41. Angle Trisection Error estimate for Mark Stark s approximate angle trisection. Pleaserefer to an angle trisection page for the description of the http://www.math.umbc.edu/~rouben/Geometry/trisect-stark-proof.html

Error estimate for Mark Stark's approximate angle trisection Please refer to An angle trisection page for the description of the problem the corresponding diagram. Let T be the measure of the angle to be trisected, that is BOA = T. Our purpose is to calculate the size of the construction error, that is, the difference between T/3 and the measure of the constructed angle E'OA. The error depends on T as well as the choice of the point E. We specify the point E in terms the measure `a' of the angle BAE. Note that a varies between (when E coincides with B) and T/2 (when E coincides with A). We write the error as e = e(T,a). Remark 1: When E corresponds to an exact trisection, the points E and E' coincide, the angle EOA is T/3, therefore the arc BE is 2T/3 hence the a = T/3. We will show that for any T and a: e(T,a) = T/6 - a/2 - arcsin((1/3)sin(T/2 - 3a/2)). Note, in particular, that e(T,T/3) = 0, which confirms the statement made in Remark 1 above. The key idea in the derivation of (1) is the observation that in the triangle OEG, the side OG is three time the side OE, therefore by the law of sines, the angle at vertex E is approximately 3 times as large as the angle at vertex G. (These angles are approximately equal to their sines, since they are pretty small.) Note: I have not drawn the line segment OE in the diagram to avoid clutter. Because of this, the triangle OEG does not really look like a closed polygon at all!) You just have to imagine that there is a line segment joining O and E.

42. Morley's Trisection Theorem well. First, it s worthwhile to review why an angle trisection cannotbe accomplished by straightedge and compass methods. Such http://www.mathpages.com/home/kmath376/kmath376.htm

Morley's Trisection Theorem In Proposition 4 of Book IV of the Elements, Euclid inscribes a circle inside an arbitrary triangle by showing that the bisectors of any two of the interior angles meet at a point equidistant from the three edges. Since there is only one such point, it follows that the bisectors of all three angles meet at the same point. Letting 2 a b , and 2 g denote the three interior angles of a triangle, the law of sines implies that the edge lengths are proportional to the sines of these angles, so we can scale the triangle to make the edge lengths equal to these sines as shown below. Since a b g p /2, the central angles are g p b p /2, and a p /2, and the law of sines gives the ratios Making use of the double-angle formula we can substitute for the sines of 2 g b and 2 a in the previous ratios and simplify to give Hence by the sine rule we see that the point of intersection is a distance of from each of the three edges, which confirms that this point is the center of the inscribed circle. Now, since the bisectors of a triangle meet at a single point (a fact which is not entirely self-evident), it seems natural to go on to consider how the tri sectors of a triangle meet. However, Euclid apparently didn't consider this question, nor did anyone else for over 2000 years. (Of course, it's impossible to trisect an arbitrary angle using Euclidean methods, i.e., by straight-edge and compass, so Euclid obviously couldn't have used trisectors in any constructions, but he could still have proven some interesting theorems about trisectors if he had wished.) It wasn't until 1899 that Frank Morley (one time was president of the American Mathematical Association) discovered that lines trisecting the angles of an arbitrary plane triangle meet at the vertices of an equilateral triangle as illustrated in the figure below, where the central triangle (in blue) is equilateral.

43. Trisection On A Budget It s well known that there is no procedure using just straightedge and compassfor trisecting an arbitrary angle in a finite number of steps. http://www.mathpages.com/home/kmath169.htm

Trisection On A Budget It's well known that there is no procedure using just straight-edge and compass for trisecting an arbitrary angle in a finite number of steps. However, we certainly can trisect an arbitrary angle to within any arbitrary precision by means of very simple straight-edge and compass operations. One approach is to simply bisect the angle, then bisect the left half-angle, then the right quarter-angle, then the left eighth-angle, and so on. The net result is 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - ... which differs from 1/3 by less than 1/2^n after n bisections. This may not be very elegant, but it's easy to remember and gives a construction with as much (finite) precision as you want. With 30 bisections the result would be within 1 part in 10^9 of a true trisection. "Hope deceives more men than cunning does." Vauvenargues, 1746 Return to MathPages Main Menu

