2. In the domain of positive real numbers, solve the system of equations

$$
\begin{aligned}
& x^{2}+2 y^{2}=x+2 y+3 z, \\
& y^{2}+2 z^{2}=2 x+3 y+4 z, \\
& z^{2}+2 x^{2}=3 x+4 y+5 z .
\end{aligned}
$$

(Patrik Bak)
SOLUTION. We prove that the problem has a unique solution $(x ; y ; z)=(10 / 3 ; 5 / 3 ; 10 / 3)$.

Notice that the right side of the second equation is the arithmetic mean of the right sides of the remaining two equations. Therefore, if we subtract from the sum of the first and third equations twice the second equation, as in the fourth homework problem, we get an equation with a zero right side

$$
\left(x^{2}+2 y^{2}\right)+\left(z^{2}+2 x^{2}\right)-2\left(y^{2}+2 z^{2}\right)=0 \text { or } 3 x^{2}-3 z^{2}=0 \text {. }
$$

Since $x$ and $z$ are positive numbers, it follows that $x=z$. Substituting $x$ for $z$, the original system of equations becomes

$$
\begin{aligned}
& x^{2}+2 y^{2}=4 x+2 y=2(2 x+y), \\
& 2 x^{2}+y^{2}=6 x+3 y=3(2 x+y) \text {, } \\
& 3 x^{2}=8 x+4 y=4(2 x+y) .
\end{aligned}
$$

By comparing the first and third equations, we get $2\left(x^{2}+2 y^{2}\right)=4(2 x+y)=3 x^{2}$, hence $x^{2}=4 y^{2}$, and with $x, y>0$ it follows that $x=2 y$. Substituting $2 y$ for $x$, the modified system of equations becomes

$$
\begin{aligned}
6 y^{2} & =10 y \\
9 y^{2} & =15 y \\
12 y^{2} & =20 y
\end{aligned}
$$

Since $y \neq 0$, each of the three equations is equivalent to $y=5 / 3$. With the derived relationships $x=2 y$ and $z=x$, it must therefore be that $(x ; y ; z)=(10 / 3 ; 5 / 3 ; 10 / 3)$. In this process, verification is not necessary.

NOTE. If we do not notice the dependence between the right sides of the given equations, we can look at the given system as a system of three linear equations with unknowns $x^{2}, y^{2}, z^{2}$ and parameters $x, y, z$. Its solution gives us the expressions

$$
x^{2}=\frac{3 x+4 y+5 z}{3}, \quad y^{2}=\frac{y+2 z}{3}, \quad z^{2}=\frac{3 x+4 y+5 z}{3} .
$$

Now it is clear that $x^{2}=z^{2}$ or $x=z$. Substituting $x$ for $z$, we get from the derived expressions

$$
x^{2}=\frac{8 x+4 y}{3}=\frac{4(2 x+y)}{3} \quad \text { and } \quad y^{2}=\frac{2 x+y}{3} .
$$

Now we see that $x^{2}=4 y^{2}$ or $x=2 y$. Therefore, the equalities $x=2 y=z$ must hold. We proceed from here in the same way as in the original solution.

Let's add to this note a mention of other consequences of the given system of equations, which are quadratic equations for two of the three unknowns $x, y, z$. For example, if we subtract the second equation from twice the third equation, we get

$$
4 x^{2}-y^{2}=2(3 x+4 y+5 y)-(2 x+3 y+4 z)=4 x+5 y+6 z .
$$

Substituting here for $z$ (or directly for 3z) from the first equation, we get for the unknowns $x$, $y$ after a simple rearrangement the quadratic equation $2 x^{2}-5 y^{2}=2 x+y$. Similarly, by subtracting the third equation from twice the first equation and using the second equation, we can obtain for the unknowns $y, z$ the equation $3 y^{2}=y+2 z$. The best result, however, is obtained by subtracting the first equation from twice the second equation, when, taking into account the third equation, we get for the unknowns $x$, $z$ the equation $x^{2}-z^{2}=0$.

For a complete solution, award 6 points. In no solution penalize the absence of a check of the found solution. In incomplete solutions, evaluate partial results as follows.

A1. Stating the solution $(x ; y ; z)=(10 / 3 ; 5 / 3 ; 10 / 3)$ (even without justification): 1 point.

A2. Deriving at least one of the quadratic equations for two unknowns, which is not a consequence of the relationships $x=2 y=z$, such as $2 x^{2}-5 y^{2}=2 x+y$ or $3 y^{2}=y+2 z: 1$ point.

B1. Deriving the relationship $x=z$ : 3 points.

B2. Deriving the relationship $x=2 y$ or $z=2 y$ : 2 points.

In total, then award $\max (\mathrm{A} 1+\mathrm{A} 2, \mathrm{~B} 1+\mathrm{B} 2)$ points.

If the solver eliminates one of the unknowns, for example $z$, by substituting the expression for $z$ from the first equation into the remaining two equations, do not award any points unless a significant progress is made in solving the resulting system of two fourth-degree equations in two unknowns, such as the relationships from A2, B1, or B2.