1 the Greek Method of Exhaustion

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The Greek Method of Exhaustion:

Leading the Way to Modern Integration

M.S. Thesis

Presented in Partial Fulfillment of the Requirements for the Degree M.S. Mathematics for

Educators in the Graduate School of The Ohio State University

Chelsea E. DeSouza

Graduate Program in Mathematics

The Ohio State University

2012

Mathematics. Committee:

Dr. Rodica Costin, Advisor. Dr. Herb Clemens, Advisor

Chelsea E. DeSouza

2012

Abstract

This paper is a critical look at the Method of Exhaustion and how it is related to some of the more modern techniques of integration. This text moves through the development of the Method of Exhaustion to some applications. These applications include estimating the area of a circle using the Method of Exhaustion, and finding the exact area of a parabolic region by Archimedes quadrature of the parabola. These ideas are extensively developed and all constructions and proofs are given in the text and appendices. Similar techniques are then suggested to the reader to extend these results to find certain volumes in the appendix. Finally, a comparison is drawn between the Method of Exhaustion and integration techniques taught in modern classroom settings.

Dedication

This document is dedicated to my family and my professors who have all helped me a great deal in my mathematical career, which would not have started without all of

your guidance.

iii

Acknowledgments

Thank you to my two advisors, Dr. Rodica Costin and Dr. Herb Clemens who were both instrumental in the development of this thesis. A big thanks also to the entire staff of the Ohio State Mathematics Department, who pushed me to be better with my mathematics.

Vita 2007-2010 ...... Undergraduate TA, Mills College

2008-2010 ...... Undergraduate Grader, Mills College

2009...... CMOP, Undergraduate Research Associate

2010...... B.S. Mathematics, Mills College

2010-2012 ...... Graduate Teaching Associate, Department

of Mathematics, The Ohio State University

2012...... M.S. Mathematics for Educators, the Ohio

Fields of Study

M.S. Mathematics for Educators

Table of Contents

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita ...... v

List of Tables ...... viii

List of Figures ...... ix

Chapter 1: Introduction ...... 1

Chapter 2: Estimating the Area of the Circle ...... 3

Chapter 3: Archimedes Quadrature of the Parabola ...... 5

Chapter 4: A Modern Integration Technique: The Trapezoidal Method ...... 19

Chapter 5: The Method Exhaustion: Leading the Way to Modern Integration ...... 22

Bibliography ...... 23

Appendix A: Some Ruler and Compass Constructions and Proofs ...... 25

Appendix B: Every isosceles can be bisected into two congruent right triangles ...... 41

Appendix C: Area of any Isosceles Triangle ...... 43

Appendix D: π ...... 44

Appendix E: Estimating Volumes using the Method of Exhaustion ...... 45

vi Appendix F: Calculus Based Explanations of Property 1, 2, and 3 ...... 46

Appendix G: Examples of Using the Quadrature of the Parabola to find the Area of the

