Jacob Rus

Ricky Reusser

Do I have it right that it's an approximate expansion that gives a very-nearly-triangular region in the diagram below?

Every one of the 48 curvilinear triangles – each has exterior angles [3π/4, π/2, 2π/3] – is congruent on the sphere. There’s no approximation involved. But if you conformally map these spherical triangles to flat triangles with exterior angles [3π/4, π/2, 3π/4], you get a flat cube; if you conformally map them to flat triangles with exterior angles [5π/6, π/2, 2π/3], you get a flat octahedron; if you map them to flat triangles with exterior angles [atan2(1, -√2), π/2, atan2(√2, -1)], you get a flat rhombic dodecahedron. Etc.

The picture below is the same as above, just zoomed in on one triangle.