Archimedes of Syracuse

Transcription

1 Chapter 8 Archimedes of Syracuse 1 The greatest of the Greek mathematicians As important and influential as Euclid s Elements has become as the epitome of Greek mathematics, it is generally acknowledged that Euclid was not the most accomplished of geometers in the ancient world. The title of greatest mathematician of the Greek era 1 has for many centuries been attached to the person of Archimedes of Syracuse ( BCE). His contributions to Greek geometry were considerable and manifold, some of which we will have a chance to see here in some detail. But Archimedes achieved fame among his contemporaries not so much for his geometrical investigations as for devoting his efforts to applying mathematics to problems of the day. And he achieved an extraordinary level of technological prowess, as witnessed by the variety of clever inventions for which he was supposedly responsible. Archimedes was the son of Phidias, an astronomer, and is also thought to have been a kinsman of Hieron II, who in Archimedes youth became the selfproclaimed king of the city-state of Syracuse on the island of Sicily. As a young man, Archimedes ventured to Alexandria in Egypt to avail himself of the best education to be found in the Greek world. There he would have been able to study the texts at the great libraries, and there he made friendships with other philosopher-mathematicians, most notably Eratosthenes of Cyrene and Conon of Samos, with whom he corresponded for many years. He is the author of a number of works of mathematics that survive today, in plane and solid geometry and arithmetic. He extended all these subjects in substantial ways and had a tremendous influence on subsequent developments in mathematics and science, especially in medieval Arabic and Persian science, and again later, in fifteenth and sixteenth century European mathematics. Archimedes eventually returned to Syracuse, where he lived the rest of his life, earning fame as an engineering consultant to the royal family of Syracuse, 1 Is such a title possible to bestow given its dependence on subjective tastes that fall into and out of favor from expert to expert and from age to age? 115

2 116 CHAPTER 8. ARCHIMEDES OF SYRACUSE designing the architecture for the city walls, and inventing defensive works for the city, like catapults and grappling hooks. Not all his inventions were for military use: he is famous for having invented the water-screw for lifting water for irrigation and domestic use; for building a planetarium, 2 a model in which small spheres mounted on a clockwork mechanism simulate the motion of the planets in the heavens; and for devising systems of tackles and pulleys for moving heavy objects with relatively little work. Still, much of the legends and stories that comes to us from the ancient world about Archimedes the man revolve around the role he played in the military siege of Syracuse in 212 BCE which ultimately led to his death. In the third and second centuries BCE, Rome was involved in a series of military conflicts (the Punic Wars) with the Greek city-state of Carthage, situated across the Mediterranean Sea on the North African coast. Caught in the middle of these conflicts was Syracuse, another city-state on the coast of the island of Sicily, which was allied under both Rome and Carthage at various points in the sequence of events. Initially allied with Carthage at the outset of the first of the Punic Wars in 263 BCE, Syracuse soon switched allegiance. Hieron managed to keep war at bay by honoring this treaty with Rome, but the situation became precarious as the Carthaginian general Hannibal was gaining the upper hand in Spain and Italy against poorly managed Roman armies. Hieron died in 215 and was succeeded by his young grandson Hieronymus, who switched his allegiance to Carthage just as the Second Punic War began. The Romans dispatched their navy to take back Syracuse by force and managed to have the 15-year-old Hieronymus assassinated. The accounts of the subsequent siege of Syracuse by the Roman general Marcellus in 213 in the military histories of Plutarch and Livy tell a fascinating story of the success enjoyed by Archimedes in the defense of the city, which managed to hold off the Romans for many months. The Roman army finally entered the city in the summer of 212 when its defenses were down during a festival held to honor the goddess Artemis. In the ensuing chaos, Archimedes was killed. The story of his death is also the subject of some ancient histories (mostly legendary). It is generally accepted however that Marcellus, despite being his adversary in the conflict, held Archimedes in high esteem for his skill in crafting the military machines that prolonged the siege and he mourned his passing with an elaborate funeral and monument, a stone plinth etched with the diagram of a sphere and its circumscribed cylinder, referring to a geometrical result that Archimedes derived in his On the Sphere and Cylinder, I.34 Cor.: Every cylinder whose base is the greatest circle in a sphere and whose height is equal to the diameter of the sphere is 3 2 of the sphere, and its surface together with its bases is 3 2 the surface of the sphere. That is, the ratio of the volumes of the circumscribing cylinder and the sphere is the same ratio of 3 : 2 as that between their total surface areas. 3 In addition to the geometric result just quoted about cylinder and sphere, Archimedes produced many other deep and far-reaching results in mechanics 2 Such a device is also called an orrery. 3 In 75 BCE, well over a century after the death of Archimedes, Cicero tells poignantly of searching in Syracuse, and eventually finding, this monument, overgrown and in disrepair.

