Connected, planar, 3-colorable graph with every face of degree 3 has an Eulerian circuit

Question

I am trying to prove that:

If G is a connected graph where every face has a degree of 3 and is 3 colourable then there exists and Euler tour.

This is what I have done:

For a graph to have an Euler tour it needs to have vertices of even degree. So we need to prove that the graph G must have even degree.

How can we prove that a 3 colourable graph, with degree of each face is 3 has vertices of even degree? (For that matter the maximum degree possible is 5, because its a planar graph so we need to show that the degree is either 2 or 4). Please help

Thanks in advance :)

Answer 1

Hint: Let $v$ be a vertex, and let $w_1, w_2, \ldots, w_k$ be the vertices it is connected to, in clockwise order. Because all faces are triangles, $w_1$ must be connected to $w_2$, and $w_3$ must be connected to $w_4$, and so on, with $w_k$ connected to $w_1$.

If the three colors are red, yellow, and blue, let $v$ be red and $w_1$ be blue, WLOG. What must the other vertices be colored, given this? What can you conclude about $w_k$?