IMO 1990 SL 10

Origin: ICE

Problem

A plane cuts a right circular cone into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the ratio of the volume of the smaller part to the volume of the whole cone.

Solution

Let r be the radius of the base and h the height of the cone. We may assume w.l.o.g. that r = 1. Let A be the top of the cone, BC the di- ameter of the circumference of the base such that the plane touches the circumference at B, O the center of the base, and H the midpoint of OA (also belonging to the plane). Let BH cut the sheet of the cone at D. By applying Menelaus’s theorem to \triangleAOC and \triangleBHO, we conclude that AD DC = CB BO \cdot OH HA = 1 2 and HD DB = HA AO \cdot OC CB = 1 4. The plane cuts the cone in an ellipse whose major axis is BD. Let E be the center of this ellipse and FG its minor axis. We have BE ED = 1 2. Let E′, F ′, G′ be radial projections of E, F, G from A onto the base of the cone. Then E sits on BC. Let h(X) denote the height of a point X with respect to the base of the cone. We have h(E) = h(D)/2 = h/3.

Hence EF = 2E′F ′/3. Applying Menelaus’s theorem to \triangleBHO we get OE′ E′B = BE EH \cdot HA AO = 1. Hence EF = 2 \sqrt 2 = \sqrt 3. Let d denote the distance from A to the plane. Let V1 and V denote the volume of the cone above the plane (on the same side of the plane as A) and the total volume of the cone. We have V1 V = BE \cdot EF \cdot d h = (2BH/3)(1/ \sqrt 3)(2SAHB/BH) h = (2/3)(1/ \sqrt 3)(h/2) h

\sqrt 3. Since this ratio is smaller than 1/2, we have indeed selected the correct volume for our ratio.