MathLabs

Problem 6

A cone of revolution has an inscribed sphere tangent to its base and sloping surface. A cylinder is circumscribed about the sphere, with its base in the base of the cone. If the volumes of the cone and cylinder are V1V_1 and V2V_2, respectively: (a) prove that V1≠V2V_1\ne V_2; (b) find the smallest possible value of V1/V2V_1/V_2, and in this case construct the half-angle of the cone.
Step 1 of 4: Express the cone dimensions from tangency
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In plain words

The sphere radius is the natural scale; only the half-angle remains as a shape parameter.

VO=r(1+1sin⁡θ),R=r(1+1sin⁡θ)tan⁡θVO=r\left(1+\dfrac1{\sin\theta}\right),\quad R=r\left(1+\dfrac1{\sin\theta}\right)\tan\theta
Detailed analysis

Let VV be the vertex, OO the sphere center, XX the center of the cone base, and rr the sphere radius. In the axial right triangle, VO=r/sin⁡θVO=r/\sin\theta because the sphere is tangent to the sloping side, while OX=rOX=r because it is tangent to the base. Thus the cone height is H=VX=r(1+1/sin⁡θ)H=VX=r(1+1/\sin\theta) and the base radius is R=Htan⁡θR=H\tan\theta.