jmc

Divide the boundary of the square into halves, thereby forming $8$ segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the $5$ segments not bordering the bottom-left segment will be distance at least $\frac{1}{2}$ apart from $A$. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least $0.5$ apart from $A$ is $\frac{0+1}{2}=\frac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.) If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be up to $\frac{1}{2}-\sqrt{0.25-x^2}$ away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is$$\frac{1}{{\left(\frac{1}{2}\right)}^2}{\int }_0^{\frac{1}{2}}\frac{1}{2}-\sqrt{0.25-x^2}dx=4\left(\frac{1}{4}-\frac{\pi }{16}\right)=1-\frac{\pi }{4}.$$ (Alternatively, one can equate the problem to finding all valid $(x,y)$ with $0<x,y<\frac{1}{2}$ such that $x^2+y^2\ge \frac{1}{4}$, i.e. $(x,y)$ is outside the unit circle with radius $\mathrm{0.5.}$) Thus, averaging the probabilities gives$$P=\frac{1}{8}\left(5+\frac{1}{2}+1-\frac{\pi }{4}\right)=\frac{1}{32}\left(26-\pi \right).$$ Our answer is $\boxed{59}$.