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REAL math stumper

C

chuckywang

Lifer
Prove that if you randomly place 5 points on a sphere, then at least 4 of those points lie in the same hemisphere.

(from the Putnam)
 
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this is a list of all the possible combinations of points with 0 being left, bottom or rear hemisphere with respect to x, y and z and 1 being right, top or front.. take any 5 of those points and 4 of them will share a set of 1's or 0's
 
Originally posted by: DeRusto
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this is a list of all the possible combinations of points with 0 being left, bottom or rear hemisphere with respect to x, y and z and 1 being right, top or front.. take any 5 of those points and 4 of them will share a set of 1's or 0's
Click to expand...

But that doesn't prove it works for the entire sphere . . . it only proves it for 8 points.
 
0 and 1 arent the actual points, they are binary representations of which hemisphere a value falls into...surely you cannot be saying that a value for x will not fall to either the left or right of its axis?
 
there is a possibility of having 2 points on 1 hemisphere, and 3 on the other or vice versa. THis contradicts what you have just mentioned.
What DeRusto said:

000
001
010
011
100
101
110
111

this is a list of all the possible combinations of points with 0 being left, bottom or rear hemisphere with respect to x, y and z and 1 being right, top or front.. take any 5 of those points and 4 of them will share a set of 1's or 0's

is kinda vague
 
Maybe my definition of hemisphere is wrong, but what if you happen to place 3 points in the southern (say) hemisphere and 2 in the northern?
 
Originally posted by: Muzzan
Maybe my definition of hemisphere is wrong, but what if you happen to place 3 points in the southern (say) hemisphere and 2 in the northern?
Click to expand...

thats what i assumed.....
 
there are at least three set of hemispheres assuming each axis of the sphere makes its own hemispheres
 
problem with that is this question asks for 5 points

edit:
thermalpaste's proof not-withstanding, does anyone disagree that any point in 3-dimensional space can be represented by one of the 8 possibilites i listed above?

now, with that in mind, is it possible to pick 5 of those possibilites and not have 4 of them that share a set of common digits? being that either the x, y or z digits will all be the same.

edit2: btw, the formula given at the end of the page on your first link for the probability that n points are in the same hemisphere is as follows: (n^2 - n + 2)/2^n

probability that 4 points fall in the same hemisphere is then (4^2 - 4 + 2)/2^4 = 14/16 = 7/8

so 7/8 points chosen will fall in the same hemisphere

there are 8 possible locations for a point, 7 of those will be in the same hemisphere, leaving only 1 possibly in a different hemisphere.

given this, you can see that it is only possible for 1 of the 5 to not share a hemisphere with the others, this still leaves 4 that DO in all cases
 
It could be true.

Edit: It has to be true. I don't have a mathmatical proof for you, but no matter how you arrange 5 point on the surface of a sphere, you can always cut that sphere in half and have 4 in the same half. What some of you are missing is the fact that there is more than one way to bisect a sphere in 3 dimensions.
 
Originally posted by: Muzzan
Maybe my definition of hemisphere is wrong, but what if you happen to place 3 points in the southern (say) hemisphere and 2 in the northern?
Click to expand...

Then you simply cut it into a eastern and western hemisphere. One way or another, you will be able to get 4 points into the same half.
 
Originally posted by: VictorLazlo
It could be true.

Edit: It has to be true. I don't have a mathmatical proof for you, but no matter how you arrange 5 point on the surface of a sphere, you can always cut that sphere in half and have 4 in the same half. What some of you are missing is the fact that there is more than one way to bisect a sphere in 3 dimensions.
Click to expand...

Exactly. Everyone is thinking of it in terms of a simple north/south (so let's put 3 in the north and 2 in the south HA!). That's not what the question is asking. You can cut the sphere in half however you want.
 
Originally posted by: VictorLazlo
It could be true.

Edit: It has to be true. I don't have a mathmatical proof for you, but no matter how you arrange 5 point on the surface of a sphere, you can always cut that sphere in half and have 4 in the same half. What some of you are missing is the fact that there is more than one way to bisect a sphere in 3 dimensions.
Click to expand...

I agree.
Those saying 3 nortth and 2 south forget there are an infinite number of hemispheres that can be viewed.
 
It could happen because you can only divide the Earth at the equator to make two hemisphere's so theres just a probability of it happening but it can.
 
Originally posted by: VictorLazlo
It could be true.

Edit: It has to be true. I don't have a mathmatical proof for you, but no matter how you arrange 5 point on the surface of a sphere, you can always cut that sphere in half and have 4 in the same half. What some of you are missing is the fact that there is more than one way to bisect a sphere in 3 dimensions.
Click to expand...

Wrong. It's not true. There's a way to arrange the dots such that no matter how you cut the sphere, they are always divided 3:2.

Here's the proof:

Draw a circle (a circle is a subset of a sphere, so it's a legitimate proof)
put dots on the circle that are evenly spaced.
Now, take bisections of the circle. See that there is in fact no way to make 4 dots be on one side of the bisection. If you take a bisection (of the sphere) that passes through all of the points, that's not legitimate because the points dont lie on either side.
 
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