Let $\triangle ABC$ be an equilateral triangle with side $a$. $M$ is a point such that $MS = d$ where $S$ is the centroid of $\triangle ABC$. Prove that the area of the triangle whose sides are of length $MA,MB,MC$ is $[\frac{\root \of3}{12}]|a^2 - 3d^2|$
This answer results in this equation: $$[ABC] = 1.5 * [T] + \frac{\root \of 3}{8}(MA^2 + MB^2 + MC^2)$$ (Where $[x]$ is the area of $x$ and $T$ is the triangle we need to find the area for.)
which on rearranging gives, $$[T] = \frac{\root \of 3}{6}(a^2 - \frac{1}{2}(MA^2 + MB^2 + MC^2))$$ So can we derive a relation between $d$ and $MA,MB,MC$?
Yes, for the centroid $G$ of any triangle $ABC$ and $M$ a point we have $$MA^2 + MB^2 + MC^2= GA^2 + GB^2 + GC^2 + 3MG^2$$
This you can show using vectors and $\vec{GA}+\vec{GB}+\vec{GC}=\vec{0}$.