exam: USAJMO

problem: Let $P(n)$ be the number of permutations $\left(a_{1}, \ldots, a_{n}\right)$ of the numbers $(1,2, \ldots, n)$ for which $k a_{k}$ is a perfect square for all $1 \leq k \leq n$. Find with proof the smallest $n$ such that $P(n)$ is a multiple of 2010 .

solution: The answer is $n=4489$. We begin by giving a complete description of $P(n)$ :  Claim - We have  $$ P(n)=\prod_{c \text { squarefree }}\left\lfloor\sqrt{\frac{n}{c}}\right\rfloor! $$   $$ S_{c}=\left\{c \cdot 1^{2}, c \cdot 2^{2}, c \cdot 3^{2}, \ldots\right\} \cap\{1,2, \ldots, n\} $$  and each integer from 1 through $n$ will be in exactly one $S_{c}$. Note also that  $$ \left|S_{c}\right|=\left\lfloor\sqrt{\frac{n}{c}}\right\rfloor . $$  Then, the permutations in the problem are exactly those which send elements of $S_{c}$ to elements of $S_{c}$. In other words,  $$ P(n)=\prod_{c \text { squarefree }}\left|S_{c}\right|!=\prod_{c \text { squarefree }}\left\lfloor\sqrt{\frac{n}{c}}\right\rfloor! $$  We want the smallest $n$ such that 2010 divides $P(n)$.  - Note that $P\left(67^{2}\right)$ contains 67 ! as a term, which is divisible by 2010 , so $67^{2}$ is a candidate. - On the other hand, if $n<67^{2}$, then no term in the product for $P(n)$ is divisible by the prime 67 .  So $n=67^{2}=4489$ is indeed the minimum.


problem: Let $n>1$ be an integer. Find, with proof, all sequences $x_{1}, x_{2}, \ldots, x_{n-1}$ of positive integers with the following three properties: (a) $x_{1}<x_{2}<\cdots<x_{n-1}$; (b) $x_{i}+x_{n-i}=2 n$ for all $i=1,2, \ldots, n-1$; (c) given any two indices $i$ and $j$ (not necessarily distinct) for which $x_{i}+x_{j}<2 n$, there is an index $k$ such that $x_{i}+x_{j}=x_{k}$.

solution: (a) $x_{1}<x_{2}<\cdots<x_{n-1}$; (b) $x_{i}+x_{n-i}=2 n$ for all $i=1,2, \ldots, n-1$; (c) given any two indices $i$ and $j$ (not necessarily distinct) for which $x_{i}+x_{j}<2 n$, there is an index $k$ such that $x_{i}+x_{j}=x_{k}$.  The answer is $x_{k}=2 k$ only, which obviously work, so we prove they are the only ones. Let $x_{1}<x_{2}<\cdots<x_{n}$ be any sequence satisfying the conditions. Consider:  $$ x_{1}+x_{1}<x_{1}+x_{2}<x_{1}+x_{3}<\cdots<x_{1}+x_{n-2} $$  All these are results of condition (c), since $x_{1}+x_{n-2}<x_{1}+x_{n-1}=2 n$. So each of these must be a member of the sequence. However, there are $n-2$ of these terms, and there are exactly $n-2$ terms greater than $x_{1}$ in our sequence. Therefore, we get the one-to-one correspondence below:  $$ \begin{aligned} x_{2} & =x_{1}+x_{1} \\ x_{3} & =x_{1}+x_{2} \\ & \vdots \\ x_{n-1} & =x_{1}+x_{n-2} \end{aligned} $$  It follows that $x_{2}=2 x_{1}$, so that $x_{3}=3 x_{1}$ and so on. Therefore, $x_{m}=m x_{1}$. We now solve for $x_{1}$ in condition (b) to find that $x_{1}=2$ is the only solution, and the desired conclusion follows.


