Given a cube $A B C D A 1 B 1 C 1 D 1$. A sphere touches the edges $A D, D D 1, C D$ and the line $B C 1$. Find the radius of the sphere, if the edge of the cube is 1.
The center $O$ of the sphere is equidistant from the lateral edges $D A, D D 1$, and $D C$ of the regular triangular pyramid $D A C D 1$ with vertex $D$, so the point $O$ lies on the line containing the height of this pyramid passing through vertex $D$, i.e., on the line $D B 1$. Let $K$ be the point of intersection of the diagonals of the square $B B 1 C 1 C$. Since the line $B C 1$ is perpendicular to the plane $A 1 B 1 C D$, then $O K \perp B C 1$, which means $K$ is the point of tangency of the sphere with the line $B C 1$ (a unique perpendicular can be dropped from point $O$ to the line $B C 1$). Drop perpendiculars $O M$ and $O N$ from the center of the sphere to the lines $B 1 C$ and $D C$ respectively. Let $r$ be the radius of the sphere. Then $O K=O N=r$, and from the similarity of triangles $D N O$ and $D C B 1$ we find that

$$
D N=D C \cdot \frac{Q_{N}}{B_{1} C}=r \cdot \frac{1}{\sqrt{2}}=\frac{r}{\sqrt{2}} .
$$

Then

$$
O M=C N=D C-D N=1-\frac{r}{\sqrt{2}}
$$

$$
\begin{gathered}
K M=|C K-C M|=|C K-O N|=\left|\frac{\sqrt{2}}{2}-r\right| \\
O M 2+K M 2=O K 2,\left(1-\frac{r}{\sqrt{2}}\right) 2+\left(\frac{\sqrt{2}}{2}-r\right) 2=r 2
\end{gathered}
$$

from which $r=\sqrt{8} \pm \sqrt{5}$, and since $D N=\frac{r}{\sqrt{2}}<1$, only $r=\sqrt{8}-\sqrt{5}$ satisfies the condition of the problem.

## Answer

$2 \sqrt{2}-\sqrt{5}$.