This is a discussion of the geometry of the Delaunay triangulation.  

Let's look at this problem on the surface of a sphere, instead of the unit square. The Delaunay triangulation of a set $P$ on the sphere is invariant under Moebius transformations (easy to verify by the definition in terms of circles), and it is equivalent to the triangulation of the convex hull of $P$. A set in the plane can be transformed to the sphere by stereographic projection.  It's Dealaunay triangulation is equivalent to a subset of the polyhedron, namely the part on the far side from the center of projection.  The problems, for the uniform distribution on the square and the uniform distribution on $S^2$, are equivalent, but the spherical setting has more symmetry.


In the projective ball model of hyperbolic space, the convex hull inherits a metric as a hyperbolic surface of finite area.   I want to describe a related metric.  Imagine there is a radar installation at each point of $P$, and that a smuggler wants to fly a small plane below the horizon of any radar.  They need to slow down when they're flying lower, at a speed proportional to the distance to $P$, so the metric will be 1/minimum distance to $P$ times spherical arc length.  Near an element of $P$, this metric is asymptotically that of a cylinder of circumference $2 \pi$, but slightly smaller near where it is attached because we're rescaling the spherical metric, rather than the Euclidean metric.

Define a "thick" part of $S^2 \setminus P$ to be the set $Q$ obtained by removing a disk around each element of $P$ whose radius is 1/3 the distance to its nearest neighbor.

[Inessential mathematical sidenote: this metric is comparable to the Poincaré metric in Q, except in situations where $P$ is confined to a tiny disk on the sphere. The definition could be modified for that situation, but it's a distraction. The smuggling metric is also comparable to the hyperbolic metric on the image of $Q$ projected to the nearest point in the convex hull of $P$.]

**Claim**: the smuggling diameter is comparable to the diameter of the Delaunay triangulation.

*Proof of Claim*: The boundary circles of $Q$ all have comparable circumference in the smuggling metric, slightly less than $2 \pi$. The Delaunay triangles all have diameter bounded above and below in the smuggling metric. This gives an easy way to transform a path in the 1-skeleton of the Delaunay triangulation to a path in $Q$ without increasing length more than a bounded factor.  Conversely, for a path in $Q$, the rate of near-encounters with elements of $P$ per smuggling distance is bounded. Therefore, you can do a simplicial approximation, pushing the path to the 1-skeleton with number of edges less than a bounded multiple of its length.

Let's now consider a variation of the problem:  A smuggler wants to go between point $X$ on the sphere and point $Y$ on the sphere, in the presence of $N$ uniformly randomly distributed radar installations.  The smuggler is conservative, and decides to never exceed spherical speed $N^{-.5}$.  This will be less than speed 1 in the smuggling metric except when there is a radar within distance $N^{-.5}$.   The density of the set of points is $4 \pi / N$ and the strip has area about $d(X,Y) * 2 N^{-.5}$, so the expected number of radars in the strip is $\Theta(\sqrt N)$.  The distribution (Poisson) has standard deviation proportional to square root of expectation, so for moderately large $N$ it is extremely unlikely that there will be more than twice the expected number within the strip: enough that even that the probability that the $\sup$ number over all $X$ and $Y$ tends rapidly to 0 with $N$.

For each radar dish that is within the strip, the smuggler merely takes a detour around the corresponding boundary circle of $Q$, adding only a constant amount of time to the trip.  This gives a path in the smuggling metric of $S^2 \setminus P$ with length $\Theta(\sqrt N)$. We want to know the length of a path in $Q$. This is now easy, since there is a distance-decreasing retraction from $S^2 \setminus P \rightarrow Q$: each cylinder can be mapped to its boundary without increasing lengths.

> Therefore, the expected maximum diameter of the Delaunay triangulation is $\Theta(\sqrt N)$

*if I have not deceived myself in haste*.

I haven't addressed the MST diameter --- maybe, with this picture, in time.

This post should really have pictures ... maybe later, or maybe someone else can supply them.