- Let $G$ be a planar graph with degrees all at least three. Suppose we can draw $G$ in a plane such that every face is a square. Prove the number of vertices of degree three is greater than the number of vertices with degree at least five.
- Let $D$ be a planar drawing of a graph such that every face is a triangle. Color the vertices of $D$ in blue, red, or green (without additional conditions). Prove the number of faces with differently colored edges is even. Hint - count in two different way the pairs $(F,e)$ such that $F$ is a face and $e$ is an edge of it with edges colored in red and blue.
I have no clue how to really approach this kind of problem.
The tools I have in mind are: Euler's formula, the inequality $3|F|<2|E|$ which is strict for triangulations, and the fact the average degree in a planar graph is strictly less than five. These look like powerful tools, but I don't know how to even think of a solution...