# 1966 IMO Problems/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.

## Solution

We will need the following lemma to solve this problem: $\emph{Lemma:}$ Suppose there is a point in a regular tetrahedron $MNOP$ such that the distances from this point to the faces $MNO$, $MNP$, $MOP$, and $NOP$ are, respectively, $x_1$, $x_2$, $x_3$, and $x_4$. Then, the value $x_1 + x_2 + x_3 + x_4$ is constant. $\emph{Proof:}$

We will compute the volume of $MNOP$ in terms of the areas of the faces and the distances from the point to the faces: $$\textrm{Volume}(MNOP) = [MNO] \cdot x_1 \cdot \frac{1}{3} + [MNP] \cdot x_2 \cdot \frac{1}{3} + [MOP] \cdot x_3 \cdot \frac{1}{3} + [NOP] \cdot x_4 \cdot \frac{1}{3}$$ $$= [MNO] \cdot \frac{(x_1 + x_2 + x_3 + x_4)}{3}$$ $$\therefore\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.$$

This value is constant, so the proof of the lemma is complete. $\emph{Proof of problem statement:}$

Let our tetrahedron be $ABCD$, and the center of its circumscribed sphere be $O$. Construct a new regular tetrahedron, $WXYZ$, such that the centers of the faces of this tetrahedron are at $A$, $B$, $C$, and $D$.

For any point $P$ in $ABCD$, $$OA + OB + OC + OD = \sum \textrm{Distances from }O\textrm{ to faces of }WXYZ$$ $$= \sum \textrm{Distances from }P\textrm{ to faces of }WXYZ \leq PA + PB + PC + PD,$$

with equality only occurring when $AP$, $BP$, $CP$, and $DP$ are perpendicular to the faces of $WXYZ$, meaning that $P = O$. This completes the proof. $\square$

~mathboy100