Difference between revisions of "2008 AMC 12B Problems/Problem 21"

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

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 $\sqrt{3}-1+\frac{1}{4} = \frac{4\sqrt{3}-3}{4} \Rightarrow \boxed{E}$

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 ==Solution 2==
-Circles centered at <math>A</math> and <math>B</math> will overlap if <math>A</math> and <math>B</math> 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 <math>A</math> to <math>B</math> will be <math>2</math>. Since <math>A</math> and <math>B</math> are separated by <math>1</math> vertically, they must be separated by <math>\sqrt{3}</math> horizontally. Thus, if <math>|A_x-B_x|<\sqrt{3}</math>, the circles intersect.
-Now, plot the two random variables <math>A_x</math> and <math>B_x</math> on the coordinate plane. Each variable ranges from <math>0</math> to <math>2</math>. The circles intersect if the variables are within <math>\sqrt{3}</math> 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 <math>\frac{(2-\sqrt{3})^2}{2}</math>. We conclude the probability the circles intersect is:<cmath>1-\frac{(2-\sqrt{3})^2}{4}=\boxed{\textbf{(E)}\frac{4\sqrt{3}-3}{4}}.</cmath>
 ==See Also==
 {{AMC12 box|year=2008|ab=B|num-b=20|num-a=22}}
 {{MAA Notice}}

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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Difference between revisions of "2008 AMC 12B Problems/Problem 21"

Revision as of 12:16, 15 July 2015

Contents

Problem

Solution 1

Solution 2

See Also