Romanian Mathematical Olympiad

The midpoint $S$ of the segment $MN$ belongs to the line $AB$, therefore $\frac{s-a}{b-a} \in \mathbb{R}$. The lines $MN$ and $AB$ are perpendicular, therefore $\frac{n-m}{b-a} \in i\mathbb{R}$.

We assume, without loss of generality, that the origin is at the circumcenter of the triangle, and $|a| = |b| = |c| = 1$. Then $\bar{a} = \frac{1}{a}$, $\bar{b} = \frac{1}{b}$, $\bar{c} = \frac{1}{c}$ and, since $s = \frac{m+n}{2}$, we have:

$$\frac{s-a}{b-a} \in \mathbb{R} \Leftrightarrow \frac{s-a}{b-a} = \frac{\bar{s}-\bar{a}}{\bar{b}-\bar{a}} \Leftrightarrow n+m=2(a+b)-ab(\bar{n}+\bar{m})$$

and

$$\frac{n-m}{b-a} \in i\mathbb{R} \Leftrightarrow \frac{n-m}{b-a} = -\frac{\bar{n}-\bar{m}}{\bar{b}-\bar{a}} \Leftrightarrow n-m = ab(\bar{n}-\bar{m}).$$

By addition, it follows that $n = a + b - ab\bar{m}$. Similarly, $p = b + c - bc\bar{m}$ and $q = c + a - ca\bar{m}$.

a.

The points $N$, $P$ and $Q$ are collinear if and only if

$$\begin{align*} \frac{n-p}{q-p} \in \mathbb{R} &\Leftrightarrow \frac{n-p}{q-p} = \frac{\bar{n}-\bar{p}}{\bar{q}-\bar{p}} \\ &\Leftrightarrow \frac{(a-c)(1-b\bar{m})}{(a-b)(1-c\bar{m})} = \frac{(\bar{a}-\bar{c})(1-b\bar{m})}{(\bar{a}-\bar{b})(1-c\bar{m})} \\ &\Leftrightarrow \frac{1-b\bar{m}}{1-c\bar{m}} = \frac{b}{c} \cdot \frac{1-\bar{b}\bar{m}}{1-\bar{c}\bar{m}} \\ &\Leftrightarrow \frac{1-b\bar{m}}{1-c\bar{m}} = \frac{b-m}{c-m} \\ &\Leftrightarrow c-b = |m|^2(c-b) \\ &\Leftrightarrow |m| = 1, \end{align*}$$

so if and only if the point $M$ belongs to the circumcircle of triangle $ABC$.

b.

Since triangles $ABC$ and $NPQ$ have the same centroid, it means that $\frac{a+b+c}{3} = \frac{n+p+q}{3} \Leftrightarrow a+b+c = \bar{m}(ab+bc+ca) \Leftrightarrow a+b+c = \bar{m}abc(\bar{a}+\bar{b}+\bar{c})$.

We apply the modulus to both members and we obtain that $|a+b+c| = |m| \cdot |a+b+c|$. The point $M$ is not located on the circumcircle of the triangle, therefore $|m| \neq 1$; it follows that $|a+b+c| = 0$. Then the centroid of the triangle $ABC$ coincides with its circumcenter, so the triangle $ABC$ is equilateral.