Let $x, y, z$ be positive real numbers such that:

$$
\begin{aligned}
& x^{2}+x y+y^{2}=2 \\
& y^{2}+y z+z^{2}=5 \\
& z^{2}+x z+x^{2}=3
\end{aligned}
$$

Determine the value of $x y+y z+x z$.
I. solution. $\mathrm{The} x, y, z>0$ unknowns cannot be equal to each other, because in the case of equality of any two, substituting and eliminating one of the unknowns from the equations leads to a contradiction. For example, in the case of $x=y$, according to (1) $x=y=\sqrt{\frac{2}{3}}$, and substituting this into (2) and (3) results in two quadratic equations for $z$ whose left sides are identical, but the right sides are 5 and 3, respectively.

Thus, we can multiply the equations by $(x-y)$, $(y-z)$, and $(z-x)$, respectively; using the known identity for the difference of cubes, we get the equations

$$
\begin{aligned}
& x^{3}-y^{3}=2(x-y) \\
& y^{3}-z^{3}=5(y-z) \\
& z^{3}-x^{3}=3(z-x)
\end{aligned}
$$

Adding these and solving for $x$:

$$
x=3 y-2 z
$$

Substituting the value of $x$ into equation (1): $(3 y-2 z)^{2}+(3 y-2 z) y+y^{2}=2$, from which

$$
13 y^{2}-14 y z+4 z^{2}=2
$$

Subtracting twice equation (2) from five times equation (5) and dividing the resulting equation by 9, we get $7 y^{2}-8 z y+2 z^{2}=0$. Solving this equation for $y$:

$$
y_{1,2}=\frac{4 \pm \sqrt{2}}{7} z
$$

Substituting these into equation (4):

$$
x_{1}=\frac{-2+3 \sqrt{2}}{7} z \quad \text { and } \quad x_{2}=\frac{-2-3 \sqrt{2}}{7} z
$$

where $x_{2}$ is definitely negative, so continuing the solution of the equation with the values of $x_{1}$ and $y_{1}$, substituting these into equation (1), after simplification: $z^{2}=\frac{19-3 \sqrt{2}}{7}$; since $z$ is a positive number, $z=\sqrt{\frac{19-3 \sqrt{2}}{7}}$ is obtained. Squaring equations (6) and (7) and substituting the value of $z^{2}$, and considering that $x$ and $y$ can only be positive numbers, we get the solutions

$$
x=\sqrt{\frac{10-6 \sqrt{2}}{7}} \quad \text { and } \quad y=\sqrt{\frac{6+2 \sqrt{2}}{7}}
$$

The roots obtained satisfy the equation.

Knowing the values of $x^{2}, y^{2}$, and $z^{2}$, summing equations (1), (2), and (3): $2\left(x^{2}+y^{2}+z^{2}\right)+(x y+y z+z x)=$ 10, from which

$$
\begin{aligned}
x y+y z+z x & =10-2\left(x^{2}+y^{2}+z^{2}\right)= \\
& =10-2 \cdot \frac{10-6 \sqrt{2}+6+2 \sqrt{2}+19-3 \sqrt{2}}{7}=2 \sqrt{2}
\end{aligned}
$$

Thus, $x y+y z+z x=2 \sqrt{2}$.

II. solution. The value of $x y+y z+x z$ can also be obtained without solving the system of equations. Notice that the given equations resemble the law of cosines. From this, we can give a geometric interpretation to the problem.

Suppose the system of equations has a solution consisting of positive numbers and let this solution be $x, y, z$. Take a point $O$ in the plane and points $A, B, C$ such that $O A=x, O B=y, O C=z, A O B \varangle=B O C \varangle=C O A \varangle=120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_773ab2a8835f402c7fd8g-2.jpg?height=572&width=415&top_left_y=126&top_left_x=844)

Then the lengths of segments $A B, B C, C A$ can be determined from the law of cosines and the system of equations:

$$
\begin{aligned}
& A B^{2}=x^{2}+y^{2}-2 x y \cos 120^{\circ}=x^{2}+y^{2}+x y=2 \\
& B C^{2}=y^{2}+z^{2}-2 z y \cos 120^{\circ}=y^{2}+z^{2}+y z=5 \\
& C A^{2}=x^{2}+z^{2}-2 x z \cos 120^{\circ}=x^{2}+z^{2}+x z=3
\end{aligned}
$$

Thus, $A B C$ is a right triangle with sides $\sqrt{2}, \sqrt{3}, \sqrt{5}$, so by the converse of the Pythagorean theorem, it is a right triangle.

Let's calculate the area of triangle $A B C$ in two ways: since it is a right triangle, its area is half the product of its legs, and also the sum of the areas of triangles $A O B, B O C, C O A$.

$$
\begin{aligned}
\frac{A B \cdot A C}{2} & =\frac{\sqrt{2} \cdot \sqrt{3}}{2}=\frac{x y \sin \left(120^{\circ}\right)}{2}+\frac{y z \sin \left(120^{\circ}\right)}{2}+\frac{z x \sin \left(120^{\circ}\right)}{2}= \\
& =\frac{\sqrt{3}}{4}(x y+y z+x z)
\end{aligned}
$$

thus

$$
\frac{\sqrt{2} \cdot \sqrt{3}}{2}=\frac{\sqrt{3}}{4}(x y+y z+x z), \quad \text { hence } \quad x y+y z+x z=2 \sqrt{2}
$$

We have thus found that if the system of equations has a solution consisting of positive numbers, then for this solution $x y+y z+x z=$ $2 \sqrt{2}$.

We still need to check whether the system of equations has a solution consisting of positive numbers.

Based on our line of thought, it is clear that the segments connecting the isogonal point of the right triangle with sides $\sqrt{2}, \sqrt{3}, \sqrt{5}$ to the vertices are a solution to the system of equations.