In the triangle $ABC$, the area $S$ and the angle $C$ are known. Find the values of the sides $a$ and $b$ so that the side $c$ is as short as possible.

## SOLUTION:



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On the one hand

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

$S=\frac{1}{2} a b \sin C \Rightarrow a b=\frac{2 S}{\sin C} . \quad$ Then,

$c^{2}=(a-b)^{2}+\frac{4 S(1-\cos C)}{\sin C}$ will be minimum when $a=b=\sqrt{\frac{2 S}{\sin C}}$.

3.