| ▲ | thaumasiotes 3 hours ago | |||||||
Well, it's easy to pick a half-n-sphere corresponding to a point on the surface. You just take the half-sphere centered at that point. For our two-dimensional circle, it would be the arc defined by the diameter that is perpendicular to the radius running between "the point" and the center of the circle. At that point you've lost the ability to say that only one such half-sphere defined by a dropped point can be a valid solution, and you've also lost the ability to say that if a valid solution exists then there must be a valid solution defined by one of the points you want to include in the half-sphere, but you can define a canonical half-sphere for any point. I was uncomfortable with the idea of picking "random points on a circle" to begin with, because of https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) , but the article doesn't even address whether the concept is well-defined. We can always choose a point on the perimeter deterministically from any chord (...that isn't a diameter), so the ill-definedness of the problem of choosing a random chord seems like it would infect the problem of choosing a random point on the perimeter. | ||||||||
| ▲ | matheist 2 hours ago | parent [-] | |||||||
Right, I was taking it as given that the problem of choosing a hemisphere canonically for a point meant "such that the argument works in the same way as for the circle". Bertrand paradox just doesn't apply here, there's a natural measure on the circle and all higher dimensional spheres. I wouldn't expect an article on this subject to need to make that clarification unless it's dealing with chords or some other situation without a natural measure. | ||||||||
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