In a triangle, one side $c$ and the two angles $\alpha$ and $\beta$ lying on it are known. What is the radius $r$ of the inscribed circle of the triangle?
The intersection point of the angle bisectors of $\alpha$ and $\beta$ will be $O$, the center of the circle. The perpendicular from $O$ to side $c$ is $OD = r$. From triangle $DOA$:

$$
r = OA \cdot \sin \frac{1}{2} \alpha
$$

From triangle $AOB$:

$$
OA = c \cdot \frac{\sin \frac{1}{2} \beta}{\sin \frac{1}{2}(\alpha + \beta)}
$$

Substituting (2) into (1), we get:

$$
r = c \cdot \frac{\sin \frac{1}{2} \alpha \sin \frac{1}{2} \beta}{\sin \frac{1}{2}(\alpha + \beta)}
$$

(Béla Grünhut, VII. grade, Pécs.)

The problem was solved by: Bernát Friedmann, S.-A.-Ujhely; Mihály Krausz, II. district, VIII. grade, Budapest; Gyula Porde, VI. grade, Szamos-Ujvár; Pál Schwarz, VIII. grade, Arad; József Suschnik, Kecskemét; István Szabó, Debrecen; László Fleischmann, Ervin Hofbauer, Nándor Kántor, Pál István Langheim, and Gyula Pósch, Budapest; Emil Geiszt, Áron Goldschmiedt, V. grade, Jenő Schiller, and Gusztáv Szabó, Győr; Leó Goldberger and Aladár Visnya, Pécs.