Problem 3

Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.

Step 1 of 6: Reduce to one pair of opposite edges

Previous step Next step

In plain words

Every vertex equidistant from $C$ and $D$ automatically sits on the plane that perpendicularly bisects $CD$ ; this geometric fact is what makes the rest of the proof work.

AC=AD,\ BC=BD \ \Rightarrow\ A,B\in\Pi

Regular tetrahedron

ABCD

Detailed analysis

Let $ABCD$ be the regular tetrahedron and let $X$ be the midpoint of edge $CD$ . Because $AC=AD$ and $BC=BD$ , both $A$ and $B$ lie on the perpendicular-bisector plane $\Pi$ of segment $CD$ , which passes through $X$ .