Show that there is a finite set of points in the plane such that for any point P in the set we can find 1993 points in the set a distance 1 from P.
Solution
The basic idea is that we can find a circle which passes through a large number of lattice points. The point (±(t2-1)/t2+1), ±2t/(t2+1)) lies on the unit circle, so the set of such points for t = 1,2,...,1993 has 1993 points on each quadrant of the unit circle. Put n = ∏11993(1+t2), then all these points can be written as m/n, where m is an integer with |m| ≤ n.
Now consider the array of points (a/n, b/n), where 0 ≤ a,b ≤ 2n. The unit circle centered at any of these points has at least one of its quadrants passing through the array and hence contains at least 1993 of the points.
© John Scholes
jscholes@kalva.demon.co.uk
8 Nov 2003
Last corrected/updated 8 Nov 03