Question 124
Let $P_{1} P_{2} P_{3} P_{4}$ be a tetrahedron in $\mathbb{R}^{3}$ and let $O$ be a point equidistant from each of its vertices. Suppose there exists a point $H$ such that for each $i$, the line $P_{i} H$ is perpendicular to the plane through the other three vertices. Line $P_{1} H$ intersects the plane through $P_{2}, P_{3}, P_{4}$ at $A$, and contains a point $B \neq P_{1}$ such that $O P_{1}=O B$. Show that $H B=3 H A$.