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Wantzel Pierre:     more detail

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1. Four Problems Of Antiquity
The problem had been settled in 1837 by pierre Laurant wantzel (18141848) whohad proven that there was no way to trisect a 60 o angle in the classical
http://www.cut-the-knot.com/arithmetic/antiquity.shtml

Extractions: Construct a square whose area equals that of a given circle. Often another problem is attached to the list: Construct a regular heptagon (a polygon with 7 sides.) The problems are legendary not because they did not have solutions, or the solutions they had were unusually hard. No, numerous simple solutions have been found yet by Greek mathematicians. The problem was in that all known solutions violated an important condition for this kind of problems, one condition imposed by the Greek mathematicians themselves: Valid solutions to the construction problems are assumed to consist of a finite number of steps of only two kinds: drawing a straight line with a ruler (or rather a straightedge as no marks are allowed on the ruler) and drawing a circle. You are referred to solutions of problems and as examples of existent solutions. That no solution exists subject to the self-imposed constraints have been proven only in the 19th century.

2. Lwf0
This problem remained unsolved until 1837, when pierre wantzel provedthat such a rulerand-compass construction is impossible.
http://geowords.com/e_/01b_plumb&edge/JohnDBarrow211.htm

Extractions: Impossible constructions (John D. Barrow, Impossibility: The Limits of Science and the Science of Limits Most architects think by the inch, talk by the yard, and should be kicked by the foot. PRINCE CHARLES The next step after the bisection of a line is to ask if it is possible to bisect an angle with these tools (Fig. 7.5). Draw any angle, then put the compass point at the corner of the angle, A; next, draw any arc that cuts the two lines. Now we just need to find the midpoint of the arc joining these two intersections. Draw two arcs, centred at B and C; then draw a line with the rule from the point where these arcs intersect to the corner A. This line bisects the angle. Fig. 7.4 Dividing a line in half by ruler and compass' constructions. Set the pair of compasses to an arc of radius greater than half the length of the horizontal line. Drawa circle of this size centred on each of the end points of the line. They will intersect at two points, above and below the line. The vertical straight line joining these two points divides the horizontal line in half. Fig. 7.5 Dividing an angle in half. Place the compass point at the corner of the angle, A. Draw any arc that intersects both lines forming the angle at B and C. Now draw two arcs of equal radius from each of the two points of intersection. The straight line from the point where these two arcs intersect to the corner of the angle divides the angle in half.

3. Johns Hopkins Magazine February 1999
after generations of mathematicians had attempted in vain to solve it, that a Frenchbridge and highway engineer named pierre Louis wantzel finally cracked the
http://www.jhu.edu/~jhumag/0299web/degree.html

Extractions: Deputy Director, National Foreign Language Center Japanese is without question the most daunting language for a native English speaker to tackle, according to Brecht. "I would like to learn Japanese but I don't have enough time in my lifetime. That's very depressing," says the linguist, whose center is based at Hopkins's Nitze School of Advanced International Studies (SAIS) . He notes that the State Department allows its students three times as long to learn Japanese as it does languages like Spanish or French. As Brecht explains it, the challenge with Japanese is threefold. First, there's the fact that the Japanese written code is different from the spoken code. "Therefore, you can't learn to speak the language by learning to read it," and vice versa. What's more, there are three different writing systems to master. The kanji system uses characters borrowed from Chinese. Users need to learn 10,000 to 15,000 of these characters through rote memorization; there are no mnemonic devices to help. Written Japanese also makes use of two syllabary systems: kata-kana for loan words and emphasis, and hira-gana for spelling suffixes and grammatical particles.

