The Farmyard Problem: Minimize the amount of ditch-digging needed to locate a straight pipe under a square piece of land Problem:

There is a farmer who has a $1\text{ mile}\times 1\text{ mile}$ square piece of land.  He knows that there is a completely straight pipe underneath some part of his property, but it could be going in any direction.  He wants to dig some ditches to find it.  He knows that if he digs all around the property we will find it, but that requires 4 miles of digging!  What choice of lines minimizes the amount of required digging, and what is that minimum amount? 

Just three lines around the perimeter will suffice, cutting the digging down to only 3 miles.  After more thought one sees that only 2 diagonal lines are needed, which is 2.82 miles of digging.
I don't think coming up with examples is that hard, it's just that proving it seems impossible.  At the moment I found something with length $\sqrt{2}+\sqrt{3/2}$ but I just can't prove it is optimal.  It is worth noting that the lines can be disconnected, and can have finitely many pieces.  The solution I had above was made up of two different pieces.
Any help is greatly appreciated!
 A: Reposting from the comments so this question will have an answer:
This problem appears to be unsolved.  This paper (Dumitrescu, Jiang, and Pach, Opaque Sets, preliminarily released in 2011, now dated May 23, 2018) provides an example with total length $\sqrt{2}+\frac{\sqrt{6}}{2} = 2.638958433764...$ which is conjectured to be optimal.  I believe this uses a Steiner tree to connect 3 vertices of the square and then a diagonal line from the 4th vertex to the center of the square.  Quoting:

The diagonal segment $[(\frac{1}{2}, \frac{1}{2}),(1, 1)]$ together with three segments connecting the corners $(0,1), (0,0), (1,0)$ to the
  ￼￼point $(\frac{1}{2}−\frac{\sqrt{3}}{6},\frac{1}{2}−\frac{\sqrt{3}}{6})$ yield a barrier of length $\sqrt{2}+\frac{\sqrt{6}}{2}$ =2.639

This is illustrated by this drawing derived from https://www.susqu.edu/brakke/opaque/opaqsq.html
 
The paper R.E.D. Jones, Opaque sets of degree α, American Mathematical Monthly, May 1964, p. 535-537, establishes a lower bound of 2.
A: The answer is a Steiner Tree. See:
http://en.wikipedia.org/wiki/Steiner_tree_problem
Of course he can stop digging when he hits the pipe so the length of the tree segments is the worse case scenario where he is unlucky enough to hit the pipe on the last scoop of dirt.
