Difference between revisions of "2009 AMC 12B Problems/Problem 22"
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− | Notice that if we "fix" the value of <math>k</math>, at, say <math>6</math>, then run through all of the values of <math>i</math>, change the <math>k</math> | + | Notice that if we "fix" the value of <math>k</math>, at, say <math>6</math>, then run through all of the values of <math>i</math>, change the value of <math>k</math> to <math>5</math>, and run through all of the values of <math>i</math> again, and so on until we exhaust all <math>49</math> combinations of <math>(k, i)</math> we get something like this: |
Revision as of 15:58, 31 July 2020
Problem
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?
Solution
Solution 1
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 .
Solution 2
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.
Solution 3
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
and then further rewritten
So there are possible sets of pairs & possible pairs possible sets of coordinates , and .
(Note: I have no idea if the above notation for summation w/ 2 indices is accurate, or even applicable in the context of this problem. If it is wrong, and someone who understands what this solution is saying could fix it, that would be great =))
See Also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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