the circles of lester, evans, parry, and their generalizations · 2010. 12. 21. · the circles of...

Forum GeometricorumVolume 10 (2010) 175–209. � �

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FORUM GEOM

ISSN 1534-1178

The Circles of Lester, Evans, Parry,and Their Generalizations

Paul Yiu

Abstract. Beginning with the famous Lester circle containing the circumcenter,nine-point center and the two Fermat points of a triangle, we survey a number ofinteresting circles in triangle geometry.

1. Introduction

This paper treats a number of interesting circles discovered by June Lester,Lawrence Evans, and Cyril Parry. We prove their existence and establish theirequations. Lester [12] has discovered that the Fermat points, the circumcenter, andthe nine-point center are concyclic. We call this the first Lester circle, and study itin §§3 – 6. Lester also conjectured in [12] the existence of a circle through the sym-median point, the Feuerbach point, the Clawson point, and the homothetic centerof the orthic and the intangent triangles. This conjecture is validated in §15. Evans,during the preparation of his papers in Forum Geometricorum, has communicatedtwo conjectures on circles through two perspectors V± which has since borne hisname. In §9 we study in detail the first Evans circle in relation to the excentralcircle. The second one is established in §14. In [9], a great number of circles havebeen reported relating to the Parry point, a point on the circumcircle. These circlesare studied in §§10 – 12.

CONTENTS

1. Introduction 1752. Preliminaries 1762.1. Intersection of a circle with the circumcircle 1762.2. Construction of circle equation 1772.3. Some common triangle center functions 1773. The first Lester circle 1784. Gibert’s generalization of the first Lester circle 1825. Center of the first Lester circle 183

Publication Date: December 21, 2010. Communicating Editor: Nikolaos Dergiades.This paper is an extended version of a presentation with the same title at the Invited Paper Session:

Classical Euclidean Geometry in MathFest, July 31–August 2, 2008 Madison, Wisconsin, USA.Thanks are due to Nikolaos Dergiades for many improvements of the paper, especially on the proofof Theorem 29.

176 P. Yiu

6. Equations of circles 1846.1. The circle F+F−G 1846.2. The circle F+F−H 1856.3. The first Lester circle 1867. The Brocard axis and the Brocard circle 1867.1. The Brocard axis 1867.2. The Brocard circle 1867.3. The isodynamic points 1878. The excentral triangle 1898.1. Change of coordinates 1899. The first Evans circle 1909.1. The Evans perspector W 1909.2. Perspectivity of the excentral triangle and Kiepert triangles 1919.3. The first Evans circle 19410. The Parry circle and the Parry point 19610.1. The center of the Parry circle 19811. The generalized Parry circles 19812. Circles containing the Parry point 19912.1. The circle F+F−G 19912.2. The circle GOK 20013. Some special circles 20113.1. The circle HOK 20113.2. The circle NOK 20214. The second Evans circle 20315. The second Lester circle 204References 209

2. Preliminaries

We refer to [15] for the standard notations of triangle geometry. Given a triangleABC , with sidelengths a, b, c, the circumcircle is represented in homogeneousbarycentric coordinates by the equation

a2yz + b2zx + c2xy = 0.

The equation of a general circle C is of the forma2yz + b2zx + c2xy + (x + y + z) · L(x, y, z) = 0 (1)

where L(x, y, z) is a linear form, and the line L(x, y, z) = 0 is the radical axis ofthe circle C and the circumcircle.

2.1. Intersection of a circle with the circumcircle . The intersections of the circleC with the circumcircle can certainly be determined by solving the equations

a2yz + b2zx + c2xy = 0,

L(x, y, z) = 0

The circles of Lester, Evans, Parry, and their generalizations 177

simultaneously. Here is an interesting special case where these intersections canbe easily identified. We say that a triangle center function f(a, b, c) represents aninfinite point if

f(a, b, c) + f(b, c, a) + f(c, a, b) = 0. (2)

Proposition 1. If a circle C is represented by an equation (1) in whichL(x, y, z) = F (a, b, c) ·

∑cyclic

b2c2 · f(a, b, c) · g(a, b, c)x, (3)

where F (a, b, c) is symmetric in a, b, c, and f(a, b, c), g(a, b, c) are triangle centerfunctions representing infinite points, then the circle intersects the circumcircle atthe points

Qf :=(

a2

f(a, b, c):

b2

f(b, c, a):

c2

f(c, a, b)

)

and

Qg :=(

a2

g(a, b, c):

b2

g(b, c, a):

c2

g(c, a, b)

).

Proof. The line L(x, y, z) = 0 clearly contains the point Qf , which by (2) isthe isogonal conjugate of an infinite point, and so lies on the circumcircle. It istherefore a common point of the circumcircle and C. The same reasoning appliesto the point Qg. �

For an application, see Remark after Proposition 11.

2.2. Construction of circle equation. Suppose we know the equation of a circlethrough two points Q1 and Q2, in the form of (1), and the equation of the lineQ1Q2, in the form L′(x, y, z) = 0. To determine the equation of the circle throughQ1, Q2 and a third point Q = (x0, y0, z0) not on the line Q1Q2, it is enough tofind t such that

a2y0z0 + b2z0x0 + c2x0y0 +(x0 + y0 + z0)(L(x0, y0, z0)+ t ·L′(x0, y0, z0)) = 0.With this value of t, the equation

a2yz + b2zx + c2xy + (x + y + z)(L(x, y, z) + t · L′(x, y, z)) = 0represents the circle Q1Q2Q. For an application of this method, see §6.3 (11) andProposition 11.

2.3. Some common triangle center functions. We list some frequently occurringhomogeneous functions associated with the coordinates of triangle centers or co-efficients in equations of lines and circles. An asterisk indicates that the functionrepresents an infinite point.

178 P. Yiu

Quartic forms

f4,1 := a2(b2 + c2) − (b2 − c2)2f4,2 := a2(b2 + c2) − (b4 + c4)f4,3 := a4 − (b4 − b2c2 + c4)f4,4 := (b2 + c2 − a2)2 − b2c2

* f4,5 := 2a4 − a2(b2 + c2) − (b2 − c2)2f4,6 := 2a4 − 3a2(b2 + c2) + (b2 − c2)2

* f4,7 := 2a4 − 2a2(b2 + c2) − (b4 − 4b2c2 + c4)

Sextic forms

f6,1 := a6 − 3a4(b2 + c2) + a2(3b4 − b2c2 + 3c4) − (b2 + c2)(b2 − c2)2* f6,2 := 2a6 − 2a4(b2 + c2) + a2(b4 + c4) − (b2 + c2)(b2 − c2)2* f6,3 := 2a6 − 6a4(b2 + c2) + 9a2(b4 + c4) − (b2 + c2)3

Octic forms

f8,1 := a8 − 2a6(b2 + c2) + a4b2c2 + a2(b2 + c2)(2b4 − b2c2 + 2c4)−(b8 − 2b6c2 + 6b4c4 − 2b2c6 + c8)

* f8,2 := 2a8 − 2a6(b2 + c2) − a4(3b4 − 8b2c2 + 3c4)+4a2(b2 + c2)(b2 − c2)2 − (b2 − c2)2(b4 + 4b2c2 + c4)

* f8,3 := 2a8 − 5a6(b2 + c2) + a4(3b4 + 8b2c2 + 3c4)+a2(b2 + c2)(b4 − 5b2c2 + c4) − (b2 − c2)4

f8,4 := 3a8 − 8a6(b2 + c2) + a4(8b4 + 7b2c2 + 8c4)−a2(b2 + c2)(4b4 − 3b2c2 + 4c4) + (b4 − c4)2

3. The first Lester circle

Theorem 2 (Lester). The Fermat points, the circumcenter, and the nine-point cen-ter of a triangle are concyclic.

