# A Hard Maths Problem



## nitrocan (Aug 6, 2008)

Since you all seem to like math problems  heres another one:
A stick is thrown to the ground and it breaks into 3 pieces. What is the possibility that these 3 pieces can form a triangle? (not when it falls, im asking if you can manage to make a triangle when you assemble them)


----------



## brunson (Aug 6, 2008)

Hmm... assuming that the end of each stick must be on a vertex.

If one of the pieces is longer than the sum of the other two, then you can't form a triangle. 

So assume the first break is not exactly in the middle of the stick, then if the longer of the two pieces breaks, a triangle can be formed. So, break a stick, choose a piece to break again, you have a 50% chance of breaking the longer of the two pieces. So I'm going with 50%, provided that the probability that either piece of the stick after the first break will be broken again is equal.

Edit:
The problem, as stated, is ambiguous.


----------



## Markus Pirzer (Aug 6, 2008)

azrian said:


> 100% of the time
> 
> EDIT: Meant to say always but accidently double-posted -.-' Sorry 'bout that



No, if the longest piece is more than 50% of the original stick you can't form a triangel, because the longest piece must be shorter than the other 2 pieces together.


----------



## nitrocan (Aug 6, 2008)

the rule for a triangle is (lets say the length of the edges are a b and c)
a-b < c < a+b

i will post the solution after some time. (i dont think its possible to guess it)


----------



## rubiksfriend (Aug 6, 2008)

Longest piece=z. yes, triangle- z>50% of stick. 
z<_50% of stick- no triangle. Right? 49%? Nescio.


----------



## not_kevin (Aug 6, 2008)

rubiksfriend said:


> 49/100=.49?



It's taking the remainder of the division. The remainder of 49/100 is 49; the remainder of 25/7 is 4 (25/7 = 3 + 4/7).


----------



## Dene (Aug 6, 2008)

azrian said:


> Crap, sorry for the double post
> 
> (Is there any way to delete a post?)



If you click "edit", there should be a "delete this post" option. When you click that it will take you to a second screen where you have to check a box, then confirm.


----------



## nitrocan (Aug 6, 2008)

as i said, the solution is complicated, but i forgot the solution  i guess i will have to wait until i meet my maths teacher again. but i remember that it is solved using the 3D cartesian corrdinate system (with x, y and z axis)


----------



## philkt731 (Aug 7, 2008)

Ok Lets say the stick is of length 1, think of it like a ruler marked from 0 to 1.
Lets say we cut it at mark x and mark y. The three lengths are x, y-x, and 1-y for the case that y>x.

From trignale inequality we get that
x<y-x+1-y ==> x<1/2
y-x<x+1-y ==> y<x+1/2
1-y<x+y-x ==> y>1/2

So 0<x<1/2 and 1/2<y<1/2+x, making the range for y's values x.

So the probability is the integral from 0 to 1/2 of x dx, which is 1/2(1/2^2-0^2)=1/8.

The proabability is also 1/8 when x>y, 

So the total probability is 1/8+1/8=1/4


Less thatn you think, huh?


----------



## blah (Aug 7, 2008)

I got something weird: 5/16. Most probably wrong  I'm just posting it here in case someone else got the same retarded answer...


----------



## hawkmp4 (Aug 7, 2008)

So, the probability of any specific, exact value (example: 1/2) in a continuous distribution is zero. So, we can assume that the stick does not break in the middle.
Then, we have one more break, and that occurs randomly. if it occurs on the short side a triangle cannot be made. if it breaks on the long side, it can.
I think its safe to say then, that the probability of a stick breaking on either side is 1/2.
Therefore, I say the probability that a triangle can be made is 1/2.


----------



## cmhardw (Aug 7, 2008)

Coolest problem I've done in a long time.

I also got 1/4 as the probability. I'm very proud of myself for solving this one, as I tried to do it without reading this thread (other than the initial problem itself) or any other math resource.

My original line of thought was to think of a line of length L, that I was going to split into 3 pieces. I tried a few non-fruitful approaches, then had the idea to think of it as 3D Cartesian coordinates (nitrocan I promise I'm not stealing your idea, if anything I just independently re-discovered the wheel).

