possibility/impossibility of pure Geometric constructions. (1 Viewer)

[who_loves_maths]

^ that's what i'm intending to find out. it makes for a good maths problem on its own.

but yes, the infinitude of its convergeance is the only uncertain thing here. otherwise everything else is fine... and yes you'll know when you actually find the diameter, since the next chord you draw will no longer be longer - it'll have the same length. this happens ONLY when the two equal lengthed 'chords' are in fact diameters. and in which case, their intersection, as i said, will be the centre of the circle.

 

[who_loves_maths]

Originally Posted by Templar
casebash is not reading everything and clearly believes he's too good for the rest of us.

well he probably is one of those very talented and ellitist ppl who attend special training, by invitation, in extra-curricular and more challenging maths. you know, those ppl who are trained to be on the reserve list for those going to olympiads and what have you...
that's why he's been around saying how boring and 'crap' HSC mathematics is.
although, on this point i very much disagree with him... however i carry no experience of mathematics outside of school to render my opinion effective.

 

[SeDaTeD]

Would a chord be the diameter only if the two chords it bisects are parallel? If so, 2 successive chords must be parallel ones, and how do you construct those, other than by chance.

Also I'm still not convinced by your/casebash's method. Is this for any triangle constructed in the circle. Because it seems to me that it only works for isosceles ones, where you join the midpoint of the two equal sides.

Edit: I see now, I thought you meant join the midpoint of the line joining 2 midpoints of the triangles sides. Didn't know you meant to extend the line and find the midpoint of the interval within the circle.

Ok, now solution 2 just comes down to finding a way to get 2 parallel lines, by not using the method in solution 1, otherwise it'll be essentially the same.

 

[who_loves_maths]

Originally Posted by SeDaTeD
Would a chord be the diameter only if the two chords it bisects are parallel? If so, 2 successive chords must be parallel ones, and how do you construct those, other than by chance.

yes, you are right except for (i) there might be other possibilities than just joining the midpoints of parallel lines. but more importantly, (ii) notice that you must not start with just two chords, but THREE - in order that the process must continue on after the first 'iteration'. in this case, as each step/iteration continues, there will be more and more choices of chords at every step from which you can choose two whose midpoints you want to join... which means there is always a growing sample space of additional chords from which two or even more lines can very well be parallel to each other at any stage of the process.

Edit: remember that the vast sample space exists because there is no restriction on how many times you can use a chord that might have been already been used in a step before!

 

[SeDaTeD]

Well, this arguement of saying constructing a bunch of chords and hoping that 2 will be parallel is along the same lines as several previous arguements, where you hope it's bloody accurate, but it's still not exact.
for your part (ii), I don't see how that impacts on what I have said. This method will arrive at finding a diameter when the two chords chosen are parallel.
It may very well be that 2 lines could be parallel, but you have no guarantee.
I could very well say let's draw a lot of random chords, there's a chance that one could be the diameter, if i drew an almost infintie number, because there's a larger sample space.
I don't dig this go on long enough and you may very well get the answer.
However, I agree with your (i), but I only made that post in relation to your method.

 

[Templar]

You can draw 2 parallel chords, as casebash has shown earlier.

And I consider this topic QED, but keep on discussing other methods.

 

[SeDaTeD]

Yes I know, but i was refering to who_loves_maths's second method, and I stated that it shouldn't use casebash's method to find parallel lines, otherwise the solutions would be essentially the same. So it's a matter of finding parallel lines a different way (and not drawing an infinite number and hoping 2 happen to be parallel).

 

[who_loves_maths]

Originally Posted by SeDaTeD
Yes I know, but i was refering to who_loves_maths's second method, and I stated that it shouldn't use casebash's method to find parallel lines, otherwise the solutions would be essentially the same. So it's a matter of finding parallel lines a different way (and not drawing an infinite number and hoping 2 happen to be parallel).

if you read my last post clearly then you'd realise that that's exactly what i'm saying too - that there might be OTHER ways of generating 2 parallel lines. and that it might occur naturally in the iterative process of the second solution.
i never said anything about drawing infinite number of lines and hoping two will be parallel, in fact it is you who said that.

so i really don't know what you are arguing to me about?!

unless you yourself can definitively prove that the second approach is not correct then all this speculation of yours is only hearsay, since you yourself do not know if or not there are other methods of drawing parallel lines... so you are really attacking me on an empty basis ?! (destructive criticism?)

 

[who_loves_maths]

Originally Posted by SeDaTeD
Well, this arguement of saying constructing a bunch of chords and hoping that 2 will be parallel is along the same lines as several previous arguements, where you hope it's bloody accurate, but it's still not exact.
for your part (ii), I don't see how that impacts on what I have said. This method will arrive at finding a diameter when the two chords chosen are parallel.
It may very well be that 2 lines could be parallel, but you have no guarantee.
I could very well say let's draw a lot of random chords, there's a chance that one could be the diameter, if i drew an almost infintie number, because there's a larger sample space.

i don't agree with what you said at all. this method is NOT like the previous methods of chance and probability if you put more thought into it. the ONLY obstacle right now is knowing whether or not each 'recursive' step of the method will lead to actually finding the diameter, otherwise this proof is fine. this is because, like i already said before, this method is structured in that you are heading always in the direction of a particular target - each new chord you draw at every step will ALWAYS be longer than the previous one, and that fact is guaranteed... so you're always approaching a limit (the diameter) at all times.
unlike your initial method where you have no guarantee that you are approaching closer and closer to drawing a 3-4-5 triangle at each successive attempt!

so there is a significant difference between this method and the previous ones.


