Suppose that for the positive numbers $x, y, z$,

$$
x^{2}+x y+y^{2}=9, \quad y^{2}+y z+z^{2}=16, \quad z^{2}+z x+x^{2}=25 .
$$

Determine the value of $x y+y z+z x$.
Transform the conditions of the problem:

$$
3^{2}=x^{2}+y^{2}-2 x y \cos 120^{\circ}, 4^{2}=y^{2}+z^{2}-2 y z \cos 120^{\circ}, 5^{2}=z^{2}+x^{2}-2 z x \cos 120^{\circ}
$$

If we consider a triangle $ABC$ with sides $AB=3$, $BC=4$, and $AC=5$, then by the converse of the Pythagorean theorem, this triangle is a right triangle. Therefore, there exists the isogonal point $P$ of the triangle $ABC$. (The point $P$ is the isogonal point of the triangle $ABC$ if and only if $\angle APB = \angle BPC = \angle CPA = 120^{\circ}$.)

Notice that if $AP=x$, $BP=y$, and $CP=z$, then the three conditions are identical to the cosine laws applied to the triangles $ABP$, $BCP$, and $CAP$. Since identical cosine laws can only be written for congruent triangles, and we have seen an example of the existence of such a triangle above, the solutions of the system of equations are: $AP$, $BP$, $CP$. Then:

$$
\begin{aligned}
6 & =\frac{3 \cdot 4}{2}=T_{ABC}=T_{APB}+T_{BPC}+T_{CPA}= \\
& =\frac{xy \sin 120^{\circ}}{2}+\frac{yz \sin 120^{\circ}}{2}+\frac{zx \sin 120^{\circ}}{2}
\end{aligned}
$$

Since $\sin 120^{\circ}=\frac{\sqrt{3}}{2}$, we have

$$
xy + yz + zx = 8 \sqrt{3}
$$

Based on the work of Zoltán Gyenes (Budapest, ELTE Apáczai Csere J. Gymnasium, 10th grade) and Gábor Horváth (Debrecen, Fazekas M. Gymnasium, 11th grade)