Solution

ID: 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 while preserving the sphere. Since every resulting average is
these averages approach six from below. Thus spherical triangulations can have average valency arbitrarily close to six.

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