Similar to AMC 10A #25

I assumed that I can just treat the point on the perimeter as if it's chosen at random along half of 1 side (from a corner to a midpoint) because no matter where the random point on the perimeter is chosen, the square can be reoriented so that it lies on the same line segment (from a corner to the midpoint of a side).

Call the distance from the corner y. From the above assumption, y ranges from [0, 0.5] Take a look at my drawing of the situation:

$\frac{dy}{0.5} = 2dy$ would be the chance of the random point being (within dy of) y

So the chance that the distance is less than 0.5 should be $\int\limits_0^{0.5} 2A(y)dy$

So now we just need to find find A(y) which is:

$A(y)=\int\limits_0^{y+0.5} \sqrt{0.25-(y-x)^2}dx$

This gives $\frac{3pi-2}{12}$

Initially, I misread the problem, so I calculated the chance that the distance is at most 0.5.... So just take 1 minus my answer to get the correct answer :)

[This is not a full solution.]

We find that the probability is determined by the integral $2\int_0^{\frac{1}{2}}\dfrac{x\sqrt{\frac{1}{4}-x^2}}{2}+\dfrac{\pi-\cos^{-1}2x}{8}\text{ d}x=\dfrac{26-3\pi}{24}$ so the answer is $26+24+3=\boxed{53}$