MST Topics in History of Mathematics Euclid’s Elements and the Works of Archimedes
Paul Yiu
Department of Mathematics Florida Atlantic University
Summer 2014
July 25 1
The works of Archimedes Recipient Propositions On the Sphere and Cylinder, Book I Dosithesus 44 On the Sphere and Cylinder, Book II Dosithesus 9 Measurement of a Circle 3 On Conoids and Spheroids Dosithesus 32 On Spirals Dosithesus 28 On the Equilibrium of Planes, Book I 15 On the Equilibrium of Planes, Book II 10 The Sand-Reckoner King Gelon Quadrature of the Parabola Dosithesus 24 On Floating Bodies, Book I 9 On Floating Bodies, Book II 10 Book of Lemmas 15 The Cattle Problem The Method Eratosthenes 15 2
Archimedes’ The Measurement of a Circle
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 to the circumference of the circle.
Proposition 2. The area of a circle is to the square on its diameter as 11 to 14.
Proposition 3. The ratio of the circumference of any circle to its diameter is less 3 1 3 10 than 7 but greater than 71 . 3
The Measurement of a circle, 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 to the circumference of the circle.
radius K
circumference 4
Euclid X.1 Two unequal magnitudes being set out, if from the greater there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out.
Given ε>0, if a sequence of magnitudes
A1,A2, ...,An,An+1, ...
A < 1 · A n is such that n+1 2 n for every , then there is some m for which Am <ε. 5
Archimedes’ use of Euclid X.1 I. If possible, let the circle be greater than K.
Inscribe a square ABCD, H AB BC CD DA bisect the arcs , , , , D then bisect (if necessary) the halves, A and so on, until the sides of the inscribed polygon whose angular points are E O the points of division subtend segments whose sum is less than the excess of the area of the circle over K. BC
Euclid X.1. Two unequal magnitudes being set out, H C − A C − K 4 and D if from the greater there is subtracted A a magnitude greater than its half, 1 C − A8 < (C − A4) 2 E O and from that which is left a magnitude greater than its half, 1 C − A16 < (C − A8) 2 BC and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out. C − A2m < C − K for some m =⇒ A2m >K. 6
Proof of Proposition 1 (continued) I. If possible, let the circle be greater than K.
. . . Thus, the area of the polygon is greater H than K. D Let AE be any side of it, and ON the A perpendicular on AE from the center O. N Then ON is less than the radius of the cir- cle and therefore less than one of the sides about E O the right angle in K. Also the perimeter of the polygon is less than the circumference of the circle, BC i.e. less than the other side about the right angle in K. Therefore, the area of the polygon is less than K; which is inconsistent with the hypothesis. Thus the area of the circle is not greater than K. 7
Proof of Proposition 1 (continued) II. If possible, let the circle be less than K.
Circumscribe a square, T G H and let two adjacent sides, D touching the circle in E, H, meet in T . A Bisect the arcs between F adjacent points of contact and draw the tangents at the points of bisection. Let A be the middle point of the arc EH, E O and FAGthe tangent at A. Then the angle TAGis a right angle. Therefore, TG > GA = GH. BC It follows that the triangle FTG is greater than half the area TEAH bounded by the arc EAH and the two tangents: (A4 − C) − (A4 − A8)=A8 − C; A − A > 1 (A − C) 4 8 2 4 . 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. 8
Proof of Proposition 1 (continued)
