Space geometry problem including centroid of a pyramid

Question

Let $VABC$ be a triangular pyramid and let $G$ be its centroid. A$\frac {MA} {MV}+ \frac {NB} {NV}+ \frac {PC} {PV}=1.$ I tried to use Menelaus' theorem, but I didn't get to any result.

Answer 1

You may assume $V=(0,0,0)$, $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Then $G=\bigl({1\over4},{1\over4}{1\over4}\bigr)$. Furthermore $M=(\mu,0,0)$, $N=(0,\nu,0)$, $P=(0,0,\pi)$, and it is assumed that $\lambda\mu\pi\ne0$. The plane $\sigma$ through the three points $M$, $N$, $P$ is given by $$\sigma:\qquad {x\over\mu}+{y\over\nu}+{z\over \pi}=1\ ,$$ hence this plane contains the point $G$ iff $${1\over\mu}+{1\over\nu}+{1\over\pi}=4\ .\tag{1}$$ On the other hand ${MA\over MV}={1-\mu\over\mu}$, and similarly for the other such quotients. It follows that $${MA\over MV}+{NB\over NV}+{PC\over PV}={1\over\mu}+{1\over\nu}+{1\over\pi}-3\ ,$$ and this is $=1$ iff $(1)$ holds, i.e., iff $G\in\sigma$.