Geometry of the icosahedron

This is an outline of the computation of the volume and the dihedral angle of an icosahedron. To follow this discussion, you should refer to a physical model of the icosahedron, not just a picture. We assume the icosahedron has edge length 1.

First we recall a couple of facts from plane geometry.

(1) The altitudes of an equilateral triangle T divide it into 6 small congruent triangles. Label the vertices of one of these triangles v, e, f, where v is a vertex of T, e is the midpoint of an edge of T, and f is the centroid of T. If T has edge length 1, then the small triangle vef has edge lengths ve = 1/2, vf = √3/3, ef = √3/6.

(2) A regular pentagon with edge length 1 has diagonal the golden ratio τ = (1 + √5)/2 ≅ 1.618. [You can also get this ratio using trig. Drop a perpendicular from a vertex of the pentagon to the diagonal between the two adjacent vertices to form a little right triangle. The segment x in the figure is 1/2 the diagonal of the pentagon. Since the angles of a regular pentagon are 108^o, we have x = sin(54^o), and so the diagonal has length 2sin(54^o) ≅ 1.618.]

Next we find the circumdiameter of the icosahedron.

Stand your model up so that a vertex A points straight up, and another vertex A' touches the table. These two vertices A and A' are exactly opposite (antipodal), and the segment between them is a circumdiameter. This segment is a diameter of the circumsphere, a sphere that passes through all 12 vertices of the icosahedron. The midpoint c of this circumdiameter is the center of the circumsphere.

Still holding your icosahedron with a vertex A pointing up, pick an edge E that ends at A, and look at the vertex B at the other end of the edge E. Find the vertex B' exactly opposite to B. Then B'A' is an edge E' of the icosahedron, and the edges E and E' are parallel. In fact the quadrilateral ABA'B' is a rectangle with opposite edges E and E'. The other two edges AB' and A'B are parallel, and they have the same length.

Now comes the tricky part. Find a vertex C of the icosahedron so that BC and CA' are edges of the icosahedron. Then BC and CA' are two edges of a planar regular pentagon made of edges of the icosahedron (the red pentagon in the figure below), and BA' is a diagonal of this pentagon. Therefore by (2) above, the length of BA' is (1 + √5)/2 (= 2sin(54^o)). So the rectangle ABA'B' is a golden rectangle!

The circumdiameter AA' is the diagonal of this rectangle. We use the Pythagorean theorem to compute that the circumdiameter is √(1² + τ²) ≅ 1.902.

Now we can find the volume of the icosahedron

We divide the icosahedron into 20 triangular pyramids, one for each face of the icosahedron. Each pyramid has base a face of the icosahedron and vertex the center c of the icosahedron. So the volume of the icosahedron is 20 times the volume of one of these pyramids: V = 20(1/3)Bh, where B is the area of the base and h is the height of the pyramid. The base is an equilateral triangle with edge length 1, so it has area (1/2)(1)(√3/2) = √3/4. Thus V = 5(√3/3)h.

To find h we use the circumdiameter we computed above. Pick a face T of the icosahedron, and consider the triangle vfc, where v is a vertex of T, f is the center of T, and c is the center of the icosahedron. This is a right triangle with right angle at f. The edge vc is half the circumdiameter, so vc = 1/2√(1² + τ²) ≅ 0.951. The edge vf = √3/3 ≅ 0.577 by (a) above. The edge fc is the height h of our pyramid. By the Pythagorean theorem h = √((0.951)² - (0.577)²) ≅ 0.756. So V ≅ 2.18. (If the icosahedron has edge length z, then the volume is 2.18 z³.)

Finally we find the dihedral angle.

Look at a face T of the icosahedron, and consider the right triangle efc, where e is the midpoint of an edge of T, f is the center of T, and c is the center of the icosahedron. The right angle is at f, and the angle θ at e is 1/2 the dihedral angle. (The dihedral angle is the angle between T and the adjacent face at e.) The leg fc of this triangle has length h ≅ 0.756. The leg ef has length √3/6 ≅ 0.289 by (1). Thus the dihedral angle is α = 2 arctan(0.756/0.289) ≅ 138.2^o. (The dihedral angle does not depend on the size of the icosahedron.)