2B. A circle with radius $r$ touches the sides of the angle $\alpha=60^{\circ}$, the second smaller circle touches the sides of the angle $\alpha$ and the first circle, the third circle touches the sides of the angle $\alpha$ and the second circle, and so on. Calculate the sum of:

a) the perimeters

b) the areas of all the circles.
Solution. Let

$\overline{A M}=r_{1}, \overline{B N}=r, \overline{O M}=a, \overline{M N}=b$.

From the similarity of triangles $O B N$ and $O A M$, we get $\frac{r}{r_{1}}=\frac{a+b}{a}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_ae7d00f0d51ae2d35791g-09.jpg?height=389&width=541&top_left_y=1822&top_left_x=941)

On the other hand, $\sin 30^{\circ}=\frac{r_{1}}{a}$, so $a=2 r_{1}$. Therefore,

$$
r=r_{1}\left(1+\frac{b}{a}\right)=r_{1}\left(1+\frac{b}{2 r_{1}}\right)=r_{1}\left(1+\frac{r+r_{1}}{2 r_{1}}\right)
$$

We get $r_{1}=\frac{r}{3}$. Similarly, $r_{2}=\frac{r_{1}}{3}=\frac{r}{9}$, and so on.

a) The sum of the perimeters is

$$
2 \pi\left(r+r_{1}+r_{2}+\ldots\right)=2 \pi\left(r+\frac{r}{3}+\frac{r}{9}+\ldots\right)=2 \pi r \frac{1}{1-\frac{1}{3}}=3 \pi r
$$

b) The sum of the areas is

$$
\pi\left(r^{2}+r_{1}^{2}+r_{2}^{2}+\ldots\right)=\pi\left(r^{2}+\frac{r^{2}}{3^{2}}+\frac{r^{4}}{3^{4}}+\ldots\right)=\pi r^{2} \frac{1}{1-\frac{1}{9}}=\frac{9}{8} \pi r^{2}
$$