Triangular Tessellation

If you tessellate a plane with irregular triangles, then it seems likely that the average degree of the resulting graph would be 6. After all, the average angle of a triangle is 60 degrees.

If you tessellate part of the plane with a finite number of triangles, then the average degree can be lower. As low as 2 for a single triangle. Can the average degree of the resulting graph ever be higher than 6?

No. Suppose that every triangle shares a side with another triangle and the graph is connected. The tessellation then forms a polygon with n sides, where n is the number exterior vertices. Let i be the number of interior vertices. If the interior of the polygon has been completely filled with triangles, then the number of edges in the tessellation graph is 3i + 2n – 3. Since the average degree is twice the number of edges divided by the number of vertices, the formula in terms of n and i is

     (6i + 4n - 6)/(i+n)

This is bounded by 6.

If the vertex density of an infinite tessellation is roughly uniform, then for expanding circles centered at the origin, i will increase proportional to the square of the circle radius, but n will increase in linear proportion to the circle radius. The i term will thus dominate for large enough a circle. In this sense infinite triangular tessellations have average degree 6. But this is vague and perhaps there is a pathological counterexample.

Update: counterexample