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

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== Solution ==
 
== Solution ==
  
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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>.
  
 
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<center><asy>
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</asy></center>
 
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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
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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>
  
 
Thus, the answer is <math>56 + 225 = \boxed{281}.</math>
 
Thus, the answer is <math>56 + 225 = \boxed{281}.</math>
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== Solution 2 ==
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First, let's figure out <math>d(P) \geq \frac{1}{3}</math> which is<cmath>\left(\frac{3}{5}\right)^2=\frac{9}{25}.</cmath>Then, <math>d(P) \geq \frac{1}{5}</math> is a square inside <math>d(P) \geq \frac{1}{3}</math>, so<cmath>\left(\frac{1}{3}\right)^2=\frac{1}{9}.</cmath>Therefore, the probability that <math>\frac{1}{5}\le d(P)\le\frac{1}{3}</math> is<cmath>\frac{9}{25}-\frac{1}{9}=\frac{56}{225}</cmath>So, the answer is <math>56+225=\boxed{281}</math>
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== Solution 3 ==
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First, lets assume that point <math>P</math> is closest to a side <math>S</math> of the square. If it is <math>\frac{1}{5}</math> far from <math>S</math>, then it should be at least <math>\frac{1}{5}</math> from both the adjacent sides of <math>S</math> in the square. This leaves a segment of <math>1 - 2 \cdot \frac{1}{5} = \frac{3}{5}</math>. If the distance from <math>P</math> to <math>S</math> is <math>\frac{1}{3}</math>, then notice the length of the side-ways segment for <math>P</math> is <math>1 - 2 \cdot \frac{1}{3} = \frac{1}{3}</math>. Notice that as the distance from <math>P</math> to <math>S</math> increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides <math>\frac{3}{5}</math> and <math>\frac{1}{3}</math> with height <math>\frac{1}{3} - \frac{1}{5} = \frac{2}{15}</math>. This trapezoid has area (or probability for one side) <math>\frac{1}{2} \cdot \left(\frac{1}{3}+\frac{3}{5}\right)\cdot \frac{2}{15} = \frac{14}{225}</math>. Since the square has <math>4</math> sides, we multiply by <math>4</math>. Hence, the probability is <math>\frac{56}{225}</math>. The answer is <math>\boxed{281}</math>.              ~Saucepan_man02
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=1|num-a=3|n=II}}
 
{{AIME box|year=2010|num-b=1|num-a=3|n=II}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:54, 7 February 2023

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

See also

2010 AIME II (Problems • Answer Key • Resources)
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