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

Take the 5 equilateral triangles that meet at one of the vertices, and cut them from the remainder of the icosahedron while still maintaining the same contact with each other. Lay this cap down on a base to form a pyramid with a regular pentagonal base. Choose any diagonal of the base; it forms a triangle with two sides of the pentagon. Let the sides of the pentagon be length 1. The angle at the vertex of the triangle that the diagonal creates is 108 degrees. Bisect this angle to form two right triangles and find that half the length of the pentagon’s diagonal is sin(54°).

Now draw a median of one of the two triangles of the pyramid directly above the chosen diagonal, from one end point of the diagonal on the pentagon to the middle of the edge of the pyramid that lies above the diagonal. Then connect that midpoint to the midpoint of the diagonal. The result is a right triangle, with half the diagonal included as a leg. The right angle is at the midpoint of the diagonal on the base, as a line directly up from the base has to be merely rotated about the diagonal as an axis, as the slant edge of the pyramid is at right angles (but skew) to the diagonal.

The leg of the right triangle is half the diagonal, that is sin(54°). The hypotenuse is the slant height of the pyramid, that is, the altitude of the equilateral triangle, which is sqrt(3)/2. The angle at the edge is in fact half the dihedral angle as the plane of the triangle is perpendicular to the edge, by symmetry. That half dihedral angle is therefore arcsin(sin(54°)/(sqrt(3)/2) = arcsin(2 sin(54°)/sqrt(3)) = 69.09484255211072 degrees. Twice this is 138.1896851042214 degrees, which is the sought dihedral angle.

Posted by Charlie on 2003-04-17 03:55:24