solution: To save the most rope, we must have $HP$ having minimum length.
  For $HP$ to have minimum length, $HP$ must be perpendicular to $CD$. [asy]
  pair C, D, H, P;
  H=(90,120);
  C=(0,0);
  D=(140,0);
  P=(90,0);
  draw(H--C--D--H--P);
  label("H", H, N);
  label("C", C, SW);
  label("D", D, SE);
  label("P", P, S);
  label("150", (C+H)/2, NW);
  label("130", (D+H)/2, NE);
  [/asy] (Among other things, we can see from this diagram that sliding $P$ away from the perpendicular position does make $HP$ longer.)
  In the diagram, $HC=150$, $HD=130$ and $CD=140$.
  Let $HP=x$ and $PD=a$.  Then $CP=140-a$.
  By the Pythagorean Theorem in $\triangle HPC$, $x^2 + (140-a)^2 = 150^2$.
  By the Pythagorean Theorem in $\triangle HPD$, $x^2+a^2 = 130^2$.
  Subtracting the second equation from the first, we obtain \begin{align*}
  (140-a)^2 - a^2 & = 150^2 - 130^2 \\
  (19600 - 280a+a^2)-a^2 & = 5600 \\
  19600 -280a & = 5600 \\
  280a & = 14000 \\
  a & = 50
  \end{align*} Therefore, $x^2 + 90^2 = 150^2$ or $x^2 = 150^2 - 90^2 = 22500 - 8100 = 14400$ so $x =120$.
  So the shortest possible rope that we can use is 120 m, which saves $130+150-120 = \boxed{160}$ m of rope.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Five points $A$, $B$, $C$, $D$, and $O$ lie on a flat field.  $A$ is directly north of $O$, $B$ is directly west of $O$, $C$ is directly south of $O$, and $D$ is directly east of $O$. The  distance between $C$ and $D$ is 140 m.  A hot-air balloon is positioned in the air at $H$ directly above $O$. The balloon is held in place by four ropes $HA$, $HB$, $HC$, and $HD$.  Rope $HC$ has length 150 m and rope $HD$ has length 130 m. [asy]
  size(250);
  pair A, B, C, D, O, H, W, X, Y, Z;
  O=(0,0);
  A=(1,1);
  D=(1.5,-.3);
  B=(-1.5,.3);
  C=(-1,-1);
  H=(0,2.5);
  W=(5/3)*(A+D);
  X=(5/3)*(A+B);
  Y=(-1)*(W);
  Z=(-1)*(X);
  draw(W--X--Y--Z--W);
  draw(A--C);
  draw(B--D);
  draw(O--H, linewidth(1));
  draw(A--H, dashed);
  draw(B--H, dashed);
  draw(C--H, dashed);
  draw(D--H, dashed);
  dot(A);
  dot(B);
  dot(C);
  dot(D);
  dot(O);
  dot(H);
  label("A", A, NE);
  label("B", B, SW);
  label("C", C, SE);
  label("D", D, NE);
  label("O", O, SE);
  label("H", H, NW);
  [/asy]
  
  To reduce the total length of rope used, rope $HC$ and rope $HD$ are to be replaced by a single rope $HP$ where $P$ is a point on the straight line between $C$ and $D$. (The balloon remains at the same position $H$ above $O$ as described above.)  Determine the greatest length of rope that can be saved.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 160
extra_info.question: Five points $A$, $B$, $C$, $D$, and $O$ lie on a flat field.  $A$ is directly north of $O$, $B$ is directly west of $O$, $C$ is directly south of $O$, and $D$ is directly east of $O$. The  distance between $C$ and $D$ is 140 m.  A hot-air balloon is positioned in the air at $H$ directly above $O$. The balloon is held in place by four ropes $HA$, $HB$, $HC$, and $HD$.  Rope $HC$ has length 150 m and rope $HD$ has length 130 m. [asy]
  size(250);
  pair A, B, C, D, O, H, W, X, Y, Z;
  O=(0,0);
  A=(1,1);
  D=(1.5,-.3);
  B=(-1.5,.3);
  C=(-1,-1);
  H=(0,2.5);
  W=(5/3)*(A+D);
  X=(5/3)*(A+B);
  Y=(-1)*(W);
  Z=(-1)*(X);
  draw(W--X--Y--Z--W);
  draw(A--C);
  draw(B--D);
  draw(O--H, linewidth(1));
  draw(A--H, dashed);
  draw(B--H, dashed);
  draw(C--H, dashed);
  draw(D--H, dashed);
  dot(A);
  dot(B);
  dot(C);
  dot(D);
  dot(O);
  dot(H);
  label("A", A, NE);
  label("B", B, SW);
  label("C", C, SE);
  label("D", D, NE);
  label("O", O, SE);
  label("H", H, NW);
  [/asy]
  
  To reduce the total length of rope used, rope $HC$ and rope $HD$ are to be replaced by a single rope $HP$ where $P$ is a point on the straight line between $C$ and $D$. (The balloon remains at the same position $H$ above $O$ as described above.)  Determine the greatest length of rope that can be saved.
extra_info.split: test