First I graphed the plane x + y + z = 1 (I am assuming the stick is of unit length). Yes I did this on paper, and not in a math program. Then I took the triangle inequalities:
x + y > z
x + z > y
y + z > x

and I graphed each of these regions. The overlap region of all of these is simply an equilateral triangle of side length sqrt(2)/2 that lies in the plane x + y + z = 1. The area of this triangle is sqrt(3)/8. The area of the triangle formed by the intercepts of x + y + z = 1 is sqrt(3)/2. 

Now take the ratio of their areas and you get 1/4. I was also surprised, I thought the probability would be higher than that.

Chris


----------



## JBCM627 (Aug 7, 2008)

hawkmp4 said:


> So, the probability of any specific, exact value (example: 1/2) in a continuous distribution is zero. So, we can assume that the stick does not break in the middle.
> Then, we have one more break, and that occurs randomly. if it occurs on the short side a triangle cannot be made. if it breaks on the long side, it can.
> I think its safe to say then, that the probability of a stick breaking on either side is 1/2.
> Therefore, I say the probability that a triangle can be made is 1/2.



I think this is correct, although I'd say its just under 50% probability, as when you have two halfs of a stick you have a line.

Edit: Just kidding, I think I see the flaw in this now... consider picking 2 random points on the stick. You want the probability that they are do not create a segment that is over half of the stick, which is 1/4 as Chris said.


----------



## hawkmp4 (Aug 7, 2008)

JBCM- the thing is, you can't have just two halves of the stick. If the stick can break anywhere along the interval (0,1), what is the probability that it breaks at exactly .5? Its 0. As it is for any value on the stick- its a continuous distribution, there's only a probability for a range of values.


----------



## JBCM627 (Aug 7, 2008)

By two halves, I meant one half being half of the stick, and the other two composing another half.

There is perhaps a 0% probability that a break will occur at a given spot, but only for a mathematically perfect stick. I'd say it should instead be called "negligable". If you say it is exactly 0, you end up with wierd things like Banach-Tarski. Maybe I'll just claim you can always form a 627-gon, as you can just duplicate your sticks.


----------



## cmhardw (Aug 7, 2008)

JBCM627 said:


> Maybe I'll just claim you can always form a 627-gon, as you can just duplicate your sticks.



haha ok this is probably dumb, but the first thing I thought of when I saw this was a second version of the problem:

"A stick falls to the ground and breaks into 3 pieces. What is the probability that it actually broke into only 2 pieces, but one of those pieces duplicated itself?"

yeah I know, it's dumb. But it made me chuckle in its dumbness ;-)

Chris


----------



## rubiksfriend (Aug 7, 2008)

philkt731, how did you know when to integrate?


----------



## Lucas Garron (Aug 7, 2008)

Has no one else in here seen this before, seriously?
It's a common example of a problem where the statement of the problem allows equally valid interpretations that give different results. See here.




JBCM627 said:


> There is perhaps a 0% probability that a break will occur at a given spot, but only for a mathematically perfect stick. I'd say it should instead be called "negligable". If you say it is exactly 0, you end up with wierd things like Banach-Tarski.


Could you explain that last sentence? I'm not sure how me saying that it's exactly 0 makes me end up with Banach-Tarski.
(By the way, I went to a talk today about the Kakeya Needle problem. We learned of a set of lines, containing one of every slope. But the total area they formed was 0.  )


----------



## hdskull (Aug 7, 2008)

Lucas Garron said:


> Has no one else in here seen this before, seriously?


Saw it at SoCal ARML.


----------



## nitrocan (Aug 7, 2008)

a long time ago, this was asked in one of my countries math olympiads. i hadn't attended that one though, my math teacher asked me this.


----------



## shadowpartner (Aug 7, 2008)

this is eye boggling for a 13 yr old like ME


----------



## mrCage (Aug 7, 2008)

Hi 

I just recall something else quite intriguing from calculus. Solids of revolution. A curve was revolved around the x-axis. The volume it confines was infinite, while the surface area was finite. Now that is not so intuitive ;-) I do not recall the function of the curve anymore. But y -> 0 as x -> + infinity. That's all i recall ...

