Given that the area of a certain triangle is $t$ and one of its angles is $C$, how should the sides $a$ and $b$ passing through the vertex $C$ be chosen so that the side $c$ opposite to $C$ is as small as possible.
Applying Carnot's theorem:

$$
c^{2}=a^{2}+b^{2}-2 a b \cos C
$$

or

$$
c^{2}=(a-b)^{2}+2 a b-2 a b \cos C
$$

$$
c^{2}=(a-b)^{2}+2 a b(1-\cos C)
$$

But

$$
1-\cos C=2 \sin ^{2} \frac{C}{2}=2 \sin \frac{C}{2} \cdot \cos \frac{C}{2} \cdot \tan \frac{C}{2}=\sin C \tan \frac{C}{2}
$$

and

$$
2 a b=\frac{4 t}{\sin C}
$$

thus

$$
2 a b(1-\cos C)=\frac{4 t}{\sin C} \cdot \sin C \tan \frac{C}{2}=4 t \tan \frac{C}{2}
$$

Substituting this into (1), we get:

$$
c^{2}=(a-b)^{2}+4 t \tan \frac{C}{2}
$$

Since $4 t \tan \frac{C}{2}$ is constant according to the problem, $c^{2}$ in (2) will be minimized when $(a-b)^{2}=0$, that is, when

$$
a=b
$$

and in this case

$$
c=2 \sqrt{t \tan \frac{C}{2}}
$$

(Pál Heimlich, Budapest, VIII. district, secondary school.)

The problem was also solved by: L. Bánó, I. Csada, I. Dömény, H. Fodor, I. Fuchs, A. Haar, Z. Harsányi, Gy. Jánosy, J. Kiss, Gy. Krampera, P. Messer, M. Pám, S. Pichler, J. Rosenberg, S. Ruvald, I. Schöffer, Gy. Schuster, Gy. Schwarz, J. Sonnenfeld, M. Werner, Mathematical Circle, Budapest, V. district, secondary school.