History of Math for the Liberal Arts CHAPTER 5 The Greeksfaculty.seattlecentral.edu/LMorales/...

2001, Lawrence Morales; MAT107 Chapter 5 – Page 1

Title Page 1 2

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History of Math 4

for the Liberal Arts 5

6

CHAPTER 5 7

8

The Greeks 9

10

11

12

Lawrence Morales 13

Seattle Central 14

Community College 15


Table of Contents 16

PART 1: Introduction................................................................................................................... 4 17

Introduction................................................................................................................................. 4 18

PART 2: Three Famous Problems .............................................................................................. 5 19

Squaring the Circle ..................................................................................................................... 5 20

Trisecting an Angle ..................................................................................................................... 5 21

Doubling a Cube ......................................................................................................................... 6 22

A “Solution” from Menaechmus................................................................................................. 6 23

Squaring Other Figures ............................................................................................................ 11 24

Fourth Century Glory ............................................................................................................... 12 25

PART 3: The Platonic Solids ..................................................................................................... 13 26

Kepler and the Platonic Solids ................................................................................................. 14 27

Discovering a Pattern ............................................................................................................... 15 28

Beauty in Form ......................................................................................................................... 15 29

PART 4: Euclid’s Elements and Geometric Algebra .............................................................. 16 30

Aristotle..................................................................................................................................... 16 31

Euclid (c. 325 – 265 B.C.E.) .................................................................................................... 16 32

Geometric Algebra.................................................................................................................... 20 33

Euclid’s Elements Book 2, Proposition 4 ................................................................................. 23 34

Solving ax = bc with Greek Construction Methods.................................................................. 25 35

Solving Quadratics with Geometric Methods ........................................................................... 29 36

Later Geometric Algebra .......................................................................................................... 31 37

PART 5: Archimedes.................................................................................................................. 32 38

Selected Work of Archimedes.................................................................................................... 33 39

The Area of a Circle.................................................................................................................. 35 40

An Approximation for π ............................................................................................................ 35 41

Areas Insides Parabolas ........................................................................................................... 37 42

Archimedes and the Law of the Lever....................................................................................... 42 43

Archimedes and the Center of Gravity...................................................................................... 48 44

PART 6: Heron's Formula......................................................................................................... 51 45


PART 7: The Golden Ratio........................................................................................................ 55 46

Summary ................................................................................................................................... 55 47

PART 8: Appendix...................................................................................................................... 57 48

PART 9: Homework ................................................................................................................... 58 49

Graphing Parabolas ................................................................................................................. 59 50

The Method of Menaechmus ..................................................................................................... 59 51

Geometric Algebra.................................................................................................................... 60 52

Measuring π and the Area of Circles........................................................................................ 60 53

Approximating the Area of a Circle.......................................................................................... 62 54

Areas of Parabolas via Archimedes.......................................................................................... 66 55

The Law of the Lever................................................................................................................. 67 56

Heron’s Formula ...................................................................................................................... 68 57

Writing ...................................................................................................................................... 69 58

PART 10: Endnotes for Chapter 5 ............................................................................................ 70 59


PART 1: Introduction 60 61 Introduction 62 63 It is difficult to discuss Greek mathematics without writing a large 64 treatise. They contributed so much to the mathematical landscape 65 that it’s impossible to provide a treatment of their contributions to 66 the field that is both short and thorough. Hence, this and many 67 other discussions on Greek mathematics are forced to look only at 68 selected highlights of what the Greeks did with mathematics. In 69 this chapter, we will look at some of the more interesting and accessible subtopics to give us a 70 feel for some of what they did. Unfortunately, we’ll leave out much more than we present (to a 71 greater degree than we have in the past). 72 73 After the Egyptians and Babylonians, the Greeks were the next major civilization in the 74 Mediterranean West to make notable inroads in mathematics. They certainly inherited many 75 empirical and intuitive rules from the Near East, but they also worked towards transforming this 76 knowledge into a very orderly, logical, and theoretical science. 77 78 In the sixth century B.C.E., classical Greek civilization was born along with the deductive 79 reasoning that led to a more formal organization of mathematics. Early in this century, many 80 Greek thinkers moved from more mythical explanations for the world and towards a more 81 rational approach to what they saw around them in nature. Calinger calls this a shift from mythos 82 (myth) to logos (word) and points out that it was “sudden, pronounced, and apparently original 83 with the Greeks.” 12 As to exactly how and why this shift may have affected mathematics, we 84 can only speculate. Some have proposed that as astronomical devices (sundials, angle-measuring 85 devices, etc) were constructed and used, and as attention was paid to harmony and proportion in 86 architecture, the initial growth of demonstrative geometry may have began.3 87 88 Thales of Miletus, mentioned in the previous chapter, emerges as the first big name in Greek 89 mathematics. As we mentioned earlier, he’s the first person to be credited with “proving” 90 mathematical statements, even though among his contemporaries he was probably better known 91 for his work in astronomy.4 92 93 The next major, well-known figure to arrive on the Greek scene is Pythagoras, who we studied 94 more closely in the last chapter. While the Pythagoreans were studying their “arithmetica,” the 95 requirements of commerce, weights, money, justice, and law were spurring changes in the laws 96 of computation. By the fifth century B.C.E., alphabets and numerical symbols were beginning to 97 emerge for use in such computations. 98 99 In the period from 480 B.C.E. to 340 B.C.E., theoretical mathematics in Greece experienced its 100 biggest spurt of growth. During this time, number theory, musical theory, methods of computing 101 area and volume, and logic all prospered. This period is also an important one for the study of 102 “nonmetrical” geometry, which included the study of three very famous problems that generated 103 a lot of interesting mathematics for many years. 104 105


PART 2: Three Famous Problems 106 107 There are three problems in Greek mathematics that are very well known for their prolonged 108 contribution to mathematics. All three involve using only an unmarked straightedge and a 109 compass (a tool that draws circles) to physically construct a solution to the problem. These are 110 the three problems: 111 112 1. Given a circle with a certain radius, how do you construct a square that has area equal to 113

the circle? This is called squaring a circle. 114 2. Given an angle, how do you trisect (cut into three equal angles) the angle? This is called 115

trisecting an angle. 116 3. Given a cube of fixed volume, how do you construct a second cube that has exactly twice 117

the volume of the original? This is called doubling a cube. 118 119 Here are visual pictures of these statements: 120 121 Squaring the Circle 122 123 124 125 126 127 128 This problem boils down to trying to construct one of the sides of the square so that the 129 corresponding area of the square was exactly that of the circle. (No approximations allowed!) 130 Once they had that, they could easily draw a square with that as its side. 131 132 Trisecting an Angle 133 134 135 136 137 138 139 140 141 142 143 For this problem, the object is to draw ray PQ so that you create an angle that is exactly one third 144 of the original. (Again, no approximations were allowed…it had to be exact, and provable.) 145

Area = πr2 Area = πr2

a a/3 P P

Q

With compass & straightedge only



146 Doubling a Cube 147 148 149 150 151 152 153 154 155 156 The goal here is to construct a line segment, AB, so that a cube with this side has exactly double 157 the volume of the original cube with side equal to x. 158 159 This last one might seem simple if you (incorrectly) assume that doubling 160 the side will double the volume, but that is not true. For example, if you 161 have a cube with side of 2 inches, then it’s volume is ( ) 82 3 ==V cubic 162 inches. If you double the side to 4 inches, you get ( ) 644 3 ==V cubic 163 inches, which is more than double the original volume of 8 cubic inches. 164 165 The Greeks did have solutions to these problems, but none of them 166 employed only the straightedge and compass. And indeed, these other techniques did help them 167 to develop a lot of interesting mathematics5. But for some reason, they really wanted a solution 168 that used only the compass and straightedge. Because many Greek philosophers held the belief 169 that the line and circle were the very basic and most perfect of curves, these tools probably 170 became the choice of Greek mathematicians. Although they tried very diligently, they could not 171 come up with solutions using just a compass and straightedge. In the 1800’s, after more than two 172 millennia of mathematical development, it was shown that each of these classic problems could 173 NOT be solved using the compass and straightedge alone. Unfortunately, the Greeks did not 174 have the tools they needed to see this impossibility, and so they persisted in their efforts. Their 175 failures led them to new discoveries and techniques in mathematics, however, so their efforts 176 were by no means a waste of time, nor were they unproductive. 177 178 A “Solution” from Menaechmus 179 180 For example, in around 350 B.C.E., Menaechmus, who was the tutor for Alexander the Great, 181 developed a process for doubling the cube by using the parabola curve that he discovered earlier. 182 We will describe his method of doubling the cube using modern algebraic notation and graphs 183 that were not available for many hundreds of years. 184 185


A B

V = x3

V = 2x3

x

Think About It

Why does doubling the side of a cube not

double its volume?


Menaechmus used a result from Hippocrates that produced the equivalent of the following two 186 equations: 187

yx

xa= 188

189 and 190

191

ay

yx

2= 192

193 In these equations, a represents the side of the original cube whose volume, a3, is to be doubled 194 to 2a3, while x and y represent numbers on the two−dimensional axes with which you are already 195 familiar. 196 197 In the first equation we can cross-multiply to get: 198 199

ayxyx

xa

=⇒= 2 200

201 Now divide both sides by a to solve for y: 202

221xy x

a a= = 203

204 In the second equation, we do the same thing and solve for x: 205 206

2

2

212

22

2

x yy a

y axyxa

x ya

=

=

=

= 207

208 209


These two equations can be graphed in the x−y 210 plane. The first is a standard parabola. (Recall 211 that an equation of the form cbxaxy ++= 2 is a 212 parabola. In this case both b and c are 0.) The 213 second is another parabola, but it is sitting on its 214 side rather than upwards or downwards. The two 215 curves are graphed here so you can see how they 216 are oriented with respect to each other. 217 218 It can be shown that the x value of the point P of 219 intersection of these two curves represents the 220 side of a cube that will have double the volume 221 of the original. In other words, the cube you want 222 will have length of x units long. Let’s do an 223 example before we prove in general that the point 224 P really does give us the length we want. 225 226 227 Example 1 228

Let a cube have a side of 4 inches. Use the method of Menaechmus to find the 229 length of a side needed to double the volume of the cube. 230

231 232 Solution 233

Since the original cube is a = 4 inches on a side, the volume is 43 = 64 cubic 234 inches. Hence, we want to double it to 128 cubic inches and determine how 235 long each side should be. Using the two equations we derived above: 236 237

2 21 14

xy xa

== 238

2 2 21 12( ) 8

12 4

yx y ya

= == 239

240 If we graph these two equations 241 together, we get the following 242 graph: 243

244 We can see that these two 245 graphs intersect around x = 5. 246 (A quick check tells us that a 247 cube of side 5 inches will have a 248 volume of 53 = 125 cubic 249 inches, which is close to the 250 desired value of 128.) In order 251 to get a more accurate answer, 252

-8 -6 -4 -2 0 2 4 6 8

-10

-5

0

5

10

-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5

2

21 xa

y =

2

21 ya

x =

P = (x,y)

Length of desired cube


we will manipulate these equations algebraically. We have the following 253 system of equations: 254 255

2

2

1418

y x

x y

= =

256

Almost any system of equations with two variables and two unknowns can be 257 solved. To do so here, we’ll substitute the first equation into the second to 258 eliminate the y variable: 259

260 261

262 We can cross multiply and proceed: 263 264

0)128(0128

1281281

3

4

4

4

=−

=−

=

=

xxxx

xx

xx

265

266 This equation gives us two solutions: 267 268

30 or 128 0x x= − = 269 270

The first one we throw out since a cube cannot have a side of 0 inches. The 271 second equation can easily be solved for x: 272 273

3

3

3

128

1280128

=

=

=−

x

xx

274

275

2

22

44

4

181 18 41 18 16 128

1 128

x y

x x

xx x

x x

=

=

= ⋅ =

⇒ =

214

y x=

You should recognize that this is what we are looking for.


