AMC12 2015 A

Answer (A): Let the square have vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$, and consider three cases. Case 1: The chosen points are on opposite sides of the square. In this case the distance between the points is at least $\frac12$ with probability $1$. Case 2: The chosen points are on the same side of the square. It may be assumed that the points are $(a,0)$ and $(b,0)$. The pairs of points in the $ab$-plane that meet the requirement are those within the square $0\le a\le 1,\ 0\le b\le 1$ that satisfy either $b\ge a+\frac12$ or $b\le a-\frac12$. These inequalities describe the union of two isosceles right triangles with leg length $\frac12$, together with their interiors. The area of the region is $\frac14$, and the area of the square is $1$, so the probability that the pair of points meets the requirement in this case is $\frac14$. Case 3: The chosen points are on adjacent sides of the square. It may be assumed that the points are $(a,0)$ and $(0,b)$. The pairs of points in the $ab$-plane that meet the requirement are those within the square $0\le a\le 1,\ 0\le b\le 1$ that satisfy $\sqrt{a^2+b^2}\ge \frac12$. These inequalities describe the region inside the square and outside a quarter-circle of radius $\frac12$. The area of this region is $1-\frac14\pi\left(\frac12\right)^2=1-\frac{\pi}{16}$, which is also the probability that the pair of points meets the requirement in this case. Cases 1 and 2 each occur with probability $\frac14$, and Case 3 occurs with probability $\frac12$. The requested probability is $$ \frac14\cdot 1+\frac14\cdot\frac14+\frac12\left(1-\frac{\pi}{16}\right)=\frac{26-\pi}{32}, $$ and $a+b+c=59$.