12.413 The lengths of the four arcs into which the entire circumference of radius $R$ is divided form a geometric progression with a common ratio of 3. The points of division serve as the vertices of a quadrilateral inscribed in this circle. Find its area.
Solution. Let points $A, B, C, D$ divide the given circle with center $O$ and radius $R$ such that $\cup A B: \cup B C: \cup C D: \cup D A=1: 3: 9: 27$.

Let $\alpha=\angle A O B$. Then $\angle B O C=3 \alpha, \angle C O D=9 \alpha, \angle D O A=27 \alpha \Rightarrow \alpha=9^{\circ}$.

$$
\begin{gathered}
S_{A B C D}=S_{O A B}+S_{O B C}+S_{O C D}+S_{O D A}= \\
=\frac{1}{2} R^{2} \sin \alpha+\frac{1}{2} R^{2} \sin 3 \alpha+\frac{1}{2} R^{2} \sin 9 \alpha+\frac{1}{2} R^{2} \sin 27 \alpha= \\
=\frac{R^{2}}{2}\left(\left(\sin 9^{\circ}+\sin 81^{\circ}\right)+\left(\sin 27^{\circ}+\sin 243^{\circ}\right)\right)= \\
=R^{2}\left(\sin 45^{\circ} \cdot \cos 36^{\circ}+\sin 135^{\circ} \cdot \cos 108^{\circ}\right)=\frac{R^{2} \sqrt{2}}{2}\left(\cos 36^{\circ}+\cos 108^{\circ}\right)
\end{gathered}
$$

But $\cos 36^{\circ}+\cos 108^{\circ}=\frac{2 \sin 36^{\circ} \cdot \cos 36^{\circ}+2 \sin 36^{\circ} \cdot \cos 108^{\circ}}{2 \sin 36^{\circ}}=$

$$
=\frac{\sin 72^{\circ}-\sin 72^{\circ}+\sin 144^{\circ}}{2 \sin 36^{\circ}}=\frac{\sin 144^{\circ}}{2 \sin 36^{\circ}}=\frac{1}{2} \Rightarrow S_{A B C D}=\frac{R^{2} \sqrt{2}}{4}
$$

Answer: $\frac{R^{2} \sqrt{2}}{4}$.