Patent Publication Number: US-2021178804-A1

Title: Angle trisector, as validated to perform accurately over a wide range of device settings by a novel geometric forming process; also capable of portraying finite lengths that only could be approximated by means of otherwise applying a compass and straightedge to a given length of unity; that furthermore functions as a level whose inherent geometry could be adapted for many other uses such as being incorporated into the design of a hydraulic car lift.

Description:
CROSS-REFERENCE TO RELATED APPLICATIONS 
     This nonprovisional patent application is a continuation application which claims the benefit of U.S. application Ser. No. 15/889,241, filed Feb. 6, 2018, now pending, and hereby is incorporated by reference. 
    
    
     BACKGROUND OF THE INVENTION 
     By now, it most certainly should be recognized that a singular angle of virtually any designated size cannot be divided into three parts when acted upon only by a straightedge and a compass. 
     Such protracted problem is considered to be so famous that herein it formally shall be referred to as the classical problem of the trisection of an angle! 
     Unfortunately, many other descriptions of such problem also exist, each conflicting in some rather subtle manner, but nevertheless having profound effect upon its interpretation. Leading examples of such differences are presented below where it is found that in certain cases throughout the twentieth century such problem of the trisection of an angle: 
     is not specified, but its solution nevertheless is posed; 
     is considered to involve the use of a ruler; and 
     is algebraically, rather than geometrically resolved. 
     A looming trisection mystery, steeped in controversy for millennia, persists right up until present day; merely because it never was examined from the proper perspective! 
     In order to crack such conundrum, not only do correct questions need to asked, but proper answers also need to be supplied. 
     In this regard, the very first of four fundamental questions is about to be posed. 
     The first riddle is: How can the classical problem of the trisection of an angle actually be solved? 
     The answer is: It cannot! 
     Granted, it is commonly known that such response has been hotly contested by many pundits over the years. 
     But, to its credit, after bearing the brunt of a constant barrage of brutal assaults by noted protagonists, such contention endures; whereby it appears that to date no solution for the classical problem of the trisection of an angle ever has been solicited that has withstood the test of time! 
     Because it is far more difficult to prove an impossibility than something that is thought to be possible, fewer proofs exist which attempt to validate that the classical problem of the trisection of an angle truly is unsolvable! 
     To this effect, page 246 appearing in  A Dictionary of Mathematics  by T. A. Millington, Barnes &amp; Noble; 1966 stated, “The classical problem of trisecting an angle by Euclidean constructions (use of straight lines and circles only) was proved to be insoluble by Wantzel in 1847”. 
     Below, my express intent is to acknowledge Wantzel&#39;s outright conclusion as being absolutely correct in such regard, in hopes of thereby identifying some distinct difficulty which prevails that otherwise prevents the classical problem of the trisection of an angle from being solved! 
     This is to be accomplished by means of delving into what before were considered to be impenetrable depths of a great enigma in order to finally unravel the very mystery that today engulfs trisection. 
     Such trisection enigma specifically shall state: If the classical problem of the trisection of an angle truly cannot be solved, then geometry, itself, must be the culprit; thereby being imperfect! 
     This author well understands that if it were learned that geometry actually is limited, marred, or otherwise flawed in some strange, presently unknown fashion, then the very perception of such form of mathematics inevitably would become altered, even to the point where it possibly might become tarnished forevermore! 
     The second riddle is: What specific, well known thermodynamics limitation now should be considered to furthermore apply to geometry? 
     The answer is: Irreversibility! 
     A commentary, as presented below, should settle any doubts, or quell any lingering suspicions, that the classical problem of the trisection of an angle might become solved. 
     A fair number of rather elaborate proofs that have gained notoriety over the years have done so by means of incorrectly claiming that the classical problem of the trisection of an angle actually can be solved. 
     Such proofs either are quite faulty in their logic, or else rely upon a basic deception; whereby their geometric solutions, although complying with a primary requirement of applying only a straightedge and compass, achieve trisection instead by having them act upon other than a singular angle of virtually any designated size! 
     In such manner, geometric solutions actually do achieve trisection, but only to their detriment by means of violating a remaining requirement that the classical problem of the trisection of an angle furthermore imposes! 
     As such, for any and all geometric solutions which might become specified in the near future claiming to succeed at trisecting an angle, perhaps it is best to offer the rather simple-minded reservation that in order to solve the classical problem of the trisection of an angle, all of its imposed requirements explicitly must be complied with! 
     In other words, the very reason why any of such actual geometric solutions should not be misinterpreted as a solution for the classical problem of the trisection of an angle is because they violate the second of its requirements, fundamentally specifying that geometric activity must proceed only from a singular angle of virtually any designated size. 
     Instead of complying with such second requirement, geometric solutions otherwise input extraneous information into such classical problem, thereby serving to corrupt it! 
     In particular, extraneous information consists of that which is not germane to such classical problem. Hence it consists of any geometric information other than that which can be derived by applying only a straightedge and compass exclusively to a singular angle of designated magnitude which is intended to be trisected. 
     Moreover, extraneous information is considered to be relevant whenever it independently can be acted upon solely by a straightedge and compass in order to pose an actual solution for some corrupted version of the classical problem of the trisection of an angle; thereby solving an entirely different problem! 
     Hence, it becomes rather obvious that such very controversy undeniably has been fueled over the years by an unrelenting confusion which has been stirred over nothing more than various cases of mistaken identity! 
     A rather elementary example of how to administer such overall theory is presented directly below; whereby for the unsolvable classical problem of the trisection of an angle of 135° designated magnitude, it is desired to specify relevant information that enables a geometric solution to be obtained. 
     For such example, the approach to obtain such relevant information is given as follows: 
     extraneous information quite readily can be determined simply by dividing such 135° angle by a factor of three. Such resulting  45  size qualifies as extraneous information because it clearly cannot be derived in any way by means of applying only a compass and straightedge to such 135° designated angle; and 
     such calculated  450  magnitude furthermore is relevant because it can be drawn solely by application of a straightedge and compass as one of the diagonals in a square of virtually any size, thereby posing a geometric solution for such corrupted classical trisection problem. 
     Such types of mishap, assuming the form of mistaken identifies, very easily can be avoided whenever any geometric solution becomes posed in the future that falsely alleges that the classical problem of the trisection of an angle actually can be solved. 
     Quite simply, such approach consists of determining whether a posed geometric solution is linked to a formally specified problem. If a corresponding problem becomes located, it then should be examined to verify that it: 
     is entirely consistent with how such classical problem of the trisection of an angle is specified above, in which case such geometric solution therefore must be in error; or 
     incorporates extraneous information that is relevant, thereby otherwise solving some corrupted version of such classical trisection problem instead. 
     On the other hand, if it is found that such geometric solution is not associated to any formally specified problem, this means that it cannot be ascertained to any certainty what such posed geometric solution even applies to! 
     Regrettably, such type mishap is well documented, thereby being known to truly have occurred in the past! 
     As such, for any literature that claims to have solved the classical problem of the trisection of an angle without indicating exactly what particular issue it has remedied, it might be a good idea to examine: 
     whether such erroneous claim was met by a great fanfare that credited its author for an outstanding discovery. 
     whether a geometric solution for some corrupted version of the classical problem of the trisection of an angle was provided instead, without expressing the exact nature of extraneous information which such solution was based upon; or 
     whether some totally unrelated type of trisection solution was identified due to the discovery of some invention; thereby only serving to expand upon the overall scope of aforementioned trisection matters as cited in connection with such attendant trisection mystery. 
     The third riddle is: What other types of trisection solutions, besides geometric solutions, are there? 
     The answer is: Motion related solutions! 
     These consist of various events, as opposed to activities where singular geometric patterns otherwise become drawn, whereby an invention becomes set to a singular angle of virtually any designated size and its associated trisector automatically becomes portrayed forthwith. 
     Such type of invention cannot solve the classical problem of the trisection of an angle because its features are quite different from a mere straightedge and compass that instead individually must be applied to a single piece of paper without otherwise violating any imposed requirements. Hence, any trisection portrayal of this nature clearly would qualify as a corrupted motion related solution of the classical problem of the trisection of an angle. 
     The fourth riddle is: Can an iteration process of successive angular bisections, as presented in complete detail later, which clearly complies with all of the requirements imposed by the classical problem of the trisection of an angle actually solve it? 
     The answer again emphatically is no! 
     In particular, this is because: 
     an infinite number of iterations would have to be performed in order to generate an exact geometric solution; thereby qualifying as a task that could never be fully completed; and 
     the resolution of the naked eye to distinguish actual drawing separations would become impaired very shortly after commencing upon such iteration process, whereby successive bisectors then would appear to run over themselves, requiring larger arcs to be drawn in order to avail added viewing relief. Eventually, the very straightedge and compass instruments themselves, along with the paper needed to avail such precision could not be manufactured due to the massive increases in their sizes which would be needed to maintain such viability. 
     In conclusion, it is impossible to solve the classical problem of the trisection of an angle when explicitly complying with all of its requirements! 
     Important forerunners of trisection, hereinafter listed chronologically by the dates of their conception include: 
     geometric construction, dating all the way back to ancient Babylonian times around the year 3000 BC, in which only a straightedge and compass are permitted for use when describing straight lines, angles, and circular arcs; 
     intersection points, as established during the same time period, in which discrete positions become completely distinguishable wherever either straight lines, arcs of circles, or straight lines in combination with the arcs of circles cross one another. Center points of circles also qualify as intersection points because they describe common locations where geometrically constructed perpendicular bisectors of respective chords of such circles meet; 
     the Pythagorean Theorem, as developed in or about the year 500 BC, in which the square of the length of the hypotenuse of a right triangle is proven to be equal to the sum of the squares of the lengths of its two sides; 
     conventional Euclidean practice, as established prior to the year 265 BC, in which definitions and rules describe the very manner in which geometric construction may be administered, and axioms identify certain geometric relationships which become evident after conducting specific geometric construction operations. Directly below, two principal definitions which further characterize such practice are specified, followed by three distinct examples of its rules: 
     its most basic rule principally states that a given set of previously defined geometric data must be furnished that specifies the locations of initial positions from which geometric construction may be launched. A designated angle of sixty degree magnitude very well could be expressed as such given data. Moreover, such definition furthermore can apply to locations that are not entirely distinguishable solely by geometric construction. For example, given data might specify a twenty degree angle; one whose magnitude only could be approximated by means of geometric construction; 
     its cardinal rule essentially stipulates that geometric construction must proceed either from a given set of previously defined geometric data, or from other locations which become distinguishable with respect to such positions by means of applying only a straightedge and compass to them; 
     at least two points must be specified in order to draw a straight line; 
     at least two points must be specified in order to draw a circle when one of those points designates its center point; and 
     at least three points must be specified in order to draw a circle when none of those points denotes its center point; 
     conventional Euclidean means, whose terminology became commonplace shortly thereafter, in which geometric construction is to be implemented in strict accordance with the specific definitions and rules stipulated by conventional Euclidean practice; 
     Euclidean commands, also taking effect during the same time period, in which various instructions describe exactly how various straight lines and circular arcs are to be geometrically constructed with respect to identified positions; 
     a sequence of Euclidean operations, as conceived during the same timeframe, in which a specific set of Euclidean commands, enumerated as a series of discrete steps become executed in consecutive order with respect to a given set of previously defined geometric data in order to distinguish various rendered positions; 
     a geometric construction pattern, as introduced during that time, in which the specific features of a given set of previously defined geometric data, in combination with any rendered positions which become established by means of executing an attendant sequence of Euclidean operations with respect to it, as well as any additional straight lines drawn between such established locations or circular extensions made to them become depicted within a single drawing; 
     an Archimedes proposition, as devised prior to 212 BC, introduced as Proposition 8 in the  Book of Lemmas , then later translated from Arabic into Latin in 1661, and finally published in the English language in 1897, in which it is stated on page 309 in  The Works of Archimedes  that if AB be any chord of a circle whose centre is 0, and if AB be produced to C so that BC is equal to the radius; if further CO meets the circle in D and be produced to meet the circle a second time in E, the arc AE will be equal to three times the arc BD; 
     an Archimedes proof for such Proposition 8, as appears on page 310 in  The Works of Archimedes , in which it is stated: 
     draw the chord EF parallel to AB, and join radius OB, OF; 
     
       
         
           
             
               
                 
                   
                     ∠ 
                      
                     
                         
                     
                      
                     COF 
                   
                   = 
                   
                     2 
                      
                     
                         
                     
                      
                     ∠ 
                      
                     
                         
                     
                      
                     OEF 
                   
                 
               
             
             
               
                 
                   
                     = 
                     
                       2 
                        
                       
                           
                       
                        
                       ∠ 
                        
                       
                           
                       
                        
                       BCO 
                     
                   
                   , 
                   
                     by 
                      
                     
                         
                     
                      
                     parallels 
                   
                   , 
                 
               
             
             
               
                 
                   
                     = 
                     
                       2 
                        
                       
                           
                       
                        
                       ∠ 
                        
                       
                           
                       
                        
                       BOD 
                     
                   
                   , 
                   
                     
                       
                         since 
                          
                         
                             
                         
                          
                         BC 
                       
                       = 
                       BO 
                     
                     ; 
                   
                 
               
             
           
         
       
     
     therefore, ∠BOF=3 ∠BOD, so that the arc BF is equal to three times the arc BD; and 
     hence the arc AE, which is equal to the arc BF, must be equal to three times the arc BD; 
     an Archimedes formulation, as evidenced only on a few sporadic occasions in the past, as typically accompanied by only partial documentation, in which a multitude of distinct Archimedes geometric construction patterns, qualifying as such because they conform to all requirements posed in such Archimedes proposition, become represented upon a single drawing. Such representation is made possible by strategically placing a Greek letter either within or alongside what later will be shown to be the given angle of such sole diagram, thereby distinguishing it to be an entire formulation in itself, instead of a mere singular geometric construction pattern. Once ascribing a specific designation, such as θ or φ at such location, it is to mean that such given angle is allowed to vary infinitesimally in size over some prescribed range of values. By means of administering a specific sequence of Euclidean operations to each of such given angles—one whose commands account for all of the individual requirements posed in such aforementioned Archimedes proposition, all of the varying shapes which emerge thereby must qualify as legitimate Archimedes geometric construction patterns; and 
     an Al-Mahani categorization, as derived prior to the year 900 AD, in which square root quantities become classified as quadratic irrational numbers, thereby distinguishing them apart from rational numbers. 
     Over the years, both mathematicians, as well as inventors alike have been somewhat awed by the spectacle of an incredibly perplexing trisection mystery whose vital secrets evidently have escaped detection. 
     Nevertheless, both parties are acutely aware that the diminution of an angle to one-third its actual size, although being truly indicative of trisection, cannot be obtained simply by reversing the sequence of Euclidean operations which governs the geometric construction of a pattern that complies with all of the requirements imposed by such Archimedes proposition. 
     Rabid speculation concerning such unexplainable dichotomy eventually gave rise to contrasting interpretations, reflective of the particular leanings of various involved personages, outlined as follows: 
     on the one hand, mathematicians even today remain splintered over how to explain trisection in terms of conventional Euclidean practice, whereby their discordant positions concerning such geometrical matters are expressed as follows: 
     one traditionalist camp contends that trisection solely via conventional Euclidean means is entirely unsolvable; whereby it becomes utterly impossible to divide an angle into three parts solely by means of applying only a straightedge and compass to it; whereas 
     another non-patronizing faction instead advocates that certain angles, in fact, can be trisected solely via straightedge and compass; whereas 
     on the other hand, inventors were the first members in society to demonstrate that trisection could be achieved by imparting certain prescribed motions. 
     The latter of such two geometrical claims, essentially alleging that certain angles can be trisected solely via straightedge and compass unfortunately has managed to gain widespread notoriety throughout the world today, thereby flourishing in the form of independent airings by a coterie of indulging sources and journalists, as well as being supplemented by a rampant proliferation of publications by self-proclaimed mathematical experts who already have accepted such premise as being generally established fact. 
     Be that as it may, nothing could be further from the truth! 
     This becomes evident by paraphrasing such latter stated mathematical claim into its only possible correct interpretation. This is accomplished by means of inserting the following bracketed italicized words to its initial substance, thereby asserting that certain angles, in fact, can be trisected solely via straightedge and compass [so long as the very magnitudes of their respective trisecting angles become disclosed beforehand]—thereby essentially solving a corrupted version of the classical problem of the trisection of an angle. 
     Such above stated clarification achieves two specific objectives, itemized as follows: 
     it completely eliminates the potential for misconstruing such latter stated mathematical claim to mean that in certain cases, trisection solely via conventional Euclidean means is entirely possible Obviously, such faulty presumption might be harmful because it very easily could be considered to refute, or contradict such former factually correct claim, as asserted above by such aforementioned traditionalist camp of mathematicians; and 
     it evidences that such latter mathematical clause really has nothing at all to do with trisection solely via conventional Euclidean means; thereby removing the intended stigma out of such statement entirely. As such, its true interpretation with such bracketed input included thereby becomes reduced to the rather insignificant equivalent meaning that all angles which can be described solely via straightedge and compass furthermore constitute respective trisecting angles of other angles whose respective magnitudes amount to exactly three times their size. Such result is of little consequence too considering that it furthermore can be broken down into the mere definition of a trisector, in concert with the understanding that certain angles effectively can be reproduced solely via straightedge and compass once having knowledge of their trigonometric properties. 
     It should be emphasized that the only way to input outside information which is relevant to the classical problem of the trisection of an angle is by means of violating the cardinal rule of conventional Euclidean practice! 
     By definition, that is because such input must consist of data that cannot be derived by means of launching geometric construction operations exclusively from a singular angle of designated magnitude which is intended to be trisected. Hence, such corrupting input must constitute that which cannot be distinguished by conventional Euclidean means. 
     As such, it is concluded that the classical problem of the trisection of an angle cannot be solved by conventional Euclidean practice; thereby exposing its very limitations! 
     Furthermore, calculating the size of a trisector merely by dividing the designated magnitude of an angle that is intended to be trisected by a factor of three is not permitted because such action cannot be duplicated by a geometric construction process which is governed by Euclidean rules. 
     Below, a composite history describing the very first important trisecting events recorded in the English speaking language is afforded; whereby trisection was observed to occur on three separate occasions as unique articulating mechanisms became invented. 
     During such first of such documented incidents, having taken place sometime during the late 1870&#39;s, Alfred Kempe discovered that so-called anti-parallelograms could be used for purposes of regulating motion! 
     One of such Kempe masterful designs, truly considered to be capable of performing trisection, is depicted as prior art in  FIG. 1A . 
     Therein, an overall formation which could have been fashioned to have either rounded or pointed corners is not displayed. Only the longitudinal centerlines of its eight linkages and radial centerlines of its eight interconnecting pivot pins are depicted instead because only these portions of such device govern trisection! 
     Such Kempe prior art features a basic fan array whose movement is controlled by three independent anti-parallelograms; the complete breakdowns of which are described as follows: 
     its fan array portion consists of four separate linkages, modeled as straight lines BA, BD, BE, and BC in  FIG. 1A ; all of which are hinged together by an interconnecting pivot pin whose shank passes through an end portion of each. In particular, the longitudinal centerlines of such four basic fan array linkages overlap one another along the radial centerline of such interconnecting pivot pin, as illustrated by a very small circle drawn located at their juncture about axis B; and 
     its control section features anti-parallelograms BFGH, BGJI, and BJLK, so denoted by identifying their diagonal compositions, rather than by enumerating their respective corners in consecutive order. 
     Such control section serves to maintain the three angles interposed between adjacent longitudinal centerlines of the four linkages, where they more particularly radiate about the hub of such basic fan array portion, at a constant magnitude; even as such sizes vary during device flexure. 
     Accordingly, such three interposed angles relate to one another according to the algebraically expressed equality ∠ABD=∠DBE=∠EBC=θ. 
     Whereas all three subtended angles in combination constitute angle ABC, its magnitude must amount to 3θ. 
     In order to trisect angle ABC, such Kempe device first must be set to a designated magnitude. In this particular case, the size of such setting is shown to be that of angle ABC, as it actually appears in  FIG. 1A . Considering that a varying magnitude of such angle ABC might have been selected instead, such drawing would have had to assume an entirely different overall shape in order to compensate for such change. 
     In effect, every time such Kempe device becomes set to a different preselected magnitude, each of such three subtended angles, nonetheless, automatically portray its actual trisector; whereby the very process of portraying an angle whose magnitude amounts to exactly one-third the size of an angle of designated magnitude is indicative of trisection! 
     In total, such Kempe device is comprised of eight linkages whose longitudinal centerlines are modeled as straight lines in  FIG. 1A , in addition to eight interconnecting pivot pins whose radial centerlines are denoted by very small circles therein. 
     Extensions made to anti-parallelogram linkage members BF and BL therein enable easier access for flexing device arm BC with respect to linkage member BA during articulation; whereas extensions made to anti-parallelogram linkages members BG and BJ enable a better viewing of trisecting members BD and BE. 
     All told,  FIG. 1A  depicts longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively constitute such Kempe mechanism. 
     Of note, various other trisecting devices exist which also are contrived of a four linkage and interconnecting pivot pin arrangement which are considered to conform to that of such basic fan array, as described in detail above. However, they each apply control mechanisms that are entirely different than the anti-parallelogram linkage arrangements, as previously developed by Mr. Kempe. 
     By means of classifying mechanisms such as these into a singular category, it becomes possible to validate the striking geometrical resemblance which exists between them. Once grouped together, their uniqueness thereby can be substantiated by means of describing how their control mechanisms differ from each other. 
     Such categorization is necessary for the very same reason that it previously has been applied in biology; namely, to suitably characterize the very diversity which exists between living things that exhibit common physical traits. 
     Likewise, by means of comparing trisecting mechanisms which exhibit similar geometries, proper conclusions can be drawn concerning both how and why they relate to one another, as well as how they fundamentally differ! 
     What should bond trisecting mechanisms together is a common geometry which they all share. 
     For example, the geometry of such basic fan array design is perhaps the simplest in all of mathematics to comprehend because it simply consists of a given angle that becomes duplicated twice, such that all three angles become grouped together at a common vertex in order to eventually geometrically construct a rendered angle whose magnitude amounts to exactly three times the size of such given angle. 
     A singular geometric construction pattern, as represented in  FIG. 1A , furthermore could qualify as an entire formulation because the sequence of Euclidean operations from which such distinctive geometric construction pattern stems furthermore could be applied to virtually any sized given angle, as algebraically denoted therein by the Greek letter θ; thereby governing the various positions which the longitudinal centerlines of linkages and radial centerlines of interconnecting pins which collectively constitute such Kempe device thereby would assume as it becomes articulated. 
     In support of such logic, it is recommended that any articulating trisecting mechanism which exhibits distinctive fan shape features should be classified as a CATEGORY I type device. Its complete inventory shall consist of: 
     CATEGORY I, sub-classification A articulating trisection devices which feature four linkages of equal length, hinged together by an interconnecting pivot pin that is passed through one end portion of each, thereby collectively constituting the array of a fan whose two inner linkages become regulated in some fashion so that their longitudinal centerlines divide the angle subtended by the longitudinal centerlines of its two outer linkages into three equal portions throughout device flexure; and 
     CATEGORY I, sub-classification B articulating trisection devices which feature three linkages of equal length, hinged together by an interconnecting pivot pin that is passed through one end portion of each, either of whose outer linkages instead could be represented by a straight line that is impressed upon a piece of paper or board, thereby collectively constituting the array of a fan whose single inner linkage becomes regulated in some fashion so that its longitudinal centerline trisects the angle subtended by the longitudinal centerlines of its two outer linkages throughout device flexure. 
     With particular regard to  FIG. 1A , each of the distinct Euclidean commands required to suitably locate straight line BC with respect to given angle ABD therein, solely by Euclidean means, is specified below, thereby together comprising the complete twenty-one steps of its sequence of Euclidean operations: 
     step 1—given angle ABD, of arbitrarily selected magnitude θ, is drawn such that its side BA is constructed to be of equal length to its other side BD; 
     step 2—point F is arbitrarily selected somewhere along side BA of given angle ABD; 
     step 3—a circle is drawn about vertex B whose radius is arbitrarily selected to be less than length BF, whereby a portion of its circumference becomes designated as the FIRST CIRCULAR ARC in  FIG. 1A ; 
     step 4—The intersection between such FIRST CIRCULAR ARC and side BD of given angle ABD becomes designated as point G; 
     step 5—a second circle is drawn about point F whose radius is set equal in length to straight line segment BG, a portion of whose circumference is designated as the SECOND CIRCULAR ARC in  FIG. 1A ; 
     step 6—a third circle is drawn about point G whose radius is set equal in length to straight line segment BF, a portion of whose circumference is designated as the THIRD CIRCULAR ARC in  FIG. 1A ; 
     step 7—the intersection point between such SECOND CIRCULAR ARC and THIRD CIRCULAR ARC is designated as point H; 
     step 8—diagonal GH and side segment FH of anti-parallelogram BFGH are drawn in order to complete its geometry; 
     step 9—an angle whose magnitude is equal to that of angle FBH is geometrically constructed with its vertex placed at point B such that its counterclockwise side becomes aligned along straight line BD, whereby the intersection of its clockwise side with diagonal GH becomes designated as point I; 
     step 10—a fourth circle is drawn about point I, whose radius is set equal in length to line segment BG, a portion of whose circumference becomes designated as the FOURTH CIRCULAR ARC in  FIG. 1A ; 
     step 11—a fifth circle is drawn about point B, whose radius is set equal in length to line segment IG, a portion of whose circumference is designated as the FIFTH CIRCULAR ARC in  FIG. 1A ; 
     step 12—the intersection point between such FOURTH CIRCULAR ARC and FIFTH CIRCULAR ARC becomes designated as point J; 
     step 13—straight line diagonal IJ and side segment BJ are drawn in order to complete an additional anti-parallelogram BGJI; 
     step 14—straight line BE is geometrically constructed to be equal in length to side BA of given angle ABD such that it aligns with side segment BJ of such additional anti-parallelogram BGJI by means of passing through point J, whereby it serves as an extension to it; 
     step 15—an angle whose magnitude is equal to that of angle GBI is geometrically constructed with its vertex placed at point B such that its counterclockwise side becomes aligned along straight line BE whereby the intersection of its clockwise side with diagonal JI becomes designated as point K; 
     step 16—a sixth circle is drawn about point K, whose radius is set equal in length to line segment BJ, a portion of whose circumference is designated as the SIXTH CIRCULAR ARC in  FIG. 1A ; 
     step 17—a seventh circle is drawn about vertex B of given angle ABD whose radius is set equal in length to line segment KJ, a portion of whose circumference is designated as the SEVENTH CIRCULAR ARC in  FIG. 1A ; 
     step 18—the intersection point between such SIXTH CIRCULAR ARC and SEVENTH CIRCULAR ARC is designated as point L; 
     step 19—straight line diagonal KL and side segment BL are drawn in order to complete the third and last of such anti-parallelograms, being duly notated as BJLK; 
     step 20—straight line BC is geometrically constructed to be equal in length to side BA of given angle ABD such that it aligns with side segment BL of anti-parallelogram BJLK by means of passing through point L, thereby serving as an extension to it; and 
     step 21—whereas side BA is constructed to be of equal length to side BD of given angle ABD, and straight lines BE and BC are geometrically constructed to be equal in length to such side BA, all four straight lines furthermore represent radii of a circle that all emanate from center point B. 
     Verification that angle ABC, as depicted in  FIG. 1A , is equal to three times the size of given angle ABD is provided in the following twenty-five step proof: 
     step 1—by construction, straight line HG is equal in length to straight line BF, and straight line GB is equal in length to straight line FH; 
     step 2—by identity, straight line BH is equal in length to straight line HB; 
     step 3—then, since the three sides represented in triangle HGB are of equal length to the corresponding sides of triangle BFH, such triangles must be congruent to each other; 
     step 4—by construction, straight line IJ is equal in length to straight line BG, and straight line JB is equal in length to straight line GI; 
     step 5—by identity, straight line BI is equal in length to straight line IB; 
     step 6—then, since the three sides represented in triangle IJB are of equal length to the corresponding sides of triangle BGI, such triangles must be congruent; 
     step 7—by construction, straight lines KL and LB respectively are equal in length to straight lines BJ and JK, and by identity, straight line BK is equal in length to straight line KB; 
     step 8—then, since the three sides represented in triangle KLB are of equal length to the corresponding sides of triangle BJK, such triangles must be congruent to each other; 
     step 9—since triangle BFH is congruent to triangle HGB, each of their corresponding angles must be of equal magnitudes, such that ∠FBH=∠GHB; 
     step 10—but, by construction ∠GBI=∠FBH; 
     step 11—then, by substitution ∠GBI=∠GHB; 
     step 12—by identity, ∠BGI=∠BGH; 
     step 13—since ∠GBI=∠GHB, in addition to the fact that ∠BGI=∠BGH, triangle BGI and triangle HGB contain two sets of angles whose respective magnitudes are of equal values, whereby such triangles must be similar to one another; 
     step 14—because triangle IJB is congruent to a triangle BGI which, in turn, is similar to triangle HGB, triangle IJB also must be similar to triangle HGB; 
     step 15—whereby angle IBJ must be equal to corresponding angle HBG; 
     step 16—because triangle BGI is congruent to triangle IJB, each of their corresponding angles must be of equal magnitudes, such that ∠GBI=∠JIB; 
     step 17—but, by construction ∠JBK=∠GBI; 
     step 18—then, by substitution ∠JBK=∠JIB; 
     step 19—by identity, ∠BJK=∠BJI; 
     step 20—since ∠JBK=∠JIB, in addition to the fact that ∠BJK=∠BJI, triangle BJK and triangle IJB contain two sets of angles whose respective magnitudes are of equal values, whereby such triangles must be similar to one another; 
     step 21—because triangle KLB is congruent to a triangle BJK which, in turn, is similar to triangle IJB, triangle KLB also must be similar to triangle IJB; 
     step 22—whereby angle KBL must be equal to corresponding angle IBJ; 
     step 23—since the whole is equal to the sum of its parts, 
       ∠ HBA+ABD=∠HBD  
 
       ∠ ABD=∠HBD−∠HBA  
 
       ∠ ABD=∠HBD−∠HBF  
 
       ∠ ABD=∠HBD−∠FBH;  
 
     step 24—whereby the following expression is obtained by substituting relevant previously determined equalities and identities ∠HBD=∠HBG, ∠IBJ=∠IBE, ∠KBJ=∠KBE, ∠KBL=∠KBC, and ∠IBG=∠IBD, and by reversing the order of the three letter designators of certain angles without influencing the values of their respective magnitudes, such that: 
     
       
         
           
             
               
                 
                   
                     
                       ∠ 
                        
                       
                           
                       
                        
                       ABD 
                     
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         HBD 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         FBH 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         HBG 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         FBH 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         IBJ 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         GBI 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         IBE 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         IBG 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         
                           ∠ 
                            
                           
                               
                           
                            
                           IBE 
                         
                         - 
                         
                           ∠ 
                            
                           
                               
                           
                            
                           IBD 
                         
                       
                       = 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         DBE 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         IBJ 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         GBI 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         KBL 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         JBK 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         KBC 
                       
                       - 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         KBJ 
                       
                     
                   
                 
               
               
                 
                   
                     
                       = 
                         
                        
                       
                         
                           
                             ∠ 
                              
                             
                                 
                             
                              
                             KBC 
                           
                           - 
                           
                             ∠ 
                              
                             
                                 
                             
                              
                             KBE 
                           
                         
                         = 
                         
                           ∠ 
                            
                           
                               
                           
                            
                           EBC 
                         
                       
                     
                     ; 
                   
                 
               
             
               
           
         
       
     
     and 
     step 25—since angle ABC is comprised of angle ABD, angle DBE, and angle EBC, the following expression demonstrates that angle ABC amounts to exactly three times the size of given angle ABD: 
     
       
         
           
             
               
                 
                   
                     
                       ∠ 
                        
                       
                           
                       
                        
                       ABC 
                     
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         ABD 
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         DBE 
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         EBC 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         ABD 
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         ABD 
                       
                       + 
                       
                         ∠ 
                          
                         ABD 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       3 
                        
                       ∠ 
                        
                       
                           
                       
                        
                       ABD 
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       3 
                        
                       
                           
                       
                        
                       
                         θ 
                         . 
                       
                     
                   
                 
               
             
               
           
         
       
     
     Notice that such above digression applies equally as well to varying magnitudes of given angle ABD since each successive step listed in such sequence of Euclidean operations and accompanying proof acts upon prior information that commences directly from such angle ABD, no matter what its initial size. That is to say, according to such proof, angle ABC still would be of magnitude 3θ, even if given angle ABD were to become varied to some degree in overall size. During such conditions,  FIG. 1A  would assume reconstituted shapes that depict actual Kempe device reconfigurations after being articulated to different positions. 
     In other words, by defining the magnitude of given angle ABD algebraically, instead of just assigning a singular value to it, such  FIG. 1A  thereby describes all of the various attitudes which such invention could assume during articulation. Such multiple number of actual drawings, thereby collectively comprising an entire formulation, essentially consists of an amalgamation of renderings which are generated solely by administering a specific sequence of Euclidean operations to a given angle ABD which is allowed to vary in size while, in each case, ultimately rendering an angle ABC which amounts to exactly three times the size of such given angle ABD. 
     As such, the three Kempe anti-parallelograms incorporated into such mechanism, as represented in  FIG. 1A , thereby serve to regulate, or strictly control the overall movement of such device, whereby trisecting angles thereby become portrayed as angle ABC becomes altered to an infinite variety of angles that are in need of being trisected. 
     A second significant trisection development also was reported in such 1897 publication entitled  The Works of Archimedes  wherein it was claimed that a marked ruler arrangement placed upon a particular geometric drawing in a prescribed manner could achieve trisection. In order to account for the enormous lapse of time which had preceded such publication, page cvi therein stipulates that some such similar operation might have contributed to the development of the conchoid; a curve considered to have been discovered by Nicomedes somewhere between 200 BC and 70 BC. 
     Such unusual application is represented as prior art in  FIG. 1B  such that straight line MR describes the longitudinal centerline of such marked ruler device. The mechanism clearly qualifies as a ruler, as opposed to just a straightedge, simply because it contains a notch etched a specific distance away from one of its tips. This is denoted by the very small circle appearing upon such drawing at point N. 
     Therein, angle QPS represents an angle that is to become trisected, suitably drawn upon a separate piece of paper and specifically sized to be of algebraically expressed  30  designed magnitude. Its sides QP and SP both are drawn to be equal in length to the distance which such notch resides away from the left endpoint of such marked ruler device. Thereafter, a circle is drawn whose center point is located at vertex P of angle QPS such that its circumference passes both through point Q and point S. Lastly, side SP is extended a sufficient length to the left, thereby completing such geometric construction pattern. 
     In order to trisect angle QPS, such marked ruler device thereby becomes indiscriminately jockeyed above such piece of paper until such time that all three of the following listed events occur: 
     its longitudinal centerline overlaps point Q; 
     its notch aligns upon the circumference of such drawn circle; and 
     its left endpoint hovers directly over straight line SP extended. 
     Therein, point N designates the position where such notch hovers above such previously drawn circle. 
     As validated below, once such device becomes set in this particular manner, angle RMP constitutes an actual trisector of angle QPS. Its respective sides consist of the longitudinal centerline of such repositioned marked ruler device, as represented by straight line MR in  FIG. 1B , along with straight line SP extended, as previously drawn upon a piece of paper that such marked ruler device now rests upon. Therein, trisector RMP is designated to be of magnitude θ, amounting to exactly one-third the size of previously drawn angle QPS. 
     As depicted in  FIG. 1B , angle RMP can be proven to be an actual trisector for angle QPS, by considering that both angles constitute respective given and rendered angles belonging to such famous Archimedes proposition. 
     The proof that  FIG. 1B  furthermore constitutes an Archimedes proposition is predicated upon the fact that it meets the description afforded in Proposition 8 in the  Book of Lemmas , properly rephrased to state that if QN be any chord of a circle whose center is designated as point P, and if QN be produced to M so that NM is equal to the radius; if further MP produced meets the circle at point S, the arc QS will be equal to three times the arc subtended between straight lines PN and PM. The accompanying proof thereby is given as: 
     
       
         
           
             ext 
             . 
             
               
                 
                   
                     
                         
                     
                      
                     
                       
                         
                           ∠ 
                            
                           
                               
                           
                            
                           PNQ 
                         
                         = 
                           
                          
                         
                           2 
                            
                           ∠ 
                            
                           
                               
                           
                            
                           NMP 
                         
                       
                       , 
                       
                         
                           by 
                            
                           
                               
                           
                            
                           
                             isos 
                             . 
                             
                                 
                             
                              
                             Δ 
                           
                            
                           
                               
                           
                            
                           NMP 
                         
                         ; 
                       
                     
                   
                 
               
               
                 
                   
                     
                       = 
                         
                        
                       
                         ∠ 
                          
                         
                             
                         
                          
                         PQN 
                       
                     
                     , 
                     
                       
                         by 
                          
                         
                             
                         
                          
                         
                           isos 
                           . 
                           
                               
                           
                            
                           ∠ 
                         
                          
                         
                             
                         
                          
                         PQN 
                       
                       ; 
                     
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         
                           ext 
                           . 
                           