A

B C

F−

F+

N

O

H

Figure 1. The first Lester circle through O, N and the Fermat points

Our starting point is a simple observation that the line joining the Fermat pointsintersects the Euler line at the midpoint of the orthocenter H and the centroid G.


Clearing denominators in the homogeneous barycentric coordinates of the Fermatpoint

F+ =(

1√3SA + S

,1√

3SB + S,

1√3SC + S

),

we rewrite it in the form

F+ = (3SBC +S2, 3SCA +S2, 3SAB +S2)+√

3S(SB +SC , SC +SA, SA+SB).

This expression shows that F+ is a point of the line joining the symmedian point

K = (SB + SC , SC + SA, SA + SB)

to the point

M = (3SBC + S2, 3SCA + S2, 3SAB + S2)

= 3(SBC , SCA, SAB) + S2(1, 1, 1).

Note that M is the midpoint of the segment HG, where H = (SBC , SCA, SAB) isthe orthocenter and G = (1, 1, 1) is the centroid. It is the center of the orthocen-troidal circle with HG as diameter. Indeed, F+ divides MK in the ratio

MF+ : F+K = 2√

3S(SA + SB + SC) : 6S2 = (SA + SB + SC) :√

3S.

With an obvious change in sign, we also have the negative Fermat point F− divid-ing MK in the ratio

MF− : F−K = (SA + SB + SC) : −√

3S.

We have therefore established

Proposition 3. The Fermat points divide MK harmonically.

A

B C

F−

F+O

H

M

G

K

Figure 2. Intersection of Fermat line and Euler line

This simple observation suggests a proof of Lester’s circle theorem by the inter-section chords theorem.

180 P. Yiu

Proposition 4. The following statements are equivalent.(A) MF+ · MF− = MO · MN .(B) The circle F+F−G is tangent to the Euler line at G, i.e., MF+ ·MF− = MG2.(C) The circle F+F−H is tangent to the Euler line at H , i.e., MF+·MF− = MH2.(D) The Fermat points are inverse in the orthocentroidal circle.

2d d d 2dH OGNM

Figure 3. The Euler line

Proof. Since M is the midpoint of HG, the statements (B), (C), (D) are clearlyequivalent. On the other hand, putting OH = 6d, we have

MO · MN = (MH)2 = (MG)2 = 4d2,see Figure 3. This shows that (A), (B), (C) are equivalent. �

Note that (A) is Lester’s circle theorem (Theorem 2). To complete its proof, itis enough to prove (D). We do this by a routine calculation.

Theorem 5. The Fermat points are inverse in the orthocentroidal circle.

O

H

M

G

F−

F+

A

B C

Figure 4. F+ on the polar of F− in the orthocentroidal circle

Proof. The equation of the orthocentroidal circle is

3(a2yz + b2zx + c2xy) − 2(x + y + z)(SAx + SBy + SCz) = 0,


equivalently, 1

−2(SAx2 +SBy2 +SCz2)+((SB +SC)yz+(SC +SA)zx+(SA +SB)xy) = 0.This is represented by the matrix

M =

⎛⎝ −4SA SA + SB SA + SCSA + SB −4SB SB + SC

SA + SC SB + SC −4SC

⎞⎠ .

The coordinates of the Fermat points can be written as

F+ = X + Y and F− = X − Y,with

X =(3SBC + S2 3SCA + S2 3SAB + S2

),

Y =√

3S(SB + SC SC + SA SA + SB

).

With these, we have

XMXt = Y MY t = 6S2(SA(SB − SC)2 + SB(SC − SA)2 + SC(SA − SB)2),and

F+MFt− = (X + Y )M(X − Y )t = XMXt − Y MY t = 0.

This shows that the Fermat points are inverse in the orthocentroidal circle. �The proof of Theorem 2 is now complete, along with tangency of the Euler line

with the two circles F+F−G and F+F−H (see Figure 5). We call the circle throughO, N , and F± the first Lester circle.

Z1 Z0

A

B C

F−

F+

N

O

H

M

G

Ki

Figure 5. The circles F+F−G and F+F−H

1It is easy to see that this circle contains H and G. The center of the circle (see [15, §10.7.2]) isthe point M = (SA(SB + SC) + 4SBC : SB(SC + SA) + 4SCA : SC(SA + SB) + 4SAB) on theEuler line, which is necessarily the midpoint of HG.

182 P. Yiu

Remarks. (1) The tangency of the circle F+F−G and the Euler line was noted in[9, pp.229–230].

(2) The symmedian point K and the Kiepert center Ki (which is the midpointof F+F−) are inverse in the orthocentroidal circle.

4. Gibert’s generalization of the first Lester circle

Bernard Gibert [7] has found an interesting generalization of the first Lester cir-cle, which we explain as a natural outgrowth of an attempt to compute the equationsof the circles F+F−G and F+F−H .

Theorem 6 (Gibert). Every circle whose diameter is a chord of the Kiepert hyper-bola perpendicular to the Euler line passes through the Fermat points.

KiZ1 Z0

A

B C

F−

F+

N

O

HM

G

Y1

Y0

Figure 6. Gibert’s generalization of the first Lester circle

Proof. Since F± and G are on the Kiepert hyperbola, and the center of the cir-cle F+F−G is on the perpendicular to the Euler line at G, this line intersects theKiepert hyperbola at a fourth point Y0 (see Figure 6), and the circle is a mem-ber of the pencil of conics through F+, F−, G and Y0. Let L(x, y, z) = 0 andL0(x, y, z) = 0 represent the lines F+F− and GY0 respectively. We may assumethe circle given by

k0((b2 − c2)yz + (c2 − a2)zx + (a2 − b2)xy) − L(x, y, z) · L0(x, y, z) = 0for an appropriately chosen constant k0.


Replacing G by H and Y0 by another point Y1, the intersection of the Kieperthyperbola with the perpendicular to the Euler line at H , we write the equation ofthe circle F+F−H in the form

k1((b2 − c2)yz + (c2 − a2)zx + (a2 − b2)xy) − L(x, y, z) · L1(x, y, z) = 0,where L1(x, y, z) = 0 is the equation of the line HY1.

The midpoints of the two chords GY0 and HY1 are the centers of the two circlesF+F−G and F+F−H . The line joining them is therefore the perpendicular bisectorof F+F−.

