Difference between revisions of "2010 AIME II Problems/Problem 2"

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== Solution ==
 
== Solution ==
  
Any point outside the square with side length <math>\frac{1}{3}</math> that have the same center as the unit square and inside the square side length <math>\frac{3}{5}</math> that have the same center as the unit square has <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math>.
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Any point outside the square with side length <math>\frac{1}{3}</math> that has the same center and orientation as the unit square and inside the square with side length <math>\frac{3}{5}</math> that has the same center and orientation as the unit square has <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math>.
  
 
<center><asy>
 
<center><asy>
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</asy></center>
 
</asy></center>
  
Since the area of the unit square is <math>1</math>, the probability of a point <math>P</math> with <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is the area of the shaded region, which is the difference of the area of two squares
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Since the area of the unit square is <math>1</math>, the probability of a point <math>P</math> with <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is the area of the shaded region, which is the difference of the area of two squares.
  
 
<math>\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}</math>
 
<math>\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}</math>

Latest revision as of 13:36, 18 March 2020

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 $56 + 225 = \boxed{281}.$

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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