A sphere of volume $V$ is inscribed in a regular tetrahedron, into which another sphere is inscribed, then a tetrahedron into that sphere, and so on to infinity. First, calculate the sum of the volumes of the spheres, and second, the sum of the volumes of the tetrahedra.
$1^{\circ}$. If $r, r_{1}, r_{2}, r_{3}, \ldots r_{n}$ are the radii of successive spheres, then according to problem 197:

$$
r_{1}=\frac{r}{3}, r_{2}=\frac{r_{1}}{3}=\left(\frac{1}{3}\right)^{2} r, \ldots r_{n}=\left(\frac{1}{3}\right)^{n} r
$$

Since the volumes of the spheres are directly proportional to the cubes of their radii, it follows that

$$
V_{1}=\frac{1}{27} V, V_{2}=\left(\frac{1}{27}\right)^{2} V, \ldots V_{n}=\left(\frac{1}{27}\right)^{n} V
$$

Thus, the volumes of the spheres form an infinite geometric progression whose first term is $V$ and common ratio is $\frac{1}{27}$; the sum of the volumes of the spheres is therefore:

$$
\frac{V}{1-\frac{1}{27}}=\frac{27}{26} V
$$

$2^{\circ}$ According to problem 197, the radius of the sphere circumscribed around a tetrahedron is:

$$
r=\frac{a}{4} \sqrt{6}
$$

from which

$$
a=r \sqrt{\frac{8}{3}}
$$

Thus, the volume of the first tetrahedron is:

$$
K=\frac{8 r^{3}}{9 \sqrt{3}}
$$

Since the volumes of the regular tetrahedra inscribed in the spheres are directly proportional to the volumes of the spheres, it follows that the volumes of the successive tetrahedra form an infinite geometric progression whose first term is $K$ and common ratio is $\frac{1}{27}$. The sum of the volumes of the tetrahedra is therefore:

$$
\frac{8 r^{3}}{9 \sqrt{3}} \frac{27}{26}=\frac{12 r^{3}}{13 \sqrt{3}}
$$

But

$$
V=\frac{4}{3} r^{3} \pi, \text{ so } r^{3}=\frac{3 V}{4 \pi}
$$

Substituting (4) into (3), we get:

$$
\frac{12 \cdot 3 V}{13 \cdot 4 \cdot \pi \sqrt{3}}=\frac{3 V \sqrt{3}}{13 \pi}
$$

(József Suschnik, 8th grade, Kecskemét.)

The problem was also solved by: Bernát Friedmann, S.-A.-Ujhely; Ervin Hofbauer, Budapest; Béla Grünhut and Aladár Visnya, Pécs.