aa r X i v : . [ m a t h . G M ] F e b Locus of Intersection for Trisection
Ramachandra BHATRajajinagar, Bengaluru,Karnataka, India - 560010.February 22, 2021
Abstract
While solving problems, if direct methods does not provide solution, indirect methodsare explored. Today, we need an indirect method to solve the problem of angle trisectionas the direct methods have been proved using algebra to be impossible by Euclidean Ge-ometry, using only straight edge and compasses. The unstoppable curiosity of Geometersand the newer advanced tools available with time have led to newer approaches to progressfurther. There is a method using Origami (paper folding) procedure to trisect an angle.The algebraic analysis of this procedure gives us a method of finding trisection using alocus of a point of intersection of two circles. The algebraic analysis and the equationfor the locus of the point of intersection of two circles leading to trisection of any givenunknown angle without any measurements is described here.
In ancient times, there were no formal measurement systems in place. While working ondesigns, Greeks defined the dimensions by considering an arbitrary length as a unit of length,for that particular design. They derived other dimensions within that design geometrically bythe addition, multiplication, subtraction and division operations. In addition, they knew theright angle and the Pythagoras theorem that helped them to get the square roots of the givenline segment. Therefore, in Euclidean Geometry, one does not measure the dimensions but theexpected designs were very precise, accurate and simple to reproduce. The only geometricaltools used in those days were a ruler (unmarked straightedge) and Compasses.With the advancement of mathematics and allied subjects, today we could actually make themeasurements to show the accuracy and precision of the Greeks and explain them through thealgebraic and trigonometric formulations. However, even today, the Three Famous Problems ofGeometry have remained unsolved. One of them being the trisecting an angle by the Euclideanprocedure.While many Geometers have been exploring to find solutions to these problems, in 1837,Pierre Wantzel proved that finding a solution to the problem of trisection of an angle isimpossible[1, 2]. This only meant, using only ruler and compasses, it is not possible to tri-sect an angle of any given value, with the tools and knowledge available at that instant of time .It should also be noted that the proof of impossibility considers primarily the constructabilityof the angle of value equivalent to one-third of the given value and not the trisectability of thegiven angle of unknown measure directly.In the literature, there are various methods described [3, 4, 5, 6, 7] to trisect an angle usingvarious additional tools (Marked ruler, trisector, Quadratrix of Hippias, hyperbola, tomahawk,linkages, etc.) and hence, are not Euclidean methods. However, the only exact and simple1rocedure for the trisection of any given angle of unknown measure demonstrated as well asproved algebraically is by Origami [8, 9, 10], i.e., the Japanese paper folding technique, byHisashi Abe. However, in this procedure, multiple conditions get satisfied simultaneously inone folding operation, by using an implicit marked ruler. The results of the algebraic analysisof this Origami method are utilized in exploring the trisection using ruler and compasses only.While doing so, it has been observed that the locus of a point of intersection of two circle givesthe exact results for trisection, though it would take a few extra steps to generate the locusgraph to implement the procedure by Euclidean Geometry.
The two approaches followed by many geometers in their exploration are:(i) to divide the given angle into three equal parts, and/or(ii) to explore the constructability of an angle of value equivalent to one-third of the givenangle.For example, some select angles such as 0, 45, 72, 90, 108 and 180 degrees were trisectable[11], by directly constructing a corresponding angle equivalent to their one-third value and notby trisecting the given angle. This means,(i) the actual value of the given angle was known (or measured) prior to attempting thetrisection, and(ii) its trisected component is constructible by Euclidean geometry.Since the measure of the given angle was unknown, Greeks could only explore the trisectabil-ity and not constructibility. In the present study, the intention was to trisect the given angleand not to find the constructability of angle of any given value. Hence, the definition by theEuclidean geometry may be presented as: “Divide the given angle (of unknown value) into three equal angles (angular parts)using only two tools, viz., (i) an unmarked straight edge (ruler) and (ii) a compasses”.
