(Projective Algebra)

Solve in $\mathbb{C}^{3}$ the system

$$
\left\{\begin{array}{l}
a^{2}+a b+c=0 \\
b^{2}+b c+a=0 \\
c^{2}+c a+b=0
\end{array}\right.
$$

## - Correction of exercises -
We must keep in mind Exercise 3: quadratic equations can be put in the form of a homography, which is also a projective transformation. We transform the first equation into $a=\frac{-c}{a+b-c}$. By setting $S=a+b+c$, we have $a=\frac{c}{c-S}=f(c)$ where $f$ is a projective transformation. Similarly, we find $b=f(a)$ and $c=f(b)$.

Let's study $f \circ f \circ f$: $c^{\prime}$ is a projective transformation that has fixed points $A, B$, and $C$. However, a projective transformation has only two distinct fixed points, so $a=b$ without loss of generality. Thus, we can solve our equation and find the triplets $(0,0,0)$ and $\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$.

## 5 Pot-pourri

## 1 IMO 2019

## - Day 1 -