From point $A$, we draw rays that enclose an angle $\alpha$ with each other. From a random point $B_{1}$ on one of the rays, we draw a perpendicular to the adjacent ray, the foot of which is $B_{1}$. From this point, we again draw a perpendicular to the next adjacent ray, and so on, infinitely. What is the sum of the perpendiculars $B_{1} B_{2}, B_{2} B_{3}, B_{3} B_{4}$, etc.? What is the sum of the areas of the triangles $A B_{1} B_{2}, A B_{2} B_{3}$, etc., if $A B_{1}=a$? What should $\alpha$ be so that the sum of the areas of all the triangles is $a^{2}$?
$1^{\circ}$.

$$
\begin{gathered}
B_{1} B_{2}=a \sin \alpha, \\
B_{2} B_{3}=A B_{2} \cdot \sin \alpha=a \cos \alpha \sin \alpha, \\
B_{3} B_{4}=A B_{3} \cdot \sin \alpha=a \cos ^{2} \alpha \sin \alpha,
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
B_{1} B_{2}+B_{2} B_{3}+B_{3} B_{4}+\ldots=a \sin \alpha\left(1+\cos \alpha+\cos ^{2} \alpha+\ldots\right)= \\
=\frac{a \sin \alpha}{1-\cos \alpha}=\frac{2 a \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \sin ^{2} \frac{\alpha}{2}}=a \operatorname{ctg} \frac{\alpha}{2}
\end{gathered}
$$

$2^{\circ}$.

$$
\begin{gathered}
t_{1}=\frac{1}{2} a^{2} \cos \alpha \sin \alpha \\
t_{2}=\frac{1}{2} a^{2} \cos ^{3} \alpha \sin \alpha \\
t_{3}=\frac{1}{2} a^{2} \cos ^{5} \alpha \sin \alpha \\
\ldots \ldots \ldots \\
t_{1}+t_{2}+t_{3}+\ldots=\frac{1}{2} a^{2} \sin \alpha\left(\cos \alpha+\cos \alpha^{2}+\cos ^{5} \alpha+\ldots\right)= \\
=\frac{a^{2} \sin \alpha \cos \alpha}{2\left(1-\cos ^{2} \alpha\right)}=\frac{a^{2} \cos \alpha}{2 \sin \alpha}=\frac{a^{2}}{2} \operatorname{ctg} \alpha
\end{gathered}
$$

$3^{\circ}$. If

$$
\frac{a^{2}}{2} \operatorname{ctg} \alpha=a^{2}
$$

then

$$
\operatorname{ctg} \alpha=2,
$$

from which

$$
\alpha=26^{\circ} 33^{\prime} 54^{\prime \prime} \text {. }
$$