<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.
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem. So, we have a hot-air balloon at point H directly above point O. There are four ropes connecting H to points A, B, C, and D. Points A, B, C, D are in the cardinal directions north, west, south, and east of O. The distance between C and D is 140 meters. Ropes HC and HD have lengths 150 meters and 130 meters, respectively. The goal is to replace ropes HC and HD with a single rope HP, where P is somewhere on the line segment CD. We need to find the maximum length of rope that can be saved by doing this.

First, let me visualize the setup.
Since H is directly above O, the balloon is at some height h above O. Points C and D are on the same horizontal line (south and east of O, respectively), but wait, no—wait, C is directly south of O, and D is directly east of O. So, actually, C and D are on different axes. Wait, maybe the field is a plane with O at the origin, A north (positive y-axis), B west (negative x-axis), C south (negative y-axis), and D east (positive x-axis). So, C is at (0, -c, 0), D is at (d, 0, 0), but since they are directly south and east of O. But the distance between C and D is 140 meters. Wait, if C is directly south of O and D is directly east of O, then the straight-line distance between C and D would be the hypotenuse of a right triangle with legs equal to the distances from O to C and O to D.

<prompt_1.1> Calculate the distance between points C and D using the Pythagorean theorem. </prompt_1.1>

The balloon is at H, which is directly above O, so H has coordinates (0, 0, h). The ropes HA, HB, HC, HD are the distances from H to each of these points. So, for example, rope HC is the distance from H to C, which is the straight line from (0, 0, h) to (0, -c, 0). The length of HC is given as 150 meters. Similarly, HD is the distance from H to D, which is from (0, 0, h) to (d, 0, 0), and that's 130 meters.

So, we can use the distance formula in 3D. For HC: the distance is sqrt[(0 - 0)^2 + (0 - (-c))^2 + (h - 0)^2] = sqrt[c^2 + h^2] = 150. Similarly, for HD: sqrt[(d - 0)^2 + (0 - 0)^2 + (h - 0)^2] = sqrt[d^2 + h^2] = 130. Also, we know that the distance between C and D is 140 meters. Since C is at (0, -c, 0) and D is at (d, 0, 0), the distance CD is sqrt[(d - 0)^2 + (0 - (-c))^2 + (0 - 0)^2] = sqrt[d^2 + c^2] = 140.

So, we have three equations:

1. sqrt[c^2 + h^2] = 150 (from HC)
2. sqrt[d^2 + h^2] = 130 (from HD)
3. sqrt[c^2 + d^2] = 140 (distance between C and D)
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, no. Wait, if C is directly south of O, then C is at (0, -k, 0) if O is at (0, 0, 0). Similarly, D is directly east of O, so D is at (m, 0, 0). Then the distance between C and D would be sqrt[(m - 0)^2 + (0 - (-k))^2] = sqrt(m^2 + k^2) = 140 meters. But we also have the lengths of HC and HD. Let's denote the height of H above O as h. Then, the length HC is the distance from H (0, 0, h) to C (0, -k, 0), which is sqrt[(0)^2 + (k)^2 + (h)^2] = sqrt(k^2 + h^2) = 150. Similarly, HD is the distance from H (0, 0, h) to D (m, 0, 0), which is sqrt[(m)^2 + (0)^2 + (h)^2] = sqrt(m^2 + h^2) = 130.

So, we have three equations:

1. sqrt(k^2 + h^2) = 150
2. sqrt(m^2 + h^2) = 130
3. sqrt(m^2 + k^2) = 140
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Our unknowns here are k, m, and h. We need to solve for these variables.

Let me square all three equations to make them easier to handle.

<prompt_2.1> Solve the equations to find the relationship between k and m. </prompt_2.1>
-----

But I need to make sure that k and m are positive real numbers.

Alternatively, maybe there's another way to approach this problem without explicitly solving for k and m.

