Conference rusure::math

    re .0,
    
    I may be wrong on this but I don't think you can have more than those 
    three regular polyhedra even with the relaxed condition.  The reason 
    being that once you start with any of the above three, the angles 
    between the triangles become fixed, and the convexity of the polyhedra
    forces you into the above one of the three.  This of course assumed
    that all the equilateral triangles have the same sizes.
    
    Eugene

Re: .2:

> 	At any edge, e, two faces meet at an angle, �.   Say that e links 
> vertices v and v'.   Then � determines the number of faces which come together 
> at v, and also the number of vertices which come together at v'.   So the
> degree of v is the same as that of v'.   Since the net of the polyhedron is
> connected, all vertices have the same degree, and the polyhedron is regular.

    Sorry, I don't grasp it.

Re: .1, .2 

    Here's an example of one such new solid. Place two pentagonal based
    pyramids base to base. This yields a ten sided non-regular equilateral
    triangle faced convex solid. There are others.

    -Franklin

    Let's see, the 3 Platonic solids.
    
    The "truncated" Dodecahedron mentioned in .3
    
    Two facing Tetrahedrons.
    
    And two Tetrahedrons with a connecting Octahedron.
    
    That should be it.
    
    --  Barry

Re .3: My brain is an ancient fermenting turnip.

	I have removed the absurd .2, but unfortunately regular readers of
this notesfile will still be able to determine the author, through the
characteristic use of the symbol �, copied into .3.   So, if I can't escape,
I'd better solve the problem.

	Firstly, some of the reckless .2 is still valid.   Two faces cannot 
share a part of an edge without sharing all of it.   Hence all the triangles 
must be of the same size.

	So now we use some graph theory, a sturdy subject where I find my
intuition much more reliable than geometry.   We use Euler's rule.   Let
vi denote the number of vertices of degree i, for i=3,4,5.

	Then: 3*v3 + 2*v4 + v5 = 12.

	The handshaking lemma says that v3+v5 is even.

	To any triangular polyhedron will correspond a triangulated planar
graph.   Since the polyhedron is triangulated, it is rigid [3n degrees of 
freedom for the vertices, with 3n-6 constraints for the edges, and 6 degrees
of freedom for the location of the rigid body in 3-space, but this is a 
hand-wave], and so no two t. polyhedra can have the same t. planar graph.

	Denote a candidate degree sequence by (v3,v4,v5), then by dink of a
certain amount of graph theoretic scratching, we get some answers.   Since the
vertices of degree 3 can never be adjacent (except in a tetrahedron) where one
exists, we can remove it and find the graphs to which a single vertex could
then be stuck.   Often, these do not exist, or where they do exist, they
do not have a face free of adjacent vertices of degree 5.

(4,0,0)		(#1) tetrahedron, T
(3,1,1)		(*) can't have deg 5 vx in graph order 5, so impossible
(3,0,3)		(*) contains K(3,3), so non-planar
(2,3,0)		(#2) T-T
(2,2,2)		(#3) T-T-T
(2,1,4)		(*) v3 vces inadj, => drop one get T-T-T, but impossible
(2,0,6)		(#4) T-O-T, T stuck on opposite faces of O.
(1,4,1)		(*) contains K(3,3), so non-planar 
(1,3,3)		(#5) T-O, the only way that the deg 3 vx can be added
(1,2,5)		(*) drop the v3 vx, get T-O or #7, but impossible
(1,1,7)		(*) drop the v3 vx, get (0,4,4), but never possible
(1,0,9)		(*) no such planar graph
(0,6,0)		(#6) octahedron, O
(0,5,2)		(#7) two degree 5 cones stuck together (the example in .3)
(0,4,4)		(?1) see below
(0,3,6)		(?2) see below
(0,2,8)		(#8) "pyramid"-"twisted square prism"-"pyramid" sandwich
(0,1,10)	(*) no such planar graph
(0,0,12)	(#9) icosahedron, I

?1: start with graph of twisted square prism, with base squares abcd & ABCD.
	Link a to c.   Link A to C.
?2: start with graph of twisted square prism, with base squares abcd & ABCD.
	Link each of a,b,c,d to a vertex e.   Link A to C.

	The next thing to do is to find out whether ?1 & ?2 correspond to
realizable polyhedra.

	The thing after that is to check whether all the 9 to 11 polyhedra
identified are convex.

	I will leave these tasks to the geometrically minded, hoping only that
I haven't missed any graphs.

Regards,
Andrew.

I have a fuzzy recollection that solids of the type described in .0 are called
deltahedrons in general.  I don't remember anything beyond that, unfortunately.

If you relax the condition that the solids are convex, can you construct an
infinite variety of them?  If not, then how many are there of this new type?
(I don't know the answer.  I'm just curious.)

Wook

�        <<< Note 1279.6 by WOOK::LEE "Wook... Like 'Book' with a 'W'" >>>
�                               -< Deltahedrons? >-
�
� I have a fuzzy recollection that solids of the type described in .0 are called
� deltahedrons in general.  I don't remember anything beyond that, unfortunately.

    This is correct.

