2002 AMC 10B 第 18 题

Question

平面上画出四个不同的圆。至少两个圆相交的点最多有多少个？

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

两个不同的圆最多相交于

2

个点。

Two distinct circles meet in at most

2

points

先数圆的配对数，再乘以

2

。

Count the pairs of circles, then multiply by

2

任意两个不同圆最多相交于

2

个点。四个圆的配对数为

\binom{4}{2} = 6

，所以最多有

6\cdot 2 = 12

个交点。

这个最大值可以达到：安排四个圆，使每一对圆都相交于两个不同的点，就会得到

12

个交点。

所以正确答案是 D。

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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