Romanian Mathematical Olympiad

Observe that $\text{tr}(A) = 0$ implies $A^2 = O_{n \times n}$ and
$$ABA \cdot ACA = O_{n \times n} = ACA \cdot ABA,$$

If $\text{tr}(A) \neq 0$, we show that $\text{rank}(A) = 1$. Let $r = \text{rank}(A)$. Then we can find matrices $X \in \mathcal{M}_{n \times r}(\mathbb{C})$ and $Y \in \mathcal{M}_{r \times n}(\mathbb{C})$ with $\text{rank}(X) = \text{rank}(Y) = r$ such that $A = XY$. Then $YX \in \mathcal{M}_r(\mathbb{C})$ and, as $\text{tr}(A) \neq 0$, we have
$$r \geq \text{rank}(YX) \geq \text{rank}(X(YX)Y) = \text{rank}(A^2) = \text{rank}(A) = r,$$
That is $\text{rank}(YX) = r$ and the matrix $YX$ is non-singular. Equality $A^2 = \text{tr}(A) \cdot A$, can be written $(X)^2 = \text{tr}(A) \cdot XY$, obtaining $(YX)^3 = \text{tr}(A) \cdot (YX)^2$. As $YX$ is non-singular, we get $YX = \text{tr}(A) \cdot I_r$. In conclusion,
$$\text{tr}(A) = \text{tr}(XY) = \text{tr}(YX) = \text{tr}(A) \cdot r,$$
giving $r = 1$. To conclude, let $B, C \in \mathcal{M}_n(\mathbb{C})$. We have $ABA = XYBXY = pXY = pA$, where $p = YBX \in \mathcal{M}_1(\mathbb{C})$. In the same way $ACA = qA$ with $q \in \mathbb{C}$, concluding the result of the problem.