Every line perpendicular to the Euler line is represented by an equation

Lt(x, y, z) := tL0(x, y, z) + (1 − t)L1(x, y, z) = 0for some real number t. Let kt := tk0 + (1 − t)k1 correspondingly. Then theequation

kt((b2 − c2)yz + (c2 − a2)zx + (a2 − b2)xy) − L(x, y, z) · Lt(x, y, z) = 0represents a circle Ct through the Fermat points and the intersections of the lineLt(x, y, z) = 0 and the Kiepert hyperbola. The perpendicular bisector of F+F− isthe diameter of the family of parallel lines Lt(x, y, z) = 0. Therefore the center ofthe circle is the midpoint of the chord cut out by Lt(x, y, z) = 0. �

Remark. If the perpendicular to the Euler line intersects it outside the segmentHG, then the circle intersects the Euler line at two points dividing the segmentHG harmonically, say in the ratio τ : 1 − τ for τ < 0 or τ > 1. In this case, theline divides HG in the ratio −τ2 : (1 − τ)2.

5. Center of the first Lester circle

Since the circumcenter O and the nine-point center N divides the segment HGin the ratio 3 : ∓1, the diameter of the first Lester circle perpendicular to the Eulerline intersects the latter at the point L dividing HG in the ratio 9 : −1. This is themidpoint of ON (see Figure 7), and has coordinates

(f4,6(a, b, c) : f4,6(b, c, a) : f4,6(c, a, b)).

As such it is the nine-point center of the medial triangle, and appears as X140 in[10].

6 2 1 3H OGN L

Figure 7. The Euler line

Proposition 7. (a) Lines perpendicular to the Euler line have infinite point

X523 = (b2 − c2, c2 − a2, a2 − b2).

184 P. Yiu

(b) The diameter of the first Lester circle perpendicular to the Euler line is alongthe line ∑

cyclic

f6,1(a, b, c)x = 0. (4)

Proposition 8. (a) The equation of the line F+F− is∑cyclic

(b2 − c2)f4,4(a, b, c)x = 0. (5)

(b) The perpendicular bisector of F+F− is the linex

b2 − c2 +y

c2 − a2 +z

a2 − b2 = 0. (6)

Proof. (a) The line F+F− contains the symmedian point K and the Kiepert center

Ki = ((b2 − c2)2, (c2 − a2)2, (a2 − b2)2).(b) The perpendicular bisector of F+F− is the perpendicular at Ki to the line

KKi, which has infinite point

X690 = ((b2 − c2)(b2 + c2 − 2a2), (c2 − a2)(c2 + a2 − 2b2), (a2 − b2)(a2 + b2 − 2c2)).�

Proposition 9. The center of the first Lester circle has homogeneous barycentriccoordinates

((b2 − c2)f8,3(a, b, c) : (c2 − a2)f8,3(b, c, a) : (a2 − b2)f8,3(c, a, b)).Proof. This is the intersection of the lines (4) and (6). �

Remarks. (1) The center of the first Lester circle appears as X1116 in [10].(2) The perpendicular bisector of F+F− also contains the Jerabek center

Je = ((b2−c2)2(b2 +c2−a2), (c2−a2)2(c2 +a2−b2), (a2−b2)2(a2 +b2−c2)),which is the center of the Jerabek hyperbola, the isogonal conjugate of the Eulerline. It follows that Je is equidistant from the Fermat points. The points Ki and Jeare the common points of the nine-point circle and the pedal circle of the centroid.

6. Equations of circles

6.1. The circle F+F−G. In the proof of Theorem 6, we take

L(x, y, z) =∑cyclic

(b2 − c2)f4,4(a, b, c)x, (7)

L0(x, y, z) =∑cyclic

(b2 + c2 − 2a2)x (8)

for the equation of the line F+F− (Proposition 8(a)) and the perpendicular to theEuler line at G. Now, we seek a quantity k0 such that the member

k0((b2 − c2)yz + (c2 − a2)zx + (a2 − b2)xy) − L(x, y, z) · L0(x, y, z) = 0


of the pencil of conic through the four points F±, G, Y0 is a circle. For this,

k0 = −3(a2(c2 − a2)(a2 − b2) + b2(a2 − b2)(b2 − c2) + c2(b2 − c2)(c2 − a2)),and the equation can be reorganized as

9(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy)

+ (x + y + z)

⎛⎝ ∑

cyclic

(b2 − c2)(b2 + c2 − 2a2)f4,4(a, b, c)x⎞⎠ = 0. (9)

The center of the circle F+F−G is the point

Z0 := ((b2 − c2)f4,7(a, b, c) : (c2 − a2)f4,7(b, c, a) : (a2 − b2)f4,7(c, a, b)).

The point Y0 has coordinates(

b2−c2b2+c2−2a2 : · · · : · · ·

).

6.2. The circle F+F−H . With the line

L1(x, y, z) =∑cyclic

(b2 + c2 − a2)(2a4 − a2(b2 + c2) − (b2 − c2)2)x = 0

perpendicular to the Euler line at H , we seek a number k1 such that

k1((b2 − c2)yz + (c2 − a2)zx + (a2 − b2)xy) − L(x, y, z) · L1(x, y, z) = 0of the pencil of conic through the four points F±, H , Y1 is a circle. For this,

k1 = 16Δ2(a4(b2 + c2 − a2) + b4(c2 + a2 − b2) + c4(a2 + b2 − c2) − 3a2b2c2),and the equation can be reorganized as

48(b2 − c2)(c2 − a2)(a2 − b2)Δ2(a2yz + b2zx + c2xy)

− (x + y + z)⎛⎝ ∑

cyclic

(b2 − c2)(b2 + c2 − a2)f4,4(a, b, c)f4,5(a, b, c)x⎞⎠ = 0.

(10)

This is the equation of the circle F+F−H . The center is the point

Z1 := ((b2 − c2)f8,2(a, b, c) : (c2 − a2)f8,2(b, c, a) : (a2 − b2)f8,2(c, a, b)).The triangle center

Y1 =(

b2 − c2f4,5(a, b, c)

:c2 − a2

f4,5(b, c, a):

a2 − b2f4,5(c, a, b)

)

is X2394.

186 P. Yiu

6.3. The first Lester circle. Since the line joining the Fermat points has equationL(x, y, z) = 0 with L given by (7), every circle through the Fermat points isrepresented by


+ (x + y + z)

⎛⎝ ∑

cyclic

(b2 − c2)(b2 + c2 − 2a2 + t)f4,4(a, b, c)x⎞⎠ = 0 (11)

for an appropriate choice of t. The value of t for which this circle passes throughthe circumcenter is

t =a2(c2 − a2)(a2 − b2) + b2(a2 − b2)(b2 − c2) + c2(b2 − c2)(c2 − a2)

32Δ2.

The equation of the circle is

96Δ2(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy)

+ (x + y + z)

⎛⎝ ∑

cyclic

(b2 − c2)f4,4(a, b, c)f6,1(a, b, c)x⎞⎠ = 0.

7. The Brocard axis and the Brocard circle

7.1. The Brocard axis . The isogonal conjugate of the Kiepert perspector K(θ) isthe point

K∗(θ) = (a2(SA + Sθ), b2(SB + Sθ), c2(SC + Sθ)),

which lies on the line joining the circumcenter O and the symmedian point K. Theline OK is called the Brocard axis. It is represented by the equation∑

cyclic

b2c2(b2 − c2)x = 0. (12)

7.2. The Brocard circle. The Brocard circle is the circle with OK as diameter. Itis represented by the equation

(a2+b2+c2)(a2yz+b2zx+c2xy)−(x+y+z)(b2c2x+c2a2y+a2b2z) = 0. (13)It is clear from

K∗(θ) = (a2SA, b2SB , c2SC) + Sθ(a2, b2, c2),

K∗(−θ) = (a2SA, b2SB , c2SC) − Sθ(a2, b2, c2)that K∗(θ) and K∗(−θ) divide O and K harmonically, and so are inverse in theBrocard circle. The points K∗

(±π3 ) are called the isodynamic points, and aremore simply denoted by J±.

