Romanian Mathematical Olympiad - District Round

a) Let $O$ be the circumcenter of the triangle $ABC$. Without loss of generality, we consider $O(0)$, while the vertices of the triangle are $A(1)$, $B(\varepsilon)$, and $C(\varepsilon^2)$, where $\varepsilon = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Because the length of the arc $NB$ is $30^\circ$, we have $AO \perp ON$, so point $N$ has the affix $i$. Moreover, $NI \perp AB$ implies the existence of $\alpha \in \mathbb{R}^*$ such that:
$$\frac{i - z_I}{1 - \varepsilon} = i\alpha \Rightarrow z_I = i - \frac{3}{2}i\alpha - \frac{\sqrt{3}}{2}\alpha,$$
where $z_I$ is the affix of the point $I$. From $|z_I| = 1$ we obtain $\alpha = 1$, so the affix of the point $I$ is $i\varepsilon$. In the same manner, from $MN \perp AC$ we obtain that the affix of the point $M$ is $i\varepsilon^2$, so the triangle $IMN$ is equilateral.

b) Using Sylvester's theorem we have the affixes of the orthocenters as follows:
$$\begin{aligned} z_{H_1} &= z_A + z_N + z_B = i + 1 + \varepsilon, \\ z_{H_2} &= z_B + z_M + z_C = \varepsilon + i\varepsilon + \varepsilon^2 = \varepsilon z_{H_1}, \\ z_{H_3} &= z_C + z_I + z_A = \varepsilon^2 + i\varepsilon^2 + 1 = \varepsilon^2 z_{H_1}, \end{aligned}$$
so the triangle $H_1H_2H_3$ is equilateral.