2009 AMC 12B Problems/Problem 22

Problem

Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there?

$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$

Solution

Solution 1

The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$ .

In our setting where $A=(0,0)$ , $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$ .

In other words, we need to count the triples of integers $(k,s,t)$ where $k>1$ , $s,t>0$ and $(k-1)st = 1,\!000,\!000 = 2^6 5^6$ .

These can be counted as follows: We have $6$ identical red balls (representing powers of $2$ ), $6$ blue balls (representing powers of $5$ ), and three labeled urns (representing the factors $k-1$ , $s$ , and $t$ ). The red balls can be distributed in ${8\choose 2} = 28$ ways, and for each of these ways, the blue balls can then also be distributed in $28$ ways. (See Distinguishability for a more detailed explanation.)

Thus there are exactly $28^2 = 784$ ways how to break $1,\!000,\!000$ into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is $784 \longrightarrow \boxed{C}$ .

Solution 2

Without the vector product the area of $ABCD$ can be computed for example as follows: If $B=(s,s)$ and $D=(t,kt)$ , then clearly $C=(s+t,s+kt)$ . Let $B'=(s,0)$ , $C'=(s+t,0)$ and $D'=(t,0)$ be the orthogonal projections of $B$ , $C$ , and $D$ onto the $x$ axis. Let $[P]$ denote the area of the polygon $P$ . We can then compute:

$\begin{align*} [ABCD] & = [ADD'] + [DD'C'C] - [BB'C'C] - [ABB'] \\ & = \frac{kt^2}2 + \frac{s(s+2kt)}2 - \frac{t(2s+kt)}2 - \frac{s^2}2 \\ & = kst - st \\ & = (k-1)st. \end{align*}$ The remainder of the solution is the same as the above.

Solution 3

We know that $A$ is $(0, 0)$ . Since $B$ is on the line $y=x$ , let it be represented by the point $(b, b)$ . Similarly, let $D$ be $(d, kd)$ . Since this is a parallelogram, sides $\overline{AD}$ and $\overline{BC}$ are parallel. Therefore, the distance and relative position of $D$ to $A$ is equivalent to that $C$ to $B$ (if we take the translation of $A$ to $D$ and apply it to $B$ , we will get the coordinates of $C$ ). This yields $C (b+d, b+kd)$ . Using the Shoelace Theorem we get

$1,000,000 = \frac{1}{2}|\left(b(b+kd) + (b+d)(kd)\right) - \left(b(b+d) + (b+kd)d\right)|$

$\Rightarrow 2,000,000 = |2kbd - 2bd|$

$\Rightarrow 1,000,000 = |kbd - bd|$

Since $k > 1, kbd > bd$ . The equation becomes

$1,000,000 = (k-1)bd$

$\Rightarrow \frac{1,000,000}{k-1} = bd$

Since $k$ must be a positive integer greater than $1$ , we know $k-1$ will be a positive integer. We also know that $bd$ is an integer, so $k-1$ must be a factor of $1,000,000$ . Therefore $bd$ will also be a factor of $1,000,000$ .

Notice that $1,000,000 = 10^6 = 2^6*5^6$ .

Let $b$ be $2^x5^y$ such that $x, y$ are integers on the interval $[0, 6]$ .

Let $d$ be $2^w5^z$ such that $w, z$ are integers, $x+w \le 6$ , and $y+z \le 6$ .

For a pair $(x, y)$ , there are $7-x$ possibilities for $w$ and $7-y$ possibilites for $z$ ( $d$ doesn't have to be the co-factor of $1,000,000$ , it just can't be big enough such that $bd > 1,000,000$ ), for a total of (7-x)(7-y) possibilities. So we want

$\sum_{k=0, i=0}^6 (7-k)(7-i)$

$\left(\text{the sum of the number of possible pairs}(w, z)\text{for all pairs}(k , i)\text{for}k[0, 6]\text{and}i[0, 6]\right)$

Notice that if we "fix" the value of k, at, say $6$ , then run through all of the values of $i$ , change the $k$ value to $5$ , and run through all of the values of $i$ again, and so on until we exhaust all $49$ combinations of $(k, i)$ we get something like this:

$1*1 + 1*2 + ... + 1*6 + 1*7 + 2*1 + 2*2 + ... + 2*6 + 2*7 + ..... + 7*1 + 7*2 + ... + 7*6 + 7*7$

which can be rewritten

$1(1+2+...+7)+2(1+2+...+7)+.....+7(1+2+...+7)$

and then further rewritten

$(1+2+...+7)(1+2+...+7)$

$\Rightarrow 28^2 = 784$

So there are $784$ possible sets of pairs $(x, y)$ & $(w, z) \Rightarrow 784$ possible pairs $(b, d) \Rightarrow 784$ possible sets of coordinates $B,$ $C$ , and $D \Rightarrow \boxed{\text{C}}$ .

(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 =))

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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