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

An Archimedean solid has a vertex configuration $\{n_1, n_2, ... , n_k\}$ if each vertex borders regular polygons with $n_i$ number of sides, $i=1,..., k$. For example, $\{3,5,3,5\}$ 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 $\{ n_1, ... , n_k\}$. We consider the internal angle of a regular polygon, which is given by $\pi - \frac{2\pi}{n}$. This means that at any vertex, the sum of internal angles of the $n_i$-gons can be expressed as $\sum_{i=1}^{k} \left( 1-\frac{2}{n_i}\right) < 2.$ since the sum of internal face angles cannot exceed $2\pi$ at each vertex. If we divide through by $\pi$ on both sides of the equation, we have the following formula,

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

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

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

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