## Task 4 - 311214

Determine all triples $(x, y, z)$ of real numbers with $x \leq y \leq z$ that satisfy the following system of equations (1), (2), (3):

$$
\begin{aligned}
x+y+z & =5 \\
x^{2}+y^{2}+z^{2} & =15 \\
x y z & =-3
\end{aligned}
$$
I. If real numbers $x, y, z$ with $x \leq y \leq z$ satisfy the system of equations (1), (2), (3), then from (1) and (2) we have:

$$
\begin{aligned}
x+y & =5-z \\
x^{2}+y^{2} & =15-z
\end{aligned}
$$

From (3) it follows that $z \neq 0$ and then

$$
x y=-\frac{3}{2}
$$

Now, $(x+y)^{2}=x^{2}+y^{2}+2 x y$; according to (4), (5), and (6), this means:

$$
\begin{aligned}
(5-z)^{2} & =15-z^{2}+2 \cdot\left(-\frac{3}{z}\right) \\
z\left(25-10 z+z^{2}\right) & =15 z-z^{3}-6 \\
2\left(z^{3}-5 z^{2}+5 z+3\right) & =0 \\
(z-3)\left(x^{2}-2 z-1\right) & =0
\end{aligned}
$$

Thus, $z=3$ or $z=1+\sqrt{2}$ or $z=1-\sqrt{2}$.

If $z=3$, then from (4) and (6) we get the system of equations

$$
x+y=2 \quad ; \quad x y=-1
$$

This leads to $x(2-x)=-1, x^{2}-2 x-1=0$ and thus $x=1+\sqrt{2}, y=1-\sqrt{2}$ or $x=1-\sqrt{2}$, $y=1+\sqrt{2}$.

If $z=1+\sqrt{2}$, then correspondingly

$$
\begin{aligned}
x+y & =4-\sqrt{2} \\
x y & =-\frac{3}{1+\sqrt{2}}=3-3 \sqrt{2} \\
x^{2}-(4 & -\sqrt{2}) x+3-3 \sqrt{2}=0
\end{aligned}
$$

with the solutions

$$
x=2-\frac{1}{2} \sqrt{2}+\sqrt{\left(2-\frac{1}{2} \sqrt{2}\right)^{2}-3+3 \sqrt{2}}=2-\frac{1}{2} \sqrt{2}+\sqrt{\frac{3}{2}+\sqrt{2}}=2-\frac{1}{2} \sqrt{2}+1+\frac{1}{2} \sqrt{2}=3
$$

and $y=1-\sqrt{2}$ or $x=1-\sqrt{2}, y=3$.

If $z=1-\sqrt{2}$, then similarly with $-\sqrt{2}$ instead of $\sqrt{2}: x=3, y=1+\sqrt{2}$ or $x=1+\sqrt{2}, y=3$.

Since $1-\sqrt{2}<1+\sqrt{2}<3$, only the triplet

$$
(x, y, z)=(1-\sqrt{2}, 1+\sqrt{2}, 3)
$$

can satisfy both the conditions (1), (2), (3) and $x \leq y \leq z$. The verification confirms the result. Thus, exactly this triplet satisfies the conditions of the problem.

Adapted from $[5]$

### 9.33.2 Second Round 1991, Class 12