2015 AMC 12A Problem 23

The second point is on the same side as the first with probability $\dfrac14,$ on the opposite side with probability $\dfrac14,$ and on an adjacent side with probability $\dfrac12.$ Opposite sides: the distance is at least $1 \ge \dfrac12$ always, probability $1.$ Same side: for points $(a, 0)$ and $(b, 0),$ the condition $|a - b| \ge \dfrac12$ has probability $\dfrac14.$ Adjacent sides: for points $(a, 0)$ and $(0, b),$ the condition $\sqrt{a^2 + b^2} \ge \dfrac12$ is the region outside a quarter-circle of radius $\dfrac12,$ with probability $1 - \dfrac14\pi\left(\dfrac12\right)^2 = 1 - \dfrac{\pi}{16}.$ The total probability is $$\begin{aligned} &\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 \\ &\quad {}+ \dfrac12\left(1 - \dfrac{\pi}{16}\right) \\ &\quad = \dfrac{26 - \pi}{32}. \end{aligned}$$ Thus $a + b + c = 26 + 1 + 32 = 59.$ Thus, the correct answer is A.