Bounded Region ...... 50

vii List of Tables

Table 1: Areas of n-gons Inscribed in a Circle ...... 4

viii

List of Figures

Figure 1: A Series of Regular Polygons Inscribed within a Circle ...... 2

Figure 2: Isosceles Triangle Inscribed in a Circle ...... 3

Figure 3: Arbitrary Isosceles Triangle ...... 4

Figure 4: Special Triangle in a Parabolic Segment ...... 5

Figure 5: The Parabolic Segment ...... 6

Figure 6: Properties of P ...... 6

Figure 7: Property 2 ...... 7

Figure 8: Adding Green Triangles ...... 8

Figure 9: Triangles of the Quadrature ...... 8

Figure 10: Points M and C ...... 9

Figure 11: Point D ...... 9

Figure 12: Triangle Congruence ...... 10

Figure 13: Triangular Height ...... 11

Figure 14: Heights of the Two Triangles ...... 11

Figure 15: Triangle P2ZM2 and M2OC ...... 12

Figure 16: External and Internal Angles of Triangles from Figure 16 ...... 12

Figure 17: Rotation of the Triangle ...... 13

Figure 18: Relation of Triangle PMB and PCB ...... 14

Figure 19: Height of PMB and PCB ...... 14

Figure 20: Unit Square ...... 16

Figure 21: Dividing S up into Four Equal Parts ...... 16

ix Figure 22: Square after the Coloring the Division Process ...... 17

Figure 23: Dividing the Unit Square up by i ...... 17

Figure 24: Trapezoid Method ...... 19

Figure 25: The Composite Trapezoidal Rule ...... 20

Figure 26: Generating Congruent Adjacent Chords in a Circle ...... 25

Figure 27: Finding Point C ...... 25

Figure 28: Perpendicularly Bisect a Segment ...... 26

Figure 29: The Perpendicular Bisector ...... 26

Figure 30: Perpendicular Bisector Proof ...... 27

Figure 31: Construct a Line through P Parallel to AB ...... 27

Figure 32: Connecting Point P to some Point O ...... 28

Figure 33: Finding Point M ...... 28

Figure 34: Finding Point D ...... 29

Figure 35: Finding Point E ...... 29

Figure 36: PE is Parallel to AB ...... 29

Figure 37: Triangles DPE and COM ...... 30

Figure 38: Finding a Side of the Equilateral Triangle ...... 31

Figure 39: The Construction of the Equilateral Triangle: Final Result ...... 31

Figure 40: The Equilateral Triangle Proof ...... 32

Figure 41: The Square ...... 33

Figure 42: Perpendicular Lines in a Circle ...... 34

Figure 43: Finding the Special Point D ...... 34

Figure 44: The Constructed Pentagon ...... 35

x Figure 45: The Special Length AD ...... 36

Figure 46: The Pentagon ...... 37

Figure 47: One of the Isosceles Triangles from the Pentagon ...... 37

Figure 48: Splitting the Isosceles Triangles in Half ...... 38

Figure 49: Finding a Side of the Regular Hexagon in the Circle ...... 39

Figure 50: The Hexagon ...... 39

Figure 51: Any Isosceles Triangle ...... 41

Figure 52: Bisecting ABC ...... 42

Figure 53: Any Isosceles Triangle can be Bisected ...... 43

Figure 54: Property 1 for the Parabola and Point P ...... 46

Figure 55: Property 3 ...... 47

Figure 56: Example 1 of the Parabolic Quadrature ...... 50

Figure 57: Example Special Triangle ...... 51

Figure 58: Finding the Area of the Special Triangle ...... 51

Figure 59: Quadrature Example 2 ...... 52

Figure 60: The Special Triangle for Example 2 ...... 53

Figure 61: Finding the Area of the Special Triangle of Ex 2 ...... 53

Chapter 1: Introduction The origins of the Infinitesimal calculus lie, not only in the significant contributions of Newton and Leibniz, but also in the centuries-long struggle to investigate area, volume, tangent and arc by purely geometric methods. To understand the important changes in proof structure, method, technique and means of presentation which emerged in the later seventeenth century it is necessary to go back to Greek sources and particularly, of course, to Archimedes. -Margaret E. Baron The benefit of looking back at the history of Calculus is the gift of hindsight. One now has the ability to make connections with the math developed by our forefathers and linearly move from one idea to the next. This paper examines the Greek’s idea of the Method of Exhaustion and how this method could have influenced the formation of integral Calculus. People from ancient times were able to study and manipulate area and volume without the modern tools of Calculus. Some of the ancient insights on area and volume can help one to more fundamentally understand the concepts of integration in modern times. As Newton and Leibnitz, the men attributed to creating modern day Calculus, had direct access to many of the achievements of ancient Greek mathematics, this is a starting point to look at the history of the concepts of Integration and more specifically, the Method of Exhaustion. This text seeks to: historically look at what the Method of Exhaustion was used for, look at how to apply the Method of Exhaustion to find useful results, and examine similarities between the Method of Exhaustion and the techniques of modern integration. The Method of Exhaustion is used to find the area of a shape by inscribing a sequence of n-sided polygons of known areas inside the shape. As n increases, the inscribed polygons tend to fill out the shape, exhausting it, hence the total area of the

1 polygons better and better approximate the area of the shape. This technique was developed by early Greek scholars and formalized by Eudoxus of Cnidus(c. 400 BC)). The Greeks lacked modern tools to construct regular polygons, so one might question if there was a limitation to the type of regular polygons that they could construct. There are records of the Greeks constructing the equilateral triangle, the square, the regular pentagon, the regular hexagon, the regular octagon, and the regular decagon. It was unknown whether of not it was possible to construct a 17-gon with a compass and straight edge until the eighteenth century. This question was finally answered by the great mathematical mind Carl Friedrich Gauss in 1796. Furthermore, it was Gauss who proved that an n-gon could only be constructed with a straightedge and compass if and only if n is the product of a power of 2 and any number of the Fermat

k primes m (numbers of the form m = 22 +1, where k is a nonnegative integer). Gauss proved the sufficient condition in Disquisitiones Arithmeticae in 1801. Pierre Wantzel proved the necessary condition in 1837 with the Gauss-Wantzel theorem. The first few ! constructible regular polygons are n: 3 = 2 +1, 4 = 22, 5 = 22 +1, 6 = 2 !(2 +1) , 8 = 23, 10 = 2" (22 +1) = 2" 5, 12 = 22 " 3, 16 = 24 , 17=24 +1. Figure 1 shows several of the constructible regular polygons! inscribed! within! a circle. !

! ! ! !

Figure 1: A Series of Regular Polygons Inscribed within a Circle Archimedes(287-212 BC) used the Method of Exhaustion to approximate the area of a circle and find an exact formula for the area enclosed by a parabolic region and a chord. A process that is similar to what Archimedes could have used is further discussed in Chapters 2 and 3.

Chapter 2: Estimating the Area of the Circle It is not difficult to use a similar method as Archimedes’ to calculate the area of a circle. Given a circle of radius r, one can inscribe in it a constructible regular polygon1. Figure 2 depicts an equilateral triangle inscribed in a circle. This triangle can be dissected into three isosceles triangles, each isosceles triangle having a vertex in the center of the circle of radius r and two congruent sides of length as depicted below in Figure 2.

Figure 2: Isosceles Triangle Inscribed in a Circle

θ1 + θ2 + θ3 = 360°

θ1 = θ2 = θ3 = 120°

The area2 A of each isosceles triangle is as follows:

1 # 2 A"AOC = A"BOC = A"COA = sin( )r 2 2 3 A = sin(120°)r2 "ABC 2

! 1 The constructions for several regular polygons inscribed in a circle are given in Appendix A. 2 Calculation of area for any isosceles triangle is given in Appendix B.

