Archimedes and the Arbelos
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Archive of SID
Archive of SID T r kh-e Elm: Iranian Journal for the History of Science, 6 (2008), pp. 1-36 Two Beautiful Geometrical Theorems by Ab Sahl K h in a 17th Century Dutch Translation Jan P. Hogendijk University of Utrecht [email protected] Abstract (received: 21/09/2008 - accepted: 15/11/2008) This article is devoted to two theorems on tangent circles, which were discovered by the Iranian geometer Ab Sahl K h (4th century A.H.). The two theorems were inspired by the Book of Lemmas (ma kh dh t) attributed to Archimedes. K h 's original treatise is lost, but the two theorems are found in Na r al-D n s 's edition of the Lemmas of Archimedes. They then appeared in Latin translations in 1659 in London, and again in 1661 in Florence, and in 1695 in a revised Dutch version in Amsterdam. The present article compares the original Arabic version of K h 's theorems (in the presentation of s ) with the revised Dutch version. Keywords: K h , s , Archimedes, geometry, circles, 17th century Dutch mathematics Introduction Waijan ibn Rustam Ab Sahl K h was an Iranian geometer and astronomer, who ßourished in the second half of the 4th century A.H./ 10th century A.D. (for biographical data and a list of works of him, see sezgin, V, 314-321, VI, 218-219; Rosenfeld and Ihsano lu, 102-105; for a general analysis of his works, see Berggrenn). K h had an outstanding reputation among his contemporaries: he was even called the Master of his Age in the Art of Geometry (the Arabic term is shaykh asrihi f in at al-handasa; see Berggren, 178). -
Squaring the Circle a Case Study in the History of Mathematics the Problem
Squaring the Circle A Case Study in the History of Mathematics The Problem Using only a compass and straightedge, construct for any given circle, a square with the same area as the circle. The general problem of constructing a square with the same area as a given figure is known as the Quadrature of that figure. So, we seek a quadrature of the circle. The Answer It has been known since 1822 that the quadrature of a circle with straightedge and compass is impossible. Notes: First of all we are not saying that a square of equal area does not exist. If the circle has area A, then a square with side √A clearly has the same area. Secondly, we are not saying that a quadrature of a circle is impossible, since it is possible, but not under the restriction of using only a straightedge and compass. Precursors It has been written, in many places, that the quadrature problem appears in one of the earliest extant mathematical sources, the Rhind Papyrus (~ 1650 B.C.). This is not really an accurate statement. If one means by the “quadrature of the circle” simply a quadrature by any means, then one is just asking for the determination of the area of a circle. This problem does appear in the Rhind Papyrus, but I consider it as just a precursor to the construction problem we are examining. The Rhind Papyrus The papyrus was found in Thebes (Luxor) in the ruins of a small building near the Ramesseum.1 It was purchased in 1858 in Egypt by the Scottish Egyptologist A. -
Archimedes of Syracuse
5 MARCH 2020 Engineering: Archimedes of Syracuse Professor Edith Hall Archimedes and Hiero II’s Syracuse Archimedes was and remains the most famous person from Syracuse, Sicily, in history. He belonged to the prosperous and sophisticated culture which the dominantly Greek population had built in the east of the island. The civilisation of the whole of ancient Sicily and South Italy was called by the Romans ‘Magna Graecia’ or ‘Great Greece’. The citis of Magna Graecia began to be annexed by the Roman Republic from 327 BCE, and most of Sicily was conquered by 272. But Syracuse, a large and magnificent kingdom, the size of Athens and a major player in the politics of the Mediterranean world throughout antiquity, succeeded in staying independent until 212. This was because its kings were allies of Rome in the face of the constant threat from Carthage. Archimedes was born into this free and vibrant port city in about 287 BCE, and as far as we know lived there all his life. When he was about twelve, the formidable Hiero II came to the throne, and there followed more than half a century of peace in the city, despite momentous power struggles going on as the Romans clashed with the Carthaginians and Greeks beyond Syracuse’s borders. Hiero encouraged arts and sciences, massively expanding the famous theatre. Archimedes’ background enabled him to fulfil his huge inborn intellectual talents to the full. His father was an astronomer named Pheidias. He was probably sent to study as a young man to Alexandria, home of the famous library, where he seems to have became close friend and correspondent of the great geographer and astonomer Eratosthenes, later to become Chief Librarian. -
Unit 3, Lesson 1: How Well Can You Measure?
