6. Determine all ordered triples $(x, y, z)$ of distinct real numbers that satisfy the set equation

$$
\{x, y, z\}=\left\{\frac{x-y}{y-z}, \frac{y-z}{z-x}, \frac{z-x}{x-y}\right\} \text {. }
$$

(J. Šimša)
Solution. If $x, y, z$ are three distinct real numbers, then the values

$$
u=\frac{x-y}{y-z}, \quad v=\frac{y-z}{z-x}, \quad w=\frac{z-x}{x-y}
$$

are clearly numbers different from $0$ and $-1$, and their product is equal to 1. Therefore, the values $x, y, z$ of any sought triplet must also have the same property. We will therefore always assume that the following conditions hold:

$$
x, y, z \in \mathbb{R} \backslash\{0,-1\}, \quad x \neq y \neq z \neq x, \quad x y z=1 .
$$

Since the given set equation is the same for the ordered triplets $(x, y, z), (z, x, y)$, and $(y, z, x)$, we will assume in addition to (2) that $x>\max \{y, z\}$, and we will distinguish two cases depending on whether $y>z$ or $z>y$. Let's also introduce the notation for the intervals $I_{1}=(0, \infty), I_{2}=(-1,0), I_{3}=(-\infty,-1)$.

Case $x>y>z$. For the fractions (1), it is clear that $u \in I_{1}, v \in I_{2}$, and $w \in I_{3}$, so $u>v>w$. The given set equation can therefore only be satisfied if $u=x, v=y$, and $w=z$. Substituting the fractions (1) and simplifying, we arrive at the equations

$$
x y+y=y z+z=z x+x, \quad \text { where } x \in I_{1}, y \in I_{2}, z \in I_{3} .
$$

According to the condition $x y z=1$ from (2), we can substitute $z x$ with $1 / y$ in the equation $x y+y=z x+x$ and further simplify:

$$
x y+y=\frac{1}{y}+x \Rightarrow x(y-1)=\frac{1-y^{2}}{y} \Rightarrow x=-\frac{1+y}{y} \Rightarrow y=-\frac{1}{1+x}
$$

(We used the fact that, given $y \in I_{2}$, $y \neq 1$.) From the last formula, it follows that the value of the first expression in the system (3) is equal to -1, so from the equality of the second expression to -1, we have

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

then the third expression in (3) is also equal to -1. Therefore, every solution to our problem (in the case where $x>y>z$) is of the form

$$
(x, y, z)=\left(t,-\frac{1}{1+t},-\frac{1+t}{t}\right)
$$

where $t \in I_{1}$ is arbitrary (since (3) holds, verification is not necessary). From the above procedure, it also follows that by choosing $t \in I_{2}$ (or $t \in I_{3}$) in formula (4), we obtain all solutions to our problem with the property $z>x>y$ (or $y>z>x$), so it is not necessary to list cyclic permutations of the triplets from formula (4) in the final answer.

Case $x>z>y$. For the fractions (1), this time we have $u \in I_{3}, v \in I_{1}$, and $w \in I_{2}$, so $v>w>u$, and the given set equation is therefore satisfied precisely when $u=y, v=x$, and $w=z$. Substituting the fractions from (1), we arrive at the system

$$
x-y=y(y-z), \quad y-z=x(z-x), \quad z-x=z(x-y) .
$$

Adding these three equations, we get

$$
0=y(y-z)+x(z-x)+z(x-y)=(y-x)(x+y-2 z),
$$

from which, given $x \neq y$, it follows that $z=\frac{1}{2}(x+y)$. Substituting back into (5), we easily find (again with $x \neq y$) that only $x=1, y=-2$, and $z=-\frac{1}{2}$ satisfy. The same triplet of numbers forms (the only) solution to the problem with the property $y>x>z$ and (the only) solution for which $z>y>x$.

Answer: The solutions to the problem are all ordered triplets (4), where $t \in \mathbb{R} \backslash\{0,-1\}$, and the three triplets $(x, y, z)$ of the form

$$
\left(1,-2,-\frac{1}{2}\right),\left(-\frac{1}{2}, 1,-2\right),\left(-2,-\frac{1}{2}, 1\right)
$$

Note. If we write out all six possible systems corresponding to the given set equation, we get, in addition to systems (3) and (5), the systems

$$
\begin{aligned}
& x-y=z(y-z), \quad y-z=y(z-x), \quad z-x=x(x-y) \\
& x-y=x(y-z), \quad y-z=z(z-x), \quad z-x=y(x-y) \\
& x-y=y(y-z), \quad y-z=z(z-x), \quad z-x=x(x-y) \\
& x-y=z(y-z), \quad y-z=x(z-x), \quad z-x=y(x-y) .
\end{aligned}
$$

The first two arise from system (5) by cyclic permutation of variables, so they can be solved by the same method as (5). Adding all three equations in each of the two remaining systems, we get the same equation

$$
x^{2}+y^{2}+z^{2}=x y+y z+z x \quad \text { or } \quad(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0
$$

which has the unique solution $x=y=z$, which are not distinct numbers.