(compact Textbooks In Mathematics) Ad Meskens, Paul Tytgat (auth.) - Exploring Classical Greek Construction Problems With Interactive Geometry Software-birkhäuser Basel (2017).pdf

Compact Textbooks in Mathematics

Ad Meskens Paul Tytgat

Exploring Classical Greek Construction Problems with Interactive Geometry Software

Compact Textbooks in Mathematics

For further volumes: http://www.springer.com/series/11225

Compact Textbooks in Mathematics This textbook series presents concise introductions to current topics in mathematics and mainly addresses advanced undergraduates and master students. The concept is to offer small books covering subject matter equivalent to 2- or 3-hour lectures or seminars which are also suitable for self-study. The books provide students and teachers with new perspectives and novel approaches. They feature examples and exercises to illustrate key concepts and applications of the theoretical contents. The series also includes textbooks specifically speaking to the needs of students from other disciplines such as physics, computer science, engineering, life sciences, finance.

Ad Meskens Paul Tytgat

Exploring Classical Greek Construction Problems with Interactive Geometry Software

Ad Meskens Department of Education and Training Artesis Plantijn University College Antwerpen, Belgium

Paul Tytgat Antwerpen, Belgium

ISSN 2296-4568 Compact Textbooks in Mathematics ISBN 978-3-319-42862-8 ISBN 978-3-319-42863-5 (eBook) DOI 10.1007/978-3-319-42863-5 Library of Congress Control Number: 2017931078 Mathematics Subject Classification (2010): 51-01, 51-03, 01A05, 1A20, 97G40 Birkhäuser © Springer International Publishing Switzerland 2017 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Cover design: deblik, Berlin Printed on acid-free paper This book is published under the trade name Birkhäuser The registered company is Springer International Publishing AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

v

Preface This book has been ten odd years in the making. Its first inception was a lecture which I was asked to give at a refresher course for mathematics teachers. The text evolved into a little booklet for the Dutch Zebra series (Epsilon publishers)1 . This is a series of 60-page booklets aimed at high school pupils preparing for their final exams. As the old saying goes, “Du choc des idées jaillit la lumière”, discussions with the editors led to a text which had more potential than the allowed 60 pages. What you have before you is a greatly expanded version of that booklet. This book is not intended as a history of mathematics, nor is it a complete overview of the problems at hand, the duplication of the cube, the trisection of an angle and the squaring of a circle. Anyone looking for new historical insights should consult other books. We only put forward our mostly educational view, sometimes new, of constructions which have been published numerous times before. The history (and sometimes mythology) of mathematics is used as an introductory story to raise students’ interest in the problem. On the other hand, neither is this a mathematics manual that observes a deductively oriented organisation. It is a well-known fact that deductive organisation is the last step in any given discovery process. Getting acquainted with a historical problem will show students that the pathway to a solution is not linear but curved, sometimes even going backwards, full of mistakes due to lack of appropriate mathematical language and symbolism. Indeed, if anything can be said about the classical Greek construction problems, it is that they are a prime example of this non-linear development. They have captured the imagination of mathematicians for thousands of years, they have led them astray, they have led them into new realms of mathematics until it turned out they were unsolvable with the restraints placed on them. The following pages are excursions, at undergraduate student level, into these three famous problems. A historian of mathematics faces a dilemma: citing and subsequently explaining an ancient text or translating the text into modern mathematical terminology. The first approach alienates the modern reader, the second betrays the ideas of the ancient author. The author of an educational exposé does not face that dilemma. He is concerned more with heritage than with history per se. Here you will find ancient problems written in modern terminology, making them accessible to the mathematical community at large. By making the point that the ancient mathematicians had fewer tools at their disposal the reader will acquire even greater admiration for their accomplishments, possibly even finding him/herself drawn to ancient texts and perhaps to exploring them in greater depth. The history of mathematics is important for teachers of mathematics. It gives them the opportunity to let their students see the motivation behind the introduction of certain concepts, it makes clear that mathematics is a non-linear endeavour and students get a glimpse of the people who created mathematics in all their human

1

Meskens and Tytgat (2015).

vi

Preface

aspects. It also makes students aware that they often come up against the same difficulties in seeking to grasp certain concepts as the inventors did. We hope that many teachers will use these chapters as enrichment material in their classes and enable their students to gain a deeper understanding of geometry. The use of dynamic geometry software packages, enables them to explore geometric relationships in an educational, enquiry-based fashion. Relationships which most likely would have remained hidden in a classic pencil and paper approach. By focusing on constructions and the use of Interactive Geometry Software or IGS for short, the reader is confronted with the same problems that ancient mathematicians once faced. The neusis construction is of particular interest to be explored with IGS, as it lets readers discover a class of interesting curves. Readers get to retrace the footsteps of Euclid, Viète and Cusanus amongst others and then, by experimenting, discover geometric relationships that far exceed their accomplishments. Over 140 exercises guide readers through methods which were developed to try and solve the problems. The exercises are at undergraduate student level and only require a command of elementary Euclidean geometry and pre-calculus algebra. These exercises are especially well-suited for students who are thinking of becoming a mathematics teacher. It may be argued that the constructions performed in IGS are not real constructions because IGS uses coordinate geometry. This is of course true, more or less. Any construction, be it on paper, or using a computer program, constitutes a particular case. IGS has the advantage that, using the Move tool, a multitude of particular cases can be checked. While a geometric assertion may not be proved using IGS it may be disproved, by finding a counterexample. IGS will not replace the deductive approach, but it adds a feature to doing geometry: experimentation. For some students, an animation showing that the medians of a triangle always intersect in one point may be more convincing than the proof itself. For the more mathematically-minded students, it is an invitation to search for that proof, showing that what IGS suggests is indeed always true. IGS is a didactic tool of which we are only seeing the beginnings of its possibilities as an incentive for engaging in proper mathematics. Writing a book like this is not possible without the help of many people and institutions. I have the pleasure of thanking Epsilon Uitgaven and NVvW (Dutch Association of Mathematics Teachers) for their permission to reuse parts of our booklet in the Zebra series. I also wish to thank the editors of the Zebra series for the useful feedback and the many valuable suggestions. Their comments made me turn a classic text into an enquiry-based and experiment-oriented exposé. Thanks are also due to VVWL (Flemish Association of Mathematics Teachers) for the use of parts of articles which appeared in their journal Wiskunde en Onderwijs. Jean Paul van Bendegem, Hiram Bollaert and Joost Bambust read through the English manuscript and also gave many valuable suggestions. I could count on Stephen Hargreaves for useful advice on English terminology. I am indebted to Artesis Plantijn University College Antwerp for its support in the form of a grant. The collegues and the staff of the department of Education and Training have supported the writing of this book in various ways. The present book would never have materialised without the support from various libraries, with particular reference to the library of the Education and Training Depart-

vii Preface

ment of Artesis Plantijn University College Antwerp, Erfgoedbibliotheek Hendrik Conscience (Antwerp), Museum Plantin Moretus (Antwerp) and the Koninklijke Bibliotheek Albert I (Brussels). Rttaalkantoor corrected the English. The longstanding collaboration with Paul Tytgat proved as invaluable as ever. We are also grateful to Yola and David Coffeen of Tesseract – Early Scientific Instruments for their permission to publish the photographs of the ellipsograph in 7 Sect. 2.4 (. Fig. 2.14). Finally, I would like to thank my wife Nicole whose support, for over thirty years, has been a continuing labour of love. In one apocryphal story, King Ptolemy wanted Euclid to teach him mathematics, but he got bored very quickly with the theorems and proofs. He asked Euclid whether there was a faster way of learning mathematics, to which Euclid replied “Sire, in mathematics there is no Royal Road.” A similar story is told about Alexander the Great and Menaechmus. You, reader, will not walk a regal road, but when you have finished the book you will feel you have been given a royal gift.

Some preliminary remarks The equivalent American style notation is given between brackets. ! 4 AB (AB) means the straight line through A and B ! 4 ŒAB (AB) means the ray with vertex A and through B 4 ŒAB (AB) means the line segment bounded by the end points A and B 4 kABk (AB) means the length of ŒAB ! 4 AB (AB or AB) means the vector represented by the line segment ŒAB and with initial point A and terminal point B a (a) means the vector represented by the line segment ŒOA and with initial 4 ! point O and terminal point A 4 ]BAC means the angle with vertex A and bounded by the rays ŒAB and ŒAC 4 †BAC (m†BAC ) means the size of the angle ]BAC (i.e. the magnitude of the smallest rotation that maps one of the rays onto the other) O in a triangle, is the angle opposite the edge a 4 A, 4 CA means the circle centered at A 4 rA means the radius of circle CA 4 4ABC Š 4DEF means that the triangles 4ABC and 4DEF are congruent, i.e. their respective sides have the same length and their respective angles have the same size. In the text at hand, “to construct” means using geometric tools with which we cannot measure. So we can use a straightedge and a compass, but we can neither use a ruler nor a protractor. This means we remain more or less faithful to the classical Greek concept of geometric construction. Diligent readers will notice that we sometimes add that the length of a line segment is a, for instance. This usually simplifies the reasoning in a proof and avoids cumbersome formulations. In these cases the construction is possible, even though the intitial line segment itself is not necessarily constructible. The bulk of the illustrations were sourced from books from three libraries, which are referred to by the following abbreviations: EHC – Erfgoedbibliotheek Hendrik Conscience, Antwerp KBR – Koninklijk Bibliotheek Albert I, Brussels MPM – Museum Plantin-Moretus/Prentenkabinet, Antwerp – Unesco-World Heritage

The text was typeset in LYX an editor for LATEX.

ix

Contents 1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.1 1.2

Ad Meskens and Paul Tytgat Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Interactive Geometry Software (IGS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 7

2

The Genesis of Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

2.1 2.2 2.3 2.4 2.5

Ad Meskens and Paul Tytgat Greeks and their Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Examples from Plato’s hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Adrian van Roomen and Apollonius’ Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Mechanical construction aids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Neusis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 12 14 19 23

3

Compass and straightedge constructions . . . . . . . . . . . . . . . . . . . . . . . . . .

27

3.1 3.2 3.3

Ad Meskens and Paul Tytgat Euclid’s Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Compass and straightedge constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . François Viète and Apollonius’ Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27 30 36

4

The Delian Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

4.1 4.2 4.3

Ad Meskens and Paul Tytgat Delos and its altar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wine gauging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Doubling the cube with a neusis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43 46 49

1.3

5

1

Trisecting an angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55

Ad Meskens and Paul Tytgat Nicomedes’ conchoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Archimedes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The first Archimedean trisection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Archimedean spiral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Flemish Jesuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hippias of Elis and the quadratrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55 59 61 63 65 71

6

Squaring the circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

6.1 6.2 6.3 6.4 6.5 6.6

Ad Meskens and Paul Tytgat Archimedes’ spiral revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dinostratus’ quadratrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hippocrates’ lunes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Franco of Liège, the demise of mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nicolas of Cusa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Archimedes’ approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 76 78 88 90 93

5.1 5.2 5.3 5.4 5.5 5.6

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Contents

Adriaan van Roomen and Ludolff van Ceulen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Kepler’s use of infinitesimal methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gregory of St Vincent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

98 99

7

Constructible numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

105

7.1 7.2 7.3

Ad Meskens and Paul Tytgat Constructing numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The theory of equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Squarable lunes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Squaring the circle is different . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

105 108 113 118

8

The Cinderella of regular polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121

8.1 8.2 8.3

Ad Meskens and Paul Tytgat The inconstructibility of the heptagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The relation with the trisection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A neusis for the heptagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121 124 125

6.7 6.8 6.9

7.4

Servicepart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

95

127 128 178 184

xi

List of Figures1 Fig. 1.1 Fig. 1.2 Fig. 1.3 Fig. 1.4 Fig. 2.12 Fig. 2.13 Fig. 2.14 Fig. 2.15 Fig. 2.18 Fig. 2.20 Fig. 2.21 Fig. 3.1 Fig. 3.9 Fig. 3.10 Fig. 4.2 Fig. 5.3 Fig. 5.4 Fig. 5.7 Fig. 5.9 Fig. 5.13 Fig. 5.14 Fig. 5.15 Fig. 5.16 Fig. 6.5 Fig. 6.15 Fig. 6.21 Fig. 6.25

1

The cubic close packing (drawings by Paul Tytgat) . . . . . . . . . . . . . . 3 The construction of Euclid 1.1 in C.a.R. (Z.u.L.) . . . . . . . . . . . . . . . . . . . . 4 Loci in GeoGebra, Cinderella and CaRMetal . . . . . . . . . . . . . . . . . . . . . 6 Dürer’s conchoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Drawing an ellipse using two tacks and piece of rope . . . . . . . . . . 19 Archimedes’ trammel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Italian ellipsograph (trammel) ca. 1700 . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Some of Frans van Schooten’s drawing devices to draw a 21 conic section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Performing a neusis (drawing by Paul Tytgat) . . . . . . . . . . . . . . . . . . . 24 An instrument to draw conchoids with (drawing by Paul Tytgat) 25 The branches of a conchoid (drawing by Paul Tytgat) . . . . . . . . . . 25 The proof of Pythagoras’ theorem in Byrne’s Euclid (1847) . . . . . . 28 The construction of a square equiareal to a rectangle ABCD from Marolois’ La Géometrie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Viète’s solution to Apollonius’ tangency problem from Opera mathematica (1646) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 A winegauger is taking a measurement with his gauging rod (left), terminology (right, drawing by Paul Tytgat) . . . . . . . . . . . . . . . 47 An instrument with which (a branch of) the conchoid can be drawn (drawing by Paul Tytgat) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Construction of the conchoid with which, for a given angle, the trisection can be accomplished (drawings by Paul Tytgat) . 57 Artistic impression of Archimedes’ claw (drawings by Paul Tytgat) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 A trisection instrument based on Archimedes’ neusis (drawing by Paul Tytgat) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Antwerp’s Carolus Borromeus Church, designed by Franciscus de Aguilon s.j. and Pieter Huyssens s.j. (Photos Ad Meskens) 66 Portrait of Gregory of St-Vincent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 Gregory of St Vincent’s generalisation of Pythagoras’ theorem 67 Frontispiece to the second book of de Aguilon’s Opticorum Libri Sex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 In all of these cases Pythagoras’ theorem holds: area B C area C D area A (drawing by Paul Tytgat) . . . . . . . . . . . . . 79 Transforming curvilinear figures into rectilinear figures, from Falco (1591) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Ludolph van Ceulen and his approximation of  . . . . . . . . . . . . . . . 96 Gregory’s figure to explain the calculation of the area between two parabolae and between a parabola and a straight line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Apart from this list of figures all other figures are made in GeoGebra by Ad Meskens

xii

List of Figures

Fig. 6.26 Fig. 7.6

Frontispiece of Gregoy of Saint Vincent’s Problema Austriacum on the squaring of the circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Part of Viète’s solution to van Roomen’s problem using chords

100 111

1

1

Introduction Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_1

1.1

Introduction

For ages, mathematicians have challenged one another by posing problems. In most cases these were generalisations or slightly adapted versions of problems that had already been solved previously. For instance, we know that every even number can be written as the sum of two uneven numbers. The proof of this is nearly self-evident: 2n D .2n  1/ C 1 D .2n  3/ C 3 D : : : Is it possible to place more rigorous demands on our question? For instance, is it possible to write each even number as the sum of two uneven prime numbers? Let’s see, we have 20 D 13 C 7, 22 D 11 C 11, 24 D 13 C 11, 26 D 13 C 13, 28 D 17 C 11, . . . Just pick an even number and you’ll see that you can always find two uneven prime numbers whose sum is your even number. Always?? That is the question! To date, no one has succeeded in proving this so-called Goldbach’s conjecture! Christian Goldbach (1690–1764) was a German mathematician, who corresponded with many famous mathematicians of his time, such as Gottfried Leibniz (1646–1716), Leonhard Euler (1707–1783), and Nicholas I Bernoulli (1687–1759). In a letter addressed to Leonhard Euler dated 7 June 1742, he proposed that: “Every integer which can be written as the sum of two primes, can also be written as the sum of as many primes as one wishes, until all terms are units.” In the margin of his letter he wrote a second conjecture: “Every integer greater than 2 can be written as the sum of three primes.” In his reply of 30 June 1742, Euler reminded him of a conversation they had had in which he had remarked that his conjecture followed from the statement: “Every even integer greater than 2 can be written as the sum of two primes.” The latter conjecture is now known as Goldbach’s conjecture1 . Another example is Fermat’s Last Theorem. We know that in a right-angled triangle with legs having lengths 3 and 4, the hypotenuse has length 5. Indeed 32 C 42 D 52 . We know a multitude of triplets .a; b; c/ with a; b; c 2 N that satisfy the equation a2 C b 2 D c 2 . These numbers are known as Pythagorean triplets. Now would it be possible to find triplets that satisfy the equation a3 C b 3 D c 3 or 4 a C b 4 D c 4 or an C b n D c n (n > 2, n 2 N). 1

Euler; Koch (2007).

2

Chapter 1  Introduction

1 Greek mathematician Diophantus (3rd century AD) had proved, amongst other theorems in number theory, that any square of a rational number can be decomposed into two other squares of rational numbers (theorem II.8). When reading the 1621 edition of Diophantus by Claude Gaspard Bachet de Méziriac (1581–1638), French mathematician Pierre de Fermat (1601 or 1606/7–1665) made marginal notes. In the margin next to theorem II.8 Fermat wrote that he had found a proof for the assertion that for all but trivial cases, no integer solutions exist for the equation an C b n D c n (n > 2, n 2 N). “Alas”, he added, “this margin is too small to contain the proof”. Nowhere among his papers the proof was found. For centuries, mathematicians tried to find a proof. Only in 1994 did Andrew Wiles succeed in proving it. The proof is incomprehensible to all but the brightest mathematicians. Wiles used the theory of elliptic curves, a subject which at first glance is light years away from number theory2 . Problems such as those cited above are the kind of problems that mathematicians love: they are easy to pose, a secondary school pupil can understand them and . . . they are very, very hard to solve. Exercise 1 Show that .2pq; p 2  q 2 ; p 2 C q 2 / is a Pythagorean triplet for all p; q 2 N. Show that .2n2 C 2n; 2n C 1; 2n2 C 2n C 1/ is a Pythagorean triplet for all n 2 N. Also show that these triplets are a special case of the triplets in the first part of the exercise. Exercise 2 Prove that .3; 4; 5/ is the only Pythagorean triplet containing three consecutive positive integer numbers. Exercise 3 Prove that no isosceles right triangle exists whose sides are integers. Exercise 4 In the Treehouse of Horror VI episode of The Simpsons (the animated TV series), one can see the equation 178212 C 184112 D 192212 . In another episode, entitled The Wizard of Waverley Terrace one can see 398712 C 436512 D 447212 . Use a pocket calculator to calculate both sides of the equations. The equations seem to hold. Use elementary number theory to disprove these equations, without actually calculating the values of both sides of the equation. (Hint: look at the divisibility of each term in these equations). Exercise 5 The Kepler conjecture was a problem posed by the astronomer Johannes Kepler (1571–1630). It says that no arrangement of equally sized spheres filling space has a greater average density than that of the cubic close packing (a packing known from material science). . Fig. 1.1 (left) shows a cubic close packing. A unit cell is the simplest structural unit which describes such a crystalline structure. The unit cell for the cubic close packing is shown in . Fig. 1.1 (right). It is a cube with one eight of a sphere at each vertex and half a sphere at each face. Show that the space taken up by the parts of the spheres is approximately 74% of the volume of the cube and hence that the total density is about 74%. The conjecture itself was proved in 2005 by Thomas Hales3 . Hint: replace 2 3

Singh (1997); Lanser (2000); Barner (2001); Meskens (2010), p. 166–169. Hales (2005).

3 1.2  Interactive Geometry Software (IGS)

1

. Fig. 1.1 The cubic close packing (also called the face-centered cubic), a crystalline structure (drawings by Paul Tytgat)

the parts of the spheres in the unit cell by the equivalent number of spheres, calculate the volume of these spheres and divide it by the volume of the unit cell. To calculate the volume of the sphere, express the radius of the spheres in terms of the edges of the cube.

The same kind of problems exists in plane geometry. This is the story of three of these problems which have haunted mathematicians for ages. They are old, they are easy to formulate and they are . . . very hard to solve. Yet with a little help, any mathematics student can get to grasp them. Consider a square. It is easy to construct, using only compass and straightedge, a square which is twice as large as a given one. One needs only to draw the diagonal and consider this as the edge of the new square. Would it be possible to construct a cube which has a double volume of a given one, using only compass and straightedge? It is easy to divide an angle into two equal angles by drawing, using only compass and straightedge, the angle bisectors. Now would it be possible to divide an angle into three equal angles, using only compass and straightedge? It is possible, using rigid transformations equivalent to using only compass and straightedge, to transform a rectangle or a polygon into a square. But is this also true for a circle? These problems are respectively referred to as the duplication of the cube, the trisection of an angle and the squaring of the circle. We shall see how mathematicians tried to solve these problems but, if limited to compass and straightedge, failed to do so.

1.2

Interactive Geometry Software (IGS)

Many of the illustrations in this book are made with GeoGebra, but they could just as easily have been made with Geometer’s Sketchpad, Cinderella or any other Interactive Geometry Software or IGS for short. Interactive Geometry Softwares are all interactive geometry, algebra, and calculus applications, intended for teachers and students. They allow one to make constructions with points, vectors, segments, lines, polygons, conic sections and functions. All of these objects can be changed dynamically afterwards. Elements can be entered and modified directly on screen. Most options can be activated by clicking an icon in the toolbar. Sometimes these icons are grouped and can be accessed through a drop down menu. The , a Lines toolbox , a Lines Properties toolbox groups typically include a Point toolbox and a Curves toolbox conic sections.

. This last toolbox holds buttons for circles, circular arcs and

4

Chapter 1  Introduction

1 . Fig. 1.2 The construction of Euclid 1.1 in C.a.R. (Z.u.L.)

Other toolboxes typically are a Transform toolbox

, allowing you to translate, rotate

, allowing you to measure the length of a line or mirror an object and a Measure toolbox segment or to determine the measure of an angle. Some of these programs also allow you to check properties such as collinearity or perpendicularity. Because of the similarity between these programs we are able to write a book using generic terms to indicate icons, without referring to a specific program. Yet the similarity of names, and in many cases also of icons, between the programs allows the reader to recognize all the names of icons in his favourite IGS. We only use straightedge and compass constructions, which are very basic geometric constructions, so they can all be carried out with the basic tools of an IGS. and click twice inside the As an example, open your IGS, click on Create Point drawing window. You will see two points appear: A and B. Now select Segment

, click

both points A and B. Click Circle with Centre and Point , click A and then B. Follow the same procedure, but reverse the order of A and B. You will see two circles. Select Intersect then click each circle – points D and C appear (. Fig. 1.2). The triangles 4ABC and 4ABD are equilateral. What you have just done is your IGS’s equivalent to Euclid’s first construction in book I of Elements (see 7 Sect. 3.1). Exercise 6 Create an IGS file in which you draw a triangle 4ABC . You can do this in one of two ways. You can draw three points and connect them with line segments. Or you can use the Polygon tool, draw three points and close the polygon by clicking the first point. 1. Determine the intersection point(s) of the perpendicular bisectors of the sides of the triangle 4ABC . 2. Determine the intersection point(s) of the medians of the triangle 4ABC . 3. Determine the intersection point(s) of the altitudes of the triangle 4ABC . 4. Determine the intersection point(s) of the bisectors of each vertex of the triangle 4ABC . In each of the above cases you will notice that the respective lines intersect in one point. Now for each of the cases, using the Move tool, move any of the vertices of the triangle around. What do you notice?

5 1.2  Interactive Geometry Software (IGS)

1

Exercise 7 Illustrate Napoleon’s theorem in IGS: if equilateral triangles are constructed on the sides of any triangle, either all outward or all inward, the centroids of those equilateral triangles themselves form an equilateral triangle. Exercise 8 A well-known tale about a treasure island is this: A pirate was stranded on a deserted island. Among the wreckage was a treasure, which he buried. The island was home to one oak tree and two birches. The pirate walked from the oak tree to one of the birches, then turned at a right angle and walked an equal number of steps. Here he drove a spike into the ground.

He returned to the oak tree and repeated the procedure with the second birch. Again, he drove a spike into the ground. He buried the treasure halfway between both spikes and then removed the spikes. A week later he was rescued. Two years later he returned to the island only to find that the coastline had receded and the oak tree had disappeared. Create an IGS file for this problem and show how the pirate will be able to find his treasure anyway (hint: choose three points (oak, birch1 and birch2) and for this configuration determine where the pirate will bury his treasure). Now move the point representing the oak around. Prove why this is the case for this configuration. Do configurations exist in which he will not be able to find his treasure?

These exercises show the great advantage of an IGS: we can experiment and see a myriad of cases by just moving one or a couple of points around. We will now take a closer look at some concepts we regularly use in the exercises: sliders and loci. A slider is a tool you can vary a parameter continuously with. Not all IGSs have these sliders, but it is not too difficult to construct one. A slider takes the form of a line segment on which a point can move indicating its the value. kAC k . First define a line segment ŒAB and a moveable point C on it. Calculate a D kABk Moving C along ŒAB will let a vary between 0 and 1. If you wish your slider to vary between kAC k . You can even let your slider take on 0 and  simply multiply the ratio by  W a D  kABk negative values. Define a line segment ŒAB and let its midpoint be M . Let C be a point on kAC k  kAM k , then a will vary between  and . In some IGSs ŒAB and define a D  kAM k some programming (“scripting”) may be necessary.

6

Chapter 1  Introduction

1

. Fig. 1.3 Left the locus generated with the command Trace, right the locus generated with the command Locus in GeoGebra (top), Cinderella (middle) and CaRMetal (bottom)

The Locus and Trace tools have similar effects, but with the first option the IGS itself draws the solution, whilst the second tool enables users to dynamically create the result. An example will clarify this immediately. Suppose a point, A (the mover), can move freely on a curve (a straight line, a circle, . . . ) called the directrix or road. Suppose the position of another point, B (the tracer), in some way depends on the position of A. It is then possible to find the locus of B as A moves along the directrix. A first method to find this locus is to select Trace and drag A along the curve using the cursor. B now traces out a line. This line is part of the locus of B. The effect is even more spectacular if you use animation. A will now automatically move along its directrix, while B traces out the locus. We can also select Locus and indicate the mover A and the tracer B. The locus immediately becomes visible. In the following paragraphs, we will call this “the locus of B with A”, meaning “the locus of B while A moves along the directrix”. Let us take a look at an example. On a straight line l select two points A and C . Now select a point D, not on l (for ease, select C on the negative x-axis, A on the positive x-axis and D on the negative y-axis). Draw the straight line AD and the straight line lC parallel to AD and

7 1.3  Polar coordinates

1

. Fig. 1.4 Dürer’s conchoid (right: Dürer (1525), EHC H 202415)

through A. Draw the line lC perpendicular to CD and through C . Determine the intersection B of lC and lA . The position of B is dependent on each of the positions of A, C and D. We will call this drawing the initial configuration. Click B and select Trace. Move C along the straight line l. We find a line which is (a part of) the locus of D as C moves along l (see . Fig. 1.3, left). Now return to the initial configuration. Select Locus, then select the mover C and the tracer B, the locus of B appears immediately (see . Fig. 1.3, right). Exercise 9 Putting A.a; 0/, C.t; 0/ and D.0; d / with t 2 R, determine the Cartesian equations of CD; lD and lA . Determine the system of equations with which you can determine the coordinates of B. Eliminate t from these equations. Prove that the resulting Cartesian equation of the locus of B with C is: y 3 C .x  a/xy C d.x  d /2 D 0

(1)

Exercise 10 Let A and B be points moving on a pair of perpendicular lines which intersect at O in such a way that kOAk C kOBk is constant. On AB, mark points C and D at a fixed distance c from A. The loci of points C and D is Dürer’s conchoid (see . Fig. 1.4). Although it is named conchoid, it is not a true conchoid (see 7 Sect. 5.1). Create an IGS file in which you draw Dürer’s conchoid. The geometric definition can be generalised analytically. Let A and B be on the x- and y-axis respectively, the condition then becomes Ax C By D b, b 2 R. Define three sliders a, b and c. Define A.a; 0/ and B.0; b  a/. Determine the points C and D on AB for which kAC k D kADk D c. Moving the slider a while Trace is on for C and D will show you Dürer’s conchoid. If you want to use Locus, the mover, slider a, and the tracer, point C or D, the loci of C and D will appear respectively. You can still move sliders b and c for other configurations.

1.3

Polar coordinates

Cartesian coordinates are not the only coordinates used in mathematics. One other coordinate system is that of polar coordinates. In polar coordinates we determine the position of a point P by a distance r to a given point O, the origin or pole, and the angle  between the positive x-axis or polar axis and ŒOP ,

8

Chapter 1  Introduction

1 . Fig. 1.5 Polar coordinates of a point P

in counterclockwise direction. We call .r;  / the polar coordinates of point P . By definition, the origin has polar coordinates.0;  / for any  . Using trigonometry, we can deduce transformation formulae to find the Cartesian coordinates if polar coordinates are known and vice versa. Take a closer look at . Fig. 1.5: in 4OPQ QO is a right angle. So we can use the formulae for right-angled triangles. Suppose P has Cartesian coordinates .x; y/ then in 4OPQ: 8
(2)

(3)

We have now proved the transformation formulae for the first quadrant. It turns out that we can use the very same sets of formulae (2) and (3) in the other quadrants. y In the first quadrant we can also use  D arctan . x Exercise 11 Prove the transformation formulae (2) and (3) for the other quadrants.

A polar equation is an equation in r and  . A solution to a polar equation is an ordered pair .R; #/ that satisfies the equation if R is substituted for r and # is substituted for  . The graph of a polar equation is the set of all points with coordinates .R; #/ which satisfy the equation. Some IGSs allow you to work in polar coordinates, but for a number of reasons it may also be more appropriate to keep working in a Cartesian coordinate system. In many cases the

9 1.3  Polar coordinates

1

polar equation can be expressed as r D f . /. First define a function r.t / D f .t / (each  in the polar equation is replaced by t ). Then let your IGS draw the curve defined by points .r.t / cos t; r.t / sin t /, you can do this in a number of ways e.g. by defining the parametric curve 8 < x D r.t / cos t :y D r.t / sin t: Here are two examples. The curve with polar equation r D 2 sin  looks like this:

We obtain a circle with radius 1 of which the centre has Cartesian coordinates .0; 1/. The curve with polar equation r D 2 C 4 cos  shows a completely different picture:

This curve is called Pascal’s limaçon, named after Etienne Pascal (1588–1651), father of the well-known mathematician Blaise. In French, “limaçon” means “little snail”. Exercise 12 (a) Use the transformation formulae to obtain the Cartesian equation of the circle with polar equation r D 2 sin . (b) Which curve is represented by the polar equation r D 2 cos ? Which Cartesian equation does this curve have? (c) Which curves are represented by r D 2a sin  and r D 2a cos ? Exercise 13 Create an IGS file in which you define a slider a. Draw the curve(s) with polar equation r D 2 C a cos . Let a vary between 0 and 10. Exercise 14 Use the transformation formulae to obtain the polar equation of the lines with Cartesian equation x D a, y D b, y D mx, y D mx C q.

11

2

The Genesis of Geometry Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_2

2.1

Greeks and their Geometry

Many elementary geometry books for secondary schools contain exercises in which a construction with compass and straightedge is called for. In many cases, these are exercises whose history goes back to Ancient Greece. In the history of western civilisation, Greek Antiquity occupies a prominent place. This civilisation is sometimes called the cradle of western civilisation, philosophy and mathematics. Greek civilisation can not be reduced to the Greek mainland however. At its zenith, it stretched from Sicily across the eastern Mediterranean to Asia Minor (now Turkey) and the Black Sea coasts (see . Fig. 2.1). Next to Athens and Sparta, the main Greek cities of the era were Alexandria on the Nile delta, Syracuse on Sicily and Milete on the coast of Asia Minor. Greek identity was defined through common trade interests, not as a nation state. Politically, the polis or the city state was the most important political entity. Despite the impression to the contrary, our knowledge about the lives of Greek mathematicians is limited to say the least. Moreover it is hard to separate fact from fiction as most biographical information about them was written several decades, sometimes even centuries, after their death. This applies even more to pre-Socratic thinkers who lived before the fifth century BC. The first Greek mathematician we can identify with some confidence is Thales of Milete (ca. 624–548/45 BC). Thales lived in Ionia (on the coast of Turkey). It was there that classical Greek science emerged, which would go on to have such a profound impact on our science. It was in Ionia that the idea took root that complex, natural phenomena can be explained by a set of basic rules. Some of the theorems that are attributed to Thales include: 1. A circle is bisected by its diameter 2. The base angles in an isosceles triangle are equal 3. When two straight lines intersect then the opposite angles are equal 4. An angle inscribed in a semicircle is a right angle (Thales’ circle theorem) 5. If two intersecting lines are cut by a two parallel lines then two similar triangles are produced (Thales’ intercept theorem)

12

Chapter 2  The Genesis of Geometry

2

. Fig. 2.1 Greek colonisation about 500 BC (after Vermaseren (1977), p. 9)

Plato (Athens ca. 427 BC–ca. 348 BC) was a philosopher, as well as a mathematician. He founded the Academy in Athens, the first institution of higher learning in the Western world. Plato, Socrates and his most famous student, Aristotle, are seen as the founders of Western philosophy and science. From the fourth century onwards, Plato’s philosophical ideas gained ground. Their influence on mathematics became visible in geometry, in which hierarchies of classes of constructions emerged. The first class was the most abstract, the third and last was the most “earthly and mechanical”: 1. constructions with only straight lines and/or circles, i.e. constructions with compass and straightedge 2. constructions in which conic sections (parabolae, hyperbolae and ellipses) are used 3. constructions with other (mechanical) construction means Of course these constructions need to be accomplished in a finite number of steps.

2.2

Examples from Plato’s hierarchy

In Plato’s hierarchy, compass and straightedge constructions are the most abstract. Here is one example of such a construction. To draw a hexagon inscribed in a circle, we observe the following procedure (see . Fig. 2.2):

»

Select the width of the compass equal to the radius of the given circle Select a point A on the circle Place the compass point in A Draw arcs which intersect the circle at B and C Place the compass point, with the same width, in B and draw an arc This arc intersects the circle at A and D

13 2.2  Examples from Plato’s hierarchy

2

. Fig. 2.2 The construction of a regular hexagon inscribed in a circle

Place the compass point, with the same width, in D and repeat the above procedure until an arc intersects the circle at C The intersection points which you have constructed are the vertices of a regular hexagon inscribed in the circle. Plato’s second category contains constructions which can be carried out with the aid of conic sections. A circle, which can be drawn with the aid of a compass, is a special type of conic section. It is therefore natural to consider the conic sections as additional construction aids. To show why they cannot be included in the first category, let us first take a look at the parabola. A parabola is defined as the locus of all points having the same distance to a given line, directrix d , and to a given point, focus F . We can construct points of a parabola with this procedure (see . Fig. 2.3):

»

Draw the perpendicular on d and through F This line intersects d in D kDF k Draw straight lines di parallel to d at regular intervals, beginning at a distance 2 to d Let the intersection of di with DF be Di For each of these lines set the compass width at kDDi k, place the compass point in F

. Fig. 2.3 The construction of a parabola with focus F and directrix d

14

Chapter 2  The Genesis of Geometry

Draw arcs which intersect di Let these intersection points be denoted by Pi;1 and Pi;2 Pi;1 and Pi;2 are points of the parabola D1 is a special point, here P1;1 and P1;2 coincide. D1 is called the vertex of the parabola, and is often referred to with the letter V . The line DF is called the axis of symmetry.

2

We notice that it is possible to construct individual points of the parabola with compass and straightedge, but that it is not possible to construct the curve itself in this way. Plato’s second category therefore is a natural extension of the first category. Exercise 15 Create an IGS file in which you define two sliders a and b both ranging from 0 to 10. Draw the straight line y D a and the point A .0; a/. Draw the straight line y D b and the circle centered at A and with radius a C b and determine their intersections B and C . For B and C select Trace and move the slider b. You will see a parabola appear. Alternatively click on Locus, select the mover, the slider b, and the tracer, B and C respectively. You will now see a parabola being traced out. What happens when you move the slider a (do NOT use Animation On)? Exercise 16 Select a system of axes x and y, in which  p x coincides with DF and y intersects ŒDF pat the ; 0 , then the Cartesian equation of d is x D  . Use midpoint. Let the coordinates of F be 2 2 the formulae for the distance between two points and the distance of a point to a straight line to prove that the Cartesian equation for the parabola in this configuration is y 2 D 2px.

2.3

Adrian van Roomen and Apollonius’ Problem

Some problems require ingenious methods to find a compass and straightedge solution, but can be easily solved using conic sections. One such problem is Apollonius’ problem, for which a solution with conic sections, i.e. belonging to Plato’s second category, is straightforward. Later we will also discuss the compass and straightedge construction (see 7 Sect. 3.3). Apollonius (ca. 262 BC–ca. 190 BC) is best known for his work on conic sections. His book Conics is preserved partly in Greek with commentaries by Eutocius and partly in an Arab translation by Thabit ibn Qurra. Apollonius’ other work has unfortunately been lost, but we know of his exploits through Pappus of Alexandria’s commentaries1 . In his book ᾿Επαφαί (Epaphaí, “Tangencies”) Apollonius posed, and solved, the problem which is now known as Apollonius’ problem: find a circle which is tangent to three given circles2 . Three given circles have at most eight circles which are tangent to all of them. Each solution encloses or excludes the circles in a different way. Adriaan van Roomen, also known as Adrianus Romanus (1561–1615) was a Flemish physician and mathematician who solved the problem using conic sections. Van Roomen became professor of mathematics and medicine first in Leuven then at Würzburg (1593). In 1604 he was ordained priest. He died in 1615 in Mainz en route to Leuven. He corresponded with the foremost mathematicians of his age, Ludolff van Ceulen and François Viète among them3 . 1

On Apollonius and Conics see Heath (1961); Fried and Unguru (2012). For a reconstruction of Apollonius’s solution see Heath (1981) vol. 2, p. 182–185. 3 Bockstaele (1976, 2009). 2

15 2.3  Adrian van Roomen and Apollonius’ Problem

2

. Fig. 2.4 Adriaan van Roomen’s challenge: to find a solution to a 45th degree polynomial equation (van Roomen (1593), MPM 8 533)

In 1593 he published Ideæ Mathematica in which he challenged his contemporaries to solve a 45th degree polynomial equation4 (see . Fig. 2.4): 45x  3795x 3 C 95 634x 5  1 138 500x 7 C 7 811 375x 9  34 512 075x 11 C 105 306 075x 13  232 676 280x 15 C 384 942 375x 17  488 494 125x 19 C 483 841 800x 21  378 658 800x 23 C 236 030 652x 25  117 679 100x 27 C 46 955 700x 29  14 945 040x 31 C 3 764 565x 33  740 259x 35 C 111 150x 37  12 300x 39 C 945x 41  45x 43 C x 45 D C s r q p C D 2C 2C 2C 2 At one point an ambassador from the Low Countries boasted to the French King Henry IV that van Roomen had listed all European mathematicians but did not mention any French mathematicians. He also referred to van Roomen’s challenge. Henry IV had Viète summoned and had him read van Roomen’s book. Viète recognised that the polynomial expresses the 45˛ ˛ given that x D chord.˛/ D 2 sin . length of a chord d45 D chord.45˛/ D 2 sin 2 2 He reported to the King that same day and gave him two solutions, of which one was 4

Bockstaele (1993).

