## Task B-3.5.

Four cities on the map determine the vertices of a quadrilateral $A B C D$ to which a circle of radius $R$ can be circumscribed. The distance between cities $A$ and $C$ is $2 R$, and the distance between cities $A$ and $B$ is equal to the distance between cities $B$ and $C$. The ratio of the distance between cities $A$ and $D$ and the distance between cities $C$ and $D$ is $\sqrt{3}$. What is the distance between cities $B$ and $D$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-21.jpg?height=580&width=603&top_left_y=1366&top_left_x=658)

Since $|A C|=2 R$, which is the diameter of the given circle, triangles $A B C$ and $A C D$ are right triangles (inscribed angles over the diameter).

Applying the Pythagorean theorem in triangle $A B C$ yields

$$
\begin{aligned}
& |A B|^{2}+|B C|^{2}=4 R^{2} \\
& |A B|=|B C|=R \sqrt{2}
\end{aligned}
$$

thus $\alpha=45^{\circ}$.

From the given ratio, we have

$$
\frac{|A D|}{|C D|}=\sqrt{3}
$$

$$
|A D|=|C D| \sqrt{3}
$$

and then from the right triangle $A D C$, we get

$$
\begin{gathered}
|A D|^{2}+|C D|^{2}=4 R^{2} \\
|C D|=R
\end{gathered}
$$

Therefore, triangle $S C D$ is equilateral, and $\beta=60^{\circ}$.

Applying the cosine rule in triangle $B C D$ yields

$$
\begin{array}{cc}
|B D|^{2}=|B C|^{2}+|C D|^{2}-2|B C| \cdot|C D| \cdot \cos (\alpha+\beta) & 1 \text{ point} \\
=2 R^{2}+R^{2}-2 \cdot R^{2} \sqrt{2} \cdot \cos \left(45^{\circ}+60^{\circ}\right) & \\
=3 R^{2}-2 \cdot R^{2} \sqrt{2} \cdot\left(\cos 45^{\circ} \cos 60^{\circ}-\sin 45^{\circ} \sin 60^{\circ}\right) & 1 \text{ point} \\
=3 R^{2}-2 \cdot R^{2} \sqrt{2} \cdot\left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) & \\
=R^{2}(2+\sqrt{3}) &
\end{array}
$$

Thus, the desired distance is $|B D|=R \sqrt{2+\sqrt{3}}$.