Equilateral Spherical Triangles


The term ‘tessellate’ means “to fit together exactly a number of identical shapes, leaving no spaces”.

Since the area of a sphere is 4piR², if n equilateral spherical triangles tessellate a sphere into n parts, the area of each triangle is:


Let us apply this to the spherical equivalent of the icosahedron on the left, and determine the spherical angle ‘a’ at each vertex of the sphere.

By central projection, the edges of the underlying icosahedron have generated a net of spherical triangles on the surface of the sphere. (I have left the edge planes in, since ‘a’ is the angle between them). The icosahedron has 20 faces, so the area A of a single spherical triangle is:


For a unit sphere, this is just:


   A = 0.62831853

The area of a spherical triangle on a unit sphere is given by:

   A =  (a + b + c) - pi       

Since a, b and c must all be equal, this becomes:

    A = 3a - pi




    a = 1.256637061 radians. To convert radians to degrees, multiply by (180 / pi)

    a = 72°

This is clearly what we expect since the total angle at each vertex must be 2pi (360°). And an icosahedron has 5 triangles meeting at each vertex, so 360° / 5 = 72°.

At this point we might ask: “How many equilateral triangles will tessellate a sphere?”

Assume we have no knowledge of the platonic solids, but can only guess how many edges meet at a single vertex. How many equilateral spherical triangles will tessellate the sphere? If e is the number of edges meeting at any given vertex, then 2pi / e will be the spherical angle for each triangle vertex.

We already know that for a unit sphere, the area A of an equilateral spherical triangle is:

   A = 3a - pi

And we also know that for a unit sphere, tessellated by n equilateral spherical triangles, the total area is:



   4pi = n(3a - pi)

Make n the subject…


Since (2pi / e) is the spherical angle for each triangle vertex, then the theoretical number of equilateral spherical triangles will be…


Tessellation only occurs when the result, n, is a whole number. As it turns out, the only valid values for e are 3, 4 and 5. As we might expect, this gives the spherical equivalents of the tetrahedron, octahedron and icosahedron.

tetra.png octa.png ico.png
tetra-sph.png octa-sph.png ico-sph.png

e = 3
n = 4
a = 120°

e = 4
n = 8
a = 90°

e = 5
n = 20
a = 72°