. Then we perform the following operation:\n\nIf $\\sum_{i=3}^{k} a_i$ is even, we increase $a_1$ by $\\frac{1}{2}\\sum_{i=3}^{k} a_i$ and replace $a_i$ with $0$ for all $i\\ge 3$ .\n\nIf $\\sum_{i=3}^{k} a_i$ is odd and $a_2\\neq 0$ , we increase $a_1$ by $\\frac{1}{2} + \\frac{1}{2}\\sum_{i=3}^{k} a_i$ , replace $a_i$ with $0$ for all $i\\ge 3$ , and reduce $a_2$ by $1$ .\n\nIf $\\sum_{i=3}^{k} a_i$ is odd and $a_2 = 0$ , we increase $a_1$ by $-\\frac{1}{2} + \\frac{1}{2}\\sum_{i=3}^{k} a_i$ , replace $a_i$ with $0$ for all $i\\ge 3$ , and replace $a_2$ with $1$ .\n\nClearly there exists a valid move sequence for the values of $\\{a_i\\}$ resulting from this operation. Moreover, the value of $A$ is not decreased, so the claim is proved.\n\nWith the claim, it is not hard to verify that the maximum value of $A$ is at most $3^{\\frac{n}{3}}$ . Since the product of the numbers on the board is initially $1$ and increases by a factor of at most $A$ , our proof of the upper bound is complete.</details>\n\n<details><summary>Lower bound</summary>We have two cases.\n\nCase 1: $2\\cdot 2^k\\le n < 3\\cdot 2^k$ for some $k$ . Choose $2\\cdot 2^k$ copies of $1$ on the board and ignore the rest (we can add them in at the end for a negligible increase). Pair off and add these numbers to create $2^k$ copies of $2$ . Then repeatedly pair off these numbers and apply the $\\text{min}(a^2,b^2)$ operation. The final number on the board is at least $2^{2^k} > 2^{\\frac{n}{3}}$ , as desired.\n\nCase 2: $3\\cdot 2^k\\le n < 4\\cdot 2^k$ for some $k$ . Choose $3\\cdot 2^k$ copies of $1$ on the board and ignore the rest (again, we can add them in at the end for a negligible increase). Group these numbers into triplets and add each triplet together to create $2^k$ copies of $3$ . Then repeatedly pair off these numbers and apply the $\\text{min}(a^2,b^2)$ operation. The final number on the board is at least $3^{2^k} = 2^{2^k\\text{log}_2(3)} > 2^{\\frac{n}{3}}$ , as desired.</details>",
"<blockquote><details><summary>What are you guys doing? This is a VERY NICE problem</summary>The key idea is to consider how the process ends, and use divide and conquer. I will prove $2^{\\frac{n+1}{3}} \\le f(n)\\le 3^{\\frac n3}$ For $2^{\\frac{n+1}{3}}\\le f(n)$ : we can check this holds for $n=2,3$ , and for $n$ odd we use the first $\\frac{n-1}{2}$ numbers to form $f(\\frac{n-1}{2})$ and other $\\frac{n+1}{2}$ numbers to form $f(\\frac{n+1}{2})$ . Then we get $\\min\\{f(\\frac{n-1}{2}), f(\\frac{n+1}{2})\\}^2\\ge 2^{2\\frac{\\frac{n-1}{2}+1}{3}}=2^{2\\frac{n+1}{6}}=2^{\\frac{n+1}{3}}$ . $n$ even is similar.\n\nFor $f(n)\\le 3^{\\frac n3}$ . Consider how the process ends. $t$ numbers contribute to one number while $n-t$ numbers contribute to the other. Clearly, $f(t)f(n-t)\\le 3^{\\frac n3}$ by i.h. Check $f(t)+f(n-t)\\le 3^{\\frac n3}$ which is boring (of course I finished all the details in the test)</details></blockquote>\n\nyep the trick for the lower bound was really ingenious, the motivation was that $2^{\\frac{n}{3}}$ I was unable to induct on, and it seemed like I was off by some factor $2^{\\frac{1}{6}}$ , so I thought about maybe changing the bound to $2^{\\frac{2n+1}{6}},$ which then lead me to think about $2^{\\frac{n+c}{3}}$ from which I saw the stronger bound.",
"We first show $f(n)\\leq 3^{\\frac{n}{3}}$ . We proceed by strong induction. The base cases are obvious. Now suppose in the last step the two numbers are $c$ and $d$ , where they are formed by $a$ copies of $1$ and $b$ copies of $1$ respectively, suppose $a\\leq b$ . $$ \\min(c^2,d^2)\\leq \\min(f(a)^2,f(b)^2)=f(a)^2\\leq f(\\lfloor\\frac{n}{2}\\rfloor)^2\\leq 3^{\\frac{n}{3}} $$ By inductive hypothesis, meanwhile, if $a\\geq 2$ then $$ c+d\\leq 3^{\\frac{a}{3}}+3^\\frac{b}{3}\\leq 2\\cdot 3^{\\frac{b}{3}}\\leq 3^{\\frac{a+b}{3}} $$ as $2\\leq 3^{\\frac{a}{3}}$ . If $a=1$ , then $c\\leq f(a)\\leq 1$ hence $$ c+d\\leq 1+3^{\\frac{b}{3}}\\leq 3^{\\frac{b+1}{3}} $$ as desired.\n\nNow we show that $f(n)>2^{\\frac{n}{3}}$ . Indeed it is easy to show that $f(1)=1$ , $f(2)=2$ and $f(3)=3$ . Now we claim that\n(i) If $2^k\\leq a<3\\cdot 2^{k-1}$ then $f(n)\\geq 2^{2^{k-1}}$ (ii) If $3\\cdot 2^{k-1}\\leq a<2^{k+1}-1$ then $f(n)\\geq 3^{2^{k-1}}$ Indeed using the fact that $f(2m+1)\\geq f(m)^2$ and the base cases this is obvious using simple induction. Now $$ 2^{2^{k-1}}> 2^{\\frac{3\\cdot 2^{k-1}-1}{3}} $$ while $$ 3^{3\\cdot 2^{k-1}}> 2^{4\\cdot 2^{k-1}}=2^{2^{k+1}} $$ so we are done.",
"What would be the MOHS difficulty for this question? ",
"I assume that $n>1$ , otherwise the problem is wrong since $f(1)=1<2^{1/3}$ .\n\nFor the lower bound, I claim that we can strengthen to $2^{(n+1)/3}$ (nonstrict). For $n \\leq 5$ just add everything. Otherwise split into $\\lfloor n/2\\rfloor$ and $\\lceil n/2\\rceil$ and reduce to having two numbers $f(\\lfloor n/2\\rfloor)$ and $f(\\lceil n/2\\rceil)$ on the board. Then apply the second operation to get at least $$ f(\\lfloor n/2\\rfloor)^2 \\geq (2^{(\\tfrac{n-1}{2}+1)/3})^2=2^{(n+1)/3}, $$ which finishes by induction.\n\nFor the upper bound, consider the last two numbers on the board, which must be at most $f(a)$ and $f(b)$ for some choice of $a,b>0$ such that $a+b=n$ . Bash out $n \\leq 4$ again to check that the upper bound holds for these $n$ . Then observe that $$ \\min(3^{2a/3},3^{2b/3}) \\leq 3^{n/3}, $$ and $$ 3^{a/3}+3^{b/3}\\leq 3^{1/3}+3^{(n-1)/3} $$ since $3^{x/3}$ is convex. It suffices to show that the RHS is at most $3^{n/3}$ for $n \\geq 5$ . This is equivalent to saying that $\\sqrt[3]{3} \\leq \\sqrt[3]{3}^{n-1}(\\sqrt[3]{3}-1) \\iff \\sqrt[3]{3}^{n-2}(\\sqrt[3]{3}-1) \\geq 1$ , so it suffices to check this for $n=3$ , in which case it is just $3\\sqrt[3]{3} \\geq 4 \\iff 81 \\geq 64$ which is true.",
"For the lower bound/odd case, why does it follow that $$ f(\\lfloor n/2\\rfloor)^2 \\geq (2^{(\\tfrac{n-1}{2}+1)/3})^2 $$ I'm pretty sure that the +1 in the exponent comes from the strengthened lower bound.\n\nWithout the strengthened bound, it seems that $$ f\\left(\\frac{n-1}{2}\\right)^2 > 2^{\\frac{n-1}{3}} $$ which isn't enough. \n\nSo, my main question is why a strengthened condition leads to the proof, while the original condition (which contains the strengthened condition (?) since $\\frac{n}{3}<\\frac{n+1}{3}$ ) can't. \n\nAnd how does one even come up with the strengthened condition, intuition wise? Is it from testing small cases like $n=1$ to $n=5$ ? Is it a general technique to prove a stronger condition of an equality to prove the statement in question? Thanks.",
"The step before constructing $f(n)$ , the resulting graph must be almost bipartite. We also have $f(a)>f(b)$ for $a>b$ , and for sufficiently large $n$ , it's optimal to split the set in half. For even $2n$ , we have $f(2n) = f(n)^2$ , and for odd $n$ , we have $f(n) = f(\\frac{n-1}{2}-1)^2 + 1$ . Induction finishes."
] |
[
"origin:aops",
"2022 Canada National Olympiad",
"2022 Contests"
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"answer_score": 144,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799970.json"
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|
Call a set of $n$ lines *good* if no $3$ lines are concurrent. These $n$ lines divide the Euclidean plane into regions (possible unbounded). A *coloring* is an assignment of two colors to each region, one from the set $\{A_1, A_2\}$ and the other from $\{B_1, B_2, B_3\}$ , such that no two adjacent regions (adjacent meaning sharing an edge) have the same $A_i$ color or the same $B_i$ color, and there is a region colored $A_i, B_j$ for any combination of $A_i, B_j$ .
A number $n$ is *colourable* if there is a coloring for any set of $n$ good lines. Find all colourable $n$ .
|
<blockquote>The answer should be $n\ge 5$ I proved that the graph of regions is bipartite and if two regions have distance $\ge 5$ then we are done.
Then I failed to prove two regions have distance $\ge 5$ .</blockquote>
If $n\le 4$ , setting $n$ parallel lines divide the plane into $n+1$ regions, so it fails.
It remains to show $n\ge 5$ works. Consider a graph where each vertex corresponds to a region and there is an edge between two vertices if and only if their corresponding regions intersect at a line. We can represent each region with a binary number with $n$ digits, where for $j=1,\cdots,n$ , a 1 on bit $j$ indicates it is on one fixed side of line $j$ , and 0 on bit $j$ indicates it is on the other side.
Claim 1: This graph is bipartite.
Proof: It suffices to show that all cycles of the graph have even length. Note two vertices are adjacent if and only if their binary numbers differ by exactly one digit. Therefore, each line must be crossed an even number of times for me to go back to the region I started with.
Claim 2: This graph has 2 vertices with distance $5$ (in fact, $n$ ).
Proof: Note $n\ge 5$ . If a line is not parallel to any of the other lines, it must intersect with all regions, so we can induct down.
Otherwise, each line is parallel to another line, so we can casework on $n=4$ and $n=5$ .
Claim 3: A coloring exists if the distance between two vertices is at least 5.
Proof: Suppose $dist(u,v)\ge 5$ . Let $S_j=\{ w | dist(u,w)=j\}$ for $j\in \mathbb{N}$ . Since the graph is bipartite, the induced subgraph of $S_j$ is an anticlique. By the definition of distance, there are no edges between $S_i$ and $S_j$ if $|i-j|\ge 2$ .
We color all vertices in $S_j$ with color $A_{(j \bmod\; 2)}, B_{(j\bmod\; 3)}$ , which works for reasons explained above.
|
[
"I'm probably misunderstanding the problem, but why are the $B_i$ colors needed? We can do a 2-coloring of the regions as in [EGMO 2017/3](https://artofproblemsolving.com/community/c6h1424941p8024557) and so all $n$ work?",
"<blockquote>I'm probably misunderstanding the problem, but why are the $B_i$ colors needed? We can do a 2-coloring of the regions as in [EGMO 2017/3](https://artofproblemsolving.com/community/c6h1424941p8024557) and so all $n$ work?</blockquote>\n\nYikes, totally forgot the key part of the problem (all of the problems I posted are from memory, as I don't have the actual problems). Should be fixed now - all color combinations must appear at least once.",
"The answer should be $n\\ge 5$ I proved that the graph of regions is bipartite and if two regions have distance $\\ge 5$ then we are done.\n\nThen I failed to prove two regions have distance $\\ge 5$ .",
"Imagine getting more progress than me\ni think i got n=3,4 doesn't work",
"bruh there are so many people better than you. I don't know the final part seems very easy but I can't do it without casework bash. Alternatively, since 5=5, we may be able to prove this claim with induction, but in contest I fakesolved because I don't understand how lines and regions behave",
"<blockquote>bruh there are so many people better than you.</blockquote>\nI am aware.\n\nalso i think induction would work\n\nn=3 clearly works. \n\nthen you can prove that each time you add a line, one of the n+1 regions you go through is split by this new line?\n\n\nedit1: in the graph theory representation pick a path such that upon removal of these points on the paths the graph becomes disconnected?\n\ncombined with extremal pinciple this might work",
"<blockquote>A ***coloring*** is an assignment</blockquote>\n\n<blockquote>A number $n$ is ***colourable*** if</blockquote>\n\nTypical Canadians being unable to decide whether to use American or British english...",
"Let $n$ be a positive integer. A set of n distinct lines divides the plane into various (possibly unbounded) regions. The set of lines is called “nice” if no three lines intersect at a single point. A “colouring” is an assignment of two colours to each region such that the first colour is from the set $\\{A_1, A_2\\}$ , and the second colour is from the set $\\{B_1, B_2, B_3\\}$ . Given a nice set of lines, we call it “colourable” if there exists a colouring such that\n(a) no colour is assigned to two regions that share an edge;\n(b) for each $i \\in \\{1, 2\\}$ and $j \\in \\{1, 2, 3\\}$ there is at least one region that is assigned with both $A_i$ and $B_j$ .\nDetermine all $n$ such that every nice configuration of $n$ lines is colourable."
