Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 17G / b / Solution

The sum of the degrees of all faces is $2e$. Each of the $t$ triangular faces contributes three. Because the graph is bridgeless and has no four-cycle, every other face has degree at least five, and therefore

No edge can border two triangular faces: two distinct triangles sharing that edge would have their other two edges form a four-cycle, while the same triangle on both sides would force the connected bridgeless graph to be that triangle, contrary to $n\ge 4$. Thus the $3t$ edge incidences belonging to triangular faces use distinct edges, soCombining the inequalities givesand hence

The Euler formula for a connected planar graph now yieldsso the triangle-pentagon planar edge bound is

Equality is possible. Cut each corner of a dodecahedron through the midpoints of its incident edges. The resulting icosidodecahedral graph haswith every edge incident to one triangular and one pentagonal face. It has no four-cycle, and