44. Angle Trisection Construction, angle trisection. Search Site map Contact us Join our mailinglist Books angle trisection. Suppose you are given an angle in the plane. http://www.mathreference.com/fld-cs,trisect.html

Construction, Angle Trisection Search Site map Contact us Join our mailing list ... Books Main Page Fields Construction Use the arrows at the bottom to step through Construction. Angle Trisection Suppose you are given an angle in the plane. Can you bisect the angle using straightedge and compass? If v is the vertex, draw a circle about v, of any radius. Let it intersect the two rays in the points s and t. Draw two more circles, same radius, about s and t. These intersect in v, but they also intersect in u, inside the angle. Now the ray from v through u bisects the angle. The Greeks wondered if an arbitrary angle could be trisected, i.e. cut into three equal pieces. Mathematicians worked on this problem for 2,000 years without success. Now we know why; it can't be done. Let's begin by showing the cube root of 2 is not constructible. If it were, it would belong to a field extension of dimension 2 m . Yet x -2 has no rational roots, and is irreducible. Hence the cube root of 2 belongs to a field extension of dimension 3. Since 3 does not divide 2 m , the cube root of 2 is not constructible. Similar reasoning holds for the 5th root of 7, and the 6th root of 38, and so on.

45. Perplexus.info :: Geometry : Angle Trisection perplexus dot info, Home Shapes Geometry angle trisection (Postedon 200206-19), How can you divide an angle into 3 equal angles? http://perplexus.info/show.php?pid=117

46. InterMath | Investigations | Algebra | Graphing Then the angle DOB is approximately the trisection of the angle AOB. Try to findand analyze another construction for approximating the trisecting an angle. http://www.intermath-uga.gatech.edu/topics/algebra/graphing/a26.htm

Search the Site Investigations Algebra Graphing ... Additional Investigations It is a well-known fact that trisecting an angle with Greek construction rules is impossible. But this does not mean that one can not provide a construction, which approximates the trisection. Use a dynamic geometry software and a spreadsheet determine how well is the construction described below. Construction: Given an angle AOB, construct the bisection AOC, with OA = OC = OB. Then construct a trisection of BC such that BD = 2CD. Then the angle DOB is approximately the trisection of the angle AOB. Is it? Explain why or why not. Click Here to get a GSP file Click Here to get a spreadsheet file Extensions Try to find and analyze another construction for approximating the trisecting an angle. Submit your idea for an investigation to InterMath

47. Singapore Science Centre ScienceNet Computer Science / IT Question No. 14557 Is there a way to TRISECT an angle? The trisection of theangle by an unmarked ruler and compass alone is in general not possible. http://www.science.edu.sg/ssc/detailed.jsp?artid=526&type=6&root=1&parent=1&cat=

48. Trisect Straightedge+compass angle trisection, given an astroid (= quadricuspidhypocycloid (x^(2/3) + y^(2/3) = 1) = red thing). Draw segment http://www.tweedledum.com/rwg/trisect.htm

Straightedge+compass angle trisection, given an astroid (= quadricuspid hypocycloid (x^(2/3) + y^(2/3) = 1) = red thing). Draw segment (blue) from center at angle A (= pi/3 in illustration) w.r.t. horizontal, of length 1/4. At its end, draw a circle (green) of radius 3/4. This will cut the astroid in eight places. The cyan, magenta, and yellow radii are at angles -A/3, (2 pi - A)/3, and (4 pi - A)/3. I have yet to work out the significance of the other five intersections. But where did we get "center" and "horizontal"? (Theological) claim: The cusps are definite, preconstructed points, giving "horizontal", "vertical" and center immediately.

