# Approximating a sphere with a regular tetrahedron

I'm studying computer graphics independently through a textbook and I've come across this (seemingly challenging) problem in chapter 1. It is stated as follows:

A different method of approximating a sphere starts with a regular tetrahedron, which is constructed from four triangles. Find its vertices, assuming that it is centered at the origin and has one vertex on the y-axis. Derive an algorithm for obtaining increasingly closer approximations to a unit sphere based on subdividing the faces of the tetrahedron.

I struggled with this problem for a while before looking up the answer. I couldn't get enough of an idea to even get started with this kind of task. After a few hours I looked up the answer in the back of the book:

We derive this algorithm in chapter 6. First we can form the tetrahedron by finding four equally spaced points on a unit sphere centered at the origin. One approach is to start with one point on the z-axis $(0,0,1)$. We can then place the other three points in a plane of constant z. One of these three points can be places on the y axis. To satisfy the requirement that the points be equidistant. The point must be $(0, 2\sqrt{2/3}, -1/3)$. The other two points can be found by symmetry to be at $(-\sqrt{6/3}, -\sqrt{2/3}, -1/3)$ and $(\sqrt{6/3}, -\sqrt{2/3}, -1/3)$.

My question(s) from this are:

1. What gave him the idea to inscribe the tetrahedron in a unit sphere? I suppose because we are trying to approximate one it makes sense to use it as a guide.

2. How does he choose the points and derive the answer? Admittedly my geometry is a little rusty, but I felt completely defeated by this question.

3. How can I properly visualize this process so I can repeat it on future questions?

• The answer looks wrong to me: point $(0,2\sqrt3, -1/3)$ is not on the y-axis. – Aretino Jun 24 '18 at 18:17
• I think the point's y coordinate you're referring to is $2\sqrt(2/3)$, right? – lolo Jun 24 '18 at 18:18
• Yes, right: $(0, 2\sqrt{2/3},-1/3)$. – Aretino Jun 24 '18 at 19:12
• It should be mentioned that the resulting triangles get smaller and also sharper. That is, one of the angles is very small. This is not a good triangulation. See this recent MSE question 2817605 "How to discretize a sphere?" for better options. – Somos Jun 24 '18 at 19:41

The question isn't so much about the maths, but the process. The tetrahedron is bounded by the sphere, and its four vertices touch the surface of the sphere.

Replace each triangular face, with three more triangular faces, with the new shared vertex on the surface of the sphere. The original three vertices remain, because they touch the sphere.

Do that again, and again until the calculation is too small to be significant.

Using a computer language, this can be done by recursion.

What gave him the idea to inscribe the tetrahedron in a unit sphere? I suppose because we are trying to approximate one it makes sense to use it as a guide.

This is the problem definition: increase the number of vertices of the polyhedron until it is almost like a sphere.

How does he choose the points and derive the answer? Admittedly my geometry is a little rusty but I felt completely defeated by this question.

He chooses a new point extended from the origin, through the centre of a face, to touch the required sphere.

How can I properly visualize this process so I can repeat it on future questions?

The tetrahedron is the simplest polyhedron. The exercise is to imagine how the centre of each of its faces can be extended out to touch the sphere, replaced by three new faces.

The process continues until the polyhedron so much resembles a sphere that it is not worth continuing.

• Is there a way I can visualize this easily in say wolfram alpha or mathematica? – lolo Jun 24 '18 at 20:17
• Sorry I can't find a video showing the expansion. The nearest static imagery I can find is this – Weather Vane Jun 24 '18 at 21:01