Let O be the center of a circle C and P_{0} a point on the circle. Take points P_{n} on the circle such that angle P_{n}OP_{n-1} = +1 for all integers n. Given that π is irrational, show that given any two distinct points P, Q on C, the (shorter) arc PQ contains a point P_{n}.

**Solution**

All the points must be distinct, for if we had P_{n} = P_{m} for n < m, then the integer m - n would be a multiple of 2π and hence π would be rational.

Choose N so that the arc length PQ < 2πR/N, where R is the radius of the circle. Divide the circle into N equal arcs length 2πR/N. Then at least two of the points P_{1}, P_{2}, ... , P_{N+1} must lie in the same arc. Suppose the points are P_{n} and P_{n+m} (with m, n positive), so that P_{n}P_{n+m} is an arc of length less than 2πR/N. Hence the arcs P_{n+m}P_{n+2m}, P_{n+2m}P_{n+3m} ... have the same length. So for some k, P_{n+km} will lie in the arc PQ.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002