IMO 1987 SL 1

Origin: AUS

Problem

Let f be a function that satisfies the following conditions: (i) If x > y and f(y) −y \geqv \geqf(x) −x, then f(z) = v + z, for some number z between x and y. (ii) The equation f(x) = 0 has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions; (iii) f(0) = 1. (iv) f(1987) \leq1988. (v) f(x)f(y) = f(xf(y) + yf(x) −xy). Find f(1987).

Solution

By (ii), f(x) = 0 has at least one solution, and there is the greatest among them, say x0. Then by (v), for any x, 0 = f(x)f(x0) = f(xf(x0) + x0f(x) −x0x) = f(x0(f(x) −x)). (1) It follows that x0 \geqx0(f(x) −x). Suppose x0 > 0. By (i) and (iii), since f(x0) −x0 < 0 < f(0) −0, there is a number z between 0 and x0 such that f(z) = z. By (1), 0 = f(x0(f(z)− z)) = f(0) = 1, a contradiction. Hence, x0 < 0. Now the inequality x0 \geqx0(f(x) −x) gives f(x) −x \geq1 for all x; so, f(1987) \geq1988. Therefore f(1987) = 1988.