10.251. Five circles are inscribed in an angle of $60^{\circ}$ such that each subsequent circle (starting from the second) touches the previous one. By what factor is the sum of the areas of all five corresponding circles greater than the area of the smallest circle?
## Solution.

Let $A$ be the vertex of the angle. $O_{i}$ and $r_{i}$ be the center and radius of the $i$-th circle $(i=1, \ldots, 5)$. Since $\frac{1}{2} \angle A=30^{\circ}$, then $A O_{i}=2 r_{i}, A O_{i-1}=2 r_{i-1}$, from which $A O_{i}=A O_{i-1}+r_{i-1}+r_{i}$ or $2 r_{i}=2 r_{i-1}+r_{i-1}+r_{i}$, i.e., $r_{i}=3 r_{i-1}$. Therefore, the radii of the circles form a geometric progression with a common ratio of 3, and the sum of the areas of the five circles will be $S=\pi r_{1}^{2}\left(1+9+9^{2}+9^{3}+9^{4}\right)=\frac{\pi r_{1}^{2}\left(9^{5}-1\right)}{9-1}=7381 \pi r_{1}^{2}$. Thus, $S: \pi r_{1}^{2}=7381$.

Answer: 7381 times.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0749.jpg?height=445&width=686&top_left_y=700&top_left_x=333)

Fig. 10.58