\begin{aligned} & {[\text { Area of a sphere and its parts }]} \\ & {[\underline{\text { Volume helps solve the task. }]}]}\end{aligned}

The base of the pyramid $S A B C$ is a triangle $A B C$, where $A B=15 \sqrt{2}, B C=20$, and the radius of the circle circumscribed around this triangle is $5 \sqrt{5}$. Three spheres are constructed on the sides of triangle $A B C$ as diameters, intersecting at point $O$. Point $O$ is the center of a fourth sphere, and the vertex of the pyramid $S$ is the point of tangency of this sphere with a certain plane parallel to the base plane $A B C$. The area of the part of the fourth sphere enclosed within the trihedral angle formed by the rays $O A, O B$, and $O C$ is $8 \pi$. Find the volume of the pyramid $S A B C$.
Since point $O$ lies on the sphere with diameter $A B$, the segment $A B$ is seen from this point at a right angle. Similarly, the sides $B C$ and $A C$ of triangle $A B C$ are seen from point $O$ at a right angle. Therefore, $O A B C$ is a trihedral angle, all of whose plane angles are $90^{\circ}$, so inside this trihedral angle, one-eighth of the sphere's surface area is enclosed. If $r$ is the radius of the sphere, then $\frac{1}{\frac{1}{b}} \cdot 4 \pi r^2 = 8 \pi$. From this, we find that $r=4$. We will prove that triangle $A B C$ is acute-angled. Let $O A=x, O B=y, O C=z$. By the Pythagorean theorem,

$$
\begin{aligned}
& \left\{\begin{array}{l}
x^{2} + y^{2} = 450 \\
y^{2} + z^{2} = 400 \\
x^{2} + z^{2} = 250
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2} = 150 \\
y^{2} = 300 \\
z^{2} = 100
\end{array}\right.\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-31.jpg?height=132&width=1127&top_left_y=2098&top_left_x=471)

Therefore, $\angle A C B < \frac{\pi}{2}$. Similarly for the other angles of triangle $A B C$. Thus, the cosines of the angles of triangle $A B C$ are positive. Let $\angle B A C = \alpha, \angle A C B = \gamma$. Suppose $R = 5 \sqrt{5}$ is the radius of the circumscribed circle of triangle $A B C$. By the sine theorem,

$$
\sin \alpha = \frac{B C}{2 R} = \frac{20}{10 \sqrt{5}} = \frac{2}{\sqrt{5}}, \quad \sin \gamma = \frac{A B}{2 R} = \frac{15 \sqrt{2}}{10 \sqrt{5}} = \frac{3}{\sqrt{10}}.
$$

$$
\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \frac{1}{\sqrt{5}}, \quad \cos \gamma = \frac{3}{\sqrt{10}},
$$

$$
\sin (\alpha + \gamma) = \sin \alpha \cos \gamma + \cos \alpha \sin \gamma = \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} + \frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}} = \frac{1}{\sqrt{2}}
$$

By the sine theorem,

$$
\begin{gathered}
A C = 2 R \sin (\pi - \alpha - \gamma) = 2 R \sin (\alpha + \gamma) = 10 \sqrt{5} \cdot \frac{1}{\sqrt{2}} = 5 \sqrt{10} \\
S_{\triangle A B C} = \frac{1}{2} A C \cdot B C \sin \gamma = \frac{1}{2} \cdot 20 \cdot 5 \sqrt{10} \cdot \frac{3}{\sqrt{10}} = 150
\end{gathered}
$$

Let $K$ be the projection of point $O$ onto the plane of triangle $A B C$. Denote $O K = h$. Expressing the volume $V$ of the pyramid $O A B C$ in two ways ($V = \frac{1}{3} S_{\triangle A B C} \cdot h$ and $V = \frac{1}{2} S_{\triangle O A B} \cdot O C$), we get

$$
h = \frac{S_{\triangle O A B} \cdot O C}{S_{\triangle A B C}} = \frac{150 x}{\frac{1}{2} \times y} = \frac{150 \cdot 10}{\frac{1}{2} \cdot \sqrt{4500} - \sqrt{300}} = 5 \sqrt{2}
$$

Let $H$ be the height of the original pyramid $S A B C$, dropped from vertex $S$. Since point $S$ lies in a plane parallel to the base $A B C$, and the sphere touches this plane at point $S$, then either $H = h + r$ or $H = h - r$.

## Therefore,

$$
V_{S A B C} = \frac{1}{3} S_{\triangle A B C} \cdot H = \frac{1}{3} S_{\triangle A B C} \cdot (h \pm r) = \frac{1}{3} \cdot 150 (5 \sqrt{2} \pm 4) = 50 (5 \sqrt{2} \pm 4)
$$

## Answer

$50 (5 \sqrt{2} \pm 4)$.