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"Automatic trisection" of an angle


Jayeskay

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well, i think you should have written "straightedge", not ruler for one thing. for another, you have in the second page constructed a vesica piscis and since it is a 120º arc it trisects the angle.

 

this is a special case however and the trisection problem specifies an arbitrary angle and using only a compass & straightedge. it is proven impossible. :(

 

angle trisection

 

Hi - I am new here.

 

Is this of interest: http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm ?

 

Jayeskay.

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hi jay. always interested to look a gift drawering in the hypotenuse. :clue:

 

under diagram 1 it says:

 

:turtle: right there @ step#2, it is wrong. Line AC=sqrt(AB^2+BC^2); pythagorean theorem.

 

Thanks -

 

That AC = 2AB is not in question as it is a 'given' of my construction.

I have been advised by an honorary maths professor that the construction is geometrically correct.

May I suggest you have a closer look at Diagram 1, then look at Diagram 2.

 

Jayeskay.

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Thanks -

 

That AC = 2AB is not in question as it is a 'given' of my construction.

I have been advised by an honorary maths professor that the construction is geometrically correct.

May I suggest you have a closer look at Diagram 1, then look at Diagram 2.

 

Jayeskay.

 

mmmm... i guess then i want to see the construction using compass & straightedge of right triangle ABC such that AC=2AB

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mmmm... i guess then i want to see the construction using compass & straightedge of right triangle ABC such that AC=2AB

 

 

The entire diagram is so constructed. It is not necessary to explain every geometric rule. For example the equation you employed in your previous comment is known by everyone at the most elementary level to automatically apply to ALL right-angled triangles regardless of the relative length of the hypotenuse and size of the two smaller angles.

 

Employing the most elementary level of geometric knowledge and procedure the diagram is demonstrating the automatic trisection of any acute angle constructed as so described, on which basis diagrams 2 and 3 proceed.

 

Jayeskay.

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The entire diagram is so constructed. It is not necessary to explain every geometric rule. For example the equation you employed in your previous comment is known by everyone at the most elementary level to automatically apply to ALL right-angled triangles regardless of the relative length of the hypotenuse and size of the two smaller angles.

 

Employing the most elementary level of geometric knowledge and procedure the diagram is demonstrating the automatic trisection of any acute angle constructed as so described, on which basis diagrams 2 and 3 proceed.

 

Jayeskay.

 

well, that's not good enough for me. as your whole demonstration hinges on a particular type of right triangle, and inasmuch as trisecting being proven impossible refers only to doing it using a compass and straightedge, you must give a compass and straightedge construction of that particular triangle in order to justify further steps. you can't just say it is a given; it is necessary to justify every step. as you imply that it ought to be elementary, then it shouldn't take you long. thnx.

 

ps perhaps you other lurking geometers would like to make your own construction for comparison. :idea: :painting:

 

pps i would lose the graph paper as it constitutes a marked rule, whether it was used as such or not.

Edited by Turtle
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well, that's not good enough for me. as your whole demonstration hinges on a particular type of right triangle, and inasmuch as trisecting being proven impossible refers only to doing it using a compass and straightedge, you must give a compass and straightedge construction of that particular triangle in order to justify further steps. you can't just say it is a given; it is necessary to justify every step. as you imply that it ought to be elementary, then it shouldn't take you long. thnx.

ps perhaps you other lurking geometers would like to make your own construction for comparison. :idea: :painting:

pps i would lose the graph paper as it constitutes a marked rule, whether it was used as such or not.

 

Turtle -

 

With respect:

 

If you are saying you do not know how to construct a line perpendicular to a given line, and a right-angled triangle with the hypotenuse twice as long as the shorter side, refer to an elementary geometry book.

 

As to my use of graph paper to construct the diagrams, you are confusing convenient draftsmanship with geometry. The diagrams but represent the geometry, in which lines and arcs are but directions without width, and points are locations without dimensions.

 

Jayeskay.

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Turtle -

 

With respect:

 

If you are saying you do not know how to construct a line perpendicular to a given line, and a right-angled triangle with the hypotenuse twice as long as the shorter side, refer to an elementary geometry book.

