# If $G$ cubic planar graph with faces only $C_5$ and $C_6$, then $G$ has $12$ pentagons and $0$ hexagons

I want to prove the questions in the title:

If $$G$$ is a connected cubic planar graph composed of only pentagons and hexagons, then it has $$12$$ pentagons and $$0$$ hexagons.

I was able to show it has $$12$$ pentagons, from Euler formula. However I am unable to proceed.

I believe it has to do with an inner structure it forms, because simply counting degrees and using Euler formula does not seem to work

• Is that a planar graph? I say that because this is a question from a book, so I expect it to be true? – MTLaurentys Jun 9 at 15:50
• What about a soccer ball? – saulspatz Jun 9 at 17:03

## 1 Answer

The Truncated icosahedral graph has 12 pentagonal faces and 20 hexagonal faces. As noted in Wikipedia, it is connected, cubic, Hamiltonian, regular and, of course, it is planar. See also the MathWorld article.

• I really do not see any problem with this. However, it is clearly contradicting the exercise in the book... I guess I should accept you answer then? – MTLaurentys Jun 9 at 16:13
• I don't know what is the context of the exercise, but it is very likely that it is wrong. About accepting my answer, there is no rush. – Somos Jun 9 at 16:46
• Okay! I will check back with the professor that is lecturing my class and come back to accept it if there is nothing else to it. – MTLaurentys Jun 9 at 16:53