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?
 A: Let $\vec{x}_1, \vec{x}_2, \vec{x}_3,\vec{x}_4$ be the 4 points on unit sphere that form the 
vertex of a regular tetrahedron. By symmetry, we have
$$\vec{x}_1 + \vec{x}_2 + \vec{x}_3 + \vec{x}_4 = \vec{0}\tag{*1}$$
Now rotate the coordinate axis so that $\vec{x}_1 = (0,0,-1)$, then the other 3 points will lie
on a plane with $z = K$ for some constant $K$. By $(*1)$, we have $-1 + 3K = 0 \implies K = \frac13$. In terms of spherical polar coordinates $(\theta, \phi)$ :
$$(x,y,z) = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$$
This plane $z = \frac13$ cut the unit sphere into two pieces. The upper piece corresponds to $\cos\theta > \frac13$ and the lower piece correponds to $\cos\theta < \frac13$. Since in polar coordinate, the surface element of the unit sphere has the form
$\sin\theta d\theta d\phi$. The area of the upper and lower pieces are
$$\int_0^{\cos^{-1}(\frac13)}\int_0^{2\pi} \sin\theta d\phi d\theta = \frac{4\pi}{3}
\quad\text{ and }\quad
\int_{\cos^{-1}(\frac13)}^{2\pi}\int_0^{2\pi} \sin\theta d\phi d\theta = \frac{8\pi}{3}
$$
A: (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.
A: 
how would I go about finding this sort of thing on my own?

Here's a sketch of a solution that uses simple geometry and no ingenuity or knowledge of coordinates. Details are kind of tedious and so are left for the reader, but feel free to ask if anything isn't clear.
Consider a tetrahedron $ABCD$ of side length $a$. Its base $ABC$ is an equilateral triangle, and by symmetry the centroid $O$ of the tetrahedron lies on the line joining $D$ to $ABC$'s centroid $P$. By elementary geometry one can find:


*

*the length of a median of $ABC$,

*the length $AP$, which is $2/3$ times the above length,

*the length $DP$, because $APD$ is a right triangle and lengths $AD$ and $AP$ are known, and

*the length $DO$, which is $3/4$ times the length $DP$ (consider dividing $ABCD$ into four congruent tetrahedra with common vertex $O$; by comparing volumes one can show that $OP=\frac14AP$).


Now you know $DO$, which is the the radius of the circumscribing sphere, and you know $OP$, which determines how far the plane is from the center of the sphere. That's everything you need to plug into the formula for the area of a spherical cap, and you should obtain that it's one-third of the area of the entire sphere.
Well, I guess this isn't entirely simple geometry because I don't know how to derive the area of a spherical cap through elementary means, so there's that.
