<s>[INST] Prove that the area of a circumscribed hexagon is twice the area of a triangle Let $\Delta ABC$ be an acute triangle and denote the circumscribed circle of $\Delta ABC$ $\Gamma$ with midpoint $O$. Let $A_1, B_1, C_1$ be the points on $\Gamma$ where the lines $AO, BO, CO$ intersect $\Gamma$. Show that the area of the hexagon $AC_1BA_1CB_1$ is twice as big as the area of $\Delta ABC$.
My first approach was to realize that the hexagon $AC_1BA_1CB_1$ contains $\Delta ABC$, which means that we have to prove that $|\Delta ABC_1| + |\Delta BA_1C| + |\Delta CB_1A| = |\Delta ABC|$. This could equivalently be written as:
$
\begin{equation}
\frac{|\Delta ABC_1| + |\Delta BA_1C| + |\Delta CB_1A|}{|\Delta ABC|} = 1.
\end{equation}
$
But since the three "outer" triangles each share a side with $\Delta ABC$, the ratio between the areas of $\Delta ABC$ and the respective "outer" triangle" will be $\frac{h_i}{H_i}$, where $h_i$ is the height of the outer triangle and $H_i$ is the height of $\Delta ABC$ (where both heights  are perpendicular to the shared side). This means that we can write the above equation as:
$
\begin{equation}
\frac{h_1}{H_1} + \frac{h_2}{H_2} + \frac{h_3}{H_3} = 1
\end{equation}
$
Now, label the points where the lines $AO, BO, CO$ intersect the opposite side of $\Delta ABC$ $P, R, Q$ respectively ($P$ on $AB$, $R$ on $AC$, $Q$ on $AB$). Since we have angles in the same segment of $\Gamma$, we get the following similar triangles: $\Delta ABC_1\sim\Delta PBC, \Delta CQA_1\sim\Delta ABQ, \Delta ARB_1\sim\Delta RBC$, which means that we can write:
$
\frac{h_1}{H_1} = \frac{|AP|}{|BP|} \\
\frac{h_2}{H_2} = \frac{|CQ|}{|BQ|} \\
\frac{h_3}{H_3} = \frac{|AR|}{|CR|}
$
Which means that we have to prove that the sum of these (above) ratios equals $1$. But from here, I can't seem to make much progress...
 [/INST] Let's assume $R$ is the circumradius of $\triangle ABC$. Then, we have:
$$S_{\triangle ABC}=\frac{1}{2} |AC||AB| \sin A=\frac{1}{2} \times 2R \sin B \times  2R\sin C \times \sin A \\ = 2R^2 \sin A\sin B \sin C.$$

Now, let's compute the area of $\triangle A_1BC$. We have:
$$S_{\triangle A_1BC}=\frac{1}{2}|A_1B||A_1C| \sin A =\frac{1}{2} \times 2R \cos C \times 2R \cos B \times \sin A \\= 2R^2 \cos C \cos B \sin A.$$
Similarly,
$$S_{\triangle B_1AC}=2R^2 \cos A \cos C \sin B \\ S_{\triangle C_1AB}=2R^2 \cos A \cos B \sin C.$$
Thus, as you observed, we need to show that:
$$\cos C \cos B \sin A+\cos A \cos C \sin B+\cos A \cos B \sin C =\sin A\sin B \sin C,$$
while $\sin A=\sin(B+C)$ and $\cos A= - \cos (B+C).$
So, we must show that:
$$\cos C \cos B \sin(B+C)- \cos (B+C) \cos C \sin B- \cos (B+C)\cos B \sin C \\= \sin(B+C)\sin B \sin C \\ \iff  sin(B+C) (\cos C \cos B - \sin B \sin C )= \cos (B+C) (\cos C \sin B + \cos B \sin C),$$
which obviously holds.
We are done.
</s>