Wait, but perhaps we need to find h? Because the total length saved is (HC + HD) - HP, where HP is the length of the new rope. So, if we can express HP in terms of h, maybe we can find the maximum saved length.

Wait, but HP is the length of the rope from H to P, where P is on CD. Since CD is the line segment between C and D, which in coordinates is from (0, -k, 0) to (m, 0, 0). So, any point P on CD can be parametrized as P = (tm, (1 - t)(-k)), where t ranges from 0 to 1. Wait, actually, parametrizing the line segment from C to D. Let's see.

<prompt_2.2> Find the optimal point P on CD to minimize HP. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
1. k^2 + h^2 = 150^2 = 22500
2. m^2 + h^2 = 130^2 = 16900
3. m^2 + k^2 = 140^2 = 19600

So, from equations 1 and 2, we can subtract equation 2 from equation 1:

(k^2 + h^2) - (m^2 + h^2) = 22500 - 16900

Which simplifies to:

k^2 - m^2 = 5600

But from equation 3, we have m^2 + k^2 = 19600.

So now, we have two equations:

a. k^2 - m^2 = 5600

b. k^2 + m^2 = 19600

We can solve these two equations for k^2 and m^2. Adding equations a and b:

(k^2 - m^2) + (k^2 + m^2) = 5600 + 19600

2k^2 = 25200

So, k^2 = 12600

Therefore, k = sqrt(12600) = sqrt(100*126) = 10*sqrt(126) ≈ 10*11.225 = 112.25 meters

Similarly, subtracting equation a from equation b:

(k^2 + m^2) - (k^2 - m^2) = 19600 - 5600

2m^2 = 14000

So, m^2 = 7000

Therefore, m = sqrt(7000) = sqrt(100*70) = 10*sqrt(70) ≈ 10*8.366 = 83.66 meters

Now, we can find h from equation 1:

k^2 + h^2 = 22500

h^2 = 22500 - 12600 = 9900

So, h = sqrt(9900) = sqrt(100*99) = 10*sqrt(99) ≈ 10*9.949 = 99.49 meters
</execution_2.1>

<execution_2.2>
If C is at (0, -k) in 2D (since we can consider the horizontal plane), and D is at (m, 0), then the line segment CD can be parametrized as (tm, -k + t(k + 0)) for t from 0 to 1? Wait, no. Let's think. To go from C (0, -k) to D (m, 0), the parametric equations would be x = m*t, y = -k + (k)*t, where t goes from 0 to 1. Because starting at t=0, x=0, y=-k (point C), and at t=1, x=m, y=0 (point D). So, yes, P = (m*t, -k + k*t) for t in [0,1].

Then, the distance HP is the distance from H (0,0,h) to P (m*t, -k + k*t, 0). So, HP = sqrt[(m*t - 0)^2 + (-k + k*t - 0)^2 + (0 - h)^2] = sqrt[(m*t)^2 + (-k + k*t)^2 + h^2].

We need to find the value of t that minimizes HP, since we want to replace HC and HD with HP where P is on CD, and the goal is to save as much rope as possible. So, the original total rope length is HC + HD = 150 + 130 = 280 meters. The new rope length is HP. So, the saved rope is 280 - HP. To maximize the saved rope, we need to minimize HP.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, the problem is to replace ropes HC and HD (total length 150 + 130 = 280 meters) with a single rope HP, where P is on CD. We need to find the length of HP for some P on CD and then see how much we can save.

But HP is a rope from H to a point P on CD. The question is, where should P be located on CD so that HP is as short as possible? Because the maximum saving would occur when HP is minimized.

<prompt_3.1> Determine the optimal point P on CD to minimize HP. </prompt_3.1>
-----

Based on execution_2.2:
But how do we find the minimum HP?

Alternatively, maybe there's a geometric interpretation. Since H is fixed, and P is on CD, the minimal HP is the distance from H to the line CD in 3D space. Wait, that's a good point. The shortest distance from point H to the line CD is the perpendicular distance. So, if we can compute the perpendicular distance from H to line CD, that would be the minimal HP.