- Per


----------



## MistArts (Aug 7, 2008)

shadowpartner said:


> this is eye boggling for a 13 yr old like ME




I am 13 too...

My answer is 25%.



Spoiler



I'll explain later.


----------



## JBCM627 (Aug 7, 2008)

Lucas Garron said:


> JBCM627 said:
> 
> 
> > There is perhaps a 0% probability that a break will occur at a given spot, but only for a mathematically perfect stick. I'd say it should instead be called "negligable". If you say it is exactly 0, you end up with wierd things like Banach-Tarski.
> ...



Consider the Banach-Tarski paradox. It claims that you can take a (mathematical) object, and can duplicate it, creating two identical figures out of one. One interpretation of this is as follows: consider the stick. You can break it up into an infinite number of points, seperate these into two (still infinite) sets, and put them back together (relying on the disputable axiom of choice) so that you now have 2 of the initial object.

This in my mind is no different than claiming the cardinality of the integers is the same as the cardinality of the even numbers.

The problem with this is that it only works for mathematically "perfect" objects; or in other words, objects able to be broken into an infinite number of pieces.

If you say that the probability is 0, this can only be implied when the stick has an infinite number of breaking points (hence, Banach-Tarski), or it is not possible to break the stick at this location. We neglect the possibility of it being impossible to break at that location. Since there are an infinite number of points, it is quite easy to create a bijection between the number of points on one stick and the number of points on two identical sticks... or to map one stick to 2 identical sticks.

In the real world however, you simply can't do this. There are a discrete number of breaking points and arrangements, so such a map would be impossible.


----------



## JBCM627 (Aug 7, 2008)

Lucas Garron said:


> Has no one else in here seen this before, seriously?
> It's a common example of a problem where the statement of the problem allows equally valid interpretations that give different results. See here.



The other visualizaton applet (and explanation) on that page seems more intuitive to me.. more obvious why it is 1/4: http://www.cut-the-knot.org/Curriculum/Probability/TriProbCartesian.shtml


----------



## philkt731 (Aug 7, 2008)

Yeah this has probably been in ARML sometime. Speaking of ARML, I can't wait till the next one haha. So far my socres have been 2,4,6,7,7. I think I need an 8


----------



## Sin-H (Aug 8, 2008)

A cousin of mine solved it this way: WRONG, look for the full solution two posts ahead

let's say, the parts of the stick are named a, b, c.

a+b < c is needed

1. case: a >= 1/2 : you cannot form a triangle
2. case: a < 1/2: then we have the cases:
2a: b >= 1/2: impossible
2b: b < 1/2: possible

50% chance for case 2 and then another 50% chance for case 2b: 0.5 * 0.5 = 0.25.
the possibility to be able to form the triangle is 1/4


----------



## nitrocan (Aug 8, 2008)

Sin-H said:


> a+b < c is needed


No, a-b < c < a+b is needed, or a+b > c


----------



## Sin-H (Aug 8, 2008)

nitrocan said:


> Sin-H said:
> 
> 
> > a+b < c is needed
> ...



You're right;

so let's do it this way:

we proved that the chance for a+b < c is 25%
it's also 25 % for a+c<b.
and also 25% for b+c <a

in none of these cases can we form a triangle.
so it's another 25% for the remaining case in which we can do it

(I am not sure about this, but my cousin is)


----------



## qqwref (Aug 8, 2008)

JBCM627 said:


> Consider the Banach-Tarski paradox. It claims that you can take a (mathematical) object, and can duplicate it, creating two identical figures out of one. One interpretation of this is as follows: consider the stick. You can break it up into an infinite number of points, seperate these into two (still infinite) sets, and put them back together (relying on the disputable axiom of choice) so that you now have 2 of the initial object.



Banach-Tarski is a result involving 3-dimensional objects: given a 3-dimensional object, you can break it up into a finite number of sets of points and then reassemble these sets into two of the original object, thus doubling the volume. Unfortunately this result does not actually hold in one or two dimensions, and since we are treating the stick as the interval [0,1] this means that the paradox is not actually applicable.


