Consider two intersecting circles of radius $r$, whose centers are a distance $O_{1} O_{2}=1$ apart. The common chord is bisected at point $O$. A ray starting from $O$ and making an angle $\alpha$ with the common chord intersects the given circles at points $A$ and $B$. What is the length of $A B$?
I. Solution: Connect point $O_{2}$ with $B$ and draw a radius $O_{1} C$ parallel to $O_{2} B$ in the other circle (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_48fae2a262c9d52aa69bg-1.jpg?height=558&width=588&top_left_y=301&top_left_x=752)

Figure 1

Since $O_{1} C = O_{2} B = r$ and $O_{1} C \parallel O_{2} B$, $O_{1} O_{2} B C$ is a parallelogram, and thus $B C \parallel O_{1} O_{2}$ and $B C = O_{1} O_{2} = 1$. If we denote the midpoint of $B C$ by $F$, then $O_{1} F \parallel O B$.

Connect point $C$ with $A$ and let the intersection of $C A$ and $O_{1} F$ be $M$. Since $C F = F B$, $C M = M A$, which means $O_{1} F$ bisects the chord $A C$. However, the line passing through the center of the circle and the midpoint of the chord is perpendicular to the chord, so $O_{1} F \perp A C$, or (since $O_{1} F \parallel O B$)

$$
A C \perp O B
$$

In the right triangle $A B C$, $\angle C = \alpha$ as the angle opposite the hypotenuse, and thus

$$
A B = B C \sin \alpha = \sin \alpha
$$

independent of the radius $r \left(>\frac{1}{2}\right)$.

László Edöcsény (Bp., XI., József Attila g. III. o. t.)

II. Solution: The labeling is shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_48fae2a262c9d52aa69bg-1.jpg?height=596&width=442&top_left_y=1632&top_left_x=820)

Figure 2

Let the extension of the ray intersect the circle centered at $O_{2}$ at point $P$, then $P$ is the central reflection of point $A$ with respect to point $O$, so $P O = O A$.

From point $P$, dilate the segment $O O_{2}$ in a $2:1$ ratio, obtaining the segment $A Q = 2 \cdot O O_{2} = 2 \cdot \frac{1}{2} = 1$ parallel to $O O_{2}$. In triangle $A B Q$, $\angle B$ is a right angle by Thales' theorem, and $\angle Q$ is equal to $\alpha$ as the angle opposite the hypotenuse. Therefore, in the right triangle $A B Q$,

$$
A B = A Q \sin \alpha = \sin \alpha
$$