16

Chapter 2  The Genesis of Geometry

2

. Fig. 2.5 Viète’s solution to van Roomen’s problem (Viète (1646), EHC G 4858) . Fig. 2.6 Circle C touches circle C1 internally and circle C2 externally

s x0 D

r 2

q p p 2 C 2 C 2 C 3, the edge of a regular 96-gon (see 7 Sect. 6.7). He

claimed that he could calculate all other solutions as well (see . Fig. 2.5). Viète published his solution in a book Ad problema quod omnibus mathematicis totius orbis construendum proposuit Adrianus Romanus, Francisci Vietae responsum (1595). In this book he challenged van Roomen to a new problem: this was nothing else than Apollonius’ problem. Viète maintained that, because all curves have a degree two it should have a compass and straightedge solution (also see 7 Sect. 3.3). van Roomen was able to solve the problem, but not with compass and straightedge. He used hyperbolae to solve the problem, his solution thus belongs to Plato’s second category5 . Suppose the circles C1 , C2 and C3 , centered at O1 , O2 and O3 and with radii r1 , r2 and r3 respectively are given. Van Roomen wants to find the locus of the centre of all circles which are tangent to C1 and C2 and the locus of the centre of all circles which are tangent to C1 and C3 . In both cases C1 has to be touched in the same way by the tangent circle, i.e. either internally or externally (see . Fig. 2.6). The intersection of these two loci then yields the centre of a circle which is tangent to the three circles and hence a solution to Apollonius’ problem. Exercise 17 Prove that if a circle C centered at O and with radius r is tangent to a circle C1 centered at O1 and with radius r1 , then kO1 Ok D r1 C r.

Suppose a circle C centered at O and with radius r is tangent to C1 and C2 (see . Fig. 2.7). We notice that kO1 Ok  kO2 Ok D .r1 C r/  .r2 C r/ D r1  r2 D constant. We therefore have to determine the locus of all points for which the differences of the distances to the centres of the circles is r1  r2 . This is a hyperbola H12 with foci O1 and O2 . All points on this hyperbola are the centre of circles which are tangent to both circles C1 and C2 . We can now do the same for C2 and C3 and we again find a hyperbola H23 . The intersections 5

Barbin and Boyé (2005), p. 10, 24–27, Bos (2001), p. 110–112.

17 2.3  Adrian van Roomen and Apollonius’ Problem

2

. Fig. 2.7 The locus of the centres of the circles tangent to two given circles is a hyperbola. kO1 Ok  kO2 Ok D .r1 C r/  .r2 C r/ D r1  r2 D constant

of these two hyperbolae yield two centres of tangent circles (see . Fig. 2.8). To see why not all intersections yield a centre for a tangent circle we need to look at the branches of the hyperbolae. For instance for H12 one branch yields all the centres of circles C which are externally tangent to C1 and C2 – i.e. C contains neither C1 nor C2 – the other branch yields the centres of circles C for which C1 and C2 are internally tangent to C (see . Fig. 2.9). Constructing e.g. H13 yields other solutions to the problem (see . Fig. 2.10). Exercise 18 Create an IGS file in which you solve Apollonius’ problem. Select three points A, B and C and draw circles CA ; CB ; CC centered at these points and with chosen radii rA ; rB ; rC . Draw a line through A and B on the one hand and B and C on the other. Determine the intersection points of these lines with the respective circles. For the line through A and B, select one of the intersection points on each circle, IA and IB and determine their midpoint MAB (see . Fig. 2.11). A and B are the foci of a hyperbola and MAB is on the hyperbola. Repeat this procedure for B and C . B and C are the foci of the second hyperbola, and MBC is on the hyperbola. Determine the intersections of the hyperbolae. Select one of the intersection points as the centre of a circle. Select the option

. Fig. 2.8 Three circles, in solid lines, are given. The hyperbolae which are the loci of the centre of circles tangent to C1 and C2 on the one hand and C2 and C3 on the other are drawn. The intersection point A is the centre of a circle externally tangent to the three circles (dotted line). The intersection point B is the centre of a circle for which the given circles are internally tangent (dashed line)

18

Chapter 2  The Genesis of Geometry

2

. Fig. 2.9 In this figure intersection point C is used to construct the tangent circle. We clearly see that one circle centered at C is externally tangent to two circles (dotted line), while another is internally tangent to two circles (dashed line). The branch of H12 through C yields all the centres of circles C which are internally tangent to C1 and C2 , the branch of H23 yields the centres of circles C which are externally tangent to C2 and C3 . Obviously a tangent circle to the three circles cannot be tangent to C2 internally and externally at the same time

. Fig. 2.10 Another solution to Apollonius’ problem. The hyperbolae which are the loci of the centres of circles tangent to C1 and C2 on the one hand, and C1 and C3 on the other, are drawn. The intersection point D is the centre of a circle externally tangent to one circle and internally tangent to two circles (dashed line). The intersection point E is the centre of a circle externally tangent to two circles and internally to one (dotted line) . Fig. 2.11 MAB is the midpoint of ŒIA ; IB . We can distinguish between four line segments for which the endpoints are on the respective circles:       ŒIA ; IB  ; IA ; IB0 ; IA0 ; IB ; IA0 ; IB0

19 2.4  Mechanical construction aids

2

Circle with centre and through point, draw the circle in such a way that it is tangent to one of the given circles. You will notice that either the circle is tangent to only two circles or that it is tangent to the three circles. Depending on which of the intersection points of the line with the circle you have chosen the tangency will either be internal or external. Exercise 19 Use the IGS file of Exercise 18 to identify all possible solutions to Apollonius’ problem. The problem has a maximum of eight solutions.

2.4

Mechanical construction aids

Although conic sections cannot be constructed with compass and straightedge, they can be constructed using mechanical means, methods from Plato’s third category. Consider for instance an ellipse. An ellipse is defined as the locus of the points of which the sum of the distances to two given points (the foci) is a constant. Figure . Fig. 2.12 shows how an ellipse can be constructed with the aid of two drawing pins and a piece of rope. Exercise 20 Create an IGS file in which you choose two numbers, a and c, c < a. Define a slider t ranging from a  c to a C c. Draw the points F 0 .c; 0/ and F .c; 0/. Draw the circles C1 , centered at F and with radius t, and C2 , centered at F 0 and with radius 2a  t. Determine the intersection points D and E of the circles C1 and C2 . Determine the locus of D and E respectively with the slider t.

There are other ways to describe an ellipse with mechanical aids (see . Fig. 2.13 right and . Fig. 2.14). One is Archimedes’ trammel for which we use a ruler, a set square, and a pencil (see . Fig. 2.13 left):

»

Draw two perpendicular lines x; y on paper; these will be the major (x) and minor (y) axes of the ellipse Mark three points A, B and C on the ruler such that kAC k is the length of the semi-major axis and kBC k is the length of the semi-minor axis With one hand, move the ruler across the paper, turning and sliding it so as to keep point A on line y, and B on line x at all times With the other hand, keep the pencil’s tip on the paper, following point C of the ruler The tip will trace out an ellipse.

. Fig. 2.12 Drawing an ellipse using two tacks and piece of rope (van Schooten (1659), KBR VH 8.040 A)

20

Chapter 2  The Genesis of Geometry

2

. Fig. 2.13 Archimedes’ trammel (right: van Schooten (1659), KBR VH 8.040 A)

. Fig. 2.14 Italian ellipsograph signed on the arm “Dominicus Lusuerg F. Romae 1700”. The Lusuerg family had some of the most remarkable craftsmen building scientific instruments of the late 17th and early 18th centuries, such as Dominicus (1669–1744) who built a wide number of mathematical instruments. Photos Tesseract – Early Scientific Instruments (Hastings-on-Hudson)

Exercise 21 Prove the procedure of Archimedes’ trammel (hint: determine the coordinates of C when AB makes an angle  with the x-axis). Prove that ŒAB is the diameter of a circle C which passes through the origin. If ŒOD is a diameter of C , prove that D traces out a circle as A moves along the x-axis (the locus of D with A). Exercise 22 Create an IGS file in which you mimic Archimedes’ trammel.

Quite a number of drawing devices with which conic sections can be drawn were proposed by Frans van Schooten in his book Exercitationum Mathematicorum libri quinque (Leiden, 1657) (see . Fig. 2.15). Frans van Schooten was the grandson of Franchois Verschooten, a Fleming who had fled the religious intolerance of the Spanish Netherlands in 1584 and had settled in the Northern Netherlands, which was fighting for independence from Spain6 . 6

van Maanen (1987), p. 212–213.

21 2.4  Mechanical construction aids

2

. Fig. 2.15 Some of Frans van Schooten’s drawing devices to draw a conic section (van Schooten (1659), KBR VH 8.040 A)

His father, Frans van Schooten sr., was a professor at the Leiden engineering school. Frans jr. followed in his footsteps. He championed the ideas of Descartes and made fundamental contributions to analytic geometry. Exercise 237 One of van Schooten’s devices can be seen in the figure (van Schooten (1659), KBR VH 8.040 A). On a long slat KL a small slat AB, with length a, can rotate about A. Another slat BE can rotate about B. On BE a point D is determined for which kBDk D kABk D a.

7

van Maanen (1987), p. 215.

22

2

Chapter 2  The Genesis of Geometry

. Fig. 2.16 The “construction” of a tractrix. After a drawing in Huygens’ manuscript on tractional motion (University Library Leiden, Ms Hug. 6, f64r)

Point D is attached to KL and can move along this slat. Put kDEk D b. As D is moved along KL E traces out an ellipse. Prove this assertion by finding the equation of the ellipse in a system in which KL is the x-axis and A the origin. Exercise 24 Create an IGS file in which you mimic van Schooten’s device of Exercise 23.

An archetypical example of a curve which is drawn mechanically is the tractrix (see . Fig. 2.16). In the seventeenth century, watches were rather heavy, and were usually kept on a chain in a pocket. The problem of which curve the watch would observe if the end of the chain was dragged along a straight line, was put by Claude Perrault (1613–1688) to Gottfried Leibniz (1646–1716) in 1676. Perrault is best remembered as the brother of Charles Perrault, author of such classic children’s tales as “Cinderella” and “Puss-in-Boots”. Trained as a physician, in 1666 Claude was invited to become a founding member of the Académie des Sciences, where he earned a reputation as an anatomist. The first known solution to Perrault’s problem was given by Christian Huygens (1629– 1695), who named the curve the tractrix from the Latin “tractus” which denotes something that is pulled along. In Huygens’ presentation, it is not a watch but a cart of which the drawbar is being drawn along a line. At the rear of the cart a pencil is attached which traces out the tractrix8 . Although solving this problem is just beyond the scope of this book we will give an outline here for those who have a knowledge of differential equations. Take a look at . Fig. 2.17. Suppose P .x; y/ is a point on the curve. Suppose C has coordinates .x; 0/, suppose D is the other end of the drawbar ŒPD, moving along the x-axis. The drawbar, which has length a, has to be tangent p to the curve. If the height is y, then using Pythagoras’ theorem, we know that kCDk D a2  y 2 . Therefore the slope of the straight y line along the drawbar is  p . On the other hand, calculus proves that the slope is 2 a  y2 dy given by . dx 8

Bos (1989), p. 10–12.

23 2.5  Neusis

2

. Fig. 2.17 The tractrix

The problem reduces to solving the differential equation y dy D p : 2 dx a  y2 This is an equation which can be solved by separating the variables to produce the following solution: ! p Za p 2 a  t2 a C a2  y 2 p 2 2 xD dt D ˙ a ln  a y : t y y

Unfortunately, the nature of this problem, involving differentials, does not allow for a construction to be created in an IGS using elementary techniques.

2.5

Neusis

A particular construction method belonging to Plato’s third category is a neusis (from the Greek νεῦσ ις from νεύειν or “neuein”, meaning “inclining or pointing towards”; plural: νεύσ εις “neuseis”). In a neusis construction, a line segment ŒAB with length a is fitted between two curves, l, the directrix, and m, the captrix (from the Latin captare, meaning “to catch”), in such a way that the straight line AB passes through a given point P , the pole of the neusis (see . Fig. 2.18). Neusis is also called verging in English. To obtain this solution one verges a ruler, on which ŒAB is marked off, through P in such a way that A follows the directrix. The solution is obtained at the position where B is on the captrix. Note that the roles of the directrix and the captrix can be interchanged. If B follows line m, then the solution is reached when A is on l (see . Fig. 2.19). When the line segment is fitted in between the curves, a solution for a particular problem is found (e.g. the trisection of an angle, the duplication of the cube, . . . ). Implicitly, by using a neusis we have defined a new curve: the locus of point B while A moves along the directrix. The question is thereby reduced to finding the intersection(s) of this locus with the captrix m. Note that we have already defined such a curve as “the locus of B with A” (see 7 Sect. 1.2).

24

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Chapter 2  The Genesis of Geometry

. Fig. 2.18 Performing a neusis (drawing by Paul Tytgat)

Exercise 25 Create an IGS file in which you select a pole P for a neusis. Draw a circle centered at O, not coincident with P , and with radius 5. Draw a straight line a which intersects the circle. Select a point A on a. Draw PA and determine the points B and C for which kABk D kAC k D 3. Determine all positions for which either ŒAB or ŒAC  can be fitted in between a and the circle by dragging A along a. Remember the point A does not have to lie within the circle. Exercise 26 Use the IGS file from Exercise 25 and draw the locus of B with A and of C with A. The intersections of both branches of the locus give the positions for either B or C when the line segment is fitted in between the straight line a and the circle. You can determine these positions by sliding A over a. Let the position of the pole P and the orientation of the line a differ. Can you find an orientation for which there are 2, 3, 4, 5, 6 intersections of the branches with the circle? Can you find a configuration with only one intersection?

We can make an instrument to draw such curves. Take a ruler with a slot in the middle. Put a cursor in the pole P and put the slot of the ruler over it. From the end of the ruler A

. Fig. 2.19 An example of a neusis. Suppose that given a pole P a line segment ŒAB has to be fitted in between the straight line and the circle. In the left figure the straight line is used as directrix, in the middle one the circle is the directrix. In the right figure (one of) the final position(s) is reached. If we use the circle as directrix A has to coincide with the intersection of the straight lines. If the straight line is used a directrix, then B has to coincide with the intersection of the line PA and the arc of a circle DF

25 2.5  Neusis

2

. Fig. 2.20 An instrument to draw conchoids with (drawing by Paul Tytgat)

. Fig. 2.21 D1 and D2 are branches of a conchoid generated by K (drawing by Paul Tytgat)

determine a point B for which kABk D a. Attach a pointer to B and a pencil to A. Let the pointer B follow the directrix l and the pencil, at A, will draw the desired curve. Of course, there are two points at a distance a from a given point. So again take the ruler and from the end A1 mark off a point B at a distance a. Then from B, mark off another point A2 at a distance a from B. Put the slot of the ruler over the cursor at P and attach a cursor to B. Attach pencils to both A1 and A2 . Let B follow the directrix l. The pencils now draw two branches of a curve, which we call a conchoid (see 7 Sect. 5.1). For directrices with polar equation r D f . /, and the pole of the neusis in the origin, the associated conchoids have a polar equation r D f . / ˙ a. Consider the following situation: the pole P for a neusis is given, a line segment has length a, the directrix is a circle with radius b, the captrix is a certain curve (not shown in . Fig. 2.22). One of the end points A of the line segment is on the circle. Which curve will the other endpoint B on ŒAP describe? In other words, what is its locus when A moves along the circle? The intersection of this line with the captrix will solve the problem. In . Fig. 2.22 we can see examples of the locus of B with A. In the figure on the left, the pole of the neusis P and the centre of the circle O coincide. The locus is a circle centered at P and with radius R D b  a. In the figure on the right, the pole and the centre do not coincide, the locus now is a closed curve. Exercise 27 Create an IGS file in which the origin is the pole P of a neusis. Draw a point A on the x-axis, which is the centre of a circle with radius 5. Select a point B on the circle. Determine points D and E on PB for which kDBk D kBEk D 3. Determine the loci of D with B and of E with B respectively. Move A along the x-axis. What do you notice?

26

Chapter 2  The Genesis of Geometry

2

. Fig. 2.22 The locus of B with A generated by a neusis in which the directrix is a circle. Left if the pole coincides with the centre of the circle, right if the pole and the centre do not coincide

Exercise 28 Create an IGS file in which the origin is the pole P of a neusis and in which the directrix is a circle with radius 5 through the origin and with its centre on the x-axis. Define a slider with variable a. Let a vary between 0 and three times the radius. Select a point B on the circle. Draw the straight line PB. Determine points D and E on PB for which kDBk D kBEk D a. Determine the loci of D with B and E with B respectively. Which kinds of curves are they? Exercise 29 Determine the polar equations of the loci you found in the previous exercise. Hint: look at the triangle with ŒBP  as one edge and the diameter through P as another. Which kind of triangle is this? Can you determine kPBk? Remember for the locus of D with B r D kPBk  kDBk.

27

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Compass and straightedge constructions Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_3

3.1

Euclid’s Elements

Of all mathematical treatises, none has had such a profound influence as Euclid’s Elements. We know practically nothing about its author Euclid (Εὐκλείδης – Eukleíd¯es lived ca. 300 BC). The apocryphal stories told about him deal more with the nature of mathematics than telling us anything about the man himself. In one story, a student asked him what profit he could derive from mathematics. To which Euclid replied: “Give the man three obols, for he wants to make a profit from everything.” In another story, King Ptolemy asked whether there was a faster way of learning mathematics, to which Euclid replied: “Sire, in mathematics there is no Royal Road”. Euclid’s Elements is arguably the most important treatise of Greek Antiquity. It is divided into 13 books (which we would call chapters), and sets out 120 definitions, 372 theorems, 93 problems, 19 porisms (corollaries) and 16 lemma’s. Euclid was not the first author to have edited such an encyclopaedic work, yet his became the standard. Book I culminates in theorems I.47 and 48, which deal with Pythagoras’ theorem and its converse. Over the ages, numerous editions of Euclid have been published. In the school book edition it survived well into the nineteenth century. Parts of our geometry course at middle school and the lower grades of secondary education still deal with Euclid’s geometry, especially the first book of Elements (see . Fig. 3.1). Elements starts out with axioms and postulates. Postulates are the rules by which new mathematical theorems can be derived. The Ancient Greeks made a distinction between axioms, which are universal truths, and postulates, which are basic truths for a particular science. In geometry these postulates are: 1. It is possible to draw a straight line between two points (line segment between two points) 2. It is possible to extend any line segment into a straight line as far as one wants (line through two points) 3. With a point as a centre and a radius it is possible to draw a circle (circle with centre and radius) 4. All right angles are equal

28

Chapter 3  Compass and straightedge constructions

3

. Fig. 3.1 The proof of Pythagoras’ theorem in Byrne’s Euclid (1847). Byrne’s edition of the first six books of Euclid’s Elements was an unusual, very appealing, as well as an educationally innovative book. Using coloured pictures he attempted to present Euclid’s proofs with as little text as possible. (Byrne (1847), Wikimedia Commons, public domain)

5. If a straight line is intersected by two other straight lines and if the sum of the interior angles is less than a straight angle then the two lines will intersect.

The first three postulates describe the basic constructions: drawing a straight line and drawing a circle. With these operations, all other constructions in Elements are performed. They also impose a restriction on a construction: only compass and straightedge are allowed as aids in the construction process. In Euclid’s view, a compass collapses when it is lifted from the paper. As a consequence, we cannot transfer lengths with it. Fortunately, it can be proved that all constructions that are performed with a collapsible compass can be performed with a modern compass. In fact, this is what Euclid’s second proposition in Book I amounts to: “To place at a given point (as an extremity) a straight line equal to a given straight line”1 . Therefore we will not distinguish between these compasses. The fourth postulate is not a construction but simply states the equality of all right angles. This postulate is necessary because it can not be proved with the previous postulates2. Yet Euclid needs this assertion for some proofs, e.g. of Pythagoras’ theorem. 1

Heath (1956) I, p. 244. The actual study of the independence of axioms of one another and of the consistency of the set of axioms did not start until the twentieth century. It uses the concept of propositional functions, which is outside the scope of this book. See e.g. Eves (1995), p. 257 ff. 2

29 3.1  Euclid’s Elements

3

. Fig. 3.2 Playfair’s axiom. m and l cannot be both parallel to BC and intersect at A

The fifth and last postulate, also known as the parallel postulate has a long history, which merits a book in its own right. Unlike the first four postulates, the parallel postulate is not self-evident and has the appearance of a theorem. Moreover proposition I.17 (In any triangle the sum of any two angles is less than two right angles) is the converse of the fifth postulate. Over the ages many attempts were made to prove the parallel postulate using Euclid’s first four postulates. Invariably some property, which turned out to be equivalent to the fifth postulate, was assumed. Ultimately the search for a proof of the fifth postulate gave rise to the discovery by Saccheri, Lobachevsky and Bolyai of what is now called non-Euclidean geometry3 . We limit ourselves to Playfair’s modern formulation of the fifth postulate: In a plane, given a straight line and a point not on it, at most one straight line parallel to the given line can be drawn through the point. In fact in his Elements of Geometry (1795) Playfair claimed that Two straight lines cannot be drawn through the same point, parallel to the same straight line without coinciding with one another. In the second edition this had become Two straight lines which intersect one another, cannot be both parallel to the same straight line4 . John Playfair (1748–1819) was a Scottish scientist, mathematician and professor of natural philosophy at the University of Edinburgh. He is now best remembered for his popularising of geologic theories. Playfair’s postulate in our formulation actually is a construction in Elements: To draw a straight line through a given point parallel to a given straight line (book I, proposition 31). To prove this construction, Euclid does not need the parallel postulate. However, the parallel line constructed in this proposition is the only one passing through the given point and this is a consequence of the parallel postulate. We use . Fig. 3.2 to see why. Suppose there are two lines, m and l, through A which are parallel to BC . Then for either m or l and BC , the sum of the interior angles on one side or the other of AD would be less than two right angles, and therefore by the parallel postulate either m or l would meet BC . But this contradicts the fact that we assumed both m and l were parallel to BC .

3

See e.g. Eves (1995), p. 219 ff. Non-Euclidean geometry is a geometry in which Euclid’s fifth postulate is replaced by another postulate in which either at least two parallel lines through a given point are allowed or none at all are allowed. For a didactical introduction see van Gulik-Gulikers (2005). 4 Playfair (1795), p. 7, Playfair (1814), p. 22.

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Chapter 3  Compass and straightedge constructions

. Fig. 3.3 Euclid I.1, the construction of an equilateral triangle

3

3.2

Compass and straightedge constructions

Euclid’s first theorem calls for a compass and straightedge construction (see . Fig. 3.3):

»

Construct an equilateral triangle on a given line segment

In Euclid’s version the solution reads5 :

»

Let AB be the given finite straight line. Thus it is required to construct an equilateral triangle on the straight line AB. With centre A and distance AB let the circle BCD be described Again, with centre B and distance BA let the circle ACE be described; and from the point C , in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. Now, since the point A is the centre of the circle CDB, AC is equal to AB Again, since the point B is the centre of the circle CAE, BC is equal to BA But CA was also proved to be equal to AB; therefore each of the straight lines CA, CB is equal to AB. And things which are equal to the same thing are also equal to one another; therefore CA is also equal to CB Therefore the three straight lines CA, AB, BC are equal to one another.

The construction can be carried out like this:

»

Let the line segment be ŒAB Position the compass point in A and set the width to kABk Draw a circle Now with the same width draw a circle centered at B Let the intersections of the circles be C1 and C2 6 Either C1 or C2 can be chosen as the third vertex of the desired equilateral triangle.

Because we have kept the compass width at kABk it follows that kABk D kBC k D kCAk and so 4ABC is equilateral. 5

Heath (1956), p. 240. Although we assume, as Euclid did, that these circles intersect, this does not follow from Euclid’s postulates! Euclid would have needed a postulate, similar to the parallel postulate, about the intersections of circles to prove this assertion. In modern geometry the existence of these points is guaranteed by some sort of continuity postulate. See e.g. Eves (1995), p. 259 ff. 6

31 3.2  Compass and straightedge constructions

3

. Fig. 3.4 The construction of a line parallel to a given one

. Fig. 3.5 Proof of the construction of a line parallel to a given one

We can also construct a straight line parallel to a given line (see . Fig. 3.4).

»

Select a point B on a given straight line a Open the compass to a chosen width Position the compass point in B and draw arcs which intersect at A and C With the previous construction construct equilateral triangles 4AD1 B and 4BD2 C , D1 and D2 being on the same side of a The straight line D1 D2 is parallel to a.

We have now described a procedure with which we can construct a parallel line. Is the line really a parallel straight line? In mathematics we have to prove our assertions, therefore we have to prove that the construction procedure is actually correct. Even though the construction may be simple, this is not necessarily true for the proof. For this construction we give two proofs (see . Fig. 3.5):

»

Consider the triangle 4D1 BD2 From the construction it follows that this triangle is isosceles, because kBD1 k D kBD2 k So †BD1 D2 D †D1 D2 B and †BD1 D2 C †D1 D2 B C †D2 BD1 is a straight angle (sum of angles of a triangle) or 2†BD1 D2 C †D2 BD1 is a straight angle (4) On the other hand it follows from the construction that 4ABD1 and 4BCD2 are equilateral triangles So †ABD1 D †D2 BC , and †ABD1 C †D1 BD2 C †D2 BC is a straight angle and so is 2†ABD1 C †D1 BD2 (5) If two straight lines are intersected by a third straight line and the corresponding angles of intersection with the transversal are congruent, then the two lines are parallel From (4) and (5) it follows that †BD1 D2 D †ABD1 These are corresponding angles of intersection So the straight lines are parallel.

An alternative proof is of a more algebraic nature:

»

Suppose the magnitude of a straight angle is a From the construction it follows that 4ABD1 and 4BCD2 are equilateral 1 From which †ABD1 D †BD1 A D †D1 AB D a 3

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Chapter 3  Compass and straightedge constructions

1 and †BCD2 D †CD2 B D †D2 BC D a 3 1 from which †D1 BD2 equals a 3 Now 4BD1 D2 is an isosceles triangle from which †BD1 D2 D †BD2 D1 The sum of the angles of a triangle is a straight angle so †BD1 D2 C †D1 BD2 C †BD2 D1 D a 1 or 2†BD1 D2 C a D a 3 1 from which †BD1 D2 D a 3 It is clear that the corresponding angles of intersection are equal, again proving the correctness of the construction.

3

Exercise 30 In the construction of a parallel line we have used equilateral triangles with sides ŒAB and ŒBC . Prove that if we use congruent isosceles triangles with base sides ŒAB and ŒBC  the third vertices define a parallel straight line. Exercise 31 If a straight line AB and a point P are given, construct a straight line through P and parallel to AB.

Another construction (6) of a straight line is this:

»

Let AB be a given line Draw a straight line l, not coincident with AB, through A Draw a circle CA centered at A and with radius kABk CA intersects l at C Draw a circle CC centered at C and with radius kABk CC intersects l at D Draw a circle CD centered at D and with radius kCBk CD intersects CC at E Then CE k AB.

Remember that in the following exercises you can only use a compass and straightedge, not a ruler or a protractor. Exercise 32 Prove construction (6). Exercise 33 If a straight line AB and a point P are given, use construction (6) to construct a straight line through P and parallel to AB. Exercise 34 Find a construction to divide a line segment into two equal parts.

33 3.2  Compass and straightedge constructions

3

. Fig. 3.6 Dividing a line segment into three equal parts

Exercise 35 Find a construction to erect a perpendicular on ŒAB through a point C on ŒAB. Exercise 36 Find a construction to erect a perpendicular on ŒAB through a point D not on ŒAB. Exercise 37 If a straight line AB and a point P are given, use Exercises 35 and 36 to construct a straight line through P and parallel to AB.

Parallel lines allow us to divide a line segment into any number of equal parts (Thales’ intercept theorem). We will describe the procedure for a division into three equal parts, but the procedure can readily be generalised for any number (see . Fig. 3.6).

»

Suppose we want to divide the line segment ŒAB into three equal parts Select a point C , not on AB, preferably such that ]CAB is an acute angle Select a point D1 on ŒAC D1 can coincide with C , but this is not necessary With the compass, construct the points D2 and D3 for which kAD1 k D kD1 D2 k D kD2 D3 k So kAD3 k D 3 kAD1 k Draw the straight line D3 B Draw the straight lines parallel to D3 B through D1 and D2 These lines intersect ŒAB in P and Q respectively and kAP k D kPQk D kQBk.

Exercise 38 1 Use the properties of similar triangles to prove that in the above procedure kAP k D kABk and 3 kAP k D kPQk D kQBk.

In the same fashion we can construct angles which obey particular demands. For instance:

»

An angle and a ray are given, construct, with the ray as one side, an angle which is congruent to the given angle (see . Fig. 3.7).

34

Chapter 3  Compass and straightedge constructions

3

. Fig. 3.7 The construction of an angle with a given ray equal to a given angle

»

Let the ray be ŒAB and the angle ]KLM Draw, with L as centre, an arc with radius equal to kLM k The arc intersects KL at a point N Draw, with the same radius, an arc centered at A This arc intersects AB at a point C Position the compass point in N and determine the length kNM k Keep this compass width, position the compass point in C and draw an arc Let the intersection of the arc centered at C and the arc centered at A be D Then †BAD is equal to †KLM . This can easily be seen: from the construction it follows that 4LMN and 4ACD are congruent (SSS).

Exercise 39 Divide, by construction, an angle into two equal parts. Prove the construction!

Using these basic constructions we can find rectilinear figures which have the same area as a given rectilinear figure. In most cases, we will “reduce” a rectilinear figure to a square by doing so. For instance:

»

Given a rectangle, construct a square of the same area (see . Fig. 3.8).

»

Suppose rectangle ABCD is given Draw the ray ŒAB Draw a circle centered at B and with radius kBC k The circle intersects ŒAB at E, which is not on ŒAB Determine the midpoint M of ŒAE (see Exercise 34) Draw a circle centered at M and with radius kMEk Draw a straight line BC , which intersects the circle at F1 and F2 We will proceed with F1 , but the argument is equally valid for F2 The square with edge ŒBF1  has the same area as the rectangle ABCD The angle ]AF1 E is subtended by a diameter and is therefore a right angle It is easy to show that 4ABF1 ; 4F1 BE and 4AF1 E are similar triangles

35 3.2  Compass and straightedge constructions

3

. Fig. 3.8 Construction of a square equiareal to a given rectangle

kF1 Bk kABk D , kF1 Bk kBEk and kF1 Bk2 D kABk  kBEk D kABk  kBC k (because kBEk D kBC k).

So we can deduce

The formula expresses that the area of the square with edge ŒF1 B is equal to the area of rectangle ABCD. For this reason, the number kF1 Bk is also called the geometric mean or mean proportional of the two numbers kABk and kBEk. Exercise 40 Create an IGS file in which you perform this construction. Define a slider a, ranging from 1 to 10. Let B be the origin. Define a point A.a; 0/ on the negative x-axis. Draw a line, m, through A and perpendicular to the x-axis. Select a point D.a; d / (d > 0) on m. Construct the rectangle ADCB. Determine E on the x-axis such that kBEk D kBC k. With ŒAE as diameter, draw the circle C , which intersects the y-axis at F1 and F2 . Draw the square BF1 GH . Draw the straight lines DC and GH and determine their intersection point L. Determine the locus of L with D. The part of the locus in the first quadrant is a parasitic part. The part of the curve in the fourth quadrant is a part of a well-known curve. Which one? Can you determine the equation of this curve? Now use Locus to draw the locus of L with B and move the slider a. What do you notice?

There are also other ways to obtain this result. Samuel Marolois (1572–1627), a Dutch mathematician and military engineer, proceeded in following fashion7 (see . Fig. 3.9):

»

7

Given a rectangle ABCD Determine the midpoint F of the edge ŒAC  Position the compass point in A and with a compass width kAF k, draw an arc The arc intersects ŒAB at E Erect the perpendicular on AB in E Position the compass point in B and draw an arc, with compass width kBF k The arc intersects the perpendicular on ŒAB at G The length of ŒGE is the length of the side of the required square with the same area as the rectangle ABCD.

Marolois (1614–1615), plate 5, van de Weyer (2011), p. 51–54.

36

3

Chapter 3  Compass and straightedge constructions

. Fig. 3.9 The construction of a square equiareal to a rectangle ABCD from Marolois’ La Géometrie (1614), EHC G 48942. Marolois’ treatise on geometry was part of his Opera Mathematica (1614) which also contained one of the most widely read treatises on perpective of its day

Exercise 418 Prove Marolois’ construction. Hint: use Pythagoras’ theorem. Put kABk D l, kAC k D b from which kAF k D

b , ... 2

Exercise 42 Consider two circles CA and CB which do not intersect. Neither circle is enclosed by the other. Construct all possible straight lines tangent to both circles. Exercise 43 Use the IGS file of Exercise 18. Let the file show only the circles CA ; CB ; CC with their centres and the tangent external circle. Define a slider R varying from  min .rA ; rB ; rC / to 0. Now redefine the circles CA ; CB ; CC as having radii rA C R; rB C R; rC C R. Let R diminish from 0 to  min .rA ; rB ; rC /. What do you notice? Now choose a situation in which the smallest circle, say CA , touches the tangent circle externally, while the others touch internally. Redefine the circles CA ; CB ; CC as having radii rA C R; rB  R; rC  R. Let R diminish from 0 to min .rA ; rB ; rC / D rA . What do you notice?

3.3

François Viète and Apollonius’ Problem

The result of Exercise 43 was used by François Viète (1540–1603) to show that there actually is a compass and straightedge solution to Apollonius’ problem. François Viète was born in Fontenay-le-Comte in 1540. After an education at the Franciscan school of Fontenay, he enrolled at the university of Poitiers in 1558 to study civil and canonical law. It took him just one year to obtain a baccalaureate and a licentiate’s degree, after which he embarked on a successful career as a lawyer in his hometown. From 1571 onwards, he held several posts both in Paris and in Brittanny. After a turbulent period in 1587–88, Viète became a counsellor and cryptographer in the service of Henry IV. As a cryptographer, he was able to decipher an intercepted letter to Philip II, King of Spain and pretender to the French throne, which helped Henry thwart Philip’s military plans9 .

8 9

van de Weyer (2011), p. 51–54. Delahaye (2005).

37 3.3  François Viète and Apollonius’ Problem

3

Viète introduced a method which, by way of an appropriate substitution, can be used to solve second, third and fourth-degree equations10. For the equation x 2 C bx D c make the substitution yDxC

b b or x D y  2 2

then y 2 D x 2 C bx C

 2 b 2

,

y2 D c C

 2 b 2

which immediately gives the value of x. For the third degree equation x 3 C bx 2 C cx C d D 0; he puts xDy

b 3

yielding the equation y 3 C py C q D 0 A second substitution yDz

p 3z

which has become the classical substitution for solving third-degree equations, yields a quadratic equation in z 3 p3 Cq D 0 27z 3  p 3 D0 z 6 C qz 3  3 z3 

)

from which z3 D 

q ˙ 2

r  p 3 3

C

 q 3 2

:

Viète only uses the positive cube root of z, but it is easily demonstrated that the six solutions for z yield three different solutions for x. 10

Meskens (2010), p. 157.

38

Chapter 3  Compass and straightedge constructions

3

. Fig. 3.10 Viète’s solution to Apollonius’ tangency problem from Opera mathematica (1646). Left for two circles externally tangent and one internally tangent to the tangent circle, right where one circle is externally tangent and the other two internally tangent to the tangent circle (Viète (1646), EHC G 4858)

An analogous substitution solves a fourth-degree equation. It goes without saying that finding these substitutions encouraged Viète and other mathematicians to search for similar substitutions for higher-degree equations, an undertaking that was, for that matter, doomed to failure. Viète found a compass and straightedge construction for Apollonius’ problem (see . Fig. 3.10). Exercise 43 shows that the tangency problem can be reduced to finding a circle tangent to two given circles and through a point. This problem can be further reduced to finding a circle tangent to a given circle and through two points11.

Suppose three circles CA , CB and CC are given.

11

Bogomolny (a); Henry.

39 3.3  François Viète and Apollonius’ Problem

3

Let rC D min.rA ; rB ; rC /. Diminish the radii of the three circles by rC . We now have two circles CA0 , CB0 and a point C .

Construct the external point of similitude P of circles CA0 and CB0 . The common tangent from P touches CA0 at D and CB0 at E. Construct the circle CDE. Let C 0 be the second intersection of CP and the circle CDE. Draw a circle which passes through C and C 0 which intersects the circle CA at F and G, e.g. the circle CDE, then G D D.

40

Chapter 3  Compass and straightedge constructions

Let H be the intersection of C C 0 and FD.

3

Construct the tangents HS and H T from H to CA .

Construct the perpendicular bisector of C C 0 . Let K and L be the intersections of the bisector with AS and AT respectively.

The circles with centres K and L respectively and radii kKSk and kLT k solve the problem.

41 3.3  François Viète and Apollonius’ Problem

3

To find the tangent circle for the original problem expand the circles again to the given circles and shrink the tangent circle by the same amount.

We have found the circle to which the three circles CA , CB and CC are externally tangent.

If we use L as the centre for the tangent circle, we again expand the radii of the circles and of the tangent circle. We have now found the tangent circle to which the circles CA , CB and CC are internally tangent.