This solution represents how long the side of the new cube should be to 276 double the volume of the original cube from 64 to 128, and we can easily see 277 that when we cube 3 128=x , we do get 128. An estimate for 278 3 128 5.03968≈ , which is close to our graphical estimate of x = 5. There is a 279 much easier way to get this answer, of course, but our goal here was to use the 280 method of Menaechmus. Keep in mind that the Greeks did not have variables, 281 equations, the root symbol, or graphs to work with.♦ 282

283 284 Check Point A 285

Let a cube have a side of 1/2 inch. Use the Method of Menaechmus to produce 286 a graph that shows the length of a side needed to double the volume of the 287 cube. 288

289 Solution See endnotes for an answer.6 290 291 292 You may be wondering why this process works. To see why, we present a proof that these 293 equations do indeed produce a cube with double the original volume. 294 295 Proof 296

We start with the system of equations that we have been using up to this point: 297 298

2

21

1

2x y

a

y xa

=

=

299

Substitute the first equation into the second to eliminate the y variable: 300 301

302

303

2

2

42

4

3

2

121

21 1

2

1

1

2

x ya

xa

x xa a

x xa

xa

=

=

= ⋅

=


We can cross multiply and proceed: 304 305

0)2(02

221

33

43

43

3

4

=−

=−

=

=

xaxxxa

xxaaxx

306

307 This equation gives us two solutions: 308 309

3 30 or 2 0x a x= − = 310 311

The first one we throw out since a cube cannot have a side of 0 inches. The 312 second equation can be solved for x: 313 314

3 3

3 3

3

3

3

2

2

02

2

a x

a

x a

x a

x

− =

=

=

= ⋅

315

316 A check is easy: If we start with side a, then the volume will be a3. If we start 317

with side 3 2⋅= ax , then its volume will be ( ) 3333 222 aaaV =⋅=⋅= , which 318 is double the original volume. ♦ 319

320 321 Squaring Other Figures 322 323 While the Greeks were unable to use a compass and straightedge to square the circle, they were 324 able to square other figures. To square an object means to take a figure and, with compass and 325 straightedge only, draw a square that has area exactly equal to it. The Greeks could easily take a 326 rectangle or triangle and square them7. Thus, any figure that was a combination of rectangles 327 and/or triangles could be squared, including the figure shown below. 328 329 The study of these area problems outlived classical Greece. In the seventeenth century, famous 330 names like Leibniz (one of the independent founders of modern 331 calculus) were working on related studies. 332 333 334

You should recognize that this is what we are looking for.


Fourth Century Glory 336 338 After two hundred years of steady development, theoretical mathematics began to 340 assume a distinct form in fourth century Greece. In particular, arithmetic and 342 geometry flourished. At Plato’s Academy, the search for rigor in proofs intensified. 344 Plato, Aristotle, and Eudoxus emphasized the need for formal definitions, postulates, 346 and axioms. They merged step-by-step, logical reasoning with technical results to 348 provide clear proofs in geometry.8 They took geometrical models and applied them 350 to astronomy. They looked at the “conic sections” like the parabola, hyperbola and 352 ellipse (probably all devised by Menaechmus) in order to study curves beyond just 354 the circle and line. These and other “philosophers” played a key role in the path that 356 mathematics traveled. 357 358 Although he probably never wrote a word, Socrates indirectly contributed to the growth of 359 mathematics by “deepening the introspective component of Greek thought.”9 His pupils 360 Xenophon and Plato helped assure that this thought was transmitted into the far future. Plato, 361 however, was the dominant intellectual of the time. He took ideas from Pythagoras, Socrates and 362 others and tried to mold them into an orderly whole. 363 364 Platonic thought maintained that there is a spiritual realm of unchanging 365 ideas or forms that is independent of the physical, material world.11 Just as 366 the Pythagoreans viewed numbers as the foundation of the universe, forms 367 were the basis of intelligent structure in the universe. The “supreme ideal 368 form” is the Good. And according to Plato, only through mathematics and 369 the Socratic dialectic could aristocratic youth come to know the Good and 370 avoid being deceived by the senses. Plato’s Academy was set up to train 371 youth in the teachings of the master. Its motto (or one of them) was “Let no 372 man ignorant of geometry enter,” which was inscribed over its gate. It was 373 one of the earliest institutions of higher learning in Europe, and remained 374 influential in Greek education well after Plato died. Furthermore, because of its emphasis on the 375 role of mathematics in the process of knowing the Good, it played an important part in the 376 development of mathematics. 377 378

Plato10

Socrates1


PART 3: The Platonic Solids 379 380 Plato believed that the triangle was the building block of the universe. In one of his books, 381 Timaues, he spoke of this belief and other ideas that he had about creation. 382 383 There are five regular Platonic Solids. These are three−dimensional objects that are sometimes 384 referred to as regular polyhedra. (“Polyhedra” is plural while “polyhedron” is singular.) They 385 are made up of faces that are polygons (square, triangles, etc); they intersect at edges; where 386 three or more edges intersect you have a vertex. Plato, who knew of the work of Pythagoras, 387 studied these polyhedra. His studies of them were significant enough that they are still known as 388 the “Platonic Solids,” even though they were known before the time of Plato. 389 390 Polyhedron is a Greek word that means "many faces." A regular polyhedron has the following 391 properties: 392 393

•It is a convex solid; that is, it takes up space. 394 •All of its faces have the same shape and size. 395 •All of its internal angles between adjacent faces are the same. 396

397 The five regular polyhedra are the Cube, Tetrahedron, Octahedron, Dodecahedron, and 398 Icosahedron. 399 400 They are pictured here:12 401 402 403

Tetrahedron Cube Octahedron Dodecahedron Icosahedron

404 By observing the pictures, you can see the following: 405

•The Cube has squares for its faces. 406 •The Tetrahedron, Octahedron and Icosahedron have equilateral triangles for their faces. 407 •The Dodecahedron has regular pentagons for its faces. 408

409 (There are cutout models available in the Appendix so you can build them if you are so inclined.) 410 411 The cube, tetrahedron, and octahedron were known in pre−Greek times, and there is some 412 archeological evidence that the dodecahedron may have been known as early as the 7th century 413 B.C.E.13 The Icosahedron was probably first discovered by Theaetetus, who proved that these 414 are the only such figures that have all of their faces congruent to each other. 415 416


Plato takes the four elements of Earth and associates them with four of the regular polyhedra. He 417 writes the following: 418 419

We must proceed to distribute the figures [the solids] we have just described between 420 fire, earth, water and air. Let us assign the cube to earth, for it is the most immobile of 421 the four bodies and most retentive of shape; the least mobile of the remaining figures 422 (icosahedron) to water; the most mobile (tetrahedron) to fire; the intermediate 423 (octahedron) to air. 424

425 As for the fifth one, the dodecahedron, Plato says: 426 427

There still remained a fifth construction, which the god used for embroidering the 428 constellations on the whole heaven. 429

430 What does this mean exactly? I’m not sure, but later Greek philosophers assigned the 431 dodecahedron to the heavens, so perhaps that is what Plato meant. (There are 12 faces on the 432 dodecahedron…perhaps that corresponds to the 12 signs of the zodiac.) 433 434 Kepler and the Platonic Solids 435 The celestial associations with the regular 436 polyhedra persisted all the way through the 437 1500’s, when Kepler published a tract called 438 The Cosmic Mystery in which he pictured the 439 universe as being made up of nested Platonic 440 solids whose inscribed spheres determine the 441 orbits of the planets. All of these were enclosed 442 in a sphere representing the outer heaven. You 443 can see a picture of that model here.14 444 445 When studying why there are only five regular 446 polyhedra, the key observation to make is that a 447 solid can only be made up of regular polygons if 448 the sum of the interior angles of all the regular 449 polygons meeting at a point (vertex) adds to less 450 than 360°. That is, if we were to lay flat on a 451 table the faces meeting at any vertex of a regular 452 polyhedron, there will always be a gap left. (You may be able to visualize this with the cut-out 453 models included in the Appendix. There is also a Lab associated with this question that you can 454 use to see what is happening here.) We also see the Platonic Solids used as inspiration for a 455 variety of artistic works throughout the ages, a topic one could study with great interest and 456 entertainment. 457 458


Discovering a Pattern 459 One of the most interesting properties of these objects is related to the number of faces, edges, 460 and vertices that each one has. Here is a table that gives the values for each of the Polyhedra: 461 462

Tetrahedron Cube Octahedron Dodecahedron Icosahedron

Vertices 4 8 6 20 12 Edges 6 12 12 30 30 Faces 4 6 8 12 20

463 It may not look like these numbers are very interesting, but if 464 you look more closely, you will find a relationship between 465 the number of vertices, edges, and faces that holds true for 466 each of the five solids. 467 468 469 470 Beauty in Form 471 There are many other solids that are not considered regular, because their faces are a mix of 472 different shapes. The following pictures are taken from the Crystallographic Polyhedra website 473 by S. Weber15 where you can view and rotate them in space to see what they look like from 474 different perspectives. 475 476

Pentagonal

Icositetrahedron Bucky Ball Triacontahedron

477 You can see in these a tremendous amount of symmetry and 478 “beauty” that spur many interesting questions about how they 479 are built and what kinds of relationships exist among “parts” of 480 these polyhedra. 481

Think About It

What is the relationship between the number of

vertices, edges and faces?

Think About It

None of these three are “regular” polyhedra. Why

not?


PART 4: Euclid’s Elements and Geometric Algebra 482 483 Aristotle 484 Aristotle was Plato’s most famous student. He enrolled at the Academy at the 485 age of seventeen and stayed there for about twenty years. (You think it’s taking 486 you long to get your degree?) Aristotle believed that pure mathematics 487 (arithmetic and geometry) is part of a “theoretical systematic knowledge in 488 philosophy, together with metaphysics and physics.”16 His science sought final 489 causes for physical effects, and so he saw mathematics, which deals with 490 abstract ideas and logical relationships, as an inferior kind of knowledge. 491 Aristotle used his work Posterior Analytics to put forth an important theory of 492 statements. This theory was a significant step towards the effort to “axiomatize” mathematics (to 493 “axiomatize” is to establish it as a field of study based on a fixed set of given statements that 494 could be built upon using logic to arrive at new and more powerful results). It was among the 495 first efforts to try to mark off starting points for proofs based on logic in classical Greece. He 496 recognized that “formal proofs must begin with basic statements that 497 are indemonstrable, for otherwise proofs would involve an infinite 498 regress.”17 This theory was a major advance for formalizing Greek 499 mathematics and was later reinforced when Euclid published his work, 500 the Elements, which was a compilation of known mathematics with an 501 axiomatic system at its core. 502 503 504 Euclid (c. 325 – 265 B.C.E.) 505 Despite the fact that the name Euclid is widely known in and out of 506 mathematical circles, very little is known about the life of this man. The 507 information we do have about him was written about 750 years after he was 508 alive (written by Proclus, 410-485 C.E.). 509 510 Most believe that he was a teacher at the great Library and Museum in 511 Alexandria. This was a place where scholars would meet to talk about 512 philosophical and literary ideas. The Alexandrian Museum became a focal 513 point for the best work by Greek scholars in the sciences and humanities. Those who were 514 employed as researchers for this establishment would also teach younger students who would 515 come to learn. 516 517 None of Euclid’s original writings are extant today, but we do have copies from later times that 518 we can examine to try to learn about what he did. Euclid’s Elements is the most important 519 mathematical text from Greek times and may even be the most important math text of all time.19 520 Elements is made up of thirteen books and is a compilation of some of the most important 521 mathematics of the time. The structure of the Elements is as follows: 522 523

Euclid18

Think About It

What does it mean for a proof to

involve “infinite regress?”


THE CONTENTS OF EUCLID’S ELEMENTS 524 Books 1 to 6: A treatment of geometry in two dimensions. 525 Books 7 to 9: Number theory. 526 Books 10: Attempts to link geometry to number theory. 527 Books 11 and 12: A study of three-dimensional geometric objects. 528 Book 13: Construction of the five regular polyhedra (Platonic Solids). 529

530 Others before Euclid had published such compilations, but when the Elements came along, it so 531 far surpassed in quality previous attempts that they have sunk into historical oblivion. Elements 532 dominated the landscape and survived where others did not. 533 534 What makes the Elements so compelling is that Euclid takes almost nothing for granted. Very 535 little is stated without being rigorously proved. To do this, Euclid adopted Aristotle’s suggestion 536 that a scientific work should begin with definitions and axioms. Everything else would be built 537 up from those starting points. 538 539 In Book 1, Euclid starts with his definitions.20 There are 23 of them given in Book 1. Some 540 examples include: 541 542

1. A point is that which has no part. 543 2. A line is “breadthless” length. 544 3. The extremities of a line are points. 545 4. A straight line is a line which lies evenly with the 546

points on itself. 547 5. A plane angle is the inclination to one another of 548

two lines in a plane which meet one another, and do 549 not lie in a straight line. 550

6. A circle is a plane figure contained by one line [the 551 radius] such that all the straight lines meeting it from 552 one point [the center] among those lying within the figure are equal to one another. 553

554 Not only are some of these difficult for us to read and understand, but also by modern standards, 555 the first few of Euclid’s definitions are really not definitions.21 Euclid is using his definitions of 556 point and line, for example, to help the reader understand the ideas involved and to claim the 557 existence of these objects in the first place. These are starting points for Euclid; they are places 558 from which to launch. 559 560

Think About It

How do Euclid’s definitions for straight lines and circles match

up with your own understanding of these

geometric objects?


