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Final Angle Trisection (first part)


CHAPTER 29

FINAL ANGLE TRISECTION
(first part)

Since reading about the trisection opus itself would require a whole book of who knows how many pages and if we were to maintain this step-by-step tempo (which many of you have requested), I have decided to partially repeat the simplest details but at the same time to insert the system of duality, therefore selecting an angle greater than 90° and smaller than 120° (at random), but of course adhering only to the compass and unmarked straightedge. I have intentionally omitted the principle of internal trisections such as the “Flower of Life” and the “Star of David” as well as the internal and external concepts of the nine-sided polygon and its multiplicative sequence, but after three pages of the “compatible” system, I elaborate the knowledge of trisection (and sectioning in general) towards the spherical (“angelic”) manner, the manner only with a compass, respectively (excluding the postulate of the enigma and its end in which the concept of the center of the equilateral triangle only with a compass will be solved). On the way we will apply the principle of figuring the given arbitrary angle only with a compass. Incidentally we must remember that we found that sectioning is a process of partitioning or dividing and that it is never done directly, but indirectly by means of codes that are natural trisections, but which were always averred to only as just some possible trisections. We often hear that the trisection problem has been solved, therefore I wonder why this has not been published and made accessible to everybody instead of making it seem so important that if published “the world would be doomed!” As a matter of fact, it is a regular geometrical creation of simple principles; so simple that any child attending the last grades of elementary school could understand them. One more fact speaks in favor of my conclusion. Every mathematician (from the teacher in elementary school to the highly educated universal mathematician – or at least our domestic or European professor) “avoids” this theme like the plague. That’s why I wish to send word to those who, now and then, pester me by assuring me that this problem is solved – yet never submit any data (according to geometrically given norm), that I don’t have time for geometrical theories and in the lifetime that I still have before me, I want to leave to the future generations as much geometrical data as possible from the domain of a “forgotten geometry”. It has been given to me to know, and my joyousness is in the fact that many all around the world will through these pages be able to share this joy with me.

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A given arbitrary angle with its arc (arc – of full circle)

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Bisector of given arbitrary angle by straight line.

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Chord of half of the given arbitrary angle.

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We divide the half into two parts with its bisector.

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This was done to find the center of the circle’s chord diameter of half of the given arbitrary angle. We divide the circle into its six parts, starting from one of the end points of the arc’s chord.

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The circle is divided by the chord (and bisector) into 4 parts. Thus, we have the right angle of the circle bisector – the chord end points and the chord has its arc in relation to the arc of the half of the given arbitrary angle, but they have common points – the end points of the chord length.

* * *

Then we establish that the “floral pattern” dividing the circle of the chord diameter by half of the angle – its external edges (second and third petal) divide the arc of the right angle into 3 parts. From these points the half-lines in the direction of the vertex of the given arbitrary angle divide the arc (its half) into three parts.

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With compass, we now simply transfer a third to the other half of the given arbitrary angle.

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In that way we have divided the given arbitrary angle (its arc) into 3 equal parts and have thereby acquired three equilateral triangles. With this principle we can divide the magnitude of any given arbitrary angle by shifting its trisection component (with compass).

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At the same time the given arbitrary angle is divided into its 6 parts, which can sometimes prove to be useful. As aforementioned: this is simply the “A” angle trisection (1st Book

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Thereafter we demonstrated that the so-called “B” system is compatible with the “A” system. The same division of half of the given arbitrary angle greater than 90° and smaller than 120°.

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By means of the circle chord diameter of half of the given arbitrary angle and four circles of same diameter that create the four-leaf pattern.

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Because the diagonals of the four-leaf pattern divide the arc of the right angle of the half of the given arbitrary angle into three parts.

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Here we first see the trisection of the right angle.

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And only from these points in the direction of the vertex half of the given arbitrary angle, rays divide the arc of the half into three equal parts.

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This surely creates three equilateral triangles of the same size.

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By applying the trisection point transfer with compass to the second half of the given arbitrary angle, it is likewise trisected into three parts …

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…and then into six parts.

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We added trisection „A“ (divided the circle into 6 parts) thus we can see the compatibility of systems „A“ and „B“ – i.e. of the spherical and rectilinear systems…

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… and the principle of transfer only with the compass.

