geometry
Description: consider a regular tetrahedron (with height 1), construct a sphere centering at one of the tetrahedron's vertex, with radius 1 also. Then what's the surface area of the sphere's portion that gets cut out by the tetrahedron? Following are links to some illustrations I made using Mathematica.
$\hskip 1.3in$ 
$$\text{a side view}$$
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$$\text{a bottom view: you are viewing the unknown area directly}$$
We know from the isoperimetric inequality that locally the surface must be a sphere (where we can include the plane as the limiting case of a sphere with infinite radius). Also, the surface must be orthogonal to the cube where they meet; if they're not, you can deform the surface locally to reduce its area. A sphere orthogonal to a cube face must have its centre on that face. You can easily show that it can't contain half the volume if it intersects only one or two faces. Thus, it must either intersect at least three adjacent faces, in which case its centre has to be at the vertex where they meet, or it has to intersect at least two opposite faces, in which case it has to be a plane.
This is supposed to be a comment but I would like to post a picture.
For any $m \ge 3$, we can put $m+2$ vertices on the unit sphere
$$( 0, 0, \pm 1) \quad\text{ and }\quad \left( \cos\frac{2\pi k}{m}, \sin\frac{2\pi k}{m}, 0 \right) \quad\text{ for }\quad 0 \le k < m$$
Their convex hull will be a $m$-gonal bipyramid which appear below.
Up to my knowledge, the largest $n$-vertex polyhedron inside a sphere is known only up to $n = 8$.
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For $n \le 8$, above configurations are known to be optimal. A proof can be found in the paper
Joel D. Berman, Kit Hanes, Volumes of polyhedra inscribed in the unit sphere in $E^3$
Mathematische Annalen 1970, Volume 188, Issue 1, pp 78-84
An online copy of the paper is viewable at here (you need to scroll to image 84/page 78 at first visit).
For $n \le 130$, a good source of close to optimal configurations can be found under N.J.A. Sloane's web page on Maximal Volume Spherical Codes. It contains the best known configuration at least up to year 1994. For example, you can find an alternate set of coordinates for the $n = 8$ case from the maxvol3.8 files under the link to library of 3-d arrangements there.
Best Answer
One useful tool in getting this kind of information is the solid angle of an object. Simply multiplying such a solid angle by $r^2$ will give you the surface area cut out from a sphere of radius $r$ centered around the apex of the angle. Since your sphere has radius $1$, you're in fact simply asking for the solid angle at the top of a tetrahedron. Wikipedia has a section on its computation. That section is for general, i.e. not neccessarily regular tetrahedra. You could apply those computations to a regular tetrahedron. Or you have a look at the article on (the regular) tetrahedron. There you find the solid angle at a vertex to be
$$\Omega = \arccos\frac{23}{27} \approx 0.55129\,\text{sr}$$
Since your radius is $1$, you can ignore the unit of sr and interpret this number directly as the surface area you asked about. Note that the size of the tetrahedron is irrelevant in all of this, as long as one of its vertices is centered at the center of the sphere and the opposite side lies completely outside the sphere.