                               
                           
                            
                           ∠ 
                         
                          
                         
                             
                         
                          
                         QPS 
                       
                       = 
                         
                        
                       
                         
                           ∠ 
                            
                           
                               
                           
                            
                           QMP 
                         
                         + 
                         
                           ∠ 
                            
                           
                               
                           
                            
                           PQM 
                         
                       
                     
                     , 
                     
                       
                         by 
                          
                         
                             
                         
                          
                         Δ 
                          
                         
                             
                         
                          
                         MPQ 
                       
                       ; 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ∠ 
                        
                       
                           
                       
                        
                       NMP 
                     
                     + 
                     
                       ∠ 
                        
                       
                           
                       
                        
                       PQN 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ∠ 
                        
                       
                           
                       
                        
                       NMP 
                     
                     + 
                     
                       2 
                        
                       ∠ 
                        
                       
                           
                       
                        
                       NMP 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     3 
                      
                     ∠ 
                      
                     
                         
                     
                      
                     NMP 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     3 
                      
                     ∠ 
                      
                     
                         
                     
                      
                     RMP 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       3 
                        
                       ∠ 
                        
                       
                           
                       
                        
                       NPM 
                     
                   
                   , 
                   
                     
                       
                         since 
                          
                         
                             
                         
                          
                         NM 
                       
                       = 
                       NP 
                     
                     ; 
                   
                 
               
             
           
         
       
     
     and 
     hence, the arc QS must be equal to three times the arc subtended between straight lines PN and PM. 
     Since the specific sequence of Euclidean operations which governs such famous Archimedes formulation furthermore distinguishes the very same geometry as now is represented in  FIG. 1B , angle RMP and angle QPS, as posed therein, also must describe its respective given and rendered angles. 
     With particular respect to  FIG. 1B , this is demonstrated as follows, wherein each of the seven steps comprising such specific sequence of Euclidean operations is stipulated directly below: 
     step 1—given acute ∠RMP, of arbitrarily selected angular size, designated therein as being of magnitude θ, and exhibiting sides of arbitrarily selected lengths, so long as RM is sufficiently longer than MP is first geometrically constructed on a piece of paper, or upon a drawing board; 
     step 2—given ∠RMP is duplicated such that its vertex is positioned at point P; whereby straight line PM denotes one of its sides, with its other side geometrically constructed so that it resides upon the same side of straight line PM as does remaining side RM of given angle RMP, such that the duplicated angle faces given angle RMP, or opens up towards it; 
     step 3—point N becomes designated as the intersection point between straight line RM and the additional side drawn by such duplicated angle, thereby completing the geometric construction of isosceles triangle MNP; 
     step 4—a circle is drawn whose origin is placed at point P which is of radius equal in length to straight line PN; 
     step 5—straight line MP is extended to point S lying upon the circumference of such formed circle; 
     step 6—the other intersection point which straight line MR makes with the circumference of such formed circle is designated as point Q; and 
     step 7—straight line PQ is drawn. 
     A three step algebraic proof, serving to verify that such geometrically constructed angle QPS is exactly three times the magnitude of given acute angle RMP, for any and all magnitudes which it otherwise might assume is provided as follows: 
     step 1—because ∠PNQ is as an exterior angle of isosceles triangle MNP, it must be equal to the sum of such triangle&#39;s equally sized included angles, denoted therein as ∠NMP and ∠NPM, which algebraically can be summed to θ+θ=2θ; 
     step 2—since angle NQP and angle PNQ reside opposite the equal length sides of isosceles triangle NPQ, with such sides furthermore representing equal length radii of such drawn circle, they must be of equal magnitude, such that ∠NQP=∠PNQ=20; and 
     step 3—because ∠QPS qualifies as an exterior angle of triangle MPQ, it thereby must be equal to the sum of such triangle&#39;s included angles QMP and MQP which, by furthermore being related in the two identities ∠QMP=∠NMP and ∠MQP=∠NQP, can be algebraically summed to amount to θ+2θ=3θ. 
     Above, it has just been proven that the magnitude of rendered angle QPS always must amount to exactly three times that of given acute angle RMP, even as such given acute angle becomes varied in size 
     Since the actual value of given acute angle RMP is not specified anywhere in the sequence of Euclidean operations that  FIG. 1B  is derived from; it can have no bearing, or direct influence upon implementing it. Hence, introducing algebraic nomenclature, as is posed in  FIG. 1B , cannot conflict in any way with its administration; thereby enabling such sequence of Euclidean operations to be applied many times over in order to generate a wide range of patterns that result as given acute angle RMP varies in size. 
     Along with such marked ruler arrangement, other trisection mechanisms whose designs could be arranged into geometric shapes that either are indicative of such famous Archimedes proposition, or particular adjuncts thereof, such as the configuration depicted in  FIG. 1B , hereinafter shall constitute CATEGORY II articulating trisection devices. 
     A third significant evolution took root during the early 1900&#39;s when an ample supply of hinged linkage assemblies, replete with interconnecting pivot pins, summarily became incorporated into a broad spectrum of up-and-coming applications. Such design practice, extending all the way up until present day, remains of paramount importance because it enables an innumerable variety of meaningful motions to become portrayed. 
     As a consequence of such progress, modern day trisection devices came into existence shortly after inventors figured out how to shape the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively constitute their mechanisms into configurations that are indicative of geometric construction patterns of rendered angles whose magnitudes amount to exactly three times the sizes of their respective given angles. 
     The very process of trisection became more sophisticated thereafter by slotting such linkages, thereby affording added degrees of freedom. This enabled such designs to travel over an ever increasing range of distinct trajectories that otherwise simply couldn&#39;t be duplicated by solid linkage designs of comparable shapes, the latter of which proved also to be both heavier and more costly. 
     A distinct example of such improved design is presented in  FIG. 1C , wherein a pair of slotted linkages are featured which can be used to actuate such device by means of rotating linkage MR in either a clockwise or counterclockwise direction relative to member MS about a hinge located at axis M. 
       FIG. 1C , was chosen to represent a typical example of such slotted linkage design for the principal reason that the very artwork which is expressed in  FIG. 1B  precisely pinpoints the particular placements of the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively comprise it. 
     Because of such association, notice that both  FIGS. 1B and 1C  denote the very same position letters, being a convenience which should permit for their easy comparison. 
     Whereas a virtually unlimited number of other geometric construction patterns which also serve to constitute such Archimedes proposition, as represented  FIG. 1B , exactly describe a series of repositioned placements of the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively comprise such mechanism, as depicted in  FIG. 1C  as it becomes actuated, the angular notations algebraically denoted as θ and  30  in  FIG. 1B  thereby also are shown to carry over into  FIG. 1C . 
     Except for these angular notations, such  FIG. 1C  merely is a truncated rendition of prior art. Because its overall geometry can be related in such manner directly to that of  FIG. 1B , and thereby furthermore can be associated to such famous Archimedes proposition, it is said to qualify as a CATEGORY II articulating trisection device. 
     This truncated phenomena becomes quite evident once realizing that such basic Archimedes proposition, as featured in  FIG. 1B , furthermore could be supplemented, or added to, merely by means of incorporating successive isosceles triangles onto it, each of whose angles of equal size thereby always would amount to the sum of the magnitudes of the non adjacent included angles of the triangle whose perimeter represents the outer envelope of the combined triangles which preceded it, thereby establishing the following progression: 
       θ+θ=2θ;
 
       θ+2θ=3θ;
 
       θ+3θ=4θ; and
 
       θ+4θ=5θ; etc.
 
     The manner in which such progression could be introduced into such prior art, as posed in  FIG. 1C , is indicated by the addition of the link, as depicted by phantom lines, which extends from axis Q therein. 
     Aside from such superfluous phantom link, upon viewing  FIG. 1C , it becomes obvious that it consists of both solid, as well as slotted linkages, along with interconnecting pivot pins, which can described in greater detail as follows: 
     solid linkages NP and PQ, each of the same length, have circular holes of the exactly the same size drilled through them located very close to each of their ends along their respective longitudinal centerlines. All four holes are located such that the distance between the respective radial centerlines of the holes drilled through linkage NP is the same as that afforded between the respective radial centerlines of the holes drilled through linkage PQ, thereby describing the same bolt hole footprints; 
     a pivot pin whose radial centerline is located at axis Q is inserted through one of the holes drilled through linkage PQ and then through the slot afforded by linkage MR which thereby constrains its movement such that axis Q always lies along the longitudinal centerline of linkage MR during device articulation; thereby furthermore conforming to the distinct geometric pattern posed in  FIG. 1B  wherein point Q is shown to reside upon straight line MR; 
     the longitudinal centerline of extended slotted linkage SP is aligned with that of linkage MR by means of drilling matching circular shaped holes at axis M, located very close to the respective endpoints of such linkages; whereby a second pivot pin then becomes inserted through such matching circular shaped holes. Hence, such axis M conforms to the location of point M, as represented in  FIG. 1B , designated as the intersection point between straight lines MR and SP extended; and 
     another hole is drilled through a portion of linkage MR which does not contain a slot, whose radial centerline both is aligned with the longitudinal centerline of linkage MR, and is offset a fixed distance away from axis M that is equal to the length between the radial centerlines of respective holes which previously were drilled through the respective ends of linkage PQ. A third pivot pin is inserted through one of the holes drilled through linkage NP, and then through the vacant hole drilled through linkage MR. Lastly a fourth pivot pin is inserted through the vacant hole drilled through linkage NP, then through the vacant hole drilled through linkage PQ, and thereafter through the slot of linkage MP. 
     Based upon such design, lengths MN, NP, and PQ all must be equal; axis N must reside along the longitudinal centerline of linkage MR and upon the circumference of a circle described about axis P of radius PN, and axis P must reside along the longitudinal centerline of slotted linkage MS; thereby conforming to the geometry posed in  FIG. 1B , wherein it is shown that corresponding straight lines MN, NP, and PQ again are all of equal length, point N resides along straight line MR and upon the circumference of a circle drawn about point P of radius PN, and point P is situated upon straight line MS. 
     In conclusion, such device, as represented in  FIG. 1C , is considered to be fully capable of portraying angle RMS as a trisector for a wide range of angle QPS designated magnitudes since each discrete setting furthermore could be fully described by a singular geometric construction pattern, such as that which is afforded in such  FIG. 1B ; as thereby generated by means of imposing a distinct sequence of Euclidean operations that conforms to that which governs such famous Archimedes proposition. 
     Therefore, once such device becomes set by means of rotating linkage MR with respect to linkage MS so that angle QPS, as posed in  FIG. 1C , becomes of particular size 30, its associated trisector, represented as an angle RMP that becomes interposed about axis M between the respective longitudinal centerlines of linkages MR and MS, automatically becomes portrayed, being of size θ. 
     Once a phantom linkage, as depicted in  FIG. 1C , becomes incorporated into such prior art, a one-to-four angular amplification thereby becomes realized with respect to angle RMS as forecasted by the progression expressed above. 
     Due to such design intricacy, such device as depicted in  FIG. 1C  becomes capable of trisecting angles of various sizes in rapid succession. Such distinct advantage clearly cannot be matched by a marked ruler arrangement that otherwise must perform the repetitious act of reproducing all of the cumbersome motions considered necessary to achieve trisection each and every time an angle of different magnitude becomes slated for trisection. 
     Mechanisms which fall under the grouping entitled, CATEGORY I, sub-classification B articulating trisection devices feature new designs, and thereby shall be fully described at a later time. 
     For the particular contingency that other trisection methods yet might become identified in the near future, consisting of different approaches than those which govern such proposed CATEGORY I and CATEGORY II groupings, it is recommended that they too should be classified into suitable categories. For example, one yet to be related method for trisecting angles consists of portraying specific contours that represent a composite of trisecting angles, or aggregate of previously established trisection points for angles whose magnitudes amount to exactly three times their respective size. It would seem only fitting, then, to group together such types of mechanisms into an entirely new CATEGORY III articulating trisection device classification. 
     All told then, just one final important question still remains largely unaddressed, being that: If a unique motion related solution is required for each and every angle of different designated magnitude that intended to be trisected, then what distinct proofs, or perhaps interrelated set of proofs, would need to be specified in order to substantiate that some newly proposed mechanism could perform trisection accurately throughout its entire range of device settings? As soon will become evident, in order to suitably answer such looming question, a novel methodology, as predicated upon an extension to a limited conventional Euclidean practice that alone is incapable of solving such famous classical problem of the trisection of an angle would need to be established; thereby making what will appear in the following pages revolutionary, rather than merely evolutionary, in nature! 
     SUMMARY OF THE INVENTION 
     A newly proposed articulating trisection invention is about to be formally introduced which consists of four distinct embodiments. Before this can occur, however, a comprehensive methodology first needs to be established that identifies specific requirements that each of their constituent designs should conform to. 
     Whereas such comprehensive methodology, in turn, then would need to rely upon new definitions, these are furnished directly below: 
     mathematically limited activity, an operation that cannot be performed when complying with all of the mathematical requirements that have been imposed upon it. The classical problem of the trisection of an angle qualifies as a very good example in this regard; 
     overlapment point, an intersection point that resides within a geometric construction pattern which, although being easily located by conventional Euclidean means with respect to its given set of previously defined geometric data, nevertheless cannot be distinguished in such manner from the lone vantage point of particular rendered information; 
     reversible geometric construction pattern, a geometric construction pattern that is entirely devoid of overlapment points. Its overall configuration can be completely reconstituted by means of launching a geometric construction activity which commences exclusively from any of its rendered information that is under current review. Reversibility proceeds because contiguous intersection points therein remain distinguishable, even with respect to such rendered information; thereby affording a pathway of return that leads all the way back to its given set of previously defined geometric data; 
     irreversible geometric construction pattern, a geometric construction pattern that harbors overlapment points. Its overall configuration cannot be completely reconstituted by means of launching a geometric construction activity which commences exclusively from certain rendered data because an availability of intrinsic overlapment points residing directly along such pathway remain impervious to detection solely by conventional Euclidean means; 
     backtrack, to distinguish intersection points or even a given set of previously defined geometric data within a geometric construction pattern by means of applying only a straightedge and compass exclusively to identified rendered information; 
     family of geometric construction patterns, an infinite number of geometric construction patterns whose overall shapes vary imperceptibly from one to the next; whereby each drawing is entirely unique due to a slight adjustment which is made to the magnitude of a given angle, denoted as θ, that appears in the very first step of a specific sequence of Euclidean operations from which all of such drawings can be exclusively derived; 
     representative geometric construction pattern, a distinct drawing which has been selected out of an entire family of geometric construction patterns to suitably characterize one relative positioning of its constituent straight lines, circular arcs, and intersection points; 
     Euclidean formulation, a practical means for representing an entire family of geometric construction patterns, all upon just a single piece of paper. Such depiction features just a singular representative geometric construction pattern that furthermore has an unmistakable double arrow notated somewhere upon it that distinguishes it apart from a singular geometric construction pattern. Such type of application could be demonstrated very easily merely by means of placing a double arrow notation upon such prior art, as posed in  FIG. 1B . For the particular case when such double arrow notation appears just above point N therein and assumes the shape of two circular arcs residing just outside of the circle drawn about point P, such convention would signify that as point N moves about the circumference of such circle, straight line NM of length equal to its radius PN would be geometrically constructed from each of such newly established N points, thereby locating respective M points somewhere along straight line SP extended; whereby corresponding Q points in turn would be geometrically located by means of extending each distinct straight line MN that becomes respectively drawn. Whereas the magnitude of angle RMP already is shown therein to be denoted algebraically by the Greek letter θ, it can assume varying sizes; unlike in the unrelated case wherein such drawing otherwise might constitute a singular geometric construction pattern, thereby requiring that such given angle RMP instead be accorded only a singular numerical magnitude; 
     animation, as it applies to the motion picture industry, furthermore also pertains to an entire family of geometric construction patterns whose distinct shapes have been organized in consecutive order for purposes of either being filmed, or quite possibly, being flapped through when inserted into a book; thereby projecting the overall illusion of motion; 
     replication, an accurate simulation of some particular motion which becomes observed as a mechanical device becomes articulated, most generally transacted by means of animating an entire family of consecutively arranged geometric construction patterns; 
     fundamental architecture, prominent portions of an articulating device which are designed to stand out more than others. Various techniques can be employed to accomplish this which include, but are not limited to, coloring such portions differently, incorporating a distinct declivity such as a groove into them, or perhaps making them more pronounced so that they protrude out beyond other device areas. Unless otherwise specified, the pathway of such fundamental architecture shall map out the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively constitute such articulating device; thereby furthermore possibly being distinguished by the representative geometric construction pattern of a Euclidean formulation; 
     static image, the projection of a solid body, as it appears at some particular point in time when viewed from a singular vantage point in space; 
     emulation mechanism, a device which features a fundamental architecture that regenerates a unique static image for each of its finite settings that furthermore can be described by a constituent drawing belonging to a particular family of geometric construction patterns; 
     trisecting emulation mechanism, a specially designed emulation mechanism in which one particular portion of the unique static image which it portrays actually trisects another portion which corresponds to such device setting; 
     geometric forming process, a novel method for geometrically describing motion! On the one hand, such newly proposed process is nothing more than an extension of conventional Euclidean practice in that each of its constituent drawings actually is a singular geometric construction pattern in itself. On the other hand, a geometric forming process remains unique in that it furthermore relates such output through a distinct association of Euclidean commands. For the condition of trisection, a particular sequence of Euclidean operations becomes specified that directs how to generate an entire family of geometric construction patterns whose rendered angles amount to exactly three times the size of their respective given angles. The overall complexity that is characteristic of a geometric forming process becomes readily apparent when considering all of the inputs which contribute to its composition, as briefly enumerated below: 
     trisection rationale, a detailed accounting of how to resolve such previously addressed trisection mystery; 
     improved drawing pretext, a method which abbreviates the rather outmoded process of otherwise unsuccessfully attempting to generate a virtually unlimited number of geometric construction patterns in order to fully describe a discrete motion which fundamentally is considered to consist of a continuum of unique shapes that instead becomes portrayed over a finite period of time in a rather effortless manner. For purposes of specifically substantiating that any static image which becomes regenerated by means of properly setting a trisecting emulation mechanism automatically portrays a motion related solution for the problem of the trisection of an angle, a corresponding geometric construction pattern whose rendered angle is of a magnitude which amounts to exactly three times its given angle can be selected from a suitable Euclidean formulation which furthermore fully describes its overall shape; thereby demonstrating that the smaller portion of such displayed static image actually trisects the larger portion which is calibrated to such setting; 
     mathematics demarcation, a natural order that can be attributed to all of mathematics; one that just now becomes recognizable as the result of partitioning conventional Euclidean practice with respect to such newly proposed geometric forming process! In effect, such categorization effort allows for any singular geometric construction pattern, thereby remaining stationary with respect to the very piece of paper it is drawn upon, to be distinguished apart from an entire group of geometric construction patterns which, not only can be related to one another through a common set of Euclidean commands, but in such manner furthermore can describe an overall outline which becomes duly portrayed by an imparted motion at some arbitrarily selected instant during its duration. Such concept actually does have a precedent; one which already was imposed upon the well known field physics years earlier, wherein: 
     statics applies to bodies which are either at rest or else are found to be moving at a constant speed, thereby signifying a specific condition which results only when forces acting upon such bodies are found to be in equilibrium; whereas 
     dynamics instead is concerned with the motions of accelerating bodies, thereby applying to a particular condition that is experienced only when forces acting upon such bodies are determined not to counterbalance one another; 
     set of rules, an accounting of how such newly proposed geometric forming process should be governed. Similar to the manner in which the very laws of motion must be interpreted differently when considering statics, as opposed to dynamics real world involvements, so too would the rules which normally apply to the conduct of conventional Euclidean practice need to be interpreted differently when instead considering the administration of such newly proposed geometric forming process. For example, when considering the varying shapes that the fundamental architecture of some particular articulating mechanism might become repositioned to over a finite period of time, design issues might arise concerning whether or not some specific interference might impede such flexure from being fully executed; and 
     probabilistic proof of mathematic limitation, an analysis that provides reasoning for how a motion related solution for the problem of the trisection of an angle can overcome a mathematic limitation which otherwise cannot be mitigated when attempting to solve the classical problem of the trisection of an angle; 
     rational number, a ratio between two integers; thereby consisting of a numerator (N) and denominator (D) which mathematically combine in order to be algebraically expressed as N/D. For any rational number (R) that furthermore is real, its actual ‘magnitude’ can be viewed! Its precise value could be obtained by means of geometrically constructing a right triangle whose two sides respectively measure N and D units in length; whereby another right triangle that is similar to it then could be drawn such that its side which corresponds to that which is D units long in such other right triangle would measure one unit in length. Hence, an established proportion N/D=x/1 would represent how the lengths of the corresponding sides of such similar triangles would relate to one another, such that the side of unknown length, x, as corresponding to that whose length is N in such other right triangle, would amount to N/D units in length; thereby being of rational value. Strictly speaking, a rational number cannot be observed, merely because it is a dimensionless fraction. However in its stead, what can be viewed is a length whose magnitude actually equals such value. For example, it reasonably could be stated that a straight line which measures 13/3 units in length is of an overall magnitude that can be expressed as a rational number. In such above given definition, notice that no indication whatsoever is afforded as to what magnitudes N and D might assume. As such, they could consist of as many digits as necessary in order to solve any given algebraic problem. Obviously, without restriction, the greater amount of digits permitted for any evaluation, the greater number of rational numbers would be contained within its overall field. For example, upon accepting numerator and denominator integers of only one digit in length, an entire field of lowest common denominator rational numbers would consist of 1, 2, 3, 4, 5, 6, 7, 8, 9, ½, 3/2, 2, 5/2, 3, 7/2, 4, 9/2, 1/3, 2/3, 4/3, 5/3, 7/3, 8/3, ½, ¾, 5/4, 7/4, 9/4, 1/5, 2/5, 3/5, 4/5, 6/5, 7/5, 8/5, 9/5, 1/6, 5/6, 7/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 8/7, 9/7, 1/8, 3/8, 5/8, 7/8, 9/8, 1/9, 2/9, 4/9, 5/9, 7/9, and 8/9, thereby comprising a total group of 58 rational numbers. However, if another rational number field instead were to be established that admits all N and D integers consisting of two or fewer digits, it naturally would consist of many more rational numbers. If such process were to be allowed to continue indefinitely to a point where another rational number field were to be permitted to grow without bounds, so would its very selectivity to a point where the actual capability of such rational number base to estimate numbers which do not belong to it would increase dramatically. Obviously, a limitless rational number grouping of this nature would enable any threshold of accuracy which might become set within an algebraic problem to be met. As a typical example to effectively demonstrate such affinity, the transcendental number π could be estimated accurately, solely as a rational number, down to a significance of two decimal places, by applying the equation π EST =R 1 =N 1 /D 1 =4T/L such that T=R 2 =N 2 /D 2 =773/1,000 and the value L=R 3 =N 3 /D 3 =985/1,000. Such rational number estimate thereby would amount to a value of π EST =4T/L=4(773/1,000)/(985/1,000)=4(773)/(985)=3.14. Such computation indicates that the rational number 3092/985=3.14 would provide an accurate estimate of the actual value of π down to a significance of two decimal places. What should stir a far greater interest, however, is that later it shall be disclosed exactly how to identify a more detailed rational number, as consisting of many more digits, that shall estimate the value of 7 down to a significance of ten decimal places. For such case, T=R 2 =N 2 /D 2 =(77,346,620,052)/100,000,000,000) and the value L=R 2 =N 2 /D 2 =(984,807,753)/(1,000,000,000). By solving the very same equation, a value π EST =4T/L=4(77,346,620,052×10 −11 )/(984,807,753×10 −9 ) would result, simplifying to π EST =4(0.77346620052)/(0.984807753)=4(7.7346620052)/(9.84807753)=3.1415926536; thereby matching the actual value of π when estimated to a ten decimal place accuracy. Hence, 309386480208/98480775300=(309,386,480,208)/(98,480,775,300)=3.1415926536 is a rational number that represents an exact value of π down to ten decimal places. Now, further suppose that such significance was not considered to be satisfactory with regard to some particular problem which instead dictated that only a rational number that can estimate the value of π down to eleven significant figures would suffice. Naturally the number determined above would not qualify, amounting to a value of 3.14159265365 down to such eleven place significance; whereby an actual value of π down to eleven decimal places of accuracy amounts to 3.14159265359. Hence, the number above could be adjusted to (309,386,480,202)/(98,480,775,300)=3.14159265359. As such, by means of continuing to apply such process, improved estimates of the value of π could be realized all the time in order to meet any accuracy requirements which might become imposed by some specific problem which needs to be solved; 
     quadratic irrational number, as first distinguished by Al-Mahani over one-thousand years ago; now stated to be the magnitude of any length which can be geometrically constructed from a given length of unity other than that which is of a rational value. Because its value cannot be a fraction, it only can be described by a string of decimal numbers that never repeats itself nor terminates, but surprisingly extends indefinitely. When algebraically expressed, a quadratic irrational number must exhibit at least one radical sign. However, it cannot feature any root which is a multiple of three, such as a cube root or even possibly an eighty-first root, because such values cannot be determined by means of applying successive Quadratic Formulas that are permitted to operate only upon either rational numbers and/or quadratic equation root values, as might become determined by such method. All that needs to be known in order to geometrically construct a square root is that upon drawing a right triangle whose sides become algebraically expressed as a and b, the altitude extending to its hypotenuse, c, will divide such base into two segments denoted respectively as s and (c−s). Hence, due to three similar right triangles which thereby become described in such manner, two residing inside of such larger initially drawn right triangle, a trigonometric relationship of the form sin θ=b/c=s/b thereby could be established. In that the proportion b/c therein identifies sides belonging to such larger right triangle, the proportion s/b would apply to corresponding sides belonging to the smaller right triangle whose hypotenuse is of length b. By multiplying each side of such resulting equation by the factor bc, the equality b 2 =cs becomes obtained. Then by taking the square root of each side, it becomes apparent that b=√{square root over (cs)}. As various rational values become substituted for c and s therein, the length of side b of such larger right triangle thereby would assume different square root magnitudes. So, if it were intended to geometrically construct side b so that it amounts to √{square root over (3)} units in length, a right triangle could be drawn whose hypotenuse, c, amounts to 3 units in length such that the altitude which lies perpendicular to it would reside a distance away from either of its ends a total of one unit of measurement; thereby setting the value of s to be one unit long. Accordingly, the value of length b would amount to √{square root over (cs)}=√{square root over (3(1))}=√{square root over (3)} units in length. In such very same manner, a fourth root of 3, as amounting to the square root of √{square root over (3)} and algebraically expressed as 
     
       
         
           
             
               
                 3 
                 
                   1 
                   / 
                   4 
                 
               
               = 
               
                 
                   
                     ( 
                     
                       3 
                       
                         1 
                         / 
                         2 
                       
                     
                     ) 
                   
                   
                     1 
                     / 
                     2 
                   
                 
                 = 
                 
                   
                     
                       3 
                       
                         1 
                         / 
                         2 
                       
                     
                   
                   = 
                   
                     
                       3 
                     
                   
                 
               
             
             , 
           
         
       
     
     thereafter could be geometrically constructed, merely by means of drawing another right triangle which this time instead exhibits dimensions of c=√{square root over (3)} and s=1, such that 
     
       
         
           
             b 
             = 
             
               
                 
                   c 
                    
                   s 
                 
               
               = 
               
                 
                   
                     
                       3 
                     
                      
                     
                       ( 
                       1 
                       ) 
                     
                   
                 
                 = 
                 
                   
                     
                       3 
                     
                   
                   . 
                 
               
             
           
         
       
     
     As another example, suppose there were an interest to geometrically construct a quadratic irrational number whose magnitude amounts to 
     
       
         
           
             
               
                 m 
                 + 
                 
                   n 
                    
                   
                     q 
                   
                 
               
             
             . 
           
         
       
     
     One way to accomplish such activity would be to obtain the square root of two straight line segments whose extremities become attached to form a longer straight line; the first of which amounts to a length that is m times the size of an arbitrarily selected length of unity, whereby such remaining segment thereby would measure n√{square root over (q)} times the size of such unit length. For example, take the case when m=2, n=¾, and q=3; whereby such first length, m, would measure 2 units long, with such second length, n√{square root over (q)}, would amount to (¾)√{square root over (3)} units in length. Yet another way to geometrically construct a length whose value amounts to 
     
       
         
           
             
               m 
               + 
               
                 n 
                  
                 
                   q 
                 
               
             
           
         
       
     
     would entail drawing a right triangle whose hypotenuse would be of a length, c, such that its sides, a and b, respectively would amount to lengths √{square root over (m)} and 
     
       
         
           
             
               
                 n 
                  
                 
                   q 
                 
               
             
             ; 
           
         
       
     
     thereby relating to one another by virtue of a Pythagorean Theorem, which unconditionally would state, 
     
       
         
           
             c 
             = 
             
               
                 
                   
                     a 
                     2 
                   
                   + 
                   
                     b 
                     2 
                   
                 
               
               = 
               
                 
                   
                     
                       
                         m 
                       
                       2 
                     
                     + 
                     
                       
                         ( 
                         
                           
                             n 
                              
                             
                               q 
                             
                           
                         
                         ) 
                       
                       2 
                     
                   
                 
                 = 
                 
                   
                     
                       m 
                       + 
                       
                         n 
                          
                         
                           q 
                         
                       
                     
                   
                   . 
                 
               
             
           
         
       
     
     Then, when m=2, n=¾, and q=3, side a thereby would amount to a value of √{square root over (m)}=√{square root over (2)} with side b being equal to 
     
       
         
           
             
               
                 n 
                  
                 
                   q 
                 
               
             
             = 
             
               
                 
                   
                     ( 
                     
                       3 
                       / 
                       4 
                     
                     ) 
                   
                    
                   
                     3 
                   
                 
               
               . 
             
           
         
       
     
     A length √{square root over (m)}=√{square root over (2)} very easily could be geometrically constructed, merely by means of drawing a right triangle whose sides each would amount to one unit in length; thereby making its hypotenuse equal to √{square root over (1 2 +1 2 )}=√{square root over (2)}. Moreover, a length of 
     
       
         
           
             
               
                 
                   ( 
                   
                     3 
                     / 
                     4 
                   
                   ) 
                 
                  
                 
                   3 
                 
               
             
             . 
           
         
       
     
     could be geometrically constructed by means of drawing yet another right triangle whose hypotenuse would be √{square root over (3)} units long such that the altitude which lies perpendicular to it would reside a distance away from either of its endpoints a total of ¾ units of measurement, such that s=¾; thereby determining a length equal to 
     
       
         
           
             
               
                 c 
                  
                 s 
               
             
             = 
             
               
                 
                   s 
                    
                   c 
                 
               
               = 
               
                 
                   
                     ( 
                     
                       3 
                       / 
                       4 
                     
                     ) 
                   
                    
                   
                     3 
                   
                 
               
             
           
         
       
     
     units in length. Thereafter, a final right triangle could be geometrically constructed whose sides are of respective lengths a=√{square root over (2)} and 
     
       
         
           
             b 
             = 
             
               
                 
                   ( 
                   
                     3 
                     / 
                     4 
                   
                   ) 
                 
                  
                 
                   3 
                 
               
             
           
         
       
     
     such that its hypotenuse would amount to a length of 
     
       
         
           
             c 
             = 
             
               
                 
                   
                     a 
                     2 
                   
                   + 
                   
                     b 
                     2 
                   
                 
               
               = 
               
                 
                   
                     2 
                     + 
                     
                       
                         ( 
                         
                           3 
                           / 
                           4 
                         
                         ) 
                       
                        
                       
                         3 
                       
                     
                   
                 
                 . 
               
             
           
         
       
     
     Lastly, for any number which resides underneath a radical sign whose value is negative, a complex number would result; thereby invalidating any possibility that such value might qualify as a quadratic irrational number, as based upon the reasoning that an imaginary number which cannot exist most certainly could not be geometrically constructed; and 
     cubic irrational number, any number whose value is neither rational nor quadratic irrational. Although such appellation is typical of Al-Mahani terminology, more specifically it is intended to signify that such types of numbers can exist only within root sets of ‘cubic’, or higher, order algebraic equations, as posed in a single variable, whose coefficients are comprised exclusively of either rational and/or quadratic irrational values. In particular, this means that cubic equations whose coefficients consist of just rational and/or quadratic irrational values can be used to convert such number types into corresponding triads of cubic irrational values. Likewise, three properly associated cubic irrational numbers could be combined mathematically to distinguish a rational or quadratic irrational number. In sharp contrast, the root set of any second order equation, as expressed in a single variable, cannot contain cubic irrational values when its coefficients consist solely of either rational and/or quadratic irrational values. One way to refute such claim would be to identify a pair of cubic irrational values, denoted as x 1  and x 2 , which could satisfy the governing requirements imposed upon such well known parabolic relationship, as algebraically assuming the form ax 2 +bx+c=0=(x−x 1 ) (x−x 2 )=x 2 −(x 1 +x 2 )x+x 1  x 2 , for the specific case when its coefficient ‘a’ amounts to a value of unity; whereby such coefficient, b, would amount to a value of −(x 1 +x 2 ), and such magnitude, c, would be equal to the product x 1 x 2 . When making such type futile attempt, however, the very difficulty which would be encountered is that whenever a pair of cubic irrational number values become selected so that the resulting magnitude of their product, x 1 x 2 , amounts to some specified rational or quadratic irrational value, such as c=1, the negative of their sums, amounting to −(x 1 +x 2 ), cannot yield yet another value equal to some stipulated rational or quadratic irrational value of b. This is because any two values belonging to a cubic irrational triad which thereafter become reduced into a single cubic irrational value, still could not be mathematically combined with such remaining value in order to calculate such hoped for result; essentially disqualifying the possibility that only two irrational number values, alone could sufficiently allow such conversion activity to take place. In other words, a quadratic equation, as posed in a single variable, cannot possibly exist which exhibits coefficients that solely are of rational value if its root set were intended to be expressed in terms of π. However, such result could be easily estimated. For example, when letting x 1 =7, and x 2 =1/π, such specific value of b would amount to −(x 1 +x 2 )=−(π+1/π). By means of simply applying the results obtained above, the value of such coefficient, b, could be approximated, as amounting to the particular value −[(309,386,480,202)/(98,480,775,300)+(98,480,775,300)/(309,386,480,202)]; whereby the magnitude of its coefficient, c, would be x 1 x 2 =(π) (1/π)=1. Above, what at face value might seem to be a rather unsupportable or even preposterous contention, when considered in combination with other prevailing claims, as rendered by various esteemed mathematics experts, who jointly concede that any cubic equation, as posed in a single variable, whose coefficients all are of rational values can contain what in their words, are ‘constructible’ roots only if accompanied by a rational root therein, leads to the extraordinary conclusion, as about to be revealed to the general public for the very first time, that each distinct algebraic equation format type associates rational, quadratic irrational, and/or cubic irrational numbers in a unique manner. That is to say, when certain types of numerical representations become related to each other within a particular problem, they can be expressed only by certain forms of algebraically equations! For example, as claimed above, no matter what rational and/or quadratic irrational values the coefficients of a parabolic equation, as posed in a single variable, might assume, it still cannot contain cubic irrational roots. Moreover, contrary to any myth which falsely might allege that a cubic irrational number cannot be geometrically constructed, it truly can! Unfortunately, a raging controversy over such matter, even today, still continues to persist! To clear this up, all that needs to be said is that it might have been overlooked on past occasions that by simply drawing an angle of arbitrarily selected size, there exists a good chance that it will exhibit trigonometric properties whose values are cubic irrational. Such claim is based merely upon the fact that most angles exhibit properties of this nature, whereby it would become highly likely that any randomly drawn angle would assume such trigonometric proportions. A more properly phrased statement, as substituted in lieu of this prefabrication, would be that a cubic irrational number is not repeatable; essentially meaning that the probability of being able to geometrically construct an angle of intended value approaches zero. Accordingly, a cubic irrational number more properly could be defined as any dimensionless value, except that which represents the very ‘magnitude’ of any length which could be geometrically constructed with respect to a given length of unity; thereby otherwise qualifying as the magnitude of a particular length which instead could be portrayed as a motion related solution for the problem of the trisection of an angle directly alongside such unit length. One of the most intriguing aspects of trisection concerns its association with any cubic equation that relates a trigonometric property of one angle to that of another whose magnitude amounts to exactly three times its size. An often ignored, but necessary starting point to account for such association would consist of explaining how some trisector actually applies to all three roots belonging to such type of cubic equation! Such perplexing concern can be easily rectified simply by divulging that when attempting to divide an angle whose designated magnitude is denoted as 3θ into three equal parts, not only does an angle, denoted as θ 1 , whose magnitude amounts to exactly one-third of its value constitute its trisector, but so would other angles denoted as θ 2 =θ 1 +120°, as well as θ 3 =θ 1 +240°. This is because when multiplying such other angles, each by a factor of three, equations would result of 3θ 2 =3θ 1 +360°=3θ, as well as 3θ 3 =3θ 1 +720°=3θ. Hence, when attempting to determine a trisector for a particular angle denoted as 3θ, it always should be kept in mind that, not one, but three angles actually meet such description; namely being, θ 1 , θ 2 =θ 1 +120°, and θ 3 =θ 1 +240°. From such knowledge, like trigonometric properties for all three of such angles could be determined whose respective values, such as cos θ 1 , cos θ 2 , and cos θ 3 , collectively would represent roots for any of such cubic equations. Accordingly, a process to obtain solutions of this nature would consist of first verifying that a particular cubic equation which is to be assessed meets such format description. For example, the famous cubic equation sin (3θ)=3 sin θ−4 sin 3  θ does so by being of cubic form, and furthermore relating the sine of an angle, denoted as θ, to that of another angle, denoted as 3θ, whose magnitude thereby would amount to exactly three times its size. Secondly, a designated magnitude of 3θ would have to become determined by means of taking the arc sine of the corresponding value which is represented in the particular equation that is to be solved. When a cubic equation of the specific form 3 sin θ−4 sin 3  θ−½=0 becomes specified, such ½ value could be equated to the sin (3θ), thereby determining that 3θ would amount to a value which is equal to the arc sine of ½, being 3θ in magnitude. Thirdly, a value of θ would be calculated, simply by dividing such determined 3θ value by three, yielding θ=3θ/3=30°/3=θ 1 =10°. Next, the equations cited above would become applied, whereby θ 2 =θ 1 +120=10°+120=130°, and θ 3 =θ 1 +240=10°+240=250°. The roots of such given equation would be sin θ 1 =sin 10°, sin θ 2 =sin 130°, and sin θ 3 =sin 250°, any of which could be substituted back into the cubic equation 3 sin θ−4 sin 3  θ=½ in order to produce the desired result of. Furthermore, the equations θ 2 =θ 1 +120° and θ 3 =θ 1 +240°, although being analogous to those which previously were determined by De Moivre in connection with his treatment of complex numbers, now have become more restricted in a sense that they furthermore are governed by additional common sense rules, such as ‘only three roots can a cubic equation contain’. Based upon such understanding, it becomes thoroughly evident that specific algebraic formulas, as listed below, that can be applied in order to suitably convert triads of cubic irrational values into either rational or quadratic irrational values, and vice versa. Below, such formulas are arranged so that they can be solved when their root set: 
     products equate to preselected rational and/or quadratic irrational numbers which can be substituted into left-hand portions of such equations: 
       cos(3θ 1 )/4=cos θ 1  cos θ 2  cos θ 3 ;
 
       sin(3θ 1 )/4=sin θ 1  sin θ 2  sin θ 3 ; and
 
       tan(3θ 1 )=tan θ 1  tan θ 2  tan θ 3 ;
 
     sums equate to preselected rational and/or quadratic irrational numbers which can be substituted into applicable left-hand portions of such equations: 
       0=cos θ 1 +cos θ 2 +cos θ 3 ;
 
       0=sin θ 1 +sin θ 2 +sin θ 3 ; and
 
       3 tan(3θ 1 )=tan θ 1 +tan θ 2 +tan θ 3 ; and
 
     sums of paired products equate to preselected rational and/or quadratic irrational numbers which can be substituted into applicable left-hand portions of such equations: 
       −¾=cos θ 1  cos θ 2 +cos θ 1  cos θ 3 +cos θ 2  cos θ 3 ;
 
       −¾=sin θ 1  sin θ 2 +sin θ 1  sin θ 3 +sin θ 2  sin θ 3 ; and
 
       −3=tan θ 1  tan θ 2 +tan θ 1  tan θ 3 +tan θ 2  tan θ 3 .
 
