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2007 AMC 12B Problems/Problem 20

Revision as of 09:57, 15 October 2008 by 1=2 (talk | contribs) (New page: ==Problem== The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y=ax+d</math>, <math>y=bx+c</math>, and <math>y=bx+d</math> has area <math>18</math>. The parallelogram bounde...)
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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

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See also

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