We write a rotational cone around a unit radius sphere. What is the minimum surface area of the cone?
I. solution. Using the notation of the figure. From the nature of the problem, it follows that $m>2$ and $FA=r>1$.

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By the Pythagorean theorem, $a=\sqrt{m^{2}+r^{2}}$, and from the similarity of triangles $ACF$ and $OCD$, we also have $a: r=(m-1): 1$, from which, in the usual way,

$$
a=r(m-1)
$$

Therefore, $\sqrt{m^{2}+r^{2}}=(m-1)r$, and thus $m^{2}+r^{2}=(m^{2}-2m+1)r^{2}$, that is, $m(m \cdot r^{2}-m-2r^{2})=0$. Since $m>0$, this can only be true if $m \cdot r^{2}-m-2r^{2}=0$, or

$$
m=\frac{2r^{2}}{r^{2}-1} \quad(r>1)
$$

From (1) and (2),

$$
a=r \cdot\left(\frac{2r^{2}}{r^{2}-1}-1\right)=\frac{r^{3}+r}{r^{2}-1}
$$

It is known that the surface area of the cone is $A=r \cdot \pi(r+a)$, from which, using (3), we obtain the one-variable expression for "A":

$$
A=r \cdot \pi \cdot\left(r+\frac{r^{3}+r}{r^{2}-1}\right)=2 \pi \frac{r^{4}}{r^{2}-1}
$$

To determine the minimum of "A", it is sufficient to determine the minimum of $\frac{r^{4}}{r^{2}-1}$, under the condition $r>1$. Since $\frac{r^{4}}{r^{2}-1}=\frac{1}{\frac{1}{r^{2}}-\frac{1}{r^{4}}}$, "A" is minimal when $\frac{1}{r^{2}}-\frac{1}{r^{4}}$ is maximal. This maximum can be determined as follows: $\frac{1}{r^{2}}-\frac{1}{r^{4}}=\left(\frac{1}{r^{2}}-\frac{1}{2}\right)^{2}+\frac{1}{4}$, and this is maximal when $\frac{1}{r^{2}}=\frac{1}{2}$, and since $r>1, r=\sqrt{2}$, and the maximum value is $\frac{1}{4}$.

Therefore, the minimum of the cone's surface area exists, and it is $8 \pi$.

II. solution. In problem F. 2595, we showed that the volume of a cone circumscribed around a sphere is at least twice the volume of the sphere.

From problem 2855 in the Collection of Geometric Problems I, we know that the measure of the surface area of a cone circumscribed around a sphere is three times the measure of the volume. Therefore, the volume and the surface area are minimal at the same time.

According to the solution of problem F. 2595 (see issue 10 of 1986), the minimum volume of a cone circumscribed around a unit sphere is $\frac{8}{3} \pi$, so the minimum surface area is $3 \cdot \frac{8}{3} \pi=8 \pi$.

Remark. The location of the minimum can also be determined using the arithmetic and geometric mean inequality or by differentiation. In the latter case, it must be proven that the zero of the derivative function is indeed a minimum point. Several submitters overlooked this. Some thought that the cone with the minimal surface area is equilateral. This is not true.