Difference between revisions of "2009 AMC 12B Problems/Problem 22"
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These can be counted as follows: We have <math>6</math> identical red balls (representing powers of <math>2</math>), <math>6</math> blue balls (representing powers of <math>5</math>), and three labeled urns (representing the factors <math>k-1</math>, <math>s</math>, and <math>t</math>). The red balls can be distributed in <math>{8\choose 2} = 28</math> ways, and for each of these ways, the blue balls can then also be distributed in <math>28</math> ways. (See [[Distinguishability]] for a more detailed explanation.) | These can be counted as follows: We have <math>6</math> identical red balls (representing powers of <math>2</math>), <math>6</math> blue balls (representing powers of <math>5</math>), and three labeled urns (representing the factors <math>k-1</math>, <math>s</math>, and <math>t</math>). The red balls can be distributed in <math>{8\choose 2} = 28</math> ways, and for each of these ways, the blue balls can then also be distributed in <math>28</math> ways. (See [[Distinguishability]] for a more detailed explanation.) | ||
− | Thus there are exactly <math>28^2 = 784</math> ways how to break <math>1,\!000,\!000</math> into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is <math>\boxed{ | + | Thus there are exactly <math>28^2 = 784</math> ways how to break <math>1,\!000,\!000</math> into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is <math>784 \longrightarrow \boxed{C}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 00:26, 4 February 2019
Contents
[hide]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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.