43

4

The Delian Problem Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_4

4.1

Delos and its altar

After defeating the Persians at Mycale (479 BC), Athens became the centre of the Greek political, commercial and intellectual world. The first philosophical Athenian school was the Sophist school. Alongside grammar and rhetoric, this school showed an interest in mathematics and in geometry in particular. One of the problems that caught their attention was the so-called Delian problem. According to legend, the island of Delos was hit by the plague. The local residents made offerings to the local oracle and asked what they had to do to stop the epidemic. The oracle answered that they should build a new cubic altar, with a size (D volume) double that of the existing one. The length of the edge should be determined by compass and straightedge methods only. The inhabitants first created an altar with an edge double that of the original. The plague did not disappear though. It became clear that to have a double volume the edge should not be doubled, but multiplied by a number different from 2. But which factor should this be? Suppose the length of the edge of the original cube is a, then the volume of the new cube should bep2a3 . If the edge of the new cube is b then b 3 D 2a3 . If b D t  a then t 3 D 2 or p 3 t D 2. 3 2 is the third root or the cubic root of 2. The problem of doubling the cube is thus reduced to constructing a line segment which is p 3 2 as long as the original edges. The problem seems easy enough. It is possible to construct a line with a length equal to any square root of a natural number, so why would this not be possible for a line with a length equal to a cubic root? The problem proved tougher than expected. A first solution was proposed by Hippocrates of Chios (fl. ca. 430 BC – not to be confused with Hippocrates of Kos, after whom the hippocratic oath is named). Hippocrates was a merchant on the isle of Chios. After suffering various misfortunes, he moved to Athens, where he became a leading mathematician. He was the first to write a systematically organised geometry textbook called Elements1 (not to be confused with Euclid’s Elements). To duplicate a cube, the equation x 3 D 2a3

1

Heath (1981) I, p. 182ff.

(7)

44

Chapter 4  The Delian Problem

needs to be solved. The equation was reduced by Hippocrates to finding two mean proportionals x and y which satisfy a x y D D : x y 2a

4

(8)

From these equations, we can deduce two equations x 2 D ay

and

y 2 D 2ax:

(9)

Note that the problem of solving a third degree cubic equation is reduced to finding two mean proportionals by solving two quadratic equations. On the one hand we need to find the geometric mean x of y and a on the other hand we need to find the geometric mean y of x and 2a. We have already seen that we can construct a geometric mean if we know the two numbers. This is not the case! In the first equation, we are unable to construct the line segment with length x because we do not know length y. In the second equation, we are unable to construct the line segment with length y because we do not know length x. It seems to be a stalemate. Exercise 44 Show that (8) can be written as the system of equations given by (9). Eliminate the parameter y and show that the resulting equation is (7). Create an IGS file in which you define a slider a, ranging from 0 to 10, and draw the curves which are given by the system of equations (9). Which curves have you drawn? Determine the coordinates of their intersection(s). (Note that when using an IGS you will only be given numerical approximations). Algebraically determine the coordinates of the intersection for any value of a.

Menaechmus (ca. 350 BC) realised that Hippocrates’ problem could be solved using conic sections. Little is known about Menaechmus, other than that he was a student of Eudoxus and that his brother was Dinostratus (see 7 Sect. 6.2). He is credited with the discovery of conic sections and some of their properties2 . Menaechmus saw that Hippocrates’ problem could be reduced to finding the intersection of two parabolae. Unfortunately, this intersection cannot be constructed using compass and straightedge alone. It is a construction from Plato’s second category. Exercise 45 Show that finding the mean proportionals of two numbers a and b, x y a D D x y b can be solved by finding the solution to either of the systems of equations 8 < ay D x 2 :bx D y 2 2

Heath (1981) I, p. 251–253, Coolidge (1945), p. 3–5, Fladt (1967), p. 5–7.

(10)

4

45 4.1  Delos and its altar

. Fig. 4.1 Plato’s construction of the mean proportional of a and 2a

or

8 < ay D x 2 :xy D ab

(11)

Which curves do these equations represent?

Other Grecian solutions belong to Plato’s third category. What they have in common is that they make use of an instrument with moving parts. The following solution is said to be Plato’s. It was used, in different guises, well into the seventeenth century (see . Fig. 4.1).

»

Consider two rods, BD and AC , perpendicular to one another, which intersect at E. Slide the rods over each other such that kEAk D a and kEDk D 2a. In the top left figure all lines represent rods. The rods F G and GH are rigidly connected to each other, so F GH is a carpenter’s square. LK is perpendicular to GH and L can slide along GH remaining parallel to F G. The cross like rods have to be placed on the other rods in such a way that A is on F G, D is on LK, the extension of DE passes through G (B does not have to coincide with G) and the extension of AE passes through L (again C does not have to coincide with L). This is done by appropriately sliding p and rotating the cross, while simultaneously sliding LK. Then kEGk is the length 3 2a of the required cube.

Exercise 46 p Use similar triangles to prove that the procedure is correct and that kEGk D 3 2a. Exercise 473 Gregory of St-Vincent (see 7 Sect. 5.5) found a construction for two mean proportionals using conic sections. To find the mean proportional between a and b, construct the rectangle ABCD 3

Sancto Vincentio (1647), p. 602.

Chapter 4  The Delian Problem

46

with sides a and b. Choose a coordinate system for which the origin coincides with A and the x- and y-axis coincide with the lines AB and AD. Then C has coordinates .a; b/. prove that the intersection of the orthogonal hyperbola through C and with the coordinate axes as asymptotes and the circumscribed circle to the rectangle intersect in a point whose coordinates are the mean proportionals between the adjacent sides of the rectangle.

4

4.2

Wine gauging

The procedure of the previous 7 Sect. 4.1 was used well into the sixteenth century, albeit in a slightly other version, by wine gaugers. Wine gaugers were employees of the local tax offices4 . From the thirteenth century onward, many European cities levied an excise on the consumption of wine. The most commonly used container type for transporting liquids was the barrel. In order to be taxed correctly the contents of the barrel had to be measured. This was the task of wine gaugers: they measured the barrel and made the necessary calculations (see . Fig. 4.2). Usually the barrel was approximated by a cylinder. To do so the diameters of the heads and the bulge were averaged. Let this averaged diameter be d and the length of the barrel l, then the volume of the approximating cylinder is V D

 2 d l: 4

Gaugers could simplify their task by using special gauging rods, such as the cubic gauging rod. This rod seems to have been primarily used in Austria. The use of this rod is based on the property that for two similar bodies B1 and B2 , which have known volumes V1 and V2 and diameters d1 and d2 respectively, the ratio of the volumes is equal to the ratio of the cubes of the diameters: V1 d3 D 13 : V2 d2 To make a measurement, the rod is inserted diagonally through the bung hole towards the lowest part of a head. With the correct graduation the volume can then be read immediately. Suppose that a rod is divided into n equal parts of a certain linear measure, the division points of which are simply called points. If the rod is introduced into the barrel in the appropriate way and at the bung hole the fifth point can be seen, then the contents will be proportional to 53 D 125 units of the cube of that linear measure. This means that the verge has to be at the 1st, 2nd, . . . , nth point. Other graduated with the numbers 1, 8, 27, . . . , n3 indicated p 3 points, in between the integer third roots such as 2, need to be constructed. In the seventeenth century, tables of cubic roots were drawn up to simplify the construction of a cubic gauging rod. In one sixteenth century manuscript we find the following procedure, closely resembling Plato’s construction5 (see . Fig. 4.3). 4 5

On winegauging see Folkerts (1974); Meskens (1994, 2013), p. 97–112, Grabiner (1996, 1998). Coignet (1576–77).

47 4.2  Wine gauging

4

. Fig. 4.2 Left: a wine gauger is taking a measurement with his gauging rod. The man at the front is constructing a quadratic gauging rod. By using a quadratic scale a gauger could immediately read the surface area of the cross section at the bung hole. With this kind of rod it was still necessary to multiply this number (or an average based on it) with the length of the barrel to find the volume (Mennher (1565), MPM A 3589), right: the parts of a barrel, terminology (drawing by Paul Tytgat)

. Fig. 4.3 An instrument with which

p 3

n can be determined (Plato-instrument, drawing by Paul Tytgat)

48

Chapter 4  The Delian Problem

4

. Fig. 4.4 Diagram of parts with which to make a Plato-instrument

»

Build an instrument consisting of a carpenter’s square F GH and a slider LK which slides over GH and is perpendicular to it Draw a horizontal straight line x and a vertical straight line y, which intersect each other at O Select A on x (to the right of O), such that kOAk D a, in which a is the length of the edge of the original cube Select D on y (under O) such that kODk D 2a Put the instrument on the drawing Make sure that the inner edge of GF always passes through A Turn and move the instrument and slide the slider LK along GH such that the inner vertex L in on x, the inner vertex G is on y, the inner edge of LK passes through D Indicate these points p with a pencil and denote them C (on x) and B (on y) respectively Then kOBk D 3 2a.

Exercise 48 p Use similar triangles to prove that the procedure is correct and that kOBk D 3 2a. Exercise 49 p Select D on y and under O such that kODk D na. Prove that that kOBk D 3 na. Exercise 50 Create an IGS file in which the construction of Exercise 9 in 7 Sect. 1.2 is made. Define a slider b ranging from 0 to 10. Select A.a; 0/ and D.0; 2a/. Determine the Cartesian equation of the locus of B with C . Determine the coordinates of the intersection with the y-axis. Explain the connection of this exercise to the wine gaugers’ method described above.

It is easy to make your own Plato instrument by cutting it out of cardboard. Take the necessary precautions when cutting the cardboard! The parts of the Plato instrument are shown in . Fig. 4.4. Take a piece of cardboard (not too thick), and draw a carpenter’s square about 3 cm wide and a ruler also 3 cm wide. Towards one end, draw two lines parallel to the edge, with a length slightly more than 3 cm. Cut the carpenter’s square and the ruler. On the ruler,

49 4.3  Doubling the cube with a neusis

. Fig. 4.5 Determining a carpenter’s square only

p 3

4

2 with

cut along the lines. You can now slide the ruler over the carpenter’s square. If you have not cut along the lines too far, the ruler should fit snugly and it should make a right angle with one leg of the square. You can now solve the problem of the duplication of the cube yourself, using this instrument. It is not really necessary to have the whole instrument to solve the problem. In the same manuscript we find a procedure using only a simple carpenter’s square (see . Fig. 4.5).

»

Draw a rectangle ABCD, such that kABk D a, in which a is the length of the edge of the original cube, and kAC k D 2a Position the angle of the carpenter’s square on the elongated line segment ŒDC  in such a way that one leg passes through B Let the intersection of the other leg with the elongated edge ŒBD be G Move and turn the square p until kAFpk D kDGk, from which kAF k D 3 2kABk D 3 2a.

Exercise 51 Prove this construction! Exercise 52 Create an IGS file in which you perform this construction. Let D be the origin. Define the points A .2a; a/ and B .2a; 0/. Construct the rectangle ABDC . Choose a point G on the x-axis. Determine the point F on AC such that kAC k D kDGk. Draw the line BF and the perpendicular ` through G. Determine the intersection point H of BF and `. Determine the locus of H with G.

4.3

Doubling the cube with a neusis

We can construct a line segment of length

»

p 3

2 using a neusis6 (see . Fig. 4.6):

Draw a regular hexagon ABCDE for which the length of the edge is k. Draw the straight lines BE and BD. Use A as the pole of a neusis and by verging fit in a line segment of length k between the two straight lines BE and BD. p 3 Suppose this is effected at G on BE, then kAGk D .1 C 2/k.

The proof draws on Menelaus’ theorem7 and the power of a point theorem.

6 7

Based on Delahaye (1997), p. 52. Menelaus’ theorem uses directed line segments. Here, we only set out the version with lengths of line segments.

50

Chapter 4  The Delian Problem

4

. Fig. 4.6 Constructing

p 3 2 with neusis

. Fig. 4.7 The power of a point P with respect to a circle

Exercise 53 Create an IGS file in which you draw a straight line PQ and a circle with radius 5. Consider two cases, one in which P is inside the circle and one in which it is outside the circle. Determine the intersection points A and B of line PQ with the circle. Determine h D kPAk  kPBk. Now, leaving P stationary, move Q about. What do you notice for h?

The power of a point theorem states that for any chord going through a point P outside a circle, the product of the distance from P to the points where the chord intersects the circle is constant. This constant is called the power of the point P with respect to the circle, and is written as h. Referring to . Fig. 4.7 this means that h D kPAk  kPBk D kP C k  kPDk D kP T k2 . It follows from these relations that h D kOP k2  r 2 , in which r is the radius of the circle. Exercise 54 Two chords ŒAB and ŒCD are drawn in a circle with radius r. The line segments ŒAB and ŒCD intersect at a point P (see . Fig. 4.8 (left)). Show that 4APD  4CPB. Deduce that kPAk  kPBk D kP C k  kPDk.

51 4.3  Doubling the cube with a neusis

4

. Fig. 4.8 The butterfly theorem (see Exercises 54 and 55)

. Fig. 4.9 Menelaus’ theorem

Exercise 55 From a point P outside a circle with radius r two straight lines which intersect the circle in A, B and C , D respectively are drawn (see . Fig. 4.8 (right)). Show that 4APD  4CPB. Deduce that kPAk  kPBk D kP C k  kPDk. Prove that this product is equal to kOP k2  r 2 . Is this expression equally valid for the situation of the previous exercise?

If a triangle 4AOG is cut by a transversal LBD then according to Menelaus’ theorem kADk kOBk kGI k   D1 kDOk kBGk kIAk (see . Fig. 4.9, left), which can also be written as kAOk  kOBk  kGI k D kODk  kBGk  kIAk: Applied to the triangle 4AOG in . Fig. 4.6, we find 2k  k  k D kkBGk  kIAk:

Chapter 4  The Delian Problem

52

Letting kBGk D x and kIAk D y, this yields 2k 2 D xy:

4

On the other hand, if we draw a circle centered at A and with radius k, we can apply the power of a point theorem. From the fact that the radius is k, we can deduce that O and B are on the circle. Suppose the line GA intersects the circle at M and N (see . Fig. 4.9, right). Then h D kGBk  kGOk D kGM k  kGN k: Now kGBk D x

and

kGOk D x C k;

kGM k D k C y  k D y

and

kGN k D k C y C k D y C 2k;

from which x .x C k/ D y .y C 2k/ :

(12)

Multiplying both sides by y 2 we find xy .xy C ky/ D y 3 .y C 2k/ : But xy D 2k 2 , so  y 3 .y C 2k/ D xy .xy C ky/ D 2k 2 2k 2 C ky D 2k 3 .2k C y/ : Cancelling out the factor 2k C y on both sides we find that y 3 D 2k 3 or p 3 kLAk D y D 2k: Exercise 56 Create an IGS file in which you construct a regular hexagon ABCDEF . Choose a point G on the straight line BE. Draw a straight line AG. Determine points H and I on AG with the property that kHGk D kGI k D kABk. Determine the locus of H and I respectively with G. The curve you see is called Nicomedes’ conchoid (see 7 Sect. 5.1). Determine point I1 on BD for which the neusis is accomplished. Determine the length of ŒAI . Exercise 57 Show that equation (12) can be rewritten as   k 2 3k 2 ; .y C k/2  x C D 2 4 which is the equation of a hyperbola. Determine the intersections of this hyperbola with the hyperbola xy D 2k 2 . Create an IGS file in which k D 1 and determine the intersections graphically.

53 4.3  Doubling the cube with a neusis

4

. Fig. 4.10 The cissoid of Diocles

Exercise 588 Let C be the circle centered at A.0; a/ and with radius a. Let t be the tangent line at T .0; 2a/. Let B be a point of t and let C be the intersection of ŒOB and the circle C . The cissoid of Diocles is the locus of all points P such that kOP k D kCBk (see . Fig. 4.10). Show that the polar equation of the cissoid is r D 2a.csc   sin / and that its Cartesian equation y3 (hint: write csc  of the polar equation as a fraction, then multiply the equation is x 2 D 2a  y with r 2 sin ). Let Q be the intersection of the cissoid and thepline x C 2y D 4a. Let R be the intersection of p OQ and t. Show that the coordinates of R are .2a 3 2; 2a/. Deduce that kTRk D 2a 3 2. Exercise 59 Create an IGS file in which you construct the cissoid of Diocles.

8

Aarts (2000), p. 194.

55

5

Trisecting an angle Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_5

The Greeks knew how to bisect an angle using compass and straightedge constructions. The question whether an angle can be trisected, or divided into n equal parts, by compass and straightedge methods then comes naturally. The question is simple enough and seems to suggest a simple solution. Here too, appearances are deceptive. It turns out that, like the duplication of the cube, the construction with compass and straightedge is impossible. We can find a construction if we allow a marked ruler and verging solutions.

5.1

Nicomedes’ conchoid

A solution to the trisection problem was proposed by Nicomedes (ca. 280 BC–210 BC), of whose life we know nothing. His solution gives rise to a curve known as Nicomedes’ conchoid.

»

Suppose the angle ]AOB has to be trisected (see . Fig. 5.1) Suppose kOBk D a Draw ŒBC  ? ŒOA , draw BD k OA Now draw a line segment OPQ, with P on ŒBC  and Q on ŒBD , such that kPQk D 2 kOBk D 2a 1 Then †AOQ D †AOB. 3

Exercise 60 Prove Nicomedes’ procedure for trisecting an angle. Exercise 61 Create an IGS file in which you draw an angle ]AOB. Draw ŒBC  ? ŒOA , draw ŒBD k ŒOA . Now draw a line OE, with E on the line segment ŒBC . Draw a circle centered at E and with radius 2 kOBk. Determine the intersection F of this circle with OE. Determine the locus of F with E (see . Fig. 5.2).

What we have done in Exercise 61 is nothing more than a neusis with O as its pole, BD as the directrix and BC as the captrix. The curve we have found as the locus of F with E is called Nicomedes’ conchoid (see . Fig. 5.2 top, in which only a very small part of the curve

56

Chapter 5  Trisecting an angle

. Fig. 5.1 pNicomedes’ neusis to construct 3 2

5

is visible). It actually is a part of one branch of a two-branched curve (see . Fig. 5.2 bottom). The name “conchoid” derives from the Greek κόγχη meaning “shell”. We can draw this branch of the curve using a neusis apparatus (see . Fig. 5.3). This neusis apparatus consists of a T-square and a moving arm. The T-square consists of a slotted arm K 0 L0 and a perpendicular arm O 0 C 0 with a cursor attached to O 0 . The other part is a slotted arm G 0 N 0 .

»

Draw the angle ]AOB which has to be trisected (see . Fig. 5.4) Position the arm O 0 over O and turn O 0 C 0 in such a way that it lies over OA Put a pencil in the slot of K 0 L0 and draw the line KL Place the slot of G 0 N 0 over O 0 and turn the arm until it is over OB

. Fig. 5.2 Top: part of the conchoid found as locus of F with E (as point on ŒBC ), kEF k D 2 kOBk. Bottom: the two branches of Nicomedes’ conchoid

57 5.1  Nicomedes’ conchoid

5

. Fig. 5.3 An instrument with which (a branch of) the conchoid can be drawn (drawing by Paul Tytgat)

Call J 0 the intersection of the slots of G 0 N 0 and K 0 L0 Now move G 0 N 0 over O 0 until kG 0 J 0 k D b D 2 kOJ 0 k D 2a Place the cursor B 0 at M 0 . Attach it to G 0 M 0 , but it should still be able to slide over K 0 L0 In other words, the length kG 0 B 0 k always equals b when B 0 slides along K 0 L0 The pencil will draw a branch of Nicomedes’ conchoid. Remove the neusis apparatus Elongate ŒOB until it intersects the straight line KL at P Draw the parallel with ŒOC through P This straight line intersects the conchoid at Q Draw the ray ŒOQ 1 Then †AOQ D †AOB. 3 c ˙ b in which  is the angle with the positive sin   0 0 x-axis and c D kO C k. In Cartesian coordinates, the equation is .y  c/2 x 2 C y 2 D b 2 y 2 .

The polar equation of the conchoid is r D

. Fig. 5.4 Construction of the conchoid with which, for a given angle, the trisection can be accomplished (drawings by Paul Tytgat)

58

Chapter 5  Trisecting an angle

5

. Fig. 5.5 Trisection of an angle by means of a conchoid

Exercise 62

kO 0 C 0 k , for any position of B 0 sin  0 0 on K L . Use this result to derive the polar equation of the conchoid. Use the properties of right-angled triangles to show that kO 0 B 0 k D

Exercise 63 Show that the Cartesian equation of the conchoid is  .y  c/2 x 2 C y 2 D b 2 y 2 : Hint: use the transformation formulae from Cartesian to polar coordinates and note that you find the polar equation.

We can trisect an angle in IGS using the following procedure:  Suppose an angle has magnitude  ( rad was chosen in . Fig. 5.5). Select a point A 6 on the negative y-axis and a point B on the positive x-axis for which kOBk D kOAk tan  (making †OAB D  ). Select a point C on the x-axis. Draw the straight line AC . Determine kABk and draw a circle centered at C and with radius 2 kABk. Determine the intersection points D and E of the circle and AC . Determine the loci of D and E respectively with C . The two branches of the conchoid become visible. Draw a parallel with the y-axis through B. Determine the intersection point F of this straight line with the branch of the conchoid defined by D with C (i.e. the branch which does not pass through A). Unfortunately not all IGSs will automatically indicate this point. However, in most IGSs, when the cursor is moved over the intersection point both the straight line and the conchoid will be highlighted. Click for a new point F at this spot. The angle ]OAF trisects ]OAB. Exercise 64 Consider the branch of Nicomedes’ conchoid determined by E with C . A is a double point (crunode) of this branch, i.e. the conchoid intersects itself at A. Using the Trace option, you will find that the conchoid passes through A twice. Prove why!

59 5.2  Archimedes

5

. Fig. 5.6 Pappus’ version of Nicomedes’ trisection

Pappus (IV.31) reduced Nicomedes’ procedure, which solved the problem by mechanical means, to a solution which used only conic sections1 . He cleverly uses Thales’ intercept theorem twice. With reference to . Fig. 5.6 draw a line segment ŒCF  with lengh 2a and parallel to ŒPQ, put kOC k D x, kP C k D y, kEQk D c, kBC k D d . Then kEQk  c  kOQk D D x kEBk kOP k

because 4OEQ  4PBQ

and   d kBC k kOQk D D y kOP k kP C k

because 4OCP  4QBP

So x d D c y

)

xy D cd

which is an orthogonal hyperbola with EO and EB as asymptotes. The problem of trisecting an angle is thereby reduced to finding the intersection of an orthogonal hyperbola with asymptotes EO and EB and a circle centered at C and with radius 2a.

5.2

Archimedes

Archimedes is undoubtedly the most important mathematician we know from Antiquity (287– 211 BC)2 . While Euclid is mainly known for one book, Archimedes’ fame rests on several smaller treatises on a variety of mathematical subjects. He was born in Syracuse, a Greek colony on Sicily. It appears that during his youth he went to (or was sent to) Alexandria, where he seems to have received a thorough mathematical education. He later returned to Syracuse, where he put himself at the service of the King. Undoubtedly the most famous story about Archimedes is his discovery of how to determine the volume of an irregular solid and, from that, how to deduce the volumetric mass 1

Sefrin-Weis (2010), p. 146ff and p. 284ff. On Archimedes see Ver Eecke (1921); Heath (1953) and for a reassessment Jaeger (2008); Paipetis and Ceccarelli (2010).

2

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Chapter 5  Trisecting an angle

5

. Fig. 5.7 Artistic impression of Archimedes’ claw (drawings by Paul Tytgat)

density of the solid. The problem posed itself when the Tyrant of Syracuse, King Hieron, asked Archimedes to determine whether the crown he had been given by the goldsmith was indeed made of solid gold. When Archimedes stepped into his bathtub, he noticed that the water level was raised. This gave him the idea for what we now call Archimedes’ law. He was so elated that he ran naked through the streets shouting: “I have found it!” (Eureka!). During Archimedes’ lifetime, he also witnessed the Second Punic War, in which Syracuse had sided with Carthage. The city was besieged by a Roman army under the command of Marcellus. Archimedes became the symbol of the heroic, if futile, resistance against Roman might. Popular lore credits Archimedes with the invention of the parabolic mirror with which he managed to destroy at least part of the Roman fleet. Parabolic mirrors concentrate the sunlight in one point, the focus3 . Roman ships coming in the range of these mirrors were targeted on their sails, which are said to have caught fire. Ships managing to sail close to the city walls were caught by Archimedes’ claw (probably a kind of lever, see . Fig. 5.7). The besieging army equally met with the effects of Archimedes’ infernal machines. How unlikely the story may be, it sheds light on the image of a mathematician to his contemporaries. Well into the sixteenth century, mathematicians simultaneously served as mathematicians, engineers, physicists and astronomers, and sometimes even as physicians. The Romans held Archimedes in the greatest esteem. When Syracuse ultimately fell, Marcellus ordered his troops to spare the life of Archimedes. Most probably he wanted to put Archimedes’ formidable genius to good use at the service of the Roman army. However, when a soldier found Archimedes, he was summoned to identify himself. Archimedes was too pen3

In German Brennpunkt, in Dutch brandpunt, literally burning point.

61 5.3  The first Archimedean trisection

5

sive about a mathematical problem and did not notice the soldier. The soldier slew Archimedes who uttered his famous last words: “Do not touch my circles” (“N¯ol¯ı turb¯are circul¯os me¯os!”, in Ancient Greek Μή μου τοὺς κύκλους τάραττε! M´¯e mou toùs kúklous táratte)4 . There are other versions of this story, including one in which Archimedes is on his way to meet Marcellus with some of his contraptions. Note the striking contradiction between a man organising, or at least making major contributions to, the defence of the city on the one hand and a pensive man trying to prove a theorem while soldiers are ransacking the city5 .

5.3

»

The first Archimedean trisection Suppose we have to trisect ]AOB (see . Fig. 5.8) Draw a circle centered at O and with radius r, for which A and B are on the circle Elongate the line segment ŒAO in the direction of A to O Draw a line through B, which intersects the circle at C and AO at D, such that kCDk D r 1 Then †ADB D †AOB. 3 Using the procedure described in 7 Sect. 3.2 (. Fig. 3.7) we can transfer the angle ]ADB to an angle having ŒOA as a leg Thereby we have trisected the angle ]AOB.

Exercise 65 Use isosceles triangles to prove Archimedes’ procedure.

We can find the points C and D using a neusis instrument called the trisector (see . Fig. 5.9). It consists of two slotted arms and one short ruler. Both the slotted arms and the ruler have two holes at the same distance r. The short ruler is mounted to both long arms with a hinge on which it turns. The distance from the hinge to the cursor on the moving arm is equal to the length of the short ruler. One end of a long arm is attached to a slot in the other arm, the directing arm. This end can thus be moved along the arm. To trisect an angle, we have to draw a semicircle of a radius equal to the distance between the hinges. With one leg on the diameter of the semicircle and with O as vertex, draw the angle you wish to trisect. Put the directing arm on the diameter. Move the other arm in such a way that it passes through the intersection of the second leg of the angle and the circle. The angle made by the two slotted arms is the desired trisector of the given angle. The neusis we have performed with the trisector can of course be described geometrically. Again consider the angle ]AOB which has to be trisected. Suppose A and B are on a circle C centered at O and with radius r. Select B as the pole, the circle as the directrix and the straight line AO as the captrix. Let the – other – intersection point of the line and the circle be C . Determine the point D outside the circle, on BC , for which kCDk D r. Determine the locus of D with C . We find a part of Pascal’s limaçon (see 7 Sect. 1.3 and . Fig. 5.10). 1 If the locus intersects the straight line OA at D0 , then †AD0 B D AOB. 3 4

Voza (2010). We refer the reader to Jaeger (2008) for a very enlightening interpretation of these stories in the light of Roman literary styles. 5

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Chapter 5  Trisecting an angle

. Fig. 5.8 Archimedes’ neusis for a trisection

5 Exercise 66 Create an IGS file in which you perform this neusis. Draw a circle with radius r, e.g. 5. Call the intersection with the positive x-axis A. Determine the point B on the circle for which †BOA D . Select a point C on the circle and draw the straight line BC . Draw a circle centered at C and with radius r. Determine the intersections D and E of the circle and the straight line BC . You can now determine the loci of D and E respectively with C . Let C move along the circle, you will meet the situations in which D is on the negative x-axis or E is on the positive x-axis or D is on the positive x-axis. In the first position measure the angle ]AD0 C , in the second the angle ]AE0 C (if it exists) and in the last the angle ]AD1 C (remember angles are measured counterclockwise). Reduce the magnitudes of these angles to the interval Œ0; 2 rad. What do you notice? E does not exist for certain positions of B, which ones?

. Fig. 5.9 A trisection instrument based on Archimedes’ neusis (drawing by Paul Tytgat)

. Fig. 5.10 Left: the part of Pascal’s limaçon found as the locus of D with C . Right: the whole curve is found by determining the locus of D and E resp. with C (with E being the other point on BC for which kCF k D r). One can find the whole limaçon as the locus of D with C if one allows C to go around the circle twice

63 5.4  The Archimedean spiral

5

. Fig. 5.11 Archimedes’ trisection  using a conchoid for an angle of 3  rad (top) and for an angle of rad 6 (bottom)

Of course, as we have already noticed, the roles of the directrix and the captrix can be interchanged. Select B as the pole and the line AO as directrix. Let the intersection of a line through B and AO be D. Determine the point C on BD for which kDC k D r, then the locus of C with D is a branch of Nicomedes’ conchoid (see 7 Sect. 5.1). 1 Let C0 be the point where the conchoid intersects the circle then †ADC0 D †AOB. 3 Exercise 67 Create an IGS file in which you perform this neusis. Draw a circle with radius r, e.g. 5. Call the intersection with the positive x-axis A. Determine the point B on the circle for which †BOA D . Select a point D on the straight line OA and draw the straight line BD. Draw a circle centered at D and with radius r. Determine the intersections C and E with the BD. Determine the locus of C and E respectively with D. Determine the intersections of this locus with the circle. You will at most find five intersections, one of which is B. Determine †CDA or †EDA at C0; C1; E0 and E1 respectively. Determine †ADC0 , †ADC1 , †ADE0 and †ADE1 and multiply by 3. Reduce the magnitudes of these angles to the interval Œ0; 2 rad. What do you notice?

5.4

The Archimedean spiral

A second Archimedean solution makes use of a curve which we call the Archimedean spiral (. Fig. 5.12). The Archimedean spiral has an equation which reads in polar coordinates r D k; k 2 R0 . We cannot construct Archimedes’ spiral using compass and straightedge. What we can do, like we have done for the parabola, is construct individual points (see . Fig. 5.12). Draw a radius of length R on the polar axis. Construct, using compass and straightedge methods, 15   3 ;:::; rad respectively with rays with initial point O which make an angle of ; ; 8 4 8 8 the polar axis.

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Chapter 5  Trisecting an angle

5 . Fig. 5.12 Construction of an Archimedean spiral

 rad with the polar axis draw a line segment with one 8 R endpoint being O and having length i , i 2 f1; : : : ; 15g. The endpoints of the line segments 16 R which are constructed thus are points of the Archimedean spiral for which k D . Start 2 out at O and draw a continuous line through these points, this line is an approximation of the Archimedean spiral. On the ray which makes an angle i

Exercise 68 15   3 ;:::; rad. Find a procedure to construct angles with magnitudes ; ; 8 4 8 8

To divide an angle into three equal parts using Archimedes’ spiral we use the following procedure. Consider the Archimedean spiral r D  (i.e. k D 1). In particular consider the first turn, i.e. 0    2 rad. Put one leg of the angle you want to trisect on the polar axis, with the vertex in the pole. Let this angle be ]AOB, with ŒOA the polar axis. The ray ŒOB intersects the spiral at C . This means that kOC k D †AOB (because r D  ). kOC k . Use the construction of 7 Sect. 3.2, . Fig. 3.6, to find a line segment with length 3 kOC k . Let the intersection with Put the compass point in O and draw a circle with radius 3 the spiral be D. Then kODk D †AOD (because r D  ). †AOB kOC k Now †AOD D kODk D D , which completes the construction of the 3 3 trisector. This is an elegant and easy solution, but again not a compass and straightedge construction. We can determine individual points of the spiral with compass and straightedge, but we cannot construct the spiral using compass and straightedge alone. With a little imagination we can see this construction as a neusis construction. We fit in a line with a given length between a point, the pole of the coordinate system, and a curve, the spiral. We can consider the point as a circle with radius 0. The spiral acts as the captrix, the point is both pole and (degenerate) directrix. The curve which is traced out by the other end point of the line segment is the circle, which we have drawn in the procedure above.

65 5.5  The Flemish Jesuits

5

Exercise 69   rad and rad into Draw an Archimedean spiral, or enlarge . Fig. 5.12 and divide an angle of 4 3 three equal parts. Try the procedure with an angle chosen at random. Exercise 70   Create an IGS file in which you draw an Archimedean spiral. Divide an angle of rad and rad 4 3 into three equal parts. Try the procedure with an angle chosen at random. Exercise 71 In the previous paragraph and in the previous exercises we have trisected an angle ]AOB using the first turn of the spiral. Use the IGS file of the previous exercise. Determine the intersections of ŒOB and the second and third turns of the spiral respectively. Perform the trisection procedure with these points. Use the option Angle to measure the magnitude of the given angle and the constructed angles. Multiply these last results with three and reduce these products to the interval Œ0; 2 rad. What do you notice?

5.5

The Flemish Jesuits

The Jesuits were an important part of intellectual life in early-seventeenth century Antwerp6 . The Jesuit College of Antwerp was founded in 1574 and flourished by the beginning of the seventeenth century. In 1615 plans for creating a school of mathematics were put into practice. Franciscus de Aguilon was to be its first mathematics professor. By the end of 1615 Gregorius a Sancto Vincentio or Gregory of St-Vincent (1584–1667), a student of Christopher Clavius, arrived at Antwerp to help de Aguilon write the curriculum (see . Fig. 5.14). By the end of 1617 the school of mathematics had opened. De Aguilon unfortunately did not live to see it open its doors. Franciscus de Aguilon was born in Brussels in January 1567, son of Pedro, a secretary to Philip II. Franciscus studied at the Colleges of Paris and Douai and became a novice at Kortrijk. On 15 September 1588 he joined the Jesuit order and became a priest in 1596. De Aguilon turned out to be not only an able mathematician but also a gifted architect. In 1615 the first stone of the Ignatius Church, now known as Carolus Borromeus Church, designed by de Aguilon was laid (see . Fig. 5.13). The impressive and richly decorated church would be finished a mere seven years later. It was not the first church Franciscus de Aguilon had built; he had already co-designed the Jesuit churches of Mons and of Tournai. Gregory was born on September 8, 1584 in Bruges. He attended secondary school in Bruges, studied philosophy in Douai and joined the Society of Jesus on October 21, 1605 at the Santo Andrea noviciate in Rome. After two years of noviciate, the Jesuit General had a post in Sicily in mind for Gregory, but Christopher Clavius managed to have him stay in Rome to study mathematics. On March 23, 1613 he was ordained as a priest in Leuven and went on to hold several posts in Brussels, ’s Hertogenbosch and Kortrijk. From 1617 to 1621, he was a mathematics teacher at the Antwerp Jesuit College. To Gregory we owe a generalisation of Pythagoras’ theorem7 . In an acute-angled triangle, erect the squares on the sides, draw the altitudes from each vertex and extend these into the 6 7

On the Jesuit mathematics school in Antwerp see Meskens (1997). Sancto Vincentio (1647), Heath (1956), I p. 404, Ostermann and Wanner (2012), p. 351.

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Chapter 5  Trisecting an angle

5

. Fig. 5.13 Antwerp’s Carolus Borromeus Church, designed by Franciscus de Aguilon s.j. and Pieter Huyssens s.j. (Photos Ad Meskens)

. Fig. 5.14 Portrait of Gregory of St-Vincent (Sancto Vincentio (1647), EHC G 4869)

squares. These altitudes divide the squares into two rectangles. If you shade these rectangles as seen in . Fig. 5.15, then the regions in the same shade of grey have the same area. Exercise 72 Prove Gregory’s generalisation of Pythagoras’ theorem.

In 1613, Jan Moretus’ widow and sons published de Aguilon’s magnum opus, Opticorum Libri Sex (see . Fig. 5.16). Despite the fact that the book is voluminous, it does not treat the whole of optics, but just those phenomena which are related to direct rays of light: no reflection, no refraction. The book was to be the first of a three volume series. de Aguilon’s untimely death prevented him from finishing this work. He had started work on his other books. Among Gregory of St Vincent’s manuscripts preserved in the Royal Library Albert I in Brussels is a manuscript by de Aguilon with preparatory writings for his other books.

67 5.5  The Flemish Jesuits

5

. Fig. 5.15 Gregory of St Vincent’s generalisation of Pythagoras’ theorem (right Sancto Vincentio (1647), EHC G 4869) . Fig. 5.16 Frontispiece to the second book of de Aguilon’s Opticorum Libri Sex. In this book de Aguilon defines the horopter plane to explain stereoscopic vision. The frontispieces were designed by Peter Paul Rubens and executed by Theodore Galle (Aguilonius (1613), EHC G 5050)

In the first part of this manuscript de Aguilon deals with trisections of an angle. In the following paragraphs we will see some examples of de Aguilon’s methods8. Exercise 73 Create an IGS file in which you draw a circle centered at O and with radius 5. Let A be the intersection of the negative x-axis with the circle. Let B be the point on the circle for which †AOB D . Define a point C on the y-axis. Draw the straight line BC and determine a point D on the circle for which kCDk D 5. Alternatively, determine a point P on BC , not on ŒCB , for which kCP k D 5 and determine its locus with C . The intersection of the locus and the circle gives you the desired point D. Extend BC until it intersects the x-axis at E. Measure †AOB and †AEB. What do you notice?

