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 3 of 6: Compare

P

with

P'

for the far pair

C,D

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In plain words

This is the same reflection trick used to find the shortest path that touches a line: unfolding the broken path $P\to C\to D$ turns the inequality into an ordinary triangle inequality.

PC+PD\ge P'C+P'D

Detailed analysis

Because $\Pi\perp CD$ at $X$ , segment $PP'$ is parallel to $CD$ , so $P,P',C,D$ are coplanar. Reflecting $C$ across line $PP'$ to $C'$ gives $PC=PC'$ ; the triangle inequality $PC'+PD\ge C'D$ is equality only when $P,C',D$ are collinear (i.e. $P=P'$ ), and since $P'\in\Pi$ gives $P'C=P'D$ , we have $C'D=P'C'+P'D=P'C+P'D$ , so $PC+PD\ge P'C+P'D$ .