Let $x, y, z$ be real numbers not equal to zero with

$$
\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y} .
$$

Determine all possible values of

$$
\frac{(x+y)(y+z)(z+x)}{x y z}
$$

(Walther Janous)

Answer. The only possible values are 8 and -1.
Let's set $\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y}=a$, then we obtain the equivalent system of equations

$$
\begin{aligned}
& x+y=a z \\
& y+z=a x \\
& z+x=a y .
\end{aligned}
$$

By adding these equations, we get

$$
2(x+y+z)=a(x+y+z),
$$

which is equivalent to

$$
(2-a)(x+y+z)=0 .
$$

If a product is zero, then one of the factors must be zero. We thus distinguish two cases:

- If $2-a=0$, i.e., $a=2$, then from the first equation we have $\frac{x+y}{2}=z$. Substituting this into the second equation and simplifying, we get $x=y$. Similarly, we get $y=z$, so we have $x=y=z$. For the desired product, we then have

$$
\frac{(x+y)(y+z)(z+x)}{x y z}=\frac{2 x \cdot 2 x \cdot 2 x}{x^{3}}=8 .
$$

This value is achieved, for example, for the tuple $(x, y, z)=(1,1,1)$.

- If $x+y+z=0$, then $x+y=-z, y+z=-x$, and $z+x=-y$. Thus, for the desired product, we have

$$
\frac{(x+y)(y+z)(z+x)}{x y z}=\frac{(-z)(-x)(-y)}{x y z}=-1 .
$$

This value is achieved, for example, for the tuple $(x, y, z)=(1,1,-2)$.