] |
[
"origin:aops",
"2022 Canada National Olympiad",
"2022 Contests"
] |
{
"answer_score": 52,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799975.json"
}
|
A pentagon is inscribed in a circle, such that the pentagon has an incircle. All $10$ sets of $3$ vertices from the pentagon are chosen, and the incenters of each of the $10$ resulting triangles are drawn in. Prove these $10$ incenters lie on $2$ concentric circles.
Note: I spent nearly no time on this, so if anyone took CMO and I misremembered just let me know.
|
Let $I_A,I_B,I_C,I_D,I_E$ be the incenters of triangles $EAB,ABC,BCD,CDE,DEA,$ respectively. Let $J_A,J_B,J_C,J_D,J_E$ be the incenters of triangles $ACD,BDE,CAE,DAB,EBC.$ **<span style="color:red">Claim:</span>** $AI_AJ_DI_BB$ is cyclic.
*Proof.* Notice \begin{align*}\angle AI_AB&=90+\tfrac{1}{2}\angle AEB&=90+\tfrac{1}{2}\angle ACB=\angle AI_BB&=90+\tfrac{1}{2}\angle ADB=\angle AJ_DB.\end{align*} $\blacksquare$ Similarly, $BI_BJ_EI_CC,$ etc, are cyclic.**<span style="color:red">Claim:</span>** $AJ_CJ_AC$ is cyclic.
*Proof.* We know $$ \angle AJ_CC=90+\tfrac{1}{2}\angle DEA=90+\tfrac{1}{2}\angle CDA=\angle AJ_AC. $$ $\blacksquare$ Similarly, $AJ_DJ_AD,$ etc, are cyclic.**<span style="color:red">Claim:</span>** $I_AI_BI_CI_DI_E$ is cyclic.
*Proof.* We see $$ II_A\cdot IA=II_B\cdot IB=II_C\cdot IC=II_D\cdot ID=II_E\cdot IE $$ so there is an inversion at $I$ that maps $A$ to $I_A$ and so on. Since $ABCDE$ is cyclic, $I_AI_BI_CI_DI_E$ is also cyclic. $\blacksquare$ **<span style="color:red">Claim:</span>** A circle concentric with $(I_AI_BI_CI_DI_E)$ circumscribes $J_AJ_BJ_CJ_DJ_E.$ *Proof.* By symmetry, it suffices to show the perpendicular bisector of $\overline{I_BI_C}$ is congruent to the perpendicular bisector of $\overline{J_DJ_A}.$ Note $$ \angle I_BI_CJ_A=360-(180-\angle I_BBC)-(180-\angle CDJ_A)=\tfrac{1}{2}\angle ABC+\tfrac{1}{2}\angle ADC=90. $$ Similarly, all other angles of $I_BI_CJ_AJ_D$ are right, so it is a rectangle. $\blacksquare$ $\square$
|
[
"I think the statement is correct",
"qwerty_ytrewq solved it in 30 minutes, but sadly I don't understand his solution\n\nDiscord links: [ page 1 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390087827496/image0.jpg)\n[ page 2 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390364647424/image1.jpg)\n[ page 3 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390607908865/image2.jpg)",
"Let the pentagon be $P_1P_2P_3P_4P_5$ with circumcircle $\\Omega$ centered at $O$ and incircle $\\omega$ centered at $I$ . Taking indices modulo $5$ , let $I_i$ and $J_i$ be the incenters of $P_{i-1}P_iP_{i+1}$ and $P_{i-2}P_iP_{i+2}$ . By Poncelet Theory, the pentagon $P_1P_3P_5P_2P_4$ has an incircle $\\omega'$ that is coaxial with $\\Omega$ and $\\omega$ , so it is centered at some point $J$ on $OI$ . By the Pingpong Lemma, there exists a point $K$ on line $OI$ such that if $\\tau_P:\\Omega\\rightarrow\\Omega$ denotes projection through $P$ , then $\\tau_O\\circ\\tau_I\\circ\\tau_J\\circ\\tau_K$ is the identity. Let $Q_i=\\tau_O\\circ\\tau_I(P_i)$ and $R_i=\\tau_J(P_i)$ , so $Q_iR_i$ passes through $K$ . Note that $Q_i$ is the midpoint of arc $P_{i-1}P_{i+1}$ and $R_i$ is the midpoint of arc $P_{i-2}P_{i+2}$ . It follows by Fact 5 that $Q_iR_i$ is the common perpendicular bisector of $I_{i-2}I_{i+2}$ and $J_{i-1}J_{i+1}$ , so $KI_{i-2}=KI_{i+2}$ and $KJ_{i-1}=KJ_{i+1}$ . Taking this over all $i$ gives the desired result.",
"<blockquote>qwerty_ytrewq solved it in 30 minutes, but sadly I don't understand his solution\n\nDiscord links: [ page 1 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390087827496/image0.jpg)\n[ page 2 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390364647424/image1.jpg)\n[ page 3 ](https://cdn.discordapp.com/attachments/927412307817005097/952021390607908865/image2.jpg)</blockquote>\n\nvery nice ideas!",
"Let $ABCDE$ be a convex pentagon such that the five vertices lie on a circle and the five sides are tangent to another circle inside the pentagon. There are ${5 \\choose 3}= 10$ triangles which can be formed by choosing $3$ of the $5$ vertices. For each of these $10$ triangles, mark its incenter. Prove that these $10$ incenters lie on two concentric circles.",
"Let $O$ be the incenter of the pentagon. Let $I_A$ be the incenter of $\\triangle EAB$ , etc. Let $M_{XY}$ be the midpoint of minor arc $XY$ of $(ABCDE)$ where $X,Y$ are arbitrary points\n\n Notice that $I_A$ and $I_B$ lie on $OA$ and $OB$ respectively, while $OI_A\\times OA=OI_B\\times OB$ as $I_A,I_B,A,B$ lie on the circle centered at the $M_{AB}$ .\n\n Hence by symmetry we have $OI_A\\times OA=OI_B\\times OB=OI_C\\times OC=OI_D\\times OD=OI_E\\times OE$ , hence an inversion \nNow let $J$ be the center of $(I_AI_BI_CI_DI_E)$ . Let $J_A$ be the incenter of $\\triangle ADC$ , etc.\n\n Notice that $M_{AE}J$ is the perpendicular bisector of $I_EI_A$ , hence $$ \\angle JM_{AE}I_A=\\frac{1}{2}\\angle I_EM_{AE}I_A=\\frac{1}{2}\\angle DM_{AE}B=\\angle M_{BD}M_{AE}B=\\angle M_{BD}M_{AE}I_A $$ Hence $J,M_{AE},M_{BD}$ are collinear. Meanwhile we notice that $$ \\angle J_BM_{BD}M_{AE}=\\angle EM_{BD}M_{AE}=\\frac{1}{2}EM_{BD}A=\\frac{1}{2}J_BM_{BD}J_D $$ Hence $M_{AE}M_{BD}$ is the perpendicular bisector of $J_BJ_D$ , which implies $JJ_B=JJ_D$ , by symmetry $J_A,J_B,J_C,J_D,J_E$ all lie on a circle centered at $J$ so we are done.",
"Is this possible to do just by angle chasing and showing that for any four points among Ia,Ib,Ic,Id,Ie it satsfies inscribed angle theroem ? And similarily for Ja,Jb... ?",
"Throughout the solution all indicates are taken modulo $5.$ Denote the pentagon as $A_1A_2A_3A_4A_5,$ as $I_i,J_i$ incenters of $A_{i-1}A_iA_{i+1},$ $A_{i-2}A_iA_{i+2}$ respectively. It's well-known that $A_iA_{i+1}I_iI_{i+1}$ are concyclic, moreover lines $A_iI_i$ pass through incenter of pentagon, so by inversion $I_1I_2I_3I_4I_5$ are concyclic with center $O.$ But well-known that $I_iI_{i+1}J_{i+2}J_{i-1}$ is rectangle, so perpendicular bisectors of $J_{i+2}J_{i-1}$ pass through $O,$ which clearly complete our proof.",
"**S**olved with **mueller.25**, **starchan**, **Siddharth03**, **AdhityaMV****<span style=\"color:#f00\">Lemma:</span>** In a cyclic quadrilateral $ABCD$ , the incenters of the four possible triangles form a rectangle.**<span style=\"color:#00f\">Proof:</span>** Let $I_1, I_2, I_3, I_4$ be the incenters of $BCD, CDA, DAB, ABC$ . First note that $DI_2I_1C$ is cyclic since $\\angle DI_2C = 90 + \\frac{\\angle DAC}{2} = 90 + \\frac{\\angle DBC}{2} = \\angle DI_1C$ . Similarly $BI_4I_1C$ is cyclic as well. So, $\\angle I_4I_1I_2 = 360 - \\angle I_4I_1C - \\angle I_2I_1C = 360 - (180 - \\angle I_4BC) - (180 - \\angle I_2DC) = \\frac{\\angle ADC + \\angle ABC}{2} = 90^\\circ$ . Similarly all angles of the quadrilateral are right, so it is a rectangle. $\\square$ Now, let $ABCDE$ be the bicentric pentagon and $I_1, I_2, I_3, I_4, I_5$ the incenters of $EAB, ABC, BCD, CDE, DEA$ and let $I$ be the incenter of the quadrilateral. Note that due to the cyclic quadrilaterals mentioned above, we get by power of point, $II_1 \\cdot IA = II_2 \\cdot IB = \\cdots = II_5 \\cdot IE$ . Considering an inversion with radius, square root of this length we see that $(ABCDE)$ swaps with $(I_1I_2I_3I_4I_5)$ , so these five incenters lie on a circle. Suppose this circle has center $O$ .\n\nLet $I_1', I_2', I_3', I_4', I_5'$ be the incenters of $ACD, BDE, CEA, DAB, EBC$ . Then note that since $I_1I_2I_5'I_3'$ is a rectangle (by the Lemma), we have that $OI_3' = OI_5'$ . Similarly we get $OI_k' = OI_{k+2}'$ for all $k$ with indices taken mod $5$ . Since $5$ is odd, this implies that $OI_k'$ is fixed across all $k$ , so the points $I_1', I_2', I_3', I_4', I_5'$ also lie on a circle with center $O$ . So the ten incenters do indeed lie on two concentric circles, as desired. $\\blacksquare$ ",
"<blockquote>\nLet the pentagon be $P_1P_2P_3P_4P_5$ with circumcircle $\\Omega$ centered at $O$ and incircle $\\omega$ centered at $I$ . Taking indices modulo $5$ , let $I_i$ and $J_i$ be the incenters of $P_{i-1}P_iP_{i+1}$ and $P_{i-2}P_iP_{i+2}$ . By Poncelet Theory, the pentagon $P_1P_3P_5P_2P_4$ has an incircle $\\omega'$ that is coaxial with $\\Omega$ and $\\omega$ , so it is centered at some point $J$ on $OI$ . By the Pingpong Lemma, there exists a point $K$ on line $OI$ such that if $\\tau_P:\\Omega\\rightarrow\\Omega$ denotes projection through $P$ , then $\\tau_O\\circ\\tau_I\\circ\\tau_J\\circ\\tau_K$ is the identity. Let $Q_i=\\tau_O\\circ\\tau_I(P_i)$ and $R_i=\\tau_J(P_i)$ , so $Q_iR_i$ passes through $K$ . Note that $Q_i$ is the midpoint of arc $P_{i-1}P_{i+1}$ and $R_i$ is the midpoint of arc $P_{i-2}P_{i+2}$ . It follows by Fact 5 that $Q_iR_i$ is the common perpendicular bisector of $I_{i-2}I_{i+2}$ and $J_{i-1}J_{i+1}$ , so $KI_{i-2}=KI_{i+2}$ and $KJ_{i-1}=KJ_{i+1}$ . Taking this over all $i$ gives the desired result.\n</blockquote>\n\nDiagram for **ABCDE** solution.\n[asy]\nimport geometry;\n size(12cm);\n point p1, p2, p3, p4, p5, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, i, o, j,\n k, q1, r1;\n\n o = (0, 0);\n\n p1 = dir(140);\n\n real r = 0.673;\n i = (0.3, 0);\n circle in = circle(i, r);\n circle ci = circle(o, 1);\n\n line tangent = tangents(in, p1)[1];\n point[] pipi1 = intersectionpoints(tangent, ci);\n p2 = pipi1[0] + pipi1[1] - p1;\n\n point nextTangent(point pi, point pin, circle in, circle ci) {\n line tangent0 = tangents(in, pi)[0];\n point[] pk = intersectionpoints(tangent0, ci);\n line tangent1 = tangents(in, pi)[1];\n point[] pl = intersectionpoints(tangent1, ci);\n\n return pk[0] + pl[0] + pk[1] + pl[1] - 2 * pi - pin;\n }\n\n p3 = nextTangent(p2, p1, in, ci);\n p4 = nextTangent(p3, p2, in, ci);\n p5 = nextTangent(p4, p3, in, ci);\n\n\n i1 = incenter(p5, p1, p2);\n i2 = incenter(p1, p2, p3);\n i3 = incenter(p2, p3, p4);\n i4 = incenter(p3, p4, p5);\n i5 = incenter(p4, p5, p1);\n\n j1 = incenter(p4, p1, p3);\n j2 = incenter(p5, p2, p4);\n j3 = incenter(p1, p3, p5);\n j4 = incenter(p2, p4, p1);\n j5 = incenter(p3, p5, p2);\n\n j = intersectionpoint(bisector(line(p1, p3), line(p2, p4), 90), bisector(line(p2, p4), line(p3, p5), 90));\n k = circumcenter(i1, i2, i3);\n\n point[] qmp1 = intersectionpoints(line(p1, i), ci);\n point qm = qmp1[0]+qmp1[1] - p1;\n point[] q1qm = intersectionpoints(line(qm, o), ci);\n q1 = q1qm[0]+q1qm[1] - qm;\n\n point[] r1p1 = intersectionpoints(line(j, p1), ci);\n r1 = r1p1[0]+r1p1[1]-p1;\n\n draw(p1--p2--p3--p4--p5--cycle, springgreen);\n draw(p1--p3--p5--p2--p4--cycle, green);\n draw(line(o,i), purple);\n draw(ci, springgreen);\n draw(in, springgreen);\n draw(circle(2*j-foot(j, p1, p3), foot(j, p1, p3)), green);\n draw(q1--r1, dashed+red);\n draw(circle(i1,i2,i3), dashed+purple);\n draw(circle(j1,j2,j3), dashed+purple);\n\n draw(i3--i4, red);\n draw(j2--j5, red);\n\n draw(p2--r1--p5, red);\n draw(p3--q1--p4, red);\n\n dot(\" $P_1$ \", p1, dir(140));\n dot(\" $P_2$ \", p2, dir(60));\n dot(\" $P_3$ \", p3, dir(10));\n dot(\" $P_4$ \", p4, dir(-30));\n dot(\" $P_5$ \", p5, dir(280));\n dot(\" $I_1$ \", i1, dir(10));\n dot(\" $I_2$ \", i2, dir(80));\n dot(\" $I_3$ \", i3, dir(230));\n dot(\" $I_4$ \", i4, 1.2*dir(90));\n dot(\" $I_5$ \", i5, dir(240));\n dot(\" $J_1$ \", j1, dir(340));\n dot(\" $J_2$ \", j2, 1.2*dir(180));\n dot(\" $J_3$ \", j3, dir(210));\n dot(\" $J_4$ \", j4, dir(100));\n dot(\" $J_5$ \", j5, dir(210));\n dot(\" $I$ \", i, dir(220));\n dot(\" $O$ \", o, dir(50));\n dot(\" $J$ \", j, dir(90));\n dot(\" $K$ \", k, dir(90));\n dot(\" $Q_1$ \", q1, dir(140));\n dot(\" $R_1$ \", r1, dir(350));\n[/asy]",
"A regular pentagon is a convex pentagon. \n"
] |
[
"origin:aops",
"2022 Canada National Olympiad",
"2022 Contests"
] |
{
"answer_score": 160,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 Canada National Olympiad/2799976.json"
}
|
Let $\triangle{ABC}$ has circumcircle $\Gamma$ , drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$ , similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$ . Prove that if $AB=DE, \angle{ACB}=60^{\circ}$ (sorry it is from my memory I can't remember the exact problem, but it means the same)
|
Let $AD \cap BC = X$ and $BE \cap CA = Y$ . The Orthocenter Reflection Lemma yields $$ XY = \frac{DE}{2} = \frac{AB}{2}. $$ Now, since $CXY \overset{-}{\sim} CAB$ , we know $$ | \cos ACB | = \frac{CX}{CA} = \frac{XY}{AB} = \frac{1}{2} $$ so $\angle ACB = 60^{\circ}$ or $\angle ACB = 120^{\circ}$ must hold. $\blacksquare$ **Remarks:** I'm pretty sure $\angle ACB = 120^{\circ}$ is possible too.