During the 19th century the French mathematician pierre wantzel proved that underthese circumstances the first two of those constructions are impossible and

Extractions: given (the radius of) a circle, construct (the side of) a square whose area equals the area of the circle. In the ancient Greek tradition the only tools that are available for these constructions are a ruler and a compass . During the 19th century the French mathematician Pierre Wantzel proved that under these circumstances the first two of those constructions are impossible and for the squaring of the circle it lasted until 1882 before a proof had been given by Ferdinand von Lindemann

5. (ROBBAR - ANGULAR UNITY)
most surprising implication of the missing theorem is that it contradicts many factsestablished by Carl Frederich Gauss and pierre Laurent wantzel about two
http://www.travelannex.com/davesafe/kafou/book/theorem.html

6. The Hundred Greatest Theorems
Karl Frederich Gauss. 1801. 8. The Impossibility of Trisecting the Angle and Doublingthe Cube. pierre wantzel. 1837. 9. The Area of a Circle. Archimedes. 225 BC. 10.
http://attila.stevens-tech.edu/~nkahl/Top100Theorems.html

Extractions: The millenium seemed to spur a lot of people to compile "Top 100" or "Best 100" lists of many things, including movies (by the American Film Institute) and books (by the Modern Library). Mathematicians were not immune, and at a mathematics conference in July, 1999, Paul and Jack Abad presented their list of "The Hundred Greatest Theorems." Their ranking is based on the following criteria: "the place the theorem holds in the literature, the quality of the proof, and the unexpectedness of the result." The list is of course as arbitrary as the movie and book list, but the theorems here are all certainly worthy results. I hope to over time include links to the proofs of them all; for now, you'll have to content yourself with the list itself and the biographies of the principals. The Irrationality of the Square Root of 2 Pythagoras and his school 500 B.C. Fundamental Theorem of Algebra Karl Frederich Gauss The Denumerability of the Rational Numbers Georg Cantor ... Pythagoras and his school 500 B.C.

7. ICT-lesvoorbeeld: Trisectie Van De Hoek
Opmerking, pierre Laurent wantzel (18141848) bewees in de 19 e eeuw dat een oplossingvan de trisectie van de hoek met passer en liniaal onmogelijk was.
http://www.aps.nl/wiskunde/Content/Lesvoorbeelden/Meetkunde/Trisectie/opdracht.h

Extractions: Trisectie van de hoek Een probleem dat al betrekkelijk vroeg in de Griekse meetkunde zijn intrede deed was: Verdeel, met passer en lineaal, een gegeven hoek ABC in drie gelijke delen. Dit naar analogie van de constructie van de bissectrice die wel gelukt was. Opdracht 1 De Grieken zochten allereerst naar een oplossing door, wat we noemen, een richt-oplossing te vinden. Bekijk de onderstaande figuur en probeer te bedenken hoe de figuur is opgebouwd, uitgaande van ABC. Hij construeerde eerst een kromme, de conchoide. Een conchoide ontstaat door B als vast punt te nemen en vervolgens op AC een punt P en op de lijn BP vanaf P een vaste afstand af te zetten. In dit geval is dat 2*AB. Als je nu door A een lijn loodrecht BC trekt, dan snijdt die de conchoide, zeg maar in F. (met de applet kun je dit nadoen). Bij die plaats van F, dus zowel op de conchoide als op de lijn door A loodrecht op BC) geldt FBC= ABC.

8. Il Teorema Di Morley
Translate this page Fu comunque solo nel 1837 che pierre wantzel (1814-1848) riuscì a dimostrare lanecessità della condizione di Gauss sui poligoni regolari e quindi anche l
http://www.lorenzoroi.net/geometria/Morley.html

9. Faq.maths : Le Fascinant Nombre Pi
Translate this page par LINDERMAN en 1882) et la règle et le compas ne perment ne tracer que des nombresalgébriques, comme l'a montré le mathématcien pierre wantzel en 1837.
http://faq.maths.free.fr/html-old/faq09.php3

Extractions: Le fascinant nombre Pi Régis Décamps Histoire - Première définition Pi est défini comme étant le rapport constant entre la circonférence et le diamètre d'un cercle Remarque : Il a déjà fallu un certain temps à l'homme pour trouver que ce rapport est constant..., et donc pour découvrir l'existence de PI. A l'origine, ce rapport est noté P. C'est Euler qui utilisa la notation de la seizième lettre de l'alphabet grec, notation gardée par la suite vue l'importance de ses travaux. Ainsi, pour tout cercle de périmètre p, de diamètre D (de rayon R), def : p = Pi * D = 2 * Pi * R Connaître l'existence d'une constante est fort intéressant, mais connaître sa valeur l'est beaucoup plus... Elle l'est d'autant plus que Pi apparait dans de très nombreux problèmes physiques et mathématiques : Par exemple, on trouve, par intégration, des formules classiques telles que :