In the previous exercise we have found another way of trisecting an angle using a neusis procedure9 . The proof is both easy and surprising. 8

Based on van Looy (1979), p. 67–76. This method can already be found in Ibn-al-Haytham’s Optica (1011–1021). Ibn-al-Haytham (ca. 965 – ca. 1040) is also known as Alhazen. See Hogendijk (1979). 9

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Chapter 5  Trisecting an angle

5

. Fig. 5.17 de Aguilon’s neusis for the trisection of an angle . Fig. 5.18 The conchoid associated with de Aguilon’s method in solid line, the conchoid associated with Archimedes’ method in dashed line

Let C be a circle centered at O and with radius r. Keep all the names of points as shown in . Fig. 5.17. To prove that this neusis construction is correct, we will prove that kDEk D r. This means that if kCDk D r, then so is kDEk.  Let †OCE D ˛ , then †CEO D  ˛. 2  4ODC is isosceles, so †COD D ˛ and †DOE D  ˛. 2 Therefore 4ODE is isosceles and kDEk D kODk D r. But we know that kCDk D r, so kDEk also equals r. De Aguilon’s method is nothing other than Archimedes’ solution (see 7 Sect. 5.3) in a different guise. Figure . Fig. 5.18 shows the conchoids resulting from the neusis using de Aguilon’s and Archimedes’s method. In another lemma, de Aguilon uses another neusislike procedure. This time, two line segments need to be of equal length, which is unknown at the start. Let O be the centre of a circle C . Let F be the point from which a straight line a is drawn.

69 5.5  The Flemish Jesuits

5

. Fig. 5.19 Another solution by de Aguilon for the trisection of an angle

This line intersects the circle at B and the line AO at E. Then it is possible to rotate the line a into such a position that kBEk D kEOk.

De Aguilon uses this result to trisect an angle. Suppose an angle ]KAC has to be trisected. Draw a circle with diameter ŒAC  which intersects AK at F . Now find the intersection E of AC and a line through F for which kBEk D kEOk. 1 Draw BO, which intersects the circle at G, then †CAG D †CAK. 3 Exercise 74 Prove that the above procedure indeed yields the trisection of an angle, i.e. that if kBEk D kEOk 1 then †CAG D †CAK. 3

In Gregory’s manuscripts kept at the Royal Library, we find a proposition which indicates that he had come to grips with what we would call the limit of a geometric series10 . Let ŒAB be a line segment one wants to divide into three equal parts. According to Gregory11 : 10 11

van Looy (1979), p. 109. Sancto Vincentio (1647), p. 111–112.

70

Chapter 5  Trisecting an angle

5

. Fig. 5.20 Trisection of an angle by infinitely many consecutive bisections

»

If one takes away from ŒAB one half ŒAC , from what remains ŒCB again one half ŒDB, from what remains ŒDC  again one half ŒCE, from what remains ŒED again ŒDF , from ŒEF  again one half ŒEG and so on, then I claim that the end of this sequence will be where ŒAB is trisected.

Gregory uses this procedure for angle trisection as well. Begin with an angle ]AOC (in . Fig. 5.20 we used a right angle to be trisected) and construct the interior bisector b1 . For the angle with sides OA and b1 construct the interior bisector b2 . Now construct the interior bisector b3 of b1 and b2 , the interior bisector b4 of b2 and b3 , the interior bisector b5 of b3 and b4 , and so on. His procedure amounts to   1 1 1 1 1 1 1 1  C  ::: D 1  C  ::: 2 4 8 16 2 2 4 8  n 1 1  1 2   D lim 1 2 n!1 1  2 1 1   D  1 2 1  2 1 2 1 D  D 2 3 3 As we will see Gregory was a mathematician able to take bold steps into new realms.

71 5.6  Hippias of Elis and the quadratrix

5

. Fig. 5.21 The quadratrix as the locus of X. The part of the quadratrix outside the circle is called the parasitic part, these points satisfy the equation but not the geometric conditions (here 0  k  R)

5.6

Hippias of Elis and the quadratrix

Hippias of Elis (late 5th century BC) found a method with which the trisection of an angle as well as the quadrature of the circle could be solved. Hippias seems to have been a talented man, sent by his fellow citizens on a diplomatic mission to Sparta. He led a peripatetic life travelling from one city to another. Plato described him as arrogant and vain12 . Hippias performed the trisection by means of a curve we now call the quadratrix. This curve is the locus of the intersection of two straight lines which simultaneously perform a certain movement (one a translation, the other a rotation). The curve itself cannot be constructed with compass and straightedge methods. Consider the square ABCD. Let a line segment ŒAF , beginning at ŒAD, rotate with constant angular velocity about A and at the same time let ŒD 0 C 0 , beginning at ŒDC , move with constant velocity towards ŒAB in such a way that both reach ŒAB at the same time. The locus of the intersection points X of ŒAF  and ŒD 0 C 0  is the curve we call the quadratrix (see . Fig. 5.21). If X 0 is the perpendicular projection of X on AB, then the equation †XAB kXX 0 k D †DAB kDAk

(13)

is always satisfied. It is clear that if ŒAF  rotates about A, then F will trace out a quarter of a circle. Let R D kAF k D kADk, †BAF D †BAX D  and suppose .x; y/ are the coordinates of X  y 2R Inserting these values in equation (13) gives D or y D . =2 R  The quadratrix therefore is the locus of the straight lines with equations y D k .0  k  R/ k   with ray ŒAB . and the ray with vertex A making an angle  D R 2 12

Heath (1981), p. 23–24.

72

5

Chapter 5  Trisecting an angle

. Fig. 5.22 The construction of points of the quadratrix. Divide the line segment ŒAD in 2n equal segments (in the figure 23 D 8). For each division point, draw the line segment ŒAi Bi  parallel to ŒAB. Select the angle with ŒAB as one leg and the first bisectrix as the second leg as the first angle. The bisectrix intersects the line segment ŒA1 B1  in K1 , a point of the quadratrix. Determine the intersections Ki with the corresponding line segments. These points are points of the quadratrix

To trisect an acute angle ]BAF with the quadratrix we follow this procedure:

»

Construct the square ABCD on ŒAB, with ŒAF  inside ]BAD Draw the quadratrix DE and call X the intersection with ŒAF  Through X draw the line segment ŒD 0 X  parallel with ŒAB and with D 0 on ŒAD 1 Construct the point H on ŒAD such that kAH k D kAD 0 k 3 Draw ŒH Y  parallel to ŒAB and Y on the quadratrix 1 From equation (13) it follows that †BAY D †BAF . 3

Exercise 75 2R . Show that the equation, in polar coordinates, of the quadratrix is r D  sin  Exercise 76 Create an IGS file in which you use the equations of the respective straight lines and the radii to draw the quadratrix as the locus of their intersection points. You will notice that the quadratrix, as drawn by IGS, has a part outside the quarter circle. We call this part the parasitic part. The points on this part of the curve satisfy the equation of the quadratrix, but they do not satisfy the geometric  condition. Here  < . 2 Exercise 77 Create an IGS file in which you construct points of the quadratrix. First draw a square ABCD. Draw the diagonal ŒAC . Draw the parallel with AB through A1 , the midpoint of ŒAD. The intersection of this parallel and the diagonal is a point of the quadratrix (see . Fig. 5.22). Now bisect the angles ]DAC and ]CAB respectively and find the midpoint of ŒA1 D and ŒAA1 . Repeat this procedure in every step (the number of angles to bisect and midpoints to find will double with every step). Exercise 78 The following method was first proposed in nineteenth century France (see . Fig. 5.23). Draw a line segment ŒPQ and divide it into three equal parts. Place the point of the compass 1 in one of the division points N or M , e.g. M , and draw a semicircle with radius kPQk. Draw 3

73 5.6  Hippias of Elis and the quadratrix

5

. Fig. 5.23 The tomahawk construction for trisecting an angle

a ray ŒNS , to the side of the semicircle, perpendicular to ŒPQ at N . The figure we have drawn is called a tomahawk. Suppose ]AOB is to be trisected. Place Q on the ray ŒOA , while the semicircle touches ray ŒOB , in such a way that the perpendicular ŒNS passes through O. The angle ]AON trisects ]AOB. The tomahawk itself can be constructed with compass and straightedge constructions, but the trisection itself is performed by rotating and translating. Again this is not a compass and straightedge construction. Prove that the tomahawk construction is correct. Exercise 79 Show that: sin 3˛ D 3 sin ˛  4 sin3 ˛ cos 3˛ D 4 cos3 ˛  3 cos ˛ tan 3˛ D

4 tan ˛  tan3 ˛ 1  tan2 ˛

Exercise 80 Colin McLaurin (1698–1746) was a Scottish mathematician, most famous for his series. He also found a way of trisecting an angle by means of a curve we now call McLaurin’s trisectrix. This curve is defined as the locus of the intersections of two moving lines, one, l, through O and another, a, through a point A.a; 0/ on the x-axis. Both lines rotate about O and A respectively with angular velocities P and 3P . If the angle of l with the x-axis is , then the angle of a with the x-axis is 3 and the angle between a and l 2. Prove that the polar equation of McLaurin’s trisectrix is a a r D ) r D .4 cos   sec /; sin 3 sin 2 2 deduce its Cartesian equation 2x.x 2 C y 2 / D a.3x 2  y 2 /. To trisect an angle ˛, draw this angle with the x-axis as one leg and with A as its vertex. ˛ Determine the intersection B with the trisectrix. Draw the line OB. OB makes an angle with the 3 x-axis.

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Chapter 5  Trisecting an angle

Exercise 81 Create an IGS file in which you draw a circle C centered at O and with radius 5. Let P be a point of the circle. Let A be a point on the x-axis. Determine the slope m.D tan ˛/: Draw a line lA through A with slope tan 3˛. Use the result of Exercise 79 to express the slope of the line through A in terms of m. Determine the intersection Q of OP and lA . Determine the locus of Q with P .

5

Exercise 82 Giovanni Ceva (1647–1734) was an Italian mathematician best remembered for the theorem named after him. He also found a method for trisecting an angle with a curve now called Ceva’s trisectrix or Ceva’s cycloid. Consider a Cartesian coordinate system. Let C1 be a circle centered at O and with radius R. Let B be a point on C1 . Let C2 be a circle centered at B and with radius R. C2 intersects the x-axis at O and D. Let C3 be a circle centered at D and with radius R. C3 intersects OB a second time in P . When B moves along the circle, P traces out Ceva’s trisectrix. Prove that the angle which DP makes with the x-axis is thrice that which OB makes with the x-axis. Prove that the polar equation of Ceva’s trisectrix is r D 3R  4R sin2  D R.1 C 2 cos 2/ and deduce its Cartesian equation. Exercise 83 Prove that the following procedure also yields Ceva’s trisectrix. Draw a circle C1 centered at O and with radius R. Let B be a point on C1 . Draw the perpendicular b to x and through B. Let D be the mirror image of O with respect to b. Draw the straight line OB. Draw the perpendicular e to OB and through D. Let P be the mirror image of B with respect to e. Then point P is a point on Ceva’s trisectrix. Create an IGS file in which you make this construction and determine the locus of P with B. Exercise 84 Ceva’s trisectrix is one curve of a class of curves defined as r D 1 C a cos 2, sometimes called botanic curves. Create an IGS file in which you define a slider a and draw some of these curves.

75

6

Squaring the circle Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_6

Constructively transforming a circle into a square is often referred to as squaring the circle. In this case squaring means to find a square with the same area as a given figure. We still refer to the square root of a number, meaning we are looking for the length of the edge of a square with the given number as area. “Squaring the circle” has become more or less a catch line for something which is impossible or unsolvable. Indeed, the problem cannot be solved with compass and straightedge methods. But as with the other problems, it has a rich history. Although we have already mentioned that the quadratrix can be used to find a square with an equal area to a circle, the nature of this problem is completely different. We will now explain some solutions that were put forward to address the problem. As in the previous chapters, we will encounter solutions to the problem which cannot be accomplished with compass and straightedge methods, but other geometrical constructions are possible. We now know that the area of a circle is given by S D r 2 . Now suppose that a square p with edge a has the same area. Then a2 D r 2 , from which a D r. The problem of p squaring the circle is thereby reduced to constructing a line segment with length , using p compass and straightedge methods. Because  is the geometric mean of  and 1, it suffices to construct a line segment with length  (see also 7 Sect. 7.1, Construction 3).

6.1

Archimedes’ spiral revisited

In 7 Sect. 5.3, we encountered a curve called Archimedes’ spiral, which we used to trisect an angle. The same curve also allows us to square the circle (see . Fig. 6.1). Suppose a circle has radius a. Draw the spiral with polar equation r D a . This spiral  intersects the perpendicular to x and through the pole at a point P , i.e. at r D a . 2 Draw a parallel to the x-axis through this point. Draw the tangents to the circle at the intersections with the x-axis. These three lines and the x-axis define a rectangle with area  s D 2a  a  D a2 , which is the area of the given circle. Once we have constructed a 2 rectangle, it is easy to construct an equiareal square (see 7 Sect. 3.2, . Fig. 3.8). Just like the case of the parabola, we note that only individual points of the spiral, but not the spiral itself, can be constructed with compass and straightedge methods.

Chapter 6  Squaring the circle

76

. Fig. 6.1 Using the Archimedean spiral to solve the circle quadrature

6

6.2

Dinostratus’ quadratrix

In a previous chapter we have encountered a curve called the quadratrix (see 7 Sect. 5.6). Dinostratus noticed that Hippias’ trisectrix also allowed the circle to be squared, hence the curve’s name Dinostratus’ quadratrix. 2 kADk 2r D . Referring to . Fig. 6.2, the length of the line segment ŒAE equals   To rigourously prove this result we need to use calculus, because this is the result of a limiting process. If X is on the quadratrix and X 0 is the perpendicular projection of X on AB, then length kXX 0 k and angle †XAB are equal to zero in the equation †XAB kXX 0 k D †DAB kDAk when X D X 0 D E. Exercise 85

2R of the quadratrix and calculate the values of r for a given value  sin  of R and some very small angles . Use the polar equation r D

Exercise 86 In a quarter circle centered at O and with radius kOSk D 1, an angle =]SOQ is drawn. Draw the tangent line t to the circle in S. OQ intersects t at R. Through Q draw the perpendicular to OS. Let the foot be P . Use area 4OPQ < area circle sector OSQ < area 4 ORS to prove that sin  cos  <  < tan 

and

cos  <

1  < sin  cos 

77 6.2  Dinostratus’ quadratrix

6

. Fig. 6.2 The quadratrix

What happens to the left-hand side and to the right-hand side respectively if  D 0 rad? What would  ? you conclude for sin 

Mathematically, we express the result of Exercise 86 as  D 1:  !0 sin  lim

This result allows us to interpret the geometry of the quadratrix. We can now say (see . Fig. 6.2) that kAEk D lim

 !0

2R 2R 2R  D lim D :  sin    !0 sin  

Dinostratus came to the same conclusion but on the basis of geometrical considerations. If you accept this step, you can construct a line segment of length , which is essential for squaring the circle. Needless to say that we cannot construct E using compass and straightedge methods. Suppose we have constructed E as shown in . Fig. 6.2. We will now proceed to construct p 2 a line segment of length . In the following let R D 1, then kAEk D (see . Fig. 6.3  left).

»

Determine the midpoint M of the line segment ŒAE, 1 then kMEk D  Draw a right-angled triangle 4EMF , right-angled in M , and choose kMF k D 1 Elongate the line segment ŒME to point G, for which kM Gk D 1 Draw a parallel straight line to FE through G, this line intersects the line FM at H Draw a semi-circle of diameter kFH k (see . Fig. 6.3) Elongate the line segment ŒME. Let the intersection with the semi-circle be K A square with edge ŒKM  will have an area S D .

78

Chapter 6  Squaring the circle

. Fig. 6.3 The construction of  using the quadratrix

6

Exercise 87 Prove that 4GMH  4EMF and deduce that 1 kGM k kMH k kMH k D ) kMH k D : D ) 1 1= kMF k kMEk p Show that kKM k D .

6.3

Hippocrates’ lunes

Hippocrates of Chios (see 7 Sect. 4.1) was one of the first mathematicians who thought he had found a way to solve the problem of squaring the circle. Hippocrates thought that if he could determine the area of curvilinear figures consisting of arcs of circles by reducing them to a rectangle it would also be possible to determine the area of a circle in the same fashion. This amounts to a compass and straightedge construction. He succeeded in squaring certain curvilinear figures, now called Hippocrates’ lunes. Consider an area of the plane bounded by two arcs of a circle, as seen in . Fig. 6.4. The shaded concave regions are called lunes (because of their resemblance to a crescent moon), the convex region is called a lens. It turns out that for five types of lunes whose area we are able to determine using compass and straightedge methods, none of the lenses are squarable. Classically, Pythagoras’ theorem reads: “[the area of] the square on the hypotenuse equals the sum [of the areas] of the squares on the perpendiculars.” Actually it does not matter which shape the figures on the sides of the triangle have, as long as they are similar Pythagoras’ theorem holds! . Fig. 6.4 Two lunes (shaded area) and a lens

79 6.3  Hippocrates’ lunes

6

. Fig. 6.5 In all of these cases Pythagoras’ theorem holds: area B C area C D area A (drawing by Paul Tytgat)

. Fig. 6.6 Archimedes’ arbelos

Exercise 88 Prove that Pythagoras’ theorem also holds if you replace square by circle or semi-circle: “[the area of] the semi-circle on the hypotenuse equals the sum [of the areas] of the semi-circles on the perpendiculars.” Exercise 89 Prove that Pythagoras’ theorem also holds if the figures on the sides are similar (see . Fig. 6.5). Exercise 901 In Greek, a shoemaker’s knife is called an arbelos. The word was used by Archimedes to describe a plane figure bounded by three semi-circles, with reference to . Fig. 6.6 SCAB , SCAC and SC CB . Draw the perpendicular to AB through C , intersecting the semi-circle SCAB at D. Then the area of the arbelos ABCD is equal to the area of the circle with diameter CD. Exercise 91 Create an IGS file in which you define a line segment ŒAB with C on it. Draw the semi-circles SCAB , SCAC and SC CB . Construct the perpendicular to AB in C , intersecting the semi-circle SCAB at D. Draw the circle with diameter ŒCD. Determine its area. Determine the area of the arbelos using the formula area arbelos D area SCAB  area SCAC  area SC CB : What do you notice? Determine the perimeters PAB ; PAC and PCB . Compare PAB to PAC C PCB . What do you notice? 1

van Lamoen (2009).

80

Chapter 6  Squaring the circle

6 . Fig. 6.7 Hippocrates’ lunes in an isosceles right-angled triangle . Fig. 6.8 Hippocrates’ lunes in a right-angled triangle

Exercise 92 Create an IGS file in which you draw a line segment ŒAC , in A erect a perpendicular l to the line segment ŒAC . Find B on l such that kABk D kAC k. 4ABC is an isoceles right-angled triangle, right-angled in A. Draw semi-circles on the sides as seen in . Fig. 6.7. Calculate the area of the shaded parts in . Fig. 6.7. To do so, use: 1 area lune D area SCAB  area quarter circle MAB C area 4ABC: 2 Determine the area of 4ABC and compare to 2.area lune/. You can move the vertices around. What do you notice? Exercise 93 Consider an isosceles right-angled triangle 4ABC , right-angled in A as in . Fig. 6.7. Use this result to determine the area of the lune on one of the perpendiculars. Show that the area of a circular sector with radius r and angle ˛ (expressed in radians) is equal 1 to S D r 2 ˛. 2 Show that the area of sector BMA is equal to the area of the semi-circle BA. Exercise 94 Create an IGS file in which you draw two perpendicular line segments ŒBA and ŒAC . 4ABC is a right-angled triangle, right-angled in A. Calculate the area of the shaded parts in . Fig. 6.8. Use 1 area lunes D area SCAB C area SCAC  area SCBC C area 4ABC: 2

81 6.3  Hippocrates’ lunes

6

. Fig. 6.9 Other properties of lunes in a right-angled triangle

Exercise 95 In a right-angled triangle 4ABC , right-angled in A, draw semi-circles with the sides as diameter as seen in . Fig. 6.8. Use Exercise 88 to show that the sum of the areas of lunes I and II (shaded areas) is equal to the area of the triangle 4ABC : area I C area II D area 4ABC Exercise 962 In a right-angled triangle 4ABC, right-angled in A, draw semi-circles as in . Fig. 6.9. Prove that in the left triangle we have area III  area I  area II D area 4ABC and in the right triangle area I C area II C area III C area IV D area 4ABC: Exercise 973 Leonardo’s claw is the grey portion of the circle in the middle figure of . Fig. 6.10. To obtain Leonardo’s Claw, cut out a lens from the big circle. The lens is made up of two equal circular segments with right angles as central angles. Now cut out a circle, with diameter along the axis of the lens, which touches the lens externally and the big circle internally. Show that Leonardo’s Claw is squarable and that its area equals the area of the square in its grasp.

The result of Exercise 93 means that we have calculated the area of a curvilinear figure by means of a rectilinear figure. Moreover the curves which determine the lune are arcs of a circle. This seems to open perspectives to do the same for a circle. Hippocrates explored similar figures hoping to find a way to square the circle. 1 Consider an isosceles trapezoid with kADk D kDC k D kCBk and kADk D p kABk 3 (see . Fig. 6.12). 2 3

Alsina and Nelsen (2010), p. 141–142. Alsina and Nelsen (2010), p. 157.

82

6

Chapter 6  Squaring the circle

. Fig. 6.10 Leonardo’s Claw

. Fig. 6.11 Constructing an isosceles trapezoid

This trapezoid can be constructed (see . Fig. 6.11). Consider a line segment ŒAB. Open a kABk compass on p . Put the compass point in A and draw a circle C1 . Now place the compass 3 point in B and draw a circle C2 . Determine the midpoint M of ŒAB. Position the compass kABk point in M and draw a circle with radius p . Let the intersection points of the circle and 2 3 ŒAB be R and S respectively. Erect perpendiculars to ŒAB in R and S. The intersection points, on the same side of ŒAB, of these perpendiculars with the circles C1 and C2 respectively determine points C and D of the trapezoid. Exercise 98 Show that the perpendicular bisectors on AD, DC and CB intersect at one point, O.

The result of Exercise 98 implies that the trapezoid can be circumscribed by a circle centered at O (see . Fig. 6.12). Construct the triangle 4ABE which is similar to triangle 4ADO. Draw a circle centered at E and with radius kABk. Together with an arc of the circumscribed circle, this circle, defines a lune AFBCD. From the similarity 4ABE  4ADO it also follows that: area 4ABE D 3 area 4ADO and area circular segment AB D 3.area circular segment AD/; because the areas of circular segments are proportional to the squares of their bases.

83 6.3  Hippocrates’ lunes

6

. Fig. 6.12 Area circular segment AB D 3.area circular segment AD/

Add the area of the trapezoid to both sides: area trapezoid ABCD C area circular segment AB D area trapezoid ABCD C 3.area circular segment AD/ area trapezoid ABCD D area trapezoid ABCD C 3.area circular segment AD/  area circular segment AB The right hand side of this equation is nothing else than the area of the lune ADCBF . Exercise 99 Show that area circular segment AB D 3.area circular segment AD/ Exercise 100 Show that the area of the lune ADCBF is equal to the area of the kite-like quadrilateral AOBE. Exercise 1014 If a regular hexagon is inscribed in a circle and six semi-circles are constructed on its edges (see . Fig. 6.13), then the area of the hexagon equals the area of the six lunes plus the area of a circle whose diameter is equal to one of the sides of a hexagon.

Hippocrates now considered an isosceles trapezoid with kADk D kDC k D kCBk

and

kADk D

1 kABk 2

(see . Fig. 6.14). Note that this trapezoid is also the half of a hexagon. 4

Alsina and Nelsen (2010), p. 140.

84

Chapter 6  Squaring the circle

. Fig. 6.13 Lunes and the regular hexagon

6 Because kAOk D kDC k and ŒAO k ŒDC  Thales’ intercept theorem says that ŒAD k ŒOC . On the other hand the line segments ŒAO and ŒDC  are parallel to one another as well, from which kADk D kOC k. Therefore A, B, C and D are on a circle centered at O and with radius kOAk. Obviously this means that area semicircle AD 1 AD 2 D D area semicircle AB AB 2 4 Because of the hypotheses we know that: area semicircle AD D area semicircle DC D area semicircle CB D area semicircle EF from which: area semicircle AD C area semicircle DC C area semicircle CB C area semicircle EF D area semicircle AB We now subtract the areas indicated by IV, V and VI from both sides and find: area semicircle EF C 3 area lune III D area trapezoid ABCD so area semicircle EF D area trapezoidABCD  3 area lune III

. Fig. 6.14 Hippocrates’ lunes in an isosceles trapezoid

(14)

85 6.3  Hippocrates’ lunes

6

Had Hippocrates succeeded in finding the area of lune III using a compass and straightedge construction, he would also have been able to find the area of the semi-circle AD and thus of the circle AD. In the previous problems it was possible to square the lunes. Unfortunately for Hippocrates, he was unable to square these lunes5 . We will return to this subject in 7 Sect. 7.3. Exercise 102 Use Exercise 101 to prove equation (14). Exercise 103 Use the formula for the area of a circular sector to determine the area of lune III. Express all angles in radians!

Hippocrates explored yet another kind of lune, the concave pentagon lune6 . It is a lune which canpbe constructed with a neusis. Consider p a concave pentagon for which kCDk D kDEk D 3a and kABk D kBC k D kEAk D 2a.

Construct a (semi) circle centered at A and with radius kABk. Erect the perpendicular bisector QP on ŒAB. Hippocrates now uses a neusis to find a certain point E on the circumference of the semi-circle.

A line through B will intersect PQ at a point D and the semi-circle at another point E. Rotate this line about B until p 6 3 2 2 D or D kDEk kDEk kEAk kEAk : 2 2 The line segment ŒDE is an edge of the concave pentagon ABCDE. 5 6

Postnikov and Shenitzer (transl.) (2000). Shelburne (2008).

86

Chapter 6  Squaring the circle

Draw a line b parallel to AB through E. Draw a circle C centered at B and with radius kABk, which intersects b at C (i.e. construct the mirror image of E about PQ). The vertices of the concave pentagon ABCDE define a lune, for which we will identify its generating circles.

6 CBAE is an isosceles trapezoid, which has a circumscribed circle centered at O1 . Draw the perpendicular bisectors on AE and BC and let their intersection be O1 . Draw the circle centered at O1 and with radius kO1 Ak. This circle also passes through B, C and E. 4EDC is a triangle, which has a circumscribed circle centered at O2 . Draw the perpendicular bisectors on ED and DC , which intersect at O2 . Draw the circle centered at O2 and with radius kO2 C k, which also passes through D and E.

In the concave pentagon AEDCB we have that kCDk D kDEk D p kBC k D kEAk D 2a. Exercise 104 Show that 4ADB and 4EAB are similar. Put a D kEAk D kABk and x D kDBk. Show that r ! r 3 11 a C : xD 2 2 2 Deduce that the neusis which Hippocrates used is not necessary.

p 3a and kABk D

6

87 6.3  Hippocrates’ lunes

We will now show that we can determine the area of the lune by compass and straightedge methods. First determine M , the intersection of AC and BO1 . Exercise 105 Prove that AC ?BO1 .

It is clear that †EO1 A D †AO1 B D †BO1 C and †EO2 F D †FO2 C . Note that †BAC D †ABC (base angles of isosceles triangle 4ABC ). and †BAC D †F CA (alternate angles of parallel lines EC and AB). Thus †F CD D †BCM . In 4DF C we have that †FDC D †O2 DC is the complementary angle of †F CD. In 4CMB we have that †MBC D †O1 BC is the complementary angle of †BCM . Triangles 4O2 DC and 4O1 BC are isosceles. Furthermore we have proved that †O1 BC D †O1 CB D †O2 DC D †O2 CD

whence †BO1 C D †DO2 C

(15)

Exercise 106 Use (15) to show that the areas of the circular sectors O1 CBAE and O2 CDE are equal.

Since the areas of both circular sectors O1 CBAE and O2 CDE are equal, the area of the lune ABCDE equals the dartlike quadrilateral EO1 CO2 . sector EO2 C D I C II sector EO1 C D II C III )

0 D I  III

)

I D III

Which proves the assertion. Exercise 107 Look closely at Jacobus Falco’s figure (. Fig. 6.15 – from Falco (1591)). Prove that the area of rectangle GM V T equals the area of the curvilinear figure QNPOSR. Prove that the way this figure is constructed in all cases results in a constructive transformation into a rectilinear figure.

88

Chapter 6  Squaring the circle

6

. Fig. 6.15 Transforming curvilinear figures into rectilinear figures kept the minds of many mathematicians occupied over the centuries. The figure shows a sixteenth century example (from Falco (1591))

6.4

Franco of Liège, the demise of mathematics

During the Middle Ages, the study of mathematics was in dire straits. The knowledge of the Roman agrimensores (surveyors) was still known, but the larger part of the Greek corpus was forgotten. Exercise 108 The following method to square the circle was used by the Roman agrimensores: draw two perpendicular diameters, divide the radii into four equal parts, elongate the radii with one fourth part. The endpoints determine an edge of the square. Determine the length of the edge of the square. Assuming the equality of the areas of the circle and the square holds, what is the value of ?7

An example of how low the study of mathematics had sunk is given by Franco of Liège (?– ca. 1083) in his treatise on the squaring of the circle8 . Geometry, as it is depicted in Franco’s treatise, had become an almost experimental science. Theorems involving equal areas are not proven but are checked, sometimes even by weighing pieces cut out of paper or cloth. Franco of Liège was one of the students of Fulbert of Chartres (ca. 960–1028). From 1066 onwards, Franco taught at the famous Cathedral School of Liège. Another student of Fulbert’s, Radolph of Liège, corresponded with Regombold, the rector p of the Cathedral School of Cologne. In one of these letters, the subject is the nature of 2. p While a line segment of length 2 is easily constructed, Radolph comes to the conclusion that 7 8

Smeur (1968), p. 16. Smeur (1968); Smeur and Folkerts (1976a,b).

89 6.4  Franco of Liège, the demise of mathematics

6

. Fig. 6.16 Franco of Liège’s attempt at the quadrature of the circle

p

2 cannot be written as a number (i.e. as what we call a rational number), although sometimes 7 17 and are used as its value. 5 12

Exercise 109 p p leads to a Suppose that p and q have no common factors. Prove that the assertion 2 D q contradiction. (Hint: a square is an even number if it is a quadruple of another square).

17 It is not unlikely that Franco knew of the correspondence. He shows that is not equal 12 p p to 2, but is larger. He succeeds in calculating a better approximation, not only for 2, but also for other irrational roots. Without a proof, he asserts that these numbers are not rational. In the following paragraphs, Franco does not prove the circle quadrature as we know it, but he tries to show that the square root of some numbers are not rational and therefore cannot be calculated exactly, but can only be constructed as the length of a line segment. For the ratio of the perimeter of a circle to its diameter (the value of ) he uses 3 71 , without questioning the correctness of the number. A circle with radius 7 therefore has a circumference of 44 and an area of 154. If the perimeter is divided into 44 equal parts, then these parts define circle sectors which can be rearranged as a rectangle (see . Fig. 6.16). The rectangle with sides 11 and 14 has the same area, 154, as the circle. To find a square with an area equal to that of the rectangle one has to transform this rectangle into a square. This should not be hard (see 7 Sect. 3.2, . Fig. 3.8) but Franco has no knowledge of the construction of mean proportionals!! In . Fig. p 6.17, EH has the same length as the diagonal of a unit square EG, from which kEH k D 2. AEF C is the square with edge 11, therefore kAEk D 11 and kEBk D 3. Exercise 110 Find K and I on ŒCD such that kHKk D kBI k D kAH k. Prove that area HBIK D p area HBDL. Calculate the distance between ŒHK and ŒBI  and compare your result to 2.

p Franco suggests, incorrectly, that the distance between ŒHK and ŒBI  is 2. He then moves the parallellogram KHBI in such a way that ŒBI  coincides with ŒAH , which is

Chapter 6  Squaring the circle

90

6 . Fig. 6.17 Franco of Liège’s attempt at the quadrature of the circle 2

possible because kBI k D kAH k. Transforming this parallellogram into a rectangle, is easy by cutting away a triangle at one end and adding it to the other p end. 2, which is not too bad an apFranco now has a square (he believes) with edge 11 C   p p 2 proximation of 154 11 C 2 D 154:11 : : : . This square, again according to Franco, has an area equal to that of a circle with radius 7.

6.5

Nicolas of Cusa

Cardinal Nicholas of Kues or Nicolas of Cusa (1401–1464), also known as Nicolas Cusanus, was a German theologian and Prince of the Church9 . He studied in Heidelberg and Padua. After his graduation, he rejected an offer made by the newly established University of Leuven. He was involved in negotiations during the Council of Florence of 1439, which sought to reunite the Eastern Orthodox Church and the Western Roman Catholic Church. In 1448 or 1449, he was made a Cardinal and bishop of Brixen in Tyrol by Pope Nicholas V. Nicolas of Cusa wrote some deeply mystical works about Christianity. He was not only interested in theology and philosophy, but equally in mathematics, astronomy and linguistics. His scientific ideas are scattered across his philosophical treatises. He advocated a reform of the Julian calendar, an idea which would lead to the Gregorian reform in the sixteenth century. Nicolas was the first to describe the use of concave lenses to treat myopia. In trying to solve the circle quadrature, Nicolas implictly put forward an approximation of . His method is a tedious one, involving trigonometry. Moreover his prose is excursive and prolix. Cusanus for instance still used a sexagesimal system to denote fractions. He stated that10 :

»

9

around the midpoint D shall be inscribed a circle EF G and a circle HI circumscribed; the straight line DE shall be so drawn, that E is the midpoint between A and B; then DB shall be drawn. Further, a straight line DK shall be drawn from D to the midpoint

Meuthen (1982). As cited in Wertz Jr. (2001).

10

91 6.5  Nicolas of Cusa

6

. Fig. 6.18 Cusanus’ attempt at the quadrature of the circle

between E and B. I maintain: DK is smaller than the radius of the circle isoperimetric 5 to the triangle, by one-fourth of the length of DK. [i.e. r D kDKk] 4 What he does is this (see . Fig. 6.18):

»

Let 4ABC be equilateral (Nicolas does not state this explicitly). Determine the centroid D of this triangle Choose a side, e.g. ŒAB and divide it into four equal line segments Let K denote the division point closest to B Draw ŒDK Divide ŒDK into four equal line segments Elongate ŒDK and draw a circle centered at K and with radius one quarter of kDKk Let the point L be the intersection of this circle and DK. Then ŒDL is the radius of a circle with a circumference equal to the circumference of the triangle.

Cusanus suspects he only gives approximations, as he himself admits: “In mathematical science, every proposition which poses the exact equality between a circle and a square is impossible [=false]” This statement also makes clear Cusanus’ reasons for his interest in the circle quadrature, in his search for a proof of God he uses geometrical analogies, e.g.11 :

»

11

Whatever is not truth cannot measure truth precisely. (By comparison, a non-circle cannot measure a circle, whose being is something indivisible.) Hence, the intellect, which is not truth, never comprehends truth so precisely that truth cannot be comprehended infinitely more precisely. For the intellect is to truth as [an inscribed] polygon is to [the inscribing] circle. The more angles the inscribed polygon has, the more similar it is to the circle. However, even if the number of its angles is increased ad infinitum, the polygon never becomes equal [to the circle], unless it is resolved into an identity with the circle. As cited in Wertz Jr. (2001).

92

Chapter 6  Squaring the circle

. Fig. 6.19 Construction lines for Cusanus’ calculation of 

6

Cusanus uses the qualitative difference in the nature of a circle and a polygon to argue that, although the human mind is created in the image of God, and is therefore finite, it cannot precisely attain the Truth itself, which is infinite12 . In the paragraphs below, we will only highlight his mathematics and refer interested readers to philosophical treatises on Nicolas of Cusa. For the given configuration (see . Fig. 6.18) Cusanus first shows that the circle with the same circumference as the perimeter of the triangle 4ABC has a radius smaller than that of the circumscribed circle, but larger than that of the inscribed circle (see . Fig. 6.19).For the  1 isoperimetric circle, he approximates the ratio of the diameter to the circumference D  p 2 12 1575 by p . 6 2700 He asserts that his approximation cannot be closer “with even a part of a minute” (this is 1 closer than ). 60 Exercise 111 In . Fig. 6.18 put the length of the edges of the triangle equal to 1, kABk D kBC k D kCAk D 1. 5 Use trigonometry to calculate the length of ŒDK and deduce the length of kDLk D kDKk. 4 Use the fact that in an equilateral triangle the centroid, orthocentre and midpoint coincide. Determine the circumference of the circle with radius kDLk and of 4ABC . What can you conclude?

p 2 12 1575 How did Cusanus arrive at the ratio p ? 6 2700 He put kDBk D 60 whence kDEk D 30. Because 4ABC is equilateral †ABC D 60ı . DB is the bisectrix, so †ABD D 30ı . In any right-angled triangle with one angle equal to 30ı , one perpendicular is half the hypotenuse. Exercise 112 If kDBk D 60, calculate kEBk, kDKk and kDLk. The circumference of the triangle equals 6 kEBk, the diameter of the circle is 2 kDLk. 12

Wertz Jr. (2001).

93 6.6  Archimedes’ approximation

6

Exercise 11313 The following procedure for finding a circle whose area was equal to that of a given square was described in the Indian Salvasutras. In square ABCD, let M be the intersection of the diagonals. Draw the circle centered at M and with radius kMAk, let kMEk be the radius of the circle perpen1 dicular to the side ŒAB and cutting ŒAB in G. Let kGN k D kGEk. Then kMN k is the radius p 3 2C 2 of the desired circle. Show that kAN k D kABk. Assuming the equality of the areas of the 3 circle and the square holds, what is the value of ?

6.6

Archimedes’ approximation

Archimedes was able to prove that the area of a circle is equal to the area of a right-angled triangle with perpendiculars equal to the radius and to the circumference of the circle respectively. Suppose a circle has an area S which is larger than the area T of a right-angled triangle with perpendiculars equal to the radius and to the circumference of the circle respectively. We can then find an integer number n for which S  area inscribed 2n -gon < S  T 1 -part of the circle. It is then clear 2n that we can find an n-gon with an area larger than the area T of the triangle. In other words, T < area 2n -gon. Now suppose that AB is an edge of the inscribed 2n -gon and ON a perpendicular from the centre of the circle to AB (so N is the middle of AB). It is clear that ON is smaller than the radius of the circle. Because the perimeter of the inscribed polygon is smaller than the perimeter of the circle we find that:   n n 1 area 2 -gon D 2 kABk  kON k 2 1 D .2n kABk/  kON k 2 1 < perimeter C  radius D T 2 But this is a contradiction. Similarly, we are able to prove that a circumscribed polygon exists for which area 2n gon < T , but this also leads to a contradiction. From this, we can conclude that the area of the circle and that of the triangle are equal. k If k is the length of the perimeter of a circle with radius 1, then its area is . 4 Now the areas of circles are to each other as the squares of their diameters, so for a circle with radius r we find that The area of a regular inscribed 2n -gon is larger than 1 

SC .2r/2 D k=4 12

and

SC D kr 2

Archimedes however was unable to calculate the value of k, but he did propose very good approximations for what we now call . 13

Katz (1998), p. 41.