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The third system is even simpler. We use the chord of the half of the given arbitrary angle and the half of the chord for correct square construction (with semicircle).

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Square construction has been described. The point where the semicircle intersects the bisector is the center of the circumscribed circle of the square, the sides of which are the magnitude of the chord.

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We can depict this, but not necessarily, since we are only interested in the arc of the circumscribed circle of the square above the chord and the part of the arc of the half of the given arbitrary angle.

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Starting from both endpoints of the chord we divide the circle into 12 parts (if this is done partially, only the radius from one endpoint or radius from the other endpoint would do.

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With the rays from these divisional poles in the direction of the given arbitrary angle we have divided the half of the given arbitrary angle into three parts.

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Everything else is a matter of compass proficiency, the skill of transferring trisection points to the other half of the given arbitrary angle. Hence, a much simpler system than the „A“ and „B“ system (therefore adapted for children). So now we want to find the magnitude of the given arbitrary angle and its components.

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Thus we have implemented for the first time in the history of our civilization an “angelic” (read: spherical system only using the compass) division of the arc of a full circle of a given arbitrary angle with the span size from the arc’s lowest point, until point by point we come back to that point regardless of the number of times this has to be done.

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Since we had no other data point except for the full arc of 360°, the following was established: 360° divided with the number of divisions times the number of divisions on the part of the arc of the circle of the given arbitrary angle gave us the magnitude of the given arbitrary angle. The procedure of this example is: 360° ÷ 29 divisions = 12,413793° x 8 divisions = 99,310344° … and that’s that! Performed again.

* * * *

AFTERWORD TO 1ST PART

At first, due to the great emotional contentment that filled me with joy, I failed to grasp what I had been given, and in the manner of human beings who by heritage of our rectilinear geometrical knowledge I got involved into the wrong conception – diving the arc with the chord. It is indeed hard to break free of a habit. I think it took me over a year to comprehend that the division of a full spherical angle should not be performed with the chord of a given arbitrary angle but with the span magnitude of its arc on the arc of the full circle. However, you will say that it is the same span as the chord. It is, and it isn’t. With rectilinear drawing we will never be able to “arouse” the new possibilities of our brain. What’s it all about? Rectilinear drawing is like the archaic belief that the Earth’s shape is a plane or disk. The next step was that the Earth is round. It was a new perception. We know that “motion or movement” in nature is never rectilinear propagation. Such a notion is rather an established custom of ours. Spherical motion belongs to nature. I don’t maintain that either one or the other is incorrect, but I wish to stress that we should familiarize ourselves with whatever is unknown; in this case the spherical or as I call it the “angelic”. I am certain that the synthesis of one and the other is the perfect form that logically brings about sequences and sequences of new scientific data. The data in the spherical domain already brings forth new unknown results of cognition. Therefore we will dedicate the next chapter to an apparently unimportant detail: construction of a triangle center only with the compass implemented on the trisection of a given arbitrary angle (directly and indirectly), once again for the first time in the history of this civilization of ours. One again this brings me satisfaction which I gladly wish to share with everybody because it was “given” to me. That is not as if somebody tells you something; instead it is an impetus of the mind, and since I am only human, it requires sequences and sequences of trials until the impetus “crystallizes”. I won’t burden you with how and what I feel (you will know when I can no longer carry on – because there will be nothing more on these pages). I only hope that in the meantime somebody will come out with a blueprint of an electronic protractor and an electronic compass to make me happy.

RIJEKA – HR May 31, 2012
Author: Tomo Perisha
English translation: S.F. Drenovac
Web Master: Slim

2 Responses to “Final Angle Trisection (first part)”

  1. Rini Greve says:

    Ik heb een vraag over het verdelen van willekurige hoeken in even en oneven delen.met behulp en een passer en een lineaal wat ik net beken heb op deze sit eis het erg omslagtig om dat even te tekenen.

    zelf doe ik het verdelen van een bepaalde hoek met diverse gelijke vlaken of dit nou in 2 of in drie moet zijn of meer. Door het plaaten van 5 cirkels deze oplossing kom in nergens tegen en is zo simpel. misschien hebben jullie er een opgave van of een tekening.

    in afwachting op uw antwoord.

    Rini Greve

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