Calculate the sides $a, b$, and $c$ of a triangle $ABC$, if the angles $A, B$, and $C$ of the triangle and the perimeter $2p$ of the triangle are given.
From the Law of Sines, it follows that

$$
b=\frac{a \sin B}{\sin A}
$$

and

$$
c=\frac{a \sin C}{\sin A}
$$

and thus

$$
2 p=a+\frac{a \sin B}{\sin A}+\frac{a \sin C}{\sin A}
$$

or

$$
2 p \sin A=a(\sin A+\sin B+\sin C)
$$

from which

$$
a=\frac{2 p \sin A}{\sin A+\sin B+\sin C}
$$

similarly

$$
\begin{aligned}
& b=\frac{2 p \sin B}{\sin A+\sin B+\sin C} \\
& c=\frac{2 p \sin C}{\sin A+\sin B+\sin C}
\end{aligned}
$$

Since

$$
\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}
$$

and

$$
\begin{aligned}
& \sin A=2 \sin \frac{A}{2} \cos \frac{A}{2} \\
& \sin B=2 \sin \frac{B}{2} \cos \frac{B}{2} \\
& \sin C=2 \sin \frac{C}{2} \cos \frac{C}{2}
\end{aligned}
$$

therefore

$$
\begin{aligned}
& a=\frac{p \sin \frac{A}{2}}{\cos \frac{B}{2} \cos \frac{C}{2}} \\
& b=\frac{p \sin \frac{B}{2}}{\cos \frac{C}{2} \cos \frac{A}{2}} \\
& c=\frac{p \sin \frac{C}{2}}{\cos \frac{A}{2} \cos \frac{B}{2}}
\end{aligned}
$$

(Armin Fried, real gymnasium VII. grade, Deva).

The problem was also solved by: Domokos Berzenczy, Deva; Béla Bolemann, Budapest; Richárd Engel, Györ; Gusztáv Grossmann, Budapest; Róbert Herczka, Budapest; Tivadar Heymann, Györ; Ármin Hirschler, real gymnasium VIII. grade, Györ; János Imre, Jr., Nyíregyháza; György Jankovics, Losoncz; Gergely Jorga, Gilad; Jenő Kis, Budapest; Elemér Meitner, Budapest; Viktor Müller, Budapest; Ignácz Pilczer, Kaposvár; Jenő Schiller, Györ; Gábor Schulhof, real gymnasium VIII. grade, Pécs; Sándor Suták, Nyíregyháza; Gusztáv Szabó, Győr; Jenő Unger, Györ; Aladár Visnya, Pécs; Ferencz Zsaborszky, real gymnasium VII. grade, Pécs.