IMO 1984 · Problem 3

In the plane two different points \(O\) and \(A\) are given. For each point \(X\) of the plane, other than \(O\), denote by \(a(X)\) the measure of the angle between \(OA\) and \(OX\) in radians, counterclockwise from \(OA\) (\(0 \le a(X) < 2\pi\)). Let \(C(X)\) be the circle with center \(O\) and radius of length \(OX + a(X)/OX\). Each point of the plane is colored by one of a finite number of colors. Prove that there exists a point \(Y\) for which \(a(Y) > 0\) such that its color appears on the circumference of the circle \(C(Y)\).

Official text of the problem. Official problem paper (PDF)

Solution

Let \(m\) be the number of colors. For every positive real number \(t\), let \(S(t)\) be the set of colors appearing on the circumference of the circle with center \(O\) and radius \(t\).

There are at most \(2^m\) possible sets \(S(t)\). Choose \(2^m+1\) distinct radii in \([1,2]\). By the pigeonhole principle, two of these radii, say \(r<s\), satisfy \[ S(r)=S(s). \]

Set \[ \theta=r(s-r). \] Since \(1\le r<s\le2\), we have \[ 0<\theta\le2<2\pi. \] Consequently, there is a point \(Y\) at distance \(r\) from \(O\) whose counterclockwise angle from \(OA\) is \(\theta\). Thus \(a(Y)=\theta>0\), and the radius of \(C(Y)\) is \[ OY+\frac{a(Y)}{OY} =r+\frac{r(s-r)}r =s. \]

The color of \(Y\) belongs to \(S(r)\), which equals \(S(s)\). Hence that color appears on the circumference of the circle with center \(O\) and radius \(s\), namely \(C(Y)\), as required. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).