2007 AMC 12B Problem 20

Two vertices of the first parallelogram lie at $(0,c)$ and $(0,d),$ and the other two have $x$-coordinates $\pm\dfrac{c-d}{b-a}.$ Its area works out to $\dfrac{(c-d)^2}{|b-a|}=18.$ The same computation for the second gives $\dfrac{(c+d)^2}{|b-a|}=72.$ So $(c-d)^2=18|b-a|$ and $(c+d)^2=72|b-a|.$ Subtracting, $4cd=54|b-a|,$ i.e. $2cd=27|b-a|.$ Thus $|b-a|$ is even, so $a+b$ is smallest with $\{a,b\}=\{1,3\};$ and $cd$ is a multiple of $27,$ so $c+d$ is smallest with $\{c,d\}=\{3,9\}.$ These satisfy all conditions, giving $a+b+c+d=1+3+3+9$ $=16.$ Thus, the correct answer is D.