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 $a+b+c+d$ ?

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

Solution

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 $c=3d$ . From the 4 corners of the parallelogram, we have that $(c-d)\left(\frac{bc-ad}{a-b}\right)=18$ . Substituting $c=3d$ ,

$\begin{eqnarray*} 2d\left(\frac{3bd-ad}{a-b}\right) &=& 18 \\ d\left(\frac{bd-ad}{a-b}+\frac{2bd}{a-b}\right) &=& 9 \\ d\left(-d+\frac{2bd}{a-b}\right) &=& 9 \\ -d^2 + \frac{2b}{a-b}d^2 &=& 9 \\ \frac{2b}{a-b}d^2 &=& 9 + d^2 \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

Taking $d=1$ , we have that $\frac{2b}{a-b} = 10$ . This gives us that $6b=5a$ . Taking the minimum values of $a=6,b=5$ we have that $a+b+c+d=6+5+3+1=15 \Rightarrow \mathrm {(C)}$

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2007 AMC 12B Problems/Problem 20

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