Problem 4. A circle is inscribed in a circular sector (the circle touches both radii and the arc of the sector). The radius of the sector is 3 times the radius of the circle. Find the ratio of the areas of the sector and the circle.
Solution. The circle $k(S, r)$ is inscribed in the circular segment $A O B$, and therefore $S$ lies on the angle bisector of $\angle A O B$. Let $C$ be the point where $k(S, r)$ touches the radius $O A$. Then triangle $OCS$ is a right triangle, and from the condition in the problem, we have $\overline{O S}=2 r$. Since the hypotenuse of triangle $O C S$ is twice the length of the leg $S C (\overline{O S}=2 r$, $\overline{S C}=r$), it follows that $\angle C O S$ is an angle of $30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_d4fd26fa5c16caf575adg-8.jpg?height=408&width=372&top_left_y=1745&top_left_x=1112)

Then $\angle A O B=60^{\circ}$, and therefore the area of the circular segment $A O B$ is $P_{1}=\frac{60^{\circ}}{360^{\circ}}(3 r)^{2} \pi=\frac{3}{2} r^{2} \pi$. The area of the circle $k(S, r)$ is $P_{2}=r^{2} \pi$, so $\frac{P_{1}}{P_{2}}=\frac{3}{2}$.