8. (10 points) In a regular hexagonal pyramid $S A B C D E F$, a point $M$ is chosen on the diagonal of the base $A D$, dividing it in the ratio $A M: M D=n: m$ $(n<m)$. A section of the pyramid is made through point $M$ by a plane parallel to the face $S A B$. Find the ratio of the area of the section to the area of triangle $S A B$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_e53c58234eff2beb6c25g-08.jpg?height=802&width=800&top_left_y=450&top_left_x=662)

We project point $S$ and our section onto the plane $A B C D E F$. Let's describe how to find the projection points of the section. Since the section is parallel to the plane $S A B$, then $K L \| A B$ (two parallel planes intersect the plane $A B C D E F$ along two parallel lines), $L N \| B S^{\prime}$ (two parallel planes intersect the plane $S B C$ along two parallel lines, the projections of these lines are also parallel), $K R \| A S^{\prime}$ (two parallel planes intersect the plane $S A F$ along two parallel lines, the projections of these lines are also parallel). To find the projection point $P$, consider three planes: our section, $A B C D E F$, and $S C D$. Their pairwise common lines intersect at one point $G$, lying on the plane $A B C D E F$. Therefore, the projections of these lines intersect at one point. Thus, points $G, N$, and $P$ lie on one line, and it is perpendicular to $A D$ ( $L N C D$ is a rhombus, so $C L \perp N G$, and it remains to note that $C L \| A D$ ). Point $Q$ is found similarly.

Note that since the planes are parallel, the ratio of the areas of the sections will be the same as the ratio of the areas of the projections onto the plane $A B C D E F$ (the areas differ by a factor of $\cos \alpha$, where $\alpha$ is the dihedral angle between the planes). The projection of the face $S B A$ is the triangle $S^{\prime} A B$, and the projection of the section is the polygon $K L N P Q R$.

Let $A M=x$ and $M S^{\prime}=y$, then it is not difficult to show that $A M=A K=K M=B H=$ $H L=B L=S^{\prime} N=S^{\prime} R=2 \cdot S^{\prime} Q=2 \cdot S^{\prime} P=x$ and $M S^{\prime}=H S^{\prime}=K R=L N=y$. The area of triangle $S^{\prime} A B$ is

$$
S_{S^{\prime} A B}=\frac{1}{2} \cdot S^{\prime} A \cdot S^{\prime} B \cdot \sin 60^{\circ}=\frac{\sqrt{3}(x+y)^{2}}{4}
$$

The area of the polygon $K L N P Q R$ is found as the sum of the areas of its parts

$$
\begin{gathered}
S_{K L N P Q R}=S_{S^{\prime} M H}+S_{S^{\prime} H L N}+S_{S^{\prime} N P}+S_{S^{\prime} P Q}+S_{S^{\prime} Q R}+S_{S^{\prime} R K M}= \\
=\frac{\sqrt{3} y^{2}}{4}+\frac{\sqrt{3} x y}{2}+\frac{\sqrt{3} x^{2}}{8}+\frac{\sqrt{3} x^{2}}{16}+\frac{\sqrt{3} x^{2}}{8}+\frac{\sqrt{3} x y}{2}=\frac{\sqrt{3}}{4}\left(y^{2}+4 x y+\frac{5}{4} x^{2}\right)
\end{gathered}
$$

We know that

$$
\frac{x}{2 y+x}=\frac{n}{m} \Longrightarrow \frac{y}{x}=\frac{m-n}{2 n}
$$

We get the ratio

$$
\begin{gathered}
\frac{S_{K L N P Q R}}{S_{S^{\prime} A B}}=\frac{4 y^{2}+16 x y+5 x^{2}}{4(x+y)^{2}}=\frac{4\left(\frac{y}{x}\right)^{2}+16\left(\frac{y}{x}\right)+5}{4\left(1+\frac{y}{x}\right)^{2}}= \\
=\frac{4\left(\frac{m-n}{2 n}\right)^{2}+16\left(\frac{m-n}{2 n}\right)+5}{4\left(1+\frac{m-n}{2 n}\right)^{2}}=\frac{(m-n)^{2}+8 n(m-n)+5 n^{2}}{(m+n)^{2}}=1+\frac{n(4 m-3 n)}{(m+n)^{2}}
\end{gathered}
$$

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 1.5568 | 1.4608 | 1.2752 | 1.5712 | 1.57 | 1.33 | 1.5328 | 1.52 | 1.48 | 1.5625 |