IMO 1968 SL 26

Origin: GDR

Problem

Let $a > 0$ be a real number and $f(x)$ a real function defined on all of $\mathbb{R}$, satisfying for all $x \in \mathbb{R}$,

$$f(x + a) = \frac{1}{2} + \sqrt{f(x) - f(x)^2}.$$

(a) Prove that the function $f$ is periodic; i.e., there exists $b > 0$ such that for all $x$, $f(x + b) = f(x)$.

(b) Give an example of such a nonconstant function for $a = 1$.

3 The problem is unclear. Presumably $n$, $i$, $j$ and the $i$th digit are fixed.

4 The problem is unclear. The correct formulation could be the following: Given $k$ parallel lines $l_1, \cdots, l_k$ and $n_i$ points on the line $l_i$, $i = 1, 2, \cdots, k$, find the maximum possible number of triangles with vertices at these points.

Solution

(a) We shall show that the period of $f$ is $2a$. From

$$\left(f(x + a) - \frac{1}{2}\right)^2 = f(x) - f(x)^2$$

we obtain

$$\left(f(x) - f(x)^2\right) + \left(f(x + a) - f(x + a)^2\right) = \frac{1}{4}.$$

Subtracting the above relation for $x + a$ in place of $x$ we get

$$f(x) - f(x)^2 = f(x + 2a) - f(x + 2a)^2,$$

which implies

$$\left(f(x) - \frac{1}{2}\right)^2 = \left(f(x + 2a) - \frac{1}{2}\right)^2.$$

Since $f(x) \geq \frac{1}{2}$ holds for all $x$ by the condition of the problem, we conclude that

$$f(x + 2a) = f(x).$$

(b) The following function, as is directly verified, satisfies the conditions:

$$f(x) = \begin{cases} \frac{1}{2} & \text{if } 2n \leq x < 2n + 1, \ 1 & \text{if } 2n + 1 \leq x < 2n + 2, \end{cases} \qquad \text{for } n = 0, 1, 2, \cdots.$$