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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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 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

Answer: B
Concepts:counting intersectionscombinationsdouble counting
Difficulty rating: 2320
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 4950300299=4351. 4950-300-299=4351.

Therefore, the correct answer is B.

← Problem 27#27
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