In parallelogram $A B C D$, the angle between diagonals $A C$ and $B D$ is $30^{\circ}$. It is known that the ratio $A C: B D=2: \sqrt{3}$. Point $B_{1}$ is symmetric to vertex $B$ with respect to line $A C$, and point $C_{1}$ is symmetric to vertex $C$ with respect to line $B D$. Find the ratio of the areas of triangle $A B_{1} C_{1}$ and parallelogram $A B C D$.
Consider two cases: $\angle A O B=30^{\circ}$ and $\angle B O C=30^{\circ}$. Apply the cosine theorem.

## Solution

Let $A O=2 a$. Then $B O=a \sqrt{3}$,

$$
S_{\mathrm{ABCD}}=\frac{1}{2} A C \cdot B D \sin 30^{\circ}=\frac{1}{2} \cdot 4 a \cdot 2 a \sqrt{3} \cdot \frac{1}{2}=2 a^{2} \sqrt{3}
$$

Let $\angle A O B=30^{\circ}$ (Fig.1). By the cosine theorem from triangle $A O B$, we find that

$$
A B^{2}=O A^{2}+O B^{2}-2 O A \cdot O B \cdot \cos 30^{\circ}=4 a^{2}+3 a^{2}-2 \cdot 2 a \cdot a \sqrt{3} \cdot \frac{\sqrt{3}}{2}=a^{2}
$$

Since $A B^{2}+O B^{2}=a^{2}+3 a^{2}=4 a^{2}=O A^{2}$, triangle $A O B$ is a right triangle, $\angle A B O=90^{\circ}$. Since $C D \parallel A B$, then $C D \perp B D$, so points $C, D$ and $C_{1}$ lie on the same line. Then $A B D C_{1}$ is a rectangle, so $A C_{1}=B D=2 a \sqrt{3}$.

Since $\angle B A B_{1}=2 \angle B A O=120^{\circ}$, then

$$
\angle B_{1} A C_{1}=\angle B A B_{1}-\angle B A C_{1}=120^{\circ}-90^{\circ}=30^{\circ}
$$

Thus,

$$
S_{\Delta \mathrm{AB}_{1} \mathrm{C}_{1}}=\frac{1}{2} \cdot A B_{1} \cdot A C_{1} \cdot \sin \angle B_{1} A C_{1}=\frac{1}{2} \cdot A B \cdot A C_{1} \cdot \sin 30^{\circ}=\frac{1}{2} \cdot a \cdot 2 a \sqrt{3} \cdot \frac{1}{2}=\frac{a^{2} \sqrt{3}}{2}
$$

Therefore,

$$
\frac{S_{\triangle A B_{1} C_{1}}}{S_{A B C D}}=\frac{\frac{a^{2} \sqrt{3}}{2}}{2 a^{2} \sqrt{3}}=\frac{1}{4}
$$

Now let $\angle B O C=30^{\circ}$ (Fig.2). Then $D B \perp B C$, points $C, B$ and $C_{1}$ lie on the same line, $D A C_{1} B$ is a rectangle, $\angle C_{1} A C=\angle B O C=30^{\circ}$,

$$
\begin{gathered}
A C_{1}=D B=2 a \sqrt{3} \\
A B_{1}=A B=\sqrt{O A^{2}+O B^{2}-2 \cdot O A \cdot O B \cdot \cos 150^{\circ}}= \\
\sqrt{4 a^{2}+3 a^{2}+2 \cdot 2 a \cdot a \sqrt{3} \cdot \frac{\sqrt{3}}{2}}=a \sqrt{13}
\end{gathered}
$$

Let $M$ be the midpoint of $B B_{1}$. From the right triangle $A M B$, we find that

$$
\sin \angle B A M=\frac{B M}{A B}=\frac{\frac{a \sqrt{3}}{2}}{a \sqrt{13}}=\frac{\sqrt{3}}{2 \sqrt{13}}
$$

Then

$$
\cos \angle B A M=\sqrt{1-\frac{3}{52}}=\frac{7}{2 \sqrt{13}}
$$

Since

$$
\angle B_{1} A C_{1}=\angle C A C_{1}+\angle M A B_{1}=\angle B O C+\angle M A B_{1}=30^{\circ}+\angle M A B_{1}
$$

then

$$
\sin \angle B_{1} A C_{1}=\sin \left(30^{\circ}+\angle M A B_{1}\right)=\frac{1}{2} \cdot \frac{7}{2 \sqrt{13}}+\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2 \sqrt{13}}=\frac{5}{2 \sqrt{13}}
$$

Thus,

$$
\begin{gathered}
S_{\triangle \mathrm{AB}_{1} \mathrm{C}_{1}}=\frac{1}{2} \cdot A B_{1} \cdot A C_{1} \cdot \sin \angle B_{1} A C_{1}=\frac{1}{2} \cdot A B \cdot B D \cdot \sin \left(30^{\circ}+\angle M A B_{1}\right)= \\
=\frac{1}{2} \cdot a \sqrt{13} \cdot 2 a \sqrt{3} \cdot \frac{5}{2 \sqrt{13}}=\frac{5 a^{2} \sqrt{3}}{2}
\end{gathered}
$$

$$
\frac{S_{\triangle A B_{1} C_{1}}}{S_{A B C D}}=\frac{\frac{5 a^{2} \sqrt{3}}{2}}{2 a^{2} \sqrt{3}}=\frac{5}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b164f828743b46f847cg-46.jpg?height=1036&width=832&top_left_y=156&top_left_x=613)

## Answer

$\frac{1}{4}$ or $\frac{5}{4}$.