IMO 1986 Problem B2 Find all functions f defined on the non-negative reals and taking non-negative real values such that: f(2) = 0, f(x) ≠ 0 for 0 ≤ x < 2, and f(xf(y)) f(y) = f(x + y) for all x, y.
f(x+2) = f(xf(2)) f(2) = 0. So f(x) = 0 for all x ≥ 2. f(y) f((2-y)f(y)) = f(2) = 0. So if y  2/(2 - y 0 ), then we can find y 1 > y 0 (and y 1 < 2) so that f(y 0 ) = 2/(2 - y 1 ). Now let x 1 = 2 - y 1 . Then f(x 1 f(y 0 )) = f(2) = 0, so f(x 1 + y 0 ) = 0. But x 1 + y 0 < 2. Contradiction. So we must have f(x) = 2/(2 - x) for all x < 2. We have thus established that if a function f meets the conditions then it must be defined as above. It remains to prove that with this definition f does meet the conditions. Clearly f(2) = 0 and f(x) is non-zero for 0 ≤ x < 2. f(xf(y)) = f(2x/(2 - y)). If 2x/(2 - y) ≥ 2, then f(xf(y)) = 0. But it also follows that x + y ≥ 2, and so f(x + y) = 0 and hence f(xf(y)) f(y) = f(x + y) as required. If 2x/(2 - y) < 2, then f(xf(y)) f(y) = 2/(2 - 2x/(2-y)) 2/(2 - y) = 2/(2 - x - y) = f(x + y). So the unique function satisfying the conditions is: f(x) = 0 for x ≥ 2, and 2/(2 - x) for 0 ≤ x < 2. s are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 27th IMO 1986 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 1998