A common design for soccer balls covers the ball with a mix of hexagons and pentagons. Instead, a British road sign directing drivers to nearby soccer stadiums incorrectly drew the soccer ball as being covered only by hexagons, as shown below. The edges of the resulting pattern, if it were possible, would form a planar graph, drawn without crossings on the ball, in which all faces are hexagons. Use the material from Monday's lecture (lecture 10a) to argue that this is impossible. (There are multiple ways of answering this, either directly using Euler's formula or indirectly using duality.)
The drawing below has one crossing. Prove that this crossing cannot be eliminated (the graph is not planar) by finding a subdivision of \(K_5\) or \(K_{3,3}\) in this graph.
A common way of drawing the complete graph on five vertices (with crossings) places its vertices at the five points of a regular pentagon, with ten edges: five edges on the sides of the regular pentagon, and five more edges on its diagonals.
The five edges on the sides of the pentagon form a cycle. Describe the flaps of this cycle, as defined in Wednesday's lecture (lecture 10b).
What is the compatibility graph of these flaps? Explain why this compatibility graph is not bipartite.
In any maximal planar graph, all faces are triangles. Is it also true that, in any maximal planar graph, all triangles are faces? Why or why not?