Squaring the Circle. A Case Study in the History of Mathematics Part II



Similar documents
Archimedes Determination of Circular Area

Postulate 17 The area of a square is the square of the length of a. Postulate 18 If two figures are congruent, then they have the same.

Grade 7 & 8 Math Circles Circles, Circles, Circles March 19/20, 2013

Discovery of Pi: Day 1

INTERESTING PROOFS FOR THE CIRCUMFERENCE AND AREA OF A CIRCLE

The Early History of Pi

Chapter 7 Quiz. (1.) Which type of unit can be used to measure the area of a region centimeter, square centimeter, or cubic centimeter?

PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures.

Finding Pi with Archimedes s Exhaustion Method

GEOMETRY. Constructions OBJECTIVE #: G.CO.12

Geometry Final Exam Review Worksheet

CK-12 Geometry: Parts of Circles and Tangent Lines

Area. Area Overview. Define: Area:

Chapters 6 and 7 Notes: Circles, Locus and Concurrence

AREA AND VOLUME WHERE DO THE FORMULAS COME FROM?

Geometry Unit 6 Areas and Perimeters

Lesson 1: Introducing Circles

History of U.S. Measurement

SA B 1 p where is the slant height of the pyramid. V 1 3 Bh. 3D Solids Pyramids and Cones. Surface Area and Volume of a Pyramid

MENSURATION. Definition

Measuring Irregular Shapes and Circles

INTRODUCTION TO EUCLID S GEOMETRY

43 Perimeter and Area

SURFACE AREA AND VOLUME

Calculating the Surface Area of a Cylinder

1.2 Chord Tables of Hipparchus and Ptolemy (Copyright: Bryan Dorner all rights reserved)

GAP CLOSING. 2D Measurement. Intermediate / Senior Student Book

88 CHAPTER 2. VECTOR FUNCTIONS. . First, we need to compute T (s). a By definition, r (s) T (s) = 1 a sin s a. sin s a, cos s a

Grade 7/8 Math Circles February 10/11, 2015 Pi

Grade 8 Mathematics Measurement: Lesson 6

THE AREA AND CIRCUMFERENCE OF A CIRCLE

CSU Fresno Problem Solving Session. Geometry, 17 March 2012

1 Solution of Homework

Geometry. Higher Mathematics Courses 69. Geometry

Geometry Notes PERIMETER AND AREA

4.2 Euclid s Classification of Pythagorean Triples

Trigonometric Functions and Triangles

Homework 2 Solutions

GEOMETRIC MENSURATION

GEOMETRY COMMON CORE STANDARDS

Circumference and Area of a Circle

16 Circles and Cylinders

Tangent Properties. Line m is a tangent to circle O. Point T is the point of tangency.

Tangent circles in the hyperbolic disk

Selected practice exam solutions (part 5, item 2) (MAT 360)

Angles that are between parallel lines, but on opposite sides of a transversal.

Section 7.2 Area. The Area of Rectangles and Triangles

Geometry Enduring Understandings Students will understand 1. that all circles are similar.

Unit 3: Circles and Volume

New York State Student Learning Objective: Regents Geometry

Student Outcomes. Lesson Notes. Classwork. Exercises 1 3 (4 minutes)

Chapter 6 Notes: Circles

Dŵr y Felin Comprehensive School. Perimeter, Area and Volume Methodology Booklet

Objectives After completing this section, you should be able to:

CIRCLE COORDINATE GEOMETRY

For the circle above, EOB is a central angle. So is DOE. arc. The (degree) measure of ù DE is the measure of DOE.

1. A student followed the given steps below to complete a construction. Which type of construction is best represented by the steps given above?

SYMBOL AND MEANING IN MATHEMATICS

Archimedes and the Arbelos 1 Bobby Hanson October 17, 2007

Problem of the Month: Circular Reasoning

Unit 10 Geometry Circles. NAME Period

Trigonometric Functions: The Unit Circle

Curriculum Map by Block Geometry Mapping for Math Block Testing August 20 to August 24 Review concepts from previous grades.

YOU MUST BE ABLE TO DO THE FOLLOWING PROBLEMS WITHOUT A CALCULATOR!

Applications for Triangles

2004 Solutions Ga lois Contest (Grade 10)

Area and Circumference

Geometry - Semester 2. Mrs. Day-Blattner 1/20/2016

2006 Geometry Form A Page 1

Charlesworth School Year Group Maths Targets

alternate interior angles

Geometry Chapter 10 Study Guide Name

Handout #1: Mathematical Reasoning

Circle Name: Radius: Diameter: Chord: Secant:

Conjectures. Chapter 2. Chapter 3

Arc Length and Areas of Sectors

Circumference CHAPTER. 1

GEOMETRY CONCEPT MAP. Suggested Sequence:

56 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 224 points.

