1st USAMO 1972 Problem 4 Let k be the real cube root of 2. Find integers A, B, C, a, b, c such that | (Ax 2 + Bx + C)/(ax 2 + bx + c) - k | < | x - k | for all non-negative rational x.
Taking the limit, we must have (Ak 2 + Bk + C) = k(ak 2 + bk + c), so A = b, B = c, C = 2a. Now notice that (bx 2 + cx + 2a) - k(ax 2 + bx + c) = (b - ak)x 2 + (c - bk)x + (2a - ck) = (x - k)( (b - ak)x + c - ak 2 ). So we require | (b - ak)x + c - ak 2 | < | ax 2 + bx + c | for all x ≥ 0. There are many ways to satisfy this. For example, take a = 1, b = c = 2. Then (b - ak)x is always positive and less than bx for positive x, and c - ak 2 is positive and less than c. 1st USAMO 1972 © John Scholes jscholes@kalva.demon.co.uk 13 Jul 2002