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LOTR Telephone

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**Risking the chastisement of repeating another forum question**
Samwise Gamgee has a square plot of land, each side being 1 unit. One day, Sam finds out that the dark Lord Sauron has a telephone line that he uses to speak with a traitor amongst the hobbits.
Gandalf informs him that the telephone line runs in a straight line parallel to the ground and passes beneath the square plot of land, but he does not know its location. Sam decides to dig up around the perimeter of his land to discover the telephone line, but Gandalf says it is not necessary to dig around the entire length of 4 units.
Sam brightens up, and says "I know what you mean. I can just dig 3 sides and still discover it. For even if the phone line runs along the fourth side, I will still detect it at the end points ! "
Gandalf shakes his head. "No, Sam. You are on the right track, but you can do better than that."
What solution does Gandalf have in mind for the optimum length of the "digging curve" ?
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Posted · Report post

attachicon.giflotr ok.png

Dig along the green lines

1 + 1 + √(0,52 + 0,52) = 2,7071

I think maybe Sam could dig even less, following the pink lines BMD + MC + AO (green), but I don't know, how to find point M.

It would be minimum for M chosen at a distance of (3 -root(3)) / 6 = 0.211 from both sides

This gives total length of digging as 2.64

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Posted · Report post

Not sure if it is the most optimum: Dig along the diagonals


The total length dug is 2*root(2)
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Posted · Report post

One approach is to dig an "X" diagonally across the square [e.g, a line from the "Top Left" corner to "Bottom Right" corner and another line from the "Top Right" corner to Bottom Left" corner ].

Note: arbitrary reference system just for explanation

The length of each line would be ~1.414 units long.

The total length of the digging would be ~ 2.828 units long [shorter than digging 3 sides]

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Posted · Report post

Not sure if it is the most optimum: Dig along the diagonals

The total length dug is 2*root(2)

I agree, we were working in parallel.

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post-54025-0-41888200-1382340758_thumb.j

The digging should be done along the red lines as shown in the figure above
AB² = x² + 0.25
BC = 1 - 2x

Total length of digging = L = 4AB + BC

L = 4(root(x² + 0.25)) + 1 - 2x

Now we need to find the minimum value of this f(x)
dL/dx = 8x/(2 root(x²+0.25) - 2

This dL/dx should be zero for the minima
This gives: 4x/root(x² + 0.25) - 2 = 0
4x² = (x² + 0.25)
x = 1/2(root3)

With this x , the total digging length is 2,732

Edited by DeGe
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Posted · Report post

Should we be digging the inside of the shape?

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post-52539-0-32838800-1382482868_thumb.p

Dig along the green lines

1 + 1 + √(0,52 + 0,52) = 2,7071

I think maybe Sam could dig even less, following the pink lines BMD + MC + AO (green), but I don't know, how to find point M.

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Posted · Report post

attachicon.giflotr ok.png

Dig along the green lines

1 + 1 + √(0,52 + 0,52) = 2,7071

I think maybe Sam could dig even less, following the pink lines BMD + MC + AO (green), but I don't know, how to find point M.

It would be minimum for M chosen at a distance of (3 -root(3)) / 6 = 0.211 from both sides

This gives total length of digging as 2.64

very elegant solution(s)! My hat is off to both of you!

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