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

[who_loves_maths]

the Jacob Steiner Theorem states that when given a fixed circle it is impossible to find the centre of that circle (throught pure constructions) using only a straightedge {ie. NO compass(es)}.

however, what about using only a ruler (ie. a straightedge, but with measurements/calibrations on it [accuracy to an infinite degree]) ?

that is to say, when given a fixed circle, is it possible to find the centre of that circle using ONLY a RULER and a pencil (which is just for drawing lines and connecting points) ?
if so, how?


any ideas and opinions (or solutions) here is appreciated, any discussions welcomed

 

[SeDaTeD]

Use a ruler to make a 3-4-5 triangle, thus obtaining a right angle. Then plonk it on the circle, extend and join to make a semicircle, and find the midpoint of that.

 

[Templar]

I think that would be the equivalent of trisecting an angle with a tomahawk.

 

[SeDaTeD]

Yeah, true, though i found the tomahawk pretty clever.

 

[who_loves_maths]

Originally Posted by SeDaTeD
Use a ruler to make a 3-4-5 triangle...

you can't do that... since that's the same as using a protractor to measure out a right angle and then draw the 3-4-5 triangle around that.

in order to just use a ruler to do it, you need to use trial and error and adjust your lines until you get to the 3-4-5 triangle, which is not a proper method to doing it in this case.

in addition, you need to draw and start all your constructions inside or on the circumference of the fixed circle... that is, the ruler is simply used to join two points at a time on/in the circle, while at the same time able to measure distances... so it doesn't have "a mind of its own" and is NOT supposed to be used in trial and error and adjusting until you (the participant) get whatever you wish.

ie. NO brute force approach plz ... if you can't guarantee that a person can draw a perfect 3-4-5 triangle using ONLY a ruler the first go, then that's the end of that method.
{also, like you said, it's the equivalent of using a tomahawk to trisect an angle, which is something you can't do with just a ruler under normal circumstances, so you still can't cheaply bypass that in this case simply by bruting your way to a 3-4-5 triangle.}

 

[Antwan23q]

wow, i thought you just had to draw a chord, and draw a perpendicular bisector of it would always lie on the centre. then u do it again to find it.

 

[who_loves_maths]

Originally Posted by antwan2bu
wow, i thought you just had to draw a chord, and draw a perpendicular bisector of it would always lie on the centre. then u do it again to find it.

no you can't do that because like i said: using ONLY a RULER... you can't make a right angle with a ruler - in mathematics, a straightedge is an abstract thing with only one infinitely long edge... in this case, the ruler is the same thing but with calibration on it so you can take measurements (but not use it to trial and error).

so you can't just build a right angle at will.

 

[Templar]

also, like you said, it's the equivalent of using a tomahawk to trisect an angle, which is something you can't do with just a ruler under normal circumstances, so you still can't cheaply bypass that in this case simply by bruting your way to a 3-4-5 triangle.

That's not the point. The point is that using a tomahawk to trisect an angle is invalid under Euclidean geometry. While your rule removes the compass and adds measurements to the straightedge, I believe this still holds in this system. Hence even using a 3-4-5 triangle and scaling it to fit the circle is not a correct solution, provided you can construct the triangle and scale it.

I personally believe it cannot be done, but I can't prove that.

 

[who_loves_maths]

Originally Posted by Templar
That's not the point. The point is that using a tomahawk to trisect an angle is invalid under Euclidean geometry. While your rule removes the compass and adds measurements to the straightedge, I believe this still holds in this system. Hence even using a 3-4-5 triangle and scaling it to fit the circle is not a correct solution, provided you can construct the triangle and scale it.

I personally believe it cannot be done, but I can't prove that.

yes, that was the point of my last post - that drawing a 3-4-5 through chancing is NOT a satisfactory solution to this problem...

in short, a construction here is one in the sense that when you make it (without braking any rules), then you've made it, and further constructions follow from that, there is 100% certainty with it; also it has to be made within the circle btw, and not on a separate piece of paper and then traced into the circle. so any 'method' which involves a degree of uncertainty and the requirement of multiple drawings with the use of a RUBBER or extra sheets of paper is not allowed, and so is not counted as a legitimate 'solution'.

 

[who_loves_maths]

just elaborating on my last post:

the use of a tomahawk AT ALL is not even the point here. SeDaTeD’s method, in its entirety, is flawed even before that part…

using chance and probability etc, is itself not illegitimate - the thing is, however, that using that method to draw the 3-4-5 triangle in the way SeDaTeD suggests is unstructured, and that’s where the problem is.

