Find the area ratio between ABC and the hatched triangle. Could you help me?
Find the area ratio between ABC and the hatched triangle.

Thanks in advance.
 A: Take the segment $AB$ as horizontal. If one moves the point $C$ along the $x$ axis, stretching the rest of the figure accordingly, every area will be preserved. If one then moves $C$ along the $y$ axis, every area will be scaled by a constant factor. Therefore, we can assume $\bigtriangleup ABC$ to be equilateral, and we can make its circumcircle the unit circle WLOG.
Let $X=CA_1\cap AB_1$, $Y=AB_1\cap BC_1$, $Z=BC_1\cap CA_1$, so that $\bigtriangleup XYZ$ is the shaded triangle. We can represent each point in our diagram as a complex number. We first calculate $$A_1=\frac{2A+B}{3},$$ $$B_1=\frac{2B+C}{3},$$ $$C_1=\frac{2C+A}{3}.$$ By Menelaus on $\bigtriangleup ABB_1$, $\frac{|B_1-X|}{|X-A|}=\frac{4}{3}$, so that subsequently $$X=\frac{4A+3B_1}{7}=\frac{4A+2B+C}{7}=A\left(4+2\omega+\omega^2\right),$$ and analogously, $$Y=B\left(4+2\omega+\omega^2\right),$$ $$Z=C\left(4+2\omega+\omega^2\right),$$ where $\omega=e^{\frac{2i\pi}{3}}$. Therefore, $\bigtriangleup XYZ$ is just $\bigtriangleup ABC$ scaled down by a factor of $|4+2\omega+\omega^2|=\sqrt{7}$, and the ratio of their areas is $\boxed{7}$.
