[Guest]

Suppose you have four points of a square on a 2D plane and your job is to connect them least possible total distance.

For our example, suppose the square would have sides of length 10.

One (obvious) solution would be to have three sides of the square, yielding a total distance of distance of 30.

Make sure to explain your solution, including the length calculations.

[unreality]

Make an X like this:

o        o

 \      /

  \    /

   \  /

    \/

    /\

   /  \

  /    \

 /      \

o        o

if a side of the square is 's', then the option you gave was 3*s. But with the X, you have sqrt(2)*s + sqrt(2)*s = 2sqrt(2)*s

which is about 2.82842712*s

I think that's the best

[Guest]

I think I have seen this one before.

O      O

 \    /

  \__/

  /  \

 /    \

O      O

[Guest]

unreality, on the right track, but there's a better one out there.

psychic_mind, you only drew a picture.

[Guest]

unreality, on the right track, but there's a better one out there.

psychic_mind, you only drew a picture.

I take no credit for this answer. Just supplying numbers that I'm certain Psychic Mind could have provided.

The solution appears as the picture that Psychic Mind drew above. The bar in the middle has length 10[1-1/sqrt(3)]. The total length is 10[sqrt(3)+1] < 20sqrt(2).

[Guest]

It can be shown that 30 degrees is the optimal angle for the diagonals with a bit of calculus. However I do not know how to prove that this formation is optimal.

Length of diagonal = 5/cos(30) = 10/sqrt(3)

4 of these = 40/sqrt(3)

Horizontal line = 10 - (2*5sin(30)) = 5

Total length = 28.094...

[Guest]

Tuckleton has the answer I found.

psychic_mind, you might want to take another look at your calculus.

[Guest]

Tuckleton has the answer I found.

psychic_mind, you might want to take another look at your calculus.

Aha. It was my trigonometry not my calculus. That sine should have been tan.