5. Find the area of the cross-section of a regular hexagonal pyramid $S A B C D E F$ 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 $S A F$, if the side of the base of the pyramid is 2, and the distance from the vertex $S$ to the cutting plane is 1. (20 points)
Solution. We will 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 $B Q$, and the point $P$ lies on the edge $F E$ and is the intersection point 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, $A Y: Y O = A Q: B O = 1: 4$, $A Y = a / 5$, $Y O = 4 a / 5$. The point $U$ is the intersection of $C P$ and $A D$, $Y U = a$, $O U = a / 5$. Triangles $C O U$ and $C F P$ are similar, $O U: F P = O C: F C = 1: 2$, $F P = 2 a / 5$, $P E = 3 a / 5$.

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_bfa14092a69dadb13b59g-11.jpg?height=537&width=348&top_left_y=768&top_left_x=957)
$H \in S R$. Then $N G: G R = S H: H R$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-11.jpg?height=489&width=491&top_left_y=1383&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, $O T: O B = O U: O Y = 1: 4$, $O T = a / 4$. Triangles $G O T$ and $G R C$ are similar, $O G: G R = O T: R C = 1: 2$, $N O: O G: G R = 3: 1: 2$. 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 $S C D$, $S H: H R = 2: 1$, then $C K$ is also a median.

The point $Z$ is the intersection of the lines $C P$ and $D E$. Triangles $P Z E$ and $U Z D$ are similar, and $P E: U D = Z E: Z D = 3: 7$, $Z E = 3 a$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-11.jpg?height=2230&width=854&top_left_y=446&top_left_x=1094)

In the plane $S D E$, the point $L$ is the intersection of the lines $Z K$ and $S E$. By Thales' theorem, we have $S L: L E = 4: 3$.

The desired section is $C K L P$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-12.jpg?height=520&width=1813&top_left_y=477&top_left_x=93)

We will calculate the area of the section using the formula $S_{c e 4} = \frac{S_{n p}}{\cos \varphi}$, where $S_{n p}$ is the area of the projection of the section onto the base plane, and $\varphi$ is the angle between the plane of the section and the base plane. We will find the area of the projection of the section onto the base plane.

The projection is a pentagon $C J X P$. The area of the projection of the section is calculated by the formula

$S_{\text {pr }} = S_{C J_{1} J} + S_{J_{1} J X X_{1}} + S_{X X_{1} P} = $

$\frac{3 a^{2} \sqrt{3}}{40} + \left(\frac{4}{7} \cdot \frac{1}{2} - \frac{1}{4} \cdot \frac{1}{5}\right) \cdot \frac{a^{2} \sqrt{3}}{4} + \left(\frac{3}{5} \cdot \frac{3}{4} - \frac{3}{7} \cdot \frac{3}{5}\right) \cdot \frac{a^{2} \sqrt{3}}{4} = \frac{51 a^{2} \sqrt{3}}{280}$,

$S_{\text {pr }} = \frac{51 \sqrt{3}}{70}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-12.jpg?height=543&width=626&top_left_y=1299&top_left_x=1269)

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

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

Using different formulas for finding the area of the triangle

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-12.jpg?height=257&width=271&top_left_y=2062&top_left_x=1555)
$D C J_{1}$, we have

$\frac{\sqrt{21} a h}{5} = \frac{4 a^{2}}{5} \cdot \frac{\sqrt{3}}{2}$, $h = \frac{2 a}{\sqrt{7}}$. Then $\cos \varphi = \sqrt{1 - \frac{7 d^{2}}{4 a^{2}}} = \frac{\sqrt{4 a^{2} - 7 d^{2}}}{2 a} = \frac{3}{4}$.

Finally, we have $S = \frac{51 a^{3} \sqrt{3}}{140 \sqrt{4 a^{2} - 7 d^{2}}}$, $S = \frac{34 \sqrt{3}}{35}$. Answer: $\frac{34 \sqrt{3}}{35}$.