     A comprehensive methodology, as presented in  FIG. 2 , evidences a critical role which new discovery plays in the development of trisecting emulation mechanisms. The main purpose of such flowchart is to provide an overall accounting of tasks that are required in order to suitably substantiate that a newly proposed invention, such as that which is about to be formally introduced herein, can perform trisection accurately over a wide range of device settings. 
     The reason why  FIG. 2  appears, even before four embodiments formally become specified, simply is because their detail designs are predicated upon such input. 
     As indicated therein, a trisecting emulation mechanism which is deemed to merit the capability to achieve trisection over a wide range of device settings must suitably demonstrate that its proposed design complies with all of the provisions specified in a prepared requirements chart. 
     Moreover,  FIG. 2  specifically itemizes pertinent ramifications which are considered to underlie the very nature of such plaguing trisection mystery. 
     To expound, by beginning at the oval shaped START symbol therein, notice that four process boxes of rectangular shape are specified in the iterative portion of such diagram just before the upper diamond shaped decision box location. 
     More specifically, they furthermore appear as entries in the very first very first column of a  FIG. 3  Trisection Mystery Iteration Processes Table under the heading entitled, PROCESS. 
     The second column of such  FIG. 3  chart provides correct responses under the heading entitled, CORRECT RESPONSE appearing therein, for each of such four listed processes, as extracted from later discussions. 
     Within this iteration portion of  FIG. 2 , the NO arrow departing such upper diamond shaped decision box signifies that a review process is needed in order to assure that for whatever identified mathematic limitation becomes proposed at a particular point in time, sufficient means are specified to overcome it. 
     Moreover, the oval shaped START symbol presented in  FIG. 2  also leads to five inputs that additionally need to be supplied, as indicated by having their names listed within parallelograms. 
     Based upon a detailed understanding of such five inputs, an overall geometric forming process is to be established from which explicit details can be gleaned in order to furnish correct responses for five rectangular shaped processes itemized to the very right of such  FIG. 2  flowchart, all leading to a second decision box cited therein. 
     The reason for preparing a requirements chart is to identify specific information that, although mostly lacking from prior art which would qualify as CATEGORY I and CATEGORY II articulating trisection devices, is needed nonetheless to suitably substantiate that the design of any newly proposed articulating invention can perform trisection accurately over its wide range of device settings. 
     Another process box located just to the left of such second decision box in  FIG. 2  discloses that an initially proposed invention might have to undergo a series of refinements in order to satisfy all of the provisions imposed by such requirements chart. 
     Whenever it can be substantiated that a newly proposed invention truly meets all of such stipulated provisions, as imposed in such cited requirements chart, it is said to thereby qualify as a trisecting emulation mechanism; thereby becoming capable of performing trisection merely by means of becoming properly set to some designated size! 
     In so doing, its constituent fundamental architecture thereby becomes reconfigured; causing the regeneration of a static image that portrays its trisector! 
     Trisection occurs because such regenerated static image must be describable by a drawing that is part of a distinct family of geometric construction patterns, each rendered angle of which is of a magnitude which amounts to exactly three times the size of its given angle; whereby the portion of such regenerated static image which corresponds to the given angle in such drawing actually portrays a trisector for its rendered angle portion, corresponding to a specific setting which such device initially becomes set to. 
     In  FIG. 4 , the four embodiments which collectively constitute such newly proposed invention are individually tabulated, each appearing as a separate line item under the heading entitled, NEWLY PROPOSED ARTICULATING TRISECTION INVENTION EMBODIMENT NAME. 
     The second column therein, accorded the heading entitled, APPLICABLE FIGURE NUMBER OF CORRESPONDING EUCLIDEAN FORMULATION, identifies a corresponding Euclidean formulation for each of such listed four embodiments, as cited in the first column therein. 
     A good starting point when referring to such Euclidean itemized formulations is to identify which angles are given and which are rendered therein. In order to expedite such activity, listings of such angles are presented in a  FIG. 9  Euclidean Formulation Rendered Angle Relation Table. First column entries cited therein under the heading entitled, APPLICABLE FIGURE NUMBER OF EUCLIDEAN FORMULATION, reiterate those listings appearing in the second column of  FIG. 4 . 
     Notice therein that each GIVEN ANGLE(S) entry, as it appears in the second column of such  FIG. 9  table, describes a particular given angle, or pair of given angles when both acute as well as obtuse angle trisection fall under consideration. Moreover, the magnitude of each cited given angle is algebraically expressed on a line appearing directly below it. 
     Clearly, the very same format applies for each entry listed in the third column therein under the heading entitled, RENDERED ANGLE(S). 
     As an example of this, for the Euclidean formulation presented in  FIG. 5 , as given angle VOO′ of algebraically expressed magnitude θ varies from zero to thirty degrees in size, acute angle VOU′ of magnitude 3θ, always amounts to exactly three times its size, thereby varying from zero to ninety degrees. 
     In this regard,  FIG. 5  represents the very first attempt to describe an entire Euclidean formulation. Its double arrow notation signifies that such improved drawing pretext characterizes an entire family of unrevealed geometric construction patterns, in addition to the lone representative geometric construction pattern that is depicted upon its very face. 
     Each and every one of such distinct drawings could be geometrically constructed merely by means of executing all of the commands which are specified in its governing fourteen step sequence of Euclidean operations, enumerated as follows: 
     step 1—given acute angle VOO′ of arbitrarily selected magnitude θ ranging anywhere from zero to thirty degrees is geometrically constructed such that its side OO′ exhibits the same length as side OV; 
     step 2—side OV of given acute angle VOO′ becomes designated as the +x-axis; 
     step 3—a +y-axis is generated orthogonally to such +x-axis, represented as a straight line drawn through vertex O of given angle VOO′ which is geometrically constructed perpendicular to such designated x-axis; 
     step 4—a portion of the circumference of a circle denoted as the FIRST CIRCULAR PORTION is geometrically constructed about center point O whose radius is equal in length to straight line OV, thereby enabling it to pass through points V and O′, both of which previously have been designated as respective termination points of angle VOO′; 
     step 5—the intersection between such FIRST CIRCULAR PORTION and such y-axis becomes designated as point T; 
     step 6—next, an angle which amounts to exactly three times the magnitude of given angle VOO′ becomes geometrically constructed, but in a completely different, yet simplified manner to the way in which it was drawn in  FIG. 1B ; one which is more in line with the fan shape, as was generated in  FIG. 1A . This is to be achieved merely by geometrically duplicating given angle VOO′ twice, and thereafter adding such result onto it in order to obtain a new angle of magnitude 3θ such that its vertex is situated at point O, its first side resides along the +x-axis, and its yet undesignated other side is orientated counterclockwise to it; 
     step 7—the intersection point between such FIRST CIRCULAR PORTION and the remaining, yet undistinguished, side of such geometrically constructed angle of size 3θ becomes designated as Point U′; 
     step 8—straight line TU′ and straight line TO become drawn, thereby completing isosceles triangle TOU′; 
     step 9—a portion of the circumference of a circle denoted as the SECOND CIRCULAR PORTION is drawn about point O′ whose radius is set equal in length to straight line OO′; 
     step 10—the yet undistinguished intersection point which such SECOND CIRCULAR PORTION makes with such previously drawn y-axis becomes designated as point T′; 
     step 11—straight line O′T′ becomes drawn; 
     step 12—a portion of the circumference of a circle denoted as the THIRD CIRCULAR PORTION is drawn about point T′ whose radius is set equal in length to straight line TU′; 
     step 13—the yet undistinguished intersection point which such THIRD CIRCULAR PORTION makes with such SECOND CIRCULAR PORTION now becomes designated as point U; and 
     step 14—straight lines T′U and O′U become drawn, thereby completing isosceles triangle T′O′U. 
     Any configuration which could be generated when implementing such sequence of Euclidean operations would exhibit a unique shape based upon the particular magnitude that becomes assigned to given angle VOO′ in its step 1. 
     The configurations that such  FIG. 5  Euclidean formulation would assume when given angle VOO′ adopts its limiting values are specified in detail as follows: 
     when given angle VOO′ is designated to be of zero degree magnitude: 
     radii OO′, OU′ and O′T′ all collapse onto the +x-axis; 
     isosceles triangle OTU′ becomes a right triangle whose hypotenuse TU′ furthermore can be described as straight line TV; 
     isosceles triangle ‘OT’U becomes a right triangle whose hypotenuse UT′ furthermore can be described as straight line UO; and 
     when given angle VOO′ is designated to be of thirty degree magnitude: 
     isosceles triangle O′OT′ becomes an equilateral triangle whose vertex T′ coincides with vertex T; 
     isosceles triangle OTU′ becomes a straight line which resides upon the +y-axis, such that its side, represented as radius OU′ collapses upon its other side, therein represented as radius OT; and 
     isosceles triangle O′T′U becomes a straight line, such that its side, represented as radius O′U emanating from a center point O′ collapses upon its other side, therein represented as radius O′T′. 
     More particularly, this means: 
     when given angle VOO′ amounts to 0°, such Euclidean formulation assumes the form of three sides of a square along with its diagonals comprised of straight lines TU′ and T′U since point T′ collapses onto point O, and points U′ and O′ collapse onto point V; and 
     when given angle VOO′ is 30, such Euclidean formulation assumes the form of an equilateral triangle one of whose sides aligns upon the +y-axis with points T′, U, and U′ collapsing onto point T. 
     Algebraically, such determination is verified below: 
     when given angle VOO′, denoted as θ in  FIG. 5 , amounts to 0°, 
       ∠ TOU′=UO′T′= 90−3θ=90°;
 
       ∠ TOO′=∠UO′O= 90−θ=90°;
 
       ∠ U′OO′=∠T′O′O= 2θ=0°;and
 
     when given angle VOO; amounts to 30°, 
       ∠ TOU′=∠UO′T′= 90−3θ=0°;
 
       ∠ TOO′=∠UO′O= 90−θ=60°;
 
       ∠ U′OO′=∠T′O′O= 2θ=60°.
 
     Accordingly, the insertion of such double arrow notation into  FIG. 5  signifies that as the magnitude θ of given angle VOO′ becomes infinitesimally increased from zero to thirty degrees in the very first step of such sequence of Euclidean operations, the overall shape of such figure will change by means of reconfiguring itself as the result of point T′ becoming displaced upwards from point O to point T along the +y-axis. Once a virtually unlimited number of geometric construction renderings which belong to such distinct family of geometric construction patterns eventually become drawn, a complete Euclidean formulation finally would be represented. 
     The particular placement of such vertical double arrow in  FIG. 5  is to signify that point T′ can reside only upon such +y-axis. Such geometric alignment easily can be confirmed once realizing that since radius O′T′ is of the same length as radius O′O, as angle VOO′ becomes varied in size, point T′ always must intersect such +y-axis at the juncture of the circumference of a circle of radius O′O which becomes drawn about point O′. 
     Each unrepresented, but differently shaped geometric construction pattern that also belongs to the family of geometric construction patterns which constitute such Euclidean formulation, as represented in  FIG. 5 , must be structured from the very same sequence of Euclidean operations whereby: 
     all unrepresented straight lines corresponding to those which appear as solid straight lines exhibited upon the representative geometric construction pattern of such Euclidean formulation, as posed in  FIG. 5 , must be equal to their respective lengths; and 
     all unrepresented internal angles that apply to such Euclidean formulation must maintain the same proportions with respect to each other as appear in its representative geometric construction pattern, as actually is posed in such  FIG. 5 . For example the magnitudes of angle O′OT′ and angle O′T′O always must remain equal to each other, even when their relative sizes become varied, since they represent angles that reside opposite the equal length sides of isosceles triangle O′T′O. 
     TU′ and T′U are depicted as phantom straight lines therein to indicate that their respective overall lengths are permitted to vary from one drawing to the next within such specific family geometric construction patterns. 
     Such Euclidean formulation, as posed in  FIG. 5 , is shown to consist of three principal portions, identified as follows: 
     isosceles triangle TOU′, as denoted by darker shading, along with +x-axis; 
     isosceles triangle T′O′U, as denoted by lighter texture; and 
     straight line OO′ which interconnects the lower vertices of isosceles triangle TOU′ and isosceles triangle T′O′U together. 
     Within  FIG. 5 , notice that vertex U′ belonging to isosceles triangle TOU′ aligns with, or superimposes directly upon, side O′U of isosceles triangle T′O′U. This is verified by the following proof: 
     since the whole is equal to the sum of its parts, 
       ∠ VOO′+∠O′OU′=∠VOU′ 
 
       θ+∠ O′OU′= 3θ
 
       ∠ O′OU′= 2θ;
 
       ∠ U′OO′= 2θ;
 
     OT=OU′=OO′ because point T, point U′ and point O′ all reside upon such FIRST CIRCULAR PORTION; 
     O′O=O′T′=O′U because point O, point T′ and point U all reside upon such SECOND CIRCULAR PORTION; 
     since OO′ is equal in length to O′O by identity, whereby via substitution OT=O′T′ and OU′=O′U; 
     T′U=TU′ by geometric construction; 
     isosceles triangle TOU′ must be congruent to isosceles triangle T′O′U since their three corresponding sides are of equal lengths; 
     since the magnitudes of angles included in isosceles triangle T′O′U must be of equal respective sizes to corresponding angles featured in its congruent triangle TOU′, then it can be said that ∠T′O′U=∠TOU′; 
     since the whole is equal to the sum of its parts, 
       ∠ VOU′+∠U′OT= 90
 
       ∠ U′OT= 90−∠ VOU′ 
 
       ∠ TOU′= 90−∠ VOU′ 
 
       ∠ TOU′= 90−3θ;
 
     since the whole is equal to the sum of its parts, and by substitution of the identities 
       ∠ O′OV=∠VOO ′=θ and ∠ O′OT=∠TOO′,  
 
       ∠ VOO′+∠O′OT= 90
 
       ∠ O′OV+∠TOO′= 90
 
       ∠ TOO′= 90−∠ O′OV  
 
       ∠ T′OO′= 90−θ;
 
     whereby the angles residing opposite the equal sides of isosceles triangle OO′T′ must be of equal magnitude, such that ∠O′T′O=∠T′OO′; 
     by substitution, the value of angle O′T′O is equal to 90−θ; 
     since the sum of the included angles in isosceles triangle OO′T′ must be 180 degrees, via substitution, 
       ∠ O′T′O+∠T′OO′+∠OO′T′= 180
 
       (90−θ)+(90−θ)+∠ OO′T′= 180
 
       ∠ OO′T′= 180−2(90−θ)
 
       ∠ OO′T′= 2θ;
 
     since the whole is equal to the sum of its parts and via substitution from above, 
       ∠ OO′T′+∠T′O′U=∠OO′U  
 
       2θ+∠ TOU′=∠OO′U  
 
       2θ+(90−3θ)=∠ OO′U  
 
       90−θ=∠ OO′U;  
 
     whereas the angles residing opposite the equal sides of isosceles triangle U′OO′ must be of equal magnitude, then ∠OO′U′=∠O′U′O; 
     since the sum of the internal angles of isosceles triangle OO′U′ must equal 180 degrees, via substitution, 
       ∠ OO′U′+∠O′U′O+∠U′OO′= 180
 
         OO′U′+∠OO′U′+∠U′OO′= 180 
       2(∠ OO′U ′)+2θ=180
 
       2(∠ OO′U ′)=180−2θ
 
       ∠ OO′U′= 90−θ;
 
     since both angle OO′U′, as well as angle OO′U are equal to a magnitude of 90−θ, they must be equal in size to each other; and 
     hence, point U′ must reside on straight line O′U. 
     For the actual representative geometric construction pattern, as is expressed upon the very face of the Euclidean formulation represented  FIG. 5 , given acute angle VOO′ amounts to exactly 16°. As such, angle VOU′ must be exactly three times that size, or 48°, and angle TOU′ must be equal to its complement, being 42°. 
     Accordingly, such algebraic proof further validates that even if such 16° given acute angle VOO′, as really is depicted in  FIG. 5 , were to have been of slightly different size, point U′ nevertheless would reside somewhere along straight line O′U. This is because the very same sequence of Euclidean operations would have governed the development of another algebraic proof for such altered case, whereby a distinct drawing of somewhat modified overall shape, but one which nonetheless belongs to the very same family of geometric construction patterns, instead would have replaced the representative geometric construction pattern that presently is depicted upon the very face of such Euclidean formulation. 
     Moreover, since such argument furthermore applies to any distinct overall shape otherwise contained within such distinct family of geometric construction patterns, such alignment of point U′ along straight line O′U thereby must pertain to any and all of such drawings which collectively comprise it. 
     Since such representative geometric construction pattern, as actually is depicted in  FIG. 5 , very easily could be appended simply by means of incorporating additional steps onto its sequence of Euclidean operations, it furthermore becomes possible to devise a never-ending assortment of Euclidean formulations which stem directly from it. 
     Accordingly, a derivative Euclidean formulation, as represented in  FIG. 6 , whose sequence of Euclidean operations builds upon that which was applied to develop such Euclidean formulation, as represented in  FIG. 5 , by means of appending another three steps onto it, thereby is said to consist of a lengthened seventeen step sequence of Euclidean operations whose last three steps are provided as follows: 
     step 15—straight line U′O is extended downwards and to the left to a position where its meets such FIRST CIRCULAR PORTION, whereby such intersection point becomes designated as point W; 
     step 16—straight line OO′ is perpendicularly bisected such that the position where its downward extension intersects such FIRST CIRCULAR PORTION becomes designated as point X; and 
     step 17—straight line OX becomes drawn. 
       FIG. 7  depicts a second derivative Euclidean formulation, thereby serving as an example of how to create others. Therein, member notations and shadings that are specified upon prior Euclidean formulations have been omitted because they no longer are needed. However, in their absence, a rectangle appears whose upper two corners align with point T′ and point U′ therein. Such drawing can be geometrically constructed by means of appending the seventeen step sequence of Euclidean operations which such derivative Euclidean formulation, as posed in  FIG. 6 , was generated from into a somewhat larger twenty-one step sequence of Euclidean operations by incorporating steps 18 through 21 onto it as follows: 
     step 18—straight line T′U′ is drawn; 
     step 19—a straight line passing through point T′ is drawn perpendicular to straight line OO′; 
     step 20—an additional straight line passing through point U′ is drawn perpendicular to straight line OO′; and 
     step 21—the intersection between radii OU′ and O′T′ is designated as point Y. 
     The proof that such newly drawn straight line T′U′ runs parallel to radius OO′ for all magnitudes which given angle VOO′ could assume, is provided directly below: 
     since angle OO′T′ and angle U′OO′, as included in triangle OO′Y, both are equal to a magnitude of 2θ, such triangle must be isosceles, whereby their opposite sides OY and O′Y must be of equal length; 
     whereas, straight line OO′ constitutes a radius belonging to both such FIRST CIRCULAR PORTION and SECOND CIRCULAR PORTION, as posed in  FIG. 7 , radii OU′ and O′T′, by being equal in length to it, must be equal in length to each other; 
     since the whole is equal to the sum of its parts, via substitution from above, 
     
       
      
       OU′=O′T′ 
      
     
     
       
      
       OY+YU′=O′Y+YT′ 
      
     
     
       
      
       OY+YU′=OY+YT′ 
      
     
     
       
      
       YU′=YT′;  
      
     
     hence, triangle T′YU′ must be isosceles: 
     whereas angle OYO′ and its vertical angle T′YU′ must be of equal magnitude, and the sum of the internal angles of a triangle must equal 180°, it can be stated for isosceles triangles OYO′ and T′YU′ that 
     
       
         
           
             
               
                 
                   
                     
                       180 
                        
                       ° 
                     
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           T 
                           ′ 
                         
                          
                         
                           YU 
                           ′ 
                         
                       
                       + 
                       
                         2 
                          
                         
                           ( 
                           
                             ∠ 
                              
                             
                                 
                             
                              
                             
                               YU 
                               ′ 
                             
                              
                             
                               T 
                               ′ 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           OYO 
                           ′ 
                         
                       
                       + 
                       
                         2 
                          
                         
                           ( 
                           
                             ∠ 
                              
                             
                                 
                             
                              
                             
                               YU 
                               ′ 
                             
                              
                             
                               T 
                               ′ 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     
                       = 
                         
                        
                       
                         
                           ∠ 
                            
                           
                               
                           
                            
                           
                             OYO 
                             ′ 
                           
                         
                         + 
                         
                           2 
                            
                           
                             ( 
                             
                               ∠ 
                                
                               
                                   
                               
                                
                               
                                 YOO 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                     
                     ; 
                   
                 
               
             
               
           
         
       
     
     hence, since ∠YU′T′ must be equal in magnitude to ∠YOO′, straight line T′U′ therefore must be parallel to radius OO′ because radius OU′, by means of acting as a transversal, distinguishes such angles to be alternate interior angles of equal magnitude with respect to each other. 
     The double arrow convention, as depicted in such second derivative Euclidean formulation posed in  FIG. 7 , indicates that point T′ intersects the +y-axis upon all drawings which belong to its distinguished family geometric construction patterns. This is because the addition of such rectangle has no bearing whatsoever upon the outcome posed by such derivative Euclidean formulation, as depicted in  FIG. 6 , because it is represented only by phantom lines which either can grow or shrink in size as the magnitude of given angle VOO′ becomes altered in variable size, denoted as θ. 
     A third derivative Euclidean formulation, as posed in  FIG. 8 , builds upon such previously described twenty-one step sequence of operations. However, because angle VOU′, while still amounting to a magnitude of 3θ therein, nevertheless is to be geometrically constructed in an entirely different manner, the following steps do not apply to  FIG. 8 : 
     steps 6-8; and 
     steps 12-21; 
     Instead, the following additional steps complete the sequence of operations for such third derivative Euclidean formulation: 
     step 22—straight line OO′ is perpendicularly bisected such that the position where its upward extension intersects straight line O′T′ becomes designated as point Y; 
     step 23—a radius is drawn which emanates from center point O, passes through point Y, and terminates at a location upon such FIRST CIRCULAR PORTION which becomes designated as point U′; 
     step 24—straight line O′T′ is extended to a position where its intersects such FIRST CIRCULAR PORTION, thereafter designated as point Z; 
     step 25—straight line OZ is drawn; 
     step 26—phantom line U′Z becomes drawn; and 
     step 27—orthogonal transformed axes x T  and YT are geometrically constructed with point O furthermore designating their origin, such that the +y T  axis superimposes upon radius OU′, as previously drawn therein. 
     An accounting of such additional steps is furnished as follows: 
     step 22 above replaces steps 16 and 21; and 
     step 23 above replaces step 7. 
     To summarize, the entire sequence of operations from which such third derivative Euclidean formulation, as posed in  FIG. 8 , was developed consists of the following steps: 
     steps 1-5; 
     steps 9-11; and 
     steps 22-27. 
     A reconciliation of the various angles appearing in  FIG. 8  is provided below: 
     whereas it previously was proven that angle OO′T′ is of magnitude 2θ whenever the magnitude of given angle VOO′ is algebraically designated to be θ, once furthermore realizing that triangle OO′Y must be isosceles because its vertex Y resides upon the perpendicular bisector of its base OO′, it must be that angle VOU′ amounts to a magnitude of 3θ because, 
     
       
         
           
             
               
                 
                   
                     
                       ∠VOU 
                       ′ 
                     
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           VOO 
                           ′ 
                         
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           O 
                           ′ 
                         
                          
                         
                           OU 
                           ′ 
                         
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           VOO 
                           ′ 
                         
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           O 
                           ′ 
                         
                          
                         OY 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           VOO 
                           ′ 
                         
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           OO 
                           ′ 
                         
                          
                         Y 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           VOO 
                           ′ 
                         
                       
                       + 
                       
                         ∠ 
                          
                         
                             
                         
                          
                         
                           OO 
                           ′ 
                         
                          
                         
                           T 
                           ′ 
                         
                       
                     
                   
                 
               
               
                 
                   
                     = 
                      
                      
                     
                       θ 
                       + 
                       
                         2 
                          
                         
                             
                         
                          
                         θ 
                       
                     
                   
                 
               
               
                 
                   
                     
                       = 
                        
                        
                       
                         3 
                          
                         
                             
                         
                          
                         θ 
                       
                     
                     ; 
                   
                 
               
             
               
           
         
       
     
     moreover, with straight line OZ being geometrically constructed to be of equal length to straight line OO′, triangle OO′Z also must be isosceles. Therefore its included angle ZOO′ must be of magnitude 180−4θ by way of the fact that ∠OO′T′=φOO′Z=∠O′ZO=2θ. Also, since angle VOO′ is equal to θ, with the whole being equal to the sum of its parts: 
     
       
         
           
             
               ∠ 
                
               
                   
               
                
               VOZ 
             
             = 
             
               
                 ∠ 
                  
                 
                     
                 
                  
                 
                   VOO 
                   ′ 
                 
               
               + 
               
                 ∠ 
                  
                 
                     
                 
                  
                 
                   O 
                   ′ 
                 
                  
                 OZ 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     ∠ 
                      
                     
                         
                     
                      
                     VOZ 
                   
                   = 
                     
                    
                   
                     
                       ∠ 
                        
                       
                           
                       
                        
                       
                         VOO 
                         ′ 
                       
                     
                     + 
                     
                       ∠ 
                        
                       
                           
                       
                        
                       
                         ZOO 
                         ′ 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     θ 
                     + 
                     
                       ( 
                       
                         180 
                         - 
                         
                           4 
                            
                           θ 
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     180 
                     - 
                     
                       3 
                        
                       
                           
                       
                        
                       
                         θ 
                         . 
                       
                     
                   
                 
               
             
           
         
       
     
     Thus, the angle which is supplementary to obtuse angle VOZ must be equal to 3θ. Once the length of straight line OZ becomes algebraically expressed by the Greek letter ρ, and the sin (3θ) becomes designated by the Greek letter η, point Z thereby must reside a vertical distance of pf above the x-axis. This same condition naturally applies for point U′ since angle VOU′ also is of magnitude 3θ with straight line OU′ also having been geometrically constructed to be of equal length to straight line OO′. 
     Hence, phantom line U′Z must remain parallel to the x-axis, even during conditions when such given angle VOO′ of designated magnitude θ varies in size, as it is capable of doing in such third derivative Euclidean formulation, as is represented in  FIG. 8 . This is because the vertical distances both being dropped from point Z and point U′ to an x-axis, furthermore represent projections of length pf. Such projections very easily furthermore could be considered as opposite equal length sides of a rectangle, thereby imposing the requirement that phantom line U′Z must remain parallel to such x-axis at all times. 
     As such, once the double arrow convention becomes applied to  FIG. 8 , such illustration may be construed to be the Euclidean formulation of an entire family of geometric construction patterns in which angle VOU′ must be of size 3θ and angle VOZ must be of magnitude 180−3θ whenever given angle VOO′ is of designated magnitude θ; such that point U′ and point Z must reside the same distance above the x-axis as for all geometric construction patterns that belong to such family. 
     As given angle VOO′ varies in size, because each intersection point Y between straight line OU′ and straight line O′T′ furthermore must reside upon the perpendicular bisector of straight line OO′, such complete family of geometric construction patterns, as depicted in  FIG. 8 , can be filmed in consecutive order for purposes of replicating the motion of a car jack as it otherwise either could be raised or lowered when attempting to change a tire. 
     Hence, it could be said that a car jack configuration of such design could regulate angle VOU′ so that its size always amounts to exactly three times the size of given angle VOO′ during conditions when its magnitude becomes varied. 
     The actual mechanics behind such activity simply is that: 
     as given angle VOO′ of designated magnitude θ becomes adjusted in size, angle OO′T′ always must be equal to double its size, amounting to a magnitude of 2θ due to the geometric construction of isosceles triangle OO′T′ thereby exhibiting two angles of (90−θ) size; such that 
     once straight line OU′ is located so that it emanates from center point O and passes through point Y, being the intersection point between straight line O′T′ and the perpendicular bisector of straight line OO′, the latter of which furthermore represents the base of isosceles triangle OO′Y, then angle O′OU′ which thereby becomes described also must be equal in size to angle OO′T′, amounting to a magnitude 2θ; such that 
     angle VOU′ must be equal to the sum of the magnitudes of such given angle VOO′ plus angle O′OU′ which calculates to θ+2θ=3θ. 
     Moreover, notice that such geometry additionally features anti-parallelogram OU′O′T′, as described by the fact that its diagonals OU′ and O′T′ are of the same length and intersect at point Y, which is located upon a perpendicular bisector of a straight line OO′ that connects two of the endpoints of such diagonals together. Proof of this lies in the understanding that such description can occur only when triangle YOT′ is congruent to triangle YO′U′, such that the other requirement of being an anti-parallelogram, being that its opposite sides OT′ and O′U′ are of equal length also becomes fulfilled. Such proof relies upon a side-angle-side (SAS) determination wherein: 
     straight line YO and straight line YO′ of isosceles triangle YO′ must be of equal length; 
     angle T′YO must be equal in magnitude to vertical angle U′YO′; and 
     straight line YT′ and straight line YU′ must be of equal length because they complete respective straight line O′T; and straight line OU′, also being of equal lengths. 
     The actual motion of such anti-parallelogram OU′O′T′, as furthermore replicated by means of animating, in consecutive order, the unique drawings which belong to the distinct family of geometric construction patterns which is represented by such third derivative Euclidean formulation, as posed in  FIG. 8 , is entirely different from that which otherwise would be portrayed by such articulating such Kempe anti-parallelogram construction, as it appears in the prior art formerly depicted in  FIG. 1A ; principally because the two opposite sides of such anti-parallelogram OU′O′T′, appearing previously as straight line segments OT′ and O′U′ in  FIG. 7 , although always remaining of equal length to each other, nonetheless must vary in size during flexure. 
     This becomes clear by further examining  FIG. 8  and noticing that as given angle VOO′ varies in magnitude, point T′ must move vertically along the y-axis whereby, as straight line segment OT′ changes in overall length, the straight line distance between point O′ and point U′ also must adjust accordingly to be equal to such length. 
     Accordingly, such car jack arrangement, while preserving the features of a previously described anti-parallelogram, by means of removing is side members enables the distances that become interposed therein to vary while remaining of equal length. 
       FIGS. 6, 7 and 8  thereby represent just three examples of how the distinct sequence of Euclidean operations which such unique Euclidean formulation was predicated upon, as posed in  FIG. 5 , furthermore could be appended and/or modified in order to establish additional Euclidean formulations of entirely different overall compositions. 
       FIGS. 5, 6, 7 and 8  essentially form a network of Euclidean formulations which allows for a wider base of unique embodiments to thereby become prescribed, all of which become capable of trisecting angles in entirely different ways. 
     Notice that such double arrow convention is notated in all of such Euclidean formulations. In  FIG. 8 , such notation is indicative of the fact that as given angle VOO′ varies in size, phantom straight line U′Z, as well as phantom straight line T′Z (being an extension of straight line O′T′) thereby represent adjustable lengths. 
     In order to provide a motion related solution for the problem of the trisection of an angle, a particular mechanism could be devised to have its fundamental architecture regenerate a static image that automatically portrays a trisector for a singular angle of any designated magnitude which such device could be set to. 
     In the event that such singular angle turns out to be acute, its magnitude algebraically could be denoted as 3θ by considering θ≤30°. Then, for a condition in which such singular angle instead turns out to be obtuse, its size would be expressed as a supplemental value, thereby becoming algebraically denoted as either 180-3θ, or 270-6θ. 
     By means of choosing a suitable Euclidean formulation from such  FIG. 9  Euclidean Formulation Rendered Angle Relation Table that refers directly to such determinable algebraic expression, a drawing could be identified out of its vast family of geometric construction patterns whose rendered angle value matches the very magnitude of such singular designated angle. 
     Then, in the event that the static image which becomes regenerated furthermore could be fully described by such identified drawing, its portion which corresponds to the given angle of such drawing automatically would portray a bona fide trisector for such device setting. 
     Serving as an example of such rather cumbersome logic, a particular device of such type is to be set to a designated magnitude of 123.3°. 
     Since 90°≤123.3°≤180°, such designated angle would qualify as being obtuse whereby: 
     in one case 180°−3θ=123.3° 
       60°−θ=41.1°
 
       −θ=41.1°−60°
 
       θ=18.9°;
 
     for the static image that becomes regenerated in such case to automatically portray a trisector of 41.1° magnitude, according to such  FIG. 9  Euclidean Formulation Rendered Angle Relation Table, the overall shape of such angle, thereby algebraically expressed therein as being of 60°−θ size, would have to match that which appears within a unique geometric construction pattern that could be drawn by commencing from a given angle of VOO′ of 18.9° magnitude, upon which becomes executed all of the remaining commands which are specified in the distinct sequence of Euclidean operations for either of such Euclidean formulations, as posed in  FIG. 6  and  FIG. 7 . Secondly, for such other case 270°−6θ=123.3° 
       90°−2θ=41.1°
 