10. Encyclopædia Britannica
pierre Laurent wantzel University of St Andrews Biographical sketch of thisFrench mathematician who solved algebraic equations by radicals.
http://www.britannica.com/search?query=pierre-simon&ct=igv&fuzzy=N&show=10&start

11. Encyclopædia Britannica
pierre Laurent wantzel University of St.Andrews Biographical sketch of this 19thcentury French mathematician known for solving equations by radicals.
http://www.britannica.com/search?query=roots solving&ct=igv&fuzzy=N&start=0&show

12. Niemo¿no¶æ Konstrukcyjnej Trysekcji Dowolnego K±ta
Te uwagi Gaussa zamienil na scisly dowód pierre L. wantzel (18141848) w 1837r. wantzel byl repetytorem w École Polytechnique w Paryzu, tej samej, w
http://ux1.math.us.edu.pl/~szyjewski/FAQ/konstruk/trysekcj.htm

Extractions: Niemo¿liwe jest podanie metody podzielenia cyrklem i linijk± dowolnego k±ta na trzy równe czê¶ci. K±ty dziel± siê na takie, które da siê podzieliæ na trzy czê¶ci cyrklem i linijk± (np. 90 ), oraz na takie takie, których cyrklem i linijk± nie da siê podzieliæ na trzy równe czê¶ci (np. 120 Jak to siê wie, to sprawa jest prosta - sprawdza siê, ¿e wszystkie punkty, które mo¿na zbudowaæ z danego odcinka jednostkowego cyrklem i linijk± odpowiadaja liczbom zespolonym, które przy ka¿dym kroku konstrukcji albo nale¿a³y do ju¿ zbudowanego cia³a, albo s± pierwiastkami trójmianu kwadratowego (o wspó³czynnikach z ju¿ zbudowanego cia³a), czyli w ka¿dym kroku konstrukcji nowe cia³o jest albo tym samym co w poprzednim kroku, albo jego rozszerzeniem stopnia 2. Dlatego stopieñ rozszerzenia (

13. MathsNet: Geometric Construction Course - Classic Problems Of Geometry
In 1837 pierre Laurent wantzel (born 1814 in Paris, France; died 1848 in Paris)published proofs on the means of deciding if a geometric problem can be
http://www.mathsnet.net/campus/construction/classic2.html

Extractions: It is impossible to construct a cube whose volume is twice that of a given volume. The reason is to do with the solutions of cubic equations. In 1837 Pierre Laurent Wantzel (born: 1814 in Paris, France; died: 1848 in Paris) published proofs on the means of deciding if a geometric problem can be solved with ruler and compasses. Gauss had stated that the problems of duplicating a cube and trisecting an angle could not be solved with ruler and compasses but he gave no proofs. In this 1837 paper Wantzel was the first to prove these results.

14. Trisection De L'angle
Translate this page Il n'est pas possible diviser un angle par construction. Démonstration en1837 par pierre Laurent wantzel (1814-1848). règle et au compas. Sites.
http://membres.lycos.fr/villemingerard/Histoire/Trisangl.htm

Extractions: 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 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 ÉQUATION 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 En généralisant Il n'est pas possible diviser un angle par construction Démonstration en 1837 par Pierre Laurent

15. Constructions Géométriques - Constructible
Translate this page pierre Laurent wantzel (1814 - 1848) Il démontre que seuls ces polygonessont constructibles. Tout nombre constructible est algébrique.
http://www.multimania.com/villemingerard/Geometri/Construc.htm

