A right circular cone has a base radius $R$; its height is $m=2R$. At what distance from the vertex must the cone be cut by a plane parallel to the base so that the total surface areas of the resulting parts are equal?
Let the generator of the whole cone be $L$, the truncated cone be $l$, and the complement cone be $l_{1}$, with the latter's height being $x$; it can be easily shown that

$$
L=R \sqrt{5}
$$

$$
l_{1}=r \sqrt{5}
$$

The surface areas of the two parts of the cone are equal, thus:

$$
R^{2} \pi+(R+r) \pi l=r \pi l_{1}
$$

from which we get

$$
l=\left(L-l_{1}\right)=(R-r) \sqrt{5}
$$

$$
\begin{aligned}
& r=\frac{R}{10} \sqrt{10(5+\sqrt{5})} \\
& x=\frac{R}{5} \sqrt{10(5+\sqrt{5})} \\
& \quad \text { Hofbauer Ervin, Budapest, 7th grade, this year's gymnasium. }
\end{aligned}
$$

Note. Concerning the construction of our $x$, we must take into account that

$$
x=R \frac{\sqrt{10+2 \sqrt{5}}}{5}=\frac{R}{\frac{1}{4} \sqrt{10-2 \sqrt{5}}}=\frac{R}{\sin 36^{\circ}}
$$

Thus, $x$ is the hypotenuse of a right-angled triangle, one of whose acute angles is $36^{\circ}$, and the opposite leg to this angle is the radius of the base of the given cone.

Friedmann Bernát, 7th grade, S.-A.-Ujhely's gymnasium.

The problem was also solved by: Szabó István, from Debrecen; Fleischmann László, Kántor Nándor, Krausz Mihály, and Pósch Gyula, from Budapest; Geist Emil, Schiller Jenő, and Szabó Gusztáv, from Győr; Goldberger Leó, Grünhut Béla, and Visnya Aladár, from Pécs.