AMC10 2002 B
AMC10 2002 B · Q18
AMC10 2002 B · Q18. It mainly tests Combinations, Circle theorems.
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
平面内画四个不同的圆，最多有多少个至少有两个圆相交的点？
(A)
8
8
(B)
9
9
(C)
10
10
(D)
12
12
(E)
16
16
Answer
Correct choice: (D)
正确答案：（D）
Solution
(D) Each pair of circles has at most two intersection points. There are $\binom{4}{2}=6$ pairs of circles, so there are at most $6\times 2=12$ points of intersection. The following configuration shows that 12 points of intersection are indeed possible:
（D）每一对圆最多有两个交点。有 $\binom{4}{2}=6$ 对圆，因此最多有 $6\times 2=12$ 个交点。下面的构型表明，确实可以达到 12 个交点：
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