Journal of Mathematics and System Science 7 (2017) 25-42 doi: 10.17265/2159-5291/2017.01.003 D DAVID PUBLISHING
The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
Kimuya M Alex Department of Mathematics, Faculty of Physics, Meru University of Science and Technology, Kenya.
Received: September 19, 2016 / Accepted: October 16, 2016 / Published: January 25, 2017.
Abstract: The objective of this paper is to provide a provable solution of the ancient Greek problem of trisecting an arbitrary angle employing only compass and straightedge (ruler). (Pierre Laurent Wantzel, 1837) obscurely presented a proof based on ideas from Galois field showing that, the solution of angle trisection corresponds to solution of the cubic equation; 3 3 1 = 0, which is geometrically irreducible [1]. The focus of this work is to show the possibility to solve the trisection of an angle by correcting some 𝑥𝑥 − 𝑥𝑥 − flawed methods meant for general construction of angles, and exemplify why the stated trisection impossible proof is not geometrically valid. The revealed proof is based on a concept from the Archimedes proposition of straightedge construction [2, 3].
Key Words: Angle trisection; Compass; Ruler (Straightedge); Classical Construction; GeoGebra Software; Greek’s geometry; Cubic equation; Plane geometry; Solid geometry
Notations sought under the Greek’s rules of Geometry’ [1, 2]. Three algebraic constraints exists which state; a length Denotes an angle Denotes a straight line and a length can only be constructible if and only if it represents a ∠ 2 Two Dimensional constructible number, an angle is constructible if and ∩����∀� 3 Three Dimensional 𝐷𝐷 only if its cosine is a constructible number, and, a 1.𝐷𝐷Introduction number is constructible if and only if it can be written in the four basic arithmetic operations and the The early Greek mathematicians were unable to extraction of the square roots but not on higher roots solve three problems of compass-ruler (straightedge) [4]. These three conditions are quite fashionable in construction; the ‘trisection of an angle’, ‘how to Euclidean constructions, and they enabled the early double the volume of a given cube’, and the problem mathematicians correctly justify the construction of all of ‘squaring a circle’. Eventually the problems were the angles multiples and sub multiples of 15 and the assumed to be unsolvable under the restrictions associated regular polygons. However, this novel imposed by the Greek mathematicians. The scope of paper presents a correct algorithm justifying that the this paper is restricted on the trisection of an arbitrary stated algebraic specifications does not apply to all the angle. The unclear proof of angle trisection plane geometrical problems, by drawing out some impossibility was established, based on ideas from limitations in the presented proof of impossibility. For Galois field of algebra, and it stated that ‘The instance, some angles such as 45°, 90° and 180° trisection of an angle corresponds to the solution of a are trisectible following the Euclidean rigor of certain cubic equation whose solution cannot be construction, but there trisection methods could not be adopted since they do not generalize for the Corresponding author: Kimuya M Alex, Department of Mathematics, Faculty of Physics, Meru University of Science partitioning of all the angles into the desired ratio [5]. and Technology, Kenya. E-mail; [email protected]. Moreover, it is geometrically possible to bisect a line 26 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex segment between two points in a plane, as well as the in one plane [8]. partitioning a line segment into equal fractions using Considering three points; , , and , not lying in the Greek’s tools of geometry [6, 7]. The presented one straight line but all connected together by straight 𝐴𝐴 𝐵𝐵 𝐶𝐶 proof of the impossibility does not show that some of lines as shown on figure (1), it is not possible for all these cases are constructible. Considering the the three points to lie on different planes. To justify trisection of and angle as a cubic equation translated this proposition, let points , , and lie on two the problem from a 2 (plane geometry) problem as distinct planes; and . Since both point and 𝐴𝐴 𝐵𝐵 𝐶𝐶 it should be sought, to a 3 (solid geometry) lie on plane , the straight line lie on plane . 𝐷𝐷 𝑀𝑀 𝑁𝑁 𝐴𝐴 𝐵𝐵 problem, which involve equations of the form 3 as Also, since points and lie on plane , the 𝐷𝐷 𝑀𝑀 𝐴𝐴𝐴𝐴 𝑀𝑀 discussed in section 1.2 . This was a serious straight line also lie on plane . Therefore, 𝑛𝑛 𝐴𝐴 𝐵𝐵 𝑁𝑁 misconstruction and due to inability to geometrically plane and plane have line in common (they 𝐴𝐴𝐴𝐴 𝑁𝑁 solve the cubic equations, and the fact that no intersect along line ). Point does not lie in line 𝑀𝑀 𝑁𝑁 𝐴𝐴𝐴𝐴 geometrical algorithm has been presented to solve the with and therefore not common for both planes. 𝐴𝐴𝐴𝐴 𝐶𝐶 partitioning of an angle into a given ratio, Thus it is not possible for points , , to lie in 𝐴𝐴𝐴𝐴 mathematicians wicked to pseudo mathematical both planes and . Considering triangle , 𝐴𝐴 𝐵𝐵 𝐶𝐶 approaches which do not redress the problem with the the whole triangle lies only in one plane. Hence, for 𝑀𝑀 𝑁𝑁 𝐴𝐴𝐴𝐴𝐴𝐴 desired degree of correctness [9]. This paper relies on any plane containing all of triangle must also a simple concept of the Archimedes theorem of contain its three vertices , , and C. This shows 𝐴𝐴𝐴𝐴𝐴𝐴 straightedge which stated “If we are in possession of a that the whole of any triangle lies only in one plane. 𝐴𝐴 𝐵𝐵 straightedge that is notched in two places, then it is Considering as the acute angle to be trisected, possible to trisect an arbitrary angle”, by revealing a it is required that the points defining the trisection ∠𝐴𝐴𝐴𝐴𝐴𝐴 geometrical solution for this ancient problem, lines lie in the curve subtending at point , contrally to the Archimedes approach of using a and that point is defined in a two dimensional ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 marked straightedge. An elementary proof of solving coordinates system. Therefore, the problem of 𝐵𝐵 the posed problem of angle trisection is then exposed trisecting an angle has to be sought following the from the construction, by trisection of 48° and 60° classical rules of Euclidean geometry. angles. 1.2 The Mistake In Pierre Laurent Wantzel’s Proof Of 1.1 To show that any three points not lying on a Angle Trisection Impossibility (1837) straight line lie in the same plane Theorem 1 illustrates how three points defining a When two rays in a plane share a common endpoint, given angle, and not collinear lie in the same plane, an angle is produced between the two rays. An and that in plane geometry two different planes cannot example is in a geometrical figure such as triangle. share all points in common.It is pellucid that algebra Any two sides of a triangle have a common point at the vertices and thus angles of some size are defined between the two sides of the figure. In this section, a brief discussion about any three points, one not collinear with the other two in a plane is presented. Consider the following theorem: Theorem 1: Any three points not lying in a straight line lie only in same plane, and every triangle lies only Fig. 1 Illustration of a plane geometry figure.