It's been stated that the probability of the stick breaking at a given point is 0. This is true, but it's because there are an infinite number of possible break points on our theoretical stick. The interesting (and kind of counterintuitive) thing is that even though the probability is 0 it can still happen: no matter where the stick does end up breaking, the situation will have probability 0. This isn't really relevant to the problem, but it's interesting to bring up.


----------



## Lucas Garron (Aug 8, 2008)

JBCM627 said:


> Lucas Garron said:
> 
> 
> > JBCM627 said:
> ...


You somewhat have to be careful with your words. But allowing a liberal interpretation of that paragraph, you're right. Although:
-Breaking it up into an infinite number of points is an odd thing to say. I can break it into one piece with an infinite number of points by doing nothing, Note that the paradox only involves decomposing into finitely many pieces. Also, th axiom of choice is more useful for showing that there is a decomposition, rather than for showing that assembly is possible for some separations.
Also, the paradox requires at least 3 dimensions. But let's say you're doing this with a 3-D cylindrical stick. 




JBCM627 said:


> This in my mind is no different than claiming the cardinality of the integers is the same as the cardinality of the even numbers.


What a mind! Y'know, if there's one thing to know about intuitions with infinity, it's not to try to apply them to analogies. Countable and uncountable sets can act very differently.
Would you mind showing me some way in which I can take apart the integers into two sets, so that each still contains every integer? And if you consider this to be different, how about halving a cube, leaving two parts which can each be scaled into the original cube like the evens into the integers?
This claim makes little sense, and Banach-Tarski still doesn't work for 1-D...



JBCM627 said:


> The problem with this is that it only works for mathematically "perfect" objects; or in other words, objects able to be broken into an infinite number of pieces.


Okay, the first line is kinda useless here. And again, Banach-Tarski does not require breaking into an infinite number of pieces *how about reading the Wikipedia entry?).



JBCM627 said:


> If you say that the probability is 0, this can only be implied when the stick has an infinite number of breaking points (hence, Banach-Tarski), or it is not possible to break the stick at this location.


Huh? That almost makes sense, if you take out "(hence, Banach-Tarski)".
How about this: You have two possible breaking points, and you pick the left only if an infinite number of coin flips are each heads. Probability 0. But you have to be careful about saying "impossible." (The left point has the same chance of being picked as any point in a continuous distribution, by the way.)




JBCM627 said:


> We neglect the possibility of it being impossible to break at that location. Since there are an infinite number of points, it is quite easy to create a bijection between the number of points on one stick and the number of points on two identical sticks... or to map one stick to 2 identical sticks.


Neglect? You don't have to neglect the possibility of a falsehood; it's just false...

What is the bijection for?




JBCM627 said:


> In the real world however, you simply can't do this. There are a discrete number of breaking points and arrangements, so such a map would be impossible.


Depends on what you mean. But can you give a physical well-defined sense in which a stick breaks into 3 pieces randomly?

Now that I've noted some Banach-Tarski things, do you still insist it is implied by probability 0 for all points?


(By the way, would you mind spell-checking words, to catch words like "negligable" and "seperate." You're much more fun to argue against than clueless people, but it's sad to see such typos. Y'know, Firefox has spell-checking in fields?  )


----------



## JBCM627 (Aug 8, 2008)

Ok, I will try to reply to this, without a quote tree...

First of all, I imagine any more generalized versions will also hold for other dimensions. See Sierpinski-Mazurkiewicz, and although this requires decomposition of an unbounded set, still works. Who ever said the stick was 1D anyway? It would be interesting to see this problem solved for a 3D cylinder or prism. Obviously the breaks (and ends) would have to be angled, as to form a sort of triangular "frame".

So point taken with the integers, pending how you define them. If you do not define an integer as a distance, then ok. Banach-Tarski is different than scaling in 1-D, but I still view it as really similar, I guess just because it deals with cutting or rearranging something infinitely complex.

So, I was incorrectly thinking of Banach-Tarski as cutting something into an infinite number of pieces, and re-assembling, or as no different than stretching/bijecting. This was incorrect, because after reading a bit more and more carefully, the theory claims you can actually cut up a sphere into a minimum of 5 pieces to duplicate it.

However, I still would like to argue that if you state there is a 0 probability at a given point, you imply this paradox. Obviously, for sake of argument, assuming you have a 3d stick.

