User:Cube26

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A4+ symmetry classification

If Hemidodecahedron (4-dimensional) has space: 4D Euclidean space and A4+ symmetry is the symmetry group of Hemidodecahedron (4-dimensional), is A4+ symmetry the 4D Euclidean symmetry group?

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My new notation for anisogons (My intention to extend the notion of a star polygon)

Let $r_{0} = 1, r_{1},...,r_{n} \in (1,+\infty)\,$and $0\leq\alpha_{0},\alpha_{1},...,\alpha_{n} \leq 2\pi\,$be "the ratio between the length of nth edge and the first one" and the "nth interior angle of the anisogon (measured in radian)".

Then $(\{r_{0},r_{1},...,r_{n}\},\{\alpha_{0},\alpha_{1},...,\alpha_{n}\})$ notates an anisogon produced by the following actions:

- Put the center $A_{1}$ and $A_{2}$ in 2D Cartesian plane.

- Draw a line from $A_{1}$ to $A_{2}$.

+ Let $i=1$.

+ Repeat the following actions but if at any step, there's a cyclic graph, terminate.

• Find $A_{i+2}\,$such that$\frac{|A_{i+1}A_{i+2}|}{|A_{1}A_{2}|}=r_{i\,(\text{mod}\,n+1)}\,$ and $\angle A_{i}A_{i+1}A_{i+2}=\alpha_{i-1\,(\text{mod}\,n+1)}$.
• Draw a line from $A_{i+1}\,$to $A_{i+2}$.
• Increment $i$
  

Note

Note that this method only depends on ratio between edge length so we need a convention on how to find and read the coordinates.

The convention is $A_{1}=(0,0)\,$and $A_{2}=(1,0)\,$as coordinates.

My new anisogon

Vertex coordinates

The vertices of the anisogon $(\{1,2\},\{\frac{\pi}{2},\frac{2\pi}{3}\})$, centered at the origin has the following vertices:

• $(0,0)$
• $(1,0)$
• $(1,2)$
• $\Big(\frac{2-\sqrt{3}}{2},\frac{5}{2}\Big)$
• $\Big(-\frac{\sqrt{3}}{2},\frac{5-2\sqrt{3}}{2}\Big)$
• $\Big(\frac{1-\sqrt{3}}{2},\frac{5-3\sqrt{3}}{2}\Big)$
• $\Big(\frac{1+\sqrt{3}}{2},\frac{7-3\sqrt{3}}{2}\Big)$
• $\Big(\frac{1+\sqrt{3}}{2},\frac{9-3\sqrt{3}}{2}\Big)$
• $\Big(\frac{\sqrt{3}-3}{2},\frac{9-3\sqrt{3}}{2}\Big)$
• $\Big(\frac{\sqrt{3}-4}{2},\frac{9-4\sqrt{3}}{2}\Big)$
• $\Big(\frac{3\sqrt{3}-4}{2},\frac{7-4\sqrt{3}}{2}\Big)$
• $\Big(2\sqrt{3}-2,4-2\sqrt{3}\Big)$
• $\Big(2\sqrt{3}-3,4-\sqrt{3}\Big)$
• $\Big(2\sqrt{3}-4,4-\sqrt{3}\Big)$
• $\Big(2\sqrt{3}-4,2-\sqrt{3}\Big)$
• $\Big(\frac{5\sqrt{3}-8}{2},\frac{3-2\sqrt{3}}{2}\Big)$
• $\Big(\frac{5\sqrt{3}-6}{2},\frac{3}{2}\Big)$
• $\Big(\frac{5\sqrt{3}-7}{2},\frac{3+\sqrt{3}}{2}\Big)$
• $\Big(\frac{3\sqrt{3}-7}{2},\frac{1+\sqrt{3}}{2}\Big)$
• $\Big(\frac{3\sqrt{3}-7}{2},\frac{\sqrt{3}-1}{2}\Big)$
• $\Big(\frac{3\sqrt{3}-3}{2},\frac{\sqrt{3}-1}{2}\Big)$
• $\Big(\frac{3\sqrt{3}-2}{2},\frac{2\sqrt{3}-1}{2}\Big)$
• $\Big(\frac{\sqrt{3}-2}{2},\frac{2\sqrt{3}+1}{2}\Big)$
• $\Big(-1,\sqrt{3}\Big)$

This notation for some polygon

Name	My notation
Star pentambus	$(\{1,1\},\{\frac{7\pi}{5},\frac{\pi}{5}\})$
Pentambus	$(\{1,1\},\{\alpha,\frac{8\pi}{5}-\alpha\})$
Dipentagon	$(\{1,a\},\{\frac{4\pi}{5},\frac{4\pi}{5}\})$
Distellagram	$(\{1,a\},\{\frac{2\pi}{5},\frac{2\pi}{5}\})\,$where $a>\frac{1+\sqrt{5}}{2}$
Stellapod	$(\{1,a\},\{\frac{2\pi}{5},\frac{2\pi}{5}\})\,$where $1<a<\frac{1+\sqrt{5}}{2}$
Complex dipentagon (Degenerate)	$(\{1,\frac{1+\sqrt{5}}{2}\},\{\frac{2\pi}{5},\frac{2\pi}{5}\})\,$ or $(\{1,\frac{1+\sqrt{5}}{2}\},\{\frac{\pi}{5},\frac{\pi}{5}\})\,$
Dipentagram	$(\{1,a\},\{\frac{3\pi}{5},\frac{3\pi}{5}\})\,$
Distellagon	$(\{1,a\},\{\frac{\pi}{5},\frac{\pi}{5}\})\,$where $a<\frac{1+\sqrt{5}}{2}$
Pentagram	$(\{1\},\{\frac{\pi}{5}\})\,$
Pentapod	$(\{1,a\},\{\frac{\pi}{5},\frac{\pi}{5}\})\,$where $a>\frac{1+\sqrt{5}}{2}$
Bowtie	$(\{1,a\},\{\text{arccos}\Big(\frac{1}{a}\Big),\text{arccos}\Big(\frac{1}{a}\Big)\})\,$
Ditetragon	$(\{1,a\},\{\frac{3\pi}{4},\frac{3\pi}{4}\})\,$
Regular n-gon	$(\{1\},\{\frac{(n-2)\pi}{n}\})\,$
Ditetragram	$(\{1,a\},\{\frac{\pi}{4},\frac{\pi}{4}\})\,$
Ditrigon	$(\{1,a\},\{\frac{2\pi}{3},\frac{2\pi}{3}\})\,$
Tripod	$(\{1,a\},\{\frac{\pi}{3},\frac{π}{3}\})\,$where $1<a<2$
Propeller tripod	$(\{1,a\},\{\frac{\pi}{3},\frac{π}{3}\})\,$where $a>2$
Hemitripod	$(\{1,2\},\{\frac{\pi}{3},\frac{π}{3}\})\,$
Rectangle	$(\{1,a\},\{\frac{\pi}{2},\frac{π}{2}\})\,$
Square	$(\{1\},\{\frac{\pi}{2}\})\,$