The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 27
has well been applied in justification of plane in , join points and , using a straight line geometric problems and the underlying concepts and add to . Given a third point in , 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 correctly verified. In this sense, degree two construct a circle with center and radius of 𝐴𝐴���𝐴𝐴� 𝐶𝐶 𝑂𝑂 𝐶𝐶 polynomials have been employed in defining the line segment joining and , and add it to 𝑂𝑂 𝐴𝐴���𝐴𝐴� constructions in a plane such as; determining the using a compass. Given two different curves and 𝐴𝐴 𝐵𝐵 𝐶𝐶 quadratic equations defining some angle bisection in (such that and are either a line or a circle 𝛾𝛾 𝑢𝑢 lines. The scope of this paper is restricted in Euclidean in ), select a point that is common to both and 𝐶𝐶 𝛾𝛾 𝑢𝑢 plane geometry of straightedge and compass and add it to . Therefore a point, line, or circle is 𝐶𝐶 𝑇𝑇 𝛾𝛾 construction. Plane geometric construction is said to be from a configuration if it can be 𝑢𝑢 𝐶𝐶 governed by many propositions and theorems, which obtained from after applying a finite number of 𝐶𝐶 greatly influenced the development of geometry. As construction steps. From these deductions, there are 𝐶𝐶 stated earlier, some few angles are geometrically only two positions in which the point can be trisectible, it is possible to bisect both an angle and a located. The point however, quadrasects angle 𝑇𝑇 straight line, as well as dividing a straight line and it must be outside the initial plane defined 𝑇𝑇 segment into the desired number of equal portions. It by radius . Trisection of an angle may not be 𝐴𝐴𝐴𝐴𝐴𝐴 is also possible to construct lines of magnitude 2 easily solved in such a linear construction, but a ���� and 5 . The existence of these well justified 𝐴𝐴𝐴𝐴 √ deeper look at the problem would obligate some constructions implies the great probability of serious exploration of the relationship between some √ trisecting a given angle, or the construction of an chords and curves in a circular plane, as presented in angle of a certain ratio. The third algebraic condition [3]. The wantzel’s proof considered the lemma; there specified in section 1.0 limits plane geometric is no power of 2 that is evenly divisible by 3 which constructions from advancing in higher orders of roots, was employed to demonstrate the angle trisection above the square roots of numbers. According to this impossibility using concepts from Galois field of article, this condition is not fashionable in restricting numbers. A proof upon this lemma can be described the extraction of higher order roots above square roots, as follows: in geometry, based on the fact that a line segment can Corollary: Let be a field, and let be an be geometrically fractioned into the required ratio [8]. extension of that is constructible out of by a 𝑄𝑄 𝑅𝑅 Depending on the application, it is quite difficult to finite order of quadratic extensions. Then does not 𝑄𝑄 𝑅𝑅 determine the magnitudes of the sliced portions of a contain any cubic extensions of . 𝑄𝑄 line segment, and algebraically, this implies the Proof: If contained a cubic extension of , 𝑉𝑉 𝑅𝑅 possibility of constructing numbers which do not then the dimension of over would be a multiple 𝑄𝑄 𝑉𝑉 𝑅𝑅 represent constructible numbers, a case contradicting of three. On the other hand, if is obtained from 𝑄𝑄 𝑅𝑅 the stated algebraic condition. This section in a simple by a tower of quadratic extensions, then the dimension 𝑄𝑄 𝑅𝑅 and brief approach discusses the angle trisection of over is a power of two. It can therefore be impossibility, to reveal how the algebraic approach stated, any point, line, or circle that can be constructed 𝑄𝑄 𝑅𝑅 turned the problem from plane geometry problem into from a configuration is definable in a field a solid geometry problem and thus the impossibility. obtained from the coefficients of all the objects in 𝐶𝐶 The impossibility proof was centered on concepts after taking a finite number of quadratic extensions, 𝐶𝐶 from the Galois field. Therefore, based on the ancient whereas trisection of an angle such as will proof, the problem can be stated as: Define a basically be definable in a cubic extension of the field ∠𝐴𝐴𝐴𝐴𝐴𝐴 configuration to be a finite collection of points, generated by the coordinates of , , . Based on lines, and circles in the Euclidean plane. Define a 𝐶𝐶 these coordinates, three plane angles , construction step to be one of the operations to enlarge 𝐴𝐴 𝐵𝐵 𝐶𝐶 and can be constructed to represent three the collection as: Given two distinct points and ∠𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐵𝐵𝐵𝐵𝐵𝐵 different planes such that; the three plane angles add ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶 𝐴𝐴
28 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex up to less than four right angles and any two of them Therefore, considering two angles, and such add up to more than the third one [6, 8]. These that = , then, , where and 𝛼𝛼 𝛽𝛽 conditions can be met in a solid geometric correspond to and respectively. Thus taking the 𝛼𝛼⁄𝛽𝛽 𝛿𝛿 𝛿𝛿 ≅ 𝑥𝑥 𝑎𝑎 𝑏𝑏 construction, when for example, the three plane angles ratios between any two given angles; = , some 𝛼𝛼 𝛽𝛽 made from isosceles triangles of equal legs meet their cases would yield a ratio from which to derive the 𝛼𝛼⁄𝛽𝛽 𝛿𝛿 vertices at a common endpoint. In analytic Euclidean relation = is geometrically difficult. plane geometry, an angle is genetically defined in a Therefore, for consistency, it is important to choose a 𝑎𝑎⁄𝑏𝑏 𝑥𝑥 two dimensions ( , - coordinates) with and as constant difference between two angles, as such the real numbers, and not in three dimensions as presented difference between the constructible angles multiples 𝑥𝑥 𝑦𝑦 𝑥𝑥 𝑦𝑦 in the Pierre Wantzel’s cubic equation of the of 15 is 15°, and multiple or sub multiples of 15°. impossibility. From this discussion, it is evident that This paper therefore considered the relation between the presented proof of impossibility dictates rotation any two angles at a difference of 10°, or a difference of objects to reach some accuracy as applied in multiple of 10 from each other in there descending solutions using ‘other methods’ of trisecting an angle order. The most significant ratio was 60° 20° = 3: 1. [24], and these operations are prohibited in Euclidean This consideration is due to the fact that the 60° ⁄ plane geometry. Therefore, the impossibility proof angle form the base for the construction of all the that an angle cannot be divided into a certain fraction, angles multiples of 15 [3]. The novelty of these or that other angles not under angles of base 60° ratios rose due to the need to generate a method in cannot be constructed has no geometric precision. which a particular considerable distance between two Thus the trisection of an angle is typically a plane points transform equally into two different points geometry problem (2D), and not a 3D problem as it without rotation or sliding of objects to locate the new has been sought. points. Consider figure (2). The resolution of this work is to geometrically ratify that the conferred 2. Hypothesis relations exists, such that: = ′ = ′ = = In a typical plane geometric construction, the , where is the radius of the circle, and that 𝐴𝐴���𝐴𝐴� 𝐴𝐴���𝐷𝐷�� 𝐷𝐷����𝐸𝐸� 𝐴𝐴���𝐴𝐴� relation between two angle can be defined by ratios; = ′ = ′. This implies 3 = 𝑟𝑟 𝑟𝑟 = , with and as the curves subtending the . larger angle to the smaller angle at a point respectively. ∠𝐵𝐵𝐵𝐵𝐵𝐵 ∠𝐸𝐸𝐸𝐸𝐷𝐷 ∠𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐵𝐵𝐵𝐵𝐵𝐵 𝑎𝑎⁄𝑏𝑏 𝑥𝑥 𝑎𝑎 𝑏𝑏 ∠𝐶𝐶𝐶𝐶𝐶𝐶
Fig. 2 Transformation of point to points ′ and geometrically.
