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Introduction to Viete’s Apollonius Gallus

Introduction to Viete’s Apollonius Gallus. Apollonius Problem.

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Introduction to Viete’s Apollonius Gallus

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  1. Introduction to Viete’s Apollonius Gallus Apollonius Problem

  2. During the 16th and 17th centuries interest in ancient works on mathematics increased and several mathematicians tried to restore and reconstruct works that had been lost, drawing upon the quotations and references in other works of ancient mathematicians.

  3. The works of Apollonius of Perga were partly unknown or lost. • The preface to the seventh book of Mathematical Collection by Pappus outlined the contents of Apollonius’s treaties On Tangencies and On Inclinations. Apollonius of Perga (262-190 B.C.)

  4. Apollonius Problem Given 3 things, each of which may be either a point, a straight line or a circle, to draw a circle which shall pass through each of the given points (so far as it is points that are given) and touch the straight lines or circles.

  5. There are 10 possible different combinations of elements, and Apollonius dealt with all 8 that had not already been treated by Euclid. • The particular case of drawing a circle to touch three given circles attracted the interest of Viete and Newton. Francois Vieta (French mathematician, 1540-1603)

  6. Adriaan van Romanus (Belgian mathematician, 1561-1615) gave a solution by means of hyperbola. • Vieta thereupon proposed a simpler construction (by means of only compass and ruler), and restored the whole treatise of Apollonius in a small work, which he entitled Apollonius Gallus (Paris, 1600).

  7. As an appendix of the book Apollonius Redivivus by Getaldi

  8. Geometric construction Mechanical drawings Doubling a cube Squaring a circle

  9. 10 problems by Pappus

  10. find more points X. CCC VII. CCL by scaling/ resizing   IX. CCP VI. CLP V. LLC  II. LPP IV. LLP VIII. CPP III. LLL I. PPP Elements IV.4 Elements IV.5

  11. Power of a Point Theorem Elements III.36, Elements III.37 = converse thm • If EZ is a tangent to the circle ABC • ECA  EBC  EC/EA = EB/EC  EAEB = EC2, which is a constant • Corollary: if another secant is drawn from E and intersect at A’, B’, then EAEB = EA’EB’ • Viete’s lemma: proof by contradiction If EAEB = EC2, then EC is tangent to the circle ABC. simplified proof

  12. II. PPL one more point • Given two points A, B and a line  • (Case 2) If AB is not parallel to  • Produce AB to  at E • Construct the point C on  by EAEB = EC2 • Construct circle ABC • It remains to prove that the circle touches  • By the lemma previously proved, the circle ABC touches  at C 

  13. IV. LLP Analysis two more points • Given a point A, two lines BC, DE • Assume there is a circle touching BC, DE at M, O respectively • The center lies on the angle bisector by symmetry • Construct angle bisector  • Construct H, L, K, I () • Construct M by HM2 = HLHA • Construct the circle ALM 

  14. IV. LLP B M H C L N  K A E I O D Proof cannot use symmetrical property; AOM; … • Construct angle bisector  • Construct H, L, K, I () • Construct M by HM2 = HLHA • Construct the circle ALM • Construct the center N • Fall perp. to O from N (Lemma) Claim: The circle touches DE at O i.e. to prove NM = NO NK=NK, NKI = 90, HK=KI  NKH  NKI  NHK = NIK, NH = NI  NHM = KHM – NHK = KIO – NIK = NIO Also, NMH = NOI = 90 NMH  NOI  MN = NO

  15. K H VIII. CPP one more point Analysis • Given a circle A and two points B, D • Assume there is a circle touching circle A at G • Join GB, GD; draw the tangent HF • GEF  GDB • EGF = EFH = FHB • D, H, F, G concyclic • BDBH = BFBG , which is constant • H can be found • Construct H (BDBH = AB2 – AK2) • Construct F (HF is tangent to A) • Construct G (produce BF) • Construct circle DBG Dealing with ratios, not geometry

  16. G K E A F D H B VIII. CPP Proof • Claim: GDB  GEF • i.e. to prove GEF = GDB • BHBD = BK2 = BFBG  D, H, F, G concyclic  GDB = HFB • HF is a tangent  EGF = EFH • GEF = 180 – EGF – GFE = 180 – EFH – GFE = HFB = GDB • GDB  GEF with parallel bases and having the same vertex G • Their circumscribed circles are mutually tangent to each other • Construct H on BD such that BHBD = AB2  AF2 (= BK2) • Construct F (tangent HF) • Construct G (produce BF) • Join DG and obtain E (Lemma)

  17. Analysis – opposed to synthesis • (Greek) the reversed solution • Viete – The Analytic Art (1591) • Zetetics (problem translated to symbols/equations) • Poristics (examination of theorems) • Exegetics (solution by derivations)

  18. Timeline • BC 260-BC 190 ApolloniusOn Tangencies • 290-350 PappusMathematical Collection • 1600 VieteApollonius Gallus • 1811 PonceletSolutions de plusieurs problemes de geometrie et de mecanique • 1816 GergonneRecherche du cercle qui en touche trois autres sur une sphere • 1879 PetersenMethods and Theories for the Solution of Problems of Geometrical Constructions • 2001 EppsteinTangent Spheres and Tangent Centers

  19. Reference • Francois Viete, Apollonius Gallus, Paris, 1600 • Thomas Heath, A History of Greek Mathematics Vol.II, Oxford, 1921 • Ronald Calinger, Vita Mathematica: Historical Research and Integration with Teaching, The Mathematical Association of America, 1997 • Euclid Elements • Google Books • Wikipedia

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