Triangulations of surface.

Question

Let $R$ be e regolar region of a surface $\Sigma$ such that $R$ is the closure `of an open set whose bourdary $\partial R$ is the union of simple closed regular curves. Let $T$ be a trangulation of $R$, $V_e$, $(V_i)$ the nunmber of external (internal) vertexes; $L_e$, $(L_i)$ the number of external (internal) sides and $S$ the number of triangles. I have to prove that the following identities hold: 1) $V_e=L_e$; 2) $3S=2L_i+L_e$; 3) If $V=V_i+V_e$ and $L=L_i+L_e$ then $V(V-1) \ge 2L$ and the equality if and only if $\partial R = \emptyset$.

Answer 1

Hint: 1. Disjoint union of n-gons.

2. For each triangle, you can count the boundary (3 edges) of that triangle.

3. Note that, $v(v-1)/2 = $$v\choose 2 $. So for each 2 vertices you can get at most 1 edge. By this you can get at most $v\choose 2$ edges.