# [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
6 months ago

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