Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Euler angles wikipedia , lookup
Multilateration wikipedia , lookup
Problem of Apollonius wikipedia , lookup
Line (geometry) wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of trigonometry wikipedia , lookup
History of geometry wikipedia , lookup
Integer triangle wikipedia , lookup
Trigonometric functions wikipedia , lookup
Compass-and-straightedge construction wikipedia , lookup
Pythagorean theorem wikipedia , lookup
Chapter 5 Hippocrates of Chios 1 The area problem The Babylonians and the Egyptians developed a rich knowledge of basic geometric concepts, like how the areas of simple plane figures – triangles, squares, rectangles, parallelograms, trapezoids, and the like – were related to the lengths of their sides, and how volumes of simple solid bodies – parallelopipeds and pyramids – related to corresponding appropriate linear dimensions.1 These versions of geometry tended to be very utilitarian, meant for use by surveyors, builders, and astronomers. The Greeks however turned geometry into a theoretical science, a new discipline that concerned itself with the properties of a very refined collection of shapes and forms – points, lines, circles, polygons and polyhedra2 ; a science founded on the deductive methods they were systematizing through philosophical dialectics. And for the first time, epistemological questions were being studied about mathematical ideas: how do we know that the geometric results we have discovered are true? Can we justify their truth so that others can come to agree to their truth independent of our authority on the matter? Are these ideas structurally interrelated? A tradition of proof through deduction was arising to organize these ideas in theoretical, rather than utilitarian, form. Geometers of the fifth and fourth centuries BCE tended to be more interested in determining what could be asserted, say, about the relationship between the lengths of the pieces into which two intersecting chords in a circle cut each 1 Indeed, every one of the great human civilizations of the world have done so, often quite independently of each other. It seems to be in the nature of human beings to geometrize in some fashion. It is important to note, however, that the geometrical relationships discovered by these early geometers were not expressed algebraically (as we do, when, for example we claim that A = 12 bh describes the area of a triangle). Algebra, the science of finding unknown numerical values from arithmetical conditions placed on them, especially in its familiar modern symbolic form, did not exist in the ancient world in this form. 2 A polyhedron is a finite solid body whose exterior is made up of plane polygons, like a cube or a pyramid. 61 62 CHAPTER 5. HIPPOCRATES OF CHIOS other than in how much land there was in some particular grower’s quadrangular pomegranate orchard. After all, once the general theorems of geometry were discovered, they could then be applied in an infinite variety of particular instances to deal with these more prosaic utilitarian issues. And while the objects of study (circles, squares, etc.) were certainly conceived of as representing the underlying forms of physical reality, their theoretical consideration was viewed by the classical Greek geometers as applying to highly abstracted and general versions of this reality. In any case, by the end of the fifth century BCE, Greek geometers had worked out a theory for the area of plane figures that successfully handled all rectilinear figures. It is this theory to which we now turn our attention. If plane geometry is the study of plane figures, then a fundamental problem in plane geometry naturally is: given a figure, determine its area. While solutions to this problem were already well known to Greek geometers by the fifth century BCE for many types of plane figures, it is the systematic way in which the problem was being addressed by them that was new. Let us follow some of this development. First of all, what is meant by area? The simple answer for us modern readers is that area is a measure of the size of a plane figure. But as soon as one attempts to make such a measure, one needs an appropriate scale to measure against. For the same figure might measure 5 square feet on one scale and 0.4645 square meters on another. Which of these measures is the correct one? Well, both are, of course. This example just illustrates that we measure area, like length and volume, relative to some unit. The value of the measure is dependent on the unit used. Now we have already mentioned (see note 9 in Chapter 2) that modern geometry is characterized by its highly arithmetized and algebraic form: in treatments of geometry given today, geometric objects are often defined and described through equations or formulas. Another reflection of this perspective more pertinent to our present discussion, however, is the modern need to quantify the chief properties of geometric objects through the measurement of their dimensions. Length, area, angle, volume, and other characteristics of geometric objects are for us numerical measures, relative to some convenient scale (feet, square meters, degrees, cubic centimeters, etc.). This way of doing geometry is quite different from that found in the work of the Greek geometers. For them, the area of a figure was not a number computed through some measurement; indeed, area was not conceived of numerically at all. Rather, it was a property attached to the object, the one having to do with its “size.” For the Greek geometer, the area of a triangle was not a number but a certain extent of two-dimensional space. In fact, in most cases, the area of the object and the object itself were not distinguished from each other. Instead, to find the area of a figure meant for them to determine the dimensions of a square having the same two-dimesional content of space as the given figure.3 This process 3 As we will see, to the Greek geometer, figures which had the same area were in fact said to be “equal,” illustrating that the figure itself was identified as one and the same with its 2. TEXT: THE QUADRATURE OF POLYGONS 63 was called quadrature (from the Latin word for square, quadratus, literally meaning four-sided ), and it manifests for us the importance of reference to the square as the most fundamental of plane figures. Notice also that quadrature of a figure requires no measure, no computation, no number; it is a purely geometric procedure. The first set of texts below is a sequence of propositions from Books I and II of Euclid’s Elements, consisting of results that were already well-known by the fifth century BCE. They systematically solve the quadrature problem for any polygonal figure. 