Difference between revisions of "2009 AMC 12B Problems/Problem 22"
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Let <math>d</math> be <math>2^w5^z</math> such that <math>w, z</math> are integers, <math>x+w \le 6</math>, and <math>y+z \le 6</math>. | Let <math>d</math> be <math>2^w5^z</math> such that <math>w, z</math> are integers, <math>x+w \le 6</math>, and <math>y+z \le 6</math>. | ||
− | For a pair <math>(x, y)</math>, there are <math>7-x</math> possibilities for <math>w</math> and <math>7-y</math> possibilites for <math>z</math> (<math>d</math> doesn't have to be the co-factor of <math>1,000,000</math>, it just can't be big enough such that <math>bd > 1,000,000</math>), for a total of (7-x)(7-y) possibilities. So we want | + | For a pair <math>(x, y)</math>, there are <math>7-x</math> possibilities for <math>w</math> and <math>7-y</math> possibilites for <math>z</math> (<math>d</math> doesn't have to be the co-factor of <math>1,000,000</math>, it just can't be big enough such that <math>bd > 1,000,000</math>), for a total of <math>(7-x)(7-y)</math> possibilities. So we want |
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− | <math>\left(\text{the sum of the number of possible pairs}(w, z)\text{for all pairs}(k , i)\text{for}k[0, 6]\text{and}i[0, 6]\right)</math> | + | <math>\left(\text{the sum of the number of possible pairs}(w, z) \text{ for all pairs}(k , i)\text{ for } k[0, 6]\text{ and } i[0, 6]\right)</math> |
− | Notice that if we "fix" the value of k, at, say <math>6</math>, then run through all of the values of <math>i</math>, change the <math>k</math> value 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: | + | 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> value 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 16:58, 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 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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