[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, \( 3\leq k \leq 5\).

Note by Tasha Kim
11 months, 1 week ago

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