Prove or disprove that there is no triangulation of n points in $\mathbb{R}^2$ where there are $\frac{n}{3}$ vertices with degree $\geq 19$.

Question

As the title says.

I know that every triangulation with $n$ vertices has $\leq 3n-6$ edges and every edge adds $2$ to the total number of degrees in the triangulation therefore the total amount of degrees should be $6n-12$. I dont know how I can get from here to the conclusion stated in the title.

Any hint/solution is appreciated!

score 0 · Accepted Answer · answered Oct 29 '17 at 20:32

0

Hint: $19 \cdot \frac{n}{3} > 6n$

answered Oct 29 '17 at 20:32

Michael Biro

13,757

1

Yeah well now I feel a little stupid. Thanks again. Stay tuned for my following questions :P – rndmusr Oct 29 '17 at 20:40

Prove or disprove that there is no triangulation of n points in $\mathbb{R}^2$ where there are $\frac{n}{3}$ vertices with degree $\geq 19$.

1 Answers1