IMO 1973 · Problem 5

\(G\) is a set of non-constant functions of the real variable \(x\) of the form \[ f(x)=ax+b,\ a\text{ and }b\text{ are real numbers,} \] and \(G\) has the following properties:

(a) If \(f\) and \(g\) are in \(G\), then \(g\circ f\) is in \(G\); here \((g\circ f)(x)=g[f(x)]\).

(b) If \(f\) is in \(G\), then its inverse \(f^{-1}\) is in \(G\); here the inverse of \(f(x)=ax+b\) is \(f^{-1}(x)=(x-b)/a\).

(c) For every \(f\) in \(G\), there exists a real number \(x_f\) such that \(f(x_f)=x_f\).

Prove that there exists a real number \(k\) such that \(f(k)=k\) for all \(f\) in \(G\).

Official text of the problem. Official problem paper (PDF)

Solution

If \(G\) is empty or every member of \(G\) is the identity function, take \(k=0\).

Otherwise, choose a nonidentity function \(f(x)=ax+b\) in \(G\). If \(a=1\), property (c) forces \(b=0\), contradicting this choice. Thus \(a\ne1\), and \(f\) has the unique fixed point \[ k=\frac{b}{1-a}. \]

Let \(g(x)=cx+d\) be any member of \(G\). The functions \(f\circ g\) and \(g\circ f\) have the same nonzero slope \(ac\). By (a) and (b), the function \[ h=(f\circ g)\circ(g\circ f)^{-1} \] belongs to \(G\). Its slope is \(1\), so \(h(x)=x+t\) for some real number \(t\). Property (c) gives a fixed point of \(h\), which forces \(t=0\). Therefore \(h\) is the identity, and hence \[ f\circ g=g\circ f. \]

Consequently, \[ f(g(k))=g(f(k))=g(k). \] Thus \(g(k)\) is a fixed point of \(f\). Since that fixed point is unique, \(g(k)=k\). As \(g\) was arbitrary, the same \(k\) is fixed by every member of \(G\). \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).