Triangle Centers and Kiepert?s Hyperbola
Except where reference is made to the work of others, the work described in this thesis is
my own or was done in collaboration with my advisory committee. This thesis does not
include proprietary or classifled information.
Charla Baker
Certiflcate of Approval:
Krystyna Kuperberg
Associate Professor
Mathematics and Statistics
Andras Bezdek, Chair
Associate Professor
Mathematics and Statistics
Pat Goeters
Associate Professor
Mathematics and Statistics
Joe F. Pittman
Interim Dean
Graduate School
Triangle Centers and Kiepert?s Hyperbola
Charla Baker
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
December 15, 2006
Triangle Centers and Kiepert?s Hyperbola
Charla Baker
Permission is granted to Auburn University to make copies of this thesis at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Charla Nicole Baker was born on December 30, 1977 in Louisville Alabama to Charles
and Ruseda Baker. She graduated from George W. Long High School and attended George
C. Wallace Community College. She entered Troy State University where she pursued
Bachelor of Science degrees in both Computer Science and Mathematics. After graduating
from Troy State University she entered graduate school at Auburn University to acquire a
Master of Science degree in Mathematics.
iv
Thesis Abstract
Triangle Centers and Kiepert?s Hyperbola
Charla Baker
Master of Science, December 15, 2006
(B.S., Troy State University, Mathematics, 2001)
(B.S., Troy State University, Computer Science, 2001)
59 Typed Pages
Directed by Andras Bezdek
In this paper, we discuss the proofs of the primary classical triangle centers and
Kiepert?s Hyperbola as a solution to Lemoine?s Problem. The deflnitions of terms which
will be used throughout the paper are presented. A brief description of well-known triangle
centers as well as complete proofs of the remaining classical triangle centers is provided.
Many of the proofs of the classical triangle centers require the use of Ceva?s Theorem.
Ceva?s Theorem is proven in the beginning prior to the introduction of the triangle cen-
ters. We also explore the proof of Kiepert?s Hyperbola as a solution to a problem posed by
Lemoine in 1868. A proof of the Nine-Point Circle is provided since the center of Kiepert?s
Hyperbola lies on the Nine-Point Circle. The trilinear coordinate system provides the basis
for the proof of Kiepert?s Hyperbola. A brief description of the system and the proofs of
its primary theorems are given. The proof of Kiepert?s Hyperbola is given along with its
properties.
v
Acknowledgments
I would like to thank my family for standing by me through the good times and the bad
and always believing in me. I would also like to thank my friends for the endless encour-
agement they have given me. Thank you to all of my professors at Auburn University. You
have given me invaluable knowledge which has allowed me to pursue this degree. Finally, I
would like to thank Dr. Bezdek for his patience, support, and knowledge over the years.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as \aums"). Bibliograpy follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (speciflcally LATEX)
together with the departmental style-flle aums.sty.
vii
Table of Contents
List of Figures ix
1 Introduction 1
2 Terms and their Definitions 3
3 Triangle Centers 6
3.1 Ceva?s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3.2 Review of Centroid, Orthocenter, Circumcenter, and Incenter . . . . . . . . 8
3.3 Gergonne Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.4 Nagel Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.5 Symmedian or Lemoine Point . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.6 Mittenpunkt or Middles-point . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.7 Spieker Center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.8 Steiner Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.9 Napoleon Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.10 Fermat Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4 Kiepert?s Hyperbola 24
4.1 Nine-Point Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 Simson Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.3 Trilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.4 Kiepert?s Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Bibliography 49
viii
List of Figures
Figure 1: Ceva?s Theorem
Figure 2: Orthocenter
Figure 3: Gergonne Point
Figure 4: Radii and Angle Bisectors
Figure 5: Nagel Point
Figure 6: Symmedian Point
Figure 7: Symmedian Line
Figure 8: Mittenpunkt
Figure 9: Excircles and Excenters
Figure 10: Symmedian Line
Figure 11: Spieker Center
Figure 12: Steiner Point
Figure 13: K on the circle
Figure 14: Napoleon Point
Figure 15: Fermat Point
Figure 16: Midpoint Lemma
Figure 17: Circumcircle, Medians, Feet of Altitudes, and Euler Points
Figure 18: H, I, G, K, L, and M are on the same circle
Figure 19: E, F, and D lie on the circle
Figure 20: Nine-Point Circle
ix
Figure 21: Simson Line
Figure 22: Trilinear Coordinates
Figure 23: Function  (t)
Figure 24: Centroid
Figure 25: Orthocenter
Figure 26: Trilinear Coordinates
Figure 27: Concurrence
Figure 28: Circumcircle
Figure 29: Kiepert?s Hyperbola
Figure 30: Concurrence at A
x
Chapter 1
Introduction
The fleld of Geometry is subdivided into many areas such as Euclidean, Non-Euclidean,
Convex, Discrete, Hyperbolic, and Algebraic. This paper explores the subject of triangle
centers as studied in Euclidean Geometry. We consider a triangle center in a general since
meaning that a triangle center is the point of concurrence of three "special" lines of a refer-
ence triangle. For many centuries, mathematicians have been discovering triangle centers.
The most well-known centers, the incenter, the centroid, the circumcenter, and the ortho-
center were discovered by the ancient Greeks thus classifying them as "classical". However,
the height of the study of triangle centers occurred during the 1800?s, where mathematicians
such as Fermat and Lemoine continued the research on triangle centers. The subject was
revisited in the early 1990?s. The study of triangle centers provides a solid foundation for
the concept of triangles and a great teaching tool for Euclidean Geometry.
In this paper, thirteen triangle centers are examined. In Chapter 2, terms and def-
initions which will be used throughout the paper are presented. Chapter 3 provides de-
scriptions and proofs for the triangle centers. A review of the properties and proofs for the
incenter, centroid, circumcenter, and orthocenter are given. Ceva?s Theorem plays an im-
portant role in many of the existence proofs of the triangle centers. The theorem is proven
and provides a framework for the existence proofs of the centroid, Nagel Point, Gergonne
Point, and the Symmedian or Lemoine Point. The Mittenpunkt or Middles-point is pre-
sented following the Symmedian Point along with the Spieker Center and Steiner Point.
