Task 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1} \mathrm{c}$ is inscribed in a sphere with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$. Segment $C D$ is the diameter of this sphere, point $K$ and $L$ are the midpoints of edge $A A_{1}$ and $A B$ respectively. Find the volume of the prism if $D L=\sqrt{2}, D K=\sqrt{3}$.
# 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 $A_1B_1C_1$, with their centers at 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. 5).

![](https://cdn.mathpix.com/cropped/2024_05_06_e9b4fbbbcf148e16432dg-25.jpg?height=700&width=606&top_left_y=221&top_left_x=725)

Fig. 5.

Draw a 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, contains the segment $OO_1$ along with point $O_1$. 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 $KA_1D$ and $LDD_1$. Given that $A_1D = AD_1 = r$ (triangle $A_1O_1D$ is equilateral), $LD_1 = \frac{r}{2}$, $DD_1 = h$, $KA_1 = \frac{h}{2}$, by the Pythagorean theorem, we obtain the system of equations

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

Solving the system, we find that $r = \frac{2\sqrt{6}}{3}$, $h = \frac{2\sqrt{3}}{3}$. Then the side of the base is $2\sqrt{2}$, its area $S = 2\sqrt{3}$, and consequently, the volume of the prism $V = S \cdot h = 4$.

Answer. 4.

## Mathematics. Variant-32 $10-11$ grade