USAMO/md/en-USAMO-1998-notes.md

USAMO 1998 Solution Notes

Evan ChEn《陳誼廷》

15 April 2024

This is a compilation of solutions for the 1998 USAMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.

These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let $\mathbb{R}$ denote the set of real numbers" are typically omitted entirely.

Corrections and comments are welcome!

Contents

0 Problems ..... 2 1 Solutions to Day 1 ..... 3 1.1 USAMO 1998/1 ..... 3 1.2 USAMO 1998/2 ..... 4 1.3 USAMO 1998/3 2 Solutions to Day 2 ..... 6 2.1 USAMO 1998/4 ..... 6 2.2 USAMO 1998/5 2.3 USAMO 1998/6 ..... 8

§0 Problems

1. 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

$$\mid a_1-b_1\mid +\mid a_2-b_2\mid +\cdots +\mid a_{999}-b_{999}\mid \backslash left|a\_\{1\}-b\_\{1\}\backslash right|+\backslash left|a\_\{2\}-b\_\{2\}\backslash right|+\backslash cdots+\backslash left|a\_\{999\}-b\_\{999\}\backslash right|$$

ends in the digit 9. 2. 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 $A M / M C$. 3. 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

$$\tan a_0\tan a_1\cdots \tan a_n\ge n^{n+1}\backslash tan\; a\_\{0\}\; \backslash tan\; a\_\{1\}\; \backslash cdots\; \backslash tan\; a\_\{n\}\; \backslash geq\; n^\{n+1\}$$

4. 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.
5. 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 $a, b \in S$.
6. 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 $n$.

§1 Solutions to Day 1

§1.1 USAMO 1998/1

Available online at https://aops.com/community/p343865.

Problem statement

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

$$\mid a_1-b_1\mid +\mid a_2-b_2\mid +\cdots +\mid a_{999}-b_{999}\mid \backslash left|a\_\{1\}-b\_\{1\}\backslash right|+\backslash left|a\_\{2\}-b\_\{2\}\backslash right|+\backslash cdots+\backslash left|a\_\{999\}-b\_\{999\}\backslash right|$$

ends in the digit 9.

Let $S$ be the sum. Modulo 2,

$$S=\sum \mid a_i-b_i\mid \equiv \sum (a_i+b_i)=1+2+\cdots +1998\equiv 1(2)S=\backslash sum\backslash left|a\_\{i\}-b\_\{i\}\backslash right|\; \backslash equiv\; \backslash sum\backslash left(a\_\{i\}+b\_\{i\}\backslash right)=1+2+\backslash cdots+1998\; \backslash equiv\; 1\; \backslash quad(\backslash bmod\; 2)$$

Modulo 5,

$$S=\sum \mid a_i-b_i\mid =1\cdot 999\equiv 4(5)S=\backslash sum\backslash left|a\_\{i\}-b\_\{i\}\backslash right|=1\; \backslash cdot\; 999\; \backslash equiv\; 4\; \backslash quad(\backslash bmod\; 5)$$

So $S \equiv 9(\bmod 10)$.

§1.2 USAMO 1998/2

Available online at https://aops.com/community/p343866.

Problem statement

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 $A M / M C$.

By power of a point we have

$$AE\cdot AF=A{B}^2=\left(\frac{1}{2}AB\right)\cdot (2AB)=AD\cdot ACA\; E\; \backslash cdot\; A\; F=A\; B^\{2\}=\backslash left(\backslash frac\{1\}\{2\}\; A\; B\backslash right)\; \backslash cdot(2\; A\; B)=A\; D\; \backslash cdot\; A\; C$$

and hence $C D E F$ is cyclic. Then $M$ is the circumcenter of quadrilateral $C D E F$.

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 $A M / M C=5 / 3$.

§1.3 USAMO 1998/3

Available online at https://aops.com/community/p343867.

Problem statement

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

$$\tan a_0\tan a_1\cdots \tan a_n\ge n^{n+1}\mathrm{.}\backslash tan\; a\_\{0\}\; \backslash tan\; a\_\{1\}\; \backslash cdots\; \backslash tan\; a\_\{n\}\; \backslash geq\; n^\{n+1\}\; .$$

Let $x_{i}=\tan \left(a_{i}-\frac{\pi}{4}\right)$. Then we have that

$$\tan a_i=\tan (a_i-{45}^{\circ }+{45}^{\circ })=\frac{x_i+1}{1-x_i}\backslash tan\; a\_\{i\}=\backslash tan\; \backslash left(a\_\{i\}-45^\{\backslash circ\}+45^\{\backslash circ\}\backslash right)=\backslash frac\{x\_\{i\}+1\}\{1-x\_\{i\}\}$$

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

$$\prod _{i=1}^n(\frac{1}{y_i}-1)\ge n^{n+1}\backslash prod\_\{i=1\}^\{n\}\backslash left(\backslash frac\{1\}\{y\_\{i\}\}-1\backslash right)\; \backslash geq\; n^\{n+1\}$$

Proof. Homogenizing, we have to prove that

$$\prod _{i=1}^n(\frac{y_0+y_1+y_2+\cdots +y_n}{y_i}-1)\ge n^{n+1}\backslash prod\_\{i=1\}^\{n\}\backslash left(\backslash frac\{y\_\{0\}+y\_\{1\}+y\_\{2\}+\backslash cdots+y\_\{n\}\}\{y\_\{i\}\}-1\backslash right)\; \backslash geq\; n^\{n+1\}$$

By AM-GM, we have

$$\frac{y_1+y_2+y_3+\cdots +y_n}{y_0}\ge n\sqrt[n]{\frac{y_1{y}_2{y}_3\dots y_n}{y_1}}\backslash frac\{y\_\{1\}+y\_\{2\}+y\_\{3\}+\backslash cdots+y\_\{n\}\}\{y\_\{0\}\}\; \backslash geq\; n\; \backslash sqrt[n]\{\backslash frac\{y\_\{1\}\; y\_\{2\}\; y\_\{3\}\; \backslash ldots\; y\_\{n\}\}\{y\_\{1\}\}\}$$

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.

§2 Solutions to Day 2

§2.1 USAMO 1998/4

Available online at https://aops.com/community/p343869.

Problem statement

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.

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 $n=6$.

Full length

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.

Proof. If equality holds, then look at the horizontal gridlines and say two gridlines are related if some rectangle has horizontal edges along both gridlines. (Hence, the graph has degree either 1 or 2 at each vertex, for equality to hold.) The reader may verify the resulting graph consists only of even length cycles and single edges, which would mean $n-1$ is even.

Hence for $n=98$ the answer is indeed 98 as claimed.

§2.2 USAMO 1998/5

Available online at https://aops.com/community/p343870.

Problem statement

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 $a, b \in S$.

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 $S$.

§2.3 USAMO 1998/6

Available online at https://aops.com/community/p150148.

Problem statement

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 $n$.

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.

Proof. If not, then we have the following diagram, where $a=A_{1} A_{2}, b=A-2 A_{3}$, $c=A_{3} A_{4}, d=A_{4} A_{5}$.

Then $A_{1} A_{4}=c+a-b$ and $A_{5} A_{2}=b+d-c$. But now

$$A_1{A}_4+A_2{A}_5=(c+a-b)+(b+d-c)=a+d=A_1{A}_2+A_4{A}_{5}A\_\{1\}\; A\_\{4\}+A\_\{2\}\; A\_\{5\}=(c+a-b)+(b+d-c)=a+d=A\_\{1\}\; A\_\{2\}+A\_\{4\}\; A\_\{5\}$$

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).