12th Putnam 1952

Let O be the center of a circle C and P₀ a point on the circle. Take points P_n on the circle such that angle P_nOP_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 P₁, P₂, ... , 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_nP_n+m is an arc of length less than 2πR/N. Hence the arcs P_n+mP_n+2m, P_n+2mP_n+3m ... have the same length. So for some k, P_n+km will lie in the arc PQ.

12th Putnam 1952

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002