[ [ correct prism $]$

The side of the base $A B C$ of a regular prism $A B C A 1 B 1 C 1$ is 1, and each of the lateral edges has a length of $\frac{1}{\sqrt{3}}$. A right circular cylinder is positioned such that point $A 1$ and the midpoint $M$ of edge $C C 1$ lie on its lateral surface, and the axis of the cylinder is parallel to line $A B 1$ and is at a distance of $\frac{1}{4}$ from it. Find the radius of the cylinder.

#
From the right triangles $A A_1 B_1$ and $A_1 M C_1$, we find that

$$
\begin{gathered}
A B_1=\sqrt{A A_1^2 + A_1 B_1^2}=\sqrt{\frac{1}{3}+1}=\frac{2}{\sqrt{3}} \\
A_1 M=\sqrt{A_1 C_1^2 + C_1 M^2}=\sqrt{1^2 + \left(\frac{1}{2 \sqrt{3}}\right)^2}=\frac{\sqrt{13}}{2 \sqrt{3}}
\end{gathered}
$$

Let $P$ and $Q$ be the projections of points $A_1$ and $M$ onto the line $A B_1$ (Fig.1). Triangle $A M B_1$ is isosceles ($M A_1 = M A = M B_1$), so its height $M Q$ is also a median, meaning $Q$ is the midpoint of $A B_1$. By the property of the median of a right triangle drawn to the hypotenuse, $Q A_1 = \frac{1}{2} A B_1 = \frac{1}{\sqrt{3}} = A A_1$. In the isosceles triangle $A A_1 Q$, the height $A_1 P$ is also a median, meaning $P$ is the midpoint of $A Q$. In the right triangle $A A_1 B_1$, it is known that

$$
\operatorname{tg} \angle A B_1 A_1 = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}}
$$

thus,

$$
\angle A B_1 A_1 = 30^\circ, \quad A_1 P = \frac{1}{2} A_1 B_1 = \frac{1}{2}
$$

Let $E$ be the midpoint of $A B$. Then

$$
M Q = C E = \frac{A B \sqrt{3}}{2} = \frac{\sqrt{3}}{2}.
$$

If $A_2$ and $M_2$ are the projections of points $A_1$ and $M$ onto the axis $l$ of the cylinder, and $R$ is the radius of the cylinder, then $A_1 A_2 = M M_2 = R$. Since the lines $l$ and $A B_1$ are parallel, the line $A B_1$ is perpendicular to two intersecting lines $A_1 P$ and $A_1 A_2$ in the plane $A_1 P A_2$, so $A_2 P \perp A B_1$ and $A_2 P \perp l$. Similarly, $M_2 Q \perp A B_1$ and $M_2 Q \perp l$, meaning $A_2 P = M_2 Q = \frac{1}{4}$. The planes $A_1 A_2 P$ and $M M_2 Q$ are parallel, as both are perpendicular to the line $l$. When parallelly translating by the vector $\vec{Q}$, point $Q$ will move to point $P$, point $M_2$ to point $A_2$, and point $M$ to some point $M_1$, such that

$$
M_1 A_2 = M M_2 = R, \quad M_1 P = M Q = \frac{\sqrt{3}}{2}, \quad A_2 M_2 = P Q = A Q - A P = \frac{\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{3}} = 2 \sqrt{2}}{2}
$$

and since the line $M Q$ is perpendicular to the plane $A A_1 B_1 B$ and $M_1 P \parallel M Q$, the line $M_1 P$ is also perpendicular to this plane, meaning $M_1 P \perp A_1 P$. Consider the plane of the right triangle $A_1 P M_1$. Suppose the ray $PA_2$ passes between the sides of the right angle $A_1 P M_1$. Let $\angle M_1 P A_2 = \alpha$. Then $\angle A_1 P A_2 = 90^\circ - \alpha$. By the cosine rule,

$$
\begin{aligned}
& R^2 = A_2 M_1^2 = P A_2^2 + P M_1^2 - 2 P A_2 \cdot P M_1 \cos \alpha = \frac{1}{16} + \frac{3}{4} - 2 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \cos \alpha \\
& R^2 = A_1 A_2^2 = P A_2^2 + P A_1^2 - 2 P A_1 \cdot P A_2 \cos (90^\circ - \alpha) = \frac{1}{16} + \frac{1}{4} - 2 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \sin \alpha
\end{aligned}
$$

Equating the right-hand sides of these equations, we get

$$
\frac{1}{2} - \frac{\sqrt{4}}{4} \cos \alpha + \frac{1}{4} \sin \alpha = 0, \quad \sqrt{3} \cos \alpha - \sin \alpha = 2, \quad \sin (60^\circ - \alpha) = 1
$$

from which $60^\circ - \alpha = 90^\circ, \alpha = -30^\circ$, which is impossible. Suppose now that the ray $PA_2$ does not pass between the sides of the angle $A_1 P M_1$ (Fig.3). Then, if $\angle M_1 P A_2 = \alpha$, then $\angle A_1 P A_2 = 90^\circ + \alpha$. By the cosine rule,

$$
R^2 = A_2 M_1^2 = P A_2^2 + M_1 P^2 - 2 P A_2 \cdot P M_1 \cos \alpha =
$$

$$
\begin{gathered}
= \frac{1}{16} + \frac{3}{4} - 2 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \cos \alpha = \frac{13}{16} - \frac{\sqrt{3}}{4} \cos \alpha \\
R^2 = A_1 A_2^2 = P A_2^2 + P A_1^2 - 2 P A_2 \cdot P A_1 \cos (90^\circ + \alpha) = \\
= \frac{1}{16} + \frac{1}{4} + 2 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \sin \alpha = \frac{5}{16} + \frac{1}{4} \sin \alpha
\end{gathered}
$$

Equating the right-hand sides of these two equations, we get

$$
\frac{13}{16} - \frac{\sqrt{3}}{4} \cos \alpha = \frac{5}{16} + \frac{1}{4} \sin \alpha \Leftrightarrow \sin \alpha + \sqrt{3} \cos \alpha = 2 \Rightarrow \sin (\alpha + 60^\circ) = 1
$$

The only solution to this equation that satisfies the problem's conditions is $-\alpha = 30^\circ$. Therefore,

$$
R^2 = A_2 M_1^2 = P A_2^2 + P M_1^2 - 2 P A_2 \cdot P M_1 \cos \alpha = \frac{1}{16} + \frac{3}{4} - 2 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \cos 30^\circ =
$$

$$
= \frac{13}{16} - \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{2} = \frac{13}{16} - \frac{3}{8} = \frac{7}{16}, \quad R = \frac{\sqrt{7}}{4}
$$

Answer

$\frac{\sqrt{7}}{4}$.