The Napoleon and Fermat Points are a result from the proof of Kiepert?s Hyperbola.
1
In 1868, Lemoine posed a problem concerning the coordinates of a given triangle if the
coordinates of the peaks of equilateral triangles raised on the sides are known. In 1869, Lud-
wig Kiepert presented a solution to the problem known as Kiepert?s Hyperbola. Kiepert?s
Hyperbola passes through several of the triangle centers presented in Chapter 3. In Chapter
4, Kiepert?s Hyperbola is examined. The chapter begins with the proof of the Nine-Point
or Feuerbach Circle since the center of Kiepert?s Hyperbola lies on the Nine-Point Circle.
The asymptotes of Kiepert?s Hyperbola are Simson Lines. Therefore, the existence proof
of Simson Lines is provided. The proof for Kiepert?s Hyperbola involves the use of trilinear
coordinates. The coordinate system is introduced in Chapter 4 and many theorems regard-
ing the system are proven. Finally, the proof of Kiepert?s Hyperbola is given along with it?s
relationship to many of the triangle centers. Kiepert?s Hyperbola not only passes through
the vertices of the given triangle, but also passes through the centroid, orthocenter, Spieker
Center, Fermat Point, and Napoleon Point[2].
2
Chapter 2
Terms and their Definitions
Altitude: The line passing through a vertex of a given triangle and perpendicular to
the opposite side.
Angle Bisector: The line which passes bisects a vertex of a given triangle.
Brocard Circle: The circle with the symmedian point and the circumcenter as its diam-
eter for a given triangle.
First Brocard Triangle: The triangle constructed by connecting the points on the
Brocard Circle where the perpendiculars through the circumcenter meet the Bro-
card Circle.
Centroid: The point of concurrence of the medians of a given triangle.
Circumcenter: The point of concurrence of the perpendicular bisectors of a given triangle.
Cyclic Quadrilateral: A quadrilateral in which all vertices lie on the same circle.
Euler Line: The line passing through the centroid, circumcenter, and orthocenter of a
given triangle.
Euler Points: The midpoints of the segments connecting the orthocenter to the vertices
of a given triangle.
Euler Triangle: The triangle connecting the three Euler Points.
Fermat Point: The point of concurrence of three lines, each passing through a vertex and
the peak of an equilateral triangle raised on the opposite side of a given triangle.
Feueurbach Circle: See Nine-Point Circle.
Gergonne Point: The point of concurrence of three lines, each passing through a vertex
3
and the point of tangency of the opposite side and the incircle of a given triangle.
Incenter: The point of concurrence of the angle bisectors of a given triangle.
Kiepert?s Hyperbola: The rectangular hyperbola formed by the point of concurrence
of three lines, each connecting a vertex of a given triangle and the peak of a similar
isosceles triangle raised on the opposite side, as the base angle varies.
Lemoine Point: See Symmedian Point.
Medial Triangle: The triangle whose vertices are the midpoints for a given triangle.
Median: The line passing through the vertex and the midpoint of the opposite side of
a given triangle.
Mittenpunkt or Middles-point: The point of concurrence of three lines, each passing
through the center of the excircle for a given side and the midpoint of that side for a
given triangle.
Nagel Point: The point of concurrence of three lines, each passing through a vertex
and the point of tangency of the opposite side and the opposite excircle of a given
triangle.
Napoleon Point: The point of concurrence of three lines, each passing through a ver-
tex and the centroid of an equilateral triangle raised on the opposite side of a given
triangle.
Nine-Point or Feuerbach Circle: The circle on which the medians, the feet of the
altitudes, and the Euler points of a given triangle lie.
Orthocenter: The point of concurrence of the altitudes of a given triangle.
Perpendicular Bisector: A line which bisects a side of a given triangle and is perpen-
dicular to that side.
4
Simson Line: A line which passes through the three feet of the altitudes from any point
on the circumcircle.
Spieker Center: The point of concurrence of the angle bisectors of the medial triangle
of a given triangle.
Steiner Point: The point of concurrence of three lines, each passing through a vertex
which parallel to the opposite side of the First Brocard Triangle for a given triangle.
Symmedian Line: For a vertex of a given triangle it is the re ection of the medial line
at that vertex across the angle bisector at that vertex.
Symmedian or Lemoine Point: The point of concurrence of the three symmedian
lines of a triangle.
Trilinear coordinates: An set of ordered triples, fi : fl :  , which are proportional to the
signed distances from a point to the sidelines of a given triangle.
Notation
Triangle ABC will be denoted as MABC.
Given MABC, a, b, and c will denote the sides BC, CA, and AB respectively.
\ABC will denote the angle with point B as its vertex as well as the measure of that angle.
The line, segment, and segment length between two points A and B will be denoted as AB.
The symbol "??!AB" will be used for vector AB.
The symbol "?=" will be used for congruence.
The symbol"s" will be used for similarity.
An altitude is understood to be the line passing through a vertex of a triangle which is
perpendicular to the opposite side.
5
Chapter 3
Triangle Centers
In this chapter, we focus on the classical triangle centers. However, a list of both
classical and recent triangle centers can be found on Clark Kimberling?s website. The
website also includes an encyclopedia of all triangle centers. Kimberling also provides more
information on the centers presented in Chapter 3 including their trilinear coordinates and
properties [3]. Before proving the existence of several triangle centers, a theorem must be
introduced. Ceva?s Theorem is useful in determining whether three lines are concurrent.
Many of the proofs in this chapter utilize Ceva?s Theorem.
