## Task B-2.5.

Circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$ as shown in the figure. Point $C$ is on circle $k_{1}$, and point $D$ is on circle $k_{2}$ such that $\varangle A C B=60^{\circ} \text{ and } \varangle B D A=30^{\circ}$. If the centers of circles $k_{1}$ and $k_{2}$ are $4 \sqrt{3} \text{ cm}$ apart, what is the length of their common chord $\overline{A B}$?

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-12.jpg?height=366&width=549&top_left_y=454&top_left_x=685)
## Second Solution.

We use the same sketch and labels as in the first solution.

Angle $\varangle A C B$ is an inscribed angle over the arc $\overparen{A B}$, so its central angle is $\varangle A S_{1} B=120^{\circ}$, or $\varangle E S_{1} B=60^{\circ}$.

Angle $\varangle B D A$ is an inscribed angle over the arc $\widehat{B A}$, so its central angle is $\varangle B S_{2} A=60^{\circ}$, or $\varangle B S_{2} E=30^{\circ}$.

Notice that triangle $S_{1} S_{2} B$ is half of an equilateral triangle with side length $a=\left|S_{1} S_{2}\right|=4 \sqrt{3} \mathrm{~cm}$.

Then, $r_{1}=\left|S_{1} B\right|=\frac{a}{2}=\frac{4 \sqrt{3}}{2}=2 \sqrt{3} \mathrm{~cm}$.

1 point

It also holds that $r_{2}=\left|S_{2} B\right|=\frac{a \sqrt{3}}{2}=\frac{4 \sqrt{3} \cdot \sqrt{3}}{2}=6 \mathrm{~cm}$. 1 point

Since triangle $S_{1} S_{2} B$ is a right triangle, its area can be calculated in two ways, so we have $P=\frac{r_{1} \cdot r_{2}}{2}=\frac{x \cdot\left|S_{1} S_{2}\right|}{2}$.

$2 \sqrt{3} \cdot 6=x \cdot 4 \sqrt{3}$

$x=3 \mathrm{~cm}$

1 point

Finally, we conclude that $|A B|=2 x=6 \mathrm{~cm}$.

Note: After noticing that triangle $S_{1} S_{2} B$ is half of an equilateral triangle with side length $a=4 \sqrt{3} \mathrm{~cm}$, the student does not need to determine $r_{1}$ and $r_{2}$, but can immediately calculate the area of the triangle $P=\frac{1}{2} \cdot \frac{a^{2} \sqrt{3}}{4}=6 \sqrt{3} \mathrm{~cm}^{2}$, for which they receive 2 points.