1976 AMC 12 Problem 28

Attempt Problem 28 of the 1976 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1976 AMC 12 solutions, or check the answer key.

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28.

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 L4n−3,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
Answer: B
Concepts:counting intersectionscombinationsdouble counting
Small hint:
Start with (1002)\binom{100}{2} intersections in general position
Big hint:
Remove the pairs among the 2525 parallel lines and collapse the pairs among the 2525 concurrent lines to one point
Solution:
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 4950−300−299=4351. 4950-300-299=4351.
Therefore, the correct answer is B.
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