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Problem 779

AMC 12 late, AIME early

Geometry Difficulty 4.3 Prove it Brazilian Mathematical Olympiad · Brazil

Given a point $p$ inside a convex polyhedron $P$ . Show that there is a face $F$ of $P$ such that the foot of the perpendicular from $p$ to $F$ lies in the interior of $F$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $F$ be the face of $P$ closer to $p$ and $p'$ be the projection of $p$ in the plane of $F$ . Suppose $p'$ lies outside of $F$ . The line through $p$ and $p'$ cuts $P$ in two points $A$ and $B$ . Let $A$ be the point between $p$ and $p'$ . $A$ belongs to a face different from $F$ and the distance from $p$ to $A$ is less than the distance between $p$ and $F$ , which contradicts the minimality of $F$ . So $p'$ lies inside $F$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.