2 Text: The quadrature of polygons From Euclid’s Elements, I.36. Parallelograms which are on equal bases and in the same parallels equal one another.4 Let ABCD and EF GH be parallelograms which are on the equal bases BC and F G and in the same parallels AH and BG.5 Figure 5.1: Elements, I.36 area. 4 Note that it is the areas of the parallelograms that is being discussed here; Euclid makes no distinction between the area of the figure and the figure itself. Further, the proposition makes it clear that the area of a parallelogram depends only on the length of its base and its corresponding height (the distance between the parallels to which reference is made in the statement of the proposition). This is obvious to the modern geometer when she writes that the formula for the area of a parallelogram with base b and height h is A = bh. But Euclid is not a modern geometer, so he sees no need to make reference to the (numerical) height of the parallelogram when the purely geometric version of the pair of parallels enclosing the figure is description enough for him. Euclid does not compute the area of the figure, he simply characterizes it in terms of its “elements.” 5 It will be obvious from the stylistic form of the writing, here and in all of the subsequent propositions drawn from the Elements, that Euclid adopts a very formal style of writing, almost like poetic verse, to lay out geometric propositions. The initial statement of the proposition, given in italics above, is called the protasis. It identifies the premises or conditions and the corresponding conclusion of the proposition at hand, often (though not in this instance) in the form “If . . . , then . . . ” The protasis is followed by a second version of the set-up of the proposition, the ekthesis, this time with letters referring to a specific, but entirely generic, illustration of the given data. The ekthesis is accompanied by a diagram to which the letters refer. This is immediately followed by the diorismos, or restatement of the conclusion, often using the words “I say that . . . ” The remaining parts of the format for Euclidean propositions will be discussed in note 7. 64 CHAPTER 5. HIPPOCRATES OF CHIOS I say that the parallelogram ABCD equals EF GH.6 Join BE and CH.7 Since BC equals F G and F G equals EH, therefore BC equals EH. But they are also parallel, and EB and HC join them. But straight lines joining equal and parallel straight lines in the same directions are equal and parallel, therefore EBCH is a parallelogram.8 And it equals ABCD, for it has the same base BC with it and is in the same parallels BC and AH with it.9 For the same reason also EF GH equals the same EBCH, so that the parallelogram ABCD also equals EF GH. Therefore parallelograms which are on equal bases and in the same parallels equal one another. Q.E.D.10 From Euclid’s Elements, I.38. Triangles which are on equal bases and in the same parallels equal one another. Let ABC and DEF be triangles on equal bases BC and EF and in the same parallels BF and AD. Figure 5.2: Elements, I.38 I say that the triangle ABC equals the triangle DEF . Produce AD in both directions to G and H. Draw BG through B parallel to CA, and draw F H through F parallel to DE. 6 The diorismos. fourth part of the propostion is called the kataskeuē (it need not appear in all propositions). It calls for any additional lines or curves that should be added to the diagram to help effect the proof. It is followed by the apodeixis, the meat of the proof, which lists the chain of reasons leading from the given premises to the final conclusion. 8 Euclid calls here on the fact, which he has proved earlier (in Proposition I.33) that in a quadrilateral, if one pair of opposite sides is parallel and equal, then so must the other pair be. 9 Euclid also cites an earlier proposition here to justify this step of the argument. But it is easy enough to see the reason directly from the diagram: since triangles ABE and BCH have equal dimensions (Figure 5.1), then they are clearly equal, and removing each of these triangles in turn from the trapezoid ABCH leaves the parallelograms ABCD and EBCH, which must therefore also be equal. 10 The final sentence of the proposition is called the sumperasma; it is simply a recapitulation of the original statement of the proposition, prefaced with a “therefore,” and typically followed by the phrase “which was to be demonstrated” (Q.E.D. See Chapter 2, note 15). 7 The 2. TEXT: THE QUADRATURE OF POLYGONS 65 Then each of the figures GBCA and DEF H is a parallelogram, and GBCA equals DEF H, for they are on equal bases BC and EF and in the same parallels BF and GH. Moreover the triangle ABC is half of the parallelogram GBCA, for the diameter AB bisects it. And the triangle F ED is half of the parallelogram DEF H, for the diameter DF bisects it. Therefore the triangle ABC equals the triangle DEF . Therefore triangles which are on equal bases and in the same parallels equal one another. Q.E.D. From Euclid’s Elements, I.42. To construct a parallelogram equal to a given triangle in a given rectilinear angle.11 Let ABC be the given triangle, and D the given rectilinear angle. Figure 5.3: Elements, I.42 It is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC. Bisect BC at E, and join AE. Construct the angle CEF on the straight line EC at the point E on it equal to the angle D.12 Draw AG through A parallel to EC, and draw CG through C parallel to EF . Then F ECG is a parallelogram. Since BE equals EC, therefore the triangle ABE also equals the triangle AEC, for they are on equal bases BE and EC and in the same parallels BC and AG.13 Therefore the triangle ABC is double the triangle AEC. But the parallelogram F ECG is also double the triangle AEC, for it has the same base with it and is in the same parallels with it,14 therefore the parallelogram F ECG equals the triangle ABC. 11 The form of the statement of this proposition is different from that of the two previous propositions we have displayed here. Whereas Propositions I.36 and I.38 are theorems, or assertions of mathematical fact, Proposition I.42 is a problem, in which the goal is to find some desired point, line, figure, or number. All of Euclid’s propositions have one of these two forms. 