Difference between revisions of "2009 AMC 12B Problems/Problem 22"
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== Problem == | == Problem == | ||
− | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Parallelogram <math>ABCD</math> has area <math>1,\!000,\!000</math>. Vertex <math>A</math> is at <math>(0,0)</math> and all other vertices are in the first quadrant. Vertices <math>B</math> and <math>D</math> are lattice points on the lines <math>y = x</math> and <math>y = kx</math> for some integer <math>k > 1</math>, respectively. How many such parallelograms are there?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Parallelogram <math>ABCD</math> has area <math>1,\!000,\!000</math>. Vertex <math>A</math> is at <math>(0,0)</math> and all other vertices are in the first quadrant. Vertices <math>B</math> and <math>D</math> are lattice points on the lines <math>y = x</math> and <math>y = kx</math> for some integer <math>k > 1</math>, respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048</math> | <math>\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048</math> | ||
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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 === | ||
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</cmath> | </cmath> | ||
The remainder of the solution is the same as the above. | The remainder of the solution is the same as the above. | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | We know that <math>A</math> is <math>(0, 0)</math>. Since <math>B</math> is on the line <math>y=x</math>, let it be represented by the point <math>(b, b)</math>. Similarly, let <math>D</math> be <math>(d, kd)</math>. Since this is a parallelogram, sides <math>\overline{AD}</math> and <math>\overline{BC}</math> are parallel. Therefore, the distance and relative position of <math>D</math> to <math>A</math> is equivalent to that of <math>C</math> to <math>B</math> (if we take the translation of <math>A</math> to <math>D</math> and apply it to <math>B</math>, we will get the coordinates of <math>C</math>). This yields <math>C (b+d, b+kd)</math>. Using the [[Shoelace Theorem]] we get | ||
+ | |||
+ | |||
+ | <math>1,000,000 = \frac{1}{2}|\left(b(b+kd) + (b+d)(kd)\right) - \left(b(b+d) + (b+kd)d\right)|</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow 2,000,000 = |2kbd - 2bd|</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow 1,000,000 = |kbd - bd|</math> | ||
+ | |||
+ | |||
+ | Since <math>k > 1, kbd > bd</math>. The equation becomes | ||
+ | |||
+ | |||
+ | <math>1,000,000 = (k-1)bd</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow \frac{1,000,000}{k-1} = bd</math> | ||
+ | |||
+ | |||
+ | Since <math>k</math> must be a positive integer greater than <math>1</math>, we know <math>k-1</math> will be a positive integer. We also know that <math>bd</math> is an integer, so <math>k-1</math> must be a factor of <math>1,000,000</math>. Therefore <math>bd</math> will also be a factor of <math>1,000,000</math>. | ||
+ | |||
+ | Notice that <math>1,000,000 = 10^6 = 2^6*5^6</math>. | ||
+ | |||
+ | Let <math>b</math> be <math>2^x5^y</math> such that <math>x, y</math> are integers on the interval <math>[0, 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 <math>(7-x)(7-y)</math> possibilities. So we want | ||
+ | |||
+ | |||
+ | <math>\sum_{k=0, i=0}^6 (7-k)(7-i)</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 <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: | ||
+ | |||
+ | |||
+ | <math>1*1 + 1*2 + ... + 1*6 + 1*7 + 2*1 + 2*2 + ... + 2*6 + 2*7 + ..... + 7*1 + 7*2 + ... + 7*6 + 7*7</math> | ||
+ | |||
+ | |||
+ | which can be rewritten | ||
+ | |||
+ | |||
+ | <math>1(1+2+...+7)+2(1+2+...+7)+.....+7(1+2+...+7)</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow (1+2+...+7)(1+2+...+7)</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow 28^2</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow 784</math> | ||
+ | |||
+ | |||
+ | So there are <math>784</math> possible sets of coordinates <math>B,</math> <math>C</math>, and <math>D \Rightarrow \boxed{\text{C}}</math>. | ||
+ | |||
+ | Note: this solution could be greatly simplified by using the Shoelace Formula on the triangle <math>ABD</math>, which we know has half the area of the parallelogram. This eliminates the need to find the coordinates of point <math>C</math>. | ||
+ | |||
+ | (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) | ||
== See Also == | == See Also == |
Latest revision as of 22:25, 9 October 2021
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? (A lattice point is any point whose coordinates are both integers.)
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
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)
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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