# Geodesic sphere using only Regular pentagons and hexagons

I know geodesic approximation to a construct a spherical dome shape needs 12 pentagons and these pentagons are regular pentagons.

However when I look closely hexagons are slightly different in their shapes and sizes. Is it mathematically possible to construct a geodesic sphere using 12 pentagons and REGULAR hexagons of the SAME SIZE? (for example like Truncated Icosahedron)

Would putting extra pentagons to force curvature between the regular hexagons solve this?

• \ I live quite close to Carbondale, Illinois, USA, where R. Buckminster Fuller built a home which incorporated a geodesic dome as its exterior, and where he taught at Southern Illinois University. – Donn Miller Oct 14 '17 at 18:07

The only possibility to have regular hexagons is the truncated icosahedron (aka. football). Whenever three regular hexagons meet at a vertex, you have three 120° angles meeting there, which makes the vertex undesireably "flat".

Using an $\ne 12$ pentagons (and otherwise only hexagons) will not give you a sphere because of Euler's polyhedron formula (unless you do not let three polygons meet at every vertex, but then your shape would be even more irregular).

If you ever are willing to accept solutions not involving regular hexagons:

The snub dodecahedron (4 regular triangles and 1 regular pentagon at every vertex) is quite close to a sphere in shape and is less "angular." It is comprised of 80 regular triangles and 12 regular pentagons. It is rigid.