Also, what is the CJMO? Personally, I only know of one [CJMO](https://artofproblemsolving.com/community/c3012738), and this problem isn't from there.
|
[
":( lucky imagine getting a doable geo\n\nour geo was so hard\n\nanyway, AB=DE implies AE || BD which implies $\\angle EAD =\\angle EBD$ which by orthocenter reflection stuff blah blah implies BHD and AHE both equilateral so we're done",
"This is my solution provided during the exam\n<details><summary>solution</summary>Connect $AD,BE$ , denote that $\\angle{EBC}=\\alpha,\\angle{AEB}=90-\\alpha$ . Since $\\widehat{AB}=\\widehat{AB},\\angle{ADB}=\\angle{AEB}=90-\\alpha$ , since $BD$ is perpendicular to $AC, \\angle{DAC}=\\alpha$ $\\widehat{CE}=\\widehat{CE},\\angle{CAE}=\\angle{CEB}=\\alpha$ , since $ED=AB,\\widehat{ED}=\\widehat{AB},\\angle{AEB}=\\angle{DAE},2\\alpha=90-\\alpha, \\alpha=30$ , $\\angle{ACB}=90-\\alpha=60^{\\circ}$ as desired</details>",
"Let $ABC$ be an acute angled triangle with circumcircle $\\Gamma$ . The perpendicular from $A$ to $BC$ intersects $\\Gamma$ at $D$ , and the perpendicular from $B$ to $AC$ intersects $\\Gamma$ at $E$ . Prove that if $|AB| = |DE|$ , then $\\angle ACB = 60^o$ .",
"Notice $\\triangle ABE\\cong\\triangle BED$ by SAS so $H=\\overline{AD}\\cap\\overline{BE}$ is the center of $\\Gamma.$ Hence, $$ 2\\angle C=\\angle AHB=180-\\angle C $$ and $\\angle C=60.$ $\\square$ ",
"Let $AD \\cap BE = H$ , $AC \\cap BE = X$ , $BC \\cap AD = Y$ . Using the fact that angles subtended by arcs of equal lengths are equal, $\\angle DBE = \\angle ADB$ so $HB = HD$ . Since $AXYB$ is cyclic, $\\angle HBY = \\angle HAX$ . By construction $ABDC$ is cyclic so $\\angle HAX = \\angle DBC$ so $BY$ is the perpendicular bisector of $\\triangle HBD$ thus $BD = HB$ . Hence $\\triangle HBD$ is equilateral and $\\angle ACB = \\angle ADB = \\angle HDB = 60^\\circ$ . ",
"Let $X$ be the foot from $A$ to $BC$ and $Y$ be the foot from $B$ to $AC$ . By reflecting the orthocenter, $XY=\\frac{1}{2}ED=\\frac{1}{2}AB$ . Since $AYXB$ is cyclic, we have $\\triangle CYX\\sim \\triangle CBA$ so $CX=\\frac{1}{2}AC$ and hence $\\cos C=\\frac{1}{2}\\implies C=60^{\\circ}$ .",
"Canada Junior Math Olympiad?",
"<blockquote>Also, what is the CJMO? Personally, I only know of one [CJMO](https://artofproblemsolving.com/community/c3012738), and this problem isn't from there.</blockquote>\n\n[Canadian Junior Math Olympiad ](https://artofproblemsolving.com/community/c1231801_canadian_junior_mathematical_olympiad)",
"Notation : $(XY)$ stands for small arc $XY$ $(DE)= (EC)+(CD)=2\\angle EAC+2<DBC=2(180^o-\\angle AEB) + 2(180^o-\\angle DBC) = 180^o +180^o-4(AB)-4(AB)=360^o-8(AB)=360^o- 4\\angle ACB=360^o-4 \\cdot 60^o=360^o -240^o=120^o =\\angle (DCE) \\Rightarrow AB= DE$ my solution might be modified to prove that the converse is also true,\n and so the problem could have been asked with **iff** condition",
"<details><summary>Sol.</summary>Let $H=AD\\cap BE$ . Then $H$ is the orthocenter of $ABC$ and thus $D,E$ are the reflections of $H$ from $BC$ and $AC$ hence $AH=AE$ . Notice that $AHB \\equiv DHE$ hence $HE=AH$ . Therefore $AHE$ is equilateral, so $\\angle AEH=60^{\\circ}$ . But $\\angle AEH=\\angle ACB$ because $AECB$ is an inscribed quadilateral q.e.d.</details>",
"Angle chase using $\\angle AEB= \\angle EBD$ and the orthocentre reflection lemma."
] |
[
"origin:aops",
"2022 Canadian Junior Mathematical Olympiad",
"2022 Contests"
] |
{
"answer_score": 122,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 Canadian Junior Mathematical Olympiad/2799974.json"
}
|
You have an infinite stack of T-shaped tetrominoes (composed of four squares of side length 1), and an n × n board. You are allowed to place some tetrominoes on the board, possibly rotated, as long as no two tetrominoes overlap and no tetrominoes extend off the board. For which values of n can you cover the entire board?
|
The answer is $\boxed{n \text{ divisible by } 4}$ . The construction is obvious by copypasting the $4 \times 4$ construction across the board. **<span style="color:#f00">Claim</span>** $4 \mid n$ .
*Proof.* It is easy to see, from say area, that $2 \mid n$ . To show that we require $4 \mid n$ consider a checkerboard coloring. Each tetrominoe covers exactly $3$ squares of one of the colors, and $1$ of the other. Thus we require an even number of tetronimoes. This implies that $8 \mid n^2$ . However it is easy to see then that we must have $4 \mid n$ as claimed. $\square$
|
[
" $4\\mid n$ only.\n\nClearly $n$ is even. If $4\\nmid n$ , then the number of T's is odd, and a checkerboard coloring gives contradiction.\n\nFor $4\\mid n$ we have \n\nTTTS\nATSS\nAABS\nABBB",
"<blockquote> $4\\mid n$ only.\n\nClearly $n$ is even. If $4\\nmid n$ , then the number of T's is odd, and a checkerboard coloring gives contradiction.\n\nFor $4\\mid n$ we have \n\nTTTS\nATSS\nAABS\nABBB</blockquote>\n\nI got the same answer",
"ye that is what I got"
] |
[
"origin:aops",
"2022 Canadian Junior Mathematical Olympiad",
"2022 Contests"
] |
{
"answer_score": 1120,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Canadian Junior Mathematical Olympiad/2800195.json"
}
|
Positive integers $a$ , $b$ , $c$ are given. It is known that $\frac{c}{b}=\frac{b}{a}$ , and the number $b^2-a-c+1$ is a prime. Prove that $a$ and $c$ are double of a squares of positive integers.
|
We have that $b^2=ac$ so by substituting it here we get $$ ac-a-c+1=(a-1)(c-1)=p $$ so one of them is equal $2$ so another one must be $2k^2$ which finishes the proof $\blacksquare$
|
[
"We have that $(a-1)(c-1)$ is prime, so one of $a$ and $c$ equals $2$ . Since $ac$ is a perfect square, it follows that the other of $a$ and $c$ is also double a perfect square.",
"This is quite simple. Simply note that $b^2=ac$ . Thus, \n\\[ac-a-c+1=(a-1)(c-1)\\]\nis a prime. This requires one of $a,c$ to be $2$ . WLOG, let $a=2$ . Note that then, $2\\mid b$ . Then, \n\\[c=\\frac{b^2}{2} = 2\\left(\\frac{b}{2}\\right)^2\\]\nThus, $a,c$ are both twice a perfect square as claimed."
] |
[
"origin:aops",
"2022 Contests",
"2022 Caucasus Mathematical Olympiad"
] |
{
"answer_score": 112,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800863.json"
}
|
In parallelogram $ABCD$ , points $E$ and $F$ on segments $AD$ and $CD$ are such that $\angle BCE=\angle BAF$ . Points $K$ and $L$ on segments $AD$ and $CD$ are such that $AK=ED$ and $CL=FD$ . Prove that $\angle BKD=\angle BLD$ .
|
$\angle BAF = \angle BCE \implies \angle EAF = \angle ECF \implies AEFC$ is cyclic. $\angle DAF = \angle ECD , \angle ADF = \angle CDE \implies ADF$ and $CDE$ are similar. $\angle BAK = \angle BCL$ and $\frac{BA}{BC} = \frac{CD}{AD} = \frac{ED}{DF} = \frac{AK}{CL}$ so $BKA$ and $BLC$ are similar so $\angle BKD = \angle BLD$ .
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Caucasus Mathematical Olympiad"
] |
{
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800865.json"
}
|
Pete wrote down $21$ pairwise distinct positive integers, each not greater than $1,000,000$ . For every pair $(a, b)$ of numbers written down by Pete, Nick wrote the number $$ F(a;b)=a+b -\gcd(a;b) $$
on his piece of paper. Prove that one of Nick’s numbers differs from all of Pete’s numbers.
|
Let WLOG $a_{21}$ be the largest number. By plugging $(a_{21}, a_i)$ for $i=1,2,...,20$ and supposing otherwise we obtain $lcm(a_1, a_2,..., a_{20})/a_{21}$ , so $2^{20} \leq a_{21}$ , size contradiction.
@below, ok you're right. We have to prove that $a_k/a_{k+1}$ (otherwise $F(a_k, a_{k+1})$ is between $a_k$ and $a_{k+1}$ )
|
[
"<blockquote>Let WLOG $a_{21}$ be the largest number. By plugging $(a_{21}, a_i)$ for $i=1,2,...,20$ and supposing otherwise we obtain $lcm(a_1, a_2,..., a_{20})/a_{21}$ , so $2^{20} \\leq a_{21}$ , size contradiction.</blockquote> $\\mathrm{lcm}$ of 20 different numbers can be less than $2^{20}$ . for example, the 30 divisors of $2^5\\cdot 3^4=2592$ ",
"imagine peter's number are a1,a2,...,a21\nWLOG:a1<a2<...<a21\nwe know that gcd(a,b) at maximum is min(a,b) so:\nF(a1,aj)=a1+aj-gcd(a1,aj)= or >aj>a1\nand\nF(ai,aj)=ai+aj-gcd(ai,aj)= or >ai>a1\nso Nick can't write a1."