<prompt_3.2> Use the perpendicular distance from H to CD to find the minimum HP. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
However, H is fixed at (0,0,h), and CD is the line segment between C (0, -k, 0) and D (m, 0, 0). So, the problem reduces to finding the point P on the line segment CD that is closest to H, and then computing the distance from H to that point P.

Once we have that minimal HP, the saved rope would be 280 - HP. Then, we need to find the greatest possible saving, which corresponds to the minimal HP.
</execution_3.1>

<execution_3.2>
But wait, in 3D space, the minimal distance from a point to a line is indeed the perpendicular distance. So, if we compute that, that would give us the minimal possible HP, which is the maximum saving.

But let me confirm. If we can find a point P on CD such that HP is perpendicular to CD, then that P would give the minimal HP. Therefore, the minimal HP is the distance from H to the line CD. Therefore, the maximum saving is 280 meters minus this minimal HP.

But we need to check if such a point P exists on the segment CD. If the perpendicular from H to line CD falls within the segment CD, then that's the point. Otherwise, the minimal distance would be the distance from H to the nearest endpoint of CD, either C or D.

So, first, let's compute the perpendicular distance from H to line CD.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
To compute the perpendicular distance from point H(0,0,h) to the line CD, which goes through points C(0, -k, 0) and D(m, 0, 0).

The formula for the distance from a point to a line in 3D is |(P - C) × direction| / |direction|, where direction is the direction vector of the line.

<prompt_4.1> Compute the perpendicular distance from H to line CD. </prompt_4.1>
-----

Based on execution_3.1:
First, let's parametrize point P on CD. Let me parametrize CD with parameter t, where t ranges from 0 to 1. When t=0, we are at point C(0, -k, 0), and when t=1, we are at point D(m, 0, 0). So, the parametric equations for CD are:

<prompt_4.2> Find the value of t that minimizes HP. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
First, let's find the direction vector of CD. From C to D: D - C = (m - 0, 0 - (-k), 0 - 0) = (m, k, 0). So, the direction vector is (m, k, 0). The vector from C to H is H - C = (0 - 0, 0 - (-k), h - 0) = (0, k, h). Then, the cross product (H - C) × direction is:

|i    j    k|
|0    k    h|
|m    k    0|

Calculating determinant:

i*(k*0 - h*k) - j*(0*0 - h*m) + k*(0*k - k*m)

= i*(-hk) - j*(-hm) + k*(-km)

= (-hk, hm, -km)

The magnitude of this cross product is sqrt[ ( -hk )^2 + ( hm )^2 + ( -km )^2 ] = sqrt( h²k² + h²m² + k²m² )

The magnitude of the direction vector is sqrt(m² + k² + 0²) = sqrt(m² + k²) = 140, as given from equation 3.

Therefore, the distance from H to line CD is sqrt( h²k² + h²m² + k²m² ) / 140.

But we can simplify the numerator:

sqrt( h²(k² + m²) + k²m² ) = sqrt( h²*19600 + k²m² )

Wait, since from equation 3, m² + k² = 19600, so k² + m² = 19600. Also, h² = 9900 from earlier. So, h²(k² + m²) = 9900*19600. Then, k²m² is (k*m)^2. Let's compute that.

But first, we need to find k*m. We have k² = 12600, m² = 7000. So, k = sqrt(12600) ≈ 112.25, m = sqrt(7000) ≈ 83.66. So, k*m ≈ 112.25*83.66 ≈ 9428.85. Therefore, (k*m)^2 ≈ (9428.85)^2 ≈ 8.89*10^8. But this might not be exact. Let's see if we can find k*m exactly.
</execution_4.1>

<execution_4.2>
x = m*t

y = -k + k*t

z = 0

So, point P is (m*t, -k + k*t, 0). Then, the distance HP is the distance from (0,0,h) to (m*t, -k + k*t, 0):

