AMC12 2002 B

AMC12 2002 B · Q14

AMC12 2002 B · Q14. It mainly tests Combinations, Geometry misc.

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

平面上画了 4 个不同的圆。至少有两个圆相交的点的最大数目是多少？

(A) 8 8

(B) 9 9

(D) 12 12

(E) 16 16

Answer

Correct choice: (D)

正确答案：（D）

Solution

For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \cdot 2 = 12$. We can construct such a situation as below, so the answer is $\boxed{\mathrm{(D)}\ 12}$. Error creating thumbnail: Unable to save thumbnail to destination

任意一对圆最多相交 $2$ 次。由于共有 ${4\choose 2} = 6$ 对圆，可能的最大交点数为 $6 \cdot 2 = 12$。如下可以构造出这种情况，因此答案是 $\boxed{\mathrm{(D)}\ 12}$。创建缩略图出错：无法将缩略图保存到目标位置

Topics