TITLE: Is there a planar graph that (almost) all its vertices have degree 6?
QUESTION [6 upvotes]: Is it true that for any $N_0\in\mathbb N$ there exists a planar graph $G=(V,E)$ on (at least) $N_0$ vertices such that at least $$|V|(1-o(1))$$ vertices has degree 6?
It is easy to show that no 6-regular planar graph exists, but is there an "almost-6-regular" planar graph?

REPLY [7 votes]: Take a regular icosahedron whose edges are unit length.
For any $n > 1$, subdivide each of its face into $n^2$ equilateral triangle with edge length $1/n$. You obtain a polyhedron with $10 n^2 + 2$ vertices, $30n^2$ edges and  $20 n^2$ faces.
Using stereographic projection from the center of one of its face,
it is easy to see the vertices and edges of this figure is a planar graph.
Aside from the original $12$ vertices, the other $10(n^2 - 1)$ vertices have degree $6$.