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 "2008 AMC 12B Problems/Problem 21"

(→‎Solution 1)
(→‎Solution 2)
Line 10: Line 10:
  
 
==Solution 2==
 
==Solution 2==
 +
Two circles intersect if the distance between their centers is less than the sum of their radii. In this problem, <math>A</math> and <math>B</math> intersect iff
 +
<math>\sqrt{(A_X-B_X)^2+(A_Y-B_Y)^2}\leq2 \Rightarrow (A_X-B_X)^2+1\leq4 \Rightarrow (A_X-B_X)\leq \sqrt{3}.</math>
 +
 +
In other words, the two chosen <math>x</math>-coordinates must differ by no more than <math>\sqrt{3}</math>. To find this probability, we divide the problem into cases:
 +
 +
1) <math>A_X</math> is on the interval <math>(0,2-\sqrt{3})</math>. The probability that <math>B_X</math> falls within the desired range for a given <math>A_X</math> is <math>A_X</math> (on the left) <math>+\sqrt{3}</math> (on the right) all over <math>2</math> (the range of possible values). The total probability for this range is the sum of all these probabilities of <math>B_X</math> (over the range of <math>A_X</math>) divided by the total range of <math>A_X</math> (which is <math>2</math>). Thus, the total probability for this interval is <cmath>\frac{1}{2}\left(\int_{0}^{2-\sqrt{3}} \frac{x+\sqrt{3}}{2}\,dx\right) =\frac{1}{2}\left(x^2/4+\frac{x\sqrt{3}}{2} \Big |_{0}^{2-\sqrt{3}}\right) </cmath> <cmath>= \frac{1}{2}(\frac{4-4\sqrt{3}+3}{4} + \sqrt{3}-\frac{3}{2}) =\frac{1}{8}.</cmath>
 +
2) <math>A_X</math> is on the interval <math>(2-\sqrt{3},\sqrt{3})</math>. In this case, any value of <math>B_X</math> will do, so the probability for the interval is simply <math>\frac{\sqrt{3}-(2-\sqrt{3})}{2} = \sqrt{3}-1</math>.
 +
 +
3) <math>A_X</math> is on the interval <math>(\sqrt{3},2)</math>. This is identical, by symmetry, to case 1.
 +
 +
The total probability is therefore <math>\sqrt{3}-1+\frac{1}{4} = \frac{4\sqrt{3}-3}{4} \Rightarrow \boxed{E}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2008|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:16, 15 July 2015

Problem

Two circles of radius 1 are to be constructed as follows. The center of circle $A$ is chosen uniformly and at random from the line segment joining $(0,0)$ and $(2,0)$. The center of circle $B$ is chosen uniformly and at random, and independently of the first choice, from the line segment joining $(0,1)$ to $(2,1)$. What is the probability that circles $A$ and $B$ intersect?

$\textbf{(A)} \; \frac {2 + \sqrt {2}}{4} \qquad \textbf{(B)} \; \frac {3\sqrt {3} + 2}{8} \qquad \textbf{(C)} \; \frac {2 \sqrt {2} - 1}{2} \qquad \textbf{(D)} \; \frac {2 + \sqrt {3}}{4} \qquad \textbf{(E)} \; \frac {4 \sqrt {3} - 3}{4}$

Solution 1

Circles centered at $A$ and $B$ will overlap if $A$ and $B$ are closer to each other than if the circles were tangent. The circles are tangent when the distance between their centers is equal to the sum of their radii. Thus, the distance from $A$ to $B$ will be $2$. Since $A$ and $B$ are separated by $1$ vertically, they must be separated by $\sqrt{3}$ horizontally. Thus, if $|A_x-B_x|<\sqrt{3}$, the circles intersect.

Now, plot the two random variables $A_x$ and $B_x$ on the coordinate plane. Each variable ranges from $0$ to $2$. The circles intersect if the variables are within $\sqrt{3}$ of each other. Thus, the area in which the circles don't intersect is equal to the total area of two small triangles on opposite corners, each of area $\frac{(2-\sqrt{3})^2}{2}$. We conclude the probability the circles intersect is:\[1-\frac{(2-\sqrt{3})^2}{4}=\boxed{\textbf{(E)}\frac{4\sqrt{3}-3}{4}}.\]

Solution 2

Two circles intersect if the distance between their centers is less than the sum of their radii. In this problem, $A$ and $B$ intersect iff $\sqrt{(A_X-B_X)^2+(A_Y-B_Y)^2}\leq2 \Rightarrow (A_X-B_X)^2+1\leq4 \Rightarrow (A_X-B_X)\leq \sqrt{3}.$

In other words, the two chosen $x$-coordinates must differ by no more than $\sqrt{3}$. To find this probability, we divide the problem into cases:

1) $A_X$ is on the interval $(0,2-\sqrt{3})$. The probability that $B_X$ falls within the desired range for a given $A_X$ is $A_X$ (on the left) $+\sqrt{3}$ (on the right) all over $2$ (the range of possible values). The total probability for this range is the sum of all these probabilities of $B_X$ (over the range of $A_X$) divided by the total range of $A_X$ (which is $2$). Thus, the total probability for this interval is \[\frac{1}{2}\left(\int_{0}^{2-\sqrt{3}} \frac{x+\sqrt{3}}{2}\,dx\right) =\frac{1}{2}\left(x^2/4+\frac{x\sqrt{3}}{2} \Big |_{0}^{2-\sqrt{3}}\right)\] \[= \frac{1}{2}(\frac{4-4\sqrt{3}+3}{4} + \sqrt{3}-\frac{3}{2}) =\frac{1}{8}.\] 2) $A_X$ is on the interval $(2-\sqrt{3},\sqrt{3})$. In this case, any value of $B_X$ will do, so the probability for the interval is simply $\frac{\sqrt{3}-(2-\sqrt{3})}{2} = \sqrt{3}-1$.

3) $A_X$ is on the interval $(\sqrt{3},2)$. This is identical, by symmetry, to case 1.

The total probability is therefore $\sqrt{3}-1+\frac{1}{4} = \frac{4\sqrt{3}-3}{4} \Rightarrow \boxed{E}$

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 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=2008_AMC_12B_Problems/Problem_21&oldid=71172"