The "double triangle" algorithm creates fractals in the following
manner:

These some of the "double triangle" fractals from Ned May's page. For
a more extensive fractal gallery and lots of other neat stuff, I recommend
you visit his website at



the larger faces of the children to the smaller faces of the parent. This model is incomplete. Each double triangle should have two arms of tetrahedrons spiraling out of them but they couldn't be constructed them because they collide with one another. 
The sides of the triangle have to be of the proportion x^0, x^1, x^2.
.618 > x >1.618, or to be more precise, 1/phi > x >phi, phi being the golden mean.
If the triangles are scalene, the tet must consist of two larger congruent triangles and two smaller congruent triangles.
It was difficult for me to explain why, but John Conway was kind enough
to help me:
On 25 Feb 1999, Hop wrote:
> I've been playing around with tetrahedrons whose 4 triangular faces
are
> all similar.
> Conjecture 1.
> The sides of the triangle have to be of the proportion x^0, x^1,
x^2.
Far from it! Indeed four copies of any acuteangled triangle
can
be used to make a tetrahedron of the special type called a "sphenoid".
You can get a sphenoid by taking alternate vertices of an arbitrary
rightangled "brick".
However, I expect you may have meant to add the requirement
that
the faces were not all congruent. In that case it seems quite
likely
that your conjectures will be easy to prove. I'll try to do so
after
I've got to the end of your message.
> Conjecture 2
> .618 > x >1.618,
> or to be more precise,
> 1/phi > x >phi, phi being the golden mean.
>
> Conjecture 3
> If the triangles are scalene, the tet must consist of two larger
> congruent triangles and two smaller congruent triangles.
Suppose the edges of the triangles are in the proportion
1:x:y with
1 < x < y, and that their smallest edges are a,b,c,d
in increasing
order, so that their entire set of edgelengths is
a < ax < ay, b < bx < by, c < cx < cy, d < dx < dy.
Then this is the set of edgelengths of the tetrahedron, but with
each edge coming twice (because it's an edge of two triangles.
But obviously the two smallest numbers in the above list
are a and b,
so we must have a = b, and the two largest numbers are cy and dy, we
must similarly have c = d, proving your Conjecture 3.
Now we see that the edgelengths of triangles numbered
1 2 3 4 are
a < ax < ay, a < ax < ay, c < cx < cy, c < cx < cy
and we know that 1 and 2 meet along their aedges, and 3
and 4
along their cyedges. So if you go through the ax and ay
edges of 1
you must get to 3 and 4, without loss of generality in that order.
This forces ax = c, ay = cx = ax^2, so the lengths are now seen to be
a < ax < ax^2 ax < ax^2 < ax^3 (each twice).
This proves your Conjecture 1. Since the longest edge of a triangle
is
less than the sum of the other two, we have ax^2 < a + ax,
which
proves your Conjecture 2.
> I am hoping someone can prove these conjectures for me (or disprove,
but
> that would be disappointing).
The fact that you made these conjectures shows that you've
obviously
thought about the situation in some detail, and I suspect that in a
way
you'd "really" proved them yourself, but didn't have enough confidence
to write out a proof for yourself. That's also a bit disappointing!
I hope my example of writing a proof will be helpful to you in the
future.
John Conway
The tetrahedrons in a fractal may be all of the same proportion but
they switch from right handed to left handed tetrahedrons with each iteration.
Shown here are the hexagons I used to cut the x=.75 tetrahedrons. I score along the inside lines to make it fold better (these lines lie on the tet's outer surface). I call the left tetrahedrons Z tets, and I call the right tetrahedrons S tets. 
 So, for example, the tets corresponding to x=.75 and x=1.33 may be the same shape, but one is left and the other is right.  It's neat that this family of fractal generating tetrahedrons are connected to the interval between (1/phi) and phi. Especially so, since the Fibonacci sequence is one Ned May's interests (Ned May's explanation of the double triangle fractal sparked the idea for these fractals). 

Hop's Pages:
Tessellating Blocks Toys that tessellate have the power to approximate any shape.
Delta Blocks (Inspired by Escher) This page discusses my toy which was inspired by M.C. Escher's print Flatworms. Flatwormsshows a structure made from the octahedron/tetrahedron tessellation. The octahedron, tetrahedron structure was also promoted by Buckminster Fuller.
Ram's Horn. A ram's horn spiral plus some spiral tessellations.
Buckminster Fuller's Octetruss This page briefly discusses Buckminster Fuller's invention based on the octahedron/tetrahedron tessellation. For more on Fuller please go to the Links to Other Folks Pages page.
Other Escher Ideas
Escher Perspective
Study This page discusses perspective,
also features one of my drawings  "Crab".
Escher Dyson Sphere To me this print by Escher looks like a Dyson Sphere! I wonder what he would have thought of this bizarre interpretation.
CurlUps as Robots? I asked a Dutch friend to translate the Escher print "Curl Ups". His comments suggest Curl Ups as robots. His translation and comments are on this page.
Miscellaneous
Fractals
from Kepler's Solids This page discusses a fractal idea I had, plus
illustrations.
Double Tetrahedron Fractal. A three dimensional version of the double triangle fractal. This model is incomplete and not very well done. However, I think it suggests an likable family of fractals. This family has an interesting connection with the golden mean
Links To Other Folk's Pages This is an indeed motley assortment of pages. Lots of good stuff!
Hop's Personal Page A page about me.
Ajo Copper News. Our business, a weekly newspaper
Gabrielle's page My sister's page