Proposition 10. K∗(±θ) are inverse in the circumcircle if and only if they are theisodynamic points.


7.3. The isodynamic points. The isodynamic points J± are also the common pointsof the three Apollonian circles, each orthogonal to the circumcircle at a vertex(see Figure 8). Thus, the A-Apollonian circle has diameter the endpoints of thebisectors of angle A on the sidelines BC . These are the points (b, ±c). The centerof the circle is the midpoint of these, namely, (b2, −c2). The circle has equation

(b2 − c2)(a2yz + b2zx + c2xy) + a2(x + y + z)(c2y − b2z) = 0.Similarly, the B- and C-Apollonian circles have equations

(c2 − a2)(a2yz + b2zx + c2xy) + b2(x + y + z)(a2z − c2x) = 0,(a2 − b2)(a2yz + b2zx + c2xy) + c2(x + y + z)(b2x − a2y) = 0.

These three circles are coaxial. Their centers lie on the Lemoine axisx

a2+

y

b2+

z

c2= 0, (14)

which is the perpendicular bisector of the segment J+J−.

O

A

B C

K

J+

J−

Figure 8. The Apollonian circles and the isodynamic points

188 P. Yiu

Proposition 11. Every circle through the isodynamic points can be represented byan equation


+ (x + y + z)

⎛⎝ ∑

cyclic

b2c2(b2 − c2)(b2 + c2 − 2a2 + t)x⎞⎠ = 0 (15)

for some choice of t.

Proof. Combining the above equations for the three Apollonian circles, we obtain

3(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy)+ (x + y + z)

∑cyclic

a2(c2 − a2)(a2 − b2)(c2y − b2z) = 0.

A simple rearrangement of the terms brings the radical axis into the form


+ (x + y + z)

⎛⎝ ∑

cyclic

b2c2(b2 − c2)(b2 + c2 − 2a2)x⎞⎠ = 0. (16)

Now, the line containing the isodynamic points is the Brocard axis given by (12). Itfollows that every circle through J± is represented by (15) above for some choiceof t (see §2.2). �Remark. As is easily seen, equation (16) is satisfied by x = y = z = 1, and sorepresents the circle through J± and G. Since the factors b2 − c2 and b2 + c2− 2a2yield infinite points, applying Proposition 1, we conclude that this circle intersects

the circumcircle at the Euler reflection point E =(

a2

b2−c2 : · · · : · · ·)

and the Parry

point(

a2

b2+c2−2a2 : · · · : · · ·)

.

This is the Parry circle we consider in §10 below.Proposition 12. The circle through the isodynamic points and the orthocenter hasequation

16Δ2 · (b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy)

+ (x + y + z)

⎛⎝ ∑

cyclic

b2c2(b2 − c2)(b2 + c2 − a2)f4,1(a, b, c)x⎞⎠ = 0.

Its center is the point

Z3 := (a2(b2 − c2)(b2 + c2 − a2)f4,1(a, b, c) : · · · : · · · ).


8. The excentral triangle

The excentral triangle IaIbIc has as vertices the excenters of triangle ABC . Ithas circumradius 2R and circumcenter I′, the reflection of I in O (see Figure 9).Since the angles of the excentral triangle are 12(B + C),

12 (C + A), and

12 (A+ B),

its sidelengths a′ = IbIc, b′ = IcIa, c′ = IaIb satisfy

a′2 : b′2 : c′2 = cos2A

2: cos2

B

2: cos2

C

2= a(b + c − a) : b(c + a − b) : c(a + b − c).

O

I

Ia

Ib

Ic

I′

A

BC

Figure 9. The excentral triangle and its circumcircle

8.1. Change of coordinates. A point with homogeneous barycentric coordinates(x, y, z) with reference to ABC has coordinates

(x′, y′, z′) = (a(b+c−a)(cy+bz), b(c+a−b)(az+cx), c(a+b−c)(bx+ay))with reference to the excentral triangle.

Consider, for example, the Lemoine axis of the excentral triangle, with equation

x′

a′2+

y′

b′2+

z′

c′2= 0.

With reference to triangle ABC , the same line is represented by the equation

a(b + c − a)(cy + bz)a(b + c − a) +

b(c + a − b)(az + cx)b(c + a − b) +

c(a + b − c)(bx + ay)c(a + b − c) = 0,

which simplifies into

(b + c)x + (c + a)y + (a + b)z = 0. (17)

190 P. Yiu

On the other hand, the circumcircle of the excentral triangle, with equation

a′2

x′+

b′2

y′+

c′2

z′= 0,

is represented by1

cy + bz+

1az + cx

+1

bx + ay= 0

with reference to triangle ABC . This can be rearranged as

a2yz + b2zx + c2ay + (x + y + z)(bcx + cay + abz) = 0. (18)

9. The first Evans circle

9.1. The Evans perspector W . Let A∗, B∗, C∗ be respectively the reflections ofA in BC , B in CA, C in AB. The triangle A∗B∗C∗ is called the triangle ofreflections of ABC . Larry Evans has discovered the perspectivity of the excentraltriangle and A∗B∗C∗.

Theorem 13. The excentral triangle and the triangle of reflections are perspectiveat a point which is the inverse image of the incenter in the circumcircle of theexcentral triangle.

X

A

BC

A∗

Ia

W

I

O

I′

Y

Y ′

I∗

Figure 10. The Evans perspector W


Proof. We show that the lines IaA∗, IbB∗, and IcC∗ intersect the line OI at thesame point. Let X be the intersection of the lines AA∗ and OI (see Figure 10). Ifha is the A-altitude of triangle ABC , and the parallel from I to AA∗ meets IaA∗at I∗, then since

II∗

2ha=

II∗

AA∗=

IaI

IaA=

Y Y ′

AY ′=

a

s,

we have II∗ = 2ahas = 4r andWI′WI =

I′IaII∗ =

2R4r . Therefore, W divides II

′ in theratio

I ′W : WI = R : −2r.Since this ratio is a symmetric function of the sidelengths, we conclude that thesame point W lies on the lines IbB∗ and IcC∗. Moreover, since I′W = RR−2r ·I ′I ,by the famous Euler formula OI2 = R(R − 2r), we haveI ′W · I ′I = R

R − 2r · I′I2 =

4RR − 2r · OI

2 =4R

R − 2r · R(R − 2r) = (2R)2.

This shows that W and I are inverse in the circumcircle of the excentral triangle.�

The point W is called the Evans perspector; it has homogeneous barycentriccoordinates,

W =(a(a3 + a2(b + c) − a(b2 + bc + c2) − (b + c)(b − c)2) : · · · : · · · )= (a((a + b + c)(c + a − b)(a + b − c) − 3abc) : · · · : · · · ).

It appears as X484 in [10].

