Geometric Probability

[ComplexZeta]

What is the probability that 11 points on a sphere lie in the same hemisphere? What about n points? (If you have solved it for 11, I can't imagine that you haven't solved it for n.) If you have seen it before, please don't give it away.

[JBL]

This is really cool. I think I'm going to start off working on a circle, though -- do you think that will give me any insight?

[zscool]

I dont think it makes it any easier to consider it on a circle first.

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Try to think about the necessary conditions for the 11 points to be on a hemisphere.

[ComplexZeta]

I think I agree with zscool. Maybe keep in mind that a sphere is (effectively) a disjoint union of two discs, two line segments, and two points. (Wow, I've been doing too much topology.) I don't know how much that helps though. By the way, this problem is really hard.

[churchilljrhigh]

is it on the sphere's surface or inside the sphere?

[Alison]

I don't know if this goes anywhere, but...
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I'm thinking convex hulls might work, because 11 points lying in an (open) hemisphere is the same thing as saying that the center of the sphere does not lie in their convex hull. For which matter, this is the same as saying that for any 4 points of the 11, the center of the sphere does not lie in the convex hull of those 4. However, this still looks messy.

[ComplexZeta]

It means on the sphere's surface. I don't think it matters though.

Interesting approach Alison. I'm also not sure if it will work, but give it a try.

[Alison]

I am, and I'm getting a messy inclusion-exclusion argument -- and that's just for five points, so I probably need to be more clever. Hm. At least I have a nice argument that the probability for 4 points is 7/8:Click to reveal hidden content

by picking one point and dividing into cases according to whether that point is inside the spherical triangle (the one with no angles > 180) formed by the other three points).

Hm. Perhaps I'll join JBL and look at circles to give my brain a rest. I'm also wondering how this generalizes to higher dimensions. I have a conjecture that the probability that the probablility that n+1 points on the surface of an n-dimensional sphere lie on a hemisphere is 1-1/(2^n), but the proof I have doesn't obviously extend.

[ComplexZeta]

7/8 is definitely correct for 4. I'll check if your conjecture is right for higher dimensions (or at least for S³, since I know how to do that fairly easily).

[gauss202]

Has anyone found a solution to this problem?

[joml88]

This problem came from the 1999 BAMM-a 50 minute 20 question multiple choice test!!! Guess who wrote those questions. None other than Paul Zeitz!

Honestly, I haven't looked at the solution yet and I don't know how to do the problem. I'm pretty much saving it for when I think I might be able to solve it...

lol, Mr. Rusczyk told me a while ago that Zeitz probably made this problem so that Gabriel Carroll wouldn't get a perfect score...it worked

O, yeah...Here's the website. It's the last question of the 1999 Test of Ingenuity.

[probability1.01]

Perhaps you can identify a hemisphere by the point in the hemisphere that is equidistant to all points on its boundary (so kinda like the north pole for the northern hemisphere on earth). Then notice that every point excludes a hemisphere of hemisphere points... if that makes any sense. 11 points share a hemisphere only if they do not collectively exclude the whole sphere. Anyway, I still have no clue about the solution .

[goodyfresh741]

[Singular]

this was also on putnam very recent.

[blahblahblah]

[Myth]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5715

[seamusoboyle]

How about saying that all the point have to be within r unit (radius) of a certain plan tangent to the sphere? Or join all the points to the centre and look at the angles between the lines....