# Geometry : Possible polygonal arrangements around a vertex

*Created on July 17, 2022. Last update on July 17, 2022.*

## Introduction

In my research about polyhedron with convex regular polygons as faces (and that do not self intersect), I have found myself going back to 2D space because everything is simpler there. The original question was about what combination of faces can one put around a single vertex ? The minimum number of faces, the valence of this vertex, is 3 which you can have in a tetrahedron, a cube or a dodecahedron. We don't consider the flat degenerate case with two of the same face one top of the other.

Regarding the upper limit for the vertex valence, since we allow concave polyhedron, there is nothing preventing putting an arbitrary large number of equilateral triangles around a single vertex, the line of the exterior edges forming a zigzag, because the faces and edges have no thickness. Though, it is not proven yet that such a vertex would exist in an actual polyhedron.

Since the problem looks not so easy to me, I have tried to tackle it in the 2D plane. The extra condition of flatness makes things much simpler. So, the question that arose was: What combination of convex polygon can one put around a vertex side to side ? This may not be so clear because the phrasing is poor, so let's have an example. I have a pentagon, a square and two equilateral triangles: can I put them together around a vertex ? The answer is yes, the configuration is illustrated in Figure 1. The total angle is 108 + 90 + 2 x 60 = 318 degrees which is smaller than 360 degrees. From this example, the set of rules for the general problem is the following:

- all polygons are convex and regular with unit length
- all polygons share a common vertex,
- two polygons are connected by a single edge,
- an edge can only belong to one or two polygons,
- any two polygons cannot cross each other or be included into one another.

## A little bit of math

The first question that comes to mind is about the relationship between
the number of vertices n (an integer larger than 3) of a polygon and its internal angle beta.
From the drawing in Figure 2, the triangle is a sector of the polygon, so we have the following
equations:
α = ^{Π}⁄_{n}
and α+^{β}⁄_{2} = ^{Π}⁄_{2}
so β = Π(n-2)/n
Note that, for all n ≥ 3, β ≤ Π and that for n tending to infinity, β tends to Π which is the angle
of the tangent of a circle. The polygon tends to a circle with an infinite radius because the edge
length is 1. Here is a table of the first few values that will be handy later:

Vertex number n | Angle β in radians | Angle β in degrees |
---|---|---|

3 | ^{1}⁄_{3} Π |
60 |

4 | ^{1}⁄_{2} Π |
90 |

5 | ^{3}⁄_{5} Π |
108 |

6 | ^{2}⁄_{3} Π |
120 |

7 | ^{5}⁄_{7} Π |
182.57... |

8 | ^{3}⁄_{4} Π |
135 |

9 | ^{7}⁄_{9} Π |
140 |

10 | ^{4}⁄_{5} Π |
144 |

11 | ^{9}⁄_{11} Π |
147.27... |

12 | ^{5}⁄_{6} Π |
150 |

13 | ^{11}⁄_{13} Π |
152.31... |

18 | ^{8}⁄_{9} Π |
160 |

20 | ^{9}⁄_{10} Π |
162 |

24 | ^{11}⁄_{12} Π |
165 |

42 | ^{20}⁄_{21} Π |
171.43... |

## Defect Angle

From the set of rules, the condition of a valid set of N polygons around a vertex is the following:
the sum of the internal angle β_{i} of the polygons P_{i} (i going from 1 to N) must be smaller or equal
than 2Π : sum(β_{i}) ≤ 2Π.

The defect angle is the difference between the full 2Π angle and the sum, it must remain positive :
D = 2Π - sum(β_{i}) ≥ 0.

For simplicity, we order the polygons by their vertex count n_{i} in ascending order so that:
n _{1} ≤ n_{2} ≤ ... ≤ n _{N-1} ≤ n _{N}

Using the previous equation and after dividing by Π, we then have:
2 - sum( (n_{i} - 2) / n_{i}) ≥ 0

At this point, we can solve this inequation for the values n_{i}. Solutions
will be noted by the sequence of vertex number between brackets: {n_{1}, n_{2}, ..., n_{N-1}, n_{N}}.
The solutions will sometimes be accompanied by the defect angle D.

## Results

There is a total of six classes from 1 to 6, the class number being the valence of the vertex.

#### Class VI : Valence of 6

Since the smallest angle we can have is that of the equilateral triangle, Π/3,
the largest valence is (2Π)/(^{Π}⁄_{3}) = 6.

