{"year":1998,"label":"1","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 $S \\equiv 9(\\bmod 10)$."}
{"year":1998,"label":"2","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 $A M \/ M C=5 \/ 3$."}
{"year":1998,"label":"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."}
{"year":1998,"label":"4","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."}
{"year":1998,"label":"5","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 $S$."}
{"year":1998,"label":"6","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)."}