REAL math stumper

[chuckywang]

Originally posted by: 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.

Hey. OP here. You proved that they can always be divided 3:2 for a circle, not a sphere. You can't just assume that if something works for a circle, it also works for a sphere.

 

[chuckywang]

One thing that everybody needs to keep in mind is that there are INFINITE ways to cut a sphere into two hemispheres. So any talk of "left", "right", "top", "bottom" hemispheres is not going to work because you can just rotate the sphere and get different "left", "right", "top", and "bottom" hemispheres.

 

[VictorLazlo]

Originally posted by: 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.

I'm going on the assumption that points on the line DO count.

Edit: If you put a dot "on" the circle, is it inside, or outside the circle?

 

[chuckywang]

Originally posted by: JustAnAverageGuy
What if those 5 random points are exactly the same point?

Then at least 4 of those points lie in the same hemisphere, trivially.

 

[chuckywang]

Originally posted by: VictorLazlo

Originally posted by: 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.

I'm going on the assumption that points on the line DO count.

That is an assumption that you can make to help you think it through. Even if points on the line don't count, you can always very very slightly move the hemisphere so that points on the line of the hemisphere are no longer on the line anymore, but still on the same side of the hemisphere. (if that makes any sense)

If it doesn't make sense, just take dots on the circle to be inside the circle.

 

[VictorLazlo]

Originally posted by: chuckywang

Originally posted by: VictorLazlo

Originally posted by: 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.

I'm going on the assumption that points on the line DO count.

That is an assumption that you can make to help you think it through. Even if points on the line don't count, you can always very very slightly move the hemisphere so that points on the line of the hemisphere are no longer on the line anymore, but still on the same side of the hemisphere. (if that makes any sense)

That's not necessary. Points on the line have to count.
If we look at the circle example, putting points "on" the circle is just putting them on a line, and clearly they count as being "in" the circle.

 

[chuckywang]

Originally posted by: VictorLazlo

Originally posted by: chuckywang

Originally posted by: VictorLazlo

Originally posted by: 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.

I'm going on the assumption that points on the line DO count.

That is an assumption that you can make to help you think it through. Even if points on the line don't count, you can always very very slightly move the hemisphere so that points on the line of the hemisphere are no longer on the line anymore, but still on the same side of the hemisphere. (if that makes any sense)

That's not necessary. Points on the line have to count.
If we look at the circle example, putting points "on" the circle is just putting them on a line, and clearly they count as being "in" the circle.

This is a point that is not important to the actual proof of the original problem. If you want, PM me and we can talk about this. No need in lengthening the thread any more.

 

[notfred]

I can't believe that someone actually said that a circle is a subset of a sphere.
A sphere is not a set. It is a sphere. You cannot take a subset of a geometric object.

Definitions:
Set: A collection of distinct elements having specific common properties.
Subset: a set whose members are members of another set; a set contained within another set

 

[JustAnAverageGuy]

Originally posted by: chuckywang

Originally posted by: JustAnAverageGuy
What if those 5 random points are exactly the same point?

Then at least 4 of those points lie in the same hemisphere, trivially.

The question was four, not at least four.

 

[chuckywang]

Originally posted by: JustAnAverageGuy

Originally posted by: chuckywang

Originally posted by: JustAnAverageGuy
What if those 5 random points are exactly the same point?

Then at least 4 of those points lie in the same hemisphere, trivially.

The question was four, not at least four.

Eh, "at least four" was sorta implied at the beginning. But I can see the confusion....*editing the original post now*

 

[Beattie]

Originally posted by: VictorLazlo

Originally posted by: 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.

I'm going on the assumption that points on the line DO count.

Edit: If you put a dot "on" the circle, is it inside, or outside the circle?

Neither. It's on the circle.

 

[Beattie]

Originally posted by: notfred
I can't believe that someone actually said that a circle is a subset of a sphere.
A sphere is not a set. It is a sphere. You cannot take a subset of a geometric object.

Definitions:
Set: A collection of distinct elements having specific common properties.
Subset: a set whose members are members of another set; a set contained within another set

Geez, you know what I meant. That it's possible to understand the relationship between the points by putting them on a circle just as it is on a sphere.

 

[chuckywang]

Originally posted by: Beattie

Originally posted by: notfred
I can't believe that someone actually said that a circle is a subset of a sphere.
A sphere is not a set. It is a sphere. You cannot take a subset of a geometric object.

Definitions:
Set: A collection of distinct elements having specific common properties.
Subset: a set whose members are members of another set; a set contained within another set

Geez, you know what I meant. That it's possible to understand the relationship between the points by putting them on a circle just as it is on a sphere.

