75th Romanian Mathematical Olympiad

If $\angle A=\angle C=\angle E=\alpha$ and $\angle B=\angle D=\angle F=\beta$, then the sum of the measures of the angles of the hexagon $ABCDEF$ is $3\alpha +3\beta =720^{\circ }$, so $\alpha +\beta =240^{\circ }$. Since $\angle B+\angle C=\alpha +\beta =240^{\circ }$, the lines $AB$ and $CD$ will intersect at a point located outside the hexagon. Denote $\{X\}=AB\cap CD$, $\{Y\}=CD\cap EF$, $\{Z\}=EF\cap AB$. The point $P$, equidistant from the sides $AB$, $CD$ and $EF$ of the hexagon, is the center of the equilateral triangle $XYZ$.

In the complex plane we consider a rectangular coordinate system with the origin at $P$. From $△BCX\sim △DEY\sim △FAZ$, there is $k\in \mathbb{R}$, $k>0$, such that $\frac{c-x}{b-x}=\frac{e-y}{d-y}=\frac{a-z}{f-z}=k\cdot \epsilon$, where $\epsilon =\cos \frac{\pi }{3}\pm i\sin \frac{\pi }{3}$. We obtain $c-x=(b-x)k\cdot \epsilon$, $e-y=(d-y)k\cdot \epsilon$, $a-z=(f-z)k\cdot \epsilon$.

Since $p=\frac{x+y+z}{3}=0$, by addition, we obtain $c+e+a=(b+d+f)k\cdot \epsilon \iff g_1=g_2\cdot k\cdot \epsilon$. From $G_1\ne G_2$ it follows that $g_1$ and $g_2$ cannot be $0$, so $G_1$, $G_2$, $P$ are three distinct points. Therefore $\frac{g_1-p}{g_2-p}=k\cdot \epsilon$, or $\angle G_{1}P{G}_2=60^{\circ }$.