2008 AMC 12B Problems/Problem 21
Contents
Problem
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 1
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
(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
Solution 2
Circles centered at and
will overlap if
and
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
to
will be
. Since
and
are separated by
vertically, they must be separated by
horizontally. Thus, if
, the circles intersect.
Now, plot the two random variables and
on the coordinate plane. Each variable ranges from
to
. The circles intersect if the variables are within
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
. We conclude the probability the circles intersect is:
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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