IMO 1961 Problem A1 Solve the following equations for x, y and z: x + y + z = a; x 2 + y 2 + z 2 = b 2 ; xy = z 2 What conditions must a and b satisfy for x, y and z to be distinct positive numbers?
A routine slog gives z = (a 2 - b 2 )/2a, x and y = (a 2 + b 2 )/4a ± √(10a 2 b 2 - 3a 4 - 3b 4 )/4a. A little care is needed with the conditions. Clearly x, y, z positive implies a > 0, and then z positive implies |b| < a. The expression under the root must be positive. It helps if you notice that it factorizes as (3a 2 - b 2 )(3b 2 - a 2 ). The second factor is positive because |b| < a, so the first factor must also be positive and hence a < √3 |b|. These conditions are also sufficient to ensure that x and y are distinct, but then z must also be distinct because z 2 = xy. s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 3rd IMO 1961 © John Scholes jscholes@kalva.demon.co.uk 19 Sep 1998 Last corrected/updated 24 Sep 2003