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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-

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12