3 2. TEXT: APPROXIMATING π 117 and hydrostatics, 4 including, for instance, the Law of the Lever, 5 and conditions under which a paraboloidal object floating in water will bob back to its vertical equilibrium if displaced away from center. 6 He is the author of more than ten books that survive to the present day, including the two from which we will read in this chapter, and an additional handful that are no longer extant. 7 Archimedes advanced the method pioneered by Antiphon for dealing with the quadrature of the circle in his work Measurement of the Circle to produce the best approximation to date to the quantity we now call π. In the text that follows, note especially in Proposition 3 how the context is dramatically different from other geometric propositions we have read up to now. Instead of working to determine a geometric relationship between objects, Archimedes is interested there in performing a careful arithmetical computation. He employs a very clever iteration procedure, whereby the results of one step of a multi-stage process are used to determine the results of the successive step in the process. The text is difficult, a technical tour-de-force in geometric and computational skill, but it provides another glimpse at the struggles taken by even the most talented mathematicians to understanding more deeply the geometry of the circle and, by extension, the elusive nature of the ubiquitous number π, which has bedeviled mathematicians throughout history. 2 Text: Approximating π From Archimedes Measurement of the Circle. 8 Proposition 1. The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other 4 Mechanics is the science of the motion of material bodies subject to forces like gravity, and hydrostatics is the similar science of bodies in water. 5 On the Equilibrium of Planes, Prop. 6-7: Magnitudes are in equilibrium at distances reciprocally proportional to the weights. In terms of modern physics, this says that if masses m and M are in balance on a lever arm so that m is a distance d from the fulcrum and the mass M is a distance D away, then m M = D d. 6 A paraboloid is the nose-cone shape formed by spinning a portion of a parabola about its axis of symmetry. 7 One book, the Method, was thought to have been lost until a single copy was discovered in 1906 by J. L. Heiberg in Constantinople (modern-day Istanbul). The manuscript vanished again for the better part of the twentieth century until reappearing in 1998, at which point it was placed under the care of the Walters Museum in Baltimore for restoration and study. The Method is vitally important for Archimedes scholarship because it was written by him expressly to explain to his colleague Eratosthenes how it was that he discovered his geometric, by means of thought experiments that balanced solids of different types against one another along a lever arm. For a more detailed exposition on the work of Archimedes, see Stein [6]; for more on the Method, see the Walters Museum website on the Archimedes Palimpsest [4]. 8 It has been argued convincingly that the shortness of this treatise by Archimedes and the fact that the Greek of the text is not in the Doric dialect that Archimedes spoke and wrote testifies that it is a copied fragment of a larger work of his on the quadrature of the circle that no longer survives.

4 118 CHAPTER 8. ARCHIMEDES OF SYRACUSE to the circumference, of the circle. 9 Let ABCD be the given circle, K the triangle described. Figure 8.1: Measurement of the Circle, Prop. 1. Then, if the circle is not equal to K, it must be either greater or less. 10 If possible, let the circle be greater than K. 11 Inscribe a square ABCD, bisect the arcs AB, BC, CD, DA, then bisect (if necessary) the halves, and so on, until the sides of the inscribed polygon whose angular points are the points of division subtend segments whose sum is less than the excess of the area of the circle over K With this result, given the impact of Elements, II.14, Archimedes has shown that the quadrature of the circle can be effected provided it is possible to construct a straight line equal in length to the circumference of the circle. Of course, this just transfers the quadrature problem from a two-dimensional one to a one-dimensional one. It is just as difficult to rectify the curve that is, straighten it as it is to square the circle. Most of us will remember from grade school that the circumference of a circle is given by the formula C = 2πR, where R is the radius. But since the area A of the circle satisfies A = πr 2, we can solve for π in the two formulas and equate the results to obtain A R 2 = C 2R, or more simply, A = 1 RC, which we recognize as the area formula for a triangle with base C 2 and height R, precisely equivalent to what Archimedes is claiming here. Of course, this is an algebraic argument, not at all the way that Archimedes could have imagined the proof. He works only geometrically, by a masterful application of Eudoxus method of exhaustion, as we will now see. 10 Notice the resemblance of the structure of this argument to that of Elements, XII.2 (Chapter 6): a double reductio ad absurdum, with contradictions generated by inscribing and circumscribing polygons inside and outside the circle. 11 The first half of the double reductio begins here. 12 Beginning with a square, Archimedes bisects the arcs between the corners of the polygon to double the number of division points around the circle and determine corner points of a (regular) polygon with double the number of sides, repeating the process until a sufficiently

5 2. TEXT: APPROXIMATING π 119 Thus the area of the polygon is greater than K. 13 Let AE be any side of it, and ON the perpendicular on AE from the centre O. Then ON is less than the radius of the circle and therefore less than one of the sides about the right angle in K. Also the perimeter of the polygon is less than the circumference of the circle, i.e., less than the other side about the right angle in K. Therefore the area of the polygon is less than K; 14 which is inconsistent with the hypothesis. Thus the area of the circle is not greater than K. 15 If possible, let the circle be less than K. Circumscribe a square, and let two adjacent sides, touching the circle in E, H, meet in T. Bisect the arcs between adjacent points of contact and draw the tangents at the points of bisection. Let A be the middle point of the arc EH, and F AG the tangent at A. 16 Then the angle T AG is a right angle. 17 Therefore T G > GA = GH. 18 It follows that the triangle F T G is greater large enough portion of the circle has been exhausted by the polygon. How large is sufficiently large? Large enough that the remaining slivers of circular segments left uncovered by the polygon comprise an area smaller than the difference in area between the circle and K, as assumed at the outset. 13 The inscribed polygon is always smaller than the circle, but it is so close in area to the circle that their difference in area is less than the difference in area between the circle and triangle K. Since the circle is also larger than K, by assumption, the area of the polygon is something between that of the circle and K. 14 Archimedes recognizes that the n-sided polygon can be decomposed into n equal triangles by radial lines emanating from the center of the circle. Each of these triangles has a height equal to ON, and base equal to one of the sides of the polygon. One can then imagine another series of n triangles (Figure 8.2), all of whose bases are equal to the side of the n-sided polygon, set side by side along a line and again sharing their opposite vertex so that they all have height equal to ON. This collection of triangles has total area equal to that of the polygon. But since this composite triangle has a base equal to the total perimeter of the polygon, which is less than the circumference of the circle, and height ON less than the radius of the circle, its area is less than that of K. 15 The first half of the double reductio argument is thus complete. 16 The technique here in the second half of the proof is entirely similar to that of the first half. Starting with a square, a succession of polygons is considered, each circumscribing the circle guaranteeing that all contain, hence are larger than, the circle and each having twice as many sides as the previously considered polygon, so as to more closely equal the circle. 17 Recall Elements, III.18 from Chapter T G > GA because the hypotenuse of right triangle T AG is longer than either of its sides. Figure 8.2: A triangle equal to the polygon.