       −2θ=41.1°-90°
 
       2θ=48.9°
 
       θ=24.45θ; and
 
     for the static image that becomes regenerated in such other case to automatically portray a trisector of 41.1° magnitude, according to such  FIG. 9  Euclidean Formulation Rendered Angle Relation Table, the overall shape of such angle, thereby algebraically expressed therein as being of 90°−2θ size, would have to match that which appears within a unique geometric construction pattern that could be drawn by commencing from a given angle of VOO′ of 24.45° magnitude upon which becomes executed all of the remaining commands which are specified in the distinct sequence of Euclidean operations for such third derivative Euclidean formulation, as posed in  FIG. 8 . 
     Although featuring unique control mechanisms, CATEGORY I sub-classification B articulating trisection devices nevertheless can be grouped together because they all feature similar fan array designs. This can be verified merely by means comparing their individual designs to one another. The results of such activity are presented in  FIG. 10 . 
     Therein, headings entitled, . . . EMBODIMENT NAME, AND . . . FIGURE NUMBER OF CORRESPONDING EUCLIDEAN FORMULATION summarize listings that appear in  FIG. 4 . 
     Radii listings, appearing in groups of three, as cited in the second column of such  FIG. 10  Category I Sub-classification B Conforming Aspect Chart, under the heading entitled, FAN PORTION RADIUS LISTINGS, align upon the longitudinal centerlines of linkages that comprise the spokes of such fan arrays. The third column therein, as headed by the words FAN PORTION COMMON INTERSECTION POINT LISTINGS, is devoted to identifying common intersection points which align upon the radial centerlines of interconnecting pivot pins that are located at the respective hubs of such fan arrays. 
     For CATEGORY I, expanded sub-classifications definitions are provided below, as are premised upon new terminology which previously was furnished at the outset of this section: 
     a CATEGORY I, sub-classification A device hereinafter shall regarded to be any articulating trisection device which features four linkages of equal length, excepting that lengths of double that size also are permissible, all hinged together about their longitudinal centerlines by an interconnecting pivot pin that is passed through one end portion of each such that its radial centerline aligns upon the common meeting point of such linkage longitudinal centerlines, or instead is passed through the center portion of a linkage which is twice such length; thereby collectively constituting the array of a fan which, in combination with the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively comprise its incorporated unique control mechanism, features a fundamental architecture that is capable of regenerating a multitude of static images over a wide range of device settings; whereby such automatically portrayed overall outlines furthermore can be described by an entire Euclidean formulation which can distinguish a central angle that amounts to the size of any designated angle which can be set into such device, as subtended between two radii of a circle which thereby becomes divided into three equal angular portions by two other radii; and 
     a CATEGORY I, sub-classification B device hereinafter shall be regarded to be any articulating trisection device which features three linkages of equal length, excepting that lengths of double that size also are permissible, all hinged together about their longitudinal centerlines by an interconnecting pivot pin that is passed through one end portion of each such that its radial centerline aligns upon the common meeting point of such linkage longitudinal centerlines, or instead is passed through the center portion of a linkage which is twice such length, thereby collectively constituting the array of a fan which, in combination with the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively comprise its incorporated unique control mechanism, features a fundamental architecture that is capable of regenerating a multitude of static images over a wide range of device settings; whereby such automatically portrayed overall outlines furthermore can be described by an entire Euclidean formulation which can distinguish a central angle that amounts to the size of any designated angle which can be set into such device, as subtended between two radii of a circle which thereby becomes trisected by another radius. 
     All in all, a total of five individual requirements, as belonging to the chart posed in the lower right hand portion of  FIG. 2 , need to be satisfied before the design of a proposed articulating invention actually can qualify as a trisecting emulation mechanism. Listed below, these consist of: 
     RQMT 1—identifying which particular settings, or range(s) thereof, such device is supposed to trisect. Providing such details should disclose whether acute, as well as obtuse angles apply; 
     RQMT 2—stating the reason the classical problem of the trisection of an angle cannot be solved. Providing such details should unmask a Euclidean limitation that needs to be mitigated; 
     RQMT 3—indicating how such device is to be operated. Providing such details should disclose whether such proposed articulating invention needs to be specifically arranged. If it does, an accompanying remark should be included for purposes of clarity stipulating that a motion related solution for the problem of the trisection of an angle can be obtained only by means of properly setting such device; 
     RQMT 4—revealing the primary function such device is supposed to perform. Providing such details should disclose whether such proposed articulating invention actually is sufficiently equipped to overcome the Euclidean deficiency of being unable to fully backtrack upon any irreversible geometric construction pattern whose rendered angle is of a magnitude which amounts to exactly three times the size of its given angle; and 
     RQMT 5—explaining why each device setting automatically portrays a unique motion related solution for the problem of the trisection of an angle. Providing such details should disclose whether all proposed articulating invention device settings were substantiated individually, or incorrectly validated by means of instead applying a particular, singular solution for all cases. 
     When a proposed articulating invention fails to meet any, or even all, of such five above itemized requirements, it is important to note that it still might be fully capable of performing trisection. However, it would become virtually impossible to substantiate that such device could perform trisection accurately throughout its entire range of device settings! 
     The detailed repercussions which would be expected to accompany such type of mishaps are delineated below. Therein, references are made to the short term notations for each of such five requirements. Accordingly, if a proposed articulating invention fails to meet: 
     RQMT 1, then a claim as to which designated magnitudes such device actually would be capable of trisecting could not be made, other than those as specifically cited within its specification or expressly depicted upon its accompanying drawing package; 
     RQMT 2, then a detailed accounting as to very manner in which such device might overcome such impediment could not be furnished; as otherwise should have been reported as a capability to fully backtrack upon any irreversible geometric construction pattern, including that of a rendered angle whose magnitude amounts to exactly three times the size of its given angle; thereby throwing serious doubt as to whether such design contains provisions that actually enable it to surpass Euclidean capabilities; 
     RQMT 3, then it could become rather difficult to decipher how to operate such device; 
     RQMT 4, then it could become incredibly difficult to logically deduce that by means of properly setting such device to a designated magnitude, a static image would become regenerated wherein overlapment points furthermore would become discernable that enable such designated magnitude to be fully backtracked upon, all the way back to its associated trisector; in effect, mitigating a Euclidean irreversibility limitation that otherwise would prevent the classical problem of the trisection of an angle from being solved and, by overcoming such difficulty, thereby automatically portray a motion related solution for the problem of the trisection of an angle; and 
     RQMT 5, then it could become quite difficult to fathom that substantiating every unique motion related solution for the problem of the trisection of an angle that possibly could be automatically portrayed by such device would entail the generation of an entire family of geometric construction patterns, all belonging to a specific Euclidean formulation. 
     Fulfilling all five requirements, as stated above, naturally would lead to a proper understanding of trisection. Such knowledge would become attained only after realizing that such listings actually work in tandem with one another. 
     For example, by acknowledging RQMT 5 to be a true statement, it would be expected that any newly proposed articulating trisection invention appropriately would account for, not just one, but many individual motion related solutions for the problem of the trisection of an angle. Hence, for any drawing which could become generated by means of executing some particular Archimedes proposition, it should be recognized that it could serve to substantiate only one motion related solution for the problem of the trisection of an angle. With regard to the representative geometric construction pattern, as expressed in  FIG. 1B , it readily should become apparent that such singular drawing could be used to substantiate only one particular solution thereof. However, if such drawing instead were to become construed to be a full blown Archimedes formulation, in itself denoted by a sufficiency of Greek letter notations, along with what should be an included double arrow located around the outside of circular arc QS, it then would describe an entire family of geometric construction patterns, each of which individually would substantiate a unique motion related solution for the problem of the trisection of an angle. 
     Obviously, such solutions would apply to different acute and/or obtuse designated magnitudes, in complete accordance with those which previously must have been specified in order to satisfy the provisions of RQMT 1. 
     Moreover, an operating procedure, as specified in order to meet the provisions stipulated in RQMT 3, thereafter could be thoroughly reviewed in order to verify that no considerable obstruction would preclude the suitable trisection of an entire range of angles, as formerly indicated in RQMT 1. 
     It is true that Wantzel and Galois, generally are credited as being instrumental in proving that an angle of designated magnitude cannot be trisected when acted upon only by a straightedge and compass. 
     However, what is quite intriguing about such work is that, while on the one hand relying rather heavily upon an analysis of various prognosticated algebraic equations, on the other hand there doesn&#39;t appear to be any tangible correlation as to how such determination, as posed in one branch of mathematics, relates to the geometric finding presented herein that the classical problem of the trisection of an angle cannot be solved due to an availability of overlapment points; thereby cause irreversibility to occur within geometric construction patterns, and making it impossible to completely backtrack from rendered angles all the way back to given angles whose respective magnitudes amount to exactly one-third their size. Inasmuch as the classical problem of the trisection of an angle requires a Euclidean solution, accounting for why it cannot be achieved requires a geometric explanation! 
     By interjecting non-geometric explanations, certain attributes that accompany trisection difficulties most certainly can be identified, but only at the risk of possibly perpetuating undesirable myths which surround such great trisection mystery; thereby preventing it from being unlocked! 
     For example, consider the rather far fetched notion that the classical problem of the trisection of an angle actually might become solved by way of obtaining a cube root, solely by conventional Euclidean means! 
     Naturally, such hypothesis would discount any possibility that unity, by posing a cube root of itself, might play a key role in any of such attempts. Nor should such cube root be confused in any way with a cubic root that, if being a real number, instead would identify the exact location where a third order curve crosses the x-axis, as displayed upon a Cartesian coordinate system. 
     Within any right triangle drawing, since the length of its hypotenuse amounts to the square root of the sum of the squares of its two sides, according to the Pythagorean Theorem, then such geometric construction pattern would be a byproduct of addition, multiplication and square root mathematical operations; which turned out to be the very basis of pursuit in Wantzel&#39;s work. In connection with such premise, as concerning the possible Euclidean extraction of a cube root, naturally a leading question which should be asked is what about cube root lengths whose ratios with respect to a given length of unity are either rational or quadratic irrational? 
     For example, the length of a straight line that is 3 inches long represents the cube root of another straight line that amounts to 27 inches in overall length; whereby such longer straight line very easily could be geometrically constructed simply by adding together nine of such 3 inch long straight lines. 
     A much needed logic that seemingly appears to be grossly lacking in such above stated scenario is that if it incorrectly were to be acceded that cube roots cannot be obtained solely by conventional Euclidean means, then it would have to follow that any geometric construction pattern whose rendered information, even when amounting to just a rendered length, is of a magnitude that amounts to the cube of any portion of its given data would have to be irreversible, not that it would present a solution of the classical problem of the trisection of an angle! This shall be further demonstrated later by means of geometrically constructing rendered lengths of cubed magnitudes. 
     In effect, Wantzel algebraically proved that addition, subtraction, multiplication, and division, as representing the various fundamental operations defined within number theory, could not be suitably applied by a straightedge and compass in any combination that could solve the classical problem of the trisection of an angle. Quite understandably such consideration would not apply to a geometric solution for the problem of the trisection of an angle! For example, the mathematical operation of performing division by a factor of four could be achieved by conventional Euclidean means merely by bisecting a straight line, and then bisecting each of its then separated portions again. As such, by means of performing such division upon the tangent of an angle whose value is 4/√{square root over (11)}, a new length could be obtained of 1/√{square root over (11)} which would be indicative of the tangent of its trisector. Obviously, Wantzel&#39;s non-geometric accounting, as briefly outlined above, couldn&#39;t possibly be expected to explain what is considered to be a Euclidean limitation; one which now, for the very first time, is to be described as an inability to fully backtrack upon any rendered angle whose magnitude amounts to exactly three times the size of its given angle due to an availability of overlapment points! 
     Whereas taking the cube root of a complex number also later on shall be shown to be synonymous with obtaining its trisector, such symbiosis represents yet another outstanding definition which could be attributed to trisection; but one which most certainly shouldn&#39;t be confused with any plausible explanation as to why trisection cannot be performed solely by conventional Euclidean means! 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1A  is a depiction of the fundamental architecture of a famous Kempe trisecting device, as shown in this particular case to be set to a designated magnitude of 143¼°; thereby divulging the whereabouts of only the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which collectively comprise such prior art, including those which are featured in its three strategically emplaced anti-parallelogram shaped control mechanisms used to strictly regulate the manner in which it such device is permitted to articulate. 
         FIG. 1B  is a prior art method for determining the trisection of an angle; except for the fact that all of the intersection points appearing therein now are denoted by different letters. It furthermore is representative of prior art, famously known as a marked ruler arrangement, in which the longitudinal centerline of a marked ruler, as denoted therein by straight line MR includes a notch at point N, being located any suitable arbitrary distance away from its tip, as located at point M along such longitudinal centerline; which furthermore sits atop a drawing of angle QPS, as algebraically expressed therein to be of 3θ designated magnitude, an added circle which is drawn about such point P at a radius which is equal in length to that of straight line segment MN, and shows its straight line SP to be extended in such a manner that the longitudinal centerline of such ruler is jockeyed about so that it passes through point Q, has its tip, M, rest somewhere upon straight line SP extended, while its notch additionally becomes located somewhere along the circumference of such drawn circle. 
         FIG. 1C  is yet another illustration of prior art; shown to have been truncated in order to apply specifically to trisection, as indicated by having its unused linkage depicted in phantom therein; which also expresses the same letter designations which appear on  FIG. 1B , thereby making it easier to compare such two drawings in order to recognize that the fundamental architecture of such device, as represented in  FIG. 1C , could be reconfigured so that it assumes the very same overall shape as that which is depicted in  FIG. 1B , thereby substantiating that, in such particular arrangement, such device would automatically portray a motion related solution for the problem of the trisection of an angle; and which additionally displays θ and 3θ algebraic angular notations for the express purpose of making it perfectly clear that such device is fully capable of trisecting, not only a specific angle of 55° designated magnitude, as actually is depicted therein, but a wide range of other device settings as well. 
         FIG. 2  is a flowchart which identifies the various elements of a comprehensive trisection methodology. 
         FIG. 3  is a Trisection Mystery Iteration Processes Table which itemizes pertinent ramifications which are considered to underlie the very nature of a plaguing trisection mystery that has persisted for millennia. 
         FIG. 4  is a Figure Number Table that cites figure numbers of Euclidean formulations and drawing packages that apply to each of the four embodiments which collective comprise such newly proposed articulating invention. 
         FIG. 5  is a Euclidean formulation, easily identified as such because it brandishes a double arrow, as well as bears algebraic angular notations upon it. 
         FIG. 6  is a derivative Euclidean formulation, as representing a geometrically constructed extension of  FIG. 5 . 
         FIG. 7  is a second derivative Euclidean formulation, thereby representing a geometrically constructed extension of  FIG. 6 . 
         FIG. 8  is a third derivative Euclidean formulation, thereby representing a geometrically constructed extension of  FIG. 7 . 
         FIG. 9  is a Euclidean Formulation Rendered Angle Relation Table that identifies acute rendered angles which appears in each of such Euclidean formulations, as cited in  FIG. 4 ; furthermore algebraically expressing each of their magnitudes, as shown therein to amount to exactly three times the size of their respective given angles. 
         FIG. 10  is a CATEGORY 1, sub-classification B Conforming Aspects Chart which identifies similarities evident within the four constituent embodiments of such newly proposed invention, as tabulated in  FIG. 4 . 
         FIG. 11  is a Mathematics Demarcation Chart, so arranged to divulge exactly which areas of mathematics can be represented only by a newly proposed geometric forming process; thereby exposing where conventional Euclidean practice actually is limited. 
         FIG. 12  is Trisecting Emulation Mechanism Flowchart that describes how a trisection emulation invention performs once a designated magnitude becomes specified. 
         FIG. 13  is a Euclidean formulation that is representative of the famous algebraic cubic function 4 sin 3  θ−3 sin θ=sin (3θ), wherein for any magnitude which given angle VOO′ might arbitrarily assume. respective lengths, algebraically expressed as 4 sin 3  θ and 3 sin θ, could be drawn solely by conventional Euclidean means, such that the difference noted between them would equal a length that thereby could be algebraically expressed as sin (3θ); in effect, enabling angle VOU′ to be geometrically constructed from such determination with its magnitude amounting to exactly three times the size of such given angle. 
         FIG. 14  is a graph of three algebraic functions; wherein the function denoted by the top legend remains continuous within the range −1≤cos θ≤+1, the function denoted by the middle legend remains continuous for all real values of cos θ except when it is equal to zero, and the function denoted by the bottom legend is entirely discontinuous in that it consists of only four discrete points, as noted within the large circles displayed therein; whereby any continuous portions of such curves furthermore could be described by a virtual unlimited number of geometric construction patterns that belong to a particular Euclidean formulation that could be developed in much the same way as that which is represented in  FIG. 13 . 
         FIG. 15  is a table of roots for the quartic equation 80 cos 4  θ−4 cos 3  θ−60 cos 2  θ+6=0, along with other supporting data, as obtained by relating the top and bottom functions denoted in such  FIG. 14  legend in order to establish the equality (4 cos 3  θ−6)/(20 cos θ)=4 cos 3  θ−3 cos θ. 
         FIG. 16  is a geometric construction pattern showing the process for geometrically solving parabolic equations of the form ax 2 +bx+c=0; merely by means of applying such famous Quadratic Formula x=(−b±√{square root over (b 2 −4ac)})/2a solely by conventional Euclidean means for the specific case when the coefficients a=−2, b=0.4, and c=0.75. 
         FIG. 17  is a geometric solution for the problem of the trisection of an angle whose designated magnitude is algebraically expressed as 3θ and whose tangent, denoted as ζ, is assigned a value of √{square root over (5)}/7. Although not representing a bona fide solution for the classical problem of the trisection of an angle, which cannot be solved, such geometric solution does succeed at resolving a quadratic equation that assumes the algebraic form z R   2 +b′z R +c′=0, as obtained by means of applying a particular abbreviated version of the Quadratic Formula z R =(½)(−b′±√{square root over (b′ 2 −4c′)}) to it for the particular case when b′=(3+γ)/(3ζ+β) and c′=(δ−ζ)/(3ζ+β), thus amounting to b′=−(105+49√{square root over (5)}))/(4√{square root over (5)}+49) and c′=(85√{square root over (5)})/(49+4√{square root over (5)}) for the particular quadratic equation which results when two cubic equations of a singular variable known to share a common root expressed, z R =tan θ, become simultaneously reduced, solely in algebraic fashion, when each is represented as: 
         ζ=tan(3θ)=√{square root over (5)}/7=(3 z   R   −z   R   3 )/(1−3 z   R   2 );and
 
           z   R   3   +βz   R   2   +γz   R +δ=0 when β=−(√{square root over (5)}+7),γ=7√{square root over (5)}+12, and δ=−12√{square root over (5)}.
 
         FIG. 18  is a diagram that indicates how an angle of arbitrarily selected designated magnitude, denoted algebraically as 3θ therein, can be trisected by means of geometrically constructing a series of properly arranged successive Euclidean bisections. 
         FIG. 19  is a Successive Bisection Convergence Chart that discloses the measure of trisection accuracy which could be obtained by means of increasing the number of properly arranged successive Euclidean bisections that take place within a particular geometric construction process; thereby indicating that just after twenty-one iterations, as indicated in the line item in which n=22 therein, trisection would be performed to an accuracy of six decimal places if the human eye were capable of detecting such activity. 
         FIG. 20  is a diagram of a complex number whose arbitrarily selected angular magnitude, algebraically denoted as θ therein, serves both as a trisector for, as well as a cube root of another complex number that becomes geometrically constructed with respect to it such that its magnitude amounts to exactly three times its size, thereby being algebraically designated as 3θ therein. 
     
    
    
     DETAILED DESCRIPTION 
     Certainly by now it should have been made quite clear that in order to unlock vital secrets, highly suspected to be hidden deep within the very recesses of a perplexing trisection mystery, a paradigm shift most definitely is warranted; one that expressly should recommend some fundamental change in overall approach concerning how to properly account for difficulties encountered when trying to solve the classical problem of the trisection of an angle. 
     Only by means of exposing such closely held secrets could the basic objective of a comprehensive trisection methodology become realized, as presented in the flowchart appearing in  FIG. 2 ; essentially being to validate that the design of some proposed invention could perform trisection accurately throughout a wide range of device settings and, in so doing, qualify as a legitimate trisecting emulation mechanism that can automatically portray various motion related solutions for the trisection of an angle. 
     Accordingly, a detailed discussion of such flowchart should precede the introduction of such newly proposed invention. In this way, any requirements posed relating to the design of its four constituent embodiments would be presented well before explaining exactly how they are to complied with. Such accounting begins with a process box entitled MATHEMATIC LIMITATION IDENTIFIED  1  therein, representing the task within such flowchart where some unknown mathematical limitation is identified that supposedly prevents the classical problem of the trisection of an angle from being solved. Obviously, since such solution must depend solely upon the communication of a straightedge and compass with respect to an angle of designated magnitude, any mathematic limitation alluded to therein must be some pronounced difficulty having to do with conventional Euclidean practice! 
     The process box referred to as UNKNOWN GEOMETRIC PROPERTY UNCOVERED  2  is where, in the course of such  FIG. 2  flowchart, an entirely new geometric property is to be uncovered which furthermore is considered to be the cause of such identified mathematic limitation. Although presently being unknown, any newly defined geometric property naturally would have to be as basic a shape as a well known straight line or circle; thereby making such trisection mystery that much more intriguing. 
     The third process box, entitled DEGREE OF IMPOSITION DELINEATED  3  is reserved for describing the extent of difficulty that such newly uncovered geometric property is anticipated to impose upon conventional Euclidean practice. 
     The process box referred to as DEVICE PRIMARY FUNCTION REVEALED  4  is where an as yet unknown capability thereby becomes revealed which assumes the form of some specially added equipment that articulating mechanisms can be fitted with that enables them to overcome, correct, or compensate for such undermining influence, as now suspected to be a mathematic limitation. 
     Next, the decision box entitled DEFICIENCY MITIGATED  5  within such  FIG. 2  flowchart serves to verify that certain equipment featured in such proposed articulating devices that are supposed to avail such suspected primary function actually are deemed to be of sufficient designs to suitably mitigate such adverse influence. If it turns out that they are not adequate to perform such identified primary function, then they require redesign. If, instead, it turns out that they perform such primary function, but do not trisect, then such suspected mathematic limitation must be an incorrect selection, and another response thereby needs to be chosen. The recourse for such noted action is indicated by the NO pathway which is shown to exit such decision box. 
     The input box entitled TRISECTION RATIONALE  6 , as shown in  FIG. 2 , is where a discussion is presented that accounts for how the correct responses, as indicated in such  FIG. 3  Trisection Mystery Iteration Processes Table, were chosen in the very first place. 
     Such trisection rationale discussion specifically directs attention to the first four processes listed in such  FIG. 2  flowchart, and proceeds by conjecturing that overlapment points residing within an irreversible geometric construction pattern elude detection from any and all Euclidean interrogations which possibly could be launched exclusively from the sole vantage point of its rendered information. 
     The very fact that overlapment points remain entirely inconspicuous in this manner furthermore evidences that it is impossible to specify a distinct set of Euclidean commands which can identify their whereabouts solely with respect to such rendered information. 
     Without such vital input, a specific sequence of Euclidean operations furthermore could not be developed that instructs how to apply a straightedge and compass in order to trace out a pathway which begins at such rendered information and leads all the way back to a given set of previously defined geometric data; whereby the very presence of overlapment points serves to circumvent reversibility! 
     Since the very concept of reversibility is entirely new with regards to conventional Euclidean practice, a validation that isosceles triangle MNP, as posed in  FIG. 1B , is a reversible geometric construction pattern is afforded directly below: 
     whereas the first three steps of a previously stipulated sequence of Euclidean operations already has accounted for how to geometrically construct isosceles triangle MNP directly from given acute angle RMP, all that is needed in order to demonstrate reversibility is to thereby geometric construct isosceles triangle MNP with respect to its rendered angle PNM instead, as is outlined in the three step sequence of Euclidean operations which follows: 
     step 1—from rendered ∠PNM, an arbitrary length NM is marked off along one of the sides with point M becoming assigned to its newly described end; 
     step 2—a circular arc is swung about point N whose radius is of length NM; and 
     step 3—point P becomes designated at the newly determined intersection of such circular arc with the other side of ∠PNM, whereby straight lines NP and PM become drawn to complete isosceles triangle MNP. 
     In order to demonstrate the actual difficulty which an intrusion of overlapment points causes, notice in  FIG. 1B  that it is impossible to geometrically construct isosceles triangle MNP solely with respect to rendered angle QPS. 
     Taking any of the specific geometric construction patterns which collectively constitute such Archimedes formulation into account, this becomes evident upon realizing that overlapment points M and N, as represented in such  FIG. 1B , never could be located solely with respect to rendered angle QPS by conventional Euclidean means. The reason for such impossibility is furnished below: 
     even though it is known that overlapment point M must reside somewhere along straight line SP extended, it cannot be determined solely via straightedge and compass exactly which of the infinite number of possible locations which resides upon it applies when commencing exclusively from rendered angle QPS; and 
     the same argument holds true for overlapment point N which is known to reside somewhere along a circle that is drawn about point P that is of radius PQ, but whose exact location cannot be precisely pinpointed exclusively with respect to rendered angle QPS solely via straightedge and compass. 
     For the particular hypothetical case when QPS amounts to exactly ninety degrees, such thirty degree trisector very easily could be geometrically constructed, simply by bisecting any angle or side of an equilateral triangle. However, the computation of dividing such ninety degree angle by a factor of three in order to arrive at the magnitude of such thirty degree trisector unfortunately cannot be duplicated solely by conventional Euclidean means. Hence, to do so only would create a corrupted version of the classical problem of the trisection of an angle; thereby solving an entirely different problem! 
     Hence, in such capacity, overlapment points function as obstructions serving to confound attempts to redefine an entire geometric construction pattern solely with respect to its rendered information. 
     Consequently, any pathway consisting of previously distinguished intersection points which originally led from given angle RMP all the way to rendered angle QPS, as depicted in  FIG. 1B , could not be retraced in complete reverse order by means of attempting to apply only a straightedge and compass with respect to such rendered angle QPS. 
     In that such discussion particularly should account for difficulties experienced when attempting to solve the classical problem of the trisection of an angle, it thereby becomes formally stipulated that it is impossible to fully backtrack upon any geometric construction pattern whose rendered angle is of a magnitude that amounts to exactly three times the size of its given angle; simply because such drawing would harbor overlapment points! 
     As such, a presence of overlapment points within such specific types of geometric construction patterns entirely thwarts attempts to generate such overall pathways in complete reverse order, solely by conventional Euclidean means; thereby preventing the classical problem of the trisection of an angle from being solved! 
     In summary, overlapment points have an affinity to impede the completion of geometric construction patterns that are replete with them for the mere reason that they cannot be entirely reconstituted solely via straightedge and compass in complete reverse order. 
     For the benefit of any remaining skeptics, it furthermore should be added that only when the magnitude of a trisected angle becomes furnished beforehand can a geometric construction pattern which specifies such trisector, in the very the form of its given angle, become fully reversible; thereby enabling some corrupted version of the classical problem of the trisection of an angle to be solved. 
     During such condition, overlapment points, by definition, then would become distinguishable intersection points with respect to such given trisecting angle; thereby making such geometric construction pattern fully reversible. However, to attempt such activity would defeat the purpose of trying to trisect an angle solely by conventional Euclidean means in the very first place; simply because the very information being sought after already has been furnished. In other words, it would be entirely senseless to generate geometric quantities such as straight lines, circles, and angles aforehand exclusively for purposes of then determining them solely via straightedge and compass. Nevertheless, a notable history of this exists which mostly has been directed towards improper attempts to trisect angles solely via straightedge and compass. 
     Such foolish endeavors stand is sharp contrast to most, if not all, other standard Euclidean procedures, such as bisection; whereby a bisector remains totally unknown until such time that it actually becomes geometrically constructed from an angle of given magnitude. 
     When only the magnitude of an angle that is intended to be trisected becomes designated, its associated geometric construction pattern remains completely unspecified. This presents a heightened problem because there virtually are a countless number of other geometric construction patterns, besides those represented in  FIGS. 1A and 1B , that also render angles whose magnitudes amount to exactly three times the size of respective given angles. Without being informed as to which particular geometric construction pattern applies in the very first place, resident overlapment points no longer become limited to specific intersection point locations upon a specific pattern. 
     Even when a specific geometric construction pattern becomes selected as a vehicle for attempting to perform trisection, such as in the case of the rendition of the Archimedes formulation, as posed in  FIG. 1B , its given angle NMP or RMP, even when designated to be of a specified size that can be duplicated solely by means of applying a straightedge and compass, still cannot be determined when launching Euclidean operations just with respect to its rendered angle QPS; principally because its resident overlapment points cannot be distinguished. 
     Such pronounced geometric construction limitation of not being able to encroach upon overlapment points when being launched from a particular direction can, in fact, be rectified rather simply; merely by affording a means for discerning overlapment points that reside within irreversible geometric construction patterns, and thereby making them entirely distinguishable with respect to rendered angles which otherwise cannot be backtracked upon! 
     Such elementary recommendation, despite its rather unsuspecting and seemingly outlandish nature, nevertheless describes exactly how a trisecting emulation mechanism can trisect virtually any designated angle which it can be set to; thereby portraying a of motion related solution for the problem of the trisection of an angle. 
     Such strange phenomena perhaps most easily can be described with respect to the motion of any CATEGORY I sub-classification A articulating trisection device because such types of devices do not first have to be specifically arranged before displaying their settings. As any of such devices becomes cycled, eventually reaching all of the settings within its entire operating range, its fundamental architecture sweeps out, or regenerates, a multitude of static images, each representing a still shot cameo of two angles, the larger of which not only amounts to exactly three times the size of the other, but furthermore is calibrated to a specific device setting. 
     The beauty of such design concept is that once any of such types of devices becomes set to a preselected designated angle, the portion of the smaller angle contained within the static image which becomes regenerated thereby automatically portrays its associated trisector. 
     In other words, by means of properly setting any trisecting emulation mechanism, its fundamental architecture becomes rearranged to a particular position such that the static image which becomes regenerated automatically portrays a motion related solution for the problem of the trisection of an angle! 
     In effect, such motion related solution distinguishes overlapment points whose availability otherwise would prevent the classical problem of the trisection of an angle from being solved! 
     Accordingly, instead of attempting to perform that which is impossible; essentially consisting of retracing a distinguishable pathway within an irreversible geometric construction pattern in complete reverse order solely by conventional Euclidean means, a trisecting emulation mechanism otherwise functions like the Dewey decimal system in a library wherein the exact name of a document that is being searched for becomes either input into a computer, or otherwise looked up in some card deck, whereby an alpha-numeric code that provides an indication of its whereabouts, thereby allows such information to forthwith become retrieved. The only glaring difference in the case of a trisecting emulation mechanism is that the magnitude of a designated angle which is slated for trisection becomes set into such device, thereby causing the regeneration of a particular static image that automatically portrays its associated trisector! 
     Accordingly, a fundamental architecture might be thought of as a mechanical means for conveniently storing a multitude of static images within the very memory of some particularly designed trisecting emulation mechanism; thereby enabling a motion related solution for the problem of the trisection of an angle of designated magnitude to be automatically portrayed at will. 
     To conclude, a unique pathway which leads from one angle all the way to another that amounts to exactly three times its size automatically becomes portrayed each and every time a static image become regenerated by means of configuring a trisecting emulation mechanism to any of its discrete device settings; thereby disclosing the actual whereabouts of nuisance overlapment points which reside along the way; simply by means of exposing them to be nothing more than commonly known intersection points. In so doing, any obstructions that otherwise normally would be encountered when attempting to solve the classical problem of the trisection of an angle, would be overcome merely by means of properly setting a trisecting emulating mechanism. 
     A basic tenet of conventional Euclidean practice is that all activity must proceed exclusively from a given set of previously defined geometric data, or else from intersection points which become located with respect to it. 
     It may well be that a purposeful adherence to such rule might explain why any serious attempt to completely retrace a geometric construction pattern exclusively from its rendered information all the way back to its given set of previously defined geometric data, solely by application of a straightedge and compass, entirely might have been overlooked in the past. 
     Moreover, only on very rare occasions, such as in the particular case of attempting to solve the classical problem of the trisection of an angle, could the prospect of possibly even engaging upon such activity arise, thereafter culminating in an avid interest to solve such classical problem without considering that a pathway leading from a rendered angle within any geometric construction pattern all the way back to a given angle whose magnitude amounts to exactly one-third of its size lies at the very heart of such difficulty! 
     Remarkably, only by means of analyzing conventional Euclidean practice from this other seldom viewed perspective could irreversibility be identified as being caused by an intrusion of overlapment points. 
     By otherwise neglecting such critical information, it would become virtually impossible to substantiate that any qualifying CATEGORY I sub-classification A or CATEGORY II articulating trisection mechanism could perform trisection accurately throughout a wide range of device settings. 
     The input box entitled IMPROVED DRAWING PRETEXT  7 , as posed  FIG. 2 , is where a new truncated drawing format is to be introduced that can represent an entire family of geometric construction patterns, all upon just a single piece of paper. 
     Whereas such  FIG. 2  flowchart is meant to apply exclusively to trisection, such improved drawing pretext, as alluded to therein, preferably should be identified as any Euclidean formulation each of whose constituent geometric construction patterns depicts a rendered angle whose magnitude amounts to exactly three times the size of its given angle. By means of suitably designing a trisecting emulation mechanism so that virtually any static image which would become regenerated as the result of its being properly set thereby would automatically portray an overall outline that furthermore could be fully described by a particular geometric construction pattern which belongs to such Euclidean formulation, then it could be substantiated that a motion related solution for the problem of the trisection of a angle could be achieved, merely by means of backtracking upon an irreversible condition that instead would have prevented the classical problem of the trisection of an angle from being solved! 
     Accordingly, the rather seemingly antiquated idea of generating singular, but unrelated geometric construction patterns thereby very easily could become dwarfed simply by means of considering the prospect that they furthermore might become linked to one another in some particular fashion through the use of an improved drawing pretext for the express purpose of geometrically describing motion! 
     The wording above is intended to infer that improved drawing pretexts, other than that of the Euclidean formulation could be devised, thereby associating their constituent drawing patterns in some distinct manner other than through specified sequences of Euclidean operations; and, upon becoming replicated might thereby describe important motions which are known to be of service to mankind! 
     Such discussion is building to the proposition that by means of properly partitioning all observed phenomena which can be described geometrically, including that of certain motions, it thereby becomes possible to envision a certain order that becomes evident within a farther reaching mathematics. 
     Such is the very purpose of the input box entitled MATHEMATICS DEMARCATION  8 , as posed in  FIG. 2  herein. Its key artifact consists of a Mathematics Demarcation Chart, as posed in  FIG. 11 , which discloses a particular partitioning which should be imposed universally in order to suitably distinguish between geometries which describe stationary patterns, as opposed to those which can quantify disparate motion related geometries. 
     As it pertains to trisection matters, the drawing pretext entry appearing in the third column of such  FIG. 11  chart, as listed directly under the cell entitled Geometric forming process, quite expectedly, turns out to be that of a Euclidean formulation; hence, limiting overall scope therein to matters in which geometric construction patterns can be associated to one another only through particular sequences of Euclidean operations. 
     Headings appearing in  FIG. 11 , are shown to run along the left side of such chart. Such arrangement enables the two principal listings appearing at the top of the second and third columns therein to serve as minor headings in themselves; thereby making it easy to differentiate between conventional Euclidean practice and a geometric forming process merely by means of comparing such two columns to one another. 
     Moreover, inasmuch as the field of geometry concerns itself with mathematically quantified depictions, algebra, on the other hand, by representing the overall language of mathematics, instead bears the biggest brunt of responsibility in validating that such alleged order truly exists; doing so by associating algebraic format types through some newly proposed equation sub-element theory! 
     One principal reference, standing as a harbinger of a newly proposed equation sub-element theory, is a relatively unknown treatise that was published in 1684; as written by one Thomas Baker and entitled,  The Geometrical Key or the Gate of Equations Unlocked . After a close affiliation with Oxford University, Mr. Baker successfully provided a solution set pertaining to biquadratic equations, perhaps more commonly referred to today as either quartic, or fourth order equations. However, it seems quite plausible that because of a serious competition among rival institutions going all the way back to that time period, Gerolamo Cardano&#39;s preceding work of 1545, as it appeared in Ars Magna, nevertheless, still managed to eclipse his later contributions. In brief, Cardano applied a transform to remove the second, or squared, term from cubic equations in order to modify them into an overall format that very easily could be resolved. However, because of such gross simplification, the all important fact that each algebraic equation is unique, in its own right, was largely ignored; hence, failing to attribute deliberate meaning to the various equation types that actually govern third order algebraic equation formats. The very stigma which such abbreviated process instilled unfortunately served to direct attention away from developing an all purpose solution that applies to all cubic equation formats, as posed in a single variable; one which obviously would lie at the very heart of any newly proposed sub-element theory; thereby not requiring that cubic equations which express second terms first become transformed in order to solve them! In retrospect, it now appears very likely, indeed, that a hit-and-miss mathematics approach of such nature most probably delayed the actual debut of a newly proposed equation sub-element theory by some four hundred years! 
     To conclude, by means of now introducing an all-purpose cubic equation solution, as presently has remained absent for all these years, the very relevancy of each format type can remain preserved so that further comparisons could be made in order to avail a more comprehensive understanding of an overall order that actually prevails within all of mathematics. 
     In such  FIG. 11  Mathematics Demarcation Chart, notice that cubic irrational numbers are listed only under the heading referred to as geometric forming process. Therein, such partitioning assignment is entirely consistent with the proposed finding that although angular portions within a regenerated static image can become automatically portrayed, even when they consist of cubic irrational trigonometric properties, nevertheless such angles cannot be geometric constructed just from a given length of unity or from another angle whose trigonometric properties are either rational or quadratic irrational! 
     That is to say, whenever the angular portion within a regenerated static image that has been calibrated to a particular device setting bears cubic irrational trigonometric properties, so must the angular portion therein which serves as its trisector. Accordingly, there is no way to relate either rational or quadratic irrational trigonometric properties of a trisector to an angle which amounts to exactly three times its size that bears cubic irrational trigonometric properties. 
     In other words, it requires, not one, but three angles that all exhibit cubic irrational trigonometric properties in order to geometrically construct an angle which exhibits either rational or quadratic irrational trigonometric properties. Such angle very well could be geometrically constructed in a manner which is analogous, or consistent with virtually any of the nine the arrangements of such products, sums, and sums of paired products, as posed in the algebraic equations previously expressed in such definition of a cubic irrational number. 
     Accordingly, any geometric construction pattern that belongs to a Euclidean formulation which furthermore is known to replicate the articulated motion of the fundamental architecture of any CATEGORY I sub-classification A trisecting emulation mechanism which thereby becomes reset every time it becomes articulated only can be approximated in size if it is meant to depict a static image either of whose two included angular portions portrays cubic irrational trigonometric properties! 
     An elementary, but nonetheless very revealing example of this concerns attempts to trisect a sixty degree angle solely by conventional Euclidean means! 
     Although such sixty angle can be distinguished merely by geometrically constructing an equilateral triangle, its associated twenty degree trisector, on the other hand, is known to exhibit transcendental trigonometric properties that cannot be geometrically constructed, when proceeding either exclusively from a given length of unity, or solely from any angle whose trigonometric properties exhibit either rational or quadratic irrational values. 
     Such explicitly stated impossibility is what actually distinguishes the realm between where angles can be portrayed which bear cubic irrational trigonometric property values, and other angles that do not which thereby can be expressed solely by conventional Euclidean means! 
     Further note in such  FIG. 11  chart that linear, as well as quadratic algebraic equation and associated function format type entries appear under both conventional Euclidean practice, as well as geometric forming process cells. This is because linear straight lines and/or second order circular arcs which remain stationary over time amidst an agitated motion would assume the very same shapes within each and every geometric construction pattern which belongs to any Euclidean formulation that furthermore could be animated in order to replicate such articulation event; thereby applying to both sides of such partitioned  FIG. 11  chart. 
     The fact that cubic equations appear only under the heading referred to as geometric forming process therein is a little more difficult to explain; having to do with the fact that by depicting actual motions, Euclidean formulations moreover can be expressed algebraically as continuums. 
     The most commonly known algebraic continuum is an infinite series whose terms become summed over some specific predetermined range of performance. 
     It naturally follows then that their integral counterparts, as realized within the field of calculus, also could apply, as well, to certain relative motions which furthermore can be geometrically described by Euclidean formulations. Quite obviously, this presumption moreover assumes that such motions actually do appear as complete continuums to any would be observer, wherein the time interval pertaining to such integral sign would approach zero; thereby confirming the very validity of yet another rather intrusive mathematical involvement. 
     Furthermore, other types of algebraic equations are considered to be continuous, beginning with that of a straight line whose linear equation of y=mx+b validates that for each and every real number x which becomes specified, a corresponding value of y truly exists. 
     With particular regard to a motion related solution for the problem of the trisection of an angle, algebraically expressed continuums relate to Euclidean formulations by well known cubic equations of a single variable in which trigonometric values of an angle of size 3θ become associated to those of an angle of size θ. 
     The key factor pertaining to such relationships is that no matter what values might be applied to either of such angles, a three-to-one correspondence nevertheless would hold between their respective angular amplitudes! 
     As an example of this, consider various motion related solutions for the problem of the trisection of an angle which could be portrayed when cycling such famous Kempe prior art from a 20 degree setting to one of 120 degrees. 
     In such case, not only would an entire Euclidean formulation with representative geometric construction pattern as fully described by  FIG. 1A  geometrically describe such three-to-one angular correspondence during device flexure, but so too would the well known algebraic cubic function which assumes the form cos (3θ)=4 cos 3  θ−3 cos θ. 
     That is to say, within such Euclidean formulation, angle ABC, when amounting to virtually any designated magnitude 3θ within the limits of 20°≤∠ABC≤120°, furthermore would algebraically relate to an angle ABD therein, of resulting size θ, by such aforementioned famous algebraic cubic function. 
     Algebraically, such relationship could be confirmed for virtually any angle within such postulated range. For example, below such functional relationship is confirmed algebraically for the particular condition when angle ABC amounts to exactly 60°: 
     
       
         
           
             
               ∠ 
                
               
                   
               
                
               ABC 
             
             = 
             
               
                 3 
                  
                 θ 
               
               = 
               
                 60 
                 ′ 
               
             
           
         
       
       
         
           
             
               θ 
               = 
               
                 
                   
                     60 
                     ′ 
                   
                   / 
                   3 
                 
                 = 
                 
                   
                     20 
                      
                     ° 
                   
                   = 
                   
                     ∠ 
                      
                     
                         
                     
                      
                     ABD 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 cos 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     ∠ 
                      
                     
                         
                     
                      
                     ABC 
                   
                   ) 
                 
               
               = 
               
                 
                   cos 
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       3 
                        
                       θ 
                     
                     ) 
                   
                 
                 = 
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     60 
                      
                     ° 
                   
                   = 
                   0.5 
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 cos 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     ∠ 
                      
                     
                         
                     
                      
                     ABD 
                   
                   ) 
                 
               
               = 
               
                 
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
                 = 
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     20 
                      
                     ° 
                   
                   = 
                   
                     0.93969262 
                      
                     
                         
                     
                      
                     … 
                   
                 
               
             
              
             
                 
             
             ; 
           
         
       
       
         
           
             
               
                 
                   
                     
                       4 
                        
                       
                           
                       
                        
                       
                         cos 
                         3 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                     - 
                     
                       3 
                        
                       
                           
                       
                        
                       cos 
                        
                       
                           
                       
                        
                       θ 
                     
                   
                   = 
                     
                    
                   
                     
                       4 
                        
                       
                         
                           ( 
                           
                             
                               0 
                               . 
                               9 
                             
                              
                             3969262 
                              
                             
                                 
                             
                              
                             … 
                           
                            
                           
                               
                           
                           ) 
                         
                         3 
                       
                     
                     - 
                     
                       3 
                        
                       
                         ( 
                         
                           
                             0 
                             . 
                             9 
                           
                            
                           3969262 
                            
                           
                               
                           
                            
                           … 
                         
                          
                         
                             
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       3.319077 
                        
                       8 
                        
                       6 
                        
                       2 
                     
                     - 
                     
                       
                         2 
                         . 
                         8 
                       
                        
                       1 
                        
                       9 
                        
                       0 
                        
                       7 
                        
                       7 
                        
                       862 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     0.5 
                     . 
                   
