Prove that a point in space has the smallest sum of the distances to the vertices of a regular tetrahedron iff it is the center of the tetrahedron.
Solution
Let the tetrahedron be ABCD and let P be a general point. Let X be the midpoint of CD. Let P' be the foot of the perpendicular from P to the plane ABX. We show that if P does not coincide with P', then PA + PB + PC + PD > P'A + P'B + P'C + P'D.
PA > P'A (because angle PP'A = 90o) and PB > P'B. P'CD is isosceles and PCD is not but P is the same perpendicular distance from the line CD as P'. It follows that PC + PD > P'C + P'D. The easiest way to see this is to reflect C and D in the line PP' to give C' and D'. Then PC = PC', and PC' + PD > C'D = P'C' + P'D = P'C + P'D.
So if P has the smallest sum, it must lie in the plane ABX and similarly in the plane CDY, where Y is the midpoint of AB, and hence on the line XY. Similarly, it must lie on the line joining the midpoints of another pair of opposite sides and hence must be the center.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
29 Sep 1998
Last corrected/updated 26 Sep 2003