The volume and surface area of the prism are:

$$
K_{1}=\frac{a^{2} m}{4} \sqrt{3}, \quad F_{1}=\frac{a^{2}}{2} \sqrt{3}+3 a m_{1}
$$

The volume and surface area of the pyramid are:

$$
K_{2}=\frac{a^{2} m}{12} \sqrt{3}, \quad F_{2}=\frac{a^{2}}{4} \sqrt{3}+\frac{3 a}{2} \sqrt{m^{2}+\frac{a^{2}}{12}}
$$

If the lateral surface area of the prism is equal to the lateral surface area of the pyramid, then

$$
3 a m=\frac{3 a}{2} \sqrt{m^{2}+\frac{a^{2}}{12}}
$$

or

$$
4 m^{2}=m^{2}+\frac{a^{2}}{12}
$$

from which

$$
m=\frac{a}{6}
$$

(Béla Bayer.)

The problem was also solved by: Zs. Benedek, J. Baumann, K. Burján, K. Czank, J. Demeter, N. Fekete, J. Filkorn, J. Frank, K. Freudenberg, J. Gellért, J. Keesz, B. Kende, T. Kerekes, D. König, B. Krausz, Gy. Krisztián, Gy. Lupsa, I. Mikuleczky, Gy. Perl, Gy. Póka, Á. Rosenberg, M. Russo, G. Sasvári, J. Sasvári, J. Scharff, H. Spitzer, E. Tézner, T. Ulmer, F. Winter.
Let one side of an equilateral triangle be $a$. Construct a right prism $\mathrm{s}$ and a right pyramid over this triangle. What is the volume and surface area of these solids if the height of each is $m$? Determine $m$ such that the lateral surface area of the prism is equal to the lateral surface area of the pyramid.