5. The cross-section of a regular hexagonal pyramid SABCDEF is formed by a plane passing through the center of the base $A B C D E F$ and parallel to the median $C M$ of the lateral face $S C D$ and the apothem $S N$ of the lateral face $S A F$. The side of the base of the pyramid is 8, and the distance from the vertex $S$ to the cutting plane is $3 \sqrt{13 / 7}$. Find the cosine of the angle between the cutting plane and the base plane. (20 points)
Solution. In the plane $S N U$ ($S U$ - the apothem of the face $S C D$) through the point $O$ draw a line $O Y$, parallel to $S N, Y \in S U, O Y$ - the midline of the triangle $2 N U$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-04.jpg?height=531&width=596&top_left_y=771&top_left_x=273)

$S K: K D=3: 5$

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-04.jpg?height=360&width=426&top_left_y=474&top_left_x=1403)
$S L: L C=3: 1$

In the plane $S C D$ through the point $Y$ draw a line $K L$, parallel to $C M, K \in S D, L \in S C$.

The medians $S U$ and $C M$ of the triangle $S C D$ intersect at point $W$ and are divided in the ratio $2: 1$, i.e., $S W: W U=2: 1$. We have $S Y: Y W: W U=3: 1: 2$. According to Thales' theorem, we arrive at the ratios $S L: L C=3: 1, S K: K D=3: 5$.

The point $V$ is the intersection point of the line $K L$ and $C D$, $K M: M D=V C: C D=1: 4$.

In the base plane, draw the line $V O$, points $R$ and $T$ are the intersection points with the sides $B C$ and $F E$ respectively. Triangles $R C V$ and $O D V$ are similar, and $S L: L C=3: 1, S K: K D=3: 5$. Let the side of the base be $a$, then $R C=F T=a / 5, B R=T E=4 a / 5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-04.jpg?height=1596&width=626&top_left_y=835&top_left_x=1269)

Let $\varphi$ be the angle between the plane of the section and the base plane. Denote the distance from point $S$ to the section plane as $d, d=3 \sqrt{13 / 7}$. The distance from point $C$ to the section is $d / 3$. In triangle $R C V$, draw the height $C G$, denote its length as $h$. Then $\sin \varphi=\frac{d}{3 h}, \cos \varphi=\sqrt{1-\frac{d^{2}}{9 h^{2}}}$. Find $R V$ using the cosine theorem:

$R V^{2}=\frac{a^{2}}{25}+\frac{a^{2}}{16}-\frac{a^{2}}{20}=\frac{21 a^{2}}{400}, R V=\frac{a \sqrt{21}}{20}=\frac{2 \sqrt{21}}{5}$. Using different

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-05.jpg?height=217&width=348&top_left_y=400&top_left_x=1525)
formulas for finding the area of triangle $R C V$, we have

$$
\begin{aligned}
& \frac{a \sqrt{21}}{20} h=\frac{a^{2}}{20} \cdot \frac{\sqrt{3}}{2}, h=\frac{a}{2 \sqrt{7}}=\frac{4}{\sqrt{7}} \text{. Then } \\
& \cos \varphi=\sqrt{1-\frac{d^{2}}{9 h^{2}}}=\sqrt{1-\frac{28 d^{2}}{9 a^{2}}}=\frac{\sqrt{9 a^{2}-28 d^{2}}}{3 a}=\sqrt{1-\frac{28 \cdot 9 \cdot 13}{9 \cdot 64 \cdot 7}}=\sqrt{1-\frac{13}{16}}=\frac{\sqrt{3}}{4} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-05.jpg?height=252&width=254&top_left_y=685&top_left_x=1603)

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