In the isosceles triangle $ABC$, we are given $a+m_{c}=s_{1}, \quad c+m_{a}=s_{2}$; what is $\gamma$? ( $c$ is the base, $m_{a}$ is the altitude to the base, $m_{c}$ is the altitude to the side.)
Let $m_{a}$ be the foot of the altitude from $A$, and $m_{c}$ be the foot of the altitude from $C$, then $\triangle A A_{1} B \sim \triangle C C_{1} B$, and thus

$$
c: m_{a}=a: m_{c}
$$

from which

$$
\frac{c+m_{a}}{a+m_{c}}=\frac{c}{a}, \text { or } \frac{s_{2}}{s_{1}}=\frac{c}{a}=2 \sin \frac{\gamma}{2}
$$

so

$$
\sin \frac{\gamma}{2}=\frac{s_{2}}{2 s_{1}}
$$

from which

$$
\cos \frac{\gamma}{2}=\sqrt{\frac{4 s_{1}^{2}-s_{2}^{2}}{4 s_{1}^{2}}}
$$

and thus

$$
\sin \gamma=\frac{s_{2}}{2 s_{1}} \sqrt{4 s_{1}^{2}-s_{2}^{2}}
$$

(Raab Rezsó, Győr.)

The problem was also solved by: Déri Zs., Deutsch I., Erdélyi I., Haar A., Kertész G., Kürti I.