Show that a planar graph with $n$ vertices and $3n-6$ edges with $\chi=3$ is Eulerian.
$\chi=3$ means there is a optimal vertex colouring with three colours. Eulerian means that the graph admits an Eulerian cycle (a cycle which contains each edge exactly once).
My thoughts on this: I know that a planar graph has at most $3n-6$ edges and that a planar graph is maximal iff each face is a triangle. So in the present case each face is a triangle. Moreover it would suffice to prove that all vertex degrees are even since this is equivalent to being Eulerian. I don't see how the information that $\chi=3$ comes in.