2019 ROMANIAN MATHEMATICAL OLYMPIAD

Let $F:[0,1]\to \mathbb{R}$, $F(x)={\int }_0^{x}f(t)dt$, and consider the continuous function $g:[0,1]\to \mathbb{R}$, $g(x)=F(x)(F(1)-F(x))$. In terms of $g$, the conclusion reads $g(a_n)=g(a_n+1/n)$, for some $a_n$ in $[0,1-1/n]$.

Notice that $g(0)=0=g(1)$. Suppose now, if possible, that $g(x)=g(x+1/n)$ for no $x$ in $[0,1-1/n]$. By continuity, either $g(x)<g(x+1/n)$, for all $x$ in $[0,1-1/n]$, or $g(x)>g(x+1/n)$, for all $x$ in $[0,1-1/n]$. Consequently, $0=g(1)-g(0)={\sum }_{k=0}^{n-1}(g(k/n+1/n)-g(k/n))\ne 0$, which is a contradiction. The conclusion follows.