] |
[
"origin:aops",
"2022 Contests",
"2022 Caucasus Mathematical Olympiad"
] |
{
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800868.json"
}
|
Do there exist 2021 points with integer coordinates on the plane such that the pairwise distances between them are pairwise distinct consecutive integers?
|
[
"Compare with [this problem](\"https://artofproblemsolving.com/community/c6h2800893_pairwise_distances_are_consecutive_numbers\")"
] |
[
"origin:aops",
"2022 Contests",
"2022 Caucasus Mathematical Olympiad"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800872.json"
}
|
|
Let $S$ be the set of all $5^6$ positive integers, whose decimal representation consists of exactly 6 odd digits. Find the number of solutions $(x,y,z)$ of the equation $x+y=10z$ , where $x\in S$ , $y\in S$ , $z\in S$ .
|
<blockquote><blockquote>Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits
We have $x_6+y_6=10$ $x_5+y_5 \geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \geq 10$ $x_3+y_3 \geq 10$ $x_2+y_2 \geq 10$ $x_1+y_1 \geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \geq 10$ has $15$ solutions
So total we have $5*15^5$ solutions.</blockquote>
Hi,good solution but i think that x1+y1 can take any value,what about you?</blockquote>
Hi, if $x_1+y_1$ less than 10, x+y can't be 7 digit number but 10z is 7 digit number
|
[
"Who can send solution of this problem",
"<blockquote>Who can send solution of this problem</blockquote>\n\nİ dont have any idea :play_ball: ",
"Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits\n\nWe have $x_6+y_6=10$ $x_5+y_5 \\geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \\geq 10$ $x_3+y_3 \\geq 10$ $x_2+y_2 \\geq 10$ $x_1+y_1 \\geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \\geq 10$ has $15$ solutions\nSo total we have $5*15^5$ solutions.",
"<blockquote>Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits\n\nWe have $x_6+y_6=10$ $x_5+y_5 \\geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \\geq 10$ $x_3+y_3 \\geq 10$ $x_2+y_2 \\geq 10$ $x_1+y_1 \\geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \\geq 10$ has $15$ solutions\nSo total we have $5*15^5$ solutions.</blockquote>\n\nHi,good solution but i think that x1+y1 can take any value,what about you?"
] |
[
"origin:aops",
"2022 Contests",
"2022 Caucasus Mathematical Olympiad"
] |
{
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800874.json"
}
|
16 NHL teams in the first playoff round divided in pairs and to play series until 4 wins (thus the series could finish with score 4-0, 4-1, 4-2, or 4-3). After that 8 winners of the series play the second playoff round divided into 4 pairs to play series until 4 wins, and so on. After all the final round is over, it happens that $k$ teams have non-negative balance of wins (for example, the team that won in the first round with a score of 4-2 and lost in the second with a score of 4-3 fits the condition: it has $4+3=7$ wins and $2+4=6$ losses). Find the least possible $k$ .
|
$k=2$
In the first round everyone gets $4-3$ points, in the second round everyone gets $4-2$ points and in the third round everyone gets $4-0$ points, and in the last round the goals difference of 2 players is not negative
|
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"origin:aops",
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"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800877.json"
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|
Point $P$ is chosen on the leg $CB$ of right triangle $ABC$ ( $\angle ACB = 90^\circ$ ). The line $AP$ intersects the circumcircle of $ABC$ at point $Q$ . Let $L$ be the midpoint of $PB$ . Prove that $QL$ is tangent to a fixed circle independent of the choice of point $P$ .
|
Let $M$ be midpoint of $AB$ . we claim that our circle is a circle with center $M$ which touches $BC$ . Note that we need to prove $M$ lies on external angle bisector of $\angle PLQ$ . from Thales Theorem we have $ML || AQ$ so $\angle MLP = \angle APB = \angle LPQ$ so we need to prove $PL = QL$ . Note that $PL = LB$ so we need to prove $\angle PQB = \angle 90$ which is true.
|
[
"We claim the desired circle is the circle whose center is the midpoint $O$ of $AB$ and that touches $BC$ .\n\nWe use complex numbers with $a=-1,b=1$ . Then by intersection formula, $p=\\frac{2cq+q-c}{c+q},l=\\frac 12(b+p)=\\frac{q(c+1)}{c+q}$ . Now the reflection of $O$ over $QL$ is given by $\\frac{\\bar lq-l\\bar q}{l-q}=\\frac{-(c+1)}{q}$ . The distance of this point to $O$ is equal to $|c+1|$ and thus fixed."
] |
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"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800880.json"
}
|
Paul can write polynomial $(x+1)^n$ , expand and simplify it, and after that change every coefficient by its reciprocal. For example if $n=3$ Paul gets $(x+1)^3=x^3+3x^2+3x+1$ and then $x^3+\frac13x^2+\frac13x+1$ . Prove that Paul can choose $n$ for which the sum of Paul’s polynomial coefficients is less than $2.022$ .
|
As stated in the title of the problem we need to show that we can choose $n$ , such that: $$ \sum_{i=0}^{n}\frac{1}{{n \choose i}} < 2.022 $$ Notice that for each $i$ , $1 < i < n$ , we have that: $$ \frac{1}{{n \choose 2}} \geq \frac{1}{{n \choose i}} $$ thus we have that: $$ LHS = 2+\sum_{i=1}^{n-1}\frac{1}{{n \choose i}} \leq 2+(n-2)\frac{2}{n(n-1)} < 2.022 $$ which implies that we must show that, there exists an $n$ such that: $$ \frac{n-2}{n(n-1)} < 0.011 $$ but notice that we have that: $$ \lim_{n \rightarrow \infty} \frac{n-2}{n(n-1)} = 0 $$ Which means that as $n$ gets bigger the fraction gets smaller and smaller, thus there exists an $n$ so that the sum is less than $2.022$
|
[
"Related to USA TST 2000/4 https://artofproblemsolving.com/community/c6h61959p371496. This result can be obtained by using the recursion in post #10 in that thread (we have to prove that the sum of the reciprocals of the coefficients has lim=2).",
"We have to prove that \n\\[\\mathop {\\lim }\\limits_{n \\to \\infty } {a_n} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\sum\\limits_{k = 0}^n {\\frac{1}{{n \\choose k}}} = 2\\]\n\nFirst, considering \n\\[{T_n} = \\frac{1}{{\\left( \\begin{array}{c}\nn\n0\n\\end{array} \\right)}} + \\frac{2}{{\\left( \\begin{array}{c}\nn\n1\n\\end{array} \\right)}} + \\frac{3}{{\\left( \\begin{array}{c}\nn\n2\n\\end{array} \\right)}} + ... + \\frac{{n + 1}}{{\\left( \\begin{array}{c}\nn\nn\n\\end{array} \\right)}} = \\sum\\limits_{k = 0}^n {\\frac{{k + 1}}{{\\left( \\begin{array}{c}\nn\nk\n\\end{array} \\right)}}} \\text{ (1)}\\]\nSince $\\[\\left( \\begin{array}{c}\nn\nk\n\\end{array} \\right) = \\left( \\begin{array}{c}\nn\nn - k\n\\end{array} \\right)\\]$ , we have\n\\[{T_n} = \\frac{{n + 1}}{{\\left( \\begin{array}{c}\nn\n0\n\\end{array} \\right)}} + \\frac{n}{{\\left( \\begin{array}{c}\nn\n1\n\\end{array} \\right)}} + \\frac{{n - 1}}{{\\left( \\begin{array}{c}\nn\n2\n\\end{array} \\right)}} + ... + \\frac{1}{{\\left( \\begin{array}{c}\nn\nn\n\\end{array} \\right)}} \\text{ (2)}\\]\n\nAdd (1) and (2) we have $$ 2T_n=(n+2)a_n $$ Also notice that $\\[\\frac{{k + 1}}{{\\left( \\begin{array}{c}\nn\nk\n\\end{array} \\right)}} = \\frac{{n + 1}}{{\\left( \\begin{array}{c}\nn + 1\nk + 1\n\\end{array} \\right)}}\\]$ , we have $$ T_n=(n+1)(a_{n+1}-1) $$ Thus we have a recursion here $$ a_n=\\frac{n+1}{2n}a_{n-1}+1 $$ From the recursion, it's easy to see that $$ a_{n+1}<a_n \\Leftrightarrow a_n>2+\\frac{2}{n} \\text{ , which is always right by induction} $$ Thus $(a_n)$ is strictly decreasing, and is lower bounded, which means $(a_n)$ converges\nFrom the recursion, we conclude that \\[\\mathop {\\lim }\\limits_{n \\to \\infty } {a_n} = 2\\]",
"It suffices to prove $\\sum_{i=0}^n \\binom{n}{i}^{-1} < 2$ . $023$ for some $n$ . Equivalently, we want $\\sum_{i=1}^{n-1} \\binom{n}{i}^{-1} < 0$ . $023$ . Each of the binomial coefficients except the two ending ones (these are $n-4$ in number) is at least as large as $\\binom{n}{2} = \\frac{n(n-1)}{2}$ , so the main sum does not exceed $\\frac{2}{n} + \\frac{2(n-4)}{n(n-1)} < \\frac{4}{n}$ . The latter can be arbitrarily small for sufficiently large $n$ (in particular, less than $0$ . $02 < 0$ . $023$ for $n \\geq 200$ ).",
"The coefficients of Paul's polynomial are $\\frac{1}{\\binom{n}{i}}$ . Therefore, we need to prove that $$ \\frac{\\sum_{i=0}^{n} i!(n-i)!}{n!} < 2.022 $$ for some $n$ . We will bound it as follows $$ \\frac{\\sum_{i=0}^{n} i!(n-i)!}{n!} \\leq \\frac{2n! + 2(n-1)! + 2!(n-2)! (n-3)}{n!} < 2 + \\frac{4}{n} $$ Take $n=1000$ to get the desired."
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Given a rectangular table with 2 rows and 100 columns. Dima fills the cells of the first row with numbers 1, 2 or 3. Prove that Alex can fill the cells of the second row with numbers 1, 2, 3 in such a way that the numbers in each column are different and the sum of the numbers in the second row equals 200.
|
<blockquote>Given a rectangular table with 2 rows and 100 columns. Dima fills the cells of the first row with numbers 1, 2 or 3. Prove that Alex can fill the cells of the second row with numbers 1, 2, 3 in such a way that the numbers in each column are different and the sum of the numbers in the second row equals 200.</blockquote> $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $x$ be the number of $1's$ , $y$ the number of $3's$ and $z$ the number of $2's$ $\Rightarrow x+y+z=100...(I)$ Let $S$ be the sum of the numbers in the second row
Let $2_i$ the $i$ -th $2's$ Let " $a\to b$ " the transformation that Alex makes, that is, if Dima places $a$ above, Alex places $b$ below $\color{red}\boxed{\textbf{If the number of 2's is even}}$ $\color{red}\rule{24cm}{0.3pt}$ Numbers of $2's=z=2k$ Alex's strategy is: $$ 2_{2i-1}\to 1, \forall 1\le i\le k $$ $$ 2_{2i}\to 3, \forall 1\le i\le k $$ $$ 1\to 2 $$ $$ 3\to 2 $$ $$ \Rightarrow S=(1)(k)+(3)(k)+(2)(x)+(2)(y) $$ $$ \Rightarrow S=4k+2x+2y $$ $$ \Rightarrow S=2x+2y+2z $$ By $(I):$ $$ \Rightarrow S=200_\blacksquare $$ $\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If the number of 2's is odd}}$ $\color{red}\rule{24cm}{0.3pt}$ Numbers of $2's=z=2k+1$ If there is at least one $1$ Alex's strategy is: $$ 2_{2i-1}\to 1, \forall 1\le i\le k $$ $$ 2_{2i}\to 3, \forall 1\le i\le k $$ $$ 2_{2k+1}\to 1 $$ $$ 1\to 2, \text{ except the last one that will become 3} $$ $$ 3\to 2 $$ $$ \Rightarrow S=(1)(k)+(3)(k)+1+(2)(x-1)+3+(2)(y) $$ $$ \Rightarrow S=4k+2+2x+2y $$ $$ \Rightarrow S=2x+2y+2z $$ By $(I):$ $$ \Rightarrow S=200_\blacksquare $$ If there is at least one $3$ Alex's strategy is: $$ 2_{2i-1}\to 1, \forall 1\le i\le k $$ $$ 2_{2i}\to 3, \forall 1\le i\le k $$ $$ 2_{2k+1}\to 3 $$ $$ 1\to 2 $$ $$ 3\to 2, \text{ except the last one that will become 1} $$ $$ \Rightarrow S=(1)(k)+(3)(k)+3+(2)(x)+1+(2)(y-1) $$ $$ \Rightarrow S=4k+2+2x+2y $$ $$ \Rightarrow S=2x+2y+2z $$ By $(I):$ $$ \Rightarrow S=200_\blacksquare $$ $\color{red}\rule{24cm}{0.3pt}$ $$ \Rightarrow \boxed{\textbf{Alex can achieve his goal}}_\blacksquare $$ $\color{blue}\rule{24cm}{0.3pt}$
|
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"answer_score": 1268,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800889.json"
}
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Let $\omega$ is tangent to the sides of an acute angle with vertex $A$ at points $B$ and $C$ . Let $D$ be an arbitrary point onn the major arc $BC$ of the circle $\omega$ . Points $E$ and $F$ are chosen inside the angle $DAC$ so that quadrilaterals $ABDF$ and $ACED$ are inscribed and the points $A,E,F$ lie on the same straight line. Prove that lines $BE$ and $CF$ intersectat $\omega$ .
|
Equivalent, but without the use of spiral similarity basic trickery (I know it is easier to figure out with it, but just for the sake of inexperienced people).
