2007 AMC 12B Problems/Problem 20

Problem

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 $a+b+c+d$ ?

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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 is $2\times$ ). The area of 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 $c = 3d$ :

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

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 is $(a,b,c,d)=(3,1,3,9)$