5. A sphere with radius $\frac{4}{9}$ lies inside a regular quadrilateral pyramid $SABCD$ with a base side of 8 and height 3. This sphere touches the base plane $ABCD$ of the pyramid and the lateral faces $SBC$ and $SCD$. The plane $\gamma$ touches the sphere, passes through point $B$, the midpoint $K$ of edge $CD$, and intersects edge $SC$ at point $M$. Find the volume of the pyramid $MBCK$. (20 points)
Solution. Since the pyramid $S A B C D$ is regular, the center $O$ of the given sphere lies in the plane $S H C$, where $S H$ is the height of the pyramid. Let $R P \| S H, R \in S C, P \in H C, O \in R P$.

Denote $S H=h, A B=a, R P=k h$. Draw $P N \| A B, N \in B C . E$ - the point of tangency of the sphere with the plane $S B C$, and let the radius of the sphere $O E=r$. Since $P C: H C=k$, then $P N=k a / 2$. Triangles $R O E$ and $R P N$ are similar, and $O E: P N=R O: R N$, or

$$
\frac{r}{k a / 2}=\frac{k h-r}{k \sqrt{h^{2}+(a / 2)^{2}}}, \quad \frac{r}{a}=\frac{k h-r}{\sqrt{4 h^{2}+a^{2}}}, \quad k=\frac{r}{h}\left(\frac{\sqrt{4 h^{2}+a^{2}}}{a}+1\right)
$$

According to the problem, $a=8, h=3, r=4 / 9$. Then $k=1 / 3$.

The point $F$ is the intersection of $A C$ and $B K$, then $F C=A C / 3$. Since $P C=k H C=H C / 3=$ $A C / 6, F P=P C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-07.jpg?height=522&width=512&top_left_y=1561&top_left_x=198)

Let $P Q \perp B K$.

Then $P Q=\frac{a}{2 \sqrt{5}}$.

A

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-07.jpg?height=717&width=1126&top_left_y=1566&top_left_x=702)

If $\alpha=\angle O Q P$, then $\operatorname{tg} \alpha=\frac{r 2 \sqrt{5}}{a}=\frac{\sqrt{5}}{9}$. The angle between the plane $\gamma$ and the base plane is $2 \alpha$. Then $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{4 \sqrt{5} a r}{a^{2}-20 r^{2}}=\frac{9 \sqrt{5}}{38}$.

Let $M G$ be the segment of the perpendicular dropped from point $M$ to the base plane $A B C$, and $M G=n h$. Then $C G=n C H$. If $G V \perp B K$, then $G V=\frac{G F}{C F} h_{0}=\frac{A C / 3-n A C / 2}{A C / 3} h_{0}=\frac{2-3 n}{2} h_{0}, h_{0}=\frac{a}{\sqrt{5}}$ - the height of triangle $В С К$, drawn from vertex $C$.

$$
\begin{gathered}
\operatorname{tg} 2 \alpha=\frac{2 n h \sqrt{5}}{(2-3 n) a}, n=\frac{4 r a^{2}}{(h+6 r) a^{2}-20 h r^{2}}=\frac{12}{37} \\
V_{M B C K}=\frac{1}{3} \cdot \frac{a^{2}}{4} n h=\frac{192}{37}
\end{gathered}
$$

Answer: $\frac{192}{37}$

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-08.jpg?height=505&width=808&top_left_y=350&top_left_x=1178)