1. In the isosceles triangle \(ABC\), given are \(\overline{AC} = \overline{BC} = 2b\) and \(\measuredangle CAB = 2\alpha\). Let \(P_1\) be the area of the inscribed circle \(k_1\) in triangle \(ABC\), \(P_2\) be the area of the circle that touches the circle \(k_1\) and the legs of triangle \(ABC\), \(\ldots\), \(P_i\) be the area of the circle that touches the circle \(k_{i-1}\) and the legs of triangle \(ABC\), and so on. Calculate the sum \(P_1 + P_2 + \ldots + P_n + \ldots\)
Solution. Let $\overline{A B}=2 a$, hence $\overline{M B}=a, \measuredangle M B O_{1}=\Varangle O_{1} B M_{1}=\alpha$. From the right triangle $O_{1} M B$ it follows that $\overline{O_{1} M}=r_{1}=a \cdot \operatorname{tg} \alpha$. Let $T$ be a point on the segment $O_{1} M_{1}$ such that $O_{2} T \perp O_{1} M_{1}$. From this, it follows that $\overline{T M_{1}}=\overline{O_{2} M_{2}}=r_{2}$. Furthermore, $\overline{O_{1} O_{2}}=r_{1}+r_{2}, \overline{O_{1} T}=r_{1}-r_{2}$ and $\angle O_{2} O_{1} T=2 \alpha$. From the right triangle $O_{1} T O_{2}$ we get $r_{1}-r_{2}=\left(r_{1}+r_{2}\right) \cos 2 \alpha$, i.e., $r_{1}(1-\cos 2 \alpha)=r_{2}(1+\cos 2 \alpha)$. From this, it follows that $r_{2}=r_{1} \operatorname{tg}^{2} \alpha$.

Similarly, $r_{3}=r_{2} \operatorname{tg}^{2} \alpha=r_{1} \operatorname{tg}^{4} \alpha$, or generally:

$$
\begin{gathered}
r_{n}=r_{1} \operatorname{tg}^{2(n-1)} \alpha, \quad n \in N \\
P_{1}=\pi r_{1}^{2}=\pi a^{2} \operatorname{tg}^{2} \alpha \\
P_{2}=\pi r_{2}^{2}=\pi r_{1}^{2} \operatorname{tg}^{4} \alpha, \ldots \\
P_{n}=\pi r_{n}^{2}=\pi r_{1}^{2} \operatorname{tg}^{4(n-1)} \alpha, \ldots
\end{gathered}
$$

Since $2 \alpha<\frac{\pi}{2}$, it follows that $\alpha<\frac{\pi}{4}$, hence $\operatorname{tg} \alpha<1$, so we can apply the formula

![](https://cdn.mathpix.com/cropped/2024_06_05_ec67a2454bbad4820097g-7.jpg?height=531&width=471&top_left_y=1393&top_left_x=990)
for the sum of an infinite geometric series

$$
\begin{aligned}
P & =P_{1}+P_{2}+\ldots+P_{n}+\ldots=\pi r_{1}^{2}\left(1+\operatorname{tg}^{4} \alpha+\operatorname{tg}^{8} \alpha+\ldots\right) \\
& =\pi r_{1}^{2} \frac{1}{1-\operatorname{tg}^{4} \alpha}=\pi a^{2} \operatorname{tg}^{2} \alpha \frac{1}{1-\operatorname{tg}^{4} \alpha}=\pi a^{2} \frac{\sin ^{2} \alpha \cos ^{2} \alpha}{\cos 2 \alpha}
\end{aligned}
$$

From the right triangle $M B C$ it follows that $a=2 b \cos 2 \alpha$, so finally we get:

$$
P=4 \pi b^{2} \cos ^{2} 2 \alpha \frac{\sin ^{2} \alpha \cos ^{2} \alpha}{\cos 2 \alpha}=\pi b^{2} \cos 2 \alpha \sin ^{2} 2 \alpha
$$