12 Euclid, in I.23, showed how to construct a copy of a given angle along a given line at any particular point on it. It is a straightforward construction procedure. See Chapter 7 for more on Euclidean constructions. 13 You recognize this as precisely the content of Proposition I.38 14 Euclid has established this in Proposition I.41. See also Exercise 7). 66 CHAPTER 5. HIPPOCRATES OF CHIOS And it has the angle CEF equal to the given angle D. Therefore the parallelogram F ECG has been constructed equal to the given triangle ABC, in the angle CEF which equals D. Q.E.F.15 From Euclid’s Elements, I.44. To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle.16 Let AB be the given straight line, D the given rectilinear angle, and C the given triangle. Figure 5.4: Elements, I.44 It is required to apply a parallelogram equal to the given triangle C to the given straight line AB in an angle equal to D. Construct the parallelogram BEF G equal to the triangle C in the angle EBG which equals D, and let it be placed so that BE is in a straight line with AB.17 Draw F G through to H, and draw AH through A parallel to either BG or EF . Join HB. Since the straight line HF falls upon the parallels AH and EF , therefore the sum of the angles AHF and HF E equals two right angles.18 Therefore the sum of the angles BHG and GF E is less than two right angles. And straight lines produced indefinitely from angles less than two right angles meet, therefore HB and F E, when produced, will meet.19 Let them be produced and meet at K. Draw KL through the point K parallel to either EA or F H. Produce HA and GB to the points L and M .20 15 Since this proposition is a problem, not a theorem, the sumperasma ends with the phrase “which was to be constructed,” in Latin, “quod erat factum.” 16 For Euclid, a “straight line” means a line segment, not an infinitely long line. Notice also the special use of the verb “apply,” meaning to build with the given conditions. 17 Here Euclid uses the result of I.42. 18 In modern geometry, we would say that angles AHF and HF E are supplementary, that is, they add up to 180◦ . But Euclid never uses the degree as a unit of angular measure. Instead, he measures all angles in reference to right angles. (See Chapter 7.) Here he needs to make use of the fact, which he proves in I.29, that alternate interior angles made by a transversal line across a pair of parallels are supplementary to each other. 19 It is clear from the diagram that these lines meet, but Euclid is careful to prove this fact, since he does not rely on the drawing for his argument. Since angle BHG is certainly smaller than AHG, the angles BHG and GF E sum to less than 180◦ . It follows then (from Euclid’s Parallel Postulate; see Chapter 7, note 7.1) that the lines HB and F E are not parallel and must meet. 20 It is curious that, after going through the trouble to argue why HB and F E must meet at a point K, Euclid does not explain how he knows that HA and GB, when produced, will 2. TEXT: THE QUADRATURE OF POLYGONS 67 Then HLKF is a parallelogram, HK is its diameter,21 and AG and M E are parallelograms, and LB and BF are the so-called complements about HK. Therefore LB equals BF .22 But BF equals the triangle C, therefore LB also equals C. Since the angle GBE equals the angle ABM , while the angle GBE equals D, therefore the angle ABM also equals the angle D. Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which equals D. Q.E.F. From Euclid’s Elements, I.45. To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle. Let ABCD be the given rectilinear figure and E the given rectilinear angle. Figure 5.5: Elements, I.45 It is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle E. Join DB. Construct the parallelogram F H equal to the triangle ABD in the angle HKF which equals E.23 Apply the parallelogram GM equal to the triangle DBC to the straight line GH in the angle GHM which equals E.24 Since the angle E equals each of the angles HKF and GHM , therefore the angle HKF also equals the angle GHM . Add the angle KHG to each. Therefore the sum of the angles F KH and KHG equals the sum of the angles KHG and GHM . eventually cut KL in points L and M . Mathematicians of the nineteenth century studied Euclid’s proof-writing techniques carefully and recognized that he wasn’t always as rigorous as he could have been. Still, the issue at this point in his proof is not a serious one. One can, if necessary, fill in a more complete proof that takes fills this small lacuna in the logic. 21 A diameter of a parallelogram corresponds to either diagonal line joining opposite corners of the figure. 22 Parallelograms, rectangles and squares are often labeled by Euclid by referring to one pair of opposite corners. In the diagram accompanying this proposition, the large parallelogram HLKF has been divided into four smaller parallelograms, two straddling the diameter HK, namely AG and M E, and two others, LB and BF , called complements by Euclid. Since the diameter of a parallelogram cuts the figure in half, into two congruent triangles, it follows that triangles HLK and KF H are equal, as are HAB and BGH, and BM K and KEB. Removing HAB and BM K from HLK leaves parallelogram LB, and removing BGH and KEB from KF H leaves parallelogram BF , so these last two are equal as well. 23 Here Euclid uses Proposition I.42. 24 Here he uses Proposition I.44. 