3 A similar process is applicable to all n-gons, whether or not they are constructible, as all n-gons can be split up into n congruent isosceles triangles, which 360° resemble the isosceles triangle depicted in Figure 3. and where" = n .

Figure 3: Arbitrary Isosceles Triangle As the n-gon is comprised of n isosceles triangles, the area of the n-gon must be n sin(")r2 . 2 With this information, the table below can be easily completed. Table 1: Areas of n-gons Inscribed in a Circle ! n 3 4 5 6

θ 120° 90° 72° 60°

3 2 5 2 2 Area sin(120°)r 2sin(90°)r2 sin(72°)r 3sin(60°)r 2 2 Aprox. Area 1.299 r2 2r2 2.378r2 2.598r2

! ! ! ! n 8 10 12 n θ 45° 36° 30° 360°/s

2 2 2 Area 4sin(45 )r 5sin(36 )r 6sin(30 )r n 2 ° ° ° sin(")r 2 2 2 2 2 Approx. Area 2.828r 2.939r 3r "r as3 n "# ! ! ! ! ! ! 3 A Calculus level proof is given of this fact in Appendix D.

Chapter 3: The Area of a Parabolic Region

Figure 4: Special Triangle in a Parabolic Segment In the 3rd century BC, Archimedes wrote a letter to his friend Dositheus that included 24 propositions about parabolas. These propositions culminated in a proof that 4 the area of a region enclosed by a parabola and a line is exactly the area of a certain 3 triangle (shown in blue in Figure 4) inscribed in the parabolic region. Examples of applying this fact to find the areas of parabolic regions are shown in Appendix G. ! The propositions proved the theorem in two different ways: one involved abstract mechanics of the lever, and the second used the Method of Exhaustion. The latter was one of the most advanced applications of the Method of Exhaustion and continued to be so until the development of Cavalieri’s quadrature formula in the 17th century. Archimedes may have proved the theorem by splitting up the region inside the parabolic segment into many special triangles. This process can be thought of as continuing on indefinitely to create an increasing number of triangles that fill up the parabolic region. Archimedes proved that the difference between the area of the parabolic region, and the sum of the triangular areas could be made as small as one pleases by constructing a large enough number of triangles in the process discussed below.

5 Take a parabolic segment enclosed by a chord AB pictured in Figure 4. There exists a point P on the parabola such that the tangent at P is parallel to the chord AB by the Mean Value Theorem. The Greeks constructed the point P as follows. Using the ruler ! and compass construction provided in Appendix A, construct a chord RQ that is parallel ! to AB. Find the midpoint of the chords AB and RQ and connect them with a line. Call P ! the point of intersection of this line and the parabola. The area of the parabolic region is ! 4 exactly A ! 3 "APB !

Figure 5: The Parabolic Segment The following properties of the parabola and of the points A, B, and P are proved in Appendix F.

Figure 6: Properties of P

6 Property 1. The straight line through P that is parallel to the axis of the parabola4

intersects AB at its midpoint M.

Figure 7: Property 2

Property 2. Every chord RQ of the parabola that is parallel to AB is bisected by PM , therefore the point P does not depend on the choice of RQ.

2 PN NQ ! ! Property 3. Using the same labeling as Figure 7, have: = 2 . PM MB ! ! Connect the points A, B, and P. This forms a triangle, pictured above in Figure 4. Repeat this process with the two remaining parabolic regions by finding the two special ! points P1 and P2 such that the tangents at P1 and P2, respectively, are parallel to AP and

PB. Connect A and P to P1, and P and B to P2 and color the two new triangles ΔAP1P

and ΔPP2B green as in Figure 9. ! ! There are now four parabolic segments that are not colored in. Find a point on each parabolic segment such that the tangent at the point is parallel to the chord enclosing it. Connect the points of the chord to their own respective new point to form a series of new triangles. Continue this process on indefinitely. The total area of all of the triangles becomes closer to the area of the main parabolic segment as the number of triangles increase.

4 The axis of the parabola is the line of symmetry of the parabola

Figure 8: Adding Green Triangles The total area of the triangles added at each step is 1/4 the total area of the triangles added at the previous step. st Proof: Let A0=Area of the 1 triangle(colored in blue), A1=Area of the two green

triangles added after the blue triangle, A2=Area of the four triangles added after the green triangles, and so on.

Will show that A1= 1/4A0. This will imply that An=1/4 An-1.

Figure 9: Triangles of the Quadrature A = A + A In Figure 9, A0 = A"APB and 1 "AP1P "PP2B . The line that passes through point P and is parallel to the parabola’s axis cuts segment AB in half at point M according to Property 1. Similarly the line that passes ! !

! 8 through P2 that is parallel to the axis of the parabola also intersects PB at its midpoint

M2 by Property 1. Extend P M so that it intersects the segment AB, call the point of intersection 2 2 ! point C. See Figure 10.

! !

Figure 10: Points M and C 1 First, will show A = A . "PP2B 4 "PMB

Draw a line that passes through P2 and is ⎜⎜ to AB. Let point D be the intersection of this line and PM , pictured below in Figure 11. ! ! !

Figure 11: Point D

DM P2C as both as parallel to the axis of the parabola and MC DP 2 (by construction).

So DPCCM is a parallelogram, therefore ! ! 9 MC = DP2 , and DM = P2C. Claim that C is the midpoint of MB. (A) ΔPBM ~ ΔM BC since PM M C, so all corresponding angles are congruent. ! ! 2 2 !

