A right truncated pyramid is inscribed around a sphere of radius $r$, with its base being a square. What is the volume of this truncated pyramid, if the diagonal of its base is $4 r$?
Let $A E F C$ be a section through the center $O$ of the sphere, perpendicular to two opposite sides of the base and the top of the pyramid; $D B$ is the height of the pyramid. Since $\angle C O D = \angle O A B$, therefore $\triangle C D O \sim \triangle O B A$, and thus

$$
A B: r = r: C D
$$

but

$$
A B = \frac{A E}{2} = r \sqrt{2}
$$

and thus

$$
C D = \frac{C F}{2} = \frac{r}{2} \sqrt{2}
$$

from which

$$
C F = r \sqrt{2}
$$

The volume of the pyramid is:

$$
K = \frac{2 r}{3}\left(\overline{A E}^{2} + \overline{C F}^{2} + A E \cdot C F\right)
$$

into which the values of $A B$ and $C F$ are substituted:

$$
K = \frac{2 r}{3}\left(8 r^{2} + 2 r^{2} + 4 r^{2}\right)
$$

or

$$
K = \frac{28 r^{3}}{3}
$$

(András Hrivnák.)

Number of solutions: 39.