2
$\begingroup$

I found the following question in a previously assigned graph theory final:

Let $G$ be a simple, planar graph with $1\leq |E(G)|< 30$. Show that $G$ has a vertex with degree at most 4.

My Attempt: this is equivalent to showing $\delta(G)\leq4$. Assume by contradiction that $\delta(G)\geq 5$. First, we know

$$5|V(G)|\leq \sum_{v\in V(G)} \text{deg}(v) \leq 2|E(G)|.$$

Also, since $G$ is planar we must have

$$|E(G)|\leq 3|V(G)|-6,$$

which implies

$$\frac{5}{2}|V(G)|\leq|E(G)|\leq 3|V(G)|-6.$$

From here I am lost. Are there any suggestions for how to continue, or is this not leading to a solution?

$\endgroup$
3
$\begingroup$

Using the fact that $|E(G)| < 30$ gives $\frac{5}{2} |V(G)| < 30$ or $|V(G)| < 12$.

Using the fact that $\frac{5}{2} |V(G)| \leq 3 |V(G)| - 6$ implies $|V(G)| \geq 12$.

Finished.

And here's a picture of a graph with exactly 30 edges that is planar and 5-regular.

https://math.stackexchange.com/a/162418/8671

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ you are right, I wrote it down incorrectly. $\endgroup$ – Holdsworth88 Sep 21 '12 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.