If the Lemoine point of triangle ABC lies on the altitude from vertex C, show that either AC = BC or angle C =90 degrees.

Question

There are so many aspects to this problem - I constructed it, but getting the Lemoine Point alone is quite a mess - that I'm not sure where to start with the proof. What aspects of the Lemoine Point should I address in order to show that either of these possibilities must be true?

Answer 1

We use barycentric coordinates. Let the Lemoine point be $L(a^2:b^2:c^2)$, so the displacement vector $\vec{CL} = (a^2:b^2:-a^2-b^2)$. Now we have $\vec{CL} \perp \vec{AB}$ iff $0=c^2(b^2-a^2)-(a^2+b^2)(b^2-a^2)$, so either $c^2=a^2+b^2$ or $a=b$ as desired.