Given a sphere of radius $r$, inscribe a cone around it such that the volume of the cone is as small as possible. What is the volume of this cone?
Let the radius of the base of the cone be $R$, the height of the cone be $m$, and the angle formed by the slant height of the cone and the radius of the base be $\alpha$. Then:

$$
R=\frac{r}{\tan \frac{1}{2} \alpha}, m=R \tan \alpha=\frac{r \tan \alpha}{\tan \frac{1}{2} \alpha}
$$

The volume of the cone is:

$$
V=\frac{R^{2} \pi m}{3}=\frac{r^{3} \pi}{3} \cdot \frac{\tan \alpha}{\tan ^{3} \frac{\alpha}{2}}
$$

or

$$
V=\frac{r^{3} \pi}{3} \cdot \frac{2}{\tan ^{2} \frac{\alpha}{2}-\tan ^{4} \frac{\alpha}{2}}
$$

Now let's find the minimum value of the function

$$
y=\frac{2}{\tan ^{2} \frac{\alpha}{2}-\tan ^{4} \frac{\alpha}{2}}
$$

$$
\tan ^{4} \frac{\alpha}{2}-\tan ^{2} \frac{\alpha}{2}+\frac{2}{y}=0
$$

from which

$$
\tan ^{2} \frac{\alpha}{2}=\frac{1}{2} \pm \sqrt{\frac{1}{4}-\frac{2}{y}}
$$

The minimum value of $y$ for which $\tan ^{2} \frac{\alpha}{2}$ is real is 8. Substituting this into (1), we get:

$$
V=\frac{3 r^{3} \pi m}{3}
$$

At the minimum value of $y$:

$$
\tan \frac{\alpha}{2}=\frac{1}{2} \sqrt{2}, R=r \sqrt{2}, m=4 r
$$

Thus, the cone with the smallest volume is obtained when the height of the cone is four times the radius of the sphere. In this case, the volume of the cone is twice the volume of the sphere.

(Nándor Kántor, VII. Evangelical Gymnasium, Budapest)

The problem was also solved by: Béla Grünhut, Pécs; Ervin Hofbauer, Budapest.