A frustum of a cone has a volume of $86256 \mathrm{dm}^{3}$, the angle between the slant height and the radius is $=80^{\circ} 25^{\prime} 22^{\prime \prime}$. At what radius of the cross-section will the surface area of the resulting frustum be equal to half the surface area of the given cone?
Let the radius of the base of the cone be $R$, its height $M$, the radius of the intersecting circle $r$, and the height of the frustum of the cone $m$.

The surface area of the cone:

$$
F_{1}=R^{2} \pi+R \pi \cdot \frac{R}{\cos \alpha}=\frac{R^{2} \pi}{\cos \alpha}(1+\cos \alpha)
$$

the surface area of the frustum of the cone:

$$
\begin{aligned}
& F_{2}=R^{2} \pi+r^{2} \pi+(R+r) \pi \cdot \frac{R-r}{\cos \alpha}= \\
& =\frac{R^{2} \pi}{\cos \alpha}(1+\cos \alpha)-\frac{R^{2} \pi}{\cos \alpha}(1-\cos \alpha)
\end{aligned}
$$

Thus, according to the problem:

$$
\begin{gathered}
\frac{R^{2} \pi}{\cos \alpha}(1+\cos \alpha)=\frac{2 R^{2} \pi}{\cos \alpha}(1+\cos \alpha)-\frac{2 R^{2} \pi}{\cos \alpha}(1-\cos \alpha) \\
r=\frac{R}{\sqrt{2} \operatorname{tg} \frac{\alpha}{2}}
\end{gathered}
$$

while the volume of the cone is:

$$
K=\frac{R^{2} \pi M}{3}=\frac{R^{3} \pi}{3} \operatorname{tg} \alpha
$$

and thus

$$
R=\sqrt[3]{\frac{3 K}{\pi \operatorname{tg} \alpha}}
$$

therefore

$$
r=\frac{1}{2} \sqrt{2} \operatorname{ctg} \frac{\alpha}{2} \sqrt[3]{\frac{3 K}{\pi \operatorname{tg} \alpha}}
$$

Substituting the given values: $r=20.11 \mathrm{dm}$.

(Jenő Filkorn, Nyitra.)

The problem was also solved by: Bender E., Burján K., Csete F. A., Czank K., Freibauer E., Kerekes T., Kiss A., Krausz B., Gy. Krisztián, Lindtner M., Gy. Lukhaub, Gy. Lupsa, Obláth R., Ovenden S., G. Sasvári, S. Szibelth, J. Szöllősy, J. Weisz.