solution: Two vertices of the first parallelogram are at $(0,c)$ and $(0,d)$.
  
  [asy]
  unitsize(0.5 cm);
  
  pair P, Q, R, S;
  
  P = (0,9);
  Q = (3,12);
  R = (0,3);
  S = (-3,0);
  
  draw(interp(P,Q,-0.4)--interp(P,Q,1.4));
  draw(interp(R,S,-0.4)--interp(R,S,1.4));
  draw(interp(P,S,-0.2)--interp(P,S,1.2));
  draw(interp(Q,R,-0.2)--interp(Q,R,1.2));
  
  label("$y = ax + c$", interp(S,R,1.4), E);
  label("$y = ax + d$", interp(P,Q,1.4), E);
  label("$y = bx + c$", interp(Q,R,1.2), SE);
  label("$y = bx + d$", interp(P,S,1.2), SE);
  
  dot("$(0,c)$", R, SE);
  dot("$(0,d)$", P, NW);
  [/asy]
  
  The $x$-coordinates of the other two vertices satisfy $ax+c=bx+d$ and $ax+d=bx+c$, so the $x$-coordinates are $\pm(c-d)/(b-a)$. Thus the parallelogram is composed of two triangles, each of which has area \[
  9=\frac{1}{2} \cdot |c-d| \cdot \left|\frac{c-d}{b-a}\right|.
  \]It follows that $(c-d)^2=18|b-a|$.
  
  By a similar argument using the second parallelogram, $(c+d)^2=72|b-a|$. Subtracting the first equation from the second yields $4cd=54|b-a|$, so $2cd = 27|b-a|$. Thus $|b-a|$ is even, and $a+b$ is minimized when $\{a,b\}=\{1,3\}$. Also, $cd$ is a multiple of 27, and $c+d$ is minimized when $\{c,d\}=\{3,9\}$. Hence the smallest possible value of $a+b+c+d$ is $1+3+3+9=\boxed{16}$. Note that the required conditions are satisfied when $(a,b,c,d)=(1,3,3,9)$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  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 $a+b+c+d$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 16
extra_info.question: 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 $a+b+c+d$?
extra_info.split: test