Introduction to Viete’s Apollonius Gallus

Apollonius Problem

• 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.

• 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.)

• 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.

Point Line Circle0 0 3 CCC

0 1 2 CCL

0 2 1 LLC

0 3 0 LLL

1 0 2 CCP

1 1 1 CLP

1 2 0 LLP

2 0 1 CPP

2 1 0 LPP

3 0 0 PPP

• 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)

• 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).

As an appendix of the book Apollonius Redivivus by Getaldi

Mechanical drawings

Doubling a cube

Squaring a circle

Geometric construction

10 problems by Pappus

V. LLC

X. CCC

VI. CLP

VII. CCL

III. LLL

IX. CCP

II. LPP

I. PPP

find more points

IV. LLP VIII. CPP

Elements IV.4 Elements IV.5

by scaling/resizing

Power of a Point Theorem• If EZ is a tangent to the circle ABC• ECA EBC

EC/EA = EB/EC EAEB = EC2, which is a constant

• Corollary: if another secant is drawn from E and intersect at A’, B’, then

EAEB = EA’EB’ • Viete’s lemma: proof by contradiction

If EAEB = EC2, then EC is tangent to the circle ABC.

Elements III.36, Elements III.37 = converse thm

simplified proof

II. PPL• 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

EAEB = EC2

• Construct circle ABC• It remains to prove that the circle

touches • By the lemma previously proved,

the circle ABC touches at C

one more point

IV. LLP• 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 = HLHA• Construct the circle ALM

Analysis

two more points

IV. LLP

Claim: The circle touches DE at Oi.e. to prove NM = NONK=NK, NKI = 90, HK=KI NKH NKI NHK = NIK, NH = NI NHM = KHM – NHK

= KIO – NIK = NIO

Also, NMH = NOI = 90NMH NOI MN = NO

Proof

• Construct angle bisector • Construct H, L, K, I ()• Construct M by HM2 = HLHA• Construct the circle ALM• Construct the center N• Fall perp. to O from N

(Lemma)

cannot use symmetrical property; AOM; …

B

CM H

N

D O I E

K

A

L

VIII. CPP • 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• BDBH = BFBG , which is constant• H can be found • Construct H (BDBH = AB2 – AK2)• Construct F (HF is tangent to A)• Construct G (produce BF)• Construct circle DBG

Analysis

H

K

Dealing with ratios, not geometry

one more point

VIII. CPP • Claim: GDB GEF• i.e. to prove GEF = GDB • BHBD = BK2 = BFBG

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 thatBHBD = AB2 AF2 (= BK2)

• Construct F (tangent HF)• Construct G (produce BF)• Join DG and obtain E

Proof

(Lemma)

H

KA

D

B

FE

G

• 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)

Timeline• BC 260-BC 190 Apollonius On Tangencies

• 290-350 Pappus Mathematical Collection

• 1600 Viete Apollonius Gallus

• 1811 Poncelet Solutions de plusieurs problemes de geometrie et de mecanique

• 1816 Gergonne Recherche du cercle qui en touche trois autres sur une sphere

• 1879 Petersen Methods and Theories for the Solution of Problems of Geometrical Constructions

• 2001 Eppstein Tangent Spheres and Tangent Centers

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