2010 AIME II Problems/Problem 2

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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 $m+n$.


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.


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 $56+225=\boxed{281}$

See also

2010 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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