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Myrtone 
Posted: May 25 2015, 02:51 PM

Casual Member Group: Members Posts: 53 Member No.: 569 Joined: 22March 10 
I have realised something about this shape. First of all, squares and equaleteral triangles have all the following properties in common:
*They all have equal side lengths as well as angles. *They can tesselate. *Unanylisable into other shapes with the same property. Any rectangle, such as an oblong, can be composed of squares, and diamonds and hexagons can be composed of triangles. Let a square be placed next to each side of a hexagon, and the gaps in between the squares can be filled with triangles makes a dozagon. So a dozagon can be composed from a combination of squares and equalateral triangles. No combination of them could make a pentagon or decagon, I think. This seems to have something to do with the divisibility of twelve. And in my country (Australia) we have a dozagon shaped 50 cent coin. 
Kodegadulo 
Posted: May 25 2015, 10:10 PM

Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
As demonstrated by my avatar (at left) , the dodecagram (gold), the complete stellation of the dodecagon (yellow), has the interesting property that it embeds four overlapping equilateral triangles (pink), three overlapping squares (green), and two overlapping regular hexagons (blue), thus highlighting the divisibility of the dozen: 10_{z} = 3 * 4 = 4 * 3 = 6 * 2.

Myrtone 
Posted: May 26 2015, 12:32 AM

Casual Member Group: Members Posts: 53 Member No.: 569 Joined: 22March 10 
I looked closely and didn't find any squares, and the pink areas appear to have four sides each.

Myrtone 
Posted: May 26 2015, 11:15 AM

Casual Member Group: Members Posts: 53 Member No.: 569 Joined: 22March 10 
The most common units of area are square units, but units of area can come in any shape, such as triangular units. A triangular unit is the area of an equilateral triangle with side lengths of one unit, just as a square unit is the area of a square with the same side length.
Both squares and oblongs have rectangular areas, as the area of them can be calculated from the length of two perpendicular sides with, in this case one of the four basic arithmetic operations, just multiply two of the sides. Right angle triangles, among some others also have rectangular areas. For example, a 5:4:3 triangle has exactly half the area of a 4:3 rectangle. In general, a C:B:A triangle has an area half that of a B:A rectangle if A^2+B^2=C^2. But there is incommensurability between the area of a equilateral triangle and that of a square, there is no ratio of integers that equals the ratio of the area of those two shapes. On the other hand, a lozenge (the shape, not cough medicine) with side lengths of one unit has exactly twice the area of an equilateral triangle with the same sidelength. And a regular hexagon with the same sidelength has twice the area of such a lozenge. Now notwithstanding incommensurability, square and triangular units can be added together. For example, the sum of one triangular unit and one square unit is one hose shaped unit. Two triangular units added to one square unit is some sort of irregular hexagonal unit, and six of them is the area of a dozagon with a length of one unit per side. So using area units of the right shape, one can calculate the area of all the above mentioned shapes with only one or more of the four basic arithmetic operations. Any irregular hexagon with the same interior angles as a regular one also has a triangular area. 
Kodegadulo 
Posted: May 26 2015, 12:02 PM


Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
The triangles, squares, and hexagons are all overlapping, so only their corner areas are exposed in color. You need to follow the lines to connect the various colored areas and imagine the shapes isolated and filled in. Four nested triangles: Three nested squares: Two nested hexagons: The dodecagon is the central figure. The hexagons represent its first stellation. The squares represent its second stellation. The triangles represent its third stellation. The dodecagram (twelvepointed star in turquoise) represents its full stellation, which can be drawn without lifting the pen. Each subsequent stellation subsumes all the preceding stellations. 

Kodegadulo 
Posted: May 26 2015, 06:31 PM

Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
Oh, by the way, it's true that "dodecagon" and "dodecagram" are Classicallyderived, but decimallybiased, names for these things, given that "dodeca" is the Greek for "twoandten". I can understand the desire to cook up a "dozenal" names instead. But "dozagon" and "dozagram" aren't very apt, because "dozen" is an English word, not a Classical Greek or Latin word. But that's where SDN comes in handy  it's our goto source for Classical (or at least quasiClassical) dozenal words. So these figures should really be called the "unquagon" and the "unquagram".