68 CHAPTER 5. HIPPOCRATES OF CHIOS But the sum of the angles F KH and KHG equals two right angles, therefore the sum of the angles KHG and GHM also equals two right angles. Thus, with a straight line GH, and at the point H on it, two straight lines KH and HM not lying on the same side make the adjacent angles together equal to two right angles, therefore KH is in a straight line with HM .25 Since the straight line HG falls upon the parallels KM and F G, therefore the alternate angles M HG and HGF equal one another. Add the angle HGL to each. Then the sum of the angles M HG and HGL equals the sum of the angles HGF and HGL. But the sum of the angles M HG and HGL equals two right angles, therefore the sum of the angles HGF and HGL also equals two right angles. Therefore F G is in a straight line with GL.26 Since F K is equal and parallel to HG, and HG equal and parallel to M L also, therefore KF is also equal and parallel to M L, and the straight lines KM and F L join them at their ends. Therefore KM and F L are also equal and parallel. Therefore KF LM is a parallelogram. Since the triangle ABD equals the parallelogram F H, and DBC equals GM , therefore the whole rectilinear figure ABCD equals the whole parallelogram KF LM . Therefore the parallelogram KF LM has been constructed equal to the given rectilinear figure ABCD in the angle F KM which equals the given angle E. Q.E.F. From Euclid’s Elements, II.14. To construct a square equal to a given rectilinear figure.27 Let A be the given rectilinear figure. Figure 5.6: Elements, II.14 It is required to construct a square equal to the rectilinear figure A. 25 While it may seem apparent from the diagram that HM is a simple extension of the line KH, it was not added to the diagram as an extension of this line, so Euclid must prove that it is so, which he does by showing that angle KHM is a 180◦ angle. 26 This mirrors the argument above to show that angle F GL is a 180◦ angle. 27 Finally, Euclid has accumulated the machinery he needs, in the guise of the propositions listed above, to solve the problem central to our discussion here, the quadrature of any polygonal figure. 2. TEXT: THE QUADRATURE OF POLYGONS 69 Construct the rectangular parallelogram BD equal to the rectilinear figure A.28 Then, if BE equals ED, then that which was proposed is done, for a square BD has been constructed equal to the rectilinear figure A. But, if not, one of the straight lines BE or ED is greater. Let BE be greater, and produce it to F . Make EF equal to ED, and bisect BF at G.29 Describe the semicircle BHF with center G and radius one of the straight lines GB or GF . Produce DE to H, and join GH. Then, since the straight line BF has been cut into equal segments at G and into unequal segments at E, the rectangle BE by EF together with the square on EG equals the square on GF .30 But GF equals GH, therefore the rectangle BE by EF together with the square on GE equals the square on GH. But the sum of the squares on HE and EG equals the square on GH,31 therefore the rectangle BE by EF together with the square on GE equals the sum of the squares on HE and EG. Subtract the square on GE from each. Therefore the remaining rectangle BE by EF equals the square on EH. But the rectangle BE by EF is BD, for EF equals ED, therefore the parallelogram BD equals the square on HE. And BD equals the rectilinear figure A.32 Therefore the rectilinear figure A also equals the square which can be described on EH. 28 He uses here Proposition I.45. Notice that he chooses the angle that determines the parallelogram to be a right angle, so that the parallelogram is in fact a rectangle. This is important to the rest of the proof. 29 One of the most elementary construction problems, which he solves in Proposition I.10, is to bisect a line segment. 30 Here Euclid is citing Proposition II.5, which we will interpret in algebraic terms here to make faster work of his argument, despite the glaring anachronism. Let a = BE and b = EF . Then since BF = a + b, GF = 21 (a + b), and we have that EG = GF − EF = 1 1 (a + b) − b = (a − b). 2 2 Now it is straightforward to verify algebraically that 1 1 ab + [ (a − b)]2 = [ (a + b)]2 , 2 2 and when we rewrite this in terms of the lines in this diagram, we get the result that BE · EF + EG2 = GF 2 , which is the algebraic form of the statement that Euclid is asserting here, “the rectangle BE by EF ” being given by multiplying its length by its width, and the squares EG and GF being obtained by squaring their sides. 31 He is citing the Pythagorean Theorem (Proposition I.47) here, of course. 32 At this point in the argument, Euclid has shown how to square a rectangle. The algebraic form of this result is a significant one: given an x × y rectangle, the square with side s of equal area satisfies the relation s2 = xy, or equivalently, xs = ys , so resolving the quadrature of the √ rectangle is equivalent to finding the square with side s = xy. In the diagram (Figure 5.6), x = EF , y = BE, and s = HE, and since HE is necessarily intermediate in length between √ EF and BE, it follows that s is between x and y in value. For this reason s = xy is also known as the geometric mean between x and y. 70 CHAPTER 5. HIPPOCRATES OF CHIOS Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilinear figure A. Q.E.F. 3 Squaring the circle? The most interesting problems with quadrature deal with figures having curved sides, the quintessential example being the quadrature of the circle. Today, we express our knowledge about the quadrature of the circle through the familiar area formula A = πr2 , or the less familiar but equivalent formula A = π4 d2 , where r represents the radius of the circle and d its diameter, but the simplicity of the formulas mask the underlying conceptual difficulties. The formula makes clear that the area of a circle is a certain multiple of the area of the square whose side is the radius (or, respectively, the diameter) of the circle, so an immediate connection to quadrature is made. This relationship can be made even clearer by expressing the area formula as A A π = π (or, alternatively, 2 = ). 