## [

 The Law of Sines
Pentagon $A B C D E$ is inscribed in a circle. It is known that $B D \| A E$ and $\angle C A E = 2 \angle C E A, \angle C B D - \angle C D B = \alpha$. Find the ratio of the perimeter of triangle $A C E$ to the radius of the circumscribed circle around it.

#
$\angle C E A=\alpha$.

## Solution

Let

$$
\angle A E C=\beta, \angle C D B=\gamma, \angle E C D=\varphi
$$

Then $\angle C A E=2 \beta$.

Let $P$ and $Q$ be the points of intersection of segment $B D$ with segments $C E$ and $A C$ respectively. Since $A E \| B D$, then

$$
\begin{gathered}
\cup A B=\cup E D, \angle A C B=\angle E C D=\varphi \\
\angle Q P C=\angle A E C=\beta, \angle P Q C=\angle C A E=2 \beta
\end{gathered}
$$

By the exterior angle theorem of a triangle,

$$
\angle Q P C=\angle P D C+\angle P C D, \text { or } \beta=\gamma+\varphi,
$$

From this, it follows that $\beta=\alpha$.

If $R$ is the radius of the given circle, then

$$
\begin{gathered}
A C+C E+A E= \\
=2 R \sin \angle C E A+2 R \sin \angle C A E+2 R \sin \angle A C E= \\
=2 R \sin \beta+2 R \sin 2 \beta+2 R \sin \left(180^{\circ}-3 \beta\right)= \\
=2 R \sin \alpha+2 R \sin 2 \alpha+2 R \sin 3 \alpha
\end{gathered}
$$

Therefore,

$$
\frac{A C+C E+A E}{R}=2(\sin \alpha+\sin 2 \alpha+\sin 3 \alpha)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-04.jpg?height=483&width=495&top_left_y=842&top_left_x=787)

## Answer

$2(\sin \alpha+\sin 2 \alpha+\sin 3 \alpha)$