HP(t) = sqrt[(m*t)^2 + (-k + k*t)^2 + h^2]

We need to find the value of t in [0,1] that minimizes HP(t). Since the square root function is monotonically increasing, minimizing HP(t) is equivalent to minimizing the square of HP(t):

f(t) = (m*t)^2 + (-k + k*t)^2 + h^2

Let's expand this:

f(t) = m²t² + ( -k + kt )² + h²

= m²t² + k² - 2k²t + k²t² + h²

Combine like terms:

f(t) = (m² + k²)t² - 2k²t + (k² + h²)

Now, to find the minimum of this quadratic function in t, we can take the derivative with respect to t and set it to zero.

df/dt = 2(m² + k²)t - 2k² = 0

Solving for t:

2(m² + k²)t = 2k²

t = k² / (m² + k²)

So, the value of t that minimizes HP is t = k² / (m² + k²)
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
We have:

k² = 12600

m² = 7000

So, k²*m² = 12600*7000 = 88,200,000

Therefore, the numerator inside the square root is:

h²(k² + m²) + k²m² = 9900*19600 + 88,200,000

Compute 9900*19600:

<prompt_5.1> Calculate the perpendicular distance from H to line CD using the above plan. </prompt_5.1>
-----

Based on execution_4.2:
Now, let's check if this t is within [0,1]. Since k² and m² are positive, t = k²/(m² + k²) is between 0 and 1. Therefore, the minimal HP occurs at this t.

So, substituting t back into the coordinates of P:

<prompt_5.2> Calculate the minimum HP and the maximum rope saved using the above plan. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
First, 100*19600 = 1,960,000, so 99*19600 = 1,960,000 - 19600 = 1,960,000 - 19,600 = 1,940,400. Then, 9900*19600 = 99*100*196*100 = (99*196)*(100*100) = (19,404)*(10,000) = 194,040,000. Wait, wait, maybe my initial calculation was wrong.

Wait, 9900*19600 = (9900*196)*100. Let's compute 9900*196:

First, 100*196 = 19,600

So, 9900 = 99*100, so 99*19,600 = (100 - 1)*19,600 = 1,960,000 - 19,600 = 1,940,400

Then multiply by 100: 1,940,400*100 = 194,040,000

Then, add 88,200,000:

194,040,000 + 88,200,000 = 282,240,000

Therefore, the numerator is sqrt(282,240,000) = sqrt(282,240,000). Let's compute that.

sqrt(282,240,000) = sqrt(28224 * 10,000) = sqrt(28224)*sqrt(10,000) = 168*100 = 16,800

Wait, sqrt(28224). Let me check: 168^2 = (170 - 2)^2 = 170^2 - 4*170 + 4 = 28900 - 680 + 4 = 28224. Yes, so sqrt(28224) = 168. Therefore, sqrt(282,240,000) = 168*100 = 16,800.

Therefore, the distance from H to line CD is 16,800 / 140 = 120 meters.
</execution_5.1>

<execution_5.2>
x = m*t = m*(k²)/(m² + k²)

y = -k + k*t = -k + k*(k²)/(m² + k²) = -k + k³/(m² + k²)

But maybe we don't need to compute x and y explicitly. Instead, let's compute HP at this t.

Alternatively, there's a formula for the minimal distance from a point to a line in 3D. Let me recall that formula.

Alternatively, since we have t, we can compute HP(t) as:

HP(t) = sqrt[ (m*t)^2 + (-k + k*t)^2 + h^2 ]

But maybe we can express this in terms of the known variables.

Alternatively, since we know t = k² / (m² + k²), let's substitute that into the expression.