49. Talk - Trisecting The Angle and discuss the mathematics that makes trisection by paper folding possible. At theend of the talk, each member of the audience trisects an angle by origami. http://www.math.ohio-state.edu/~goldstin/exposition/tri.html

Trisecting the Angle, or Better Math through Cheating Sample Abstract Although there are those who still attempt to solve this classical problem, it is now a well established mathematical fact that it is impossible to trisect an angle using only a compass and a straightedge. But this doesn't mean that the ancient Greeks didn't know how to trisect an angle. In this talk, we will discuss why the compass and straightedge construction is impossible and show two simple methods for trisecting an angle with additional equipment: a compass and ruler construction known to Archimedes, and a more recent technique using origami. Description The mathematics behind this talk comes from the discussion of compass and straightedge constructions in What is Mathematics by Courant and Robbins and from the discussion of origami constructions on Thomas Hull's Origami Mathematics Page . After sketching a proof of the impossibility of trisection by compass and straightedge, we perform the Archimedes compass-and-ruler trisection and discuss the mathematics that makes trisection by paper folding possible. At the end of the talk, each member of the audience trisects an angle by origami. Level High School.

50. The Regular Nine-gon And Angle Trisection angle trisection and the regular ninegon. We know that it is impossibleto trisect an angle using only a straight-edge and compass. http://www.nevada.edu/~baragar/geom/nine.html

Angle trisection and the regular nine-gon We know that it is impossible to trisect an angle using only a straight-edge and compass. Since Geometer's Sketchpad mimics such constructions, one cannot write a script or sketch that trisects an arbitrary angle (using only the buttons and construction pull down menu in sketchpad.) However, one can create a sketch that mimics Archimede's trisection using a notched straight-edge. Such a sketch is below. The one step that is not a valid construction must be done by hand. In the sketch at the right, select the (acute) angle CAB to be trisected by moving the point C . Now, move P so that the line PQ goes through C . The angle CQB is one third of the angle CAB Sorry, this page requires a Java-compatible web browser. The step that must be done by hand moving P so that PQ goes through C is the step which is not a valid construction. The regular pentagon is constructible. Thus, one can write a sketch which produces a regular pentagon inscribed in a given circle (see below). The regular nine-gon, on the other hand, cannot be constructed using only a straight-edge and compass. But, one can use Archimedes' construction. This is done below. Again, one can adjust the circle in which a regular nine-gon is to be constructed by moving A and B . Try doing this. Note how the figure is distorted. Now, adjust

51. Trisection From MathWorld trisection from MathWorld trisection is the division of a quantity, figure, etc. into three equal parts, i.e., kmultisection with k = 3. See also angle trisection, Bisection, Multisection, http://rdre1.inktomi.com/click?u=http://mathworld.wolfram.com/Trisection.html&am

52. Trisection -- From MathWorld into three equal parts, ie, kmultisection with k = 3. angle trisection,Bisection, Multisection, Trisected Perimeter Point, Trisectrix. search. http://mathworld.wolfram.com/Trisection.html

INDEX Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics ... Alphabetical Index ABOUT THIS SITE About MathWorld About the Author DESTINATIONS What's New MathWorld Headline News Random Entry ... Live 3D Graphics CONTACT Email Comments Contribute! Sign the Guestbook MATHWORLD - IN PRINT Order book from Amazon Geometry Geometric Construction Geometry ... Angles Trisection Trisection is the division of a quantity, figure, etc. into three equal parts, i.e., k multisection with k Angle Trisection Bisection Multisection Trisected Perimeter Point ... search Eric W. Weisstein. "Trisection." From MathWorld A Wolfram Web Resource. http://mathworld.wolfram.com/Trisection.html Wolfram Research, Inc.

53. Euclid Challenge - Trisection Of Any Angle By Straightedge And Compass - Page 7 Page 7 trisection of Any angle by Straightedge and Compass. Turn the “chartpage” 90° to the left, from “portrait” to “ “landscape”. http://www.euclidchallenge.org/pg_07.htm

EUCLID CHALLENGE Successful Response by Milton Mintz May 10, 2002 Page 7: Trisection of Any Angle by Straightedge and Compass Turn the “chart page” 90 to the left, from “portrait” to “ “landscape”. Grasp with the left hand a compass leg above the legpoint. (Identify as leg R). Place legpoint of leg R on point D’. Grasp with the right hand the other compass leg above the legpoint. (Identify as leg S). Place legpoint of leg S on point D. Move both legpoints at a uniform rate; legpoint of leg R along arc D’ C’ towards point C’; and legpoint of leg S along arc DC towards point C. When the legpoint of leg R reaches point P’, stop both legpoints. At the location of the legpoint of leg S on arc DC, mark point T Trisection point *. From point D, on arc AD, length arc DT, mark point T’. Arc TT’ trisects Angle ABC by straightedge and compass. Confirmation of the “Uniform Rate” : The maximum travel on arc D’C’ = 5.625 . (1/8 of 45 and on arc DC = 7.5 (1/6 of 45 or 1/4 of 30 ). These two angles, 5.625