The pieces from Banach-Tarski are not smooth pieces, but are rather "jagged" and infinitely complex themselves. For these pieces to be infinitely "jagged", or some of them at least, there still needs to be an infinite number of points on the stick. The object still needs to be mathematical, not physical. So 0% probability implies an infinitely dense object ("non-empty", as otherwise put), which means generalized Banach-Tarski applies. A physical stick will have a limited number of breaking points, but a mathematical one will not. So if the stick has a 0% probability of being broken at a certain point, as stated before, I can disassemble it and reassemble it into 2 sticks. And make my wonderful 627-gon.



Lucas Garron said:


> Depends on what you mean. But can you give a physical well-defined sense in which a stick breaks into 3 pieces randomly?


Yep... consider a very small stick with a few tree cells, end to end. Or we can consider an atomic stick, or perhaps smaller, to avoid ambiguity... consider a string of electrons going 'round and 'round in a particle accelerator in a nice orderly fashion. Say we have 10 electrons. There are 9 breaking points now (Although, I suppose the whole thing would fall apart when the accelerator was shut off, so pretend it is quasi stable and only breaks at 2 points.) anyway, there is a non-0 probability for this physical stick to break at a given point, and is extendable for any physical stick composed of indecomposable particles.



Lucas Garron said:


> (By the way, would you mind spell-checking words?)
> ...
> Also, th axiom
> ...
> *how about reading the Wikipedia entry?)


Check yourself, too then 

At my 3D stick problem, would it be 0%? Because there would only be one possible arrangement of angles that would work for a given set of stick lengths, so probability of that angle occurring is 1/Infinity = 0% (well, again, for a mathematically perfect stick )

Another interesting extension is to consider the probability of forming a triangle of stick length integer n where the stick is only breakable at integer intervals.

This post has been spell checked.

Btw, very off topic, Lucas, I was never able to confirm this re: random walk, have you heard of a proof?:


JBCM627 from WCA forum said:


> Actually, from what I can tell, not backtracking will generate the same probability distributions as allowing backtracking. I went up to a 3x3 grid looking at various maps... and I can't find a proof of this or anything online. So I'm learning towards it not being an issue.


----------



## nitrocan (Aug 8, 2008)

so guys, should i keep putting up math problems, or is it getting boring?


----------



## cmhardw (Aug 8, 2008)

nitrocan said:


> so guys, should i keep putting up math problems, or is it getting boring?



I for one always love doing math problems. I simply know too few to post many myself without having to really think about or try to find one from somewhere.

Chris


----------



## qqwref (Aug 12, 2008)

JBCM627 said:


> The pieces from Banach-Tarski are not smooth pieces, but are rather "jagged" and infinitely complex themselves. For these pieces to be infinitely "jagged", or some of them at least, there still needs to be an infinite number of points on the stick. The object still needs to be mathematical, not physical. So 0% probability implies an infinitely dense object ("non-empty", as otherwise put), which means generalized Banach-Tarski applies. A physical stick will have a limited number of breaking points, but a mathematical one will not. So if the stick has a 0% probability of being broken at a certain point, as stated before, I can disassemble it and reassemble it into 2 sticks. And make my wonderful 627-gon.



Excuse me, what? This sounds like just mathobabble to me (and if that isn't a word it should be). You can't dispute that a mathematical stick has an infinite number of points and that it has a probability of 0 of being able to break anywhere. But, look, the stick HAS to be one-dimensional, or else we have to take into account the possibility that the stick can break into any three partitions! I've never dropped a stick and had it break across the long way before, which is why we were all assuming that you can describe the breaking point by a distance alone. If you want to talk about a three-dimensional stick and think about breaking it into any possible subsets (and this is where Banach-Tarski could come in) you have to provide a very rigorous definition of what it means to form the pieces into a triangle. If any subset is possible and you're considering a three-dimensional theoretical stick, pieces have a probability of 0 of having a length (in the intuitive sense) shorter than the length of the stick (because there are an uncountably infinite number of points that must NOT be present in a given subset for it to be shorter), so the probability of making a triangle (in the intuitive sense) is 1 and the problem becomes boring.

