Making a regular tetrahedron out of concrete

Question

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.

Answer 1

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"

https://www.youtube.com/watch?v=YGzqbEeVWhs

Answer 2

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

Answer 3

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!