We have $\angle DFE = 180^{\circ} - \angle AFD = \angle ABD = \angle BCD$ from the cyclic $ABDF$ and the tangency around $\omega$ ,
similarly $\angle AED = \angle ACD = \angle DBC$ . Hence $\triangle DBC \sim \triangle DEF$ . Hence $\frac{BD}{DC} = \frac{DE}{DF}$ , i.e. $\frac{BD}{DE} = \frac{DC}{DF}$ and also $\angle BDC = \angle EDF$ , i.e. $\angle BDE = \angle CDF$ . Hence $\triangle BDE \sim \triangle CDF$ , so $\angle DBE = \angle DCF$ , thus after extending $BE$ and $CF$ to intersect $\omega$ , the respective arcs would be equal and so the intersection points would coincide, done.
|
[
"Again easy P4 (Like lsast year's P4). But anyway, it's better than last year's P4.\nInvert around circle $(A,AB)$ and assume $X'$ is image of $X$ under this inversion. We get $D'=AD\\cap \\omega, E'=AE\\cap CD', F'=AF\\cap BD'$ and let $T=BE\\cap CF$ .\nSince $AE\\cdot AE'=AC^2$ we get $(CEE')$ tangents to $AC \\implies \\angle E'EC=\\angle E'CA=\\angle D'BC \\implies BCEF'$ is cyclic. Similarly $BCFE'$ is cyclic. So $\\angle TEF=\\angle BEF'=\\angle BCF'$ and $\\angle F'FC=\\angle F'CA \\implies \\angle BTC=\\angle TEF + \\angle EFT = \\angle BCF'+\\angle F'CA=\\angle BCA \\implies T\\in \\omega$ . So we are done!",
"Well, almost trivial. Note that there is a spiral similarity taking $DEF$ to $DBC$ (more formally, we have that triangles $DBE$ and $DCF$ are also similar, so if $BE$ intersects $CF$ at $P$ , then $BCDP$ is cyclic).",
"Easy for P4... $\\angle DBC = \\angle DCF = \\angle 180 - \\angle ACD = \\angle 180 - \\angle AED = \\angle DEF$ and $\\angle BCD = \\angle 180 - \\angle ABD = \\angle EFD$ so $DBC$ and $DEF$ have spiral similarity. Let $BE$ meet $CF$ at $X$ . It's well known that $DEXF$ and $DBCX$ are cyclic so $X$ lies on $\\omega$ .",
"Plot twist: only angle chasing suffices! (Credits to @africanboy)\n\nWe have $\\angle DFE = 180^{\\circ} - \\angle AFD = \\angle ABD = \\angle BCD$ from the cyclic $ABDF$ and the tangency around $\\omega$ , similarly $\\angle AED = \\angle ACD = \\angle DBC$ and hence $\\angle BDC = \\angle EDF$ . Now define $T = BE \\cap \\omega$ . Then $\\angle DTE = \\angle BTD = \\angle BCD = \\angle DFE$ , hence $DFTE$ is cyclic. From here $\\angle ETF + \\angle CTE = (180^{\\circ} - \\angle FDE) + \\angle BTC = (180^{\\circ} - \\angle BDC) + \\angle BTC = 180^{\\circ}$ , so $CE$ passes through $T$ and we are done."
] |
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"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800890.json"
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Prove that infinitely many positive integers can be represented as $(a-1)/b + (b-1)/c + (c-1)/a$ , where $a$ , $b$ and $c$ are pairwise distinct positive integers greater than 1.
|
$a=(2k+1)×k$ $b=(2k+1)×k-1$ $c=2k+1$ $(a-1)/b + (b-1)/c + (c-1)/a=k+1$ $k$ is a positive integer greater than 1
|
[] |
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"2022 Contests",
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"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800891.json"
}
|
Do there exist 100 points on the plane such that the pairwise distances between them are pairwise distinct consecutive integer numbers larger than 2022?
|
[
"Compare with [this problem](\"https://artofproblemsolving.com/community/c6h2800872_pairwise_distances_of_lattice_points_are_consecutive_numbers\")"
] |
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"2022 Contests",
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800893.json"
}
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|
Let $ABC$ be an acute triangle. Let $P$ be a point on the circle $(ABC)$ , and $Q$ be a point on the segment $AC$ such that $AP\perp BC$ and $BQ\perp AC$ . Lot $O$ be the circumcenter of triangle $APQ$ . Find the angle $OBC$ .
|
Let $H$ be orthocenter of $ABC$ , $S$ be foot of $A$ on $BC$ and $QB$ meet $APQ$ at $K$ . we have $\angle APK = \angle 90 = \angle ASB \implies BS || KP$ . It's well known that $P$ is reflection of $H$ across $BC$ so $HB = BK$ so $B$ lies on perpendicular bisector of $KP$ so $OB \perp KP || BC \implies OB \perp BC \implies \angle OBC = \angle 90$ .
|
[
"Let $H$ be orthocenter of $\\triangle ABC$ and $D$ be foot of $A \\text{-altitude}$ in $\\triangle ABC$ . Define $E=BQ\\cap (APQ)$ , where $E\\ne Q$ . $\\angle EQA=90 \\implies O\\in EA$ . $\\angle APE=\\angle AQE=90\\implies BD||EP$ and since $DH=DP$ we get $BH=BE$ . So $OB$ is $E\\text{-midline}$ in $\\triangle AEH \\implies OB||AH\\perp BC \\implies OB\\perp BC \\implies \\angle OBC=90.$ ",
"Sniped, I was just writing that. I got motivated for this construction (of $E$ ) by taking the reflection of $H$ wrt $B$ and doing PoP."
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"2022 Contests",
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800894.json"
}
|
There are $n > 2022$ cities in the country. Some pairs of cities are connected with straight two-ways airlines. Call the set of the cities {\it unlucky}, if it is impossible to color the airlines between them in two colors without monochromatic triangle (i.e. three cities $A$ , $B$ , $C$ with the airlines $AB$ , $AC$ and $BC$ of the same color).
The set containing all the cities is unlucky. Is there always an unlucky set containing exactly 2022 cities?
|
The answer is negative. We will construct a graph with 2023 vertices such that the whole graph is unlucky but every 2020 subgraph is not so. Note that if we add some empty vertices to the graph a counterexample for all $n$ is built.
<span style="color:#f00">**Observation 1.** </span> If we color $K_5$ properly every vertex has 2 edges of each color; otherwise 3 edges have same color and the triangle forming by their uncommon vertices cannot be colored properly.
<span style="color:#f00">**Observation 2.** </span> $K_5$ is colored quite uniquely. <details><summary>Proof</summary>Take a vertex $v_1$ and let it connect with color blue to $v_2, v_3$ and color red to $v_4, v_5$ . $v_4v_5$ must be blue and $v_2v_3$ must be red. WLOG assume that $v_2v_4$ is red, then $v_3v_4$ must be blue, so $v_3v_5$ must be red and $v_2v_5$ must be blue.</details>
<span style="color:#f00">**Observation 3.** </span> If we connect one vertex to four vertices of a properly colored $K_5$ , the edges of that vertex must be colored exactly like the missing vertex of $K_5$ . <details><summary>Proof</summary>This is because every one of four vertices due to observation 1 misses an specific color to be complete and hence the edges of any vertex connecting to these four has a unique way of coloring.</details>
It is now enough to make a cycle of $K_5$ s. Put 2023 vertices on a circle and connect each vertex to all 4 nearest points in each direction. Assume for the sake of contradiction that the said graph could be properly colored. Then consider 5 consecutive points on the circle. They form a $K_5$ . Now the next 5 points on the circle form a $K_5$ that only differs in one vertex and due to observation 3 must be colored with the same pattern. We can conclude that the edges that connect two consecutive points is colored in a repeating pattern of length 5. But 2023 is not divisible by 5 and hence such pattern cannot exist. The contradiction means the said graph with $2023$ vertices cannot be properly colored. Meanwhile every subgraph of size 2022 can be colored properly. We only need to color a $K_5$ and just as described color other $K_5$ s one by one in each direction.
<details><summary>Motivation</summary>Sticking to $K_5$ came to mind after noting that $R(3,3) = 6$ and so every $K_6$ cannot be properly colored, this suggest that a weak $K_6$ could be bound to some restriction. After noting that $K_5$ is restricted, working with a cycle to abuse the restriction is quite natural.</details>
|
[
"bump this gem"
] |
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"2022 Contests",
"2022 Caucasus Mathematical Olympiad"
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{
"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Caucasus Mathematical Olympiad/2800897.json"
}
|
Let $a$ and $b$ be two positive real numbers, and $AB$ a segment of length $a$ on a plane. Let $C,D$ be two variable points on the plane such that $ABCD$ is a non-degenerate convex quadrilateral with $BC=CD=b$ and $DA=a$ . It is easy to see that there is a circle tangent to all four sides of the quadrilateral $ABCD$ .
Find the precise locus of the point $I$ .
|
@2 above since $ABCD$ is not degenrate, not all points on this circle work. My solution is same with at 2 above, but I find the locus as arc of this circle.
Let circle with center $B$ and radius $b$ be $\Gamma$ and let Homothety with center $A$ and radius $\frac{a}{a+b}$ maps $\Gamma$ to $\omega$ . Then it's obvious that $B$ lies on $\omega$ . Let $AX_1$ and $AX_2$ tangent to $\omega$ at $X_1,X_2$ , rescpectively. Then locus of $I$ is $\text{arc}$ $X_1BX_2$ of $\omega $ , except points $X_1,X_2$ and $B$ .
|
[
"Is a,b fixed?",
"<blockquote>Is a,b fixed?</blockquote>\n\nYes.",
"Since $ABCD$ is a deltoid, $I$ lies on its axis of symmetry $AC$ . Therefore, by bisector theorem on $BCA$ , we have $AB:BC=AI:IC$ , which implies $AI=\\frac{a}{a+b}AC$ . Since $C$ lies on a fixed circle of radius $b$ and center $B$ , it follows that the locus of $I$ is this circle scaled with center $A$ and factor $\\frac{a}{a+b}$ (which in fact passws trough $B$ ).",
"Well not exactly. First of all the problem says “non degenerate convex quadrilateral” but not all points on the circle satisfy this property.",
"<blockquote>Well not exactly. First of all the problem says “non degenerate convex quadrilateral” but not all points on the circle satisfy this property.</blockquote>\n\nYes I omitted the fact that the two diametrically opposite on segment $AB$ don't work",
"By the way,I am curious about why the title is 2022 National Olympiad.Isn't that 2021?",
"Clearly $I$ lies on $AC$ . Also, from the angle bisector theorem, $\\frac{AI}{AC} = \\frac{a}{a+b}$ , so its obvious that the locus is the result of taking the locus of $C$ and performing a homothety $H\\left(A, \\frac{a}{a+b}\\right)$ on it. As the locus of $C$ is clearly half of a circular arc, we are done.",
"Not a single complete solution...\n\nAfter obtaining $I$ is on that circular arc, as not all points must work, we need to still give some discussion.\n\nThen, we have to split into multiple cases, and I summarize the result:\nIf $a=b$ , the entire circle except two diametrically opposite points is the locus.\nIf $a<b$ , we can see there are two arcs with central angle $\\cos^{-1} \\frac ab$ If $a>b$ , we can see there are two arcs with central angle $\\cos^{-1} \\frac ba$ I hope somebody can provide a good solution: without a complete discussion, one can only obtain 15 marks, which the committee correctly noted was too high, but this mark was given due to the fact that locus is not often tested.\n\nIndeed, the average mark is 18.6, for silver is >19, but still not 20 even for the training camp.",
"<blockquote>By the way,I am curious about why the title is 2022 National Olympiad.Isn't that 2021?</blockquote>\n\nHow can aops change the year, It's been 2025\n",
"This result generalizes to any tangential quadrilateral. This is shown in the paper \"On a Circle Containing the Incenters of Tangential Quadrilaterals\", by Albrecht Hess.\n\nLinks to paper:\n[https://web.archive.org/web/20141214205151/http://forumgeom.fau.edu/FG2014volume14/FG201437.pdf](https://web.archive.org/web/20141214205151/http://forumgeom.fau.edu/FG2014volume14/FG201437.pdf)\n[https://cjhb.site/Files.php/Books/Off%20Topic/FG201437.pdf](https://cjhb.site/Files.php/Books/Off%20Topic/FG201437.pdf)"
] |
[
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"2022 Contests",
"2022 China National Olympiad"
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{
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2022 Contests/2022 China National Olympiad/2742283.json"
}
|
Find the largest real number $\lambda$ with the following property: for any positive real numbers $p,q,r,s$ there exists a complex number $z=a+bi$ ( $a,b\in \mathbb{R})$ such that $$ |b|\ge \lambda |a| \quad \text{and} \quad (pz^3+2qz^2+2rz+s) \cdot (qz^3+2pz^2+2sz+r) =0. $$
|
Take $p=q=r=s=1$ ,we got $\lambda \leq \sqrt3$ To prove $\lambda = \sqrt 3$ ,we suppose the roots of the two equations are $x_1,x_2,x_3;y_1,y_2,y_3$ we have(Viete) $\sum x_i \sum {y_i}=\sum\frac{1}{x_i}\sum\frac{1}{y_i}=4$ but one can easily prove that $\sum x_i \sum \frac{1}{x_i},\sum y_i \sum \frac{1}{y_i}>4$ when $|b|<\sqrt3 |a|$ <details><summary>reason</summary>When $x_i$ are reals this is just Cauchy (notice $x_i<0$ ),otherwise suppose $x_1$ is real we have $(x_2+x_3)(\frac{1}{x_2}+\frac{1}{x_3})=\frac{4x^2}{x^2+y^2}>1(x_{2,3}=x\pm yi)$ ,use Cauchy again.
