It seems only fair that I now post my own no-paper-required solution… - delta_november
Feb. 6th, 2013
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
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.
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)
[x y 1/3] dot [0 0 1] = cos(theta)
cos(theta) = 1/3
theta = acos(1/3) which is a hair more than 70 degrees