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 2 of 6: Compare

P

with its projection on

\Pi

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In plain words

Projecting onto a plane can only shorten the distance to points that already lie in that plane, and $A,B\in\Pi$ by step 1.

PA\ge P'A,\quad PB\ge P'B

Detailed analysis

For an arbitrary point $P$ , let $P'$ be the foot of the perpendicular from $P$ to $\Pi$ . Since $\angle PP'A=\angle PP'B=90°$ , the Pythagorean theorem gives $PA^2=PP'^2+P'A^2\ge P'A^2$ and similarly for $B$ , with equality exactly when $P=P'$ .