IMO 1968 · Problem 5

Let \(f\) be a real-valued function defined for all real numbers \(x\) such that, for some positive constant \(a\), the equation \[ f(x+a)=\frac12+\sqrt{f(x)-[f(x)]^2} \] holds for all \(x\).

(a) Prove that the function \(f\) is periodic (i.e., there exists a positive number \(b\) such that \(f(x+b)=f(x)\) for all \(x\)).

(b) For \(a=1\), give an example of a non-constant function with the required properties.

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

Answer

(a) A period is \(b=2a\).

(b) One example is \[ \boxed{f(x)=\frac12+\frac12\left|\sin\frac{\pi x}{2}\right|.} \]

Solution

(a) Since the square root is real, we have \[ f(x)-[f(x)]^2\ge 0, \] so \(0\le f(x)\le1\). Moreover, applying the given equation at \(x-a\) gives \[ f(x)=\frac12+\sqrt{f(x-a)-[f(x-a)]^2}\ge\frac12. \] Thus \[ \frac12\le f(x)\le1\qquad\text{for every real }x. \]

The given equation now yields \[ \begin{aligned} f(x+a)-[f(x+a)]^2 &=\frac14-\left(f(x+a)-\frac12\right)^2\\ &=\frac14-\bigl(f(x)-[f(x)]^2\bigr)\\ &=\left(f(x)-\frac12\right)^2. \end{aligned} \] Applying the equation once more, we obtain \[ \begin{aligned} f(x+2a) &=\frac12+\sqrt{f(x+a)-[f(x+a)]^2}\\ &=\frac12+\left|f(x)-\frac12\right|\\ &=f(x), \end{aligned} \] where the last equality uses \(f(x)\ge\frac12\). Since \(2a>0\), this proves periodicity.

(b) Define, for every real \(x\), \[ f(x)=\frac12+\frac12\left|\sin\frac{\pi x}{2}\right|. \] Then \[ f(x)-[f(x)]^2 =\frac14\left(1-\sin^2\frac{\pi x}{2}\right) =\frac14\cos^2\frac{\pi x}{2}. \] Consequently, \[ \begin{aligned} \frac12+\sqrt{f(x)-[f(x)]^2} &=\frac12+\frac12\left|\cos\frac{\pi x}{2}\right|\\ &=\frac12+\frac12\left|\sin\frac{\pi(x+1)}{2}\right|\\ &=f(x+1). \end{aligned} \] Thus the required equation holds for all real \(x\) when \(a=1\). Finally, \(f(0)=\frac12\) and \(f(1)=1\), so \(f\) is non-constant.

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