1992 AMC 12 Problem 22

Attempt Problem 22 of the 1992 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 1992 AMC 12 solutions, or check the answer key.

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

22.

Ten points are selected on the positive xx-axis, X+,\mathrm{X}^+, and five points are selected on the positive yy-axis, Y+.\mathrm{Y}^+. The fifty segments connecting the ten selected points on X+\mathrm{X}^+ to the five selected points on Y+\mathrm{Y}^+ are drawn. What is the maximum possible number of points of intersection of these fifty segments that could lie in the interior of the first quadrant?
250250
450450
500500
12501250
25002500
Answer: B
Concepts:combinatorial geometryintersectionsbinomial coefficient
Small hint:
An interior crossing is determined by choosing two points from each axis
Big hint:
For each selected pair on each axis, exactly one pair of the four connecting segments crosses
Solution:
Choose two of the 1010 points on the xx-axis and two of the 55 points on the yy-axis. Among the four connecting segments, the two with reversed endpoint order cross exactly once. The points can be chosen so no three segments meet at one interior point, so the maximum is (102)(52)=45⋅10=450. \binom{10}{2}\binom{5}{2}=45\cdot10=450.
Thus the correct answer is B.
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