94

Chapter 6  Squaring the circle

6

. Fig. 6.20 Left the inscribed and circumscribed equilateral triangles of a circle, right: the inscribed and circumscribed regular hendecagons or 11-gons

So if we know the value of k (D ), we may be able to find a way to construct a square with the same area as the circle. To approximate this number k, Archimedes used the fact that Spn < SC < SPn , in which pn is the regular n-gon inscribed in the circle and Pn the regular n-gon circumscribed about the circle. Archimedes was able to calculate the areas of the inscribed and circumscribed 96-gon and < k < 3 71 or 3:1408 < k < 3:1429. arrived at 3 10 71 22 These are very good approximations and 3 71 or still survives as a value for  in the 7 lower grades. It is clear that the larger n becomes, the closer the areas of the inscribed and circumscribed polygons will approximate the area of the circle. Exercise 114 Create an IGS file in which you define a slider a ranging from 3 to 100 in steps of 1. Draw a circle centered at A and with radius e.g. 5. Define a point B on the circle with coordinates 2 2 ; R sin . R cos a a Determine the intersection points C and D of this circle and the x-axis. Choose C on the positive x-axis and define a regular polygon with ŒCB as the edge and with a vertices. You will first see a triangle 4CBE. Define the tangents in C , B and E respectively and determine the intersections F ,G and H of the tangents, taken pairwise. Define a regular polygon with edge ŒF G and a vertices. This is the circumscribed polygon (see . Fig. 6.20). Define the numbers b D .circumference polygon1/=.2radius/

and

c D .circumference polygon1/=.2radius/

[alternatively define .area polygon2/=.radius/2 and .area polygon2/=.radius/2 ]. Use the slider to increase the number of edges. What do you notice for the values of b and c?

95 6.7  Adriaan van Roomen and Ludolff van Ceulen

6

Exercise 115 If pn and Pn denote the perimeters of the regular n-gons inscribed in and circumscribed about the same circle, and if an and An are their areas, show that pn  D cos Pn n

6.7

and

 an D cos2 : An n

Adriaan van Roomen and Ludolff van Ceulen

Archimedes’ method was revived in the late sixteenth century by Adriaan van Roomen (see 7 Sect. 2.3) in his book Ideæ Mathematica (1593)14 and later in In Archimedis (1597), in which he published a value for  to sixteenth decimal places  D 3:1415926535897931. He used Archimedes’ method for regular polygons having 230 edges, by consecutively doubling the number of vertices of a regular polygon. His result would soon be superseded by the work of his friend Ludolff van Ceulen (1540–1610). In August 1593, Josephus Justus Scaliger (1540–1609) was appointed professor at the University of Leiden. The next year he published two books Cyclometrica Elementa duo (1594) on the squaring of the circle and Mesolabium (1594) on the duplication of the cube. His work was already being circulated in the form of manuscript copies, in which he claimed to have solved the three classical geometry problems of Antiquity. For this, he was vehemently criticised by François Viète. p Scaliger claimed that the ratio of the circumference of the circle to the diameter was 10. He also criticised Archimedes for using an arithmetical approach to the squaring of the circle. Following Aristotle’s advice that one “cannot pass from one genus to another”, he was opposed to solving a geometric problem by means of arithmetic15 . The first to read his book was Ludolff van Ceulen, who advised him not to distribute it. Van Roomen read the book in the autumn of 1594 and in his correspondence he was very critical of Scaliger. He wrote a devastating answer to Scaliger’s claims. In Archimedis was published in 1597. The book contains the Greek text with a Latin translation of Archimedes’ On the measurement of the circle, an Apology of Archimedes, refutations of Scaliger’s claims and Finnaeus’ and Ursus’ quadratures of the circle. Ludolff van Ceulen was a German-Dutch reckoning master (see . Fig. 6.21 left), who went from Hildesheim to Antwerp, where he stayed for some time with his brothers. Around 1562, he settled in Delft, where he made his fame as a fencing and reckoning master. Around the mid-1580s, Ludolff van Ceulen got involved in a dispute. At the request of his collegue Adriaan Anthonisz, he proved that Simon van der Eycke’s squaring of the circle was wrong. Van der Eycke had published a book in 1584 in which he claimed the value for the 69 . ratio of the circumference of a circle to its diameter was 3 484 In his 1596 book Vanden Circkel (About the Circle), van Ceulen published his ten-year research on the area of the circle and the areas of in- and circumscribed polygons (see . Fig. 6.21 right). In the same book, he also criticised Scaliger, without mentioning his name. Van Ceulen was buried in the Pieterskerk in Leiden. On his tombstone, the 35 decimals of  he had calculated were engraved. The original has long disappeared, but a new commemo14

van Roomen (1593), f (**iv) v, also Bockstaele (1976, 1993, 2009). Aristotelean dogma had it that one could not use algebra for solving geometric problems (Yu (2003), p. 44). “It follows that we cannot in demonstrating pass from one genus to another. We cannot, for instance, prove geometrical truths by arithmetic. [. . . ] Nor can the theorem of any one science be demonstrated by means of another science, unless these theorems are related as subordinate to superior. [. . . ] ” (Aristotle, Posterior Analytics, Book I, section 7). 15

96

Chapter 6  Squaring the circle

6

. Fig. 6.21 Left: portrait of Ludolff van Ceulen, right: Ludolph van Ceulen’s approximation of , a lower limit in the upper half of the circle, an upper limit in the lower half of the circle (van Ceulen (1619), EHC G 4867)

rative stone reminds us of his feat. In German  is sometimes referred to as die Ludolphsche Zahl (Ludolff’s number). Van Ceulen used the same method as van Roomen to calculate . Whereas van Roomen’s 1593 book puts forward “recipes” without formulae to calculate the edge, van Ceulen’s 1596 book explains the method in a symbolism which is still quite readable. We will determine the area of an inscribed and circumscribed regular polygon. By constructing a 2n-gon from an n-gon we can give a better approximation of the area. We can then construct a 4n-gon etc16 . To begin with, draw an inscribed and circumscribed hexagon in a circle with radius 1 (see . Fig. 6.22). The edge of the inscribed hexagon is 1 and the circumference is p6 D 6. Exercise 116 1 Referring to . Fig. 6.22, if a hexagon is inscribed in a circle with radius 1, prove that kAC k D 2 and calculate kOC k ; kCDk and kEDk.

p We find that kEDk D

3 . 3

p p 2 3 The edge of the circumscribed polygon thus is and its circumference is P6 D 4 3. 3 p p Now pn < 2 < Pn from which 6 < 2 < 4 3 and 3 <  < 2 3.

Exercise 117 Prove that .2 sin ˛/ D 2  2

q

r 4  .2 sin 2˛/ and .2 sin ˛/ D 2  2

2

2C

Deduce a formula to calculate the length of the edges of a 2i n-gon. Hint: In . Fig. 6.23 kABk D 2 sin 2˛ and kA0 B 0 k D 2 sin ˛. 16

Based on Wepster (2010a,b).

q 4  .2 sin 4˛/2 .

97 6.7  Adriaan van Roomen and Ludolff van Ceulen

6

. Fig. 6.22 The inscribed and circumscribed hexagon of a circle

It was this formula that allowed van Roomen and van Ceulen to “quickly” calculate the edges of a 2i n-gon, in a circle with radius 1, knowing the edge of the n-gon and to improve the approximation step by step. Putting of a regular hexagon equal to x, then the edge pthe edge p of a regular dodecagon equals s12 D 2  4  x 2 . Beginning with a hexagon, the formula for a 192-gon (D 25  6-gon) reads as

s192

v v u s u u r u q u t p t D 2  2 C 2 C 2 C 2 C 4  x2

Exercise 118 In a circle with radius 1, prove that if Sn is the edge of the circumscribed regular polygon and sn 2sn the edge of the inscribed polygon, that Sn D p . (Hint: calculate tan ˛) 4  sn2 Exercise 119 Use the van Roomen-van Ceulen formula to calculate, to eight decimals, the circumference of the inscribed and circumscribed 24, 48, 96 and 192-gons.

. Fig. 6.23 The inscribed hexagon and inscribed dodecagon of a circle

Chapter 6  Squaring the circle

98

Exercise 120 Use a spreadsheet and make a table in which you iteratively calculate the values of the circumference and circumscribed polygons beginning with the hexagon. Define the cell A.i C 1/ qof inscribed p as 2  4  A.i/2 : Define cell A .1/ as the edge of the inscribed hexagon (e.g. 1). In another column do the same for the circumscribed polygons. Calculate the circumferences of the inscribed and circumscribed polygons in the other columns. What do you notice?

6.8

6

Kepler’s use of infinitesimal methods

By the beginning of the seventeenth century, mathematicians started to explore completely new ideas concerning the sums of areas: infinitesimal methods. Let’s start with a simple idea. Consider a rectangle, which we divide intofour new  rectanB B gles, each having length L and width . Then the area of the rectangle is 4 L  D LB. 4 4 Suppose we make the parts even smaller and divide the rectangle into n smaller rectB angles. Then every rectangle has length L and width . The area then of course still is n   B D LB. n L n Now suppose n can become infinitely large and the rectangle is cut into infinitely many rectangles, which have zero width. These rectangles have become lines. We would write   B lim n L  D lim L  B D LB: n!C1 n!C1 n But infinitesimal calculus was still in its infancy, the limit concept would not be introduced until well into the nineteenth century. Mathematicians were still trying to come to grips with these new concepts. The fundamental question they had to answer was: “How can one add an infinity of lines to find an area?” Sometimes they made astounding and correct discoveries, sometimes they made fundamental errors. One of the first to use such infinitesimal methods was the astronomer Johannes Kepler. He put forward the following solution to the circle quadrature. Consider a circle, with radius r and perimeter P , which is cut up, like a cake, into n sectors. Take the sectors from the circle and arrange them as shown in . Fig. 6.24. This method seems similar to Franco of Liège’s method (see 7 Sect. 6.4), but Kepler takes a very different approach. Now consider a very small interior angle for each sector, i.e. let n become very large. Consider one such circular sector OAB and also 4OAB (see . Fig. 6.24). Kepler states that the height h of 4OAB is nearly equal to the radius r of the circular sector OAB, h  r. P P or kABk D xi  . Similarly, the base of 4OAB is nearly equal to n n Now the area of the parallellogram, which is made up of the triangles, is SD

n X 1 i D1

2

hxi 

n X 1 i D1

2

1 X 1 xi D rP r 2 i D1 2 n

rxi D

  2r D  r D r 2 : 2

99 6.9  Gregory of St Vincent

6

. Fig. 6.24 Kepler’s transformation of a circle into a parallellogram

Which is the same result as that of Archimedes (see 7 Sect. 6.6), but instead of using a triangle for comparison to the circle, Kepler uses a parallellogram. While Archimedes’ proof is cumbersome, but based on a solid geometric and logical basis, Kepler’s proof is quite easy but not rigourous even by the standards of his time. Kepler does not build his proof on a finite number of sectors, he boldly states that there are an infinite number. In that case, he argues, the difference between sector and triangle is negligible. He considers this kind of reasoning to be legitimate.

6.9

Gregory of St Vincent

The most famous mathematician to tackle the squaring of the circle problem with infinitesimal methods was Flemish Jesuit Gregory of St-Vincent (see also 7 Sect. 5.5). In 1615, Gregory was sent to Antwerp to teach mathematics at the Jesuit college. Here Gregory developed methods similar to those of his contemporary Kepler. He was one of the first to use the sum of rectangles to approximate the area between curves. He would then put the width of the rectangles equal to zero, in effect making them into lines. Being the first mathematician to venture into this territory, he had to build up the mathematics and its axioms involved in these processes, sometimes erring along the way. In . Fig. 6.25 we can see how he uses rectangles to approximate the area between two parabolae and between a parabola and a straight line. By making the width of the rectangles infinitely small and by adding the “areas” of these rectangles, he was able to calculate the area between the parabola and the straight line. By 1625, Gregory thought he could solve the problem of squaring the circle, considered as determining the exact area of a circle (not as a construction problem). However, before they were allowed to publish any of their work, Jesuits had to seek permission from the SuperiorGeneral in Rome. In Gregory’s case, the General forwarded his manuscript to Christopher Grienberger s.j., the mathematics professor at the Jesuit College in Rome, the famous Collegium Romamum. Grienberger was not convinced by Gregory’s results and summoned him to Rome. Gregory’s journey proved to be in vain, he was unable to convince Grienberger and failed to get permission to publish. His book entitled Problema Austriacum would not be published until 1647 (see . Fig. 6.26). By that year, Girolamo Cardano had already published a book explaining a very similar method. Grienberger had objected to Gregory’s method because it was not logically sound, i.e. it failed to live up to the strict classical Greek exigences of accuracy. Incidentally, Gregory’s

100

Chapter 6  Squaring the circle

6 . Fig. 6.25 Gregory’s figure to explain the calculation of the area between two parabolae and between a parabola and a straight line (from Sancto Vincentio (1647), EHC G 4869) . Fig. 6.26 Frontispiece of Gregoy of Saint Vincent’s Problema Austriacum on the squaring of the circle. Notice the angel holding a square window and the sun’s rays passing through it casting a circular spot on the ground (from Sancto Vincentio (1647), EHC G 4869)

treatise did contain a major flaw and a fundamental error. However, this error was not discovered until the treatise was published in 1647. It was Christian Huygens, one of the greatest mathematical minds of his generation, who found the error. It damaged Gregory’s reputation considerably. Yet we should not be too harsh on Gregory. He had boldly gone where no man had gone before, seeking out strange new mathematical worlds. To Gregory we owe (the equivalent of) Zb 1 the formula dx D ln b  ln a. A formula he proved using his rectangle method17. x a

17

Actually, Gregory proved that the area under a rectangular hyperbola and one of its asymptotes (also acting, in our a c D . b d

notation, as x-axis) is the same over the segment Œa; b as over the segment Œc; d  if

101 6.9  Gregory of St Vincent

6

. Fig. 6.27 Using the rectangle method to calculate the area between a straight line y D x and the x-axis over the interval Œ0; 1

In the paragraphs below, we will use infinitesimal methods to determine the area between a straight line and a parabola, a problem which Gregory also solved18. Exercise 121 Use the method of induction to prove that n X

iD

iD1

1 n .n C 1/ : 2

(16)

Suppose we want to find the area under the straight line y D ax (a > 0) between x D 0 1 1 1 and x D b (> 0). Obviously this is a triangle and its area is S D b y.b/ D b ab D ab 2 . 2 2 2 We will now, as an example, use infinitesimal methods to arrive at the same result. b Consider our triangle. We divide the leg along the x-axis into n pieces, with length . n Draw rectangles as in . Fig. 6.27. ib aib Then the height of a rectangle with vertex at x D is y D . n n 2 b aib iab Therefore the area of this rectangle is Sr D  D 2 . n n n The area of the triangle can be approximated by SD

n X iab 2 i D1

D

n2

n ab 2 X i n2 i D1

ab 2 1 n .n C 1/ n2 2 nC1 1 D ab 2 2 n

D

in which we have used the result of Exercise 121. 18 Gregory’s method was more complicated. For instance instead of using rectangles with equal bases, in his method the bases are in a geometric progression.

Chapter 6  Squaring the circle

102

. Fig. 6.28 Using the rectangle method to calculate the area between a parabola y D x 2 and the x-axis over the interval Œ0; 1

6

We notice that if n becomes very large nC1 Mathematically lim D 1, so n!1 n SD

1 2 ab : 2

nC1  1. n

(17)

If you are lucky your IGS will be able to calculate a Riemann sum. A Riemann sum is a mathematical tool you can approximate the area between a curve and the x-axis with. Basically it is the sum of the areas of rectangles with a base on the x-axis and one vertex on the curve. Depending on which vertex is on the curve, we distinguish between a left Riemann sum and a right Riemann sum. In our example the right vertices are on the curve so we have calculated the right Riemann sum for the area between the straight line y D x and the x-axis between x D 0 and x D 1. Some IGSs allow you to calculate these Riemann sums, in others you have to revert to tricks, such as calculating the area of each rectangle in a list and using a spreadsheet to calculate their sum. Draw n rectangles with base Œai ; bi  and height f .bi / on the interval Œa; b between the x-axis and the function y D f .x/, as in . Fig. 6.27 and . Fig. 6.28. Calculate the sum of these areas, call this RightRiemannSum(f; a; b; n). LeftRiemannSum(f; a; b; n) is similar: draw n rectangles with base Œai ; bi  and height f .ai / on the interval Œa; b between the x-axis and the function y D f .x/ and calculate the sum of their areas. Exercise 122 Create an IGS file in which you draw a line y D x, by defining a function f .x/ D x. Define a slider n which varies from 3 to 100. Now calculate RightRiemannSum(f; 0; 1; n). What is the value of this sum when n D 3 and when n D 100? Compare this number to the result of 17 when a D 1 and b D 1. Now define a slider a > 0 and define f .x/ D ax. Again define a slider n which varies from 3 to 100 and calculate RightRiemannSum(f; 0; 1; n). What is the value of this sum when n D 3 and when n D 100? Compare this number to the result of (17).

103 6.9  Gregory of St Vincent

6

. Fig. 6.29 The area between a parabola y D x 2 and a straight line yDx

Exercise 123 Use the method of induction to prove that n X iD1

i2 D

1  n.n C 1/.2n C 1/ D n 2n2 C 3n C 1 : 6 6

Use this result to calculate the area under the parabola y D x 2 and above the x-axis, between x D 0 and x D b.> 0/. Use a similar method as the one above. i 2 b2 ib b i 2 b2 is y D 2 , so the area of a rectangle is Sr D  2 . Now Hint: the height at x D n n n n 2n C 1 D 2. summate all the rectangle areas. Hint: lim n!1 n Exercise 124 Create an IGS file in which you draw a parabola y D x 2 by defining a function f .x/ D x 2 . Define a slider n which varies from 3 to 100. Now calculate RightRiemannSum(f; 0; 1; n). What is the value of this sum when n D 3 and when n D 100? Compare this number to the result you found in Exercise 123 when a D 1 and b D 1. Define a slider t > 0 and calculate RightRiemannSum(f; 0; t; n). Deduce that the area between t3 the x-axis and the parabola f .x/ D x 2 over the interval Œ0; t is equal to . 3 Exercise 125 Use equation (17) and the result of Exercise 123 to prove that the area between the parabola y D x 2 1 and the straight line y D x, between x D 0 and x D 1, equals (see . Fig. 6.29). 6

To square a circle one would have to calculate the area under a circle with radius R and above the line x D 0. This cannot readily be done using Gregory’s rectangle method. It can be done once we know the concept of Riemann sum, which produces an integral. In this case, to obtain the area under Ra circle p with centre O and radius R and above the line x D 0 one has R to calculate the integral R R2  x 2 dx.

104

Chapter 6  Squaring the circle

Exercise 126 Create an IGS file in which you draw a semi-circle with radius 1 by defining a function f .x/ D p 1  x 2 . Define a slider n which varies from 3 to 100. Now calculate RightRiemannSum(f; 0; 1; n). What is the value of this sum when n D 3 and when n D 100? Compare this number to .

Although Gregory did not succeed in determining the area of a circle, his student Joannes della Faille (1597–1652) did. Ironically, della Faille succeeded in determining the centroid of an ellipse and deduced the area, and hence also the area of a circle, using classical geometric methods19.

6

19

Meskens (2005).

105

7

Constructible numbers Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_7

7.1

Constructing numbers

In the previous chapters we have taken a look at some classical Greek geometric compass and straightedge construction problems. These constructions are a consequence of Euclid’s first three postulates: 1. It is possible to draw a straight line between points 2. It is possible to elongate any segment in a straight line as far as one wants 3. With a point as the centre and a radius it is possible to draw a circle We will call a number a constructible if we can construct a line segment with length a in a finite number of steps1 .

Axiom 1 If A and B are two different points, then a one-to-one correspondence exists between the set of real numbers R and straight line AB, such that A corresponds to 0 and B corresponds to 1.

The axiom tells us that if we choose two different points A and B, then we actually turn the straight line into a ruler with ŒAB as unit line segment. Every point on the line will correspond to a real number x such that if C is to the left of D then xC < xD . Definition 1 If the line segment ŒAB is given and A and B correspond to 0 and 1 respectively, then a point is called constructible if it can be constructed using compass and straightedge methods in a finite number of steps. The corresponding number is called a constructible number.

Note that this gauge can be transferred to any straight line through A or B using Thales’ intercept theorem. After we have shown that parallel lines exist and can be constructed (see 7 Sect. 3.2) we can transfer this gauge to any straight line. 1 For a thorough algebraic exposé on constructible numbers as an extension field of the rational numbers, see e.g. Lorenz (2006), p. 1–13, Finston and Morandi (2014), p. 93–102.

106

Chapter 7  Constructible numbers

. Fig. 7.1 Dividing a segment into three equal segments

. Fig. 7.2 The construction of the sum and the difference of two given constructible numbers

7

In the following constructions, you will note that we can construct a multitude of numbers. But can we construct all of them? i Construction 1 Divide a line segment into n equal segments, or, if a number a is constructible, then so is

a (see . Fig. 7.1). n

We have already proved this construction in 7 Sect. 3.2 (. Fig. 3.6). i Construction 2 If a and b are constructible numbers, then a C b and a  b .a > b/ are also constructible (see . Fig. 7.2).

»

Let the length of ŒOA be a and let that of ŒOB be b Draw the straight line OA and draw a circle with radius b and centre O The circle intersects the line at two points C and C 0 Suppose C 0 lies outside ŒOA then kAC k D a  b and kAC 0 k D a C b.

i Construction 3 If a and b are constructible numbers, then ab is also constructible (see . Fig. 7.3).

»

On a straight line through O determine the points C and D such that kOC k D 1 and kODk D a Draw a random straight line l through O not coinciding with CD Determine A on l such that kOAk D b Draw a line through A and C Now draw a straight line through D parallel to AC , this line intersects l at B and kOBk D ab as a consequence of Thales’ intercept theorem.

Exercise 127 Prove that if a and b are constructible numbers, then ab is also constructible. Deduce a method to construct a2 , a3 and a4 .

107 7.1  Constructing numbers

7

. Fig. 7.3 The construction of the product of two given constructible numbers

Exercise 128 a Prove that if a and b are constructible numbers, then is also constructible. b

From the foregoing, it is obvious that all (positive) rational numbers can be constructed. Can other non-rational numbers be constructed? Well, yes some can. i Construction 4 If a is a constructible number, then

»

p

a is also constructible (see . Fig. 7.4).

Choose three collinear points A, B and C such that kABk D a and kBC k D 1 Draw the circle with ŒAC  as diameter The perpendicular to AC and through B intersects the circle at D It is obvious that 4ADB  4DCB kDBk kABk D so kDBk kBC k p a kDBk D and from which kDBk D a. 1 kDBk

Note that this construction is also the construction of the geometric mean of a and 1 (see 7 Sect. 3.2, . Fig. 3.8). These five constructions are the basic constructions. All other constructions for constructing numbers can be reduced to a composition of these five constructions. This means that p n all positive rational numbers and all numbers of the kind 2 a, with a constructible, are constructible. Moreover all sums, differences, products, quotients and square roots of constructible numbers are constructible. s p p 10 8 7C p is constructible. For instance, the number 2 C 1 C 16 3 We will denote the set of constructible numbers as K. . Fig. 7.4 The construction of the square root of a constructible number

108

Chapter 7  Constructible numbers

Exercise 129 If a and b are constructible numbers, find a way to construct p (a) ab p (b) a n 1 (c) .abcd / 4 Exercise 130 Some constructible numbers have strange shapes and yet are so simple. Try these: p p p p Prove that 4 4  2 3 C 97  56 3 is an integer number. 6 6 Prove that p C p is an integer number. 3 3 3C 3

7

Exercise 131 (a) Consider the quadratic equation x 2  bx C c D 0 (b; c > 0). Draw a line segment ŒAB of length b. From the midpoint of ŒAB, drop the perpendicular bisector. Find point C on the p perpendicular bisector for which kM C k D c. Draw a circle centered at C and with radius b , this circle intersects ŒMB at D. Then x D kADk D kAM k C kMDk is a solution to the 2 quadratic equation. Prove this assertion. (b) Consider the equation x 2 C bx  c D 0 (b; c > 0). Draw a line segment ŒAB of length b. From the midpoint of ŒAB drop the perpendicular bisector. Find point C on the perpendicular p b bisector for which kM C k D c. Draw a circle centered at A and with radius , this circle 2 intersects ŒAC  at E. Then kEC k is a solution to the quadratic equation. Prove this assertion. (c) Use this result to prove that you can divide a line segment of length a, using only straightedge and compass, into two parts such that the difference of their squares is equal to their product. Show that the longer segment is the mean proportional between the shorter segment and the whole line. This ratio is called the golden ratio or golden section (sectio aurea). Exercise 132 Let the coordinates of A, B, C and D be constructible numbers. Show that the coordinates of the intersections of AB and CD, AB and a circle centered at C and with radius kCDk, and the circles centered at A and C and with radii kABk and kCDk respectively, are also constructible.

With our understanding of constructible numbers, we can now take another look at the Greek construction problems.

7.2

The theory of equations

The three classical problems we have focused on were solved, in the negative, in the nineteenth century. The duplication of the cube and the trisection of the angle were proved to be impossible by Pierre Wantzel (1814–1848). He proved that a constructible number is the solution to a polynomial equation of degree 2n with rational coefficients.

Theorem 1 p 3 2…K

109 7.2  The theory of equations

7

The proof uses two theorems from the theory of equations, which we will not prove. i Lemma 1 A cubic equation with rational coefficients, which does not have a rational solution, does not have a constructible solution.

i Lemma 2 If the coefficients of the polynomial equation an x n C : : : C a1 x C a0 D 0 are

p integers, then every rational solution to the equation can be written as with p a factor of q a0 and q a factor of an .

If we consider the general cubic equation a3 x 3 C a2 x 2 C a1 x C a0 D 0 with ai 2 Z, then p the last theorem states that every rational solution can be written as with p a factor of a0 q and q a factor of a3 . For example: x 3  1 D 0, it is clear that if there is a rational solution it has to be x D ˙1. ˙1 are the only factors of 1 and a3 D a0 D 1. ˙1 . A rational solution therefore can be written as ˙1 3 2 Note that x  1 D .x  1/.x C x C 1/ D 0, the equation has x D 1 as a solution, but no other real (and therefore no other constructible) solutions. This does not contradict Lemma 1. Lemma 1 does not guarantee extra constructible solutions if there is a rational solution. It only states that the lack of a rational solution implies that there are no constructible solutions. x 3  2x 2  5x C 6 D 0. a3 D 1 and a0 D 6. We only need to look for factors of 6, which are ˙1; ˙2; ˙3; ˙6. After some trial and error, we find .x  1/.x C 2/.x  3/ D 0, having solutions x D 1; x D 2; x D 3, and these obviously are constructible numbers. 2x 3 13x 2 30x 9 D 0. a3 D 2 and a0 D 9, we have to consider all fractions of which the denominator is a factor of 2 and the nominator a factor of 9. The factors of 2 are ˙1; ˙2, 3 those of 9 are ˙1; ˙3; ˙9. After some trial and error, we find that x D  is a solution to 2 p p 8 ˙ 76 the equation. The other solutions are x1;2 D D 4 ˙ 19, and these are obviously 2 constructible. p We know that 3 2 is a solution to the equation x 3  2 D 0. The factors of a3 D 1 are ˙1, those of a0 D 2 are ˙1; ˙2. Possible rational solutions are ˙1 en ˙2. None of these numbers satisfies the equation. There are no rational solutions!! As a consequence the equation x 3  2 D 0 has no constructible solutions. The duplication of the cube is impossible if we can find at least one cube for which we cannot find a cube twice as large using compass and straightedge methods. From the foregoing, it is clear that the edge of a cube which has double the volume as a cube of which the edge has length 1 is not constructible. More generally, if a line segment p has a constructible length a then it is impossible to construct a line segment with length 3 2a using compass and straightedge methods. We can treat the problem of the trisection of an angle in a similar way. To see why, consider the following figures (in . Fig. 7.5). Let the length of the chord ŒCB, which is subtended by 3˛, be d3 and let x be the length of the chord subtended by ˛. sin 2 sin 3˛ Using the sine rule in 4ABC , which is right-angled in C , we find from D 2R d3 which d3 D 2R sin 3˛ and x D 2R sin ˛. Now we know from the inscribed angle theorem that an angle # inscribed in a circle is half the central angle 2# that subtends the same arc on the circle. So, referring to . Fig. 7.5

110

Chapter 7  Constructible numbers

. Fig. 7.5 Determining an equation for the trisection of an angle using inscribed angles (left) and central angles (right)

7

ˇ 3ˇ ˇ and d3 D 2R sin and x D 2R sin . This result can readily be generalised 2 2 2 nˇ . to dn D 2R sin 2 We can now generalise this result for the division of an angle into n C 2 parts. Let R D 1, let the chord subtended by the angle .n C 2/ which has to be divided into n C 2 parts be denoted by dnC2 .D 2 sin .n C 2/  / and let x be the chord subtended by  . From the sum-to-product identities we know: right, ˛ D

sin ..n C 2/  /  sin ..n  2/  / D 2 sin n cos 2 which after some manipulations: sin ..n C 2/  / D 2 sin n cos 2  sin ..n  2/  /  D 2 1  sin2  sin n  sin ..n  2/  / ˇ we find 2       .n  2/ ˇ nˇ 2 ˇ sin .n C 2/ D 2 1  sin sin  sin ˇ 2 2 2 2    x 2 d dnC2 dnC2 n D2 12  2 2 2 2   1 2 dnC2 D 2 1  x dn  dn2 2

Putting  D

Now put d1 D 2 sin .˛/ D 2 sin ˛ D x, which allows us to draw up a table with the lengths of the chords in consideration of x D 2 sin ˛. d1 D x d3 D 3x  x 3 d5 D 5x  5x 3 C x 5 d7 D 7x  14x 3 C 7x 5  x 7 :: :

111 7.2  The theory of equations

. Fig. 7.6 Part of Viète’s solution to van Roomen’s problem using chords (Viète (1646), EHC G 4858)

7

112

Chapter 7  Constructible numbers

. Fig. 7.7 Nicomedes’ solution for the angle trisection

Viète noticed that the left-hand side of van Roomen’s equation was the expression for d45 , the right-hand side therefore was the length of this chord (see 7 Sect. 2.3) (see . Fig. 7.6). The expression for d3 can be written as 2 sin 3˛ D 3x  x 3 ;

7

(18)

which is a third degree equation. On the other hand, if we put y D sin ˛ and c D sin 3˛ then 4y 3  3y C c D 0:

(19)

Exercise 133 If y D sin ˛ and c D sin 3˛ then 4y 3  3y C c D 0, prove this assertion! Also show that this equation is equivalent with equation (18). Hint: first prove that sin 3˛ D 3 sin ˛  4 sin3 ˛ (see Exercise 79). Exercise 1342 . Fig. 7.7 shows Nicomedes’ solution to the trisection problem. Put kOC k D b, kBQk D x, kOP k D y. Prove that x 3  3x  2b D 0: Hint: first prove that kEQk D

(20) y C 3a . 2

The trisection of an angle is impossible if an ˛ exists for which equation (19) has no solution which can be written as a C b sin ˛, a; b 2 K. Exercise 135 If a line segment is given that has a length sin ˛ (which does not need to be constructible), then ˛ can be constructed.

The trisection of an angle in general is impossible if at least one angle can be found which cannot be trisected using compass and straightedge methods.  Consider the angle  D 3˛ D rad. 6 1 Equation (19) now becomes 4x 3  3x C D 0 ) 8x 3  6x C 1 D 0. 2 The factors of 8 are ˙1; ˙2; ˙4; ˙8, those of 1 are ˙1. 2

Bunt et al. (1988), p. 118–119.

113 7.3  Squarable lunes

7

1 1 1 We have to check whether any one of ˙1; ˙ ; ˙ or ˙ is a solution to the equation. 2 4 8 Substituting the values into the equation shows that none of these are solutions. Therefore the equation has neither rational nor constructible solutions.   An angle of magnitude rad cannot be constructed, and therefore the angle rad cannot 18 6 be trisected. Because we have found one counterexample, we can decide that the trisection of an angle in general is not possible. Remember the question was whether we can trisect any angle using compass and straightedge methods. Exercise 136  Use equations (18) and (20) to prove that an angle of rad cannot be trisected. 6 Exercise 137  Show that an angle of rad can be trisected. 2

7.3

Squarable lunes

In 7 Sect. 6.3 we encountered lunes for which we were able to determine the area by constructing a polygon with an equal area. We will now investigate whether and how we can construct other lunes, and whether we can express the area of these lunes as a constructible number. Consider . Fig. 7.8 (left) in which we need to determine the area of the shaded part. The lune is bounded by two arcs of a circle. For each arc we will determine the centres of their circles, calculate the area of the segments and subtract the triangles with the centres and endpoints of the arcs as vertices (see . Fig. 7.8 (right)).

. Fig. 7.8 A lune defined by two angles

114

Chapter 7  Constructible numbers

area CEDF D area CFDG  area CEDG area CEDG D area circular sector ACED  area triangle 4ACD 2˛ area ACED D R2 D ˛R2 2 2ˇ 2 r D ˇr 2 area BCFD D 2     1 1 area CEDF D ˇr 2  r 2 sin 2ˇ  ˛R2  r 2 sin 2˛ 2 2  1 r 2 sin 2ˇ  R2 sin 2˛ D ˇr 2  ˛R2  2 To simplify the calculations, we will assume that

7

ˇr 2 D ˛R2 :

(21)

Hippocrates used two assumptions to find constructible lunes: 1. Two circular sectors (with radius r and R) corresponding to the lunes’ arcs have the same area 2. The central angles (2˛; 2ˇ) of the two circular arcs are commensurable. These assumptions are equivalent to equation (21). It can be shown that these relations hold for all squarable lunes. The proof however goes beyond the scope of this book3. Exercise 1384 Show that Hippocrates’ assumptions are equivalent to equation (21). Express algebraically that 2˛ and 2ˇ, and hence also ˛ and ˇ, are commensurable. Let  be the greatest common denominator of 2˛ and 2ˇ. Express the areas of the two sectors in terms of  and deduce that r2 ˛ D 2: ˇ R

Using equation (21) we find that: area CEDF D

1 2 R sin 2˛  r 2 sin 2ˇ : 2

˛R2 and it follows that r2   1 ˛R2 area CEDF D S D R2 sin 2˛  r 2 sin 2 2 : 2 r

Now from equation (21) we also have that ˇ D

Let m D 3

(22)

R2 r2

See e.g. Rothe (2014), p. 696 ff. The proof uses algebraic numbers and some fundamental theorems of number theory. Algebraic numbers are real numbers which are the solution to a polynomial equation with integer coefficients. 4 Rothe (2014), p. 696 ff.

115 7.3  Squarable lunes

7

Now

) ) )

kC Gk D R sin ˛ D r sin ˇ R sin ˇ D sin ˛ r R R2 sin 2 ˛ D sin ˛ r r p sin m˛ D m sin ˛

(23)

we can rewrite equation (22) as   1 ˛R2 2 2 SD R sin 2˛  r sin 2 2 2 r ! 1 2 R2 ˛R2 D r sin 2˛  sin 2 2 2 r2 r 1 2 r .m sin 2˛  sin 2m˛/ 2 1 D r 2 .2m sin ˛ cos ˛  2 sin m˛ cos m˛/ 2   p p D r 2 m sin ˛ 1  sin2 ˛  sin m˛ 1  sin2 m˛   p p p D r 2 m sin ˛ 1  sin2 ˛  m sin ˛ 1  m sin2 ˛ p p  p p D r 2 m sin ˛ m 1  sin2 ˛  1  m sin2 ˛

D

(from (23)) (24)

To know whether the lune is squarable, we need to be able to determine whether sin ˛ is constructible, i.e. that we can draw a line segment with length sin ˛ using compass and straightedge methods. It is obvious that this will only be the case for certain values of m. A lune is squarable if for certain constructible values of m the equation sin m˛ D p m sin ˛ yields an expression for sin ˛ which can be constructed. If sin ˛ is constructible, then so is expression (24). It turns out that only five values for m will yield a constructible result5 . These numbers are 3 5 m D 2; 3; ; 5; . 2 2 p If m D 2 then sin 2˛ D 2 sin ˛, this is Hippocrates’ isosceles right triangle case. sin 2˛ D ) ) )

2 sin ˛ cos ˛ D

p p

2 sin ˛

2 sin ˛ p 2 _ sin ˛ D 0 cos ˛ D 2 p 2 sin ˛ D 2

Obviously sin ˛ D 0 is not a valid solution. 5

See Postnikov and Shenitzer (transl.) (2000).

