Okay, let's try to solve this problem. So, we have a hot air balloon at point H directly above point O. There are four ropes: HA, HB, HC, and HD connecting the balloon to points A, B, C, and D respectively. The points A, B, C, D are located around O such that A is north, B is west, C is south, and D is east. The distance between C and D is 140 meters. 

The problem says that ropes HC and HD have lengths 150 m and 130 m respectively. We need to replace these two ropes with a single rope HP, where P is a point on the straight line between C and D. The goal is to determine the greatest length of rope that can be saved by making this replacement. 

First, I need to visualize the setup. Points A, B, C, D are all around O, which is the origin. A is north, so directly above O. B is west, so directly to the left of O. C is south, directly below O. D is east, directly to the right of O. So, the coordinates might be something like A(0, y), B(-x, 0), C(0, -z), D(w, 0), but maybe we can assume O is at (0,0). 

But since the distance between C and D is 140 meters, and assuming that the field is flat, CD is a straight line. If C is south of O and D is east of O, then CD is a diagonal line from south to east. Wait, but if C is directly south and D is directly east, then the line CD would be from (0, -k) to (m, 0), but maybe in the diagram, they are arranged such that CD is a horizontal or vertical line? Wait, the asy code shows that C is at (-1,-1) and D is at (1.5,-0.3). Hmm, but maybe the exact coordinates aren't necessary. 

Wait, the problem states that the distance between C and D is 140 meters. Since C is directly south of O and D is directly east of O, the coordinates of C would be (0, -c) and D would be (d, 0). The distance between C and D would then be sqrt((d - 0)^2 + (0 - (-c))^2) = sqrt(d^2 + c^2) = 140. So, d^2 + c^2 = 140^2. 

But maybe we don't need to find c and d. The problem gives the lengths of HC and HD as 150 m and 130 m. Since H is directly above O, the coordinates of H would be (0, 0, h), where h is the height above O. Then, the lengths of the ropes HA, HB, HC, HD can be calculated using the distance formula in 3D. 

For HC, which is from H(0,0,h) to C(0, -c, 0), the distance is sqrt((0 - 0)^2 + (-c - 0)^2 + (0 - h)^2) = sqrt(c^2 + h^2) = 150. Similarly, HD is from H(0,0,h) to D(d, 0, 0), so distance is sqrt(d^2 + h^2) = 130. 

Therefore, we have two equations:

1. c^2 + h^2 = 150^2
2. d^2 + h^2 = 130^2

And we also know that sqrt(d^2 + c^2) = 140, so:

3. d^2 + c^2 = 140^2

So, we have three equations here. Let me write them down:

1. c^2 + h^2 = 22500
2. d^2 + h^2 = 16900
3. d^2 + c^2 = 19600

If we subtract equation 2 from equation 1, we get:

(c^2 + h^2) - (d^2 + h^2) = 22500 - 16900
c^2 - d^2 = 5600

But from equation 3, we know that c^2 + d^2 = 19600. So, we have:

c^2 - d^2 = 5600
c^2 + d^2 = 19600

Adding these two equations:

2c^2 = 25200
c^2 = 12600
c = sqrt(12600) = sqrt(126 * 100) = 10*sqrt(126) ≈ 10*11.225 ≈ 112.25 m

Then, substituting back into equation 3:

d^2 = 19600 - c^2 = 19600 - 12600 = 7000
d = sqrt(7000) ≈ 83.666 m

Then, from equation 1:

h^2 = 22500 - c^2 = 22500 - 12600 = 9900
h = sqrt(9900) ≈ 99.499 m

So, the height h is approximately 99.5 meters. 

Now, the problem is to replace ropes HC and HD with a single rope HP, where P is a point on the line CD. The goal is to minimize the total length of the ropes, which currently is HC + HD = 150 + 130 = 280 meters. By replacing them with HP, the new rope will be just HP. The amount saved will be (150 + 130) - HP = 280 - HP. So, to maximize the saving, we need to minimize HP. Therefore, we need to find the position P on CD such that HP is minimized. 

Wait, but HP is a single rope, so the length of HP must be less than or equal to the sum of HC and HD. But in reality, since P is on CD, the length HP can be found using the coordinates of H and P. 

