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 $\dfrac12$ is $\dfrac{a - b\pi}{c},$ where $a,$ $b,$ and $c$ are positive integers and $\gcd(a, b, c) = 1.$ What is $a + b + c?$

$59$

$60$

$61$

$62$

$63$

Solution:

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 $\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 + \dfrac12\left(1 - \dfrac{\pi}{16}\right) = \dfrac{26 - \pi}{32}.$ Thus $a + b + c = 26 + 1 + 32 = 59.$

Thus, the correct answer is A.

2015 AMC 12A Problem 23

Solution:

Problem 23 in Other Years