The parallelogram bounded by $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $50$. If $a,b,c,d$ are positive integers, what is the smallest possible value of $c+d$?

1. Let $k=|a-b|$. The two area conditions give $\dfrac{(c-d)^2}{k}=18$ and $\dfrac{(c+d)^2}{k}=50$.
2. Dividing these equations shows $\dfrac{|c-d|}{c+d}=\dfrac35$, so $\{c,d\}=\{4m,m\}$ for some positive integer $m$.
3. Then $k=\dfrac{(c-d)^2}{18}=\dfrac{9m^2}{18}=\dfrac{m^2}{2}$. Since $k$ is an integer, $m$ must be even, so the least possible $c+d=5m$ is $10$.
4. This value is attained with $(c,d)=(8,2)$ and $(a,b)=(1,3)$, giving areas $36/2=18$ and $100/2=50$. The minimum is $\boxed{10}$.