A strucured series of constructions with a certain "target", but whose process depends on the intermediate steps taken can be allowed since the only variance is the length of the process each time, the target does not change and will/can always be reached because the steps follow a well laid out path/logic (eg. iterative or recursive methods.)...

this is unlike SeDaTeD’s method because IF the first attempt at a 3-4-5 triangle fails, then the next new line can be started, placed and drawn at ANY other angle, so there is no guarantee that the target triangle will ever be constructed (since there are an infinite number of different angles).
Hence, it’s unstructuredness is what denies it as a definitive proof.

 

[SeDaTeD]

Haha, I think you took my post too seriously.

 

[who_loves_maths]

Originally Posted by SeDaTeD
Haha, I think you took my post too seriously.

lol, on hindsight, yes i probably did. sorry if i was a bit too "aggressive" about it, didn't mean it that way...

but honestly, when i first saw your method, i thought it was pretty neat and subtle. but just unfortunately not rigorous enough for the particular restrictions of this problem.

 

[SeDaTeD]

Hehe, it's cool. I knew my method was fudgy anyway. I am aware of the problems of trying to contstcut a right angle that way, and it may turn out that you wont get it. Just threw it in, might've convinced some people, obviosuly not you or Templar.

 

[who_loves_maths]

Originally Posted by SeDaTeD
Hehe, it's cool. I knew my method was fudgy anyway. I am aware of the problems of trying to contstcut a right angle that way, and it may turn out that you wont get it. Just threw it in, might've convinced some people, obviosuly not you or Templar.

haha... well, do you know or suspect of any other possible methods of doing this problem?
because the Jacob Steiner Theorem EXPLICITLY says that one cannot find the centre with a straightedge ---> which tends to lead ppl on to believe/think that perhaps it's possible with something like a calibrated ruler?... otherwise, the theorem might as well just says that the construction is not possible for not only a straightedge, but also a ruler...

 

[Templar]

I believe the comment about straightedge is that rulers, by convention, does not exist in mathematics.

Just to clear it up, my point is that even if you can constructed a 3-4-5 triangle properly (let's just say with ruler and compass), and you can stretch it (however that is done mathematically...), by superpositioning this is mathematically invalid, which you mentioned. The reference to the tomahawk was that what he was doing is somewhat similar, and we all know that the tomahawk is not a mathematical method to trisect an angle (not that there is one).

 

[who_loves_maths]

Originally Posted by Templar
I believe the comment about straightedge is that rulers, by convention, does not exist in mathematics.

Just to clear it up, my point is that even if you can constructed a 3-4-5 triangle properly (let's just say with ruler and compass), and you can stretch it (however that is done mathematically...), by superpositioning this is mathematically invalid, which you mentioned. The reference to the tomahawk was that what he was doing is somewhat similar, and we all know that the tomahawk is not a mathematical method to trisect an angle (not that there is one).

^ it doesn't matter if the ruler is officially mathematical object or not... don't be so close-minded; mathematics is a subject of problem-solving, and in this particular problem we have a ruler, not a straightedge. and we're simply asked to solve that problem. (you can say it's indeterminate if you like, that would count as a 'solution'.)

also, there is no "superpositioning", it is in fact a 'projection'. and i thought that Geometric Projection is a mathematically viable concept. (isn't that what they in fact used to prove that you can't find the centre of a circle with just a straightedge? or at least one of the proofs.)

 

[casebash]

Easy, provided the marking are actually small enough for you to mark around the circle. draw a line AB with A clockwise from B. Let AC be a 1 unit chord with C clockwise from A, and BD be 1 unit with D anticlockwise from B. Let EC be a 1 unit chord with E clockwise from C, and FD be 1 unit with F anticlockwise from D. Join CB and AD at X. Join BE and AF at Y. XY joins to give a diamater. Now construct a second diameter.

 

[Templar]

Read previous threads please. Then you'll realise you can't draw unit chords.

 

[SeDaTeD]

Hmm, what about constructing 3 tangents to the circle, creating a triangle with sides of lenght a, b and c. Then the area of the triangle can be found by Heron's formula then equated with A=rs, where r is the radius of the circle and s = (a+b+c)/2. Then r can be found. Then you can do a bit of island hopping around the circumference with chords of length r, then joining up the opposite ones, which should interset at the centre. But then this method won't please Templar either so I'll leave it at that.

 

[Templar]

Can you even draw a tangent? I'm not sure how you can do so. But good thinking.