We divide the height of a cone into three equal parts, and draw planes parallel to the base through the division points. How do the volumes of the resulting solids relate to each other?
Let the height of the cone be $3 m$; if the radius of the top cross-section is $r$, then the middle one is $2 r$, and the bottom one is $3 r$.

The volume of the top cone:

$$
V_{1}=\frac{r^{2} \pi m}{3}
$$

The volume of the middle frustum of the cone:

$$
V_{2}=\frac{m \pi}{3}\left(4 r^{2}+2 r^{2}+r^{2}\right)=\frac{7 r^{2} \pi m}{3}
$$

The volume of the bottom frustum of the cone:

$$
\begin{gathered}
V_{3}=\frac{m \pi}{3}\left(9 r^{2}+6 r^{2}+4 r^{2}\right)=\frac{19 r^{2} \pi m}{3} \\
V_{1}: V_{2}: V_{3}=\frac{r^{2} \pi m}{3}: \frac{7 r^{2} \pi m}{3}: \frac{19 r^{2} \pi m}{3}
\end{gathered}
$$

$$
V_{1}: V_{2}: V_{3}=1: 7: 19
$$

(Nándor Kántor, VIII. class of the Budapest State Gymnasium)

The problem was also solved by: Béla Bálint, Hugó Berger, Mór Feuer, Antal Freund, Bernát Friedmann, Emil Geist, Zsigmond Goldstein, Károly Goldziher, Béla Grünhut, Ervin Hofbauer, Mór Klein, Frigyes Riesz, István Szabó.