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Difference between revisions of "2010 AIME II Problems/Problem 2"
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==Problem==
 
== Problem 2 ==
 
== Problem 2 ==
 
A point <math>P</math> is chosen at random in the interior of a unit square <math>S</math>. Let <math>d(P)</math> denote the distance from <math>P</math> to the closest side of <math>S</math>. The probability that <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is equal to <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
A point <math>P</math> is chosen at random in the interior of a unit square <math>S</math>. Let <math>d(P)</math> denote the distance from <math>P</math> to the closest side of <math>S</math>. The probability that <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is equal to <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.

Revision as of 14:46, 9 August 2018

Problem

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 have the same center as the unit square and inside the square side length $\frac{3}{5}$ that have the same center 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 $\boxed{281}$

See also

2010 AIME II (Problems • Answer Key • Resources)
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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