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"Automatic trisection" of an angle


Jayeskay

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Thank you again Phillip:     I was interested in your response, and would ask how many angles you drew and subjected to the construction..   In defense of my work I drew some 220 angles yielding some 660 sectors(? if this is not the rightterm for the outcomes of trisection please excuse)   When the results of these trisections wer plotted against the original size, the points clustered round the line y  =  x/3,  with some 95% lying between the lines y = (x/3)  +  2  and y  =  (x/3)  -2.  When tested for chi-squared against y  =  x/3, the total value for 660 points came to 65: this was well within the table value for 400 degrees of freedom.    I allowed 400 degrees of freedom, because in every angle subjected to "Trisection"  once two were measured, the fact that the third one must complete the value for the original angle, meant that in each angle there were only 2 degrees of freedom: I may have been overly restrictive, but you will applreciate the more stringent, the more arguably the result holds.   apropos of this , in e.g an anglr o6 69deg where we would expect 3 x 23 degree products,  then if two were 24, we have only 22deg left for the thirs angle, hence I allowed the toerance of 2 deg in the trisection products.  , then there is only   Now I know that in Maths this is not proof: I have also developed a set of Euclidean type theorems, which prove:           FG6:    In any circle centre O with points A,B,C on the circumference s.t. arc AB  =  arc BC,, then for any point P outside the circle s.t AP cuts the circle at R, BP cuts the circle at S  and CP cuts the circle at T  then arc RS  =  arc ST           FG7:        In any circle given 2 equal adjacent arcs then the angles at any point P outside the circle  then the equal arcs will subtend 2 equal adjacent angles at theat point            FG8:     If in any circle givn three adjacent ,equal arcs, then for any point P outside the circle,  these arcs will subtend 3, adjacent, equal angles.    It is the last theorem which gices the clue to the construction:    The semi circle centre E, is then cut into three equal arcs by the well established construction of setting up chords equal to PE/SE; These three arcs will now by theorem G* subtend three equal angles at any point in the plane, so we join them to B, the original vertex.   Further in my defense, I would  comment that our line unlike Euclid's line will have

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Dear Phillip,

           I really wonder what you are looking for as proof: I am looking at my 220 drawings as “ confirmation”  My Proof is built up through the series of theorems which I have  worked through extending the basic circle equality theorems  of Euclid: My research has led me to a series of equality theorems concerning the equality of angles on equal arcs of the circle ,  subtended to a point or points outside the circle: The crucial theorems are listed in my last, catalogues as FGn.  They are new, I did not find them anywhere in the literature, but the key one FG8 states that  Given three equal, adjacent arcs on the circumference of a circle, then the three adjacent angles at any point on the plane, subtended by these arcs will be equal.  Again the proof is built up from the construction of the angles off the same arcs at the circumference of the circle, then using the property that the external angle of a triangle is equal to the two internal opposite angles ( SUM OF)

That result proved, and returning to the actual description of the construction,the three equal arcs on the semicircle centre E through P and S, will subtend three equal angles at any point in the plan3e, thus at point B where the threeangles have their vertex BPAn  and BPC being the outide arms of the angles is trisected.

Now I think we should terminate this exchange, because I amk not going to unveil the proofs until I have absolutely ensured the security of my copyright: no need for sneery comments: I have made noe to you, and enlightened minds should be above them

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galet even assuming that picture is true that's a 90 degree angle, which is already known how to trisect. you have to trisect an arbitrary angle.

grogorous, i wish i could say i wish you luck but i truth i do not. you do not seem to comprehend basic math have have already posted erroneous proofs here. I'm not even a  serious mathematician and i spotted your errors. probably for the best we end the conversation here. its been proven rigorously you cant trisect an arbitrary angle with a compass and straight edge. 

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  • 1 year later...

In this way we can make any angle trisection using only ruler and compass

This proves that the trisection angle is possible only with a ruler and compass

 

                                                                                                                                                                                                                                                                                                                                                                              Dane Gacesa 2306936710262

Edited by galet
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This is a device for angles trisection

In this way we can make any angle trisection using only ruler and compass

This proves that the trisection angle is possible only with a ruler and compass

At first glance at your sketch, Galet, I got a “hey, this could work” thrill. :)

 

Let me expand on your “possible with straightedge and compass” remark.

 

It’s well known that an arbitrary angle can be trisected using various tools beyond the straightedge and compass, such as tomahawk. The “device” you describe, though, isn’t a movable “tool” like a straightedge, compass, or tomahawk – it can be constructed give an angle Q-P-R we want to trisect, as follows:

  • Draw a circle centered at P intersecting P-Q at Q and P-R at R.
  • Construct lines A-B and A-C intersecting at a 90o angle at A.
  • Draw a circle centered at A, intersecting the lines A-B at B and A-C at C.
  • At the bisection of A-B, draw a line intersecting the circle at X. C-A-X is a 30o angle, the trisection of 90[o]o[/o] angle C-A-B.
  • Draw a line C-Q.
  • Draw a line B-R intersecting C-Q at E.
  • Draw a line X-E intersecting the circle passing through Q and R at Y

    (Note that point D is sketched, but not required.

post-1347-0-59710400-1477849690_thumb.png

We suggest that angle Q-P-Y is the trisection of Q-P-R

 

:( Unfortunately, a rough protractor measurement shows it isn’t.

 

I think this construction looks deceptively promising because it resembles a sketch of a 3-dimensional object with parallel planes containing C-A-B, Q-P-R and D-E. If we were allowed to do this, with C-E intersecting the 3 plane orthogonally (at 90o angles), the sketched tool would work. But the trisection problem is required to be done on a single, 2-D plan, not in 3-D space.

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