6 120 CHAPTER 8. ARCHIMEDES OF SYRACUSE than half the area T EAH. 19 Similarly, if the arc AH be bisected and the tangent at the point of bisection be drawn, it will cut off from the area GAH more than one-half. 20 Thus, by continuing the process, we shall ultimately arrive at a circumscribed polygon such that the spaces intercepted between it and the circle are together less than the excess of K over the area of the circle. Thus the area of the polygon will be less than K. 21 Now, since the perpendicular from O on any side of the polygon is equal to the radius of the circle, while the perimeter of the polygon is greater than the circumference of the circle, it follows that the area of the polygon is greater than the triangle K; 22 which is impossible. Therefore the area of the circle is not less than K. 23 Since then the area of the circle is neither greater nor less than K, it is equal to it. Proposition 2. The area of a circle is to the square on its diameter as 11 to By area T EAH, Archimedes means that portion of the triangle T EH which lies outside the circle. 20 The need to ensure that more than half of these triangular areas are used up is vital for the success of the method of exhaustion. Recall that in the middle of the proof of Elements, XII.2 in Chapter 6 a crucial reference to Proposition X.1 is made; this proposition states that if more than half of any area is removed, again and again in succession, eventually the area remaining will be smaller than any preassigned amount; that is, it will eventually become exhausted. This is Archimedes goal: to ensure that this successive doubling of the sides of the circumscribed polygon will eventually result in one so close in area to that of the circle that the difference between it and the circle is less than the difference between K and the circle. 21 Since both the circumscribed polygon and K are larger than the circle, but the difference between the polygon and the circle is less than the difference between K and the circle, then K must be larger than the polygon. 22 This is the mirror conclusion to that in the first part of the double reductio argument: by decomposing the polygon into n equal triangles by lines radiating from O to its corners, we can see that the polygon is equal in area to a triangle composed of n triangles with bases each equal to a side of the polygon and height equal to the radius of the circle, as in Figure 8.2. But the composite triangle will have a base equal to the perimeter of the polygon, which is greater than the circumference of the circle and height equal to the radius of the circle, so its area must be less than that of K. 23 This nicely completes the second part of the double reductio. 24 Recalling that the area A of a circle with radius R and diameter D = 2R is given by the formula A = πr 2 = π «D 2 = π 2 4 D2, we see that the ratio of the area of a circle to the square on its diameter is exactly π 4. If Archimedes were to be taken literally here, we would be able to claim that π 4 = which implies that π = 22. While there is close agreement between these numbers and the use of 7 the fraction 22 is a common and very good approximation to π, it is not exactly true. As Heath 7 notes in his 1897 English edition of this work of Archimedes [5], The text of this proposition is not satisfactory. Clearly, if Archimedes did indeed state this proposition (it could have been inserted by a later editor), he must have meant it in the sense of an approximation to the value of the ratio of circle and square on diameter. In addition, Heath goes on to say that Archimedes cannot have placed it before Proposition

7 2. TEXT: APPROXIMATING π 121 Proposition 3. The ratio of the circumference of any circle to its diameter is less than but greater than Let AB be the diameter of any circle, O its center, AC the tangent at A; and let angle AOC be one-third of a right angle. 26 3, as the approximation depends on the result of that proposition. Indeed, it does follow as a corollary directly from Proposition 3 (see the Exercises at the end of this chapter), so no separate proof was deemed necessary. 25 We will quote Heath again here, who finds the editorial comments of the sixth century CE commentator, Eutocius of Askalon, quite useful: In view of the interesting questions arising out of the arithmetical content of this proposition of Archimedes, it is necessary, in reproducing it, to distinguish carefully the actual steps set out in the text as we have it from the intermediate steps (mostly supplied by Eutocius) which it is convenient to put in for the purpose of making the proof easier to follow. Accordingly all the steps not actually appearing in the text have been enclosed in square brackets, in order that it may be clearly seen how far Archimedes omits actual calculations and only gives results. We, too, will retain most of the editorial comments that Eutocius/Heath supplied, as Archimedes is difficult to follow here for a reader not steeped in the milieu of third century Greek geometry. He only shows the finished product of his computations, or rather, the editions of his work that survive for us have this characteristic; it may be that his original manuscripts, now lost to us, were more forthcoming with the details. The reader must work hard to fill in the gaps left in the proof to understand how he was able to make the next claim, and luckily Eutocius (and Heath as well) have done this for us. In some cases, it is impossible to reconstruct his steps absolutely because multiple paths can be offered to arrive at the same conclusions. The proof is structured in two parts, first to show that the ratio of circumference to diameter of a circle and we will continue to call this ratio π despite the fact that Archimedes would not have done so is less than 22 by estimating measurements of the perimeter of circumscribed 7 polygons to the circle; and then to show that the ratio is more than 223 by estimating 71 measurements of the perimeter of inscribed polygons. 26 Archimedes begins by considering the geometry of the circumscribed regular hexagon. In the diagram (Figure 8.3), C is one of the corners of this hexagon and A is the point of tangency of one of the sides having C as endpoint, hence it is also the midpoint of that side of the polygon. The opposite corner of this side is beyond the point H and does not appear in the diagram; let us call it C. Then angle C OC is the central angle of a regular hexagon, and so it measures (360 6 =) 60, whence angle AOC is 30, or as Archimedes says onethird of a right angle. But as angle ACO is also 60, we see that triangle AOC must be a triangle. Therefore, a simple application of the Pythagorean theorem shows (see the Exercises) that the sides of this triangle are in the proportions OC : OA : AC = 2 : 3 : 1. This is Archimedes next assertion, except that he does this in the form of an inequality by replacing 3 with a very good rational approximation, the slighty-too-large ratio 265 : 153. How did Archimedes come up with this approximation? While no one can say for sure, but the best explanation seems to be that he used a number of anthyphairetic steps (see the discussion in Chapter 4 on anthyphairesis) with the segments OA and AC to find this good approximation. (One of this chapter s exercises will duplicate this computation.) Once the ratio OA : AC is computed, the ratio OC : AC, known to be equal to 2 : 1, is easy to express in similar terms.