Euclid then moves on to state several postulates. (We will consider axioms and postulates to be 561 the same thing.) These are statements that are to be accepted as true and do not need to be 562 proven. “They just are!” (Most students think that most of the rules in math work “just because.” 563 In reality, very few rules behave this way and the only things that “just are” tend to be 564 mathematical definitions or axioms.) Here are few of Euclid’s postulates: 565 566 Euclid’s Five Postulates 567

1. [It is possible] to draw a straight line from any point to any [other] point. 568 2. [It is possible] to draw a finite straight line continuously in a straight line. (A ray) 569 3. [It is possible] to construct a circle with any center and radius. 570

4. All right angles are equal to each other. 571 5. If a straight line intersecting two straight lines 572

make the interior angles on the same side less 573 than two right angles, the two straight lines, if 574 produced indefinitely, meet on that side on 575 which the angles are less than two right angles. 576

577 The first four seem pretty straightforward. The last, called 578 the parallel postulate, is the most intriguing of the 579 postulates. It claims that if line c crosses the lines a and b, 580 making the sum of angles 1 and 2 less than two right angles 581 (180o), then the lines a and b will eventually cross in the direction of A. It’s essentially a 582 roundabout way of stating that two parallel lines do not ever intersect. 583 584 Not everyone has agreed that this should be a postulate and have tried to show that it could 585 actually be proved using the first four postulates. All such attempts have failed, however. (It is 586 interesting to note that if you choose a different set of axioms or postulates for your geometry, 587 you can get a different geometry. The geometry that we are used to working with is called 588 “Euclidean Geometry” because it’s based on the axioms/postulates of Euclid. But there are other 589 valid geometries such as hyperbolic and spherical geometry that also can be used.) 590 591 Next, Euclid includes common notions, which are truths that are common to all sciences. They 592 include: 593 594

1. Things which are equal to the same thing are equal to each other. 595 (If A=C and B=C, then A=C) 596 2. If equals are added to equals, the wholes are equal. 597 (If A=B, then A+C=B+C…or, you can add the same thing to both sides of an 598

equation.) 599 3. If equals are subtracted from equals, the remainders are equal. 600 (If A=B, then A-C=B-C…or, you can subtract the same thing from both sides of an 601

equation.) 602 603 These first three common notions are used regularly in basic algebra. 604 605

1

2

a

b

c

A


After laying a foundation of definitions, postulates, and common notions, Euclid then presents 606 his propositions. These are statements that tell the reader how to go about accomplishing a task 607 or are statements of mathematical fact that must be proved using the definitions, postulates, and 608 common notions. There are dozens of propositions in the Elements, and many of them are 609 challenging to follow if you have not carefully read through the previous propositions. We’ll 610 present just the very first proposition here.22 It shows how to build an equilateral triangle if you 611 start with the length of one side. The process is described and then reasoning is given to explain 612 why it works. The wording used is reflective of a translation of Euclid’s words into English, 613 which accounts for some of its awkwardness. 614 615 Proposition 1 To construct an equilateral triangle on a given finite straight line.

Let AB be the given finite straight line. It is required to construct an equilateral triangle on the straight line AB.

Describe the circle BCD with center A and radius AB. Again describe the circle ACE with center B and radius BA. Join the straight lines CA and CB from the point C at which the circles cut one another to the points A and B.

Post.3 Post.1

Now, since the point A is the center of the circle CDB, therefore AC equals AB. Again, since the point B is the center of the circle CAE, therefore BC equals BA.

I.Def.15

But AC was proved equal to AB, therefore each of the straight lines AC and BC equals AB.

And things which equal the same thing also equal one another, therefore AC also equals BC. C.N.1

Therefore the three straight lines AC, AB, and BC equal one another.

Therefore the triangle ABC is equilateral, and it has been constructed on the given finite straight line AB. ♦

616 Notice that Euclid is very careful to describe what he is doing and presents a logical argument 617 for each step. The postulates, definitions and common notions are referenced in the column to 618 the right so that the reader can go back and look up what they say. All of the Elements is written 619 in this structured way and so it provides a valuable training in deductive reasoning and logic. 620 (This is a skill that seems to have been lost in the recent development of mathematics education 621 in this country, which is sad. Everyone can benefit from learning how to construct and follow a 622 logical argument.) 623 624


Besides the Elements, Euclid wrote about a dozen other works that covered many branches of 625 mathematics, but they are rarely discussed in any length in history of math treatments.23 They 626 are simply overshadowed by his immensely popular, still-in-print masterpiece. 627 628 Geometric Algebra 629 One of the most interesting aspects of Greek mathematics is the fact that they could do the 630 equivalent of a lot of our algebra even though they did not have variables like x or y, equations, 631 equal signs, or any of the modern symbols that we do. Instead, they interpreted everything in 632 terms of geometry. This shows up prominently in the Elements. 633 634 To the Greeks, a number was represented by a straight line segment with the length of the 635 segment representing the number. 636 637 For example, to represent the number five, the Greeks did not have the symbol “5.” Instead, they 638 would visualize and work with a straight line segment that was five units long. This, to them, 639 was the number 5. 640 641 642 643 644 Because of this geometric interpretation for number, what we would think of as an unknown 645 variable x, they would think of as a line segment with length to be determined. If their problem 646 involved the square of an unknown number, ( 2x for example) then they would represent this as a 647 square with the unknown quantity being the side of the square: 648 649 650 651 652 653 654 655 656 657 658 659 660 661 If the Greeks wanted to cube a number, they would envision a literal cube with the number to be 662 cubed represented as one of its sides. 663

x

x

x

The square of x is the area of this square, x2

5

Think About It

Have you ever thought about why, when we take a number and multiply it by itself, we say that we are “squaring the number?” Or

when we take a number and multiply it by itself two more times (5×5×5), we say we are cubing that number? The approach of the

Greeks should answer that question for you!


This is the “square on the whole.”

In the Elements, Euclid says the following in Book 2, Proposition 4: 664 665 If a straight line is cut at random, the square on the whole is equal to the squares on the 666 segments and twice the rectangle contained by the segments. 667 668 669 Let’s interactively explore this a little. Here is a line segment that has been cut “at random” at the 670 point P. 671 672 673 674 675 676 677 This proposition tells us something about the 679 “square on the whole.” The square on the whole 681 is the square that is formed with each side equal 683 to the whole length of segment AB. We’ll call it 685 ABCD. 687 689 This square (ABCD) has some area. According to 691 this proposition, it’s equal to the area of the 693 695

“squares on the segments” 697 and 699

“twice the rectangle” 701 703

The segments being referred to are segments AP 705 and PB…. they’re formed when the line segment 707 AB is cut at point P. 709 711 You’ll notice that we can use these segments to cut the “square on the whole” into four regions; 712 two squares and two rectangles, labeled 1 through 4 in the picture. 713 714 The two rectangles are regions and ; the squares are regions and . 715 716 Let the length of AP = a and let the length of PB = b. Use this information to label the rest of the 717 segments on the outer edge of the “square on the whole” with either a or b. 718 719 In terms of a and b, what are the areas of rectangular regions and ?24 720 721 In terms of a and b, what are the areas of square regions and ?25 722 723 When you add up the four regions that make up the square on the whole, what do you get in 724 terms of a and b?26 725 726

P BA

b a


What is the area of the “square on the whole,” in terms of a and b?27 Make sure that your 727 expression is completely simplified. 728 729 With all of these facts in hand, we can know see what Euclid was proposing. He’s simply saying 730 that ( ) 222 2 bababa ++=+ , where the left side of the equation represents the square on the 731 whole, while the right side represents the total sum of the pieces that make up this square. When 732 you read his statement again, it should now make sense. 733 734

If a straight line is cut at random, the square on the whole is equal to the squares on the 735 segments and twice the rectangle contained by the segments. 736

737 For us, this is a relatively easy fact to see since we are familiar with the algebraic notation we’ve 738 used here…something Euclid did not have available to him. Euclid stated this fact and then went 739 on to prove it in rigorous detail. If you would like to look at what his proof looks like (after 740 being translated into English), see the next page. It’s not easy to follow, but it might be 741 interesting to look at to see how detailed Euclid gets. 742


Euclid’s Elements Book 2, Proposition 4 743

744 If a straight line is cut at random, then the square on the whole equals the 745 sum of the squares on the segments plus twice the rectangle contained by 746 the segments. 747 748 Proof 749 750 Let the straight line AB be cut at random at C. 751 752 I say that the square on AB equals the sum of the squares on AC and CB 753 plus twice the rectangle AC by CB. 754 755 Describe the square ADEB on AB. Join BD. Draw CF through C parallel to either AD or EB, 756 and draw HK through G parallel to either AB or DE. (I.46 and I.31) 757 758 Then, since CF is parallel to AD, and BD falls on them, the exterior angle CGB equals the 759 interior and opposite angle ADB. (I.29) 760 761 But the angle ADB equals the angle ABD, since the side BA also equals AD. Therefore the angle 762 CGB also equals the angle GBC, so that the side BC also equals the side CG. (I.5,I.6) 763 764 But CB equals GK, and CG to KB. Therefore GK also equals KB. Therefore CGKB is 765 equilateral. (I.34) 766 767 I say next that it is also right-angled. Since CG is parallel to BK, the sum of the angles KBC and 768 GCB equals two right angles. (I.29) 769 770 But the angle KBC is right. Therefore the angle BCG is also right, so that the opposite angles 771 CGK and GKB are also right. (I.34) 772 773 Therefore CGKB is right-angled, and it was also proved equilateral, therefore it is a square, and 774 it is described on CB. For the same reason HF is also a square, and it is described on HG, that is 775 AC. Therefore the squares HF and KC are the squares on AC and CB. (I.34) 776 777 Now, since AG equals GE, and AG is the rectangle AC by CB, for GC equals CB, therefore GE 778 also equals the rectangle AC by CB. Therefore the sum of AG and GE equals twice the rectangle 779 AC by 780 CB. (I.43) 781 782 But the squares HF and CK are also the squares on AC and CB, therefore the sum of the four 783 figures HF, CK, AG, and GE equals the sum of the squares on AC and CB plus twice the 784 rectangle AC by CB. 785 But HF, CK, AG, and GE are the whole ADEB, which is the square on AB. Therefore the square 786 on AB equals the sum of the squares on AC and CB plus twice the rectangle AC by CB. 787


Therefore if a straight line is cut at random, the square on the whole equals the squares on the 788 segments plus twice the rectangle contained by the segments. 789 790 The proof is done. ♦ 791 792 If you tried to get through the previous page and had difficulty, you’re not alone. Studying 793 Euclid can be a challenging task when you first encounter his style…it takes time to adjust to it. 794 795 We (and the Greeks) can represent other algebraic equations with geometric drawings. 796 797 798 Example 2 799 800

Consider the figure shown. What algebraic expression is represented by this 801 figure? 802

803 Solution 804

To answer this question, we must think like the Greeks did. To them, numbers 805 were line segments. When multiplying or squaring numbers, the Greeks would 806 envision rectangles or squares with the areas of these objects representing the 807 product they were interested in computing. In this figure, we see a large 808 rectangle that is broken up into two smaller rectangles. The area of the largest 809 rectangle, ABCD, is equal to the following: 810

Area of ABCD = (Length)(Width) 811 = (a)(b+c) 812 = ab + ac 813 814

On the other hand, the large rectangle is 815 made up of two smaller rectangles. One has 816 length equal to a and width equal to b. Thus 817 its area is ab. The other rectangle has length 818 equal to c and width equal to a. Thus its area 819 is ca. Adding these together gives a total 820 area of ab+ac. This is exactly what we got 821 above. 822

823 Hence, what we see in this picture is a geometric demonstration of the 824

distributive property of multiplication that you have used many times 825 before in your mathematical studies. 826

