8.1. On the lateral edge $S B$ of a regular hexagonal pyramid $S A B C D E F$, a point $P$ is chosen such that $S P$ is 10 times greater than $P B$. The pyramid is divided into two parts by a plane passing through point $P$ and parallel to the edges $S E$ and $F E$. Find the ratio of the volume of the larger of these parts to the volume of the smaller part. If necessary, round the answer to the nearest hundredth.
Answer: 116.44. Solution. Let $S P=c x, P B=b x$ (for the original problem $c=10$, $b=1$). Draw a line parallel to $E F$ from point $P$, and denote the intersection of this line with edge $S C$ as $Q$. Through points $P$ and $Q$, draw planes perpendicular to $P Q$ (Fig. 1). These planes are obviously perpendicular to the base plane.

The cut-off body consists of a prism (Fig. 1) and two pyramids (Fig. 2).

Denote: $H$ and $H_{p}$ - the perpendiculars dropped from points $S$ and $P$ to the base, respectively; $S_{\perp}$ and $P_{\perp}$ - the bases of these perpendiculars; $B_{1}$ - the intersection of $B F$ with the cutting plane; $a$ - the length of the base edge.

![](https://cdn.mathpix.com/cropped/2024_05_06_2cb18ec48844428a3b16g-10.jpg?height=719&width=806&top_left_y=1348&top_left_x=268)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_2cb18ec48844428a3b16g-10.jpg?height=691&width=779&top_left_y=1362&top_left_x=1118)

Fig. 2

Then we get (notations see Figs. 1-3):

$H_{P}=\frac{b}{b+c} H$ (from the similarity of $\triangle B P P_{\perp}$ and $\triangle B S S_{\perp}$);

$B B_{1}=\frac{b}{b+c} \cdot B F=\frac{b}{b+c} \cdot \sqrt{3} a$ (from the similarity of $\triangle B P B_{1}$ and $\triangle B S F$);

$A_{1} B_{1}=\frac{B B_{1}}{B G} \cdot A G=\frac{2 b}{b+c} \cdot A G=\frac{a b}{b+c}$ (from the similarity of $\triangle A_{1} B B_{1}$ and $\triangle A B G$);

$P Q=\frac{c}{b+c} B C=\frac{a c}{b+c}$ (from the similarity of $\triangle S P Q$ and $\triangle S B C$).

![](https://cdn.mathpix.com/cropped/2024_05_06_2cb18ec48844428a3b16g-11.jpg?height=609&width=665&top_left_y=478&top_left_x=750)

Fig. 3

The volume of the original pyramid is: $V=V_{\text {ABCDEF }}=\frac{1}{3} \cdot \frac{6 \cdot a^{2} \sqrt{3}}{4} \cdot H=\frac{\sqrt{3} a^{2}}{2} H$;

The volume of the prism (see Fig. 1) is: $V_{\text {prism }}=P Q \cdot S_{P P_{1} P_{2}}=P Q \cdot \frac{H_{P}}{2} \cdot B B_{1}=\frac{b^{2} c}{2(b+c)^{3}} \cdot \sqrt{3} a^{2} H$.

Since $S_{A_{1} B B_{1} P_{1} P_{2}}=S_{A_{1} B B_{1}}+S_{B_{1} B P_{1} P_{2}}=B B_{1}\left(\frac{A_{1} B_{1}}{2}+B P_{1}\right)=B B_{1}\left(\frac{A_{1} B_{1}}{2}+\frac{a-P Q}{2}\right)$ $=\left(\frac{b}{b+c}\right)^{2} \cdot \sqrt{3} a^{2}$, the volume of each of the two pyramids (see Fig. 2) with vertices at points $P$ and $Q$ respectively is: $V_{\text {pyr. }}=\frac{1}{3} S_{A_{1} B P_{1} P_{2}} \cdot H_{P}=\frac{1}{3}\left(\frac{b}{b+c}\right)^{3} \cdot \sqrt{3} a^{2} H$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2cb18ec48844428a3b16g-11.jpg?height=237&width=1242&top_left_y=2052&top_left_x=356)
Denoting $\frac{c}{b}=n$, we get the desired ratio: $\frac{(1+n)^{3}-n-\frac{4}{3}}{n+\frac{4}{3}}$. Given $n=10$, the answer is: $\frac{3959}{34}=116.44117 \ldots$