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2002 AMC 12B Problems/Problem 14

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Problem

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 9\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 12\qquad\mathrm{(E)}\ 16

Solution

For any given pair of circles, they can intersect at most 2 times. Since there are {4\choose 2} = 6 pairs of circles, the maximum number of possible intersections is 6 \cdot 2 = 12. We can construct such a situation as below, so the answer is \mathrm{(D)}.

Image:2002_12B_AMC-14.png

See also

2002 AMC 12B (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
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