Let us consider the addition rule for Sin(A+B). Let A = θ and B = 2 ∗ θ and draw the diagramfor the same. Next, consider A = 2 ∗ θ and B = θ and draw the corresponding diagram withthe same unit length and angle value θ . Overlay the two diagrams (on same scale). This ispresented in an overladen square ABCD in Fig. 1.Let the given angle ∠ BAF = 3 θ . When this angle is trisected, let us consider the threeequal components as α, β and γ , i.e., 3 θ = α + β + γ .In addition, let us draw a circle with radius = 0.5 unit of length with J (= mid point ofAF) as the center. K and L are the intersection points of the circle with AH and AG, the linesof trisected angles. Now, from the diagram, it could be found that segments FK = KL = LE= Sin ( θ ).Hence, as exprected, the angle trisection corresponds to the trisection of an arc of a circleand the length of cords corresponding to each section is given by sine of the trisected componentof the given angle [ sin ( θ )]. 2igure 1: Trigonometric addition formula diagram and TrisectionFrom the diagram, it may be noted that ∠ BAF = 3 θ and 3 θ = α + β + γ and α = β = γ ∠ BAG = α ; ∠ GAH = β ; ∠ HAF = γ .AE = Cos (3 θ ); EF = Sin (3 θ ); AF = 1 = AG.BG = Sin ( θ ); GF = 2 ∗ Sin ( θ ); AF = 1 = AG.In the Origami procedure, BG corresponds to the first fold. The geometrical picture of the Origami procedure is given in Fig. 2.Figure 2: Hisashi Abe’s Origami diagram for TrisectionLet us assume the given angle ∠ AOB = 3 θ (for convenience). Considering the similarisosceles triangles DCO and HOC, it has been proved already that the three angles α, β and γ are all mutually equal [8]. Hence, 3 θ = α + β + γ = 3 ∗ θ , i.e., to say, α = β = γ = θ , and hencethe trisection.Here, segments OH = OC and OS = SD = HG = GC = CP.If we define OH = OC as the unit of length (=1), then OS = CP = sin( θ ) and OP = cos( θ ).From the Origami method, it is clear that the two unknowns in the procedure for trisectionare the unit of length (= 1) and the fold length (= sin( θ )) and they are interdependent. While3olding the paper, the value for the fold length (= OS = CP = sin( θ )) is arbitrarily assumedand the corresponding value for the unit length (= 1 = OH) is determined. While doing so,the segment OD is matched with CH such that OH = OC = 1. Here, the segment OD servesas the marked ruler to identify the location of point G (corresponding to the point S on OD)which is the advantage over the Euclidean method. From the Origami procedure and the addition formula for the trigonometric function [Sin(A+B)],it is clear that we need to explore a method to find the value of either unit length (segmentlength=1) or the value for sin( θ ) to achieve the trisection, while we could assume the value ofone of these two interdependent parameters arbitrarily. Now, the options available for the ex-ploration of trisection of a given angle is summarized in the following GFR (Given-Find-Report)table. GFR Table for the solution explorationOptions Given { & Assumed } Parameter to Find ReportFirst 3 θ , Paper, { & Sinθ } θ ) Origami procedureSecond 3 θ , S & C, { & Sin θ } θ ) Euclidean procedure*Third 3 θ , S & C, { & 1 } Sin θ (or Cos θ ) Euclidean procedure*Fourth 3 θ , S & C, { & Sin θ } Cosθ ) Locus method*** Using only S & C (Straightedge and Compassess); ** Present study
In the present study, the locus of a point of intersection of two circles is made use of to findthe exact dimension of the unit length segment, while assuming a value for sin ( θ ). Let the given angle be ∠ AOB = 3 θ (= 3* θ ). Similar to the Origami procedure, let us markthe two folds as lines parallel to OA (base segment of the given angle) at C and D. The lengthOC = CD corresponds to Sin θ (assumed value) for the given angle.Figure 3: Diagram for the locus of the intersectionNow, to find the corresponding value of the unit length, draw the first circle with O asthe center and OD as the radius. Let the circle cut the first fold line at J. Mark the point of4rojection of point J on OA as K. At this stage, KJ = OC and OJ = OD = 2 * OC. Hence, ∠ KOJ = 30 ◦ and OK = √ θ ), where θ is the valueof the trisected angle, i.e., ∠ AOB = 3* ∠ AOJ. Here, JN = 2 * KJ = OD, same as in Origamiprocedure. Hence, the proof of trisection is similar to that for the Origami procedure.Figure 4: Locus of intersection and the trisection
