2007 AMC 12B Problems/Problem 20


The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?

$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution 1

Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which gives $c=3d$ (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$, it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$, so substituting $c = 3d$:

\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\]

Thus $3|d$, and we verify that $d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$. Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$.

Solution 2

The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-c}{b-a}$, $\frac{(d-c)^2}{b-a}=18$, and the area contained by the latter is $\frac{(c+d)^2}{b-a}=72$. Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$. $c^2=18(b-a)$ so $c$ is a multiple of $3$. Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$, so the sum is $\boxed{\textbf{(D)} 16}$.

Solution 3

Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$. Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is $\frac{bh}{2}$, but since a parallelogram is two such triangles, the area becomes $bh$.

Let $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$, or $(c-d)$. Also notice that $h_1$ is the distance from the $x$-axis to the intersection of lines $ax+c$ and $bx+d$. This is equivalent to the value of the $x$-coordinate of intersection, so we solve for $x$:


$\Rightarrow bx-ax = c-d$

$\Rightarrow (b-a)x = c-d$

$\Rightarrow x = \frac{c-d}{b-a}$.

The area of the smaller parallelogram is $b_1*h_1$, or

$(c-d) * \frac{c-d}{b-a}$

$\Rightarrow \frac{(c-d)^2}{b-a}$.

$b_2$ is the distance from $(0, -d)$ to $(0, c)$, or $(c+d)$. $h_2$ is the $x$-coordinate of the intersection of the lines $ax+c$ and $bx-d$. Again, we solve for x:


$\Rightarrow bx-ax = c+d$

$\Rightarrow (b-a)x = c+d$

$\Rightarrow x = \frac{c+d}{b-a}$.

The area of the larger parallelogram is $b-1*h_1$, or

$(c+d) * \frac{c+d}{b-a}$

$\Rightarrow \frac{(c+d)^2}{b-a}$.

The areas of the parallelograms are given to us: $18$ and $72$. Therefore we can set up a ratio:

$\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}$

$\Rightarrow 18(c+d)^2 = 72(c-d)^2$

$\Rightarrow (c+d)^2 = 4(c-d)^2$

$\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2$

$\Rightarrow 3c^2 - 10cd +3d^2 = 0$

$\Rightarrow (3c-d)(c-3d)=0$

$\Rightarrow c=3d, c=\frac{d}{3}$

We established earlier that $c>d$, so $c=3d$. Plugging this into the intial equations yields

$\frac{16d^2}{b-a} = 72$


$\frac{4d^2}{b-a} = 18$

Solving for $d$, we get

$d = 3\sqrt{\frac{b-a}{2}}$

We want the sum of $a$, $b$, $c$, and $d$. We can now rewrite this

$a + b + 12\sqrt{\frac{b-a}{2}}$

We are told that $a$, $b$, $c$, and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this

$\frac{b-a}{2} = k^2$

where $k$ is some positive integer.

Rearranging, we get

$b = a + 2k^2$

Now we can rewrite the sum as

$a + a + 2k^2 +12k$.

Since both $a$ and $k$ must be at least $1$, the minimum value is

$1 + 1 + 2(1)^2 + 12(1) = 16 \Rightarrow \boxed{\text{D}}$.

See also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions

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