IMO 1984

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Problem A3

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 + (∠AOX)/OX, where ∠AOX is measured in radians in the range [0, 2π). 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

Suppose the result is false. Let C1 be any circle center O. Then the locus of points X such that C(X) = C1 is a spiral from O to the point of intersection of OA and C1. Every point of this spiral must be a different color from all points of the circle C1. Hence every circle center O with radius smaller than C1 must include a point of different color to those on C1. Suppose there are n colors. Then by taking successively smaller circles C2, C3, ... , Cn+1 we reach a contradiction, since each circle includes a point of different color to those on any of the larger circles.

 


Solutions are also available in     Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

 

25th IMO 1984

© John Scholes
jscholes@kalva.demon.co.uk
19 Oct 1998