1st APMO 1989 Problem 5 f is a strictly increasing real-valued function on the reals. It has inverse f -1 . Find all possible f such that f(x) + f -1 (x) = 2x for all x. Solution
: f(x) = x + b for some fixed real b. Suppose for some a we have f(a) ≠ a. Then for some b ≠ 0 we have f(a) = a + b. Hence f(a + b) = a + 2b (because f( f(a) ) + f -1 ( f(a) ) = 2 f(a), so f(a + b) + a = 2a + 2b ) and by two easy inductions, f(a + nb) = a + (n+1)b for all integers n (positive or negative). Now take any x between a and a + b. Suppose f(x) = x + c. The same argument shows that f(x + nc) = x + (n+1)c. Since f is strictly increasing x + c must lie between f(a) = a + b and f(a+b) = a + 2b. So by a simple induction x + nc must lie between a + nb and a + (n+1)b. So c lies between b + (x-a)/n and b + (a+b-x)/n or all n. Hence c = b. Hence f(x) = x + b for all x. If there is no a for which f(a) ≠ a, then we have f(x) = x for all x. 1st APMO 1989 © John Scholes jscholes@kalva.demon.co.uk 23 Sep 2002