IMO 1984 SL 13

Origin: BUL

Problem

Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume 1 does not exceed 3\pi.

Solution

Let Z be the given cylinder of radius r, altitude h, and volume \pir2h = 1, k1 and k2 the circles surrounding its bases, and V the volume of an inscribed tetrahedron ABCD. We claim that there is no loss of generality in assuming that A, B, C, D all lie on k1 \cupk2. Indeed, if the vertices A, B, C are fixed and D moves along a segment EF parallel to the axis of the cylinder (E \ink1, F \ink2), the maximum distance of D from the plane ABC (and consequently the maximum value of V ) is achieved either at E or at F. Hence we shall consider only the following two cases: (i) A, B \ink1 and C, D \ink2. Let P, Q be the projections of A, B on the plane of k2, and R, S the projections of C, D on the plane of k1, respectively. Then V is one-third of the volume V ′ of the prism ARBSCPDQ with bases ARBS and CPDQ. The area of the quadri- lateral ARBS inscribed in k1 does not exceed the area of the square inscribed therein, which is 2r2. Hence 3V = V ′ \leq2r2h = 2/\pi. (ii) A, B, C \ink1 and D \ink2. The area of the triangle ABC does not exceed the area of an equilateral triangle inscribed in k1, which is \sqrt 3r2/4. Consequently, V \leq \sqrt 4 r2h = \sqrt 4\pi < 3\pi.