2009 AMC 12B Problems/Problem 22
From AoPSWiki
Contents |
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 ![[P]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/5/c/4/5c49b1bdcece7d212a29112164e298ae6b88c2ae.gif)
denote the area of the polygon 
. We can then compute:
![\begin{align*}[ABCD] & = [ADD'] + [DD'C'C] - [BB'C'C] - [ABB']\\& = \frac{kt^2}2 + \frac{s(s+2kt)}2 - \frac{t(2s+kt)}...](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/0/1/001f5b89b5af5a65be0d5dd04ff4fd7f46c88e60.gif)
See Also
| 2009 AMC 12B (Problems • 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 | ||
