5. The cross-section of a regular hexagonal pyramid SABCDEF is formed by a plane passing through the vertex $C$ of the base $A B C D E F$ and parallel to the median $B M$ of the lateral face $S A B$ and the apothem $S N$ of the lateral face $S A F$. The side length of the base of the pyramid is 2, and the distance from the vertex $S$ to the cutting plane is 1. Find the cosine of the angle between the cutting plane and the base plane. (20 points)
Solution. Construct the section of the pyramid. In the plane $S A F$, through the point $M$, draw a line $M Q$ parallel to $S N$, where $Q$ lies on the line $A F$, and $M Q$ is the midline of the triangle $S A N$. Given $A F = a$, $A Q = Q N = \frac{a}{4}$, where $a$ is the side length of the base of the pyramid.

The plane $S Q B$ is parallel to the plane of the section. Through the point $C$, draw a line $C$ parallel to $B Q$, where the point $P$ lies on the edge $F E$ and is the intersection of this edge with the plane of the section.

The point $Y$ is the intersection of $B Q$ and $A D$. Triangles $Y A Q$ and $Y O B$ are similar, with $A Y: Y O = A Q: B O = 1: 4$, $A Y = a / 5$, $Y O = 4 a / 5$. The point $U$ is the intersection of $C P$ and $A D$, with $Y U = a$, $O U = a / 5$. Triangles $C O U$ and $C F P$ are similar, with $O U: F P = O C: F C = 1: 2$, $F P = 2 a / 5$, $P E = 3 a / 5$.

In the plane $S N R$ (where $S R$ is the apothem of the face $S C D$), through the point $G$ (the intersection of $C P$ and $N R$), draw a line $G H$ parallel to $S N$,

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-12.jpg?height=534&width=346&top_left_y=818&top_left_x=958)
$H \in S R$. Then $N G: G R = S H: H R$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-12.jpg?height=497&width=491&top_left_y=1422&top_left_x=882)

The point $T$ is the intersection of $C$ and $B E$. Triangles $O T U$ and $O B Y$ are similar, with $O T: O B = O U: O Y = 1: 4$, $O T = a / 4$. Triangles $G O T$ and $G R C$ are similar, with $O G: G R = O T: R C = 1: 2$, $N O: O G: G R = 3: 1: 2$. Then $S H: H R = 2: 1$.

The point $K$ is the intersection of $C H$ and $S D$. Since $S R$ is the median of $S C D$, and $S H: H R = 2: 1$, then $C K$ is also a median.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-12.jpg?height=505&width=332&top_left_y=2209&top_left_x=1093)

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-12.jpg?height=1047&width=563&top_left_y=499&top_left_x=1389)

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-12.jpg?height=937&width=497&top_left_y=1599&top_left_x=1453)

Let the distance from point $S$ to the plane of the section be $d$, with $d = 1$. The distance from point $D$ to the section is also $d$. In triangle $D C J_{1}$, draw the height $D V$ and denote its length as $h$. Then $\sin \varphi = \frac{d}{h}$, $\cos \varphi = \sqrt{1 - \left(\frac{d}{h}\right)^{2}}$. Find $C J_{1}$ using the cosine rule:

$$
C J_{1}^2 = \frac{16 a^2}{25} + a^2 - \frac{4 a^2}{5} = \frac{21 a^2}{25}, \quad C J_{1} = \frac{\sqrt{21} a}{5}
$$

Using different formulas to find the area of triangle $D C J_{1}$, we have

$$
\frac{\sqrt{21} a h}{5} = \frac{4 a^2}{5} \cdot \frac{\sqrt{3}}{2}, \quad h = \frac{2 a}{\sqrt{7}}
$$

Then $\cos \varphi = \sqrt{1 - \frac{7 d^2}{4 a^2}} = \frac{\sqrt{4 a^2 - 7 d^2}}{2 a} = \frac{3}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-13.jpg?height=257&width=260&top_left_y=777&top_left_x=995)

Answer: $\frac{3}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-13.jpg?height=545&width=623&top_left_y=367&top_left_x=1279)