1984 IMO Problems/Problem 3


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 $C(X)$.


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]

See Also

1984 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
All IMO Problems and Solutions