A cone of revolution has an inscribed sphere tangent to the base of the cone (and to the sloping surface of the cone). A cylinder is circumscribed about the sphere so that its base lies in the base of the cone. The volume of the cone is V1 and the volume of the cylinder is V2.
(a) Prove that V1 ≠ V2;
(b) Find the smallest possible value of V1/V2. For this case construct the half angle of the cone.
Solution
Let the vertex of the cone be V, the center of the sphere be O and the center of the base be X. Let the radius of the sphere be r and the half-angle of the cone θ.
Then the the cone's height is VO + OX = r(1 + 1/sin θ), and the radius of its base is r(1 + 1/sin θ) tan θ. Hence V1/V2 = (1/6) (1 + 1/sin θ)3 tan2θ = (1 + s)3(6s(1 - s2)), where s = sin θ.
We claim that (1 + s)3(6s(1 - s2)) ≥ 4/3. This is equivalent to 1 + 3s + 3s2 + s3 ≥ 8s - 3s3 or 1 - 5s + 3s2 + 9s3 >= 0. But we can factorise the cubic as (1 - 3s)2(1 + s). So we have V1/V2 ≥ 4/3 with equality iff s = 1/3.
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.
2nd IMO 1960
© John Scholes
jscholes@kalva.demon.co.uk
18 Sep 1998
Last corrected/updated: 24 Sep 2003