Olympiad calculus problem This problem is from a qualifying round in a Colombian math Olympiad, I thought some time about it but didn't make any progress. It is as follows.
Given a continuous function $f : [0,1] \to \mathbb{R}$ such that $$\int_0^1{f(x)\, dx} = 0$$ Prove that there exists $c \in (0,1) $ such that $$\int_0^c{xf(x) \, dx} = 0$$
I will appreciate any help with it.

Let $G(x) = \int_{0}^{x} t f(t) \, \mathrm{d}t$, and suppose that $G(c) \neq 0$ for all $c \in (0, 1)$. Then by IVT, either

*

*(Case 1) $G > 0$ on $(0, 1)$, or

*(Case 2) $G < 0$ on $(0, 1)$.

However,
\begin{align*}
0
= \int_{0}^{1} f(x) \, \mathrm{d}x
= \int_{0}^{1} \frac{G'(x)}{x} \, \mathrm{d}x
= G(1) + \int_{0}^{1} \frac{G(x)}{x^2} \, \mathrm{d}x.
\end{align*}
This is $ > 0 $ in Case 1 and $ < 0$ in Case 2, a contradiction.