𝑫𝑫 𝑫𝑫 𝑬𝑬
Journal of Mathematics and System Science 7 (2017) 25-42 doi: 10.17265/2159-5291 3. Materials and Methods out the following steps of construction using a ruler and compass. 3.1 Materials (1) Draw a straight line between two points and The required mathematical tools in solving this . 𝐴𝐴 problem include; (2) Mark a point equidistant from both 𝐵𝐵 Compass and and draw an arc centered at . 𝐶𝐶 𝐴𝐴���𝐴𝐴� 𝐴𝐴 Ruler (straightedge) (3) Join the chord , and construct its bisection 𝐵𝐵 𝐵𝐵𝐵𝐵 𝐴𝐴 Piece of a drawing paper through a point to cut curve at . 𝐶𝐶���𝐶𝐶� Pencil (4) Construct the bisection of to cut 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 Computer curve at . ∠𝐸𝐸𝐸𝐸𝐸𝐸 GeoGebra Software installed in the computer. (5) Further, construct the bisection of to cut 𝐵𝐵𝐵𝐵 𝐹𝐹 curve at . 3.2 Methodology ∠𝐸𝐸𝐸𝐸𝐸𝐸 (6) Position the compass at point and make an 𝐹𝐹𝐹𝐹 𝐺𝐺 This section has improved on a construction presented arc of length to cut curve at . 𝐸𝐸 for the classical construction of angles in general [3]. (7) With the compass at point , mark point of �𝐷𝐷��𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 The present theorem begins by depicting how one can distance between carve . 𝐻𝐻 𝐼𝐼 construct angles multiples of 2 by constructing a (8) Construct the bisector of to cut curve 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 32° angle using the usual Greek’s tools of geometry, at point . ∠𝐼𝐼𝐼𝐼𝐼𝐼 𝐵𝐵𝐵𝐵 and then the construction of angles multiples of 5, (9) With the compass at point , mark an arc of 𝐽𝐽 and 10 or their sub multiples, by constructing an radius to cut curve at . = 48°. 𝐼𝐼 angle of 40° , and the methods are justified by (10) Reflect point about to produce point ’. 𝐵𝐵𝐵𝐵 𝐼𝐼𝐼𝐼 𝐾𝐾 ∠𝐾𝐾𝐾𝐾𝐾𝐾 trisecting angles of 48° and 60° respectively. (11) Again using chord , place the compass at 𝐵𝐵 𝐴𝐴 𝐵𝐵 3.2.1 Construction of ° angle using a ruler ’ and mark a point ’ on the arc ’ . ’ is an 𝐵𝐵𝐵𝐵 (straightedge) and compass only. image of point . 𝟑𝟑𝟑𝟑 𝐵𝐵 𝐽𝐽 𝐵𝐵 𝐶𝐶 𝐽𝐽 (1) Draw a straight line between two points and . (12) Draw a straight line from point through ’ 𝐽𝐽 (2) Mark a point equidistant from both to meet the produced diameter ’ externally at a 𝐴𝐴 𝐵𝐵 𝐾𝐾 𝐽𝐽 and and draw an arc centered at . point . 𝐶𝐶 �𝐴𝐴��𝐴𝐴� 𝐴𝐴 𝐵𝐵𝐵𝐵 (3) Join the chord , and construct its bisection 3.2.3 Proof 𝐵𝐵 𝐵𝐵𝐵𝐵 𝐴𝐴 𝑀𝑀 through a point to cut curve at . Consider figure (4). 𝐶𝐶���𝐶𝐶� (4) Construct the bisection of to cut 3.2.3.1 To show = = 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 curve at . This part of proof employs the compass equivalence ∠𝐸𝐸𝐸𝐸𝐸𝐸 𝑴𝑴𝑴𝑴�����′ �𝑨𝑨��𝑨𝑨�′ 𝑨𝑨𝑨𝑨���� (5) Further, construct the bisection of to cut theorem to justify that; length ’ is equal to the 𝐵𝐵𝐵𝐵 𝐹𝐹 curve at . radius of the original (blue) circle, by drawing a circle ∠𝐸𝐸𝐸𝐸𝐸𝐸 𝑀𝑀𝑀𝑀 (6) Position the pair compass at point and make of radius ′, and center ’. If ′ = ′, then the 𝐹𝐹𝐹𝐹 𝐺𝐺 an arc of length to cut curve at . circumference of the circle with radius ′ has to 𝐸𝐸 �𝑀𝑀���𝑀𝑀� 𝐽𝐽 �𝑀𝑀���𝑀𝑀� �𝐴𝐴��𝐴𝐴� (7) With the compass at point , mark point of pass through point as shown in figure (5). 𝐷𝐷���𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 �𝑀𝑀���𝑀𝑀� distance between carve . = 32°. Thus, the circle in blue circumference is congruent 𝐻𝐻 𝐼𝐼 𝐴𝐴 3.2.2 Trisection of ° Using Compass-ruler to the circle defined by the circumference in orange, 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 ∠𝐼𝐼𝐼𝐼𝐼𝐼 (Straightedge) Construction to Proof the Construction implying: ′ = ′ = , circles radii. Therefore, ∠𝟒𝟒𝟒𝟒 of ° this proof indicates that if a trisection is not precisely 𝑀𝑀����𝑀𝑀� �𝐴𝐴��𝐴𝐴� 𝐴𝐴���𝐴𝐴� Figure (4) depicts the results obtained after carrying correct, the two circles would not be congruent. ∠𝟑𝟑𝟑𝟑
Journal of Mathematics and System Science 7 (2017) 25-42 doi: 10.17265/2159-5291
Fig. 3 Construction of ° angle.