Figure 12: Triangle Congruence

And M2 is the midpoint of PB by proposition 1. CB 1 M C (1) = = 2 MB 2 PM ! 1 So, CB = MC = MB 2 ! 1 (2) M C = PM 2 2 ! 2 PD DP2 Applying property 3 to Figure 12, we get = 2 PM MB ! (3) PM = 4 PD, by the fact that MC = DP2 , and (2)

DM = 3PD , by (3) and the fact that PM! = PD + DM (4) P C = 3PD, since DM = P C ! ! 2 ! 2 (5) by (2) AND (3) ! M2C = 2PD ! ! ! (6) P C = P M + M C ! 2 !2 2 2 (6) P M = PD by (4) , (5), and (6). ! 2 2 Claim the height of ΔPCB is twice the height of ΔPP B, where the heights of both ! 2 triangles are with respect to the shared side PB. !

! 10

Figure 13: Triangular Height

Let Z be the point of intersection of the height of ΔPP2B with respect to side PB, and the line PB, and O be the point of intersection of PB and the height of ΔPCB with respect to PB. ! ! ! !

Figure 14: Heights of the Two Triangles

Look at the triangles ΔZP2M2 and ΔM2CO.

Figure 15: Triangle P2ZM2 and M2OC

! !

Figure 16: External and Internal Angles of Triangles from Figure 16

The extension of ZO, pictured above in Figure 16, form corresponding angles with the extension of the line segments ZP2 and OC . ZP2 OC by Euclid’s Elements

Book 1 Proposition! 27. So, we can rotate ΔZP2M2 so that it is inscribed inside of

ΔM2CO so that the two triangles look like Figure 17. ! ! !

Figure 17: Rotation of the Triangle

Have,

⇒ΔZP2M2 ~ ΔM2CO

ZP 2 MP 1 = 2 = by (5) and (6) OC MC 2

(7) OC = 2ZP2

Recall OC is the height of PCB and ZP is the height of PP2B with respect to ! Δ 2 Δ 1 ! PB. So then have A = PC# ZP , by (7). "PCB 2 2 ! 1 ! (8) A A "PP2B = "PCB ! 2 Now! compare the two triangles ΔPCB and ΔPMB pictured below in Figure 18.

Figure 18: Relation of Triangle PMB and PCB

Look at the height of both of these triangles with respect to the shared side PB. These are pictured below in Figure 19 in green. Let U be the point of intersection of the

height of ΔPMB with respect to side PB, and the line PB, and recall that! O is the point of intersection of the height of ΔPCB with respect to PB, and PB.

! ! ! !

Figure 19: Height of PMB and PCB

MU ⎜⎜CO by Euclid’s Elements Book 1 proposition 27. ⇒ΔMUB~ΔCOB by an argument that is similar to (A) CO MC 1 ! ! = = by (1) MU MB 2 ⇒ MU = 2CO 1 ! (9) A = A "PCB 2 "PMB ! !

14 ! 1 A = A by (8) and (9). "PP2B 4 "PMB 1 A similar argument can be used to show A = A . "PP1A 4 "PMA ! A"APB = A"PMB + A"PMA = A0 1 So have, A = A . ! 1 4 0 ! ! ! 1 This argument implies that A = A . Let A be the total area of all of the n "1 4 n !triangles added after an indefinite amount of time.

A=A0+A1+A2+…. ! 1 1 1 A = A ;A = A ;A = A and so on. 1 4 0 2 4 1 3 4 2 " 1% n (1 A = $ ' A , where n is the number of times more triangles have been added to n # 4& 0 ! the quadratic region " 1% " 1% 2 " 1% 3 " 1% 4 ! So A = A +$ ' A +$ ' A +$ ' A +$ ' A + ... 0 # 4& 0 # 4& 0 # 4& 0 # 4& 0 ( " 1% " 1% 2 " 1% 3 " 1% 4 + (10) A = *1 +$ ' +$ ' +$ ' +$ ' + ...- A0 ) # 4& # 4& # 4& # 4& , !

(" 1% " 1% 2 " 1% 3 " 1% 4 + The Greeks could evaluate the series C = *$ ' +$ ' +$ ' +$ ' +- ! )# 4& # 4& # 4& # 4& , geometrically with a unit square pictured below in Figure 20.

Figure 20: Unit Square

Divide this square up into four even square parts. Label these squares S , where i i j is equal to the number of times a division has taken place, and j={k⎜k=1,2,3,4}. j depends on the location of each individual square. j=1 if the square is in the upper right ! corner, 2 if the square is in the upper left corner, 3 if the square is in the lower left corner, and 4 if the square is in the lower right corner.

Figure 21: Dividing S up into Four Equal Parts

Color in S and divide S into four equal square parts and color in S . Continue 13 11 23 the process of dividing the top square in fourths and coloring in the bottom left square indefinitely. The square looks like Figure 21 after this process. ! ! !

Figure 22: Square after the Coloring the Division Process Each colored square represents a term from the series C, as every square has an 1 area that is always equal to a power of . The unit square S is collection of the set of all 4 squares S . i j S = S + S + S + S + S + S + S + S + S + ... 12 13 14 2!2 23 24 32 33 34 This sum can be grouped by i, the number of times a division has occurred. !