Besides, as I think I've stated before, Banach-Tarski is completely unrelated to the situation at hand, and I don't even know why it's been brought up. Just because an object has an infinite number of points doesn't mean you can apply Banach-Tarski! It says you can break up a (three-dimensional) stick into a finite number of subsets that you can rearrange them into two sticks. But that's not relevant because we're taking one stick and breaking it up into three parts, once. You don't get to reassemble them however you want, you don't get to break the stick anywhere near enough to get 627 of them, and you don't even get enough parts to use the paradox because (as has been previously mentioned) you need five pieces to reassemble them into two of the original object. Even if the stick dropped and shattered into a million random partitions, you'd still have a probability of 0 of them being just the right pieces to reassemble into two sticks.

The thing is, Banach-Tarski is not a paradox in the sense of "something that causes a contradiction", but one in the sense of "something that goes completely against intuition and is hard to believe even after you prove it". There's nothing wrong with assuming a 3-dimensional theoretical stick has an infinite number of points. Banach-Tarski has to do with the tricky notions of volume and density, where if you break up a sphere into three pieces which each (somehow) contain 1/3 of the points then each piece has 1/3 of the density of the original, despite not containing any 'less' points. The fact that you talk about Banach-Tarski as if it makes sense tells me that you don't understand it too well, since in my experience the more you understand what the paradox means the more it seems impossible to believe. Hell, I've looked through a book that PROVES the Banach-Tarski theorem, and I still think it's crazy! Believe me, it's not an obvious result at all, and you can't explain it by analogy or by simple reasoning. Take a look at the original theorem written in mathematical notation, and you won't believe it either.


----------



## JBCM627 (Aug 13, 2008)

qqwref said:


> But, look, the stick HAS to be one-dimensional


I initially assumed it was 3-dimensional; there is nothing wrong with this. You assume we are working in a Euclidean space with Euclidean geometry; there is nothing wrong with this, either. These differing assumptions will probably be the root of any and all disagreement about this topic.



qqwref said:


> or else we have to take into account the possibility that the stick can break into any three partitions!


Yes, and I find this quite a bit more interesting than considering a 1D object. And you can't break them into just _any_ partitions (see A below).



qqwref said:


> If any subset is possible and you're considering a three-dimensional theoretical stick, pieces have a probability of 0 of having a length (in the intuitive sense) shorter than the length of the stick (because there are an uncountably infinite number of points that must NOT be present in a given subset for it to be shorter), so the probability of making a triangle (in the intuitive sense) is 1 and the problem becomes boring.


What? A) there are certainly restrictions on the subsets, such as the need for them to be fully connected to themselves and not interlocking with others, and B) There are certainly also uncountably many ways to partition a stick so that each segment is shorter than the initial length. C) as you said before, this depends on how you define your ability to form a triangle. As I noted in an earlier post, another consideration of this results in a 0% chance of forming a triangle.



qqwref said:


> It says you can break up a (three-dimensional) stick into a finite number of subsets that you can rearrange them into two sticks. But that's not relevant because we're taking one stick and breaking it up into three parts, once.


Obviously I am interpreting this problem quite liberally.



qqwref said:


> you need five pieces to reassemble them into two of the original object.


I think this lower bound is only for spheres.



qqwref said:


> Even if the stick dropped and shattered into a million random partitions, you'd still have a probability of 0 of them being just the right pieces to reassemble into two sticks.


I’d break it myself so that they could be reassembled.



qqwref said:


> The thing is, Banach-Tarski is not a paradox in the sense of "something that causes a contradiction"


 So technically, its not even a paradox.



qqwref said:


> "something that goes completely against intuition and is hard to believe even after you prove it".


 Actually, the more in depth I look at it, the more and more it does make sense.



qqwref said:


> Banach-Tarski has to do with the tricky notions of volume and density, where if you break up a sphere into three pieces which each (somehow) contain 1/3 of the points then each piece has 1/3 of the density of the original, despite not containing any 'less' points.


