# Number of triangle regions in a connected planar simple graph

Suppose a connected planar simple graph G with v vertices such that all its regions are triangles (cycle with 3 edges).

I'm trying to figure out into how many regions does a representation of the planar graph G.

So here's my thinking: If G is a connected planar simple graph, with e edges and v>=3 , then e<= 3v-6 . Plus if we add all degrees of all regions, the sum must be equal to 2e. Since we know that the regions are triangles, we know that the degree of each region is 3. Also, r <= e – v + 2

In the end, here's what we have : r <= e – v + 2 r <= 3v –6 - v +2 r <= 2(v+1)

So r = 2(v+1)?

Something must be wrong in my reasoning.

Let $$n=|V|$$, $$e=|E|$$, and the number of faces be $$f$$.

In a triangulation (a plane graph where every face boundary is a 3-cycle) $$G$$, we know that $$e=3n-6$$.

We substitute this value for $$e$$ into Euler's Formula for planar surfaces;

Euler's Formula

$$n-e+f=2$$

$$n-(3n-6)+f=2$$

$$f=2-n+(3n-6)$$

$$f=2n-4$$