For a real number $a$, let $\lfloor a \rfloor$ denote the [greatest integer](https://artofproblemsolving.com/wiki/index.php/Ceiling_function) less than or equal to $a$. Let $\mathcal{R}$ denote the region in the [coordinate plane](https://artofproblemsolving.com/wiki/index.php/Coordinate_plane) consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$. The region $\mathcal{R}$ is completely contained in a [disk](https://artofproblemsolving.com/wiki/index.php/Disk) of [radius](https://artofproblemsolving.com/wiki/index.php/Radius) $r$ (a disk is the union of a [circle](https://artofproblemsolving.com/wiki/index.php/Circle) and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$, where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$.
The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\left(\frac12,\frac12\right)$. The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$, is $\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.$ The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$, is $\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.$ The latter is the larger, and is $\frac {\sqrt {130}}2$, giving an answer of $130 + 2 = \boxed{132}$.