2 r d 4 This brings forward the fact that the ratio between the area of the circle and the area of the square on its radius (or its diameter) is a constant value, independent of how large or small the circle is. This constant ratio between area of the circle and the area of the square on its radius we have recognized to be so important a constant that we have given it a special name, π.33 But giving the number a special symbol gives no indication of its value. What is this number π? Ancient pre-Greek mathematicians dealt with the area problem for the circle in various ways, but nearly all of these presented solutions that were at best good approximations to an exact quadrature. An Egyptian papyrus, for example, records that a circle is equal to the square on 89 its diameter. This is equivalent to saying that π = 256/81 = 3.16049 . . . , a bit larger than it should be. In the Hebrew Scriptures, in the first Book of Kings, we read about the construction of the ceremonial bowl that was erected at Solomon’s Temple in Jerusalem (1 Kgs 7:23) that it was to be 10 cubits across and 30 cubits around, giving a value of π = 3, which is too small. This latter was a very commonly used approximation, no doubt for its simplicity, and was used in Babylonia as well. In the Indian Sulvasutras, the main source for ancient Indian mathematics prior to the seventh century BCE, we find the following verse: If you wish to turn a circle into a square, divide the diameter into 8 parts, and again one of these eight parts into 29 parts; of these 29 parts, remove 28, and moreover the sixth part (of the one part left) less the eighth part (of the sixth part). 33 It may be surprising to learn that, despite the fact that we use a Greek letter to represent this number, this notation does not come to us from the Greeks. The first use of the symbol π for this number (the first letter of the Greek word periphereia, meaning circumference, from the related fact that it is also the ratio of the circumference to the diameter of any circle; see Chapter 8). dates to the early eighteenth century in Britain, and was popularized by the noted Swiss mathematician Leonhard Euler (see Chapter ???). 4. TEXT: THE QUADRATURE OF A LUNE 71 The Brahmin author means to say that the area of the circle is equal to the area of a square whose side is a certain fraction of the diameter of the circle, that fraction being ! " 1 1 1 7 + − − , 8 8 × 29 8 × 29 × 6 8 × 29 × 6 × 8 which is equivalent to having π = 3.0883 . . . The problem of the quadrature of the circle is an old and thorny problem, so much so that, to this day, we use the phrase “squaring the circle” to refer to any intractable problem. We will return to a discussion of the quadrature of the circle later on (see Chapter 6), for work on this problem helped to fuel the development of ideas that are central to calculus. But for now, we will look at the quadrature of another figure, called a lune, that it was thought might provide a fruitful inroad towards resolution of the problem of squaring the circle. A lune is the shape formed by two arcs of intersecting circles, the region that is interior to one of the circles but exterior to the other (Figure 5.7). Figure 5.7: A lune. Toward the end of the fifth century BCE, Hippocrates of Chios34 , a geometer of considerable prowess, lived and taught in Athens. As for many other figures in the ancient world, we know very little about his life. Over 800 years later, the historian Proclus reported that Hippocrates was the first to write a work on elements of geometry. Indeed, Hippocrates may have been the source for much of what eventually was included by Euclid in his Elements some 100 years or so later. Our main interest in Hippocrates here is his discovery of the quadrature of certain lunes. We will learn how he did this by consulting two short excerpts from other historians of geometry who wrote some centuries after Hippocrates’ life; none of his own writings survive today. 4 Text: The quadrature of a lune From Joannes Philoponus, In Aristotelis Physica.35 34 Do not confuse him with Hippocrates of Cos, a comtemporary, but renowned as a physician. It is this other Hippocrates for whom the famous oath of physicians, to above all do no harm to their patients, is named. Both Cos and Chios are islands in the Aegean Sea. 35 Philoponus was a sixth century CE philosopher and Christian theologian who studied the works of Greek philosophers, especially those of Aristotle, and wrote extensive commentaries 72 CHAPTER 5. HIPPOCRATES OF CHIOS Hippocrates of Chios was a merchant who fell in with a pirate ship and lost all his possessions. He came to Athens to prosecute the pirates and, staying a long time in Athens by reason of the indictment, consorted with philosophers, and reached such proficiency in geometry that he tried to affect the quadrature of the circle. He did not discover this, but having squared the lune he falsely thought from this that he could square the circle also. For he thought that from the quadrature of the lune the quadrature of the circle could also be calculated.36 From Simplicius, In Aristotelis Physica.37 Eudemus38 , however, in his History of Geometry says that Hippocrates did not demonstrate the quadrature of the lune on the side of a square39 but generally, as one might say.40 For every lune has an outer circumference equal to a semicircle or greater or less, and if Hippocrates squared the lune having an outer circumference equal to a semicircle and greater and less, the quadrature would appear to be proved generally.41 I shall set out what Eudemus wrote word for word, adding only for the sake of clearness a few things taken from Euclid’s Elements on account of the summary style of Eudemus, who set out on them. This excerpt comes from a commentary on Aristotle’s Physics. A native of Alexandria in Egypt (a city about which much will be said later), he lives at the end of the period of Greek progress in the sciences. 36 We will analyze in detail Hippocrates’ quadrature of one type of lune in the next text below. It is enough for now to understand the point Philoponus makes here concerning the quadrature of the circle. For Hippocrates’ work represented the first time that someone had determined the area of a figure with curved sides – in fact, with circular sides – and it was thought that the techniques he used for the quadrature of the lune might lead to a resolution to the quadrature of the circle. Alas, this was not to be. 37 Simplicius was another sixth century commentator on early Greek texts, notably on the works of Aristotle and Euclid. He was born in Cilicia, a Roman province which is today part of Turkey. He studied in Alexandria under a pupil of Proclus, and at Athens at Plato’s Academy. 