827 828

a a

b c A B

D C


Check Point B 829 Consider the figure shown. What 830

algebraic fact is represented with this 831 picture? 832

833 Solution See the endnotes for an answer. 28 834 835 836 837 838 839 Today, we are very used to expressing these relationships with variables and modern notation. 840 The Greeks, on the other hand, viewed these algebraic facts and properties in highly geometric 841 ways. In fact, they could even solve many “equations” with geometric figures. Instead of doing 842 algebra with symbols like x and y, they would use the compass and straightedge to “build” a 843 solution. We start with a simple class of such equations: ax bc= . 844 845 846 Solving ax = bc with Greek Construction Methods 847 By Euclid’s time, Greek geometric algebra had progressed far enough that it could solve some 848 basic equations with unknown quantities. The equations were interpreted geometrically and then 849 solved with methods of construction (with compass and straightedge). For example, take the 850 linear equation ax = bc. This was viewed as a statement that two areas were equal. The Greeks 851 would solve this equation for x as follows (with all steps being done with a compass and 852 straightedge): 853 854 1. Construct a rectangle ABCD with sides AB = b, BC = c. This rectangle has area equal to 855

bc. 856 857

b

c

A B

CD 858 859 2. Extend AB to the left by a units to point E 860

b

c

a A B

CD

E

861 862

p

q

r s


3. Construct a line segment through point D that extends far enough so that it meets BC (at 863 the point F) when it is extended downwards. 864

865 b

c

a A B

CD

E

F 866 867 4. Complete rectangle EBFH. The desired unknown quantity x is the line segment CF = KH. 868 869 870

b

c

a

xx

a

A B

CD

E

FH

K

G

Area = ax

Area = bc

871 872 873 To see why CF = KH is the value of x that satisfies the equation, we first note that the area of 874 rectangle KDGH is its length times its width, or ax. Also, the area of rectangle ABCD is bc. If 875 we can show that these two areas are equal, then x will satisfy the equation ax = bc and we’d be 876 done. To do so, we now view the rectangle as two equal triangles, ∆EHF and ∆EBF. 877 878

879 880


Each of these two triangles is made up of three pieces: 881 882

∆EHF = ∆EKD + ∆DGF + KDGH 883 ∆EBF = ∆EAD + ∆DCF + ABCD 884

885 Since these two triangles’ areas are equal, we have: 886 887

∆EKD + ∆DGF + KDGH = ∆EAD + ∆DCF + ABCD 888 889

However, ∆EKD = ∆EAD and ∆DGF = ∆DCF (Why?). So, we can cancel all of these from the 890 equations and we’re left with: 891

KDGH = ABCD 892 893

But in terms of areas, this means that ax = bc, as these are the areas of these two triangles. ♦ 894 895 896 This method gave the Greeks a way to solve any linear equation of the form ax = bc using a 897 compass and straightedge. When they implemented it, they didn’t end up with a numerical 898 symbol like we would. Instead, they created an actual line segment whose length x is the desired 899 quantity. 900 901 902 Example 3 903 For example, suppose we had the equation that we wanted to solve with 904

“geometric algebra”: 905 906

43

23

21

×=x 907

908 Solution 909

910

First notice that the equation is in the form cbax ×= . We have 21

=a , 23

=b 911

and 43

=c . We want to build a figure with these lengths, so we convert to 912

decimals to more easily measure the lengths we are going to draw. Hence, we 913 have a = 0.5, b = 1.5, and c = 0.75. 914 915

The first step is to draw rectangle ABCD with 916

length of 5.123==b and width of 917

75.043==c . 918

919 920

A B

C D

b =1.5

c = 0.75


A B

C D

b

c

a E

A B

C D

b a E

F

c

Then line segment EA is extended 921

from AB with length of 5.021==a . 922

923 924 925 926 The next step is to extend a line 928 segment through points E and D that will 930 cross line segment BC extended downwards. 932 934 936 938 940 942 944 946 948 950 952

With that done, the larger rectangle 953 EBFG can be constructed and the 953 solution CFx = can be measured. It 953 should be very close to the actual answer 953 that you would get with “modern” algebra. 953 (This picture’s scale is a little off but you will 953 get relatively accurate results if you are careful 953 with your graph paper or ruler.) 953

970 972 974 975

Solve the equation above for x with regular algebra. What do you get for x? Is 976 it the same as the measurement of line segment CF? Explain any 977 discrepancies. ♦ 978

979 980

A B

C D

b =1.5 a = 0.5

c = 0.75

E

x = 2.25

G


Check Point C 981 Build a geometric solution to the following equation: 453 ×=x . Use a ruler 982

to measure carefully. 983 984 Solution 985

Check your own solution with regular algebra. Work with a ruler using 986 centimeters to make the drawing a reasonable size. How close do you get on 987 your drawing to the actual algebraic answer? 988 989

990 991 992 993 994 995 996 997

998 999 1000 Solving Quadratics with Geometric Methods 1001 When faced with a quadratic equation, Euclid would reduce the 1002 equation so that it looked like one of the following: 1003

2

2

2

)()()(

bxaxbaxxbaxx

=−

=−

=+

1004

1005 Applying results he already knew about areas to the equation at 1006 hand, he could solve each of these. 1007 1008 1009 For example, in Book I, Proposition 6 of the Elements, Euclid shows how to solve 2)( baxx =+ , 1010 which we will write as 22 baxx =+ . His method would essentially be to take a given line 1011 segment AB = a, construct the rectangle AQKF of known area b2 on AB in such a way that it 1012 will exceed (the rectangle ABFH on the whole segment AB) by a square figure, say x2. This 1013 would produce a figure like the one shown below: 1014 1015

x

x

xy

a QB

KH

A

F

ax x^2

1016 1017

x2

Think About It

What happens to your solution and your

drawing if you approach the problem by looking at

3 10 2x = × ?

Think About It

How is this approach similar to the

Babylonian approach for solving quadratics?

(See Chapter 3)


With such a construction, the line segment BQ would be the solution to the equation because 1018 under these conditions, you would have: 1019 1020

Rectangle AQKF = Rectangle ABHF + Square BQKH 1021 or 1022

22 xaxb += 1023 1024

Of course, how you construct rectangle AQKF is another matter into which we will not delve. 1025 (It’s not straightforward!) The point is that Euclid, and others before him, had specific methods 1026 for solving such equations and these methods were based on geometric approaches rather than 1027 the use of variables (which were not available to them at the time). The Babylonians could solve 1028 equations like these as well, and their approaches end up being pretty equivalent to these (when 1029 you translate everything to modern algebraic notation), even though they may have looked 1030 different on tablets. As we saw in Chapter 3, the Babylonians used step-by-step algorithms to 1031 solve their quadratics. While the methods of the Greeks may seem cumbersome to us due to our 1032 unfamiliarity with them, it’s reasonable to expect that mathematicians familiar with these 1033 techniques could probably solve quadratic equations with their compass and straightedge as fast 1034 (or faster) than we can with our modern quadratic formula. 1035 1036


Later Geometric Algebra 1037 1038 In later centuries, other mathematicians also had geometric solutions to 1039 quadratics. Notably, Al−Khwarizmi (780−850 C.E.)29 had a method of 1040 “completing the square” that has survived to this day, albeit couched in 1041 modern algebraic terms. 1042 1043 Al−Khwarizmi is best known for his work on algebra and he wrote a book on 1044 the subject entitled treatise Hisab al-jabr w'al-muqabala (“The Condensed 1045 Book on Calculation by Restoring and Balancing”). From the title you see the 1046 work al−jabr, which is where our word algebra comes from. (Aha! Now you 1047 know who’s responsible for algebra...or at least the word.) The word al−jabr 1048 basically means to restore an equation by eliminating negative quantities. 1049 Hence, the process of al−jabr would take the equation 22 85 xxx −= and 1050 eliminate the 28x− by adding 28x+ to both sides, giving xx 59 2 = . 1051 1052 Al−muqabala is the act of balancing an equation by subtracting positive values of the same 1053 power from both sides, starting on the side of the largest power. This would be a typical 1054 al−muqabala operation: 1055 1056

xx

xx

10212929102950

2

2

=+−=−+=+

1057

1058 These methods are familiar to us. We are very used to adding and subtracting to or from both 1059 sides of an equation to balance and eventually solve it. The chapter (as yet unwritten) on non-1060 Western math will include a more thorough treatment of his work as well as examinations of 1061 mathematics from the Islamic, Indian, and Chinese traditions. 1062 1063


PART 5: Archimedes 1064 1065 Shortly after Euclid, Archimedes of Syracuse (287-212 B.C.E.) 1066 emerged on the mathematical scene. He is widely considered to be the 1067 most prolific and significant mathematician of antiquity. Scholars 1068 have recently categorized his work and writings into three main 1069 categories:30 1070 1071 1. Pure geometry with measurements of plane and solid figures. 1072 2. Inquiries into hydrostatics and (geometrical) mechanics. 1073 3. Other works on topics like arithmetic. 1074 1075 We know more about Archimedes31 than we do any other ancient 1076 mathematician, but that doesn’t mean there isn’t a lot of legend surrounding his name. From 1077 what we do know, his father probably taught him geometry, and he apparently visited Egypt, 1078

studying under students of Euclid.32 While there, he is said to 1079 have invented a device used for irrigation (although not all 1080 scholars agree that he actually did this). A long cylinder with a 1081 spiral running its entire length33 was placed into the Nile and as 1082 the spiral was rotated, water would be drawn up and over the 1083 levees. It is now known as Archimedes’ screw. This particular 1084 picture34 is one of many portrayals of this device that have been 1085 made over the years. 1086 1087

After his studies were over, he returned to Syracuse (where he was born and raised). There, he 1088 concentrated on work in geometry and its applications. Due to his studies, he was able to create 1089 some very clever inventions and solve some interesting problems. 1090 1091 One of the most famous stories about Archimedes is one in which King Heiro asks him to 1092 determine whether or not a wreath he had commissioned was made of pure gold, as he had 1093 requested. Archimedes was to determine this without hurting the wreath. (Common legend has it 1094 being a crown, but we’re pretty certain it was a wreath instead.) After a while, it is reported that 1095 Archimedes, while bathing in a public bathroom, came upon a way to do it. He was so excited 1096 that… 1097 1098

He did not delay, but in his joy leapt out of the tub, and rushing naked towards his home, 1099 he cried out in a loud voice that he had found what he sought. As he ran he repeatedly 1100 shouted in Greek, “Eureka! Eureka” (“I have found it! I have found it!”)35 1101 1102

How did he find his solution? While stepping into a full bathing pool, he noticed that his body 1103 caused water to spill over the sides. He concluded that his submerged body would correlate to 1104 the volume and weight of the displaced water. He figured he could then get equal weights of gold 1105 and silver, observe what water each displaced, and use the results to determine if the wreath was 1106 pure gold or not. It is doubtful that something this simple would get Archimedes excited enough 1107 to run naked through he streets. This discovery and solution would have been a minor feat 1108 compared to many of his other accomplishments, but it is hard to beat back myth and legend. 1109


1110 Archimedes was also said to have accomplished great feats with levers and pulleys. One of the 1111 most famous quotes attributed to him is “Give me a place to stand on [and a fulcrum]! And I will 1112 move the entire Earth.” When King Hiero actually challenged him on this, Archimedes 1113 supposedly rigged up a system that could move a fully loaded ship weighing 4,200 tons along the 1114 beach (out of water)36. 1115 1116 When Macrellus attacked Syracuse, many of the devices invented by Archimedes helped to 1117 deflect these attacks. We have no archeological evidence of these devices, but literary stories 1118 suggest he oversaw the use of battering rams, cranes, catapults, ballistic weapons, and even 1119 complicated pulleys used to lift ships from the sea and then drop them from a height sufficient to 1120 cause them considerable damage.37 Roman soldiers were fearful of such weapons and would 1121 often flee when confronted by them. 1122 1123 (Despite these wonderful stories, it is not exactly certain how he felt about mechanical 1124 inventions. However, very few of his writings dealt with such work, so we know little about how 1125 he viewed these devices compared to his other work.) 1126 1127 Archimedes died during the Second Punic War in 212 B.C.E. when the Romans finally breached 1128 the walls of Syracuse. During the looting, a Roman soldier is said to have killed Archimedes. 1129 Accounts of his death range in their dramatic appeal. Many accounts have him engrossed in a 1130 problem when the Roman soldier appears. Legend has it (among others) that he told the soldier 1131 “Do not spoil my circles” (which he was drawing in the sand) and, insulted, the soldier killed 1132 him. 1133 1134 A few years ago, the oldest known copy of Archimedes work was auctioned off for two million 1135 dollars. This is the most ever paid for a mathematical text. It is now in a museum in Baltimore 1136 being restored for future study. 1137 1138 Selected Work of Archimedes 1139 Archimedes did so much it’s difficult to catalog. His work touched on geometry, arithmetic, 1140 statics, optics, astronomy, engineering, hydrostatics, and much more. Of all his mathematical 1141 achievements, he seems to have been most proud of On the Sphere and Cylinder.38 This work 1142 consisted of two books with 53 propositions making up its contents. Samples of the results he 1143 states and prove include the following: 1144 1145 1. The surface area of a sphere is four times the area of a great circle of the sphere. This 1146

translates to the equation for the surface area of sphere as 24 rS π= . 1147 2. If about a sphere there is circumscribed a cylinder whose height is equal to the diameter of 1148

the sphere, then the volume of the cylinder is three halves of the volume of the sphere; and 1149 the surface of the circumscribing cylinder, including its bases, is three halves of the surface 1150 of the sphere. 1151