𝟑𝟑𝟑𝟑
Fig. 4 Geometrical Proof of Trisecting ° Angle.
𝟒𝟒𝟒𝟒
The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 31
Fig. 5 To Justify that ′ = ′ = .
3.2.3.2 To show that𝑴𝑴𝑴𝑴��� �� �𝑨𝑨��𝑨𝑨� =�𝑨𝑨𝑨𝑨��� subject; = 180° ( ′ ′ + ′ ). (3) From figure (4), = = , implying triangle Substituting equations (1) and (2) in equation (3); ∠𝑲𝑲𝑲𝑲𝑲𝑲 𝟑𝟑∠𝑱𝑱𝑱𝑱𝑱𝑱 ∠𝐾𝐾𝐾𝐾𝐾𝐾 − ∠𝐵𝐵 𝐴𝐴𝐽𝐽 ∠𝐽𝐽 𝐴𝐴𝐴𝐴 is equilateral. = 30°. The goal of this = 180° ( + 180° 4 ) 𝐴𝐴���𝐴𝐴� �𝐵𝐵��𝐵𝐵� �𝐶𝐶��𝐶𝐶� section of proof is to show that = 32° . = + 4 = 3 . (4) 𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐸𝐸𝐸𝐸𝐸𝐸 ∠𝐾𝐾𝐾𝐾𝐾𝐾 − ∅ − ∅ Applying a concept of transformations, the point is Thus from (4), = 3 . ∠𝐼𝐼𝐼𝐼𝐼𝐼 ∠𝐾𝐾𝐾𝐾𝐾𝐾 −∅ ∅ ∅ reflected on the circumference of the circle, on the Since = 48°, = 16° (a bisector of 𝐽𝐽 ∠𝐾𝐾𝐾𝐾𝐾𝐾 ∠𝐽𝐽𝐽𝐽𝐽𝐽 plane ′ to yield point ′ as shown above. ), implying = 32°. ∠𝐾𝐾𝐾𝐾𝐾𝐾 ∠𝐽𝐽𝐽𝐽𝐽𝐽 Let = . Figure (6) demonstrates the presented proof of 𝐵𝐵𝐵𝐵𝐵𝐵 𝐽𝐽 ∠𝐼𝐼𝐼𝐼𝐼𝐼 ∠𝐼𝐼𝐼𝐼𝐼𝐼 Triangle ′ ′ , by the property . It trisecting 48° using the GeoGebra software. ∠𝐽𝐽𝐽𝐽𝐽𝐽 ∅ follows that, ′ " = . (1) 𝐵𝐵 𝐴𝐴𝐴𝐴 ≡ 𝐵𝐵𝐵𝐵𝐵𝐵 𝑆𝑆𝑆𝑆𝑆𝑆 3.3 Construction∠ of ° angle using compass and Since: ′ = ′ , ′ ′ = ′ ′ = (Base ∠𝐵𝐵 𝐴𝐴𝐴𝐴 ∅ ruler (straightedge) only. angles of isosceles triangle ′ ). 𝟒𝟒𝟒𝟒 𝑀𝑀����𝑀𝑀� �𝐴𝐴��𝐴𝐴� ∠𝐽𝐽 𝑀𝑀𝑀𝑀 ∠𝐵𝐵 𝐴𝐴𝐴𝐴 ∅ Again, ′ = 2 (Sum of two interior angles is Carrying out the following steps of a construction 𝐴𝐴𝐴𝐴 𝑀𝑀 equal to the size of the opposite exterior angle of the would help to construct an angle of 40° using only a ∠𝐴𝐴𝐴𝐴 𝐾𝐾 ∅ triangle). We also have; ′ = ′ = 2 , (Base compass and a ruler, as illustrated in figure (7). angles of isosceles triangle ′ ). (1) Draw a straight line between two points and ∠𝐴𝐴𝐽𝐽 𝐾𝐾 ∠𝐽𝐽 𝐾𝐾𝐾𝐾 ∅ From these deductions, ′ = 180° 4 . (2) . 𝐽𝐽 𝐴𝐴𝐴𝐴 𝐴𝐴 Consider; ′ ′ + ′ + = 180° (2) Mark a point equidistant from both ∠𝐽𝐽 𝐴𝐴𝐴𝐴 − ∅ 𝐵𝐵 (Angles on a straight line). Making the and and draw an arc centered at . ∠𝐵𝐵 𝐴𝐴𝐽𝐽 ∠𝐽𝐽 𝐴𝐴𝐾𝐾 ∠𝐾𝐾𝐾𝐾𝐾𝐾 𝐶𝐶 �𝐴𝐴��𝐴𝐴� 𝐴𝐴 ∠𝐾𝐾𝐾𝐾𝐾𝐾 𝐵𝐵 𝐵𝐵𝐵𝐵 𝐴𝐴
32 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
Fig. 6 To illustrate the Trisection of ° angle.