38 Eudemus of Rhodes lived in the late fourth century BCE and was the first historian of mathematics. He was a fellow student with Aristotle in Athens and authored histories of geometry, arithmetic and astronomy, none of which survive today. What we know of his work is based on references and quotes in other works like this commentary on Aristotle’s Physics by Simplicius. Here Simplicius quotes from Eudemus’ history of geometry. Because Eudemus is writing so much closer in time to Hippocrates and the other Greek geometers, he is considered a valuable historical source. 39 By this, Simplicius means the quadrature of the lune constructed as follows: given a circle and an inscribed square, build the circle whose diameter is one of the sides of the square. The lune being described here is the one inside the second smaller circle and outside the larger. 40 Simplicius seems to believe that Hippocrates had claimed to have successfully squared the circle as a result of his work on lunes, but the argument he gives does not accomplish this, so Simplicius is not to be trusted on this account. 41 A lune has two boundary arcs; the line segment that joins the endpoints of these arcs forms a chord in each of the two circles. Simplicius is here describing the situation in which the arc of the outer circle is a semicircle, or is less than a semicircle, or more than a semicircle. If Hippocrates had successfully worked out the quadrature of the lune in all these cases, then by arranging that the outer arc of the lune increases to its greatest possible extent and the inner arc shrinks to a point , then the lune becomes a full circle and the quadrature of the circle would be resolved! 4. TEXT: THE QUADRATURE OF A LUNE 73 his proofs in abridged form in conformity with the ancient practice. He writes thus in the second book of the History of Geometry: The quadratures of lunes, which seemed to belong to an uncommon class of propositions by reason of the close relationship to the circle, were first investigated by Hippocrates, and seemed to be set out in correct form; therefore we shall deal with them at length and go through them. He made his starting-point, and set out as the first of the theorems useful to this purpose, that similar segments of circles have the same ratios as the squares on their bases.42 And this he proved by showing that the squares on the diameters have the same ratios as the circles.43 Having first shown this he described in what way it was possible to square a lune whose outer circumference was a semicircle. He did this by circumscribing about a right-angled isosceles triangle a semicircle and about the base a segment of a circle similar to those 42 A segment of a circle is any portion of the circle cut of by a single line. Segments are similar if the arcs that bound them come from equal central angles in their respective circles (the shaded areas labeled s and S in Figure 5.8). Thus, to say that “similar segments of circles have the same ratios as the squares on their bases” means that the ratio of the areas of the segments is the same as the ratio of the squares of the lengths of their straight line bases. In symbols, if one segment s has base b and the other segment S has base B, then s b2 = 2. S B 43 This result is the key idea behind this quadrature. We will carefully study a proof of it in the next chapter. For now, however, we will take it as given. The result says that if one circle c has diameter d and the other circle C has diameter D, then c d2 = . D2 C Hippocrates uses this to derive the result about segments of circles stated above. While the details of this proof are unknown to us, here is one plausible reconstruction. Draw the diameter d through one of the endpoints of b, then complete the triangle two of whose sides are b and d. (See Figure 5.8.) Similarly, in circle C, draw diameter D through one of the endpoints of B, then complete the triangle two of whose sides are B and D. Then, on the one hand, similarity of the segments means that segment s is the same fraction of circle c that segment S is of circle C, or s S = , c C or alternately, s c = . S C On the other hand, since the triangles in the two circles and have equal corresponding angles, they are similar, so their corresponding sides are proportional: b d = . B D Combining these last three proportions yields s c d2 b2 = = 2 = 2, S C D B hence the desired result. 74 CHAPTER 5. HIPPOCRATES OF CHIOS cut off by the sides.44 Since the segment about the base is equal to 44 The figure described here is displayed in Figure 5.9. The right angle is at B. The semicircle circumscribed around it forms the outer circumference of the lune. The inner circumference is formed by the arc of a second circle that makes a segment with the base P R of the triangle similar to the two segments between the semicircle and the other two sides P Q and QR of the triangle. Each of these three segments cut off a 90o arc from its circle. Simplicius described this lune as one ”on the side of a square”. To see that he is talking about the same lune as the one described by Eudemus, consider the entire circle that forms the inner circumference of this lune (Figure 5.10). The base of the semicircle that forms the outer circumference of the lune is one side of a square that can be inscribed in this inner circle. Figure 5.8: Similar segments. Figure 5.9: Hippocrates’ lune. Figure 5.10: Hippocrates’ lune, expanded view. 75 5. SCHOLIUM: THE DEFINITE INTEGRAL the sum of those about the sides45 , it follows that when the part of the triangle above the segment about the base is added to both the lune will be equal to the triangle.46 Therefore the lune, having been proved equal to the triangle, can be squared. In this way, taking a semicircle as the outer circumference of the lune, Hippocrates readily squared the lune. 5 Scholium: The definite integral At the other end of the timeline of the trajectory of ideas that began with the area problem in ancient Greece is the modern calculus concept of the definite integral. Loosely speaking, the integral can be viewed geometrically as an area