Thus, by continuing the process, T G H we shall ultimately arrive at a circum- D scribed polygon such that A the spaces intercepted between it and the F circle are together less than the excess of K over the area of the circle. Euclid X.1 applied to E O A4 − C and K − C =⇒ A2m 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; which is impossible. Therefore the area of the circle is not less than K. This completes the proof of Archimedes’ Proposition 1. 9 Euclid VI.3: Angle bisector theorem If an angle of a triangle is bisected and the straight line cutting the angle cuts the base also, E the segments of the base will have the same ratio of the remaining sides of the triangle; and if the segments of the base have the same ratio as A the remaining sides of the triangle, the straight line joined from the vertex to the point of the section will bisect the angle of the triangle. B D C 10 1 Proposition 3A: Upper bound π<37 Let AB be the diameter of any cir- cle, O its centre, AC the tangent at A; and let the angle AOC be one- C third of a right angle. Then √ D OA : AC= 3:1> 265 : 153, and A OB OC : CA=2:1= 306 : 153. First, draw OD bisecting the angle AOC and meeting AC in D. Now Eucl. VI.3, CO : OA = CD : DA, so that CO + OA = OA : AD. Therefore, OA : AD > 571 : 153. Hence, 1OD2 : AD2= OA2 + AD2 : AD2 > 5712 + 1532 : 1532 > 349150 : 23409, OD : DA > 591 1 : 153 so that 8 . 11 1 Proposition 3A: Upper bound π<37 (continued) Secondly, let OE bisect the angle AOD, meeting AD in E. ... C 1 OE : EA > 1172 : 153. D 8 E F G A Thirdly, OB let OF bisect the angle AOE and meet AE in F . ... 1 OF : FA > 2339 : 153. 4 Fourthly, let OG bisect the angle AOF , meeting AF in G. ... 1 OA : AG > 4673 : 153. 2 12 1 Proposition 3A: Upper bound π<37 (continued) GH is one side of a regular poly- gon of 96 sides circumscribed to the given circle. C And since OA : AG > 4673 1 : 153 2 , while AB =2OA, GH =2AG, it follows that G A H OB AB : perimeter of polygon of 96 sides 1 > 4673 : 153 × 96 2 1 > 4673 : 14688. 2 But, 14688 667 1 1 =3+ 2 < 3 . 4673 1 4673 1 7 2 2 Therefore the circumference of the circle (being less than the perimeter of the polygon) 1 3 1 AB is a fortiori less than 7 times the diameter . 1Archimedes, On the Sphere and Cylinder I, Proposition 1. 13 10 Proposition 3B: lower bound π>371 Let AB be the diameter of a circle, and let AC, meeting the circle in C, C make the angle CAB equal to one - third of a right angle. Join BC. D d Then √ AC : CB = 3:1< 1351 : 780. B OA First, let AD bisect the angle BAC and meet BC in d and the circle in D. Join BD. Then ∠BAD = ∠dAC = ∠dBD, and the angles at D, C are both right angles. It follows that the triangles ADB,[ACd], BDd are similar. Therefore, AD : DB = BD : Dd = AC : Cd = AB : Bd (Euclid V I.3) = AB + AC : Bd + Cd = AB + AC : BC, or BA + AC : BC = AD : DB. But AC : CB < 1351 : 780, from above, while BA : BC = 2 : 1 = 1560 : 780. Therefore, AD : DB < 2911 : 780. Hence, AB2 : BD2 < (29112 + 7802) : 7802 = 9082321 : 608400. 3 Thus, AB : BD < 3013 4 : 780. 14 10 Proposition 3B: lower bound π>371 (continued) Secondly, let AE bisect the angle BAD, meeting the circle in E; and C let BE be joined. Then we prove in the same way as D before, that . . . E F G 9 B AB : BE < 1838 : 240. OA 11 Thirdly, let AF bisect the angle BAE, meeting the circle in F . ... 1 AB : BF < 1009 :66. 6 Fourthly, let the angle BAF be bisected by AG meeting the circle in G. ... 1 BG : AB > 66 : 2017 . 4 BG is a side of a regular inscribed polygon of 96 sides. It follows that 1 1 : AB > 96 × 66 : 2017 > 6336 : 2017 . perimeter of polygon 4 4 6336 > 3 10 And 2017 1 71 . 4 3 10 Much more then is the circumference of the circle greater than 71 times the diam- eter. 15 Reformulation of Archimedes method Let an and bn respectively denote the perimeters of a circumscribing regular 6 · 2n−gon and an inscribed regular 6 · 2n−gon of a circle of diameter 1. Then √ a0 =2 3, b0 =3, and 2anbn an+1 = ,bn+1 = an+1bn. an + bn This was the basis of practically all calculations of π before the 16th century. Correct to the first 34 decimal places, π =3.1415926535897932384626433832795028 ······ 1 Quadrature of the parabola in Archimedes’ Method Method, Proposition 1 Let ABC be a segment of a parabola bounded by the straight line AC and the parabola ABC, and let D be the middle point of AC. Draw the straight line DBE (DB) parallel to the axis of the parabola and join AB, BC. ABC 4 ABC Then shall the segment be 3 of the triangle . B A D C 2 Method, Proposition 1 (continued) H F K E B A D C From A draw AKF parallel to DE (DB), and let the tangent to the parabola at C meet DBE in E and AKF in F . Produce CB to meet AF in K, and again produce CK to H, making KH equal to CK. Consider CH as the bar of a balance, K being its middle point. 