                 
               
             
           
         
       
     
     Additionally, a specific nature that is found to be evident within algebraic continuums furthermore shall become addressed, wherein: 
     a Euclidean formulation, each of whose constituent geometric construction patterns exhibits a rendered angle whose magnitude amounts to exactly three times the size of its given angle, is to become obtained by means of having the value of its sine described by a length of 3 sin θ−4 sin 3  θ; thereby conforming to the famous cubic function 3 sin θ−4 sin 3  θ=sin (3θ); and 
     a graph is to become developed that distinguishes between the continuity of such well known cubic function 4 cos 3  θ−3 cos θ=cos (3θ) and the discontinuity that clearly is evident within a function that otherwise assumes the form (4 cos 3  θ−6)/(20 cos θ)=cos (3θ). 
     Note that in this presentation such issue is addressed even before a more important detailed discussion that shall describe the very designs of such four newly proposed embodiments. 
     One method of algebraically relating a quadratic equation to two independent cubic functions that share a common root, wherein each function is limited only to a singular variable, is to link their respective coefficients together by means of what commonly is referred to as a simultaneous reduction process. 
     Since such common root, as denoted as z R  below, occurs only when the value y in such functions equals zero, the following second order parabolic equation, thereby assuming the well known form ax 2 +bx+c=0, can be derived from the following two given cubic equations: 
         y   1 =0= z   R   3 +β 1   z   R   2 +γ 1   z   R +δ 1 ;
 
         y   2 =0= z   R   3 +β 2   z   R   2 +γ 2   z   R +δ 2 ;
 
         z   R   3 +β 1   z   R   2 +γ 1   z   R +δ 1 =0= z   R   3 +β 2   z   R   2 +γ 2   z   R +δ 2 ;
 
       β 1   z   R   2 +γ 1   z   R +δ 1 =0=β 2   z   R   2 +γ 2   z   R +δ 2 ;
 
       0=(β 2 −β 1 ) z   R   2 +(γ 2 −γ 1 ) z   R +(δ 2 −δ 1 ); and
 
       0= az   R   2   +bz   R   +c.    
     Therein, whenever coefficients a, b, and c become specified, a straight line of length equal to such common root z R  can be determined solely by conventional Euclidean means, simply by developing a geometric construction pattern that is representative of the famous Quadratic Formula z R =(−b±√{square root over (b 2 −4ac)})/2a. Since such approach is not germane just to trisection, but nevertheless is relevant to a proper understanding of the dichotomy which exists between cubic functions of a single variable and an algebraically related famous parabolic equation, such geometric construction approach is to be described later on; after the four embodiments of such newly proposed invention first become formally introduced. Moreover, such particular resolution shall pertain to the specific circumstance when the coefficients in such well known parabolic equation, assuming the particular form az R   2 +bz R +C=0=ax 2 +bx+c become assigned the respective values of a=−2, b=0.4, and c=0.75, thereby later being described by the second order equation of a single variable of the particular form −0.2x 2 +0.4x+0.75=0. 
     In such  FIG. 11  Mathematics Demarcation Chart, algebraic equations and their associated functions are addressed interchangeably. Such association between them easily can be recognized when considering that by reformatting the function stipulated above into equation format, its overall content in no way changes, but only becomes perceived from a completely different perspective, such that: 
         z   3   +βz   2   +γz+δ=y ; and 
         z   3   +βz   2   +γz +(δ− y )=0
 
     In such first case, the variable z can change in value, thereby promoting a new corresponding value for y. 
     However, in such second case, generally a specific value of z is being sought after based upon the particular values which are assigned to its second order coefficient β, its linear coefficient γ, and its scalar coefficient δ−y. Notice that in such particular later reformatting, no attention whatsoever is directed to the fact that such value y also signifies a particular height above an x-axis within an orthogonal coordinate system at which a horizontal line passes through the curve that can be algebraically expressed as z 3 +z 2 +γz+δ=y at three specific locations whose corresponding values away from the y-axis amount to the respective magnitudes of z. Such perceived distinctions also suitably should be accounted for, in order to serve as yet other rudimentary elements, as contained within an all-encompassing newly proposed equation sub-element theory. 
     In such  FIG. 11  chart, it further is indicated that only certain real numbers can reside within specific algebraic equations types, as well as their associated functions; thereby even further evidencing an overall order that exists within a farther reaching mathematics! 
     Such relationships are further addressed in section 9.3, as entitled  Cubic Equation Uniqueness Theorem , also appearing within such above cited treatise; wherein it is stated that with respect to equation formats of singular variable, “Only cubic equations allow solely rational and quadratic irrational numerical coefficients to co-exist with root sets comprised of cubic irrational numbers”. 
     Such technical position doesn&#39;t address higher order equations merely because they represent byproducts of cubic relationships which are fashioned in a singular variable. 
     Neither does such contention dispute, nor contradict the fact that cubic irrational root pairs can, and do exist within quadratic equations of singular unknown quantity. 
     An example of this follows with respect to the parabolic equation presented below, followed by an associated abbreviated form of the Quadratic Formula: 
     
       
         
           
             
               
                 a 
                  
                 
                   x 
                   2 
                 
               
               + 
               
                 b 
                  
                 x 
               
               + 
               c 
             
             = 
             0 
           
         
       
       
         
           
             
               
                 x 
                 2 
               
               + 
               
                 
                   b 
                   a 
                 
                  
                 x 
               
               + 
               
                 c 
                 a 
               
             
             = 
             0 
           
         
       
       
         
           
             
               
                 
                   x 
                   2 
                 
                 + 
                 
                   
                     b 
                     ′ 
                   
                    
                   x 
                 
                 + 
                 
                   c 
                   ′ 
                 
               
               = 
               0 
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 
                   x 
                   = 
                     
                    
                   
                     
                       
                         - 
                         b 
                       
                       ± 
                       
                         
                           
                             b 
                             2 
                           
                           - 
                           
                             4 
                              
                             
                                 
                             
                              
                             ac 
                           
                         
                       
                     
                     
                       2 
                        
                       a 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           - 
                           b 
                         
                         / 
                         a 
                       
                       ± 
                       
                         
                           ( 
                           
                             1 
                             / 
                             a 
                           
                           ) 
                         
                          
                         
                           
                             
                               b 
                               2 
                             
                             - 
                             
                               4 
                                
                               
                                   
                               
                                
                               ac 
                             
                           
                         
                       
                     
                     
                       2 
                        
                       
                         a 
                         / 
                         a 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           - 
                           b 
                         
                         / 
                         a 
                       
                       ± 
                       
                         
                           
                             
                               b 
                               2 
                             
                             / 
                             
                               a 
                               2 
                             
                           
                           - 
                           
                             4 
                              
                             
                                 
                             
                              
                             
                               ac 
                               / 
                               
                                 a 
                                 2 
                               
                             
                           
                         
                       
                     
                     2 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           - 
                           b 
                         
                         / 
                         a 
                       
                       ± 
                       
                         
                           
                             
                               ( 
                               
                                 b 
                                 / 
                                 a 
                               
                               ) 
                             
                             2 
                           
                           - 
                           
                             4 
                              
                             
                               c 
                               / 
                               a 
                             
                           
                         
                       
                     
                     2 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           - 
                           
                             b 
                             ′ 
                           
                         
                         ± 
                         
                           
                             
                               b 
                               ′2 
                             
                             - 
                             
                               4 
                                
                               
                                   
                               
                                
                               
                                 c 
                                 ′ 
                               
                             
                           
                         
                       
                       2 
                     
                     . 
                   
                 
               
             
           
         
       
     
     After examining such abbreviated Quadratic Formula, it becomes obvious that the only way in which such roots can be of cubic irrational value is when either coefficient b′ and/or c′ also turns out to be cubic irrational. 
     As such, a corollary furthermore states, “Cubic irrational root pairs which appear in parabolic equations or their associated functions require supporting cubic irrational coefficients”. 
     Just as in the general case of conventional Euclidean practice where stringent rules apply, so to should they be specified in support of a geometric forming process. With respect to such flowchart, as posed in  FIG. 2 , such entries pertain to the input box entitled SET OF RULES  9 . 
     A few of the very simple rules which apply to geometric forming are elicited directly below. Their intent is to simplify the overall administration of such process by means of requiring fewer lines in any attendant substantiation. As duly furnished below, some of them might appear to be rather straightforward, even to the point where they may be considered as being somewhat obvious such that: 
     one principal rule is that the overall length of a linkage which belongs to any trisecting emulation mechanism is considered to remain constant throughout device flexure. Naturally, such rule applies so long as the linkage under consideration remains totally inelastic and intact during device flexure. From such rule, a wide variety of relationships thereby can be obtained, a small portion of which are listed as follows: 
     when two straight solid linkages of equal length become attached along their longitudinal centerlines at a common end by an interconnecting pivot pin which situated orthogonal to it, such three piece assembly thereby shall function as an integral hinged unit, even during conditions when one of such linkages becomes rotated respect to the other about the radial centerline of such interconnecting pivot pin; and 
     whenever one free end of such integral three piece unit thereby becomes attached along its longitudinal centerline to the solid end of another straight slotted linkage along its longitudinal centerline by means of inserting an second interconnecting pivot pin through a common axis which is orthogonal to such longitudinal centerlines, and thereafter the remaining unattached end of such initial integral three piece hinged unit has a third interconnecting pivot pin inserted orthogonally through its longitudinal axis whose radial centerline lies equidistant away from the radial centerline of its hinge as does the radial centerline of such added second interconnecting pivot pin, whereby such third interconnecting pivot pin furthermore passes through the slot of such slotted linkage, the longitudinal centerlines of such three linkages, together with the radial centerlines of such three interconnecting pivot pins collectively shall describe an isosceles triangle shape in space, even during device flexure. For example, when viewing prior art, as posed in  FIG. 1C , notice that the triangle whose vertices are described by axis M, axis N, and axis P must remain isosceles no matter what magnitude becomes applied to angle RMS. Such is the case because the constant distance between axis N and axis M always must be equal to that which lies between axis N and axis P therein; 
     a second rule which more particularly pertains to trisection is that the various shapes that collectively comprise an entire family of geometric construction patterns all bear a distinct geometric relationship to one another based upon the fact that they all stem from the very same sequence of Euclidean operations. A few examples of how such rule can be administered are presented below: 
     the radial centerlines of interconnecting pivot pins which become constrained within trisecting emulation mechanism linkage slots must remain aligned along the straight line, or even curved paths of their respective longitudinal centerlines during device flexure; as based upon the design principle that the constant width of such slot, whose longitudinal centerline also remains coincidental with that of such slotted linkage, is just slightly larger than the diameter of the shanks of the interconnecting pins which are constrained within it; 
     static images which become regenerated whenever a CATEGORY I sub-classification A trisecting emulation mechanism becomes cycled over its wide range of device settings automatically portray a virtually unlimited number of unique overall shapes which furthermore fully can be fully described by a Euclidean formulation; more particularly meaning that for any discrete device setting, the longitudinal centerlines of linkages and radial centerlines of interconnecting pivot pins which constitute its fundamental architecture furthermore can be described by the respective straight lines and intersection points of a geometric construction pattern which belongs to such Euclidean formulation; and 
     as a CATEGORY I sub-classification A trisecting emulation mechanism becomes cycled over a wide range of device settings, any change which can be observed in the magnitude of the intrinsic angles of its fundamental architecture furthermore fully can be described by those which become exhibited between corresponding straight lines within a Euclidean formulation which describes the overall shapes of its portrayed static images; 
     another rule is that both rational, as well as quadratic irrational numbers can be algebraically equated to specific sets of cubic irrational numbers. The procedure for accomplishing this consists of first selecting a specific rational or quadratic irrational number that is to be characterized and then setting it equal to the left-hand side of one of nine equations presented in the preceding definition of a cubic irrational number, wherein: 
     for the particular case when a rational number of ⅛ is to be further characterized, the first of such nine equations can be applied in order to determine a value for 3θ 1  as follows wherein each concluding three dot notation indicates that such number extends an infinite number of decimal places to the right, thereby being indicative of an actual cubic irrational number: 
     
       
         
           
             
               
                 
                   cos 
                    
                   
                     ( 
                     
                       3 
                        
                       
                           
                       
                        
                       
                         θ 
                         1 
                       
                     
                     ) 
                   
                 
                 / 
                 4 
               
               = 
               
                 
                   1 
                   / 
                   8 
                 
                 = 
                 
                   cos 
                    
                   
                       
                   
                    
                   
                     θ 
                     1 
                   
                    
                   
                       
                   
                    
                   cos 
                    
                   
                       
                   
                    
                   
                     θ 
                     2 
                   
                    
                   
                       
                   
                    
                   cos 
                    
                   
                       
                   
                    
                   
                     θ 
                     3 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               cos 
                
               
                 ( 
                 
                   3 
                    
                   
                     θ 
                     1 
                   
                 
                 ) 
               
             
             = 
             
               1 
               / 
               2 
             
           
         
       
       
         
           
             
               
                 
                   
                     3 
                      
                     
                       θ 
                       1 
                     
                   
                   = 
                     
                    
                   
                     arc 
                      
                     
                         
                     
                      
                     cosine 
                      
                     
                         
                     
                      
                     
                       1 
                       / 
                       2 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       60 
                        
                       ° 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     θ 
                     1 
                   
                   = 
                     
                    
                   
                     60 
                      
                     
                       ° 
                       / 
                       3 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     20 
                      
                     ° 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       1 
                     
                   
                   = 
                     
                    
                   
                     cos 
                      
                     
                         
                     
                      
                     20 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       0.93969262 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     θ 
                     2 
                   
                   = 
                     
                    
                   
                     
                       θ 
                       1 
                     
                     + 
                     
                       120 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       20 
                        
                       ° 
                     
                     + 
                     
                       120 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       140 
                        
                       ° 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       2 
                     
                   
                   = 
                     
                    
                   
                     cos 
                      
                     
                         
                     
                      
                     140 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         
                           0 
                           . 
                           7 
                         
                       
                        
                       66044443 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       θ 
                       3 
                     
                     = 
                       
                      
                     
                       
                         θ 
                         1 
                       
                       + 
                       
                         240 
                          
                         ° 
                       
                     
                   
                   ; 
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       20 
                        
                       ° 
                     
                     + 
                     
                       240 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     260 
                      
                     ° 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                   = 
                     
                    
                   
                     cos 
                      
                     
                         
                     
                      
                     260 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         
                           0 
                           . 
                           1 
                         
                       
                        
                       73648177 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                   and 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     
                       
                         ( 
                         
                           3 
                            
                           
                             θ 
                             1 
                           
                         
                         ) 
                       
                       / 
                       4 
                     
                   
                   = 
                     
                    
                   
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       1 
                     
                      
                     
                         
                     
                      
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       2 
                     
                      
                     
                         
                     
                      
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         cos 
                          
                         
                             
                         
                          
                         
                           θ 
                           
                             1 
                              
                             
                                 
                             
                           
                         
                          
                         cos 
                          
                         
                             
                         
                          
                         
                           θ 
                           2 
                         
                       
                       ) 
                     
                      
                     cos 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           - 
                           
                             0 
                             . 
                             7 
                           
                         
                          
                         1984631 
                          
                         
                             
                         
                          
                         … 
                       
                        
                       
                           
                       
                       ) 
                     
                      
                     
                       ( 
                       
                         
                           - 
                           
                             0 
                             . 
                             1 
                           
                         
                          
                         73648177 
                          
                         
                             
                         
                          
                         … 
                       
                        
                       
                           
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       1 
                       / 
                       8 
                     
                   
                   ; 
                 
               
             
           
         
       
     
     when a rational number of magnitude 0, −¾, or −3 is to be further characterized, then the fourth, fifth, seventh, eighth, and ninth of such nine equations would apply; whereby those that pertain to the sin θ are validated for the particular case when a value of 34.3° becomes assigned to θ 1  as follows: 
     
       
         
           
             
               
                 
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       1 
                     
                   
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     34.3 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       0.563526048 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     θ 
                     2 
                   
                   = 
                     
                    
                   
                     
                       θ 
                       1 
                     
                     + 
                     
                       120 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       34.3 
                        
                       ° 
                     
                     + 
                     
                       120 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       154.3 
                        
                       ° 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       2 
                     
                   
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     154.3 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       0.433659084 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       θ 
                       3 
                     
                     = 
                       
                      
                     
                       
                         θ 
                         1 
                       
                       + 
                       
                         240 
                          
                         ° 
                       
                     
                   
                   ; 
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       34.3 
                        
                       ° 
                     
                     + 
                     
                       240 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       274.3 
                        
                       ° 
                     
                   
                   ; 
                   and 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     274.3 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         
                           0 
                           . 
                           9 
                         
                       
                        
                       97185133 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   0 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         1 
                       
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         2 
                       
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         3 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       0.563526048 
                        
                       
                           
                       
                        
                       … 
                     
                     + 
                     
                       0.4336590 
                        
                       
                           
                       
                        
                       … 
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         3 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       0.997185133 
                        
                       
                           
                       
                        
                       … 
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         3 
                       
                     
                   
                 
               
             
             
               
                 
                   
                     
                       = 
                         
                        
                       0.997185133 
                     
                     ) 
                   
                   - 
                   
                     0.997185133 
                      
                     
                         
                     
                      
                     … 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     0 
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       
                         
                           - 
                           3 
                         
                         / 
                         4 
                       
                       = 
                         
                        
                       
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             2 
                           
                         
                         + 
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             3 
                           
                         
                         + 
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             2 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             3 
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                        
                       
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             2 
                           
                         
                         + 
                         
                           
                             ( 
                             
                               sin 
                                
                               
                                   
                               
                                
                               
                                 θ 
                                 1 
                               
                                
                               sin 
                                
                               
                                   
                               
                                
                               
                                 θ 
                                 2 
                               
                             
                             ) 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             3 
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                        
                       
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             2 
                           
                         
                         + 
                         
                           
                             ( 
                             
                               
                                 - 
                                 sin 
                               
                                
                               
                                   
                               
                                
                               
                                 θ 
                                 3 
                               
                             
                             ) 
                           
                            
                           
                               
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             3 
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                        
                       
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             2 
                           
                         
                         - 
                         
                           
                             sin 
                             2 
                           
                            
                           
                               
                           
                            
                           
                             θ 
                             3 
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                        
                       
                         
                           0.24437819 
                            
                           
                               
                           
                            
                           … 
                         
                         - 
                         
                           
                             ( 
                             
                               
                                 - 
                                 
                                   0 
                                   . 
                                   9 
                                 
                               
                                
                               97185133 
                                
                               
                                   
                               
                                
                               … 
                             
                              
                             
                                 
                             
                             ) 
                           
                           2 
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                        
                       
                         
                           0.24437819 
                            
                           
                               
                           
                            
                           … 
                         
                         - 
                         
                           
                             ( 
                             0.99437819 
                              
                             
                                 
                             
                             ) 
                           
                            
                           
                               
                           
                            
                           … 
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                        
                       
                         
                           0.24437819 
                            
                           
                               
                           
                            
                           … 
                         
                         - 
                         
                           ( 
                           
                             
                               
                                 0 
                                 . 
                                 2 
                               
                                
                               4 
                                
                               4 
                                
                               3 
                                
                               7 
                                
                               8 
                                
                               1 
                                
                               9 
                             
                             + 
                             
                               3 
                               / 
                               4 
                             
                           
                           ) 
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         = 
                           
                          
                         
                           
                             - 
                             3 
                           
                           / 
                           4 
                         
                       
                        
                       
                           
                       
                       ; 
                       
                           
                       
                        
                       
                         and 
                          
                         
                             
                         
                          
                         as 
                          
                         
                             
                         
                          
                         a 
                          
                         
                             
                         
                          
                         check 
                       
                     
                   
                 
               
                
               
                 
 
               
               - 
               
                 sin 
                  
                 
                     
                 
                  
                 
                   
                     ( 
                     
                       3 
                        
                       
                         θ 
                         1 
                       
                     
                     ) 
                   
                   / 
                   4 
                 
               
             
             = 
             
               sin 
                
               
                   
               
                
               
                 θ 
                 1 
               
                
               
                   
               
                
               sin 
                
               
                   
               
                
               
                 θ 
                 2 
               
                
               
                   
               
                
               sin 
                
               
                   
               
                
               
                 θ 
                 3 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         3 
                          
                         
                           θ 
                           1 
                         
                       
                       ) 
                     
                   
                   = 
                     
                    
                   
                     
                       - 
                       4 
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       
                         1 
                          
                         
                             
                         
                       
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       2 
                     
                      
                     
                         
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       + 
                       
                         0 
                         . 
                         9 
                       
                     
                      
                     74761194 
                      
                     
                         
                     
                      
                     … 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     3 
                      
                     
                       θ 
                       1 
                     
                   
                   = 
                     
                    
                   
                     arc 
                      
                     
                         
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         0.974761194 
                          
                         
                             
                         
                          
                         … 
                       
                        
                       
                           
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       77.1 
                        
                       ° 
                     
                   
                   , 
                   
                     or 
                      
                     
                         
                     
                      
                     102.9 
                      
                     ° 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       θ 
                       1 
                     
                     = 
                       
                      
                     
                       77.1 
                        
                       
                         ° 
                         / 
                         3 
                       
                        
                       
                           
                       
                        
                       or 
                     
                   
                   , 
                   
                     102.9 
                      
                     
                       ° 
                       / 
                       3 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       25.7 
                        
                       ° 
                     
                   
                   , 
                   
                     
                       or 
                        
                       
                           
                       
                        
                       34.3 
                        
                       ° 
                     
                     ; 
                   
                 
               
             
           
         
       
     
     and 
     lastly, an example is afforded for the particular case when a quadratic irrational number is to be further characterized, such that when: 
     
       
         
           
             
               
                 tan 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     3 
                      
                     
                       θ 
                       1 
                     
                   
                   ) 
                 
               
               = 
               
                 - 
                 
                   3 
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 3 
                  
                 
                   θ 
                   1 
                 
               
               = 
               
                 arctan 
                  
                 
                   ( 
                   
                     - 
                     
                       3 
                     
                   
                   ) 
                 
               
             
             ; 
           
         
       
       
         
           
             
               3 
                
               
                 θ 
                 1 
               
             
             = 
             
               
                 - 
                 60 
               
                
               ° 
             
           
         
       
       
         
           
             
               θ 
               1 
             
             = 
             
               
                 - 
                 20 
               
                
               ° 
             
           
         
       
       
         
           
             
               
                 
                   
                     tan 
                      
                     
                         
                     
                      
                     
                       θ 
                       1 
                     
                   
                   = 
                     
                    
                   
                     tan 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         
                           - 
                           20 
                         
                          
                         ° 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         
                           0 
                           . 
                           3 
                         
                       
                        
                       63970234 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     θ 
                     2 
                   
                   = 
                     
                    
                   
                     
                       θ 
                       1 
                     
                     + 
                     
                       120 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         - 
                         20 
                       
                        
                       ° 
                     
                     + 
                     
                       120 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       100 
                        
                       ° 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     tan 
                      
                     
                         
                     
                      
                     
                       θ 
                       2 
                     
                   
                   = 
                     
                    
                   
                     tan 
                      
                     
                         
                     
                      
                     100 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         
                           5 
                           . 
                           6 
                         
                       
                        
                       7128182 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       θ 
                       3 
                     
                     = 
                       
                      
                     
                       
                         θ 
                         1 
                       
                       + 
                       
                         240 
                          
                         ° 
                       
                     
                   
                   ; 
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         - 
                         20 
                       
                        
                       ° 
                     
                     + 
                     
                       240 
                        
                       ° 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     220 
                      
                     ° 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     tan 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                   = 
                     
                    
                   
                     tan 
                      
                     
                         
                     
                      
                     220 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       0.839099631 
                        
                       
                           
                       
                        
                       … 
                     
                   
                    
                   
                       
                   
                   ; 
                   and 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       - 
                       tan 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         3 
                          
                         
                             
                         
                          
                         
                           θ 
                           1 
                         
                       
                       ) 
                     
                   
                   = 
                     
                    
                   
                     tan 
                      
                     
                         
                     
                      
                     
                       θ 
                       1 
                     
                      
                     
                         
                     
                      
                     tan 
                      
                     
                         
                     
                      
                     
                       θ 
                       2 
                     
                      
                     
                         
                     
                      
                     tan 
                      
                     
                         
                     
                      
                     
                       
                         θ 
                         3 
                       
                       . 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         tan 
                          
                         
                             
                         
                          
                         
                           θ 
                           1 
                         
                          
                         
                             
                         
                          
                         tan 
                          
                         
                             
                         
                          
                         
                           θ 
                           2 
                         
                       
                       ) 
                     
                      
                     
                         
                     
                      
                     tan 
                      
                     
                         
                     
                      
                     
                       θ 
                       3 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         2.0641777 
                          
                         
                             
                         
                          
                         … 
                       
                        
                       
                           
                       
                       ) 
                     
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         0.8390996 
                          
                         
                             
                         
                          
                         … 
                       
                        
                       
                           
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     1.732050808 
                      
                     
                         
                     
                      
                     … 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       3 
                     
                   
                   ; 
                 
               
             
           
         
       
     
     and 
     conversely, whenever trigonometric values of triads θ 1 , θ 2 , and θ 3  become afforded as given quantities, geometric construction patterns can be approximated which are analogous to the above equations. For example, a unit circle can be drawn which exhibits three radii that emanate from its origin describing angles of θ°, (θ+120°), and (θ+240°) with respect to its x-axis and terminate upon its circumference. Accordingly, from the equation below, the sum of their three ordinate values always must be equal to zero, verified algebraically as follows: 
     
       
         
           
             
               
                 
                   0 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         1 
                       
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         2 
                       
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         3 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         1 
                       
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           
                             θ 
                             1 
                           
                           + 
                           
                             1 
                              
                             2 
                              
                             0 
                           
                         
                         ) 
                       
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           
                             θ 
                             1 
                           
                           + 
                           
                             2 
                              
                             4 
                              
                             0 
                           
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                         1 
                       
                     
                     + 
                     
                       ( 
                       
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           
                               
                           
                            
                           cos 
                            
                           
                               
                           
                            
                           120 
                         
                         + 
                         
                           cos 
                            
                           
                               
                           
                            
                           
                             θ 
                             1 
                           
                            
                           sin 
                            
                           
                               
                           
                            
                           120 
                         
                       
                       ) 
                     
                     + 
                   
                 
               
             
             
               
                 
                     
                    
                   
                     ( 
                     
                       
                         sin 
                          
                         
                             
                         
                          
                         
                           θ 
                           1 
                         
                          
                         
                             
                         
                          
                         cos 
                          
                         
                             
                         
                          
                         240 
                       
                       + 
                       
                         cos 
                          
                         
                             
                         
                          
                         
                           θ 
                           1 
                         
                          
                         
                             
                         
                          
                         sin 
                          
                         
                             
                         
                          
                         240 
                       
                     
                     ) 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       θ 
                     
                     + 
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           
                             
                               - 
                               1 
                             
                             / 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       cos 
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           
                             
                               3 
                             
                             / 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                   
                 
               
             
             
               
                 
                     
                    
                   
                     
                       
                         sin 
                          
                         
                             
                         
                          
                         
                           θ 
                            
                           
                             ( 
                             
                               
                                 - 
                                 1 
                               
                               / 
                               2 
                             
                             ) 
                           
                         
                       
                       + 
                       
                         cos 
                          
                         
                             
                         
                          
                         
                           θ 
                            
                           
                             ( 
                             
                               
                                 3 
                               
                               / 
                               2 
                             
                             ) 
                           
                         
                       
                     
                     ; 
                     
                       such 
                        
                       
                           
                       
                        
                       that 
                     
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   0 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           
                             1 
                             - 
                             
                               1 
                               / 
                               2 
                             
                             - 
                             
                               1 
                               / 
                               2 
                             
                           
                           ) 
                         
                       
                     
                     + 
                     
                       cos 
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           
                             
                               
                                 3 
                               
                               / 
                               2 
                             
                             - 
                             
                               
                                 3 
                               
                               / 
                               2 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       sin 
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           0 
                           ) 
                         
                       
                     
                     + 
                     
                       cos 
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           0 
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   0. 
                 
               
             
           
         
       
     
     Before even trying to solve the classical problem of the trisection of an angle, either the designated magnitude of an angle which is intended to be trisected or some geometric construction pattern which fully describes it first needs to be furnished! 
     To the contrary, if such information instead were to be withheld, then the exact size of an angle which is intended to be trisected would not be known; thereby making it virtually impossible to fulfill the task of dividing into three equal parts. 
     In effect, such provision of an a priori condition performs the very important role of identifying exactly which classical problem of the trisection of an angle is to be solved out of a virtually infinite number of possible forms it otherwise could assume depending upon which designated magnitude comes under scrutiny! 
     For example, attempting to trisect a sixty degree angle solely by conventional Euclidean means poses an entirely different problem than trying to trisect a seventeen degree angle by means of applying the very same process. 
     From an entirely different point of view, whenever a motion related solution for the problem of the trisection of an angle becomes portrayed, it signifies that an actual event has taken place. Such is the case because some period of time must elapse in order to reposition a trisecting emulation mechanism to a designated setting. 
     If this were not the case, specifically meaning that an element of time would not be needed in order to effect trisection, then a motion related solution for the problem of the trisection of an angle thereby could not occur; simply because without time, there can be no motion! 
     In support of such straightforward line of reasoning, however, it surprisingly turns out that a trisecting emulation mechanism furthermore can portray a stationary solution for the problem of the trisection of an angle, as well; not as an event, but by sheer coincidence; meaning that such portrayed solution materializes before time can expire! 
     The only way this could occur is by having such solution be portrayed before an a priori condition becomes specified; thereby suggesting that such solution becomes posed even before defining the full extent problem which it already has solved. 
     Essentially, such stationary solution for the problem of the trisection of an angle consists of a condition in which the designated magnitude of an angle which is intended to be trisected just so happens to match the particular reading that a trisecting emulation mechanism turns out to be prematurely set to before such activity even commences. 
     The only problem with such stationary solution scenario is that its probability of occurrence approaches zero; thereby negating its practical application. Such determination is computed as such singular reading selection divided by the number all possible readings which such device could be set to, generally comprised of a virtually unlimited number of distinct possibilities, and thereby amounting to a ratio which equates to 1/∞→0. 
     The input box appearing in such  FIG. 2  flowchart, entitled PROBABILISTIC PROOF OF MATHEMATIC LIMITATION  10 , refers to the specific results which can be obtained by realizing that a given angle within a geometric construction pattern furthermore must serve the dual role of also being a trisector for any rendered angle therein whose magnitude amounts to exactly three times its size; thereby signifying that a trisection event successfully has been performed solely by conventional Euclidean means! 
     Unfortunately although posing a legitimate solution for the classical problem of the trisection of an angle, such rather elementary approach also proves to be entirely impractical; simply because there is no way of assuring that such generated rendered angle matches the designated magnitude of an angle which is intended to be trisected; as had to be specified as an a priori condition even before attempting to generate such solution! 
     Since such a priori condition might have specified any of an infinite number of possible designated magnitudes, the probability of such geometric construction activity proving successful approaches zero, as again calculated by the ratio 1/∞→0. 
     Therefore, the practicality of actually attempting to solve such classical problem of the trisection of an angle solely by conventional Euclidean means now easily can be evaluated; whereby any singular geometric construction pattern which could be generated in such manner that the magnitude of its rendered angle amounts to exactly three times the size of a given angle, as well as turns out to be equal to a designated magnitude which previously was identified, because it bears a probability that approaches zero percent of posing a legitimate solution for such classical trisection problem, pretty much should be considered to be an impossible avenue for obtaining such solution! 
     Another interpretation is that an angle could be divided into three equal parts by means of applying only a straightedge and compass to it, but only under the highly unusual condition that an unlimited number of opportunities become extended, thereby assuring success. Unfortunately, such alternate approach also should be viewed to be quite unacceptable because it would take forever to complete. 
     To follow through with such discussion, it should be mentioned, however, that an approach to solve such classical problem of the trisection of an angle in this very manner already was discovered. As copyrighted in chapter six of my never before published 1976 treatise entitled,  Trisection, an Exact Solution , as filed under copyright registration number TXu 636-519, such infinite point solution can trisect in a precise manner by means of performing a multitude of consecutive angular bisections, all geometrically constructed upon just a single piece of paper. Since such solution was authored more than forty years ago, it is included herein for purposes of being shared with the general public for the very first time, but only after formally introducing the four embodiments of such newly proposed articulating trisection invention first. 
     In  FIG. 2 , notice that all five of such parallelogram shaped listed input boxes, along with the non-iterative, or YES output portion of such DEFICIENCY MITIGATED  5  decision box, all funnel into a downstream process box which is entitled GEOMETRIC FORMING PROCESS DEVISED  11 . 
     Within such flowchart, although such process box is limited basically to trisection matters, a geometric forming process nevertheless is indicative of a whole gamut of improved drawing pretexts, besides that Euclidean formulations, which could be developed in order to chart certain other distinct motions which lie outside of its presently discussed purview, or very narrow scope which hereinafter is to be addressed in this presentation. Accordingly, it is important to note that such overall process, at some future date, furthermore could prove to be the source of countless other discoveries which would require either a motion related geometric substantiation, and/or an analogous higher order algebraic solution; thereby evidencing the enormity of a geometric forming capability with regard to its profound influence upon other forms of mathematics. 
     In 1893, Thomas Alva Edison at long last showcased his kinetoscope. Obviously, such discovery spurred on the development of a cinematic projector by the Lumiere brothers shortly afterwards. Unfortunately, many instances can be cited in human history in which follow-up inventions of far larger practical importance succeed earlier landmark cases. Ironically, such type of mishap befell Edison on another occasion, as well; being when he developed a direct current capability which thereafter became improved upon by Tesla during such time that he introduced alternating current. Accordingly, one fitting way to suitably address such above described disparity would be to unequivocally state that due to a series of ongoing technical developments, an entire motion picture industry eventually became ushered in; whereby a great fanfare finally arose, as caused by a rather unsuspecting audience who became more and more accustomed to witnessing the actual footages of world events at the cinema, as opposed to just reading about them in the newspapers. Over time, the general public began to welcome viewing news in a more fashionable setting. In retrospect, Kempe&#39;s attempts to disclose how to articulate an anti-parallelogram linkage assembly for the express purpose of performing trisection most certainly appeared to receive far less critical attention. Whether or not there existed a large interest in such subject matter is hard to fathom, for just consider: A full fifteen years prior to Lumiere&#39;s actual cinematic projector debut, dating back all the way to the late 1880&#39;s, it obviously would have been very difficult, if not impossible, to reveal in sufficient detail to any awaiting crowd, and that much less to one that might have been gathered some distance way, just how to articulate an anti-parallelogram linkage device in order to satisfactorily perform trisection. Moreover, consider: Had a presentation to this effect successfully been pulled off at that very time period, it more fittingly might have been mistaken for some sort of magic act! Be that as it may, had there also been a considerable demand levied beforehand, for example by some predisposed mathematics party who might have expressed an interest in viewing such purported trisection capability, it evidently would have had very little effect in the overall scheme of things. As it were, way back in the 1880&#39;s, with such industrial community seriously lagging behind in development, as least in comparison to what actually had become accomplished just ten to fifteen years later, fewer news organizations would have been available to disseminate important technical information of that kind. In sharp contrast, only rather recently has it truly become possible to pictorially describe just how a Kempe anti-parallelogram trisection device actually functions. In today&#39;s technology, a presentation very easily could be made, merely by means of simulating the relative movement of such Kempe anti-parallelogram device within a modem day computer. However, without being predisposed to such type of information, or even to a lesser extent, thoroughly apprised of such professed trisection capabilities, it most certainly would be very difficult, indeed, to foresee that the overall technique used to create the very illusion of motion all those prior years, merely by means of animating some ragtag assortment of pictures, or possibly even some collection of photographs whose overall shapes would have been known to differ imperceptibly from one to the next, furthermore could have been applied to replicate an observed motion by means of instead animating an entire family of related geometric construction patterns! Hover, had such association truly been made those many years before, it well might have contributed to substantiating that some articulating prior art mechanism actually could perform trisection effectively throughout its wide range of device settings. 
     Another possible reason for such noticeable omission could be a reticence, or complacency stemming from the fact that, not only had conventional Euclidean practice proved entirely satisfactory for use on most prior occasions, but moreover that, up until now, generating a singular drawing pattern was the preferred way to pictorially display various aspects of mathematics. 
     Unfortunately, as it just so happens to turn out, one of the very few instances in which a singular conventional Euclidean practice approach should not be applied, just for the very reasons expressed above, is when attempting to provide the solution for the classical problem of the trisection of an angle! 
     As such, it might well be that a recommendation never before was raised, thereby proposing to extend conventional Euclidean practice into a geometric forming process that is fully capable of describing certain motions, simply because such aforementioned complacency very well by now actually might have escalated into a full blown reluctance on the part of a seemingly silent majority of mathematical authoritarians to overcome the crippling Euclidean limitation of not being able to backtrack upon irreversible geometric construction patterns! 
     With regard to the particular damage levied upon trisection matters over the years by not otherwise adopting a formal geometric forming process, consider the very first English language trisection involvement, tracing all the way back to a particular drawing which appears on page 309 of such 1897  The Works of Archimedes . Inasmuch as such drawing is accompanied by a complete accounting of such previously referred to Archimedes proposition, as well as a suitable algebraic proof needed to substantiate it, the apparent problem is that such drawing only is a singular geometric construction pattern, thereby applying only to the specific chord length which appears within its depicted circle. In order for such drawing depiction to be fully consistent with such Archimedes proposition and supporting algebraic proof, it should be represented by an entire Euclidean formulation, replete with an infinite number of other chord lengths which furthermore could be described within such circle, and which such Archimedes proposition and supporting algebraic proof also apply to. Without such incorporation, such drawing remains quite adequate for substantiating the arbitrarily selected chord pattern which is illustrated therein, but nonetheless remains grossly impractical because it cannot represent such infinite number of other chord shapes and attendant sizes with its circle, and thereby also remain subject to the very requirements posed by such included proposition. Whereas such drawing evidently was presented as a convention of the time, it must be presumed that it was provided merely as an example of all of the other possible geometric construction patterns which also could have been drawn while still satisfying all of the requirements of such proposition. Unfortunately, the key element that never was stated therein is that all of such other possible geometric construction patterns furthermore must stem from the very same sequence of Euclidean operations that governs such singular drawing, as is represented therein. 
     Based upon such prior trisection rationale discussion, it becomes apparent that a singular geometric construction pattern can depict only one event which takes place during an entire articulation process, thereby representing only a momentary viewing which neither can provide an indication of where a particular motion might have originated from, nor where it might have ended up. 
     Accordingly, such singular drawing format remains somewhat deficient from the standpoint that it cannot even define all of the various geometries needed to characterize an entire articulated motion! 
     As such, a singular geometric construction pattern can be likened to a still photograph. Whereas the latter gave birth to the motion picture industry, it seems only appropriate that the former should serve as the basis for an improved geometric approach that becomes capable of characterizing motion! 
     Such newly proposed geometric forming process capitalizes upon the novel prospect that it requires an entire family of geometric construction patterns to adequately represent all of the unique shapes needed to represent a complete articulation event. 
     Accordingly, Euclidean formulations can be of service in motion related problems which cannot be fully interpreted by a singular geometric construction pattern. 
     With particular regard to trisection matters, the magnitude of at least one rendered angle exhibited within any constituent geometric construction pattern that belongs to a substantiating Euclidean formulation, quite obviously would need to amount to exactly three times the size of its given angle. 
     Hence, by means of verifying that its outline matches the overall shape of a corresponding regenerated static image that becomes automatically portrayed once a trisecting emulation mechanism becomes properly set, its smaller static image portion thereby could be substantiated to qualify as an associated trisector for such device setting. 
     As such, a Euclidean formulation, recognizable by its double arrow notation, could dramatically simplify the overall process needed to substantiate that some proposed invention has been designed so that it can perform trisection accurately over a wide range of device settings and, in so doing, thereby become referred to as a bona fide trisecting emulation mechanism; as duly is depicted in the lower right hand portion of such  FIG. 2  flowchart. 
     Hence, applying such novel geometric forming process in this respect thereby validates that overlapment points, normally considered to be detrimental because they remain inconspicuous, can be supplanted with intersection points that become fully distinguishable as regenerated static images become automatically portrayed by means of properly setting trisecting emulating mechanisms 
     In closing, it should be mentioned that when imposing a controlled motion, it becomes possible to discern overlapment points; whereby such Euclidean limitation of otherwise not being able to distinguish them by means of backtracking exclusively from a rendered angle within an irreversible geometric construction becomes rectified! 
     Recapping, an overall explanation just has been afforded for the very first time which maintains that a discernment of overlapment points leads to trisection. Hence, it couldn&#39;t possibly have been referred to in any prior art. 
     Moreover, since such explanation alone accounts for how a motion related solution for the problem of the trisection of an angle can be portrayed, prior art couldn&#39;t possibly have rendered a differing substantiation that actually accounts for such professed capabilities. 
     Any further discussion concerning specific amounts of time which may be needed to arrange trisecting emulation mechanisms to particular device settings are omitted herein because such input is irrelevant when attempting to substantiate a motion related solution for the problem of the trisection of an angle; especially when considering that such times obviously would vary depending upon a user&#39;s dexterity, as well as the varying distances encountered when going from where such device might be temporarily positioned to a particular device setting. 
     In conclusion, if the logic proposed in such  FIG. 3  Trisection Mystery Iteration Processes Table turns out to be entirely true, meaning that an inability to solve the classical problem of the trisection of an angle results because it impossible to backtrack upon any irreversible geometric construction pattern, then it would be utterly senseless to attempt drawing any type of singular geometric construction pattern whatsoever, solely by conventional Euclidean means, in an effort to achieve such ends! 
     Moreover, when considering that it is necessary to exert a motion in order to properly set any trisecting emulation mechanism, such warranted flexure could not, in any way, be fully described solely by a singular geometric construction pattern! 
     The process box entitled CLASSICAL PROBLEM OF THE TRISECTION OF AN ANGLE SOLUTION DISCREDITED  12  is to serve as the principal focal point within such flowchart, as represented in  FIG. 2 , where two distinct, independent Euclidean trisection approaches are to be discredited. Below, it should become rather obvious that such second listed approach is an entire reversal of the first: 
     not only is it impossible to fully backtrack upon any rendered angle within a geometric construction pattern whose magnitude amounts to exactly three times the size of its given angle, thereby explaining why the classical problem of the trisection of an angle cannot be solved; but 
     conversely, the probability that the magnitude of a rendered angle matches the designated magnitude of an angle that is intended to be trisected approaches zero whenever such rendered angle becomes geometrically constructed such that its magnitude amounts to exactly three times the arbitrarily selected size of a given angle. 
     The fact that a duration of time is needed in order to effect a motion related solution for the problem of the trisection of an angle eliminates the possibility that such form of solution potentially might double as a solution for the classical problem of the trisection of an angle. This is because any geometric construction pattern, once drawn, cannot be modified just by applying time to it; thereby affording a probability that still approaches zero that its overall outline just might happen to superimpose upon that which otherwise could be automatically portrayed whenever a static image becomes regenerated by means of properly setting any trisecting emulation mechanism. 
     Moreover, when investigating whether a geometric solution furthermore might qualify as a solution for the classical problem of the trisection of an angle, it should be remembered that if extraneous information were to become introduced into such problem that turns out to be relevant to determining its solution, then only a solution for some corrupted version of the classical problem of the trisection of an angle could be obtained; thereby solving an entirely different problem and, in so doing, discrediting any potential claims that might incorrectly allege that the classical problem of the trisection of an angle has been solved. 
     Lastly, for those remaining skeptics who otherwise would prefer to believe that a solution for the classical problem of the trisection of an angle might yet be specified, all they need to do is disprove that an availability of overlapment points actually prevents backtracking upon a rendered angle within any geometric construction pattern all the way back to a given angle whose magnitude amounts to exactly one-third of its size! 
     In other words, to dispute the new theory that is presented herein, it is now up to them to identify some as yet unidentified geometric construction pattern which would enable an angle of virtually any designated magnitude they might decide upon to be trisected; when neither violating the rules which pertain to conventional Euclidean practice, not introducing any extraneous information which could be considered to be relevant to its solution! 
     Over time, as such ascribed overlapment attribution finally becomes acknowledged to be the real cause for being unable to solve the classical problem of the trisection of an angle, ongoing analysis thereby could be performed in order to confirm, beyond any shadow of doubt, that trisection of an angle of any magnitude cannot be performed solely by means of applying only a straightedge and compass to it! 
     The process box entitled SINGULAR DRAWING SOLUTION DISPELLED  13  is included in such  FIG. 2  flowchart to address the fact that although a singular drawing solution can be described for any regenerated static image that automatically becomes portrayed once a trisecting emulation mechanism becomes properly set, designing a device of that type which has only one discrete setting would be entirely impractical! 
     Conversely, any proposed articulating trisection invention that only specifies a singular motion related solution for the trisection of angle couldn&#39;t possibly substantiate a trisection capability for its remaining wide range of settings! 
     The process box described as SUPPLEMENTAL DEVICE CAPABILITIES SPECIFIED  14  is the principal location in such  FIG. 2  flowchart where information pertaining to such MATHEMATICS DEMARCATION  8  input box contributes to an understanding that trisecting emulation mechanisms additionally have the affinity to portray exact lengths that only could be approximated when performing geometric construction upon a given length of unity. 
     Such fact is duly reflected in such  FIG. 11  Mathematics Demarcation Chart wherein cubic irrational real number types appear only in its third column, as headed by the geometric forming process cell; thereby indicating that deliberate motions must be imparted in order portray them. They can appear either as the ratios of portrayed lengths with respect to a given length of unity, or as trigonometric properties inherent within trisecting angles which become portrayed during certain trisection events. 
     For the particular case of the fourth embodiment of such newly proposed invention, a supplemental device leveling capability also is to be thoroughly described. 
     Within a right triangle, if the ratio between the length of one of its sides to that of its hypotenuse is cubic irrational, so must be the other. In other words, if one trigonometric property of a right triangle is cubic irrational, so must be all of its trigonometric properties! 
     It then logically would follow that for any right triangle that exhibits cubic irrational trigonometric properties whose hypotenuse amounts to one unit in length, the lengths of its constituent sides each would have to be of a cubic irrational value. 
     Such association enables the lengths of the sides of such right triangle to compensate for each other. With regard to the Pythagorean Theorem, this means that only the sum of the squares of two cubic irrational values can equal a value of one; thereby avoiding the common pitfall of otherwise attempting to equate such rational unitary value to the square of a cubic irrational value added to the square of either a rational or quadratic irrational value! 
     The reason that a right triangle which exhibits cubic irrational trigonometric properties truly can be geometrically constructed is because of the large number of geometric construction patterns which exist, all meeting such criteria; whereby the probability of drawing just one of them out of sheer coincidence increases dramatically. 
     Attempting to reproduce any one of them just be conventional Euclidean means, however, nevertheless would prove fruitless, resulting only in a mere approximation thereof; one which might prove suitable when being considered as a duplicate rendering, but not when taking into account differences between them which possibly only would become discernable well beyond what the capabilities of the human eye could detect. 
     By finally acknowledging that angles which exhibit cubic irrational trigonometric properties actually can be portrayed, their exact measurements would become revealed for the very first time, despite the fact that their real values can be described only by decimal patterns that are never-ending. Perhaps such new found capability very well might become perceived as an uncharted gateway that unfortunately was overlooked time and time again in the past! 
       FIG. 1B  presents a very good example of such capability to portray angles which exhibit cubic irrational trigonometric properties. Therein, angle QPS amounts to exactly sixty degrees. Such sixty degree angle QPS was chosen because, as stated earlier, its magnitude can be represented exactly by any of the included angles within an equilateral triangle, and thereby can be drawn solely by conventional Euclidean means. 
     Its associated trisector ∠NMP=∠QMP=∠RMP must be equal to exactly one-third of its size, amounting to a value which computes to 60°/3=20°. 
     Upon interpreting  FIG. 1B  to be representative of a famous as a marked ruler arrangement, angle NMP would be twenty degrees in magnitude. 
     Moreover consider that the notch appearing in its ruler resides away from its endpoint, M, one unit of measurement. 
     In isosceles triangle NMP, since length MN=length NP=1, it logically follows that twice the cosine of angle NMP would amount to the ratio between length MP length MN, whereby the following relationship thereby could be obtained: 
     
