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?

$8$

$9$

$10$

$12$

$16$

答案：D

知识点：交点计数数对计数

难度评级：1280

解答：

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.

2002 AMC 10B 第 18 题

答案：D

难度评级：1280

解答：

其他年份的第 18 题