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Jeffery Mewtamer

Post subject: Some questions about polyhedra. Posted: Thu Jul 04, 2013 10:59 am 

Joined: Sun Nov 23, 2008 2:18 am

Over the last few months, I have started amassing a collection of pieces for the polydron geometric building set, and playing around with it has gotten thinking about various polyhedra, and some questions have bee nagging at my mind that I cannot find solutions to via Google.
1. I constructed a polyhedron with the following properties. 12 pentagonal faces and 10 triangular faces. the pentagonal faces are arranged into 2 groups of six similiar to a regular dodecahedron. the two "halfdodecahedron" are positioned such that they touch only at the vertexes of the pentagons. The spaces between the two "halfdodecahedra" are filled by an equatorial band of triangles, in five pairs. All edges have dihedral angles less than 180. The vertexes are as follows: 10 x 5,5,5; 10 x 5,5,3,3; and 5 x 5,3,5,3.
I have three main questions about this polyhedra: Does it have a special name? It appears to be a Johnson solid, but does not appear to match any of the Johnson solids, leading me to believe that their is a slight distortion from regular in either the triangles or pentagons. What is this distortion? If one wanted to replace each pair of triangles with a rhombus, what rhombus would make for the least deviation from all the pentagons being regular?
2. Polydron features 4 different triangle pieces: A equalateral triangle with unit edge length. a equalateral triangle with edge length sqrt(2). A right triangle with sides 1:1:sqrt(2). An isoceles triangle with sides 1:sqrt(2):sqrt(2).
Using these triangles, how many different tetrahedra can be constructed and how many of each triangle would be needed to construct all of them simultenously?
I suspect that there are 11 tetrahedra constructible with such a set of triangles, from the following logic: 2 regular tetrahedron: 1 small, 1 large Treating all edges as equvalent and ignoring orientation: There is one way have having one edge of irregular length, and the irregular length can be either longer or shorter than the regular length. There are 2 ways of picking egges to have irregular length(either an adjacent pair, or a nonadjacent pair), and again two options for the irregular length compared to the regular length. The are three ways of picking three edges(3 meeting at a vertex, three forming a triangle, three making a zigzag pattern), with the odd length being longer equalvalent to the odd length being shorter. for 2+2+4+3=11.
3. Not related to polydron since it lacks the pieces needed to construct trapazohedra in general, but I was wondering: is the triangular trapazohedron the only one whose faces can be rhombic?
_________________ Just so you know, I am blind.
I pledge allegiance to the whole of humanity, and to the world in which we live: one people under the heavens, indivisible, with Liberty and Equality for all.
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jaap

Post subject: Re: Some questions about polyhedra. Posted: Thu Jul 04, 2013 11:54 am 

Joined: Wed Mar 15, 2000 9:11 pm Location: Delft, the Netherlands

Jeffery Mewtamer wrote: 1. I constructed a polyhedron with the following properties. 12 pentagonal faces and 10 triangular faces. the pentagonal faces are arranged into 2 groups of six similiar to a regular dodecahedron. the two "halfdodecahedron" are positioned such that they touch only at the vertexes of the pentagons. The spaces between the two "halfdodecahedra" are filled by an equatorial band of triangles, in five pairs. All edges have dihedral angles less than 180. The vertexes are as follows: 10 x 5,5,5; 10 x 5,5,3,3; and 5 x 5,3,5,3.
I have three main questions about this polyhedra: Does it have a special name? It appears to be a Johnson solid, but does not appear to match any of the Johnson solids, leading me to believe that their is a slight distortion from regular in either the triangles or pentagons. What is this distortion? If one wanted to replace each pair of triangles with a rhombus, what rhombus would make for the least deviation from all the pentagons being regular?
I don't think it will have a name, because it is indeed not a Johnson solid. If the edge length of the pentagons is 1, then the length of the edge shared by the two triangles needs to be about 1.0514622. (The exact value is 2f / sqrt(4f+5) where f is the golden ratio.) Jeffery Mewtamer wrote: 2. Polydron features 4 different triangle pieces: A equalateral triangle with unit edge length. a equalateral triangle with edge length sqrt(2). A right triangle with sides 1:1:sqrt(2). An isoceles triangle with sides 1:sqrt(2):sqrt(2).
Using these triangles, how many different tetrahedra can be constructed and how many of each triangle would be needed to construct all of them simultenously?
I suspect that there are 11 tetrahedra constructible with such a set of triangles, I agree with that, though it could be 12 because that zigzag one is asymmetric and you could include its mirror image in the count. Jeffery Mewtamer wrote: 3. Not related to polydron since it lacks the pieces needed to construct trapazohedra in general, but I was wondering: is the triangular trapazohedron the only one whose faces can be rhombic? Yes, because the hexagon is the only regular polygon with a radius equal to its edge length (consider what the top view of the trapezohedron would look like if the faces were rhombi).
_________________ Jaap
Jaap's Puzzle Page: http://www.jaapsch.net/puzzles/


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Jeffery Mewtamer

Post subject: Re: Some questions about polyhedra. Posted: Tue Jul 30, 2013 11:17 am 

Joined: Sun Nov 23, 2008 2:18 am

Revisiting the above mentioned Johnson nearmiss based on a Dodecahedron, I realized that any number of rows of triangles can be placed between the two halves of the dodecahedra. With two rows, the triangles appear to form a ring of ten rhombuses around the equator of the resulting solid forming what George w. Hart describes as a Zonish Polyhedra(specifically, a regular dodecahedra with one zone added), but is the diamond the correct rhombus for such a transformation of the Dodecahedron, or is the actual rhombus of different angles and too small a defect to be noticeable with the materials I am using. With 3 or more rows, the ten equalatorial faces appear to just degenerate into a straight namond where n is the number of rows of triangles, but are they proper namonds, or does the light defect in the n=1 case persist for higher n? Also, I am interested in the family of "square barrels" mention here and here, and the generalizations of the spenocorona, sphenomegacorona, and hebesphenomegacorona mentioned her but the pages in question lack sufficient text for me to get a grasp of the shapes being described, and my vision is far too poor for me to attempt looking at the graphical representations. Any help in wrapping my mind around these shapes without my vision would be much appreciated.
_________________ Just so you know, I am blind.
I pledge allegiance to the whole of humanity, and to the world in which we live: one people under the heavens, indivisible, with Liberty and Equality for all.
My Shapeways Shop


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Gus

Post subject: Re: Some questions about polyhedra. Posted: Tue Jul 30, 2013 4:12 pm 

Joined: Sun Mar 15, 2009 12:00 am Location: Jarrow, England

Have you tried looking at this website? If you select a solid and then click on "coordinates" at the bottom of the 3D java window it gives you the formulae for calculating the vertices and faces.
_________________ My Shapeways Shop: http://www.shapeways.com/shops/gus_shop


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