P.S. also, i suspect that this method is similar in nature to something like Newton's Method. where the convergeance toward a close approximation of a root is dependent on your INITIAL choice. similarly, whether this method will succeed or not looks like it might depend on the choice of chords at each step or choice of placement of initial chords.
However, i do not hear/see you complaining about Newton's Method as an illegitimate iterative method???
most iterative algorithms are of this similar uncertain nature - it doesn't mean that they don't work!

 

[SeDaTeD]

Point taken, but how do you guarantee that you will always find an exact diameter?
"the ONLY obstacle right now is knowing whether or not each 'recursive' step of the method will lead to actually finding the diameter, otherwise this proof is fine" I totally agree. What i said for the recursive method to obtain the exact diameter, is that for the final step , the two chords chosen must be parallel. (check) The thing i'm iffy on is that how do you know that you can get parallel lines?
It's true that each step will arrive at a much longer chord, and approach the diameter, but you want to find the exact one don't you?
And yes, i don't know/ can't prove if the second method will or will not arrive at parallel lines. But do you know if it will of will not? I'm saying that it relies on eventually obtaining a pair of parallel lines. If it does occur naturally and you can prove it, then I have nothing to argue about. (Actually am I arguing or trying to point things out? Come to think of it, I don't think I am arguing with you, just looking at a loose point in the second proof, whihc you too are aware of)
Actually, I think I'm agreeing with most of your last couple of points. I just didn't like the idea that it may only just approach the solution, not the exact thing.
And i've got nothing against iterative methods. Just that this question wants the EXACT centre/ diameter. I don't think newton's method actually obtains the exact root, but it gets bloody close.

 

[who_loves_maths]

Originally Posted by SeDaTeD
Point taken, but how do you guarantee that you will always find an exact diameter?
"the ONLY obstacle right now is knowing whether or not each 'recursive' step of the method will lead to actually finding the diameter, otherwise this proof is fine" I totally agree. What i said for the recursive method to obtain the exact diameter, is that for the final step , the two chords chosen must be parallel. (check) The thing i'm iffy on is that how do you know that you can get parallel lines?
It's true that each step will arrive at a much longer chord, and approach the diameter, but you want to find the exact one don't you?
And yes, i don't know/ can't prove if the second method will or will not arrive at parallel lines. But do you know if it will of will not? I'm saying that it relies on eventually obtaining a pair of parallel lines. If it does occur naturally and you can prove it, then I have nothing to argue about. (Actually am I arguing or trying to point things out? Come to think of it, I don't think I am arguing with you, just looking at a loose point in the second proof, whihc you too are aware of)
Actually, I think I'm agreeing with most of your last couple of points. I just didn't like the idea that it may only just approach the solution, not the exact thing.
And i've got nothing against iterative methods. Just that this question wants the EXACT centre/ diameter. I don't think newton's method actually obtains the exact root, but it gets bloody close.

okay i understand what you're trying to say, but what i'm trying to say is that i am, as you said, aware of all this fact as Templar has already pointed them out - the problem with convergeance. and i have already said in a previous post that that's currently what i am endeavouring to find out (and hopefully i can in the not too distant future ).

so i apologise to you if i seemed ignorant in my last post to you, but plz realise that i'm merely pissed off because it seems to me like you haven't read any of my responses to Templar on the same issue which you are concerned about right now - telling me that i have no guarantee that i will end up with the diameter, after i have already expressed to Templar that that's exactly what i am investigating atm because i too am concerned about it, is rather annoying. i just don't want to repeat myself that's all.
ie. yes i already KNOW that there is no guarantee of it because of the convergeance issue - the question now is whether or not a guarantee exists.

retrospectively and honestly, i am also also pissed off because i can't answer in fullness both your concerns over this method as of yet

 

[SeDaTeD]

Ok, I apologise for bringing up the same thing in a much more long winded way haha. Anyways good luck on trying to find out if it works or not.

 

[who_loves_maths]

^ thanks. maybe you'd like a crack at it too ?

 

[casebash]

who_loves_maths is correct, i did make to many mistakes. Even if we don't count the first, which i maintain is due to an ambiguity, my accuracy is still quite poor. The complaint about my excessive amount of errors was quite reasonable considering my unreasonable response to a request for a proof. As someone said, anyone can trig bash the coordinates of the point which is why i callled it "obvious'. I realise now that my phrasing implied that I thought you were an idiot and i am sorry for that. BTW, when i said 'get' i didn't mean construct, but find the coordinates of. I specifically mentioned at the end of the post detailing how that it was not a geometric method, as i did read the post pointing out that my method was not geometric. If you checked it and pretty sure you would find it is a valid method for finding the coordinates of the point.

 

[who_loves_maths]

Originally Posted by casebash
who_loves_maths is correct, i did make to many mistakes. Even if we don't count the first, which i maintain is due to an ambiguity, my accuracy is still quite poor. The complaint about my excessive amount of errors was quite reasonable considering my unreasonable response to a request for a proof. As someone said, anyone can trig bash the coordinates of the point which is why i callled it "obvious'. I realise now that my phrasing implied that I thought you were an idiot and i am sorry for that.

you needn't apologise at all casebash. we have all come across this type of situation before to know better than get too overly irate over it {except for the rare spur-of-the-moment things } and i'm sure i speak for SeDaTeD and Templar in saying that any discussions on mathematical problems (as long as its stimulating and not repetitive) are beneficial for all parties involved, in retrospect.

in the end, it was you who solved the problem of this thread, so once again congratulations. and thanks for instigating progress into this thread when it was at its most stale.