A right circular cone has a volume $V$; determine the volume and surface area of the triangular pyramid circumscribed around the cone, if the base of the pyramid is an equilateral triangle.
The volume of a cone:

$$
V=r^{2} \pi \frac{m}{3}, \text { or } 3 V=r^{2} \pi m
$$

The height of an equilateral pyramid: $m=r \sqrt{3}$, and thus

$$
3 V=r^{3} \pi \sqrt{3}
$$

from which

$$
r=\sqrt[3]{\frac{V \sqrt{3}}{\pi}}
$$

Knowing the radius of the base circle, we can calculate the area of the pyramid's base. If one edge of the base of the pyramid is $a$, then

$$
a=2 r \sqrt{3}
$$

Thus, the area of the base is:

$$
t_{1}=3 r^{2} \sqrt{3}
$$

Therefore, the volume of the pyramid will be:

$$
K=3 r^{3}=\frac{3 V \sqrt{3}}{\pi}
$$

To calculate the surface area of the pyramid, we also need to determine the area of the lateral faces. The base of the lateral faces is $a$, and the height is the slant height of the cone, which is $2 r, \text { so }$ the area of one lateral triangle is

$$
t_{2}=2 r \sqrt{3} \cdot r=2 r^{2} \sqrt{3}
$$

Thus, the total surface area of the pyramid will be:

$$
F=3 r^{2} \sqrt{3}+6 r^{2} \sqrt{3}=9 r^{2} \sqrt{3}
$$

or

$$
F=9 \sqrt{3} \sqrt[3]{\frac{3 V^{2}}{\pi^{2}}}
$$

(Note. Let's also calculate the area of the cross-section of the cone; the base is $2 r$, the height is $r \sqrt{3}, \text { so }$:

$$
t_{3}=r^{2} \sqrt{3}
$$

Comparing (5), (7), and (9), we see that the areas of the cross-section of the cone, one lateral face of the pyramid, and the base of the pyramid are in the ratio $1: 2: 3$; that is,

$$
t_{3}: t_{2}: t_{1}=1: 2: 3
$$

The problem was also solved by: Friedmann Bernát, S.A.-Ujhely; Geist Emil, Győr; Hofbauer Ervin and Kántor Nándor, Budapest; Krausz Mihály, főreál. VIII. o. t., Budapest, II. district; Messinger Ábrahám, S.-A.-Ujhely; Suschnik József, Kecskemét; Visnya Aladár, Pécs; Goldstein Zsigmond, Nyíregyháza.