Extractions: Accueil Dictionnaire Rubriques Index ... M'écrire Édition du: Rubrique: CONSTRUCTIONS Constructible Bissection Sommaire de cette page CONSTRUCTIBLE: Théorème de Gauss CONSTRUCTION DES POLYGONES TYPES DE CONSTRUCTIONS Pages voisines Allumettes Centre du cercle Cycloïde Géométrie ... Triangles CONSTRUCTIBILITÉ Peut-on construire rigoureusement une figure géométrique en utilisant des outils précis: règle, compas, marque sur la règle  CONSTRUCTIBLE: Théorème de Gauss Historique Les Grecs connaissaient de nombreuses possibilités de construction Il a fallu 2000 ans pour que Gauss (1796) démontre le théorème suivant Théorème de Gauss (formulation n°1) Il est possible de diviser la circonférence en un nombre impair de parties égales si, et seulement si, le nombre est un nombre premier de Fermat Théorème de Gauss (formulation n°2) Le polygone n'est constructible avec une règle et un compas que si : m étant entier et n premier Ou si n est formé exclusivement de combinaisons d'un nombre premier de Fermat (comme 3, 5, 17, 257, 65 537) et de puissances de deux. Théorème de Gauss (formulation n°3) Un polygone régulier à n côtés est constructible à la règle et au compas si n est de la forme r p p ...p k où les p i sont des nombres premier de Fermat distincts Contient les cas r = et k= Les nombres premiers de Fermat sont de la forme p o et alors p est une puissance de 2.

16. Duplication Du Cube
Translate this page Abordé par Descartes en 1637*. Puis par Gauss (1777 - 1855)*. C'est wantzel (1814- 1848). Panoplie du constructible pierre Delezoïde - Lycée Buffon - Paris XV.
http://www.multimania.com/villemingerard/Histoire/Duplcube.htm

Extractions: 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 DUPLICATION DU CUBE HISTORIQUE ÉQUATION Pages voisines Doubler le carré Règle et compas Transcendant Histoire ... Hilbert DUPLICATION DU CUBE ou Problème de Délos Problème délien ou déliaque Problème de l'autel d'Apollon DUPLICATION DU CUBE Duplication du cube Trouver le volume doublé du cube revient à déterminer la racine cubique de deux. Cette opération est impossible avec la règle et le compas En cherchant (et en trouvant) des solutions utilisant les intersections de cônes, de cylindres et de tores, les Grecs découvrirent les sections coniques, paraboles et hyperboles et même les conchoïdes et les cissoïdes. HISTORIQUE Athènes 430 av. J.-C. Les habitants de l'île de Délos souffrent de fièvre Il veulent que cesse l'épidémie de peste ruinant le pays L'oracle leur recommande de doubler le volume de leur autel cubique, dédié à Apollon De nombreuses tentatives furent des échecs Et, la peste redoublait

17. The Classical Greek Problems
In 1837 pierre wantzel proved that the classical Greek problem of a squaring a cubecould not be solved with the restriction of using only straight lines and
http://www.math.rutgers.edu/courses/436/Honors02/classical.html

Extractions: The Classical Greek Problems Patricia DiJoseph There were three problems that the ancient Greeks (600BC to 400AD) tried unsuccssfully to solve by Euclidean methods, all of which were proven unsolvable by these means as much as two thousand years later, as a result of progress in algebra, and the idea of analytic geometry in the sense of Descartes. The Greeks wanted to solve these problems using only a Euclidean constructions, or as they themselves called them, "plane" methods. Though they were never able to do so ( as they cannot be done this way, they did find a series of remarkably clever constructions using more powerful techniques, involving so-called "solid" and "mechanical" methods, as well as a technique called "verging". Then, in the 19th century, the impossibility of finding purely Euclidean constructions for these problems was finally proved. The three classical Greek problems were problems of geometry: doubling the cube, angle trisection, and squaring a circle. Duplication of the cube is the problem of determining the length of the sides of a cube whose volume is double that of a given c ube. A cube by definition is a three dimensional shape comprised of a height, width, and depth all of the same magnitude s. To find its volume, one multiplies the length (s) by the width (s) and then by the depth (s): the volume is s(s(s or s3. Diagram not converted, here and below

18. The Problem Of Angle Trisection In Antiquity
The first person to prove its impossibility results was pierre Laurent wantzel, whopublished his proofs in a paper called Research on the Means of Knowing If
http://www.math.rutgers.edu/courses/436/436-s00/Papers2000/jackter.html

