$\left.\begin{array}{ll}{[\quad \text { The Law of Sines }}\end{array}\right]$

Quadrilateral $A B C D$ is inscribed in a circle, and chord $D E$, parallel to $A B$, intersects $B C$. It is known that $\angle A C E=60^{\circ}$ and $2 \angle B D C+3 \angle C B D=\alpha$. Find the ratio of the radius of the circle inscribed in triangle $B C D$ to the radius of the circle circumscribed around this triangle.
Express the angles $B D C$ and $C B D$ in terms of $\alpha$.

## Solution

Let $\angle B D C=\beta, \angle C B D=\gamma$.

From triangle $B C D$, we find that $\beta+\gamma=120^{\circ}$, and by the given condition $2 \beta+3 \gamma=\alpha$. From the obtained system, we find that

$$
\gamma=\alpha-240^{\circ}, \beta=360^{\circ}-\alpha
$$

Let $R$ and $r$ be the radii of the circumcircle and incircle of triangle $B D C$, respectively, $S$ be its area, and $P$ be its perimeter. Express $S$ in two ways:

$$
\begin{gathered}
S=\frac{1}{2} B C \cdot C D \sin \angle B C D=\frac{1}{2} \cdot 2 R \sin \beta 2 R \cdot \sin \gamma \sin 60^{\circ}= \\
=\sqrt{3} R^{2} \sin \left(360^{\circ}-\alpha\right) \sin \left(\alpha-240^{\circ}\right)=\sqrt{3} R^{2} \sin \alpha \sin \left(\alpha+120^{\circ}\right)
\end{gathered}
$$

and

$$
\begin{aligned}
S & =\frac{P r}{2}=\left(R \sin \beta+R \sin \gamma+R \sin 60^{\circ}\right) r= \\
& =R r\left(-\sin \alpha+\sin \left(\alpha+120^{\circ}\right)+\sin 60^{\circ}\right)
\end{aligned}
$$

By equating the right-hand sides of the obtained equations, we find that

$$
\frac{r}{R}=\frac{\sqrt{3} \sin \alpha \sin \left(\alpha+120^{\circ}\right)}{\sin \alpha-\sin \left(\alpha+120^{\circ}\right)-\sin 60^{\circ}}=\cos \left(\alpha+60^{\circ}\right)-\frac{1}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-05.jpg?height=460&width=552&top_left_y=1256&top_left_x=753)

## Answer

$\frac{\sqrt{3} \sin \alpha \sin \left(\alpha+120^{\circ}\right)}{\sin \alpha-\sin \left(\alpha+120^{\circ}\right)-\sin 60^{\circ}}=\cos \left(\alpha+60^{\circ}\right)-\frac{1}{2}$.