10,11

Given a cube $A B C D A 1 B 1 C 1 D 1$ with bases $A B C D$ and $A 1 B 1 C 1 D 1$. Point $M$ is the midpoint of edge $A B$, and point $K$ is the midpoint of edge $C D$. Find the radius of the sphere passing through points $M, K, A 1, C 1$, if the edge of the cube is $\sqrt{41}$.
Let the edge of the cube be denoted by \(a\). The center \(O\) of the sphere passing through points \(M, K, A_1\), and \(C_1\) is equidistant from points \(A_1\) and \(C_1\), so point \(O\) lies in the plane \(\alpha\), which is perpendicular to the segment \(A_1 C_1\) and passes through its midpoint, i.e., in the plane \(B D D_1 B_1\). Point \(O\) is equidistant from points \(M\) and \(K\), so it lies in the plane \(\beta\), which is perpendicular to the segment \(M K\) and passes through its midpoint. The planes \(\alpha\) and \(\beta\) intersect along the line \(P Q\), where \(P\) and \(Q\) are the centers of the squares \(A B C D\) and \(A_1 B_1 C_1 D_1\) respectively. Let \(O Q = x\), \(O M = O K = O A_1 = O C_1 = R\). By the Pythagorean theorem from the right triangles \(O P M\) and \(O Q A_1\), we find that

\[
\begin{gathered}
R^2 = O A_1^2 = O Q^2 + Q A_1^2 = x^2 + \left(\frac{a}{\sqrt{2}}\right)^2 = x^2 + \frac{a^2}{2} \\
R^2 = O M^2 = O P^2 + P M^2 = (a - x)^2 + \left(\frac{a}{2}\right)^2 = (a - x)^2 + \frac{a^2}{4}
\end{gathered}
\]

From the equation

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-05.jpg?height=83&width=386&top_left_y=1299&top_left_x=839)

we find that \(x = \frac{3}{8} a\). Therefore,

\[
R^2 = \sqrt{x^2 + \frac{a^2}{2}} = \sqrt{\frac{a^2}{\rho_1^2} + \frac{a^2}{2}} = \frac{a \sqrt{11}}{8}.
\]

If \(a = \sqrt{41}\), then

\[
R = \sqrt{41} \cdot \frac{\sqrt{11}}{8} = \frac{11}{8}
\]

## Answer

\(\frac{11}{8}\)