11.5. Given an acute angle O. On one of its sides, we take a point $A_{1}$ and drop a perpendicular $A_{1} A_{2}$ to the other side of the angle. Then from point $A_{2}$, we drop a perpendicular to $O A_{1}$, obtaining point $A_{3}$, and so on. How can we use a compass and a straightedge to construct a segment equal to the length of the infinite broken line $L=A_{1} A_{2}+A_{2} A_{3}+\ldots$? Provide the construction and justify that the result is a segment of the required length.
Solution: Let the measure of angle $O$ be $\alpha$, and the length of segment $A_{1} A_{2}=m-$ see the figure. From the similarity of triangles $O A_{1} A_{2}$ and $A_{1} A_{2} A_{3}$, we find that $\angle A_{1} A_{2} A_{3}=\alpha$, and $A_{2} A_{3}=m \cos \alpha$. Next, from the similarity of triangles $O A_{2} A_{3}$ and $A_{2} A_{3} A_{4}$, we find that $\angle A_{2} A_{3} A_{4}=\alpha$, and $A_{3} A_{4}=m \cos ^{2} \alpha$. Similarly, we sequentially obtain (a rigorous proof is conducted using mathematical induction):

$$
A_{4} A_{5}=m \cos ^{3} \alpha, \quad A_{5} A_{6}=m \cos ^{4} \alpha, \quad \ldots, \quad A_{k} A_{k+1}=m \cos ^{k-1} \alpha, \quad \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_de0eff2f5c97fa664b16g-08.jpg?height=511&width=603&top_left_y=161&top_left_x=138)

To the solution of problem 11.5,

first stage

![](https://cdn.mathpix.com/cropped/2024_05_06_de0eff2f5c97fa664b16g-08.jpg?height=346&width=619&top_left_y=798&top_left_x=136)

To the solution of problem 11.5 second stage

Then, using the formula for the sum of an infinite decreasing geometric series, we get:

$$
L=m+m \cos \alpha+m \cos ^{2} \alpha+\ldots=\frac{m}{1-\cos \alpha}
$$

and it remains to show how to construct a segment of length

$$
x=\frac{m}{1-\cos \alpha}
$$

given the angle $\alpha$ and a segment of length $m$. This can be done, for example, as follows.

Rewrite the last equality as

$$
\frac{x}{m}=\frac{m}{m-m \cos \alpha}
$$

The segment of length $m-m \cos \alpha=A_{1} A_{2}-A_{2} A_{3}$ is easily constructed. Now take an arbitrary angle with vertex $K$, on one of its sides choose points $B$ and $C$, and on the other side choose point $D$ such that $K B=K D=m$, $K C=m-m \cos \alpha$. Now draw a line through point $B$ parallel to line $C D$, and denote the point of intersection of this line with the second side of the angle as $E$—see the figure.

The segment $K E$ is the desired one, which follows from the similarity of triangles $K C D$ and $K B E$.

## Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct construction and correct proof that the constructed segment has the required length | 7 points |
| Correctly found the length of the required segment, provided a correct construction, but did not prove that the constructed segment has the required length | 6 points |
| Correctly found the length of the required segment, but the construction is missing or incorrect | 4 points |
| The problem is reduced to finding the sum of a geometric series, but due to an arithmetic error, the length of the required segment is found incorrectly; the construction is missing or incorrect | 2 points |
| Incorrect answer; the construction is missing or incorrect | 0 points |