Points on a Sphere Part 3
 

I posted this problem a while back but I found it very interesting that I wanted to spend a bit more time on it .

The problem is as follows :

What is the probability that n randomly chosen points on a sphere will all lie on a single hemisphere ? Although there are several solutions to this problem , I have yet to encounter a combinatorial argument . In other words , show that the probability in question is (n^2-n+2)/2^n .

Before we solve this problem , it may be a good idea to solve the analogous case in R^2 ; that is , what is the probability that n points on a circle lie on a semicircle ?

It turns out that the answer in R^2 is n/2^(n-1) . There is a straightforward combinatorial argument to this question which goes as follows :

There are 2^n pairs of antipodes . There are also 2n semi-circles that have the property that all the points lies on that side . Therefore , the probability is simply 2n/2^n or n/2^(n-1) .

In R^3 , we know that there are 2^n pairs of antipodes , but it isn't so obvious that there are n^2-n+2 ways of having n points on a hemisphere .

I'll let others think about it for a bit and then I'll post my answer . Try to be as thorough as possible with your solution .

Good luck !!

Re: Points on a Sphere Part 3 [spoiler]
 

Each point/antipode pair determines a great circle of points of distance pi/2 radians away from both, e.g., the north pole/south pole pair corresponds to the equator.

Generically, n great circles divide the sphere into n^2-n+2 regions (for n>0). Each region corresponds to a choice of point or antipode such that all chosen points are within pi/2 radians of every point in the region, hence contained within all hemispheres centered in that region.

This is analogous with what happens for the circle, and it's just an interpretation of what happens in this more general proof.

Re: Points on a Sphere Part 3 [spoiler]
 

Very nice Pzhon !

I think the easiest way to show that the sphere divides into n^2-n+2 regions is to notice that 2nc2+2 = n^2-n+2 .

ie, take 3 points , say a,b,c and their antipodes a',b',c' . There are three equators or large circles created by each pair of antipodes . This divides the sphere into the following regions : ab,ac,bc, a'b',a'c',b'c' + both sides of aa' . This is simply 2*3c2 +2 ways of partitioning the sphere so that all the points are in that region .

Also note that ab is just the intersection in surface area of the center of hemisphere a with the center of hemisphere b .