10.346. A circle with center $O_{1}$ is divided by the diameter $A B$ into two semicircles. In one of them, two new semicircles are constructed, based on $O_{1} A$ and $O_{1} B$ as diameters. A circle is inscribed in the curvilinear figure bounded by the contours of these three semicircles. How many times smaller is its area compared to the area of the given circle?
## Solution.

Let $O_{2}$ be the midpoint of $A O_{1}$ - the center of one semicircle, $O_{3}$ be the midpoint of $B O_{1}$ - the center of the second semicircle, $O_{4}$ - the center of the circle inscribed in the curvilinear figure mentioned in the problem (Fig. 10.129).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0806.jpg?height=436&width=517&top_left_y=886&top_left_x=757)

Fig. 10.129

Let $R$ be the radius of the circle with center $O_{1}$, and $S_{1}$ be its area, $r$ be the radius of the circle with center $O_{4}$, and $S_{2}$ be its area. We have: $O_{1} O_{2}=O_{1} O_{3}=\frac{R}{2}$, $O_{2} O_{4}=O_{3} O_{4}=\frac{R}{2}+r, O_{1} O_{4}=R-r$. In triangle $O_{2} O_{4} O_{3}: O_{2} O_{4}=O_{3} O_{4}$, point $O_{1}$ is the midpoint of $O_{2} O_{3}$. Therefore, $O_{1} O_{4} \perp O_{2} O_{3}$, then

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0807.jpg?height=537&width=1167&top_left_y=77&top_left_x=93)

Fig. 10.130

Fig. 10.131

$O_{2} O_{4}^{2}=O_{1} O_{2}^{2}+O_{1} O_{4}^{2}$, i.e., $\left(\frac{R}{2}+r\right)^{2}=\left(\frac{R}{2}\right)^{2}+(R-r)^{2} . R^{2}=3 R r, R=3 r$.

Therefore, $\frac{S_{1}}{S_{2}}=\frac{R^{2}}{r^{2}}=9$.

Answer: 9 times.