9.2. Perspectivity of the excentral triangle and Kiepert triangles.

Lemma 14. Let XBC and X′IbIc be oppositely oriented similar isosceles trian-gles with bases BC and IbIc respectively. The lines IaX and IaX ′ are isogonalwith respect to angle Ia the excentral triangle (see Figure 11).

X

A

B C

Ia

Ib

Ic

X′

θ

θ

Figure 11. Isogonal lines joining Ia to apices of similar isosceles on BC and IbIc

192 P. Yiu

Proof. The triangles IaBC and IaIbIc are oppositely similar since BC and IbIcare antiparallel. In this similarity X and X′ are homologous points. Hence, thelines IaX and IaX ′ are isogonal in the excentral triangle. �

We shall denote Kiepert perspectors with reference to the excentral triangle byKe(−).Theorem 15. The excentral triangle and the Kiepert triangle K(θ) are perspectiveat the isogonal conjugate of Ke(−θ) in the excentral triangle.Proof. Let XY Z be a Kiepert triangle K(θ). Construct X′, Y ′, Z ′ as in Lemma14 (see Figure 11).

(i) IaX ′, IbY ′, IcZ ′ concur at the Kiepert perspector Ke(−θ) of the excentraltriangle.

(ii) Since IaX and IaX ′ are isogonal with respect to Ia, and similarly for thepairs IbY , IbY ′ and IcZ and IcZ ′, the lines IaX, IbY , IcZ concur at the isogonalconjugate of Ke(−θ) in the excentral triangle. �

Ia

Ib

Ic

V−

A

B C

C′′

A′′

B′′

Figure 12. Evans’ perspector V− of K(−π

3

)and excentral triangle

We denote the perspector in Theorem 15 by V (θ), and call this a generalizedEvans perspector. In particular, V

(π3

)and V

(−π3 ) are the isodynamic pointsof the excentral triangle, and are simply denoted by V+ and V− respectively (see


Figure 12 for V−). These are called the second and third Evans perspectors respec-tively. They are X1276 and X1277 of [10].

Proposition 16. The line V+V− has equation∑cyclic

(b − c)(b2 + c2 − a2)x = 0. (19)

Proof. The line V+V− is the Brocard axis of the excentral triangle, with equation∑cyclic

b′2 − c′2a′2

· x′ = 0

with reference to the excentral triangle (see §7.1). Replacing these by parameterwith reference to triangle ABC , we have

∑cyclic(b− c)(b + c− a)(cy + bz) = 0.

Rearranging terms, we have the form (19) above. �

Proposition 17. The Kiepert triangle K(θ) is perspective with the triangle of re-flections A∗B∗C∗ if and only if θ = ±π3 . The perspector is K∗(−θ), the isogonalconjugate of K(−θ).

This means that for ε = ±1, the Fermat triangle K(ε · π3 ) and the triangle ofreflections are perspective at the isodynamic point J−ε (see Figure 13 for the caseε = −1).

J+

A

B C

C′′

A′′

B′′

A∗

B∗

C∗

Figure 13. K (−π3

)and A∗B∗C∗ perspective at J+

194 P. Yiu

9.3. The first Evans circle. Since V± are the isodynamic points of the excentraltriangle, they are inverse in the circumcircle of the excentral triangle. Since Wand I are also inverse in the same circle, we conclude that V+, V−, I , and W areconcyclic (see [1, Theorem 519]). We call this the first Evans circle. A strongerresult holds in view of Proposition 10.

Theorem 18. The four points V (θ), V (−θ), I , W are concyclic if and only ifθ = ±π3 .

A

B

C

A∗

B∗

C∗

Ia

Ic

Ib

I

W

V−

V+

I′

Figure 14. The first Evans circle

We determine the center of the first Evans circle as the intersection of the per-pendicular bisectors of the segments IW and V+V−.

Lemma 19. The perpendicular bisector of the segment IW is the line

bc(b + c)x + ca(c + a)y + ab(a + b)z = 0. (20)

Proof. If M ′ is the midpoint of IW , then since O is the midpoint of II′, fromthe degenerate triangle II′W we have OM ′ = I

′W2 =

R2

OI . This shows that the


midpoint of IW is the inverse of I in the circumcircle. Therefore, the perpendicularbisector of IW is the polar of I in the circumcircle. This is the line

(a b c

) ⎛⎝ 0 c2 b2

c2 0 a2

b2 a2 0

⎞⎠

⎛⎝xy

z

⎞⎠ = 0,

which is the same as (20). �

Remark. M ′ = (a2(a2−b2+bc−c2) : b2(b2−c2+ca−a2) : c2(c2−a2+ab−b2))is the triangle center X36 in [10].

Lemma 20. The perpendicular bisector of the segment V+V− is the line

(b + c)x + (c + a)y + (a + b)z = 0. (21)

Proof. Since V+ and V− are the isodynamic points of the excentral triangle, theperpendicular bisector of V+V− is the polar of the symmedian point of the excen-tral triangle with respect to its own circumcircle. With reference to the excentraltriangle, its Lemoine axis has equation

x′

a′2+

y′

b′2+

z′

c′2= 0.

Changing coordinates, we have, with reference to ABC , the same line representedby the equation

a(b + c − a)(cy + bz)a(b + c − a) +

b(c + a − b)(az + cx)b(c + a − b) +

c(a + b − c)(bx + ay)c(a + b − c) = 0,

which simplifies into (21). �

Proposition 21. The center of the first Evans circle is the point(

a(b − c)b + c

:b(c − a)c + a

:c(a − b)a + b

).

Proof. This is the intersection of the lines (20) and (21). �

Remark. The center of the first Evans circle is the point X1019 in [10]. It is alsothe perspector of excentral triangle and the cevian triangle of the Steiner point.

Proposition 22. The equation of the first Evans circle is

(a − b)(b − c)(c − a)(a + b + c)(a2yz + b2zx + c2xy)

−(x + y + z)⎛⎝ ∑

cyclic

bc(b − c)(c + a)(a + b)(b + c − a)x⎞⎠ = 0. (22)

196 P. Yiu

A

B C

J+

J−

G

O

K

E

X111

X352

X353

X351

X23

X187

Figure 15. The Parry circle

10. The Parry circle and the Parry point

The Parry circle CP, according to [9, p.227], is the circle through a number ofinteresting triangle centers, including the isodynamic points and the centroid. Weshall define the Parry circle as the circle through these three points, and seek toexplain the incidence of the other points.

First of all, since the isodynamic points are inverse in each of the circumcircleand the Brocard circle, the Parry circle is orthogonal to each of these circles. Inparticular, it contains the inverse of G in the circumcircle. This is the trianglecenter

X23 = (a2(a4−b4 +b2c2−c4), b2(b4−c4 +c2a2−a4), c2(c4−a4 +a2b2−b4)).(23)

The equation of the Parry circle has been computed in §7.3, and is given by (16).Applying Proposition 1, we see that this circle contains the Euler reflection point

E =(

a2

b2 − c2 ,b2

c2 − a2 ,c2

a2 − b2)

(24)


and the point

P =(

a2

b2 + c2 − 2a2 ,b2

c2 + a2 − 2b2 ,c2

a2 + b2 − 2c2)

(25)

which we call the Parry point.

