75th 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:

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

and

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

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

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

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 $\frac{a+b+c}{3}=\frac{n+p+q}{3}\iff a+b+c=\bar{m}(ab+bc+ca)\iff a+b+c=\bar{m}abc(\bar{a}+\bar{b}+\bar{c})$.

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 $|m|\ne 1$; it follows that $|a+b+c|=0$. Then the centroid of the triangle $ABC$ coincides with its circumcenter, so the triangle $ABC$ is equilateral.