In triangle $ABC$, the midpoints of sides $AB$ and $AC$ are $C_{1}$ and $B_{1}$, respectively, and the center of the circumcircle is $O$. At what angle do the circles passing through points $A, B_{1}, C_{1}$ and points $B, C, O$ intersect?
The angle in question is the angle between the tangents at the common point of the two circles. This angle can be determined at any common point, because the common points, and their tangents, are pairwise mirror images of each other with respect to the line connecting the two centers; furthermore, it is sufficient to determine any one of the angles between the two tangents, the other being the supplementary angle to $180^{\circ}$.

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The circle $k_{1}$ defined by the points $A, B_{1}, C_{1}$ is a scaled-down version of the circumcircle $k$ of the original triangle from point $A$, because the triangle $A B_{1} C_{1}$ is obtained by such a scaling from the triangle $A C B$. Thus, $k_{1}$ passes through $O-\mathrm{n}$, so it is convenient to determine the angle between $k_{1}$ and the circle $k_{2}$ passing through points $B, C, O$ at point $O$.

The angles between the two tangents are equal to the angles between the lines of the radii of the two circles at point $O$, since the radius is perpendicular to the tangent, and the angles with perpendicular sides are either equal or supplementary to $180^{\circ}$. The line of the radius of $K_{2}$ at $O$ is the central line of $k$ and $k_{2}$, so it is the perpendicular bisector of the common chord $B C$, and in $k_{1}$ the radius at $O$ lies on the line $O A$, because $O A$ is a diameter, as it is seen perpendicularly from the midpoints of the sides used. This reduces our task to determining the angles between the line $O A$ and the perpendicular bisector $f$ of side $B C$.

Now consider the ray of $f$ starting from $O$ that passes through the midpoint $F$ of the arc $B C$ not containing $A$, and choose the labeling so that the angles of the triangle are denoted in the usual way with $\beta \geqq \gamma$. If $\beta>\gamma$, the arc $F B A$ is not larger than a semicircle, so by bisecting the angle $F O A$ with the ray $O B$, and using the theorems of central and inscribed angles,

$$
\begin{gathered}
\angle F O A = \angle F O B + \angle B O A = 2 \cdot \angle F A B + 2 \cdot \angle B C A = \angle C A B + 2 \gamma = \\
= \alpha + 2 \gamma = 180^{\circ} - (\beta - \gamma).
\end{gathered}
$$

Accordingly, the other ray of $f$ forms an angle of $\beta - \gamma$ with the ray $O A$, and the sought angle is equal to this. - If, however, the given triangle is isosceles with $A B = A C$, or $\beta = \gamma$, it is immediately clear that the circles in question are tangent to each other, and their angle is $0^{\circ}$, since their centers lie on the axis of symmetry of the triangle.

Remark. If specifically $\alpha = 90^{\circ}$, then $O$ lies on the hypotenuse $B C$, and $k_{2}$ does not exist, since a circle can have at most 2 points on a straight line. It is easy to show that our result can be interpreted for this case as the angle between the line of the hypotenuse $B C$ and the circle $k_{1}$ being $|\beta - \gamma|$.