$\left.\begin{array}{c}\text { Cube } \\ \text { [Vector product }]\end{array}\right]$

Given a cube $A B C D A 1 B 1 C 1 D 1$ with edge $a$. Points $M$ and $K$ are the midpoints of edges $A B$ and $C D$ respectively. Find the radius of the sphere passing through points $M, K, A 1$ and $C 1$.
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 $\alpha$, perpendicular to the segment $A 1 C 1$ and passing 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 $\beta$, perpendicular to the segment $M K$ and passing 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

$$
\begin{gathered}
R 2=O A 21=O Q 2+Q A 21=x 2+\left(\frac{a}{(\sqrt{2}}\right) 2=x 2+\frac{a^{2}}{2} \\
R 2=O M 2=O P 2+P M 2=(a-x) 2+\frac{\left.\frac{a}{2}\right)}{(2)}=(a-x) 2+\frac{a^{2}}{4}
\end{gathered}
$$

From the equation

$$
x 2+\frac{\mathbf{a}^{2}}{2}=(a-x) 2+\frac{\mathbf{a}^{2}}{4}
$$

we get $x=\frac{3 a}{3}$. Therefore,

$$
R=\sqrt{x^{2}+\frac{a^{2}}{2}}=\sqrt{\frac{9 a^{2}}{64}+\frac{a^{2}}{2}}=\frac{a \sqrt{41}}{8} .
$$

## Answer

$\frac{a \sqrt{41}}{8}$