 

As to my use of graph paper to construct the diagrams, you are confusing convenient draftsmanship with geometry. The diagrams but represents the geometry, in which lines and arcs are but directions without width, and points are locations without dimensions.

 

Jayeskay.

 

i am saying i want to see your construction. as to the graph paper it is an unecessary confusion.

 

anyway, i drew the construction myself & have it attached below. by all means let me know if i have erred in it. with that settled, we can proceed to the rest of your constructions step by step.

 

ps already i have a question: how is point D located/chosen in Diagram 1?

 

edit: i made some clarifications in the text of my drawing; attachment #2

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i am saying i want to see your construction. as to the graph paper it is an unecessary confusion.

 

anyway, i drew the construction myself & have it attached below. by all means let me know if i have erred in it. with that settled, we can proceed to the rest of your constructions step by step.

 

ps already i have a question: how is point D located/chosen in Diagram 1?

 

 

Line AB = AD = AE = DC = DH.

Line AB = distance BD.

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Line AB = AD = AE = DC = DH.

Line AB = distance BD.

 

roger. that should be a statement in your construction steps. after putting my compass to your drawing just now, i made that assumption that AD=AB and proceeded to follow your steps in diagram 1 and make the construction. first attachment below. - note that i chose a different H for comparative purpose.

 

as you might expect, i have some issues. :doh: :lol: first, you don't show the justifications in step #8 for "automatically trisected", as in the form of theorems such as "opposite interior angles are equal" or some such a matter as applies to a proof of the statement.

 

secondly, and most importantly, presuming that angle ICJ is proven 1/3 of angle IHJ, this does not in fact trisect angle IHJ. you need to actually construct the 2 lines that trisect IHJ. in the second attachment i have added eye-balled points L & M and lines LH & MH to show what i would expect to see when i read "trisection of an angle".

:cap:

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roger. that should be a statement in your construction steps. after putting my compass to your drawing just now, i made that assumption that AD=AB and proceeded to follow your steps in diagram 1 and make the construction. first attachment below. - note that i chose a different H for comparative purpose.

 

as you might expect, i have some issues. :doh: :lol: first, you don't show the justifications in step #8 for "automatically trisected", as in the form of theorems such as "opposite interior angles are equal" or some such a matter as applies to a proof of the statement.

 

secondly, and most importantly, presuming that angle ICJ is proven 1/3 of angle IHJ, this does not in fact trisect angle IHJ. you need to actually construct the 2 lines that trisect IHJ. in the second attachment i have added eye-balled points L & M and lines LH & MH to show what i would expect to see when i read "trisection of an angle".

:cap:

 

Automatic trisection:

1. In isosceles triangle CDH, angle DHC = angle DCH.

2. In isosceles triangle DHJ, angle HDJ = angle HJD.

3. External angle HDJ to isos Triangle CDH is twice the size of angles DHC and DCH.

4. Relatively in the diagram as a whole:

Angles DCH and DHC = 1.

Angles JDH and DJH = 2.

Angle IHJ, being an external angle to triangle JHC = 3.

5. Therefore the angle ICJ trisects the angle IHJ.

 

 

The trisecting line:

Step 6 creates the randomly chosen trisecting line to C, and the automatically trisected angle IHJ.

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roger. that should be a statement in your construction steps. after putting my compass to your drawing just now, i made that assumption that AD=AB and proceeded to follow your steps in diagram 1 and make the construction. first attachment below. - note that i chose a different H for comparative purpose.

 

as you might expect, i have some issues. :doh: :lol: first, you don't show the justifications in step #8 for "automatically trisected", as in the form of theorems such as "opposite interior angles are equal" or some such a matter as applies to a proof of the statement.

 

secondly, and most importantly, presuming that angle ICJ is proven 1/3 of angle IHJ, this does not in fact trisect angle IHJ. you need to actually construct the 2 lines that trisect IHJ. in the second attachment i have added eye-balled points L & M and lines LH & MH to show what i would expect to see when i read "trisection of an angle".

:cap:

 

 

Auto trisection-

Relatively, within the diagram:

Angles DHC and DCH each = 1.

Angles DJH and external JDH each = 2.

External angle IHJ = 3.

