exam: USAMO

problem: Suppose that the set $\{1,2, \ldots, 1998\}$ has been partitioned into disjoint pairs $\left\{a_{i}, b_{i}\right\}$ $(1 \leq i \leq 999)$ so that for all $i,\left|a_{i}-b_{i}\right|$ equals 1 or 6 . Prove that the sum  $$ \left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{999}-b_{999}\right| $$  ends in the digit 9.

solution: Let $S$ be the sum. Modulo 2,  $$ S=\sum\left|a_{i}-b_{i}\right| \equiv \sum\left(a_{i}+b_{i}\right)=1+2+\cdots+1998 \equiv 1 \quad(\bmod 2) $$  Modulo 5,  $$ S=\sum\left|a_{i}-b_{i}\right|=1 \cdot 999 \equiv 4 \quad(\bmod 5) $$  So $S \equiv 9(\bmod 10)$.


problem: Let $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ be concentric circles, with $\mathcal{C}_{2}$ in the interior of $\mathcal{C}_{1}$. From a point $A$ on $\mathcal{C}_{1}$ one draws the tangent $A B$ to $\mathcal{C}_{2}\left(B \in \mathcal{C}_{2}\right)$. Let $C$ be the second point of intersection of ray $A B$ and $\mathcal{C}_{1}$, and let $D$ be the midpoint of $\overline{A B}$. A line passing through $A$ intersects $\mathcal{C}_{2}$ at $E$ and $F$ in such a way that the perpendicular bisectors of $\overline{D E}$ and $\overline{C F}$ intersect at a point $M$ on line $A B$. Find, with proof, the ratio $A M / M C$.

solution: By power of a point we have  $$ A E \cdot A F=A B^{2}=\left(\frac{1}{2} A B\right) \cdot(2 A B)=A D \cdot A C $$  and hence $C D E F$ is cyclic. Then $M$ is the circumcenter of quadrilateral $C D E F$. ![](https://cdn.mathpix.com/cropped/2024_11_19_4d9e3ecba99734fe13c2g-4.jpg?height=806&width=792&top_left_y=1179&top_left_x=638)  Thus $M$ is the midpoint of $\overline{C D}$ (and we are given already that $B$ is the midpoint of $\overline{A C}$, $D$ is the midpoint of $\overline{A B}$ ). Thus a quick computation along $\overline{A C}$ gives $A M / M C=5 / 3$.


problem: Let $a_{0}, a_{1}, \ldots, a_{n}$ be numbers from the interval $(0, \pi / 2)$ such that $\tan \left(a_{0}-\frac{\pi}{4}\right)+$ $\tan \left(a_{1}-\frac{\pi}{4}\right)+\cdots+\tan \left(a_{n}-\frac{\pi}{4}\right) \geq n-1$. Prove that  $$ \tan a_{0} \tan a_{1} \cdots \tan a_{n} \geq n^{n+1} $$

solution: $  Let $x_{i}=\tan \left(a_{i}-\frac{\pi}{4}\right)$. Then we have that  $$ \tan a_{i}=\tan \left(a_{i}-45^{\circ}+45^{\circ}\right)=\frac{x_{i}+1}{1-x_{i}} $$  If we further substitute $y_{i}=\frac{1-x_{i}}{2} \in(0,1)$, then we have to prove that the following statement:  Claim - If $\sum_{0}^{n} y_{i} \leq 1$ and $y_{i} \geq 0$, we have  $$ \prod_{i=1}^{n}\left(\frac{1}{y_{i}}-1\right) \geq n^{n+1} $$   $$ \prod_{i=1}^{n}\left(\frac{y_{0}+y_{1}+y_{2}+\cdots+y_{n}}{y_{i}}-1\right) \geq n^{n+1} $$  By AM-GM, we have  $$ \frac{y_{1}+y_{2}+y_{3}+\cdots+y_{n}}{y_{0}} \geq n \sqrt[n]{\frac{y_{1} y_{2} y_{3} \ldots y_{n}}{y_{1}}} $$  Cyclic product works.  Remark. Alternatively, the function $x \mapsto \log (1 / x-1)$ is a convex function on $(0,1)$ so Jensen inequality should also work.


problem: A computer screen shows a $98 \times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result, the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.

solution: The answer is 98 . One of several possible constructions is to toggle all columns and rows with even indices.  In the other direction, let $n=98$ and suppose that $k$ rectangles are used, none of which are $n \times n$ (else we may delete it). Then, for any two orthogonally adjacent cells, the edge between them must be contained in the edge of one of the $k$ rectangles.  We define a gridline to be a line segment that runs in the interior of the board from one side of the board to the other. Hence there are $2 n-2$ gridlines exactly. Moreover, we can classify these rectangles into two types:  - Full length rectangles: these span from one edge of the board to the other. The two long sides completely cover two gridlines, but the other two sides of the rectangle do not. - Partial length rectangles: each of four sides can partially cover "half a" gridline.  See illustration below for $n=6$. ![](https://cdn.mathpix.com/cropped/2024_11_19_4d9e3ecba99734fe13c2g-6.jpg?height=321&width=327&top_left_y=1601&top_left_x=522) ![](https://cdn.mathpix.com/cropped/2024_11_19_4d9e3ecba99734fe13c2g-6.jpg?height=315&width=326&top_left_y=1604&top_left_x=865)  Full length ![](https://cdn.mathpix.com/cropped/2024_11_19_4d9e3ecba99734fe13c2g-6.jpg?height=350&width=318&top_left_y=1601&top_left_x=1212)  Partial length  Since there are $2 n-2$ gridlines; and each rectangle can cover at most two gridlines in total (where partial-length rectangles are "worth $\frac{1}{2}$ " on each of the four sides), we immediately get the bound $2 k \geq 2 n-2$, or $k \geq n-1$.  To finish, we prove that: Claim - If equality holds and $k=n-1$, then $n$ is odd.   Hence for $n=98$ the answer is indeed 98 as claimed.


problem: Prove that for each $n \geq 2$, there is a set $S$ of $n$ integers such that $(a-b)^{2}$ divides $a b$ for every distinct $a, b \in S$.

solution: This is a direct corollary of the more difficult USA TST 2015/2, reproduced below. Prove that for every positive integer $n$, there exists a set $S$ of $n$ positive integers such that for any two distinct $a, b \in S, a-b$ divides $a$ and $b$ but none of the other elements of $S$.


problem: Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$ ) such that there exists a convex $n$-gon $A_{1} A_{2} \ldots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i} A_{i+1} A_{i+2} A_{i+3}$ have an inscribed circle, where indices are taken modulo $n$.

solution: The main claim is the following: Claim - We can't have both $A_{1} A_{2} A_{3} A_{4}$ and $A_{2} A_{3} A_{4} A_{5}$ be circumscribed.  ![](https://cdn.mathpix.com/cropped/2024_11_19_4d9e3ecba99734fe13c2g-8.jpg?height=621&width=892&top_left_y=1049&top_left_x=585)  Then $A_{1} A_{4}=c+a-b$ and $A_{5} A_{2}=b+d-c$. But now  $$ A_{1} A_{4}+A_{2} A_{5}=(c+a-b)+(b+d-c)=a+d=A_{1} A_{2}+A_{4} A_{5} $$  but in the picture we have an obvious violation of the triangle inequality. This immediately gives an upper bound of $\lfloor n / 2\rfloor$. For the construction, one can construct a suitable cyclic $n$-gon by using a continuity argument (details to be added).