## Task B-4.4.

The area of triangle $A B C$ is $P=3+\sqrt{3}$. Calculate the area of the circle circumscribed around triangle $A B C$, if the lengths of the arcs $\widehat{A B}, \widehat{B C}$ and $\overparen{C A}$ are in the ratio $5: 3: 4$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_b0f5f68256ba1552c492g-16.jpg?height=497&width=529&top_left_y=1785&top_left_x=752)

The lengths of the arcs are in the ratio $l_{1}: l_{2}: l_{3}=5: 3: 4$, i.e., $l_{1}=5k, l_{2}=3k, l_{3}=4k$. From $l_{1}+l_{2}+l_{3}=2\pi$ it follows that

$k=\frac{2\pi}{12}=\frac{\pi}{6} \Rightarrow l_{1}=\frac{5\pi}{6}, l_{2}=\frac{\pi}{2}, l_{3}=\frac{2\pi}{3}$.

Thus, $\varangle ASB=150^{\circ}, \varangle BSC=90^{\circ}, \varangle ASC=120^{\circ}$.

According to Thales' theorem, the inscribed angle is half the central angle over the same chord. Therefore,

$\gamma=\frac{1}{2} \cdot 150^{\circ}=75^{\circ}, \beta=\frac{1}{2} \cdot 90^{\circ}=45^{\circ}, \alpha=\frac{1}{2} \cdot 120^{\circ}=60^{\circ}$.

The lengths of the sides are calculated using the Law of Sines:

$$
\begin{aligned}
& a=2R \sin 45^{\circ}=R \sqrt{2} \\
& b=2R \sin 60^{\circ}=R \sqrt{3}
\end{aligned}
$$

$\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}=\frac{\sqrt{2}}{4}(1+\sqrt{3})$.

The area of the triangle is $P=\frac{1}{2} a b \sin 75^{\circ}=\frac{1}{2} R \sqrt{2} R \sqrt{3} \frac{\sqrt{2}}{4}(1+\sqrt{3})=\frac{R^{2} \sqrt{3}}{4}(1+\sqrt{3})$.

If $P=3+\sqrt{3}$, then from $\frac{R^{2} \sqrt{3}}{4}(1+\sqrt{3})=3+\sqrt{3}$ it follows that $R^{2}=4$.

The area of the circle is $P=4\pi$.