Romanian Mathematical Olympiad

Consider an orthonormal coordinate system with the origin at $O$. For any point $Z$ in the plane, we denote its complex coordinate by $z$.

a.
From Sylvester's relation, we have $h_1 = m + a + b$, $h_2 = m + b + c$, and $h_3 = m + c + a$. Also, let $h = a + b + c$ be the complex coordinate of the orthocenter of triangle $ABC$. Therefore, we obtain:
$$s(M) = 6 + |h|^2 + 2m\overline{h} + 2\overline{m}h.$$
If triangle $ABC$ is equilateral, then $h = a + b + c = 0$, hence $s(M) = 6$.

b.
Assume, by contradiction, that triangle $ABC$ is not equilateral, which is equivalent to $h \neq 0$. Then, since $s(M_1) = s(M_2)$, we have: $|h|^2 + 2m_1\overline{h} + 2\overline{m}_1h = |h|^2 + 2m_2\overline{h} + 2\overline{m}_2h \Leftrightarrow \overline{h}(m_1 - m_2) + h\frac{m_2 - m_1}{m_1m_2} = 0 \Leftrightarrow m_1m_2 = \frac{h}{\overline{h}}$.

Similarly, we obtain $m_1m_3 = \frac{h}{\overline{h}}$. Since $m_1 \neq 0$, we get $m_2 = m_3$, which is a contradiction with $M_2 \neq M_3$. Therefore, triangle $ABC$ is equilateral.