Figure 23: Dividing the Unit Square up by i

17 1 Figure 23 shows that the colored square of each layer has the total area of each 3 1 layer. Hence, the total area of all of the colored squares is S. And by design, S has an 3 area of 1. ! (" 1% " 1% 2 " 1% 3 " 1% 4 + 1 (11) C = *$ ' +$ ' +$ ' +$ ' + ...- = ! )# 4& # 4& # 4& # 4& , 3 Recall that A is the limiting area of the triangles. ( " 1% " 1% 2 " 1% 3 " 1% 4 + 4 ! A = *1 +$ ' +$ ' +$ ' +$ ' + ...- A0 = A0 by (10) and (11) ) # 4& # 4& # 4& # 4& , 3 A is also the area of the parabolic region, so the parabolic must also have an area 4 that is A . Examples of finding the area of a parabolic region with this technique are ! 3 0 given in Appendix G.

Chapter 4: A Modern Integration Technique: The Trapezoidal Method

Figure 24: Trapezoid Method The Trapezoidal Method is a technique employed in modern Calculus courses and is similar to the Method of Exhaustion. In the Trapezoidal Method, the area under the graph of a function f(x) on a finite interval (a,b) is approximated by a trapezoidal area. This technique is easy to implement because it is very easy to calculate the area of a trapezoid. Fig. 24 above shows the graph of f together with its approximation by a trapezoid. The trapezoid pictured above in Figure 24 has an area A, where f (b) " f (a) A = (b " a) . The area underneath the graph of f(x) is approximately equal to 2 A. b f (b) # f (a) ! $ f (x)dx " (b # a) a 2 This method is very similar to the Method of Exhaustion as the Trapezoidal Method uses shapes whose areas can be readily calculated to approximate the area of the ! graph. The Method of Exhaustion, however is a limiting method that inscribes a sequence of shapes in the region, whereas the original Trapezoidal Method is not. There is only one polygon being used to estimate the area under the graph. In fact, one can see

19 that the one trapezoid’s area is not a very good approximation for the area under the curve.

Figure 25: The Composite Trapezoidal Rule Better approximations can be obtained by breaking up the interval (a,b) into several subintervals, pictured above in Fig. 25; the procedure is called the Composite Trapezoidal Rule. The total area of all of the smaller trapezoids then approximates the area of the f(x). This process of breaking the interval up into several subintervals reduces the error because the smaller trapezoids do not omit or over count the areas that the one large trapezoid did. It is to be expected that the sum of the trapezoidal areas become better approximates to the area of f(x) as the number of subintervals increase. The Composite Trapezoidal Method is a direct application of the Method of Exhaustion, as this method inscribes a series of trapezoids within the curve of the function and the x–axis. As the number of intervals increases, the number of trapezoids increases as well and the total area of the trapezoids can be made to become arbitrarily close to the area of the function. M h T( f ,h) ( f (x ) f (x )) The area of f(x) can then be approximated as , = # k"1 + k 2 k=1 b " a where M is the number of subintervals, and h = . M The Trapezoidal Method and the Composite! Trapezoidal Method are intuitive and are similar to the Method of Exhaustion. The methods estimate the area and exhaust ! lower bounds by increasing the number of subintervals, which is analogous to increasing the number of polygons to estimate the area of the circle or increasing the number of

20 triangles to find the area of a parabolic segment. The Trapezoidal Method is a lesser used method in introductory Calculus courses, but is quite similar to the widely taught Riemann Sum Method as they both use reasoning similar to the Method of Exhaustion.

Chapter 5: The Method Exhaustion: Leading the Way to Modern Integration The ideas behind the Method of Exhaustion are accessible to people of all levels of mathematics. Tackling a complex situation by splitting it into simpler steps is something that mankind has done all throughout history to better understand our surroundings. In the particular problem of calculating areas, the Greeks modeled different properties of a circle and a parabolic region by using approximations via shapes that were previously understood. The Greeks were knowledgeable about the properties of constructible regular n-gons, and in particular about triangles, and were able to use this knowledge to better understand the concept of the area of a circle and the area of a parabolic region. Although the term “limit” was never used in any of the Greek texts recovered, the Method of Exhaustion used the idea of limiting the area of the object in question by constructing shapes with closer and closer areas to the area of the shape. The difference of the area of the shape and the area of the polygon could be made as small as one chose, by making an appropriate choice for the polygon. The Composite Trapezoidal Method, which is similar to the more popular Riemann Sum Method, is, at its essence, an application of the Method of Exhaustion. The limiting total trapezoidal area of the interval becomes closer to the area under the graph of f(x) as the number of trapezoids increase. One can make the difference between the two areas as small as one pleases by increasing the number of subintervals. One could also conjecture that the Method of Exhaustion directly influenced both Newton and Leibnitz. The Composite Trapezoidal Method and Riemann Sums are directly related to the Method of Exhaustion, and one could conjecture that both Newton and Leibnitz were directly influenced by the Method of Exhaustion, which is why so many techniques used to estimate the definite integral use a process that is similar to the Method of Exhaustion. The topics of Mathematics are forever building upon the achievements of the past and this is evident with modern day integration and the Method of Exhaustion.

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Appendix A: Some Ruler and Compass Constructions and Proofs A.1 Generating Congruent Adjacent Secants in a Circle

Given a circle C of radius r. Let AB be a secant to C, pictured below in Figure 26.

Figure 26: Generating Congruent Adjacent Chords in a Circle

1. Starting at A, create a circle with center A and radius AB.

2. Mark the point of intersection between the two circles. Call this point C. Connect

A to C, depicted in Figure 27.

