Question

Let $S$ be a square of side length $1.$ Two points are chosen independently at random on the sides of $S.$ The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c},$ where $a,$ $b,$ and $c$ are positive integers with $\gcd(a,b,c)=1.$ What is $a+b+c?$

$59$

$60$

$61$

$62$

$63$

Solution:

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 $\frac14\cdot\frac14+\frac12\left(1-\frac{\pi}{16}\right)+\frac14=\frac{26-\pi}{32}.$ Hence $a+b+c=26+1+32=59$ .

Thus, A is the correct answer.

2015 AMC 10A Problem 25

Solution:

Problem 25 in Other Years