A regular ten-sided pyramid has a height of $39 \mathrm{~cm}$; in the triangular faces, the angle at the apex is $18^{\circ}$. What is the volume of the pyramid?
Let the area of the base be $A$, the height of the pyramid $m$, the slant height $l$, the base edge $a$, and the radius of the circle circumscribed around the base $r$. Then the volume of the pyramid is:

$$
K=\frac{A m}{3}=\frac{m}{3} \cdot \frac{5}{2} \cdot a^{2} \operatorname{ctg} 18^{\circ}=\frac{5}{6} m a^{2} \operatorname{ctg} 18^{\circ}
$$

But

$$
m^{2}+r^{2}=l^{2}
$$

$$
r=\frac{a}{2 \sin 18^{\circ}}
$$

$$
l=\frac{a}{2 \sin 9^{\circ}}
$$

Squaring (3) and (4) and substituting into (2), we get:

$$
a^{2}=\frac{4 m^{2} \sin ^{2} 18^{\circ}}{1+2 \cos 18^{\circ}}
$$

Substituting this into (1), we get:

$$
K=\frac{5 m^{3} \sin 36^{\circ}}{3\left(1+2 \cos 18^{\circ}\right)}
$$

Substituting the value of $m$, we get:

$$
K=20023 \mathrm{~cm}^{3}
$$

(Julius Kovács, Budapest.)

The problem was also solved by: L. Brichta, N. Ehrenfeld, P. Füstös, P. Hajdú, V. Horti, Gy. Keller, E. Kirchknopf, E. Kiss, Gy. Léber, E. Mellinger, S. Pichler, P. Sárközy, Gy. Schuster, O. Szilas.