My friend finished making dodecahedrons, and her next project is to make regular icosahedrons (20-sided dice). Again she wants to know: what is the dihedral angle between any two adjacent faces?

Perhaps this can be solved without the use of spherical trig? ;P

Each side of the die is an equilateral triangle. Consider two adjacent sides sharing one edge, and on each triangle draw an altitude from the middle of the shared edge to the corner opposite. If we connect the ends of these two lines with a third line, and find the lengths of all three lines, we can calculate the angle X between the two triangular sides using the law of cosines.
Set the length of each edge of the icosahedron arbitrarily to 1. The height of the altitude on each triangle is then sqrt(3)/2.

Every vertex of the icosahedron consists of five triangles meeting at one point. The perimeter created by the outer edges of these five triangles is a regular pentagon with length of sides = 1. For any two adjacent triangles in this group, one end of the shared edge falls on a corner of the pentagon, and the corners of the pentagon immediately on either side are coincident to the corners of the triangles opposite their shared edge. A line between these two corners will have a length L where

L^2 = 1^2 + 1^2 – 2*1*1*cos(108) = 2.61803

Plugging in the lengths of the three sides of our mystery triangle,

L^2 = [sqrt(3)/2]^2 + [sqrt(3)/2]^2 – 2* sqrt(3)/2* sqrt(3)/2*cos(X)
X = arccos [(L^2 – 3/4 – 3/4) / (-2*3/4)]
X = arccos(-0.74535)
X = 138.189 degrees

Posted by Tim Axoy on 2003-04-18 12:55:41