8 122 CHAPTER 8. ARCHIMEDES OF SYRACUSE Figure 8.3: Measurement of the Circle, Prop. 3: a portion of the circumscribed polygons. Then 27 OA : AC [ = 3 : 1] > 265 : 153 (8.1) and OC : CA [ = 2 : 1] = 306 : 153. (8.2) First, draw OD bisecting the angle AOC and meeting AC in D. 28 Now [by Elements VI.3 29 ] CO : OA = CD : DA 27 Beginning here and throughout the rest of this text are displayed a number of equations and inequalities involving the ratios with which Archimedes is working. It is necessary to emphasize that this is is to make Archimedes proof intelligible to a modern reader without undue effort and not because Archimedes would have written his mathematics in this way. The use of signs like +,, and even = and <, is extremely useful for modern readers, but none of these symbols was available to Archimedes. Instead, he would have written out in words the content of all these relations, making the language very difficult to parse in this form. 28 The point D will be one corner of the regular dodecagon (having 12 sides) that circumscribes the circle; the opposite corner of this side of the dodecagon whose midpoint is A is also beyond the point H and does not appear in the diagram; let us call it D. Then angle D OD is a central angle of this polygon, and so it measures ( =) 30, whence angle AOD is 15, or half of angle AOC. 29 This proposition states that the angle bisector of any angle in a triangle cuts the opposite side into two parts that have a ratio equal to the ratio of the sides of the original triangle which are on the corresponding sides of the bisector. In the context of the triangle AOC, OD bisects angle AOC, so the proposition allows to Archimedes to make this next assertion. (A proof of Elements VI.3 is in the Exercises.)

9 2. TEXT: APPROXIMATING π 123 Hence 31 so that 30 [CO + OA : OA = CA : DA or] CO + OA : CA = OA : DA. Therefore [by adding (1) and (2)] OA : AD > 571 : 153. (8.3) OD 2 : AD 2 [= ( OA 2 + AD 2) : AD 2 > ( ) : ] > : so that 32 OD : DA > : 153. (8.4) Secondly, 33 let OE bisect the angle AOD, meeting AD in E. [Then DO : OA = DE : EA 30 Expressed as an equation, the previous proportion is Adding one to both sides, we obtain which on combining the fractions yields CO + OA OA CO OA = CD DA. CO OA + OA OA = CD DA + DA DA + OA CD + DA = + DA OA DA DA. 31 This next computation is an application of the Pythagorean Theorem to triangle AOD. 32 Archimedes will now evaluate a square root. The number is a perfect square so that causes no difficulty, but is not a perfect square: he needs to find a useful approximation. He can work out that the square root of is very close to and so he settles on as a convenient lower bound Before we move on, compare lines (8.1) and (8.2) with (8.3) and (8.4). The first set gives estimates of measures of the sides of triangle AOC, a portion of the circumscribed hexagon. The second set gives estimates of corresponding measures for triangle AOD, a portion of the circumscribed dodecagon. In both sets of relations we have ratios between OA, the radius of the circle, and either the half-length of the side of the circumsribed polygon (AC in line (8.1) and AD in line (8.3)), or the distance from the center to a corner of the polygon (OC in line (8.2) and OD in line (8.4)). Archimedes has completed the first stage of the computational analysis. From line (8.1), he could determine the ratio of the perimeter of the hexagon (which is 12 times the length of AC) to the diameter of the circle (twice the radius OA); this would give a crude overestimate to the ratio of the circumference of the circle to its diameter. In order to achieve better accuracy, Archimedes leaves aside the hexagon and concentrates instead on measuring the dodecagon, whose perimeter is closer in length to that of the circle s circumference. But since lines (8.3) and (8.4) bring him to the same point with the dodecagon as did lines (8.3) and (8.4) for the hexagon, he sees that he can carry this same iterative procedure forward to obtain measures for the corresponding line segments in the circumscribed 24-gon obtained by bisecting the