Let me consider the coordinates. Let's set O at (0,0,0). Then, H is at (0,0,h) where h ≈ 99.5 m. Points C and D are at (0, -c, 0) and (d, 0, 0), respectively. So, the line CD goes from (0, -c, 0) to (d, 0, 0). 

We can parametrize point P on CD. Let’s parameterize CD with a parameter t, where t=0 corresponds to C and t=1 corresponds to D. So, the coordinates of P would be:

P = C + t*(D - C) = (0, -c, 0) + t*(d, c, 0) = (td, -c + tc, 0)

So, the coordinates of P are (td, c(t - 1), 0). 

Then, the length of HP is the distance from H(0,0,h) to P(td, c(t - 1), 0). 

Calculating HP:

HP = sqrt[(td - 0)^2 + (c(t - 1) - 0)^2 + (0 - h)^2]
= sqrt[(t^2 d^2) + (c^2 (t - 1)^2) + h^2]

We need to find the value of t in [0,1] that minimizes this expression. 

To minimize HP, we can minimize HP^2, which is:

HP^2 = t^2 d^2 + c^2 (t - 1)^2 + h^2

Let’s denote S(t) = t^2 d^2 + c^2 (t - 1)^2 + h^2

To find the minimum, take derivative of S(t) with respect to t and set to zero.

dS/dt = 2 t d^2 + 2 c^2 (t - 1) = 0

So,

2 t d^2 + 2 c^2 (t - 1) = 0

Divide both sides by 2:

t d^2 + c^2 (t - 1) = 0

Expand:

t d^2 + c^2 t - c^2 = 0

Factor t:

t (d^2 + c^2) - c^2 = 0

Then,

t = c^2 / (d^2 + c^2)

But from equation 3, we know that d^2 + c^2 = 140^2 = 19600. Therefore,

t = c^2 / 19600

But earlier, we found c^2 = 12600. So,

t = 12600 / 19600 = 126 / 196 = 9/14 ≈ 0.6429

So, the point P that minimizes HP is located at t = 9/14 along the line CD from C to D. 

Now, let's compute HP at this t. 

First, let's compute HP^2:

HP^2 = t^2 d^2 + c^2 (t - 1)^2 + h^2

We can substitute t = 9/14:

HP^2 = (9/14)^2 d^2 + c^2 (9/14 - 1)^2 + h^2

Simplify:

= (81/196) d^2 + c^2 (-5/14)^2 + h^2
= (81/196) d^2 + (25/196) c^2 + h^2

Factor out 1/196:

= (81 d^2 + 25 c^2)/196 + h^2

But we know from earlier that d^2 + c^2 = 19600, so 81 d^2 + 25 c^2 = 81 d^2 + 25 c^2. Let's compute this:

81 d^2 + 25 c^2 = 25(d^2 + c^2) + 56 d^2 = 25*19600 + 56 d^2

Wait, maybe not. Let's compute 81 d^2 + 25 c^2. Since d^2 = 7000 and c^2 = 12600, then:

81 d^2 = 81 * 7000 = 567,000
25 c^2 = 25 * 12,600 = 315,000
Total = 567,000 + 315,000 = 882,000

So, HP^2 = 882,000 / 196 + h^2

Calculate 882,000 / 196:

Divide numerator and denominator by 14: 882,000 ÷ 14 = 63,000; 196 ÷14=14. So, 63,000 /14 = 4,500

Therefore, HP^2 = 4,500 + h^2

But h^2 is 9900, so:

HP^2 = 4,500 + 9,900 = 14,400

Therefore, HP = sqrt(14,400) = 120 meters

So, the length of the new rope HP is 120 meters. Therefore, the original ropes HC and HD sum to 150 + 130 = 280 meters, and the new rope is 120 meters. Therefore, the rope saved is 280 - 120 = 160 meters.

Wait, that seems straightforward, but let me verify the calculations again.

