2002 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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Concepts:counting intersections counting pairs

Difficulty rating: 1280

18.

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

$8$

$9$

$10$

$12$

$16$

Solution:

Any two distinct circles intersect in at most $2$ points. There are $\binom{4}{2} = 6$ pairs of circles, giving at most $6\cdot 2 = 12$ intersection points.

This maximum is achievable by a configuration where every pair of circles crosses twice, so the answer is $12.$

Thus, the correct answer is D.

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