10 124 CHAPTER 8. ARCHIMEDES OF SYRACUSE so that Therefore DO + OA : DA = OA : AE.] [It follows that OA : AE [ > ( ) : 153 by (8.3) and (8.4) ] > : 153 (8.5) OE 2 : EA 2 > ({ } ) : > ( ) : > : ] Thus 34 OE : EA > : 153. (8.6) Thirdly, 35 let OF bisect the angle AOE and meet AE in F. We thus obtain the result [corresponding to (8.3) and (8.5) above] that [Therefore OA : AF [ > ( ) : 153 ] > : 153 (8.7) OF 2 : F A 2 > ({ } ) : > : ] Thus 36 OF : F A > : 153 (8.8) Fourthly, 37 let OG bisect the angle AOF, meeting AF in G. We have then OA : AG [ > ( ) : 153 by means of (8.7) and (8.8) ] > : 153 Now the angle AOC, which is one-third of a right angle, has been bisected four times, and it follows that AOG = 1 48(a right angle). central angles of the dodecagon. This is precisely what he does. Indeed, he will not stop there, continuing on to obtain measures for the corresponding line segments in the circumscribed 48-gon, and then again for the circumscribed 96-gon, where he finally pauses to give his final highly accurate estimate. 34 Once more Archimedes requires the computation of a convenient lower bound for the square root of a number, , which is not a perfect square Lines 8.5 and 8.6 are for the 24-gon what lines 8.3 and 8.4 are for the dodecagon. Now Archimedes moves on to step four of this five step iteration, the measurement of the circumscribed 48-gon. 36 This is the last of the square root estimates. 37 This is the fifth and final step of the iteration, to measure the half-side of the circumscribed 96-gon.

11 2. TEXT: APPROXIMATING π 125 Make the angle AOH on the other side of OA equal to the angle AOG, and let GA produced meet OH in H. Then GOH = 1 24(a right angle). Thus GH is one side of a regular polygon of 96 sides circumscribed to the given circle. And, since OA : AG > : 153 while AB = 2OA, GH = 2AG, it follows that But AB : (perimeter of polygon of 96 sides) [ > : ] = > : [ < ] 1 2 < Therefore the circumference of the circle (being less than the perimeter of the polygon) is a fortiori less than times the diameter AB. 38 Figure 8.4: Measurement of the Circle, Prop. 3: working with inscribed polygons. Next let AB be the diameter of a circle, and let AC, meeting the circle in C, make the angle CAB equal to one-third of a right angle. Join BC. 39 Then 40 AC : CB[ = 3 : 1 ] < 1351 : Notice that in the last line of computation above, Archimedes has inverted the ratio he has been working with, which forces the inequalities to invert as well: (circumference) : AB < (perimeter of 96-gon) : AB < : < This ends the first part of the analysis. Now, Archimedes will consider making estimates of ratios of lines having to do with the inscribed regular hexagon, dodecagon, 24-gon, 48-gon, and 96-gon. These will allow him to give an accurate underestimate for the ratio of circumference to diameter of the circle. 39 Recall from Chapter 6, note 6, that according to Elements, III.20, angle CAB is half as large as the central angle COB (not drawn in the figure). So if angle CAB is 30 (one-third of a right angle), then angle COB is 60. Therefore BC is one side of an inscribed regular hexagon. 40 Recall (Exercise 7.4) that the side of the inscribed hexagon is equal to the radius of the cir-

12 126 CHAPTER 8. ARCHIMEDES OF SYRACUSE First, 41 let AD bisect the angle BAC and meet BC in d and the circle in D. Join BD. Then 42 BAD = dac = dbd and the angles at D, C, are both right angles. It follows that the triangles ADB, [ACd,] BDd are similar. Therefore 43 AD : DB = BD : Dd [ = AC : Cd ] = AB : Bd [by Elements, VI.3] = (AB + AC) : (Bd + Cd) or BA + AC : BC = AD : DB. (8.9) [But AC : CB < 1351 : 780 from above, while BA : BC = 2 : 1 = 1560 : 780. ] Therefore [Hence AD : DB < 2911 : 780 (8.10) AB 2 : BD 2 < ( ) : = : ] Thus 44 AB : BD < : 780 (8.11) cle, so AB : BC = 2 : 1. Also, by Elements III.20, angle ACB is right. So by the Pythagorean Theorem, AC : CB = 3 : 1. Once again, therefore, Archimedes needs a convenient approximation this time an underestimate to 3. By extending the anthyphairetic procedure that allowed him to get the estimate in line (8.1) above, he arrives at the approximation 1351 : 780. (See the chapter Exercises for details.) 41 The computations of the ratios of the sides of triangle ABC give Archimedes the first set of measurements for this part of the analysis. Here BC is the side of the inscribed hexagon; in successive stages of the iteration, he will compute estimates for the corresponding ratios in the inscribed dodecagon, 24-gon, 48-gon, and 96-gon. Each of these is obtained simply by bisecting the angle of the triangle at A. 42 Since angle CAB measures 30, angle DAB measures 15, so angle DOB (not drawn) is 30 and BD is the side of the inscribed dodecagon. Triangles dac and dbd have equal angles at d and both have right angles: one at C, the other at D. Therefore, the third angle in each triangle has the same measure: angle dac = angle dbd. 43 The first two lines below rely on the similarity of the triangles just mentioned. The third line has a citation for Elements, VI.3. See note 29 above. The proposition is being applied to triangle ABC with Ad as angle bisector. Finally, the last step is equivalent to the arithmetical fact that if a b = c d, then a b = a+c b+d = c, whose verification is left to the chapter Exercises. 44 d Lines (8.10) and (8.11) give the estimated measures for the sides of triangle ABD, which apply to the inscribed dodecagon.