       
         
           
             
               
                 MP 
                 _ 
               
               / 
               
                 MN 
                 _ 
               
             
             = 
             
               2 
                
               
                   
               
                
               
                 cos 
                  
                 
                   ( 
                   
                     ∠ 
                      
                     
                         
                     
                      
                     NMP 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   MP 
                   _ 
                 
                 / 
                 1 
               
               = 
               
                 2 
                  
                 
                     
                 
                  
                 cos 
                  
                 
                     
                 
                  
                 20 
                  
                 ° 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 
                   
                     MP 
                     _ 
                   
                   = 
                     
                    
                   
                     2 
                      
                     
                       ( 
                       
                         
                           0 
                           . 
                           9 
                         
                          
                         3969262 
                          
                         
                             
                         
                          
                         … 
                       
                        
                       
                           
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     1.879385242 
                      
                     
                         
                     
                      
                     … 
                   
                 
               
             
           
         
       
     
     Hence, a cubic irrational value 1.879385242 . . . must be the exact length of base MP of isosceles triangle NMP; whereby the three dots notated after such number indicates that such decimal pattern extends on indefinitely. 
     Since the cosine of twenty degrees furthermore is a transcendental, number, the above procedure also could distinguish such number types, thereby constituting a subset of cubic irrational numbers. 
     Once having devised a suitable geometric forming process, it thereby becomes possible to verify that device candidates which wish to qualify as trisecting emulation mechanisms conform to the various elements which funnel into such process box. For example, all devices must be shown to be fully capable of performing the primary function of regenerating static images, or be bound by the same set of rules. Devices which meet such criteria, but thereafter are found to share common design traits, should be categorized as such in order to assure that each item appearing within any particular group features some fundamental performance difference which qualifies it as being individually unique. The TRISECTION INVENTIONS CATEGORIZED  15  process box represents the location within such  FIG. 2  flowchart where associations of this nature are to be carried out. 
     The process box therein entitled REQUIREMENTS CHART PREPARED  16  is intended to distinguish that, although CATEGORY I and CATEGORY II prior art devices actually can perform trisection over a wide range of device settings, certain aspects of such capability never before were completely substantiated. The remainder of such  FIG. 2  flowchart, including the decision box entitled DESIGN REQUIREMENTS MET  18 , have been added for the express purpose of specifying that all of such posed requirements must be satisfied in order for a proposed design, as described by the process box entitled PROPOSED INVENTION DESIGN REFINEMENT  17 , to fully qualify as a trisection emulation mechanism, as itemized in the TRISECTING EMULATION MECHANISM SUBSTANTIATED  19  process box described therein. 
     In closing, a novel geometric forming process just has been proposed which suitably explains how to rectify a major Euclidean limitation, essentially consisting of an incapability to distinguish overlapment points; as achieved simply by means of imposing a controlled motion which makes it possible to discern them! 
     Although trisection today can be performed because of such identified motion related compensation, were such deleterious behavior otherwise to remain unchecked, then trisection, as sought after by countless futile attempts to solve the famous classical problem of the trisection of an angle still would remain a very illusive problem! 
     Accordingly it is concluded that a geometric forming process thereby eclipses a rather limited conventional Euclidean practice that has been in vogue for millennia! 
     Now that new definitions have been provided, and a resulting comprehensive methodology, as presented in  FIG. 2  has been suitably described, it is due time to account for exactly how a trisecting emulation mechanism operates. 
       FIG. 12  has been prepared just for this purpose. Such flowchart commences by means of supplying details to an input box, as entitled DESIGNATED ANGLE SPECIFIED  120  therein. Such specific activity consists merely of selecting the designated magnitude of an angle that is intended to become trisected. 
     The decision box entitled DEVICE NEEDS TO BE SPECIFICALLY ARRANGED  121  is where it is to be determined which particular embodiment is to be utilized to perform such anticipated trisection; whereby: 
     if either such first, second, or fourth embodiment were to be chosen, then the YES route would apply, thereby leading to a process box entitled DEVICE IS SPECIFICALLY ARRANGED  122  which is where such device is to be specifically arranged in accordance with applicable provisions; or 
     if such third embodiment were to be chosen, then the NO route would apply, thereby leading to a process box entitled, DEVICE IS SET  123 . 
     At this stage in the flowchart, such chosen device now should be properly set to a magnitude which matches the designated magnitude which first was specified. 
     The next process box entitled, STATIC IMAGE BECOMES REGENERATED  124  refers to the fact that by having properly set such device, a specific static image became regenerated, a particular portion of which assumed the overall outline of an actual trisector for such device setting; thereby automatically portraying a motion related solution for the problem of the trisection of an angle. 
     Activities which appear inside of the large square shaped dotted line are those which are to be performed exclusively by any trisecting emulation mechanism which might be placed into use, thereby being considered as properties that are intrinsic to it. 
     Outside of such trisecting emulation mechanism dotted box, the process box entitled, TRISECTOR AUTOMATICALLY PORTRAYED  125  is where such motion related solution for the trisection of an angle thereafter can be witnessed. 
     In connection with such input box entitled MATHEMATICS DEMARCATION  8 , as posed in  FIG. 2  herein, it previously was mentioned that a Euclidean formulation, each of whose constituent geometric construction patterns exhibits a rendered angle whose magnitude amounts to exactly three times the size of its given angle, is to become obtained by means of having the value of the sine of any of such rendered angles described by a length of 3 sin θ−4 sin 3  θ; thereby conforming to a famous function expressed as 3 sin θ−4 sin 3  θ=sin (3θ). 
     With regard to the very limited scope of trisection covered in this presentation, it should suffice to say that discussions below are to begin by significantly pointing out that the pretext of a Euclidean formulation just so happens to be conducive to physically describing various equations which have an infinite number of solutions! 
     Perhaps the most relevant of these, as specified below, assume the form of three very famous cubic expressions which address trisection by means of relating trigonometric properties of one angle of variable size to another whose magnitude always amounts to exactly three times its size: 
     
       
         
           
             
               
                 cos 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     3 
                      
                     θ 
                   
                   ) 
                 
               
               = 
               
                 
                   4 
                    
                   
                     cos 
                     3 
                   
                    
                   θ 
                 
                 - 
                 
                   3 
                    
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 sin 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     3 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
               
               = 
               
                 
                   3 
                    
                   
                       
                   
                    
                   sin 
                    
                   
                       
                   
                    
                   θ 
                 
                 - 
                 
                   4 
                    
                   
                       
                   
                    
                   
                     sin 
                     3 
                   
                    
                   
                       
                   
                    
                   θ 
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               tan 
                
               
                   
               
                
               
                 ( 
                 
                   3 
                    
                   
                       
                   
                    
                   θ 
                 
                 ) 
               
             
             = 
             
               
                 
                   
                     3 
                      
                     tan 
                      
                     
                         
                     
                      
                     θ 
                   
                   - 
                   
                     
                       tan 
                       3 
                     
                      
                     θ 
                   
                 
                 
                   1 
                   - 
                   
                     3 
                      
                     
                         
                     
                      
                     
                       tan 
                       2 
                     
                      
                     θ 
                   
                 
               
               . 
             
           
         
       
     
     Whenever the magnitude of an angle that is algebraically denoted to be of size 3θ becomes supplied as a given quantity in any of such three cubic expressions, then such algebraic relationship truly would typify trisection! 
     This is because, a corresponding magnitude of θ, being an exact trisector of such given 3θ value, then could be computed simply by means of dividing such given value by a factor of three; thereby enabling a determination of the constituent trigonometric properties, as specified above. 
     For example, for the particular condition when it is given that: 
     
       
         
           
             
               3 
                
               θ 
             
             = 
             
               75 
                
               ° 
             
           
         
       
       
         
           
             
               
                 
                   θ 
                   = 
                     
                    
                   
                     75 
                      
                     
                       ° 
                       / 
                       3 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     25 
                      
                     ° 
                   
                 
               
             
           
         
       
       
         
           
             
               cos 
                
               
                   
               
                
               θ 
             
             = 
             0.906307787 
           
         
       
       
         
           
             
               3 
                
               
                   
               
                
               cos 
                
               
                   
               
                
               θ 
             
             = 
             
               2.718 
                
               9 
                
               23361 
             
           
         
       
       
         
           
             
               
                 4 
                  
                 
                     
                 
                  
                 
                   cos 
                   3 
                 
                  
                 
                     
                 
                  
                 θ 
               
               = 
               
                 2.97 
                  
                 7 
                  
                 742406 
               
             
             ; 
           
         
       
       
         
           
             
               
                 
                   
                     cos 
                      
                     
                       ( 
                       
                         3 
                          
                         θ 
                       
                       ) 
                     
                   
                   = 
                     
                    
                   
                     
                       4 
                        
                       
                         cos 
                         3 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                     - 
                     
                       3 
                        
                       
                           
                       
                        
                       cos 
                        
                       
                           
                       
                        
                       θ 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       2.9777 
                        
                       4 
                        
                       2 
                        
                       406 
                     
                     - 
                     
                       2.7 
                        
                       1 
                        
                       8 
                        
                       9 
                        
                       2 
                        
                       3 
                        
                       361 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     0.258819095 
                   
                    
                   
                       
                   
                   ; 
                 
               
             
           
         
       
     
     and 
     As a check, 3θ=75°√ 
     Conversely, if an infinite number of magnitudes of θ were to become supplied as given values instead, each of such three algebraic relationships thereby could be suitably represented by means of developing a newly established Euclidean formulation that fully could distinguish it! 
     This is because all three of such above cited cubic expressions are continuous and their respective right-hand terms furthermore are geometrically constructible. 
     To aptly demonstrate this, a Euclidean formulation, as posed in  FIG. 13 , has been developed to suitably represent such famous cubic relationship sin (3θ)=3 sin θ−4 sin 3  θ; wherein any geometric construction pattern belonging to thereby would exhibit a discrete value of sin (3θ) for each and every selected real sin θ value existing within the range of −1 to +1. 
     The governing sequence of Euclidean operations for such new Euclidean formulation is specified as follows: 
     given angle VOO′ is geometrically constructed of an arbitrarily selected magnitude that algebraically is denoted as θ such that its side OO′ exhibits the same length as its side OV; 
     side OV is designated to be the x-axis; 
     a y-axis is drawn, hereinafter represented as a straight line which passes through vertex O of given angle VOO′ and lies perpendicular to such x-axis; 
     a UNIT CIRCLE ARC becomes geometrically constructed, hereinafter to be represented as a portion of the circumference of a circle drawn about center point O whose radius is set equal in length to OV, thereby enabling it to pass through points V and O′, both of which previously have been designated as respective termination points of angle VOO′; 
     point T thereafter becomes designated as the intersection between such UNIT CIRCLE ARC and such geometrically constructed y-axis; 
     a straight line which passes through point O is drawn at forty-five degree angle counterclockwise to such x-axis; 
     another straight line which passes through point O is drawn making a three-to-one slope with the +x-axis; 
     a horizontal straight line is drawn which passes through point O′ and thereby lies parallel to the x-axis; 
     the juncture between such horizontal straight line and the y-axis becomes designated as “sin θ”, thereby denoting its vertical distance above such x-axis; 
     a vertical straight line is drawn so that it remains parallel to the y-axis while passing through the intersection made between such forty-five degree straight line and such horizontal straight line; 
     the horizontal distance such vertical straight line resides to the right of such y-axis also thereby is to be designated as “sin θ” along such x-axis; 
     a second vertical straight line is drawn which passes through coordinate point V, thereby being tangent to such previously drawn UNIT CIRCLE ARC; 
     a slanted straight line is drawn which originates at point O and passes through the intersection point made between such second vertical straight line and such horizontal straight line; 
     the angle which such slanted straight line makes with the x-axis becomes designated as “ω”, not to be confused with angle VOO′ amounting to a slightly larger magnitude of θ; 
     a second horizontal straight line is draw which passes through the intersection point made between such slanted straight line and such vertical straight line; 
     the juncture of such second horizontal straight line with the y-axis becomes designated as “h 1 ”, thereby denoting its unknown vertical distance above point O; 
     a second slanted straight line is drawn which extends from point O to the intersection point made by such second horizontal straight line with such second vertical straight line; 
     the angle which such second slanted straight line makes with the x-axis thereafter becomes designated as “φ”; 
     a third horizontal straight line is drawn so that it passes through the intersection point made between such second slanted straight line and such vertical straight line; 
     the juncture of such third horizontal straight line with the y-axis becomes designated as “h 2 ”, thereby denoting its unknown vertical distance above point O; 
     a fourth horizontal straight line is drawn so that it passes through the intersection point made between such straight line which exhibits a 3:1 slope with respect to the x-axis and such vertical straight line; 
     the juncture which such fourth horizontal straight line makes with the y-axis becomes denoted as “3 sin θ”, thereby distinguishing its vertical distance above point O; 
     a fifth horizontal straight line is drawn at a distance directly below such fourth horizontal straight line which measures four times the height which such third horizontal straight line resides above such x-axis, algebraically denoted therein as “4h 2 ”; 
     the juncture which is made between such fifth horizontal straight line and the y-axis becomes designated as “sin (3θ)”, thereby denoting its vertical distance above point O; and 
     the intersection point of such fifth horizontal straight line with such UNIT CIRCLE ARC becomes designated as point U′. 
     The proof for such  FIG. 13  Euclidean formulation is provided below: 
     
       
         
           
             
               
                 
                   
                     
                       tan 
                        
                       
                           
                       
                        
                       ω 
                     
                     = 
                     
                       
                         
                           
                             h 
                             1 
                           
                           / 
                           sin 
                         
                          
                         
                             
                         
                          
                         θ 
                       
                       = 
                         
                        
                       
                         sin 
                          
                         
                             
                         
                          
                         
                           θ 
                           / 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   
                     
                       
                         h 
                         1 
                       
                       = 
                         
                        
                       
                         
                           sin 
                           2 
                         
                          
                         
                             
                         
                          
                         θ 
                       
                     
                     ; 
                   
                 
               
             
              
             
                 
               
                 
 
               
                
               
                 
                   
                     
                       
                         
                           
                             
                               tan 
                                
                               
                                   
                               
                                
                               ϕ 
                             
                             = 
                             
                               
                                 
                                   
                                     h 
                                     2 
                                   
                                   / 
                                   sin 
                                 
                                  
                                 
                                     
                                 
                                  
                                 θ 
                               
                               = 
                                 
                                
                               
                                 
                                   h 
                                   1 
                                 
                                 / 
                                 1 
                               
                             
                           
                         
                       
                       
                         
                           
                             = 
                               
                              
                             
                               
                                 sin 
                                 2 
                               
                                
                               
                                   
                               
                                
                               
                                 θ 
                                 / 
                                 1 
                               
                             
                           
                         
                       
                     
                      
                     
                       
 
                     
                      
                     
                       h 
                       2 
                     
                   
                   = 
                   
                     
                       
                         sin 
                         3 
                       
                        
                       
                           
                       
                        
                       θ 
                        
                       
                         
 
                       
                        
                       4 
                        
                       
                         h 
                         2 
                       
                     
                     = 
                     
                       4 
                        
                       
                           
                       
                        
                       
                         sin 
                         3 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                   
                 
                 ; 
                 
                   
 
                 
                  
                 
                   
                     
                       
                         
                           sin 
                            
                           
                               
                           
                            
                           
                             ( 
                             
                               3 
                                
                               θ 
                             
                             ) 
                           
                         
                         = 
                           
                          
                         
                           
                             3 
                              
                             
                                 
                             
                              
                             sin 
                              
                             
                                 
                             
                              
                             θ 
                           
                           - 
                           
                             4 
                              
                             
                                 
                             
                              
                             
                               sin 
                               3 
                             
                              
                             
                                 
                             
                              
                             θ 
                           
                         
                       
                     
                   
                   
                     
                       
                         
                           = 
                             
                            
                           
                             
                               3 
                                
                               sin 
                                
                               
                                   
                               
                                
                               θ 
                             
                             - 
                             
                               4 
                                
                               
                                 h 
                                 2 
                               
                             
                           
                         
                         ; 
                       
                     
                   
                 
               
             
           
         
       
     
     and 
     since point U′ lies upon such UNIT CIRCLE ARC and exhibits a sin (3θ) ordinate value, radius OU′ must reside at an angle of 3θ with respect to the x-axis. 
     Accordingly,  FIG. 13  distinguishes an entire family of geometric construction patterns, all generated from the very same sequence of Euclidean operations as stipulated above; with the only exception being that the respective magnitudes of given angle VOO′ becomes slightly altered each time a new geometric construction pattern becomes drawn. 
     Based upon a reasoning that such famous cubic relationship sin (3θ)=3 sin θ−4 sin 3  θ actually can be fully distinguished by an entire family of geometric construction patterns which together comprise such newly proposed Euclidean formulation, as posed in  FIG. 13 , it theoretically might become possible to devise yet another rather crude, or cumbersome, trisecting emulation mechanism which, due to a considerable increase in its number of overall working parts, obviously would be considered to lie far beyond the very scope of this presentation. In order to become feasible, however, a newly fashioned device of such type would have to be designed so that when it becomes articulated by means of rotating its axis U′ circumferentially about axis O in accordance with such double arrow notation as expressed in  FIG. 13 , such motion additionally could be replicated by means of animating the conglomeration of geometric construction patterns which belong to such Euclidean formulation in successive order. 
     In conclusion, any algebraic determination that can be made by means of relating like trigonometric properties that exist between one value and another that amounts to exactly three times its magnitude, as specified in such three cited famous cubic expressions, furthermore can be fully described by a geometric construction pattern which belongs to one of three Euclidean formulations which could be developed to characterize them. 
     For example, if a particular value of 1.119769515 radians were to be accorded to θ, then an algebraic determination could be made, as follows of 3θ, which furthermore fully could be described by a singular geometric pattern which belongs to such newly proposed Euclidean formulation, as posed in  FIG. 13 : 
     
       
         
           
             θ 
             = 
             
               
                 1 
                 . 
                 1 
               
                
               19769515 
                
               
                   
               
                
               radians 
             
           
         
       
       
         
           
             
               
                 sin 
                  
                 
                     
                 
                  
                 θ 
               
               = 
               0.9 
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         3 
                          
                         θ 
                       
                       ) 
                     
                   
                   = 
                     
                    
                   
                     
                       3 
                        
                       sin 
                        
                       
                           
                       
                        
                       θ 
                     
                     - 
                     
                       4 
                        
                       
                         sin 
                         3 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       3 
                        
                       
                         ( 
                         
                           0 
                            
                           .9 
                         
                         ) 
                       
                     
                     - 
                     
                       4 
                        
                       
                         
                           ( 
                           0.9 
                           ) 
                         
                         3 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     2.7 
                     - 
                     
                       4 
                        
                       
                         ( 
                         
                           
                             0 
                             . 
                             7 
                           
                            
                           2 
                            
                           9 
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     2.7 
                     - 
                     2.916 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         0 
                       
                        
                       .216 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     3 
                      
                     
                         
                     
                      
                     θ 
                   
                   = 
                     
                    
                   
                     π 
                     + 
                     
                       
                         0 
                         . 
                         2 
                       
                        
                       1 
                        
                       7 
                        
                       7 
                        
                       1 
                        
                       5 
                        
                       891 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     3.3593 
                      
                     0 
                      
                     8 
                      
                     545 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     3 
                      
                     
                       
                         ( 
                         
                           
                             1 
                             . 
                             1 
                           
                            
                           1 
                            
                           9 
                            
                           7 
                            
                           6 
                            
                           9515 
                         
                         ) 
                       
                       . 
                     
                   
                 
               
             
           
         
       
     
     Such above furnished overall detailed accounting explains exactly why all three of such previously cited famous cubic expressions remain incredibly important! 
     More particularly, this is because each of such three expressions can be considered to be a distinctive format type, in itself, one that furthermore can be broken down into an infinite number of unique relationships that have three cubic roots each. 
     Such scenario is far different than what transpires with respect to discontinuous functions, as are about to be discussed in detail next. 
     Also in connection with such input box entitled MATHEMATICS DEMARCATION  8 , as posed in  FIG. 2 , it previously was mentioned that a graph is to become developed that distinguishes between the continuity of such well known cubic function 4 cos 3  θ−3 cos θ=cos (3θ) and the discontinuity that very clearly accompanies the function (4 cos 3  θ−6)/(20 cos θ)=cos (3θ). 
       FIG. 14  is intended to make clear such distinction. 
     Its top legend identifies the path charted by a curve for such first famous cubic function, algebraically expressed as y=4 cos 3  θ−3 cos θ=cos (3θ) wherein: 
     abscissa values in x signify cos θ magnitudes; and 
     ordinate values in y signify cos (3θ) magnitudes. 
     Such well known curve is shown to be continuous within the specific range of −1≤x≤+1, thereby accounting for all real number values of cos θ. 
     The second legend therein identifies the particular function y=(4 cos 3  θ−6)/(20 cos θ) wherein abscissa values in x again signify cos θ magnitudes. Such curve also is shown to be continuous in the same range, except for the fact that it is discontinuous at x=0. Notice that as the value of x, or cos θ, nears zero from a negative perspective, the corresponding value of y approaches positive infinity, and as it nears zero from the positive side, the corresponding value of y approaches negative infinity; thereby maintaining a one-to-one relationship between x and y values all along its overall path. 
     Where the curves identified by such first and second legends intersect, they can be equated due to the fact that they exhibit both x values of equal magnitude, as well as y values of equal size. Algebraically this can be expressed by the equation y=(4 cos 3  θ−6)/(20 cos θ)=cos (3θ), as typified by the third legend, as displayed in  FIG. 14 . 
     Hence, such intersection points, shown to be positioned at the centers of such four large circles drawn therein, locate positions where (4 cos 3  θ−6)/(20 cos θ)=cos (3θ). 
     By then substituting 4 cos 3  θ−3 cos θ for cos (3θ), as shown below, the following fourth order equation can be obtained, along with a determination of the four associated roots for cos θ and other relevant quantitative details: 
     
       
         
           
             
               
                 
                   4 
                    
                   
                     cos 
                     3 
                   
                    
                   θ 
                 
                 - 
                 
                   3 
                    
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
               
               = 
               
                 
                   
                     4 
                      
                     
                       cos 
                       3 
                     
                      
                     θ 
                   
                   - 
                   6 
                 
                 
                   20 
                    
                   
                       
                   
                    
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
               
             
             ; 
           
         
       
     
     and 
     via cross multiplication, 
       (4 cos 3 θ−3 cos θ)(20 cos θ)=4 cos 3 θ−6;
 
       80 cos 4 θ−60 cos 2 0=4 cos 3 θ−6;
 
       80 cos 4 θ−4 cos 3 θ−60 cos 2 θ+6=0; and
 
       cos 4 θ− 1/20 cos 3 θ−¾ cos 2 θ+ 3/40=0.
 
     Values of the roots of such quartic equation are provided in  FIG. 15 . The first column therein, as headed by the term VALUE, contains various entries of algebraic significance. For each of such five listed entries, corresponding values are cited each of the four the roots θ 1 , θ 2 , θ 3 , and θ 4  which appear as headings in the following four columns. Notice that for each of such particular values of θ, as specified in the second line item therein, a respective value of cos (3θ) appears in the fifth line item therein which is equal to the value of (4 cos 3  θ−6)/(20 cos θ), as it appears in the sixth line item therein. 
     In conclusion, the cos (3θ)=(4 cos 3  θ−6)/(20 cos θ) quartic function clearly qualifies as being discontinuous because it consists of only four distinct points, as are identified by circles appearing in such of  FIG. 14 . 
     With particular regard to the two continuous curve representations drawn in  FIG. 14 , a Euclidean formulation could be generated, whereby each of the singular geometric construction patterns which belong to it can be algebraically determined; three examples of which are presented directly below: 
     
       
         
           
             
               
                 at 
                  
                 
                     
                 
                  
                 x 
               
               = 
               
                 
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
                 = 
                 1 
               
             
             ; 
           
         
       
       
         
           
             
               
                 
                   y 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           4 
                            
                           
                               
                           
                            
                           
                             cos 
                             3 
                           
                            
                           
                               
                           
                            
                           θ 
                         
                         - 
                         6 
                       
                       ) 
                     
                     / 
                     
                       ( 
                       
                         20 
                          
                         
                             
                         
                          
                         cos 
                          
                         
                             
                         
                          
                         θ 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       [ 
                       
                         
                           4 
                            
                           
                             ( 
                             1 
                             ) 
                           
                         
                         - 
                         6 
                       
                       ] 
                     
                     / 
                     
                       [ 
                       
                         2 
                          
                         0 
                          
                         
                           ( 
                           1 
                           ) 
                         
                       
                       ] 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         4 
                         - 
                         6 
                       
                       ) 
                     
                     / 
                     20 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       - 
                       2 
                     
                     / 
                     20 
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         1 
                       
                       / 
                       10 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 at 
                  
                 
                     
                 
                  
                 x 
               
               = 
               
                 
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
                 = 
                 
                   1 
                   / 
                   2 
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 
                   y 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           4 
                            
                           
                               
                           
                            
                           
                             cos 
                             3 
                           
                            
                           
                               
                           
                            
                           θ 
                         
                         - 
                         6 
                       
                       ) 
                     
                     / 
                     
                       ( 
                       
                         20 
                          
                         
                             
                         
                          
                         cos 
                          
                         
                             
                         
                          
                         θ 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     
                       = 
                         
                        
                       
                         [ 
                         
                           
                             
                               ( 
                               4 
                               ) 
                             
                              
                             
                               
                                 ( 
                                 
                                   1 
                                   / 
                                   2 
                                 
                                 ) 
                               
                               3 
                             
                           
                           - 
                           6 
                         
                         ) 
                       
                     
                     ] 
                   
                   / 
                   
                     [ 
                     
                       ( 
                       
                         20 
                          
                         
                           x 
                            
                           
                             ( 
                             
                               1 
                               / 
                               2 
                             
                             ) 
                           
                         
                       
                       ] 
                     
                   
                 
               
             
             
               
                 
                   
                     
                       = 
                         
                        
                       
                         [ 
                         
                           
                             
                               ( 
                               4 
                               ) 
                             
                              
                             
                               ( 
                               
                                 1 
                                 / 
                                 8 
                               
                               ) 
                             
                           
                           - 
                           6 
                         
                         ) 
                       
                     
                     ] 
                   
                   / 
                   10 
                 
               
             
             
               
                 
                   
                     
                       = 
                         
                        
                       
                         ( 
                         
                           
                             1 
                             / 
                             2 
                           
                           - 
                           6 
                         
                         ) 
                       
                     
                     ] 
                   
                   / 
                   10 
                 
               
             
             
               
                 
                   
                     
                       
                         = 
                           
                          
                         
                           - 
                           5.5 
                         
                       
                       ) 
                     
                     ] 
                   
                   / 
                   10 
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         0 
                       
                        
                       .55 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 at 
                  
                 
                     
                 
                  
                 x 
               
               = 
               
                 
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
                 = 
                 
                   
                     - 
                     1 
                   
                   / 
                   2 
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 
                   y 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           4 
                            
                           
                               
                           
                            
                           
                             cos 
                             3 
                           
                            
                           
                               
                           
                            
                           θ 
                         
                         - 
                         6 
                       
                       ) 
                     
                     / 
                     
                       ( 
                       
                         20 
                          
                         
                             
                         
                          
                         cos 
                          
                         
                             
                         
                          
                         θ 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     [ 
                     
                       
                         
                           ( 
                           4 
                           ) 
                         
                          
                         
                           
                             ( 
                             
                               
                                 - 
                                 1 
                               
                               / 
                               2 
                             
                             ) 
                           
                           3 
                         
                       
                       - 
                       
                         6 
                         / 
                         
                           [ 
                           
                             ( 
                             
                               20 
                                
                               
                                 x 
                                  
                                 
                                   ( 
                                   
                                     
                                       - 
                                       1 
                                     
                                     / 
                                     2 
                                   
                                   ) 
                                 
                               
                             
                             ] 
                           
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     
                       = 
                         
                        
                       
                         [ 
                         
                           
                             
                               ( 
                               4 
                               ) 
                             
                              
                             
                               ( 
                               
                                 
                                   - 
                                   1 
                                 
                                 / 
                                 8 
                               
                               ) 
                             
                           
                           - 
                           6 
                         
                         ) 
                       
                     
                     ] 
                   
                   / 
                   
                     - 
                     10 
                   
                 
               
             
             
               
                 
                   
                     
                       
                         = 
                           
                          
                         
                           ( 
                           
                             
                               
                                 - 
                                 1 
                               
                               / 
                               2 
                             
                             - 
                             6 
                           
                           ) 
                         
                       
                       ] 
                     
                     / 
                     
                       - 
                       1 
                     
                   
                    
                   0 
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       - 
                       
                         ( 
                         
                           6 
                            
                           .5 
                         
                         ) 
                       
                     
                     / 
                     
                       - 
                       10 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     0.65 
                     . 
                   
