In the right-angled triangle $A B_{0} C_{0}$, which is right-angled at $B_{0}$. From $B_{0}$, we draw a perpendicular to $A C_{0}$; from the foot of this perpendicular, $C_{1}$, we draw a perpendicular to $A B_{0}$, and from the foot of this perpendicular, $B_{1}$, we again draw a perpendicular to $A C_{0}$, and so on, indefinitely. The feet of the perpendiculars drawn to $A C_{0}$ and $A B_{0}$ are $C_{1}, C_{2}, \ldots C_{i}, \ldots$, and $B_{1}, B_{2}, \ldots B_{i}, \ldots$ respectively. The following sums are to be calculated (from $i=0$ to $i=\infty$):

$1^{0} . \sum B_{i} C_{i} ; \quad 2^{0} \cdot \sum A B_{i} ; \quad 3^{0} \cdot \sum A C_{i} ; \quad 4^{0} \cdot \sum C_{i+1} B_{i}$.
$1^{0}$. Examine the parts of the broken line $C_{0} B_{0} C_{1} B_{1} C_{2} B_{2} \ldots C_{i} B_{i} \ldots$ and express these parts using $C_{0} B_{0}$ (or $B_{0} C_{0}$) and $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_65940067176997c2e9cdg-1.jpg?height=328&width=512&top_left_y=229&top_left_x=796)

$$
\text { I. } \begin{cases}B_{0} C_{1}=B_{0} C_{0} \cos \alpha, & B_{1} C_{1}=B_{0} C_{1} \cos \alpha=B_{0} C_{0} \cos ^{2} \alpha \\ B_{1} C_{2}=B_{0} C_{0} \cos ^{3} \alpha, & B_{2} C_{2}=B_{0} C_{1} \cos ^{4} \alpha \text { and so on. }\end{cases}
$$

Accordingly, $B_{0} C_{0}, B_{1} C_{1}, B_{2} C_{2}, \ldots, B_{i} C_{i}, \ldots$ form a geometric sequence whose first term is $B_{0} C_{0}$, and the common ratio is $\cos ^{2} \alpha$ with $0<\cos ^{2} \alpha<1$. Therefore,

$$
\begin{gathered}
\lim _{i \rightarrow \infty} \sum^{i=0} B_{i} C_{i}=\frac{B_{0} C_{0}}{1-\cos ^{2} \alpha}=\frac{B_{0} C_{0}}{\sin ^{2} \alpha}=\frac{B_{0} C_{0} \cdot A C_{0}^{2}}{B_{0} C_{0}^{2}}= \\
=A C_{0} \cdot \frac{A C_{0}}{B_{0} C_{0}}=A C_{0} \operatorname{cosec} \alpha
\end{gathered}
$$

$4^{0}$. Similarly, among the terms of the sequence under I., there are $B_{0} C_{1}, B_{1} C_{2}, \ldots$, and so on. These form a geometric sequence whose first term is $B_{0} C_{1}=B_{0} C_{0} \cos \alpha$ and the common ratio is $\cos ^{2} \alpha$, so that

$$
\lim _{i \rightarrow \infty} \sum^{i=0} C_{i+1} B_{i}=\frac{B_{0} C_{0} \cos \alpha}{1-\cos ^{2} \alpha}=A C_{0} \cos \mathrm{ec} \alpha \cos \alpha=A C_{0} \operatorname{cotg} \alpha=A B_{0} \cos \mathrm{ec} \alpha
$$

$2^{0} . A B_{0}=B_{0} C_{0} \operatorname{cotg} \alpha, \quad A B_{1}=B_{1} C_{1} \operatorname{cotg} \alpha=B_{0} C_{0} \cos ^{2} \alpha \operatorname{cotg} \alpha, \quad A B_{2}=B_{2} C_{2} \operatorname{cotg} \alpha=B_{0} C_{0} \cos ^{4} \alpha \operatorname{cotg} \alpha$, and so on. The distances $A B_{i}$ form a geometric sequence whose first term is $B_{0} C_{0} \operatorname{cotg} \alpha$, and the common ratio is $\cos ^{2} \alpha$; thus

$$
\lim _{i \rightarrow \infty} \sum^{i=0} A B_{i}=\frac{B_{0} C_{0} \operatorname{cotg} \alpha}{1-\cos ^{2} \alpha}=\frac{A B_{0}}{\sin ^{2} \alpha}=A B_{0} \operatorname{cosec}^{2} \alpha
$$

$3^{0} . \mathrm{The} A C_{0}, A C_{1}, A C_{2}, \ldots, A C_{i}$ terms of the sequence are: $A C_{0}=\frac{B_{0} C_{0}}{\sin \alpha}$.

$$
\begin{aligned}
& A C_{1}=B_{0} C_{1} \operatorname{cotg} \alpha=B_{0} C_{0} \cos \alpha \operatorname{cotg} \alpha=\frac{B_{0} C_{0}}{\sin \alpha} \cos ^{2} \alpha \\
& A C_{2}=B_{1} C_{2} \operatorname{cotg} \alpha=B_{0} C_{0} \cos ^{3} \alpha \operatorname{cotg} \alpha=\frac{B_{0} C_{0}}{\sin \alpha} \cos ^{4} \alpha \text { and so on. }
\end{aligned}
$$

Accordingly, we now have a geometric sequence that arises from the one in $1^{0}$ if we multiply its terms by $\frac{1}{\sin \alpha}=\cos$ ecc $\alpha$; thus

$$
\lim _{i \rightarrow \infty} \sum^{i=0} A C_{i}=\left(\lim _{i \rightarrow \infty} \sum^{i=0} B_{i} C_{i}\right) \operatorname{cosec} \alpha=A C_{0} \operatorname{cosec}^{2} \alpha
$$

The sums of the four infinite series are respectively:

$$
A C_{0} \operatorname{cosec} \alpha, \quad A B_{0} \operatorname{cosec}^{2} \alpha, \quad A C_{0} \operatorname{cosec}^{2} \alpha, \quad A B_{0} \operatorname{cosec} \alpha
$$