10th Chinese 1995 Problem B1 Four solid balls radii 3, 3, 2, 2 touch each other. What is the radius of the solid ball which touches all four balls?
6/11 Solution Let the large spheres have centers C, D and the two spheres radius 2 have centers A, B. Let the small sphere have center O and radius r. Let M, N be the midpoints of AB, CD respectively. Evidently the lines AB, CD and MN are mutually perpendicular. Obviously CN = 3, AM = 2, OC = r+3, and OB = r+2. Looking at the triangle AMC, we have CM = √21. Hence from the triangle CNM, we have MN = √12. From CNO, ON = √(r 2 + 6r) and from AMO, OM = √(r 2 + 4r). Since NO + OM = MN, we have √(r 2 + 6r) + √(r 2 + 4r) = √12. Squaring: r 2 + 6r = r 2 + 4r + 12 - 2√12√(r 2 + 4r). Squaring again, (r - 6) 2 = 12(r 2 + 4r), so 11r 2 + 60r - 36 = 0. Conveniently, this factorises on sight to (11r - 6)(r + 6) = 0. Since r must be positive, it is 6/11. 10th Chinese 1995 © John Scholes jscholes@kalva.demon.co.uk 5 Jan 2003