calculator. Given a line or curve in thexy-plane with equation y = f (x) (here, f (x) represents some formula in x), there is a region bounded by this curve and the x-axis. If we assume for the moment that the curve lies above the axis (that is, that all values of y = f (x) are positive), then the region has this curve as its top edge and the x-axis as bottom edge (Figure 5.11). If we specify two numbers a < b on the x-axis and erect the vertical lines perpendicular to the axis at these points, then these lines can serve as left and right edges of the region (these lines have equations x = a and x = b). We therefore have a region bounded on three sides by straight lines and on #b the fourth by a curve whose equation we know. We use the notation a f (x) dx to represent the area of this region. This is read “the integral of f (x) between a # and b.” The integral symbol was first used by Leibniz in 1686. (We will catch up to this in Chapter ???.) The values placed as subscript and superscript next to the integral symbol are respectively called the lower limit and upper limit of integration; the expression that follows is called the integrand ; and the symbol dx to the right of the integrand is called a differential. The notation gives Thus, the lune can be seen to rest on neighboring vertices of a square inscribed in the circle that forms its inner circumference. 45 This is determined by combining the result on areas of similar segments with the Pythagorean theorem. Referring to Figure 5.9, the Pythagorean theorem says that (P Q)2 + (QR)2 = (P R)2 . But b = P Q = QR is the common length of the base of the two small segments, and B = P R is the length of the base of the larger segment, so we can express this relation as b2 + b2 = B 2 , or in terms of ratios, b2 b2 + 2 = 1. 2 B B But by the result on similar segments, we can transform this into a relationship between the corresponding segments themselves: s s + = 1, S S or more simply, s + s = S. This shows that “the segment about the base is equal to the sum of those about the sides.” 46 The lune in Figure 5.9 is comprised of both of the two small circular segments and the isosceles right triangle. Since the sum of the two small segments is equal to the one larger one, and replacing the two smaller areas with the larger one produces the triangle P QR, Hippocrates has therefore shown that the lune equals the triangle. 76 CHAPTER 5. HIPPOCRATES OF CHIOS Figure 5.11: The definite integral as area calculator. all the necessary information for describing the region whose area we wish to determine: the integral symbol indicates that an area above the x-axis is being calculated, the limits of integration indicate the left- and right-side bounds of the region, the integrand determines the upper boundary of the region, and the differential points to the x as the underlying input variable. For example, the integral $ 5 2x + 1 dx 2 represents the area of the trapezoid bounded by the x-axis, the vertical lines x = 2 and x = 5, and the line y = 2x + 1. The reader is urged to sketch a graph of this region (perhaps with the aid of a graphics calculator) to see which trapezoid we are considering here. Notice that the very same region is represented by the integral $ 5 2t + 1 dt 2 as well, since the only thing that has changed is the name of the input variable.47 Since there is a large class of plane figures whose areas we can calculate from basic geometry, there is then a large class of integrals that we can compute, even without the benefit of sophisticated machinery from calculus. For instance, $ 2 2 dx = 8, −2 since the integral asks for the area of a rectangle (the graph of y = 2 is a horizontal line) 4 units wide by 2 units tall. This generalizes to produce areas of any rectangle that sits on the x-axis: if k is any constant, $ b k dx = (b − a)k. a We also have $ 8 3x dx = 96 0 47 You can imagine the same geometric picture as being displayed on a ty-plane instead of an xy-plane. 77 6. SUMMARY since the region we are considering here is a triangle with base 8 units long and height 3 × 8 = 24 units tall, whence the area of the triangle is 21 (8 × 24) = 96. Other simple figures can be represented by not-so-simple looking integrals. For instance, $ 4 [8 − 2|x − 1| − 2|x − 3|] dx 0 asks for the area of a trapezoid, and $ 4 −4 % 16 − x2 dx the area of a semicircle. (Check these out with a graphics device.) As we consider the development of more tools for handling the area problem, we will be able to extend the class of integrals we can compute even further. The exercises at the end of this chapter include some more explorations of these ideas. 6 Summary • By the fifth century BCE, Greek geometry had become a theoretical science whose truths were tested through deductive logic applied to abstracted shape and form. • A natural and important geometric problem was the area problem: given a figure, is it possible to find (that is, construct a diagram of) the square with the same area? • Propositions from Books I and II of Euclid’s Elements, culminating in Proposition II.14, show how the area problem had been resolved for polygonal figures. • Euclid organized his mathematics into theorems, assertions of some mathematical truth, and problems, procedures for drawing a geometrical object. Both types of proposition were laid out in a formal style consisting of protasis (initial statement of the entire proposition), ekthesis (clear enunciation of the givens or hypotheses), diorismos (declaration of the conclusion), kataskeuē (additional machinery needed to execute the proof), apodeixis (the main argument), and sumperasma (recapitulation). Propositions proved earlier in the text can be used in the course of proving later propositions. • Babylonian, Egyptian and Greek geometers all knew that the area of a circle had a fixed ratio with the square on its diameter (or radius), or numerically, with the square of its diameter (or radius). The use of the symbol π to denote this ratio, however, dates only to the early eighteenth century CE. The value of this ratio was approximated to varying degrees 78 CHAPTER 5. HIPPOCRATES OF CHIOS of accuracy in the ancient world, but could not be determined exactly. This computational