67th Romanian Mathematical Olympiad

a) If the triangle $OAB$ is oriented counterclockwise, then $m(\angle AOB)=\arg \frac{b}{a}$, otherwise $m(\angle AOB)=\arg \frac{a}{b}$. Then $$\sin (\angle AOB)=\left|\frac{|a|}{2|b|}\left(\frac{b}{a}-\frac{\bar{b}}{\bar{a}}\right)\right|=\frac{|\overline{ab}-\overline{ab}|}{2|a||b|}$$ and since $[OAB]=\frac{1}{2}OA\cdot OB\cdot \sin (\angle AOB)$, we obtain the conclusion.

b) Let $\epsilon =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$. We may assume that the complex coordinates of the points $A$, $B$ and $C$ are $1$, $\epsilon$, and ${\epsilon }^2$, respectively. Let $P$ be a point with complex coordinate $p$, in the interior of the circle. Then $(p-1)+\epsilon (p-\epsilon )+{\epsilon }^2(p-{\epsilon }^2)=0$. Consider the points $D(p-1)$, $E(\epsilon (p-\epsilon ))$ and $F({\epsilon }^2(p-{\epsilon }^2))$. Observe that, $OD=PA$, $OE=PB$ and $OF=PC$. Let $J$ be a point such that $ODJF$ is a parallelogram. We obtain that the complex coordinates of points $J$ and $F$ are opposite numbers. Then $OJ=PC$, and the side lengths of $ODJ$ are $PA$, $PB$, and $PC$. But $$\begin{array}{rl}[ODJ]& =[ODE]=\frac{1}{4}|(\bar{p}-1)\cdot \epsilon \cdot (p-\epsilon )-(p-1)\cdot \bar{\epsilon }\cdot (\bar{p}-\bar{\epsilon })|\\ & =\frac{1}{4}((\epsilon -{\epsilon }^2)|p{|}^2-(\epsilon -{\epsilon }^2))=\frac{\sqrt{3}}{4}|p{|}^2-1\end{array}$$ Since $|p|<1$, we obtain $[ODE]=S(P)=\frac{\sqrt{3}}{4}(1-|p{|}^2)$, from which the conclusion follows easily.