GRADE 7 MATHEMATICS NAME DATE PERIOD Unit 3, Lesson 1: How Well Can You Measure? 1. Estimate the side length of a square that has a 9 cm long diagonal. 2. Select all quantities that are proportional to the diagonal length of a square. A. Area of a square B. Perimeter of a square C. Side length of a square 3. Diego made a graph of two quantities that he measured and said, “The points all lie on a line except one, which is a little bit above the line. This means that the quantities can’t be proportional.” Do you agree with Diego? Explain. 4. The graph shows that while it was being filled, the amount of water in gallons in a swimming pool was approximately proportional to the time that has passed in minutes. a. About how much water was in the pool after 25 minutes? b. Approximately when were there 500 gallons of water in the pool? c. Estimate the constant of proportionality for the number of gallons of water per minute going into the pool. Unit 3: Measuring Circles Lesson 1: How Well Can You Measure? 1 GRADE 7 MATHEMATICS NAME DATE PERIOD Unit 3: Measuring Circles Lesson 1: How Well Can You Measure? 2 GRADE 7 MATHEMATICS NAME DATE PERIOD Unit 3, Lesson 2: Exploring Circles 1. Use a geometric tool to draw a circle. Draw and measure a radius and a diameter of the circle. 2. Here is a circle with center and some line segments and curves joining points on the circle. Identify examples of the following. -
Archimedes and Pi
Archimedes and Pi Burton Rosenberg September 7, 2003 Introduction Proposition 3 of Archimedes’ Measurement of a Circle states that π is less than 22/7 and greater than 223/71. The approximation πa ≈ 22/7 is referred to as Archimedes Approximation and is very good. It has been reported that a 2000 B.C. Babylonian approximation is πb ≈ 25/8. We will compare these two approximations. The author, in the spirit of idiot’s advocate, will venture his own approximation of πc ≈ 19/6. The Babylonian approximation is good to one part in 189, the author’s, one part in 125, and Archimedes an astonishing one part in 2484. Archimedes’ approach is to circumscribe and inscribe regular n-gons around a unit circle. He begins with a hexagon and repeatedly subdivides the side to get 12, 24, 48 and 96-gons. The semi-circumference of these polygons converge on π from above and below. In modern terms, Archimede’s derives and uses the cotangent half-angle formula, cot x/2 = cot x + csc x. In application, the cosecant will be calculated from the cotangent according to the (modern) iden- tity, csc2 x = 1 + cot2 x Greek mathematics dealt with ratio’s more than with numbers. Among the often used ratios are the proportions among the sides of a triangle. Although Greek mathematics is said to not know trigonometric functions, we shall see how conversant it was with these ratios and the formal manipulation of ratios, resulting in a theory essentially equivalent to that of trigonometry. For the circumscribed polygon We use the notation of the Dijksterhuis translation of Archimedes. -
The Arbelos in Wasan Geometry: Chiba's Problem
Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.8, (2019), Issue 2, pp.98-104 THE ARBELOS IN WASAN GEOMETRY: CHIBA’S PROBLEM HIROSHI OKUMURA Abstract. We consider a sangaku problem involving an arbelos with two congruent circles, and give several conditions where the two congruent circles appear. Also we show several pairs of congruent circles and Archimedean circles. 2010 Mathematical Subject Classification: 01A27, 51M04, 51N20. Keywords and phrases: arbelos, Archimedean circle, non-Archimedean congruent circles. 1. INTRODUCTION We consider the arbelos appeared in Wasan geometry, and consider an arbelos formed by three semicircles α, β and γ with diameters AO , BO and AB , respectively for a point O on the segment AB . The radii of α and β are denoted by a and b, respectively. We consider the following sangaku .(in 1880 [5] (see Figure 1 (نproblem proposed by Chiba ( ઍ༿ӫ࣏҆ δ ε γ α h β B O H A Figure 1. Problem 1. For a point H on the segment AO, let h be the perpendicular to AB at H. Let δ be the circle touching α and β externally and h, and let ε be the incircle of the curvilinear triangle made by α, γ and h. If δ and ε touch and δ is congruent to the circle of diameter AH, find the radius of ε in terms of a and b. Circles of radius rA = ab /(a + b) are said to be Archimedean. In this paper we generalize the problem and consider two circles touching at a point on a perpendicular to the line AB . -