The only solution is : {3, 3, 3, 3, 3, 3}, D = 0.

#### Class V : Valence of 5

There are five solutions, that I classify in two categories:

- 4 solutions of the form {3, 3, 3, 3, n} with 3 ≤ n ≤ 6, D = 0 for n = 6.
- 1 solution: {3, 3, 3, 4, 4}, D = 0.

#### Class IV : Valence of 4

There are six categories, one of them being an infinite family:

- an infinite family of solutions of the form {3, 3, 3, n} with n ≥ 3.
- 9 solutions of the form {3, 3, 4, n} with 4 ≤ n ≤ 12, D = 0 for n = 12.
- 3 solutions of the form {3, 3, 5, n} with 5 ≤ n ≤ 7, D = 2/15 Π for n = 7.
- 1 solution: {3, 3, 3, 6, 6}, D = 0.
- 3 solutions of the form {3, 4, 4, n} with 4 ≤ n ≤ 6, D = 0 for n = 6.
- 1 solution: {4, 4, 4, 4}, D = 0.

#### Class III : Valence of 3

There are 18 categories, five of them being an infinite family:

- an infinite family of solutions of the form {3, 3, n} with n ≥ 3.
- an infinite family of solutions of the form {3, 4, n} with n ≥ 4.
- an infinite family of solutions of the form {3, 5, n} with n ≥ 5.
- an infinite family of solutions of the form {3, 6, n} with n ≥ 6.
- 9 solutions of the form {3, 7, n} with 7 ≤ n ≤ 42, D = 0 for n = 42.
- 3 solutions of the form {3, 8, n} with 8 ≤ n ≤ 24, D = 0 for n = 24.
- 9 solutions of the form {3, 9, n} with 9 ≤ n ≤ 18, D = 0 for n = 18.
- 9 solutions of the form {3, 10, n} with 10 ≤ n ≤ 15, D = 0 for n = 15.
- 9 solutions of the form {3, 11, n} with 11 ≤ n ≤ 13, D = 0.007323... rad for n = 13.
- 1 solution: {3, 12, 12}, D = 0.
- an infinite family of solutions of the form {4, 4, n} with n ≥ 4.
- 9 solutions of the form {4, 5, n} with 5 ≤ n ≤ 20, D = 0 for n = 20.
- 9 solutions of the form {4, 6, n} with 6 ≤ n ≤ 12, D = 0 for n = 12.
- 9 solutions of the form {4, 7, n} with 7 ≤ n ≤ 9, D = 0.02493... rad for n = 9.
- 1 solution: {4, 8, 8}, D = 0.
- 9 solutions of the form {5, 5, n} with 5 ≤ n ≤ 10, D = 0 for n = 10.
- 9 solutions of the form {5, 6, n} with 6 ≤ n ≤ 7, D = 0.0190... rad for n = 7.
- 1 solution: {6, 6, 6}, D = 0.

#### Class II : Valence of 2

For one polygon the internal angle cannot be larger than Π then any
pair of polygons is a solution.

The defect angle is:
D = 2Π - Π(n_{1} - 2)/n_{1} - Π(n_{2} - 2)/n_{2} = 2Π(n_{1} + n_{2})/(n_{1} n_{2})

#### Class I : Valence of 1

Any single polygon is a valid solution. The defect angle is:
D = Π(n_{1} + 2)/n_{1}

## Application to 3D space

Now, that we now all of that, what about 3D ? First, with the solutions we already have when the valence is 3 or more, by moving the shape to 3D and closing the gap between polygons where the defect angle lies, we form a local convex shape at this vertex. Consequently, any convex polyhedron (even those not strictly convex) with regular faces must have their vertices in this set. For instance, the icosahedron only has {3,3,3,3,3} vertices.

Second, there is more degrees of freedom in 3D than in 2D. In the 2D study, we discarded the cases where the polygons would overlap. In these cases, we can define the defect angle to be negative. When moving the shape to 3D, it becomes possible to glue the edges of the polygons at the two extremities together resulting in a vertex where faces necessarily form both mounts and valleys, folds and creases. In my future work, I will study what kind of configurations exists at a 3D vertex which implies explaining dihedral and solid angles.

## References

Nothing found on the subject, please let me know if you have anything.