1.4. Find all real solutions of the system

$$
\begin{aligned}
a x+b y & =(x-y)^{2} \\
b y+c z & =(y-z)^{2} \\
c z+a x & =(z-x)^{2}
\end{aligned}
$$

where $a, b, c$ are given positive real numbers.

## 2nd BALKANIAN MATHEMATICAL OLYMPIAD

SOFIA, Bulgaria, May 1985.
1.4. Let us introduce new variables

$$
s=x-y, \quad t=y-z
$$

Then the system becomes

$$
\begin{aligned}
a(s+y)+b y & =s^{2} \\
b y+c(y-t) & =t^{2} \\
c(y-t)+a(s+y) & =(s+t)^{2}=s^{2}+2 s t+t^{2} \\
& =2 s t+[a(s+y)+b y]+[b y+c(y-t)]
\end{aligned}
$$

The last relation gives $y=-s t / b$. After substituting, from the first two equations we obtain the following system in $s$ and $t$ :

$$
\begin{aligned}
{[a b-(a+b) t] s } & =s^{2} \\
{[-(b+c) s-b c] t } & =t^{2}
\end{aligned}
$$

We have three possibilities.

a) If $s=0$, then $-c t=t^{2}$ and we have either $t=0$, i.e.

$$
(x, y, z)=(0,0,0)
$$

or $t=-c$, i.e.

$$
(x, y, z)=(0,0, c)
$$

b) In a similar way, for $t=0$ we obtain $s=a$ and:

$$
(x, y, z)=(a, 0,0)
$$

c) If $s \neq 0$ and $t \neq 0$, then

$$
\begin{aligned}
a b-(a+b) t & =b s \\
-b c-(b+c) s & =b t
\end{aligned}
$$

As $a, b, c>0$, the determinant of this system is different from 0. Solving it for instance by elimination, we get $s=-b, t=b$ and one more solution:

$$
(x, y, z)=(0, b, 0)
$$

## Figure 2.1.