problem is related to the geometric problem of the quadrature of the circle, a problem that remained unresolved in Greek mathematics. • Hippocrates of Chios, an accomplished geometer of the mid-fifth century, discovered how to work the quadrature of a certain type of lune. His discovery was based on knowing that circles are to each other as the squares on their diameters. This caused a certain amount of excitement because it was thought that knowing how to square the lune could lead to a method for squaring the circle. #b • The definite integral, denoted a f (x) dx, is used in modern calculus to represent the area of the region bounded by the x-axis, the vertical lines x = a, x = b, and the graph of the equation y = f (x). Know how to... • find the geometric mean √ xy between two given values x and y • describe the quadrature of Hippocrates’ lune • interpret and calculate definite integrals that represent the areas of familiar figures from plane geometry 7 Exercises 1. A trapezoid is a four-sided figure with just one pair of parallel opposite sides, called its bases. (In a parallelogram, both pairs of opposite sides are parallel.) We typically orient a trapezoid so that its bases are horizontal with the longer base below the shorter one, whence the height of the trapezoid is the distance between the bases. Draw a trapezoid, label its longer base B, its shorter base b, and its height h. By dissecting the trapezoid into smaller figures, derive the area formula A = 12 (B + b)h. 2. (a) In note 30 (page 69), the algebraic identity ab + ! a−b 2 "2 = ! a+b 2 "2 appears. By multiplying through by 4 and expanding both sides of the equation, show that the identity is correct. (b) Draw a square and mark off a point on the top edge somewhere to the right of the midpoint. Label the piece of this edge to the left of your point a, and the remaining piece to the right b. (So a > b.) Now working clockwise around the square, divide the other three 7. EXERCISES 79 sides into pieces of length a and b so that line segments of length a and b alternate around the perimeter of the square. In each corner of the square, these line segments determine four a × b rectangles; insert the necessary lines to display these four rectangles. Why must the remaining quadrilateral in the center of the diagram be a square, and how big is this square? (c) Show that the diagram in part (b) provides a geometric verification of the algebraic identity described in part (a). 3. What is the area of an equilateral triangle whose side length is one unit? (Hint: Drawing in the altitude of the triangle divides the equilateral triangle into two congruent right triangles; since you know the length of the short side and the hypotenuse of one of these right triangles, the Pythagorean theorem will determine the length of the other side.) 4. What is the area of a regular hexagon whose side length is one unit? By regular, we mean that all sides are equal and all interior angles are equal. 5. What is the arithmetic mean of the numbers 24 and 54? What is their geometric mean? 6. Find two pairs of whole numbers (like 24 and 54) whose arithmetic mean and geometric mean are both whole numbers. 7. Evaluate each of these integrals by sketching graphs of the corresponding regions and using area formulas from plane geometry: #9 (a) −1 5 dx ' #7& (b) 0 2 − 72 x dx #5 (c) −3 (4 − |x − 1|) dx #4 (d) 0 (8 − 2|x − 1| − 2|x − 3|) dx #4√ (e) 1 −x2 + 8x − 7 dx 8. The “lune on the side of the equilateral triangle” has a quadrature which is resolved by a procedure similar to the one Hippocrates gives for the “lune on the side of the square.” Inscribe an equilateral triangle ABC in a circle with center O and build semicircles on the sides of the triangles. This creates three equal lunes, numbered 1, 2, 3 in the diagram below. Also, let D be the point opposite B on the diameter of the large circle. (a) What are the measures of the angles in triangle OAD? How can you tell? (b) What are the measures of the angles in triangle BAD? How can you tell? (c) Show that AD = 21 BD. 80 CHAPTER 5. HIPPOCRATES OF CHIOS Figure 5.12: The lune on the side of an equilateral triangle. (d) Let x = AB be the length of the side of the equilateral triangle ABC, and let d = BD be the diameter of the large circle. Then AD = d2 . We can apply the Pythagorean Theorem to the right triangle BAD. √ Show from this that x = 23 d. (e) It follows that the ratio of the areas of the semicircle on diameter AB to the semicircle on diameter BD is 34 . Why? (f) Use the diagram to argue that, in terms of area, 3(semicircle onAB)+(triangleABC) = 3(lune)+2(semicircle onBD), and from this show that lune = 1 1 (triangleABC) + (semicircleBD). 3 12 The quadrature of this lune is dependent on finding the quadrature of the circle that forms its inner edge. Therefore, unlike the quadrature of Hippocrates’ lune, the quadrature of this lune is unresolved. 9. Hippocrates is said to have attempted the quadrature of another lune, the “lune on the side of the hexagon,” which we repeat here. Inscribe a regular hexagon (regular means that all the sides and interior angles are equal) in a circle centered at C. Drawing in the diagonals that join opposite vertices of the hexagon dissects the hexagon into six triangles. On each side of the hexagon build a semicircle. This constructs six lunes, marked 1 through 6 in the diagram (Figure 5.13). (a) In triangle ABC, like the other five triangles into which the hexagon is dissected, is equilateral. Why? (b) The semicircles containing lunes 1, 2, 3, and 4 have area equal to the upper half of the large circle. Why? 7. EXERCISES 81 Figure 5.13: The lune on the side of a regular hexagon. (c) The three lunes 1, 2, and 3, together with the semicircle containing lune 4 have total area equal to the upper half of the hexagon (which is a trapezoid). Why? (d) It follows that the area of one of these lunes equals one-third the difference between the area of the trapezoid and the area of the semicircle that contains the lune. Why? Because this quadrature is dependent on that of the circle that forms the inner edge of the lune (just as in the previous exercise), the quadrature of this lune is unresolved. 82 CHAPTER 5. HIPPOCRATES OF CHIOS