problem: Let $A X Y Z B$ be a convex pentagon inscribed in a semicircle of diameter $A B$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $A X, B X$, $A Z, B Z$, respectively. Prove that the acute angle formed by lines $P Q$ and $R S$ is half the size of $\angle X O Z$, where $O$ is the midpoint of segment $A B$.

solution: 【 First approach using angle chasing. Define $T=\overline{P Q} \cap \overline{R S}$. Also, let $2 \alpha, 2 \beta, 2 \gamma, 2 \delta$ denote the measures of arcs $\overparen{A X}, \widehat{X Y}, \widehat{Y Z}, \widehat{Z B}$, respectively, so that $\alpha+\beta+\gamma+\delta=90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_11_19_a4f5a16d632e1d6ed723g-5.jpg?height=812&width=763&top_left_y=1210&top_left_x=652)  We now compute the following angles:  $$ \begin{aligned} & \angle S R Y=\angle S Z Y=90^{\circ}-\angle Y Z A=90^{\circ}-(\alpha+\beta) \\ & \angle Y Q P=\angle Y X P=90^{\circ}-\angle B X Y=90^{\circ}-(\gamma+\delta) \\ & \angle Q Y R=180^{\circ}-\angle(\overline{Z R}, \overline{Q X})=180^{\circ}-\frac{2 \beta+2 \gamma+180^{\circ}}{2}=90^{\circ}-(\beta+\gamma) \end{aligned} $$  Hence, we can then compute  $$ \begin{aligned} \angle R T Q & =360^{\circ}-\left(\angle Q Y R+\left(180^{\circ}-\angle S R Y\right)+\left(180^{\circ}-\angle Y Q P\right)\right) \\ & =\angle S R Y+\angle Y Q P-\angle Q Y R \\ & =\left(90^{\circ}-(\alpha+\beta)\right)+\left(90^{\circ}-(\gamma+\delta)\right)-\left(90^{\circ}-(\beta+\gamma)\right) \end{aligned} $$  $$ \begin{aligned} & =90^{\circ}-(\alpha+\delta) \\ & =\beta+\gamma \end{aligned} $$  Since $\angle X O Z=\frac{2 \beta+2 \gamma}{2}=\beta+\gamma$, the proof is complete.


solution: 【 Second approach using Simson lines, ignoring the diameter condition. In this solution, we will ignore the condition that $\overline{A B}$ is a diameter; the solution works equally well without it, as long as $O$ is redefined as the center of $(A X Y Z B)$ instead. We will again show the angle formed by lines $P Q$ and $R S$ is half the measure of $\widehat{X Z}$.  ![](https://cdn.mathpix.com/cropped/2024_11_19_a4f5a16d632e1d6ed723g-6.jpg?height=820&width=732&top_left_y=912&top_left_x=665)  Now it's straightforward to see $A P Y R T$ is cyclic (in the circle with diameter $\overline{A Y}$ ), and therefore  $$ \angle R T Y=\angle R A Y=\angle Z A Y $$  Similarly,  $$ \angle Y T Q=\angle Y B Q=\angle Y B X $$  Summing these gives $\angle R T Q$ is equal to half the measure of arc $\widehat{X Z}$ as needed.


problem: A triangle is called a parabolic triangle if its vertices lie on a parabola $y=x^{2}$. Prove that for every nonnegative integer $n$, there is an odd number $m$ and a parabolic triangle with vertices at three distinct points with integer coordinates with area $\left(2^{n} m\right)^{2}$ 。

solution: A triangle is called a parabolic triangle if its vertices lie on a parabola $y=x^{2}$. Prove that for every nonnegative integer $n$, there is an odd number $m$ and a parabolic triangle with vertices at three distinct points with integer coordinates with area $\left(2^{n} m\right)^{2}$.  For $n=0$, take instead $(a, b)=(1,0)$. For $n>0$, consider a triangle with vertices at $\left(a, a^{2}\right),\left(-a, a^{2}\right)$ and $\left(b, b^{2}\right)$. Then the area of this triangle was equal to  $$ \frac{1}{2}(2 a)\left(b^{2}-a^{2}\right)=a\left(b^{2}-a^{2}\right) . $$  To make this equal $2^{2 n} m^{2}$, simply pick $a=2^{2 n}$, and then pick $b$ such that $b^{2}-m^{2}=2^{4 n}$, for example $m=2^{4 n-2}-1$ and $b=2^{4 n-2}+1$.


