## Task B-4.7.

In a right-angled triangle $ABC$, the length of the hypotenuse $\overline{AB}$ is $c$, and at vertex $A$ there is a larger acute angle of measure $\alpha$. The perpendicular to the hypotenuse at point $A$ intersects line $BC$ at point $D$. The perpendicular at point $D$, to the previously drawn perpendicular $AD$, intersects line $AC$ at point $E$. In the obtained point $E$, we draw another perpendicular to the previous perpendicular $DE$, and its intersection with line $DC$ is marked as $F$. If we continue to draw "perpendiculars to perpendiculars" as described, we will obtain a sequence of segments $\overline{AD}, \overline{DE}, \overline{EF}, \ldots$ Determine the sum of their lengths.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-31.jpg?height=805&width=697&top_left_y=311&top_left_x=565)

1 point

Let's denote the lengths of the obtained segments as follows:

$$
x_{1}=|A D|, \quad x_{2}=|D E|, \quad x_{3}=|E F|, \quad \ldots
$$

In triangle $A C D$, the angle at vertex $D$ is $\alpha$ (angles with perpendicular sides), in triangle $C D E$, the angle at vertex $E$ is also $\alpha, \ldots$

All the obtained segments are hypotenuses in these right triangles.

It follows that $\sin \alpha=\frac{b}{x_{1}}=\frac{|C D|}{x_{2}}$, so $x_{2}=\frac{x_{1} \cdot|C D|}{b}$.

On the other hand, $\operatorname{tg} \alpha=\frac{b}{|C D|}$, so $|C D|=\frac{b}{\operatorname{tg} \alpha}$.

Thus, $x_{2}=\frac{x_{1}}{\operatorname{tg} \alpha}$ and similarly $x_{3}=\frac{x_{2}}{\operatorname{tg} \alpha}, \ldots$

The lengths of the drawn segments form a geometric sequence with the common ratio $\frac{1}{\operatorname{tg} \alpha}$.

Since $\alpha$ is greater than the two acute angles in a right triangle, $\alpha>45^{\circ}$, so $\operatorname{tg} \alpha>1$ and $\frac{1}{\operatorname{tg} \alpha}<1$.

The sum of the lengths of all observed segments is

$$
\begin{aligned}
x_{1}+x_{2}+x_{3}+\ldots & =\frac{x_{1}}{1-\frac{1}{\operatorname{tg} \alpha}} & & 1 \text{ point} \\
& =\frac{x_{1} \operatorname{tg} \alpha}{\operatorname{tg} \alpha-1}=\frac{\frac{b}{\sin \alpha} \cdot \frac{a}{b}}{\operatorname{tg} \alpha-1}=\frac{c}{\operatorname{tg} \alpha-1} . & & 1 \text{ point}
\end{aligned}
$$