A sphere of radius $\sqrt{41}$ passes through the vertices $B, C, C_1$ and the midpoint of the edge $A_1 D_1$ of the cube $A B C D A_1 B_1 C_1 D_1$ ( $A A_1$ || $B B_1$ || $C C_1$ || $D D_1$ ). Find the surface area of the cube.
Let $M$ be the midpoint of edge $A_1D_1$. The center $O$ of the sphere is equidistant from the endpoints of segment $BC$, so point $O$ lies in the plane perpendicular to edge $BC$ and passing through its midpoint. Similarly, point $O$ lies in the plane perpendicular to edge $CC_1$ and passing through the midpoint of $CC_1$. These two planes intersect along a line passing through the centers $P$ and $Q$ of the faces $AA_1D_1D$ and $BB_1C_1C$ respectively. Point $O$ is equidistant from the endpoints of segment $MC_1$, so point $O$ lies in the plane perpendicular to segment $MC_1$ and passing through its midpoint $E$. This plane intersects the line $PQ$ at a point equidistant from points $B, C, C_1$, and $M$, i.e., at point $O$. Consider the orthogonal projection of the cube onto the plane passing through the line $PQ$, parallel to the face $ABCD$. Let $A', B', C'$, and $D'$ be the projections of vertices $A, B, C$, and $D$ respectively, and $E'$ be the projection of point $E$. Denote the edge of the cube by $x$ and the radius of the sphere by $R$. From the right triangles $PQC', POE'$, and $POM$, we find that

$$
\begin{aligned}
& P C'=\sqrt{P D'^2 + C' D^2}=\sqrt{\frac{x^2}{4} + x^2}=\frac{x \sqrt{5}}{2}, P E'=\frac{1}{2} P C'=\frac{x \sqrt{5}}{4} \\
& \operatorname{tg} \angle QPC'=\frac{C'Q}{PQ}=\frac{1}{2}, \cos \angle QPC'=\frac{1}{\sqrt{1+\operatorname{tg}^2 \angle QPC'}}=\frac{2}{\sqrt{5}} \\
& P O=\frac{P E'}{\cos \angle QPC'}=\frac{\frac{x \sqrt{5}}{4}}{\frac{2}{\sqrt{5}}}=\frac{5}{8} x \\
& 41=R^2=OM^2=PO^2+PM^2=\frac{25}{64} x^2 + \frac{x^2}{4}=\frac{41}{64} x^2
\end{aligned}
$$

from which $x^2=64$. Therefore, the total surface area of the cube is $6x^2=384$.

## Answer

384.00