IMO 1975 SL 9

Origin: NET

Problem

Let f(x) be a continuous function defined on the closed interval 0 \leqx \leq1. Let G(f) denote the graph of f(x): G(f) = {(x, y) \inR2 | 0 \leq x \leq1, y = f(x)}. Let Ga(f) denote the graph of the translated function f(x −a) (translated over a distance a), defined by Ga(f) = {(x, y) \in R2 | a \leqx \leqa + 1, y = f(x −a)}. Is it possible to find for every a, 0 < a < 1, a continuous function f(x), defined on 0 \leqx \leq1, such that f(0) = f(1) = 0 and G(f) and Ga(f) are disjoint point sets?

Solution

Suppose n is the natural number with na \leq1 < (n + 1)a. If a function f with the desired properties exists, then fa(a) = 0 and let w.l.o.g. f(a) > 0, or equivalently, let the graph of fa lie below the graph of f. In this case also f(2a) > f(a), since otherwise, the graphs of f and fa would intersect between a and 2a. Continuing in this way we are led to 0 = f(0) < f(a) < f(2a) < \cdot \cdot \cdot < f(na). Thus if na = 1, i.e., a = 1/n, such an f does not exist. On the other hand, if a ̸= 1/n, then we similarly obtain f(1) > f(1 −a) > f(1 −2a) > \cdot \cdot \cdot > f(1 −na). Choosing values of f at ia, 1 −ia, i = 1, . . . , n, so that they satisfy f(1 −na) < \cdot \cdot \cdot < f(1 −a) < 0 < f(a) < \cdot \cdot \cdot < f(na), we can extend f to other values of [0, 1] by linear interpolation. A function obtained this way has the desired property.