# It seems only fair that I now post my own no-paper-required solution… - delta_november

## Feb. 6th, 2013

### 10:35 pm

It seems only fair that I now post my own no-paper-required solution to the tetrahedral angle question.

Consider unit vectors normal (and pointing out) from the four sides. Call them A, B, C and D. It is clear from inspection that:

A+B+C+D = 0

thus:

A+B+C = -D

Let's sit the tetrahedron on its base such that:

D = [0 0 -1]

It is clear from inspection that the vertical components of vectors A, B and C are equal. Since the three vertical components sum to unity, each one must be 1/3.

So:

A = [x y 1/3] where x and y are at this point unknown, and irrelevant to this problem.

Since these are all unit vectors, their dot products are equal to the cosine of the angles between them. The angle between side A and the vertical (-D) is:

A dot -D = cos(theta)

thus:

[x y 1/3] dot [0 0 1] = cos(theta)

thus:

cos(theta) = 1/3

thus:

theta = acos(1/3) which is a hair more than 70 degrees

kiwano(Link)delta_november(Link)The interesting thing about this proof is that while it does not extend to other platonic solids, I believe it does work for all dimensions.

An equilateral triangle has angle acos (1/2).

A tetrahedron has angle acos (1/3).

I believe that for higher dimension N, the angle should be acos (1/N).

While it breaks my brain to visualize, the vectors don't care. There are still more of them adding to make unity, so the fraction decreases. In four dimensions we're talking about a vector normal to a solid rather than a surface, but dot products still obey their usual rules.

kiwano(Link)The higher-dimension generalization you point out looks correct. I'd probably refer to the vectors as being normal to a hypersurface or hyperplane rather than a solid, but that's probably because I'm still amused by that quirk of nomenclature where a hypersomething is like a something, but with more dimensions than you'd expect..