(3) Join the chord , and construct𝟒𝟒𝟒𝟒 its bisection trisecting an angle of 60° using GeoGebra. Consider through a point to cut curve at . figure (8) generated after performing the following 𝐶𝐶���𝐶𝐶� (4) Construct the bisection of to cut curve steps of construction using the geometry software: 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 at . (1) Draw a straight line between two points and . ∠𝐸𝐸𝐸𝐸𝐸𝐸 (5) Further, construct the bisection of to cut (2) Mark a point equidistant from both 𝐵𝐵𝐵𝐵 𝐹𝐹 𝐴𝐴 𝐵𝐵 curve at . and and draw an arc centered at . ∠𝐸𝐸𝐸𝐸𝐸𝐸 𝐶𝐶 𝐴𝐴���𝐴𝐴� 𝐴𝐴 (6) Position the pair compass at point and make (3) Join the chord , and construct its bisection 𝐹𝐹𝐹𝐹 𝐺𝐺 𝐵𝐵 𝐵𝐵𝐵𝐵 𝐴𝐴 an arc of length to cut curve at . through a point to cut curve at . 𝐸𝐸 �𝐶𝐶��𝐶𝐶� (7) With the compass at point , mark point of (4) Construct the bisection of to cut curve 𝐷𝐷���𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 distance between carve . = 32. at . 𝐻𝐻 𝐼𝐼 ∠𝐸𝐸𝐸𝐸𝐸𝐸 (8) Construct the bisection of angle to cut (5) Further, construct the bisection of to cut 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 ∠𝐼𝐼𝐼𝐼𝐼𝐼 𝐵𝐵𝐵𝐵 𝐹𝐹 curve at . curve at . 𝐼𝐼𝐼𝐼𝐼𝐼 ∠𝐸𝐸𝐸𝐸𝐸𝐸 (9) Bisect at on curve . (6) Position the pair compass at point and make 𝐵𝐵𝐵𝐵 𝐽𝐽 𝐹𝐹𝐹𝐹 𝐺𝐺 (10) Placing the compass at , mark a point , an arc of length to cut curve at . ∠𝐽𝐽𝐽𝐽𝐽𝐽 𝐾𝐾 𝐵𝐵𝐽𝐽 𝐸𝐸 distance on curve . = 40°. (7) With the compass at point , mark point of 𝐼𝐼 𝐿𝐿 �𝐷𝐷��𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 3.3.1 Trisection of ° angle to proof the distance between carve . �𝐵𝐵��𝐵𝐵� 𝐼𝐼𝐼𝐼 ∠𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻 𝐼𝐼 construction of ° angle (8) Construct the bisection of to cut curve 𝟔𝟔𝟔𝟔 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 This section presents an illustrative construction of at . 𝟒𝟒𝟒𝟒 ∠𝐼𝐼𝐼𝐼𝐼𝐼 𝐵𝐵𝐵𝐵 𝐽𝐽 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 33
Fig. 7 Construction of ° angle.
(9) Bisect at 𝟒𝟒𝟒𝟒 on curve . an arc of radius to cut curve at . (10) Placing the compass at , mark a point , (7) With the compass at point , mark point of ∠𝐽𝐽𝐽𝐽𝐽𝐽 𝐾𝐾 𝐵𝐵𝐵𝐵 𝐷𝐷���𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 distance on curve . distance between carve . 𝐼𝐼 𝐿𝐿 𝐻𝐻 𝐼𝐼 (11) Construct the bisection of to cut curve (8) Construct the bisection of to cut curve �𝐵𝐵��𝐵𝐵� 𝐼𝐼𝐼𝐼 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 at . at . ∠𝐿𝐿𝐿𝐿𝐿𝐿 ∠𝐼𝐼𝐼𝐼𝐼𝐼 From the algebra window, by default the software (9) Bisect at on curve . 𝐵𝐵𝐵𝐵 𝑀𝑀 𝐵𝐵𝐵𝐵 𝐽𝐽 awards the symbols , , for , , and (10) Placing the compass at , mark a point , ∠𝐽𝐽𝐽𝐽𝐽𝐽 𝐾𝐾 𝐵𝐵𝐵𝐵 respectively. It is clearly shown that, distance on curve . 𝛼𝛼 𝛽𝛽 𝛾𝛾 ∠𝐶𝐶𝐶𝐶𝐶𝐶 ∠𝐿𝐿𝐿𝐿𝐿𝐿 𝐼𝐼 𝐿𝐿 = = = 20°. (11) Construct the bisection of to cut curve ∠𝑀𝑀𝑀𝑀𝑀𝑀 �𝐵𝐵��𝐵𝐵� 𝐼𝐼𝐼𝐼 3.3.2 A Geometrical Proof that it is Possible to at . ∠𝐶𝐶𝐶𝐶𝐶𝐶 𝐿𝐿𝐿𝐿𝐿𝐿 ∠𝑀𝑀𝑀𝑀𝑀𝑀 ∠𝐿𝐿𝐿𝐿𝐿𝐿 Trisect ° Angle (12) Reflect point about point to get point 𝐵𝐵𝐵𝐵 𝑀𝑀 Consider the following steps of construction and the ’ as shown below. 𝟔𝟔𝟔𝟔 𝐵𝐵 𝐴𝐴 produced diagram; (13) With the compass at ′, mark a point ′such 𝐵𝐵 (1) Draw a circle with radius , and its dimeter that, ′ ′ = . Draw a straight line from point 𝐵𝐵 𝑀𝑀 extended outside as depicted below. through ′ to meet the extended diameter �𝐴𝐴��𝐴𝐴� �𝐵𝐵��𝑀𝑀��� �𝐵𝐵��𝐵𝐵�� 𝐶𝐶 (2) Mark a point on the circumference ’externally at a point . 𝑀𝑀 equidistant from both and . 3.3.3 Theorem 2: Based on the classical Greek’s 𝐶𝐶 𝐵𝐵𝐵𝐵 𝑁𝑁 (3) Join the chord , and construct its bisection at rules of construction it can be deduced that: “If the 𝐴𝐴���𝐴𝐴� 𝐴𝐴 𝐵𝐵 a point to cut curve at . terminal point of the curve or the chord subtending 𝐶𝐶𝐶𝐶 (4) Construct the bisection of to cut curve the “trisection angle” at the center of the circle is 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 at . reflected on the opposite side at circumference in the ∠𝐸𝐸𝐸𝐸𝐸𝐸 (5) Further, construct the bisection of to cut same plane, and a line drawn from the terminal of the 𝐵𝐵𝐵𝐵 𝐹𝐹 curve at . “trisected angle” through the mirrored point to ∠𝐸𝐸𝐸𝐸𝐸𝐸 (6) Position the pair compass at point and make intersect the diameter outside the circle , such that, 𝐹𝐹𝐹𝐹 𝐺𝐺 𝐸𝐸
