5. Inside a regular quadrilateral pyramid $S A B C D$ with base $A B C D$, there is a regular quadrilateral prism $K L M N K_{1} L_{1} M_{1} N_{1}$, the base $K L M N$ of which lies in the plane $A B C$. The center of the base $K L M N$ of the prism is located on the segment $A C$, $K L\|A C$, $K N\|$ $B D$ (points $K$ and $B$ lie on the same side of $A C$), the side of the base of the prism is 2, and the lateral edge $K K_{1}$ of the prism is 1. The vertices $L_{1}$ and $M_{1}$ of the upper base of the prism $K L M N K_{1} L_{1} M_{1} N_{1}$ belong to the lateral faces $S B C$ and $S C D$ of the pyramid $S A B C D$ respectively. The plane $\gamma$ passes through the points $B, K_{1}$, and $N_{1}$. Find the volumes of the parts into which the plane $\gamma$ divides the pyramid $S A B C D$, if the side of the pyramid is $8 \sqrt{2}$, and its height is 4. (20 points)
Solution. Plane $\gamma$ contains $BD$.

Given $AB = a = 8\sqrt{2}$, $SO = h = 4$, $KL = b = 2$, $KK_1 = h_0 = 1$.

$PR \parallel AC \parallel KL$, $RF \parallel SO$, $RF = h_0$.

$\triangle RCF \sim \triangle SCO$, $\frac{CF}{CO} = \frac{RF}{SO} = \frac{h_0}{h} = \frac{1}{4}$, $CF = \frac{CO}{4}$.

Plane $PQR$ is parallel to $ABC$.

$GR \perp M_1L_1$, $GR = \frac{b}{2} = 1$. $KE \perp BD$,

$KE = CO - CF - GR - KL = \frac{3CO}{4} - \frac{3b}{2} = 6 - 3 = 3$.

Let $\alpha$ be the angle of inclination of plane $\gamma$ to the base plane $ABC$. Then $\tan \alpha = \frac{h_0}{KE} = \frac{1}{3}$.

Let point $H$ be the intersection of plane $\gamma$ and edge $SC$. Then $\angle HOC = \alpha$. If $HU$ is

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-14.jpg?height=811&width=893&top_left_y=608&top_left_x=1173)
the perpendicular dropped to plane $ABC$, then $\frac{HU}{OU} = \tan \alpha = \frac{1}{3}$.

$\Delta HCU \sim \Delta SCO$, $\frac{HU}{SO} = \frac{CU}{CO}$, $\frac{HU}{h} = \frac{CO - OU}{CO}$, $\frac{HU}{h} = \frac{CO - 3HU}{CO}$, $\frac{HU}{4} = \frac{8 - 3HU}{8}$, $20HU = 32$, $HU = \frac{8}{5}$.

Volume of pyramid $HBCD: V_1 = \frac{1}{3} \cdot \frac{BD \cdot CO}{2} \cdot HU = \frac{512}{15}$. Volume of pyramid $SABCD: V = \frac{1}{3} a^2 h = \frac{512}{3}$.

Volume of the second part $V_2 = \frac{512}{3} - \frac{512}{15} = \frac{2048}{15}$.

Answer: $\frac{512}{15}, \frac{2048}{15}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-14.jpg?height=660&width=742&top_left_y=1686&top_left_x=1108)