On two parallel lines, the distinct points $A_{1}, A_{2}, A_{3}, \ldots$ respectively $B_{1}$, $B_{2}, B_{3}, \ldots$ are marked in such a way that $\left|A_{i} A_{i+1}\right|=1$ and $\left|B_{i} B_{i+1}\right|=2$ for $i=1,2, \ldots$ (see Figure). Provided that $\angle A_{1} A_{2} B_{1}=\alpha$, find the infinite sum $\angle A_{1} B_{1} A_{2}+\angle A_{2} B_{2} A_{3}+\angle A_{3} B_{3} A_{4}+\ldots$

![](https://cdn.mathpix.com/cropped/2024_04_17_fa027955eeb5a8b983dbg-02.jpg?height=220&width=391&top_left_y=1250&top_left_x=339)
Answer: $\pi-\alpha$.

Let $C_{1}, C_{2}, C_{3}, \ldots$ be points on the upper line such that $\left|C_{i} C_{i+1}\right|=1$ and $B_{i}=C_{2 i}$ for each $i=1,2, \ldots$ (see Figure 2 ). Then for any $i=1,2, \ldots$ we have

$$
\angle A_{i} B_{i} A_{i+1}=\angle A_{i} C_{2 i} A_{i+1}=\angle A_{1} C_{i+1} A_{2}=\angle C_{i+1} A_{2} C_{i+2}
$$

Hence

$$
\begin{aligned}
& \angle A_{1} B_{1} A_{2}+\angle A_{2} B_{2} A_{3}+\angle A_{3} B_{3} A_{4}+\ldots= \\
& \quad=\angle C_{2} A_{2} C_{3}+\angle C_{3} A_{2} C_{4}+\angle C_{4} A_{2} C_{5}+\ldots=\pi-\alpha .
\end{aligned}
$$