3 Method, Proposition 1 (continued) H F K E B A D C Now, since CE is a tangent to the parabola and CD the semi-ordinate, EB = BD; “for this is proved in the Elements [of Conics].” 2 See also QP, Proposition 2. 2Heath’s footnote: i.e. the works on conics by Aristaeus and Euclid. 4 Archimedes QP: Proposition 1 Let P be a point on a parabola, and QQ a chord parallel to the tangent at Q P . (i) The line through P parallel to the axis of the parabola bisects QQ . (ii) Conversely, if the line through P paral- P V lel to the axis bisects QQ , then the chord is parallel to the tangent at P . Q 5 Archimedes QP: Proposition 2 If in a parabola QQ be a chord parallel to the tangent at P , and if a straight line be drawn through P which is either itself the axis or parallel to the axis, and which meets QQ in V and the tangent at Q to the parabola in T , then PV = PT. Q V T P Q 6 Archimedes QP: Proposition 5 Let Qq be the base of any segment of a parabola. If the diameter of the parabola through any point R meets Qq in O and the tangent at Q in E, then QO : Oq = ER : RO. Q O E R q 7 Method, Proposition 1 (continued) H F M K E N P B A O D C Let MO be any straight line parallel to ED, and let it meet CF, CK, AC in M, N, O, and the curve in P . Since FA, MO are parallel to ED, it follows that FK = KA and MN = NO. Now, by the property of the parabola, “proved in a lemma,” (By QP, Proposition 5, MP : PO = CO : OA.) MO : OP = CA : AO = CK : KN = HK : KN. 8 Method, Proposition 1 (continued) H F M K E N P B A O D C MO : OP = HK : KN. Take a straight line TGequal to OP, and place it with its center of gravity at H, so that TH = HG; then, since N is the center of gravity of the straight line MO, and MO : TG= MO : OP = HK : KN, it follows that TGat H and MO at N will be in equilibrium about K.[On the Equilibrium of Planes, I. 6,7] 9 Method, Proposition 1 (continued) H F M K E N P B A O D C Similarly, for all other straight lines parallel to DE and meeting the arc of the parabola, (1) the portion intercepted between FC, AC with its middle point on KC and (2) a length equal to the intercept between the curve and AC placed with its center of gravity at H will be in equilibrium about K. 10 Method, Proposition 1 (continued) H F M K E N P B A O D C Therefore K is the center of gravity of the whole system consisting (1) of all the straight lines as MO intercepted between FC, AC and placed as they actually are in the figure, and (2) of all the straight lines placed at H equal to the straight lines as PO intercepted between the curve and AC. 11 Method, Proposition 1 (continued) H F M K E N P B A O D C And, since the triangle CFA is made up of all the parallel lines like MO, and the segment CBA is made up of all the straight lines like POwithin the curve, it follows that the triangle, placed where it is in the figure, is in equilibrium about K with the segment CBA placed with its center of gravity at H. 12 Method, Proposition 1 (continued) H F K E W B A D C Divide KC at W so that CK =3WK; then W is the center of gravity of the triangle ACF , “for this is proved in the books on equilibrium”. (I. 15). Therefore, ACF : segment ABC = HK : KW =3:1. ABC = 1 ACF Therefore, segment 3 . But, ACF =2ACK =4ABC. ABC = 4 ABC Therefore, segment 3 . 13 Archimedes’ postscript Now the fact here stated is not actually demonstrated by the argument used; but that argument has given a sort of indication that the conclusion is true. Seeing then that the theorem is not demonstrated, but at the same time suspecting that the conclusion is true, we shall have recourse to the geometrical demonstration which I myself discovered and have already published. 