3.1 Ceva?s Theorem
Theorem 3.1 (Ceva?s Theorem) If D, E, and F are points on the sides c, a, and b respec-
tively of a given triangle, MABC, then the lines AE, BF, and CD concur at a point K if
and only if ADDB ? BEEC ? CFFA = 1.
Proof.
A
F
E
D
C
BK
Figure 1: Ceva's Theorem
6
First, assume that the segments AE, BF, and CD concur at a point K.(Figure 1) Triangles
CKE and BKE share altitude hK. Therefore
area(MCKE) = CE ?hk2 and area(MBKE) = EB ?hk2 .
Therefore
CE
EB =
area(M CKE)
area(M BKE).
In the same manner,
CE
EB =
area(M BAE)
area(M CAE).
Since MAKB = MBAE - MBKE and MAKC = MCAE - MCKE then
CE
EB =
area(M CKA)
area(M AKB).
Likewise,
AF
FC =
area(M AKB)
area(M CKB) and
BD
DA =
area(M CKB)
area(M CKA).
Therefore,
CE
EB ?
AF
FC ?
BD
DA =
area(M CKA)
area(M BKA)?
area(M AKB)
area(M CKB)?
area(M BKC)
area(M AKC) = 1.
Now, assume that for MABC with any points D, E, and F on it?s sides,
AD
DB ?
BE
EC ?
CF
FA = 1.
Let the lines AE and BF intersect at a point K. Then there is some point, X, on side AB
such that CX passes through K.
Then
AX
XB ?
BE
EC ?
CF
FA = 1.
7
Therefore, by setting the two equations equal to each other,
AD
DB =
AX
XB.
So D = X. Thus, the lines AE, BF, and CD are concurrent at a point. ?
3.2 Review of Centroid, Orthocenter, Circumcenter, and Incenter
The centroid, orthocenter, circumcenter, and incenter are perhaps the most well known
triangle centers. Euler found that the incenter, circumcenter, and orthocenter all lie on the
same line known as the Euler Line. This section provides a brief review of their proofs and
properties.
Centroid: The centroid is the point of concurrence of the medians of a given triangle.
The centroid is the center of mass for the given triangle. The existence of the centroid is
easily proven using Ceva?s Theorem.
Orthocenter: The orthocenter of a given triangle is the point of concurrence of the altitudes.
We will provide the proof which utilizes simple vector calculus.
Theorem 3.2 Given MABC, the three altitudes of the triangle concur at a point called the
orthocenter.
Proof.
8
O
P
M
A
BC
bc
a
Figure 2: Orthocenter
Given MABC let vectors a, b, and c be vectors from the circumcenter O to the vertices A,
B, and C respectively.(Figure 2) Let the circumcenter be O. Now Let OP be the vector from
the circumcenter O to the centroid P. Then
P = a+b+c3 .
Let ??!OM be the vector with same direction which is three times the length of ??!OP. Then
??!OM = a + b + c.
Now ??!AB = b - a and
??!MC = c - (a + b + c) = -a - b.
Therefore,
??!AB ? ??!MC = (b - a) ? (-a - b) = -b2 + a2 = 0
9
since b = a. Therefore, ??!MC is perpendicular to ??!AB. In the same manner, ??!MA is perpen-
dicular to ??!CB and ??!MB is perpendicular to ?!AC. Thus, M is the point of concurrence of the
altitudes of triangle ABC. ?
Circumcenter: The circumcenter of a given triangle is the point of concurrence of the
perpendicular bisectors of the triangle. The existence proof for the circumcenter follows
from the fact that the point lies equal distance from each vertex.
Incenter: The incenter of a given triangle is the point of concurrence of the angle bisectors
of the triangle. The existence proof shows that the incenter is equal distance from the
sidelines of the triangle.
3.3 Gergonne Point
Theorem 3.3 (Gergonne?s Theorem) Given MABC, let the incircle have center I and be
tangent to the sides c, a, and b at the points D, E, and F respectively. Then the lines AE,
BF, and CD concur at a point K called the Gergonne Point.(Figure 3)
I
E
D
F
C B
A
K
Figure 3: Gergonne Point
10
Proof.
I
E
D
F
C B
A
Figure 4: Radii and Angle Bisectors
I lies on the angle bisectors of the triangle and the radii of the incircle, ID, IE, and
IF are perpendicular to the sides AB, BC, and CA.(Figure 4) Therefore, three pairs of
congruent triangles are formed such that BE = BD, CF = CE, and AD = AF then
AD
DB ?
BE
EC ?
CF
FA = 1.
By Ceva?s Theorem the lines AE, BF, and CD are concurrent.
?
3.4 Nagel Point
Theorem 3.4 (Nagel?s Theorem) Given MABC let E, I, and L be the points on the sides
c, a, and b respectively which are tangent to the side?s excircles. Then the lines AI, BL,
and CE concur at a point called the Nagel Point.(Figure 5)
Proof.
11
C
J
K
M
D
L
I
E
B
A
H
F
Figure 5: Nagel Point
Since BM = BK, AM = AL, and CL = CK then BK = s where s is the semiperimeter
of MABC. Likewise CF = s. Therefore,
CL = EB = s - a.
Also
AE = CI = s - b and AL = IB = s - c.
Therefore,
12
AE
EB ?
IB
CI ?
CL
AL =
s?b
s?a ?
s?c
s?b ?
s?a
s?c = 1.
So by Ceva?s Theorem the lines AI, BL, and CE concur. ?
3.5 Symmedian or Lemoine Point
Another application of Ceva?s Theorem is a proof involving the symmedian lines of a
triangle. A symmedian line of a given triangle at a vertex is the line re ection of the median
of that vertex across the angle bisector. The three symmedian lines of a given triangle are
concurrent by Ceva?s Theorem.
Theorem 3.5 Given MABC, the symmedian lines AS, BS?, and CS" are concurrent at a
point called the Symmedian or Lemoine Point.(Figure 6)
A
C S
S'
B
S"
Figure 6: Symmedian Point
Proof.
