IMO 1973 SL 17

Origin: POL

Problem

Let F be a nonempty set of functions f : R \toR of the form f(x) = ax + b, where a and b are real numbers and a ̸= 0. Suppose that F satisfies the following conditions: (1) If f, g \inF, then g ◦f \inF, where (g ◦f)(x) = g[f(x)]. (2) If f \inF and f(x) = ax + b, then the inverse f −1 of f belongs to F (f −1(x) = (x −b)/a). (3) None of the functions f(x) = x + c, for c ̸= 0, belong to F. Prove that there exists x0 \inR such that f(x0) = x0 for all f \inF.

Solution

Let f1(x) = ax+b and f2(x) = cx+d be two functions from F. We define g(x) = f1◦f2(x) = acx+(ad+b) and h(x) = f2◦f1(x) = acx+(bc+d). By the condition for F, both g(x) and h(x) belong to F. Moreover, there exists h−1(x) = x−(bc+d) ac , and h−1 ◦g(x) = acx + (ad + b) −(bc + d) ac = x + (ad + b) −(bc + d) ac belongs to F. Now it follows that we must have ad + b = bc + d for every f1, f2 \inF, which is equivalent to b a−1 = d c−1 = k. But these formulas exactly describe the fixed points of f1 and f2: f1(x) = ax + b = x ⇒x = b a−1. Hence all the functions in F fix the point k.