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?
$\endgroup$ 12 Answers
$\begingroup$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).
$\endgroup$ $\begingroup$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.
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