4th Brazil 1982 Problem 6 Five spheres of radius r are inside a right circular cone. Four of the spheres lie on the base of the cone. Each touches two of the others and the sloping sides of the cone. The fifth sphere touches each of the other four and also the sloping sides of the cone. Find the volume of the cone.
(1/3)πr 3 (2√2+1) 3 . Solution The left-hand diagram shows the four spheres on the base. Evidently AC = 2r√2. The right-hand diagram shows a vertical section through A, C and the center O of the top sphere. P is the apex of the cone and QR is a diameter of its base. Evidently PQR is similar to OAC and its sides are parallel and a distance r outside the corresponding sides of OAC. Also AOC is congruent to ABC, so ∠AOC = 90 o . Hence ∠QPR = 90 o also and so OP = r√2. The altitude from O in AOC has length AC/2 = r√2. Hence the altitude from P in PQR has length OP + r√2 + r = (2√2+1)r. Thus radius of the cone's base is also (2√2+1)r and its volume is (1/3)πr 3 (2√2+1) 3 . 4th Brazil 1982 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 2003 Last corrected/updated 24 Oct 03