                 
               
             
           
         
       
     
     Naturally any geometric construction pattern which possibly could be drawn which belongs to such Euclidean formulation would identify just a single point which lies upon the two curve potions represented by the second legend in  FIG. 14 . 
     Above, the length (½) 3  would be geometrically constructed in much the same fashion as was the sin 3  θ in  FIG. 13 . The development of such envisioned Euclidean formulation would encompass first generating a length which is equal to (½) 2 , solely by conventional Euclidean means; produced in similar manner to length h 1 , as it appears therein. From such length, another length representative of the algebraic expression (½) 3  would become drawn, similar to h 2 , as it appears therein. 
     From the above calculations, it should become rather clear that an entire family of geometric construction patterns could be drawn for the function y=(4 cos 3  θ− 6 )/(20 cos θ). The corresponding sequence of Euclidean operations needed to conduct such activity could be obtained merely by administering the formula represented on the right hand side of the equation given above, thereby represented as (4 cos 3  θ−6)/(20 cos θ); whereby only the value of cos θ would be altered in during such development. 
     Each respective length of the ordinate value y then could be drawn by way of the proportion y/1=(4 cos 3  θ−6)/(20 cos θ), thereby producing such length ‘y’ by means of applying only a straightedge and compass. 
     As such, the function y=(4 cos 3  θ−6)/(20 cos θ) could be fully described by yet another entirely separate Euclidean formulation. Even though each of such generated geometric construction patterns belonging to such Euclidean formulation most certainly would not relate trigonometric values of angles to those of angles which amount to exactly one-third their respective size, it nevertheless would be possible to design an entirely new invention whose distinctive flexure, maybe even being a harmonic motion, could be replicated by means of animating the entire family of geometric construction patterns which belong to such newly devised Euclidean formulation in successive order. 
     Obviously, such types of involvements inevitably should serve as building blocks for mathematics! 
     More specifically stated, a novel assortment of sundry mechanical devices that exhibit capabilities well beyond those of trisecting emulation mechanisms whose fundamental architectures during flexure regenerate static images that automatically portray overall geometries that furthermore can be fully described by Euclidean formulations additionally can be quantified algebraically! 
     In this vein, prior claims made in connection with such  FIG. 11  Mathematics Demarcation Chart, now are to be somewhat bolstered by theorizing that the very formats expressed by algebraic equations give clear indication of the types of geometric construction practices they support. 
     Such explanation begins with what clearly is known concerning any linear function of the form y=mx+b. 
     Its geometric construction counterpart consists merely of locating a second point which lies a magnitude that algebraically is denoted as ‘b’ either directly above or below a first point, depending upon the sign placed in front of such coefficient. For example, in the equation y=6x−3, such second point would be situated exactly three units of measurement below such first point. In order to complete such singular geometric construction pattern, a straight line next would need to be drawn which passes through such second described point and furthermore exhibits a slope, ‘m’, whose rise and run values could be depicted as the sides of a right triangle, the ratios of whose mutual lengths amount to such magnitudes. 
     Second order functions of a singular variable cannot be fully described by a geometric construction process, thereby necessitating instead that they be fully charted by means of plotting a y value that appears upon a Cartesian coordinate system that becomes algebraically determined for each x value belonging to such function. 
     However, conventional Euclidean practice can be of assistance in determining the roots of quadratic functions. For example, consider an entire set of parabolic functions whose overall format type thereby could be expressed as ax 2 +bx+c=y. 
     For any specific values which its coefficients might be respectively assigned, a singular algebraic function belonging to such format type would become specified. Its roots would indicate where such singular curve crosses the x-axis; but only could when the variable ‘y’ within such function amounts to zero; hence becoming representative of a quadratic equation which instead would belong to another simplified format type, algebraically expressed as ax 2 +bx+c=0 which would typify a subset of such parabolic function format type. 
     By means of referring back to the previous discussion regarding such input box entitled MATHEMATICS DEMARCATION  8 , as posed in  FIG. 2 , note that it was mentioned that a geometric construction pattern that is representative of the famous Quadratic Formula z R =(−b±√{square root over (b 2 −4ac)})/2a would be created to resolve the parabolic equation of −0.2x 2 +0.4x+0.75=0 belonging to such ax 2 +bx+c=0 format type. 
     Herein,  FIG. 16  represents such very solution. 
     The very sequence of Euclidean operations from which such singular geometric construction pattern is derived is provided directly below: 
     a square each whose sides is of unit length is drawn; 
     a right triangle is inscribed within it such that: 
     its first side begins at one of the corners of such square, extends a length of 0.75, or ¾ of a unit from it, and becomes drawn so that it aligns upon a side of such square, thereafter becoming algebraically denoted as being of length ‘c’ therein; 
     its second side, drawn at a right angle away from the endpoint of such first side, is to be of unit length also such that its endpoint resides somewhere along the opposite side of such previously drawn square; and 
     its hypotenuse then is to become drawn; 
     a straight line of length of 0.8 units which extends from a point which resides somewhere upon the first side of such previously drawn right triangle that is parallel to its second side, and terminates somewhere along its hypotenuse is to be drawn as follows: 
     a straight line reference becomes drawn that lies parallel the first side of such previously drawn right triangle and resides 0.8 units in length above it; 
     from the intersection point of such straight line reference and the hypotenuse of such previously drawn right triangle, another straight line is drawn that is perpendicular to such straight line reference; 
     such 0.8 units in length which spans the distance between the first side of such previously drawn right triangle and such straight line reference is to be algebraically denoted as ‘−4a’ therein; and 
     the span of the first side of such previously drawn right triangle which extends from its beginning point to where it intersects such straight line which was drawn to be of 0.8 units in length thereby can be algebraically denoted to be of a length ‘−4ac’ due to the fact that it represents a corresponding side belonging to another right triangle which is similar such previously drawn right triangle, thereby meeting the proportion c/1=−4ac/−4a; 
     a semicircle is drawn whose diameter aligns upon the side of such square that the first side of such previously drawn right triangle also aligns with whose circumferential portion lies outside of such square; 
     such 0.8 unit straight line next is to be extended below the side of such square until it meets such previously drawn circumferential portion, from which two more straight lines are to be drawn, each terminating at a lower corner of such square, thereby describing a second right triangle whose hypotenuse then can be denoted as √{square root over (−4ac)}, since is squared value is equal to the area of the rectangle inscribed in such square whose sides are of unit and −4ac respective lengths by virtue of the Pythagorean Theorem; 
     the remaining side of such newly drawn right triangle, as appearing within such previously drawn semicircle, becomes extended a distance that amounts to 0.4 units in length such that the circumference of a whole circle can be drawn about its new endpoint, being of a radius that thereby can be algebraically denoted to be of length ‘b’ therein; 
     a straight line then is drawn which extends from the beginning of the first side of such previously drawn right triangle that terminates at the center point of such whole circle, thereby being algebraically denoted to be of length √{square root over (b 2 −4ac)} as determined by Pythagorean Theorem, once realizing that it represents the hypotenuse of yet another right triangle whose respective sides are of lengths b and √{square root over (−4ac)}; 
     such newly drawn straight line then becomes extended until it reaches the far circumference of such circle, thereby to become algebraically denoted to be of overall length b+√{square root over (b 2 −4ac)}; 
     its span extending from the beginning of the first side of such previously drawn right triangle to the near circumference of such circle thereby becomes algebraically denoted to be of length −b+√{square root over (b 2 −4ac)}; 
     another straight line then is drawn which passes through the corner of such previously drawn square upon which the vertex of such previously drawn right triangle was geometrically constructed, and its first side began, which furthermore lies perpendicular to the diameter of such newly drawn circle which is shown, being a total length of unity such that 0.4 units of such overall length resides to right side of such diameter, thereby becoming algebraically denoted to be of length −2a; 
     with respect to such last drawn straight line: 
     a straight line is drawn perpendicular to its left termination point; and 
     two more straight lines are drawn emanating from its rightmost termination point, each of which passes through respective locations where the diameter drawn for such circle intersects its circumference; 
     the longer cutoff made upon such lastly drawn perpendicular straight line thereby is algebraically denoted to be of length x 1 , signifying an overall length whose magnitude is equal to the value of the first root of such given quadratic function −0.2x 2 +0.4x+0.75=y, as determined by the respective sides of two right triangles that establish the proportion x 1 /1=(b+√{square root over (b 2 −4ac)})/−2a, therefore amounting to x 1 =(−b−√{square root over (b 2 −4ac)})/2a; and 
     the shorter cutoff made upon such lastly drawn perpendicular straight line thereby is algebraically denoted to be of length −x 2 , signifying an overall length whose magnitude is equal to the negative value of the second root of such given quadratic, as determined by the respective sides of two right triangles that establish the resulting proportion −x 2 /1=(−b+√{square root over (b 2 −4ac)})/−2a, thus amounting to x 2 =(−b+√{square root over (b 2 −4ac)})/2a. 
     Likewise, a cubic functions of a single variable also cannot be fully described by a single geometric construction pattern, but instead requires an entire Euclidean formulation to describe what otherwise would need to become fully plotted by means of algebraically determining a value of y for each x value belonging to such function; as is the case for the either of the continuous cubic curves which are charted in  FIG. 14 . 
     Notice that when interpreting such continuous cubic function y=(4 cos 3  θ−6)/(20 cos θ): 
     when reading from right to left, it indicates an entire family of unique geometric construction patterns, each of which can be generated by means of applying the very same sequence of Euclidean operations, whereby only the magnitude of its given value, cos θ, becomes slightly altered; but 
     when otherwise going from left to right, it becomes indicative of a certain motion which could be imparted by some mechanical device whose fundamental architecture during flexure can be replicated by means of animating a Euclidean formulation which could fully describe its constituent overall shapes. That is to say, a geometric forming process which should be incorporated into the fold of mathematics can characterize trisection for virtually any of the equations contained within the three very famous cubic curves expressed above! 
     As such, a sequel, or follow-on development, being one that presently is considered to be well beyond the very limited scope postulated herein, might entail placing parameters of time within continuous algebraic cubic functions, thereby opening up an entirely new gateway for mathematical investigation; principally because motion cannot transpire without it. 
     It is in this area of discussion that perhaps the greatest confusion abounds concerning trisection! 
     In order to suitably avoid its pitfalls, it becomes necessary to pose one last riddle which finally should fully expose any disturbing myths that yet might be perpetuated by such great trisection mystery. 
     The last riddle is: Can the classical problem of the trisection of an angle actually be solved after gaining an understanding of the role which algebraic expressions play in the determination of the magnitude of a trisector for an angle of virtually any designated magnitude? 
     Again, such answer, most emphatically, turns out to be a resounding no! 
     Such above proposed determination can be substantiated by examining the proceedings associated with a cubic equation containing a single variable which becomes resolved by means of simultaneously reducing it with respect to another cubic equation of a single variable which harbors a common root, whereby such algebraic process enables vital information to be converted into second order form. 
     Naturally, such algebraic approach cannot solve the classical problem of the trisection of an angle! 
     However, it can serve to justify that there is a certain order within mathematics that most certainly should be exposed for the benefit of mankind! 
     As a relevant example of this, one of the three famous cubic functions cited above is to undergo such simultaneous reduction process, wherein ζ is to denote the particular value of the tangent of a designated magnitude of an angle, 3θ, that is about to be trisected; thereby becoming algebraically expressed as tan (3θ). Since such famous cubic equations can track trigonometric relationships which exist between various given angles and those amounting to exactly three times their respective sizes, such previously mentioned common root, denoted as z R , is to represent corresponding values of tan θ, thereby enabling the following algebraic cubic equation expressions to be reformatted as follows: 
     whereas, tan (3θ)=(3 tan θ−tan 3  θ)/(1−3 tan 2 θ); 
       then, ζ=(3 z   R   −z   R   3 )/(1−3 z   R   2 )
 
       ζ(1−3 z   R   2 )=3 z   R   −z   R   3  
 
         z   R   3 =3 z   R −ζ(1− 3   z   R   2 ).
 
     In order to perform such simultaneous reduction, a generalized cubic equation format type of the form z 3 +βz 2 +γz+δ=0 now is to become introduced, as well. 
     In order to determine what common root values any of such equations which belong to such generalized cubic equation format type share in common, in such above equation: 
         z   3   +βz   2   +γz+δ= 0; 
         z   R   3   +βz   R   2   +γz   R +δ=0; and
 
         z   R   3 =−(β z   R   2   +γz   R +δ).
 
     Such format type is to be referred to as the generalized cubic equation because its accounts for virtually every possible equation that a cubic equation of a single variable could possibly assume! 
     Since such famous tangent cubic function can be arranged as z R   3 −3ζz R   2 −3z R +ζ=0, it must be a subset of such generalized cubic equation for the specific case when coefficient β=−3ζ, γ=−3, and δ=ζ. 
     As, I&#39;m sure the reader by now must have guessed, the significance of such association is that both equation formats thereby must bear a common root! 
     Moreover, the term format, as addressed above, applies to a whole family of equations that exhibit identical algebraic structures, but differ only in respect to the particular values of the algebraic coefficients they exhibit! 
     Such mathematical phenomenon occurs because the uncommon roots of each particular equation belonging to such generalized cubic equation format, when arranged in certain combinations with common roots, z R , which they share with respective equations that belong to such famous tangent cubic equation format, actually determine such other coefficient values, as will be more extensively explained below. 
     By equating z R   3  terms, the following quadratic equation relationships can be obtained by means of removing mutual cubic parameters: 
     
       
         
           
             
               
                 3 
                  
                 
                   z 
                   R 
                 
               
               - 
               
                 ζ 
                  
                 
                   ( 
                   
                     1 
                     - 
                     
                       3 
                        
                       
                         z 
                         R 
                         2 
                       
                     
                   
                   ) 
                 
               
             
             = 
             
               
                 - 
                 
                   ( 
                   
                     
                       βz 
                       R 
                       2 
                     
                     + 
                     
                       γ 
                        
                       
                         z 
                         R 
                       
                     
                     + 
                     δ 
                   
                   ) 
                 
               
               = 
               
                 
                   
                     
                       
                         z 
                         R 
                         3 
                       
                        
                       
                         
 
                       
                       ( 
                       
                         
                           3 
                            
                           ζ 
                         
                         + 
                         β 
                       
                       ) 
                     
                      
                     
                       z 
                       R 
                       2 
                     
                   
                   + 
                   
                     
                       ( 
                       
                         3 
                         + 
                         γ 
                       
                       ) 
                     
                      
                     
                       z 
                       R 
                     
                   
                   + 
                   
                     ( 
                     
                       δ 
                       - 
                       ζ 
                     
                     ) 
                   
                 
                 = 
                 0 
               
             
           
         
       
       
         
           
             
               
                 
                   a 
                    
                   
                       
                   
                    
                   
                     z 
                     R 
                     2 
                   
                 
                 + 
                 
                   b 
                    
                   
                       
                   
                    
                   
                     z 
                     R 
                   
                 
                 + 
                 c 
               
               = 
               0 
             
             ; 
             
               
                 
                   
                     and 
                      
                     
                       
 
                     
                     ( 
                     
                       
                         3 
                          
                         ζ 
                       
                       + 
                       β 
                     
                     ) 
                   
                    
                   
                     z 
                     R 
                     2 
                   
                 
                 + 
                 
                   
                     ( 
                     
                       3 
                       + 
                       γ 
                     
                     ) 
                   
                    
                   
                     z 
                     R 
                   
                 
                 + 
                 
                   ( 
                   
                     δ 
                     - 
                     ζ 
                   
                   ) 
                 
               
               = 
               0 
             
           
         
       
       
         
           
             
               
                 z 
                 R 
                 2 
               
               + 
               
                 
                   
                     3 
                     + 
                     γ 
                   
                   
                     
                       3 
                        
                       ζ 
                     
                     + 
                     β 
                   
                 
                  
                 
                   z 
                   R 
                 
               
               + 
               
                 
                   δ 
                   - 
                   ζ 
                 
                 
                   
                     3 
                      
                     ζ 
                   
                   + 
                   β 
                 
               
             
             = 
             0 
           
         
       
       
         
           
             
               
                 z 
                 R 
                 2 
               
               + 
               
                 
                   b 
                   ′ 
                 
                  
                 
                     
                 
                  
                 
                   z 
                   R 
                 
               
               + 
               
                 c 
                 ′ 
               
             
             = 
             0. 
           
         
       
     
     Such last alteration, amounting to the division of each contained coefficient by a factor of ‘a’, gives an indication of how to further manipulate algebraic equation results in order to realize their geometric solutions in a more efficient manner, leading to an abbreviated Quadratic Formula of the form z R =(−b+√{square root over (b 2 −4ac)})/2a=[−b′+√{square root over (b′ 2 −4(1)(c′))}]/2(1)(½)(−b′±√{square root over (b′ 2 −4c′)}). 
     Obviously, such abbreviated Quadratic Formula then applies only to quadratic equations of a singular variable whose squared term coefficients are equal to unity! 
     In order to simultaneously reduce two cubic equations in a single variable which share a common root, their remaining root values must be different. 
     To demonstrate how this works, a generalized cubic equation is to be determined whose uncommon roots, for the sake of simplicity exhibit values of z S =3 and z T =4. 
     For the example which is about to be presented below, a common root value of z R =√{square root over (5)} is to be assigned because it is of quadratic irrational magnitude, and thereby can be geometrically constructed directly from a given length of unity, thereby representing the length of the hypotenuse of a right triangle whose sides are of lengths 1 and 2, respectively. 
     As such, the magnitude of (could be determined merely by means of computing the overall value associated with (3z R −z R   3 )/(1−3z R   2 ) (3√{square root over (5)}−5√{square root over (5)})/(1−3×5)=√{square root over (5)}/7. 
     Notice that such calculation furthermore must be of quadratic irrational magnitude, thereby enabling such length to be represented as the very starting point within an upcoming geometric construction process. 
     Accordingly, such famous cubic relationship in a single variable z R   3 −3ζz R   2 −3z R +ζ=0 would assume the particular form z R   3 −3 (√{square root over (5)}/7) z R   2 −3z R +√{square root over (5)}/7=0. 
     As for such generalized cubic equation, since it can be stated that: 
         z−z   R =0; 
         z−z   S =0; and 
         z−z   T =0. 
     By thereafter multiplying such three equations together, the following algebraic expression could become obtained: 
       ( z−z   R )( z−z   S )( z−z   T )=0; or 
         z   3 −( z   R   +z   S   +z   T ) z   2 +( z   R   z   S   +z   R   z   T   +z   S   z   T ) z−z   R   z   S   z   T =0; and
 
         z   3   +βz   2   +γz+δ= 0. 
     By equating coefficients of like terms, the following three relationships can be determined: 
     
       
         
           
             
               
                 
                   β 
                   = 
                     
                    
                   
                     - 
                     
                       ( 
                       
                         
                           z 
                           R 
                         
                         + 
                         
                           z 
                           S 
                         
                         + 
                         
                           z 
                           T 
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     - 
                     
                       ( 
                       
                         
                           5 
                         
                         + 
                         3 
                         + 
                         4 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       - 
                       
                         ( 
                         
                           
                             5 
                           
                           + 
                           7 
                         
                         ) 
                       
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   γ 
                   = 
                     
                    
                   
                     
                       
                         z 
                         R 
                       
                        
                       
                         z 
                         S 
                       
                     
                     + 
                     
                       
                         z 
                         R 
                       
                        
                       
                         z 
                         T 
                       
                     
                     + 
                     
                       
                         z 
                         S 
                       
                        
                       
                         z 
                         T 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         ( 
                         
                           5 
                         
                         ) 
                       
                        
                       
                         ( 
                         
                           3 
                           + 
                           4 
                         
                         ) 
                       
                     
                     + 
                     
                       3 
                        
                       
                         ( 
                         4 
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         7 
                          
                         
                           5 
                         
                       
                       + 
                       12 
                     
                   
                   ; 
                   and 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   δ 
                   = 
                     
                    
                   
                     
                       - 
                       
                         z 
                         R 
                       
                     
                      
                     
                       z 
                       S 
                     
                      
                     
                       z 
                       T 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       - 
                       
                         ( 
                         
                           5 
                         
                         ) 
                       
                     
                      
                     
                       ( 
                       
                         3 
                          
                         
                           ( 
                           
                             ( 
                             4 
                             ) 
                           
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       - 
                       12 
                     
                      
                     
                       
                         5 
                       
                       . 
                     
                   
                 
               
             
           
         
       
     
     Such generalized cubic equation format would be z 3 −(√{square root over (5)}+7) z 2 +(7√{square root over (5)}+12) z−12√{square root over (5)}=0. 
     Accordingly: 
     
       
         
           
             
               
                 
                   
                     b 
                     ′ 
                   
                   = 
                     
                    
                   
                     
                       3 
                       + 
                       γ 
                     
                     
                       
                         3 
                          
                         ζ 
                       
                       + 
                       β 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       3 
                       + 
                       
                         ( 
                         
                           
                             1 
                              
                             2 
                           
                           + 
                           
                             7 
                              
                             
                               5 
                             
                           
                         
                         ) 
                       
                     
                     
                       
                         3 
                          
                         
                           ( 
                           
                             
                               5 
                             
                             / 
                             7 
                           
                           ) 
                         
                       
                       - 
                       
                         ( 
                         
                           
                             5 
                           
                           + 
                           7 
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         1 
                          
                         5 
                       
                       + 
                       
                         7 
                          
                         
                           5 
                         
                       
                     
                     
                       - 
                       
                         ( 
                         
                           
                             4 
                              
                             
                                 
                             
                              
                             
                               
                                 5 
                               
                               / 
                               7 
                             
                           
                           + 
                           7 
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       - 
                       
                         ( 
                         
                           
                             105 
                             + 
                             
                               49 
                                
                               
                                 5 
                               
                             
                           
                           
                             
                               4 
                                
                               
                                 5 
                               
                             
                             + 
                             49 
                           
                         
                         ) 
                       
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     b 
                     
                       ′ 
                       2 
                     
                   
                   = 
                     
                    
                   
                     
                       
                         1 
                          
                         0 
                          
                         
                           5 
                           2 
                         
                       
                       + 
                       
                         2 
                          
                         1 
                          
                         0 
                          
                         
                           ( 
                           49 
                           ) 
                         
                          
                         
                           5 
                         
                       
                       + 
                       
                         4 
                          
                         
                           9 
                           2 
                         
                          
                         
                           ( 
                           5 
                           ) 
                         
                       
                     
                     
                       
                         1 
                          
                         6 
                          
                         
                           ( 
                           5 
                           ) 
                         
                       
                       + 
                       
                         8 
                          
                         
                           ( 
                           49 
                           ) 
                         
                          
                         
                           5 
                         
                       
                       + 
                       
                         4 
                          
                         
                           9 
                           2 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         
                           23 
                            
                           
                             , 
                           
                            
                           0 
                            
                           3 
                            
                           0 
                         
                         + 
                         
                           10 
                            
                           
                             , 
                           
                            
                           290 
                            
                           
                             5 
                           
                         
                       
                       
                         
                           2 
                            
                           
                             , 
                           
                            
                           4 
                            
                           8 
                            
                           1 
                         
                         + 
                         
                           3 
                            
                           9 
                            
                           2 
                            
                           
                             5 
                           
                         
                       
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     c 
                     ′ 
                   
                   = 
                     
                    
                   
                     
                       δ 
                       - 
                       ζ 
                     
                     
                       
                         3 
                          
                         ζ 
                       
                       + 
                       β 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           - 
                           1 
                         
                          
                         2 
                          
                         
                           5 
                         
                       
                       - 
                       
                         
                           5 
                         
                         / 
                         7 
                       
                     
                     
                       
                         3 
                          
                         
                           ( 
                           
                             
                               5 
                             
                             / 
                             7 
                           
                           ) 
                         
                       
                       - 
                       
                         ( 
                         
                           
                             5 
                           
                           + 
                           7 
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         8 
                          
                         5 
                          
                         
                           5 
                         
                       
                       
                         
                           4 
                            
                           
                             5 
                           
                         
                         + 
                         49 
                       
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       - 
                       4 
                     
                      
                     
                       c 
                       ′ 
                     
                   
                   = 
                     
                    
                   
                     
                       - 
                       
                         ( 
                         
                           
                             340 
                              
                             
                               5 
                             
                           
                           
                             49 
                             + 
                             
                               4 
                                
                               
                                 5 
                               
                             
                           
                         
                         ) 
                       
                     
                      
                     
                       ( 
                       
                         
                           49 
                           + 
                           
                             4 
                              
                             
                               5 
                             
                           
                         
                         
                           49 
                           + 
                           
                             4 
                              
                             
                               5 
                             
                           
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       - 
                       
                         ( 
                         
                           
                             
                               6 
                                
                               8 
                                
                               0 
                                
                               0 
                             
                             + 
                             
                               16 
                                
                               
                                 , 
                               
                                
                               660 
                                
                               
                                 5 
                               
                             
                           
                           
                             
                               2 
                                
                               
                                 , 
                               
                                
                               4 
                                
                               8 
                                
                               1 
                             
                             + 
                             
                               3 
                                
                               9 
                                
                               2 
                                
                               
                                 5 
                               
                             
                           
                         
                         ) 
                       
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             b 
                             2 
                           
                           - 
                           
                             4 
                              
                             
                               c 
                               ′ 
                             
                           
                         
                         = 
                           
                          
                         
                           
                             
                               ( 
                               
                                 
                                   23 
                                    
                                   
                                     , 
                                   
                                    
                                   0 
                                    
                                   3 
                                    
                                   0 
                                 
                                 + 
                                 
                                   10 
                                    
                                   
                                     , 
                                   
                                    
                                   290 
                                    
                                   
                                     5 
                                   
                                 
                               
                               ) 
                             
                             - 
                             
                               ( 
                               
                                 6800 
                                 + 
                                 
                                   16 
                                    
                                   
                                     , 
                                   
                                    
                                   660 
                                    
                                   
                                     5 
                                   
                                 
                               
                               ) 
                             
                           
                           
                             
                               ( 
                               
                                 49 
                                 + 
                                 
                                   4 
                                    
                                   
                                     5 
                                   
                                 
                               
                               ) 
                             
                             2 
                           
                         
                       
                     
                   
                   
                     
                       
                         
                           = 
                             
                            
                           
                             
                               
                                 16 
                                  
                                 
                                   , 
                                 
                                  
                                 2 
                                  
                                 3 
                                  
                                 0 
                               
                               - 
                               
                                 6 
                                  
                                 
                                   , 
                                 
                                  
                                 370 
                                  
                                 
                                   5 
                                 
                               
                             
                             
                               
                                 ( 
                                 
                                   49 
                                   + 
                                   
                                     4 
                                      
                                     
                                       5 
                                     
                                   
                                 
                                 ) 
                               
                               2 
                             
                           
                         
                         ; 
                       
                     
                   
                 
                  
                 
                   
 
                 
                 ± 
                 
                   
                     
                       b 
                       2 
                     
                     - 
                     
                       4 
                        
                       
                         c 
                         ′ 
                       
                     
                   
                 
               
               = 
               
                 
                   ± 
                   
                     
                       
                         16 
                          
                         
                           , 
                         
                          
                         230 
                       
                       - 
                       
                         6 
                          
                         
                           , 
                         
                          
                         370 
                          
                         
                           2 
                         
                       
                     
                   
                 
                 
                   49 
                   + 
                   
                     4 
                      
                     
                       5 
                     
                   
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 
                   
                     z 
                     R 
                   
                   = 
                     
                    
                   
                     
                       
                         - 
                         
                           b 
                           ′ 
                         
                       
                       ± 
                       
                         
                           
                             b 
                             
                               ′ 
                               2 
                             
                           
                           - 
                           
                             4 
                              
                             
                               c 
                               ′ 
                             
                           
                         
                       
                     
                     2 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         1 
                          
                         0 
                          
                         5 
                       
                       + 
                       
                         
                           49 
                            
                           
                             5 
                           
                         
                         ± 
                         
                           
                             
                               16 
                                
                               
                                 , 
                               
                                
                               230 
                             
                             - 
                             
                               6 
                                
                               
                                 , 
                               
                                
                               370 
                                
                               
                                 5 
                               
                             
                           
                         
                       
                     
                     
                       
                         9 
                          
                         8 
                       
                       + 
                       
                         8 
                          
                         
                           5 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         1 
                          
                         0 
                          
                         5 
                       
                       + 
                       
                         
                           49 
                            
                           
                             5 
                           
                         
                         ± 
                         
                           
                             
                               ( 
                               
                                 
                                   - 
                                   65 
                                 
                                 + 
                                 
                                   49 
                                    
                                   
                                     5 
                                   
                                 
                               
                               ) 
                             
                             2 
                           
                         
                       
                     
                     
                       
                         9 
                          
                         8 
                       
                       + 
                       
                         8 
                          
                         
                           5 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         1 
                          
                         0 
                          
                         5 
                       
                       + 
                       
                         
                           49 
                            
                           
                             5 
                           
                         
                         ∓ 
                         
                           ( 
                           
                             
                               6 
                                
                               5 
                             
                             - 
                             
                               49 
                                
                               
                                 5 
                               
                             
                           
                           ) 
                         
                       
                     
                     
                       
                         9 
                          
                         8 
                       
                       + 
                       
                         8 
                          
                         
                           5 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         
                           4 
                            
                           0 
                         
                         + 
                         
                           98 
                            
                           
                             5 
                           
                         
                       
                       
                         
                           
                             9 
                              
                             8 
                           
                           + 
                           
                             8 
                              
                             
                               5 
                             
                           
                         
                          
                         
                             
                         
                       
                     
                   
                   ; 
                   
                     170 
                     
                       98 
                       + 
                       
                         8 
                          
                         
                           5 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         ( 
                         
                           
                             5 
                           
                           
                             5 
                           
                         
                         ) 
                       
                        
                       
                         [ 
                         
                           
                             
                               8 
                                
                               
                                 
                                   ( 
                                   
                                     5 
                                   
                                   ) 
                                 
                                 2 
                               
                             
                             + 
                             
                               9 
                                
                               8 
                                
                               
                                 5 
                               
                             
                           
                           
                             
                               9 
                                
                               8 
                             
                             + 
                             
                               8 
                                
                               
                                 5 
                               
                             
                           
                         
                         ] 
                       
                     
                   
                   ; 
                   
                     170 
                     
                       98 
                       + 
                       
                         8 
                          
                         
                           5 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         
                           
                             
                               5 
                             
                              
                             
                               ( 
                               
                                 8 
                                  
                                 
                                   5 
                                 
                               
                               ) 
                             
                           
                           + 
                           98 
                         
                         ) 
                       
                       
                         
                           
                             9 
                              
                             8 
                           
                           + 
                           
                             8 
                              
                             
                               5 
                             
                           
                         
                          
                         
                             
                         
                       
                     
                   
                   ; 
                   
                     170 
                     
                       98 
                       + 
                       
                         8 
                          
                         
                           5 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       5 
                     
                   
                   ; 
                   
                     
                       
                         1 
                          
                         7 
                          
                         0 
                       
                       
                         
                           9 
                            
                           8 
                         
                         + 
                         
                           8 
                            
                           
                             5 
                           
                         
                       
                     
                     . 
                   
                 
               
             
           
         
       
     
     Naturally, the last of such three famous continuous cubic equations, as stipulated above, alternatively could have been resolved algebraically without having to resort to such cumbersome simultaneous reduction process. 
     This could be achieved simply by realizing that once a value of (becomes designated, an angle of 3θ magnitude that it is representative of very easily could be determined trigonometrically; whereby, a value for z R  which corresponds to its trisector, computed as being one-third of such value, and thereby algebraically expressed merely as θ, thereafter also could be trigonometrically determined. 
     Unfortunately, the pitfall that accompanies such shortened algebraic process is that such common root, z R , does not become identified solely by conventional Euclidean means! 
     The method to do so would be to draw straight lines whose lengths are of magnitudes which are equal to the value of roots belonging to such abbreviated Quadratic Formula z R =(½)(−b′±√{square root over (b′2−4c′)}), much in the same manner as was employed earlier when quadratic roots first were determined by means of geometric construction in  FIG. 16 . 
     For such algebraic determination, as made above, the magnitude of a trisector for an angle whose tangent is of a designated magnitude √{square root over (5)}/7 could be geometrically constructed by means of applying the following sequence of Euclidean operations; thereby rendering a particular pattern, as is depicted in  FIG. 17 : 
     two right triangles are drawn in the lower right corner which share a common side of length (49+4√{square root over (5)})/100, and whose other mutual sides are of respective lengths: 
       (105+49√{square root over (5)})/100; and
 
       85√{square root over (5)}/100;
 
     such common side is extended to a unit length; 
     a perpendicular straight line is drawn above the newly formed endpoint of such extension; 
     the hypotenuses appearing in such two previously drawn right triangles are extended until they intersect such newly drawn perpendicular straight line, thereby depicting two more similar right triangles; 
     whereby, the lengths of the unknown sides of such two newly drawn right triangles can be determined by virtue of the proportions established between the known lengths of corresponding sides of their respective similar right triangles and their common side of unit length, thereby enabling designations of −b′ length and c′ to be notated upon such drawing to reflect the following determinations: 
     
       
         
           
             
               b 
               ′ 
             
             = 
             
               
                 
                   - 
                   
                     ( 
                     
                       
                         
                           1 
                            
                           0 
                            
                           5 
                         
                         + 
                         
                           49 
                            
                           
                             5 
                           
                         
                       
                       
                         
                           4 
                            
                           
                             5 
                           
                         
                         + 
                         49 
                       
                     
                     ) 
                   
                 
                  
                 
                   
 
                 
                 - 
                 
                   b 
                   ′ 
                 
               
               = 
               
                 
                   ( 
                   
                     
                       
                         1 
                          
                         0 
                          
                         5 
                       
                       + 
                       
                         49 
                          
                         
                           5 
                         
                       
                     
                     
                       
                         4 
                          
                         
                           5 
                         
                       
                       + 
                       49 
                     
                   
                   ) 
                 
                  
                 
                   ( 
                   
                     
                       
                         1 
                         / 
                         1 
                       
                        
                       0 
                        
                       0 
                     
                     
                       
                         1 
                         / 
                         1 
                       
                        
                       0 
                        
                       0 
                     
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   - 
                   
                     b 
                     ′ 
                   
                 
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         1 
                          
                         0 
                          
                         5 
                       
                       + 
                       
                         49 
                          
                         
                           5 
                         
                       
                     
                     ) 
                   
                   / 
                   100 
                 
                 
                   
                     ( 
                     
                       49 
                       + 
                       
                         4 
                          
                         
                           5 
                         
                       
                     
                     ) 
                   
                   / 
                   100 
                 
               
             
             ; 
           
         
       
       
         
           
             
               c 
               ′ 
             
             = 
             
               
                 
                   85 
                    
                   
                     5 
                   
                 
                 
                   
                     4 
                      
                     
                       5 
                     
                   
                   + 
                   49 
                 
               
                
               
                 ( 
                 
                   
                     1 
                     / 
                     100 
                   
                   
                     1 
                     / 
                     100 
                   
                 
                 ) 
               
             
           
         
       
       
         
           
             
               
                 
                   c 
                   ′ 
                 
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       8 
                        
                       5 
                        
                       
                         5 
                       
                     
                     ) 
                   
                   / 
                   100 
                 
                 
                   
                     
                       ( 
                       
                         49 
                         + 
                         
                           4 
                            
                           
                             5 
                           
                         
                       
                       ) 
                     
                     / 
                     1 
                   
                    
                   0 
                    
                   0 
                 
               
             
             ; 
           
         
       
     
     next, a square whose sides are of length −b′ is to be drawn, as indicated in the lower left-hand corner of  FIG. 17 ; 
     a rectangle then becomes drawn whose base of unit length is to align along the lower side of such square and whose left lower corner is to share the very position which the left hand lower corner of such square occupies; 
     a straight line then is to become drawn which extends from such newly identified common corner, passes through an intersection point which is made between the upper side of such previously drawn square and the right side of such newly drawn rectangle, and thereafter continues as a large diagonal until it intersects with the right side of such previously drawn square; 
     the distance between such newly determined intersection point above the lower side of such square of base dimension −b′ is to become denoted as b′ 2 , as determined by the proportion established between the corresponding sides of two new similar right triangles whose respective hypotenuses align upon such just drawn long diagonal, whereby such proportion becomes calculated as b′2/−b′=−b′/1; 
     a horizontal line next is set off a distance of b′ 2  above the based of such previously drawn square; 
     another horizontal line of is set off a distance of 4c′ above the based of such previously drawn square; 
     the intervening length existing between them, amounting to a magnitude of b′ 2− −4c′, must constitute the entire area of the small rectangle they furthermore describe, as bounded by the two opposite side of such previously drawn rectangle whose base is equal to a length of unity; 
     a second square of unit base dimension then becomes described such that its lower portion aligns directly upon such previously described rectangle of area equal to b′ 2− −4c′; 
     a semicircle thereby can be drawn to the right of such square whose diameter aligns upon its left side; 
     straight lines thereafter are drawn from the respective ends of such semicircle diameter to the point residing upon its circumference which intersects the horizontal straight line which resides at a distance of b′ 2  above the base of such previously drawn square whose respective sides each are ‘b’ in length; 
     by virtue of the Pythagorean Theorem, such lower straight line, as drawn from the lower extremity of the diameter of such semicircle and extending to a point lying upon its circumference, must amount to a length which is equal to the square root of the b′ 2 −4c′ area of such previously described rectangle; 
     such length thereafter is reproduced as an extension to the horizontal straight line previously drawn which resides a distance of 4c′ above the base of the previously drawn square whose sides each equal −b′ in length; 
     such new straight line extension is notated as being of overall length 2z R =−b′+√{square root over (b′ 2 −4c′)}, as is indicated both at the very the top and very bottom of such drawing; and 
     such overall length thereafter becomes bisected in order to distinguish and thereby designate a length z R  which amounts to one-half such magnitude. 
     Obviously, such geometric construction approach cannot pose a solution for the classical problem of the trisection of an angle; simply because the generalized cubic equation format that contributes to its very determination, specifically being z 3 −(√{square root over (5)}+7)z 2 +(7√{square root over (5)}+12)z−12√{square root over (5)}=0, could not be derived without a prior awareness of the very solution itself. 
     A second less complicated example demonstrating that it is possible to apply algebraic information in order to create a geometric solution for the problem of the trisection of an angle pertains to a generalized cubic equation whose coefficients β and γ are set to zero, and whose coefficient δ amounts to a value of +1, thereby establishing the specific cubic equation z R   3 +1=0. 
     From such information, the following details can be gleaned: 
     
       
         
           
             
               
                 z 
                 R 
                 5 
               
               + 
               1 
             
             = 
             0 
           
         
       
       
         
           
             
               z 
               R 
               5 
             
             = 
             
               - 
               1 
             
           
         
       
       
         
           
             
               
                 
                   
                     z 
                     R 
                   
                   = 
                     
                    
                   
                     
                       - 
                       1 
                     
                     3 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     - 
                     1 
                   
                 
               
             
           
         
       
       
         
           
             
               tan 
                
               
                   
               
                
               θ 
             
             = 
             
               - 
               1 
             
           
         
       
       
         
           
             
               
                 
                   θ 
                   = 
                     
                    
                   
                     arc 
                      
                     
                         
                     
                      
                     tan 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         - 
                         1 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       135 
                        
                       ° 
                     
                   
                   ; 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     3 
                      
                     θ 
                   
                   = 
                     
                    
                   
                     3 
                      
                     
                       ( 
                       θ 
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     3 
                      
                     
                       ( 
                       
                         135 
                          
                         ° 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       405 
                        
                       ° 
                     
                   
                   ; 
                   
                     and 
                      
                     
                         
                     
                      
                     as 
                      
                     
                         
                     
                      
                     a 
                      
                     
                         
                     
                      
                     check 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   ζ 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           3 
                            
                           
                             z 
                             R 
                           
                         
                         - 
                         
                           z 
                           R 
                           3 
                         
                       
                       ) 
                     
                     / 
                     
                       ( 
                       
                         1 
                         - 
                         
                           3 
                            
                           
                             z 
                             R 
                             2 
                           
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           - 
                           3 
                         
                         + 
                         1 
                       
                       ) 
                     
                     / 
                     
                       ( 
                       
                         1 
                         - 
                         3 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       - 
                       2 
                     
                     / 
                     
                       - 
                       2 
                     
                   
                 
               
             
           
         
       
       
         
           
             
               tan 
                
               
                 ( 
                 
                   3 
                    
                   
                       
                   
                    
                   θ 
                 
                 ) 
               
             
             = 
             
               + 
               1 
             
           
         
       
       
         
           
             
               
                 
                   
                     3 
                      
                     θ 
                   
                   = 
                     
                    
                   
                     arc 
                      
                     
                         
                     
                      
                     tan 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         + 
                         1 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       ( 
                       
                         
                           3 
                            
                           6 
                            
                           0 
                         
                         + 
                         45 
                       
                       ) 
                     
                      
                     ° 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     405 
                      
                     
                       ° 
                       . 
                     