Therefore angle ICJ trisects angle IHJ.

 

Trisection line-

Line CI automatically constructs the trisecting line and the randomly constructed angle IHJ.

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Automatic trisection:

1. In isosceles triangle CDH, angle DHC = angle DCH.

2. In isosceles triangle DHJ, angle HDJ = angle HJD.

3. External angle HDJ to isos Triangle CDH is twice the size of angles DHC and DCH.

4. Relatively in the diagram as a whole:

Angles DCH and DHC = 1.

Angles JDH and DJH = 2.

Angle IHJ, being an external angle to triangle JHC = 3.

5. Therefore the angle ICJ trisects the angle IHJ.

 

 

The trisecting line:

Step 6 creates the randomly chosen trisecting line to C, and the automatically trisected angle IHJ.

 

you honestly can't bring yourself to actually draw it? if you leave your diagram 1 unchanged, it is no geometric proof. i'll look over your steps tomorrow. extraordinary claims require extraordinary proof.

 

looking ahead to your diagram 2, those measures you marked out on the left must be derived by construction; you cannot use the graph paper as a measure as that is using a ruler. :naughty:

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Not much.

 

First of all I, unlike Turtle, would find no difficulty in constructing the triangle ABC which is a classic exercise. However, I agree that the last step in each case is not properly argued and I won't spend too much of my time looking for proof or disproof of it.

 

Mostly, it is not a construction starting from the angle to be trisected, as a given. This is the main objection. There are many ways to triple a given angle, heck its actually trivial, but to say that one can simply find an angle such that tripling it matches the given angle just doen't qualify as a solution to the ancient problem. On this account, you are proposing three new and complicated ways to do something that has long been done much more simply.

 

What's more, some dude proved that a SE&C solution would amount to an algebraic method for solving a certain kind of equation, which was already a proven impossibility.

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Alright, to your credit, I found how to argue for step 8 in the first diagram, it didn't even take that long once I fetched meself paper and pencil.

 

In doing it, it dawned on me that you could well construct the IHJ angle without any need for triangle ABC, all you need to do is set a compass in C and mark off D, then H, then J and it's simpler, eh?

 

So it remains: Given IHJ, how do you go about finding the right point C for trisecting it? :scratchchin:

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Not much.

 

First of all I, unlike Turtle, would find no difficulty in constructing the triangle ABC which is a classic exercise. However, I agree that the last step in each case is not properly argued and I won't spend too much of my time looking for proof or disproof of it.

...

 

well, i only had some dificulty seeing the construction in my minds eye. i had no problem actually doing the construction and posted my drawing in post #13. no doubt that there are other constructions than mine. by all means let me know if i erred there as i like my own errors even less than someone else's. :agree:

 

do you also agree with me that diagram 1, or any drawing of a trisection, requires points and lines inside the angle to be trisected that correspond to a trisection? after all, a construction is a drawing, isn't it? if you askered a student to do a construction & got no drawering, would you accept that as a proper response?

 

always glad to hear from you though prof! :smart: :thumbs_up

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Sorry, I hadn't followed your exchanges quite exactly, so in the light of this subtlety let me put it as follows: I wouldn't require this guy to spell it out in detail. If a student uses something already known, in spelling out a method, one may consider it an independent thing. Unless of course it is an actual execution that is required. When the aim is to argue a point, one can certainly use standard things. For instance, a student won't always demonstrate Pythagoras' theorem each time it comes into demostrating something else. Even less does a researcher do so when publishing a novel finding.

 

Your construction has no error. It can be done more simply though; once the segment of length two is done there's your hypotenuse and, without changing the compass, you can make the intersection between arcs from the midpoint and one end.

 

do you also agree with me that diagram 1, or any drawing of a trisection, requires points and lines inside the angle to be trisected that correspond to a trisection? after all, a construction is a drawing, isn't it? if you askered a student to do a construction & got no drawering, would you accept that as a proper response?
I'm not sure if I read your intent. Do you mean he should trace a parallel to his line AC through H and then remaining line through H? As far goes proving his point, this is superfluous, I rest that his real shortcoming is that he would need to give a construction for the reverse of what he does.
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