October 2nd, 2007, 6:03 pm
Sorry, for the weird reasoning, I read: "probability of 3 points on a sphere that can be contained in a hemi-sphere." Apologies.But here's a contructive proof for the actual problem I think you'll like:Take the 3 points --Look at the possible displacements. (Displacement = moving the point across the diagonal to its mirror image spot on the other side)As there are 3 mirror image spots to be filled (one for each point) -- there can be 2^3 = 8 displacements.So for instance: for (0,120,240) -> (0,120,240), (180,120,240), (0,300,240), ... ,(180,300,60)You will notice that there are exactly 2 displacements when the semi-circle does not fit.Thus for any configuration, 6 out of 8 displacements will fit on a semi-circle.So prob = 6/8 = 3/4EXCEPTIONs:When the points are EXACTLY on a semi-circle, this does not hold -- however we can safely ignore these points as they are zero-content (or something to that effect)