Remark. The line EP also contains the symmedian point K .

Lemma 23. The line EG is parallel to the Fermat line F+F−.

Proof. The line F+F− is the same as the line KKi, with equation given by (5).The line EG has equation∑

cyclic

(b2 − c2)f4,2(a, b, c)x = 0. (26)

Both of these lines have the same infinite point

X542 = (f6,2(a, b, c) : f6,2(b, c, a) : f6,2(c, a, b)).

�

Proposition 24. The Euler reflection point E and the centroid are inverse in theBrocard circle.

A

B C

O

K G

E

F+

F−

H

M

Figure 16. E and G are inverse in the Brocard circle

Proof. Note that the line F+F− intersects the Euler line at the midpoint M of HG,and G is the midpoint of OM . Since EG is parallel to MK , it intersects OK atits midpoint, the center of the Brocard circle. Since the circle through E, G, J± isorthogonal to the Brocard circle, E and G are inverse to each other with respect tothis circle. �

198 P. Yiu

The following two triangle centers on the Parry circle are also listed in [9]:(i) the second intersection with the line joining G to K , namely,

X352 = (a2(a4 − 4a2(b2 + c2) + (b4 + 5b2c2 + c4)), · · · , · · · ),(ii) the second intersection with the line joining X23 to K , namely,

X353 = (a2(4a4 − 4a2(b2 + c2) − (2b4 + b2c2 + 2c4)), · · · , · · · ),which is the inverse of the Parry point P in the Brocard circle, and also the inverseof X352 in the circumcircle.

10.1. The center of the Parry circle.

Proposition 25. The perpendicular bisector of the segment GE is the linex

b2 + c2 − 2a2 +y

c2 + a2 − 2b2 +z

a2 + b2 − 2c2 = 0. (27)Proof. The midpoint of EG is the point

Z4 := ((b2+c2−2a2)f4,5(a, b, c) : (c2+a2−2b2)f4,5(b, c, a) : (a2+b2−2c2)f4,5(c, a, b)).By Lemma 23 and Proposition 8(b), the perpendicular bisector of EG has infinitepoint X690. The line through Z4 with this infinite point is the perpendicular bisectorof EG. �Proposition 26. The center of the Parry circle CP is the point

(a2(b2 − c2)(b2 + c2 − 2a2), · · · , · · · ).Proof. This is the intersection of the line (27) above and the the Lemoine axis(14). �Remark. The center of the Parry circle appears in [10] as X351.

11. The generalized Parry circles

We consider the generalized Parry circle CP(θ) passing through the centroid andthe points K∗(±θ) on the Brocard axis. Since K∗(θ) and K∗(−θ) are inverse inthe Brocard circle (see §7.2), the generalized Parry circle CP(θ) is orthogonal tothe Brocard circle, and must also contain the Euler reflection point E. Its equationis

3(b2 − c2)(c2 − a2)(a2 − b2)(16Δ2 sin2 θ − (a2 + b2 + c2)2 cos2 θ)(a2yz + b2zx + c2xy)+ (x + y + z)

∑cyclic

b2c2(b2 − c2)(f6,2(a, b, c) sin2 θ + f6,3(a, b, c) cos2 θ)x = 0.

The second intersection with the circumcircle is the point

Q(θ) =(

a2

f6,2(a, b, c) sin2 θ + f6,3(a, b, c) cos2 θ, · · · , · · ·

).

The Parry point P is Q(θ) for θ = π3 . Here are two more examples.(i) With θ = π2 , we have the circle GEO tangent to the Brocard axis and with

centerZ5 := (a2(b2 − c2)(b2 + c2 − 2a2)f4,4(a, b, c) : · · · : · · · ).


It intersects the circumcircle again at the point(a2

f6,2(a, b, c),

b2

f6,2(b, c, a),

c2

f6,2(c, a, b)

). (28)

This is the triangle center X842.(ii) With θ = 0, we have the circle GEK tangent to the Brocard axis and with

center

Z6 := (a2(b2 − c2)(b2 + c2 − 2a2)((a2 + b2 + c2)2 − 9b2c2) : · · · : · · · ).It intersects the circumcircle again at the point

Z7 :=(

a2

f6,3(a, b, c):

b2

f6,3(b, c, a):

c2

f6,3(c, a, b)

). (29)

A

B C

G

K

O

E

Z7

X842

Z6

Z5

Figure 17. The circles GEO and GEK

12. Circles containing the Parry point

12.1. The circle F+F−G. The equation of the circle F+F−G has been computedin §6.1, and is given by (9). Applying Proposition 1, we see that the circle F+F−Gcontains the Parry point and the point

Q =(

1(b2 − c2)f4,4(a, b, c) :

1(c2 − a2)f4,4(b, c, a) :

1(a2 − b2)f4,4(c, a, b)

).

200 P. Yiu

This is the triangle center X476 in [10]. It is the reflection of the Euler reflectionpoint in the Euler line. 2

A

B C

O

K G

E

J+

J−

P

F+

F−

X23

X476

Figure 18. Intersections of F+F−G and the circumcircle

12.2. The circle GOK . Making use of the equations (13) of the Brocard circle and(12) of the Brocard axis, we find equations of circles through O and K in the form

(a2+b2+c2)(a2yz+b2zx+c2xy)−(x+y+z)⎛⎝ ∑

cyclic

b2c2((b2 − c2)t + 1)x⎞⎠ = 0

(30)for suitably chosen t.

With t = −a4+b4+c4−b2c2−c2a2−a2b23(b2−c2)(c2−a2)(a2−b2) , and clearing denominators, we obtain theequation of the circle GOK .

3(b2 − c2)(c2 − a2)(a2 − b2)(a2 + b2 + c2)(a2yz + b2zx + c2xy)

+ (x + y + z)

⎛⎝ ∑

cyclic

b2c2(b2 − c2)(b2 + c2 − 2a2)2x⎞⎠ = 0.

2To justify this, one may compute the infinite point of the line EQ and see that it is X523 =(b2 − c2 : c2 − a2 : a2 − b2). This shows that EQ is perpendicular to the Euler line.


A

BC

HK OG

E

X691

P

X842

X112

Figure 19. The circles GOK and HOK

This circle GOK contains the Parry point P and the point

Q′ =(

a2

(b2 − c2)(b2 + c2 − 2a2) : · · · : · · ·)

,

which is the triangle center X691. It is the reflection of E in the Brocard axis.3

Remark. The line joining P to X691 intersects(i) the Brocard axis at X187, the inversive image of K in the circumcircle,(ii) the Euler line at X23, the inversive image of the centroid in the circumcircle.

13. Some special circles

13.1. The circle HOK . By the same method, with

t = −a4(c2 − a2)(a2 − b2) + b4(a2 − b2)(b2 − c2) + c4(b2 − c2)(c2 − a2)

16Δ2(b2 − c2)(c2 − a2)(a2 − b2)

3This may be checked by computing the infinite point of the line EQ′ as X512 = (a2(b2 −c2), b2(c2 − a2), c2(a2 − b2), the one of lines perpendicular to OK.

202 P. Yiu

in (30), we find the equation of the circle HOK as

16Δ2(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy)

+(x + y + z)

⎛⎝ ∑

cyclic

b2c2(b2 − c2)(b2 + c2 − a2)f6,2(a, b, c)x⎞⎠ = 0.

