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.

This is a streamlined version of Thomas Andrews' proof:
Put $F(x):=\int_0^x f(t)dt$ and consider the auxiliary function $\phi(x)={1\over x}\int_0^x F(t)dt$. Then $\phi(0)=0$, $\ \phi(1)=\int_0^1 F(t)dt=:\alpha$, and by partial integration one obtains
$$\phi'(x)=-{1\over x^2}\int_0^xF(t)dt +{1\over x}F(x)={1\over x^2}\int_0^x t f(t)dt\ .$$
The mean value theorem provides a $\xi\in(0,1)$ with $\phi'(\xi)=\alpha$. If $\alpha$ happens to be $0$ we are done. Otherwise we invoke $F(1)=0$ and conclude that $\phi'(1)=-\alpha$. It follows that there is a $\xi'\in(\xi,1)$ with $\phi'(\xi')=0$.
