Prove or disprove: If $n < 20$ and the length of a smallest cycle in $G$ is $5$, then $G$ has a vertex of degree $2$ or less.

Question

I'm working in the following graph theory excercise:

Let $G$ be a connected plane graph of order $n≥5$ and size $m$, prove or disprove that If $n < 20$ and the length of a smallest cycle in $G$ is $5$, then $G$ has a vertex of degree $2$ or less.

By this question I think is false because each face has at least $5$ edges adjacent to it, how should I proceed to check if the desired grade of degree $2$ exists? Any hint or help will be really appreciated.

Answer 1

Since $m \le \frac{5}{3}(n-2)$ then $\delta(G) \le \frac{2 m}{n} = \frac{10}{3}(1- \frac {2}{n}) $. But since $n \lt 20$, then $1- \frac {2}{n} \lt 1 - \frac{2}{20} = \frac{9}{10}$

Therefore $\delta(G) \lt \frac{10}{3}\frac {9}{10} = 3$. So $\delta(G) = 0, 1, 2$