[Geometry] Condition on the Vertex Configurations for an Archimedean Solid

An Archimedean solid has a vertex configuration {n1,n2,...,nk}\{n_1, n_2, ... , n_k\} if each vertex borders regular polygons with nin_i number of sides, i=1,...,ki=1,..., k. For example, {3,5,3,5}\{3,5,3,5\} will give a vertex belonging to four distinct faces, where each will be alternating triangles and pentagons.

Let's consider an Archimedean solid have a vertex configuration {n1,...,nk}\{ n_1, ... , n_k\}. We consider the internal angle of a regular polygon, which is given by π2πn \pi - \frac{2\pi}{n}. This means that at any vertex, the sum of internal angles of the nin_i-gons can be expressed as i=1k(12ni)<2.\sum_{i=1}^{k} \left( 1-\frac{2}{n_i}\right) < 2. since the sum of internal face angles cannot exceed 2π2\pi at each vertex. If we divide through by π\pi on both sides of the equation, we have the following formula,

i=1k(12ni)<2. \sum_{i=1}{k}\left( 1-\frac{2}{n_i} \right) <2.

The smallest regular polygon we can get is a triangle, with internal angle 60 degrees, 2π6\frac{2\pi}{6} radians at each corner. If we suppose that k=6k=6, the sum of internal angles of the nin_i-gons therefore, is a minimum, of 2π66=2π\frac{2\pi}{6}\cdot 6 = 2\pi, but the sum of face angles at a vertex must not exceed 2π2\pi or be equal to 2π2\pi since this would create a flat plane and we cannot construct a solid. Hence k cannot be greater than or equal to 6, meaning k5.k \leq 5.

Additionally, we see that k3k\geq 3 because if there are any two polygons glued together at any vertex or side, the solid will fall flat and fold into itself.

Therefore, 3k5 3\leq k \leq 5.

Note by Bright Glow
3 years, 1 month ago

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