259. If 0 is the center of the circle, $L$ is the projection of $N$ onto the plane of the base, then the point $M$, since $M$ is the closest point to $N$ on the circle, must lie on the segment LO. On the other hand, since $N$ is the closest point to $M$ on the diagonal of the face, $MN$ is perpendicular to this diagonal, and therefore $KN$ is also perpendicular to this diagonal, where $K$ is the projection of $M$ onto the face containing this diagonal (Fig. 53).

Let $|A L|=a x, \triangle A N K$ be an isosceles right triangle, so $|I \cdot K|=|A L|=a x$,

$$
\begin{aligned}
& |M K|=|O D| \frac{|L K|}{|L D|}=\frac{a x}{1-2 x} \\
& |K D|=\frac{a}{2}(1-4 x)
\end{aligned}
$$

Writing the Pythagorean theorem for $\triangle M O E$ ( $M E$ is parallel to $A D$ ), we get the equation for $x$

$$
\begin{aligned}
& \frac{(1-4 x)^{2}}{4}+\left(\frac{1}{2}-\frac{x}{1-2 x}\right)^{2}=\frac{25}{144} \Leftrightarrow \\
& \quad \Leftrightarrow[6(1-4 x)(1-2 x)]^{2}+[6(1-4 x)]^{2}=[5(1-2 x)]^{2}
\end{aligned}
$$

Replacing $5^{2}=3^{2}+4^{2}$ in the right part and moving it to the left, we get

$$
\begin{aligned}
& {[6(1-4 x)(1-2 x)]^{2}-} \\
& -[3(1-2 x)]^{2}+[6(1-4 x)]^{2}-[4(1-2 x)]^{2}=0 \Leftrightarrow \\
& \Leftrightarrow 9(1-2 x)^{2}(1-8 x)(3-8 x)+4(5-16 x)(1-8 x)=0 \Leftrightarrow \\
& \Leftrightarrow(1-8 x)\left[9(1-2 x)^{2}(3-8 x)+4(5-16 x)\right]=0
\end{aligned}
$$

; It is easy to see that the point $K$ must be to the left of point $D$, i.e., $0<x<1 / 4$, so the expression in the square brackets is not equal to zero, $x=1 / 8$.

Answer: $a \frac{\sqrt{34}}{24}$.
259. Given a cube with edge $a$. Let $N$ be a point on the diagonal of a side face, $M$ be a point on the circle lying in the plane of the base, with the center at the center of the base and radius $\frac{5}{12} a$. Find the minimum value of the magnitude $|M N|$.