problem: Two permutations $a_{1}, a_{2}, \ldots, a_{2010}$ and $b_{1}, b_{2}, \ldots, b_{2010}$ of the numbers $1,2, \ldots, 2010$ are said to intersect if $a_{k}=b_{k}$ for some value of $k$ in the range $1 \leq k \leq 2010$. Show that there exist 1006 permutations of the numbers $1,2, \ldots, 2010$ such that any other such permutation is guaranteed to intersect at least one of these 1006 permutations.

solution: A valid choice is the following 1006 permutations:  | 1 | 2 | 3 | $\cdots$ | 1004 | 1005 | 1006 | 1007 | 1008 | $\cdots$ | 2009 | 2010 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 2 | 3 | 4 | $\cdots$ | 1005 | 1006 | 1 | 1007 | 1008 | $\cdots$ | 2009 | 2010 | | 3 | 4 | 5 | $\cdots$ | 1006 | 1 | 2 | 1007 | 1008 | $\cdots$ | 2009 | 2010 | | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | | 1004 | 1005 | 1006 | $\cdots$ | 1001 | 1002 | 1003 | 1007 | 1008 | $\cdots$ | 2009 | 2010 | | 1005 | 1006 | 1 | $\cdots$ | 1002 | 1003 | 1004 | 1007 | 1008 | $\cdots$ | 2009 | 2010 | | 1006 | 1 | 2 | $\cdots$ | 1003 | 1004 | 1005 | 1007 | 1008 | $\cdots$ | 2009 | 2010 |  This works. Indeed, any permutation should have one of $\{1,2, \ldots, 1006\}$ somewhere in the first 1006 positions, so one will get an intersection.  Remark. In fact, the last 1004 entries do not matter with this construction, and we chose to leave them as $1007,1008, \ldots, 2010$ only for concreteness.  Remark. Using Hall's marriage lemma one may prove that the result becomes false with 1006 replaced by 1005 .


problem: Let $A B C$ be a triangle with $\angle A=90^{\circ}$. Points $D$ and $E$ lie on sides $A C$ and $A B$, respectively, such that $\angle A B D=\angle D B C$ and $\angle A C E=\angle E C B$. Segments $B D$ and $C E$ meet at $I$. Determine whether or not it is possible for segments $A B, A C, B I$, $I D, C I, I E$ to all have integer lengths.

solution: The answer is no. We prove that it is not even possible that $A B, A C, C I, I B$ are all integers. ![](https://cdn.mathpix.com/cropped/2024_11_19_a4f5a16d632e1d6ed723g-9.jpg?height=427&width=504&top_left_y=923&top_left_x=776)  First, we claim that $\angle B I C=135^{\circ}$. To see why, note that  $$ \angle I B C+\angle I C B=\frac{\angle B}{2}+\frac{\angle C}{2}=\frac{90^{\circ}}{2}=45^{\circ} . $$  So, $\angle B I C=180^{\circ}-(\angle I B C+\angle I C B)=135^{\circ}$, as desired. We now proceed by contradiction. The Pythagorean theorem implies  $$ B C^{2}=A B^{2}+A C^{2} $$  and so $B C^{2}$ is an integer. However, the law of cosines gives  $$ \begin{aligned} B C^{2} & =B I^{2}+C I^{2}-2 B I \cdot C I \cos \angle B I C \\ & =B I^{2}+C I^{2}+B I \cdot C I \cdot \sqrt{2} . \end{aligned} $$  which is irrational, and this produces the desired contradiction.