Double Tetrahedron Fractal

Ned May's Description of a double triangle fractal led me to the idea of a double tetrahedron fractal.
The text and illustration below are taken from May's web site.
"Double Triangle" Fractals

The "double triangle" algorithm creates fractals in the following manner: Begin with line segmentAB, and add point C to form a triangle. Repeat the same process by forming similar triangles on line segments AC and CB, creating points D, and E. Once again, repeat the process of creating similar triangles on the new line segments AD, DC, CE, and EB, giving rise to points F, G, H, and I
Continue in this manner until such time as a line segment becomes shorter than a pre-defined length (usually less than one pixel in the display), and then plot a point at the midpoint of that line. Beak Claw Ferns Mica

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
http://luna.moonstar.com/~nedmay/chromat/fractals.htm

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My first thought was "Is there a 3D version of this?"
A triangle is a simplex in two dimensions, so it seemed reasonable to try the 3-D simplex, the tetrahedron.
- After a few iterations this model is starting to look like a four legged animal. My daughter thinks is the bull dog who is always saving Tweety Bird from the nefarious Sylvester.
- This is made by shrinking tetrahedrons and attaching 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.
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To clone and shrink a tetrahedron, You would need the larger triangle faces to be similar to the smaller triangle faces.
Playing around with this three things became evident to me:

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 acute-angled 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
right-angled "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  a-edges, and  3  and 4
along their  cy-edges.  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

> 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

Conway 's a powerful mind. Getting a letter from him is no distinction for he is respectful and friendly to the clueless as well as the savy. Regardless, I still get happy every time I see his name in my mailbox. 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. If you start with x^0 and move counter clock wise to x^1 and then to the side x^2, all numbers x such that (1/phi) Here is the first few iterations of the double triangle fractal from a 45 degree Isosceles triangle. Note that the triangles also spiral into themselves. But this is much easier to render in 2 dimensions than 3. A possibility is to use a 3-D printer like those used in rapid prototyping. The space occupied by 1, 3, or an odd number of tetrahedrons could be a clear resin. That space occupied by an even number of tetrahedrons could be a tinted clear resin. I believe this would be a beautiful and hypnotic object.

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.

Curl-Ups 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