Therefore, the circle HOK intersects the circumcircle at(

a2

(b2 − c2)(b2 + c2 − a2) ,b2

(c2 − a2)(c2 + a2 − b2) ,c2

(a2 − b2)(a2 + b2 − c2))

,

which is the triangle center X112, and X842 given by (28).

Remarks. (1) The circle HOK has center

Z8 := (a2(b2 − c2)f4,2(a, b, c)f4,3(a, b, c) : · · · : · · · ).

(2) X112 is the second intersection of the circumcircle with the line joining X74with the symmedian point.

(3) X842 is the second intersection of the circumcircle with the parallel to OKthrough E. It is also the antipode of X691, which is the reflection of E in theBrocard axis.

(4) The radical axis with the circumcircle intersects the Euler line at X186 andthe Brocard axis at X187. These are the inverse images of H and K in the circum-circle.

13.2. The circle NOK . The circle NOK contains the Kiepert center because bothO, N and K, Ki are inverse in the orthocentroidal circle.

A

B C

F−

F+

N

O

H

K

Ki

M

Figure 20. The circle NOK


This circle has equation

32(a2 − b2)(b2 − c2)(c2 − a2)(a2 + b2 + c2)Δ2(a2yz + b2zx + c2xy)+

∑cyclic

b2c2(b2 − c2)f8,4(a, b, c)x = 0.

Its center is the point

Z9 := (a2(b2 − c2)f8,1(a, b, c) : · · · : · · · ).

14. The second Evans circle

Evans also conjectured that the perspectors V± = V (±π3 ) and X74, X399 areconcyclic. Recall that

X74 =(

a2

f4,5(a, b, c):

b2

f4,5(b, c, a):

c2

f4,5(c, a, b)

)

is the antipode on the circumcircle of the Euler reflection point E and X339, theParry reflection point, is the reflection of O in E.

We confirm Evans’ conjecture indirectly, by first finding the circle through V±and the point

X101 = (a2(c − a)(a − b) : b2(a − b)(b − c) : c2(b − c)(c − a))on the circumcircle. Making use of the equation (22) of the first Evans circle, andthe equation (19) of the line V+V−, we seek a quantity t such that

(a − b)(b − c)(c − a)(a + b + c)(a2yz + b2zx + c2xy)

−(x + y + z)⎛⎝ ∑

cyclic

(bc(b − c)(c + a)(a + b)(b + c − a) + t(b − c)(b2 + c2 − a2))x⎞⎠ = 0,

represents a circle through the point X101. For this, we require

t = −abc(a2(b + c − a) + b2(c + a − b) + c2(a + b − c) + abc)

(b + c − a)(c + a − b)(a + b − c) ,

and the equation of the circle through V± and X101 is

16Δ2(a − b)(b − c)(c − a)(a2yz + b2zx + c2xy)

− (x + y + z)⎛⎝ ∑

cyclic

b2c2(b − c)f4,5(a, b, c)x⎞⎠ = 0.

It is clear that this circle does also contain the point X74.The center of the circle is the point

Z10 = (a2(b − c)((b2 + c2 − a2)2 − b2c2), · · · , · · · ).Now, the perpendicular bisector of the segment OE is the line∑ x

a2f4,4(a, b, c)= 0,

204 P. Yiu

which clearly contains the center of the circle. Therefore, the circle also containsthe point which is the reflection of X74 in the midpoint of OE. This is the same asthe reflection of O in E, the Parry reflection point X399.

Theorem 27 (Evans). The four points V±, the antipode of the Euler reflection pointE on the circumcircle, and the reflection of O in E are concyclic (see Figure 21).

A

B

C

Ia

Ic

Ib

V−

V+

I′

O

E

X399

X74

Figure 21. The second Evans circle

15. The second Lester circle

In Kimberling’s first list of triangle centers [8], the point X19, the homotheticcenter of the orthic triangle and the triangular hull of the three excircles, was calledthe crucial point. Kimberling explained that this name “derives from the name ofthe publication [13] in which the point first appeared”. In the expanded list in [9],this point was renamed after J.W. Clawson. Kimberling gave the reference [2], and


commented that this is “possibly the earliest record of this point”.4

Cw =(

a

b2 + c2 − a2 ,b

c2 + a2 − b2 ,c

a2 + b2 − c2)

. (31)

Proposition 28. The Clawson point Cw is the perspector of the triangle boundedby the radical axes of the circumcircle with the three excircles (see Figure 24).

O

A

B

C

A′

B′C′

Cw

Figure 22. The Clawson point

Proof. The equations of the excircles are given in [15, §6.1.1]. The radical axeswith circumcircle are the lines

La := s2x + (s − c)2y + (s − b)2z = 0,Lb := (s − c)2x + s2y + (s − a)2z = 0,Lc := (s − b)2x + (s − a)2y + s2z = 0.

These lines intersect at

A′ = (0 : b(a2 + b2 − c2) : c(c2 + a2 − b2)),B′ = (a(a2 + b2 − c2) : 0 : c(b2 + c2 − a2)),C ′ = (a(c2 + a2 − b2) : b(b2 + c2 − a2) : 0).

It is clear that the triangles ABC and A′B′C ′ are perspective at a point whosecoordinates are given by (31). �

4According to the current edition of [10], this point was studied earlier by E. Lemoine [11].

206 P. Yiu

Apart from the circle through the circumcenter, the nine-point center and theFermat points, Lester has discovered another circle through the symmedian point,the Clawson point, the Feuerbach point and the homothetic center of the orthic andthe intangents triangle. The intangents are the common separating tangents of theincircle and the excircles apart from the sidelines. These are the lines

L′a := bcx + (b − c)cy − (b − c)bz = 0,L′b := −(c − a)cx + cay + (c − a)az = 0,L′c := (a − b)bx − (a − b)ay + abz = 0.

These lines are parallel to the sides of the orthic triangles, namely,

−(b2 + c2 − a2)x + (c2 + a2 − b2)y + (a2 + b2 − c2)z = 0,(b2 + c2 − a2)x − (c2 + a2 − b2)y + (a2 + b2 − c2)z = 0,(b2 + c2 − a2)x + (c2 + a2 − b2)y − (a2 + b2 − c2)z = 0.

The two triangles are therefore homothetic. The homothetic center is

To =(

a(b + c − a)b2 + c2 − a2 ,

b(c + a − b)c2 + a2 − b2 ,

c(a + b − c)a2 + b2 − c2

). (32)

A

B

C

To

Figure 23. The intangents triangle

Theorem 29 (Lester). The symmedian point, the Feuerbach point, the Clawsonpoint, and the homothetic center of the orthic and the intangent triangles are con-cyclic.


A

B

C

To

Cw

A′

B′

C′

KFe

Z11

Figure 24. The second Lester circle

There are a number of ways of proving this theorem, all very tedious. For ex-ample, it is possible to work out explicitly the equation of the circle containingthese four points. Alternatively, one may compute distances and invoke the inter-secting chords theorem. These proofs all involve polynomials of large degrees. Wepresent here a proof given by Nikolaos Dergiades which invokes only polynomialsof relatively small degrees.

