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 6 of 6: Conclude the center is the strict minimizer

Previous step Next step

In plain words

The whole argument only ever produces $\ge$ ; strictness appears exactly when $P$ is forced off one of the three symmetry planes, which happens for every $P\ne O$ .

OA+OB+OC+OD < PA+PB+PC+PD \quad (P\ne O)

Detailed analysis

For any $P\ne O$ , $P$ fails to lie in at least one of the three planes from step 5, so the corresponding inequality from step 4 is strict for that plane. Hence the sum of distances from $O$ to the four vertices is strictly less than the sum from any other point $P$ .