Problem:
Parallelogram ABCD has area 1,000,000. Vertex A is at (0,0) and all other vertices are in the first quadrant. Vertices B and D are lattice points on the lines y=x and y=kx for some integer k>1, respectively. How many such parallelograms are there?
Answer Choices:
A. 49
B. 720
C. 784
D. 2009
E. 2048
Solution:
Let B=(b,b) and D=(d,kd), so C=(b+d,b+kd). Let E=(b+d,0) and F=(0,b+kd). Rectangle AECF is the disjoint union of parallelogram ABCD, two rectangles with length d and height b, two isosceles right triangles with leg length b, and two right triangles with leg lengths d and kd. It follows that the area of ABCD is
(b+d)(b+kd)−2bd−b2−kd2=(k−1)bd
Therefore each parallelogram with the required properties determines, and is determined by, an ordered triple (k−1,b,d) of positive integers whose product is 1,000,000=2656. The number of ways to distribute the six factors of 2 among the three integers k−1,b, and d is (3−16+3−1)=(28)=28. The six factors of 5 can also be distributed in 28 ways, so there are 282=784 parallelograms with the required property.
OR
The area of a triangle with vertices (x1,y1),(x2,y2),(x3,y3) is
21∣∣∣∣∣∣∣x1y11x2y21x3y31∣∣∣∣∣∣∣=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Thus the area of △ABD is 21(k−1)bd and the area of △CBD is the same. Then proceed as in the first solution.
The problems on this page are the property of the MAA's American Mathematics Competitions