Archimedes Palimpsest a Brief History of the Palimpsest Tracing the Manuscript from Its Creation Until Its Reappearance Foundations...The Life of Archimedes
Archimedes Palimpsest A Brief History of the Palimpsest Tracing the manuscript from its creation until its reappearance Foundations...The Life of Archimedes Birth: About 287 BC in Syracuse, Sicily (At the time it was still an Independent Greek city-state) Death: 212 or 211 BC in Syracuse. His age is estimated to be between 75-76 at the time of death. Cause: Archimedes may have been killed by a Roman soldier who was unaware of who Archimedes was. This theory however, has no proof. However, the dates coincide with the time Syracuse was sacked by the Roman army. The Works of Archimedes Archimedes' Writings: • Balancing Planes • Quadrature of the Parabola • Sphere and Cylinder • Spiral Lines • Conoids and Spheroids • On Floating Bodies • Measurement of a Circle • The Sandreckoner • The Method of Mechanical Problems • The Stomachion The ABCs of Archimedes' work Archimedes' work is separated into three Codeces: Codex A: Codex B: • Balancing Planes • Balancing Planes • Quadrature of the Parabola • Quadrature of the Parabola • Sphere and Cylinder • On Floating Bodies • Spiral Lines Codex C: • Conoids and Spheroids • The Method of Mechanical • Measurement of a Circle Problems • The Sand-reckoner • Spiral Lines • The Stomachion • On Floating Bodies • Measurement of a Circle • Balancing Planes • Sphere and Cylinder The Reappearance of the Palimpsest Date: On Thursday, October 29, 1998 Location: Christie's Acution House, NY Selling price: $2.2 Million Research on Palimpsest was done by Walter's Art Museum in Baltimore, MD The Main Researchers Include: William Noel Mike Toth Reviel Netz Keith Knox Uwe Bergmann Codex A, B no more Codex A and B no longer exist. -
Archimedes and the Arbelos1 Bobby Hanson October 17, 2007
Archimedes and the Arbelos1 Bobby Hanson October 17, 2007 The mathematician’s patterns, like the painter’s or the poet’s must be beautiful; the ideas like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics. — G.H. Hardy, A Mathematician’s Apology ACBr 1 − r Figure 1. The Arbelos Problem 1. We will warm up on an easy problem: Show that traveling from A to B along the big semicircle is the same distance as traveling from A to B by way of C along the two smaller semicircles. Proof. The arc from A to C has length πr/2. The arc from C to B has length π(1 − r)/2. The arc from A to B has length π/2. ˜ 1My notes are shamelessly stolen from notes by Tom Rike, of the Berkeley Math Circle available at http://mathcircle.berkeley.edu/BMC6/ps0506/ArbelosBMC.pdf . 1 2 If we draw the line tangent to the two smaller semicircles, it must be perpendicular to AB. (Why?) We will let D be the point where this line intersects the largest of the semicircles; X and Y will indicate the points of intersection with the line segments AD and BD with the two smaller semicircles respectively (see Figure 2). Finally, let P be the point where XY and CD intersect. D X P Y ACBr 1 − r Figure 2 Problem 2. Now show that XY and CD are the same length, and that they bisect each other. -
Solutions Manual to Accompany Classical Geometry
Solutions Manual to Accompany Classical Geometry Solutions Manual to Accompany CLASSICAL GEOMETRY Euclidean, Transformational, Inversive, and Projective I. E. Leonard J. E. Lewis A. C. F. Liu G. W. Tokarsky Department of Mathematical and Statistical Sciences University of Alberta Edmonton, Canada WILEY Copyright © 2014 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representation or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. -
Chapter 6 the Arbelos