34 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
Fig. 8 Trisection of ° angle. the distances between𝟔𝟔𝟔𝟔 the point of intersection and the triangles ′ and ′ are both isosceles. reflected point is equal to the radius of the circle, then Applying the approach used in the previous proof and 𝑁𝑁𝑁𝑁 𝐴𝐴 𝑀𝑀 𝐴𝐴𝐴𝐴 it is geometrically possible to trisect a given angle.”. letting = , then ′ = ′ = . A proof elaborating this proposition is presented Thus ′ = 2 (‘sum of two interior angles in a ∠𝑀𝑀𝑀𝑀𝑀𝑀 𝜃𝜃 ∠𝑀𝑀 𝑁𝑁𝑁𝑁 ∠𝑀𝑀 𝐴𝐴𝐴𝐴 𝜃𝜃 using the following diagram. The point is the triangle equal the size of the opposite exterior angle of ∠𝐴𝐴𝑀𝑀 𝐶𝐶 𝜃𝜃 intersection between the diameter of the circle and a the triangle’). It follows that ′ = ′ = 2 𝑁𝑁 line from the terminal of the arc (point ) of the acute (base angles of an isosceles triangle). Again, ∠𝑀𝑀 𝐶𝐶𝐶𝐶 ∠𝐴𝐴𝑀𝑀 𝐶𝐶 𝜃𝜃 , through ′, where ′ is a reflection of point ′ = 180° 4 , so that we have: 𝐶𝐶 . ′ + ′ + = 180° ∠𝐶𝐶𝐶𝐶𝐶𝐶 𝑀𝑀 𝑀𝑀 ∠𝐶𝐶𝐶𝐶𝑀𝑀 − 𝜃𝜃 Let the arbitrary angle to be trisected be the acute (Angles in a straight line). (5) 𝑀𝑀 ∠𝑁𝑁𝑁𝑁𝑀𝑀 ∠𝐶𝐶𝐶𝐶𝑀𝑀 ∠𝐶𝐶𝐶𝐶𝐶𝐶 . From the construction steps it is clear that By substituting for ′ = 180° 4 we have; triangle is equilateral since = = . = 180° ′ + ′ ∠𝐶𝐶𝐶𝐶𝐶𝐶 ∠𝐶𝐶𝐶𝐶𝑀𝑀 − 𝜃𝜃 Therefore this proof is to show the trisection of a 60° = 180° ( ′ + 180° 4 ) (6) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴���𝐴𝐴� �𝐵𝐵��𝐵𝐵� �𝐶𝐶��𝐶𝐶� ∠𝐶𝐶𝐶𝐶𝐶𝐶 − ∠𝑁𝑁𝑁𝑁𝑀𝑀 ∠𝐶𝐶𝐶𝐶𝑀𝑀 angle. Consider, ′ = ′ = (radii). ′ is a = + 4 = 3 ∠𝐶𝐶𝐶𝐶𝐶𝐶 − ∠𝑁𝑁𝑁𝑁𝑀𝑀 − 𝜃𝜃 reflection of about point . Thus triangle Therefore, �𝑁𝑁��𝑁𝑁��� 𝑀𝑀����𝐴𝐴� 𝐴𝐴���𝐴𝐴� 𝑀𝑀����𝐴𝐴� ∠𝐶𝐶𝐶𝐶𝐶𝐶 −𝜃𝜃 𝜃𝜃 𝜃𝜃 ′ ′ , by property . It implies that = 3 . (7) �𝐴𝐴��𝐴𝐴�� 𝐴𝐴 𝑀𝑀𝑀𝑀𝑀𝑀 ≡ 𝑀𝑀 𝐴𝐴𝐴𝐴 𝑆𝑆𝑆𝑆𝑆𝑆 ∠𝐶𝐶𝐶𝐶𝐶𝐶 𝜃𝜃 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 35
Fig. 9 Angle Trisection Proof.
Equation (7) implies that, = 3 ′ = 4.1 Construction of a Pentagon to justify the construction of a ° angle using only a ruler and 3 1 = ∠, as𝐶𝐶𝐶𝐶𝐶𝐶 required.∠𝑀𝑀 𝑁𝑁𝑁𝑁 3 compass 𝟖𝟖 ∠3.3.4𝑀𝑀𝑀𝑀𝑀𝑀 Use⟹ of� GeoGebra∠𝐶𝐶𝐶𝐶𝐶𝐶 software∠𝑀𝑀𝑀𝑀𝑀𝑀 to justify the Proof (1) Draw a circumference of radius . The use of GeoGebra in this section is to exemplify (2) Mark a point on the circumference, two important aspects from the provided proof. First is 𝐴𝐴���𝐴𝐴� equidistant from both and centered at . to show line ′ lie on such that ′ = 𝐶𝐶 (3) Join the chord , and construct its bisection ′ and line ′ is a radius of the circle as discussed 𝐴𝐴���𝐴𝐴� 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝑁𝑁���𝑁𝑁��� �𝑁𝑁��𝑁𝑁� 𝑁𝑁���𝑁𝑁��� through a point to cut curve at . earlier. The other aspect verified is that; ′ = 𝐶𝐶���𝐶𝐶� 𝑀𝑀����𝐴𝐴� 𝑀𝑀����𝐴𝐴� (4) Construct the bisection of to cut curve ′ = 2 . 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 ∠𝑀𝑀 𝐶𝐶𝐶𝐶 at . From figure (10), it can be visualized that the ∠𝐸𝐸𝐸𝐸𝐸𝐸 ∠𝐴𝐴𝑀𝑀 𝐶𝐶 𝜃𝜃 (5) Further, construct the bisection of to cut presented cases ′ = ′ = ′ = 5 (radii), 𝐵𝐵𝐵𝐵 𝐹𝐹 curve at . and ′ = ′ = where = 20° are ∠𝐸𝐸𝐸𝐸𝐸𝐸 �𝑁𝑁��𝑁𝑁��� 𝑀𝑀����𝐴𝐴� �𝐵𝐵���𝐴𝐴� 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 (6) Position the pair compass at point and make absolutely correct. The methodology made it possible 𝐹𝐹𝐹𝐹 𝐺𝐺 ∠𝑀𝑀 𝑁𝑁𝑁𝑁 ∠𝑀𝑀 𝐴𝐴𝐴𝐴 𝜃𝜃 𝜃𝜃 an arc of length to cut curve at . to trisect the 60°, and this implies the possibility to 𝐸𝐸 (7) With the compass at point , mark point of construct all angles multiples and sub-multiples of 10. �𝐷𝐷��𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 distance between carve . 𝐻𝐻 𝐼𝐼 4. Application (8) Construct the bisection of to cut curve 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 at . In this section, a rationalization of the presented ∠𝐼𝐼𝐼𝐼𝐼𝐼 (9) Bisect at on curve . proofs is made by construction of some regular 𝐵𝐵𝐵𝐵 𝐽𝐽 (10) Placing the compass at point , make an arc of polygons; (Pentagon and Nonagon), to represent some ∠𝐽𝐽𝐽𝐽𝐽𝐽 𝐾𝐾 𝐵𝐵𝐵𝐵 length to cut curve at . of the regular polygons which could not be 𝐼𝐼 (11) With the compass at , mark a point on geometrically solved. �𝐵𝐵��𝐵𝐵� 𝐼𝐼𝐼𝐼 𝐿𝐿 the circumference using chord as shown below. 𝐿𝐿 𝑀𝑀
𝐵𝐵𝐵𝐵 36 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
Fig. 10 Justification of the Angle Trisection Proof.
