In the square $A B C D$ with an area of 1, side $A D$ is extended beyond point $D$ and a point $O$ is taken on the extension such that $O D=3$. From point $O$, two rays are drawn. The first ray intersects segment $C D$ at point $M$ and segment $A B$ at point $N$, the second ray intersects segment $C D$ at point $L$ and segment $B C$ at point $K, O N=a, \angle B K L=\alpha$. Find the area of the polygon $B K L M N$.
Triangle $O D M$ is similar to triangle $O A N$, and triangle $K C L$ is similar to triangle $O D L$.

## Solution

The similarity coefficient of triangles $O D M$ and $O A N$ is $3 / 4$, so $S_{O D M}=9 / 16 S_{A O N}$ and $S_{A N M D}=\frac{7}{16} S_{A O N}=\frac{7}{32} A O \cdot A N=\frac{7}{8} \sqrt{a^{2}-16}$

Since $\angle K O D=180^{\circ}-\alpha$, then $L D=O D \operatorname{tg}\left(180^{\circ}-\alpha\right)=-3 \operatorname{tg} \alpha, C L=C D-L D=1+3 \operatorname{tg} \alpha$.

From the similarity of triangles $K C L$ and $O D L$, it follows that

$S_{K L C}=\left(\frac{C L}{L D}\right)^{2} S_{O D L}=\frac{(1+3 \operatorname{tg} \alpha)^{2}}{9 \operatorname{tg}^{2} \alpha} \cdot\left(-\frac{9}{2} \operatorname{tg} \alpha\right)=\frac{(1+3 \operatorname{tg} \alpha)^{2}}{2 \operatorname{tg} \alpha}$.

Therefore, $S_{B K L M N}=1-\frac{7}{8} \sqrt{a^{2}-16}+\frac{(1+3 \operatorname{tg} \alpha)^{2}}{2 \operatorname{tg} \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ef0b80dacb6d98eff5ag-35.jpg?height=306&width=898&top_left_y=1656&top_left_x=586)

## Answer

$1-\frac{7}{8} \sqrt{a^{2}-16}+\frac{(1+3 \operatorname{tg} \alpha)^{2}}{2 \operatorname{tg} \alpha}$