Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AIME II Problems/Problem 2"
(→‎Solution)
Line 25: Line 25:
 
Thus, the answer is <math>56 + 225 = \boxed{281}.</math>
 
Thus, the answer is <math>56 + 225 = \boxed{281}.</math>
  
−
== Solution ==
+
== Solution 2 ==
 
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>
 
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>
  

Revision as of 21:04, 28 August 2021

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 $56+225=\boxed{281}$

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_2&oldid=161136"