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The icosidodecahedral graph is the skeleton of the icosidodecahedron. It has 30 vertices, 60 edges, 20 triangular faces, and 12 pentagonal faces. Every edge separates a triangle from a pentagon, so it attains equality in the triangle-pentagon planar edge bound.

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