Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere of radius 6. Segment $C D$ is a diameter of this sphere. Find the volume of the prism if $A D=4 \sqrt{6}$.
# Solution.

The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.

It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (Fig. 3).

![](https://cdn.mathpix.com/cropped/2024_05_06_e9b4fbbbcf148e16432dg-15.jpg?height=672&width=623&top_left_y=138&top_left_x=725)

Fig. 3.

Draw the diameter $C_1D$ of the circle centered at point $O_1$ through point $C_1$. We will show that $CD$ is the diameter of the sphere. Indeed, the plane $CC_1D$ is perpendicular to the planes of the bases and, therefore, together with point $O_1$ contains the segment $OO_1$. Since $C_1D = 2DO_1$, the line $CD$ intersects the segment $OO_1$ at its midpoint, i.e., at the center $M$ of the given sphere.

Let $D_1$ be the projection of point $D$ onto the plane of the base $ABC$, the height of the prism is $h$, and the radii of the circles centered at $O$ and $O_1$ are $r$. Consider the triangles $CC_1D$ and $ADD_1$. Given that $C_1D = 2r$, $AD_1 = r$ (triangle $AOD_1$ is equilateral), $CC_1 = DD_1 = h$, by the Pythagorean theorem, we obtain the system of equations:

$$
\left\{\begin{array}{l}
h^2 + 4r^2 = 12^2 \\
h^2 + r^2 = (4\sqrt{6})^2
\end{array}\right.
$$

Solving the system, we find that $r = 4$, $h = 4\sqrt{5}$. Then the side of the base is $4\sqrt{3}$, its area $S = 12\sqrt{3}$, and consequently, the volume of the prism $V = S \cdot h = 48\sqrt{15}$.

Answer. $48\sqrt{15}$.