REAL math stumper

[chuckywang]

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)

 

[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

 

[CarpeDeo]

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

But that doesn't prove it works for the entire sphere . . . it only proves it for 8 points.

 

[DeRusto]

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?

 

[thermalpaste]

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

 

[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?

 

[thermalpaste]

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?

thats what i assumed.....

 

[DeRusto]

there are at least three set of hemispheres assuming each axis of the sphere makes its own hemispheres

 

[thermalpaste]

IT is possible if the 4 points are equidistant from each other.........

 

[thermalpaste]

CHECK THIS OUT

 

[thermalpaste]

AND THIS......

 

[DeRusto]

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

 

[DeRusto]

would SOMEONE please reply to this thread and say whether i am an idiot or a genius?

 

[Beattie]

It's not true.

 

[Ilmater]

It's absolutely not true.

 

[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.

 

[VictorLazlo]

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?

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.

 

[silverpig]

Hmm, that's a thinker.

 

[silverpig]

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.

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.

 

[Lonyo]

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.

I agree.
Those saying 3 nortth and 2 south forget there are an infinite number of hemispheres that can be viewed.

 

[thermalpaste]

Originally posted by: DeRusto
would SOMEONE please reply to this thread and say whether i am an idiot or a genius?

:thumbsdown:

 

[Zero Plasma]

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.

 

[edro]

My head hurts.

 

[Beattie]

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.

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.

 

[JustAnAverageGuy]

What if those 5 random points are exactly the same point?