(Notice $x<0$ ,otherwise $2x+x_1<0,x^2+y^2+2xx_1>0(Viete) \rightarrow 3x^2 \leq y^2$</details> $\square$
|
[
"Not a difficult problem! The answer is sqrt(3).\nTo prove it we can simply assume the contradictory. Using the fact that these two equality have a solution in R, assume them out as x1, x2 to get another two quadratic ones, like x^2+A1x+B1. Then by the assumption, A1^2>B1. Use p/q and q/p,r/s and s/r to establish two equations, and the problem will be down by Cauchy. The inequality holds when p=q=r=s=1.",
"<details><summary>复仇 (Revenge)</summary>The answer is $\\lambda=\\sqrt{3}$ , obtained at $p=q=r=s=1$ . It remains to show it works.\n\nSuppose $(x^2+ax+b)(px+t)=px^3+2qx^2+2rx+s$ $(x^2+cx+d)(qx+v)=qx^3+2px^2+2sx+r$ Here all variables except for $x$ are real. All other than $a,c$ are guaranteed positive.\n\nWTS: Either $d\\ge c^2$ or $b\\ge a^2$ .\n\nFirst, let's write some equations. $ap+t=2q, at+bp=2r, bt=s$ $cq+v=2p, cv+dq=2s, dv=r$ Step 1: Eliminating $v,t,s,r,p$ .\n\nWe can eliminate $r$ with the statement $2dv=at+bp$ Similarly, $2bt=cv+dq$ Furthermore, we can assume $p=1$ , we get $$ 2d(2-cq)=a(2q-a)+b $$ $$ 2b(2q-a)=c(2-cq)+d $$ Solving yields $$ d=\\frac{c(2-cq)+2a(2q-a)^2}{4(2-cq)(2q-a)-1} $$ $$ b=\\frac{2c(2-cq)^2+a(2q-a)}{4(2-cq)(2q-a)-1} $$ Assume for contradiction $d<c^2$ and $b<a^2$ .\n\nWe know $d,b>0$ and the numerator is positive. Therefore, it is safe to consider the following inequalities: $$ c^2>\\frac{c(2-cq)+2a(2q-a)^2}{4(2-cq)(2q-a)-1} $$ $$ a^2>\\frac{2c(2-cq)^2+a(2q-a)}{4(2-cq)(2q-a)-1} $$ WLOG, $q\\le 1$ (this is equivalent to $q\\le p$ unscaled). Expanding and cancelling, $$ 4c^2(2-cq)(2q-a)-c^2>2c-c^2q+2a(2q-a)^2 $$ $$ 4c^2(2-cq)(2q-a)>2c+2a(2q-a)^2 $$ $$ 4a^2(2-cq)(2q-a)-a^2>2c(2-cq)^2+2a-a^2 $$ $$ 2c^2(2-cq)(2q-a)>c+a(2q-a)^2 $$ $$ 2a^2(2-cq)(2q-a)>c(2-cq)^2+a $$ Taking their product yields $$ 4a^2c^2(2-cq)^2(2q-a)^2 > c^2(2-cq)^2+ca+ca(2-cq)^2(2q-a)^2+a^2(2q-a)^2 $$ Observe the geometric mean of RHS is $c(2-cq)a(2q-a)$ So $a^2c^2(2-cq)^2(2q-a)^2 > c(2-cq)a(2q-a)$ Since all terms are positive, $ c(2-cq)a(2q-a)>1$ $ qc(2-cq)a(2q-a)>q$ Note $qc(2-cq)\\le 1, a(2q-2)\\le q^2$ , so $q^2>q$ , contradicting our assumption that $q<1$ . \n\nI can't believe this problem is just bash.</details>",
"<blockquote>Not a difficult problem! The answer is sqrt(3).\nTo prove it we can simply assume the contradictory. Using the fact that these two equality have a solution in R, assume them out as x1, x2 to get another two quadratic ones, like x^2+A1x+B1. Then by the assumption, A1^2>B1. Use p/q and q/p,r/s and s/r to establish two equations, and the problem will be down by Cauchy. The inequality holds when p=q=r=s=1.</blockquote>\n\nI'm sorry but could you please explain it more clearly i couldn't catch up with you that well :-(",
"I think my solution is along the lines of what he is doing, except for I used AM-GM to finish.",
"A post for storage~\n\n(1) When $p=q=r=s$ , the equation turns to $x^3+2x^2+2x+1=0$ , which means $x=-1$ or $x=\\frac{-1\\pm \\sqrt{3}i}{2}$ , so $\\lambda \\leq \\sqrt{3}$ .\n\n(2) Suppose that when $\\lambda = \\sqrt{3}$ the result isn’t true.\nLet $px^3+2qx^2+2sx+r=0$ has roots $a_1+b_1i, a_1-b_1i, r_1$ and the other equation has $a_2+b_2i, a_2-b_2i, r_2$ as roots (2 complex roots and 1 real root), so $|b_1|<\\sqrt{3} |a_1|$ and $|b_2|<\\sqrt{3} |a_2|$ .\nBy Vieta,\n\\begin{align*}\np:-2q:2s:-r &= 1:2a_1+r_1:2a_1 r_1+{a_1}^2+{b_1}^2:{a_1}^2 r_1+{b_1}^2 r_1 \nq:-2p:2r:-s &= 1:2a_2+r_2:2a_2 r_2+{a_2}^2+{b_2}^2:{a_2}^2 r_2+{b_2}^2 r_2,\n\\end{align*}\nso\n\\begin{align*}\n\\left(\\frac{2a_1}{{a_1}^2+{b_1}^2}+\\frac{1}{r_1}\\right) \\left(\\frac{2a_2}{{a_2}^2+{b_2}^2}+\\frac{1}{r_2}\\right) &= 4 \\text{ }(1)\n(2a_1+r_1)(2a_2+r_2) &= 4 \\text{ }(2)\nr_1, r_2 &< 0 \\text{ }(3).\n\\end{align*}\n(just multiply out the ratios)****** When $a_1 \\geq 0$ , from $0 < 2a_1 r_1+{a_1}^2+{b_1}^2$ and $2a_1+r_1<0$ we get ${b_1}^2 > 3{a_1}^2$ . The same goes for $a_2$ .****** When $a_1, a_2 < 0$ , the first one gives \n\\begin{align*}\n4 &> \\left( \\frac{2a_1}{{a_1}^2+3{a_1}^2}+\\frac{1}{r_1} \\right) \\left( \\frac{2a_2}{{a_2}^2+3{a_2}^2}+\\frac{1}{r_2} \\right) \n&= \\left(\\frac{1}{2a_1}+\\frac{1}{r_1}\\right) \\left(\\frac{1}{2a_2}+\\frac{1}{r_2}\\right),\n\\end{align*}\nso by (2) and Cauchy there’s a contradiction!",
"<details><summary>my solution</summary>We claim that the answer is $\\sqrt 3$ . Not hard to see that $\\lambda = \\sqrt 3$ works because $p=q=r=s=1$ gives us that the roots of the polynomial are $-1, \\frac{1 \\pm \\sqrt{3}}{2}$ . Assume the contrary, let $\\exists ~p,q,r,s \\in \\mathbb R^+$ such that all roots $z$ of $(px^3+2qx^2+2rx+s) \\cdot (qx^3+2px^2+2sx+r) =0$ have $\\frac{Im(z)}{Re(z)} < \\sqrt{3}$ .\n\nSince $px^3+2qx^2+2rx+s$ must have a real root, it can be factorized to $(x+k_1)(c_2x^2+c_1x+c_0)$ for $r_1,c_0,c_1,c_2\\in \\mathbb R$ .[rule]**<span style=\"color:#00f\">Claim:</span>** $c_1^2 > c_0c_2$ and $k_1,c_0,c_1,c_2 \\in \\mathbb R^+$ \n<span style=\"color:#f00\">Proof:</span> suppose that $c_1^2 \\leq c_0c_2$ , we get that the roots of the polynomial are $-k_1$ and $(-c_1 \\pm \\sqrt{c_1^2 - 4c_0c_2})/(2c_2)$ (which are definitely not real). Therefore, $$ \\frac{\\sqrt{4c_0c_2 - c_1^2}}{|c_1|} \\geq \\frac{\\sqrt{3c_1^2}}{|c_1|} = \\sqrt{3}. $$ Contradict to what we assume at first.\nFor the second part, it's obvious that $k_1 \\in \\mathbb R^+$ since the real root is clearly negative. By checking the coefficient, $c_0,c_2 \\in \\mathbb R^+$ . If $c_1 \\leq 0$ , we have $0 \\leq 2q = c_2 + rc_1, 0 \\leq 2r = c_1 + rc_0$ . Hence, $c_2 \\leq -rc_1, c_1 \\leq -rc_0$ . Multiplying those inequality leads to the contradiction. Hence, $c_0 \\in \\mathbb R^+$ [rule]\nBy the claim and multiplying $(x+k_1)(c_2x^2+c_1x+c_0)$ back, there exist $\\epsilon_1,\\epsilon_2 \\in \\mathbb R^+$ and $m_1,n_1 \\in \\mathbb R^+_0$ such that $px^3 +2qx^2 + 2rx + s = (px^3 + (\\sqrt{pm_1} + \\epsilon_1)x^2 + m_1x) + (n_1x^2 + (\\sqrt{n_1s} + \\epsilon)x + s)$ and $m_1n_1 = ps$ . Hence, \\begin{align*} 2q &>\\sqrt{pm_1} + n_1 2r &> \\sqrt{sn_1} + m_1 \\implies 4qr &> \\sqrt{psm_1n_1} + \\sqrt{pm_1^3} + \\sqrt{sn_1^3} + m_1n_1 & \\overset{\\text{AM-GM}}{\\geq} ps + 2ps + ps = 4ps.\\end{align*} Hence, $qr > ps$ .\n\nThe roots of $qx^3 + 2px^2 + 2sx + r$ also similarly imply that $ps > qr$ . Contradiction.</details>",
"Thank you,Mr.He."
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Find all positive integers $a$ such that there exists a set $X$ of $6$ integers satisfying the following conditions: for every $k=1,2,\ldots ,36$ there exist $x,y\in X$ such that $ax+y-k$ is divisible by $37$ .
|
Clearly $37 \nmid a$ . We will take the elements of $X$ as residues $\pmod{37}$ . Clearly all of them must be nonzero and distinct.
So if $\omega$ is any primitive $37$ th root of unity, then it is necessary and sufficient for $$ \left(\sum_{n\in X} \omega^{an}\right)\left(\sum_{n\in X} \omega^{n}\right)=\left(\sum_{n\in X} \omega^{a^2n}\right)\left(\sum_{n\in X} \omega^{an}\right) =\sum_{n=1}^{36} \omega^n = -1. $$ Consider the polynomial $\sum_{n \in X} x^{a^2n}-\sum_{n\in X} x^{n}$ . For each term $x^m$ in the polynomial, replace it with $x^{m\text{'s remainder mod 37}}$ . Then this polynomial has degree less than $37,$ but is divisible by $x^{37}-1.$ So it's the zero polynomial, and $a^2X \equiv X \pmod{37}$ . Thus, $ord_{37}(a^2) \mid 6.$ If $a^3 \equiv -1,$ $$ \left(\sum_{n\in X} \omega^{an}\right)\left(\sum_{n\in X} \omega^{n}\right)=\left(\sum_{n\in X} \omega^{-\frac{n}{a^2}}\right)\left(\sum_{n\in X} \omega^{n}\right)=\left(\sum_{n\in X} \omega^{-n}\right)\left(\sum_{n\in X} \omega^{n}\right) $$ which is the magnitude of $\sum_{n\in X} \omega^{n}$ squared, which cannot be $-1.$ If $a^3 \equiv 1,$ then $aX \equiv X \pmod{37}$ . So $$ \left(\sum_{n\in X} \omega^{n}\right)^2=\left(\sum_{n\in X} \omega^{an}\right)\left(\sum_{n\in X} \omega^{n}\right)=\sum_{n=1}^{36} \omega^n $$
for any primitive $37$ th root of unity $\omega$ . Take the polynomial $\left(\sum_{n\in X} x^{n}\right)^2-\sum_{n=1}^{36} x^n,$ and for each term $x^m$ in the polynomial, replace it with $x^{m\text{'s remainder mod 37}}$ . Then the result has degree less than $37$ and is divisible by $\frac{x^{37}-1}{x-1}$ and $x,$ so it is a zero polynomial, but it's not hard to see that this is impossible.
So $ord_{37}(a^2) = 2$ or $6.$ In the former case, $\boxed{a \equiv 6 \pmod{37}}$ works if $X = \{16,17,18,19,20,21 \}$ . Note that in this case, $$ \left(\sum_{n\in X} \omega^{6n}\right)\left(\sum_{n\in X} \omega^{n}\right) = \omega^{16} \cdot \frac{\omega^{6}-1}{\omega-1} \cdot \omega^{16 \cdot 6} \cdot \frac{\omega^{36}-1}{\omega^{6}-1} = \omega \cdot \frac{\omega^{36}-1}{\omega-1}=-1 $$ for any primitive $37$ th root of unity $\omega$ . $\boxed{a \equiv 31 \pmod{37}}$ works as well by an identical argument.