54. Euclid Challenge - Trisection Of Any Angle By Straightedge And Compass - Page 4 May 10, 2002. Page 4 trisection of Any angle by Straightedge and Compass.Note 1 3. 45º 90º, Bisect angle, 22½º - 45º, trisection X 2. http://www.euclidchallenge.org/pg_04.htm

EUCLID CHALLENGE Successful Response by Milton Mintz May 10, 2002 Page 4: Trisection of Any Angle by Straightedge and Compass Note 1: Basic range of angles Adjustment before trisection Range of angles after adjustment Adjustment after trisection Between: Between: Add 22 Trisection minus 7 None None Bisect angle Trisection X 2 Take ¼ of angle Trisection X 4 Note 2: EXAMPLE ANGLE: 30 Since 60º is a frequent test angle, the above 30º example was used so that the resulting trisection could be doubled. Previous Page Top of Page Next Page

55. Trisection De L'angle trisection DE L angle. ÉQUATION pour la trisection. Trisectingthe angle by Steven Dutch. § angle trisection The Geometry Center. http://membres.lycos.fr/villemingerard/Histoire/Trisangl.htm

NOMBRES - Curiosités, théorie et usages Accueil Dictionnaire Rubriques Index ... M'écrire Édition du: Rubrique: HISTOIRE ANTIQUITÉ Introduction Duplication du cube Trisection de l'angle Quadrature du cercle ... Heptagone Sommaire de cette page ÉQUATION pour la trisection CONSTRUCTION à l'équerre Pages voisines Règle et compas Transcendant Histoire Hilbert ... Bissection Trisection Découper un angle quelconque en deux parts égales Découper un angle quelconque en trois parts égales Bissection Découper un angle quelconque en deux parts égales est facile Pourquoi est-ce si difficile pour trois? ÉQUATION POUR trois Idée de la démonstration avec un angle de 20° Calculons en général cos(3a) = cos(a)cos(2a) - sin(a)sin(2a) = cos(a)(cos (a) - sin (a)) - 2sin (a)cos(a) = cos(a)(2cos (a) - 1) - 2(1 - cos (a))cos(a) (a) - 3cos(a) Prenons le cas particulier de a o cos(3a) = cos(60 o L'équation, dans ce cas, devient (a) - 3cos(a) (a) - 6cos(a) - 1 En remplaçant cos(a) = x Avec v = 2x = v Voir Équation Solutions rationnelles ? Supposons que Oui, alors v = p/q fraction minimale (simplifiée) En remplaçant dans l'équation = (p/q) - 3(p/q) - 1 En multipliant par q = p - q En reformulant q = p = p (p² - 3q²) On déduit que p est divisible par q Conséquence p est divisible par q Impossible p/q est une fraction irréductible par hypothèse Et en factorisant avec p p + q = q (3p + q²) On déduit que q est divisible par p Conséquence q est divisible par p Impossible p/q est une fraction irréductible par hypothèse La supposition est fausse v n'est par rationnel

56. Les Problèmes Impossibles De L'antiquité angle. · Construction de l heptagonerégulier. angle trisection. · Constructing a regular heptagon. http://membres.lycos.fr/villemingerard/Histoire/Hisantiq.htm

