1992 AHSME Problems/Problem 22

Problem

Ten points are selected on the positive $x$ -axis, $X^+$ , and five points are selected on the positive $y$ -axis, $Y^+$ . The fifty segments connecting the ten points on $X^+$ to the five points on $Y^+$ 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?

$\text{(A) } 250\quad \text{(B) } 450\quad \text{(C) } 500\quad \text{(D) } 1250\quad \text{(E) } 2500$

Solution

$\fbox{B}$ We can pick any two points on the $x$ -axis and any two points on the $y$ -axis to form a quadrilateral, and the intersection of its diagonals will thus definitely be inside the first quadrant. Hence the answer is $10C2 \times 5C2 = 450$ .

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1992 AHSME Problems/Problem 22

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