What are the coordinates of the vertices of a regular tetrahedron, relative to its centroid?

Question

I am trying to draw an equilateral/regular tetrahedron in Processing (subset of Java), so I have to define 4 triangles that meet at the 4 vertices. I have been able to find the coordinates for the vertices assuming 1 of the vertices is at the origin.

However, the tetrahedron rotates, and I would like its rotation to be about its centroid. This where I get stuck, I need to define the 4 planes of the tetrahedron when the centroid is at the origin.

I hope my question is clear. :/ Thanks in advance.

Answer 1

According to this Wikipedia article, if the origin is at the centre, then you can take the vertices to be

$$(\pm 1, 0, -1/\sqrt{2}),\,\,\,\,\, (0, \pm 1, 1/\sqrt{2}).$$

The article does also give some alternatives such as $(1, 1, 1, ), (1, -1, -1), \dots $ which might be easier to work with.

Answer 2

I happened to have a suitable image on my laptop (for my freshman course), so I couldn't resist

It is supposed to be a methane molecule as opposed to a tetrahedron, so think of the hydrogen atoms (green) as vertices of the tetrahedron. As you see, this is the second arrangement in Old John's (+1) answer. One vertex is at the point $(1,1,1)$ and the other three at the points $(\pm1,\pm1,\pm1)$ with exactly two minus signs occuring. In other words, four selected corners of a cube.

I second Old John's opinion that this is very well suited for 3D-rotations, because you surely know the effect of a rotation on the cube...