[The triangle formed by the bases of two altitudes and a vertex] The Law of Sines

In an acute triangle $A B C$, altitudes $C C_{1}$ and $A A_{1}$ are drawn. It is known that $A C=1$ and $\angle C_{1} C A_{1}=\alpha$. Find the area of the circle circumscribed around triangle $C_{1} B A_{1}$.
Triangle $C_{1} B A_{1}$ is similar to triangle $C B A$ with a coefficient of $\cos \angle B=\cos \left(90^{\circ}-\alpha\right)=\sin \alpha$, so $A_{1} C_{1}$ $=A C \sin \alpha=\sin \alpha$.

Let $R$ be the radius of the circumcircle of triangle $C_{1} B A_{1}$. Then $R=\frac{A_{1} C_{1}}{2 \sin \angle B}=\frac{2 \sin \alpha}{2 \cos \alpha}=\frac{1}{2} \operatorname{tg} \alpha$, and the area of the circle is

$\pi R^{2}=\pi / 4 \operatorname{tg}^{2} \alpha$

## Answer

$\pi / 4 \operatorname{tg}^{2} \alpha$