For real numbers $x, y$, the following holds:

$$
\begin{aligned}
& |x+y|=1-z, \\
& |y+z|=1-x, \\
& |z+x|=1-y .
\end{aligned}
$$

Determine the values that the sum $x+y+z$ can take. For each valid sum, provide an example of corresponding numbers $x, y, z$. \quad (Mária Dományová \& Patrik Bak)
SOLUTION. If $x+y \geq 0$, then according to the first given equation, we have $x+y=1-z$, so the sought sum $x+y+z$ is equal to 1. We reach the same conclusion if $y+z \geq 0$ or $z+x \geq 0$. It remains to assess the case where all three sums $x+y, y+z$, $z+x$ are negative. Then the given system can be rewritten as

\[
\begin{aligned}
& -x-y=1-z \\
& -y-z=1-x \\
& -z-x=1-y
\end{aligned}
\]

By adding these three equations, we get $x+y+z=-3$. The sum $x+y+z$ thus always takes either the value 1 or the value -3. We will show that both of these values are achievable on the solutions of system (1).

Finding the necessary triples $(x, y, z)$ will be easy. As it turns out, it is sufficient to limit ourselves to triples of the same numbers $x=y=z$. Then the system (1) reduces to a single equation $|2x|=1-x$. This equation clearly has two solutions $x=-1$ and $x=\frac{1}{3}$. These correspond to the solutions $(-1,-1,-1)$ and $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$ of the original system (1), which already confirm that both sums $x+y+z=-3$ and $x+y+z=1$ are achievable.

NOTE. The equality $x+y+z=-3$ is satisfied by the only solution $(-1,-1,-1)$ of system (1), because this is the only solution of system (2).** The equality $x+y+z=1$ is satisfied by infinitely many solutions $(x, y, z)$ of system (1); all are determined by the condition

\[
(x \leq 1) \wedge(y \leq 1) \wedge(z \leq 1) \wedge(x+y+z=1).
\]

The first three inequalities express that the right-hand sides of the equations in system (1) are non-negative; if they are satisfied, then from the equality $x+y+z=1$ it follows that the sums $x+y, y+z, z+x$ (equal to $1-z, 1-x, 1-y$ respectively) are also non-negative, so $(x, y, z)$ is a solution of system (1).[^0]