Prove that for every positive integer m we can find a finite set S of points in the plane, such that given any point A of S, there are exactly m points in S at unit distance from A.

**Solution**

Take a_{1}, a_{2}, ... , a_{m} to be points a distance 1/2 from the origin O. Form the set of 2^{m} points ±a_{1} ±a_{2} ± ... ±a_{m}. Given such a point, it is at unit distance from the m points with just one coefficient different. So we are home, provided that we can choose the a_{i} to avoid any other pairs of points being at unit distance, and to avoid any degeneracy (where some of the 2^{m} points coincide).

The distance between two points in the set is |c_{1}a_{1} + c_{2}a_{2} + ... + c_{m}a_{m}|, where c_{i} = 0, 2 or -2. So let us choose the a_{i} inductively. Suppose we have already chosen up to m. The constraints on a_{m+1} are that we do not have |c_{1}a_{1} + c_{2}a_{2} + ... + c_{m}a_{m} + 2a_{m+1}| equal to 0 or 1 for any c_{i} = 0, 2 or -2, apart from the trivial cases of all c_{i} = 0. Each | | = 0 rules out a single point and each | | = 1 rules out a circle which intersects the circle radius 1/2 about the origin at 2 points and hence rules out two points. So the effect of the constraints is to rule out a finite number of points, whereas we have uncountably many to choose from.

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

8 Oct 1998

Last updated/corrected 14 Mar 03