61st Putnam 2000 Problem A1 k is a positive constant. The sequence x i of positive reals has sum k. What are the possible values for the sum of x i 2 ? Solution
: any value in the open interval (0, k 2 ). Since the terms are positive we have 0 < ∑ x n 2 < ( ∑ x n ) 2 = k 2 . So certainly all sums must lie in the interval (0, k 2 ). We show how to get any value in the interval by explicit construction. We may represent the value as hk 2 where 0 < h < 1. Take c = (1 - h)/(1 + h). Take the geometric sequence with first term k(1 - c) and ratio c ( k(1 - c), kc(1 - c), kc 2 (1 - c), ... ). We have ∑ x n = k as required, and ∑ x n 2 = k 2 (1 - c)/(1 + c) = hk 2 as required. 61st Putnam 2000 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2001