Chapter 7  Constructible numbers

116

So sin 2˛ D

p

p 2 2 D 1; 2

the area then is S Dr

p 2

! p   2 p 2 1 2 2 2 2 0 Dr D R 2 2 2 p

If m D 3 then sin 3˛ D

7

p

3 sin ˛;

this the isosceles trapezoid case. We know sin 3˛ D 3 sin ˛  4 sin3 ˛ (see Exercise 133) So p sin 3˛ D 3 sin ˛ p ) 3 sin ˛  4 sin3 ˛ D 3 sin ˛ p ) 3  4 sin2 ˛ D 3 p ) 4 sin2 ˛ D 3  3 s s p p 3 3 3 3 _ sin ˛ D  ) sin ˛ D 4 4 q p 1 3 3 ) sin ˛ D 2 Obviously the sine cannot be negative in this case, so 1 0 v  u s p  p u q 3 3 3 C p 1 p Bp 3 3 t C S D r2 3 3 3B 3 1  1   A @ 2 4 4 ! p q p q q p 2 p p 3 3 1 D 3  3r 43C 3 49C3 3 2 2 2 p q q   q p 2 p p p 3 D 3  3r 3 1 C 3  5 C 3 3 4 p q q q  p 2 p p 3 D 3  3r 3 C 3 3  5 C 3 3 4

(25)

117 7.3  Squarable lunes

3 3 If m D then sin ˛ D 2 2 ˛ Put ˇ D then 2

r

7

3 sin ˛. 2 r

3 sin 2ˇ 2 r 3 2 sin ˇ cos ˇ D 2 p D 6 cos ˇ r 3 D 2 sin ˇ cos ˇ 2 p D 6 cos ˇ p q D 6 1  sin2 ˇ

sin 3ˇ D )

3 sin ˇ  4 sin3 ˇ

)

3  4 sin2 ˇ

)

3 sin ˇ  4 sin3 ˇ

)

3  4 sin2 ˇ

)

3  4 sin2 ˇ

)

9  24 sin ˇ C 16 sin ˇ D 6  6 sin ˇ

)

16 sin4 ˇ  18 sin2 ˇ C 3 D 0

2

4

(26)

2

p p 9 ˙ 33 18 ˙ 132 D sin ˇ D 32 16 q q p p 1 1 sin ˇ D 9  33 ^ cos ˇ D 7 C 33 4 4

)

2

)

which are constructible numbers. 2 2 Now equation (26) pimplies 3 4 sin ˇ > 0. Inserting the possible values for sin ˇ reveals 9 C 33 that if sin2 ˇ D then 3  4 sin2 ˇ < 0. 16 p 1p 9  33, from which Therefore we only retain sin ˇ D 4 q q q p p p 1 1 1 9  33  7 C 33 D 30 C 2 33: sin ˛ D 2 sin ˇ cos ˇ D 2  4 4 8 Also p ˛ 1  cos ˛ 1 C 33 sin ˇ D sin D ) cos ˛ D : 2 2 8 2

2

Exercise 139 p p p  p p p 1  Prove that 34  2 33 D 1 C 33 and that 38  6 33 D p 9  33 . 3 Exercise 140 Prove that the area of this lune is SD

p  1 2 r 3 C 33 32

q p 30 C 2 33:

118

Chapter 7  Constructible numbers

Exercise 141 Prove that cos 3˛ D 4 cos3 ˛  3 cos ˛ (see also Exercises 79 and 133). Use this result to prove that sin 5˛ D 16 sin5 ˛  20 sin3 ˛ C 5 sin ˛

(27)

Hint: sin 5˛ D sin .3˛ C 2˛/ D : : : p Use equation (27) to solve sin 5˛ D r5 sin ˛ for sin ˛ (this is the case m D 5). 5 5 5 sin ˛ for sin ˛ (this is the case m D ). Use equation (27) to solve sin ˛ D 3 3 3 ˛ Hint: put ˇ D 3 In both cases you will find a biquadratic equation, in sin ˛ and sin ˇ respectively.

7

In the last exercise you have identified the two other squarable lunes. Now you can also determine their area. Exercise 142 (For the extremely brave) In Exercise 79 and 133 you have expressed sin 3˛ in terms of sin ˛ and in Exercise 141 you have done the same for sin 5˛. Now use the fact that sin 45˛ D sin .5 .3 .3˛/// D sin .3 .3 .5˛/// to show that van Roomen’s x polynomial (see 7 Sect. 2.3) is equivalent to expressing sin 45˛ in terms of D sin ˛. 2 You can also use de Moivre’s theorem for complex numbers and expand the binomial with the binomial theorem.

7.4

Squaring the circle is different ...

The third classical Greek problem, squaring the circle, could not be solved using Wantzel’s methods. Algebraically squaring the circle is equivalent to solving the equation x 2   D 0. We know that a quadratic equation with constructible coefficients yields constructible solup tions. This equation therefore leads to an infernal circle: to prove that  and thus  is constructible, we need to know whether the constant term of the equation, , is constructible, which is exactly what we need to prove. It was suspected in the nineteenth century that  was not an algebraic number. Algebraic numbers are numbers which are the solution to a polynomial equation with integer coefficients. By definition, constructible numbers are also algebraic, but not all algebraic numbers are constructible.p For instance, 3 2 is an algebraic number, because it is the solution to the equation x 3 2 D 0, but as we have seen it is not a constructible number.  however is not the solution to such an equation. We say that  is transcendental. Although it was clear that transcendental numbers must exist, it was only in 1844 that Joseph Liouville (1809–1882) succeeded in constructing such a transcendental number, thereby identifying the first such number: 1 1 1 1 C 2Š C 3Š C : : : C nŠ C : : : D 0:11000100000000000000000100:: : 10 10 10 10

119 7.4  Squaring the circle is different . . .

7

He also tried, in vain, to prove that the numbers e and  are transcendental as well. It was Charles Hermite (1822–1901) who succeeded in proving that the number e was transcendental. Building on his work, Ferdinand Lindemann (1852–1939) was able to prove that p1 e x1 C p2 e x2 C : : : C pn e xn cannot equal zero if all xi and pi are distinct algebraic numbers. Now it is proven in complex analysis that e i  C 1 D 0. This implies that i is not algebraic. But we know that i is algebraic, because it is a solution to the equation x 2 C 1 D 0, therefore  is transcendental. Since  is not algebraic, it is a forteriori not constructible. As is often the case in mathematics, the solution to one problem leads to a multitude of other problems. E.g., is   algebraic or transcendental? Since the nineteenth century, we have learned that if a is an algebraic number, not equal to 0 or 1, and if b is an irrational algebraic 3 p  p p 2 3 number, then ab is transcendental6 . For instance, 2 2 or 2 are transcendental numbers. Because  is not algebraic, the question on the nature of   remains open. Why don’t you have a go at the proof?

6

Gelfond (1934).

121

8

The Cinderella of regular polygons Ad Meskens and Paul Tytgat © Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5_8

8.1

The inconstructibility of the heptagon

Among the regular polygons, the regular heptagon most certainly is Cinderella. Contrary to a triangle, a square, a pentagon or a hexagon, the heptagon is not constructible using compass and straightedge methods. In what follows, we will guide you through the proof of this nonconstructibility1 . Exercise 143 Consider . Fig. 8.1. Show that if †BAC D ˛, then: †ACB D ˛ †CBD D †CDB D 2˛ †DCE D †CED D 3˛ †EDF D †EFD D 4˛ and so on, if the initial angle allows it. Hint: this configuration is a generalisation of the configuration used for the trisection of an angle shown in . Fig. 5.8.

We call this configuration Viète’s ladder. Viète used this configuration to calculate the edge of a regular heptagon. Exercise 144 Show that in a regular n-gon, the internal angle in a vertex is equal to Let ABCDEF G be a regular heptagon with edge equal to 1. Determine †BAG and from this deduce a value for †BAC . Show that †BAC D †CAD D †DAE D †EAF D †FAG.

Draw the diagonals from A and let kAC k D kAF k D x kADk D kAEk D y 1

Inspired by Ostermann and Wanner (2012), p. 162–164.

.n  2/ rad. n

122

Chapter 8  The Cinderella of regular polygons

. Fig. 8.1 Viète’s ladder

Cut the pentagon ABCDE. Fold the figure along diagonal AD. Turn and fold along diagonal AC . You will find a figure as seen in . Fig. 8.2. The right figure is Viète’s Ladder2 . To make a distinction we have added an accent to the letters in the right figure. It is obvious that 4ADE Š 4A0 D 0 E 0 .

8

Exercise 145 Show that 4ACF  4C 0 B 0 D 0 and 4ADE  4D 0 E 0 C 0 . y 1 Deduce that kB 0 D 0 k D and kC 0 E 0 k D x y Use these results to show that y D1C

y x

and

yDxC

1 : y

(28)

Exercise 146 Show that, if you eliminate y from these equations, you obtain x 3  x 2  2x C 1 D 0:

(29)

Which equation do you obtain when you eliminate x from the system of equations? Use Lemma 2 in 7 Sect. 7.1 to identify the possible rational solutions to equation (29). Are these numbers in effect solutions to the equation? Which conclusion can you draw?

The line segments ŒAC  and ŒAD cannot be constructed using compass and straightedge. So the regular seven-pointed stars ACEGBDF and ADGCFBE cannot be constructed with the aid of only compass and straightedge. The edges of these stars are also the diagonals of the regular heptagon. This means that the regular heptagon ABCDEF G cannot be constructed using just compass and straightedge either. Should this be the case, then the seven-pointed stars could be constructed, which we have proved to be impossible. The conclusion seems strange. The edge with length 1 is constructible, so why cannot the heptagon be constructed? Because you need two parameters to construct a regular polygon, the length of the edge and the measure of the internal angle at a vertex.

2

A similar “folding” method was used by Arabic mathematician Ab¯u’l-J¯ud. In answer to a question posed by AlB¯ır¯un¯ı, he proved that the edge of a heptagon is not equal to half that of an equilateral triangle inscribed in the same circle. Heron of Alexandria had already stated that half the edge of an equilateral triangle inscribed in the same circle as the heptagon is a very good approximation for the edge of the heptagon. See Hogendijk (1987); Berggren (2007), p. 580ff.

123 8.1  The inconstructibility of the heptagon

8

. Fig. 8.2 Folding a regular heptagon into an equilateral triangle . Fig. 8.3 Archimedes’ pseudo neusis to determine 2 cos 7

Exercise 147  Show that cos is not constructible. 7

5  rad nor angle rad is con7 7 structible using compass and straightedge. On one leg we would be able to determine a line segment with length 1. We can then construct the perpendicular on the other leg. In doing so,   we have constructed a right-angled triangle with edges cos and sin . But we have shown 7 7   that cos cannot be constructed, so neither can rad. 7 7 From the latter assertion it follows that neither angle

Exercise 1483 The following pseudo-neusis method was attributed to Archimedes by the Arabs (see . Fig. 8.3). Create an IGS file in which you perform this operation. Draw a unit square ABCD, draw the diagonal ŒAC . Let E be a point on ŒAC . Draw line DE and determine the intersection points F and G of DE with ŒBC  and with AB respectively. Use the Polygon tool to define the triangles 4DEC and 4FBG. Move E along ŒAC  until area 4DEC D area 4FBG. Determine the length  kAGk and compare this length to 2 cos . 7  Knowing the value of cos allows us to construct a heptagon with an edge of given length. 7 3

Hartshorne (2000), p. 270, Martin (1998), p. 136.

Chapter 8  The Cinderella of regular polygons

124

8.2

The relation with the trisection

As unlikely as it seems, there is a connection between the construction of the heptagon and the trisection of an angle. Exercise 149

1 Consider the equation (29): x 3  x 2  2x C 1 D 0. Use Viète’s substitution x D t C (see 3 7 Sect. 3.3) and prove that the resulting equation is 7 7 D 0: t3  t C 3 27

(30)

We know that cos 3˛ D 4 cos3 ˛  3 cos ˛ (see Exercise 136). Putting z D cos ˛ we find that

8

3 1 z 3  z  cos 3˛ D 0 4 4

(31)

This is an equation of the same form as equation (30). If we put t D nz equation (30) becomes z3 

7 z 7 C D 0: 3 n2 27n3

(32)

We now want to determine n in such a way that the coefficients of equations (31) and (32) are equal. p 7 1 2 7 3 Then  2 D  ) n D and 3n 4 3 1 7  cos 3˛ D 4 27n3 )

47  27

1 p !3 2 7 3 p 7 1 D p D 14 2 7

cos 3˛ D 

We know the value of cos 3˛, using the result of Exercise 135 we know that an angle 3˛ can be constructed. To find z D cos ˛, we need to determine ˛, which means we have to trisect 3˛. We can trisect 3˛ by any of the constructions described in 7 Chapter 5. Obviously it is this step which prevents a compass and straightedge construction. Once we know z, we can backtrack to t and x again using compass and straightedge methods.

125 8.3  A neusis for the heptagon

8

. Fig. 8.4 A neusis for the heptagon

8.3

A neusis for the heptagon

 rad using neusis. Try performing this neusis in an IGS 7 file (see . Fig. 8.4). Draw a square ABCD with the length of the edge equal to the length of the edge of the regular heptagon. Draw the diagonal ŒBD. Draw an arc of a circle C centered at B and with radius kBDk. Define the perpendicular bisector m on ŒDC . Choose a point E on m. Draw the straight line AE. Define a circle centered at E and with radius kABk. Determine the intersection points F and G of the circle and the straight line AE. Determine the locus of F with E. Which kind of curve is this? Let H be the intersection point of the locus and the circle. Let the line segment ŒEF  D ŒEH  in this position be ŒEN FN . ŒEN FN  is the edge of the regular heptagon, in the desired orientation. Define the perpendicular bisector of ŒEN FN  and determine its intersection point O with straight line m. Point O is the centre of the circumscribed circle. Draw this circle. Draw the circle centered at EN and with radius kEN FN k .D kABk/. Determine the intersection

It is possible to construct an angle of

126

Chapter 8  The Cinderella of regular polygons

5 rad 7 (alternatively, you can mirror ŒEN FN  about m, the image of FN is I ). It is now easy to determine the other points of the heptagon, as we know the circumscribed circle. Repeat the procedure, but instead of choosing E on the straight line, choose E on the circle C . Which kind of curve do you obtain as locus of F with E?

point I of this circle and the circumscribed circle. Angle ]FN EN I has a magnitude

Exercise 150 Use the method of Exercise 144 in 7 Sect. 8.1 to show that the enneagon (9-gon or nonagon) is not constructible. Why is it not possible to decide whether the hendecagon (11-gon) is constructible using this method? Can you determine the lengths of the diagonals in these regular polygons?

8

127

Servicepart Solutions – 128 References – 178 Index – 184

© Springer International Publishing Switzerland 2017 A. Meskens, P. Tytgat, Exploring Classical Greek Construction Problems with Interactive Geometry Software, Compact Textbooks in Mathematics, DOI 10.1007/978-3-319-42863-5

128

Solutions Exercise 1

2  4p 2 q 2 C p 4  2p 2 q 2 C q 4 D p 4 C 2p 2 q 2 C q 4 D p 2 C q 2 : For the second part of the exercise put p D n C 1 and q D n. Exercise 2 Let .n  1; n; n C 1/ be a Pythagorean triplet. Then

.n  1/2 C n2 D .n C 1/2

)

n2  4n D 0

)

nD0_nD4

Only for n D 4 do we have three consecutive positive integers: .3; 4; 5/. Exercise 3 Let the sides of the isosceles triangle be given by the triplet .a; a; b/ 2 N 3 , then

a2 C a2 D b 2 , 2a2 D b 2

)

p

2a D b:

Therefore, if a 2 N then b … N. Exercise 4 In the equation 178212 C 184112 D 192212 , it is clear that the left-hand side is

uneven (because 1782 is even and 1841 is uneven) and the right-hand side is even (because 1922 is even). So the equation cannot hold. In the equation 398712 C 436512 D 447212 both 3987 and 4365 are divisible by 7, and hence 398712, 436512 and 398712 C 436512 are divisible by 7, while 4472 is not and therefore neither is 447212. Exercise 5 We divide the cubic close packing into unit cells, such as the one in the figure. We

see that a unit cube consists of half a sphere at each face and one eight of a sphere at each vertex.

1 1 The unit cube therefore consists of 6  C 8  spheres or the equivalent of 4 spheres. We 2 8 now try to determine how much space they take up relative to the unit cube. Suppose the edge

129 Solutions

of the sphere has length a and the radius of the spheres is r. We see that the length of apface 2a, diagonal of the cube equals four radii of the spheres. The face diagonal ŒAC  has length p p 2a . The because it is the hypotenuse of the triangle 4ABC . Therefore 4r D 2a or r D 4 volume of one sphere then is p !3 p p 2a 2 3 4 3 4 4 2 2 3 Vs D r D  a : D  3 a D 3 3 4 3 4 24 The volume of the cube of course is Vc D a3 . The volume taken up by the spheres relative to the cubes is p 2 3 p 4 a 2 4Vs 24 Vrel D D D  0:74: Vc a3 6 This means that 74% of the volume of the cube is filled with spheres. Exercise 8

Let C be the oak tree and P and Q the birches. For P there are two points which satisfy conditions P1 and P2 , for Q these are Q1 , and Q2 . Let M11 be the midpoint of P1 and Q1 , M12 of P1 and Q2 , M21 of P2 and Q1 and M22 of P2 and Q2 . We notice that as we move C around M11 and M22 remain stationary, while M12 and M21 move about. So the property only holds if one angle is made clockwise, the other counterclockwise. If both right angles are made in the same direction the property does not hold. We assume the pirate is clever enough to choose the solution which is drawn in the righthand figure. Select the x-axis through Q and P and the y-axis, perpendicular to x, through C . It is easy to prove that 4COP Š 4PBP1 and 4COQ Š 4QOQ 1 . Let  the coordinates of C , Q and P be as shown in the figure, then P1 p C cy ; p  cx and Q1 q  cy ; q C cx . Therefore     p C cy  q  cy p  cx C q C cx pq pCq ; or M11 ; ; M11 2 2 2 2 which is independent of cx and of cy . Therefore M11 will remain stationary. The proof for M22 is analogous.  The coordinates for P2 are P2 p  cy ; cx  p yielding   p  q  2cy q  p C 2cx M21 ; 2 2 which is neither independent of cx nor of cy and therefore will not remain stationary when C is moved around. An analogous calculation can be made for M12 .

130

Solutions

Exercise 9

d d t AC W y  d D  xI lC W y D  .x  a/I lA W y D .x  t /: t t d The coordinates of D are found by determining the intersection of lA and lC . We can solve the d cartesian equation of lC for t to find that t D  .x  a/. Inserting this value in the cartesian y equation of lA gives the desired equation. Exercise 12 (a) Multiply both sides of the equation with r to obtain r 2 D 2r sin  . Using the

transformation formulae this becomes: x 2 C y 2 D 2y

,

x 2 C y 2  2y D 0

,

x 2 C .y  1/2 D 1:

,

x 2 C y 2  2y C 1 D 1

Which is the cartesian equation of a circle centered at .0; 1/ and with radius 1. (b) Again multiply both sides of the equation with r to obtain r 2 D 2r cos  . This now gives us the cartesian equation .x  1/2 C y 2 D 1, which is the equation of a circle centered at .1; 0/ and with radius 1. (c) The first equation is the equation of a circle centered at .0; a/ and with radius a, the second of a circle centered at .a; 0/ with radius a. Exercise 13 These are all limaçons. Depending on the value of a you find a curve which has an

inner loop, or one which is bulging in. The curves with equations r D 2 C a cos  and r D 2  a cos  are each others mirror image with respect to the straight line perpendicular to the polar axis and through the pole. For those who want to prove this: consider r1 D 2 C a cos 1 and r2 D 2  a cos 2 ; in which we have introduced indices to be able to make a distinction, then: 8 <x2 D .2  a cos 2 / cos 2 :y D .2  a cos  / sin  : 2 2 2 Now put 2 D   3 , 3 is nothing more than the supplementary angle of 2 . Then x2 D .2  a cos .  3 // cos .  3 / D 2 cos .  3 /  a cos2 .  3 / D 2 cos 3  a cos2 3 D .2 C a cos 3 / cos 3 : This last expression is of the same kind as x1 D r1 .1 / cos 1 whence x2 D x1 .

131 Solutions

Exercise 14 The straight line with cartesian equation x D a has polar equation r cos  D a,

a or r D a sec  . Analogously the line with cartesian equation cos  y D b has polar equation r D b csc  . y D mx represents a straight line through the origin. Using the transformation formulae we find

this is often written as r D

r sin  D mr cos  ) sin  D m cos  ) m D tan  )  D arctan m i  h yielding the equation  D 0 . for some value 0 in  ; 2 2 In general the line with cartesian equation y D mx C q has polar equation r sin  D mr cos  C q

)

r .sin   m cos  / D q

)

rD

q : sin   m cos 

x2 . A smaller magnitude of a 2a gives the graph a more closed appearance, sharply curved at the vertex.

Exercise 15 The equation of the parabola is x 2 D 2ay or y D

Exercise 16 A point A.x; y/ is on the parabola P if the distance of A to d is equal to the

distance of A to F . Or: A2P

, , ,

r ˇ p ˇˇ p 2 ˇ C y2 x ˇx C ˇ D 2 2

d.A; d / D d.A; F / , ˇ p ˇˇ2  p 2 ˇ C y2 ˇx C ˇ D x  2 2 p2 p2 D x 2  px C C y2 x 2 C xp C 4 4

,

2px D y 2

Exercise 17 Let circles C and C1 be tangent in T , then the tangent line t in T to C is also the tangent line in T to C1 . Now OT ?t and O1 T ?t , so OT k O1 T , so OT and O1 T coincide.

Therefore kO1 Ok D r1 C r.

Exercises 18 and 19 If we denote an external tangency to the first circle as 1E, and an internal

tangency as 1I, we find the following possibilities: 1E2E3E;

1E2E3I;

1E2I3E;

1E2I3I;

1I2E3E;

1I2E3I;

Some configurations of the initial circles do not allow all 8 solutions.

1I2I3E;

1I2I3I:

132

Solutions

Exercise 21

It is obvious from the figure that 8 ˆ x D .a  b/ cos  C b cos  ˆ ˆ < D a cos  ˆ ˆ ˆ :y D b sin  which are the parameter equations of an ellipse with major axis a and minor axis b. 4AOB is a right-angled triangle. In a right-angled triangle the midpoint of the hypotenuse, here ŒAB, is also the centre of the circumscribed circle. Therefore O is on the circle for which ŒAB is a diameter. Obviously M is the midpoint of both ŒOD and ŒAB and ]AOB is a right angle. Therefore AOBD is a rectangle. If A .x; 0/ and B .0; y/ then D .x; y/. Whence 8 < x D l cos ˛ :y D l sin ˛ which reduces to x 2 C y 2 D l 2 , which is the equation of a circle centered at O and with radius l. Exercise 22 Select a point A on the y-axis, draw a circle centered at A and with radius a  b

and determine the intersection B with the x-axis. On the ray ŒAB determine the point C at a distance b of B and draw the locus of C with A. Exercise 23 Put A .0; 0/, †DAB D ˛ and E .x; y/. 4BAD is isosceles, so †BDA D

†DAB D ˛. It is obvious that B .a cos ˛; a sin ˛/. Because D is the mirror image about the perpendicular on the x-axis and through B we find that D .2a cos ˛; 0/. The coordinates for E therefore are: 8 8 x ˆ
whence  2  y 2 x C  D1  b C 2a b

,

x2 .b C 2a/

2

C

y2 D 1: b2

133 Solutions

Exercise 24 Draw a circle centered at the origin A and with radius a (say 3), choose a point D

on the x-axis and draw a circle centered at D and with radius a. Determine the intersection points of the circles, B and C . Draw the ray ŒBD , draw a circle centered at D and with radius b and determine the intersection point E with the ray ŒBD . Now determine the locus of E with A.

Exercises 25 and 26 You can find 2 to 6 configurations for which the line can be fitted in. If

pole P is on the straight line a, there is a degenerate solution if the length of the chord which is subtended equals 3 (D r). In the figures below, we set out a few possible configurations with the intersections of the locus with the circle.

Exercise 27

134

Solutions

In general we find two curves. When the pole of the neusis is on the circle, we see that both loci merge into one curve, Pascal’s limaçon (figure under left). Exercises 28 and 29 Let the second intersection of the diameter through P and the x-axis be Q.

We know 4PBQ is a right-angled triangle, with right angle in B (Thales’ circle theorem). Let the angle of ŒPB with the x-axis be  . Therefore kPBk D kPQk cos  D 2R cos  D 10 cos : For the locus of D with B, we find that r D 10 cos   a, and for the locus of E with B that r D 10 cos  C a. These are the polar equations of limaçons.

Exercise 30 In the proof we have never used the property that the triangles are equilateral. We did use the fact that kBD1 k D kBD2 k. For this assertion it suffices that we have congruent isosceles triangles. Exercise 31 We follow a procedure which is similar to that of the construction of a parallel

straight line to a given straight line. Draw a circle C centered at P and of a chosen radius, larger than d.P; AB/. Let the intersections of the circle and AB be C1 and C2 respectively. Draw a circle centered at C2 and with radius kC1 C2 k. Let the second intersection of this circle and AB be C3 . Draw circles with radii kP C1 k and centered at C2 and C3 respectively. Let the intersection of both circles (on the same side of AB as P ) be P1 . PP1 is parallel to AB. The proof runs as follows: Let the magnitude of a straight angle be a. It is clear that 4P C1 C2 Š 4P1 C2 C3 , moreover because of the construction these are equilateral triangles, so †P C1 C2 D †P C2 C1 D †P1 C2 C3 D †P1 C3 C2 :

(1)

We know from the construction that C1; C2 and C3 are collinear, so †P C2 C1 C †P C2 P1 C †P1 C2 C3 D a:

(2)

In triangle 4C1 P C2 we have that †C2 C1 P C †C1 P C2 C †P C2 C1 D a:

(3)

From (1), (2) and (3) we have †P C2 C1 D †C1 P C2 . From the SAS property 4P C1 C2 Š 4P C2 P1 , so †P C2 C1 D †C2 PP1 , which proves the construction. An alternative construction: Draw a circle C centered at P and of a chosen radius, larger than d.P; AB/. Let the intersection of this circle and AB be C1 and C2 . Draw a circle with the same compass width and the compass point in C1 . This circle intersects AB at C3 . With the same compass width and the compass point in C3 , draw an arc which intersects the first circle C , on the same side of AB as P , at Q. Then PQ is parallel to AB. Find the proof yourself. Hint: kP C1 k D kC1 C3 k D kC3 Qk D kPQk from which 4P C1 C3 Š 4P C3 Q. Exercise 32 From the construction it is immediately clear that 4ACB Š 4CDE, therefore

†BAC D †ECD. If two straight lines are intersected by a third straight line and the corresponding angles of intersection with the transversal are congruent, then the two lines are parallel. Therefore AB k CE.

135 Solutions

Exercise 33 Draw AP and draw a circle C1 centered at A and with radius kAP k. C1 intersects AB at C . Draw a circle C2 centered at P and with radius kAP k. C2 intersects AP a second time at D. Draw a circle C3 centered at D and with radius kP C k. C3 intersects C2 in Q. Then

PQ k AB.

Exercise 34 Let ŒAB denote the line segment. Draw a circle C1 centered at A and with radius kABk and a circle C2 centered at B and with radius kABk. The two circles intersect each

other at two points C1 and C2 . Draw the straight line C1 C2 . This line intersects ŒAB at the midpoint, and is therefore called the perpendicular bisector of ŒAB. The distance of any point P of this line to A always equals the distance of P to B. We now prove the construction. We know that the triangles 4ABC1 and 4ABC2 are isosceles, so kAC1 k D kBC1 k D kAC2 k D kBC2 k. Construct the bisectrix of †AC1 B in C1 (this is the line C1 C2 – check this!) and let D be the intersection of the bisectrix and ŒAB. Now consider the triangles 4AC1 D and 4BC1 D. We know that kAC1 k D kBC1 k and both triangles have C1 D in common as an edge. Moreover because C1 D is the bisectrix †AC1 D D †BC1 D. From the congruence property SAS we can conclude that 4AC1 D Š 4BC1 D. In other words kADk D kDBk, so D is the midpoint of ŒAB. To prove that C1 C2 ?AB we can easily see that †C1 AC2 D †C1 BC2 and †AC1 B D †AC2 B, from which we conclude that AC1 BC2 is a rhombus. In a rhombus, the diagonals intersect at right angles, so AB?C1 C2 . Exercise 35 Place the point of the compass in C . Open the compass at will. With this opening

draw a circle, which intersects the straight line AB at A1 and B1 . Obviously C is the midpoint of ŒA1 B1 . Construct the perpendicular bisector on ŒA1 B1 . This line passes through C and is perpendicular to AB. The proof of this construction is a corollary of the proof in Exercise 34. Exercise 36 Place the point of the compass in D, choose an opening of the compass in such a way that the circle will intersect AB. Let the points of intersection of the circle and AB be A1 and B1 respectively. Then kDA1 k D kDB1 k D R, so D is on the perpendicular bisector of ŒA1 B1 . Construct this perpendicular bisector. The proof of this construction is a corollary of the proof in Exercise 34. Exercise 37 Construct l through P , such that l?AB, construct m through P such that m?l, then m k AB. Exercise 38 We know that 4AD1 P  4AD3 B and 3 kAD1 k D kAD3 k. From the similarity

of the triangles it follows that kAD3 k kAD1 k D kAP k kABk

and

3 kAD1 k kAD1 k D ; kAP k kABk

which proves the equality. The other equalities are corollaries. Exercise 39 Consider an angle ]ABC . Place the compass point in B and describe an arc of

a circle with a chosen radius. The arc intersects the straight lines AB and BC at D and E respectively. With the same compass width draw circles centered at D and E respectively. Let

136

Solutions

the second intersection of these circles be F (the first one is B). Then 4BDF Š 4BEF (SSS) and †DBF D †FBE. BF bisects ]ABC . Exercise 40

x2 The locus of L with B is part of a parabola with equation x 2 D ay or y D  . A smaller a magnitude of a gives the graph a more closed appearance, sharply curved at the vertex. Exercise 41 Put

kABk D kCDk D l

kAC k D kBDk D b;

and

then kAF k D kF C k D kAEk D

b 2

and

kEBk D l 

b : 2

In the right-angled triangle 4FAB we have  2 b b2 2 2 2 2 D l2 C : kBF k D kFAk C kABk D l C 2 4 Since point G is on a circle centered at B and with a radius kBF k we find that s b2 kBGk D l 2 C : 4 In the right-angled triangle 4GEB we have   b2 b2 b2 b 2 D l2 C  l  l 2 C lb  D lb: kGEk D kBGk  kEBk D l C 4 2 4 4 2

2

2

2

We see that the square of p kGEk equals the area of the rectangle, so GE is the edge of the square we are looking for. lb is sometimes called the geometric mean of l and b. Exercise 42 There are two pairs of solutions. Let CA and CB be the given circles and let rB < rA . Construct the circle CAB centered at A

and with radius rA  rB . Determine the midpoint M between B and A. Draw an arc of a circle

CM centered at M and with radius kMAk. Let C be one of the intersections of this arc with

137 Solutions

the circle CAB . Draw the straight line AC , which intersects CA at D. Draw the line BC . BC is at right angles with the radius CA, because of Thales’ circle theorem in CM . Therefore BC is tangent to CAB . Construct the parallel line t to BC through D. Repeat the procedure with the other intersection of the arc of the circle CM with the circle CAB .

To find the other pair of tangents, consider the circle CAB centered at A and with radius rA CrB and repeat the above procedure. Exercise 43 In both cases you will notice that the radius of the tangent circle increases or diminishes with the same amount as the radii of the circles CA , CB , CC .

a x y D D is shorthand for two equations: x y 2a a x x y D and D x y y 2a a y ( D can also be used as an alternative for one of these equations). It is obvious that these x 2a equations can be transformed into x 2 D ay and y 2 D 2ax. The two equations x 2 D ay and y 2 D 2ax represent parabolae, with vertex in the origin, but with the y- and x-axis as axes of symmetry. We find the intersection by solving the system of equations 8 2  2 2 < x 2 D ay ) x D y x ) D 2ax: a : 2 a y D 2ax p 3 Solving this last equation pfor xpyields  x D 2a. The intersection points of the parabolae 3 3 therefore are .0; 0/ and 2a; 22 a . Exercise 44

a x y D D is shorthand for a system of two equations, e.g. x y b a x x y D and D or x y y b a x a y D and D : x y x b

Exercise 45

The first system yields equations (10) which represent two parabolae, the second system yields equations (11), which represent a parabola and an (orthogonal) hyperbola.

138

Solutions

Exercise 46 4LED; 4GEL and 4AEG are all right-angled triangles, with right angle in E.

Moreover †ELD D †EGL D †GAE

and †EDL D †GLE D †AGE:

Therefore 4LED  4GEL  4AEG and kELk kGEk kEDk D D kELk kEGk kEAk

or

y x 2a D D ; y x a

which can be rewritten as equations (8), which were solved in Exercise 44. Exercise 47 The equation of an orthogonal hyperbola having the coordinate axes as asymptotes is xy D k. The equation of the orthogonal hyperbola passing through C therefore is xy D ab. The equation of the circle circumscribed about the rectangle is    a 2 b 2 a2 C b 2 x C y D : 2 2 4

Therefore we have to solve the system of equations 8 <x 2  ax C y 2  by D 0 :

xy D ab

Which yields the equation: a2 b 2 ab b D0 x2 x ) x 4  ax 3  ab 2 x C a2 b 2 D 0  3 ) x  ab 2 .x  a/ D 0 p 3 ) x D ab 2 _ x D a p  p 3 3 Therefore the intersections are .a; b/ and ab 2 ; a2 b . The latter coordinates are the mean proportionals between a and b. p p  3 Choosing a D 1, b D 2 we find 22 ; 3 2 . x 2  ax C

Exercise 48 When the slats are in their final position, we see three similar right-angled triangles

4AOB; 4BOC and 4COD. Then kOAk kOBk kOC k D D kOBk kOC k kODk

or

a kOBk kOC k D D ; 2a kOBk kOC k

these are the same equations as equations (8), which were solved in Exercise 44. From these, after some arithmetical manipulations, we find kOBk3 D 2a3 , which is what we had to prove.

139 Solutions

Exercise 49 Repeat the procedure of the previous exercise with kODk D na. Exercise 50 We use equation (1) (see 7 Sect. 1.2) and put c D 2b, then

y 3 C .x  b/xy  2b.x  b/2 D 0: The y-axis is intersected for x D 0, so y 3 C 0  2b.0  b/2 D 0

,

y 3  2b 3 D 0

,

yD

p 3

2b:

The configuration of the straight lines when D is on the y-axis is the same as those of the slats of the winegauger’s instrument. Exercise 51 Again, we turn to use the properties of similar triangles. We know that

4BAF  4EDB  4GDE

)

kBAk kEDk kGDk D D kAF k kDBk kDEk

from which kDEk2 D kDBk  kGDk )

kDBk2  kBAk2 kAF k2

and

kAF k  kEDk D kBAk  kDBk

D kDBk  kGDk

)

Now kDBk D 2 kBAk

and

kGDk D kAF k

from which 2 kBAk3 D kAF k3

)

p 3

2 kBAk D kAF k :

Exercise 52

The locus of H with G is a curve with equation y 3 D  .x C 2a/ .xy  a .x C 2a// :

kDBk  kBAk2 D kAF k2  kGDk :

140

Solutions

Exercise 54

]AP C and ]BPD are vertically opposite angles, therefore †AP C D †BPD. ]CAB is subtended by the circular arc CB. ]CDB is subtended by the circular arc CB. Therefore †CAB D †CDB D †PDB. 4AP C and 4BPD have two congruent angles therefore 4AP C  4BPD. In similar triangles the ratios of the corresponding sides are equal, so kPDk kPAk D kP C k kPBk )

kPAk  kPBk D kP C k  kPDk

Exercise 55

Obviously †APD D †BPD. ]ADC is subtended by the circular arc AC . ]ABC is subtended by the circular arc AC . Therefore †ADC D †ABC . 4PAD and 4P CB have two congruent angles, therefore 4PAD  4P CB. In similar triangles, the ratios of the corresponding sides are equal, so kPDk kPAk D kP C k kPBk ) kPAk  kPBk D kP C k  kPDk We now consider a ray through the centre O, which intersects the circle at E and F respectively.

141 Solutions

Now kPEk D kOP k  R and kPF k D kOP k C R. We know that for any intersection points A and B of a straight line and the circle kPAk  kPBk D kPEk  kPF k D .kOP k  r/  .kOP k C r/ D kOP k2  R2 Alternatively: kPAk  kPBk D kP T k2 D kOP k2  kOT k2 D kOP k2  R2 The formula is not valid if P is in the inside of the circle, however then kPEk  kPF k D R2  kOP k2 (proof as an exercise). Exercise 57

x .x C k/ D y .y C 2k/ , , ,

,

x C kx C k 2   k 2 3k 2 xC C 2 4   k 2 .y C k/2  x C 2 2  k xC .y C k/2 2  3k 2 3k 2 4 4 2

D y 2 C 2ky C k 2 D .y C k/2 D

3k 2 4

D 1

142

Solutions

The of this hyperbola with the hyperbola xy D 2k 2 are .k; 2k/ and p intersections  p 3 3 22 k; 2k . Exercise 58 The polar equation of the circle C is r D 2a sin  (see Exercise 11), the equation

of the line t is r D 2a csc  , the line OP is given by  D  for some value of i straight  h 2  ; . Thus in polar coordinates, we find C .2a sin ; / and B .2a csc ; /. Hence 2 2 kCBk .D kOP k/ D 2a.csc   sin /: Let kOP k D r and  D  and the polar equation of the cissoid becomes r D 2a.csc   sin  /: y3 Using the transformation formulae, we find the cartesian equation x 2 D : 2a  y   1  sin  r D 2a.csc   sin  / ) r D 2a sin  ) )

r 3 sin  D 2ar 2  2ar 2 sin2   2  x C y 2 y D 2a x 2 C y 2  2ay 2 ;

which yields the desired result. Substituting x D 4a  2y into the equation of the cissoid yields: y3 .4a  2y/ D 2a  y 2

)

4 .2a  y/ D y 3

3

)

p 2 3 4a yD p : 1C 3 4

And xD

4a p : 3 4

1C

The slope of OQ is p 2 3 4a p p 3 4 1 1C 3 4 mOQ D D D p 3 4a 2 2 p 1C 3 4 1 and its cartesian equation y D p x. It immediately follows that the coordinates of R, the 3 2   p intersection of OQ and t are 2a 3 2; 2a . kTRk is equal to the horizontal distance to the y-axis and thus to the absolute value of the x-coordinate of R. Exercise 59 Draw a line t with equation y D 2a, now draw the point A.0; a/ and the circle C centered at A and with radius a. Choose a point B on t and draw the straight line OB. Determine the instersection point C of OB and C . Draw the circle C2 centered at O and with radius kCBk. Determine the intersection point P of C2 and OB. The locus of P with B is the

cissoid of Diocles.