1152


This latter proposition is a mouthful, but graphics might help. In this first picture39, you 1153 see a sphere that is inscribed in a cylinder…or a cylinder circumscribed around a 1154 sphere. Let’s call the diameter of the sphere d, and its radius r. That means that the 1155 radius of the base of the cylinder is also r, and the height of the cylinder is d. The first 1156 part of statement (b) above is: 1157 1158

Volume of the cylinder = Three halves the volume of the sphere 1159 2 33 4

2 3r h rπ π= × 1160

1161 We get the equation above from the standard formulas for the volume of a cylinder 1162

( )hr 2π and sphere

3

34 rπ . It’s important to keep in mind that Archimedes did not 1163

have use of these equations…they are more modern. He’s giving (and proving!) 1164 relationships between these two geometric objects with relatively few 1165 sophisticated formulas to use. Substituting the d for h, and then 2r for d, above, 1166 this equation becomes: 1167 1168

Volume of the cylinder = Three halves the volume of the sphere 1169 1170

2 3

2 3

2 2

3 3

3 42 32

(2 ) 22 2

r h r

r d rr r r

r r

π π

π π

π π

π π

= ×

=

=

=

1171

1172 From this last statement we can confirm that Archimedes’ claim is correct. Archimedes proof, 1173 however, is not nearly as simple. 1174 1175

h=d

d=2r


The Area of a Circle 1176 1177 In another work called Dimension of the Circle, Archimedes’ first proposition is the following: 1178 1179 The area of every circle is equal to the area of a right triangle where the radius of the circle is 1180 one leg of the right triangle and the other leg of the triangle equals the circumference of the 1181 circle. 1182 1183 In visual terms, Archimedes is equating the following two areas: 1184 1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 This is actually a pretty interesting result because Archimedes has taken a circle and transformed 1198 it into a figure (a triangle) with equal area. Furthermore, a triangle is one of the easiest 1199 geometrical figures to draw. We know that the Greeks knew how to take a triangle and square it 1200 (with compass and straightedge), so this is one step away from being able to square the circle. 1201 Unfortunately, Archimedes did not give a way to construct such a triangle. He only proved that if 1202 you had a triangle that met the given conditions, it would be equal (in area) to the given circle. 1203 1204 An Approximation for π 1205 1206

Another of Archimedes’ great achievements is his proof that the number π is between 713 and 1207

71103 , which appeared in his work called Measurement of a Circle.40 That is: 1208

71103

713 <<π 1209

(Note that 722

713 = , which is often given as a good approximation to π.) 1210

1211 Archimedes used what is called the “method of exhaustion” to prove this inequality. The method 1212 of exhaustion is one in which, by successive approximations, a desired value is estimated to 1213 within an acceptable level of accuracy. It is not easy to describe without seeing an example of it 1214 at work. 1215 1216

r r

C

=

Circle with radius r and circumference C

Right triangle with legs of r and C


To find an approximation of π, Archimedes took a circle (with the radius 1217 equal to ½ so that its perimeter/circumference, 2πr = 2π(1/2) = π) and 1218 inscribed (inside) and circumscribed regular polygons of 6,12,24,48 and 96 1219 sides. You can see the first step illustrated here. Note the hexagons (six-sided 1220 figures) that are inscribed and circumscribed in and around the circle. The 1221 hexagon inside the circle has a perimeter that is smaller than that of the 1222 circumference of the circle. The hexagon on the outside has a perimeter 1223 larger than that of the circumference of the circle. Calculating the perimeters 1224 of the hexagons was something Archimedes could do (with some work) and so it gave him a 1225 crude estimate for π. 1226 1227 The next step was to double the number of sides in the hexagons from 6 to 12. 1228 For the inscribed polygon with 12 sides, you can see that its perimeter is much 1229 closer to the perimeter of the circle than its predecessor (shown with dotted lines). 1230 By subsequently circumscribing a 12-sided hexagon on the outside of the circle, 1231 and computing its perimeter, Archimedes could (and did) get an even smaller 1232 range within which π must lie. (If you think the calculations for such perimeters 1233 are easy, try them!) 1234 1235 Doubling to 24, then 48, and then 96 sides, Archimedes gets close to “exhausting” the entire 1236 circle (hence the name of the method) as the polygons look more and more like circle. In the 1237 process, he arrives at his famous inequality and we end up with the most common fractional 1238 estimate of π used in modern times. This is the first recorded proof that actually computed an 1239 estimate for π. (Most civilizations knew that the ratio of the circumference of a circle to its radius 1240 gave you a constant number, and many even tried to find the value of this number, which we 1241 now call π. But they simply gave the estimate and did not prove or show how they got it.) 1242 1243 This method of exhaustion is often recognized by some to be very similar to “taking a limit,” in 1244 modern terms, and is connected with ideas related to calculus (Chapter 7). Although Archimedes 1245 was not using or inventing calculus, his methods share a theme with the later discovery of 1246 calculus that transformed the world. 1247 1248 1249


Areas Inside Parabolas 1250 1251 Another great accomplishment of Archimedes is 1252 simple, elegant, and powerful. In Quadrature of a 1253 Parabola, Archimedes uses the method of 1254 exhaustion to find the area trapped inside a 1255 parabola by a “chord” that cuts it off. His 1256 ingenious result states that the area trapped inside 1257 the parabola is four thirds of the area of the 1258 inscribed triangle. 1259 1260

AreaParabola = 43

AreaTriangle 1261

1262 1263 For example, consider the graph of 2xy = and the 1264 horizontal line 6=y , which acts as a chord here. The horizontal line creates an area trapped 1265 inside the parabola (shown as shaded). 1266 1267

The shaded area is 34 as large as the area inside of the triangle inscribed in the trapped area. (The 1268

inscribed triangle will be taken to be the one formed with one of its vertices at the parabola’s 1269 vertex and the other two vertices at the points where the chord intersects the parabola.) 1270 1271 Once again, this is an example of taking a non-standard shape (parabola) and reducing its area to 1272 one that is very familiar (the triangle). Since the area of a triangle is easy to compute, the area 1273 trapped inside a parabola is easy to compute as well. 1274 1275

-4 -2 0 2 40

2

4

6

8

10Chord


Example 4 1276 Find the area inside of the parabola as shown in the given picture. (The chord 1277

cutting off the parabola is the x axis.) 1278 1279 Solution We first just note that Archimedes did not have use of the x-y plane…we are 1280

using it here for convenience. 1281 1282

The inscribed triangle has a height of 1283 25 and a base of 10 (since it extends 1284 from –5 to +5), so the area of the 1285 triangle is 1286

1287

Area = 125251021

=×× 1288

1289 Thus, the area trapped inside the 1290 parabola is 1291

1292

32166

3500125

34

==× . ♦ 1293

1294 1295

Check Point D 1296 Find the area inside of the parabola as 1297

shown in the given picture. (The 1298 chord cutting off the parabola is the x 1299 axis.) The vertex of the parabola is at 1300 (1,9) in case you are having trouble 1301 reading it. 1302

1303 1304 Solution See endnotes for answer.41 1305 1306 1307 1308 1309 Most of us have simply never even thought about the area of such a figure. We’ve been limited 1310 to areas of rectangles, squares, and circles (or combinations of these figures), and yet it is 1311 possible to exactly compute areas of multitudes of regions. Using ancient methods of 1312 Archimedes, or the modern methods of calculus, our geometric horizons can be extended. 1313 1314

-2 0 2 40

2

4

6

8

10

-5 0 5 10

-30

-20

-10

0

h

b


Example 5 1315 Find the area trapped inside the parabola 1162 +−= xxy and the line y = 6. 1316 1317 Solution This will require a healthy dose of review of quadratic equations and their 1318

graphs 1319

Recall that the vertex of a parabola has an x coordinate of ab

2− . For this 1320

particular parabola, we have the values a = 1, b = −6, and c = 11. So the x 1321 coordinate of the vertex is: 1322

2( 6)2(1)

623

bxa

x

x

x

−=

− −=

=

=

1323

1324 To find the y coordinate of the vertex, we plug this into the equation: 1325

1326 23 6(3) 11

9 18 112

yyy

= − += − +=

1327

We thus have a vertex of (3,2). 1328 1329

If we want to know where 1330 this parabola intersects the 1331 line y = 6, we set the equation 1332 for the parabola equal to 6 1333 and then solve for x: 1334

)1)(5(0

5601166

2

2

−−=+−=

+−=

xxxxxx

1335

1336 This yields the points x = 5 1337 and x = 1 as points of 1338 intersection. We can now 1339 sketch a pretty good graph of the parabola and the line y = 6, and see what 1340 area we are after. 1341

1342

0 2 4 60

2

4

6

h

b


The inscribed triangle has a height of 6−2 = 4. Its base is 4 units long (since 1343 the intersection points are at x = 1 and x = 5, which is a distance of 4). So the 1344

inscribed triangle has area = ( )( ) 84421

= . 1345

Therefore the parabola has area of 3210

332)8(

34

== ♦ 1346

1347 1348 Check Point E 1349 Find the area trapped inside the parabola 542 +−= xxy and the line y = 5. 1350 1351 Solution See the endnotes for an answer.42 1352 1353 1354 1355 1356 Example 6 1357 Find the area of the 1358

shaded region shown. 1359 It may seem like a 1360 strange area to want to 1361 compute, but it turns 1362 out that these kinds of 1363 computations are 1364 common in calculus 1365 and have very specific 1366 applications to a 1367 variety of disciplines. 1368

1369 Solution For this problem, we 1370

will consider the 1371 regions A, B, C, and D, 1372 as labeled. We want 1373 the area A+B+C as our final answer. 1374

1375 Area A is not too hard to find using Archimedes’ formula: 1376 1377

364

1634

)4)(8(21

34A Area

=

×=

×=

1378

1379

y = 5

The inscribed triangle’s area

-10 -8 -6 -4 -2 0 2

-4

-2

0

2

4

6

y=5

Region D

Region B Region C

Region A


To get Area B + Area C, we will first get Area D. To do this, we look at the 1380 parabola with a chord of y = 5 cutting it off. The enclosed area is Area A + 1381 Area D and is computed with Archimedes’ formula: 1382 1383

723

216

5434

)9)(12(21

34 D Area A Area

==

×=

×=+

1384

Therefore we can find Area D: 1385 1386

3152D Area

364

3216 D Area

3216 D Area

364

3216 D Area A Area

=

−=

=+

=+

1387

1388 Now that we have Area D, we can find Area B + Area C. Notice that regions 1389 B, C, and D form a rectangle that is 12 units long and 5 units high. Thus, its 1390 area is 60. We use this to find Area B + Area C: 1391 1392

328

3152

3180

315260 C Area B Area

603

152 C Area B Area

60 D Area C Area B Area

=

−=

−=+

=++

=++

1393

1394

The inscribed triangle’s area


Finally, we can find the shaded region by adding up the following: 1395 1396

3230

392

328

364 C Area B AreaA Area

=

=

+=++

1397

This gives us our final area. ♦ 1398 1399 1400

Archimedes and the Law of the Lever 1401 One of Archimedes’ greatest and simplest discoveries is the Law of the Lever. The law tells you 1402 how to balance two weights that are on a lever by moving a fulcrum to a particular point on the 1403 lever. Here is a picture that will be useful for us during this discussion. 1404 1405 1406 1407 1408 1409 1410 1411 1412 1413 1414 1415 1416 In this drawing, we will call point P the point where the fulcrum is located. The distance of the 1417 weight w to the fulcrum is denoted by d. The distance of the second weight W to the fulcrum is 1418 D. 1419 1420 Archimedes pondered the question, “If W is a distance D from the fulcrum, and w is a distance d 1421 from the fulcrum, where should the fulcrum be in order for the weights to balance at point P?” 1422 1423 In order to determine what was needed to balance these weights, Archimedes began his treatment 1424 of the subject by giving three assumptions that he would work with. They are given as follows:43 1425 1426 Assumption 1 1427 Equal weights that are equal distances from the fulcrum will balance. If equal 1428

weights are unequal distances from the fulcrum, they will not balance and the 1429 weight at the greater distance will cause its side of the lever to move down. 1430

