Vertices $A, B$ and $D 1$ of the cube $A B C D A 1 B 1 C 1 D 1$ lie on the lateral surface of a cylinder, the axis of which is parallel to the line $D C 1$. Find the radius of the base of the cylinder, if the edge of the cube is $a$.

#
Consider the orthogonal projection of a cylinder onto the plane $B C D 1 A 1$. This plane is perpendicular to the line $D C 1$, and therefore it is perpendicular to the axis of the cylinder. The orthogonal projections of points $B$ and $D 1$ onto this plane are the points $B$ and $D 1$ themselves, while the orthogonal projection of point $A$ is point $F$, the intersection of the diagonals of the square $A B B 1 A 1$. The orthogonal projection of the circle at the base of the cylinder is a circle circumscribed around the triangle $BFD1$. Its radius $r$ is equal to the radius of the base of the cylinder. In the rectangle $B C D 1 A 1$, it is known that

$$
\begin{gathered}
A 1 D 1=a, A 1 B=a \sqrt{2}, A 1 F=\frac{a \sqrt{2}}{2}, B D 1=a \sqrt{3} \\
F D 1=\sqrt{A_{1} D^{2}+A_{1} F^{2}}=\sqrt{a^{2}+\frac{a^{2}}{2}}=a \sqrt{\frac{3}{2}} \\
\sin \angle F B D 1=\frac{A_{1} D_{1}}{B D_{1}}=\frac{1}{\sqrt{3}}
\end{gathered}
$$

Therefore,

$$
r=\frac{F D_{1}}{2 \sin \angle F B D_{1}}=\frac{a \sqrt{\frac{3}{2}}}{2 \cdot \frac{1}{\sqrt{3}}}=\frac{3 a \sqrt{2}}{2} .
$$

## Answer

$\frac{3 a \sqrt{2}}{2}$.