Figure 27: Finding Point C

25 Then AB is congruent to AC.

A.2 Perpendicularly Bisect a Segment (this is also a method to find the ! ! midpoint of a line segment)

Start with a line segment AB.

1. Create a circle centered at A of radius r, where r is larger than ½ length of AB

2. Create a circle centered at B of the same radius r, pictured below in Figure 28.

Figure 28: Perpendicularly Bisect a Segment

3. Connect the intersection of the two circles. This new line, pictured in Figure 29,

is the perpendicular bisector of AB.

Figure 29: The Perpendicular Bisector Proof: Denote the two points of intersection of the two circles points C and D, shown

below in Figure 30. Use the known fact that for any two intersecting circles(not

26 necessarily congruent) the line connecting their centers if perpendicular to their common

chord.

Figure 30: Perpendicular Bisector Proof

AB connect the two circle’s centers and CD is the common chord.

Let O be the point of intersection of AB and CD. ! ! The right triangles, ΔACO and ΔBCO are congruent, as AC = CB = r, and CM is ! ! common. ! ! AO = OB

A.3 Construct a Line through a Given Point Parallel to a given Line

1. Start with a given line AB and the point P that we would like new parallel line to

pass through, pictured below in Figure 31.

Figure 31: Construct a Line through P Parallel to AB

27 2. Draw a line that passes through P and some point O on AB(where O is neither A nor

B), pictured below in Figure 32. !

Figure 32: Connecting Point P to some Point O

3. Draw a circle centered at O with radius less than ½ OP . Label the point of

intersection of the circle and OP as point C, and the point of intersection of the circle ! with AB as point M, pictured below in Figure 33. !

Figure 33: Finding Point M

4. Extend the line OP past the point P. Draw a circle centered at P of radius OC . Label

D the point of intersection of this circle and the line OP, which is not on the segment ! ! OP., pictured in Figure 34. !

Figure 34: Finding Point D

5. Draw a circle centered at D of radius CM . Label E on of the points of intersection

of the two circles, shown is Figure 35. !

Figure 35: Finding Point E

PE is parallel to AB. Connect the points P, E and extend.

! !

Figure 36: PE is Parallel to AB

29 Proof: Look "DPE and "COM , pictured below in Figure 37.

DP = CO = OM = PE , and DE = CM , by construction. ! !

! !

Figure 37: Triangles DPE and COM Implies and all the statements are related by “implies”

"DPE = "COM

< DPE =< COM

! < DPE and < COM are congruent alternate angles.

! PE is parallel to OM by Euclid’s Elements: Book 1, proposition 27. ! ! A.3 Equilateral triangle inscribed within a circle ! ! 1. Draw the given circle C1 having radius r with center O1.

2. Pick a point on C1 and label this point O2 Draw a circle C2 of radius r that passes

through O1, shown in Figure 38. Label the point of intersection of the two circles, A

and B. Connect pts A and B.

Figure 38: Finding a Side of the Equilateral Triangle

3. AB is a side of the equilateral triangle inscribed in C1. Using the construction for creating secants in a circle of equal length, create two more secants of length AB that intersect on the circle C1.

Figure 39: The Construction of the Equilateral Triangle: Final Result

31 Proof of Construction: To show that AB is a side of the equilateral triangle

inscribed inside of C. First show that < !!!! = 120°: !

Figure 40: The Equilateral Triangle Proof

Connect A and B to O1 and O2, respectively.

A01 = O1B = r, and AO2 = O2B = r , by construction.

So ΔAO2O1 and ΔBO2O1 are equilateral triangles. ! ! "AO2O1 = "BO2O1

< AO1O2 =< BO1O2 = 60° ! Implies < AO1B =120° ! ! A B =1/3 of the circle. AB = AD = BD, hence ΔABD is an equilateral triangle. ! A.4 Square Inscribed within a Circle ! ! 1. Draw a circle C of radius r with center O.

2. Draw a line through O intersecting C at points A and B ( AB is a diameter).

3. Create a perpendicular bisector through the line AB. (See construction for ! perpendicular bisector.) !

32 4. Label the points of intersection of this new line and the circle C and D. Connect

the points A, C, B, and D to the two adjacent points, pictured below in Figure 41.

Figure 41: The Square

ABCD is a square inscribed in a circle C.

Proof:CO = DO = AO = BO = r .

CD is the perpendicular bisector of AB. !

! So ACBD is a rhombus. ! The four triangles ΔAOC, ΔCOB, ΔBOD, and ΔDOA are right isosceles triangles, hence

are 90°, hence ABCD is a square.

A.5 Regular Pentagon Inscribed in a Circle

1. Draw a circle C of radius r with center O, and draw a line through the diameter of

the circle. Label one of the points of intersection of the circle and the line A.

33 2. Draw a perpendicular bisector through the line through the diameter (See the construction in A2). Label the point of intersection between the new line and the circle B, pictured below in Figure 42.

Figure 42: Perpendicular Lines in a Circle

3. Use construction given to find the midpoint of OB , mark this point C.

4. Create a new circle with center at C, radius AC . Mark the point of intersection of ! the new circle extension of line BO . Call this point D, pictured below in Figure 43. !

Figure 43: Finding the Special Point D

34 5. AD has the same length as the sides of the pentagon. Using the construction for

creating secants in a circle of equal length, create five connected secants of length ! AD. Connect the five points on circle C. This is the regular pentagon inscribed within

a circle, pictured below in Figure 44.

