1976 AMC 12 第 28 题

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28.

直线 L1L_1L2L_2\ldotsL100L_{100} 两两不同。所有直线 L4nL_{4n}(其中 nn 为正整数)彼此平行。所有直线 L4n3L_{4n-3}(其中 nn 为正整数)都经过给定点 AA。完整集合 {L1,L2,,L100}\{L_1,L_2,\ldots,L_{100}\} 中任意两条直线的交点数最大为

Lines L1,L_1, L2,L_2, ,\ldots, L100L_{100} are distinct. All lines L4n,L_{4n}, nn a positive integer, are parallel to each other. All lines L4n3,L_{4n-3}, nn a positive integer, pass through a given point A.A. The maximum number of points of intersection of pairs of lines from the complete set {L1,L2,,L100}\{L_1,L_2,\ldots,L_{100}\} is

43504350

43514351

49004900

49014901

98519851

答案:B
知识点:交点计数组合双重计数
难度评级:2320
小提示:

先从一般位置下的 (1002)\binom{100}{2} 个交点开始

Start with (1002)\binom{100}{2} intersections in general position

大提示:

去掉 2525 条平行线之间的线对,并把 2525 条共点直线之间的所有线对合并成一个点

Remove the pairs among the 2525 parallel lines and collapse the pairs among the 2525 concurrent lines to one point

解答:

能被 44 整除的下标有 2525 个,与 1(mod4)1\pmod4 同余的下标也有 2525 个。从 (1002)=4950\binom{100}{2}=4950 个交点出发,平行线组不产生交点,因此减去 (252)=300\binom{25}{2}=300。共点线组的 300300 对直线只产生一个点,而不是 300300 个不同的点,因此再减去 299299。其余交点都可以选成互不相同,所以最大值为 4950300299=4351 4950-300-299=4351\text{。}

所以正确答案是 B

There are 2525 indices divisible by 44 and 2525 congruent to 1(mod4).1\pmod4. Starting from (1002)=4950,\binom{100}{2}=4950, the parallel group contributes no intersections, removing (252)=300.\binom{25}{2}=300. The concurrent group’s 300300 pairs all give one point rather than 300300 distinct points, removing another 299.299. All remaining intersections can be chosen distinct, so the maximum is 4950300299=4351. 4950-300-299=4351.

Therefore, the correct answer is B.

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