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What possible connected planar graphs are there that satisfy the following properties?

  1. Every face (including unbounded face) is triangular, i.e. is bounded by exactly 3 edges.
  2. For every vertex, the set of faces not incident to that vertex is either singleton or empty.

So far I can only think of two genuinely distinct such graphs, namely $K_3$ and $K_4$. I'm hoping that these are the only possibilities.

2 Answers 2

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If you allow loops (as well as both sides of an edge counting as distinct sides of the same face), there's this degenerate case:

graph with degree sequence 4,1,1

Otherwise there's only $K_3$ and $K_4$. When at least two faces meet at each vertex, any two vertices must share a face, and because the faces are triangular, this means that they are neighbors. So the graph is complete! But a complete graph with more than $4$ vertices is not planar.

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    I'm not allowing loops, as I want my graph to correspond to a non-degenerate polyhedron. But thanks! – 2017-02-22
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    @Heiro: Note that the text **below** the figure is a much shorter argument for your conclusion than Simon's in the non-degenerate case. – 2017-02-22
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    Ahh yes, good point. – 2017-02-22
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    At least you got 2 different proofs for your question. – 2017-02-22
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Let $V$, $E$ and $F$ denote the amount of vertices, edges and faces of such a graph. The first condition states that $E=\frac{3F}{2}$, hence $2$ divides $F$. The second condition states that the degree of a vertex is either $F$ or $F-1$. Now use Euler's formula to get $V=2+E-F=2 + \frac{3F}{2}-F=2+\frac{F}{2}$. Now since the sum of the degree of the vertices is exactly $2E$ we have the inequalities:

$$V(F-1) \leq 2E \leq VF$$ Substituting $E=\frac{3F}{2}$ and $V=2+\frac{F}{2}$ we get $$(2+\frac{F}{2})(F-1) \leq 3F \leq (2+\frac{F}{2})F$$ hence $$(4+F)(F-1) \leq 6F \leq (4+F)F$$ This means that $F^2-3F-4 \leq 0$ and $0 \leq F^2-2F$. This means that $2 \leq F \leq 4$. Then you can easily check graphs with $2$ or $4$ faces ($3$ faces is not possible, since $2 \nmid 3$). So the only graphs are indeed $K_3$ and $K_4$.

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    Perfect, this is exactly the solution I came to in the interim! – 2017-02-22