maximizing distance in a polyhedron

Question

How can I prove that the point that is maximally distant to a specific point in a convex polyhedron must be in a vertex of the polyhedron?

Answer 1

Let be $P$ that specific point. Let be $Q$ any point in the polyhedron that is not a vertex. Then, there exist a real number $\varepsilon>0$ and a nonzero vector $\vec v$ such that the segment that connects the points $Q'=Q+\varepsilon\vec v$ and $Q''=Q-\varepsilon\vec v$ is contained in the polyhedron.

Then the segment $PQ$ is the median of the triangle $PQ'Q''$; thus, one of the segments $PQ'$ and $PQ''$ are longer than $PQ$.