Cut a sphere with three parallel planes so that the resulting spherical segments and spherical caps form a geometric progression with a common ratio of $q$.
Our task is solved if we determine the heights of the spherical caps and segments, which we will denote as $m_{1}, m_{2}, m_{3}$, and $m_{4}$. The surface areas of the spherical caps and segments are then $2 r \pi m_{1}, 2 r \pi m_{2}, 2 r \pi m_{3}$, and $2 r \pi m_{4}$. If $m_{1}, m_{2}, m_{3}, m_{4}$ form a geometric progression with a common ratio $q$, then the surface areas of the spherical caps and segments also form such a geometric progression. But

$$
m_{1}+m_{2}+m_{3}+m_{4}=m_{1} \frac{q^{4}-1}{q-1}=2 r
$$

so

$$
m_{1}=2 r \frac{q-1}{q^{4}-1}
$$

(József Glück, Debrecen.)

The problem was also solved by: Blum J., Dömény I., Fekete M., Fodor H., Földes R., Haar A., Heimlich P., Jánosy Gy., Krampera Gy., Kräuter F., Messer P., Schuster Gy., Sonnenfeld J., Tandlich E.