[ The inscribed angle is half the central angle ]

In an acute-angled triangle $A B C$, altitudes $C H$ and $A H_{1}$ are drawn. It is known that $A C=2$, and the area of the circle circumscribed around triangle $H B H_{1}$ is $\pi / 3$. Find the angle between the altitude $C H$ and the side $B C$.

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Let $P$ be the point of intersection of the altitudes of triangle $ABC$. Points $H$ and $H_{1}$ lie on the circle with diameter $BP$.

Let the radius of this circle be $R$. Then

$\pi R^{2}=\pi / 3$. Hence, $R=\frac{1}{\sqrt{3}}$

First method. From the solution of problem $\underline{54814}$, it follows that $2 R=AC \operatorname{ctg} \angle B=2 \operatorname{ctg} \angle B$. Therefore, $\angle B=60^{\circ}$.

Thus, $\angle B C H=90^{\circ}-\angle B=30^{\circ}$.

Second method. Let $O$ be the center of the circumcircle of triangle $ABC$, and $M$ be the midpoint of side $AC$.

The distance from point $O$ to side $AC$ is half the distance from point $P$ to vertex $B$ (see problem $\underline{53528}$), and

since $BP=\frac{2}{\sqrt{3}}$, then $OM=\frac{1}{\sqrt{3}}$.

From the right triangle $OMC$, it is clear that $\angle MOC=60^{\circ}$.

By the inscribed angle theorem, $\angle B=1 / 2 \angle AOC=\angle MOC=60^{\circ}$. Therefore, $\angle B C H=90^{\circ}-\angle B=30^{\circ}$.

## Answer

$30^{\circ}$.