                   
                 
               
             
           
         
       
     
     Such algebraic determination, as made above, thereby enables the trisection of an angle to be geometrically constructed as follows: 
     from a designated value of =tan (3θ)=+1, an angle designated as 3θ which amounts to exactly 450 in magnitude first becomes geometrically constructed with respect to the +x-axis; and 
     from an algebraically determined common root value of z R =−1, a trisecting angle designated as θ which amounts to exactly 135° in magnitude thereafter becomes geometrically constructed with respect to the +x-axis. 
     Needless to say, such geometric construction, as posed above, although representing geometric solution for the problem of the trisection of an angle, nevertheless does not pose a solution for the classical problem of the trisection of an angle. This is because a value for such common root z R  cannot be ascertained solely by means of a geometric construction which proceeds exclusively from a given value of (=tan (3θ)=+1. 
     Although a straight line of slope z R =−1 could be geometrically constructed rather easily from another line of given slope (=+1, such geometric construction pattern represents just one out of an infinite number of straight line possibilities which otherwise could be distinguished geometrically from a given value of (=+1. 
     Hence, the sequence of Euclidean operations which governs such trisection can be completed with certainty only by incorporating such algebraic determination that z R =+1, or else simply by algebraically dividing such geometrically constructed  405  angle by a factor of three. 
     In either case, since both of such algebraic results are tied only to such 135° trisector of slope z R =−1, the only way to determine such information solely via straightedge and compass from a geometrically constructed  450  angle would be to distinguish them from the results of a Euclidean trisection which has not yet been performed. 
     Such process entails knowledge of the results of a geometric construction before it actually becomes conducted, thereby violating the rules of conventional Euclidean practice which require that geometric construction can proceed only from a given set of previously defined geometric data. 
     In order to further emphasize just how the use of aforehand knowledge inadvertently creeps into conventional Euclidean practice, thereby grossly violating its very rules, a last rather telling example is afforded below whereby given angle NMP, as depicted in  FIG. 1B , is to be of the very size which actually appears in such figure; thereby very closely amounting to twenty degrees. As such: 
     angle QPS, being geometrically constructed to three times that size, must be exactly sixty degrees. It becomes very easy to draw such rendered angle because the internal angle of a geometrically constructed equilateral triangle is that same size; 
     the next step is to determine whether or not Euclidean operations can be launched exclusively from such designated angle QPS in order to locate the correct positions of points M and N; and 
     as it turns out, intersection points M and N cannot be distinguished solely via straightedge and compass solely from such rendered angle QPS. That is to say, there is absolutely no geometric construction that can be performed with respect to such sixty degree angle QPS which can locate points M and N, short of having aforehand awareness of their respective locations. 
     Such above analysis reveals that with respect to the particular geometry represented in such famous  FIG. 1B  Archimedes Euclidean formulation, when commencing only from angle QPS of designated sixty degree magnitude, points M and N truly qualify as overlapment points. 
     Were this above assertion not to be true, it would be tantamount to trisecting such sixty degree angle QPS solely by means of applying a straightedge and compass to it; thereby solving the classical problem of the trisection of an angle without having any other predisposed knowledge and, in so doing, accomplishing a feat that is entirely impossible! 
     With regard to a prior discussion concerning the input box entitled PROBABILISTIC PROOF OF MATHEMATIC LIMITATION  10 , it was mentioned that trisection can be achieved by means of performing a multitude of consecutive angular bisections, all geometrically constructed upon just a single piece of paper. 
     Such approach generates a geometric construction pattern that is indicative of a geometric progression whose: 
     constant multiplier, “m”, is set equal to −½; and 
     first term, “f”, is algebraically denoted as 3θ. 
     Moreover, the overall sum, “s”, of such geometric progression consisting of an “n” number of terms can be reresented by the common knowledge formula: 
     
       
         
           
             
               
                 
                   
                     s 
                     = 
                       
                      
                     
                       
                         f 
                          
                         
                           ( 
                           
                             
                               m 
                               n 
                             
                             - 
                             1 
                           
                           ) 
                         
                       
                       / 
                       
                         ( 
                         
                           m 
                           - 
                           1 
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       3 
                        
                       
                           
                       
                        
                       
                         
                           θ 
                            
                           
                             ( 
                             
                               
                                 
                                   - 
                                   1 
                                 
                                 / 
                                 
                                   2 
                                   n 
                                 
                               
                               - 
                               1 
                             
                             ) 
                           
                         
                         / 
                         
                           ( 
                           
                             
                               
                                 - 
                                 1 
                               
                               / 
                               2 
                             
                             - 
                             1 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     
                       = 
                         
                        
                       
                         
                           - 
                           2 
                         
                          
                         
                             
                         
                          
                         
                           θ 
                            
                           
                             ( 
                             
                               
                                 
                                   - 
                                   1 
                                 
                                 / 
                                 
                                   2 
                                   n 
                                 
                               
                               - 
                               1 
                             
                             ) 
                           
                         
                       
                     
                     ; 
                   
                 
               
             
               
           
         
       
     
     whereby 
     for an infinite number of terms, such equation thereby reduces to, 
     
       
         
           
             
               
                 
                   
                     s 
                     = 
                       
                      
                     
                       
                         - 
                         2 
                       
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           
                             
                               
                                 - 
                                 1 
                               
                               / 
                               
                                 2 
                                 ∞ 
                               
                             
                             - 
                             1 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       
                         - 
                         2 
                       
                        
                       
                           
                       
                        
                       
                         θ 
                          
                         
                           ( 
                           
                             0 
                             - 
                             1 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       2 
                        
                       
                           
                       
                        
                       
                         θ 
                         . 
                       
                     
                   
                 
               
             
               
           
         
       
     
     Such result indicates that after conducting an infinite number of successive bisection operations, it becomes possible to geometrically construct an angle that amounts to exactly ⅔ the size of an angle of designated 3θ magnitude, whereby their difference then must distinguish its trisector. 
     Below, a method is furnished which describes how to geometrically construct the first five terms appearing in such governing geometric progression; and in so doing thereby assuming the form 3θ−3θ/2+3θ/4−3θ/8+3θ/16=33θ/16. 
     In such development, the value of the first term, algebraically denoted as 3θ, can be set equal to virtually any designated magnitude that is intended to be trisected. By inspection, it furthermore becomes apparent that the numerical value of each succeeding term is equal to one-half the magnitude of its predecessor. As such, values for such diminishing magnitudes can be geometrically constructed merely by means of bisecting each of such preceding angles. 
     Lastly, wherein positive values could applied in a counterclockwise direction, negative magnitudes would appear in a completely opposite, or clockwise direction, with respect to them. 
     The specific details which pertain to a  FIG. 18  drawing of this nature are itemized as follows: 
     an angle of magnitude 3θ is drawn such that its vertex aligns upon the origin of an orthogonal coordinate system with its clockwise side residing along its +x-axis; 
     such given angle, being of magnitude 3θ, becomes bisected, whereby such bisector resides at an angle relative to such +x-axis that amounts to ½(3θ)=3θ/2; 
     the upper portion of such bisected angle, amounting to a size of 3θ/2, then itself becomes bisected, whereby a determination made as to the location of such second bisector would place it at an angle of 3θ/2+3θ/4=9θ/4 with respect to the +x-axis; 
     the angle formed between such first bisector and second bisector next becomes bisected, whereby a determination made as to the location of such third bisector would place it at an angle of 9θ/4−3θ/8=15θ/8 with respect to the +x-axis; and 
     the angle formed between such second bisector and third bisector then itself becomes bisected, whereby a determination made as to the location of such fourth bisector would place it at an angle of 15θ/8+3θ/16=33θ/16 with respect to the +x-axis. 
     Quite obviously, it remains possible to continue such activity until such time that the naked eye no longer could detect a bisector for an arc that invariably becomes smaller and smaller with each subsequent bisection operation. 
     In this regard, the resolution of the naked eye is considered to be limited to about one minute of arc, thereby amounting to 1/60 th  of a degree, whose decimal equivalent is 0.01667°. 
     Once the human eye no longer can detect gradations resulting from such bisectors process, they could be located erroneously or even superimposed upon prior work. 
     Since the use of a microscope might increase such perception capabilities, it might enable a few additional bisections to become accurately determined. However, being that an infinite number of bisections are needed in order to generate a precise trisector in this manner, such enhancement only would serve to slightly improve upon the overall approximation of any trisector which becomes produced. 
     The Successive Bisection Convergence Chart, as presented in  FIG. 19 , describes the results produced by such geometric progression as the number of terms is shown to increase in its first column, as headed by the term n. 
     The second column therein is devoted to calculations which apply to such geometric progression, based upon the number of terms it contains. In each line item, the last value provided indicates the overall size of the angle which would become geometrically constructed by means of conducting such successive bisection process. 
     Notice that  FIG. 19  is discontinued at a value of n=22. This is because, at this point in such overall geometric construction process, an accuracy of six decimal places, amounting to (2.000000)θ would become realized. 
     Since the only time that a bisection operation is not conducted is when n=1, each successive line item within such  FIG. 19  chart depicts a geometric construction pattern that could be generated by means of performing a total of n−1 bisection operations. 
     Hence, an accuracy of one-millionth could be obtained by means of conducting twenty-one successive bisections. 
     The analysis presented below discloses that for a 20° trisector, such above summarized process of successive angular bisections would have to be disbanded during the twelfth bisection operation due to the naked eye no longer being able to discern the exact placement of its bisector. 
     As such, the number of terms this condition would apply to, as indicated in such  FIG. 19  chart, would be when n=13. 
     From such  FIG. 19  chart, the separation needed to be distinguished when performing such twelfth bisection is calculated to be 
     
       
         
           
             
               
                 
                   
                     
                       
                         2.000244 
                          
                         
                             
                         
                          
                         θ 
                       
                       - 
                       
                         1.999512 
                          
                         
                             
                         
                          
                         θ 
                       
                     
                     = 
                       
                      
                     
                       0.000732 
                        
                       θ 
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       0.000732 
                        
                       
                         ( 
                         
                           20 
                            
                           ° 
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   
                     = 
                       
                      
                     
                       0.01464 
                        
                       
                         ° 
                         . 
                       
                     
                   
                 
               
             
               
           
         
       
     
     Therefore, since such 0.01464° needed separation clearly is smaller than the 0.01667° which the naked eye is capable of perceiving; it means that such twelfth bisector could be located erroneously. 
     When referring to  FIG. 18 , notice that an angle of size 3θ whose vertex is placed at the origin of a Cartesian Coordinate System such that its clockwise side aligns upon its +x-axis is indicative of such geometric progression for the particular condition when n=1. 
     Additionally, four subsequent bisections are depicted, each of which is considered to have been performed solely by conventional Euclidean means. 
     The purpose of the shading therein is to suitably distinguish between each of such bisection activities as follows: 
     such angle of magnitude +3θ is bisected in order to distinguish two separate arcs, each being of 3θ/2 size; 
     with the upper portion of such bisected angle, amounting to a size of 3θ/2, then itself becoming bisected, the determination made as to the location of such second bisector would place it at an angle of 3θ/4 counterclockwise of such first bisector position; 
     with the angle formed between such first bisector and second bisector, amounting to a size of 3θ/4, then itself becoming bisected, the determination made as to the location of such third bisector would place it at an angle of 3θ/8 clockwise of such second bisector position, ad denoted by the minus sign notation; and 
     with the angle formed between such second bisector and third bisector, amounting to a size of 3θ/8, then itself becoming bisected, the determination made as to the location of such fourth bisector would place it at an angle of 3θ/16 counterclockwise of such third bisector position. 
     As to the role which cube roots could play in a geometric solution of the problem of the trisection of an angle, below it is shown how to determine the length of a straight line, half which amounts to its cube root value, whereby it could be algebraically stated that: 
     
       
         
           
             
               
                  
                 3 
               
               = 
               
                  
                 / 
                 2 
               
             
             ; 
           
         
       
     
     such that by cubing both sides; 
         =   3 /8 
       8 =   3    
       4(2)=   2    
       2√{square root over (2)}= 
 
       √{square root over (2)}= /2; and
 
     relevant information then is to be introduced in the form of an angle whose complement furthermore turns out to be its trisector, algebraically determined as follows: 
     
       
         
           
             θ 
             = 
             
               
                 90 
                  
                 ° 
               
               - 
               
                 3 
                  
                 
                     
                 
                  
                 θ 
               
             
           
         
       
       
         
           
             
               
                 3 
                  
                 θ 
               
               + 
               θ 
             
             = 
             
               90 
                
               ° 
             
           
         
       
       
         
           
             
               4 
                
               
                   
               
                
               θ 
             
             = 
             
               90 
                
               ° 
             
           
         
       
       
         
           
             θ 
             = 
             
               22.5 
                
               ° 
             
           
         
       
       
         
           
             
               2 
                
               
                   
               
                
               θ 
             
             = 
             
               45 
                
               ° 
             
           
         
       
       
         
           
             
               
                 3 
                  
                 θ 
               
               = 
               
                 67.5 
                  
                 ° 
               
             
             ; 
           
         
       
       
         
           
             
               sin 
                
               
                   
               
                
               
                 ( 
                 
                   3 
                    
                   θ 
                 
                 ) 
               
             
             = 
             
               
                 3 
                  
                 
                     
                 
                  
                 sin 
                  
                 
                     
                 
                  
                 θ 
               
               - 
               
                 4 
                  
                 
                     
                 
                  
                 
                   sin 
                   3 
                 
                  
                 
                     
                 
                  
                 θ 
               
             
           
         
       
       
         
           
             
               cos 
                
               
                   
               
                
               
                 ( 
                 
                   90 
                   - 
                   
                     3 
                      
                     
                         
                     
                      
                     θ 
                   
                 
                 ) 
               
             
             = 
             
               sin 
                
               
                   
               
                
               θ 
                
               
                   
               
                
               
                 ( 
                 
                   3 
                   - 
                   
                     4 
                      
                     
                         
                     
                      
                     
                       sin 
                       2 
                     
                      
                     
                         
                     
                      
                     θ 
                   
                 
                 ) 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     cos 
                      
                     
                         
                     
                      
                     θ 
                   
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                        
                       
                           
                       
                       [ 
                       
                         
                           
                             ( 
                             2 
                             ) 
                           
                            
                           
                             ( 
                             
                               1 
                               - 
                               
                                 2 
                                  
                                 
                                     
                                 
                                  
                                 
                                   sin 
                                   2 
                                 
                                  
                                 
                                     
                                 
                                  
                                 θ 
                               
                             
                             ) 
                           
                         
                         + 
                         1 
                       
                       ] 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                        
                       
                           
                       
                       [ 
                       
                         
                           2 
                            
                           cos 
                            
                           
                               
                           
                            
                           
                             ( 
                             
                               2 
                                
                               
                                   
                               
                                
                               θ 
                             
                             ) 
                           
                         
                         + 
                         1 
                       
                       ] 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     θ 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         
                           2 
                            
                           
                               
                           
                            
                           cos 
                            
                           
                               
                           
                            
                           45 
                            
                           ° 
                         
                         + 
                         1 
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                       [ 
                       
                         
                           
                             ( 
                             2 
                             ) 
                           
                            
                           
                             ( 
                             
                               
                                 2 
                               
                               2 
                             
                             ) 
                           
                         
                         + 
                         1 
                       
                       ] 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       θ 
                        
                       
                         ( 
                         
                           
                             2 
                           
                           + 
                           1 
                         
                         ) 
                       
                     
                   
                 
               
             
           
         
       
       
         
           
             
               1 
               
                 
                   2 
                 
                 + 
                 1 
               
             
             = 
             
               tan 
                
               
                   
               
                
               θ 
             
           
         
       
       
         
           
             
               
                 1 
                 
                   
                     2 
                   
                   + 
                   1 
                 
               
                
               
                 ( 
                 
                   
                     
                       2 
                     
                     - 
                     1 
                   
                   
                     
                       2 
                     
                     - 
                     1 
                   
                 
                 ) 
               
             
             = 
             
               tan 
                
               
                   
               
                
               θ 
             
           
         
       
       
         
           
             
               
                 
                   2 
                 
                 - 
                 1 
               
               
                 2 
                 - 
                 1 
               
             
             = 
             
               tan 
                
               
                   
               
                
               θ 
             
           
         
       
       
         
           
             
               
                 2 
               
               - 
               1 
             
             = 
             
               tan 
                
               
                   
               
                
               θ 
             
           
         
       
       
         
           
             
               
                 2 
               
               - 
               1 
             
             = 
             
               1 
               
                 tan 
                  
                 
                   ( 
                   
                     3 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     tan 
                      
                     
                       ( 
                       
                         3 
                          
                         θ 
                       
                       ) 
                     
                   
                   = 
                     
                    
                   
                     
                       1 
                       
                         
                           2 
                         
                         - 
                         1 
                       
                     
                      
                     
                       ( 
                       
                         
                           
                             2 
                           
                           + 
                           1 
                         
                         
                           
                             2 
                           
                           + 
                           1 
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         2 
                       
                       + 
                       1 
                     
                     
                       2 
                       - 
                       1 
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         2 
                       
                       + 
                       1 
                     
                   
                   ; 
                 
               
             
           
         
       
     
     it therefore becomes possible to geometrically construct a right triangle whose sides amount to respective lengths of 1 and 1+√{square root over (2)} such that its tangent, ζ, amounts to a value of 1+√{square root over (2)}; 
     whereby such √{square root over (2)} length is drawn as the hypotenuse of a 45θ right triangle, and such 1+√{square root over (2)} thereby represents the addition of its side added to such hypotenuse length; and 
     such hypotenuse of length √{square root over (2)} after becoming doubled and thereby amounting to 2√{square root over (2)}, being its cubed value, thereafter can be bisected in order to arrive at its cube root. 
     The algebraic cubic equation which correlates to this geometric construction process assumes the form of z R   3 +3z R   2 +3z R +(3−2ζ)=0; as determined below: 
       tan(3θ)=√{square root over (2)}+1=ζ
 
       √{square root over (2)}=ζ−1; and
 
       tan θ= z   R =√{square root over (2)}−1
 
         z   R +1=√{square root over (2)}
 
       ( z   R +1) 3 =(√{square root over (2)}) 3  
 
       ( z   R +1) 3 =2√{square root over (2)}
 
       ( z   R +1) 3 =2(ζ−1)
 
       ( z   R   3 +3 z   R   2 +3 z   R +1)−2(ζ−1)=0
 
         z   R   3 +3 z   R   2 + 3   z   R +(3−2)=0.
 
     To finalize a discussion raised earlier,  FIG. 20  relates one complex number to another which serves both as its trisector, as well its cube root. 
     To elaborate upon this, complex numbers typically are represented geometrically as straight lines which appear upon an xy plane known as the complex plane. 
     Each straight line featured therein commences from the origin of a rectilinear coordinate system, and contains an arrow at its termination point to express direction. 
     The convention used to specify a complex number is first to indicate its real numerical magnitude, followed by its imaginary component. Such imaginary aspect is represented by an Arabic letter, i, used to denote an imaginary term √{square root over (−1)}, followed by its magnitude. 
     As such, the coordinate values of complex number termination points designate their respective imaginary and real number magnitudes; thereby fully describing them. 
     In  FIG. 20 , such two complex numbers are shown to be expressed as cos (3θ)+i sin (3θ), and cos θ+i sin θ. 
     Conversely, since the ratio between the magnitudes of the real and imaginary portions of such first complex number is (sin 3θ)/(cos 3θ)=tan 3θ, the straight line which represents it, by exhibiting such slope, thereby must pass through the origin while forming an angle of 3θ with such x-axis. 
     Likewise, the straight line which represents such second complex number, by exhibiting a slope of tan θ, thereby must pass through the origin while instead forming an angle of θ with respect to the x-axis and, in so doing, trisecting such angle of 3θ magnitude. 
     The fact that the complex number cos θ±i sin θ also turns out to be the cube root of the first complex number cos (3θ)+i sin (3θ) furthermore is to be verified algebraically by applying the binomial expansion (A+B) 3 =A 3 +3A 2 B+3AB 2 +B 3  for the express condition when the A=cos θ, and B=i sin θ as follows: 
     
       
         
           
             
               
                 A 
                 3 
               
               + 
               
                 3 
                  
                 
                   A 
                   2 
                 
                  
                 B 
               
               + 
               
                 3 
                  
                 
                   AB 
                   2 
                 
               
               + 
               
                 B 
                 3 
               
             
             = 
             
               
                 
                   cos 
                   
                     3 
                      
                     
                         
                     
                   
                 
                  
                 θ 
               
               + 
               
                 3 
                  
                 
                   ( 
                   
                     
                       cos 
                       2 
                     
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
                  
                 
                   ( 
                   
                     i 
                      
                     
                         
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
               
               - 
               
                 3 
                  
                 
                   ( 
                   
                     cos 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
                  
                 
                   ( 
                   
                     
                       sin 
                       2 
                     
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
               
               + 
               
                 
                   ( 
                   
                     i 
                      
                     
                         
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
                 3 
               
             
           
         
       
       
         
           
             
               
                 ( 
                 
                   A 
                   + 
                   B 
                 
                 ) 
               
               3 
             
             = 
             
               
                 
                   cos 
                   3 
                 
                  
                 
                     
                 
                  
                 θ 
               
               + 
               
                 3 
                  
                 
                   ( 
                   
                     1 
                     - 
                     
                       
                         sin 
                         2 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     i 
                      
                     
                         
                     
                      
                     sin 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
               
               - 
               
                 3 
                  
                 
                   ( 
                   
                     cos 
                      
                     
                         
                     
                      
                     θ 
                   
                   ) 
                 
                  
                 
                   ( 
                   
                     1 
                     - 
                     
                       
                         cos 
                         2 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                   
                   ) 
                 
               
               - 
               
                 i 
                  
                 
                     
                 
                  
                 
                   sin 
                   3 
                 
                  
                 
                     
                 
                  
                 θ 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   ( 
                   
                     
                       cos 
                        
                       
                           
                       
                        
                       θ 
                     
                     + 
                     
                       i 
                        
                       
                           
                       
                        
                       sin 
                        
                       
                           
                       
                        
                       θ 
                     
                   
                   ) 
                 
                 3 
               
               = 
               
                 
                   cos 
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       3 
                        
                       θ 
                     
                     ) 
                   
                 
                 + 
                 
                   i 
                    
                   
                       
                   
                    
                   sin 
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       3 
                        
                       
                           
                       
                        
                       θ 
                     
                     ) 
                   
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   cos 
                    
                   
                       
                   
                    
                   θ 
                 
                 ± 
                 
                   i 
                    
                   
                       
                   
                    
                   sin 
                    
                   
                       
                   
                    
                   θ 
                 
               
               = 
               
                 
                   
                     
                       cos 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           3 
                            
                           θ 
                         
                         ) 
                       
                     
                     ± 
                     
                       i 
                        
                       
                           
                       
                        
                       sin 
                        
                       
                           
                       
                        
                       
                         ( 
                         
                           3 
                            
                           θ 
                         
                         ) 
                       
                     
                   
                   3 
                 
                 . 
               
             
           
         
       
     
     Lastly, one final justification is about to be put forth, essentially claiming that only an availability of overlapment points can fully account for why the classical problem of the trisection of an angle cannot be solved! 
     Public sentiment on this topic, as highly influenced by the earlier discoveries of Wantzel and Galois dating all the back to the mid 1800&#39;s, instead generally leans to attributing an inability to geometrically construct cube roots as being the principal cause which prevents trisection. 
     Moreover, at the very heart of this matter lies a fundamental issue of constructability. 
     To openly dispute such issue, upon drawing an angle of arbitrarily selected magnitude, there is a good chance that its trigonometric properties will turn out to be cubic irrational. This is because a far greater number of angles exist which exhibit cubic irrational trigonometric properties than do other angles whose trigonometric properties are of rational and quadratic irrational value. 
     From such initial angle, an entire geometric construction pattern could be generated which belongs to the Euclidean formulation, as posed in  FIG. 13 . Therein, such singular drawing would depict just how a given angle VOO′ actually relates to rendered angle VOU′, amounting to exactly three times its size, by virtue of specific trigonometric properties which are inherent to each of such angles, as characterized by the famous cubic equation sin (3θ)=3 sin θ−4 sin 3  θ. 
     The basic problem with such scenario is that such drawing, although fully constructible by a process of sheer random selection, never could be repeated; thereby becoming relegated to approximation when attempting to reproduce it. 
     More particularly stated, although the likelihood of drawing an angle which exhibits cubic irrational trigonometric properties is quite high, as due to a substantial availability of them, the probability of geometrically constructing a specific angle, even one which might feature a particular transcendental trigonometric property such a pi for example, nevertheless approaches zero; being entirely consistent with the previously stipulated premise that absolutely no cubic irrational length can be geometrically constructed, but only approximated, from a given unit length. 
     To further emphasize this outstanding difficulty, consider the largely unknown fact that even the rarified transcendental number, π, can be approximated by means of geometric construction well beyond what the naked eye could detect. 
     To demonstrate this, a rational number very easily can be described by the ratio of two cubic irrational numbers by an algebraic manipulation such as: 
     
       
         
           
             
               
                 
                   1 
                    
                   3 
                 
                 9 
               
               = 
               
                 
                   
                     1 
                      
                     3 
                   
                   9 
                 
                  
                 
                   ( 
                   
                     
                       tan 
                        
                       
                           
                       
                        
                       20 
                        
                       ° 
                     
                     
                       tan 
                        
                       
                           
                       
                        
                       20 
                        
                       ° 
                     
                   
                   ) 
                 
               
             
             ; 
             whereby 
           
         
       
       
         
           
             
               
                 
                   
                     
                       1 
                        
                       3 
                     
                     9 
                   
                   = 
                     
                    
                   
                     
                       13 
                       9 
                     
                      
                     
                       
                         tan 
                          
                         
                             
                         
                          
                         20 
                          
                         ° 
                       
                       
                         tan 
                          
                         
                             
                         
                          
                         20 
                          
                         ° 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       4.7316130455 
                        
                       
                           
                       
                        
                       … 
                     
                     
                       3.2757321084 
                        
                       
                           
                       
                        
                       … 
                     
                   
                 
               
             
           
         
       
     
     Similarly, the actual transcendental value of π can be multiplied to the sin 80° in order to produce another transcendental length as follows: 
       π sin 80°=3.093864802 . . . ; and
 
       π(0.9848077530 . . . )=4(0.77346620052 . . . ).
 
     Moreover, all of the stated values in such above equation, except for l, furthermore very closely could be approximated as actual rational numbers, down to a significance of at least ten decimal places; being well beyond the accuracy of what the naked eye could detect. 
     Such estimated result is furnished directly below, whereby all constructible rational numbers thereby could be algebraically expressed as follows: 
     
       
         
           
             
               
                 
                   π 
                    
                   
                     ( 
                     
                       
                         984 
                          
                         
                           , 
                         
                          
                         807 
                          
                         
                           , 
                         
                          
                         7 
                          
                         5 
                          
                         3 
                       
                       
                         1 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         000 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   4 
                    
                   
                     ( 
                     
                       
                         77 
                          
                         
                           , 
                         
                          
                         346 
                          
                         
                           , 
                         
                          
                         620 
                          
                         
                           , 
                         
                          
                         0 
                          
                         5 
                          
                         2 
                       
                       
                         100 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         0 
                          
                         0 
                          
                         0 
                       
                     
                     ) 
                   
                 
               
                
               
                 
 
               
                
               
                 
                   π 
                    
                   
                     ( 
                     
                       
                         984 
                          
                         
                           , 
                         
                          
                         807 
                          
                         
                           , 
                         
                          
                         7 
                          
                         5 
                          
                         3 
                       
                       
                         1 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         000 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   4 
                    
                   
                     ( 
                     
                       
                         19 
                          
                         
                           , 
                         
                          
                         336 
                          
                         
                           , 
                         
                          
                         655 
                          
                         
                           , 
                         
                          
                         0 
                          
                         1 
                          
                         3 
                       
                       
                         25 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         000 
                          
                         
                           , 
                         
                          
                         00 
                          
                         0 
                       
                     
                     ) 
                   
                 
               
                
               
                 
 
               
                
               
                 
                   π 
                    
                   
                     ( 
                     L 
                     ) 
                   
                 
                 = 
                 
                   4 
                    
                   
                     ( 
                     T 
                     ) 
                   
                 
               
             
             ; 
             or 
           
         
       
       
         
           
             
               π 
                
               
                   
               
                
               L 
             
             = 
             
               4 
                
               
                 T 
                 . 
               
             
           
         
       
     
     Notice that such above described rational lengths 4, T, and L now can be geometrically constructed from an arbitrarily applied, or given length of unity. 
     In the above example, there is little need to attempt to reduce the rational length T any further than is indicated. This is because it is necessary only to know that a rational length of T=19,336,655,013/25,000,000,000 could be made use of to geometrically construct another length that very closely approximates the actual value of pi. 
     From such equation πL=4T, as determined above, the proportion 
     
       
         
           
             
               π 
               T 
             
             = 
             
               4 
               L 
             
           
         
       
     
     readily could be established; whereby a very close estimation of the length pi thereby could be identified from the geometric construction of two similar right triangles whose sides respectively consist of drawn rational lengths 4, T, and L. Understandably, the level of accuracy attributed would amount to only three, or perhaps four at the very most, significant digits. 
     To conclude, since transcendental lengths describe decimal sequences which are considered to continue on indefinitely, they cannot be exactly geometrically constructed from any long-hand division computation that is indicative of a pair of rational numbers whose quotients begin to repeat themselves. 
     In the past, such difficulty merely was bypassed by means of considering only geometric construction patterns which could be redrawn. 
     Such process simply entails selecting a given angle whose trigonometric properties are either rational or quadratic irrational. For example, upon considering a given angle VOO′ whose sine is equal to ⅓, the following algebraic relationship could be obtained: 
     
       
         
           
             
               sin 
                
               
                   
               
                
               θ 
             
             = 
             
               1 
               / 
               3 
             
           
         
       
       
         
           
             
               θ 
               = 
               
                 1 
                  
                 9.47122063 
                  
                 ° 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 
                   
                     sin 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         3 
                          
                         θ 
                       
                       ) 
                     
                   
                   = 
                     
                    
                   
                     
                       3 
                        
                       
                           
                       
                        
                       sin 
                        
                       
                           
                       
                        
                       θ 
                     
                     - 
                     
                       4 
                        
                       
                           
                       
                        
                       
                         sin 
                         3 
                       
                        
                       
                           
                       
                        
                       θ 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       3 
                        
                       
                         ( 
                         
                           1 
                           / 
                           3 
                         
                         ) 
                       
                     
                     - 
                     
                       4 
                        
                       
                         
                           ( 
                           
                             1 
                             / 
                             3 
                           
                           ) 
                         
                         3 
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     23 
                     / 
                     27 
                   
                 
               
             
           
         
       
       
         
           
             
               
                 
                   
                     3 
                      
                     θ 
                   
                   = 
                     
                    
                   
                     5 
                      
                     8 
                      
                     .4136619 
                      
                     ° 
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     3 
                      
                     
                       
                         ( 
                         
                           1 
                            
                           9 
                            
                           .47122063 
                            
                           ° 
                         
                         ) 
                       
                       . 
                     
                   
                 
               
             
           
         
       
     
     Obviously the sin (3θ) also must be a rational value because it amounts to the sum of three times such selected rational value of ⅓ plus four times the value of its cube; meaning that all coefficients within such resulting equation 23/27=3 sin θ−4 sin 3  θ very handily would consist of only rational numbers! 
     Accordingly, an associated geometric solution for the problem of the trisection of an angle very easily could be drawn merely geometrically constructing an angle whose sine equals ⅓. 
     Notice, however, that such particular drawing would remain entirely irreversible, despite being characterized by that very geometric construction pattern, as just described, belonging to the Euclidean formulation, as posed in  FIG. 13 ; thereby specifically depicting a given angle VOO′ which would exhibit a sine value of exactly ⅓. In other words, the claim that the classical problem of the trisection of an angle cannot be solved becomes further bolstered, even for an angle whose sine value amounts to 23/27; as predicated upon the fact that an availability of overlapment points must remain at work which prevents such drawing from fully being backtracked upon. Naturally, in such specific case, relevant data, as previously stipulating that the sine of the trisector for such angle would amount to exactly ⅓, only would qualify as extraneous information, whereby its use would violate the very Euclidean requirements which just so happen to be levied upon such problem. 
     Next, the issue of attempting to extract cube roots is to be addressed. In order to do this, consider that some Euclidean formulation someday might become devised, each of whose constituent geometric construction patterns would be fully reversible, as well as exhibit a rendered length that amounts to the cube of its given length. In so doing, it naturally would follow that for each of such singular drawings, a cube root of such rendered length value thereby could be geometrically constructed without having to introduce any additional relevant information. 
     Now, if a Euclidean formulation of such nature truly could be devised, an overriding question then would be whether such capability could in some way overcome the irreversible nature of any geometric construction pattern in which the magnitude of a rendered angle amounts to exactly three times the size of its given angle. For instance, could such magical Euclidean cube root capability enable angle VOU′, as appearing upon the irreversible representative geometric construction pattern for such Euclidean formulation, as posed in  FIG. 13 , to be fully backtracked upon all the way to given angle VOO′ in order to solve the classical problem of the trisection of an angle? 
     Naturally, an activity of this nature would be severely limited in that some far-fetched reversible Euclidean cube root capability only could be applied to any known aspect of such rendered angle VOU′. Such is the case because when attempting to solve the classical problem of the trisection of an angle, other lengths in  FIG. 13 , such as sin 3  θ, still would remain unknown. Since it is impossible to take the cube root of an unknown value, such very difficulty would thwart any attempts to fully backtrack from rendered angle VOU′ all the way back to given angle VOO′. 
     Accordingly, it is conjectured that some as yet undeveloped Euclidean capability to extract cube roots would have little to no impact whatsoever upon enabling the classical problem of the trisection of an angle to become solved; as based upon the fact that such hypothetical cube root development couldn&#39;t possibly offset the irreversibility of such  FIG. 13  representative geometric construction pattern. Once recognizing that it otherwise must be an availability of overlapment points which actually prevents a backtracking activity of this nature from being accomplished, it becomes rather obvious that an introduction of any professed Euclidean cube root capability couldn&#39;t possibly rectify a plaguing Euclidean irreversibility limitation which instead actually prevents the classical problem of the trisection of an angle from actually being solved! 
     In closing, it is important to note that vital input leading to the very discovery of significant findings, as presented herein, never even would have been obtained had it not been for one strange incident which occurred in 1962. It was then, that my high school geometry teacher informed me that it was impossible to perform trisection solely by conventional Euclidean means. Her disclosure moved me greatly. I become intrigued; thereby fueled with a relentless curiosity to ascertain secrets needed to unlock a trisection mystery that had managed to baffle mathematicians for millennia! 
     Naturally, during such prolonged fifty-five year investigation, certain critical aspects pertaining to trisection became evident well ahead of others. For example, I realized that a general perception of geometry dating back all the way to the time of Archimedes perhaps might be better served by means of now considering a much needed extension to it; one that would transcend beyond the confines of conventional Euclidean practice, and amplify even upon Webster&#39;s own definition of such word; whereby from an availability of straight lines, intersection points, circles, triangles, rectangles and parallelograms, leading to an overall profusion of spheres, prisms and even pyramids, eventually would emerge the far greater understanding that any visualization which could be mathematically interpreted diagrammatically should be considered to be of a geometric nature! 
     Such enhanced perception would apply to real world events wherein certain articulating mechanisms, even those capable of performing trisection, would be credited for accomplishing specific geometric feats that otherwise could not be matched solely by conventional Euclidean means. Certain famous convolutions then would comprise known geometric shapes, such as the Conchoid of Nicomedes, the Trisectrix of Maclaurin, the catenary or hyperbolic cosine, the elliptical cone, the parabola, the Folium of Decartes, the Limacon of Pascal, the Spiral of Archimedes, the hyperbolic paraboloid, as well as logarithmic and even exponential curves; as previously were considered to be taboo within an otherwise limited realm of conventional Euclidean practice. 
     Revolutionary material, as presented herein, consists largely of a wealth of information that can be traced directly to a newly established methodology that, in turn, is predicated upon a proposed extension to conventional Euclidean practice. In order to succeed at developing such rather unconventional output, it became essential to take good notes over extended periods of time. Moreover, copyrights conveniently served to document dates pertaining to significant discoveries. 
     Many concepts, as expressed herein, stem from a far broader pretext which previously was referred to as equation sub-element theory Upon reading my unfinished treatise entitled,  The Principles of Equation Sub - element Theory ; United States Copyright Number TXu 1-960-826 granted in April of 2015, it would become apparent that such purported new field of mathematics unfortunately only is in its embryonic stage of development. By no means should it be considered to be complete! In fact, such document already was amended under United States Copyright Number TXu 1-976-071 during August of 2015, and presently is undergoing yet another revision in order to keep abreast with recent findings, some of which are to be disseminated to the public for the very first time herein. Such copyright process permits premature theories to become documented, and thereafter revised without difficulty in order to suitably become refined into viable output. 
     Any prior art issue which might arise concerning the concurrent preparation of two documents which might contain somewhat similar, or even closely related information could be reconciled by means of controlling which becomes published and/or disseminated first. 
     In this regard, such above described copyrighted material should pose no problem because it never before was published, nor even disseminated to the general public in any manner whatsoever. Hence, there is no compelling reason to suspect that information contained therein might qualify as prior art material. Such position is predicated upon one basic understanding; being, that because the exclusive right granted by such copyrights to reproduce and/or distribute never before was exercised, it becomes impossible for anyone to be aware of the very nature of such material. 
     Conversely, if the argument that such copyrighted material actually should qualify as prior art otherwise were to persist in some thoroughly unabated manner, it then would require a review by some expert who, by gaining access in some surreptitious manner to undisclosed information, thereby independently only would collaborate that such unfinished copyrighted information is seriously flawed. For example, such hypothetical review would reveal that the term transcendental was used inappropriately throughout such copyright and amendment thereto. Today such mistake can be easily explained by mentioning that a thorough understanding of Al-Mahani&#39;s work was gained only after such copyrighted information first became amended. Therefore, the correct replacement term, being cubic irrational, couldn&#39;t possibly have appeared in earlier forms of such copyrights. Moreover, had such copyrighted information been released to the public, well before it completion, then inaccurate information stating that only transcendental values, as consisting of a limited subset of all cubic irrational numbers, could be automatically portrayed by means of performing trisection; thereby contradicting correct details as presented herein. 
     Regarding the 2½ year interim which elapsed between the granting of such two 2015 copyrights and the present day completion of this disclosure, such period of time is indicative of an expected turnaround needed to effectively update information that well should be construed to include complex revolutionary material, thereby exceeding that of evolutionary projects by some considerable degree; whereby more leniency should be extended for their proper update. 
     By means of documenting what might appear to be similar theory concurrently in dual records, a process of leap frog would unfold, whereby what might have seemed to be credible information appearing in a copyrighted document, when worked upon earlier, soon would become outdated by a subsequent accounting, such as this one; thereby necessitating yet another revision of such copyrighted document to be completed before its release in order to remain totally consistent with refinements now incorporated herein. 
     Accordingly, by means of publishing the contents of this disclosure well ahead of any portion of such, as yet undisclosed 500+ page copyrighted treatise, this document shall be the first to become disseminated anywhere on earth. Lastly, whereas such copyrights, as identified directly above, evidently do not appear to qualify as prior art, it thereby should not be necessary to furnish a copy of them along with the submittal of this patent disclosure.