Smallest dilation of a quadrilateral? What is the smallest dilation of a quadrilateral in $\mathbb{R}^d$?
This may be an open problem;
my question is: Is this indeed open?
It will take me some time to explain the terms.
The notion of dilation derives from Gromov, as far as I know
(He defines a version in
Metric Structures for Riemannian and Non-Riemannian Spaces,
p.11,
although he called it distortion).
I came upon it myself via $t$-spanners.
The version in which I am interested is this.
Let $P$ be a polygon (its boundary, not its interior), and $x,y$ two points on
$P$.  You can think of $P$ in $\mathbb{R}^2$, but also $\mathbb{R}^3$ and $\mathbb{R}^d$
for $d>3$ are interesting.
Define $\delta(x,y)$ as the maximum (supremum) of $d_P(x,y) / | x y |$,
where $d_P(x,y)$ is the distance between $x$ and $y$ following
$P$ (the shortest path staying on the closed path that consitutes $P$),
and $|xy|$ is the Euclidean distance in $\mathbb{R}^d$.
Thus $\delta(x,y)$ measures how much $P$ dilates w.r.t. Euclidean distance.
I am interested in the minimum value $\delta(P)$
of $\delta(x,y)$ over all
$x,y \in P$, for all $n$-gons $P$, for fixed $n$.

Example 1.
If $P$ is a unit square, then $\delta(x,y)$
for $x,y$ opposite corners is $\sqrt{2}$, but
$\delta(P)=2$ because with $x,y$ midpoints of
opposite sides, $\delta(x,y)= 2/1$.

Example 2.
If $P$ is an equilateral triangle, $\delta(P)=2$, as shown in the figure.
In fact, the dilation of any triangle is $\ge 2$ [Lemma 7 in the 2nd paper below].
            

Example 3.
It is known the the dilation of any closed curve $C$
satisfies $\delta(C) \ge \pi/2$, with equality achieved
only by the circle. [Corollary 23 in the first paper below.] This is (apparently) due to Gromov.

So I finally come to my question.  By reading these two
papers,
"Geometric Dilation of Closed Planar Curves: New Lower Bounds,"
and
"On Geometric Dilation and Halving Chords,"
it appears to me that the minimum dilation of a quadrilateral
in $\mathbb{R}^2$ (and $\mathbb{R}^d$) is not known.
I had heard this was the case three years ago in a seminar
in Brussels, but (a) I didn't quite believe it,
(b) it was hearsay, and (c) it is now out of date.
I am trying to clarify with the authors of these papers, but in parallel
I would appreciate any information on the status of
this question.
The latter paper cited above proves a lower bound of $4 \tan(\pi/8) \approx 1.66$
(if I have interpreted it correctly).
Addendum. I don't want to close-out this question, but I have heard from one of the authors of
the above cited papers, and indeed it appears that the dilation of a planar quadrilateral is unknown
[as of July 2010, the original posting date].
So I have tentatively tagged this as an open-problem, and I will update if new information surfaces.  Thanks for everyone's interest and input!
 A: 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.
Update
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
correctly,
here is his quadrilateral with the (conjectured) smallest dilation:
$h=0.896158$, $w=0.25552$:

          