Extractions: Rutgers, Spring 2000 The problem of trisecting an angle was posed by the Greeks in antiquity. For centuries mathematicians sought a Euclidean construction, using "ruler and compass" methods, as well as taking a number of other approaches: exact solutions by means of auxiliary curves, and approximate solutions by Euclidean methods. The most influential mathematicians to take up the problem were the Greeks Hippias, Archimedes, and Nicomedes. The early work on this problem exhibits every imaginable grade of skill, ranging from the most futile attempts, to excellent approximate solutions, as well as ingenious solutions by the use of "higher" curves [Hobson]. Mathematicians eventually came to the empirical conclusion that this problem could not be solved via purely Euclidean constructions, but this raised a deeper problem: the need for a proof of its impossibility under the stated restriction. The trisection of an angle, or, more generally, dividing an angle into any number of equal parts, is a natural extension of the problem of the bisection of an angle, which was solved in ancient times. Euclid's solution to the problem of angle bisection, as given in his Elements , is as follows: To bisect a given rectilineal angle: Let the angle BAC be the given rectilineal angle. Thus it is required to bisect it. Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined. I say that the straight line AF has bisected the angle BAC. For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively. And the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. Therefore the given rectilineal angle BAC has been bisected by the straight line AF

19. Trisectie Van Een Hoek
eerste echte bewijs van de onoplosbaarheid van de eerste twee problemen werd gegevendoor de weing bekende Franse wiskundige pierre Laurent wantzel (18141848
http://www.pandd.demon.nl/trisect.htm

Extractions: Tot de meetkundige problemen die ook door de Grieken niet konden worden opgelost (niet onbegrijpelijk, overigens), behoren Trisectie van de hoek Gegeven een hoek, verdeel die hoek met behulp van passer en liniaal (we schrijven in hetgeen volgt "penl") Verdubbeling van de kubus Gegeven een lijnstuk met lengte 1, construeer een lijnstuk met lengte 2 met behulp van penl (het zogenoemde " Delisch probleem ") Kwadratuur van de cirkel Gegeven een lijnstuk met lengte 1, construeer een vierkant met oppervlakte p met behulp van penl Deze drie problemen zijn inderdaad onoplosbaar (er is een verschil tussen "niet kunnen oplossen" en "onoplosbaar").

20. Geometriske Konstruktioner Med Passer Og Lineal
Først i 1837 blev det af pierre Laurent wantzel (1814 1848) bevist, at det erumuligt at tredele en generel vinkel og fordoble en terning med passer og
http://www.matematiksider.dk/klassisk.html

Extractions: Det var grækerne, der førte geometrien frem til en høj grad af fuldkommenhed. Her spillede den såkaldte Pythagoræer-skole en helt central rolle. Dette religiøst filosofiske broderskab, der var blevet grundlagt af navnkundige Pythagoras (6. århundrede f. Kr.), forsøgte at udtrykke alle forhold i naturen ved forhold mellem naturlige tal. Opdagelsen af de såkaldte usigelige tal , altså de tal, vi i dag kalder irrationale , var så stor en skuffelse for pythagoræerne, at de i høj grad forlod aritmetikken til fordel for geometrien. I geometrien kan man for eksempel konstruere en længde, der er lig med kvadratroden af 2, hvorimod tallet ikke kan skrives som en brøk mellem hele tal, idet det jo ikke er rationalt. Længden kan konstrueres som diagonalen i en retvinklet trekant, hvor begge kateter har længden 1. Herefter begyndte en lang æra med geometrien i centrum. Da den rette linje og cirklen hører til de mest ædle geometriske figurer er det ikke svært at forstå, at man begyndte at foretage konstruktioner med passer og lineal. Og opgaverne var ikke af praktisk art, som tilfældet havde været med ægypterne. Grækerne betragtede geometrien for matematikkens egen skyld. Man var ikke interesseret i tilnærmede løsninger. Geometrien blev betragtet som en ophøjet åndelig disciplin, som skulle besjæle eleverne med moralsk kraft, så de kunne begå handlinger til gavn for almenheden. Platon (427 - 347 f.Kr.) var en af geometriens store fortalere. Han mente, at man skulle lære at trække tankeindholdet, selve idéen, ud fra det konkrete.

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