Triangulation proofs

Question

How can I prove that:

$a)$ In every triangulation with $n$ nodes there is a node Degree $\leq 5$

$b)$ Each triangulation with $n$ nodes has at least $\frac{n}{27}$ pairs of non-adjacent nodes of degree $\leq 8$.

Let me know if you need further information.

Answer 1

a) is a consequence of the following theorem (see e.g. https://en.wikipedia.org/wiki/Planar_graph#Other_planarity_criteria):

For every planar graph $(V,E)$, if $|V| \geq 3$, then $|E| \leq 3 |V|-6$

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Assume (towards a contradiction) that every node $v$ has degree $deg(v) \geq 6$. If $|V|<3$, the contradiction is immediate. Otherwise, we have $|E| = 1/2 \sum_{v \in V} deg(v) \geq 1/2 \sum_{v \in V} 6 = 3 |V|$. But (since every triangulation is planar, and $|V| \geq 3$), we have $|E| \leq 3|V|-6$ - a contradiction.