Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 14E / b / ii / Solution

Subdividing one face in the stated way adds one vertex, three edges, and two net faces. Starting from any spherical triangulation and repeating this operation makes $V\to \infty$ while preserving the sphere. Since every resulting average isthese averages approach six from below. Thus spherical triangulations can have average valency arbitrarily close to six.