Question

A point $P$ is chosen at random in the interior of a unit square $S.$ Let $d(P)$ denote the distance from $P$ to the closest side of $S.$ The probability that $\frac{1}{5} \le d(P) \le \frac{1}{3}$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

The points with $d(P) \ge t$ form a concentric square of side $1 - 2t.$ So $d(P) \ge \frac{1}{5}$ puts $P$ inside the concentric square of side $\frac{3}{5},$ and $d(P) \le \frac{1}{3}$ keeps $P$ outside the open concentric square of side $\frac{1}{3}.$

Since the unit square has area $1,$ the probability is the area between those two squares: $\left(\frac{3}{5}\right)^2 - \left(\frac{1}{3}\right)^2 = \frac{9}{25} - \frac{1}{9} = \frac{81 - 25}{225} = \frac{56}{225}.$ Thus $m + n = 56 + 225 = 281.$

2010 AIME II Problem 2

Solution:

Problem 2 in Other Years