(12) Mark equal intervals of length along the Thus = 72°. circumference to produce figure (11). 4.1.1 Construction of a Nonagon to justify the �𝐵𝐵��𝐵𝐵�� 𝜃𝜃 From the construction it is observed that, the chord construction of a ° angle using only a ruler and equally stroked 5 times along the circumference compass 𝟏𝟏𝟏𝟏 to produce the regular Pentagon. Let the subtended (1) Draw a circumference of radius . �𝐵𝐵��𝐵𝐵�� angle be and the be . Since (2) Mark a point on the circumference, 𝐴𝐴���𝐴𝐴� = (radii of circle), triangle is an isosceles equidistant from both and centered at . 𝑀𝑀𝑀𝑀𝑀𝑀 𝜃𝜃 ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 𝐴𝐴���𝐴𝐴� 𝐶𝐶 and therefore = = (base angles of (3) Join the chord , and construct its bisection �𝐴𝐴��𝐴𝐴�� 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴���𝐴𝐴� 𝐴𝐴 𝐵𝐵 𝐴𝐴 an isosceles triangle). Triangle by through a point to cut curve at . ∠𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 𝐶𝐶���𝐶𝐶� property; . It follows that = = . (4) Construct the bisection of to cut curve 𝑁𝑁𝑁𝑁𝑁𝑁 ≡ 𝑀𝑀𝑀𝑀𝑀𝑀 𝐷𝐷 𝐵𝐵𝐵𝐵 𝐸𝐸 Thus = 2 (interior angle of the regular at . 𝑆𝑆𝑆𝑆𝑆𝑆 ∠𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 ∠𝐸𝐸𝐸𝐸𝐸𝐸 pentagon). The size of the angle can be found (5) Further, construct the bisection of to cut ∠𝐵𝐵𝐵𝐵𝐵𝐵 𝛼𝛼 𝐵𝐵𝐵𝐵 𝐹𝐹 by applying the expression; curve at . ∠𝐵𝐵𝐵𝐵𝐵𝐵 ∠𝐸𝐸𝐸𝐸𝐸𝐸 90° (2 4) = 2 (Size of one interior angle (6) Position the pair compass at point and make 𝐹𝐹𝐹𝐹 𝐺𝐺 of a regular pentagon), where is the number of an arc of length to cut curve at . 𝑛𝑛 − ⁄𝑛𝑛 𝛼𝛼 𝐸𝐸 sides of the regular polygon (7) With the compass at point , mark point of 𝑛𝑛 �𝐷𝐷��𝐷𝐷� 𝐸𝐸𝐸𝐸 𝐻𝐻 Since = 15, it follows; (8) distance between carve . 𝐻𝐻 𝐼𝐼 90° (10 4) 5 = 2 (9) (8) Construct the bisection of to cut curve 𝑛𝑛 𝐷𝐷���𝐷𝐷� 𝐻𝐻𝐻𝐻 Therefore, 2 = 108° and = 54° (10) at . − ⁄ 𝛼𝛼 ∠𝐼𝐼𝐼𝐼𝐼𝐼 = = 180° 2 = 180° 108° = 72° (11) (9) Bisect at on curve . 𝛼𝛼 𝛼𝛼 𝐵𝐵𝐵𝐵 𝐽𝐽 ∠𝑀𝑀𝑀𝑀𝑀𝑀 𝜃𝜃 − 𝛼𝛼 − ∠𝐽𝐽𝐽𝐽𝐽𝐽 𝐾𝐾 𝐵𝐵𝐵𝐵 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 37
Fig. 11 Classical Construction of a Pentagon.
Fig. 12 Geometrical Construction of a Nonagon.
38 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
(10) Placing the compass at , mark a point through a point R to cut curve PQ at S. distance on curve . (4) Draw the bisector of angle to cut the 𝐼𝐼 𝐿𝐿 (11) Make equal intervals of length along the curve at . �𝐵𝐵��𝐵𝐵� 𝐼𝐼𝐼𝐼 𝑆𝑆𝑆𝑆𝑆𝑆 circumference to produce the figure below. (5) Place your compass at point and make a �𝐼𝐼�𝐼𝐼� 𝑃𝑃𝑃𝑃 𝑇𝑇 In this case, the chord equally marked 9 small arc of length to cut curve at . 𝑄𝑄 intervals along the circumference to produce the (6) Join the point to and there you have 𝐿𝐿𝐿𝐿 �𝑅𝑅��𝑅𝑅� 𝑆𝑆𝑆𝑆 𝑈𝑈 regular nonagon. Let the subtended be and = 50° . 𝑈𝑈 𝑂𝑂 also let be . Since lines and are The steps were carried out for; compass and ruler ∠𝐿𝐿𝐿𝐿𝐿𝐿 𝜃𝜃 ∠𝑈𝑈𝑈𝑈𝑈𝑈 equal (circle radii), triangle is an isosceles and construction, and also using the GeoGebra software as ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 𝐴𝐴���𝐴𝐴� �𝐴𝐴��𝐴𝐴� therefore = = (base angles of an illustrated in figures (13) and (14). 𝐿𝐿𝐿𝐿𝐿𝐿 isosceles triangle). Triangle by property; From figure (13), the author did not provide a proof ∠𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 . It follows that = = . Thus that the method would provide an angle of exactly10°. 𝑀𝑀𝑀𝑀𝑀𝑀 ≡ 𝐿𝐿𝐿𝐿𝐿𝐿 = 2 (interior angle of the regular nonagon). In figure (14), it is clear that = 49.78° and 𝑆𝑆𝑆𝑆𝑆𝑆 ∠𝐴𝐴𝐴𝐴𝐴𝐴 ∠𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 The size of the angle can be found by not = 50° as stated in the article. This ∠𝐵𝐵𝐵𝐵𝐵𝐵 𝛼𝛼 ∠𝑈𝑈𝑈𝑈𝑈𝑈 applying the expression for calculating the interior implies that the generated chord subtends an ∠𝐵𝐵𝐵𝐵𝐵𝐵 ∠𝑈𝑈𝑈𝑈𝑈𝑈 angles size as; angle of 10.22° at point , and not 10° as �𝑅𝑅��𝑅𝑅� 90° (2 4) = 2 discussed. 𝑂𝑂 (Size of one interior angle of a regular nonagon), 5.1.1 Use of GeoGebra software to illustrate the 𝑛𝑛 − ⁄𝑛𝑛 𝛼𝛼 where is the number of sides of the regular mistake in construction of ° from the given polygon. (12) method. 𝑛𝑛 𝟖𝟖 Here, = 9 and therefore using in equation In his work, the author described how to produce an (12); angle of ° using the traditional drafting tools of 𝑛𝑛 𝑛𝑛 90° (18 4) 9 = 2 = 140°. (13) Greek’s geometry. The following is the procedure of 𝟖𝟖 Therefore construction presented in the work; − ⁄ 𝛼𝛼 = 70°. (14) (1) Draw a straight line between two points and = = 180° 2 = 180° 140° = 40°. (15) . 𝛼𝛼 𝑂𝑂 Thus = 40° as it is expected. ∠𝐿𝐿𝐿𝐿𝐿𝐿 𝜃𝜃 − 𝛼𝛼 − 𝑃𝑃 5.0 Revised𝜃𝜃 Errors in the Classical construction of Angles in General Paper [3]
