1, Could nine faces occur? The combinatorics (i.e. counting argument) of the Euler formula do not prohibit it. Here is a method for construction a combinatorial polyhedron with nine faces, all of which are quadrilaterals (and with 18 edges and 11 vertices). Start with two tetrahedra and “glue” them together to make a polyhedron with six triangles. Along with the inside triangle of this polyhedron (where you glued faces together) find the mid-points of the three edges and then cut off the vertices up to these midpoints (this will be some sort of curvy slice). What you cut off will give three new “quadrilateral faces” where we put quotes around these words because you cannot physically cut them with planes – they are two trianglesl in space that you can pretend are quadrilaterals (and therefore the combinatorics work). Also, the six original faces are now cut in a way so they are quadrilaterals. Draw a net for this “almost polyhedron”. Extra Credit: Could you really make this polyhedron with nine quadrilateral faces?
2, An interesting question that a student brought up is whether all of the edges of the mystery polyhedron were either an edge between two hexagons or one pentagon and one hexagon. Is this breakdown of edges 50/50? That is, must there be an equal number of edges adjacent one pentagon and one hexagon as there are edges between two hexagons? If claim so, justify your claim and if you believe there is another ratio, explain why this ratio must be correct.