42nd Swedish 2002 Problem 6 A tetrahedron has five edges length 3 and circumradius 2. What is the length of the sixth edge?
6√(3/7) Solution Let the tetrahedron be ABCD. Let AD be the sixth edge and O the circumcenter. Let the foot of the perpendicular from O to the plane ABC be H. H must be the center of the equilateral triangle ABC. So AH = √3. OA = 2, so by Pythagoras, OH = 1. Let M be the midpoint of BC. HM = (√3)/2, so OM = (√7)/2. By symmetry the line OM passes through the midpoint of AD. So (AD/2)/AM = OH/OM. Hence AD = 6√(3/7). Thanks to Hector Mommaerts for pointing out an error in the original solution. 42nd Swedish 2002 © John Scholes jscholes@kalva.demon.co.uk 21 December 2003 Last corrected/updated 28 Mar 04