Figure 44: The Constructed Pentagon

Proof: Claim that AD is the length of a side of a regular pentagon inscribed within

circle of radius r. !

Figure 45: The Special Length AD 1 1 OC = OB = r , and OA = r, by construction. 2 2

5 AC = r , and! AC = DC by construction. ! 2

5 DC = ! r ! 2

5 1 OD = DC " OC = r " r ! 2 2 )10 " 2 5 AD = ! 2

To show that AD is equivalent to the length of the regular pentagon inscribed in the ! circle of radius r, pictured below in Figure 46: !

Figure 46: The Pentagon

Let a be a side of the regular pentagon. The pentagon can be split up into 5 congruent isosceles triangles of the form of Figure 47:

Figure 47: One of the Isosceles Triangles from the Pentagon

Each of these isosceles triangles can be bisected into two congruent right triangles

(proof in Appendix B), pictured in Figure 48.

Figure 48: Splitting the Isosceles Triangles in Half

Can use trig to calculate a in terms of r.

)10 " 2 5 a=2rsin(36°). Note, 2sin(36°)= . So indeed the length of AD is the length 2 of the sides of the pentagon inscribed inside of the circle. ! 6. Regular hexagon! inscribed in a circle

1.Draw a circle C of radius r and center O1.

2. Pick a point on C and label this point 02.

3. Draw a circle C2 of radius r that passes through the center of O. Label the point of intersection of the two circles, A and B, pictured below in Figure 49.

Figure 49: Finding a Side of the Regular Hexagon in the Circle

4. AO2 and O2B are sides of the regular hexagon inscribed in the circle. Use the

construction before mentioned to create four more connected secants with length ! ! AO2 .

Figure 50: The Hexagon

Proof: Follows from the construction of the equilateral triangle.

7. Regular octagon inscribed in a circle

1. Use same construction for constructing the square in a circle. See Figure 41.

2. Bisect

CE = AE =side of the octagon !

! ! 39 3. Use the construction before mentioned to create six more connected secants with

length AE .

Note: Can now construct any n-gon, where n = k2m , where k=3,4,5,6,8 and m is any ! natural number by this bisection process. !

Appendix B: Every isosceles can be bisected into two congruent right triangles Start with any Isosceles ΔABC, pictured below in Figure 52.

Figure 51: Any Isosceles Triangle

Bisect < ABC and extend the ray down to the line AC in the following manner. Draw a circle C1 centered at B. Let D and E the points of the intersection of ΔABC and C1. ! Draw a circle C2 centered at D with the same radius as C1. Draw a circle C3 centered at E with the same radius as C1. Label the intersection of C2 and C3 point F. Connect

B to F and extend. Let D be the intersection of ΔABC and the extension of BF

Figure 52: Bisecting ABC

By construction

(1) ΔABD = ΔCBD by SAS

To show ΔABD and ΔCBD are right triangles:

Appendix C: Area of any Isosceles Triangle Previously showed in Appendix B that every isosceles can be dissected into two

congruent right triangles, depicted below in Figure 54.

Figure 53: Any Isosceles Triangle can be Bisected

Can find the height of each right triangle by trigonometry.

" 1 " (1) Height = rcos( ); a = rsin( ) 2 2 2

1" 1 % So the area of each right triangle is: $ a'( Height) 2# 2 & ! " 1 % ( ( So the area of the isosceles triangle =$ a'( Height) = r2 sin( )cos( ), by (1) # 2 & 2 2 ! 1 = r2 sin(") by the double angle formula. 2 !

Appendix D: π Claim: lim ! !"#! ! = !, where ! = °. !→! ! !

1 360° = lim !"#!( ) 2 !→! !

This is an indeterminate form

360° 1 sin ( ) 360° = lim 360° ! = = 180° = ! 2 !→! 360° 2 !

Appendix E: Estimating Volumes using the Method of Exhaustion The Greeks were able to extend the idea of the Method of Exhaustion for calculating the area of planar objects to calculating the volume of three-dimensional objects. They were able to calculate the volume of a sphere by inscribing a series of polyhedron of known volume into the shape. The volumes of each tetrahedron were known as each polyhedron of n faces could be divided up into n pyramids with a base comprised of n sides. The Greeks knew the formula for the volume of a pyramid by the

Egyptians, as many Greek scholars studied in Alexandria and learned from the Egyptians.

The Method of Exhaustion was also used to find the volume produced when rotating a two-dimensional object around an axis of rotation. One can fill the space with a sequence of shapes that better and better approximate the volume.

Appendix F: Calculus Based Explanations of Property 1, 2, and 3

Figure 54: Property 1 for the Parabola and Point P Proof of Property 1: The straight line through P that is parallel to the axis of the

parabola intersects AB at its midpoint M.

Shall prove this using a modern Calculus technique. ! Without loss of generality, let the parabola in question be defined by y = ax 2 , where

a " 0 (by choosing an appropriate coordinate system). The axis of symmetry is the y-

! 2 2 axis. Let the point A=(x0 , y0) and B=(x1, y1), where y0 = ax0 and y1 = ax1 . The slope ! y " y ax 2 " ax 2 a(x 2 " x 2) of AB is given by 1 0 = 1 0 = 1 0 = a(x + x ).There exists a point x x x x x x 1 0 1 " 0 1 " 0 ! 1 " 0 !