143 Solutions

Exercise 60

Let R be the midpoint of ŒPQ. Let †POC D ˛, then †PQB D ˛ (alternate angles). Let †OP C D ˇ, then †BPQ D ˇ (opposing angles). 4BRQ is isosceles, in a rightangled triangle the length of the median on the hypotenuse is equal to half the length of the hypotenuse. Therefore †QBR D ˛ (equal base angles in an isosceles triangle) and †BRP D 2˛ (exterior angle is equal to the sum of the opposing interior angles). Furthermore 4BRP and 4BOR are isosceles. Now †BOC D †BOP C †POC D †BOR C †POC D 2˛ C ˛ D 3˛: Therefore the line OPQ is the trisectrix of ]BOC . Exercise 62  is the angle ]L0 B 0 G 0 . Depending on the position of the slat O 0 B 0 †O 0 B 0 C 0 D 

or †O 0 B 0 C 0 D    . In the first case kO 0 C 0 k D kO 0 B 0 k sin ;

in the second kO 0 C 0 k D kO 0 B 0 k sin .   / D kO 0 B 0 k sin : To find the polar equation we have to determine the length r D kO 0 P k in which P is a point of the conchoid. It is clear that kO 0 P k D kO 0 B 0 k C kB 0 P k D

c C b: sin 

Exercise 63 Use the transformation formulae

8 < x D r cos  :y D r sin 

which yields .r sin   c/2 r 2 D b 2 r 2 sin2 

,

.r sin   c/2 D b 2 sin2 

,

r sin   c D ˙b sin 

,

rD

c ˙b sin 

.if sin  ¤ 0/:

144

Solutions

Exercise 64 This branch (as well as the other) is defined as the locus of a neusis with pole A and the x-axis as the directrix. For each point on C on the x-axis, there is a point on the straight line AC for which kEC k D 2 kABk. Draw the circle centered at A and with radius 2 kABk. The circle intersects the x-axis at C1 and C2 . For these points, the point on AC at a distance 2 kABk is A itself. Therefore A is twice on the conchoid. Exercise 65 Because kDC k D kCOk D kOBk D r, 4DCO and 4COB are isosceles. From

which †ODC D †COD and †OCB D †CBO. Now the magnitude of an outer angle of a triangle equals the sum of both opposing inner angles, from which †OCB D †ODC C †COD

and

†AOB D †ODC C †CBO D †ODC C †OCB D †ODC C †ODC C †COD D 3†ODC D 3†ADB: Exercises 66 and 67 If you want to trisect an angle  you will find angles with magnitude

  2  4 ; C ; C rad in the intersections. Multiply these angles by 3 and reduce them to 3 3 3 3 3 Œ0; 2 and you will find  .

Exercise 68 On the x-axis, construct a perpendicular through O. For the construction, we

confine ourselves to the first quadrant, the constructions in the other quadrants are analogous. Construct the bisector of a right angle. Now bisect the angles defined by the bisector and the x-axis and the perpendicular respectively. Exercise 71 If you use Archimedes’ spiral to trisect an angle  you will find angles with mag-

  2  4 ; C ; C rad, using the first, second and third turn respectively. Multiply 3 3 3 3 3 these angles by 3 and reduce them to Œ0; 2 and you will find  .

nitude

Exercise 72

All shaded triangles in the figures have the same area, which proves Gregory’s generalisation. The method used here is the same method as the one Euclid used when proving Pythagoras’ theorem. Consider the triangles in figures (a) and (b). They have the same area because they have the same base ŒAB (the edge of the square) and they have the same altitude kDEk. The triangles in figures (b) and (c) are congruent: kDAk D kABk, kALk D kAC k, †DAC D †BAL (SAS-property). The triangles in figures (c) and (d) have the same area because they have the same base ŒAL and the same altitude kLKk.

145 Solutions

Now repeat this procedure for the rectangles in the other shades of gray. Exercise 74

kBEk D kEOk so 4BED is isosceles. †EOB D †EBO, now ]EOB is a central angle and ]FBG is an inscribed angle. _

_

†FOG D 2†FBG D 2†EOB ) F G D 2AB. _ _ _ _ Now AB D GC and F G D 2GC . Therefore †CAF D 3†CAG. Exercise 75 The quadratrix is defined as the curve for which

†BAX kXX 0 k : D †BAD kDAk Put kDAk D R, we know †DAB D

 . If †XAB D  and kAX k D r then 2

kXX 0 k D r sin : The equation becomes  r sin  D ; =2 R which yields the desired equation. Exercise 75 Draw a circle with radius R (e.g. 5). Draw a point D 0 on the y-axis and draw the

 y kAD 0 k  D it follows that  D , therefore the slope =2 R 2R   0 kAD k  and its equation is given by y D mx. of the line AF is given by m D tan 2R 0 0 Determine the intersection point X of D C and AF . Determine the locus of X with D 0 , this locus is the quadratrix. horizontal line D 0 C 0 . Since

146

Solutions

Exercise 78

The names of the points are indicated on the figure. Draw the radius ŒMD. We know that the tangent segments ŒOD and ŒON  have the same length because 4ONM Š 4ODM . Also 4ONA Š 4ONM . This implies that †AON D †NOM D †MOD, from which †AOD D †AOB D 3†AON: Exercise 79

sin 3˛ D sin .˛ C 2˛/ D sin ˛ cos 2˛ C cos ˛ sin 2˛ D : : : D 3 sin ˛  4 sin3 ˛ cos 3˛ D cos .˛ C 2˛/ D cos ˛ cos 2˛  sin ˛ sin 2˛ D : : : D 4 cos3 ˛  3 cos ˛ sin 3˛ 3 sin ˛  4 sin3 ˛ D cos 3˛ 4 cos3 ˛  3 cos ˛ 3 sin ˛  4 sin3 ˛ sin ˛ 3 3  4 tan3 ˛ 3˛ cos cos ˛ D D 1 4 cos3 ˛  3 cos ˛ 3 2 C 4 3 cos ˛ cos ˛ 1 sin ˛    4 tan3 ˛ 3 3 tan ˛ 1 C tan2 ˛  4 tan3 ˛ 2˛ cos ˛ cos D D 1 3 .1 C tan2 ˛/ C 4 3  C 4 2 cos ˛ 3 tan ˛  tan3 ˛ D 1  3 tan2 ˛

tan 3˛ D

Exercise 80

147 Solutions

From the left figure it is obvious that, using the sine rule, r a D sin .  3 / sin 2

)

r a D sin 3 sin 2

which, using the results of Exercise 79, yields:

)

r a D 3 2 sin  cos  3 sin   4 sin  r a D 2 cos  3  4 sin2 

! 3 4 sin2   cos  cos  !  4 1  cos2  3  cos  cos    1 4 cos   cos 

)

a rD 2

)

rD

)

a 2 a r D .4 cos   sec  / 2

)

rD

a 2

To find the cartesian equation, we use the transformation formulae   a 1 rD 4 cos   2 cos   a r  ) r2 D 4r cos   2 cos    r2 a ) r2 D 4r cos   2 r cos    2 x C y2 a 2 2 4x  ) x Cy D 2 x  2   ) 2x x C y 2 D a 4x 2  x 2 C y 2   ) 2x x 2 C y 2 D a 3x 2  y 2 Exercise 81 See figure in Exercise 80 right.

The equation of the straight line through O is y D mx, in which m D tan  . The equation of the straight line through A.a; 0/ is y D n.x  a/, in which n D tan 3 D

3m  m3 1  3m2

(see Exercise 79). The intersection of these lines is given by the system: 8
148

Solutions

Eliminating m from these equations yields: y  y 3 3  y D x  x 2 .x  a/ y 13 x 3yx 2 y 3  3 3 x .x  a/ ) y D x3 x  3y 2 x x3 2 3yx  y 3 ) yD 3 .x  a/ x  3y 2 x   3 ) y x  3y 2 x D 3yx 2  y 3 .x  a/   ) yx x 2  3y 2 D y 3x 2  y 2 .x  a/ ) )

2x 3 C 2xy 2  3x 2 a C y 2 a D 0   2x x 2 C y 2 D a 3x 2  y 2

The latter equation is the equation of MacLaurin’s trisectrix. Exercise 82

4OBD and 4BDP are isosceles. The configuration is the same as in the proof of Archimedes’ trisection (see proof of Exercise 65).

) ) )

r R D sin .  3 / sin  R sin 3 rD sin  3 sin   4 sin3  r DR sin   r D R 3  4 sin2 

To find the cartesian equation, we use the transformation formulae  r D R 3  4 sin2   r 3 D R 3r 2  4r 2 sin2   2   p x C y2 x 2 C y 2 D R 3 x 2 C y 2  4y 2   2 p x 2 C y 2 D R 3x 2  y 2 x C y2 3  2  2 x C y 2 D R2 3x 2  y 2

149 Solutions

Exercise 83

Draw circles C1 centered at O and with radius R and C2 (not shown in the figure), centered at B and with radius R. C2 intersects the x-axis in D. 4OBD is isosceles .kOBk D kBDk D R/ Therefore the median through B and the altitude a on ŒOD and through B coincide. Therefore D is the mirror image of O with respect to a. Draw circle C3 centered at D and with radius R. Let P be the intersection of C3 and OB. Draw the perpendicular e to OB and through D. 4BDP is isosceles so P is the mirror image of B with respect to e. Exercise 84 Some of the curves which you can find.

Exercise 85 The smaller you choose  , the closer  D

2r 2r will be to the number .  sin  

Exercise 86

kOP k D cos  I

kQP k D sin  I

kRSk D tan 

1 1 kOP k kQP k D cos  sin  2 2 1  area circular sector QOS D kORk  D 2 2 1 1 area 4ORS D kORk kRSk D tan  2 2 area 4OPQ D

150

Solutions

From the figure it is obvious that area 4OPQ < area circular sector QOS < area 4ORS )

1  1 cos  sin  < < tan  2 2 2

)

cos  sin  <  < tan 

Dividing these inequalities by sin  yields cos  <

 1 1 < tan  D sin  sin  cos 

If  D 0 both the left and right-hand sides are equal to 1, which would tempt one to conclude  that D 1, which is of course incorrect, because one cannot divide by 0. sin     1 or mathematically lim D 1. We can however say that if   0 then  !0 sin  sin  Exercise 87 4GMH  4EMF because GH and AF are parallel lines by construction. Parallel lines intersect other straight lines at the same angles. From the similarity we deduce that

kGM k kMH k D kMF k kMEk

)

1 kMH k D 1 1=

)

kMH k D :

In any right-angled triangle the altitude on the hypotenuse is the mean proportional of the two segments of the hypotenuse. Hence p kFM k  kMH k D kKM k2 ) 1   D kKM k2 ) kKM k D : Exercise 88 Pythagoras’ theorem for a right-angled triangle 4ABC , right-angled A, states:

kBC k2 D kBAk2 C kAC k2 . The area of the semi-circle on BC is   kBC k 2 SCBC D  ; 2

the area of the semi-circles on BA and AC are resp.     kBAk 2 kAC k 2 SCBA D  and SCAC D  : 2 2 Now

     kBAk 2 kAC k 2 C D kBAk2 C kAC k2 2 2 4 4    2 2 2 D kBAk C kAC k D kBC k D SCBC : 4 4 

SCBA C SCAC D 

Exercise 89 The ratio between the areas of similar figures is equal to the square of the ratio of corresponding lengths (e.g. a diameter) of those figures. Here we take the lengths of the sides as corresponding lengths and obtain:

A B C D 2 D 2 a2 b c

)

B CC D

b2 c2 b2 C c2 AC 2A D A D A: 2 a a a2

151 Solutions

Exercise 90

We use the semi-circular version of Pythagoras’ theorem as proved in the previous exercise. In figure (a) we have that A C A1 C A2 D B1 C B2 , in figure (b) B1 D A1 C C1 and in figure (c) B2 D A2 C C2 . Substituting the values for B1 and B2 in the first equation yields A C A1 C A2 D A1 C C1 C A2 C C2 ; which simplifies to A D C1 C C2 D C: Exercise 93 To show that the area of a circle sector equals S D

1 2 r ˛, we use the rule of three: 2

2 ˛ D : 2 r S In the left-hand side of the equation we have used a special sector: a full circle. The area of the circle sector BMA is equal to !2 p 2  1  1 2 D kBAk2 ; S D kBM k .†BMA/ D kBAk 2 2 2 2 8 which is equal to the area of the semi-circle on ŒBA. The area of the lune on AB is given by: area lune D SCAB  sector BMA C 4BMA D D

1 kABk2   kBAk2 C kBM k2 8 8 2

1 kBAk2 : 4

Exercise 95 Referring to . Fig. 6.8 we know from the previous exercise that

SCAB C SCAC D SCBC ;

We deduct areas III and IV from both sides (these are the areas defined by the perpendiculars and the semi-circle on the hypotenuse): SCAB C SCAC D SCBC

area III C area IV D area III C area IV area I C area II D area 4ABC

152

Solutions

Exercise 96

Referring to the figure, let T1 denote the area of the right-angled triangle to the left of the altitude on the hypotenuse, and T2 that of the area of the triangle to the right, so that T1 CT2 D T . Again, we turn to use Pythagoras’ semi-circular theorem. In the left triangle II C a C T1 and I C b C T2 are the areas of the semi-circles on the legs of a right-angled triangle, so they sum to the area of the semicircle on the hypotenuse A C a C b, from which II C a C T1 C I C b C T2 D III C a C b; or T1 C T2 D III  II  I; which had to be proved. In the right-angled triangle, we use the results of Exercise 95 and find that .II C III/ C .I C IV/ D T1 C T2 D T: Exercise 97 Let R denote the radius of the original circle and S, C and L the areas of the circular segment to the left of the dashed chord, the white circle and Leonardo’s Claw, respectively. Then

SD

R2 R2  ; 4 2

i.e. the difference between the area of the circular sector and the isosceles triangle (see figure (a)). p 2R R The altitude in the isosceles triangle is equal to p D . 2 2

153 Solutions

 p p  2R D 2  2 R. Subtracting from the dip ameter ofpthe large circle, we find that the diameter of the small circle is 2R; therefore its 2R radius is (see figure (b)). 2 The area of Leonardo’s Claw becomes:   R2 R2 1  D R2 L D R2  C  2S D R2  R2  2 2 4 2 p The diagonal of the square equals 2R, therefore the side of the square is R and its area R2 (see figure (c)). Therefore the thickness of the lens is 2R 

Exercise 98 The trapezoid ABCD is isosceles and DC k AB. Let the perpendicular bisectors be mAD , mDC and mCB and the midpoints MAD , MDC and MCB . The bisector mCD is an axis of symmetry (mirror reflection) of the trapezoid (check for yourself that †A D †B and deduce that mCD is an axis of symmetry). mAD intersects mDC at a point O. The mirror image of mAD is a line l through O. The mirror image of MAD with respect to mDC is MCB . So line l passes through MCB . The property “at right angles” is invariant under a mirror reflection, so image mAD ?image AD or l?CB. This means that line l is the perpendicular bisector mCB of CB. Therefore the perpendicular bisectors mAD ; mDC and mCB intersect at one point O, which is the centre of the circumscribed circle. Exercise 99 4AOD  4AED ) †AOD D †AEB. Let †AOD D ˛, then

area circular segment AD D area circular sector AOD  area 4AOD D

1 1 kOAk2 ˛  kOAk kODk sin ˛ 2 2

D

1 kOAk2 .˛  sin ˛/ 2

154

Solutions

On the other hand area circular segment AB D area circular sector AEB  area 4AEB 1 1 kEBk2 ˛  kEBk kEAk sin ˛ 2 2 1 D kEBk2 .˛  sin ˛/ 2 3 D kAOk2 .˛  sin ˛/ 2

D

D 3 area circular segment AD Exercise 100

Because 4ABE  4ADO, †AOD D †AEB and †AOB D 3†AOD we have †AOB D 3†AEB (all angles are expressed inpradians). Furthermore kAEk D 3 kAOk. Therefore area circle sector AOD D

1 3 kAOk2 .†AOD/ D kAOk2 .†AEB/ 2 2

and 2 1 1 p 3 kAOk .†AEB/ kAEk2 .†AEB/ D 2 2 3 D kAOk2 .†AEB/ 2

area circle sector AEB D

sector AOB D I C II sector AEB D II C III )

0 D I  III

)

I D III

155 Solutions

Exercise 101 The radius of the small circles is half that of the large circles, therefore four times

their area equals the area of the large circle. Exercise 102 From Exercise 84 we deduce that

area semi-circle AD D area semi-circle DC D area semi-circle CB D area semi-circle EF and 4 area semi-circle AD D area semi-circle AB which we can also write as 3 area semi-circle AD C area semi-circle EF D area semi-circle AB or area semi-circle EF D area semi-circle AB  3 area semi-circle AD semi-circle EF D area trapezoid AB C 3 area VI  3 area semi-circle AD semi-circle EF D area trapezoid AB  .3 area VI  3 area semi-circle AD/ semi-circle EF D area trapezoid AB  3 area lune III Exercise 103 Referring to . Fig. 6.14

S D area semicircle CB  area VI D area semicircle CB  .area circular sector COB  area 4COB/     kOBk 2 kOBk2  kOBk2  C sin  SD 2 2 2 3 2 3 p 3   D kOBk2  kOBk2 C kOBk2 8 6! 4 p 3  D  C kOBk2 24 4

156

Solutions

We notice that the area of the lune is a function of . Therefore we do not know, and neither did Hippocrates, whether we can construct this lune with compass and straightedge because we do not know whether we can construct a line segment of length  (see 7 Chapter 7 and esp. 7 Sect. 7.3 and 7 Sect. 7.4 for an explanation). Exercise 104 We will first suppose that the pentagonal figure is already constructed. From this,

we will deduce the lengths of certain line segments, which will then allow us to actually draw the pentagonal figure.

E, D, B are collinear, therefore †EBA D †DBA. Furthermore 4ADB and 4EAB are isosceles. Two isosceles triangles are similar r if they have equal r base angles. 3 3 We have a D kEAk D kABk, kDEk D a and x D kDBk. kEAk D 2 2 kEAk kDBk D . From 4ADB  4EAB it follows that kABk kEBk r 3 Now kEBk D x C a so 2 r r ! a 3 3 x 2 2 D a Da )x C ax  a2 D 0 r )x xC a 2 2 3 xC a 2 which has solutions r r ! 3 11 a xD ˙ ; 2 2 2 of which in this geometrical context r r ! 3 11 a xD C 2 2 2 satisfies the condition of being positive. A line segment with length x!can easily be constructed. Draw a circle centered at B and r r 3 11 a with radius x D C . One intersection of this circle and the line PQ is the 2 2 2 vertex D.

157 Solutions

Above, we have only used quadratic equations, the solutions to which are constructible with compass and straightedge (see Exercise 131). Therefore we do not need a neusis. Exercise 105

4AO1 B Š 4BO1 C

)

†ABO1 D †O1 BC

)

†ABM D †MBC:

Furthermore kABk D kBC k and †CAB D †ACB (4ABC is isosceles) )

4AMB Š 4CMB

)

†AMB D †BM C and ]AM C is a straight angle

)

]AMB and ]BM C are right angles

Exercise 106 Let

†DO2 C D ˛:

p

(1)

The sector O2 EDC has a radius 3a and an interior angle †EO2 D D 2˛, therefore 1 p 2 3a  2˛ D 3a2 ˛: area O2 CDE D 2 p From (1) and (15) †BO1 C D ˛. The circular sector O1 CBAE has radius 2a and †EO1 C D 3˛; therefore 1 p 2 area O1 CBAE D 2a  3˛ D 3a2 ˛: 2 Exercise 107 Falco constructs a curvilinear figure in which the two curves QNP and RSO are

a translation of each other along the other (straight) edges. A consequence is that the piece which looks to have been cut from the rectangle is added on the other side.

The total area therefore remains unchanged. In Falco’s figure, PNM and OSV have the same area. OSV has been “cut” from the rectangle and added again on the top. The same goes for the figures QGN and RT S. The curvilinear figure RQP V therefore has the same area as the rectangle T GM V . Exercise 108

158

Solutions

5 It is clear that half a diagonal of the square is equal to R, therefore to find an edge E we can 4 apply Pythagoras’ theorem  E2 D

5 R 4

2

 C

5 R 4

2

)

ED

5 p R 2: 4

If we assume that the area of the square and the circle are equal then S D E 2 D S

) p

25 2 R D R2 8

)

D

25 D 3 18 : 8

p , and p; q 2 N. Furthermore p and q have no common factors, q in other words the fraction is irreducible. If we square both sides of the equation we find Exercise 109 Suppose

p2 D2 q2

)

2D

2q 2 D p 2 :

So p 2 is even, and so is p. We can write 2p 0 D p. The equation becomes .2p 0 /2 D2 q2

)

2q 2 D 4p 02

)

q 2 D 2p 02 :

Then q 2 is even, and so is q. We can write 2q 0 D q. The equation becomes p p0 2p 0 D : D q 2q 0 q0 p But we started out with the assumption that was irreducible, so we have a contradiction. In q p other words, 2 can not be written as a rational number. Exercise 110

HBDL is a rectangle, KHBI is a parallellogram. For a parallellogram, the area is produced by base  height. For KHBI ŒHB is a base and ŒBD the height, from which:  p  (1) area KHBI D kHBk  kBDk D 3  2  11 D area HBDL:

159 Solutions

On the other hand, we can also consider BI as the base. In which case area KHBI D kHZk  kKH k  p  D kHZk 11 C 2  p  D 11 3  2 from (1)   p  p  p  p 7  11 5  2 2 3  2 11  2 3 2 D kHZk D 11  p D 11  121  2 7  17 11 C 2 p D 1:409 : : :  2 2

2

2

Exercise 111 Let kABk D kBC k D kCAk D 1 also kCEk C kEBk D kCBk from which

p kCEk D

3 : 2

4ABC is equilateral, so the respective altitudes, medians and angle bisectors coincide. The centroid therefore is also orthocentre, circumcentrepand intersection of the angle bisectors. 3 1 Since D is the centroid kDEk D kCEk D . 3 6 1 We also know kEKk D kKBk D . 4 7 So kDKk2 D kDEk2 C kEKk2 D and 48 r p p 5 7 21 5 21 D and kDLk D kDKk D : kDKk D 48 12 4 48 According to Nicolas of Cusa kDLk is the radius of the circle isoperimetric to the triangle. So 2 kDLk D 3 and p p 3 144 144 21 62 21 D D p D D : 2 kDLk 210 105 10 21 Exercise 112 4ABC is equilateral, so the respective altitudes, medians and angle bisectors

coincide and †EBD D In 4DEB:

 . 6

p p 3  D 30 3 kEBk D kDBk cos D 60 6 2  1 kDEk D kDBk sin D 60  D 30: 6 2

Because kEBk D

and

1 1 kABk and kEKk D kKBk D kABk: 2 4

kEKk D kKBk D

p kEBk D 15 3: 2

160

Solutions

4DEK is right-angled in E, from which kDKk2 D kDEk2 C kEKk2 D 1575 D 7:225 and

p kDKk D 15 7:

Furthermore kDLk D

5 75 p 7: kDKk D 4 4

According to Nicolas of Cusa 75 p p p 7 2 kDLk 1 75 7 5 7 2 D D p D p D p  6 kEBk 6  30 3 12  30 3 24 3 At first glance, this expression does not resemble Cusa’s own result in 7 Sect. 6.5, but after some manipulation we find: p p p p p 2 12 1575 2:5 7  225 2:5  15 7 5  15 7 5 7 p D p D p D p D p 6 2700 6 3:900 6  30 3 12  30 3 24 3 which is, of course, the same value. Exercise 113

p

2 Let kABk D d . Using Pythagoras’ theorem, it is clear that kMAk D d and therefore 2 p d 2 d : kEGk D 2 2 p

So kMN k D

21 d dC D 6 2

p

2C2 d: 6

If the circle and the square have the same area, then p 

2C2 d 6

!2

 D d2

)

D

6 p 2C2

2

0  p  12 6 2 2 A  3:08831 : : : D@ 2

Exercise 114 The values of b and c converge to a number with the same decimals as  and

b <  < c.

161 Solutions

Exercise 115

Let the radius of the circle be r and suppose we have an inscribed regular n-gon pn and a circumscribed n-gon Pn . In triangle 4Oab we have that 2 : kOak D kObk D r and †aOb D n  Since 4Oab is isosceles, 4Oma is right-angled, †aOm D therefore n   and kabk D 2r sin : kamk D r sin n n The total circumference thus is  pn D 2nr sin : n  Triangle 4OF G is isosceles and kOf k D r therefore kf Gk D r tan and the circumfern  ence is Pn D 2nr tan . n It immediately follows that  2nr sin pn  n D  D cos n : Pn 2nr tan n Similarly we find for the area:  1  An D n  r  2r tan D nr 2 tan 2 n n and     an D n  r cos  2r sin D nr 2 cos sin n n n n from which    nr 2 cos sin cos an n n D n D cos2  D  1 An n nr 2 tan  n cos n

162

Solutions

Exercise 116

p    3 1 D sin , from which kOC k D D cos and so We know that kAC k D 2 6 2 6 p 3 : kCDk D 1  2 We see that 4OCA  4ODE from which kEDk kAC k D kOC k kODk so 1 p 3 1 kEDk 2 p D and kEDk D p D : 1 3 3 3 2 Exercise 117

2

q p 4  .2 sin 2˛/2 D 2  4  4 sin2 2˛ D 2  2 cos 2˛  D 2  2 1  2 sin2 ˛ D 2  2 C 4 sin2 ˛ D .2 sin ˛/2

Using this formula:

q .2 sin 2˛/ D 2  4  .2 sin 4˛/2 q .2 sin ˛/2 D 2  4  .2 sin 2˛/2 s   q 2 D 2  4  2  4  .2 sin 4˛/ 2

r D2

2C

q 4  .2 sin 4˛/2

(33)

163 Solutions

This is the basis for a proof by induction. v v u s u u r u q u t t 2 .2 sin ˛/ D 2  2 C 2 C 2 C : : : 2 C 4  .2 sin 2n ˛/2 with n  1 2s under the square roots. The proof of the induction step is essentially the same as proving equation (33). Alternatively, one can also write: v v u s u u r u q u   t ˛ 2 t 2 sin n D 2  2 C 2 C 2 C : : : 2 C 4  .2 sin ˛/2 2 The edge of a regular n-gon inscribed in a circle with radius 1 equals 2 sin ˛, in which ˛ is the central angle subtending the edge as its chord. Exercise 118 With reference to the figure of Exercise 116, we see that:

1 1 sn Sn 2 ; , s 2 D 2 1 1 sn 1 2 2sn . which after simplification yields Sn D p 4  sn2 kAC k kEDk D kOC k kODk

Exercise 120 Use B.i/ D ROOT.2  ROOT.4  .B.i  1//O2//

2i n

length edge

perimeter



1

6

3

12

0:51763809 : : :

6:211657 : : :

3:105829 : : :

24

0:261052384 : : :

6:265257 : : :

3:132629 : : :

48

0:130806258 : : :

6:2787 : : :

3:141032 : : :

96

0:065438166 : : :

6:282064 : : :

3:141452 : : :

192

0:032723463 : : :

6:282905 : : :

3:141584 : : :

384

0:016362279 : : :

6:283115 : : :

3:141590 : : :

6

Exercise 121 We will use the method of complete induction.

We notice that the assertion holds for small values of n. If n D 1

then

1 X

i D 1 and

1 1  .1 C 1/ D 1: 2

i D 3 and

1 2  .2 C 1/ D 3: 2

i D 6 and

1 3  .3 C 1/ D 6: 2

i D1

If n D 2

then

2 X i D1

If n D 3

then

3 X i D1

164

Solutions

Now suppose the assertion holds for n D k, will the assertion also hold for n D k C 1? kC1 X i D1

k X

1 k.k C 1/ C .k C 1/ 2 i D1     1 k C2 D .k C 1/ k C 1 .k C 1/ D 2 2 1 1 D .k C 2/.k C 1/ D .k C 1/ ..k C 1/ C 1/ 2 2

iD

i C .k C 1/ D

This implies that the assertion holds for all positive integers. Exercise 123 We will use the method of complete induction.

We notice that the assertion holds for small values of n. If n D 1

then

1 X

i 2 D 1 and

123 D 1: 6

i 2 D 5 and

235 D 5: 6

i 2 D 14 and

347 D 14: 6

i D1

If n D 2

then

2 X i D1

If n D 3

then

3 X i D1

Suppose that the assertion holds for n D k, will the assertion also hold for n D k C 1? kC1 X

k X

k .k C 1/ .2k C 1/ C .k C 1/2 6 i D1  .k C 1/ 2k 2 C 7k C 6 .k C 1/ .k .2k C 1/ C 6 .k C 1// D D 6 6 .k C 1/ .k C 2/ .2k C 3/ D 6 .k C 1/ ..k C 1/ C 1/ .2 .k C 1/ C 1/ D 6

.i C 1/2 D

i D1

i 2 C .k C 1/2 D

which concludes the proof. The area under the parabola is approximated by: n n X i 2b3 b3 X 2 b 3 n .n C 1/ .2n C 1/ D i D  n3 n3 i D1 n3 6 i D1

D

b 3 n C 1 2n C 1   6 n n

So b 3 n C 1 2n C 1 b3 b3   D 12D : n!1 6 n n 6 3 lim

165 Solutions

1 between the straight 2 1 line and the x-axis. Inserting b D 1 in the results of Exercise 123 yields an area between the 3 parabola and the x-axis. Therefore the area between the parabola and the straight line equals 1 1 1  D . 2 3 6

Exercise 125 Inserting a D 0 and b D 1 in equation (17) yields an area

Exercise 127

Obviously 4OAC  4OBD ) ) )

kOAk kOBk D kOC k kODk b kOBk D 1 a kOBk D ab

which proves the assertion.

To construct a2 , a3 and a4 , draw two intersecting straight lines l and m, let kOAk D 1 and kOBk D a. Draw a circle with centered at O and with radius a. The circle intersects m in P . Draw line AP and the parallel line through B. The intersection of this parallel line with m is point Q. Now by the construction kOQk D a2 . Draw a circle centered at O and with radius a2 , this circle intersects line l in C . Draw a parallel to AP through C , this line intersects m in R. Now by the construction kORk D a3 . To construct a4 , draw the line AQ and the parallel to it through C , this line intersects b at T . Now by the construction kOT k D a4 .

166

Solutions

Exercise 128

Select points D, A and B in such a way that kODk D a, kOAk D 1 and kOBk D b. Connect B and D, draw a parallel l line to BD through A. Then kOC k D a=b in which C is the intersection point of l and OD. Exercise 130

p 2 p p 42 3D 32 3C1 D 31 ;  p p 2 p and 97  56 3 D 49  56 3 C 48 D 7  4 3 q q   p p p p 3  1 C 7  4 3 D 3: 4 4  2 3 C 97  56 3 D 4   p  p  6 3C 3 6 3 3 6 6 p C p D C D 6: 93 93 3 3 3C 3

Exercise 131 (a) We know that kABk D b, kAM k D kMBk D

p b and kM C k D c. 2

b . Therefore kCDk2 D kCM k2 C kMDk2 or 2 p b 2  4c : kMDk D 2

By construction we also know that kCDk D kMDk2 D

b2 b 2  4c c D 4 4

So kADk D

bC

and

p

b 2  4c 2

and

kDBk D

b

p

b 2  4c ; 2

which are the solutions to the quadratic equation x 2  bx C c D 0.

167 Solutions

(b) Again we know that kABk D b, kAM k D kMBk D the point on ŒAC  for which kAEk D kAM k.

p b and kM C k D c. Let E be 2

p b 2 C 4c By construction we also know that kAC k D , therefore 2 p p b 2 C 4c b b 2 C 4c  b  D ; kEC k D 2 2 2 which is a solution to the quadratic equation x 2 C bx  c D 0. The second solution is negative and therefore cannot be constructed. (c) To divide a line segment into two segments, denote the parts as x and a  x. The condition can be written as the equation x 2  .a  x/2 D x .a  x/

or x 2 C ax  a2 D 0;

which can be solved using the method of (b). To show that the longer segment is the mean proportional between the shorter segment and the whole line, we rewrite the equation x 2 C ax  a2 D 0 as x ax x 2 D a2  ax , x 2 D a.a  x/ , D ; ax x which proves the assertion. Exercise 132 If the coordinates of the points are constructible, then the cartesian equations of

the straight lines through them have constructible coefficients. Let y D m1 x C q1 be the equation of AB and y D m2 x C q2 the equation of CD. The intersection of these lines is given by the system of equations: 8
168

Solutions

Let x 2  2ax C y 2  2by C c D 0 be the equation of the circle centered at C and with radius kCDk. The coefficients in this equation are constructible because they are rational expressions of constructible numbers. To find the intersection with AB, we need to solve the system 8 < y D m1 x C q1 (34) :x 2  2ax C y 2  2by C c D 0 Substituting the value of y from the first equation into the second equation yields the equation ux 2 C vx C w D 0:

(35)

This is a quadratic equation in which u, v and w are rational expressions of constructible numbers and which therefore are constructible. The solution is either a rational expression or a rational expression which contains a square root. So this equation yields a constructible number. The coordinates of the intersections of two circles (if they exist) are given by the system of quadratic equations: 8 <x 2  2a1 x C y 2  2b1 y C c1 D 0 :x 2  2a x C y 2  2b y C c D 0 2 2 2 Subtracting the second equation from the first equation yields: 8 <2.a1  a2 /x  2.b1  b2 /y C c1  c2 D 0 :

x 2  2a2 x C y 2  2b2 y C c2 D 0

This system is of the same kind as the system (34), therefore the conclusion is the same. Exercise 133 See also Exercise 79

sin 3˛ D sin .˛ C 2˛/ D sin ˛ cos 2˛ C cos ˛ sin 2˛ D : : : D 3 sin ˛  4 sin3 ˛: or, using the assumption, c D 3y  4y 3 . d3 D 3x  x 3

Exercise 134

)

2 sin 3˛ D 3 .2 sin ˛/  .2 sin ˛/3

)

2 sin 3˛ D 6 sin ˛  8 sin3 ˛

)

sin 3˛ D 3 sin ˛  4 sin3 ˛:

169 Solutions

From Exercise 65 we know that a D kOBk D kPRk D kQRk : Moreover 4PBQ, 4P CO and 4BEQ have two congruent angles in common, the right angle and †POC D †PQB D †BQE. Consequently 4PBQ  4P CO  4BEQ; which yields the continued proportion kBQk kCOk kEQk D D : kPQk kPOk kBQk 4ORB is isosceles, so 1 1 kERk D kORk D .y C a/ ; 2 2 and y C 3a 1 : kEQk D kERk C kRQk D .y C a/ C a D 2 2 Whence x b y C 3a D D : 2a y 2x Put a D 1 then b D cos 3˛ and 8 <xy D 2b ) x 3  3x  2b D 0 : x2 D y C 3 Exercise 135 Suppose the line segment ŒOA has a length sin ˛. Draw a circle centered at O and

with radius 1. Erect the perpendicular to ŒOA at A. This perpendicular intersects the circle at B. Then †ABO D ˛. Suppose †ABO D ˛ is given. Draw a circle centered at B and with radius 1. This circle intersects leg AB at C . From C , construct the perpendicular on BO, let the foot be D. The triangle 4BCD is a right-angled triangle in which the legs have length cos ˛ and sin ˛.  rad and R D 1 then 1 D x 3 3x ) x 3 3x1 D 0. 6 Potential solutions are ˙1 neither of which satisfies the equation, therefore the equation has neither rational nor constructible solutions. p p 3  3 3 D a. Select a D 3 then b D and x  3x  2b D 0, now b D a cos 6 2 2 3 x  3x  3 D 0, with potential solutions ˙3, neither of which satisfies the equation, therefore the equation has neither rational nor constructible solutions. Exercise 136 d3 D x 3 3x, if ˛ D

1  1  D and cos D . These are rational numbers and thus 6 2 3 2   constructible. Exercise 135 allows us to construct the angles rad and rad. Therefore a 6 3 right angle can be trisected. For a right angle equation (19) becomes: 4x 3  3x C 1 D 0. It is clear that x D 1 is a solution. All solutions to this equation are by consequence constructible. Exercise 137 We know that sin

170

Solutions

Exercise 138 Hippocrates used two assumptions to find constructible lunes:

(1) Two circular sectors (with radii r and R respectively) corresponding to the lunes’ arcs have the same area (2) The central angles (2˛, 2ˇ) of the two circular arcs are commensurable ˛ n r2 D D 2. ˇ m R Let 2˛ be the central angle of the arc centered at A. Let 2ˇ be the central angle of the arc centered at B. It is obvious that ˇ > ˛. Let  be the greatest common denominator of 2˛ and 2ˇ which, by assumption (2), are commensurable. n ˛ Therefore ˛ D n and ˇ D m for some m > n 1 and D . ˇ m Subdivide SA D ACED into n congruent sectors of central angle  and centered at A. Subdivide SB D BCFD into m congruent sectors of central angle  and centered at B. Now by assumption (1), the two circular sectors have the same area. Therefore the n smaller sectors centered at A and with radius R taken together and the m smaller sectors centered at B and with radius r taken together have the same area. R2 Now the area of one of the n sectors is given by , while one of the m sectors has 2 2 r . area 2 Therefore

We will show that

nR2 m r 2 n r2 D ) D 2: 2 2 m R From which we conclude that ˛ n r2 D D 2 ˇ m R

and ˇr 2 D ˛R2 :

We can prove that this relation holds for all squarable lunes. Exercise 139

q

p

q

34  2 33 D q

q

r p 2 p 1 C 33 D 1 C 33: 33  2 33 C 1 D p

q p 1 38  6 33 D 27  6 33 C 11 D p 81  18 33 C 33 3 r p  p 2 1 9  33 D p 9  33 D p : 3 3 p

p

171 Solutions

Exercise 140

p ! r r p q p p  3 1  3 34  2 33 S Dr 30 C 2 33   1   30 C 2 33 2 8 2 64 ! r r r p q p  p 3 1 3 1 C 33 3 1  2 Dr    1   30 C 2 33 30 C 2 33 2 8 2 8 2 64 s 0r r q p 1 p p 1 3 3 33 38  6 33 A 1 C 1 30 C 2 33 @ D r2   8 2 2 8 8 2 r q    p  p p  3 1 p 1  21 30 C 2 33 p 3  1 C 33  p 9  33 see Exercise 139 Dr 8 2 8 2 3 q   p p 1 30 C 2 33 12 C 4 33 D r2 128 q p  p  r2 D 30 C 2 33 3 C 33 32 r

2

3 1  2 8

Exercise 141 See also Exercises 79 and 133

cos 3˛ D cos .˛ C 2˛/ D cos ˛ cos 2˛  cos ˛ cos 2˛ D : : : D 4 cos3 ˛  3 cos ˛ sin 5˛ D sin .2˛ C 3˛/ D sin 2˛ cos 3˛ C cos 2˛ sin 3˛    D 2 sin ˛ cos ˛ 4 cos3 ˛  3 cos ˛ C 1  2 sin2 ˛ 3 sin ˛  4 sin3 ˛ :: : D 16 sin5 ˛  20 sin3 ˛ C 5 sin ˛ If m D 5 then sin 5˛ D

p p

5 sin ˛

16 sin ˛  20 sin ˛ C 5 sin ˛ D 5 sin ˛  p  16t 5  20t 3 C 5  5 t D 0  p  16t 4  20t 2 C 5  5 D 0 _ t D 0 5

3

The last equation is a biquadratic equation with solutions: p p p p 20 ˙ 4 5 C 4 5 5˙ 5C4 5 2 t D D 32 8 p p p p 5 5C4 5 5C 5C4 5 2 2 Only t D is a valid solution because t D > 1. 8 8 So s r p p q p 5 5C4 5 1 10  2 5 C 4 5 sin ˛ D D 8 4

172

Solutions

If m D

5 , then 3

r

5 sin ˛ 3 r 5 sin 3ˇ sin 5ˇ D 3 r 5 16 sin5 ˇ  20 sin3 ˇ C 5 sin ˇ D 3 sin ˇ  4 sin3 ˇ 3  p  p  4 15  15 sin2 ˇ C 5  15 D 0 _ sin ˇ D 0 16 sin4 ˇ  3 5 sin ˛ D 3

The biquadratic equation has solutions p p  4p 4 p p p 60 C 6 15 15  15 ˙ 15  15 ˙ 60 C 6 15 2 3 3 sin ˇ D D 32 24 Now  sin ˛ D sin 3ˇ D sin ˇ 3  4 sin2 ˇ s p p p p p p ! 15  15 ˙ 60 C 6 15 15  15 ˙ 60 C 6 15 D  34 24 24 s p p p p p ! p 15  15 ˙ 60 C 6 15 15  15 ˙ 60 C 6 15 D  3 24 8 Exercise 142

sin .5 .3 .3˛/// D 16 sin5 .3 .3˛//  20 sin .3 .3˛// C 5 sin .3 .3˛//  5  3 D 16 3 sin 3˛  4 sin3 3˛  20 3 sin 3˛  4 sin3 3˛  C 5 3 sin 3˛  4 sin3 3˛    3 5 D 16 3 3 sin ˛  4 sin3 ˛  4 3 sin ˛  4 sin3 ˛    3 3  20 3 3 sin ˛  4 sin3 ˛  4 3 sin ˛  4 sin3 ˛    3  C 5 3 3 sin ˛  4 sin3 ˛  4 3 sin ˛  4 sin3 ˛ !    x 3   x 3 3 5 x x D 16 3 3  4 4 3 4 2 2 2 2 !    x 3   x 3 3 3 x x  20 3 3  4 4 3 4 2 2 2 2 !     x 3  x 3 3 x x C5 3 3 4 4 3 4 2 2 2 2 :: :

173 Solutions

Remember that .i sin ˛/2k 2 R. sin 45˛ D Im .cos 45˛ C i sin 45˛/   D Im .cos ˛ C i sin ˛/45 ! ! 45 X 45 k 45k D Im .i sin ˛/ cos ˛ k kD0 ! ! 22 45 1 X 2kC1 442k D cos ˛ .i sin ˛/ 2k C 1 i kD0 ! 22 X 45 D .1/k sin2kC1 ˛ cos442k ˛ 2k C 1 kD0 ! 22 X  22k 45 .1/k sin2kC1 ˛ 1  sin2 ˛ D 2k C 1 kD0

:: : D

22 X j D0

.1/

j

" j X kD0

45 2k C 1

!

