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

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


Solution. If $x, y, z$ are three mutually distinct real numbers, then

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

are clearly numbers different from 0 and -1 whose product is equal to 1 . This property must therefore be possessed also by the values $x, y, z$ from any such triple. We will thus assume from now on that

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

Since the given set relation is the same for each of the ordered triples $(x, y, z)$, $(z, x, y)$ and $(y, z, x)$, we will assume in addition to (2) that $x>\max \{y, z\}$, and will distinguish two cases, according as $y>z$ or $z>y$. Let us introduce the following notation for intervals: $I_{1}=(0, \infty), I_{2}=(-1,0), I_{3}=(-\infty,-1)$.

The case of $x>y>z$. For the fractions (1) we clearly have $u \in I_{1}, v \in I_{2}$ and $w \in I_{3}$, so $u>v>w$. The given set equation can thus be fulfilled only when $u=x, v=y$ and $w=z$. Upon substituting from (1) and an easy manipulation we arrive at the equations

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

In view of the condition $x y z=1$ from (2) we can replace the term $z x$ in the equation $x y+y=z x+x$ by $1 / y$. This leads to

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

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

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

But then also the third expression in (3) is equal to -1 . Any solution of our problem (in the current case of $x>y>z$ ) must therefore be of the form

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

where $t \in I_{1}$ is arbitrary (in view of (3), we do not need to worry about checks). From the procedure used it also follows that taking $t \in I_{2}$ (or $t \in I_{3}$, respectively) in the formula (4) we get all solutions of our problem satisfying $z>x>y(y>z>x)$, so that it is not necessary to list the cyclic permutations of the triples from (4) in our final answer below.

The case of $x>z>y$. Now we have for the fractions (1) $u \in I_{3}, v \in I_{1}$ and $w \in I_{2}$, so $v>w>u$, and the set equation in question is satisfied only if $u=y$, $v=x$ and $w=z$. Upon 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 up these three equations yields

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

which in view of $x \neq y$ implies that $z=\frac{1}{2}(x+y)$. Substituting this back into (5) we find (taking again into account that $x \neq y$ ) that the only solution is $x=1, y=-2$ and $z=-\frac{1}{2}$. The same triple also forms the (unique) solution satisfying $y>x>z$, as well as the (unique) solution for which $z>y>x$.

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

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

## Czech-Slovak-Polish Match  Bílovec, June 25-26, 2007

$\mathbb{N} / \mathbb{1}^{6}$
