10. A sphere touches all the edges of the pyramid $S A B C$, and the lateral edges $S A, S B$, and $S C$ at points $A^{\prime}, B^{\prime}$, and $C^{\prime}$. Find the volume of the pyramid $S A^{\prime} B^{\prime} C^{\prime}$, if $A B=B C=S B=5$ and $A C=4$.
Answer: $\frac{2 \sqrt{59}}{15}$.

Solution. Since the sphere touches all edges, the pyramid has the property that

$$
B C+A S=A B+S C=A C+S B=9
$$

from which $A S=S C=4$, so $\triangle A S C$ is equilateral.

![](https://cdn.mathpix.com/cropped/2024_05_06_948de1dbe9747fa89f7eg-3.jpg?height=551&width=640&top_left_y=1392&top_left_x=1299)
Then $\triangle A^{\prime} S C^{\prime}$ is also equilateral, so $S A^{\prime}=S C^{\prime}=A^{\prime} C^{\prime}=2$. Draw $S H-$ the height of the pyramid $S A B C$ and $S D \perp A C$ (then $B D \perp A C$ by the theorem of three perpendiculars). From $\triangle A B C$ we find $B D=\sqrt{21}$, from which $S_{A B C}=2 \sqrt{21}$. From $\triangle S A C$ we find $S D=2 \sqrt{3}$. Next, from $\triangle S D B$ we find

$$
\cos \angle S B D=\frac{S B^{2}+B D^{2}-S D^{2}}{2 S B \cdot B D}=\frac{21+25-12}{10 \sqrt{21}}=\frac{17}{5 \sqrt{21}}
$$

Therefore, $\sin \angle S B D=\sqrt{1-\frac{289}{525}}=\frac{2 \sqrt{59}}{5 \sqrt{21}}$ and $S H=S B \cdot \sin \angle S B D=\frac{2 \sqrt{59}}{\sqrt{21}}$.

We find the volume of the pyramid $S A B C: V_{S A B C}=\frac{1}{3} \cdot S H \cdot S_{A B C}=\frac{4 \sqrt{59}}{3}$.

Since $\frac{V_{S A^{\prime} B^{\prime} C^{\prime}}}{V_{S A B C}}=\frac{S A^{\prime} \cdot S B^{\prime} \cdot S C^{\prime}}{S A \cdot S B \cdot S C}$, we finally get

$$
V_{S A^{\prime} B^{\prime} C^{\prime}}=\frac{1}{10} V_{S A B C}=\frac{2 \sqrt{59}}{15} .
$$