A World of Polyhedra
A cube can be defined as a convex polyhedron where at each vertex three squares meet each other. We may write this briefly as .
An truncated icosahedron is a 32-faces convex polyhedron, known from the pattern of a soccer ball. At each vertex, one regular pentagon and two regular hexagons meet, so that .
Using this notation, which of the following arrangements cannot represent a convex polyhedron made of regular polygons?
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1 solution
The sum of the angles at the vertex should remain less than 3 6 0 ∘ . The internal angles of a regular n -gon may be calculated as α = 1 8 0 ∘ − n 3 6 0 ∘ .
Thus we find ( 3 , 3 , 3 , 3 , 4 ) → 4 ⋅ 6 0 ∘ + 9 0 ∘ = 3 3 0 ∘ ; ( 3 , 5 , 3 , 5 ) → 2 ⋅ 6 0 ∘ + 2 ⋅ 1 0 8 ∘ = 3 3 6 ∘ ; ( 4 , 6 , 8 ) → 9 0 ∘ + 1 2 0 ∘ + 1 3 5 ∘ = 3 4 5 ∘ ; ( 3 , 1 0 , 1 0 ) → 6 0 ∘ + 2 ⋅ 1 4 4 ∘ = 3 4 8 ∘ ; ( 3 , 6 , 3 , 6 ) → 2 ⋅ 6 0 ∘ + 2 ⋅ 1 2 0 ∘ = 3 6 0 ∘ .
The last situation results in a total angle of 3 6 0 ∘ , making the vertex flat: this situation corresponds to a tessalation of regular hexagons and equilateral triangles: