8,9

In a triangular pyramid $P A B C$, the lateral edge $P B$ is perpendicular to the plane of the base $A B C$, $P B=6, A B=B C$ $=\sqrt{15}, A C=2 \sqrt{3}$. A sphere with center $O$ lying on the face $A B P$ touches the planes of the other faces of the pyramid. Find the distance from the center $O$ of the sphere to the edge $A C$.
Let $M$ be the midpoint of $A C$. Then $B M \perp A C$ and $P M \perp A C$, so $B M P$ is the linear angle of the dihedral angle between the faces $A B C$ and $A P C$. Denote $\angle B M P=\alpha$. From the right triangle $A M B$, we find that

$$
B M=\sqrt{A B^{2}-A M^{2}}=\sqrt{15-3}=\sqrt{12}=2 \sqrt{3}
$$

therefore

$$
\operatorname{tg} \alpha=\frac{B P^{\prime}}{B M}=\frac{6}{2 \sqrt{3}}=\sqrt{3}
$$

so, $\alpha=60^{\circ}$, and $M P=2 B M=4 \sqrt{3}$. By connecting the center $O$ of the sphere to the vertices of the pyramid $P A B C$, we divide the pyramid $P A B C$ into three triangular pyramids with a common vertex $O$ and bases $A P C, B P C$, and $A B C$. The height of each of them, drawn from the vertex $O$, is equal to the radius $r$ of the sphere. Then

$$
\begin{aligned}
V_{P A B C} & =\frac{1}{3} S_{\triangle A P C} \cdot r+\frac{1}{3} S_{\triangle B P C} \cdot r+\frac{1}{3} S_{\triangle A B C} \cdot r= \\
& =\frac{1}{3} r\left(S_{\triangle A P C}+S_{\triangle B P C}+S_{\triangle A B C}\right)
\end{aligned}
$$

$$
r=\frac{3 V_{P A B C}}{S_{\triangle A P C}+S_{\triangle B P C}+S_{\triangle A B C}}=
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-18.jpg?height=120&width=546&top_left_y=154&top_left_x=756)

$=\frac{\frac{1}{2} \cdot 2 \sqrt{3} \cdot 2 \sqrt{3} \cdot 6}{\frac{1}{2} \cdot 2 \sqrt{3} \cdot 4 \sqrt{3}+\frac{1}{2} \cdot \sqrt{15} \cdot 6+\frac{1}{2} \cdot 2 \sqrt{3} \cdot 2 \sqrt{3}}=$

$=\frac{36}{12+3 \sqrt{15}+6}=\frac{36}{18+3 \sqrt{15}}=\frac{12}{6+\sqrt{15}}$.

Since the sphere is inscribed in the dihedral angle formed by the faces $A B C$ and $A C P$, its center $O$ lies in the bisector plane of this dihedral angle. Let $K$ be the foot of the perpendicular dropped from point $O$ to the line $A C$ - the edge of the considered dihedral angle. Then

$$
O M=\frac{r}{\sin \frac{\pi}{2}}=\frac{r}{\sin 30^{\circ}}=2 r=\frac{24}{6+\sqrt{15}}
$$

## Answer

$\frac{24}{6+\sqrt{15}}$