"Automatic trisection" of an angle

[Rade]

Jayeskay...I have a question. If it was possible to trisect 'any' constructed chance acute angle of HIJ in your figure 1, then there should be a line length along IH and IJ that would be equal to a line constructed between JH, such that a perfect isosceles triangle was formed with angle JIH = 60 degrees. This isosceles triangle should then be trisected by angle DCI, however I am not able to construct any such isosceles triangle from your figure 1 :( ...any suggestions on the length along line CD that would allow such an isosceles triangle to be formed ?

[Turtle]

Thanks for the detailed response. I knew that the trisection of an angle impossible with ruler and compass - it should not even try, but I think that my

"T" divider elegant tool for practical use. With a few of these tools is possible n - section.

 

See here for paper claiming arbitrary acute angle trisection using ruler and compass:

 

snip ...

 

:doh: no, no, no. not a ruler. a straight-edge people. it is trisection of arbitrary angle by straightedge and compass that is proven impossible. that's Proven. Period...end of story. dismissing the proof is as good as dismissing geometry itself.

[Rade]

no, no, no. not a ruler. a straight-edge people. it is trisection of arbitrary angle by straightedge and compass that is proven impossible.

Well, this site does claim to make the trisection by STRAIGHTEDGE and COMPASS, as seen from the introduction by the author copied below. He shows a ruler on the title page which adds confusion.

 

http://larrysimpson.zymichost.com/aaat.pdf

 

"Introduction: A method for the classic problem of arbitrary angle trisection constructible by straightedge and compass was developed by the author during December, 2012."

Edited April 9, 2013 by Rade

[Turtle]

Well, this site does claim to make the trisection by STRAIGHTEDGE and COMPASS, as seen from the introduction by the author copied below. He shows a ruler on the title page which adds confusion.

 

snip...

 

"Introduction: A method for the classic problem of arbitrary angle trisection constructible by straightedge and compass was developed by the author during December, 2012."

 

yes well, in spite of your mistake i understood your implication and so there was no need to read the paper. what part of "proven impossible" do you not understand? what's the point of adding yet another fairy tale? you only add to the confusion. good grief.

 

 

EDIT: >> allow me to elucidate. Why Trisecting the Angle is Impossible

Edited April 9, 2013 by Turtle

[Rade]

yes well, in spite of your mistake i understood your implication and so there was no need to read the paper. what part of "proven impossible" do you not understand? what's the point of adding yet another fairy tale? you only add to the confusion.

Well since you ask, the part of "proven impossible" that I understand is that working in GEOMETRY ALONE it is impossible to prove that an acute angle cannot be trisected using the methods given in the OP. If I error, please along a geometric proof that does not require algebra, such as the cos(2) argument related to 60 degree angle.

 

Also, why was the OP allowed to linger so long with so many responses given that it has been proven impossible via algebra ?

Edited April 10, 2013 by Rade

[Turtle]

yes well, in spite of your mistake i understood your implication and so there was no need to read the paper. what part of "proven impossible" do you not understand? what's the point of adding yet another fairy tale? you only add to the confusion. good grief.

 

 

EDIT: >> allow me to elucidate. Why Trisecting the Angle is Impossible

 

Well since you ask, the part of "proven impossible" that I understand is that working in GEOMETRY ALONE it is impossible to prove that an acute angle cannot be trisected using the methods given in the OP. If I error, please along a geometric proof that does not require algebra, such as the cos(2) argument related to 60 degree angle.

 

Also, why was the OP allowed to linger so long with so many responses given that it has been proven impossible via algebra ?

 

you are some piece of work rade.talk, talk, talk. :blahblahblah:

 

i won't even hazard a guess what "GEOMETRY ALONE" means or why suddenly you introduce acute angles. (and please do not tell me.) if you have a construction to present, present it in a new thread.

 

i rather suspect from your response that you didn't read the material at the link i gave above. perhaps you missed my adding it in an edit, but read it you should as it goes to some length to point out common errors in trisection claims. what i'm quoting below is just a summary of such errors and no substitute for reading the paper.

 

...

Proving It Can't Be Done

 

Follow the Rules

Many people who "solve" the angle trisection problem inadvertently violate the rules:

• You can only use a compass and a plain straightedge.
• You cannot use the straightedge for measuring, or put marks on it.
• You can only use the compass for drawing arcs around a fixed center. You cannot slide the pivot.
• You cannot use a straightedge and compass to construct some other tool.
• You cannot use a straightedge and compass to construct some other curve.
• All loopholes and evasions of these restrictions violate the rules....

 
i will however hazard a guess on the lingering topic. i believe the thinking goes 'if "we" don't allow -if not encourage- bul...erhm...nonsense , then there will be nothing to talk about and the board will die' or some such a matter along those lines. i'll leave it to the staff to correct me or not. :coffee_n_pc:

[Jayeskay]

My link at < http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm >

has a new suggested trisection for comment.

[Rade]

My link at < http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm >has a new suggested trisection for comment.

Is angle ABE trisected by NDA at exactly 4 graph paper squares to the left of point B along line AB ? Again, how exactly do you determine a perfect trisection of ABE without using the graph paper as a ruler, which violates rule of using only compass and straightedge, and not ruler of any type ? It might be better that you redo graphs with plain white background, not graph paper ?

