Let's calculate the volume of the tetrahedron that can be circumscribed around four mutually tangent, equal-radius spheres.

(Kövi Imre, teacher at the Fógymnasium, Iglón).
The centers of the spheres determine a tetrahedron whose faces are parallel to the faces of the circumscribed tetrahedron of the spheres. If we calculate the radius of the sphere inscribed in the smaller tetrahedron $(\rho) \mathrm{s}$ and add to it the radius of the given spheres $(r)$, we obtain the radius of the sphere inscribed in the larger tetrahedron $(R)$; thus:

$$
R=\rho+r .
$$

The radius of the sphere inscribed in a tetrahedron is:

$$
\rho=\frac{a}{12} \sqrt{6}
$$

if we denote the edge of the tetrahedron by $a$ (Problem 1.197); but $a=2 r$, and thus

$$
\begin{gathered}
\rho=\frac{r}{6} \sqrt{6} \\
R=\frac{r}{6}(6+\sqrt{6})
\end{gathered}
$$

The edge of the tetrahedron circumscribed around the sphere of radius $R$ (2) is:

$$
b=\frac{12 R}{\sqrt{6}}=\frac{2 r}{\sqrt{6}}(6+\sqrt{6})
$$

or

$$
b=2 r(\sqrt{6}+1)
$$

Thus, the volume of the tetrahedron is:

$$
V=\frac{b^{3}}{12} \sqrt{2}=\frac{4 r^{3}(\sqrt{6}+1)^{3}}{3 \sqrt{2}}
$$

or

$$
V=\frac{r^{3}}{3}(2 \sqrt{3}+\sqrt{2})^{3} .
$$

(Béla Grünhut, real gymnasium VIII. class, Pécs.)

The problem was also solved by:

Bernát Friedmann, István Szabó.