Solution
= 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 is
$$
\overline d=6-\frac{12}{V},
$$
these averages approach six from below. Thus spherical triangulations can have average valency arbitrarily close to six.