Question

对于平面上的四条互不相同的直线，恰有 $N$ 个不同的点同时位于其中两条或更多条直线上。所有可能的 $N$ 值之和是多少？

For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N?$

$14$

$16$

$18$

$19$

$21$

答案：D

知识点：交点计数分类讨论

难度评级：1380

小提示：

当直线处于一般位置时，最大值是 $\binom{4}{2} = 6$ 。

The maximum is $\binom{4}{2} = 6$ when the lines are in general position

大提示：

按平行类和共点情况枚举构型；检查从 $0$ 到 $6$ 中哪些计数能实现

Enumerate configurations using parallel classes and concurrences; check which counts from $0$ to $6$ are achievable

解答：

当四条直线全部平行或全部共点时，交点数分别为 $0$ 和 $1$ 。三条平行线被第四条截时有 $3$ 个交点。三条共点直线加上一条不经过公共点的直线时有 $4$ 个交点。一对平行且没有三线共点时有 $5$ 个交点；四条直线处于一般位置时有 $6$ 个交点。

数值 $2$ 不可能。两条直线在 $P$ 点相交后，第三条若不经过 $P$ ，就至少产生一个新交点；第四条不同的直线还必须产生另一个新交点。如果其余直线都经过 $P$ ，则只有一个交点。因此可实现的值为 $0,1,3,4,5,6$ ，它们的和为 $19$ 。

因此，正确答案是 D。

The values $0$ and $1$ occur when all four lines are parallel or all four are concurrent. Three parallel lines crossed by a fourth give $3$ points. Three concurrent lines together with a fourth line not through their common point give $4.$ One parallel pair with no three concurrent gives $5,$ and four lines in general position give $6.$

The value $2$ is impossible. Once two lines meet at $P,$ a third line not through $P$ creates at least one new intersection; a fourth distinct line must then create another new point. If every remaining line passes through $P,$ there is only one intersection point instead. Thus the achievable values are $0,1,3,4,5,6,$ whose sum is $19.$

Thus, the correct answer is D.

2019 AMC 12A 第 8 题

答案：D

难度评级：1380

小提示：

大提示：

解答：

其他年份的第 8 题