The perimeter of a triangle is $2 s=42 \mathrm{~cm}$ and two of its angles are $\alpha=53^{\circ} 7^{\prime} 50^{\prime \prime}$ and $\beta=59^{\circ} 29^{\prime} 22^{\prime \prime}$. The base of an equilateral cylinder is the circle inscribed in this triangle. What is the surface area and volume of this cylinder?
$$
\begin{aligned}
& \tan \frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \\
& \tan \frac{\beta}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}
\end{aligned}
$$

$$
\begin{aligned}
& \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{(s-c)}{s} \\
& s-c=s \tan \frac{\alpha}{2} \tan \frac{\beta}{2}
\end{aligned}
$$

The radius of the inscribed circle of the triangle:

$$
\rho=(s-c) \tan \frac{\gamma}{2}
$$

Substituting (3):

$$
\rho=s \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}
$$

The surface area and volume of the right circular cylinder:

$$
F=6 \rho^{2} \pi, V=2 \rho^{3} \pi
$$

Substituting the value of $\rho$ into these formulas:

$$
F=6 \pi\left(s \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}\right)^{2} \text { and } V=2 \pi\left(s \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}\right)^{3}
$$

Substituting the given values:

$$
F=301.6 \text{ cm}^{2}, \quad V=402.1 \text{ cm}^{3}
$$

(Goldstein Zsigmond, gym. VIII. grade, Nyíregyháza.)

The problem was also solved by: Bálint Béla, Feuer Mór, Freund Antal, Friedmann Bernát, Geist Emil, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Riesz Frigyes.