First, coordinates:

- O is (0,0,0)
- H is (0,0,h) with h = sqrt(9900) ≈ 99.4987
- C is (0, -c, 0) with c = sqrt(12600) ≈ 112.2497
- D is (d, 0, 0) with d = sqrt(7000) ≈ 83.666

Line CD goes from (0, -c, 0) to (d, 0, 0). Parametric equation: P(t) = (td, -c + tc, 0), t ∈ [0,1]

We found t = c^2 / (c^2 + d^2) = 12600 / 19600 = 9/14 ≈ 0.6429

Then, HP^2 = t^2 d^2 + (1 - t)^2 c^2 + h^2

Wait, let's compute that again:

HP^2 = (td)^2 + ( ( -c + tc )^2 ) + h^2

= t^2 d^2 + ( -c(1 - t) )^2 + h^2

= t^2 d^2 + c^2 (1 - t)^2 + h^2

Which is the same as before.

Plugging in t = 9/14:

= (81/196) d^2 + (25/196) c^2 + h^2

= (81 d^2 + 25 c^2)/196 + h^2

But:

81 d^2 + 25 c^2 = 81*7000 + 25*12600 = 567,000 + 315,000 = 882,000

Divide by 196: 882,000 / 196 = 4,500

Add h^2 (9900): 4,500 + 9,900 = 14,400

Square root: 120. Correct.

Therefore, the total rope saved is 280 - 120 = 160 meters.

But let me think again. The problem says "the greatest length of rope that can be saved." So, replacing two ropes with one, the maximum saving is 160 meters. Is there a possibility that there's a different point P that could result in a shorter HP? But since we found the minimum of HP^2 by taking derivative, which gives the minimal HP. So, this should be the minimal length of HP, hence the maximum saving.

Alternatively, maybe there's a reflection principle here. Since H is fixed, and we need to go from H to P on CD. The minimal HP would correspond to reflecting H over CD and finding the straight line distance. Wait, that might be another way to approach it.

Yes, reflection method. To find the shortest path from H to line CD, you can reflect H across CD and then the minimal distance is the straight line from H_reflected to C or D? Wait, no. Wait, reflection over the line CD. Let me recall.

The shortest path from a point to a line is the perpendicular distance. But here, since CD is a straight line in 3D space, the minimal HP is the perpendicular distance from H to CD. But wait, H is not in the plane of CD. CD is in the z=0 plane, and H is at (0,0,h). So, the perpendicular distance from H to line CD would be the minimal distance, which would be sqrt( (distance from O to CD)^2 + h^2 ). Wait, maybe not. Let me think.

Alternatively, the minimal HP is the minimal distance from H to any point on CD. To compute this, we can parametrize CD as we did before and find the minimum distance. Which is exactly what we did, resulting in HP = 120 meters. 

Alternatively, using calculus, we found that the minimal distance is 120. Therefore, the greatest length of rope saved is 280 - 120 = 160. 

Therefore, the answer should be 160 meters.

But let me check if there is a miscalculation. Let's verify HP again.

Given HP^2 = 4,500 + 9,900 = 14,400. So HP = sqrt(14,400) = 120. That seems correct.

Alternatively, maybe we can compute the coordinates of P when t = 9/14. Let's compute P:

P = (td, c(t - 1)) = ( (9/14)*sqrt(7000), sqrt(12600)*(9/14 - 1) )

Compute sqrt(7000) ≈ 83.666, so (9/14)*83.666 ≈ 54.000

sqrt(12600) ≈ 112.2497, so (9/14 - 1) = -5/14 ≈ -0.3571

Therefore, c(t - 1) ≈ 112.2497*(-0.3571) ≈ -40.000

Therefore, P ≈ (54, -40, 0)

Then, H is (0, 0, 99.4987). So, distance HP is sqrt(54^2 + (-40)^2 + 99.4987^2)

Calculate 54^2 = 2916, 40^2 = 1600, 99.4987^2 ≈ 9900

So, HP^2 ≈ 2916 + 1600 + 9900 = 14416

sqrt(14416) ≈ 120.066, which is approximately 120. So, that checks out. The slight discrepancy is due to rounding. Therefore, HP is indeed 120 meters.

Therefore, the greatest length of rope that can be saved is 280 - 120 = 160 meters.

**Final Answer**
The greatest length of rope that can be saved is \boxed{160} meters.
Thus, the greatest length of rope that can be saved by replacing the ropes $HC$ and $HD$ with the single rope $HP$ is $\boxed{160}$ meters.
</think>