Nah, it's gotta be -10
:D

Its been a long time, been a long time,
Been a long lonely, lonely, lonely, lonely, lonely time.
Yes it has.

It's like this:
$-2 \text {ln}Y$

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: $f(x)=0$, for all its domain.

In which case the distance from vertex would be the circumradius.
If the midpoint means midpoint of side, which I don't think, then the distance becomes 1/2.