4.085 In an angle containing $60^{\circ}$, five circles are inscribed such that each subsequent circle, starting from the second, touches the previous one; How many times greater is the sum of the areas of all five corresponding circles compared to the area of the smallest circle?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-093.jpg?height=475&width=663&top_left_y=171&top_left_x=416)

$\angle O_{1} A B_{1}=30^{\circ} \Rightarrow A O_{1}=2 a, A O_{2}=2 b$. But $A O_{2}=b+a+A O_{1}=3 a+b \Rightarrow \frac{b}{a}=3$.

Therefore, if $r_{i}$ is the radius of the $i$-th inscribed circle, then $\frac{r_{i}}{r_{i-1}}=3$, i.e., $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$ form a geometric progression with a common ratio of 3. If $S_{i}$ is the area of the $i$-th circle, then $S_{1}+S_{2}+S_{3}+S_{4}+S_{5}=\pi r_{1}^{2}\left(1+3^{2}+3^{4}+3^{6}+3^{8}\right)=S_{1} \cdot 7381$.

## Answer: 7381 times.

LOGARITHMS. EXPONENTIAL AND LOGARITHMIC EQUATIONS

Simplify the expressions (7.295 - 7.299):