I was told this question by a friend, who said that their friend had thought about it on and off for six months without any luck. I have then had it for a while without any luck either. It is in the style of a contest question, but it is unlike any other, as really nothing seems to gain any ground.
Here's the question:
Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function satisfying the following property: If $ABC$ is the equilateral triangle with side lengths 1, we have for any point $P$ inside $ABC$, $f(\overline{AP})+f(\overline{BP})+f(\overline{CP})=0$ where $\overline{AP}$ is the distance from point P to vertex $A$. (Example: by taking $P=A$ we see $f(0)=-2f(1)$.)
Prove or disprove: $f$ must be identically zero on its domain
Notice that continuity is critical since otherwise I could just take the function $f(x)=0$ whenever $x\in (0,1)$ and $f(0)=1$, $f(1)=\frac{-1}{2}$.