1431 Assumption 2 1432 Assume two weights do balance on a lever. If we add something to one of 1433

those weights, they will no longer balance and, in fact, the side where we add 1434 more weight will go down. 1435

w W

Fulcrum

d D

P


1436 Assumption 3 1437 Assume two weights do balance on a lever. If we take something away from 1438

one of those weights, they will no longer balance and, in fact, the side that was 1439 not changed will go down. 1440

1441 This set of assumptions serves as Archimedes’ starting point, and everything else that he 1442 develops from here is based on these ideas. By applying these assumptions, he arrived at new 1443 results, which we will call Propositions. Propositions must be proved, in general, and 1444 Archimedes obliged in each case. 1445 1446 Proposition 1 1447 Weights that balance at equal distances from the fulcrum must be equal. 1448 1449

Proof: 1450 1451 This is a proof by contradiction. (See Chapter 4.) We start with the 1452

assumption that the two weights balance. What if are not equal (in weight)? 1453 Then simply remove from the greater weight the difference between them so 1454 that they are equal. You now have two equal weights at equal distances. But 1455 Assumption 3 says that when we have weights that balance and then remove 1456 something from one (which we just did), they no longer balance. But the 1457 weights are now equal (since we removed the difference), and they are still 1458 equal distances from the fulcrum, but they do not balance. However, 1459 Assumption 1 says that equal weights at equal distances do balance. This is a 1460 contradiction and tells us our original assumption that they were not equal 1461 does not hold water. Hence, the weights are indeed equal. 1462

1463 Proposition 2 1464 Unequal weights and equal distances from the fulcrum do not balance. The 1465

side with the greater weight will tilt down. 1466 1467

Proof: 1468 1469 Remove from the greater weight the difference between it and the lighter 1470

weight (so that they are now equal). By Assumption 1 the weights that remain 1471 must balance (because they are equal and equidistant from the fulcrum). But 1472 if we put back the weight that we just removed, then Assumption 2 tells us 1473 that they will no longer balance and that the heavier weight tilts down. 1474

1475


Proposition 3 1476 Weights that are not equal will balance at unequal distances from the fulcrum. 1477

The heavier weight will be a shorter distance from the fulcrum than the lighter 1478 weight will be. 1479

1480 Proof: 1481

1482 Archimedes assumes there is some balancing point (which could be 1483

dangerous…why?) and that it is located somewhere between the two weights. 1484 By using a series of mini proofs by contradiction, Archimedes’ established 1485 this proposition as well, but we will not include it here.44 1486

1487 In Archimedes’ next proposition, he talked about the “center of gravity” of an object but did not 1488 really define what he meant by that phrase. Ideas from calculus can help us today to form a good 1489 definition of this term, but for our purposes, we will rely a bit on a more fuzzy “definition.” 1490 What we do is picture a weight concentrated at a single point in space. That is, the weight is 1491 viewed as if it were all located at a single point. We’ll call this point the weight’s center of 1492 gravity. With this idea in mind, he could then state his next proposition, which was, and is, 1493 crucial to his Law of the Lever. 1494 1495 1496 Proposition 4 1497 If two equal weights have different centers of gravity, then the center of 1498

gravity of the two weights combined is the midpoint of the line segments that 1499 join their centers of gravity. 1500

1501 1502 1503 1504 1505 1506 1507 1508

Proof: 1509 1510 To construct this proof, we consider the picture above. The points S and T (on 1511

the lever) to be considered as being on the lever and represent the centers of 1512 gravity for the two weights shown. P is the midpoint between points S and T. 1513 We start by (falsely) assuming that the weights do not balance at the midpoint 1514 and instead balance at some point C. We then look for some contradiction. 1515 First of all, notice that distance SC is not equal to distance TC, since C is not 1516 the midpoint. By Assumption 1, the two weights do not balance at C (as long 1517 as it is not point P). This is a blatant contradiction to our assumption that they 1518 balance at C. So they do indeed balance at point P, which is the midpoint. ♦ 1519

1520

w P

w C

S T


Finally, Archimedes was ready to give and prove the Law of the Lever, using Proposition 4 (and 1521 related results that we do not state here) to complete his task. Archimedes considered the two 1522 separate cases where the weights were commensurable and when the weights were 1523 incommensurable. (Now why would he do that?) 1524 1525 Proposition 5 1526 Weights that are commensurable balance at distances (from the fulcrum) that 1527

are inversely proportional to the magnitudes of the weights. 1528 1529 While you may be used to the notion of proportionality, the idea of inverse proportionality may 1530 not be as familiar to you. IF he had said that the distances were proportion to the weights, we 1531 would get the following: 1532 1533

wW

dD= 1534

1535 However, to say they are inversely proportional means the following: 1536 1537

Ww

wW

dD

==/1/1 1538

1539 That is: 1540

Ww

dD= 1541

1542 You can almost think of w and W as swapping roles, which makes sense since this is an inverse 1543 proportion. In any case, this is the great result he’s looking for. We notice that we can write this 1544 equality in other forms. Specifically, if we cross multiply, we get the following: 1545 1546 1547 1548 1549

1550 1551

This is not something Archimedes would have done for he would not have thought to multiply a 1552 weight and a distance.45 But it may be easier for us to remember and we generally don’t worry 1553 about these matters as much as the ancient Greeks did. In the one last case he had to consider, 1554 Archimedes stated the following: 1555 1556 Proposition 6 1557 Weights that are incommensurable balance at distances (from the fulcrum) 1558

that are inversely proportional to the magnitudes of the weights. 1559 1560 When you read this statement, it appears to be almost identical to Proposition 5, with the 1561 exception of one word. But we have to remember that Archimedes and the Greeks thought of 1562 commensurable and incommensurable numbers as very different, so he has two cases. While we 1563

Archimedes’ Law of the Lever

dwDW =


can think of the commensurables and incommensurables as comprising the real numbers, and 1564 treat them as one big set, that notion had not yet been fully developed. Anyway, his proof of 1565 Proposition 6 uses a consequence of Proposition 5, but again, we won’t state the proof here due 1566 to its complexity. 1567 1568 The application of the Law of the Lever is much easier than its derivation, thankfully. The 1569 previous discussion, however, does paint a picture of the logical rigor that Archimedes, and the 1570 Greeks in general, were interested in maintaining. Let’s now look at some simple examples. 1571 1572 Example 7 1573 A 25-pound weight is placed 5 feet from a fulcrum. How far on the other side 1574

of the fulcrum, must a 50-pound weight be placed to balance the weights? 1575 1576 Solution 1577 Here’s a picture of what we have. 1578

1579 1580 1581 1582 1583 1584 1585 1586 1587 1588 1589

Using the Law of the Lever, we have the following: 1590 1591

5.250

12512550

)25)(5()50(

==

==

=

D

DD

dwDW

1592

Therefore, the 50-pound weight must be placed 2.5 feet from the fulcrum. ♦ 1593 1594 Check Point F 1595 A 70-pound weight is placed 4 feet from a fulcrum. How far on the other side 1596

of the fulcrum, must a 200-pound weight be placed to balance the weights? 1597 1598 Solution 1599 See the endnote for the answer.46 1600 1601 1602

25 50

Fulcrum

5 ft D

P


Example 8 1603 A 30-pound weight is placed at one end of a 10-foot plank of wood (the 1604 lever). A 50-pound weight is placed at the other end. Where should a fulcrum 1605 be placed to balance the weights? 1606

1607 1608 1609 Solution 1610 A picture might help here: 1611 1612 1613 1614 1615 1616 1617 1618 1619 1620 1621 1622

1623 1624 The goal here is to find the distances d and D. Using the Law of the Lever, 1625 we have the following: 1626

1627

dDdD

dwDW

3050)30()50(

==

= 1628

1629 This equation cannot be solved since it has two variables, so we need more 1630

information. However, we also know that: 1631 1632

dDdD

−==+

1010 1633

1634 We can use substitute this into our previous result to eliminate one of the 1635

variables, as follows: 1636 1637

25.6425

80500305050030)10(50

3050

==

==−=−

=

d

ddddd

dD

1638

30 50

Fulcrum

d D

P


1639 Once we know d, we know D, which is 75.325.610 =− (ft). 1640 1641 A quick check shows the following: 1642

1643

5.1875.187)25.6(30)75.3(50

== 1644

1645 Thus, the fulcrum should be placed 3.75 feet from the 50-pound weight, 1646

which is 6.75 feet from the 30-pound weight.♦ 1647 1648 1649 Check Point G 1650

A 40-pound weight is placed at one end of a 12-foot plank of wood (the 1651 lever). A 70-pound weight is placed at the other end. Where should a fulcrum 1652 be placed to balance the weights? 1653

1654 Solution 1655 Check the endnotes for the solution.47 1656 1657 1658 Archimedes and the Center of Gravity 1659 If we know the center of gravity for each of the two objects on a lever, as in the previous section, 1660 we can think of the law of the lever as describing the center of gravity of the two objects 1661 combined.48 The fulcrum point where they balance is their combined center of gravity. As we 1662 said earlier, the phrase "center of gravity" is not one that Archimedes ever really defined, but he 1663 uses this general idea to work with the centers of gravity for a variety of different solid objects or 1664 flat objects such as triangles, parabolas, disks, parallelograms, etc. 1665 1666 For the center of gravity of a flat object, we will define it here for our purposes as the point 1667 where the flat object will balance on a vertical pole. (Alternatively, it's the point where the object 1668 "can be suspended and remain motionless, no matter in which position it is placed." 49) 1669 1670 1671 1672 1673 1674 1675 1676 1677 1678 1679 Archimedes explores the centers of gravity for a variety of objects, as we said before, but there is 1680 one that is worth mentioning here. If you were given a cardboard triangle that was uniformly 1681 dense, at what point inside the triangle could you place a vertical needle or pole so that the 1682 cardboard triangle remained perfectly balanced? 1683

P

A

C

B


This is the result that Archimedes gives as an answer to that question. We start be looking at the 1684 triangle from above, so that it is flat as we look down on it. 1685 1686 1687 1688 1689 1690 1691 1692 1693 1694 1695 1696 1697 1698 1699 Now, on each side of the triangle we mark off the midpoint of that side. 1700 1701 1702 1703 1704 1705 1706 1707 1708 1709 1710 1711 1712 1713 1714 1715 Now connect each of the midpoints with the vertex opposite to that side. 1716 1717 1718 1719 1720 1721 1722 1723 1724 1725 1726 1727 1728 1729

A

B

C

A

B

C

MC MA

MB

A

B

C

MC MA

MB

P


1730 You will note that the three lines, called medians of the triangle, all seem to go through one point 1731 inside the triangle, called P in the diagram above. It can be shown that these three lines do 1732 indeed pass through the exact same point, and Archimedes proved that this point, P, is the center 1733 of gravity. There are many such interesting points in the triangle. One of the most interesting 1734 studies related to a triangle is the construction of the Nine Point Circle. The endnotes have a list 1735 of web sites that you may want to explore that describe this interesting geometrical 1736 phenomenon.50 1737 1738 1739 1740 1741


PART 6: Heron's Formula 1742 1743 The Greek mathematician Heron (circa 10 C.E. to 75 C.E.), sometimes known as Hero, was a 1744 geometer who is now best known for a formula that gives the area of a triangle. You are 1745 undoubtedly familiar with the formula: 1746 1747

bhA21

= 1748

1749 1750 1751 1752 1753 1754

where b is the base of a triangle and h is its height. A is, of course, the area of the triangle. The 1755 height of the triangle is the distance from a vertex to the opposite side. In the figure above, the 1756 dotted line with length h is perpendicular to the base, b. 1757 1758 This is fine when you know the height of a triangle, but what if you do not? For example, what if 1759 you have a triangle and you know the length of its three sides, but you do not know the height? 1760 Finding the height is possible, but requires results from trigonometry, and that is material that 1761 you may have not seen before. It would be helpful if we had a formula for determining the area 1762 of a triangle that could be found by knowing the lengths of the three sides and which does not 1763 require knowing the height of the triangle. 1764 1765 Heron's formula does just that. Let's start with a triangle that has sides with length a, b, and c, as 1766 shown below. 1767 1768 1769 1770 1771 1772 1773 1774 1775 1776 1777 We will define the semiperimeter of the triangle to be half the perimeter. Thus, if we let s be the 1778 semiperimeter, then: 1779 1780