A
C S BM
Figure 7: Symmedian Line
13
For MABC let the median from vertex A intersect BC at the point M. Let the sym-
median line from vertex A intersect BC at the point S.(Figure 7) Then by computing the
areas of triangles CAM and BAS,
area(M BAS)
area(M CAM) =
BS
CM
since triangles BAS and CAM share a common height h.
Then using an alternative area formula,
2?area(BAS) = AB?AS sin(SAB)
and
2?area(CAM) = AM?AC sin(CAM).
Since \SAB = \CAM then,
area(M BAS)
area(M CAM) =
AB ?AS
AM ?AC =
BS
CM.
For triangles ASC and AMB,
area(M ASC)
area(M AMB) =
AC ?AS
AM ?AB =
CS
BM.
Since CM = BM, dividing the two equations gives
BS
CM
CS
BM
= BSCS = AB
2
AC2.
For the other vertices,
CS0
AS0 =
BC2
AB2 and
AS"
BS" =
AC2
BC2
14
with S? and S" being the point of intersection of the symmedian lines from vertex B and
vertex C respectively.
Now multiplying the three equalities yields
BS
CS ?
CS0
AS0 ?
AS"
BS" =
AB2
AC2 ?
BC2
AB2 ?
AC2
BC2 = 1.
By Ceva?s Theorem the symmedian lines AS, BS?, and CS" are concurrent.
?
3.6 Mittenpunkt or Middles-point
Theorem 3.6 (Mittenpunkt or Middles-point) Given MABC with Oa, Ob, and Oc as the
excenters and A?, B?, and C? as the midpoints of sides a, b, and c respectively, the lines
OaA0, ObB0, and OcC0 are concurrent at a point called the Mittenpunkt.(Figure 8)
15
A
Oc
Oa
Ob
BC
L
M
N
   Figure 8: Mittenpunkt
S
R
Proof. Let L and M be points on the excircle with center Ob tangent to BA and BC
respectively. Let N be the point on the excircle with center Oc tangent to CA. (Figure 9)
The lines ObA and OcA bisect the external angles of vertex A.
16
A
C'B'
Oc
Oa
Ob
BC
A'
L
M
N
S
R
Figure 9: Excircles and Excenters
Then
\ObAC = \LAOb = fi.
In the same manner,
\BCOa = \OaCS =  
where S is the point on tangency of the line AC with the excircle with center Oa.
Also
\CBOa = \OaBR = fl
where R is the point of tangency of the line AB to the excircle with center Oa.
So
17
\CAB = ? - 2fi, \ABC = ? - 2fl, and \BCA = ? - 2 .
Therefore,
fi = ? ?\CAB2 , fl = ? ?\ABC2 , and  = ? ?\BCA2 .
Then
\BOaC = ? ?
 ? ?\BCA
2
?
?
 ? ?\ABC
2
?
= \BCA2 + \ABC2 = ? ?\CAB2 = fi.
Likewise,
\CObA = fl
and
\BOcA =  .
Therefore, MAOcB is similar to M OcObOa. Let line OcT be the angle bisector of \AOcB.
Re ectingMAOcB about line OcT we get thatMAOcB andMOcObOa are homothetic. Thus,
the image of the median OcC? is the median OcC0r of M OcObOa. Therefore, line OcC? is a
symmedian line for M OaOcOb.(Figure 10)
18
Oc
Oa
Ob
T
C'
A
B
     Figure 10: Symmedian Line
Ar
Br
C'r
Likewise the lines ObB? and OaA? are symmedian lines for M OaOcOb. Therefore, by
Theorem 3.5 the lines OaA0, ObB0, and OcC0 are concurrent.
?
3.7 Spieker Center
Theorem 3.7 Given MABC, let D, E, and F be the midpoints of the sides c, a, and b
respectively. Then the angle bisectors of the medial triangle, MDEF concur at a point K
known as the Spieker Center.(Figure 11)
19
A
C B
Figure 11: Spieker Center
F
E
D
Proof. Since the angle bisectors of any given triangle concur at a point, the angle bisectors
of medial triangle, MDEF, concur at a point. ?
3.8 Steiner Point
Theorem 3.8 GivenMABC with circumcircle having center X and Brocard Triangle,Ma?b?c?,
then the lines through the vertices MABC which are parallel to the "opposite" sides of the
Brocard triangle concur at a point known as the Steiner Point.(Figure 12)
20
A
X
B
C
S
b'
a'
c'
K
Figure 12: Steiner Point
Proof. Let the line through vertex B and parallel to a?c? and the line through vertex C
and parallel to a?b? concur at a point K.(Figure 13)
A
X
B
C
S
b'
a'
c'
K
Figure 13: K on the circle
Then,
21
\BKC = \BAC
since both angles cut ofi the same chord of the circumcircle with center X. Therefore, K
lies on the circumcircle of MABC. In the same manner, let the line through vertex B and
parallel to ac and the line through vertex A and parallel to cb concur at a point M.
Then,
\BMA = \ACB
since both angles cut ofi the same chord of the circumcircle with center X. Therefore, M
also lies on the circumcircle.
Since line BK can only pass through a circle in at most two points, then
K = M.
Thus the lines BK, CK, and AK are concurrent at the point K on the circumcircle ?
3.9 Napoleon Point
Theorem 3.9 Given MABC with equilateral triangles raised on it?s sides, the lines ,each
passing through a vertex and the centroid of the opposite equilateral triangle, concur at a
point called the Napoleon Point.(Figure 14)
22
A
C
B
Figure 14: Napoleon Point
Proof. Refer to the proof of Kiepert?s Hyperbola in Chapter 4. ?
3.10 Fermat Point
Theorem 3.10 Given MABC with equilateral triangles raised on it?s sides, the lines ,each
passing through a vertex and the peak of the opposite equilateral triangle, concur at a point
called the Fermat Point.
