Applied Mathematics in Electrical and Computer Engineering A Possible Solution of Trisection Problem SIAVASH H. SOHRAB Robert McCormick School of Engineering and Applied Science Department of Mechanical Engineering Northwestern University, Evanston, Illinois 60208 [email protected] http://www.mech.northwestern.edu/web/people/faculty/sohrab.php Abstract: - A solution of the ancient Greek problem of trisection of an arbitrary angle employing only compass and straightedge and its algebraic proof are presented. It is shown that while Wantzel’s [1] theory of 1837 concerning irreducibility of the cubic x3 3x 1 0 is correct it does not imply impossibility of trisection of arbitrary angle since rather than a cubic equation the trisection problem is shown to depend on the quadratic equation y2 3 c 0 where c is a constant. The earlier formulation of the problem by Descartes the father of algebraic geometry is also discussed. If one assumes that the ruler and the compass employed in the geometric constructions are in fact Platonic ideal instruments then the trisection solution proposed herein should be exact. Key-Words: The trisection problem, angle trisection, Wantzel theory, regular polygons, heptagon. 1 Introduction The classical trisection problem requires trisecting the algebraic proof of its validity will be presented. an arbitrary angle employing only a compass and a Historically, it was proven by Wantzel [1] that the straightedge or unmarked ruler. The general rules trisection of an arbitrary angle by only compass and concerning the construction instruments and straightedge is impossible if such a construction acceptable solution of the problem are most requires the existence of rational roots of the cubic eloquently described by Dunham [2] equation Indeed, Greek geometers performed trisection (1) by introducing auxiliary curves like the quadratrix of Hippias or the spiral of As described by Dunham [2]: Archimedes, but these curves were not themselves constructible with compass and (a) If we can trisect the general angle with straightedge and thus violated the rules of the compass and straightedge, game. It is rather like reaching the top of o (b) Then we can surely trisect a 60 angle, Everest by helicopter: It achieves the end by an unacceptable means. For a legitimate trisection, (c) So, we can find a constructible solution of only compass and straightedge need apply. , The second rule is that the construction must require only a finite number of steps. (d) So, we can find a rational solution for There must be an end to it. An “infinite , construction,” even if it has trisection as a (e) And this rational solution must be either c/d = 1 limiting outcome, is no good. Construction that goes on forever may be the norm for interstate or c/d = 1. highways, but it is impermissible in geometry. Finally we must devise a procedure to when x is a rational number denoted by the ratio x = trisect any angle. Trisection a particular angle, c/d. Since by Wantzel’s [1] proof (1) is irreducible or even a thousand particular angles, is and (d) is not true then one must conclude that (a) insufficient. If our solution is not general, it is cannot be true. The algebraic equation (1) in not a solution.” Wantzel’s theory [1] originates from the trigonometric equation In this study an unexpectedly simple solution 3 of the ancient Greek trisection problem along with cos 4cos ( / 3) 3cos( / 3) (2) ISBN: 978-1-61804-064-0 277 Applied Mathematics in Electrical and Computer Engineering that when applied to the angle = /3 with similarity between (1) and (6) the failure of Wantzel x 2cos( / 3) results in (1). A cubic equation of [1] to reference the work of Descartes is unfortunate. the form Finally, if parallel to (5) instead of full angle in (2) one applies the trigonometric identity for half angle 3 x qx r 0 (3) 3 cos( / 2) = 4cos ( / 6) 3cos( / 6) (7) was also employed by Descartes [3] in connection and considers = 60 = 3 with one to the trisection problem [4] y2 cos( / 6) arrives at “Descartes dealt with the problem of angle trisection by reducing the problem to a third- y3 3y 3 0 (8) degree equation and constructing it via intersection of circle and parabola” that also does not possess any rational roots. It is interesting to note that cubic equation of the type Descartes proposed a solution of trisection problem by employment of a parabola, a non-constructible x3 b 2 x b 2 c (9) hence transcendental curve, as shown in Fig.1 reproduced from his book of geometry [3] that was first solved by Omar Khayyam using intersection of conics [5] also reduces to (1) and (6) when ( b i 3 , c = 1/3) and ( , c = 1/3), respectively. According to Wantzel’s theory of 1837 [1] only rational numbers x c / d that are roots of algebraic equations