5.1.0 Use of GeoGebra software to illustrate the fault in construction of ° in the given method.
According to the𝟏𝟏 𝟏𝟏article ‘Classical construction of angles in general’, the author presented the following steps of constructing an angle of 50° [3]; (1) Draw a straight line and mark two points and on the line. 𝑂𝑂 (2) Mark a point equidistant from both 𝑃𝑃 and and draw an arc centered at . 𝑄𝑄 𝑂𝑂���𝑂𝑂� 𝑂𝑂 Fig. 13 Wrong illustration of constructing ° angle for (3) Join the chord , and draw its bisector 𝑃𝑃 𝑃𝑃𝑃𝑃 𝑂𝑂 compass-ruler construction. 𝟓𝟓𝟓𝟓 𝑄𝑄���𝑄𝑄�
The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 39
Fig. 14 Illustration that the size of = . ° and not ° using GeoGebra Software
(2) Mark a point equidistant∠𝑼𝑼𝑼𝑼𝑼𝑼 from𝟒𝟒𝟒𝟒 𝟕𝟕 both𝟕𝟕 𝟓𝟓chord𝟓𝟓 would always subtend an angle of 8° at a and and draw an arc centered at . point. Contrary to figure (16), it is evident from the 𝑄𝑄 �𝑂𝑂��𝑂𝑂� 𝑂𝑂 �𝑅𝑅��𝑅𝑅� (3) Join the chord , and construct its bisector construction that the equivalent angle of chord is 𝑃𝑃 𝑃𝑃𝑃𝑃 𝑂𝑂 through a point to cut curve at . 8.22° and not exactly 8° as presented. 𝑄𝑄���𝑄𝑄� �𝑅𝑅��𝑅𝑅� (4) Construct the bisector of angle to cut 𝑅𝑅 𝑃𝑃𝑃𝑃 𝑆𝑆 curve at . 𝑆𝑆𝑆𝑆𝑆𝑆 (5) Further, construct the bisector of angle to 𝑃𝑃𝑃𝑃 𝑇𝑇 cut curve at . 𝑆𝑆𝑆𝑆𝑆𝑆 (6) Position the pair compass at point and make 𝑆𝑆𝑆𝑆 𝑈𝑈 an arc of length to cut curve at . 𝑃𝑃 = 8° 𝑅𝑅���𝑅𝑅� 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑉𝑉 The construction was performed for ruler-compass ∠𝑉𝑉𝑉𝑉𝑉𝑉 construction and also using the GeoGebra software and the results presented in figures (15) and (16) respectively. From figure (16), the obtained results does not Fig. 15 The incorrect presentation of constructing ° correctly agree with the statement that the generated angle for compass-ruler construction. 𝟖𝟖
40 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
Fig. 16 Illustration that the size of = . ° and not ° using GeoGebra software
6. Results and Discussion ∠𝑽𝑽𝑽𝑽𝑽𝑽 𝟖𝟖 𝟐𝟐𝟐𝟐 𝟖𝟖 revealing a correct geometric theorem of trisecting an arbitrary angle, and its precision confirmed by the This paper has presented a flawless methodology of trisection of 48° and 60° angles. A proposition solving the ancient problem of angle trisection. governed by use of compass and ruler is presented Through the ages, mathematicians sought the contrary to the Archimedes theorem of having a trisection of an arbitrary angle but no geometrical marked straightedge notched in two places [24]. proof has been found by this day. An ambiguous proof Through this work, it has been shown that, the general of the angle trisection impossibility closed the doors consideration of angle trisection solution as a cubic in solving this crucial problem by assuming it equation solution, genetically corresponds to solving impossible [1, 2, 4, 12-15]. However, mathematicians the trisection of an angle in solid geometry. and other practitioners still crack their minds to have Geometrically, an angle is defined by two rays with a the problem solved under the stated restrictions of common endpoint, and only a solid angle can be Greek’s geometry. It has well been defined that there solved in a 3 consideration. Some algebraic exists a considerable ration between any two angles at irrationalities are constructible in plane geometry as 𝐷𝐷 a difference of 10° from each other, but this novel stated before, and the fact that a straight line segment reflection was drawn in an erratic manner with no can be proportionally fragmented and its proof did not justified proof [3]. This present proof concerns concern degree three polynomials, shows the
The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex 41 uncertainty in the Pierre Wantzel’s proof of Acknowledgments impossibility. Thus the cubic equation 3 3 1 = The author of the paper gratefully acknowledges the 0 is not geometrically precise. GeoGebra 5.0 software 𝑥𝑥 − 𝑥𝑥 − advice and the constructive suggestion by the advisors; was used to exemplify the correctness of the proposed Dr. E. Mwenda (Department of Mathematics), Mr. method, and also to show the consistency of the Stephen Karanja (Department of Mathematics) and Mr. construction for both Euclidean constructions and the Daniel Maitethia (Physics), in the organization of the computer aided design (CAD) approaches. The choice paper. Heartfelt thanks to my family and friends for of using the open source GeoGebra as one of the their prayers. interactive geometry software was because of its good geometry environment (Toolbox) compared to some References other software, and its ease in application. [1] Wantzel, P.L., Recherches sur les moyens de reconnaitre si un problème de géométrie peut se résoudre avec la 7. Conclusion règle et le compass. Journal de Mathematiques pures et The problem of trisecting an angle has pondered in appliques, (1837), Vol. 2, pp.366-372. [2] Ruth Helgerud, Trisection of the Angle ,Texas A&M the minds of mathematicians since the antiquity, but University, April 4, 2011 no geometric algorithm has been made to solve the [3] Kimuya .M. Alex, Classical Construction of Angles in problem. Most of the presented methodologies bend General, International Journal of Mathematical Archive the stated rules, and none has met the desired level of (IJMA), 2016,Vol.7, Issue 3: 47-51, March. [4] University of Toronto, The Three Impossible accuracy [22]. This novel work presents a method of Constructions of Geometry, trisecting an angle using the traditional Greek’s tools
42 The Possibility of Angle Trisection (A Compass-Straightedge Construction) Kimuya M Alex
[15] H.Dorrie, 100 Great Problems of Elementary