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## Question

Four towns, A, B, C and D, are located such that their centers form the vertices of a square 1 mile on a side. Town planners want to build a set of roads that connect the four town centers while minimizing the cost, which can be considered to increase linearly with road length.

What set of roads minimizes that cost?

A               B

C               D

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I think...

Spoiler

Let a diagonal road connect each of the four towns to the endpoints of a vertical road that runs through the center of the square. Call the length of the vertical road x. Then the total length of all of the roads, geometrically, is

T = x + 2√(x2 -2x +2)

Then, dT/dx = 1 + (2x -2) / √(x2 -2x +2), which has a zero at x = 1 - (1/√3). The total road length, by the above formula, is then ~2.73205.

Special thanks to Cygnet for suggesting there might be a minimum hidden in such a configuration.

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Spoiler

Total length: (2*sqrt(5) + 1) / = 2.7361

It would look something like this:

\__/
/  \

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Make an X? Length would be 2*sqrt2 mi. Seems too easy though...

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11 minutes ago, Thalia said:

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Make an X? Length would be 2*sqrt2 mi. Seems too easy though...

You're correct. It can be a little bit shorter than that.

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8 hours ago, ThunderCloud said:

I think...

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Let a diagonal road connect each of the four towns to the endpoints of a vertical road that runs through the center of the square. Call the length of the vertical road x. Then the total length of all of the roads, geometrically, is

T = x + 2√(x2 -2x +2)

Then, dT/dx = 1 + (2x -2) / √(x2 -2x +2), which has a zero at x = 1 - (1/√3). The total road length, by the above formula, is then ~2.73205.

Special thanks to Cygnet for suggesting there might be a minimum hidden in such a configuration.

I assumed my answer was along the right lines, but I was too lazy to actually find the true minimum with it...thanks for doing the dirty work!

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Another approach:

Spoiler

Let the small angle in the diagram be theta. Total length then is 1 + 2 sec(theta) - tan(theta), whose derivative is (2 sin(theta) - 1)/cos2(theta) which is zero for theta=30o. The three lines intersect, nicely enough, at 120o. Using 30o for theta, the length is 1 + (4-1)/sqrt(3) = 1 + sqrt(3), which is ThunderCloud's result

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