    There are a total of 8 convex deltahedra (proven by van der Waerden?)
    (I don't know what to call some of them.)

    o tetrahedron
    o octahedron
    o icosahedron
    o triangular dipyramid
    o pentagonal dipyramid
    o a 12 faced convex deltahedron
    o a 14 faced convex deltahedron
    o a 16 faced convex deltahedron

    I'll have to look at Andrew's posting to compare these results.

    I don't know how many non-convex deltahedra there are, but it
    is probably known exactly how many there are.

    - Jim

    Hmmm, I missed the T+O deltahedron (got the T+O+T) and called the
    Icosehdron a Dodecahedron (I always get theri names reversed).
    
    Re .7:
    
    There are an infinite number of non-convex deltahedrons.
    
    --  Barry

	Jim's list agrees with .5.

	The shapes with 12,14,16 faces correspond to ?1,?2 & #8, which are
all derived from the net of the twisted square prism, by either sticking a
square pyramid over a square face, or drawing a diagonal.   I would still be
interested to see what ?1 & ?2 look like.

	The shapes left over are T-O, T-O-T, T-T-T, which are presumably
non-convex.   It's easy to see that T-T-T is not convex, by imagining a
line segment joining the two extreme vertices.   Similarly, a pleasant
moment's surd-juggling shows that the point of the tetrahedron in T-O cannot
access any of the three distance corners of the octahedron.  This follows from
the surprising fact that the unit tet is taller than the unit oct, when both 
are resting on a face.   

	So none of these three candidates is convex.

	If we abandon the convexity constraint, then it seems obvious that we
can glue tetrahedra together to make fairly arbitrary shapes.   Attempting
to build a helical pattern seems to be the best way to get a shape that is
infinitely extensible without possibility of self-intersection.

Regards,
Andrew.

>	If we abandon the convexity constraint, then it seems obvious that we
>can glue tetrahedra together to make fairly arbitrary shapes.

	There's a rough analogy between these shapes and the alkanes,
organic compounds of carbon and hydrogen, without multiple bonds or loops.
The analogy is not perfect because the 4 covalent bonds that each carbon
atom has are not located at the vertices of a regular tetrahedron, but are
offset slightly according to electrostatic and interatomic forces.   Methane,
ethane, propane and butane are the first four non-branching members of this
family, but there is no shortage [yet!] of naturally occuring alkane compounds 
with many more carbons in the chain.

Regards,
Andrew.

Re: .5:

    Excellent job with the graph theory Andrew considering you
    did'nt build any of these little beasties! 

	    I did ... ALL of them!

    with the aid of an incredible reasonably priced, high quality, massly 
    distributed (e.g., Toys-R-Us) ZAKS set from Ohio Art (courtesy of 
    my 8-year).

> (4,0,0)		(#1) tetrahedron, T

    This is #1

> (2,3,0)		(#2) T-T

    This is #2

> (2,2,2)		(#3) T-T-T

    Not a deltahedron.

> (2,0,6)		(#4) T-O-T, T stuck on opposite faces of O.

    This contains coplanar faces and therefore is not a deltahedron.
    In fact, its a parallelopiped comprised of six congruent parallograms 
    of angles 60 and 120 degrees.

> (1,3,3)		(#5) T-O, the only way that the deg 3 vx can be added

    This also contains coplanar faces.

> (0,6,0)		(#6) octahedron, O

    This is #3 or equivalently a triangular based triangular (twisted) prism

> (0,5,2)		(#7) two degree 5 cones stuck together (the example in .3)

    This is #4

> (0,2,8)		(#8) "pyramid"-"twisted square prism"-"pyramid" sandwich

    This is #5


> (0,0,12)		(#9) icosahedron, I

    This is #6 or equivalently, a pentagonal based triangular (twisted) 
    prism sandwiched between two pentagonal based pyramids.


> (0,3,6)		(?2) see below

    This is #7. 

    As Andrew's (?2) notation indicates, this one of the two peculiar ones.
    However it atleast admits a simple description, namely, it consists 
    of stellating each of the three square sides of a prism of triangular 
    base with square based pyramids.

> (0,4,4)		(?1) see below

    This is #8 and the last.

    Collect $200 and advance to Boardwork!

    This is by far the most unique of all the deltahedrons. It is so 
    peculiar in that it does not admit a simple description as 
    sandwiched/stellated prisms (as far as I know) that it's 
    analysis warrants further study. Here's a template of it: 

		     ________
		     \  /\  /
		      \/__\/
		      /\  /\
		     /__\/__\____      
			 \  /\  /
			  \/__\/ 
			  /\  /\ 
			 /__\/__\      


    The top and bottom halves of this non-regular dodecahedron are congruent 
    and 180 degrees out of phase. If you build this little goody (I encourage 
    you all to do so strongly) it appears to be the 1st (however crude) 
    prototype of the famous Faberge' Egg. Also notice that the angle between	
    the edges of the joining faces between the top and bottom halves is
    almost 180. Also each half separately has one degree of motion to 
    form an open and closing mouth.


Re: .8, .7

    As mentioned above T+O+T ( (2,2,2) in Andrew's notation ) is not a 
    parallelopiped and not a deltahedron.

Re: .9

    Another way to generate an infinite family of convex deltrahedrons 
    BUT allowing coplanar faces is by iteratively doubling lengths 
    of the faces of the T+O+T parallelopiped.


    -Franklin