# Mutually disjoint triangles in certain planar graph

Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.

Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?

(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)

• For my own edification.. what is an odd face ? – T. Fo Dec 30 '18 at 5:04
• Face with odd number of sides. – Zachary Hunter Dec 30 '18 at 5:09
• I think "those bounded by an odd length cycle" is a sensible definition. – Finallysignedup Dec 30 '18 at 5:11

Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.

(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)

• Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question. – Finallysignedup Dec 30 '18 at 6:03

Take two cycles of the same order 2n+1, $$C_a$$ and $$C_b$$ and number each in clockwise fashion.

"Glue together" $$v_{1a}$$ and $$v_{1b}$$, and then do the same for $$v_{2a}$$ and $$v_{2b}$$, so that there is just one edge between $$v_1$$ and $$v_2$$.

Then, for each $$i > 2$$ create an edge between $$v_{ia}$$ and $$v_{ib}$$.

Finally, subdivide the edge between $$v_1$$ and $$v_2$$.

• Thanks. Sorry that I could not tick both answers. – Finallysignedup Dec 30 '18 at 6:02
• Understandable, it is the fault of my laziness in creating diagrams. – Zachary Hunter Dec 30 '18 at 6:04