5. Given an arbitrary convex quadrilateral $A B C D$. Points $K$ and $M$ are the midpoints of sides $A B$ and $C D$ respectively. Point $P$ is the intersection of lines $K C$ and $B M$, point $N$ is the intersection of $A M$ and $K D$. Find the area of quadrilateral $K P N M$, if angles $C B P$ and $N D A$ are 30 degrees, $B P C-105^{0}$, $D A N-15^{0} . B P=2 \sqrt{2}, N D=\sqrt{3}$.
# Solution and Answer.

$S_{K P M N}=S_{A M B}-S_{A N K}-S_{K P B}=S_{A D K}+S_{K C B}-S_{A N K}-S_{K P B}=S_{A D N}+S_{P C B}$

By the Law of Sines

$$
\begin{aligned}
& \frac{A N}{\sin 30}=\frac{N D}{\sin 15} \Rightarrow A N=\sqrt{3} \cdot \frac{1}{2} \cdot \frac{2 \sqrt{2}}{\sqrt{3}-1}=\frac{\sqrt{6}(\sqrt{3}+1)}{2} \Rightarrow \\
& S_{A N D}=\frac{1}{2} \sqrt{3} \frac{\sqrt{6}(\sqrt{3}+1)}{2} \sin 135=\frac{1}{2} \sqrt{3} \frac{\sqrt{6}(\sqrt{3}+1)}{2} \cos 45=\frac{3(\sqrt{3}+1)}{4}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_009f2643c33564a3e4a2g-5.jpg?height=403&width=511&top_left_y=912&top_left_x=1452)

Similarly, $\frac{C B}{\sin 105}=\frac{P B}{\sin 45} \Rightarrow B C=2 \sqrt{2} \cos 15 \cdot \frac{2}{\sqrt{2}}=\frac{4(\sqrt{3}+1)}{2 \sqrt{2}} \Rightarrow$

$$
S_{C P B}=\frac{1}{2} 2 \sqrt{2} \frac{2(\sqrt{3}+1)}{\sqrt{2}} \sin 105=\frac{2(\sqrt{3}+1)}{1} \cos 45=\sqrt{2}(\sqrt{3}+1)
$$

$$
S_{K P M N}=\frac{3(\sqrt{3}+1)}{4}+\frac{4 \sqrt{2}(\sqrt{3}+1)}{4}=\frac{(\sqrt{3}+1)(4 \sqrt{2}+3)}{4}
$$