A
C
B
Figure 15: Fermat Point
Proof. Refer to the proof of Kiepert?s Hyperbola in Chapter 4. ?
23
Chapter 4
Kiepert?s Hyperbola
In 1868, Lemoine posed the following problem [2]: Construct a triangle, given the
peaks of the equilateral triangles constructed on the sides (p. 188). Ludwig Kiepert posed
a solution to the problem in 1869. Kiepert generalized the problem to isosceles triangles.
Kiepert?s solution is stated as follows[2]: If three triangles, MA?BC, MAB?C, and MABC?,
with equivalent base angles are constructed on the sides of a given triangle, MABC, then the
lines AA?, BB?, and CC? concur. The locus of the point P as the base angle of the isosceles
triangles varies forms the equation
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 = 0;
or equivalently,
bc(b2 ?c2)
fi +
ca(c2 ?b2)
fl +
ab(a2 ?b2)
 = 0(p:189):
He showed that the lines, each passing through a vertex of the given triangle and a peak
of the constructed isosceles triangle, concur at a point. As the base angle of the isosceles
triangles vary the locus of the point of concurrence creates a rectangular hyperbola. The
hyperbola passes through the vertices of the given triangle and several triangle centers
including the centroid, orthocenter, Spieker center, Napoleon Point, Fermat Point, and
the Brocard point [5]. The Simson Lines of the given triangle are the asymptotes for
the rectangular hyperbola. The existence proof for Simson Lines is given in section two.
The proof of Kiepert?s Hyperbola requires the use of trilinear coordinates. Several theorems
involving trilinear coordinates will be proven prior to the completion of Kiepert?s Hyperbola
24
theorem. The center of Kiepert?s Hyperbola lies on the Nine-Point Circle. The Nine-Point
Circle proof is well known. However, for completeness the proof is given in the flrst section
of this chapter.
4.1 Nine-Point Circle
The Nine-Point Circle or Feuerbach Circle passes through the medians, the feet of the
altitudes, and the Euler points of a given triangle. A lemma is needed to prove that all nine
points lie on the same circle.
Lemma 1 Let X be any point on a circle with center O and radius r. Let T be any flxed
point. Then the locus of M, the midpoint of segment TX, is on a circle with radius r2 and
center N which is the midpoint of segment OT.
Proof.
O
X
T
M
N
Figure 16: Midpoint Lemma
By Figure 16, MOXT is similar to MNMT. Therefore MN = 12 XO or MN = r2. Thus
M lies on a circle with radius r2 and center N. ?
25
Theorem 4.1 For a given triangle the midpoints, the feet of the altitudes, and the Euler
points lie on a circle called the Nine-Point Circle.
Proof. Let a, b, and c be sides BC, CA, and AB respectively. For a given triangle, MABC,
let the midpoints of the sides a, b, and c be D, E, and F respectively. Let the feet of the
altitudes to sides a, b, and c be G, H, and I respectively. Let J be the orthocenter of MABC
and let K, L, and M be the midpoints of AJ, BJ, and CJ where K, L, and M are the Euler
Points of the triangle. Also, let O be the circumcenter of MABC with a circumcircle of
radius r. Let X be the midpoint of OJ.(Figure 17)
G
H
I
A
B
C D
E FJ
O
K
LM
Figure 17: Circumcircle, Medians, Feet of  Altitudes, and 
                  Euler Points
X
\ACJ ?= \ABH since both angles are complements of \CAB. Let J be the point on
which the line BH touches the circle. Then \ABN ?= \ACN since they cut ofi arc NA.
26
G
H
I
A
B
C D
E FJ
O
K
LM
N
X
Figure 18: H, I, G, K, L, and M are on the same circle
Therefore, \ACN ?= \ACJ and \ABN ?= \ACJ. Since \JHC ?= \NHC then MHCN
?= MHCJ. Thus JH = HN. By Lemma 1 J lies on the circle with X as the center and radius
r
2. Likewise H and I lie on the circle. L is the midpoint of JB so L is on the circle with
center X and radius r2 by Lemma 1. Likewise M and K lie on the circle. Therefore H, I, J,
L, M, and K lie on the circle with center X and radius r2.(Figure 18)
27
G
H
I
A
B
C D
E FJ
O
K
LM
N
Figure 19: E, F, and D lie on the circle
X
 
Y
Let line BO meet the circumcircle at the point Y.(Figure 19) Then YCJA is a parallelogram.
Therefore YJ and CA intersect at E since E is the midpoint of CA. Therefore E is also the
midpoint of YJ. By Lemma 1, E lies on the circle with center X and radius r2 as do D and
F.(Figure 20)
28
G
H
I
A
B
C D
E FJ
O
K
LM
Figure 20: Nine-Point Circle
X
?
4.2 Simson Line
Since the asymptotes for Kiepert?s Hyperbola are Simson Lines, the existence proof of
Simson Lines will be given.
Theorem 4.2 Given MABC and a point P on it?s circumcircle then the feet of the perpen-
diculars, W, U , and V ,to the sides BC, AC, and AB respectively are collinear. The line
passing through the feet of the altitudes is called a Simson Line.(Figure 21)
29
A
B
C
P
W
U
V
Figure 21: Simson Line
Proof. Quadrilateral CAPB is cyclic therefore
\WCA + \WPB + \APW = 180?.
Likewise, quadrilateral CUPW is cyclic so
\WCA + \UPA + \APW = 180?.
So
\WPB = \UPA.
And since quadrilateral PVWB is also cyclic then
\WPB = \WVB
since they cut ofi the same angle. Quadrilateral PVAU is cyclic therefore,
\UPA = \UVA.
30
Therefore,
\WVB = \UVA.
Since \WVB and \UVA are vertical angles the U, V, and W are collinear. ?
4.3 Trilinear Coordinates
Many coordinate systems are used to determine triangle properties. In cartesian
coordinates the distances from two given perpendicular axes are assigned to each point.
Polar coordinates are more useful for certain analytic computations. If the signed distance
from the origin O to a point P and the signed angle OP to the x-axis are assigned to P then
polar coordinates are being used. Barycentric coordinates are a set of ordered triples
of masses, w1 : w2 : w3, deflned for a point P inside a given triangle such that P lies on the
centroid of the triangle. Trilinear coordinates can be very useful when exploring various
properties of triangles and their centers. Trilinear coordinates are a set of ordered triples,
fi : fl :  , of numbers which are proportional to the signed distances from a point to the
sidelines of a given triangle.(Figure 22)
P
A
BC
b c
a
??
?
Figure 22: Trilinear Coordinates
In trilinear coordinates the ratio of the distances are important so that
31
fi : fl :  = k fi : k fl : k  
for any nonzero constant k.
Actual trilinear coordinates give the actual directed distances from a point to the sides
of a given triangle. Actual trilinear coordinates are given as
kfi : kfl : k 
where k = 2 Mafi+bfl +c [3]. More information on trilinear coordinates can be found in the
works of Kimberling[3][4] and Coxeter[1].
To determine collinearity of three points in trilinear coordinates a determinant must be
evaluated.
Theorem 4.3 Given three points fi1 : fl1 :  1, fi2 : fl2 :  2, and fi3 : fl3 :  3, the three
points are collinear if
flfl
flfl
flfl
flfl
flfl
fl
fi1 fl1  1
fi2 fl2  2
fi3 fl3  3
flfl
flfl
flfl
flfl
flfl
fl
= 0:
Proof. Let MABC be the reference triangle. Let P and P? be points such that P = 0 and
P? = 1 and let t be a point on the line passing through P and P? such that 0 < t < 1.(Figure
23)
32
1
0
t
?1
?1
?1
1-t
?2??1
x
Figure 23: Function ?(t)
From Figure 23, the distance from any point t between P and P? is given by the function
 = t 1 + (1 - t) 2.
In the same manner,
fi = tfi1 + (1 - t)fi2,
fl = tfl1 + (1 - t)fl2.
Therefore,
2
66
66
64
fi
fl
 
3
77
77
75 = (1?t)
2
66
66
64
fi2
fl2
fi2
3
77
77
75+t
2
66
66
64
fi1
fl1
 1
3
77
77
75:
Thus, [fi, fl,  ] is a linear combination of [fi1, fl1,  1] and [fi2, fl2,  2].
Therefore,
33
flfl
flfl
flfl
flfl
flfl
fl
fi fl  
fi1 fl1  1
fi2 fl2  2
flfl
flfl
flfl
flfl
flfl
fl
= 0:
?
A line in trilinear coordinates is deflned in terms of three parameters: l, m, and n. Using
the theorem for collinearity, we can derive the equation of a line in trilinear coordinates.
Theorem 4.4 Given two points fi1 : fl1 :  1 and fi2 : fl2 :  2 a line through these two
points in trilinear coordinates has the form
l fi + m fl + n  = 0
with
l = fl1 2 -  1fl2
m =  1fi2 - fi1 2
n = fi1fl2 - fl1fi2.
Proof. Using the collinearity theorem and replacing one set of coordinates with fi : fl :  
we get,
flfl
flfl
flfl
flfl
flfl
fl
fi fl  
fi1 fl1  1
fi2 fl2  2
flfl
flfl
flfl
flfl
flfl
fl
=
34
(fl1 2 -  1fl2)fi + ( 1fi2 - fi1 2)fl + (fi1fl2 - fl1fi2) = 0.
Therefore, the equation of the line passing through fi1 : fl1 :  1 and fi2 : fl2 :  2 is given by
lfi + mfl + n = 0
where
l = fl1 2 -  1fl2
m =  1fi2 - fi1 2
n = fi1fl2 - fl1fi2.
?
Now that we have determined the equation for a line in trilinear coordinates, a theorem
is introduced regarding line concurrence.
Theorem 4.5 Three trilinear lines
l1 fi + m1 fl + n1  = 0,
l2 fi + m2 fl + n2  = 0,
l3 fi + m3 fl + n3  = 0
concur at one point if
flfl
flfl
flfl
flfl
flfl
fl
l1 m1 n1
l2 m2 n2
l3 m3 n3
flfl
flfl
flfl
flfl
flfl
fl
= 0:
35
Proof. If three lines,
l1fi + m1fl + n1 = 0,
l2fi + m2fl + n2 = 0,
l3fi + m3fl + n3 = 0
are concurrent, then they meet at one point. Therefore, the system of equations has a
solution. Since trilinear coordinates deflne distances to the sides of a reference triangle,
then at most two of the coordinates can be zero. However, the trivial solution 0 : 0 : 0 is
also a solution of the system. Thus, two solutions exist.
Therefore,
flfl
flfl
flfl
flfl
flfl
fl
l1 m1 n1
l2 m2 n2
l3 m3 n3
flfl
flfl
flfl
flfl
flfl
fl
= 0:
Since
flfl
flfl
flfl
flfl
flfl
fl
l1 m1 n1
l2 m2 n2
l3 m3 n3
flfl
flfl
flfl
flfl
flfl
fl
= 0
then
36
flfl
flfl
flfl
flfl
flfl
fl
l1 m1 n1
l2 m2 n2
l3 m3 n3
flfl
flfl
flfl
flfl
flfl
fl
=
l1
flfl
flfl
flfl
fl
m2 n2
m3 n3
flfl
flfl
flfl
fl
+m1
flfl
flfl
flfl
fl
l2 n2
l3 n3
flfl
flfl
flfl
fl
+n1
flfl
flfl
flfl
fl
l2 m2
l3 m3
flfl
flfl
flfl
fl
=
(m2n3 ?m3n2)l1 +(l2n3 ?l3n2)m1 +(l2m3 ?l3m2)n1 = 0
The point of concurrence is given as
P = m2n3 ?m3n2 : l2n3 ?l3n2 : l2m3 ?l3m2.
?
Remark: The line at inflnity in trilinear coordinates is given as
afi + bfl + c = 0
since the distances from the line at inflnity to the vertices of the reference triangle become
equal in the limit.
4.4 Kiepert?s Hyperbola
For the proof of Kiepert?s Hyperbola, we assume that the given triangle is scalene.
We will, however, discuss the cases in which the given triangle is an equilateral or isosceles
triangle.
37
Equilateral Triangle: If the given triangle, MABC, is equilateral then as the base angle of
the three isosceles triangles raised on the sides of the given triangle varies the lines, each
passing through a vertex and the peak of the isosceles triangle raised on the opposite side,
always concur at the centroid of the triangle.
Isosceles Triangle: If the given triangle, MABC, is an isosceles triangle with side AB equal
to side CA, then the point of concurrence lies on the perpendicular bisector of side BC. But
in the limit case where ` is approaching ?2 the lines, each passing through a vertex and the
peak of the isosceles triangle raised on the opposite side, are parallel.
Theorem 4.6 If three triangles, MA?BC, MAB?C, and MABC?, with equivalent base angles
are raised on the sides of a given triangle, MABC, then the lines AA?, BB?, and CC? concur.
The locus of the point P as the base angle of the isosceles triangles varies forms a hyperbola
with equation
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 = 0;
or equivalently,
bc(b2 ?c2)
fi +
ca(c2 ?b2)
fl +
ab(a2 ?b2)
 = 0:
Before we prove Theorem 4.6, note that some of the special cases have already been proven.
Case 1: If ` = 0 then the lines AA?, BB?, and CC? are the medians of MABC. Thus, the
lines concur at the centroid of MABC.(Figure 24)
38
C
B
A
A'
B' C'
Figure 24: Centroid
Case 2: In the limit case, as ` approaches ?2, the lines AA?, BB?, and CC? get closer
to the altitudes of MABC. Therefore, in the limit case, the hyperbola passes through the
orthocenter but the orthocenter, is not a point of concurrence.(Figure 25)
B'
A'C
A
B
C'
P
Figure 25: Orthocenter
Proof. In the general case, let MABC be acute and let the constructed isosceles triangles
lie outside the given triangle. Let ` represent the base angle of the constructed triangles.
39
Also let A, B, and C represent the angle at vertex A, vertex B, and vertex C respectively.
In trilinear coordinates, let
A = 1 : 0 : 0,
B = 0 : 1 : 0,
C = 0 : 0 : 1.
Letting x be the length of side A?C then the distance from vertex A? to side BC is x
sin(`).(Figure 26)
A
BC
B' C'
A'
??
Figure 26: Trilinear Coordinates
A
C B
x
Since the incenter of 4ABC and A? lie on opposite sides of BC, the trilinear coordinates
for the distance is -x sin(`). Likewise, the distance from A? to side AC and side AB is
x sin(C +`) and x sin(B +`) respectively. Therefore,
A?= -x sin(`) : x sin(C +`) : x sin(B +`)
or equivalently
A? = -sin(`) : sin(C +`) : sin(B +`).
40
In the same manner,
B? = sin(C +`) : -sin(`) : sin(A+`)
and
C? = sin(B +`) : sin(A+`) : -sin(`).
By the deflnition of a trilinear line, the line passing through the points A and A? have
parameters l, m, and n as follows: l = 0, m = -sin(B +`), and n = sin(C +`). Therefore,
the equation for line AA? is -sin(B + `)fl + sin(C + `) = 0. The equations for lines BB?
and CC? can be obtained in the same way so that
AA? = -sin(B +`)fl + sin(C +`) = 0,
BB? = -sin(C +`) + sin(A+`)fi = 0,
CC? = -sin(A+`)fi + sin(B +`)fl = 0.
The lines AA?, BB?, and CC? concur if the matrix of their coe?cients has a determinant of
0.
flfl
flfl
flfl
flfl
flfl
fl
0 ?sin(B +`) sin(C +`)
sin(A+`) 0 ?sin(C +`)
?sin(A+`) sin(B +`) 0
flfl
flfl
flfl
flfl
flfl
fl
=
0 ((sin(B +`)(sin(C +`))?(?sin(B +`)((?sin(A+`)(sin(C +`))+(sin(C +
`))((sin(A+`)(sin(B +`)) = 0
41
Therefore the lines AA?, BB?, and CC? concur. The point, P, of concurrence is given by the
trilinear coordinates
sin(B +`)sin(C +`) : sin(A+`)sin(C +`) : sin(A+`)sin(B +`).(Figure 27)
A
BC
B' C'
A'
??
Figure 27: Concurrence
P
Finally, notice that the same linear algebra computation holds in the case of the given
triangle being an obtuse triangle and for the isosceles triangles being constructed on the
interior of the given triangle. In these cases, signed distances are used.
As the base angle ` varies, the locus of the point P forms a curve with the equation,
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 = 0;
or equivalently,
bc(b2 ?c2)
fi +
ca(c2 ?b2)
fl +
ab(a2 ?b2)
 = 0:
To show that the point P of concurrence is a solution to the equation above, we substitute
the trilinear coordinates of point
42
P = sin(B = `)sin(C +`) : sin(A+`)sin(C +`) : sin(A+`)sin(B +`)
in the formula
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 = 0
for fi, fl, and  respectively.
