$[\quad$ Prism

The lower base of the prism $A B C A 1 B 1 C 1$ with lateral edges $A A 1, B B 1, C C 1$ is an equilateral triangle $A B C$ with side $2 a$. The projection of the prism onto the base plane is a trapezoid with side $A B$ and an area twice the area of the base. The radius of the sphere passing through the vertices $A$, $B, A 1, C 1$ is $2 a$. Find the volume of the prism. (Find all solutions).
Let $A_2, B_2$ and $C_2$ be the orthogonal projections of the vertices $A_1, B_1$ and $C_1$ onto the base plane $ABC$. The projection of the prism onto the plane $ABC$ is a trapezoid with side $AB$, if either points $A_2$ and $C_2$ lie on the ray $AC$ (Fig.1), or points $B_2$ and $C_2$ lie on the ray $BC$ (Fig.2). Consider the first of these cases. Denote $BB_2 = x$. The quadrilaterals $AB B_2 A_2$ and $B B_2 C_2 C$ are parallelograms, so $AA_2 = BB_2 = CC_2 = x$. The height of the trapezoid $AB B_2 C_2$ is equal to the height of the equilateral triangle $ABC$, the semi-sum of the bases of this trapezoid is $\frac{1}{2}(x + x + 2a) = x + a$, and since the area of the trapezoid is twice the area of the triangle $ABC$, we have $x + a = 2a$, from which $AA_2 = x = a$, i.e., the projection of vertex $A_1$ is the midpoint of side $AC$ of the base $ABC$. The sphere passes through points $A$ and $B$, so its center lies in the plane $\alpha$ passing through the midpoint $E$ of segment $AB$ perpendicular to $AB$. Similarly, the center of the sphere lies in the plane $\beta$ passing through the midpoint $F$ of segment $A_1 C_1$ perpendicular to $A_1 C_1$. Since $EC \perp AB$ and $FC \perp A_2 C_2$ and $C$ is the midpoint of $A_2 C_2$, these two planes intersect along the line $FC$. Therefore, the center of the sphere lies on the line $FC$, and since the distances from point $C$ to vertices $A$ and $B$ are $2a$, i.e., the radius of the sphere, the center of the sphere is point $C$. Then $CC_1 = CA_1 = 2a$ and

\[
\begin{gathered}
CF = \sqrt{CC_1^2 - C_1 F^2} = \sqrt{4a^2 - a^2} = a \sqrt{3} \\
V_{ABC A_1 B_1 C_1} = S_{\triangle ABC} \cdot FC = \frac{(2a)^2 \sqrt{3}}{4} \cdot a \sqrt{3} = a^3
\end{gathered}
\]

Consider the second case. Similarly to the previous one, we find that $AA_2 = BB_2 = CC_2 = a$. The center of the sphere lies on the line $M M_1$ of intersection of planes $\alpha$ and $\beta$, perpendicular to the plane $ABC$, so the line $M M_1$ is also perpendicular to the plane $ABC$. Let point $M$ lie in the plane $ABC$, and point $M_1$ in the plane $A_1 B_1 C_1$. Note that point $M$ lies at the intersection of the heights $CE$ and $B_2 F_2$ of the equilateral triangles $ABC$ and $A_2 B_2 C_2$. In this case, $EF_2$ is the midline of the trapezoid $ABC_2 A_2$, so $EF = \frac{1}{2}(BC_2 + AA_2) = \frac{1}{2}(3a + a) = 2a$. From the similarity of triangles $CMB_2$ and $EMF_2$, it follows that $\frac{MF_2}{MB_2} = \frac{B_2 C}{EF_2} = 2$. Therefore,

\[
OM_1 = OM_2 = \frac{1}{2} MM_1, \quad MF_2 = \frac{2}{3} B_2 F_2 = \frac{2}{3} \cdot \frac{2a \sqrt{3}}{2} = \frac{2a \sqrt{3}}{3}
\]

Let $O$ be the center of the sphere. Then $OC_1 = OA = 2a$, and since $M M_1 C_1 C_2$ is a rectangle, $M_1 C_1 = M C_2 = M B = M A$, so the right triangles $OM_1 C_1$ and $OMA$ are equal by the leg and hypotenuse. Therefore, $MO = OM_1 = \frac{1}{2} MM_1$. From the right triangle $MC_2 F_2$ we find that

\[
MC_2 = \sqrt{MF_2^2 + F_2 C_2^2} = \sqrt{\left(\frac{2a \sqrt{3}}{3}\right)^2 + a^2} = \frac{a \sqrt{21}}{3}
\]

Then $M_1 C_1 = M C_2 = \frac{a \sqrt{21}}{3}$ and

\[
OM_1 = \sqrt{OC_1^2 - M_1 C_1^2} = \sqrt{(2a)^2 - \left(\frac{2 \sqrt{21}}{3}\right)^2} = \frac{a \sqrt{5}}{\sqrt{3}}
\]

Therefore,

\[
V_{ABC A_1 B_1 C_1} = S_{\triangle ABC} \cdot MM_1 = \frac{(2a)^2 \sqrt{3}}{4} \cdot 2 \cdot \frac{a \sqrt{5}}{\sqrt{3}} = 2a^3 \sqrt{5}
\]

## Answer

$3a^3, 2\sqrt{5}a^3$.