Two random points on a square's edge over unit apart

Two points are chosen independently and uniformly on the perimeter of a square with side $1$. The probability that they are more than $1$ apart equals $\tfrac{c-\pi}{d}$. Find $\tfrac cd$.

$3/4$. Given the side of the first point, the second is on the same side ($\tfrac14$, never more than $1$ apart), the opposite side ($\tfrac14$, always more than $1$ apart almost surely) or an adjacent side ($\tfrac12$, apart iff $x^2+y^2>1$, probability $1-\tfrac\pi4$). The total is $\tfrac34-\tfrac\pi8=\tfrac{6-\pi}{8}$.