G is a set of non-constant functions f. Each f is defined on the real line and has the form f(x) = ax + b for some real a, b. If f and g are in G, then so is fg, where fg is defined by fg(x) = f(g(x)). If f is in G, then so is the inverse f-1. If f(x) = ax + b, then f-1(x) = x/a - b/a. Every f in G has a fixed point (in other words we can find xf such that f(xf) = xf. Prove that all the functions in G have a common fixed point.
Solution
f(x) = ax + b has fixed point b/(1-a). If a = 1, then b must be 0, and any point is a fixed point. So suppose f(x) = ax + b and g(x) = ax + b' are in G. Then h the inverse of f is given by h(x) = x/a - b/a, and hg(x) = x + b'/a - b/a. This is in G, so we must have b' = b.
Suppose f(x) = ax + b, and g(x) = cx + d are in G. Then fg(x) = acx + (ad + b), and gf(x) = acx + (bc + d). We must have ad + b = bc + d and hence b/(1-a) = c/(1-d), in other words f and g have the same fixed point.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 1998