We construct pyramids perpendicular to each face of a cube. What is the volume of the resulting solid if all its vertices lie on the surface of a sphere and each edge of the cube is $a$?
If the edge of a cube is $a$, then the radius of the sphere circumscribed around it is $\frac{a}{2} \sqrt{3}$, and thus the height of each pyramid is $\frac{a}{2} \sqrt{3} - \frac{a}{2} = \frac{a}{2}(\sqrt{3}-1)$. Therefore, the volume of the entire solid is:

$$
K=a^{3}+6 \cdot \frac{a^{3}}{6}(\sqrt{3}-1)=a^{3} \sqrt{3}
$$

(Pál Pilczer, Kaposvár.)

The problem was also solved by: Bayer B., Burján K., Demjén E., Filkorn J., Holzmann M., Kertész F., Gy. Krisztián, Lázár L., Gy. Lukhaub, Gy. Lupsa, Zs. Moskovits, Gy. Perl, Gy. Póka, Á. Rosenberg, M. Russo, J. Schwarz, K. Smodics, Gy. Sümegi, H. Szmodics, A. Weisz.