22nd USAMO 1993 Problem 1 n > 1, and a and b are positive real numbers such that a n - a - 1 = 0 and b 2n - b - 3a = 0. Which is larger? Solution
: a > b. Note that a n = a + 1 > 1 (since a is positive). Hence a > 1. So a 2n = (a + 1) 2 = a 2 + 2a + 1. Put a = 1 + k, then a 2 = 1 + 2k + k 2 > 1 + 2k, so a 2 + 1 > 2 + 2k = 2a. Hence a 2n > 4a. So (b/a) 2n = (b + 3a)/a 2n < (b + 3a)/4a. If b/a ≥ 1, then (b + 3a)/4a ≤ (b + 3b)/4a = b/a, so (b/a) 2n < b/a. Contradiction. Hence b/a < 1. 22nd USAMO 1993 © John Scholes jscholes@kalva.demon.co.uk 23 Aug 2002