Task 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. Segment $C D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $A A_{1}$. Find the volume of the prism if $C K=2 \sqrt{3}, D K=2 \sqrt{2}$.
# Solution.

The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along circles circumscribed around 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-20.jpg?height=702&width=605&top_left_y=120&top_left_x=814)

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 a 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 triangles $CAK$ and $K A_1D$. Given that $A_1D = AD_1 = r$ (triangle $A_1O_1D$ is equilateral), $AC = r\sqrt{3}$, $AK = K A_1 = \frac{h}{2}$, by the Pythagorean theorem, we obtain the system of equations:

$$
\left\{\begin{array}{l}
\frac{h^2}{4} + 3r^2 = (2\sqrt{3})^2 \\
\frac{h^2}{4} + r^2 = (2\sqrt{2})^2
\end{array}\right.
$$

Solving the system, we find that $r = \sqrt{2}$, $h = 2\sqrt{6}$. Then the side of the base is $\sqrt{6}$, its area $S = \frac{3\sqrt{3}}{2}$, and, consequently, the volume of the prism $V = S \cdot h = 9\sqrt{2}$.

Answer. $9\sqrt{2}$.