Updated on 25th July -- see below
Make an isosceles trapezium by starting with a rectangle of height 12 and width 66/13, and attaching Pythagorean (5,12,13) triangles to each side. Then the perimeter P is 600/13, and the height h is 12; and the numbers have been selected so that the width w at the equator is 12 too (where the equator is the horizontal line that divides the perimeter into two halves of equal length). So the dilation is P/2h = 25/13, which is less than 2.
We can try this with a general right-angled triangle. It is convenient to let the angle at the base of the trapezium be 2θ, so in our example we have sin 2θ = 12/13. We find that the dilation is equal to 1 + sin 2θ. However, if θ is too small, then we can achieve a larger dilation simply by cutting across the corner; this dilation is 1/sin θ. So we get the smallest dilation for an isosceles trapezium when 1/sin θ = 1 + sin 2θ. I had to resort to numerical methods to solve this; I got θ = 0.5555166235227462... radians, for a dilation of 1.89615765267304...
We can't improve on this by using a non-isosceles trapezium, but a smaller dilation might be achieved by a general non-trapezoidal quadrilateral.
Sorry if this all looks a bit provisional. If I get a full answer, I will put more effort into drawing some nice pictures and stuff.
I have carried out a computer search for the smallest dilation, as follows.
Without loss of generality, we can restrict the search to quadrilaterals ABCD such that:
- A=(0,0), D=(1,0);
- B and C lie above the x-axis;
- all side lengths are <= 1.
Step 1: Evaluate the dilation of all such quadrilaterals with x- and y-coordinates a multiple of 0.01. Save the 10000 quadrilaterals with the smallest dilation to file.
Step 2: For each B,C in the file, evaluate the dilation for the 10000 quadrilaterals with coordinates differing from B,C by a multiple of 0.001 between -0.005 and +0.004. (In other words, decrease the grid size by a factor of 10.) Of the resulting 100000000 quadrilaterals, save the 10000 with the smallest dilation to file.
Step 3: Repeat Step 2 with the grid size decreased by a factor of 10. And so on.
This procedure converges on the isosceles trapezium described above. So while this is not a proof, it is likely that the smallest possible dilation of a quadrilaterlal is indeed 1.89615765267304... (which is the real root of the polynomial $x^5 - x^4 - 4x - 4$).
Edit by J.O'Rourke (16Aug10). If I've followed Tony's description in the comment below
here is his quadrilateral with the (conjectured) smallest dilation: