A pyramid has a volume $V$, and the area of its base is $A$; at what distance from the base should we cut the pyramid with a plane parallel to the base so that the volume of the resulting frustum is $V_{1}$?
Let the total height of the pyramid be $m$, and the height of the complementary pyramid with volume $V_{2}$ be $x$.

According to the problem,

$$
\begin{aligned}
V_{2} & =V-V_{1} \\
V & =\frac{A m}{3}
\end{aligned}
$$

from which

$$
m=\frac{3 V}{A}
$$

It is known that

$$
V: V_{2}=m^{3}: x^{3}
$$

Substituting (1) and (2) into this:

$$
V: V-V_{1}=\frac{27 V^{3}}{A^{3}}: x^{3}
$$

from which

$$
x=\frac{3}{A} \sqrt[3]{V^{2}\left(V-V_{1}\right)}
$$

and thus the distance from the base

$$
m-x=\frac{3}{A}\left(V-\sqrt[3]{V^{2}\left(V-V_{1}\right)}\right)
$$

(Probst Elemér, m. kir. teng. akad. növendék, Fiume.)

The problem was also solved by: Bojedain F., Dénes A., Devecis M., Erdős A., Freibauer E., Goldziher K., Kármán T., Kertész T., Spitzer Ö., Szabó I., Szabó K., Weisz J.