13 2. TEXT: APPROXIMATING π 127 Secondly, 45 let AE bisect the angle BAD, meeting the circle in E; and let BE be joined. Then we prove, in the same way as before, 46 that [Hence Therefore AE : EB [ = BA + AD : BD < ( ) : 780 by (8.9) and (8.10) ] = : 780 = : = 1823 : 240. (8.12) AB 2 : BE 2 < ( ) : = : ] AB : BE < : 240 (8.13) Thirdly, 47 let AF bisect the angle BAE, meeting the circle in F. Thus [It follows that Therefore AF : F B [ = (BA + AE) : BE < : 240 by (8.12) and (8.13) ] = : = 1007 : 66 (8.14) AB 2 : BF 2 < ( ) : 66 2 = : 4356 AB : BF < : 66 (8.15) Fourthly, 48 let the angle BAF be bisected by AG, meeting the circle in G. Then AG : GB [ = (BA + AF ) : BF ] < : 66 [by (8.14) and (8.15). And AB 2 : BG 2 < ({ } ) : 66 2 = : ] 45 Now, in step three of the iterated computations, Archimedes concentrates on measuring the inscribed regular 24-gon. 46 The proportion in the first line of the next computation follows in exactly the same way for the 24-gon in the same way that line (8.9) was derived for the dodecagon. 47 Next come the estimates for the ratio measures corresponding to the 48-gon. 48 And the final step obtains estimates for the ratio measures corresponding to the 96-gon.

14 128 CHAPTER 8. ARCHIMEDES OF SYRACUSE Therefore AB : BG < : 66 whence BG : AB > 66 : (8.16) [Now the angle BAG which is the result of the fourth bisection of the angle BAC, or of one-third of a right angle, is equal to one-fortyeighth of a right angle. Thus the angle subtended by BG at the centre 49 is 1 24(a right angle). Therefore BG is a side of a regular inscribed polygon of 96 sides. It follows [from (8.16)] that And (perimeter of 96-gon) : AB [ = : ] > = 6336 : Much more then is the circumference of the circle greater than times the diameter. Thus the ratio of the circumference to the diameter < The conic sections but > We have seen how diagrams drawn with straightedge and compass provided the medium whereby Greek geometry was realized (literally), in spite of the fact that the objects of its attention seem to be non-physical abstractions of reality. The first three postulates of Euclid s Elements (see Chapter 7) indicate the importance of these basic tools of the geometers for making diagrams. Indeed, much of the compelling nature of the problem of the quadrature of the circle derived from the inability of geometers to devise a construction to square the circle similar to the construction in Elements, II.14 that resolved this problem for polygons with straight sides (Chapter 5). Archimedes was well aware of this conundrum, and his work in Measurement of the Circle can be interpreted as an admission that, as the actual quadrature appeared impossible, an approximate one was the best that geometers could hope for. The intractability of the problem of the quadrature of the circle was not the only unsolved problem in Greek geometry. Another famous problem that resisted solution was the problem of the duplication of the cube. As various apocryphal stories relate it, the worshipers at a Greek temple are instructed to double the size of a cubical altar; when some calamity befalls after they have rebuilt the altar by doubling the lengths of all the sides, they realize that instead of doubling the volume of the altar, they have increased it by a factor of 8 = 2 2 2, neglecting to account for the three-dimensionality of the solid. The duplication of the cube (Figure 8.5) asks for the dimension of the altar required to correctly double the volume of a cube, and moreover, it asks for a 49 This is another application of Elements III.20 (see note (6)).

15 3. THE CONIC SECTIONS 129 construction that will produce the proper line segment of the side of the new cube given the side of the original. 50 Figure 8.5: The duplication of the cube. The first work towards a solution of this problem was carried out by Hippocrates of Chios. He showed that the problem would be solved if it were possible to construct two mean proportionals between the given side and its double. Recall from Chapter 5 (note 32) that a geometric mean between two given line segments a and b is the line x that satisfies a : x = x : b, or in equation form, a x = x b. (8.17) For this reason, the geometric mean is also called the mean proportional between the segments a and b. To find two mean proportionals between a and b is to find a pair of segments x and y so that a : x = x : y = y : b, or again in equation form, a x = x y = y b. (8.18) Now if x is the (single) mean proportional between a and b, then from (8.17) we deduce that x = ab. This illustrates algebraically how the mean proportional identifies the side of the square whose area equals that of the rectangle with sides a and b. Indeed, Elements VI.13 provides the construction of this mean proportional: referring back to Exercise 8 on page 114, if a and b are set side by side along the same line and the circle with this longer segment as diameter is then drawn, the mean proportional x between a and b is constructed by erecting the perpendicular at the point where a and b meet the circle. In the case of two mean proportionals x and y between a and b, we note that since the three fractions in (8.18) are equal, then the product of the three of them equals the cube of any one of them: a b = a x x y y ( a ) 3 b =. x 50 This problem would not be definitively resolved until the nineteenth century, in an 1837 paper by Pierre Laurent Wantzel, a 23-year-old Frenchman, in which he proved, using sophisticated algebraic arguments that no straightedge and compass construction would ever solve the duplication problem, since the class of line segments which were so constructible does not contain the type of line required by this problem.

