1984 IMO Problems/Problem 3

Problem

Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\angle AOX\over OX}$ , where $\angle AOX$ is measured in radians in the range $[0,2\pi)$ . Prove that we can find a point $X$ , not on $OA$ , such that its color appears on the circumference of the circle $C(X)$ .

Solution

Let $a_{n},\ n\ge 1$ be a sequence of positive reals such that $\left(\sum_{n\ge 1}a_{n}\right)^{2}<2\pi\ (*)$ . For each $n\ge 1$ , let $\mathcal C_{n}$ be the circle centered at $O$ with radius $\sum_{i=1}^{n}a_{i}$ .

Because of $(*)$ , we can find points $x_{i,j}\in\mathcal C_{i},\ i,j\ge 1$ such that for all $i,j\ge 1$ we have $\mathcal C(x_{i,j})=\mathcal C_{i+j}$ . We now forget about all the other points, and work only with the matrix $M=(x_{i,j})$ .

Suppose we use $n$ colors. There must be one, $c_{1}$ , which appears infinitely many times on the first row of $M$ , in, say, points $x_{1,j_{1}},x_{1,j_{2}},\ldots$ . Then $c_{1}$ cannot appear on the lines $j_{k}+1$ , $k\ge 1$ . Next, there is a color $c_{2}$ which appears infinitely often among the points $x_{j_{1}+1,j_{k}-j_{1}}$ , $k\ge 2$ . But then $c_{2}$ cannot appear on the lines $j_{k}+1$ for such $k$ . Repeating this procedure, we reach a stage where we have a row of $M$ (infinitely many actually) on which none of our $n$ colors $c_{1},\ldots,c_{n}$ can appear. This is a contradiction.

This solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]

1984 IMO (Problems) • Resources
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1984 IMO Problems/Problem 3

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