P=(x2,y2) on the parabola whose tangent is parallel to AB by the Mean Value Theorem. ! Since y’=2ax! , then !

46 y1 " y0 2ax2 = = a(x1 + x0 ) x1 " x0 , hence

x1 + x0 x2 = ! 2

So x2 is the midpoint of x0 and x1. Recall that we ensured that M is the intersection of ! AB and the y-axis. So the x-component of the point M is x2.

M=(x2, yM). ! To show yM is the midpoint of y0 and y1:

The equation of the line AB is y = a(x1 + x0)(x " x0 ) + y0

x1 + x0 Can solve for yM by plugging in x = into the above equation. 2 2 ! ! y + y Solving this algebraically, get y = 1 0 , so indeed M is the midpoint of AB. M 2 !

! !

Figure 55: Property 3

Property 2 follows from Property 1 since RQ is parallel to AB and the point P is

dependent on the slope of AB, not the particular points A and B. Can then take AB to be ! ! RQ and apply Property 1 to get that RQ is bisected by the line through P that is parallel ! ! to the axis of the parabola

! !

47 Proof of Property 3: The parabola can still be assumed to be of the form y=ax2.

2 PN NQ Take any chord RQ of the parabola that is parallel to AB. Then = 2 . PM MB

Let A-(xA, yA), B=(xB, yB), M=(xM, yM), N=(xN, yN), Q=(xQ, yQ), and P=(xP, yP), where 2 y(xA,B,P!)=a xA,B,P ! ! yA " yB Then AB has slope m = = a(xA + xB ). xA " xB

xA + xB So, 2axp=m, which implies x p = . ! 2 ! M is the midpoint of AB by property 1, so xM=xP. Also have xp=xN by Property 2. y y m(x x ) ax 2 x(x x )(x x ) M = B + M "! B = B + A + B M " B 2 yN = yQ + m!(x N " xQ ) = a(xQ ) + a(xA + xB )(xN " xQ )

2 ! PN NQ Want to show = 2 . PM MB ! # 1 & a% (x )2 + (x + x ) x " x + (x + x )2( PN y " y $ Q A B ( N Q ) 4 A B ' = N P = ! 2 PM yM " yP a(xA " xB ) 4

xA + xB 2 2 2 2 4(xQ " ( )) 4(xQ ) " 4(xA + xB )xQ + (xA + xB ) (2xQ " (xA + xB )) 2 = 2 = 2 = 2 ! (xA " xB ) (xA " xB ) (xA " xB )

2 2 ! # xA + xB & # # xA + xB && 2 2 2 % " xQ ( +% a(xA + xB )% " xQ (( NQ (xN " xQ ) + (yN " yQ ) $ 2 ' $ $ 2 '' And = = 2 (x x )2 (y y )2 2 2 MB M " B + M " B # xA + xB & # # xA + xB && % " xB ( +% a(xA + xB )% " xB (( $ 2 ' $ $ 2 '' 2 2 # x + x & 2 # x + x & % A B " x ( ) 1+ a(x + x ) % A B " x ( $ 2 Q ' ( ( A B ) ) $ 2 Q ' = 2 = 2 ! # xA + xB & 2 # xA + xB & % " x ( ) 1+ a(x + x ) % " x ( $ 2 B ' ( ( A B ) ) $ 2 B ' 2 "" xA + xB % % 4$$ ' ( xQ ' ## 2 & & = 2 ! x x ( A ( B )

! 48 2 PN NQ So indeed, = 2 . PM MB

Appendix G: Examples of Using the Quadrature of the Parabola to find the Area of the Bounded Region Ex. 1 Look at the specific example y=-x2 + 4 with the points A=(-2,0) and

B=(2,0).

Figure 56: Example 1 of the Parabolic Quadrature

The slope of AB is 0. Can use Calculus to solve for the point P whose tangent has slope

0. This point is (0,4). !

Figure 57: Example Special Triangle 4 The Area of the parabolic region enclosed by AB is A . A can be found using 3 "APB "APB several techniques. The simplest technique is to inscribe the triangle in a rectangle ! ! shown below in Figure 60. !

Figure 58: Finding the Area of the Special Triangle The rectangle is comprised of ΔAPB, and two right triangles. These points and shapes are in the xy-coordinate system, so can easily find the height and base measurement of all

51 three right triangles. ΔAPB must then have an area that is equal to the area of the

rectangle minus the areas of the right triangles.

!∆!"# = 16 − 4 − 4 = 8

Hence, the area of the quadratic region enclosed by y=-x2 + 4 and AB is

4 32 8 = 3 3 ! Ex. 2 Look at the same function y=-x2 + 4, but now let A=(-2,0) and B=(0,4). Then the

slope of AB=2.

Figure 59: Quadrature Example 2 Can find point P by Calculus again by differentiating the function to find the point whose

tangent is parallel to AB, so P=(-1,3).

Figure 60: The Special Triangle for Example 2 Using a similar technique as in example 1, can easily find the area of the special triangle.

Now the larger rectangle is comprised of 3 right triangles, the special triangle and one small rectangle.

Figure 61: Finding the Area of the Special Triangle of Ex 2

! = 8 − 1 − ! − ! − 4 = 1 ∆!"# ! !

4 Hence the area of the parabolic region enclosed by the graph of y=-x2 + 4 and AB is 3 .

! 53