Advanced Euclidean Geometry

In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data.

Conjectures for Geometry for Math 70 By I. L. Tse

Angle - a figure formed by two rays or two line segments with a common endpoint called the vertex of the angle; angles are measured in degrees

Incenter Circumcenter

Lesson 18: Looking More Carefully at Parallel Lines

Solutions to Practice Problems

How To Solve The Pythagorean Triangle

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:

Session 7 Circles and Pi (π)

Circles: Circumference and Area Lesson Plans

Geometry Course Summary Department: Math. Semester 1

Pythagoras Theorem. Page I can identify and label right-angled triangles explain Pythagoras Theorem calculate the hypotenuse

Algebra Geometry Glossary. 90 angle

Week 1 Chapter 1: Fundamentals of Geometry. Week 2 Chapter 1: Fundamentals of Geometry. Week 3 Chapter 1: Fundamentals of Geometry Chapter 1 Test

6.1 Basic Right Triangle Trigonometry

CHAPTER 8, GEOMETRY. 4. A circular cylinder has a circumference of 33 in. Use 22 as the approximate value of π and find the radius of this cylinder.

Lies My Calculator and Computer Told Me

Geometry Regents Review

Circumference of a Circle

Transcription:

Squaring the Circle A Case Study in the History of Mathematics Part II

π It is lost in the mists of pre-history who first realized that the ratio of the circumference of a circle to its diameter is a constant. All the ancient civilizations knew this fact. Today we call this ratio π and express this relationship by saying that for any circle, the circumference C and the diameter d satisfy: C = πd. The use of the symbol π for this ratio is of relatively recent origin; the Greeks did not use the symbol. π was first used by the English mathematicians Oughtred (1647), Isaac Barrow (1664) and David Gregory (1697) to represent the circumference of a circle. The first use of π to represent the ratio of circumference to diameter was the English writer William Jones (1706). However, it did not come into common use until Euler adopted the symbol in 1737. 1

Archimedes Euclid XII.2 says that the ratio of the area of any circle to the square of its diameter is also a constant, but does not determine the value of this constant. It was Archimedes (287 212 B.C.) who determined the constant in his remarkable treatise Measurement of a Circle. There are only three propositions in this short work (or at least, that is all of that work that has come down to us) and the second proposition is out of place indicating that what we have is probably not the original version. 2 We shall look at the first and third proposition.

Archimedes A few preliminary ideas: The area of a regular polygon is easily determined if you know that the area of a triangle = ½bh (b the length of a side, h the length of the altitude drawn to that side). In a regular polygon of n sides (all sides equal, all angles equal), draw the lines from the center to each of the vertices creating n congruent triangles. apothem The area is thus n times the area of one of the triangles = n(½sa) where s is the length of a side and a the length of an apothem (line drawn from center perpendicular to a side). Since ns is the perimeter (Q) of the polygon we get : A = ½aQ.

Archimedes Consider a regular polygon inscribed in a circle. Let K be the difference in the areas (area of circle area of polygon). If you now double the number of sides of the polygon, the area you have added to the original polygon is more than ½K. By repeating the procedure, you can make the area difference as small as you like, or in other words, for any positive number K, you can find a regular polygon (with enough sides) inscribed in a circle so that the area difference is less than K. This is Eudoxus' method of exhaustion. It also works for circumscribed polygons.

Archimedes Proposition 1: The area of any circle equals the area of a right triangle one of whose sides is a radius of the circle and the other has length equal to the circumference of the circle. Pf: Let C be the length of the circumference of our circle and r its radius. Denote by A the area of the circle. The area of the triangle is T = ½rC. Case I: Assume that A > T. K = A T is a positive number. By the method of exhaustion, one can find an inscribed regular polygon so that A (area of the inscribed polygon) < A T. From this we see that T < area of the inscribed polygon. Since the polygon is inscribed, its perimeter Q is less than C and its apothem a is less than r, so area of the inscribed polygon = ½aQ < ½rC = T, which gives a contradiction.

Archimedes Proposition 1: The area of any circle equals the area of a right triangle one of whose sides is a radius of the circle and the other has length equal to the circumference of the circle. Pf (cont.): Case II : A < T Now T A is a positive number and we can find a regular polygon circumscribed about the circle so that (area of circumscribed polygon) A < T A. or, more simply area of circumscribed polygon < T. Since the polygon is circumscribed, the length of its apothem is r and its perimeter is greater than C so, area of circumscribed polygon = ½rQ > ½rC = T. Another contradiction!

