3. In a right-angled triangle $ABC$, with the right angle at vertex $B$, the angle $\alpha$ at vertex $A$ and the altitude from $B$ to the hypotenuse $BA_1 = h_1$ are given. From point $A_1$, a perpendicular $A_1 B_1$ is drawn to $BC$, from point $B_1$ a perpendicular $B_1 A_2$ is drawn to $AC$, from point $A_2$ a perpendicular $A_2 B_2$ is drawn to $BC$, and so on. Then, in each of the triangles $ABA_1$, $BA_1B_1$, $A_1B_1A_2$, $B_1A_2B_2$, etc., an inscribed circle is drawn.

a) Determine the sum of the areas of all the inscribed circles.

b) Determine the angle $\alpha$ such that this sum is maximized.
Solution. a) It is easily shown that the radius of the inscribed circle in a right-angled triangle with hypotenuse $c$ and acute angle $\alpha$ is equal to $\frac{c}{2}(\sin \alpha + \cos \alpha - 1)$. The given triangles

$$
A B A_{1}, B A_{1} B_{1}, A_{1} B_{1} A_{2}, B_{1} A_{2} B_{2}, \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_118f91fe8bf1788f2605g-06.jpg?height=312&width=471&top_left_y=1495&top_left_x=990)

are right-angled with acute angle $\alpha$ and hypotenuses respectively equal to

$$
\frac{h_{1}}{\sin \alpha}, h_{1}, h_{1} \sin \alpha, h_{1} \sin ^{2} \alpha, \ldots
$$

Therefore, the desired sum of the areas of their inscribed circles is

$$
\begin{aligned}
S & =\frac{\pi}{4} h_{1}^{2}(\sin \alpha+\cos \alpha-1)^{2}\left(\frac{1}{\sin ^{2} \alpha}+1+\sin ^{2} \alpha+\ldots\right) \\
& =\frac{\pi}{4} h_{1}^{2}\left(\frac{\sin \alpha+\cos \alpha-1}{\sin \alpha \cos \alpha}\right)^{2}
\end{aligned}
$$

b) The expression in the last parentheses on the interval $\alpha \in\left(0, \frac{\pi}{2}\right)$ has a maximum for $\alpha=\frac{\pi}{4}$ and this maximum is equal to $4(3-2 \sqrt{2})$, so $S=\pi h_{1}^{2}(3-2 \sqrt{2})$.