Show the $\frac{1}{a} + \frac{3}{b} > 1$ for this planar graph.

Question

The connected planar graph, G satisfies the $(1)$ ~ $(3)$.

• $(1)$ The number of the vertex of the $G$ is $n$.

• $(2)$ The degree of all the vertices are $4$.

• $(3)$ Each vertex of G lies on exactly one face of degree $a$ and exactly three faces of degree $b$ (Here the $d(f)$ is a degree of the face and $a< b$)

Show the $\frac{1}{a} + \frac{3}{b} > 1$.

In the solution it claimed like the below.

First, $4n = \sum d(v) = 2e$

The number of the faces whose degree is the $a$ ; $\frac{n}{a}$ (*)

The number of the faces whose degree is the $b$ ; $\frac{3n}{b}$ (**)

Hence, $v-e+f = n-2n +({\frac{n}{a}}+\frac{3n}{b})=2$. From this it deducted the results $\frac{1}{a} + \frac{3}{b} > 1$.

But My question is why are the (*) and (**) hold? Is there any particular reason for claiming those?

p.s.) Thanks for the kabenyuk

Answer 1

This is a standard argument that shows up in many places, so people get very sloppy about expressing it.

Here is a very formal version. Let $A$ be the set of all pairs $(f,v)$ where $f$ is a face of degree $a$ and $v$ is a vertex of that face. Then:

• $|A| = n$, because each of the $n$ vertices is part of exactly one such pair, by condition (3).
• If there are $k$ faces of degree $a$, then $|A| = ak$, because each of the faces of degree $a$ is part of exactly $a$ such pairs. (Any vertex on any such face gives us a pair.)

Therefore $ak = |A| = n$, so $k = \frac na$.

We can proceed similarly if we define $B$ to be the set of all pairs $(f,v)$ where $f$ is a face of degree $b$ and $v$ is a vertex of that face. The only difference is that by condition (3), $|B| = 3n$: for every vertex $v$, there are $3$ pairs $(f,v)$ where $f$ has degree $b$ and $v$ lies on $f$. So we will get $\frac{3n}{b}$ such faces instead of $\frac nb$.