# 2010 AIME II Problems/Problem 2

## Problem 2

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 $m+n$.

## Solution

Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$. $[asy] unitsize(1mm); defaultpen(linewidth(.8pt)); draw((0,0)--(0,30)--(30,30)--(30,0)--cycle); draw((6,6)--(6,24)--(24,24)--(24,6)--cycle); draw((10,10)--(10,20)--(20,20)--(20,10)--cycle); fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray); fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white); [/asy]$

Since the area of the unit square is $1$, the probability of a point $P$ with $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares. $\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$

Thus, the answer is $56 + 225 = \boxed{281}.$

## Solution 2

First, let's figure out $d(P) \geq \frac{1}{3}$ which is $$\left(\frac{3}{5}\right)^2=\frac{9}{25}.$$Then, $d(P) \geq \frac{1}{5}$ is a square inside $d(P) \geq \frac{1}{3}$, so $$\left(\frac{1}{3}\right)^2=\frac{1}{9}.$$Therefore, the probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is $$\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$$So, the answer is $56+225=\boxed{281}$

## Solution 3

First, lets assume that point $P$ is closest to a side $S$ of the square. If it is $\frac{1}{5}$ far from $S$, then it should be at least $\frac{1}{5}$ from both the adjacent sides of $S$ in the square. This leaves a segment of $1 - 2 \cdot \frac{1}{5} = \frac{3}{5}$. If the distance from $P$ to $S$ is $\frac{1}{3}$, then notice the length of the side-ways segment for $P$ is $1 - 2 \cdot \frac{1}{3} = \frac{1}{3}$. Notice that as the distance from $P$ to $S$ increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides $\frac{3}{5}$ and $\frac{1}{3}$ with height $\frac{1}{3} - \frac{1}{5} = \frac{2}{15}$. This trapezoid has area (or probability for one side) $\frac{1}{2} \cdot \left(\frac{1}{3}+\frac{3}{5}\right)\cdot \frac{2}{15} = \frac{14}{225}$. Since the square has $4$ sides, we multiply by $4$. Hence, the probability is $\frac{56}{225}$. The answer is $\boxed{281}$. ~Saucepan_man02

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 