The Very First Functional Equation at IMO

Problem

The Very First Functional Equation at IMO

Solution 1

First note that, for all $x\in\mathbb{R},\,$ $\displaystyle 0\le f(x)\le 1\,$ and, moreover $\displaystyle \frac{1}{2}\le f(x)\le 1.\,$ Squaring gives

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

i.e.,

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

Letting here $x-a\,$ for $x\,$ gives

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

Now we subtract the two equations:

$\displaystyle f^2(x+a)-f(x+a)=f^2(x-a)-f(x-a)$

or

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

from which we conclude that $f(x+a)=f(x-a),\,$ implying that $f\,$ has a period $2a.\,$

To produce an example of such a function, note that if $\displaystyle v=\frac{1}{2}+\sqrt{u-u^2}\,$ then also $\displaystyle u=\frac{1}{2}+\sqrt{v-v^2},\,$ implying that we can choose arbitrary $u,\,\displaystyle u\in\left[\frac{1}{2},1\right]\,$ and define $f\,$ piecewise:

$f(x)\begin{cases} u,&x\in [0,1)\\ v,&x\in [1,2) \end{cases}$

and expand the function periodically with period $2.\,$ The simplest such function is obtained with $u=1,\,$ so that then $\displaystyle v=\frac{1}{2}.$

Solution 2

$\displaystyle \begin{align} f(x+a)&=\frac{1}{2}+\sqrt{f(x)-f^2(x)}\\ f(x+2a)&=\frac{1}{2}+\sqrt{f(x+a)-f^(x+a)}\\ &=\frac{1}{2}+\sqrt{\left(\frac{1}{2}+\sqrt{f(x)-f^2(x)}\right)-\left(\frac{1}{2}+\sqrt{f(x)-f^2(x)}\right)^2}\\ &=\frac{1}{2}+\sqrt{\left(\frac{1}{2}+\sqrt{f(x)-f^2(x)}\right)-\left(\frac{1}{4}+\sqrt{f(x)-f^2(x)}+f(x)-f^2(x)\right)}\\ &=\frac{1}{2}+\sqrt{\frac{1}{4}-f(x)+f^2(x)}\\ &=\frac{1}{2}+\sqrt{\left(\frac{1}{2}-f(x)\right)^2}\\ &=\frac{1}{2}-\left(\frac{1}{2}-f(x)\right),\,because f(x)\ge\frac{1}{2}\\ &=f(x). \end{align}$

Hence, $2a\,$ is a period.

Solution 3

For the argument of the square-root to be non-negative and the function to evaluate to $1/2$ plus a non-negative term, $f(x)\in[1/2,1]$.$

$\displaystyle \begin{align} &f(x+a)=\frac{1}{2}+\sqrt{f(x)-f^2(x)} \\ \Rightarrow &\left[f(x+a)-\frac{1}{2}\right]^2=f(x)-f^2(x) \\ \Rightarrow &f^2(x+a)-f(x+a)+\frac{1}{8}=- \left[ f^2(x)-f(x)+\frac{1}{8} \right] \\ \Rightarrow &f^2(x+2a)-f(x+2a)=f^2(x)-f(x) \\ \Rightarrow &f(x+2a)=f(x)~\text{or}~f(x+2a)=1-f(x). \end{align}$

From the bounds on $f(x)$, $f(x+2a)=1-f(x)$ only if $f(x+2a)=f(x)=1/2$. Thus, in general, $f(x+2a)=f(x)$.

To find an example function, let $g(x)=f^2(x)-f(x)+1/8$. From the range of $f(x)$, $g(x)\in[-1/8,1/8]$. We also need $g(x+1)=-g(x)$. Thus, let

$\displaystyle g(x)=\frac{\cos \pi x}{8}.$

Thus,

$\displaystyle \begin{align} &f^2(x)-f(x)+\frac{1-\cos\pi x}{8}=0 \\ &f^2(x)-f(x)+\frac{1}{4}\sin^2\left(\frac{\pi x}{2}\right)=0 \end{align}$

The root of this quadratic equation that satifies the bounds for $f(x)$ is

$\displaystyle f(x)=\frac{1}{2}+\frac{1}{2}\sqrt{1-\sin^2\left(\frac{\pi x}{2}\right)}.$

Acknowledgment

This is Problem 5 from the 1968 IMO (10^{th}), see IMO Compendium (1st edition, p 49). Solution 2 is by N. N. Taleb; Solution 3 is by Ami Itagi.

 

Cyclic inequalities in three variables

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