Archimedes Proposition 1: The area of any circle equals the area of a right triangle one of whose sides is a radius of the circle and the other has length equal to the circumference of the circle. Pf (cont.): Since A > T and A < T have both led to contradictions, the only other possibility, A = T, must be true. QED The proof method is called a double reductio ad absurdum, exactly the same method used in Euclid XII.2. The method of exhaustion is used in both proofs to obtain the contradictions. The only difference in method is that Euclid (or Eudoxus) reduces the second case to the first case, so uses inscribed polygons in both parts of his proof, while Archimedes switches to circumscribed polygons for the second case. [Dunham is a bit too harsh on Euclid's result while he dumps praise on Archimedes!] 3

Archimedes Archimedes has proved that for any circle, A = ½rC, and since we know that C = πd, we get A = ½r(πd) = ½r(π2r) = πr 2 our familiar high school formula. Even though Archimedes showed the equivalence of a circle with a rectilinear figure, easily converted to a square, this is not a solution of the quadrature problem. The proof is indirect, it does not give a means for constructing the triangle with straightedge and compass. Before looking at Proposition 3, let's consider one of several methods known to the Greeks of using curves to perform a quadrature of the circle however, the curves used in this way can not themselves be constructed with straightedge and compass!

Quadratrix of Hippias Hippias of Elis (ca. 425 B.C.) is credited with inventing a curve which he used to trisect angles. Dinostratus (ca. 350 B.C.) is usually given credit for using Hippias' curve to find a quadrature of the circle. He and his followers referred to the curve as the Quadratrix ( squarer ) and this name stuck. Quadratrix I Quadratrix II 4 5 The defining property implies that : arc BED = (AB) 2 /AH which expresses the arc in terms of straight line segments, thus permitting a quadrature of the circle.

Archimedes In proposition 3 Archimedes turns his attention to the circumference of a circle. Again using inscribed and circumscribed regular polygons, their perimeters provide upper and lower bounds for the circumference of the circle. This gives him a means of calculating bounds for the number π. Proposition 3: The ratio of the circumference of any circle to its diameter is less than 3 1/7 (22/7) but greater than 3 10/71 (223/71). (3.140845... < π < 3.142857...) What is remarkable about this result is not the underlying idea, but rather the skill of Archimedes in carrying out the computations. He started with inscribed and circumscribed hexagons, then doubled the size, and again, and again and yet again, ending with 96-sided polygons. At each step he calculates the perimeters. This involves approximating radicals which is where he shows his genius.

The Computation of π Early Phase There has always been an interest in the precise value of π. As we have seen, the ancient Egyptians used π = 3.1604938.... Other ancient civilizations were not as precise, generally using π = 3. This can be seen in Babylonian clay tablets and in the Bible (I Kings: 7:23) Then He made the molten sea (circular), ten cubits from brim to brim, while a line of 30 cubits measured it around. After Archimedes improvements were made by taking larger and larger polygons (except for the Romans not very concerned with precision, they dropped back to the value 3 1/8) 6.

The Computation of π Early Phase 7 ca. 150 A.D. - The first improvement over Archimedes values was given by Claudius Ptolemy of Alexandria in the Almagest, the most famous Greek work on astronomy. Ptolemy gives a value of π = 377/120 = 3.1416... ca. 480 The early Chinese worker in mechanics, Tsu Ch'ung-chih, gave the approximation π = 355/113 = 3.1415929... correct to 6 decimal places. ca. 530 The early Hindu mathematician Āryabhata gave π = 62,832/20,000 = 3.1416... as an approximation. It is not known how this was obtained, but it could have been calculated as the perimeter of a regular inscribed polygon of 384 sides.

The Computation of π Early Phase ca. 1150 The later Hindu mathematician Bhāskara gave several approximations. He gave 3927/1250 as an accurate value, 22/7 as an inaccurate value, and 10 for ordinary work. 1429 Al-Kashi, astronomer royal to Ulugh Beg of Samarkand, computed π to sixteen decimal places using perimeters. 1579 The eminent French mathematician François Viète found π correct to nine decimal places using polygons having 393,216 sides. 1593 Adriaen van Roomen, more commonly known as Adrianus Romanus, of the Netherlands, found π correct to 15 places using polygons having 2 30 sides.

The Computation of π Early Phase 1610 Ludolph van Ceulen of the Netherlands computed π to 35 decimal places using polygons having 2 62 sides. He spent a large part of his life on this task, and his achievement was considered so extraordinary that his widow had the number engraved on his tombstone (now lost). To this day, the number is sometimes referred to as the Ludophine number. 1621 The Dutch physicist Willebrord Snell, best known for his discovery of the law of refraction, devised a trigonometric improvement of the classical method so that from each pair of bounds given by the classical method he was able to obtain considerably closer bounds. By his method, he was able to get van Ceulen's 35 places using a polygon with only 2 30 sides.

The Computation of π Early Phase 1630 Grienberger, using Snell's refinement, computed π to 39 decimal places. This was the last major attempt to compute π by the method of Archimedes.