1952 AMC 12 第 31 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

31.

平面内有 1212 个点,其中任意三点不共线。这些点所确定的直线条数为:

Given 1212 points in a plane no three of which are collinear, the number of lines they determine is:

2424

5454

120120

6666

以上答案均不正确

None of these

答案:D
知识点:组合基本计数
难度评级:1150
小提示:

每条直线由这些点中的两个确定

Each line is determined by choosing two of the points

大提示:

任意三点不共线保证不同的点对确定不同的直线

The condition that no three are collinear ensures that different pairs determine different lines

解答:

每一对点确定一条直线;又因为任意三点不共线,所以没有一条直线会由多于一对点重复计数。因此,直线条数为 (122)=12112=66 \binom{12}{2}=\frac{12\cdot11}{2}=66\text{。}

因此,正确答案是 D

Every pair of points determines one line, and no line is counted by more than one pair because no three points are collinear. Therefore the number of lines is (122)=12112=66. \binom{12}{2}=\frac{12\cdot11}{2}=66.

Thus, the correct answer is D.

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