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The icosidodecahedron has 20 triangular and 12 pentagonal faces, with the faces alternating around each of its 30 vertices. It can be obtained by cutting a dodecahedron through the midpoints of its edges.

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