Chapter 6 The arbelos 6.1 Archimedes’ theorems on the arbelos Theorem 6.1 (Archimedes 1). The two circles touching CP on different sides and AC CB each touching two of the semicircles have equal diameters AB· . P W1 W2 A O1 O C O2 B A O1 O C O2 B Theorem 6.2 (Archimedes 2). The diameter of the circle tangent to all three semi- circles is AC CB AB · · . AC2 + AC CB + CB2 · We shall consider Theorem ?? in ?? later, and for now examine Archimedes’ wonderful proofs of the more remarkable§ Theorems 6.1 and 6.2. By synthetic reasoning, he computed the radii of these circles. 1Book of Lemmas, Proposition 5. 2Book of Lemmas, Proposition 6. 602 The arbelos 6.1.1 Archimedes’ proof of the twin circles theorem In the beginning of the Book of Lemmas, Archimedes has established a basic proposition 3 on parallel diameters of two tangent circles. If two circles are tangent to each other (internally or externally) at a point P , and if AB, XY are two parallel diameters of the circles, then the lines AX and BY intersect at P . D F I W1 E H W2 G A O C B Figure 6.1: Consider the circle tangent to CP at E, and to the semicircle on AC at G, to that on AB at F . If EH is the diameter through E, then AH and BE intersect at F . Also, AE and CH intersect at G. Let I be the intersection of AE with the outer semicircle. Extend AF and BI to intersect at D. -
On Archimedes' Measurement of a Circle, Proposition 3
On Archimedes’ Measurement of a circle, Proposition 3 Mark Reeder February 2 1 10 The ratio of the circumference of any circle to its diameter is less than 3 7 but greater than 3 71 . Having related the area of a circle to its perimeter in Prop. 1, Archimedes next approximates the circle perimeter with circumscribed and inscribed regular polygons and then finds good rational estimates for these polygon perimeters, thereby approximating the ratio of circumference to diameter. The main geometric step is to see how the polygon perimeter changes when the number of sides is doubled. We will consider the circumscribed case. Let AC be a side of a regular circumscribing polygon, and let AD be a side of a regular polygon with the number of sides doubled. C D θ O A θ B To make Archimedes’ computation easier to follow, let x = AC, y = AD, r = OA, c = OC, d = OD. We want to express the new ratio y/r in terms of the old ratio x/r. But these numbers will be very small after a few subdivisions, so they will be difficult to estimate. Instead, we will express r/y in terms of r/x. These are big numbers, which can be estimated by integers. 1 From Euclid VI.3, an angle bisector divides the opposite side in the same ratio as the other two sides of a triangle. Hence CD : DA = OC : OA. In our notation, this means x − y c x c + r r r c = , or = , or = + . y r y r y x x From Euclid I.47, we have r c r2 = 1 + , x x2 so that r r r r2 = + 1 + (1) y x x2 Thus, the new ratio r/y is expressed in terms of the old ratio r/x, as desired. -
Pappus of Alexandria: Book 4 of the Collection
Pappus of Alexandria: Book 4 of the Collection For other titles published in this series, go to http://www.springer.com/series/4142 Sources and Studies in the History of Mathematics and Physical Sciences Managing Editor J.Z. Buchwald Associate Editors J.L. Berggren and J. Lützen Advisory Board C. Fraser, T. Sauer, A. Shapiro Pappus of Alexandria: Book 4 of the Collection Edited With Translation and Commentary by Heike Sefrin-Weis Heike Sefrin-Weis Department of Philosophy University of South Carolina Columbia SC USA [email protected] Sources Managing Editor: Jed Z. Buchwald California Institute of Technology Division of the Humanities and Social Sciences MC 101–40 Pasadena, CA 91125 USA Associate Editors: J.L. Berggren Jesper Lützen Simon Fraser University University of Copenhagen Department of Mathematics Institute of Mathematics University Drive 8888 Universitetsparken 5 V5A 1S6 Burnaby, BC 2100 Koebenhaven Canada Denmark ISBN 978-1-84996-004-5 e-ISBN 978-1-84996-005-2 DOI 10.1007/978-1-84996-005-2 Springer London Dordrecht Heidelberg New York British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Control Number: 2009942260 Mathematics Classification Number (2010) 00A05, 00A30, 03A05, 01A05, 01A20, 01A85, 03-03, 51-03 and 97-03 © Springer-Verlag London Limited 2010 Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licenses issued by the Copyright Licensing Agency.