[ Auxiliary area. The area helps to solve the task]

Positive numbers $a, b, c, x, y$, such that

$$
\begin{aligned}
& x^{2}+x y+y^{2}=a^{2} \\
& y^{2}+y z+z^{2}=b^{2} \\
& x^{2}+x z+z^{2}=c^{2}
\end{aligned}
$$

Express the value $x y+y z+x z$ in terms of $a, b, c$.
The equation $x^{2}+x y+y^{2}=a^{2}$ can be interpreted as the Law of Cosines in a triangle with sides $x, y, a$ and an angle of $120^{\circ}$.

## Solution

Draw three rays from point $O$ with angles of $120^{\circ}$ between them, and mark segments $O A=z, O B=x, O C=y$ on them. According to the Law of Cosines, the sides of triangle $A B C$ are $a, b$, and $c$. Furthermore, $1 / 2(x y+y z+x z) \sin 120^{\circ}=S_{\text {OBC }}+$ $S_{O A C}+S_{O A B}=S_{A B C}$, which means $x y+y z+x z=\frac{4}{\sqrt{3}} S_{A B C}$.

## Answer

$x y+y z+x z=4 \sqrt{\frac{p(p-a)(p-b)(p-c)}{3}}$, where $p=1 / 2(a+b+c)$.