31. Describe a regular octahedron around a sphere of radius $R$ and find its volume.
31. The three diagonals of an octahedron are mutually perpendicular. Let \( a \) be half the length of a diagonal, \( r \) be the radius of the inscribed sphere, and \( O \) be the common center of the octahedron and the inscribed sphere. If we take the diagonals as coordinate axes and point \( O \) as the origin, then each face of the octahedron will intercept segments of length \( a \) on the axes. Therefore, the distance \( r \) from the face to the origin is given by the formula:

$$
\frac{1}{a^{2}}+\frac{1}{a^{2}}+\frac{1}{a^{2}}=\frac{1}{r^{2}} ; a^{2}=3 r^{2} ; a=r \sqrt{3}(\mathrm{c} \text{m. problem } 57 \text{ from } \S 5) .
$$

(Another way to obtain this relationship is the formula \( V_{0}=\frac{1}{3} S_{0} \cdot r \), where \( V_{0} \) and \( S_{0} \) are the volume and the surface area of the octahedron.)

The length \( a \) is not difficult to construct (for example, as the side length of an equilateral triangle inscribed in a circle of radius \( r \)). Finally, to describe an octahedron around a given sphere, draw three mutually perpendicular lines through the center of the sphere \( O \) and mark segments of length \( a \) from point \( O \) on these lines. This will give us all 6 vertices of the octahedron. The volume of the octahedron is:

$$
V=4 \cdot \frac{1}{3} a^{2} a=\frac{4}{3} a^{3}=\frac{4}{3} r^{3} \cdot 3 \sqrt{3}=4 r^{3} \sqrt{3}
$$