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​∣∣∣∣∣∣∣​x1​y1​1x2​y2​1x3​y3​1​∣∣∣∣∣∣∣​=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