In a sphere of radius $r$, a diameter $A B$ and three equal chords $A C, A D$ and $A F$ are drawn at an angle $\alpha$ to each other. Find the volume of the body bounded by the planes of triangles $A C D, A D F, A C F, B C D, B D F$ and $B C F$.
Isosceles triangles $C A D, C A F$ and $D A F$ are equal by two sides and the angle between them, so $C D=C F$ $=D F$. Therefore, triangle $C D F$ is equilateral, and since the lateral edges of the triangular pyramid $A C D F$ with vertex $A$ are equal, the pyramid $A C D F$ is regular. Its height $A H$ passes through the center of the equilateral triangle $C D F$. Since the given sphere passes through points $C, D$ and $F$, its center $O$ lies on the line $A H$, so the vertex $B$ of the pyramid $B C D F$ also lies on the line $A H$. Therefore, the pyramid $B C D F$ is regular. Let $C D=D F=C F=a$. From the isosceles triangle $C A D$, we find that $A C=\frac{a}{2 \sin \frac{\pi}{2}}$. Consider the section of the sphere by the plane passing through points $A, C$ and $B$. We get a right triangle $A B C$, in which from the vertex of the right angle $C$ the height $C H$ is drawn. Further, we have:

$$
C H=\sqrt{A B^{2}-A C^{2}}=\sqrt{4 r^{2}-\frac{a^{2}}{16 \sin^{2} \frac{\pi}{2}}}, A C \cdot B C=A B \cdot C H
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_41a6f93ed2bb2ffd440eg-24.jpg?height=117&width=408&top_left_y=542&top_left_x=825)

from which

$$
a^{2}=\frac{16}{3} r^{2} \sin^{2} \frac{\pi}{4}\left(3-4 \sin^{2} \frac{\pi}{4}\right)
$$

The geometric body given in the condition consists of two regular pyramids $A C D F$ and $B C D F$ with a common base $C D F$, and the sum of the heights of these pyramids is equal to the diameter of the sphere. Let $V$ be the desired volume. Then

$$
\begin{aligned}
& V=V_{A C D F}+V_{B C D F}=\frac{1}{3} S_{\triangle C D F} \cdot A H+\frac{1}{3} S_{\triangle C D F} \cdot B H= \\
& =\frac{1}{3} S_{\triangle C D F} \cdot A B=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot 2 r=\frac{8}{3 \sqrt{3}} r^{3} \sin^{2} \frac{\pi}{4}\left(3-4 \sin^{2} \frac{\pi}{4}\right)
\end{aligned}
$$

## Answer

$\frac{8}{3 \sqrt{3}} r^{3} \sin^{2} \frac{\pi}{4}\left(3-4 \sin^{2} \frac{\pi}{4}\right)$