P r o b l e m 1. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 1), with an edge length of 1. A sphere is drawn through the vertices $A$ and $C$ and the midpoints $F$ and $E$ of the edges $B_{1} C_{1}$ and $C_{1} D_{1}$, respectively. Find the radius $R$ of this sphere.
Solution. The center $M$ of the sphere is equidistant from points $A$ and $C$, so it lies in the plane of symmetry of these points, i.e., the plane $B D D_{1}$. For the same reason, point $M$ also lies in the plane $A_{1} C_{1} C$ (the plane of symmetry of points $F$ and $E$). Therefore, point $M$ lies on the line $O O_{1}$, which is the intersection of the planes $A_{1} C_{1} C$ and $B D D_{1}$.

We introduce a rectangular coordinate system. We do not know where on the line $O O_{1}$ the point $M$ is located. Therefore, $M(0.5 ; 0.5 ; z)$. But $M E^{2}=M C^{2}$, i.e.,

$$
\begin{aligned}
& (0.5-0.5)^{2}+(0.5-1)^{2}+(z-1)^{2}= \\
& =(1-0.5)^{2}+(1-0.5)^{2}+(0-z)^{2}
\end{aligned}
$$

From this, $z=\frac{3}{8}$ and $k^{2}=M C^{2}=(1-0.5)^{2}+(1-0.5)^{2}+$ $+\left(0-\frac{3}{8}\right)^{2}=\frac{41}{64}, R=\frac{\sqrt{41}}{8}$.