)(21 cbas ++= 1781

1782

b

h

b

a c

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Heron.html


With this definition in hand, we can now state Heron's formula: 1783 1784 The area of a triangle with sides of length a, b, and c is given by: 1785 1786

1787 1788 1789 1790 1791 1792

When you first look at this, it may not seem all that odd. It’s just another formula, after all. 1793 However, when you think about it for a bit, it becomes a bit surprising. First of all, why would 1794 the perimeter (or part of it) be related to the area of a triangle? We're told over and over that area 1795 and perimeter are two very different quantities, and they are, so why do they get linked like this 1796 in this formula? Second, where and why does the square root appear? Both of these questions are 1797 answered when one looks at the proof that Heron gave, which we will only explore part of, but 1798 before then, let's look at a straightforward example. 1799

1800 Example 9 1801

Suppose a triangle has sides 17, 25, and 26. What is its area?51 1802 1803

Solution 1804 We will first compute the semiperimeter: 1805 1806

( )

34

6821

)262517(21

)(21

=

=

++=

++= cbas

1807

1808 Now we can compute the area of the triangle: 1809 1810

( )( )( )

20441616

26342534173434

==

−−−=A

1811

1812 Hence, the are of the given triangle is 204 square units. ♦ 1813 1814

Heron’s Formula for the Area of a Triangle

))()(( csbsassA −−−=


Check Point H 1815 Suppose a triangle has sides 30, 20, and 35. What is its area? 1816 1817

Solution 1818 See the endnote for the answer.52 1819 1820 1821

We now turn our attention to the questions of why the perimeter is related to this formula and 1822 why the square root appears. The first question concerning the perimeter we'll try to answer here. 1823 The second one, pertaining to the square root, will be omitted due to its complexity. 1824 1825 Let's start by looking at a typical triangle: 1826 1827 1828 1829 1830 1831 1832 1833 1834 1835 1836 1837 Heron's first step in finding his formula is to inscribe a circle into the triangle. This move is a bit 1838 strange, but as we'll see, it gets the job done. The inscribed circle looks like this: 1839 1840 1841 1842 1843 1844 1845 1846 1847 1848 1849 1850 In this figure, capital letters represent points. Point E, for example, is the point where the triangle 1851 and circle touch. The same is true for points D and F. The letters a, b, and c represent the lengths 1852 of the sides. Point O is the center of the circle. Segments OE, OD, and OF are perpendicular to 1853 BC, AB, and AC, respectively, and are also radii of the circle. The dotted lines are angle 1854 bisectors; that is, they cut each of angles A, B, and C in half. 1855 1856

b

a c

b

a c

A

B

C

E

F

D

O


With all of this in place, we can examine the areas of several of the triangles that are created in 1857 this figure.53 1858 1859

Area ∆AOB = ( ) crODABheightbase21

21)(

21

=×= 1860

Area ∆BOC = ( ) arOEBCheightbase21

21)(

21

=×= 1861

Area ∆COA = ( ) brOFACheightbase21

21)(

21

=×= 1862

1863 1864 Note that triangles ∆AOB, ∆BOC, and ∆COA, when combined, give the area of ∆ABC. 1865 Therefore, we have the following: 1866 1867

Area of ∆ABC = brarcr21

21

21

++ 1868

1869

When we factor out a 21 and the variable r, we get the following: 1870

1871

Area of ∆ABC = ( )cbar ++21 1872

1873 But, )( cba ++ is the perimeter of the original triangle, and so one half of that is its 1874 semiperimeter, s. We therefore have: 1875 1876

Area of ∆ABC = ( )cbar ++21 1877

Area of ∆ABC = rs 1878 1879 1880 We have thus shown how the area of a triangle is directly related to (a.) the perimeter of the 1881 triangle and (b.) the radius of a circle that is inscribed inside the triangle. Hence, when we look at 1882

Heron’s interesting formula, ))()(( csbsassA −−−= , where )(21 cbas ++= , we get some 1883

feel for where this formula is coming from. 1884 1885 There is still the whole question of where the square root comes from. Since we now know that 1886 the area can be related to the radius of the inscribed circle, a good guess would be that the square 1887 root comes from this geometric fact. However, the derivation of this part of Heron’s formula 1888 goes a little beyond where we want to be, so we will skip it and trust that the reader will believe 1889 that everything works out nicely. 1890

1891 1892


PART 7: The Golden Ratio 1893 This section is to be added at a later date. You may be asked to read an accompanying Lab for 1894 background information on this topic. 1895 1896 1897 1898 Summary 1899 There were many other Greek mathematicians that we did not get a chance to talk about. To do 1900 so would require volumes. But what we have seen is that the Greeks were instrumental in 1901 broadening the horizons beyond practical and utilitarian needs. They pushed the limits of the 1902 subject and introduced deductive reasoning, logic, and formal proof into mathematics, forever 1903 changing the landscape of the field. To this very day, an overwhelming majority of 1904 mathematicians do not believe that a piece of mathematics is valid unless it is proven with formal 1905 logic. Even in the age of computers, machines that can compute millions of instructions per 1906 second take a back seat to good old-fashioned proof. For that, we have the Greeks to thank for 1907 this legacy. 1908


Blank page1909


PART 8: Appendix 1910 Cut out each series of figures, folding on internal lines to create the four of the five Platonic 1911 solids. 1912 1913 Equilateral Triangles 1914 1915 1916 1917 1918 1919 1920 1921 1922 1923 Squares 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 Pentagons 1934 1935 1936 1937 1938 1939 1940 1941 1942 1943 1944 1945 1946 1947 Hexagons 1948 1949 1950 1951 1952 1953 1954 1955


Blank page1956


PART 9: Homework 1957 1958 Graphing Parabolas 1959 1960 Graph the following the following parabolas. Make sure you find the vertex, y intercept, and x 1961 intercepts, if there are any. (Recall that the x coordiate of the vertex of a parabola can be found 1962

with the equation abx

2−

= , its y-intercept can be found by letting x = 0 and solving for y, and its 1963

x-intercept(s) can be found by setting y = 0 and then solving for x.) 1964 1965 1) 2 2 24y x x= + − 1966 2) 2 12 20y x x= + + 1967 3) 202 ++−= xxy 1968 4) 352 +−= xxy 1969

1970 The Method of Menaechmus 1971 1972 Recall that the method of Menaechmus uses following two equations: 1973 1974

2 2

and2

x yy xa a

= = 1975

1976 For problems (5) through (8), use the method of Menaechmus to construct a graphical solution to 1977 the case where a cube with side a is to have its volume doubled. Your graph should be 1978 constructed with graph paper and the scales should be carefully chosen. (A blank page of graph 1979 paper is included on the next page…make copies if you need more.) In each case, compare your 1980 graphical solution to the actual answer and comment on how close your graphical solution is to 1981 the actual answer. 1982 1983 5) Let a = 4 6) Let a = 6 1984 1985 7) Let a = 8 8) Let a = 10 1986 1987 1988


Geometric Algebra 1989 1990 Give at least two different drawings that show the solution to the following linear problems (in 1991 the form ax = bc) using the Greek methods of construction described in this chapter. (See 1992 Example 3.) All drawings should either be done using graph paper to measure distances or use 1993 centimeters as the unit of measure. Use a ruler, as neatness counts here. Label all relevant 1994 information: 1995 1996 9) 4x = 12 10) 5x = 24 1997 1998 11) 3x = 16 12) 6x = 18 1999 2000 13) Use the diagram to provide a geometric explanation of why the formula 2001

( )( )2 2a b a b a b− = − + is true. You may have to redraw the figure with pieces rearranged. 2002 Explain clearly in words and/or with algebra why your drawing “proves” the difference of 2003 squares formula. 2004

2005 2006 2007 2008 2009 2010

2011 2012 2013 2014 2015 2016 2017 2018 2019

2020 Measuring π and the Area of Circles 2021 2022 The Greeks were certainly not the first to try to estimate the value of π. A Babylonian tablet 2023

found in 1936 asserts that the formula for the area of a circle is A = 2112

C , where C is the 2024

circumference of the circle. Use this formula to approximate the area of circles with the given 2025 dimensions. Your answers should be given as reduced fractions and in decimal form (to 5 2026 decimal places). For each case, also compute the actual area of that circle and compare it to the 2027 approximate value. (Recall that the circumference of a circle is rC π2= .) 2028 2029 14) Circumference = 5 15) Circumference = 10 2030 2031 16) Circumference = 20 17) Circumference = 100 2032 2033

a

a

bb


18) When the Babylonians used 2112

C to approximate the area of a circle, they were actually 2034

using a value of 3 for π. Prove this fact. (Hint: Take 2C rπ= and substitute it into the 2035 Babylonian equation. Set this equal to the known formula for the area of a circle, 2A rπ= 2036 and then solve for π.) 2037

_____ 2038 2039 19) When a more accurate approximations of area is required by the Babylonians, the tablet 2040

indicated that the 1/12 in the formula 2112

C should be multiplied be the base-60 number 2041

0;57,36. Convert the multiplying factor of 0;57,36 to a base-10, reduced fraction. 2042 2043 For Problems (20) to (23), use the formula and answer from Number (19) to approximate the 2044 area of circles with the following measurements. Your answers should be given as reduced 2045 fractions and in decimal form (to 5 decimal places). For each case, also compute the actual area 2046 of that circle and compare it to the approximate value. (Recall that the circumference of a circle 2047 is rC π2= .) 2048 2049 20) Circumference = 5 21) Circumference = 10 2050 2051 22) Circumference = 20 23) Circumference = 100 2052 _____ 2053 2054 24) When the Babylonians used this multiplication factor of 0;57,36, what value of π were they 2055

effectively using? Write your answer as a reduced fraction, showing all work. When they 2056 used this multiplying factor, how close were they to the actual value of π? 2057

_____ 2058 2059 Our friends the Egyptians also had a method of calculating the area of a circle. Their method is 2060

equivalent to the formula 28

9A d =

, where d is the diameter of the circle. For problems (25) to 2061

(28) use this formula to approximate the area of circles with the following measurements. Your 2062 answers should be given as reduced fractions and in decimal form (to 5 decimal places). For 2063 each case, also compute the actual area of that circle and compare it to the approximate value. 2064 2065 25) Diameter = 10 26) Diameter = 20 2066 2067 27) Diameter = 50 28) Diameter = 100 2068 2069 2070 29) When the Egyptians used this formula for the area of a circle, what value of π were they 2071

utilizing? Show all work. Hint: d = 2r. 2072 _____ 2073 2074


In his book, Measurement of a Circle, Archimedes stated: The area of a circle is to the square 2075 of its diameter as 11 is to 14. This is essentially a proportional statement that is equivalent to 2076

1411

2 =dA . For problems (30) to (33), use this proportion to approximate the area of circles with 2077

the following measurements. Your answers should be given as reduced fractions and in decimal 2078 form (to 5 decimal places). For each case, also compute the actual area of that circle and 2079 compare it to the approximate value. 2080 2081 30) Diameter = 10 31) Diameter = 20 2082 2083 32) Diameter = 50 33) Diameter = 100 2084 2085 34) Show that Archimedes proportion (used in the previous four problems) leads to 22/7 for the 2086

value of π. Show all work. Hint: rd 2= and 2rA π= . 2087 _____ 2088 2089 35) The sixth century Hindu mathematician Aryabhata had the following procedure for finding 2090

the area of a circle. Half the circumference multiplied by half the diameter is the area of 2091 the circle. How accurate is this rule and how does it compare to the other methods given 2092 above? Show all work and steps. 2093

2094 2095 Approximating the Area of a Circle 2096 2097 36) The circle shown here has 12 equally sized sectors marked off. Assume the 2098

radius is 1. 2099 2100 a. What is the perimeter of the circle? (Ignore the 12 segments for now and use 2101

rC π2= .) 2102 2103 b. What is the area of the circle? (Again, ignore the segments and use 2rA π= ) 2104 2105