Accueil Dictionnaire Rubriques Index ... M'écrire Édition du: Rubrique: Histoire antiquité Introduction Duplication du cube Trisection de l'angle Quadrature du cercle ... Heptagone Sommaire de cette page TROIS ou QUATRE PROBLÈMES MÉTHODES DE CONSTRUCTION TRANSCENDANT Pages voisines Règle et compas Transcendant Histoire Hilbert ... Équations - Glossaire TROIS ou QUATRE PROBLÈMES LES TROIS CÉLÈBRES PROBLÈMES DE L'ANTIQUITÉ + 1 assimilé Quadrature du cercle Duplication du cube Trisection de l'angle Construction de l'heptagone régulier Squaring a circle Doubling the Cube Angle trisection Constructing a regular heptagon Il est impossible de résoudre ces problèmes Duplication du cube Trisection de l'angle Construction de l'heptagone régulier Quadrature du cercle Ces trois problèmes sont de la même catégorie Il s'agit de la résolution d'équations du troisième degré Leurs racines ne sont pas constructibles Problème un peu différent, mais tout aussi impossible L'impossibilité résulte du fait que p est transcendant , c'est-à-dire non-algébrique L'aire du cercle de rayon 1 est p Il n'est pas possible de construire un carré d'aire p Constructible que l'on peut construire géométriquement que l'on peut construire à la règle et au compas MÉTHODES DE CONSTRUCTION Les trois méthodes de Pappus Plane (2D) En Volume (3D) Linéaire construction avec des droites et des cercles dites "à la règle et au compas"

57. Trisection Selon Nicomède angle selon Nicomède. Nicomède proposa une solutionapprochée de la trisection de l angle par la construction http://www.sciences-en-ligne.com/momo/chronomath/anx3/trisection.html

Trisection de l'angle selon prop trisection de l'angle par la construction d'une trisectrice : ^xOA est l'angle que nous voudrions trisecter l'angle ^AOB = ^OJA = 2t, c'est dire que ^xOA = 3t. Vous voyez se construire point par point la branche ( G ) de la trisectrice G trisection de l'angle ^xOA. r = OP + PP' = OK/cos t + 2a avec t p C'est une (prononcer ) de la droite (AK). D'une façon générale, une G r = 1/cos t + 2 p . Les portions ( G G 2) et ( G p p p p p Autres trisections usant de courbes : Morley : Pour en savoir plus Ed. Hermann, Paris - 1989

58. La Trisection De L'angle angle rosesoit tangent au cercle. Alors la trisection apparaitra aussitôt. http://perso.wanadoo.fr/therese.eveilleau/pages/truc_mat/textes/trisection.htm

L e trisecteur et les trisectrices... Manipulons L'instrument Exp l ications On pourrait se faire angle Et, sinon vivre au calme, Attaquer l'entourage, Se reposer ensuite Guillevic 1967 M anipulons L e trisecteur F A L' P ositionner le point S L 'instrument M ode d'emploi : E xplic ations N AI = IJ = JB et l'angle AIS est droit.

59. Trisection D'un Angle angle par pliage 1) Tracer une droite jaune parallèle à (AB)2) D est le symétrique du point A 3) Il faut trouver la droite rouge axe du http://perso.wanadoo.fr/math.lemur/hub2d/anglepli.htm

La trisection d'un angle par pliage 1) Tracer une droite jaune parallèle à (AB) 2) D est le symétrique du point A 3) Il faut trouver la droite rouge axe du pliage qui envoie D sur la droite (AC) et A sur la droite jaune D'où connaissance des points K et J La trisectrice de Maclaurin Tracer une droite passant par A parallèle à la droite (OM) Tracer une droite passant par A parallèle à la droite (OM')

60. Trisection De L'Angle Avec Geoplan-W angle avec Geoplan-W. La démonstration de ce quiest observé à l’issue de la construction est accessible en http://www2.ac-lille.fr/math/trisection_de_l'angle_avec_geoplan-w.htm

Trisection de l'Angle avec Geoplan-W La démonstration de ce qui est observé à l’issue de la construction est accessible en collège dès la classe de 4 ème , et peut constituer un réinvestissement de géométrie en classe de 2 nde Télécharger l'activité au format word (25 ko) Télécharger l'activité au format word zippé (5 ko) Contexte historique : Ce problème tint longtemps en haleine les géomètres grecs et leurs successeurs, qui cherchèrent à construire à la règle non graduée et au compas le tiers d’un angle donné quelconque. La réponse arriva seulement au 19 ème siècle : une telle construction fut alors définitivement prouvée impossible, tout comme celles de la Quadrature du Cercle et de la Duplication du Cube. Cependant Pappus d’Alexandrie y est arrivé par une autre méthode, qui utilise la règle graduée pour reporter une longueur imposée. Construction : Créer les trois points O, A, B définissant l’angle en A à trisecter Créer les droites (OA), (AB)

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