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

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

$$
S_{c e u}=\frac{S_{n p}}{\cos \varphi}=\frac{S_{A, B K L D_{1}}}{\cos \varphi}=\frac{a^{2}-a^{2} / 12}{\cos \varphi}=\frac{11 a^{2}}{12 \cos \varphi}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_03a96cebf8326f3655efg-2.jpg?height=657&width=756&top_left_y=1148&top_left_x=1141)

$A_{1} P \cdot K L=2 S_{A_{1} K L}=2\left(a^{2}-a^{2} / 4-a^{2} / 12-a^{2} / 3\right)=2 a^{2} / 3$.

$A_{1} P=2 a^{2} /(3 K L)=4 a / \sqrt{13}, \cos \varphi=A_{1} P / N P=\frac{4 a}{\sqrt{13} \cdot \sqrt{a^{2} / 4+16 a^{2} / 13}}=\frac{8}{\sqrt{77}}$.

$S_{c e u}=\frac{11 a^{2} \sqrt{77}}{96}$

Answer: $\frac{11 a^{2} \sqrt{77}}{96}$.

## Solution Variant #2

1 In a basketball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game with each other team. The ratio of the number of victories won by teams from city $B$ to the number of victories won by teams from city $A$ is $14: 19$. Find $n$, given that there were no draws in the tournament.

Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ victories. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n-1) n$ victories. The number of matches between teams from city $A$ and teams from city $B$ is $2 n^{2}$. Let $m$ be the number of victories in these matches won by teams from city $A$, then teams from city $B$ won $2 n^{2}-m$ victories in these matches. In total, teams from city $A$ won $\frac{(n-1) n}{2}+m$ victories, and teams from city $B$ won $(2 n-1) n+2 n^{2}-m$ victories. According to the condition, we have $\frac{\frac{(n-1) n}{2}+m}{(2 n-1) n+2 n^{2}-m}=\frac{19}{14}$, $7 n^{2}-7 n+14 m=76 n^{2}-19 n-19 m, 33 m=69 n^{2}-12 n, m=\frac{23 n^{2}-4 n}{11}, \frac{23 n^{2}-4 n}{11} \leq 2 n^{2}, n^{2}-4 n \leq 0$. The number $n$ can be $1,2,3,4$. By substituting into the equation $m=\frac{23 n^{2}-4 n}{11}$, we get that $m$ is an integer only when $n=4$.

Answer: 4.