# A Problem From the Romanian Mathematical Competitions (2016)

### Solution

Introduce the function $\displaystyle g:\,[0,1]\to\mathbb{R}\,$ with

$\displaystyle g(y)=\begin{cases} \frac{1}{y}\int_0^yf(x)dx)-\frac{1}{y^2}\int_0^yxf(x)fx, & y\in(,1]\\ 0,& y=0. \end{cases}$

By l'Hôpital's rule, $\displaystyle \lim_{y\to 0^{+}}=0,\,$ implying that $g\,$ is continuous on $[0,1]\,$ and differentiable on $(0,1).\,$ Therefore, according to Rolle's theorem, there is $c\in (0,1)\,$ such that $g'(c)=0.\,$ Fort that $c,\,$ $\displaystyle \int_0^cxf(x)dx=\frac{c}{2}\cdot\int_0^cf(x)dx.$

### Acknowledgment

This is from the Romanian Mathematical Competitions magazine (2016, the Putnam for Seniors section). The problem is by Cezar Lupu and Leonard Giugiuc. I am grateful to Leo for bringing the problem to my attention, along with the author's solution.