A right circular cone has a base radius of $25 \mathrm{~cm}$ and a slant height of $42 \mathrm{~cm}$; this cone needs to be cut into two parts by a plane parallel to the base so that the volumes of the two resulting solids are equal; the question is: what is the radius of the circle of intersection, and what is the height of the smaller cone?
If the radius of the base of the cone is $r$, the height is $m$, the slant height is $l$, the radius of the intersecting circle is $\rho$, and the height of the cone is $x$, then according to the problem:

$$
r^{2} m=2 \rho^{2} x, \quad \text { or } \quad \frac{r^{2}}{\rho^{2}}=\frac{2 x}{m}
$$

but

$$
r^{2}: \rho^{2}=m^{2}: x^{2}
$$

thus

$$
\frac{m^{2}}{x^{2}}=\frac{2 x}{m}
$$

from which

$$
x=m \sqrt[3]{\frac{1}{2}}=\sqrt{l^{2}-r^{2}} \cdot \sqrt[3]{\frac{1}{2}}
$$

and

$$
\rho=r \sqrt[3]{\frac{1}{2}}
$$

Substituting the given values:

$$
x=\sqrt{1139} \cdot \sqrt[3]{\frac{1}{2}} \quad \text { and } \quad \rho=25 \sqrt[3]{\frac{1}{2}}
$$

(Ede Freubauer.)

The problem was also solved by: Csete F. A., Czank K., Dolowschiák M., Faith F., Filkorn J., Kárf J., Kerekes T., Kiss A., Klein A., Kornis Ö., Krausz B., Krisztián Gy., Lukhaub Gy., Neumann J., Obláth R., Oltay K., Perl Gy., Petrogalli G., Pollák N., Rehberger Z., Sasvári G., Szabó J., Szibelth S., Weisz J.