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2008 AMC 12B Problems/Problem 21

Revision as of 13:16, 15 July 2015 by Daniellionyang (talk | contribs) (→‎Solution 2)

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

Solution 2

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

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