Lemma 30. The equation of the circle passing through three given points P1 =(u1 : v1 : w1), P2(u2 : v2 : w2) and P3 = (u3 : v3 : w3) is

a2yz + b2zx + c2xy − (x + y + z)(px + qy + rz) = 0where

p =D(u1, u2, u3)

s1s2s3D(1, 2, 3), q =

D(v1, v2, v3)s1s2s3D(1, 2, 3)

, r =D(w1, w2, w3)

s1s2s3D(1, 2, 3),

with

s1 = u1+v1+w1, s2 = u2+v2+w2, s3 = u3+v3+w3, D(1, 2, 3) =

∣∣∣∣∣∣u1 v1 w1u2 v2 w2u3 v3 w3

∣∣∣∣∣∣ ,

208 P. Yiu

and

D(u1, u2, u3) =

∣∣∣∣∣∣a2v1w1 + b2w1u1 + c2u1v1 s1v1 s1w1a2v2w2 + b2w2u2 + c2u2v2 s2v2 s2w2a2v3w3 + b2w3u3 + c2u3v3 s3v3 s3w3

∣∣∣∣∣∣ ,

D(v1, v2, v3) =

∣∣∣∣∣∣s1u1 a

2v1w1 + b2w1u1 + c2u1v1 s1w1s2u2 a

2v2w2 + b2w2u2 + c2u2v2 s2w2s3u3 a

2v3w3 + b2w3u3 + c2u3v3 s3w3

∣∣∣∣∣∣ ,

D(w1, w2, w3) =

∣∣∣∣∣∣s1u1 s1v1 a

2v1w1 + b2w1u1 + c2u1v1s2u2 s2v2 a

2v2w2 + b2w2u2 + c2u2v2s3u3 s3v3 a

2v3w3 + b2w3u3 + c2u3v3

∣∣∣∣∣∣ .

Proof. This follows from applying Cramer’s rule to the system of linear equations

a2v1w1 + b2w1u1 + c2u1v1 − s1(pu1 + qv1 + rw1) = 0,a2v2w2 + b2w2u2 + c2u2v2 − s2(pu2 + qv2 + rw2) = 0,a2v3w3 + b2w3u3 + c2u3v3 − s3(pu3 + qv3 + rw3) = 0.

�Lemma 31. Four points Pi = (ui : vi : wi), i = 1, 2, 3, 4, are concyclic if andonly if

D(u1, u2, u4)D(u1, u2, u3)

=D(v1, v2, v4)D(v1, v2, v3)

=D(w1, w2, w4)D(w1, w2, w3)

=s4D(1, 2, 4)s3D(1, 2, 3)

.

Proof. The circumcircles of triangles P1P2P3 and P1P2P4 have equations

a2yz + b2zx + c2xy − (x + y + z)(px + qy + rz) = 0,a2yz + b2zx + c2xy − (x + y + z)(p′x + q′y + r′z) = 0

where p, q, r are given in Lemma 30 above and p′, q′, r′ are calculated with u3,v3, w3 replaced by u4, v4, w4 respectively. These two circles are the same if andonly if p = p′, q = q′, r = r′. The condition p = p′ is equivalent to D(u1,u2,u4)D(u1,u2,u3) =s4D(1,2,4)s3D(1,2,3)

; similarly for the remaining two conditions. �

Finally we complete the proof of the second Lester circle theorem. For

P1 = K = (a2 : b2 : c2),

P2 = Fe = ((b − c)2(b + c − a) : (c − a)2(c + a − b) : (a − b)2(a + b − c),P3 = Cw = (aSBC : bSCA : cSAB),

P4 = To = (a(b + c − a)SBC : b(c + a − b)SCA : c(a + b − c)SAB),we have

D(u1, u2, u4)D(u1, u2, u3)

=(b + c − a)(c + a − b)(a + b − c)

a + b + c=

s4D(1, 2, 4)s3D(1, 2, 3)

.

The cyclic symmetry also shows that

D(v1, v2, v4)D(v1, v2, v3)

=D(w1, w2, w4)D(w1, w2, w3)

=(b + c − a)(c + a − b)(a + b − c)

a + b + c.


It follows from Lemma 31 that the four points K, Fe, Cw, and To are concyclic.This completes the proof of Theorem 29.

For completeness, we record the coordinates of the center of the second Lestercircle, namely,

Z11 := (a(b − c)f5(a, b, c)f12(a, b, c) : · · · : · · · ),wheref5(a, b, c) = a

5 − a4(b + c) + 2a3bc − a(b4 + 2b3c − 2b2c2 + 2bc3 + c4) + (b − c)2(b + c)3,f12(a, b, c) = a

12 − 2a11(b + c) + 9a10bc + a9(b + c)(2b2 − 13bc + 2c2)− a8(3b4 − 2b3c − 22b2c2 − 2bc3 + 3c4) + 4a7(b + c)((b2 − c2)2 − b2c2)− 10a6bc(b2 − c2)2 − 2a5(b + c)(b − c)2(b2 − 4bc + c2)(2b2 + 3bc + 2c2)+ a4(b − c)2(3b6 + 2b5c − 19b4c2 − 32b3c3 − 19b2c4 + 2bc5 + 3c6)− 2a3(b + c)(b − c)2(b6 + 2b5c − 3b4c2 − 2b3c3 − 3b2c4 + 2bc5 + c6)+ a2bc(b4 − c4)2 + a(b + c)(b − c)4(b2 + c2)2(2b2 + 3bc + 2c2)− (b − c)4(b + c)2(b2 + c2)3.

References

[1] N. Altshiller-Court, College Geometry, second edition, Barnes and Noble, 1952; Dover reprint,2007.

[2] J. W. Clawson and M. Goldberg, Problem 3132, Amer. Math. Monthly, 32 (1925) 204; solution,ibid., 33 (1926) 285.

[3] L. S. Evans, A rapid construction of some triangle centers, Forum Geom., 2 (2002) 67–70.[4] L. S. Evans, A conic through six triangle centers, Forum Geom., 2 (2002) 89–92.[5] L. S. Evans, Some configurations of triangle centers, Forum Geom., 3 (2003) 49–56.[6] L. S. Evans, A tetrahedral arrangement of triangle centers, Forum Geom., 3 (2003) 181–186.[7] B. Gibert, Hyacinthos message 1270, August 22, 2000.[8] C. Kimberling, Central points and central lines in the plane of a triangle, Math. Magazine, 67

(1994) 163–187.[9] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998)

1–285.[10] C. Kimberling, Encyclopedia of Triangle Centers, available at

http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.[11] E. Lemoine, Quelques questions se rapportant à l’étude des antiparallèles des côtés d’un trian-

gle, Bulletin de la Société Mathématique de France, 14 (1886) 107–128.[12] J. Lester, Triangles III: Complex triangle functions, Aequationes Math., 53 (1997) 4–35.[13] R. Lyness and G.R. Veldkamp, Problem 682 and solution, Crux Mathematicorum 9 (1983)

23–24.[14] P. Yiu, Hyacinthos message 1726, November 3, 2000.[15] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes,

2001.

Paul Yiu: Department of Mathematical Sciences, Florida Atlantic University, 777 Glades Road,Boca Raton, Florida 33431-0991, USA

E-mail address: [email protected]

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