2002 AMC 12B Problems/Problem 14

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 ${4\choose 2} = 6$ circles, the maximum number of possible intersections is $6 \cdot 2 = 12$ . We can construct such a situation as below, so the answer is $\mathrm{(D)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2002 AMC 12B Problems/Problem 14

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