If you assume that each sector of the circle is approximately a triangle, you can estimate the 2106 area of the circle by finding the area of one triangle and then multiply by how many 2107

triangles you have. Recall that the area of a triangle is given by bhA21

= . 2108

2109 c. Assuming that the circle has twelve sectors as shown, what would be a good approximation 2110

for the length of the base of each sector/triangle? Explain or show how you get your 2111 answer. 2112

2113 d. Use your answer to part (c.) to find the area of this circle. Explain your answer. 2114 2115


37) Repeat the process of Problem (36) for a circle with radius 4 that is cut into 20 equal 2116 sectors. What is your result? Show all steps. 2117

2118 38) This process of Problems (35) and (36) seems to work really well, but what is wrong with 2119

it, fundamentally? Explain. 2120 2121 39) The more sectors that a circle has, the better this process is for approximating the area of a 2122

circle. Explain why this is true. You should be careful to carefully justify any claims you 2123 make. 2124

2125 40) We can extend this to a more general case where we have an arbitrary number of sectors in 2126

the circle. Let’s say we have n sectors in the circle and that there are many of them. Let’s 2127 also assume that we don’t know what the radius is, so we will call it r . Since we don’t 2128 know the radius’ length, we also don’t know the circumference of the circle, so let’s say the 2129 circumference is C. 2130

2131 a. What is the approximate length of the base of each sector/triangle in the circle? Your 2132

answer should be given in terms of C, r, and n. 2133 2134 b. If you have n sectors, use your result from part (a.) to find the area of the circle. Your final 2135

answer should be in terms of C and r. 2136 2137 c. How is your result similar to that of the Hindu mathematician Aryabhata? (See Problem 2138

(35) above.) 2139 2140 d. Based on your answer to Problem (35) what can you say about this method? 2141 2142 e. Does this method give you the exact area of the circle? Explain. 2143 _____ 2144 2145


41) Another way to use these circle sectors to approximate the area of a circle is to cut them up 2146 and rearrange them to form a parallelogram. A parallelogram looks like this: 2147

2148 2149 2150 2151 2152 2153 2154 2155 The base is labeled b, and the height is labeled h on the figure. (The height forms a right 2156

angle with the base.). The area of a parallelogram is given by hbA ×= . It’s essentially the 2157 same formula for the area of a rectangle. 2158

2159 Returning to the circle with the sectors, we cut the sectors up and rearrange them so that 2160

they look like this: 2161 2162 2163 2164 2165 Note that this has a shape that is very close to a parallelogram: 2166 2167 2168 2169 2170 2171 2172 2173

This means that the area of a circle can be approximated with the area of a parallelogram. 2174 2175

Let’s say we have n sectors in the circle and that there are many of them. Let’s also 2176 assume that we don’t know what the radius is, so we will call it r . Since we don’t know 2177 the radius’ length, we also don’t know the circumference of the circle, so let’s say the 2178 circumference is C. 2179 2180

a. If you align the sectors up to look like a parallelogram, what is the length of the base of the 2181 resulting parallelogram? Your answer should be in terms of C. 2182

2183 b. What is a reasonable value to take for the height of the parallelogram? Explain your 2184

answer. 2185 2186 c. Find the area of the parallelogram using the formula for the area of a parallelogram 2187

( hbA ×= , where b is the base and h is the height) and your answers from parts (a.) and 2188 (b.) Your answer should be in terms of C and r. 2189

2190

h

b


d. How is your result similar to that of the Hindu mathematician Aryabhata? (See problem 2191 (35) above). What do you conclude? 2192

2193 e. Explain any “problems” with this method of approximating the area of a circle. Please try 2194

to be very clear in your explanation. 2195 2196 2197

42) Archimedes’ great work on this topic says that The area of a circle is equal to the area of a 2198 right triangle where one leg is equal to the radius of the circle and the other leg is equal to 2199 the circumference of the circle. How does his treatment compare to that of the Hindu 2200 mathematician Aryabhata? (See problem (35) above.) Explain how you get your 2201 conclusion. 2202

2203 2204


Areas of Parabolas via Archimedes 2205 2206 Find the areas of the following shaded regions using the formula of Archimedes: 2207 2208 2209 2210

43)

-4 -2 0 2 4 60

5

10

15y=14

44)

-6 -4 -2 0 2

-6

-4

-2

0

2

4

6

y=-4

The horizontal line is y = 14 The horizontal line is y = −4

45)

-30 -20 -10 0 10 20

-400

-200

0

200y=225

46)

0 2 4 6 8 10 12 14

-6

-4

-2

0

2

4

The line is y = 225. The parabola is 2 400y x= −

The parabola is 2x y= . Shade the area trapped between this parabola and the line x = 3. What is that area?

2211 47) Find the area of the region enclosed by the line y = 10 and 2 10 26y x x= − + using the 2212

formula of Archimedes. Include a graph of your region. 2213 2214


48) Find the area of the region enclosed by the line y = −10 and ( )25 6y x= − + + using the 2215 formula of Archimedes. Include a graph of your region. 2216

2217 49) Find the area of the region enclosed by the line y = 16 and 2 8 16y x x= + + using the 2218

formula of Archimedes. Include a graph of your region. 2219 2220 50) Find the area of the region enclosed by the line y = 0 and 2 5 3y x x= − + − using the 2221

formula of Archimedes. Include a graph of your region. 2222 2223

51) Find the area of the shaded region 2224 shown. Take the vertex to be 2225

−12,

21 and the x intercepts to be 2226

−3 and +4. The horizontal line is 2227 given by the equation y = 8. 2228

2229 2230 2231 2232 2233 2234 2235 2236 2237 2238 2239

2240 The Law of the Lever 2241 2242 52) A 30-pound weight is placed 8 feet from a fulcrum. How far on the other side of the 2243

fulcrum, must a 50-pound weight be placed to balance the weights? 2244 2245 53) A 60-pound weight is placed 6 feet from a fulcrum. How far on the other side of the 2246




fulcrum, must a 50-pound weight be placed to balance the weights? 2256

-4 -2 0 2 4 6

-10

-5

0

5

10y=8


2257 57) A 65-pound weight is placed 7 feet from a fulcrum. How far on the other side of the 2258

fulcrum, must a 50-pound weight be placed to balance the weights? 2259 _____ 2260 2261 58) 40-pound weight is placed at one end of a 7-foot plank of wood (the lever). A 20-pound 2262

weight is placed at the other end. Where should a fulcrum be placed to balance the 2263 weights? 2264

2265 59) 60-pound weight is placed at one end of a 3-foot plank of wood (the lever). A 60-pound 2266










2285 Heron’s Formula 2286 2287 64) Use Heron’s Formula to find the area of a triangle if it has sides with lengths 3, 6, and 8. 2288

Round your answer to the nearest tenth of a square unit. 2289 2290 65) Use Heron’s Formula to find the area of a triangle if it has sides with lengths 15, 9, and 20. 2291

Round your answer to the nearest tenth of a square unit. 2292 2293 66) Use Heron’s Formula to find the area of a triangle if it has sides with lengths 12, 50, and 2294

55. Round your answer to the nearest tenth of a square unit. 2295 2296 67) Use Heron’s Formula to find the area of a triangle if it has sides with lengths 3, 4, and 5. 2297

Round your answer to the nearest tenth of a square unit. When you are done, explain 2298 another, more straightforward way to find the area of this triangle. Explain your reasoning 2299 carefully. 2300

2301


68) Use Heron’s Formula to try to find the area of a triangle if it has sides with lengths 5, 7, 2302 and 12. What numerical result do you get? Explain why the formula does not work in this 2303 case. 2304

2305 69) Use Heron’s Formula to try to find the area of a triangle if it has sides with lengths 5, 7, 2306

and 14. What numerical result do you get? Explain why the formula does not work in this 2307 case. 2308

2309 2310 Writing 2311 2312 Write a short essay on the given topic. It should not be more than one page and if you can type it 2313 (double−spaced), I would appreciate it. If you cannot type it, your writing must be legible. 2314 Attention to grammar is important, although it does not have to be perfect grammatically…I just 2315 want to be able to understand it. 2316 2317 70) Use the internet or library to research some aspect of the “Golden Ratio” and, specifically, 2318

its role in Greek mathematics. Give a brief presentation on what you find. Cite all of your 2319 sources. 2320

2321 71) Use the internet or library to research an important aspect of Greek civilization that was not 2322

covered in this text or in this course. Give a brief presentation on what you find. Cite all of 2323 your sources. 2324

2325 72) Use the internet or library to research what is called the “Triangle Inequality.” Explain it in 2326

your own words, including as many examples as is necessary to illustrate what it says. 2327 Then discuss how this problem will help explain what is happening with Problems (68) and 2328 (69). 2329

2330 73) Many believe that when it came to mathematics, the Greeks were heavily influenced by the 2331

Egyptians and/or the Babylonians. Use the internet or library to research what evidence 2332 may exist for this claim and present your results in a brief essay. 2333

2334 74) In this homework section, we’ve seen a variety of methods for computing π or finding the 2335

area of a circle. Which method do you think is the most creative or interesting? Explain 2336 your opinion. Use at least one unique example of the method that you pick in your 2337 write−up. 2338


PART 10: Endnotes for Chapter 5 2339 2340 1 Calinger, page 61 2 3 Calinger, page 64 4 Calinger, page 65 5 Bunt, page 89 6 Solution to Check Point A a = ½ so the equations are 22y x= and 2x y= . The graphs look like the following:

x

y

0 0.2 0.4 0.6 0.8

-0.5

0

0.5

1

7 Dunham, pages 11−17 8 Calinger, page 96 9 Calinger, page 98 10 http://www.windows.umich.edu/cgi-bin/tour_def/people/ancient_epoch/plato.html; Courtesy of Corbis-Bettmann. From sculpture in the Vatican, Rome 11 Calinger, page 102 12 http://prtr-13.ucsc.edu/sro/tt/html/regular_polyhedra.html 13 Katz, page 94 14 http://www.dartmouth.edu/~matc/math5.geometry/unit6/unit6.html#Platonic Solids 15 http://www.nirim.go.jp/~weber/POLYHEDRA/p_01.html 16 Calinger, page 106 17 Calinger, page 107 18 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Euclid.html 19 Katz, page 58 20 The web has a fantastic online, interactive version of the Elements at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html . It should not be missed. 21 Katz, page 61 22 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html 23 Including this work. Calinger, pages 147-8 have a discussion of these works. 24 Each of these areas are a×b 25 Region 2 has area = b2, while region 3 has area = a2

26 You get 2222 2 bababaabab ++=+++ 27 The square on the whole has area ( ) 222 2 bababa ++=+ 28Solution to Check Point B (p + q)(r + s) = pr + ps + qr + qs The area of the whole is equal to the area of the two rectangles that make it up. 29 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Al-Khwarizmi.html


30 Calinger, page 150 31 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Archimedes.html 32 Calinger, page 151 33 http://www.mcs.drexel.edu/~crorres/Archimedes/Screw/ScrewEngraving.html 34 http://www.mcs.drexel.edu/~crorres/Archimedes/Screw/ScrewSlide19.jpg 35 Vitruvius’ De Architectura as quoted by Calinger, page 152 36 Calinger, page 153 37 Calinger, page 154 38 Burton, page 205 39 http://www.mcs.drexel.edu/~crorres/Archimedes/Tomb/Cicero.html 40 Measurement of a Circle; See Classics of Mathematics by Calinger, page 137 41 Solution to Check Point D The area is (4/3)(1/2)(6)(9) = 36 42 Solution to Check Point E The area is 10.6666 43 Stein, page 8. 44 Stein, page 9-10. 45 Stein, Sherman. Archimedes: What Did He Do Besides Cry Eureka? MAA, 1999. Page 7. 46 Solution to Check Point F

About 1.4 feet from the fulcrum. 47 Solution to Check Point G

About 7.63 feet from the 40-pound weight, or 4.36 feet from the 70-pound weight. 48 Stein, page 15 49 Stein, page 15 50 Nine Point Circle Web Sites The Nine Point Circle: http://www.csm.astate.edu/Ninept.html Nine Point Circle: http://jwilson.coe.uga.edu/EMT669/Student.Folders/McFarland.Derelle/Nine/9ptcircle.html 51 Dunham, William; Journey Through Genius, page 118. 52 Solution to Check Point H

About 299.413 square units 53 Dunham, page 122.


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