I don't consider volume and density very tricky. But herein lies an intuitive understanding of the paradox: the initial object has an infinite density, and so these partitioned objects still also have an “infinite density”. 1/3 of infinity is still infinity.



qqwref said:


> Hell, I've looked through a book that PROVES the Banach-Tarski theorem, and I still think it's crazy! Believe me, it's not an obvious result at all, and you can't explain it by analogy or by simple reasoning. Take a look at the original theorem written in mathematical notation, and you won't believe it either.


I don’t believe you, and I’m sorry if any “mathematical notation” confuses you so much that you believe it even less. If anything, the detail of the proof should allow it to make more sense. I will admit I don’t understand everything in the proof itself, but I do have a grasp of the flow of the proof, and do understand at least some of the details.

As for it intuitively making sense, http://www.kuro5hin.org/story/2003/5/23/134430/275… scroll down to “how can this ever be intuitive”. That’s a good enough analogy for me, and is very simple reasoning. And instead of just looking through a book, you can go here: http://www.jstor.org/stable/2321514?seq=5.


----------



## JBCM627 (Aug 13, 2008)

Lucas Garron said:


> JBCM627 said:
> 
> 
> > This in my mind is no different than claiming the cardinality of the integers is the same as the cardinality of the even numbers.
> ...



Btw, I guess I'm not the only one with such a mind...


from that one link in the previous post I made said:


> In fact, if you think about it, this is not any stranger than how we managed to duplicate the set of all integers, by splitting it up into two halves, and renaming the members in each half so they each become identical to the original set again



I mean, think about it... 1 group of infinity = 2 groups of infinity? That's all BT really claims at the end, which is the same claim |*N*| = |*N*-even| makes. And I really don't want to post in this thread any more, so will probably refrain from doing so...


----------



## Lucas Garron (Aug 13, 2008)

JBCM627 said:


> qqwref said:
> 
> 
> > you need five pieces to reassemble them into two of the original object.
> ...


That made me laugh very hard. 

I've given up on this thread, because I'd be responding to something false you state every other line. I still respect you, I just don't want to have to keep dispelling actually false presumptions (not just alternate interpretations).

(Just to make this post more useful, simple proof that [knowing that 5 is necessary for a sphere] 5 necessary for a rod: There are lots of spherical regions in the rod, and any of them requires 5.)


----------



## qqwref (Aug 13, 2008)

JBCM627 said:


> You assume we are working in a Euclidean space with Euclidean geometry; there is nothing wrong with this, either.


Excuse me, what? Please explain where I assumed this.



JBCM627 said:


> A) there are certainly restrictions on the subsets, such as the need for them to be fully connected to themselves and not interlocking with others,


How do you define "interlocking"? And what in the problem itself tells you that the pieces have to be connected and non-interlocking?



JBCM627 said:


> B) There are certainly also uncountably many ways to partition a stick so that each segment is shorter than the initial length


That's not relevant to probability. There are uncountably many real numbers and uncountably many complex numbers, but that doesn't mean that a random complex number (however you choose it) has a nonzero chance of being real.



JBCM627 said:


> Obviously I am interpreting this problem quite liberally.


Oh, sorry, I thought we were doing math.



JBCM627 said:


> I’d break it myself so that they could be reassembled.


Doctor: Your bone is fractured in four pieces!
JBCM: *crack* Look, I broke it again, so now it can be reassembled into two bones!
(but seriously, this is not a part of the problem, and thus completely irrelevant)



JBCM627 said:


> I don't consider volume and density very tricky. But herein lies an intuitive understanding of the paradox: the initial object has an infinite density, and so these partitioned objects still also have an “infinite density”. 1/3 of infinity is still infinity.


That's why I said the concept is tricky, because the definition you're using is completely useless unless you're talking about finite sets of points. One useful way to define density might be the probability that a random point inside a set's bounding set (i.e. the smallest continuous, 'nice' set that contains it) is in the set. So if you create a set which contains exactly 1/2 of the points (chosen randomly) in the cube [0,1]x[0,1]x[0,1], the set will have density 1/2. You run into definitions like this in measure theory and related fields.



JBCM627 said:


> I’m sorry if any “mathematical notation” confuses you so much that you believe it even less.


It's easy to know something but not really believe it... But seriously, what I mean is that when the result is written down explicitly in mathematical terms, it is just as counterintuitive and unbelievable as before, if not more so. Of course I know it's true, but that's different from intuitively believing it.



