7 Added link to code, and description of further ideas for illustrating and analyzing

The Mathematica notebook containing these computations is here. The code in the notebook itself is brief, since algorithms for Delaunay triangulations and minimum spanning trees and graph diameter are provided in packages in the Mathematica distribution. It took me more effort to make it work than it should have, because functions in these auxiliary packages are poorly documented. Here are some simple ideas for better showing what's happening, and analyzing further:

• Draw the Delaunay triangulation in a different color, along with the tree
• Indicate the increasing clumps accessible with changing step size by edge thickness and/or color coding edges of the spanning tree. One idea is to use random colors for short edges within clumps, then use a saturated average color (with average weighted by clump size) when new edges make clumps collide and merge. Delaunay triangles could also be color coded to indicate clumping.
• Make a 3-dimensional plot of graph distance from a randomly selected member of $P$ to other elements of the tree, using the TriangularSurfacePlot function from the ComputationalGeometry package. Also: try showing distance from basepoint by color coding, perha
• Do bigger experiments, and make plots showing the actual data: e.g. make a ListPlot[] of the log of the actual diameter for trees of sizes something like `Floor[Exp[ Range[2,6,.01]]]', but start with a much more modest range and aim for a more ambitious range.
• Draw contour plots or 3-D plots of Delaunay graph distance from a randomly selected vertex. How quickly do the contour lines begin to look like circles as the size of $P$ increases?
• The following is an earlier answer, which essentially shows that Delaunay "arc length" converges to a multiple of Euclidean arc length. Missing point: how much might larger detours around radars

6 added 404 characters in body

After mulling over this question for a few days, I think it's inconsistent with the stated facts stated that the diameter of the Euclidean MST is $\Theta(\sqrt n)$. Here's why.

Let's first interpret the theorem cited mentioned in the question, : that the diameter of the Delaunay triangulation for $P$ has expected diameter $\Theta(\sqrt n)$. For each $n$, the Delaunay triangualations for $P$ define a probability measure on the space of metrics on $n$-element subsets of the unit square. We can interpolate this metric to a metric on the unit squareby making each triangle of the Delaunay triangulation an equilateral triangle, and rescale it by $1/\sqrt n$ to get a measure on the space of metrics on the square. As $n$ goes to infinity, these metrics have uniformly bounded $\epsilon$ complexity (=minimum number of balls of size $\epsilon$ needed to cover). The set of metrics of $\epsilon$-complexity bounded by some fixed function of $\epsilon$ is compact , in the Gromov-Hausdorff topology, so the space of probability measures on these metric spaces is compact in the weak topology, so there exists at least a subseqeunce of $n$ such that the scaled Delaunay metrics converge to a measure on metric spaces.

The metric spaces are a.s. Lipschitz equivalent to the standard metric on the unit square, using the theorem about diameter (which can be used to deduce that the Delaunay distance between two random elements $p, q \in P$ is $O(\sqrt n)$). These metrics are path-metric spaces. The shortest paths are rectifiable arcs in the Euclidean sense as well as in the particular metric, since the two metrics are Lipschitz equivalent. A rectifiable arc in the plane has a tangent line almost everywhere, so there are rescaled limits where it looks like a almost all limiting Delaunay geodesics are straight linelines.

Now consider the MST. The Euclidean MST for a Delaunay triangulation is as a metric tree.

As a further test, I calculated the diameter of the Euclidean MST's for 25 uniform pseudo-ranodm pseudo-random distributions of $2^k$ points, with for each $k$ ranging from 2 through 8. The sample mean diameters are {2.64, 5.8, 10.32, 18.32, 29.88, 49.12, 78.88}. The best linear fit toto the log of the diameter as a function of $k$ is $.03 + .55 k$. The sample standard deviations for the log of the diameter were nearly constant, all about .15. (that is, the diameters fluctuate on a multiplicative scale, typically about $\pm 16 \%$). It's curious that $\approx n^{\sqrt n^{]log(\sqrt 3 / 2}$2)}$, but this could be coincidence, since the numerical experiment was modest in size. Here's the plot of log of mean sample diameter vs.$k$: The actual asymptotic behavior of diameter seems intricately tied with percolation, a subject that I do not understand very well. That is: you can think of building up the MST by successively adding edges of length$t$if they join points which are not already on different connected components. This gives an increasing union of equivalence relations on elements of$P$, consisting of clumps that can be connected by steps not greater than$t$. One would expect that at for a$t$greater than some critical length distance$t$, t(n)$ that is approximately a constant time $1/\sqrt n$, large-scale clusters are expectedlikely. Paths between points in such the a cluster clusters must stay in the cluster. The geometry of the increasing clusters should enable one to estimate the Hausdorff dimension of tree geodesics, which should in turn give the an exponent of growth for the diameter of the Euclidean MST.

As a further test, I calculated the diameter of the Euclidean MST's for 25 uniform pseudo-ranodm distributions of $2^k$ points, with $k$ ranging from 2 through 8. The sample mean diameters are {2.64, 5.8, 10.32, 18.32, 29.88, 49.12, 78.88}. The best linear fit toto the log of the diameter as a function of $k$ is $.03 + .55 k$. It's curious that$Exp[.55] = 1.733$, close to $\sqrt 3 = 1.732$. This suggests a power law that asymptotic diameter $\approx n^{\sqrt 3 / 2}$, but this could be coincidence, since the numerical experiment was modest in size. Here's the plot of log of mean sample diameter vs. $k$: