Content uploaded by Miklós Hoffmann
Author content
All content in this area was uploaded by Miklós Hoffmann
Content may be subject to copyright.
The Gergonne Conic
Sonja Gorjanc
Department of Mathematics, Faculty of Civil Engineering
University of Zagreb
Kaˇci´ceva 26, 10000 Zagreb, Croatia
sgorjanc@grad.hr
Mikl´os Hoffmann
Institute of Mathematics and Computer Science
K´aroly Eszterh´azy College
Le´anyka str. 4., H-3300 Eger, Hungary
hofi@ektf.hu
June 17, 2011
Abstract
The notion of Gergonne point was generalized in several ways during the
last decades, including the work of Konecny [5]. Given a triangle V1V2V3, a
point Iand three arbitrary directions qi, we find a distance x=I Q1=I Q2=
IQ3along these directions, for which the three cevians ViQiare concurrent.
If Iis the incenter, qiare the direction of the altitudes, and xis the radius of
the incenter, the point of concurrency is the Gergonne point. For arbitrary
directions qi, it is shown that each point Igenerally yields two solutions, and
points of concurrency lie on a conic, which can be called the Gergonne conic.
Key words: Gergonne point, conics, projectivity, pencil of conics
1 Introduction
The well-known center of the triangle is the Gergonne point, the intersection of
three cevians defined by the contact points of the inscribed circle [4]. As almost
every triangle center, it appears in several configurations (see e.g. [3]), and the
point is also generalized by several authors, for example for higher dimensions in
[2] or by embedding in a more general context in [6]. Gergonne points is also
generalized by [5], applying circles concentric to the inscribed circle.
The first question naturally arises from this point of view: if the radius of the
circle xis altered, what will be the path of the point G(x). Boyd et al. computed
1
the convex coordinates of G(x) in [1], which obviously yields that the path is a
hyperbola.
The general problem can be formulated as follows: given a triangle V1V2V3, a
point Iand three arbitrary directions qi, find a distance x=IQ1=IQ2=IQ3
along these directions, for which the three cevians ViQiare concurrent. Normally
these lines will not meet in one point: instead of one single center Gwe have
three different intersection points: G12 =V1Q1∩V2Q2,G23 =V2Q2∩V3Q3and
G31 =V3Q3∩V1Q1. In this paper we find all the solutions and consequences of
this generalized problem.
2 The general solution
In [3] the authors proved the following theorem:
Theorem 1 Let V1, V2, V3and Ibe four points in the plane in general position.
Let q1, q2, q3be three different oriented lines through I(Vi6∈ qi). Then there exist
at most two values x∈R\{0}such that finding points Qialong the lines qias
IQ1=IQ2=IQ3=x, the lines ViQiare concurrent.
It was shown that altering the value x, the points G12,G23 and G31 separately move
on three conics ci,(i= 1,2,3), defined in the following way: For a real number xand
i= 1,2,3,let Qi(x) be a point on qithrough Ifor which IQi(x) = x, (i= 1,2,3).
The correspondences Qi(x)↔Qj(x) define perspectivities (qi)[(qj), (i6=j). Let
li(x) be a line connecting Viand Qi(x). The correspondences li(x)↔lj(x) define
projectivities (Vi)⊼(Vj), (i6=j). The intersection points of corresponding lines of
these projectivities lie on three conics:
(V1)⊼(V2)⇒c3
(V1)⊼(V3)⇒c2
(V2)⊼(V3)⇒c1.(1)
It was shown in [3] that these conics have three common points: Iand two points
G1,G2that can be real and different, imaginary or coinciding. These common
points G1and G2are solutions of the general problem and thus can be considered
as generalized Gergonne points, naturally depending on the triangle, the directions
and the point I.
Remark 1 According to the projective principles in the proof, the statement re-
mains valid if we substitute the condition IQ1=I Q2=I Q3=xwith the more
general case, when only the ratios of these lengths have to be preserved.
2
3 The Gergonne conic
According to [3], for given triangle, directions and point Ithere are at most two
generalized Gergonne points. Now for arbitrary Iwe collect all these points with
respect to a fixed triangle and directions.
To formulate the main result, we need some further notations and lemmas. Let
the line qibe parallel to qithrough Vi(c.f. Fig. 1). Finding the points
C3=q1∩q2, C2=q1∩q3, C1=q2∩q3,
the conic ciwill pass through Ci, because Ciis the intersection point of qjand qk
that are the corresponding lines in the pro jectivities given by (1). Thus, for every
Ithe conic paths ciare uniquely determined by five points, separately:
c1= (I, C1, G23 , V2, V3), c2= (I , C2, G31 , V3, V1), c3= (I, C3, G12, V1, V2),
which have 3 common points: I,G1,G2, where G1,G2are the solutions of the
problem.
Figure 1: Points Ci, (i=1,2,3) are defined by lines qiparallel to qithrough Vi
Lemma 1 The vertex V1(V2or V3) is the solution for a point Iif and only if
I∈t1(t2or t3), where t1(t2or t3) is the line which contains the points at equal
distance from V1V2(V2V3or V3V1) and V1V3(V2V1or V3V2) measuring along the
directions q2(q3or q1) and q3(q1or q2), respectively.
Proof. The statement follows directly from the definition of the generalized Ger-
gonne point and the properties of a central similarity (see Fig. 2).
3
I
q1
3
q
2
q
V
1V2
V3t1
Figure 2: Vertex V1of the triangle can also be a generalized Gergonne point for
points Ibeing in special position
Lemma 2 If a point I6=Vilies on the line V2V3(V3V1or V1V2), then the conic
c1(c2or c3) splits into the line V2V3(V3V1or V1V2) and another line which passes
through the solutions.
Proof. If I∈V2V3, then the first correspondence given by (1) is the perspectivity
(V1)[(V2). Thus, the conic c1splits into the axes of this projectivity and the self
corresponding ray V2V3. According to the proof of theorem 1 [3], the axes of the
projectivity (V1)[(V2) cuts the conics c2and c3into the solutions G1,G2. This
proof is also valid for the circular shift of index.
Lemma 3 If a point I6=Cilies on the line C2C3(C3C1or C1C2), then each of
the conics c2,c3(c3,c1or c1,c2) splits into the line C2C3(C3C1or C1C2) and
another line which passes through the corresponding vertex V3,V2(V1,V3or V2,
V1) and one solution G1.
Proof. Let I∈C2C3, and let x1be the distance between the points Iand V1. In this
case the previous described correspondence l1(x)↔l2(x) leads to the following:
- the line C2C3corresponds with all lines l2(x)∈(V2), x∈R,
- the line l2(x1) corresponds with all lines l1(x)∈(V1), x∈R.
Thus, the conic c3splits up into the lines C2C3and V2Q2, where IQ2=x1and
the line IQ2is parallel to q2. In the same way: the conic c2splits up into the lines
C2C3and V3Q3, where IQ3=x1and the line IQ3is parallel to q3. It is clear that
the intersection point of the lines V2Q2and V3Q3lies on the conic c1and is the
solution G. The second solution G16=Iis the intersection point of the conic c1
and the line C2C3, see Fig. 3. This proof is also valid for the circular shift of index.
4
Iq1
3
q
2
q
V
1
V2
V3
G
Q3
C1
c1
3
c
2
c
2
C
3
C
2
Q
1
G
Figure 3: If point Iis on the line C2C3then conics c2and c3split up into pairs of
lines
Lemma 4 If Gis the solution for I, then it is the solution for every Ij∈l, where
lis the line through Iand G. Consequently if Gis the solution for I1and I2as
well, then the points G,I1and I2are collinear.
Proof. The statement follows directly from the definition of the generalized Ger-
gonne point and the properties of a central similarity (see Fig. 4).
V
1
V2
V3
G
l
Ij
I
Figure 4: If Gis the solution for I, then it is the solution for every point Ijon the
line IG.
By these lemmas, we can prove our main result, that for given triangle and
directions, all the generalized Gergonne points lie on one conic, which can be called
Gergonne conic.
Theorem 2 For given V1, V2, V3, q1, q2, q3, all solutions lie on one conic Γwhich
passes through V1, V2, V3.
Proof. Let I,G1and lbe as in lemma 4 (see Fig. 5). For every Ij∈lwe consider
the conics cj
2= (V1, V3, C2, G1, Ij) and cj
3= (V1, V2, C3, G1, Ij). These conics have
5
4 common points: V1, G1, Ijand (according to the previous theorem) the other
solution Gj
2. The correspondence cj
2↔cj
3is (1,1) correspondence between the
pencils of conics (V1, V3, C2, G1) and (V1, V2, C3, G1). According to the Chasles’s
formula [8], the intersection points of the corresponding conics lie on one plane
curve of the order 1 ·2 + 1 ·2 = 4. In our case this quartic splits up into the lines
land C2C3and one conic Γ. Namely, for Ic=C2C3∩lconics cc
2and cc
3split up
into C2C3and the lines V3G1,V2G1(according to lemma 3); every Ij∈lis the
part of cj
2∩cj
3; all solutions Gj
2lie on one residual conic Γ which passes through V1
because it is the intersection point for all pairs (cj
2, cj
3). Since Γ is also the result
of the correspondences cj
1↔cj
2and cj
1↔cj
3, we can conclude that it also passes
through V2and V3.
If G6=G1is any other solution for any other point I6=I, then, according to
lemma 4, Gis the solution for the intersection point I=l∩l. According to the
previous consideration, Galso lies on the conic Γ.
q1
3
q
2
q
V
1
V2
V3
C1
2
C
3
C
1
Gl
IcI1I2
I3
I4
1
G2
4
G23
G2
2
G2
1
2
c2
2
c
3
2
c
4
2
c
c
2
c
2
3
c
3
3
c
4
3
c
c
3
c
1
3
c
Figure 5: The conic Γ contains all the solutions for the given triangle and directions.
Further details in the proof of theorem 2. (This figure is computed and plotted by
the software Mathematica)
Lemma 5 On every line l=IG1there exists a unique point I∆for which the
solutions coincide, i. e. G∆
2=G1.
Proof. We consider the pro jectivity (l)⊼(G1)1where corresponding elements are
1In the proof of theorem 2, with Ij↔Gj
2, we defined projective correspondence between the
range (l1) and the 2nd order range (Γ). Since the cross ratio of the points on one conic can be
defined as the cross ratio of lines which join them with any other point on this conic, we can
conclude that Ij↔G1Gj
2defines the projectivity between (l1) and (G1).
6
Ij↔G1Gj
2. The point I∆which corresponds with the tangent line of the conic Γ
at its point G1has coinciding solutions in G1.
Theorem 3 For given V1, V2, V3, q1, q2, q3, all lines l=IG (where G is the solution
for I) form one envelope conic ∆with tangent lines CiCj.
Proof. Let G1and G2be two different solutions for Iand l1=IG1,l2=I G2.
For every point Gj∈Γ let Ij
1∈l1and Ij
2∈l2be the points for which Gjis
the solution. (Ij
1and Ij
2are uniquely determined: Ij
1=l1∩(V2, V3, C1, G1, Gj),
Ij
2=l1∩(V2, V3, C1, G2, Gj).) According to lemma 4 points Ij
1,Ij
2and Gjare
collinear.
There are two projective correspondences (l1)⊼(Γ) and (l2)⊼(Γ) which induce
the projectivity (l1)⊼(l2) with corresponding pairs Ij
1↔Ij
2. This projectivity
determines one envelope conic ∆ and it is clear, according to lemma 4, that every
line which joins any point with its solution is the tangent line of it (see Fig. 6).
According to lemma 3 and lemma 4: if I∈CiCj, one solution always lies on
the line CiCjand it is the solution for every I∈CiCj,I6=Ci,I6=Cj.
The touching point of ∆ and l1(l2) corresponds with I∈l2(I∈l1), thus this
touching point is I∆
1(I∆
2) from lemma 5, i. e. the point with coinciding solutions
G1∈l1(G2∈l2).
q1
3
q
2
q
V
1
V2
V3
C1
2
C
3
C
1
G
2
G
G1
1
I2
l1
l2
G3
G2
2
I2
3
I2
I2
1
I1
2
I1
3
I1
I1
I
Figure 6: The lines connecting any point Iwith its solutions G1and G2are tangent
to the conic ∆ at I∆
1and I∆
2, respectively
7
Corollary 1 The conic ∆separates the possible positions of Iin the way men-
tioned in the proof of theorem 1 in [3].
Proof. Since lines which join points Iwith their solutions are the tangent lines of
∆ it is clear that solutions are real and different for I∈Ext∆, real and coinciding
for I∈∆ and imaginary for I∈I nt∆.
4 Construction of conics Γand ∆
Since V1,V2,V3lie on Γ it is enough to construct only two other points on it.
Arbitrary number of points G∈Γ can be constructed according to lemma 3. Here
we will describe the construction of the points Gi∈Γ, Gi6=Vi,Gi∈qi. They
can be constructed as the solutions for the points C1,C2or C3. For example: If
I=C2=q1∩q3the conic c2splits into the lines q1,q3. The conic c1splits into
the lines q3and V2Q′′
2, where Q′′
2is on the same distance from C2as the point V3
measuring along the direction q2. The conic c3splits into the lines q1and V2Q′
2,
where Q′
2is on the same distance from C2as the point V1measuring along the
direction q2. The intersection points of c1,c2,c3are G1and G3. The point G2
can be constructed, in the same way, as one solution for the point C1. According
to lemma 4 the point G1(G2or G3) is a common solution for the points C2,C3
(C3,C1or C1,C2), see Fig. 7a.
Since the lines qiare the tangent lines of ∆ it is enough to construct only two
other tangent lines of it. According to theorem 3 the tangent lines of the conic
∆ join points with their solutions. Thus, the construction of the lines tiwhich is
described in lemma 1 determines the conic ∆. The lines tican be constructed in
an elementary way, see Fig. 7b.
5 Special cases
So far we considered that the directions qiand the point Iare in general position.
This way, according to theorem 1, the number of solutions is at most two. But at
the previously mentioned result of Konecny, if the directions are perpendicular to
the corresponding sides of the triangle and Iis the incenter, there are infinitely
many solutions. To overcome this problem the following lemma and theorem clarify
the special cases.
Lemma 6 For given V1, V2, V3, q1, q2, q3there exists a point Iwith more than
two solutions if and only if the points Cicoincide with one point C(the lines qiare
concurrent). In this case there are infinitely many solutions for the point I. These
solutions lie on the conic ΓIwhich passes through the points V1, V2, V3, I and C.
For every point Ij6=I, point Cis one solution and the other solution Gjis the
intersection point of conic ΓIand the line II j.
8
x
ab
q1
3
q
2
q
V
1
V2
V3
C1
2
C
3
C
1
t
2
t
3
t
C1
2
C
3
C
2
Q
2
Q
1
G
2
G
3
G
V
1V2
V3
x
x
Figure 7: Construction of conics Γ and ∆
Proof. If there are three solutions G1, G2, G3for I, then the conics cicoincide
because each pair of them passes through the same five points. Thus there exists
one conic ΓIwhich passes through seven points G1, G2, G3, V1, V2, V3and I, and
every point on it is a solution for I. Since the conic ΓIcontains points Ci(Ci∈ci)
we can conclude that the points Cimust coincide because in the opposite case
the line C1C2which passes through V3would cut ΓIin three points, which is
impossible.
If the lines qiare concurrent with the intersection point C, then Cis the solution
for every Ij(Ccorresponds with the points at infinity on the lines qithrough Ij
in the projectivities given by (1)). Let I1, I 2be two different points with different
solutions G16=C, G26=C. According to lemma 4, G1, G2are the solutions for
the intersection point Iof the lines I1G1,I2G2. Thus, there are three solutions
for I(G1,G2,C), and according to the previous consideration there are infinitely
many solutions for Iwhich lie on the conic ΓI.
Now, it is clear that for every point Ij6=Ione solution is Cand the other Gj
must lie on the line IjIand the conic ΓI.
Since in this case the lines which join any point with its solutions pass through
Iand C, the envelope conic ∆ splits up into the pencils of lines (I) and (C).
9
Let the lines V1V2,V1V3,V2V3be denoted by a3,a2,a1, respectively, and let
db(P, a) denote the distance from a point Pto a line ameasuring along a direction
b.
Theorem 4 Let V1, V2, V3and Cbe four points in the plane in general position
and let three oriented lines qicoincide with the lines CVi,i= 1,2,3. Then there
exists a point Isuch that
dqi(I, aj) = dqj(I , ai), i 6=j, i, j = 1,2,3 (2)
and every point G∈ΓI(where ΓIis the conic through the points V1, V2, V3, C, I)
is a generalized Gergonne point for I. For every other point Pthere exist two
generalized Gergonne points, GP
1=Cand GP
2= ΓI∩P I .
Proof. According to lemma 6 in such a case the envelope conic ∆ splits into the
pencils of lines (C) and (I), where Iis the point with infinitely many general-
ized Gergonne points on the conic ΓI. Thus, according to theorem 3 the lines
V1I, V2I , V3Iare the lines t1, t2, t3from lemma 1, respectively, and their intersec-
tion point Ihas the property (2). The other statements follow directly from lemma
6.
1
V
3
V
q1
q2q3
C
I
I
3
t
2
t
1
t
G2
A
A
G1
A
=
2
V
Figure 8: Construction of the point Iwith infinitely many solutions for given
C, V1, V2, V3and Gergonne points for arbitrary point A.
Since that for every qi,i= 1,2,3, there are two different orientations, there
are 23possibilities for the choosing of the oriented lines qi. But, these choices in
pairs determine the same way for the finding generalized Gergonne points (if the
orientations are opposite for every i= 1,2,3). Thus, for the given non-oriented
lines qithere are four different possibilities for finding the generalized Gergonne
points.
Corollary 2 If the point Cfrom theorem 4 is the orthocenter Hof △V1V2V3, then
the point Iis the incenter or the excenter Ji,i= 1,2,3, of △V1V2V3. The conics
ΓI,ΓJiare Feuerbach hyperbolas.
10
q1
q2
q3
I
C
1
V
3
V
2
V
I
3
t
2
t
1
t
Figure 9: Construction of the point Iand the conic ΓIin the case when the
fundamental elements C,Vi,qidiffer from those in Fig. 8 only in the orientation
of q2.
Proof. If the lines qi,i= 1,2,3, are the altitudes of △V1V2V3, the line tiis the
internal angle bisector bior the external angle bisector b′
iof △V1V2V3, depending
on the orientation of qi. The proof of this statement is elementary and is based on
triangle congruence theorems. Thus, the point Ifrom theorem 4 is the incenter or
excenter of △V1V2V3. The conics ΓIand ΓJipass through the ortocenter Hare
rectangular hyperbola which are called Feuerbach hyperbolas.
An even more specialized situation is when Iand C, from theorem 4, coincide.
This is the equilateral triangle V1V2V3with the center Cand the positive orien-
tation on qigiven by −−→
CVi. In this case, for any point Pthere are infinitely many
generalized Gergonne points which lie on the line P C .
6 Further research
Conics Γ, ∆ and ci,(i= 1,2,3) play central role in the construction. The affine
types of these conics however, may only be determined by analytical approach or
by closer study of the type of projective pencils determined by cevians. It is also a
topic of further research how the types of solutions depend on the ratios mentioned
in Remark 1. The exact representation of the length of the radius by the given
data can also be discussed analytically in a further study.
References
[1] J.N. Boyd and P.N. Raychowdhury, The Gergonne point generalized through
convex coordinates, Int. J. Math. Math. Sci.,22 (1999), 423-430.
11
I
H
1
V
3
V
2
V
3
b
2
b1
b
3
J
3
b
,
2
b
,
1
b
,
1
J
2
J
I
1
J
2
J
3
J
P
Figure 10: For an arbitrary point Pthe generalized Gergonne point is the inter-
section of the corresponding hyperbola Γ and the line P I or P Ji.
[2] M. Hajja, P. Walker, The inspherical Gergonne center of a tetrahedron, Journal
for Geometry and Graphics,8(2004), 23-32.
[3] M. Hoffmann and S. Gorjanc, On the generalized Gergonne point and beyond,
Forum Geom.,8(2008) 151-155.
[4] C. Kimberling, Gergonne Point.
http://faculty.evansville.edu/ck6/tcenters/class/gergonne.html.
[5] V. Konecny, Problem department, Math. Mag.,69 (1990), 130-131.
[6] B. Odehnal, Beitr¨age zur Algebra und Geometrie, 51 (2010) 477-491.
[7] O. Veblen and J. W. Young, Projective Geometry, Boston, MA, Ginn, 1938.
[8] H. Wieleitner, Theorie der ebenen algebraischen Kurven h¨oherer Ordnung, G.J.
G¨oschensche Verlagshandlung, Leipzig, 1905.
[9] B. Wojtowicz, Desargues’ Configuration in a Special Layout, Journal for Ge-
ometry and Graphics,7(2003), 191-199.
12
