2009 AMC 12B Problems/Problem 22
Parallelogram has area . Vertex is at and all other vertices are in the first quadrant. Vertices and are lattice points on the lines and for some integer , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
The area of any parallelogram can be computed as the size of the vector product of and .
In our setting where , , and this is simply .
In other words, we need to count the triples of integers where , and .
These can be counted as follows: We have identical red balls (representing powers of ), blue balls (representing powers of ), and three labeled urns (representing the factors , , and ). The red balls can be distributed in ways, and for each of these ways, the blue balls can then also be distributed in ways. (See Distinguishability for a more detailed explanation.)
Thus there are exactly ways how to break into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is .
Without the vector product the area of can be computed for example as follows: If and , then clearly . Let , and be the orthogonal projections of , , and onto the axis. Let denote the area of the polygon . We can then compute:
The remainder of the solution is the same as the above.
We know that is . Since is on the line , let it be represented by the point . Similarly, let be . Since this is a parallelogram, sides and are parallel. Therefore, the distance and relative position of to is equivalent to that of to (if we take the translation of to and apply it to , we will get the coordinates of ). This yields . Using the Shoelace Theorem we get
Since . The equation becomes
Since must be a positive integer greater than , we know will be a positive integer. We also know that is an integer, so must be a factor of . Therefore will also be a factor of .
Notice that .
Let be such that are integers on the interval .
Let be such that are integers, , and .
For a pair , there are possibilities for and possibilites for ( doesn't have to be the co-factor of , it just can't be big enough such that ), for a total of possibilities. So we want
Notice that if we "fix" the value of , at, say , then run through all of the values of , change the value of to , and run through all of the values of again, and so on until we exhaust all combinations of we get something like this:
which can be rewritten
So there are possible sets of coordinates , and .
Note: this solution could be greatly simplified by using the Shoelace Formula on the triangle , which we know has half the area of the parallelogram. This eliminates the need to find the coordinates of point .
(Notational note: I'm not sure if the notation for double index summation is correct or even applicable in the context of this problem. If someone could fix the notation so that it is correct, or replace it without changing the general content of this solution, that would be great. If the notation is correct, then just delete this footnote)
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