The base of the right parallelepiped $A B C D A 1 B 1 C 1 D 1$ is a square $A B C D$ with a side length of 4, and the length of each lateral edge $A A 1, B B 1, C C 1, D D 1$ is 6. A right circular cylinder is positioned such that its axis lies in the plane $B B 1 D 1 D$, and the points $A 1, C 1, B 1$ and the center $O$ of the square $A B C D$ lie on the lateral surface of the cylinder. Find the radius of the cylinder (find all solutions).
1. Suppose the axis $l$ of the cylinder intersects the line $OB_1$ at some point $K$ (Fig.1), and the line $BB_1$ at point $F$. Let $A_2, B_2, C_2$, and $O_2$ be the projections of points $A, B, C$, and $O$ onto the line $l$, respectively. Then the point $A_2$ coincides with the point $C_2$, and since points $O$ and $B_1$ are equidistant from the axis of the cylinder, $B_1B_2 = OO_2 = R$ (where $R$ is the desired radius of the cylinder), so point $K$ is the midpoint of the segment $OB_1$. Draw a line through point $K$ parallel to $BD$. Let this line intersect the edges $BB_1$ and $DD_1$ at points $N$ and $M$, respectively. Then $N$ and $M$ are the midpoints of these edges. Denote $\angle FKN = \alpha, \angle B_1KN = \beta$. We will assume that $\alpha > 0$ and $\cos \alpha > 0$. Squaring both sides of the obtained equation:

$$
9 \sin^2 \alpha \cos^2 \alpha = 2.9 \sin^2 \alpha (1 - \sin^2 \alpha) = 2
$$

$$
9 \sin^4 \alpha - 9 \sin^2 \alpha + 2 = 0, \left(\sin^2 \alpha - \frac{2}{3}\right)\left(\sin^2 \alpha - \frac{1}{3}\right) = 0
$$

from which $\sin \alpha = \sqrt{\frac{2}{3}}$ or $\sin \alpha = \frac{1}{\sqrt{3}}$. In the first case,

$$
R = \sqrt{2} \sin \alpha + 3 \cos \alpha = \sqrt{2} \cdot \sqrt{\frac{2}{3}} + 3 \cdot \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}
$$

in the second case,

$$
R = \sqrt{2} \sin \alpha + 3 \cos \alpha = \sqrt{2} \cdot \frac{1}{\sqrt{3}} + 3 \cdot \sqrt{\frac{2}{3}} = 4 \sqrt{\frac{2}{3}}
$$

2. Suppose the axis $l$ is parallel to the line $OB_1$ (Fig.2). Then $OB_1$ is a generatrix of the cylinder. Let $A_2, B_2, C_2$, and $O_2$ be the projections of points $A, B, C$, and $O$ onto the line $l$, respectively, and $O_1$ be the center of the square $A_1B_1C_1D_1$. Then the point $A_2$ coincides with the point $C_2$, and since points $O$ and $B_1$ are equidistant from the axis of the cylinder, $B_1B_2 = OO_2 = R$ (where $R$ is the desired radius of the cylinder), the line $O_1C_2$ is perpendicular to the axis $l$ of the cylinder. If the lines $O_1C_2$ and $OB_1$ intersect at point $E$, then $O_1E \perp OB_1$. Denote $\angle OO_1E = \angle BOB_1 = \beta$. Then

$$
OB_1 = \sqrt{BB_1^2 + B_1O_1^2} = \sqrt{36 + 8} = 2\sqrt{11}, \sin \beta = \frac{BB_1}{OB_1} = \frac{6}{2\sqrt{11}} = \frac{3}{\sqrt{11}}
$$

$$
O_1E = OO_1 \cos \beta = 6 \sqrt{\frac{2}{11}}, C_2E = B_1B_2 = R, C_2O_1 = C_2E - O_1E = R - 6 \sqrt{\frac{2}{11}}
$$

From the right triangle $C_1OC_2$ we find that $C_1C_2^2 = O_1C_2^2 + O_1C_1^2$, or

$$
R^2 = \left(R - 6 \sqrt{\frac{2}{11}}\right)^2 + 8
$$

from which $R = \frac{20}{3} \sqrt{\frac{2}{11}}$.

## Answer

$\frac{5}{\sqrt{3}}, 4 \sqrt{\frac{2}{3}}, \frac{20}{3} \sqrt{\frac{2}{11}}$