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
Apollonius of Perga (262-190 B.C.)
Inclinations.
• 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
Circle
0
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
Geometric
construction
Mechanical
drawings
Doubling a
cube
Squaring a
circle
10 problems
by Pappus
find more
points
X. CCC
VII. CCL


V. LLC
by scaling/
resizing
IX. CCP
VI. CLP

IV. LLP
II. LPP
III. LLL
I. PPP
Elements IV.4
Elements IV.5
VIII. CPP
Power of a Point Theorem
Elements III.36,
Elements III.37
= converse thm
• If EZ is a tangent to the circle ABC
• ECA  EBC
simplified
proof
 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.
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
•
•
•
•
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
B
M
H
L
•
•
•
•
•
•
IV. LLP
C
Proof
cannot use symmetrical property; AOM; …
Claim: The circle touches DE at O
N
K

i.e. to prove NM = NO
NK=NK, NKI = 90, HK=KI
A
 NKH  NKI
E
I
O
 NHK = NIK, NH = NI
D
 NHM = KHM – NHK
Construct angle bisector 
= KIO – NIK
(Lemma)
Construct H, L, K, I ()
= NIO
2
Construct M by HM = HLHA
Also, NMH = NOI = 90
Construct the circle ALM
NMH  NOI  MN = NO
Construct the center N
Fall perp. to O from N
one more point
VIII. CPP
Analysis
K
H
Dealing with ratios,
not geometry
• 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
VIII. CPP
Proof
G
E
A
K
F
D
H
B
• Construct H on BD such that
BHBD = AB2  AF2 (= BK2)
• Construct F (tangent HF)
• Construct G (produce BF)
• Join DG and obtain E
• 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
(Lemma)
•
•
•
•
•
•
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