In the latter case, $X=\{x,a^2x,a^4x,\cdots,a^{10}x\}$ for some nonzero residue $x$ . We can assume WLOG $x=1,$ then this forces $X=\{1,27,26,36,10,11\}, aX=\{8,31,23,29,6,14\}.$ Note $26+6\equiv 1+31 \pmod{37}$ which is contradiction. $\square$
|
[
"<details><summary>Solution</summary>Throughout the solution, a and elements of X are in $\\mathbb{Z}_{37}$ .\n\nThe answer is $a\\equiv \\pm 6$ only. \n\nConstruction: $X=\\{16,17,18,19,20,21\\}$ Proof of optimality: Let $\\omega$ be any primitive $37$ th root of unity. Then $\\left(\\sum_{t\\in X} \\omega^{at}\\right)\\left(\\sum_{t\\in X} \\omega^{t}\\right)=-1$ is sufficient and necessary.\n\nIn particular, $\\left(\\sum_{t\\in X} \\omega^{a^2t}\\right)\\left(\\sum_{t\\in X} \\omega^{at}\\right)=-1$ , so $\\sum_{t\\in X} \\omega^{a^2t}-\\omega^t=0$ . This polynomial is divisible by $x^{37}-1$ . \n\nI claim this implies $a^2X=X$ in $\\mathbb{Z}_{37}$ . Sort the remainders of $a^2X, X$ into increasing order, say $0\\le y_1<y_2<\\cdots<y_6\\le 36$ for $a^2X$ and $0\\le x_1<x_2<\\cdots<x_6 \\le 36$ . We can see $x_6=y_6, x_5=y_5, x_4=y_4$ , and so on, implying $a^2X=X$ . This implies if $t\\in X, a^2t\\in a^2X$ , so $a^2t\\in X$ .\n\nIf we consider a permutation $\\pi: [6]\\rightarrow [6]$ such that $x_j\\cdot a^2=x_{\\pi(j)}$ . Letting \nlet $d=ord_{37}(a^2)$ , the permutation is a disjoint union of cycles of length $d$ . Therefore, $d\\in \\{1,2,3,6\\}$ .\n\n\nCase 1: $d=1$ . If $a\\equiv 1(\\bmod\\; 37)$ then $(x+y)_{x,y\\in X}$ cannot be pairwise distinct.\n\nIf $a\\equiv -1(\\bmod\\; 37)$ then $-1=\\left(\\sum_{t\\in X} \\omega^{at}\\right)\\left(\\sum_{t\\in X} \\omega^{t}\\right)=|\\sum_{t\\in X} \\omega^{t}|^2$ , contradiction.\n\nCase 2: $d=3$ . If $a^3\\equiv -1(\\bmod\\; 37)$ , we get $\\left(\\sum_{t\\in X} \\omega^{at}\\right)\\left(\\sum_{t\\in X} \\omega^{t}\\right)=|\\sum_{t\\in X} \\omega^{t}|^2$ again. If $a^3\\equiv 1(\\bmod\\; 37)$ , we can get $X=\\{c,a^2c,a^4c,d,a^2d,a^4d\\}$ so $aX=X$ , so $\\frac{x^{37}-1}{x-1} \\mid (\\sum\\limits_{t\\in X} x^t)^2+1$ , which fails by the same reasoning as $a\\equiv 1$ .\n\nCase 3: $d=2$ . Then $a$ can be 6 or 31 mod 37. Check $\\{16,17,18,19,20,21\\}$ works.\n\nCase 4: $d=6$ , then $X=\\{t,a^2t,a^4t,\\cdots,a^{10}t\\}$ . We need $aX+X=\\{1,\\cdots,36\\}$ .\n\nClearly, $t\\ne 0$ . If we divide all elements of $X$ by $t$ (mod 37) then we have that $X$ is the set of $6$ th powers mod 37 and $aX$ is the set of cubes that are not 6th powers mod 37. We can find $X=\\{1,27,26,36,10,11\\}, aX=\\{8,31,23,29,6,14\\}$ . Note $1+6\\equiv 36+8(\\bmod\\; 37)$ , so this case fails.</details>",
"My solution. I replace the set X with S. \nWhen a=6, if I let S={1,-1,m,-m,n,-n} then it is easy to see that (m^2+1)(n^2+1)=1 (mod 37). SInce m can't be 2, I let m=3 and see that n =5 satifies the modulo condition. The rest is just check.\n\n**Attachments:**\n\n[CHINA MO 2022 P3.docx](https://cdn.artofproblemsolving.com/attachments/0/7/eb42b600e2fe164f67b9badf29f680b74eaf48.docx)"
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A conference is attended by $n (n\ge 3)$ scientists. Each scientist has some friends in this conference (friendship is mutual and no one is a friend of him/herself). Suppose that no matter how we partition the scientists into two nonempty groups, there always exist two scientists in the same group who are friends, and there always exist two scientists in different groups who are friends.
A proposal is introduced on the first day of the conference. Each of the scientists' opinion on the proposal can be expressed as a non-negative integer. Everyday from the second day onwards, each scientists' opinion is changed to the integer part of the average of his/her friends' opinions from the previous day.
Prove that after a period of time, all scientists have the same opinion on the proposal.
|
<details><summary>Wrong solution</summary>Solved with **bora_olmez**
The problem says the graph is connected and non-bipartite. Call the given operation *averaging* and denote it by $f$ . I claim that for a graph $G$ if $f^k(G) = G$ then all the vertices have the same number. Note that the sum of opinions cannot increase since the sum is now the floor of some sums. So since the sum has to eventually return to the same thing, the sum stays constant and for this, every vertex has all its neighbours as the same number. So since between any two vertices, we may find a path with an even number of edges between them (because its not bipartite), they must have the same number, so every vertex has the same number, as claimed. $\square$ .
So suppose eventually the graph is unchanged, then because there are only finitely many possibilities, it must cycle back to the same thing. But from the previous paragraph, this implies that averaging leaves this graph unchanged, as desired. $\blacksquare$</details>
|
[
"Consider the graph of friendships is connected. The condition of having two scientists in different groups is equivalent to saying the graph is connected (partition connected components). The condition of having two scientists in the same group is equivalent to saying the graph is not 2-colorable/bipartite (by definition of bipartite).\n\nNotice the opinion of all scientists is bounded above by the maximal initial opinion. Therefore the set of options takes a finite number of states and is deterministic, so it eventually enters a cycle. Note that maximal opinion is weakly decreasing, so by periodicity, it is eventually constant. Notice if a maximal vertex is adjacent to a nonmaximal one, it is nonmaximal the next turn. Once it is constant, notice the number of nonmaximal to nonmaximal edges is weakly increasing, so by periodicity it is eventually constant. Also, the number of maximal to maximal edges is weakly decreasing, so by periodicity, it is eventually constant. Since maximal to maximal edges cannot disappear, if there are any then any bordering vertex is always maximal, and so on implying the entire graph is maximal. This is as desired. So consider the case when no maximal to maximal edges exist. Since nonmaximal to nonmaximal edges cannot disappear, no other types of edges can become nonmaximal to nonmaximal edges. This means for any nonmaximal to maximal edge, the nonmaximal vertex must become maximal. So the nonmaximal vertex only borders maximal vertices. Assume some nonmaximal vertex borders only nonmaximal vertices, then by our previous result all of those nonmaximal vertices only border nonmaximal vertices, and so on, implying the entire graph is nonmaximal. This is a contradiction, so all nonmaximal vertices border only maximal vertices. This implies that the set of maximal and nonmaximal vertices are a bipartition of the graph, giving contradiction. So we are done.",
"Ignore $$ $$ ",
"Take the obvious graph interpretation.\n\n<details><summary>Addressing bora's solution</summary>Inconsistent is correct. However, there is a different monovariant.\n\n\nConsider an alternate progress where the value of each vertex (call $x_v$ ) is replaced by the average of all its neighbours. Then I claim $\\sum\\limits_{v\\in G} \\deg(v)x_v$ is constant.\n\nProof: Let $y_v$ be the new values. In the new graph, $\\deg(v)y_v=\\sum\\limits_{uv \\text{ edge}} x_u$ Therefore, for the new graph, $\\sum\\limits_{v\\in G} \\deg(v)y_v=\\sum\\limits_{uv \\text{ edge}} (x_u+x_v) = \\sum\\limits_{v\\in G} \\deg(v)x_v$ , as needed.</details>\n\n<details><summary>Solution</summary>I claim if $x_k$ is nonconstant, then the maximal $x_v$ eventually decreases. Suppose $x_t<x_k$ , and we will show that eventually $x_k$ will decrease. To prove this, note the graph is not bipartite and connected. This means we can upper bound $x_k$ by $c_1x_{v_1}+\\cdots+c_nx_{v_n}$ where $\\sum c_i=1$ (by replacing value of each vertex with the average of values of neighbours without $\\lfloor \\rfloor$ ). It suffices to show that eventually, $c_i>0$ for all $i$ , which implies $x_k$ cannot be of this value and must decrement by at least 1, bringing the floor back.\n\nAs $t$ tends to infinity, $c_j>0$ if and only if there exists a path from $v_j$ to $v_k$ of length $t$ . A path of length $l$ with $l\\le t$ and $2|l-t$ is okay because I can just travel an edge forward and back $\\frac{t-l}{2}$ times. Since the graph is connected and not bipartite, it has an odd cycle, so we are done as any vertex can go to the odd cycle.\n\nSince the minimal $x_v$ cannot decrease, we are done.</details>",
"A similar solution:\n\nLet our graph be $G$ . It’s given that $G$ is non-bipartite and connected, so there’s an odd cycle $C$ in $G$ .\nThe main observation is that the maximum opinion is weakly decreasing, and since it has a lower bound (e.g. $0$ ), it’s eventually a constant $M$ . From this point, we say that a node “is not M” iff at the time its opinion isn’t $M$ .\nIt’s clear that if a node isn’t M, all its neighbours aren’t M on the following day.\n1) Therefore, starting from any vertex $v_0$ , its neighbours won’t be M on the following day, and so do its neighbours’ neighbours. We’ll eventually reach a node $v_1$ in the odd cycle and make it not M (due to connectivity).\n2) Knowing that at some point there’s a node in $C$ that’s not M, we can see that the two neighbours of $v_1$ in $C$ won’t be M on the following day, and the neighbours of neighbours won’t be M on the day after...\n3) At some point, we’ll arrive at two adjacent nodes in $C$ that are not M at the same time (since $C$ is odd).\n4) If there are two “not M” adjacent nodes, their direct neighbours and themselves won’t be M later. These “not M” nodes expand outward and remain to be “not M” since every one of them is connected to at least one “not M” node. Eventually all nodes aren’t M, which is a contradiction!\nThis implies that if the maximum reaches a constant, all opinions are the same (they’re all $M$ ).",
"I found this video helpful: https://youtu.be/EuLaHqFyTtI",
"Consider the obvious Graph interpretation. Let the given graph be $G$ .\nWe know by the conditions that $G$ is connected and not bipartite, and that the maximum vote M cannot increase. Note that after the first day, if a vertex at any given time has not the maximum vote, then it will never:\nindeed, for a vertex $v$ to have the maximum vote it needs to have all the adjecent nodes' votes M, which will never happen since all its neighbours cannot increase their votes to M because of the presence of $v$ . On the converse, due the connectivity of $G$ we know that there exists a vertex $u$ whose vote is M that has a neighbour with a vote numerically less than M, thus $u$ 's vote will deacrease. In particular the number of nodes with the maximum vote is strictly decreasing and since M is bounded by $0$ we will reach a state in which all the nodes have the maximum vote, thus the same vote. \nRemark: I didn't use the fact that $G$ is not bipartite so I'm not sure about this solution.",
"<blockquote>Consider the obvious Graph interpretation. Let the given graph be $G$ .\nWe know by the conditions that $G$ is connected and not bipartite, and that the maximum vote M cannot increase. Note that after the first day, if a vertex at any given time has not the maximum vote, then it will never:\nindeed, for a vertex $v$ to have the maximum vote it needs to have all the adjecent nodes' votes M, which will never happen since all its neighbours cannot increase their votes to M because of the presence of $v$ . On the converse, due the connectivity of $G$ we know that there exists a vertex $u$ whose vote is M that has a neighbour with a vote numerically less than M, thus $u$ 's vote will deacrease. In particular the number of nodes with the maximum vote is strictly decreasing and since M is bounded by $0$ we will reach a state in which all the nodes have the maximum vote, thus the same vote. \nRemark: I didn't use the fact that $G$ is not bipartite so I'm not sure about this solution.</blockquote>\n\nConsider the graph $K_{3,3}$ . We obtain that $G$ must be not bipartite.",
"Here is my solution:- \n\nConvert this to a graph $\\mathcal{G}$ where scientists are the vertices two friends are joined by edges. \n\nAs per condition $\\mathcal{G}$ is a connnected non biparite graph\n\nAssuem the contrary at each step color the vertex with the maximum valued opinion as \"red\" others as blue\n\nIt is easy to see after a point the values at \"red\" vertices is fixed. and also we assume number of red vertices are always fixed.\n\nIt is easy to see that each step for any vertex $v$ all of its neighours are red it becomes red , otherwise blue next day\n\n\nClaim: Two blue vertices can never be joined by an egde. \n\nProof: Assume the contrary $u$ and $v$ are two vertices joined like this. \n\nLemma: None of them will ever turn red after they were together blue.\n\nProof: Say it does happen , It is easy to see both cant turn red for first time on same day after they were blue together say WLOG $u$ turns red first but for $u$ to turn red it neighours must all be red before hand that implies $v$ is red but thats contradiction! \n\n\nNow we prove entually all vertices will become non red ,we induct on the distance between the vertex and $u$ from which it is easy to see . \n\nEasy to see any red vertex will always atleast once will turn to blue\n\n\nNow say we have a odd cycle in the graph, $v_1 \\to v_2 .... \\to v_k \\to v_1$ in the graph which clearly has to exist. now observe eventually one vertex turns to blue and as the blue color spreads to its neighours its count grows largely and will not platue eventually cuasing an edge between two blue vertices.\n"
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"path": "Contest Collections/2022 Contests/2022 China National Olympiad/2742829.json"
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|
On a blank piece of paper, two points with distance $1$ is given. Prove that one can use (only) straightedge and compass to construct on this paper a straight line, and two points on it whose distance is $\sqrt{2021}$ such that, in the process of constructing it, the total number of circles or straight lines drawn is at most $10.$
Remark: Explicit steps of the construction should be given. Label the circles and straight lines in the order that they appear. Partial credit may be awarded depending on the total number of circles/lines.
