2008 AMC 12B Problems/Problem 21
Problem 21
Two circles of radius 1 are to be constructed as follows. The center of circle is chosen uniformly and at random from the line segment joining
and
. The center of circle
is chosen uniformly and at random, and independently of the first choice, from the line segment joining
to
. What is the probability that circles
and
intersect?
Solution
Two circles intersect if the distance between their centers is less than the sum of their radii. In this problem, and
intersect iff
In other words, the two chosen X-coordinates must differ by no more than . To find this probability, we divide the problem into cases:
1) is on the interval
. The probability that
falls within the desired range for a given
is
(on the left)
(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
(over the range of
) divided by the total range of
. Thus, the total possibility for this interval is
.
2) is on the interval
. In this case, any value of
will do, so the probability for the interval is simply
.
3) is on the interval
. This is identical, by symmetry, to case 1.
The total probability is therefore , which is answer choice 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 |