In a finite connected planar graph $G$, such that $\delta(G)\geq 3$, show that at least $2$ faces have at most $5$ edges. I have tried writing out degree sum from Handshaking lemma and then substituting in Euler's formula but I am stuck there.
Planar graph, number of faces, minimum vertex degree 3
graph theoryplanar-graphs
Related Solutions
We will count the outer face as its own face. So a cycle has two faces, for example.
The idea is to take each connected component $C_1$, $C_2$, etc., and connect them by drawing a bridge between $C_1$ and $C_2$, $C_2$ and $C_3$, etc. We will thus add $n-1$ edges (the restriction is if you contract the connected components we had previously, you must form a tree).
- $F = f$ remains the same.
- $V = n$ since no vertices were added.
- $E = m + k - 1$ since $k-1$ edges were added.
Now apply Euler's formula.
The discharge method works here. (You can forget this proof, but try to remember the method. It can be very powerful. It belongs in your toolbox next to the pigeonhole principle and the principle of induction).
Assume $G$ is a counterexample with a minimum number of vertices ($n$). We may also assume that we have maximized the number of edges ($e$). So $G$ has minimum degree at least three and $d(u)+d(v)>13$ for every edge $uv$. The edge maximization implies that if there are any non-adjacent $u$ and $v$ with $d(u)+d(v)>13$, then $G+uv$ is not planar.
Now we assign a charge to each vertex: a vertex $v$ gets charge $6-d(v)$. Applying the formula $e\leq 3n-6$ (equality only holds for triangulations), we find the total charge $\sum_v(6-d(v))=6n-2e\geq6n-(6n-12)=12$.
Now we discharge: a vertex $v$ with a positive charge distributes its charge evenly among its neighbours. We do this for all vertices simultaneously.
Since no charge is lost, there must still be vertices with a positive charge.
Let $v$ be a vertex with a positive charge.
Case 1: $d(v)\leq 8$. Since a vertex can only obtain a positive charge from a neighboring vertex $u$ with degree at most 5, we find an edge $uv$ with $d(u)+d(v)\leq 13$. Contradiction.
Case 2: $d(v)=9$. $v$ has no neighbours of degree 3 or 4, so its positive charge can only be obtained from neighbours of degree 5 and each of them contributes a charge of $\frac 15$ (they started with charge 1 and sent an equal part, i.e. $\frac 15$ to each of their neighbours). Since $v$ started out with charge $-3$ it would need 16 neighbours to end up with a positive charge. Contradiction.
Case 3: $d(v)=10$. As case 2: no neighbours of degree 3, and each neighbour of degree 4 or 5 contributes at most charge $\frac 12$. Since $v$ started with charge $-4$, it needs at least 9 neighbours of degree 4 or 5. Let $v_1,\ldots,v_{10}$ be the neighbours of $v$, arranged clockwise. Without loss of generality we may assume that $v_{10}$ is the only vertex that (possibly) has a degree larger than 5. Since $v_1$ and $v_2$ cannot be adjacent (since $d(v_1)+d(v_2)\leq10$), the face containing $v,v_1,v_2$ contains at least one other neighbour $u$ of $v_1$. The degree of $u$ must be at least 9 (or $d(v_1)+d(u)\leq13$). If $u$ is not adjacent to $v$, we can add edge $uv$, contradicting the edge-maximality. So $u$ is another neighbor of $v$, which implies $u=v_{10}$. We have shown that $v_1$ is adjacent to $v_{10}$. Exactly the same argument shows that $v_2,\ldots,v_9$ are adjacent to $v_{10}$. Now consider the adjacent "4-gons" $vv_2uv_1$ and $vv_2uv_3$. Since $d(v_2)\geq4$, $v_2$ must have neighbours that are inside one of these 4-gons, and not neighbours of $v$. Without loss of generality we may assume that $v_2$ has clockwise neighbours $w_0=v,w_1,\ldots,w_t,w_{t+1}=v_2$ where $t\geq1$ and $w_1,\ldots,w_t$ are inside $vv_2uv_1$. Since $d(w_1)\geq3$ we can now add edge $vw_1$ and again contradict edge-maximality.
Case 4: $d(v)\geq11$. Each neighbour of degree <6 contributes at most 1. $v$ starts with charge $6-d(v)$, so it needs at least $d(v)-5$ neighbours of degree less than 6 to end up with a positive charge. Since $d(v)-5>\frac{d(v)}2$ we again must find two consecutive neighbours of low degree and, like in case 3, these must either be adjacent or expose a high-degree neighbour that can be connected to $v$ without breaking planarity. This final contradiction finishes the proof.
For the counterexample we can repeat the same arguments and find that there must be at least 12 vertices of degree 10, each of them surrounded by 5 vertices of degree 3.
This immediately brings up the idea of the icosahedron and indeed, if you take the icosahedron, and add a vertex in the center of each triangle face and connect it to the three corners, you get a very symmetric planar graph with 20 vertices of degree 3 and 12 of degree 10. No two vertices of degree 3 are adjacent, so this is our desired counterexample.
Edit to add: a stellated icosahedron:
Best Answer
Using the fact $\delta(G)\geq 3$, one could prove each edge is part of a cycle, and therefore bound exactly two faces.
Denote by $e,v,f$ the number of edges, vertices and faces in $G$ respectively.
Intuition
Since the proof is quite technical, I'll try to sum up the proposition I use:
Proof
Now assume by contradiction the statement is false, and therefore at least $f-1$ of the faces have at least $6$ edges.
This implies: $$2e=\sum\limits_{i=1}^{f} \text{#edges of the face } f_i\geq 6(f-1)\implies \\ 2e\geq 6(f-1)\implies (f-1)\leq \frac{e}{3}$$
Using Euler formula $v-e+f = 2$, we get $$ 2 = v-e+f\leq v-e+\frac{e}{3} +1=v-\frac{2e}{3} + 1$$
Since
$$2e = \sum\limits_{u\in V(G)}deg(u) \geq 3v\implies \frac{2e}{3}\geq v$$
and finaly we get a contradiction: $$v-\frac{2e}{3}\leq 0 \implies v-\frac{2e}{3} + 1 \leq 1$$ and because $$v-e+f \leq v-\frac{2e}{3} + 1 $$ we get
$$2 = v-e+f \leq 1 $$
which is a contradiction.