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Pretty easy problem
Construction brings to mind the methods to multiply a segment. To multiply a segment directly by a factor $a$ , we must construct $a-1$ circles. Note that we can 'copy the segment' by marking points a unit away from each other.
The length $\sqrt{2021}$ immediately gives us the idea of using the Pythagorean theorem. The most obvious way to do this is note $2021 = 45^2 - 2^2$ . Copy the unit segment $45$ times and $2$ times, such that we now have marked a distance $A=45$ with endpoints $A_1, A_2$ and $B=2$ .
Now construct the perpendicular bisector of $A$ and set it to the center of a circle with diameter $45$ . At this point we have constructed $2$ lines and $4$ circles. Now draw a circle with radius $2$ at $A_1$ and let its intersection with the larger circle be $X$ . Then $A_2X$ will have our desired length of $\sqrt{2021}$ and we are done! $\blacksquare$
|
[
"Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?",
"<blockquote>Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?</blockquote>\n\nIf so, then $5$ lines/circles suffice. On the line $l_1$ with the two points $A_0$ and $A_1$ ,copy the length $2021$ times in order to obtain the points $A_2,...,A_{2022}$ such that $A_iA_{i+1}=1$ for all $i=1,...,2021$ . Now construct the circle $C_1$ of center $A_{2020}$ and radius $A_{2020}A_{2022}$ and the circle $C_2$ of center $A_{2022}$ and radius $A_{2020}A_{2022}$ . By intersecting these two circles we get the points $P_1$ and $P_2$ , we get two points lying on a new line $\\ell_2$ , which is perpendicular to $\\ell_1$ and passes through $A_{2021}$ . Now take the circle $C_3$ of center $A_{1011}$ and radius $A_0A_{1011}=1011$ , which is in other words the circle with diameter $A_0A_{2022}$ . Intersecting it with $\\ell_2$ we get two points $Q_1$ and $Q_2$ . Since $A_0Q_1A_{2022}$ is right-angled and $A_{2021}$ is its projection on $\\ell_1$ , it follows that $A_{2021}Q_1=\\sqrt{A_0A_{2021}\\cdot A_{2021}A_{2022}}=\\sqrt{2021}$ , for a total of $5$ things used.\nHowever if we can't copy distances as I did then I think it would be quite difficult, since you probably can create an integer in logarithmic time (for example $2^n$ in at least $n$ steps, I think).",
"<details><summary>Click to expand</summary>Denote by $B(x,r)$ the circle of radius $r$ centered at point $x$ . Assume the two given points are $(0,0)$ and $(1,0)$ .\n(1) Connect the two given points with a line.\n(2) Draw $B((1,0),1)$ and find $(2,0)$ .\n(3) Draw $B((2,0),1)$ and find $(3,0)$ .\n(4) Connect the two intersections of $B((1,0),1)$ and $B((2,0),1)$ and find $(1.5,0)$ .\n(5) Draw $B((3,0),3)$ and find $(6,0)$ .\n(6) Draw $B((6,0),6)$ and find $(12,0)$ .\n(7) Draw $B((12,0),10.5)$ and find $(22.5,0)$ .\n(8) Draw $B_1=B((22.5,0),22.5)$ and find $A_1=(45,0)$ .\n(9) Draw $B_2=B((0,0),2)$ and assume $B_1$ intersects $B_2$ at $A_2$ .\n(10) Connect $A_1,A_2$ . The distance between $A_1A_2$ is $\\sqrt{45^2-2^2}=\\sqrt{2021}$</details>",
"<details><summary>Solution</summary>WLOG the two points of distance 1 are horizontal. Call the points $(0,0)$ and $(0,1)$ .\n\nBy positioning the compass and pencil such that their distance is 1, we can mark $(0,2),\\cdots,(0,45)$ with one extra line passing through $(0,0), (0,1)$ . (tell me if this is not allowed)\n\nDraw a circle centered at $(0,22)$ with radius 22 and a circle centered at $(0,23)$ with radius 22. Say they intersect at P, Q . Draw PQ and let K be the intersection of PQ and the x-axis. Note $K=(0,\\frac{45}{2})$ .\n\nNow draw a circle centered at $K$ passing through $(0,0)$ .\n\nDraw a circle centered at $(0,45)$ with radius 2. The intersection of this point with the circle centered at $(0,\\frac{45}{2})$ has distance $\\sqrt{2021}$ from the origin.</details>",
"<blockquote><blockquote>Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?</blockquote>\n\nIf so, then $5$ lines/circles suffice. On the line $l_1$ with the two points $A_0$ and $A_1$ ,copy the length $2021$ times in order to obtain the points $A_2,...,A_{2022}$ such that $A_iA_{i+1}=1$ for all $i=1,...,2021$ . Now construct the circle $C_1$ of center $A_{2020}$ and radius $A_{2020}A_{2022}$ and the circle $C_2$ of center $A_{2022}$ and radius $A_{2020}A_{2022}$ . By intersecting these two circles we get the points $P_1$ and $P_2$ , we get two points lying on a new line $\\ell_2$ , which is perpendicular to $\\ell_1$ and passes through $A_{2021}$ . Now take the circle $C_3$ of center $A_{1011}$ and radius $A_0A_{1011}=1011$ , which is in other words the circle with diameter $A_0A_{2022}$ . Intersecting it with $\\ell_2$ we get two points $Q_1$ and $Q_2$ . Since $A_0Q_1A_{2022}$ is right-angled and $A_{2021}$ is its projection on $\\ell_1$ , it follows that $A_{2021}Q_1=\\sqrt{A_0A_{2021}\\cdot A_{2021}A_{2022}}=\\sqrt{2021}$ , for a total of $5$ things used.\nHowever if we can't copy distances as I did then I think it would be quite difficult, since you probably can create an integer in logarithmic time (for example $2^n$ in at least $n$ steps, I think).</blockquote>\n\nWell but for “copying”, shall we count them as steps?",
"No, because it doesn't create a new circle or line.",
"<blockquote>No, because it doesn't create a new circle or line.</blockquote>\n\nInstead I would say that we either count it as $0$ steps, because in practice you don't need to draw the whole circle or as $2021$ steps because technically you need to draw the entire circles to copy the distance, but surely not a single step.",
"Or you can think of drawing a line cost one dollar, drawing a circle cost one dollar, nothing else costs money, and you have 10 dollars.",
"I’m not sure but I think the intention was to only draw “a line through two points” and “a circle with a certain center and radius”. The ruler is assumed to have no markings on it (and, perhaps, the compass can’t be used to transfer distance).\n\nThe main idea is basically divide & conquer. But it really takes time to reduce $11-20$ steps to $10$ . \n\n<details><summary>Construction</summary>Let $P_0, P_1$ be the two given points. In the following construction, $R-S$ denotes drawing a line through $R, S$ (using the ruler once) and $C(R, RS)$ means to draw a circle with center $R$ and radius $RS$ (using the compass).\n\n1) $P_0-P_1$ . Call this line $l$ .\n2) $C(P_1, P_1P_0)$ and call this $C_1$ .\n....Let $C_1$ intersects $l$ at $P_0, P_2$ , so $P_1P_2=1$ .\n3) $C(P_2, P_2P_0)$ and call this $C_2$ .\n.....Let $C_2$ intersects $l$ at $P_0, P_3$ , so $P_2P_3=2$ .\n4) $C(P_3, P_3P_1)$ and call this $C_3$ .\n.....Let $C_3$ intersects $l$ at $P_1, P_4$ , so $P_3P_4=3$ .\n5) $C(P_4, P_4P_1)$ and call this $C_4$ .\n.....Let $C_4$ intersects $l$ at $P_1, P_5$ , so $P_4P_5=6$ .\n6) $C(P_5, P_5P_2)$ and call this $C_5$ .\n.....Let $C_5$ intersects $l$ at $P_2, P_6$ , so $P_5P_6=11$ .\n7) $C(P_6, P_6P_1)$ and call this $C_6$ .\n.....Let $C_6$ intersects $l$ at $P_1, P_7$ , so $P_6P_7=23$ , meaning that $P_2P_7=45$ .\n* 8) $C(P_1, P_1P_4)$ and call this $C_7$ .\n.....Let $C_7$ intersects $C_4$ at $A, B$ .\n* 9) $A-B$ and call this $l’$ .\n.....This is a line perpendicular to $l$ at $P_3$ .\n10) $C(P_2, P_2P_7)$ and name this $C_8$ .\n.....Let $C_8$ meet $l’$ at $X$ .\n\nNow $XP_3=\\sqrt{2021}$ as $2021 + 2^2=45^2$ .</details>",
"There are many solutions that interpret this as number of operations instead of number of circles and lines drawn... congratulations for proving this stronger statement!",
"<blockquote>I’m not sure but I think the intention was to only draw “a line through two points” and “a circle with a certain center and radius”. The ruler is assumed to have no markings on it (and, perhaps, the compass can’t be used to transfer distance).\n\nThe main idea is basically divide & conquer. But it really takes time to reduce $11-20$ steps to $10$ . \n\n<details><summary>Construction</summary>Let $P_0, P_1$ be the two given points. In the following construction, $R-S$ denotes drawing a line through $R, S$ (using the ruler once) and $C(R, RS)$ means to draw a circle with center $R$ and radius $RS$ (using the compass).\n\n1) $P_0-P_1$ . Call this line $l$ .\n2) $C(P_1, P_1P_0)$ and call this $C_1$ .\n....Let $C_1$ intersects $l$ at $P_0, P_2$ , so $P_1P_2=1$ .\n3) $C(P_2, P_2P_0)$ and call this $C_2$ .\n.....Let $C_2$ intersects $l$ at $P_0, P_3$ , so $P_2P_3=2$ .\n4) $C(P_3, P_3P_1)$ and call this $C_3$ .\n.....Let $C_3$ intersects $l$ at $P_1, P_4$ , so $P_3P_4=3$ .\n5) $C(P_4, P_4P_1)$ and call this $C_4$ .\n.....Let $C_4$ intersects $l$ at $P_1, P_5$ , so $P_4P_5=6$ .\n6) $C(P_5, P_5P_2)$ and call this $C_5$ .\n.....Let $C_5$ intersects $l$ at $P_2, P_6$ , so $P_5P_6=11$ .\n7) $C(P_6, P_6P_1)$ and call this $C_6$ .\n.....Let $C_6$ intersects $l$ at $P_1, P_7$ , so $P_6P_7=23$ , meaning that $P_2P_7=45$ .\n* 8) $C(P_1, P_1P_4)$ and call this $C_7$ .\n.....Let $C_7$ intersects $C_4$ at $A, B$ .\n* 9) $A-B$ and call this $l’$ .\n.....This is a line perpendicular to $l$ at $P_3$ .\n10) $C(P_2, P_2P_7)$ and name this $C_8$ .\n.....Let $C_8$ meet $l’$ at $X$ .\n\nNow $XP_3=\\sqrt{2021}$ as $2021 + 2^2=45^2$ .</details></blockquote>\n\nThe problem asks for a line and two points on it, so I think you need one more line $XP_3$ .",
" $l’$ (Line $XP_3$ ) is already constructed in the process.",
"<blockquote>Just curious. Can one “copy a distance” using a compass? A bit ambiguous there..?</blockquote>\n\nBased on the official solution, copying distance without drawing circle is not allowed. You are only allowed two operations: (1) draw a line through two pre-constructed points; (2) draw a circle with centre being a pre-constructed point and radius being the distance between two pre-constructed points. New pre-constructed points can only be form via intersection of such circles and lines. For those curious, the original question has an additional line at the start making reference to the straightedge and compass problem, so it assumes the reader knows the standard rules of the game :p\n\nDid a video solution here: [https://www.youtube.com/watch?v=n1DOunWAyyk](https://www.youtube.com/watch?v=n1DOunWAyyk)",
"China MO is wack.\n\nCall our points $(0,0)$ and $(1,0)$ .\n1. Draw the x-axis\n2. Draw circle centered $(1,0)$ with radius $1$ to mark point $(2,0)$ .\n3. Draw circle centered $(2,0)$ with radius $1$ to mark point $(3,0)$ .\n4. Draw circle centered $(3,0)$ with radius $3$ to mark point $(6,0)$ .\n5. Draw circle centered $(6,0)$ with radius $6$ to mark point $(12,0)$ .\n6. Draw circle centered $(12,0)$ with radius $12$ to mark point $(24,0)$ .\n7. Draw circle centered $(12,0)$ with radius $10$ to mark point $(22,0)$ .\n8. Draw the radical axis of the circles in steps 2. and 3. to mark $(1.5,0)$ 9. Draw the circle centered at $(1.5,0)$ with radius $22.5$ to mark $(-21,0)$ 10. Draw the circle centered at $(24,0)$ with radius $2$ and intersect with the previous circle to get a point that is $\\sqrt{2021}$ away from $(-21,0)$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 China National Olympiad"
] |
{
"answer_score": 138,
"boxed": false,
"end_of_proof": true,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 China National Olympiad/2742832.json"
}
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