Planar graph with triangular faces in which the link of every vertex bounds a face What possible connected planar graphs are there that satisfy the following properties?


*

*Every face (including unbounded face) is triangular, i.e. is bounded by exactly 3 edges.

*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.
 A: 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:

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.
A: 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 $K_4$.
