# A Functional Equation from 1968 IMO

Here is Problem 5 from the 1968 International Mathematical Olympiad [Compendium, p. 49]:

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

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

1. Prove that the function $f$ is periodic; i.e., there exists $b\gt 0$ such that for all $x,$ $f(x+b)=f(x).$
2. Give an example of such a nonconstant function for $a=1.$

Solution

### References

1. D. Djukic et al, The IMO Compendium, Springer, 2011 (Second edition)

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

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

1. Prove that the function $f$ is periodic; i.e., there exists $b\gt 0$ such that for all $x,$ $f(x+b)=f(x).$
2. Give an example of such a nonconstant function for $a=1.$

### Solution

Note at the outset that, for all $x,$ $f(x)\ge\frac{1}{2},$ and, since it is defined for all real $x,$ $f(x)\le 1$ (to keep the expression under the square root non-negative.)

Set $g(x)=f(x)-\frac{1}{2}.$ Then $g$ satisfies a simplified equation

$g(x+a)=\sqrt{\frac{1}{4}-g^2(x)}$

which is equivalent to $g^2(x+a)+g^2(x)=\frac{1}{4}.$ This problem is related to the one posed by Dan Sitaru and Leo Giugiuc. Replace $x$ with $x-a$ to obtain $g^2(x)+g^2(x-a)=\frac{1}{4}.$ From the two equations, $g(x-a)=g(x+a),$ for all $x,$ making $g$ periodic with period $2a.$

The simplest example of such a function is the one defined piecewise:

$g(x)=\begin{cases} \frac{1}{2}, & x\in [0,1)\\ 0, & x\in [1,2) \end{cases}$

and extended periodically for $x$ outside $[0,2).$