14 Chapter 25: Archimedes’ quadrature of the parabola Propositions 1-5 elementary propositions in conics 6-17 investigation by mechanics 18-21 elementary propositions in conics 22-24 proof by method of exhaustion 15 Archimedes QP: Proposition 1 If from a point on a parabola a straight line be drawn Q which is either itself the axis or parallel to the axis, as PV, and if QQ be a chord parallel to P V the tangent to the parabola at P and meeting PV in V , then QV = VQ . Q Conversely, if QV = VQ , the chord QQ will be parallel to the tan- gent at P . 16 Archimedes QP: Proposition 2 If in a parabola QQ be a chord parallel to the tangent at P , Q and if a straight line be drawn through P which is either itself the axis or parallel to the axis, and which meets QQ in V and the V T P tangent at Q to the parabola in T , then PV = PT. Q 17 Tangent of a parabola A parabola has equation y2 =4ax. A typical point on the parabola is P (at2, 2at) P =(at2, 2at). The tangent at P has slope dy y 2a 1 = = = . dx x 2at t Equation of tangent: y − 2at = 1 (x − at2) t , or x − ty + at2 =0. 18 Verifying QP 1 2 2 Let Q =(at1, 2at1) and Q =(at2, 2at2) be two points on the parabola. Q a(t2+t2) V = 1 2 ,a(t + t ) Their midpoint is 2 1 2 . The line joining them has slope 2at − 2at 2 1 2 = . P V 2 2 at1 − at2 t1 + t2 Therefore, the tangent at P =(at2, 2at) is QQ parallel to 1 = 2 Q if and only if t t1+t2 , i.e., t + t t = 1 2 . 2 This is the case when PV is parallel to the axis of the parabola. 19 Verifying QP 2 If in a parabola QQ be a chord parallel to the tangent at P , Q and if a straight line be drawn through P which is either itself the axis or parallel to the axis, and which meets QQ in V and the V T P tangent at Q to the parabola in T , then PV = PT. 2 The tangent at Q is the line x − t1y + at1 =0. Q It intersects the line PV at the point T =(at1t2,a(t1 + t2)). at2+at2 at t + 1 2 2 1 2 2 = a t1+t2 Since 2 2 , P is the midpoint of TV. Note: the tangent at Q also passes through the same point T . 20 Archimedes QP: Proposition 3 Q If from a point on a parabola a straight line be drawn which is ei- Q ther itself the axis or parallel to the axis, as PV, and if from two other points Q, Q on the parabola straight lines be drawn parallel to P V the tangent at P and meeting PV V in V , V respectively, then PV : PV = QV 2 : Q V 2. P =(at2, 2at) 1 Proof. If , its tangent has slope t . The line through Q =(as2, 2as) parallel to the tangent at P has equation y−2as = 1 (x − as2) y =2at V =(a(t2 +(s − t)2), 2at) t . It intersects the line at the point . Note that QV 2 =[a((s − t)2]2 +(2a(s − t))2 =4a2(s − t)2(1 + t2). QV 2 = 4a2(s−t)2(1+t2) =4a(1 + t2) s Therefore, PV a(s−t)2 , which is independent of . Therefore, for another point Q with corresponding V on PV, we have QV 2 : QV 2 = PV : PV . 21 Archimedes QP: Proposition 4 If Qq be the base of any segment of a parabola, and P the vertex of the segment, and if the diameter through any other point R meets Qq in O and QP (produced if necessary) in F , then QV : VO= OF : FR. Q Q K R O F K P V P V W W F O R q q Draw the ordinate RW to PV, meeting QP in K. Then, PV : PW = QV 2 : RW 2 (QP 3), whence, by parallels, PQ : PK= PV : PW = QV 2 : RW 2 = QV 2 : OV 2 = PQ2 : PF2. In other words, PQ, PF, PK are in continued proportion; 22 Archimedes QP: Proposition 4 (continued) Q Q K R O F K P V P V W W F O R q q therefore, PQ : PF = PF : PK = PQ± PF : PF ± PK = QF : KF. Hence, by parallels, QV : VO= QP : PF = QF : KF = OF : FR. 23 Archimedes QP: Proposition 5 If Qq be the base of any segment of a parabola, P the vertex of the segment, and Q PV its diameter, and if the diameter of the parabola through any other point R meet Qq in O and the tangent at Q in E, then T V P QO : Oq = ER : RO. O E F R Proof. Let the diameter through R meet QP in F . q Then by Proposition 4, QV : VO= OF : FR. Since QV = Vq, it follows that QV : qO = QV : QV − VO= OF : OF − FR = OF : OR. (1) Also, if VP meet the tangent in T , by QP 2, PT = PV, and therefore EF = OF. Accordingly, doubling the antecedents in (1), we have Qq : Oq = OE : OR, whence, QO : Oq = Qq − Oq : qO = OE − OR : OR = ER : RO.