Well, either way....properties that are true for circles are not necessarily true for spheres.

 

[RayH]

It only takes 3 distinct points to form a plane which cuts the sphere into 2 sections.
Any additional distinct points will be in one of those 2 sections.
3 points forming a plane which divides a sphere into hemispheres is just one case of the above.

 

[chuckywang]

Originally posted by: RayH
It only takes 3 distinct points to form a plane which cuts the sphere into 2 sections.
Any additional distinct points will be in one of those 2 sections.
3 points forming a plane which divides a sphere into hemispheres is just one case of the above.

You are very close. But it is not necessarily true that some 3 points out of the 5 will be able to form a plane that cuts the sphere into hemispheres.

 

[desteffy]

choose two points, they will let you draw a hemisphere. (by cutting the sphere by the plane intersection those two points and the center of the sphere)

Then of the remaining 3 points, at least two must be on one half of the circle by the pigeon hole principal, giving you 4 on one hemisphere

 

[chuckywang]

Originally posted by: desteffy
choose two points, they will let you draw a hemisphere. (by cutting the sphere by the plane intersection those two points and the center of the sphere)

Then of the remaining 3 points, at least two must be on one half of the circle by the pigeon hole principal, giving you 4 on one hemisphere

THE WINNAR!

 

[kcthomas]

Originally posted by: desteffy
choose two points, they will let you draw a hemisphere. (by cutting the sphere by the plane intersection those two points and the center of the sphere)

Then of the remaining 3 points, at least two must be on one half of the circle by the pigeon hole principal, giving you 4 on one hemisphere

that would only guarentee 3 on each side. there is 1 point on each half sphere, then 2 extras = 3 total.

 

[kcthomas]

Originally posted by: 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.

think of a circle as a cross section of the sphere. if you cut a really really small layer of the sphere out down the middle you would have a circle. if all the points lie on this layer then they are all along the cut of the original sphere, which can be said to be in the same hemisphere.

 

[desteffy]

Originally posted by: kcthomas

Originally posted by: desteffy
choose two points, they will let you draw a hemisphere. (by cutting the sphere by the plane intersection those two points and the center of the sphere)

Then of the remaining 3 points, at least two must be on one half of the circle by the pigeon hole principal, giving you 4 on one hemisphere

that would only guarentee 3 on each side. there is 1 point on each half sphere, then 2 extras = 3 total.

No, we are dealing with closed (including the border) hemispheres here. so, 2 on the border + 2 more that are either on the border or on the one side = 4

 

[desteffy]

Originally posted by: chuckywang

Originally posted by: desteffy
choose two points, they will let you draw a hemisphere. (by cutting the sphere by the plane intersection those two points and the center of the sphere)

Then of the remaining 3 points, at least two must be on one half of the circle by the pigeon hole principal, giving you 4 on one hemisphere

THE WINNAR!

woot! No fair though since I saw it when I took the putnam so I already figured it out a while ago.

 

[TuxDave]

Another puzzle using the same idea.

You have an infinitely wide plane, a neverending bucket of blue paint and a neverending bucket of red paint. Assuming you manage to paint every point on the plane either blue or red, prove to me that there will exist 2 points 1 mile apart that will be the same color.

 

[desteffy]

Originally posted by: TuxDave
Another puzzle using the same idea.

You have an infinitely wide plane, a neverending bucket of blue paint and a neverending bucket of red paint. Assuming you manage to paint every point on the plane either blue or red, prove to me that there will exist 2 points 1 mile apart that will be the same color.

Choose 3 points that form a triangle with all sides equal to one mile. Since there are 3 points, at least two of them are the same color.
Can you do this for 3 colors

 

[TuxDave]

Originally posted by: desteffy

Originally posted by: TuxDave
Another puzzle using the same idea.

You have an infinitely wide plane, a neverending bucket of blue paint and a neverending bucket of red paint. Assuming you manage to paint every point on the plane either blue or red, prove to me that there will exist 2 points 1 mile apart that will be the same color.

Choose 3 points that form a triangle with all sides equal to one mile. Since there are 3 points, at least two of them are the same color.
Can you do this for 3 colors

Yeah you got it. It was the only creative problem I saw using the pidgeon hole principle. What are you asking. If the plane was colored 3 different colors prove that 2 points 1 mile apart will be the same color?

 

[desteffy]

Originally posted by: TuxDave


Yeah you got it. It was the only creative problem I saw using the pidgeon hole principle. What are you asking. If the plane was colored 3 different colors prove that 2 points 1 mile apart will be the same color?

yes

Given a plane RxR, color each point with one of three colors. Is it necessairly true that there will be two points exactly one mile apart of the same color.