JBCM627 said:


> As for it intuitively making sense, http://www.kuro5hin.org/story/2003/5/23/134430/275… scroll down to “how can this ever be intuitive”. That’s a good enough analogy for me, and is very simple reasoning.


What strikes me about that explanation is that there's nothing about it that wouldn't apply to one- or two-dimensional objects, and it's known that Banach-Tarski is false for smaller dimensions, so that explanation can't be completely correct. Crystalline structures appear in two dimensions as well, and they compare it to the one-dimensional case of splitting up the integers. By the way - try it in one dimension. Just try splitting up an uncountable number of objects into two uncountable and dense everywhere subsets - you'll find that it's trickier than you think.



JBCM627 said:


> And instead of just looking through a book, you can go here: http://www.jstor.org/stable/2321514?seq=5.


I quote: "This article is available for purchase from the publisher for $12.00 USD.". No thank you, I'll stay with my library.


----------



## DavidWoner (Aug 13, 2008)

philkt731 said:


> Ok Lets say the stick is of length 1, think of it like a ruler marked from 0 to 1.
> Lets say we cut it at mark x and mark y. The three lengths are x, y-x, and 1-y for the case that y>x.
> 
> From trignale inequality we get that
> ...





cmhardw said:


> Coolest problem I've done in a long time.
> 
> I also got 1/4 as the probability. I'm very proud of myself for solving this one, as I tried to do it without reading this thread (other than the initial problem itself) or any other math resource.
> 
> ...



this is seriously how you guys did this? graphing? integrating? there is no need for calculus or planar geometry. it is the simplest of probability problems, with just a splash of 4th grade geometry. there is no disrespect intended, but it seems that your brains are getting in the way. here is the simple, unambiguous intuitive solution:

we have already covered the whole in-order-to-make-a-triangle-the-sum-of-any-two-sides-must-be-greater-than-the-third thing. the only way for this to not be true is if both of the breaks occur on the same half of the stick. so:

we are assuming that there is an equal probability of a break occurring at any given point, and that the breaks occur independent of one another. so its just the probability of one break occuring on one side(.5) time the probability of the second break happening on that same side(.5)

so .5 * .5 = .25.

its a problem that is much easier to solve visually.


----------



## cmhardw (Aug 13, 2008)

Vault312 said:


> this is seriously how you guys did this? graphing? integrating? there is no need for calculus or planar geometry. it is the simplest of probability problems, with just a splash of 4th grade geometry. there is no disrespect intended, but it seems that your brains are getting in the way. here is the simple, unambiguous intuitive solution:



With no disrespect intended, you did not account for the case when both breaks are more than half the stick apart from each-other.

I think this is probably what you intended to write?

P(no triangle) = P(both breaks occur on the same half) + P(both breaks more than half the stick apart)

P(both breaks on same half) = 0.5
P(both breaks more than half the stick apart = 0.25

P(no triangle) = 0.25 + 0.5 = 0.75

1-P(no triangle) = P(triangle) = 0.25

--edit--
by the way this is the method one of my good friends from high school used when I told him this problem. So no credit to me for the idea.
--edit--


Chris


----------



## DavidWoner (Aug 13, 2008)

cmhardw said:


> Vault312 said:
> 
> 
> > this is seriously how you guys did this? graphing? integrating? there is no need for calculus or planar geometry. it is the simplest of probability problems, with just a splash of 4th grade geometry. there is no disrespect intended, but it seems that your brains are getting in the way. here is the simple, unambiguous intuitive solution:
> ...



ah. yes. that. well doing the wrong math and still getting the right answer is sort of my specialty. still, i think using probabilities is the simplest way to solve this.

i realized that there was a hole in my math because the .5 from the first break shouldnt have been there, since the position of the first break is unimportant(at least for what i was doing). much like rolling the same number on two dice, the first is can be anything, the second just has to match. however, i just decided to roll with it. although having the two breaks more than half the stick away from each other seems to fill this hole. sorry this is literally the first non-arithmetic math that i have done in over 3 months. i will think before speaking next time.

i should get out my old textbooks and practice before i go to college...


----------

