2015 AMC 10A Problem 25

Fix one of the two points. The second point is on the same side with probability $\frac14$, on an adjacent side with probability $\frac12$, and on the opposite side with probability $\frac14$. On the same side, two coordinates $a,b\in[0,1]$ are at distance at least $\frac12$ when $|a-b|\ge\frac12$. This region consists of two right triangles with total area $\frac14$. On adjacent sides, the distance has the form $\sqrt{a^2+b^2}$. The failing region is a quarter circle of radius $\frac12$, so the success probability is $1-\frac{\pi}{16}$. On opposite sides, the distance is always at least $1$, so the success probability is $1$. Therefore the desired probability is $$\begin{aligned} &\frac14\cdot\frac14 \\ &\quad {}+\frac12\left(1-\frac{\pi}{16}\right) \\ &\quad {}+\frac14=\frac{26-\pi}{32}. \end{aligned}$$ Hence $a+b+c=26+1+32=59$. Thus, A is the correct answer.