16 130 CHAPTER 8. ARCHIMEDES OF SYRACUSE Cross multiplication leads to ax 3 = a 3 b, or more simply, x 3 = a 2 b, whence x = 3 a 2 b. This allows us to see the connection with Hippocrates idea for solving the problem of the duplication of the cube: if a is the side of the given cube and b = 2a is its double, then the first of the two mean proportionals between a and b will satisfy x 3 = a 2 b = 2a 3, showing that x is the side of the cube whose volume is twice that of the cube on a. But while this does algebraically solve the duplication problem, it does not geometrically solve it. Therefore, short of exhibiting a construction that will produce a line of length x, Greek geometers would not consider the problem fully solved. A student of Eudoxus named Menaechmus was responsible for the next advance in solving this problem by the use of curved lines known as conic sections. In the mid-fourth century BCE, the Greek geometers conceived of a cone as generated by spinning a right triangle about one of its two legs (Def.18 in Elements, XI). If the resulting cone was sliced by a plane perpendicular to the hypotenuse of this spinning triangle, the resulting section (where the plane met the surface of the cone), was a curve called a conic section. Three types of conic section were distinguished by whether the angle at the apex of the cone was acute, right, or obtuse (Figure 8.6), and the resulting curves had intrinsic distinguishing features as well. For instance, the section of the acute-angled cone was a closed curve while the other two were not. 51 Figure 8.6: Sections of the acute-, right-, and obtuse-angled cone. The reader with a little knowledge of these curves may recognize that the conic sections are the curves we now call ellipse, parabola, and hyperbola. While this is true, it is also the case that in the mid-fourth century BCE, these terms had not yet been associated with these curves. That took place late in the third century BCE when Apollonius of Perga wrote a highly influential treatise on conic sections that reoriented the way geometers thought of these curves. But until then, the curves were identified with the cumbersome terms section of the acute-angled cone (ellipse), section of the right-angled cone (parabola), and section of the obtuse-angled cone (hyperbola). Acknowledging the anachronism in the context of Menaechmus work, we will nevertheless use the familiar and more compact Apollonian terms for these three types of curve in our discussion here. 51 They had their endpoints on the base of the cone.

17 3. THE CONIC SECTIONS 131 Menaechmus approached the duplication problem by beginning with Hippocrates reduction to two mean proportionals. From (8.18) above, it is clear that x is the (single) mean proportional between a and the unknown second line y, while y is the (single) mean proportional between the unknown x and b = 2a. In an effort to obtain a geometric representation of these quantities, Menaechmus may have carried out the following analysis. Using the well-known construction for the mean proportional between two lines that was preserved in Elements, VI.13 (see Exercise 8 in Chapter 7, and especially the accompanying Figure (7.9)), a single diagram was drawn to display the mean proportionals x between the fixed line a and a number of possible values of y (Figure 8.7). The collection of these lines, when translated from their positions inside the circles that produced them to become tangents to these circles determine from their endpoints a locus of points (the collection of a typically infinite number of points that satisfy a particular condition) that traces out the parabola. Figure 8.7: Menaechmus parabola. Some decades later, a certain Aristaeus, about whom not much else is known, wrote a book called Solid Loci on conic sections. This work may have been the source of another book called Conics, written by Euclid, but no copies remain to the best of our knowledge. 52 We speculate from comments on these works by later Greek writers that Aristaeus and Euclid wrote about the connection between the curve identified above by Menaechmus and the section of the rightangled cone (Figure 8.8) Even in ancient times, the success of Apollonius Conics may have contributed to a loss of status for works written earlier on the conic sections whose treatments were much less comprehensive than Apollonius. We will have more to say on this later. 53 The following exposition is taken from Dijksterhuis reconstruction of the early work on conics in [2].

18 132 CHAPTER 8. ARCHIMEDES OF SYRACUSE Figure 8.8: The symptom of the parabola as section of the right-angled cone. In Figure 8.8, in every circular section of the right-angled cone parallel to the base of the cone, W Z is the geometric mean between segments XW and W Y. Two such sections are displayed in the figure along with the parabola formed by slicing the cone by the plane containing points V, W, and Z and which is perpendicular to the side AY of the cone. So by Elements, VI.13, XW : W Z = W Z : W Y, and since XW = T V = 2UV, we can write this proportion in the form 2UV : W Z = W Z : W Y. (8.19) But since triangles UV A and V W Y are both triangles, they are similar, so AV : UV = W Y : W V, or more conveniently, On multiplying together (8.19) and (8.20), we get 2AV : 2UV = W Y : W V. (8.20) 2AV : W Z = W Z : W V. (8.21) Therefore, if a fixed line of length a can be made to correspond to the fixed length 2AV while lines of variable length x and y can be made to correspond to variable lines W Z and W V respectively, (8.21) is transformed into a; x = x : y, or ay = x 2. (8.22) These two equivalent relationships, one a proportion, another the statement of equality 54 between a rectangle of sides a and y and a square of side x, would both be referred to later by Apollonius as the symptom of the parabola. This symptom describes the relation between the ordinate x = W Z and the abscissa It is worth reemphasizing that in an age before algebra, these would not have been viewed as symbolic equations but as relationships between lines. 55 This is a Latin word, from the same root as the English word scissors, which translates the phrase from Apollonius Conics in which he describes it as the line cut off from the axis V W by the ordinate.

