IMO 1960 - Problem 7

Problem: A sphere is inscribed in a regular cone. Around the sphere a cylinder is circumscribed so that its base is in the same plane as the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ the volume of the cylinder.
(a) Prove that $V_1=V_2$ is impossible.
(b) Find the smallest $k$ for which $V_1=kV_2$, and in this case construct the angle at the vertex of the cone.
Solution: Let $A$ be the vertex of the cone, $O$ the center of the sphere, $S$ the center of the base of the cone, $B$ a point on the base circle, and $r$ the radius of the sphere. Let $\angle{SAB}=\alpha$. We easily obtain $AS=\frac{r(1+sin(\alpha))}{sin(\alpha)}$ and $SB=\frac{r(1+sin(\alpha)tan(\alpha)}{sin(\alpha)}$ and hence $V_1=\pi.SB^2.\frac{SA}{3}=\pi.r^3.\frac{(1+sin(\alpha))^2}{3sin(\alpha)(1-sin(\alpha))}$. We also have $V_2=2\pi.r^3$ and hence $k=\frac{(1+sin(\alpha))^2}{6sin(\alpha)(1-sin(\alpha))}\implies (1+6k)sin^2(\alpha)+2(1-3k)sin(\alpha)+1=0$.
The discriminant of this quadratic must be nonnegative: $(1-3k)^2-(1+6k)\ge 0\implies k\ge \frac{4}{3}$. Hence we cannot have $k=1$. For $k=\frac{4}{3}$ we have $sin(\alpha)=\frac{1}{3}$, whose construction is elementary.