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.
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.
Answer:
Let's think step by step.

user
160
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.
test