19 3. THE CONIC SECTIONS 133 y = W V of the point Z on the curve. The line segment of length a = 2AV is called the orthia and the line extending V W is called the diameter or axis of the parabola. In the same way, Menaechmus developed a second parabola to represent the other proportion in (8.18), with b = 2a: x : y = y : 2a, or 2ax = y 2. (8.23) This time, the parabola treats y as ordinate, x as abscissa, and 2a as orthia. But the two parabolas are relating the same pair of unknown line segments x and y, so if both parabolas could be combined in the same diagram, as in Figure 8.9, where the vertical axis is the diameter of the first parabola and the horizontal axis is the diameter of the second parabola, then the ordinates to the two diameters of the point where the curves meet would then be the desired segments that solve the condition of the double mean proportional (8.18), and consequently, the segment x so produced solves the problem of the duplication of the cube. Figure 8.9: Menaechmus two parabola solution of the duplication of the cube. It is important to observe that this was not a solution in the orthodox sense of the term, one that could provide the desired line segments using the standard tools of straightedge and compass. In Menaechmus solution, the two parabolas, neither of which could be drawn with straightedge and compass alone, were instrumental in producing the sought-for segments x and y. Still, this clever piece of geometry illustrates that the fourth century was a time in which geometric research was expanding to include a wider collection of objects of investigation, and the conic sections would be a rich source of mathematical study for the next century, culminating in Apollonius Conics, a monumental work in eight books 56 representing the epitome of Greek investigation into this family of curves and much revered by mathematicians in later 56 The first four books have been preserved in their original Greek, but Books V-VII are known only in their later ninth century CE Arabic translation and Book VIII is now lost.

20 134 CHAPTER 8. ARCHIMEDES OF SYRACUSE centuries, who referred to Apollonius as the Great Geometer on account of this work. Another of the problems that was solved by the time Aristaeus or Euclid wrote on the conic sections was the problem of the tangent line: given a point on a conic curve, determine the tangent line there. We know this because Archimedes refers to finding this result in the elements of conics. We are interested in the problem of the tangent line to the parabola, but as the works of Aristaeus and Euclid are lost to us, the first clear enunciation of this theorem to which we have access, however, is from the Conics of Apollonius. We present its statement and proof in full here below. Conics, I.33. If in a parabola some point is taken, and from it an ordinate is dropped to the diameter and to the straight line cut off by it on the diameter from the vertex, a straight line in the same straight line from its extremity is made equal, then the straight line joined from the point thus resulting to the point taken will touch the section. Figure 8.10: Conics, I.33. Let there be a parabola (Figure 8.10) whose diameter is the straight line AB, and let the straight line CD be dropped ordinatewise, and let the straight line AE be made equal to the straight line ED, and let the straight line AC be joined. I say that the straight line AC produced will fall outside the section. For if possible, let it fall within, as the straight line CF, 57 and let the straight line GB be dropped ordinatewise. And since 58 BG 2 : CD 2 > F B 2 : CD 2, 57 This is the reason for the funny drawing of the curve in Figure 8.10; to show that AC is tangent to the parabola, Apollonius shows indirectly that the line AC cannot dip inside the curve. 58 Since BG > F B, so must BG 2 > F B 2.

21 3. THE CONIC SECTIONS 135 but 59 F B 2 : CD 2 = BA 2 : AD 2, and 60 BG 2 : CD 2 = BE : DE, therefore BE : DE > BA 2 : AD 2. But BE : DE = 4(BE)(EA) : 4(DE)(EA); therefore also 4(BE)(EA) : 4(DE)(EA) > AB 2 : AD 2. Therefore alternately 4(BE)(EA) : AB 2 > 4(DE)(EA) : AD 2 ; and this is absurd, for since AE = DE, hence 61 4(DE)(EA) = AD 2. But, 62 4(BE)(EA) < AB 2, for E is not the midpoint of AB. Therefore the straight line AC does not fall within the section; therefore it touches it. Now Archimedes also made considerable contributions to the study of the conic sections, devoting an entire work to the presentation of one such advance entitled Quadrature of the Parabola, a portion of which is the next text that we will consider. In it, he proves a theorem he discovered that determines the quadrature of a parabolic segment, the region enclosed between the curve and any secant line that cuts it in two points. These theorems show us how quadrature problems for plane regions and tangent problems for curves were favorites of the Greek geometers. 63 Archimedes 59 By the similarity of triangles F BA and CDA, F B : CD = BA : AD. 60 The symptom of the parabola states that BG 2 = (BE)(o) and CD 2 = (DE)(o) where o is the orthia of the curve. Forming ratios between these statements gives the next assertion made here. 61 Since E is the midpoint of AD, we have 4(DE)(EA) = (2DE)(2EA) = (AD)(AD). 62 Citing a result from Euclid equivalent to the algebraic statement that, given any two unequal numbers x and y, (x + y) 2 4xy = (x 2 + 2xy + y 2 ) 4xy = x 2 2xy + y 2 = (x y) 2 > 0, whence (x+y) 2 > 4xy. In the present context, it says that AB 2 = (AE+EB) 2 > 4(BE)(EA), in direct contradiction to the earlier statement that 4(BE)(EA) : AB 2 is a greater ratio than equal to equal. 63 To these we can also add cubature problems (volumes) and surface area problems for solids, and to a lesser degree, rectification problems for curves (lengths along the curves).