show/hide this revision's text 3 Addressed the remark n European vs US time.

I am not sure what you are asking in your first question (what is a "unit triangulation" ?) but one can prove the following:

There is no geodesic triangulation of the plane with all vertices of degree >6 and such that the following conditions holds:

b) the angles of triangles are bounded from below.

Condition b) means that each triangle is bi-Lipshitz to an equilateral triangle, so this is probably what you ask. And the reason you mention is valid.

Examples of geodesic triangulation with vertices of degree 7 are given in the previous answer, and it follows from, b) that some triangles in these examples are very thin.

Edited on 9/9/12 at 9:30 a.mlater. My first answer also had condition

a) all triangles are bounded.

I claimed that b) can be replaced with a).

This condition CANNOT be used to replace b). Here is an example of a geodesic trianglation of the plane with all vertices of arbitrarily high degree, and bounded triangles.

Begin with an equilateral triange inscribed in the unit circle. From each vertex draw segments from the unit circle to the concentric circle of radius 2, so that the degree of each vertex on the unit circle becomes large. Add some chords of the circle of radius 2, connecting the endpoints of these segmnts. Draw additional segments to make this a triangulation of some polygon inscribed in the circle of radius 2. All inner vertices of this triangulation have large degree.

Now repeat this construction. We obtain a sequence of larger and larger polygons $P_n$ each inscribed in a circle of radius $n$, and each polygon is triangulated so that the degrees of inner vertices is large. Continuing this indefinitely, we obtain a triangulation of the plane.
show/hide this revision's text 2 added an example, corrected a mistake in the previous ans
one of the following conditions holds:

a) the diameters of the triangles are bounded, or

and it follows froma), b) that some triangles in these examples are very thin.

Edited on 9/9/12 at 9:30 a.m. My first answer also had condition

a) all triangles are bounded.

I claimed that b) can be replaced with a).

This condition CANNOT be used to replace b).Here is an example of a geodesic trianglation of the plane with allvertices of arbitrarily high degree, and longbounded triangles.

Begin with an equilateral triange inscribed in the unit circle. From each vertex draw segmentsfrom the unit circle to the concentric circle of radius 2, so that the degree of each vertexon the unit circle becomes large. Add some chords of the circle of radius 2, connecting theendpoints of these segmnts. Draw additional segments to make this a triangulation of somepolygon inscribed in the circle of radius 2. All inner vertices of this triangulationhave large degree.

Now repeat this construction. We obtain a sequence of larger and larger polygons $P_n$ eachinscribed in a circle of radius $n$, and each polygon is triangulated so that thedegrees of inner vertices is large. Continuing this indefinitely, we obtain a triangulation ofthe plane.
show/hide this revision's text 1

I am not sure what you are asking in your first question (what is a "unit triangulation" ?) but one can prove the following:

There is no geodesic triangulation of the plane with all vertices of degree >6 and such that one of the following conditions holds:

a) the diameters of the triangles are bounded, or

b) the angles of triangles are bounded from below.

Condition b) means that each triangle is bi-Lipshitz to an equilateral triangle, so this is probably what you ask. And the reason you mention is valid.

Examples of geodesic triangulation with vertices of degree 7 are given in the previous answer, and it follows from a), b) that some triangles in these examples are very thin and long.