22  k j k

!#

 x 2j C1 2

D 45x  3795x 3 C    C x 45 Exercise 143 We know †BAC D ˛, 4ABC is isosceles so †BAC D †ACB D ˛.

†CBD D 2˛ (in a triangle the exterior angle equals the sum of the opposite angles). Now consider 4BCD. 4BCD is an isosceles triangle so †CBD D †BDC D 2˛. We can now repeat the procedure with 4BCD and 4CDE and so on.

Exercise 144

Divide the regular polygon into triangles as is done for the heptagon in the figure. It is clear 2 rad. The other angles are base angles that for these isosceles triangles the vertex angle is n and are equal to ˇ.

174

Solutions

The sum of the angles of a triangle is a straight angle so ˇ C ˇ C

2 D  rad, whence n

2 .n  2/ D rad. n n The base angle is half of this result, but the interior angle of the polygon is equal to the sum of the base angles of two adjacent triangles. Therefore we can conclude that the interior angle .n  2/ is equal to rad. n 2ˇ D  

The angle BAG is an interior angle of a regular heptagon, so 5 †BAG D rad. 7 The angles †BAC; †CAD; †DAE; †EAF; †FAG are angles inscribed in a circle, subtending a chord of length 1. Therefore they subtend equal arcs and are therefore equal, from which 5 rad, †BAC C †CAD C †DAE C †EAF C †FAG D 7 5  and 5†BAC D rad, so †BAC D rad. 7 7   0 0 0 0 0 0 so †B C A D rad. †B A C D rad, 7 7 Now in 4B 0 C 0 D 0 : 2 †D 0 B 0 C 0 D †C 0 D 0 B 0 D rad (exterior angle triangle). 7 On the other hand in 4ACF : †ACF C †CFA C †FAC D  rad

(sum of the angles of a triangle)

and †ACF D †CFA (isosceles triangle), so 2†ACF D  

7 3 4 3 D  D rad 7 7 7 7

or †ACF D

2 rad. 7

175 Solutions

Exercise 145 Two isosceles triangles with equal base angles are similar so

4ACF  4C 0 B 0 D 0 : We now know that that kB 0 D 0 k D

kCF k kAC k D x, so D x, and, using kCF k D kADk D y, we find 0 0 kB C k kB 0 D 0 k

y : x

The proof for 4ADE  4D 0 E 0 C 0 is analogous. Exercise 146 From the construction we know that kADk D kA0 D 0 k so y D 1 C

y . x

1 On the other hand kAEk D kA0 E 0 k so y D x C y 8 y ˆ
)

)

x D x .x  1/ C .x  1/

)

x 3  x 2  2x C 1 D 0

2

2

2

x2 .x  1/ )

2

D

x2 C1 x1

x 2 D x 3  x 2 C x 2  2x C 1

.29/

If we want to eliminate x we solve the second equation for x and find x D this result into the first equation:  2  y 1 y2  1 y3  y y2  1 y2 y D Cy ) D C y y y y y )

y3  y D y2  1 C y2

)

y2  1 , and insert y

y 3  2y 2  y C 1 D 0

Possible rational solutions for equation (29) are x D ˙1. Inserting into the equation yields: for x D 1W

1  1  2 C 1 D 1 ¤ 0I

for x D 1W

1  1 C 2 C 1 D 1 ¤ 0:

The equation has no rational solution and therefore no constructible solution. Neither ŒAC  nor ŒAF  can be constructed using compass and straightedge methods. Exercise 147 In 4CBA draw the altitude from B. Call the foot H , then 4ABH is right-angled

in H . Moreover, because 4CBA is isosceles kCH k D kHAk D

x : 2

176

Solutions

Furthermore †BAH D †BAC D cos .†BAH / D

 rad so 7

kAH k kBAk

therefore cos

 x=2 x D D 7 1 2

 . 7  Consequently (see Exercise 133) an angle of magnitude rad cannot be constructed using 7 compass and straightedge methods.

Now x is not a constructible number so neither is cos

Exercise 150

Fold the enneagon in the same fashion as we did with the heptagon. Thales’ intercept theorem then leads to the equations kAC k D x; kADk D y;

z ; kCEk D 1; x kAEk D kAF k D z;

kBDk D

kDF k D

1 ; x

1 z D y, x C 1 D z and y C D z. x z Eliminating x and y from these equations yields:

and 1 C

z 3  3z 2 C 1 D 0

(36)

If this equation has rational solutions these are z D ˙1. Obviously neither is a solution to equation (36), therefore there are no constructible solutions to the equation. If z is not a constructible number, then neither is x. Using the same procedure as in the  proof of the inconstructibility of the heptagon, we find that angle rad is not constructible 9 and therefore the enneagon is not constructible using compass and straightedge methods. Fold the hendecagon in the same fashion as we did with the heptagon and the enneagon. This gives us the following relations: y D 1C

z ; x

zDxC

w ; y

wDyC

y z

and w D z C

1 w

177 Solutions

Eliminating x, y and z yields w 5  3w 4  3w 3 C 4w 2 C w  1 D 0:

(37)

This is a fifth degree equation, so we can draw no conclusions about the constructibility of the roots. Lemma 1 only allows us to draw conclusions about the constructibility of solutions to cubic equations. Solving equations (36) and (37) gives us the length of the longest diagonal. Unfortunately no general algorithm exists to solve a fifth degree equation algebraically, so we have to revert to numerical methods. For the enneagon z D 2:879 : : : , for the hendecagon z D 3:513 : : :

178

References Primary sources1 Franciscus Aguilonius. Opticorum libri sex philosophis iuxtà ac mathematicis utiles. Jan I Moretus, Antwerpen, 1613. EHC G 5050. Aristotle (Trans. G. R. G. Mure). Posterior analytics. URL http://classics.mit.edu//Aristotle/posterior.1.i.html. Retrieved 25-4-2015. Oliver Byrne. The First Six Books of the Elements of Euclid. William Pickering, London, 1847. http://commons.wikimedia.org/wiki/Category:Byrne http://publicdomainreview.org/collections/the-first-six-books-of-the-elements-of-euclid-1847/ http://babel.hathitrust.org/cgi/pt?id=gri.ark:/13960/t9766b543;view=1up;seq=7. Michiel Coignet. Est de Michaelis Coigneti 1576, 1576–77. part of the manuscript written 1603 and later; Bibliothèque Nationale Paris, Ms Néer 56. Albrecht Dürer. Underweysung der Messung. s.n., Nürenberg, 1525. EHC H 202415. Leonhard Euler. Correspondence, letter 765 & 766. URL http://eulerarchive.maa.org/. Retrieved 24-8-2014. Jacobus Falco. Jacobus Falco Valentinus, miles ordinis Montesiani, hanc circuli quadraturam invenit. Petrus Bellerus I, Antwerpen, 1591. MPM 8 499. Johannes Kepler. Ausszug auss der Uralten MesseKunst Archimedis. Johann Planck, Linz, 1616. EHC G 86948. Samuel Marolois. Opera mathematica, ou: Œuvres mathematiques traictans de geometrie, perspective, architecture, et fortification. H. Hondius, Hagae-Comitis (Den Haag), 1614–1615. EHC G 48942. Valentin Mennher. Practicque pour brievement apprendre a ciffrer, & tenir livre de comptes. Gillis I Coppens van Diest, Antwerpen, 1565. MPM A 3589. John Playfair. Elements of Geometry. Bell & Bradfute and G.G. & J. Robinson, Edinburgh-London, 1795. URL http://lib.ugent.be/catalog/rug01:001104161. University of Ghent. John Playfair. Elements of Geometry (4th ed.). Bell & Bradfute, Edinburgh, 1814. https://books.google.be/. John Playfair. Elements of Geometry (9th ed.). Bell & Bradfute, Edinburgh, 1836. https://books.google.be/. Gregorius à Sancto Vincentio. Opus geometricum quadraturae circuli et sectionum coni, decem libris comprehensum. Joannes et Jacobus Meursius, Antwerpen, 1647. EHC G 4869. Ludolph van Ceulen. De circulo et adscriptis liber. Colster, Lugdunum Batavorum (Leiden), 1619. EHC G 4867. Adriaan van Roomen. Ideæ mathematicae pars prima, sive Methodus polygonorum. Jan I van Keerberghen, Antwerpen, 1593. MPM 8 533. Frans van Schooten. Mathematische oeffeningen begrepen in vijf boecken: waer by gevougt is een tractaet handelende van reeckening in speelen van geluck, door Christianus Hugenius. Gerrit van Goedesbergh, Amsterdam, 1659. KBR VH 8.040 A. Frans van Schooten. Eerste [- vijfde] bouck der mathematische oeffeningen. Gerrit van Goedesbergh, Amsterdam, 1659–1660. KBR VB 4.820a B. François Viète. Opera mathematica. Bonaventura et Abraham Elzevir, Lugdunum Batavorum (Leiden), 1646. EHC G 4858.

1

Abbreviations used to refer to libraries.

EHC – Erfgoedbibliotheek Hendrik Conscience, Antwerp KBR – Koninklijke Bibliotheek Albert I, Brussels MPM – Museum Plantin-Moretus/Prentenkabinet, Antwerp – Unesco-World Heritage

179 References

Bibliography J.M. Aarts. Meetkunde. Facetten van planimetrie en stereometrie. Epsilon Uitgaven, Utrecht, 2000. Roger C. Alperin. Trisections and Totally Real Origami, 2004. URL arXiv:math/0408159v1 [math.HO] 11 Aug 2004. Claudi Alsina and Roger B. Nelsen. Charming Proofs; A Journey into Elegant Mathematics, volume 42 of Dolciani Mathematical Expositions. MAA, 2010. Kirsti Andersen. The Geometry of an Art: The History of the Mathematical Theory of Perspective from Alberti to Monge. Sources and Studies in the History of Mathematics and Physical Sciences. Springer Science & Business Media LLC, New York, 2007. W.S. Anglin. Mathematics: a Concise History and Philosophy. Springer-Verlag, New York – Berlin, 1994. Eric Bainville and Bernard Genevés. Constructions using Conics. The Mathematical Intelligencer, 22(3): 60–72, 2000. Adnan Baki, Temel Kosa, and Bulent Guven. A comparative study of the effects of using dynamic geometry software and physical manipulatives on the spatial visualisation skills of pre-service mathematics teachers. British Journal of Educational Technology, 42(2):291–310, 2011. Évelyne Barbin and Anne Boyé. François Viète, Un mathématicien sous la Renaissance. Vuibert, Paris, 2005. Klaus Barner. How Old did Fermat become? NTM International Journal of History & Ethics of Natural Sciences, Technology & Medicine, 9:209–228, 2001. J. Lennart Berggren. The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook, chapter Mathematics in Medieval Islam, pages 515–676. Princeton University Press, Princeton – Woodstock, 2007. Frits Beukers. Pi, volume 6 of Zebra. Epsilon Uitgaven, Utrecht, 2000. Paul Bockstaele. The correspondence of Adriaan van Roomen. Lias, 3:85–129 and 249–299, 1976. Paul Bockstaele. A Challenge to the Mathematicians of the University of Leuven as a New Year’s Gift for 1639. In M. Folkerts & J.P. Hogendijk, editor, Vestigia Mathematica, pages 15–28. Rodopi, 1993. Paul Bockstaele. Between Viète and Descartes: Adriaan van Roomen and the Mathesis Universalis. Archive for the History of Exact Sciences, 63(4):433–470, 2009. Alexander Bogomolny. Francois Viete’s Reduction of CCC to PCC from Interactive Mathematics Miscellany and Puzzles, a. URL http://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/CCC-Viete.shtml. Retrieved 8-8-2014. Alexander Bogomolny. Delian problem from Interactive Mathematics Miscellany and Puzzles, b. URL http:// www.cut-the-knot.org/do%20you%20know/Delian.shtml. Retrieved 17-11-2013. Benjamin Bold. Famous Problems of Geometry and how to solve them. Dover Publications, New York, 1982. Karl Bopp. Die Kegelschnitte des Gregorius a St Vincentio in vergleichender Bearbeitung, volume XX of Abhandlungen zur Geschichte der mathematischen Wissenschaften. Teubner Verlag, Leipzig, 1907. Henk J.M. Bos. Recognition and Wonder: Huygens, tractional motion and some thoughts on the history of Mathematics. Tractrix, 1:3–20, 1989. Henk J.M. Bos. Johann Molther’s Problema Deliacum, 1619. In M. Folkerts & J.P. Hogendijk, editor, Vestigia Mathematica, pages 29–46. Rodopi, 1993. Henk J.M. Bos. Redefining Geometric Exactness. Springer-Verlag, New York, 2001. Lucas N.H. Bunt, Philip S. Jones, and Jack D. Bedient. The Historical Roots of Elementary Mathematics. Dover Publications, New York, 1988. Karine Chemla, editor. History of Science, History of Text, volume 238 of Boston Studies in the Philosophy of Science. Springer, Dordrecht, 2004. Constantinos Christou, Nikos Mousoulides, Marios Pittalis, and Demetra Pitta-Pantazi. Proofs through Exploration in Dynamic Geometry Environments. International Journal of Science and Mathematics Education, 2(3):339–352, 2004. Julian L. Coolidge. A History of the Conic Sections and Quadric Surfaces. Clarendon Press, Oxford, 1945. Gregg De Young. Translating Playfair’s Geometry into Arabic: Mathematics and Missions. In Nathan Sidoli and Glenn Van Brummelen, editors, From Alexandria Through Baghdad: Surveys and Studies in the Ancient Greek and Medieval Mathematical Sciences in Hounour of J.L. Berggren, pages 503–527. Springer Verlag, New York – Heidelberg, 2014.

180

References

Jean-Pierre Delahaye. Le fascinant nombre . Belin-Pour la Science, Paris, 1997. Jean-Pierre Delahaye. Viète et les codes secrets. In Barbin and Boyé (2005). David Dennis. Historical Perspectives for the Reform of Mathematics Curriculum: Geometric Curve Drawing Devices. Master’s thesis, Cornell University, 1995. Heinrich Dörrie. 100 Great Problems of Elementary Mathematics. Dover Publications, New York, 1965. Howard Eves. From Five Fingers to Infinity, chapter Omar Khayyam’s Solution of Cubic Equations, pages 302–303. In Swetz (1994), 1994. Howard Eves. College Geometry. Jones and Bartlett Publishers, Boston – London, 1995. Julie Fink and Nicholas Molbert. Sectioning Angles using Hyperbolic Curves. Rose-Hulman Undergraduate Mathematics Journal, 14:183–194, 2013. David R. Finston and Patrick J. Morandi. Abstract Algebra; Structure and Application. Undergraduate Texts in Mathematics and Technology. Springer International Publishing Switzerland, Cham – Heidelberg – New York - Dordrecht – London, 2014. Kuno Fladt. Geschichte und Theorie der Kegelschnitte und der Flächen zweiten Grades. Ernst Klett Verlag, Stuttgart, 1967. Menso Folkerts. Die Entwicklung und Bedeutung der Visierkunst als Beispiel der praktischen Mathematik der frühen Neuzeit. Humanismus und Technik, 18/1:1–41, 1974. Michael. N. Fried and Sabetai Unguru. Apollonius of Perga’s Conica : text, context, subtext. Brill, Leiden, 2012. Aleksandr O. Gelfond. Sur le septième Problème de Hilbert. Bulletin de l’Académie des Sciences de l’URSS. Classe des sciences mathématiques, VII:623–634, 1934. Stephen Gersh. Rewriting the Proslogion; Nicholas of Cusa’s Transformation of Anselm of Canterbury’s Proof of the Existence of God. Epekeina, 1:65–89, 2012. Judith V. Grabiner. The Origins of Cauchy’s Rigorous Calculus. MIT Press, Cambridge, Mass., 1981. Judith V. Grabiner. A Mathematician among the Molasses Barrels: MacLaurin’s unpublished Memoir on Volumes. Proceedings of the Edinburgh Mathematical Society, 39:193–240, 1996. Judith V. Grabiner. ’Some Disputes of Consequence’: McLaurin among the Molasses Barrels. Social Studies of Science, 28:139–168, 1998. René Grothmann. The Geometry Program C.a.R. International Journal of Computer Discovered Mathematics, 1(1):45–61, 2016. Iris Gulikers and Klaske Blom. ’A Historical Angle’, a Survey of Recent Literature on the Use and Value of History in Geometrical Education. Educational Studies in Mathematics, 47:223–258, 2001. Thomas C. Hales. A Proof of the Kepler Conjecture. Annals of Mathematics, 162(3):1065–1185, 2005. Vagn L. Hansen. Shadows of the Circle. World Scientific, Singapore – River Edge, NJ – London, 1998. Robin Hartshorne. Geometry: Euclid and Beyond. Springer-Verlag, New York, 2000. Thomas L. Heath. The works of Archimedes. Dover Publications, New York, 1953. Thomas L. Heath. Euclid; The Thirteen Books of the Elements (3 vols). Dover Publications, New York, 1956. Thomas L. Heath. Treatise on Conic Sections: with Introductions, including an Essay on the Earlier History of the Subject. Heffer, Cambridge, 1961. Thomas L. Heath. A History of Greek Mathematics (2 vols). Dover Publications, New York, 1981. Philippe P. A. Henry. La solution de François Viète du problème d’Adriaan van Roomen. URL http://egg.epfl. ch/~phenry/VieteRoomen.pdf. Retrieved 18-8-2014. Jan P. Hogendijk. On the Trisection of an Angle and the Construction of a Regular Nonagon by means of Conic Sections in Medieval Islamic Geometry. Preprint 113, University of Utrecht, Department of Mathematics, 1979. Jan P. Hogendijk. How Trisections of the Angle were transmitted from Greek to Islamic Geometry. Historia Mathematica, 8:417–438, 1981. Jan P. Hogendijk. Abu’l-Jud’s answer to a question of al-Biruni concerning the regular heptagon. In D.A. King and G. Saliba, editors, From Deferent to Equant: a volume of studies in the ancient and medieval Near East in honor of E.S. Kennedy, volume 500 of Annals of the New York Academy of Sciences, pages 175–184. New York Academy of Sciences, New York, 1987. Markus Hohenwarter, Judith Hohenwarter, Yves Kreis, and Zsolt Lavicza. Teaching and Learning Calculus with Free Dynamic Mathematics Software GeoGebra, 2008. URL http://tsg.icme11.org/document/get/666. Retrieved 15-08-2015.

181 References

Mary Jaeger. Archimedes and the Roman Imagination. The University of Michigan Press, Ann Arbor, 2008. Victor J. Katz. A History of Mathematics: an Introduction. Addison-Wesley, Reading, Mass., 2nd edition, 1998. Victor J. Katz, editor. The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook. Princeton University Press, Princeton-Woodstock, 2007. James King and Doris Schattschneider, editors. Geometry Turned On: Dynamic Software in Learning, Teaching, and Research, number 41 in Notes, Washington, DC, 1997. The Mathematical Association of America. Dick Klingens. Kwadreerbare maantjes, 01 2008. URL http://www.pandd.nl/downloads/kwadreerbaar.pdf. Retrieved 26-8-2014. Wilbur R. Knorr. The Ancient Tradition of Geometric Problems. Dover Publications, New York, 1986. Wilbur R. Knorr. The Ancient Tradition of Geometric Problems. Dover Publications, New York, 1993. Helmut Koch. Der Briefwechsel von Leonhard Euler und Christian Goldbach. Birkhäuser Verlag, Basel, 2007. Ulrich Kortenkamp, Axel M. Blessing, Christian Dohrmann, Yves Kreis, Paul Libbrecht, and Christian Mercat. Interoperable interactive geometry for europe – first technological end educational results and future cchallenge of the intergeo project. Proceedings of CERME 6, January 28th-February 1st 2009, 2010. URL http://ife.ens-lyon.fr/editions/editions-electroniques/cerme6/. Steven G. Krantz. The Proof is the Pudding; The Changing Nature of Mathematical Proof. Springer Science & Busines Media LLC, New York, NY, 2011. Colette Laborde. Dynamic Geometry Environments as a Source of Rich Learning Contexts for the Complex Activity of Proving. Educational Studies in Mathematics, 44(1):151–161, 2000. Peter Lanser. De laatste stelling van Fermat, volume 7 of Zebra. Epsilon, Utrecht, 2000. J. Dennis Lawrence. A Catalog of Special Plane Curves. Dover books on advanced mathematics. Dover Publications, New York, NY, 1972. Falko Lorenz. Algebra, Volume I: Fields and Galois Theory. Universitext. Springer, New York, NY, 2006. Arman Margaryan and Nerses Aramian. Compass constructions, 10 2014. URL http://arxiv.org/abs/1410. 3042. Ramón Marrades and Ángel Gutiérrez. Proofs produced by Secondary School Students learning Geometry in a Dynamic Computer Environment. Educational Studies in Mathematics, 44(1):87–125, 2000. George E. Martin. Geometric Constructions. Springer-Verlag, New York, 1998. J. P. McCarthy. The Cissoid of Diocles. The Mathematical Gazette, 25(263):12–15, February 1941. Ad Meskens. Wine gauging at late Sixteenth and early Seventeenth Century Antwerp. Historia Mathematica, 21:121–147, 1994. Ad Meskens. The Jesuit Mathematics School in Antwerp in the early Seventeenth Century. The Seventeenth Century, 12:11–22, 1997. Ad Meskens. Joannes della Faille s.j.: Mathematics, Modesty and Missed Opportunities, volume 53 of Belgisch historisch instituut te Rome. Bibliotheek. Belgisch Historisch Instituut te Rome, Brussel-Rome, 2005. Ad Meskens. Travelling Mathematics: the fate of Diophantos’ Arithmetic, volume 41 of Science Networks. Historical Studies. Birkhäuser Verlag, Basel, 2010. Ad Meskens. Practical Mathematics in a Commercial Metropolis: Mathematical Life in late 16th century Antwerp, volume 31 of Archimedes. Springer Science & Business Media B.V., Dordrecht, 2013. Ad Meskens and Paul Tytgat. Met passer, liniaal en neusislat. Number 41 in Zebra. Epsilon Uitgaven, Amsterdam, 2015. Erich Meuthen. Nikolaus von Kues, 1401 – 1464: Skizze einer Biographie. Aschendorff, Münster, 1982. Dermot Moran. Platonism at the Origins of Modernity: Studies on Platonism and Early Modern Philosophy, chapter Nicholas of Cusa (1401–1464): Platonism at the Dawn of Modernity, pages 9–29. Springer Netherlands, Dordrecht, 2008. Pier Daniele Napolitani. Between Myth and Mathematics: the Vicissitudes of Archimedes and his Work. Lettera Matematica, 1:105–112, 2013. R.W.D. Nickalls. Viète, Descartes and the Cubic Equation. The Mathematical Gazette, 90:203–208, 2006. John J. O’Connor and Edmund F. Robertson. The MacTutor History of Mathematics archive. URL http:// www-history.mcs.st-and.ac.uk/. Retrieved 25-8-2014.

182

References

Alexander Ostermann and Gerhard Wanner. Geometry by Its History. Undergraduate Texts in Mathematics. Springer Verlag, Dordrecht, 2012. Stephanos A. Paipetis and Marco Ceccarelli, editors. The Genius of Archimedes – 23 Centuries of Influence on Mathematics, Science and Engineering, volume 11 of History of mechanism and Machine Science, Dordrecht – Heidelberg – London – New York, NY, 2010. Springer Science & Busines Media B.V. Elena Pastuchová, Jana Galanová, and Alfonz Smola. Application of atypical curves for the formation surfaces and reflectors. Journal of Electrical Engineering, 53:336–338, 2002. George M. Phillips. Mathematics is not a Spectator Sport. Springer Science & BusinesSpringer Inc., New York, NY, 2005. M.M. Postnikov and Abe Shenitzer (transl.). The Problem of Squarable Lunes. The American Mathematical Monthly, 107:645–651, 2000. Jürgen Richter-Gebert and Ulrich H. Kortenkamp. The Cinderella.2 Manual. Springer Verlag, Berlin – Heidelberg, 2012. Franz Rothe. Several Topics in Geometry, 01 2014. URL http://math2.uncc.edu/~frothe/. Retrieved 15-2-2015. Ken Saito. Doubling the Cube: A New Interpretation of Its Significance for Early Greek Geometry. Historia Mathematica, 22:119–137, 1995. C. Schoy. Drei planimetrische Aufgaben des arabischen Mathematikers Abu’l=Jud Muhammad ibn al=Lith. ISIS, 7:5–8, 1925. Heike Sefrin-Weis. Pappus of Alexandria. Book 4 of the Collection. Springer Verlag, London, 2010. Brian J. Shelburne. The Five Quadrable (Squarable) Lunes, 04 2008. URL http://www4.wittenberg.edu/ academics/mathcomp/bjsdir/TheFiveLunes120408.pdf. Retrieved 26-8-2014. Eugene V. Shikin. Handbook and Atlas of Curves. CRC Press, Boca Raton – New York, NY – London – Tokyo, 1995. Simon Singh. Fermat’s Enigma: the Quest to solve the World’s Greatest Mathematical Problem. Walker, New York, 1997. Matthijs H. Sitters. Sybrandt Hansz Cardinael; rekenmeester en wiskundige. Verloren, Hilversum, 2008. Alfons J.E.M. Smeur. De verhandeling over de cirkelkwadratuur van Franco van Luik van omstreeks 1050. Koninklijke Vlaamse Academie, Brussel, 1968. Alfons J.E.M. Smeur and Menso Folkerts. A Treatise on the Squaring of the Circle by Franco of Liège, of about 1050. I. Archives Internationales d’Histoire des Sciences, 26:59–105, 1976a. Alfons J.E.M. Smeur and Menso Folkerts. A Treatise on the Squaring of the Circle by Franco of Liège, of about 1050. II. Archives Internationales d’Histoire des Sciences, 26:225–253, 1976b. John Stillwell. The Four Pillars of Geometry. Springer Science & BusinesSpringer Inc., New York, NY, 2005. John Stillwell. Mathematics and Its History. Undergraduate Texts in Mathematics. Springer Science & Busines Media LLC, New York, NY, 2010. Dirk J. Struik. From Five Fingers to Infinity, chapter Omar Khayyam, Mathematician, pages 297–301. In Swetz (1994), 1994. Frank J. Swetz, editor. From Five Fingers to Infinity. Open Court, Peru, Ill., 1994. Nikolaj Tschebotaröw. Über quadrierbare Kreisbogenzweiecke. I. Mathematische Zeitschrift, 39(1):161–175, 1935. ISSN 0025-5874. C.J. Vamvacas. The Founders of Western Thought – The Presocratics, Boston Studies in the Philosophy of Science, volume 257 of Boston Studies in the Philosophy of Science, chapter Thales of Miletus (ca. 625 – 546 B.C.), pages 2 – 33. Springer Science+Business Media B.V., Dordrecht, 2009. Glen Van Brummelen. Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry. Princeton University Press, Princeton, 2012. Glen Van Brummelen and Michael Kinyon, editors. Mathematics and the Historian’s Craft; The Kenneth O. May Lectures. CMS Books in Mathematics. Canadian Mathematical Society – Societé mathématique du Canada, New York, NY, 2005. Delphi van de Weyer. Opera Mathematica: La Géometrie, S. Marolois. Master’s thesis, Vrije Universiteit Brussel, Brussel, 2011. Iris van Gulik-Gulikers. Geschiedenis van de niet-Euclidische meetkunde, volume 21 of Zebra. Epsilon, Utrecht, 2005.

183 References

Floor van Lamoen. Passen en meten met cirkels: de arbelos van Archimedes, volume 30 of Zebra. Epsilon Uitgaven, Utrecht, 2009. Herman van Looy. Chronologie en analyse van de mathematische handschriften van G. a Sancto Vincentio. PhD thesis, K.U. Leuven, 1979. Jan A. van Maanen. Aangeslibd land, kegelsnedetrekkers en oneindige bierglazen. In Thomas Coppens, editor, Voordrachten en werkgroepen van jhet vijfde congres van de Vlaamse Vereniging Wiskunde Leraars Neerpelt juli 1987, volume 8 of Monografieën van de VVWL, pages 203–220, Wilrijk, 1987. VVWL. Paul Ver Eecke. Les œuvres complètes d’Archimède. Desclée de Brouwer, Paris, 1921. B.A. Vermaseren. Atlas der algemene en vaderlandse geschiedenis. Wolters-Noordhoff, Groningen, 1977. Cettina Voza. The Death of Archimedes: a Reassessment. In Paipetis and Ceccarelli (2010), pages 507–514. Steven Wepster. In de ban van de cirkel. Euclides, 85/3:98–100, 2009. Steven Wepster. Ludolph van Ceulen (1540–1610), Meester der rekenmeesters. Pythagoras, 49(3):12–15, 2010a. Steven Wepster. Hoe van Ceulen  insloot. Pythagoras, 49(3):26–28, 2010b. Steven Wepster and Marjanne de Nijs. Meester Ludolphs koordenvierhoek, volume 31 of Zebra. Epsilon Uitgaven, Utrecht, 2010. William F. Wertz Jr. Nicolaus of Cusa’s On the Quadrature of the Circle, 2001. URL http://www. schillerinstitute.org/fid_97-01/012_Cusa_quad_circ.html. Retrieved 26-8-2014. Jiyuan Yu. The Structure of Being in Aristotle’s Metaphysics, volume 52 of The new synthese historical library: texts and studies in the history of philosophy. Kluwer, Dordrecht, 2003.

184

Index

Index

A Ab¯u’l-J¯ud 122 agrimensores 88 Al-B¯ır¯un¯ı 122 Alexandria 11, 59 Anthonisz, Adriaan 95 Antwerp 65, 66, 95, 99 Apollonius of Perga 14 – Apollonius’ problem 14, 16–19, 36, 38 – Conics 14 – Tangencies 14 Archimedes 59–61, 79, 93–95, 99, 123 – arbelos 79 – Archimedes’ spiral 63, 64, 75, 144 – Archimedes’ trisection 63, 148 Aristotle 12, 95 Asia Minor 11 Athens 11, 12, 43

B Bachet de Méziriac, Claude Gaspard 2 Bernoulli, Nicholas I 1 Bolyai, János 29 Bruges 65 Brussels 65, 66

C Ceva, Giovanni 74 Clavius s.j., Christopher 65 curve – Archimedes’ spiral 63–65, 75, 144 – botanic curve 74 – Ceva’s trisectrix 74 – cissoid of Diocles 53, 142 – conchoid 7, 25, 56–58, 63, 68, 143, 144 – Dinostratus’ quadratrix 71, 72, 75–78, 145 – Dürer’s conchoid 7

– ellipse 12, 19, 22, 104, 132 – hyperbola 12, 16–18, 46, 52, 59, 100, 137, 138, 142 – limaçon 130, 134 – McLaurin’s trisectrix 73 – Nicomedes’ conchoid 52, 55–57, 63 – parabola 12–14, 44, 63, 75, 99–103, 131, 136, 137, 164, 165 – Pascal’s limaçon 9, 61, 62, 134 – tractrix 22, 23 Cusanus, Nicolas 90–92, 159, 160

D de Aguilon s.j., Franciscus 65–69 – Opticorum Libri sex 66, 67 Delft 95 della Faille s.j., Joannes 104 Delos, island 43 – oracle 43 Dinostratus 44, 76, 77 – quadratrix 71, 72, 75–78, 145 Diocles – cissoid 53, 142 Diophantus 2 Dürer – conchoid 7

E equation – quadratic 37 Euclid 27–30, 43, 59, 105, 144, vii – Elements 4, 27–29, 43 – parallel postulate 28–30 Eudoxus 44 Euler, Leonhard 1 Eutocius 14

F Fermat, Pierre de 2 – Fermat’s Last Theorem 1 Fontenay-le-Comte 36 Franco of Liège 88–90, 98 Fulbert of Chartres 88

G Goldbach, Christian 1 – Goldbach’s conjecture 1 Grienberger s.j., Christopher 99

H Hales, Thomas 2 Henry IV, King of France 15, 36 Hermite, Charles 119 Heron of Alexandria 122 Hildesheim 95 Hippias of Elis 71, 76 Hippocrates of Chios 43, 44, 78, 81, 83, 85, 86, 114, 115, 156, 170 – Elements 43 – Hippocrates’ lunes 78, 80, 84 Huygens, Christian 22, 100

I Ibn Qurra, Thabit 14 Ibn-al-Haytham 67

J jesuit church – Antwerp, Carolus Borromeus 65 – Mons 65 – Tournai 65 jesuit college – ’s Hertogenbosch – Antwerp 65 – Brussels 65 – Douai 65

65

185 Index

– Kortrijk 65 – Paris 65

K Kepler, Johannes 2, 98, 99 – Kepler conjecture 2 Kortrijk 65

L Leibniz, Gottfried 1, 22 Leiden 95 – Engineering school 21 – University 95 Leonardo’s Claw 81, 82, 152, 153 Leuven 14, 65 – University 14, 90 Lindemann, Ferdinand 119 Lobachevsky, Nikolai 29 locus of B with A; definition 6

M Mainz 14 Marolois, Samuel 35, 36 – Opera Mathematica 36 McLaurin, Colin 73 – trisectrix 73 Menaechmus 44, vii Menelaus – Menelaus’ theorem 49, 51 Milete 11

N Napoleon’s theorem

5

neusis 23–26, 49, 50, 52, 55, 56, 61–64, 67, 68, 85, 86, 125, 134, 144, 157 Nicomedes 55, 59, 112 – conchoid 52, 55–57, 63 non-Euclidean geometry 29

T

P Paris 36 Pascal, Blaise 9 Pascal, Etienne 9 Perrault, Claude 22 Philip II, King of Spain 36, 65 Philip III, King of Spain 36 Pirate’s dilemma 5 Plato 12, 44–46, 71 Playfair, John 29 Ptolemy, King of Egypt 27, vii Pythagoras – Pythagoras’ theorem 22, 27, 28, 36, 65–67, 78, 79, 144, 150–152, 158, 160 – Pythagorean triplet 1, 2, 128

R Radolph of Liège

Sicily 11, 59, 65 slider, definition 5 Socrates 12 Sparta 11, 71 Syracuse 11, 59, 60

88

S Saccheri, Girolamo 29 Salvasutras 93 Sancto Vincentio s.j., Gregorius a 45, 65–67, 69, 70, 99–101, 103, 104, 144 – Problema Austriacum 99, 100 Scaliger, Joseph Justus 95

Thales of Milete 11 – circle theorem 11, 134, 137 – intercept theorem 11, 33, 59, 84, 105, 106, 176 The Simpsons (TV-series) 2

V van Ceulen, Ludolff 14, 95–97 – Vanden Circkel 95 van Roomen, Adriaan 14–16, 95–97, 111, 112, 118 – IdeæMathematica 15, 95 – In Archimedis 95 van Schooten, Frans 20–22 – Exercitationum Mathematicorum 20 vander Eycke, Simon 95 Viète, François 14–16, 36–38, 95, 111, 112, 121, 124 – Ad Problema 16 – Opera Mathematica 38 – Viète’s Ladder 121, 122

W Wantzel, Pierre 108 Wiles, Andrew 2 Wine gauger 46, 47 Würzburg 14