Kodegadulo 
Posted: May 26 2015, 07:50 PM

Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
To create an unquagon, you arrange a dozen vertices evenly around a circle, then draw an edge from each vertex to the next vertex.
To create the first stellation of the unquagon, you draw an edge from each vertex to the vertex two down from it, skipping one vertex in between. However, because 10_{z} Ã· 2 = 6 with no remainder, you wind up returning to the same vertex after six edges, having created a regular hexagon. So in order to include all the vertices, you need to create two overlapping hexagons. To create the second stellation of the unquagon, you draw an edge from each vertex to the vertex three down from it, skipping two vertices in between. However, because 10_{z} Ã· 3 = 4 with no remainder, you wind up returning to the same vertex after four edges, having created a square. So in order to include all the vertices, you need to create three overlapping squares. To create the third stellation of the unquagon, you draw an edge from each vertex to the vertex four down from it, skipping three vertices in between. However, because 10_{z} Ã· 4 = 3 with no remainder, you wind up returning to the same vertex after three edges, having created an equilateral triangle. So in order to include all the vertices, you need to create four overlapping triangles. To create the fourth (and final) stellation of the unquagon, you draw an edge from each vertex to the vertex five down from it, skipping four vertices in between. Since 5 is not a divisor of 10_{z}, you wind up visiting all the vertices before returning to the first one. There is no further stellation, because if you draw an edge from each vertex to the vertex six down from it, you simply wind up drawing six diameters through the center of the circle, and these cannot be considered polygons. 
Shaun 
Posted: May 27 2015, 10:32 AM

Dozens Disciple Group: Admin Posts: 1,116 Member No.: 3 Joined: 2August 05 
You can see all the shapes in this version:
Dodecagon, hexagon,square and triangle 
dgoodmaniii 
Posted: Jun 1 2015, 12:15 AM


Dozens Demigod Group: Admin Posts: 1,927 Member No.: 554 Joined: 21May 09 
You can also see the process of constructing an unquagon in the plane with compass and straightedge here, courtesy of Wikipedia: Constructing the Unquagon 

Myrtone 
Posted: Jul 9 2015, 12:33 PM

Casual Member Group: Members Posts: 53 Member No.: 569 Joined: 22March 10 
On a hyperboliods all regular polygons including Unquagons can be made to tesselate. Consider both order 3 and order 4 tiling, as both orders are factors of twelve.

Kodegadulo 
Posted: Feb 24 2016, 07:15 AM

Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
I cooked up this animated gif of my unquagram avatar:
Wait a few seconds and it will start. 
Kodegadulo 
Posted: Feb 24 2016, 02:13 PM

Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
Here is another animation of my unquagram that makes a fairly good "trice timer": It cycles once per Primel 'trice (triciaday) (Dometric's "minette"), pausing for 1 Primel 'lull (quadciaday) per point,

Kodegadulo 
Posted: Feb 21 2018, 11:28 PM


Obsessive poster Group: Moderators Posts: 4,190 Member No.: 606 Joined: 10September 11 
Thanks, but there's nothing particularly mystical or religious about my avatar, hotdog8. I don't go for that. It's just an ordinary unquagram (dodecagram), the full stellation of the unquagon (dodecagon), as already explained in the thread above. It's colorized to demonstrate the high divisibility of the dozen, highlighting the 1 unquagon (yellow), 2 hexagons (blue), 3 squares (red), and 4 triangles (green) embedded within the unquagram (gold). These figures, respectively, representing the 0th, 1st, 2nd, 3rd, and 4th stellations of the unquagon. The animation in the secondtolast post should make this clear. Each greater stellation contains within it all the lesser ones. The unquagram itself contains all the stellations. It's true that in theory one could draw the unquagram in a single stroke without lifting the pen, just like the pentagram, but in practice it's very difficult to do so successfully by hand, because of the need to exactly intersect all those internal points making up _all_ the stellations. A different animation, as shown in the previous post acts as a handy Primel ′trice·timer: It cycles with a period of 1 tricia·day = 1 ′triqua·timel = 1 ′trice = 42_{z}=50_{d} seconds, filling in another one of its points once every 1 quadcia·day = 1 ′biqua·timel = 1 ′lull = 4.2_{z} seconds. This is the animation I've used as my avatar. Other people have used different arrangements of these shapes as symbols for dozenal divisibility as well, as mentioned above. 