[Jayeskay]

Is angle ABE trisected by NDA at exactly 4 graph paper squares to the left of point B along line AB ?

Again, how exactly do you determine a perfect trisection of ABE without using the graph paper as a ruler, which violates rule of using only compass and straightedge, and not ruler of any type ? It might be better that you redo graphs with plain white background, not graph paper ?

 

The graph paper is a drafting convenience and no measurements are made using it.

The drafting represents the geometry, which is described in the notes.

The diagram is constructable using only a straight-edge and compass.

The geometry is a suggested true trisection.

Edited April 27, 2013 by Jayeskay

[Dane]

[Dane]

[phillip1882]

just wanted to briefly comment, no, the relationshp df/fe = bf/fc is no constant. for example if the angle you are trying to trisect is 90 degrees, 1/3 is 30 degrees, making traingle FEB a 30 60 90 triangle with the simple ratio FB = sqrt(3) and FE = 2. it will definately not be this ratio for other angles.

[Dane]

You're right.

 

I'm very angry at myself, but my accessories for angle trisection is good

I was not in control. I apologize to everyone.

You especially like to thank. Can not forgive myself

[grogosaurus]

Dear Sir,   Please have the courtesy to read the following: As a graduate Mathematician of St Andrews University, and having trained in research methods for four years after that,  I have approached the problem in this way:  You accept the circle equality theorems : equal arcs  if and only if  equal chords             if and only if  equal angles at the centre  if and only if   equal (but half size) angles at the circumference.                                     I have developed new theoremd extending to a pair of concentric circles

Theorem:      Given two equal arcs on the inner circle they will subtend equal angles on the circumference of the outer circle

 Theorem     Given two adjacent and equal arcs on the inner circle, they will subtend equal and adjacent angles at the circumference of the outer circle

Theorem: Given three equal and contiguous arcs on the inner circle, they will subtend three mutually adjacent angles at any point on               outer circle

These I have proved with Euclidean rigour       Following on this I have used these newly proved theorems to support the folloowing

(1|)       given any angle L ABC

              (2)        Centre B and a suitable radius, draw an arc  to cut AB at P and BC at S                                                                      (3)        Join PS

              (4)        Centre and a suitable radius, draw a seeking arc between AB & BC                                                                              (5)        Centre S and the same radius draw  a defining arc to cut the previous arc centre P at D (6) align the straight edge along BD and mark E the mid point of PS (7) Centre   E  and radius PE draw a semi circle through P and S (8) Centre   P and radius, draw an arc to cut semicircle centre E at Q (9) Centre S and radius PE draw an arc to cut semicircle centre E radius PE, at R These chord/arcs will subtend three equal angle s of 60º at E             (10) Join BQ, BR,

 

these lines will create three equal angles subtended by arc PQ,  QR,  QR

Nd there fore these lines will create three equal angles subtended by arc PQ,  QR,  QR

Therefore LPBQ  =  LQBR   =   L RBS

The theory outlined in the new theorems, a bank of some 220 angles in the range 0 <  q < 180 yielding some 660 angles  which when graphed against  the original size clustered around y  =   x with 90% within the lines, y = x  +2,  and y  =  x  -2, and a chi  squared value of 65 to the nearest whole number well within the limit for 440 degrees of freedom.

 I therefore contend that with the theorems predicting the construction, and the empirical results confirming it, that this represents the nearest  we can get to trisecting the general angle;  Sunbsequent testing confirmed that it works for reflea angles also.

 

If you do not read this then please note it is already copyright in the UK for 10 yrs

[phillip1882]

um, this isn't really trisecting an abitrary angle. this is constructing a 60 degree angle off a given one and trisecting that.

[grogosaurus]

Dear Phillip   Thankyou for your comment  I think things went awry about item 5:

(6) align the straight edge along BD and mark E the mid point of PS

(7) Centre   E  and radius PE draw a semi circle through P and S

(8)should read: Centre   P and radiusPE, draw an arc to cut semicircle centre E at Q

(9) Centre S and radius PE draw an arc to cut semicircle centre E radius PE, at R

 COMMENT:These chord/arcs will subtend three equal angle s of 60º at E            

(10) Join BQ, BR, Now you are joining the ends of these equal arcs to the original vertex B, forming angles  ABQ,  QBR, R$B

OR the angles PBQ,  QBR,  RBS   which, since equal contiguopus ( or if you prefer adjacent) arcs on a circle ( or semi circle, will subtend equal angles at any point on the plane ( one of my newly researched theorems above) so effectively we have a construction:  Space does not allow me to  put in the three developmental theorems which build up to the construction, but as soon as I can get a website up and running with the geometry drawing facility I will refer you to it.

[phillip1882]

okay it was actually step 10 that i was stuck on, it still doesnt work however. the two edge angles PBQ and RBS are equal, but they don't equal QBR, or only do in the special case where ABC is a 60 degree angle. i constructed this diagram and measured the angles myself. i hope the copywriters of this work didn't make a profit off it.
 

Edited September 3, 2015 by phillip1882