6. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $a$, point $K$ is the midpoint of edge $B_{1} C_{1}$, point $L$ lies on edge $C_{1} D_{1}$, such that $D_{1} L=3 C_{1} L$, and point $N$ is the midpoint of edge $A A_{1}$. Construct the section of the cube by the plane passing through points $K, L, N$. Describe the algorithm for constructing the section. Find the area of the resulting section.

(20 points)
# Solution:

Algorithm:

1) Connect $K$ and $L$. $E$ is the intersection point of lines $K L$ and $A_{1} B_{1}$.
2) Connect $E$ and N. F is the intersection point of lines $E N$ and $B B_{1}$.
3) $G$ is the intersection point of lines $K L$ and $A_{1} D_{1}$.
4) $M$ is the intersection point of lines $G N$ and $D D_{1}$.
5) $F K L M N$ is the desired section.

$\varphi$ is the angle of inclination of the section plane to the plane $A_{1} B_{1} C_{1} D_{1}$.

$$
S_{c e \varphi}=\frac{S_{n p}}{\cos \varphi}=\frac{S_{\mathrm{A}_{\mathrm{A},}, L D_{1}}}{\cos \varphi}=\frac{a^{2}-a^{2} / 16}{\cos \varphi}=\frac{15 a^{2}}{16 \cos \varphi}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_03a96cebf8326f3655efg-5.jpg?height=605&width=825&top_left_y=1171&top_left_x=1072)

$A_{1} P \cdot K L=2 S_{A_{1} K L}=2\left(a^{2}-a^{2} / 4-3 a^{2} / 8-a^{2} / 16\right)=5 a^{2} / 8$.

$A_{1} P=5 a^{2} /(8 K L)=a \sqrt{5} / 2, \cos \varphi=A_{1} P / N P=\frac{5 a}{2 \sqrt{5} \cdot \sqrt{a^{2} / 4+5 a^{2} / 4}}=\frac{5}{\sqrt{30}}$.

$S_{c e 4}=\frac{3 a^{2} \sqrt{30}}{16}$.

Answer: $\frac{3 a^{2} \sqrt{30}}{16}$.

## Solution for Variant 8