A cone of height $m$ is divided into three parts by two planes parallel to the base, such that the areas of the lateral surfaces of the parts are equal. What are the distances of the cutting planes from the apex of the cone?
Let the radii of the sections from the vertex to the base be: $r_{1}, r_{2}, r_{3}$; the corresponding side edges: $l_{1}, l_{2}, l_{3}$; the corresponding heights: $m_{1}, m_{2}, m$.

According to the problem:

$$
\begin{aligned}
r_{1} \pi l_{1} & =\frac{1}{3} r_{3} \pi l_{3} \\
r_{2} \pi l_{2} & =\frac{2}{3} r_{3} \pi l_{3}
\end{aligned}
$$

from which:

$$
\frac{r_{3}}{r_{1}} \cdot \frac{l_{3}}{l_{1}}=3, \frac{r_{3}}{r_{2}} \cdot \frac{l_{3}}{l_{2}}=\frac{3}{2}
$$

From the corresponding similar triangles, it follows that

$$
\frac{r_{3}}{r_{1}}=\frac{l_{3}}{l_{1}}=\frac{m}{m_{1}}, \frac{r_{3}}{r_{2}}=\frac{l_{3}}{l_{2}}=\frac{m}{m_{2}}
$$

Substituting these values into (3), we get:

$$
\frac{m^{2}}{m_{1}^{2}}=3 \text { and } \frac{m^{2}}{m_{2}^{2}}=\frac{3}{2}
$$

from which

$$
m_{1}=\frac{m}{3} \sqrt{3} \text { and } m_{2}=\frac{m}{3} \sqrt{6}
$$

The problem was solved by: Friedmann Bernát, S.-A.-Ujhely; Galter János, Székely-Udvarhely; Grünhut Béla, Pécs; Hofbauer Ervin and Kántor Nándor, Budapest; Oberle Károly, VIII. o. t., Budapest, Piarist High School; Szabó István, Debrecen; Visnya Aladár, Pécs; Zemplén Győző, Fiume.