Area of a spherical cap formed by the plane containing any side of an inscribed regular tetrahedron I was trying to think about this problem today and realized that practically all of my high school geometry has deserted me, so "how to find it" answers would be greatly appreciated. As to the actual problem: Imagine that a unit sphere has a regular tetrahedron inscribed in it. The plane containing any arbitrary side of the tetrahedron cuts the sphere's surface into two parts with different areas. What are the areas of the two parts, and how would I go about finding this sort of thing on my own?

(Another hint) Find the distance from the center of the inscribed tetrahedron to the center of one of its faces, call that $r$ [Once one can get the coordinates of some regular tetrahedron's vertices and center, after rescaling one can get this $r$.]
Once that $r$ is known, there is likely an available formula on-line for the area of a spherical cap given the distance of the cutting plane from the sphere origin. Alternately maybe this can be set up as a surface integral and done directly.