are acceptable solution to the trisection problem. This is because the criterion of geometric constructability based on Descartes’s analytic geometry only admits rational operations of addition, subtraction, multiplication, division, and extraction of square roots thus requiring existence of rational roots of polynomials of various degrees [1- 8]. Over three decades after Wantzel’s work Hermite Fig.1 Descartes solution of trisection of an arbitrary (1873) and then Lindemann (1882) respectively angle NOP employing a parabola GAF [3]. proved the existence of transcendental numbers e and [2]. It is known that concerns about basing By geometric construction based on the trisected geometric constructability only on the application of angle shown in Fig.1 Descartes arrived at the cubic geometrical (algebraic) curves and not mechanical equation [3] (transcendental) curves were raised by ancient Greek mathematicians as well as Newton [4]. Clearly, the z3 3z q 0 (4) question of rational versus transcendental numbers could be connected to the admissibility of algebraic where z = NQ and q = NP. It is now clear that for versus transcendental curves as means of geometric = 60 = 3 and a circle of unity radius NO = 1 construction just mentioned. Indeed, in view of the one has NP = q = NO = 1 and both (4) as well as fact that trisecting an angle involves and as was the trigonometric relation emphasized by Gauss transcendental numbers by far are more abundant than rational numbers, it is sin( / 2) = 3sin( / 6) 4sin3 ( / 6) (5) reasonable to anticipate that trisection problem may involve transcendental numbers that are not solutions with the definition z 2sin( / 6) lead to of polynomials with integer coefficients. Therefore Wantzel’s proof may not be applicable to the 3 z 3z 1 0 (6) resolution of trisection problem because this problem involves transcendental numbers and not rational that like (1) has no rational roots. One notes that numbers. In other words, the assumption that only the chord NQ in Fig.1 of Descartes becomes the rational numbers x = c/d should be allowed as unknown NQ = . In view of the acceptable constructible solutions made in steps (c)- ISBN: 978-1-61804-064-0 278 Applied Mathematics in Electrical and Computer Engineering (d) above may not be valid. Such a situation will 2 A Solution of Trisection Problem bex somewhat 2cos( / 3)similar to von Neumann’s proof of Before discussing the trisection problem however, let impossibility of hidden variables in quantum us first examine two different geometric mechanics [9] and the fact that it was later found to configurations that one encounters in the construction be inapplicable to quantum mechanics not because of regular polygons shown schematically in Figs.2a of an error in the theory but rather due to invalid and 2b that were helpful in arriving at the suggested assumptions made in its axiomatic foundation. resolution of trisection problem to be discussed in the Because of the overwhelming abundance of sequel. In Fig.2a the edges of an arbitrary angle AG transcendental numbers there is no reason to are located at points of tangency A and G of the require the root x = 1.879385242… of equation (1) small circles Ri with origins at the vertices of regular to be a rational number. polygons. On the other hand, in Fig.2b the edges of Another important observation concerning the angle GF are located at the position of origins of the classical proof (a)-(e) discussed above is that small circles at points G and F. the term cos on the left hand side of (2) is treated as a constant by substitution for cos( = /3) = ½ while on the right hand side of (2) is treated as the unknown variable that is to be determined. Clearly, such a practice disregards the fact that both the left as well as the right hand sides of the trigonometric equation (2) are in fact functions of the true independent variable of the problem that is the angle . To show how such a practice could influence judgments about constructability one expresses (7) as cos( / 2) (a) 4cos2 ( / 6) 3 (10) cos( / 6) If in the process of geometric construction the ratio on the left hand side of (10) becomes a constant cos( / 2) / cos( / 6) = c for a given angle then one arrives at c y2 3 (11) where y 2cos( / 6). Now, as opposed to the cubic equation (1) the quadratic equation (11) only (b) involves extraction of square root thus making trisection of angle constructible. In fact, once the Fig.2 Two configurations of dividing finite angles into ratio is determined through regular polygons: (a) Integral number of full circles construction the sine of the trisected angle can be (t-o-t-o-t-o-t) versus (b) Two half circles at the directly related to this ratio by noting that (7) could boundaries (o-t-o-t-o-t-o).