## T-1 A

Determine all triples $(a, b, c)$ of real numbers satisfying the system of equations

$$
\begin{aligned}
& a^{2}+a b+c=0 \\
& b^{2}+b c+a=0 \\
& c^{2}+c a+b=0
\end{aligned}
$$

Answer. The solutions are

$$
(a, b, c) \in\left\{(0,0,0),\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)\right\}
$$


Solution. If one of the numbers $a, b$ and $c$ is equal to zero, it is easy to see that the other two numbers also have to be equal to zero, which gives us the solution $(0,0,0)$.

Now assume that $a, b, c \neq 0$.

If all three numbers are positive, then the left-hand side of each equation is positive, while the right-hand sides are equal to zero, which is impossible.

Let us assume that only one of the numbers is positive, and without loss of generality let it be $a$. Since $b, c0$, which is a contradiction.

It remains to consider the two following cases:

(a) All three numbers are negative.

We substitute $a=-x, b=-y$ and $c=-z$, where $x, y, z>0$. The original system transforms into

$$
\begin{aligned}
& x^{2}+x y=z \\
& y^{2}+y z=x \\
& z^{2}+z x=y
\end{aligned}
$$

The system is cyclic, so we can assume that $x \leqslant y$ and $x \leqslant z$. Now we have

$$
\begin{aligned}
& x^{2}+x y=z \geqslant x \Longrightarrow x+y \geqslant 1 \\
& y^{2}+y z=x \leqslant y \Longrightarrow y+z \leqslant 1
\end{aligned}
$$

From the previous two inequalities we conclude that

$$
x+y \geqslant 1 \geqslant y+z, \quad \text { i.e. } \quad x \geqslant z
$$

On the other hand $x \leqslant z$, so we get $x=z$.

Now, from equation (1) it follows that $x+y=1$, while from equation (2) it follows that

$$
x=y^{2}+y z=y^{2}+y x=y(y+x)=y
$$

Thus $x=y=z$ and from $x+y=1$ we see that $x=y=z=1 / 2$.

We easily verify that $(a, b, c)=\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$ is indeed a solution.

(b) Exactly one of the numbers is negative.

Without loss of generality we can assume that $c$ is negative, while $a$ and $b$ are positive. From the second equation we conclude that $b(b+c)=-a0$.

Adding $a+b$ to the first equation and cyclic permutation yields

$$
a+b+c=(1-a)(a+b)=(1-b)(b+c)=(1-c)(c+a)
$$

The last product is positive. This implies that $1-a>0$ and $1-b<0$ by our above considerations. Therefore $0<a+c<1+c<b+c<0$, a contradiction.
