# Making a regular tetrahedron out of concrete

I'm trying to make the following tetrahedron made of concrete just for fun: Each edge is a beam with a triangular cross section. I imagine the easiest way is to make 6 identical truncated triangular prisms and glue them. Identical because I would need to make only one mold.

The problem I'm having is figuring out the angles to make the mold. Currently I have the following equilateral triangle prism I made just for testing: Equilateral because it could be rotated to whatever edge it would be placed, but I tried recreating it on Autocad and the pieces wouldn't fit together.

What I want is to find out what are the angles I need at the end of each prism and build a wooden piece to put into the mold to make the final piece.

• Practically, it might be easier to make two tetrahedral shells (3 equilateral triangles each) and put one inside the other with triangular prism spacers and then do the concrete with one pour. – Michael Biro May 5 '18 at 2:17
• Note that your image shows roughly rectangular beams as edges. For triangular edge beams, the intersection at a vertex will be quite complex, I think. Perhaps use e.g. OpenSCAD to procedurally generate the shape (with triangular beams) first? – Nominal Animal May 5 '18 at 2:27
• You will need to know some of the angles given on the Wikipedia page. en.wikipedia.org/wiki/Tetrahedron – John Wayland Bales May 5 '18 at 3:36
• I found a page that does this in wood. I think you can see that each of the six final wooden pieces has 7 sides. The "inner" face of your triangular prism becomes a hexagon sbebuilders.blogspot.com/2013/03/… – Will Jagy May 5 '18 at 15:42
• I put an answer, including a cardboard version of one seven sided piece. With Pictures! – Will Jagy May 5 '18 at 20:47

It is convenient to place the image of the tetrahedron with vertices at every other vertex of a cube. I am imagining the cube of side 10 units, whatever they might be.

The seven sides of one of the six pieces, the one going between $(0,0,10)$ and $(10,10,10)$ are given by:

Long hexagon: $$z=8$$ Two long trapezoids: $$x-y+z = 10$$ $$-x+y+z = 10$$ Four triangles, invisible after final assembly: $$x = z$$ $$y = z$$ $$y+z = 10$$ $$x+z = 10$$

The edges bounding the polygons, if made from thin cardboard as I did as a child, come from pairs of intersecting planes; the vertices from triples of intersecting planes.

I made jpegs of these, if it lets me load them... you can print them out, cut out around the outline, make folds where needed, then use some tape to place the two figures together. Hmmmm. The paper is won't stay flat, very hard to tape together. Better to glue these to some thin cardboard, then cut out those pieces, then put together. Still, I can tell already from the paper version that i got it right.

Version by user Nominal Animal: Alright, did with cardboard, then lightly scored some of the lines for folding, rather than cutting into all separate pieces. Came out well. I do have a digital camera, so you can get some idea.

Someday in the far future, cryptoarchaeologists will happen upon this site, see that I tried to convince the OP that he ought to build his shape in something easy first, as it has seven sides, and they will say, "It was good"

• Very nice! I shamelessly stole your plans, and put an SVG version of the net here, in case anyone wants to create one from a single piece of cardboard, or play with it in Inkscape. I included small triangular tabs for gluing, too. Did you notice the cross section is not an equilateral triangle? (Side lengths are $2$, $\sqrt{3}$, and $\sqrt{3}$. No, OP did not ask for equilateral triangular prisms, I just wanted to point it out.) Now, whose 3D printer can I borrow? :) – Nominal Animal May 6 '18 at 21:01
• @NominalAnimal, good. You should find some way to call this to the OP's attention. With your version, he might actually go ahead and build six out of cardboard or paper and see how they really fit together. Note how the dihedral angles come out, for instance between two of the triangles we get $120^\circ.$ And yes, The answer currently below this pointed out that one dihedral angle of the triangular cross section is $\arccos \frac{1}{3}.$ The way i labeled the figures, the triangular cross section sides are $\{ 2, \sqrt 3, \sqrt 3 \}$ – Will Jagy May 6 '18 at 22:19
• @WillJagy Wow! I'm very impressed how far you went. This is so awesome! Thank you very much! – Yuri F Becker May 7 '18 at 1:09

I think you are after the dihedral angle of the regular tetrahedron which is $\cos^{-1}(1/3)$ or about 70.53 degrees.

• Wow, thanks, you're right. I thought all prisms would be equilateral. So I have to make 1 angle 70.53 and the other two 54.733. – Yuri F Becker May 5 '18 at 4:17

With your help I was able to make a 3D model. Here are the results for a tetrahedron inside a cube of side 10 and height of the triangular cross section = 2.

First slicing the planes through the solid. The final piece is this   Now subtracted the piece from a triangular prism. Thank you all!

• Now that you have gotten this far: please place all six pieces in the same cubical shell. If that is too much, at least place three meeting at a single corner, so you see how that works out. If you place three pieces at (10,10,10) the equations for the defining planes are the same as mine but, let's see, cyclically permute the variables $x,y,z.$ That was what I had in mind, seeing the final product before making something out of concrete, which is heavy and crumbles. – Will Jagy May 9 '18 at 16:23
• Oh: as you have noticed, the choice of $z=8$ was arbitrary. At first i set it to $z=9$ which is fine for cardboard or wood. Then I thought concrete is not usually made long and thin, and the ends of these pieces will already be very pointy and fragile. – Will Jagy May 9 '18 at 16:27