Let OC = JK = a and OK = b. The coordinates of the points are O = (0, 0); D = (0, 2a); J= (b, a). ∴ The equation for Circle 1: ( x + y ) = OJ = OQ = ( b + a ).and the equation for Circle 2: ( x − b ) + ( y − a ) ) = J Q = (2 a ) = 4 a ).At Q - the point of intersection,( x + b − bx ) + ( y + a − ay ) = 4 a .( x + y ) − bx + ay ) + ( b + a ) = 4 a .( b + a ) − bx + ay ) + ( b + a ) = 4 a . ∴ ( bx + ay ) = ( b − a ) or y = ( − ba ) ∗ x + ( b − a ) a .The value of b is variable (not constant) and varies with the values of x and y. Hence, theequation for the locus corresponds to a curve and not a straight line. Every point Q on thelocus curve corresponds to ∠ AOQ (= 3 δ ) for which ∠ AOJ = δ is its trisected angle. Hence,the procedure gives the trisection for the spectrum of angles from ∠ AOD (= 90 ◦ ) till ∠ AOB(= given angle = 3 θ ). Therefore, when Q lies on the line OB, it gives the trisection for thegiven angle ∠ AOB.When Q coincides with D (in the beginning), x = 0, y = 2a, b = √ sin ( θ ) and b = cos ( θ ). Then, x = cos (3 θ ) and y = sin (3 θ ).Hence, the value of b varies between √ cos ( θ ) while the value of x varies between 0and cos (3 θ ) for the locus of the point Q between point D and Point N. The proof for the trisection of a given unknown angle by the Origami method (Abe’s method)is well established. Using the same concept of the dimensions and the intersections, trisectionprocedure using straight edge and compass is explored in the present study. The locus of thepoint of intersection of two circles establishes the trisection procedure. To draw the completelocus line, we need to find the points at various locations using compass. Once the concept isunderstood, it would be easy to do so. When the locus line is very close to the intersectionpoint with the angle line (OB), it is expected to be linear at that very short length. Hence,by finding two points, one before and one after the line (OB), it is possible to find the exactunit length and hence the trisection. This locus graph is a universal plot and can be madeseparately and used for any angle between 0 and 90 ◦ . Hence, it is possible, not impossible, totrisect an angle with only unmarked ruler and compasses. It only needs extra effort to find theexact point of intersection of the line of locus with the the angle line (OB). The support and cooperation of my family during the exploration is gratefully appreciated. Inthe early stage of this study (Mar-2016), while exploring suitable software to verify the accuracyof the ideas, the suggestion of Mr. John Page ([email protected]) to use GeoGebra [12]was of great help and the same is acknowledged here.All the explorations of different options were done using the GeoGebra software. Hence,a special thanks for the creator of the GeoGebra software who made it so user-friendly andavailable freely on the Internet [12].
Conflict of Interest
There is no conflict of interest.
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Angle Trisection
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An interesting example of angle trisection by paper folding
Angle Trisection By Paper Folding
Alumni of th B Main Road, 1. R V High School, Itgi,6th Block, Rajajinagar, Taluk: Siddapur (Uttara Kannada),Bengaluru - 560 010. Karnataka, INDIA.Karnataka, INDIA. 2. Sarada Vilas College, Mysore.Mobile: +91-9902461175. 3. Yuvaraja’s College, Mysore.Email: rs bhat @ yahoo.com
4. Indian Institute of Technology, Bombay.