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2008 AMC 12B Problems/Problem 21

Revision as of 13:16, 15 July 2015 by Daniellionyang (talk | contribs) (→‎Solution 1)

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

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AMC 12 Problems and Solutions

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