2008 AMC 12B Problems/Problem 21

Problem 21

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

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 $A_X (2)$ . Thus, the total possibility for this interval is $\frac{1}{2}(\int_{0}^{2-\sqrt{3}} \frac{x+\sqrt{3}}{2}\,dx\}) =\frac{1}{2}(x^2/4+\frac{x\sqrt{3}}{2} \Big |_{x=0}^{x=2-\sqrt{3}}) = \frac{1}{2}(\frac{4-4\sqrt{3}+3}{4} + \sqrt{3}-\frac{3}{2}) = .125$ .

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 $\sqrt{3}-1+.25 = \frac{4\sqrt{3}-3}{4}$ , which is answer choice E.

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2008 AMC 12B Problems/Problem 21

Problem 21

Solution

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