8,9

In a sphere of radius 9, three equal chords $A A 1, B B 1$, and $C C 1$ are drawn through a point $S$ such that $A S=4, A 1 S=8, B S < B 1 S, C S < C 1 S$. Find the radius of the sphere circumscribed about the pyramid $S A B C$.
The products of the segments of intersecting chords are equal, so $B S \cdot S B 1=A S \cdot S A 1=4 \cdot 8=32$. In addition, $B B 1=A A 1=A S+S A 1=4+8=12$. From the system

$$
\left\{\begin{array}{l}
B S \cdot S B_{1}=32 \\
B S+S B_{1}=12
\end{array}\right.
$$

and the condition $B S<B 1 S$, we find that $B S=4$ and $S B 1=8$. Similarly, $C S=4$ and $S C 1=8$. Let $A 2, B 2$, and $C 2$ be the feet of the perpendiculars dropped from the center $O$ of the sphere to the chords $A A 1, B B 1$, and $C C 1$, respectively. Then the points $A 2, B 2$, and $C 2$ are the midpoints of these chords. Note that the triangular pyramid $S A B C$ is similar to the triangular pyramid $S A 2 B 2 C 2$, since these pyramids are homothetic with the center of homothety $S$ and the coefficient

$$
k=\frac{A S}{S A_{2}}=\frac{A S}{-A A_{2}-A S}=-\frac{4}{-6-4}=-2
$$

Therefore, the radius $R$ of the sphere circumscribed around the pyramid $S A B C$ is twice the radius $r$ of the sphere circumscribed around the pyramid $S A 2 B 2 C 2$. The segment $O S$ is seen from the points $A 2, B 2$, and $C 2$ at a right angle, meaning that these points lie on the sphere with diameter $O S$, i.e., $O S$ is the diameter of the sphere circumscribed around the pyramid $S A 2 B 2 C 2$. From the right triangles $A O A 2$ and $O S A 2$, we find that

$$
\begin{gathered}
O A 22=O A 2-A A 22=O A 2-\left(\frac{1}{2} A A 1\right) 2=92-62=45 \\
2 r=O S=\sqrt{S A_{2}^{2}+O A_{2}^{2}}=\sqrt{4+45}=7 \\
\text { Therefore, } R=2 r=7
\end{gathered}
$$

## Answer

7.00

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