smc

WLOG, let the first point be on the bottom side of the square. The points where the second point could exist are outside a circle of radius 0.5 centered on the first point. The parts of the square that lie in this circle are the distance from the point to the closest side of the square $n$, the distance from the point to the outside of the circle (the radius $0.5$), and any portion of the nearest side that lies within the circle as represented by the Pythagorean Theorem $\sqrt{\frac{1}{4}-n^2}$. Thus, the total length the second point can exist in can be represented by $f(n)=4-(0.5+n+\sqrt{\frac{1}{4}-n^2})$. Distributing, $f(n)=3.5-n-\sqrt{\frac{1}{4}-n^2}$. Then, we can find the average of this function through calculus (wow more calc?). This formula is as follows, $$\frac{1}{a_f-a_i}{\int }_{a_i}^{a_f}f(x)dx$$ For this case, the limits of integration are $0$ and $0.5$ ($0\le n\le 0.5$). Then, we have, $$2{\int }_0^{\frac{1}{2}}3.5-n-\sqrt{\frac{1}{4}-n^2}dn$$ $$2({\int }_0^{\frac{1}{2}}3.5dn-{\int }_0^{\frac{1}{2}}ndn-{\int }_0^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}dn)$$ Even if you don't know how to integrate, as long as you know the idea of integration, you can figure these out. Graphing the first two, you can see that the first is a rectangle of length $0.5$ and width $3.5$. The second is an isosceles right triangle of leg length $0.5$. $${\int }_0^{\frac{1}{2}}3.5dn=\frac{7}{4},{\int }_0^{\frac{1}{2}}ndn=\frac{1}{8}$$ Recognize that the third integral is a semicircle of radius $0.5$ and centered at the origin. This is where $\pi$ comes in. From $0$ to $0.5$, the integral is simply a quarter circle. $\frac{{\frac{1}{2}}^2\pi }{4}=\frac{\pi }{16}$. $${\int }_0^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}dn)=\frac{\pi }{16}$$ If you want me to actually integrate these, look below. Do note that this is for those that have a limited knowledge of integration or those that have little time but are being very clever. Overall, where second point could lie to satisfy the problem is a length of $2(\frac{7}{4}-\frac{\pi }{16}-\frac{1}{8})=\frac{26-\pi }{8}$. By contrast, the total length where it could lie is the perimeter of the square $4$. So the possible points that the second point could be make up $\frac{\frac{26-\pi }{8}}{4}=\frac{26-\pi }{32}$ of the square's perimeter. Obviously, $\gcd (32,26,1)=1$. $32+26+1=59⟹\boxed{59}$ Sorry for the long explanation!