28th CanMO 1996 Problem 2 Find all real solutions to the equations x = 4z 2 /(1 + 4z 2 ), y = 4x 2 /(1 + 4x 2 ), z = 4y 2 /(1 + 4y 2 ).
It is immediate from the equations given that 0 <= x, y, z < 1. Rearranging, we have 4z 2 = x/(1 - x), 4x 2 = y/(1 - y), 4y 2 = z/(1 - z). Multiplying gives (xyz) 2 (1 - x)(1 - y)(1 - z) = xyz/64. If x = 0, then from the first equation in the question, z = 0 and hence also y = 0. That is one solution. Otherwise we have xyz non- zero and hence x(1 - x) y(1 - y) z(1 - z) = 1/64. But x(1 - x) = 1/4 - (x - 1/2) 2 <= 1/4 with equality iff x = 1/2. Similarly for y and z, so the only other solution is x = y = z = 1/2. 28th CanMO 1996 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002