12th Putnam 1952

Problem A7

Let O be the center of a circle C and P0 a point on the circle. Take points Pn on the circle such that angle PnOPn-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 Pn.



All the points must be distinct, for if we had Pn = Pm 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, ... , PN+1 must lie in the same arc. Suppose the points are Pn and Pn+m (with m, n positive), so that PnPn+m is an arc of length less than 2πR/N. Hence the arcs Pn+mPn+2m, Pn+2mPn+3m ... have the same length. So for some k, Pn+km will lie in the arc PQ.



12th Putnam 1952

© John Scholes
5 Mar 2002