Thus,
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 =;
sin(B ?C)
sin(B +`)sin(C +`) +
sin(C ?A)
sin(C +`)sin(A+`) +
sin(A?B)
sin(A+`)sin(B +`) =;
sin(B ?C)sin(A+`)+sin(C ?A)sin(B +`)+sin(A?B)sin(C +`)
sin(A+`)sin(B +`)sin(C +`) :
Using trigonometric identities, it can be shown that
sin(B ?C)sin(A+`)+sin(C ?A)sin(B +`)+sin(A?B)sin(C +`)
sin(A+`)sin(B +`)sin(C +`) = 0:
Therefore, the locus of the point P as the base angle ` varies is given by,
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 = 0:
Next we show that the equation
sin(B;?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 = 0;
is equivalent to the equation
bc(b2 ?c2)
fi +
ca(c2 ?b2)
fl +
ab(a2 ?b2)
 = 0:
First, construct the circumcircle of MABC. By moving vertex A along the circumcircle a
right triangle, MABC, is constructed with AB having a length of twice the radius of the
circumcircle and no change in \A.(Figure 28)
43
A
r
r
a
A
A
B
C
Figure 28: Circumcircle
Therefore, sinA = a2r. The angles sinB and sinC can be found in a similar manner such
that
sinA = a2r, sinB = b2r, and sinC = c2r.
By the law of cosines
cosA = b
2 +c2 ?a2
2bc ; cosB =
a2 +c2 ?b2
2ac ; cosC =
a2 +b2 ?c2
2ab :
And sin(B ?C) = sinBcosC +sinC cosB =
b
2r
a2 +b2 ?c2
2ab +
c
2r
a2 +c2 ?b2
2ac =
1
4r
flfl
flfla2 +b2 ?c2
a ?
a2 +c2 ?b2
a
flfl
flfl abc
abc =
1
4abcr(2bc(b
2 ?c2)) =
1
2rabcbc(b
2 ?c2):
44
sin(C ?A) and sin(A?B) can be derived in the same way so that,
sin(B ?C) = 12rabc(b2 ?c2)bc;
sin(C ?A) = 12rabc(c2 ?a2)ca;
sin(A?B) = 12rabc(c2 ?a2)ab:
Therefore,
sin(B ?C)
fi +
sin(C ?A)
fl +
sin(A?B)
 =
1
2rabc(b
2 ?c2)bc
fi +
1
2rabc(c
2 ?a2)ca
fl +
1
2rabc(c
2 ?a2)ab
 =
bc(b2 ?c2)
fi +
ca(c2 ?a2)
fl +
ab(c2 ?a2)
 = 0:
Now we must show that the equation
bc(b2 ?c2)
fi +
ca(c2 ?a2)
fl +
ab(c2 ?a2)
 = 0
is the equation of a conic section.
Using a difierent notation let,
fi = da,
fl = db,
 = dc,
where da, db, and dc are the distances to sides a, b, and c respectively. Also let the vertices
have the following coordinates,
45
A = (x1, y1),
B = (x2, y2),
C = (x3, y3).
The vector perpendicular to ??!AB is
v =
flfl
flfl
flfl
fl
y2 ?y1
?(x2 ?x1)
flfl
flfl
flfl
fl
:
Let ?!PA be the vector from any point P = (x,y) to A and be given by
r =
flfl
flfl
flfl
fl
x?x1
y ?y1
flfl
flfl
flfl
fl
:
Then the distance from P to side AB is given by projecting r onto v which gives
dc = j(x?x1)(y2 ?y1)?(x2 ?x1)(y ?y1)jc .
In the same manner,
da = j(y3 ?y2)(x?x2)?(x3 ?x2)(y ?y2)ja ,
db = j(y1 ?y3)(x?x3)?(x1 ?x3)(y ?y3)jb .
Therefore,
46
fi = j(y3 ?y2)(x?x2)?(x3 ?x2)(y ?y2)ja ,
fl = j(y1 ?y3)(x?x3)?(x1 ?x3)(y ?y3)jb ,
 = j(x?x1)(y2 ?y1)?(x2 ?x1)(y ?y1)jc .
Plugging these values in for
bc(b2 ?c2)
fi +
ca(c2 ?b2)
fl +
ab(a2 ?b2)
 = 0:
we get
(b2?c2)
(y3?y2)(x?x2)?(x3?x2)(y?y2) +
(c2?b2)
(y1?y3)(x?x3)?(x1?x3)(y?y3) +
(a2?b2)
(x?x1)(y2?y1)?(x2?x1)(y?y1) = 0:
By getting a common denominator, the equation becomes a quadratic involving x2, y2, and
xy terms. This quadratic is the equation of a conic section. The equation actually deflnes a
rectangular hyperbola. The proof that this conic section is a rectangular hyperbola is given
by Eddy and Fritsch [2].
A
BC
O
I
Figure 29:  Kiepert's Hyperbola
47
Kiepert?s Hyperbola is a solution to Lemoine?s problem. Therefore, the values of ` are
given which will provide the coordinates of the given triangle. The point of concurrence,
P, lies on the vertex A when ` = -A and \A is acute. When \A is acute and ` = -A the
lines BB? and CC? pass through vertex A. In Figure 30, BB?, CC?, and AA? intersect at the
vertex A.
A
C B
B'
A'C'
Figure 30: Concurrence at A 
If \A is obtuse, then ` = 180 - A when P = A. For vertices B and C, P = B and P = C
when ` = -B and ` = -C respectively. ?
48
Bibliography
[1] H. S. M. Coxeter, Some applications of trilinear Coordinates, Linear Algebra Appl.,
226-228 (1995), 375-388.
[2] R. H. Eddy and R. Fritsch, The Conics of Ludwig Kiepert: A Comprehensive Lesson
in the Geometry of the Triangle, Math. Mag., 67 (1994), no. 3, 188-205.
[3] C. Kimberling, Central Points and Central Lines in the Plane of a Triangle, Math.
Mag., 67 (1994), no. 3, 163-187.
[4] C. Kimberling, Major Centers of Triangles, The Amer. Math. Month., 104 (1997), no.
5, 431-438.
[5] C. Kimberling, Triangle Centers and Central Triangles, Congres. Numer, 125, 1-295,
1998.
49