75th Romanian Mathematical Olympiad

Let $(f,g)$ be a pair of functions satisfying the given condition. We shall show that $f^{\prime \prime }$ is constant. Suppose that $f^{\prime \prime }$ is not constant. Then, because $(f(x)-g(0))\cdot (f^{\prime }(x)-g^{\prime }(0))\cdot (f^{\prime \prime }(x)-g^{\prime \prime }(0))=0$, for all $x\in \mathbb{R}$, and $f^{\prime \prime }$ is continuous, there is $a\in \mathbb{R}$ and $r>0$ such that $f^{\prime \prime }(x)\notin \{0,g^{\prime \prime }(0)\}$, for any $x\in (a-r,a+r)$. It follows that $(f(x)-g(0))\cdot (f^{\prime }(x)-g^{\prime }(0))=0$, for $x\in (a-r,a+r)$.

Two cases are possible. Case 1. There exists $b\in (a-r,a+r)$ such that $f^{\prime }(b)\ne g^{\prime }(0)$. By continuity of $f^{\prime }$, there is $s>0$ such that $(b-s,b+s)\subset (a-r,a+r)$ and $f^{\prime }(x)\ne g^{\prime }(0)$, for all $x\in (b-s,b+s)$, implying $f(x)=g(0)$ for $x\in (b-s,b+s)$. We get $f^{\prime \prime }(x)=0$ for $x\in (b-s,b+s)$, a contradiction.

Case 2. $f^{\prime }(x)=g^{\prime }(0)$, for any $x\in (a-r,a+r)$. We obtain $f^{\prime \prime }(x)=0$, for all $x\in (a-r,a+r)$, a contradiction again.

Thus $f^{\prime \prime }$ is constant on $\mathbb{R}$. Let $m\in \mathbb{R}$ such that $f^{\prime \prime }(x)=2m$, for any $x\in \mathbb{R}$. As $(f^{\prime }(x)-2mx{)}^{\prime }=0$, for all $x\in \mathbb{R}$, there is $n\in \mathbb{R}$ such that $f^{\prime }(x)-2mx-n=0$, for $x\in \mathbb{R}$. As a consequence there exists $p\in \mathbb{R}$ such that $f(x)=m{x}^2+nx+p$, for all $x\in \mathbb{R}$. In the same way, $g^{\prime \prime }$ is constant on $\mathbb{R}$, implying the existence of $m^{\prime },n^{\prime },p^{\prime }\in \mathbb{R}$ such that $g(x)=m^{\prime }{x}^2+n^{\prime }x+p^{\prime }$, for $x\in \mathbb{R}$. In the case $m\ne m^{\prime }$, as the equation $(f(x)-g(x))\cdot (f^{\prime }(x)-g^{\prime }(x))=0$ has at most three real solution, the condition of the problem is not possible. Thus $m=m^{\prime }$, so the pairs of functions satisfying the given conditions are of the form $f(x)=m{x}^2+nx+p$ and $g(x)=m{x}^2+n^{\prime }x+p^{\prime }$, for any $x\in \mathbb{R}$.