First, compute m*t = m*(k²)/(m² + k²)

Similarly, -k + k*t = -k + k*(k²)/(m² + k²) = -k + k³/(m² + k²) = [ -k(m² + k²) + k³ ] / (m² + k²) = [ -k*m² - k³ + k³ ] / (m² + k²) = -k*m² / (m² + k²)

Therefore, HP(t) = sqrt[ (m²k^4)/(m² + k²)^2 + (k²m^4)/(m² + k²)^2 + h^2 ]

Factor out 1/(m² + k²)^2:

= sqrt[ (m²k^4 + k²m^4)/(m² + k²)^2 + h^2 ]

Factor numerator inside the sqrt:

= sqrt[ (k²m²(k² + m²))/(m² + k²)^2 + h^2 ]

Simplify:

= sqrt[ (k²m²)/(m² + k²) + h^2 ]

But from earlier, we have k² + m² = 19600, and h² = 9900. So:

= sqrt[ (k²m²)/19600 + 9900 ]

But we need to compute k²m².

We have k² = 12600, m² = 7000, so k²m² = 12600*7000 = 88,200,000

Therefore:

= sqrt[ 88,200,000 / 19600 + 9900 ]

Calculate 88,200,000 / 19600:

Divide numerator and denominator by 100: 882000 / 196

196*4500 = 882000 (since 196*4000=784000, 196*500=98000; 784000+98000=882000)

Therefore, 882000 / 196 = 4500

Therefore:

= sqrt[ 4500 + 9900 ] = sqrt[14400] = 120 meters

So, HP_min = 120 meters.

Therefore, the minimal HP is 120 meters. Therefore, the maximum rope saved is 280 - 120 = 160 meters.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
But wait, let me confirm this.

<prompt_6.1> Verify the calculations for the saved rope and the final answer using the above plan. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Earlier, we found that the perpendicular distance from H to line CD is 120 meters, which is indeed the minimal HP. Therefore, replacing HC (150) and HD (130) with HP (120) saves 150 + 130 - 120 = 160 meters. Therefore, the greatest length of rope that can be saved is 160 meters.

But let me verify once more. The key steps were:

1. Find the minimal HP by calculating the perpendicular distance from H to CD, which is 120 meters.
2. Subtract this from the original 280 meters to get 160 meters saved.

But let's check if the perpendicular distance is indeed achievable. Since t = k² / (m² + k²) ≈ 12600 / 19600 ≈ 0.6428, which is between 0 and 1, so yes, the closest point P lies on the segment CD. Therefore, this is valid.

Alternatively, if the perpendicular from H to CD falls outside the segment CD, then the minimal distance would be the distance to the nearest endpoint, either C or D. But in this case, since t is between 0 and 1, the minimal distance is indeed 120 meters.

Therefore, the answer is 160 meters saved.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
But let me check once more with the coordinates.

<prompt_7.1> Perform a final verification of the solution using the above plan. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Suppose we compute the coordinates of P.

Given t = k² / (m² + k²) = 12600 / 19600 = 126/196 = 9/14 ≈ 0.6429.

Therefore, x = m*t = sqrt(7000)*(9/14)

Similarly, y = -k + k*t = -sqrt(12600) + sqrt(12600)*(9/14) = sqrt(12600)*(1 - 9/14) = sqrt(12600)*(5/14)

Then, HP would be sqrt(x² + y² + h²)

Compute x²:

(sqrt(7000))^2*(9/14)^2 = 7000*(81/196) = 7000*(81)/196 = (7000/196)*81 = (35.714... ) *81 ≈ 2892.857... 

Similarly, y²:

(sqrt(12600))^2*(5/14)^2 = 12600*(25/196) = 12600/196 *25 = 64.2857... *25 ≈ 1607.142...

Then, x² + y² ≈ 2892.857 + 1607.142 ≈ 4500

Adding h² = 9900, total inside sqrt is 4500 + 9900 = 14400, sqrt of that is 120. So, correct.

Therefore, HP is 120 meters, original total was 280, so saved 160 meters. Therefore, the answer is 160 meters.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
**Final Answer**
\boxed{160}
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

<Final_answer>
Therefore, the greatest length of rope that can be saved is $\boxed{160}$ meters.
</Final_answer>
</think>