12.408 In an acute-angled triangle $ABC$ with side $AC$ equal to $\boldsymbol{b}, \angle \boldsymbol{A}=\alpha, \angle B=\beta$ radians. A circle is drawn through the orthocenter (the point of intersection of the altitudes) and the feet of the altitudes dropped onto sides $AB$ and $BC$. Find the area of the common part of the triangle and the circle.
Solution. Let $P$ be the orthocenter of the given $\triangle ABC$, and $AE, CD$ be its altitudes.

The circle passing through points $P, E, D$ will also pass through vertex $B$ (Theorem 8). By the same Theorem 8, $BP$ is the diameter of this circle ($\angle BDP = 90^\circ$).

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-292.jpg?height=424&width=363&top_left_y=85&top_left_x=104)

Let $O$ be the center of the circle, and $x$ be its radius. Since $\angle DOE = 2\beta$ (the central angle subtending arc $DE$, which is equal to $2\beta$ - Theorem 6), the area $S_c$ of sector $DOE$ is $x^2 \beta$.

$$
S_{BOD} = \frac{1}{2} S_{BPD} = \frac{BD \cdot PD}{4} = \frac{BP^2 \cdot \sin \alpha \cdot \cos \alpha}{4} =
$$

$$
\begin{gathered}
= \frac{x^2}{2} \cdot \sin 2\alpha \left(\angle DBP = 90^\circ - \alpha\right), S_{BOE} = \frac{1}{2} S_{BPE} = \frac{BE \cdot PE}{4} = \\
= -\frac{BP^2 \sin (\alpha + \beta) \cos (\alpha + \beta)}{4} = -\frac{x^2}{2} \cdot \sin (2\alpha + 2\beta) \left(\angle PBE = 90^\circ - \angle C = \alpha + \beta - 90^\circ\right)
\end{gathered}
$$

In the end, the area of the common part of circle $O$ and $\triangle ABC$ is

$$
S_c + S_{BOD} + S_{BOE} = x^2 (\beta - \sin \beta \cdot \cos (2\alpha + \beta))
$$

Now, let's determine $x$. $\triangle ADC \sim \triangle BPD \Rightarrow \frac{BP}{AC} = \frac{BD}{CD} = \cot \beta \Rightarrow x = \frac{1}{2} b \cdot \cot \beta$.

From this, we get the answer.

Answer: $\frac{1}{4} b^2 \cot^2 \beta (\beta - \sin \beta \cdot \cos (2\alpha + \beta))$.