5th Mexico 1991 Problem A3 4 spheres of radius 1 are placed so that each touches the other three. What is the radius of the smallest sphere that contains all 4 spheres?
√(3/2) + 1 Solution The centers of the 4 spheres form a regular tetrahedron side 2. The center of the fifth sphere must obviously be at the center of the tetrahedron. An altitude of the tetrahedron will have one end at the centroid of a face, so its length is √(2 2 - ((2/3)√3) 2 ) = √(4 - 4/3) = 2√(2/3). So the distance from the center of the fifth sphere to one of the other centers is (3/4)2√(2/3) = √(3/2). Hence the radius of the fifth sphere is √(3/2) + 1. 5th Mexican 1991 © John Scholes jscholes@kalva.demon.co.uk 28 February 2004 Last corrected/updated 28 Feb 04