A continuous center point of a convex spherical polygon In discrete geometry, the center point $c$ of a discrete set $S$ of $n$ points in the plane is such that any half plane containing $c$ contains (roughly) $n/3$ points of $S$. (Such a center point always exists.)

My problem is similar but in a continuous setting.
I have a convex spherical polygon $P$ on the unit sphere. In this context an "area-center" point $c$ is such that any half sphere that contains $c$ also contains a part of $P$ whose area is at least a constant fraction of the area of $P$.

The definition can be similarly phrased with convex polygons on the plane and half-planes instead of half-spheres. It can also be generalised to higher dimension.

My questions are:


Does this definition has a well-known name? ("area-center point"?)
Do we know algorithm to compute such an area-center point? (exact or approximate)


(My motivation for finding such a point is for doing logarithmic search on the unit sphere.)

I ask this question here because searching the web only leads me to centroids or discrete center point, or the Tverberg generalization, but nothing related to the specific definition given above. (I have already asked question 30 minutes ago on math SE, but realised it is probably more relevant to comp. geom. hence its posting here)

Thank you !
For your second question...

For any distribution, if you take a sample $R$ of size $O(1/epsilon^2 )$ of it, and compute its center point, it is going to be a $\geq (1/3-\epsilon)$-center point, with probability $\geq 1-\epsilon^{O(1)}$ by the $\epsilon$-net theorem (well, more precisely the $\epsilon$-sample theorem). You can apply this to a polygon - you just need to sample points uniformly inside the polygon. This works in any constant dimension.

Similar ideas were used in the paper "Approximating Center Points with Iterated Radon Point". Or so I believe... ;)

Computing the exact center point for a general polygon is nasty as far as I know. There was a paper in the 90s by Dan Halperin on maintaining the area bisector of a polygon, and it was painful, if my memory serves me right.