Forum Geometricorum Volume 10 (2010) 175–209. 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 of interesting 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 their equations. Lester [12] has discovered that the Fermat points, the circumcenter, and the nine-point center are concyclic. We call this the first Lester circle, and study it in §§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 center of the orthic and the intangent triangles. This conjecture is validated in §15. Evans, during the preparation of his papers in Forum Geometricorum, has communicated two conjectures on circles through two perspectors V± which has since borne his name. In §9 we study in detail the first Evans circle in relation to the excentral circle. The second one is established in §14. In [9], a great number of circles have been reported relating to the Parry point, a point on the circumcircle. These circles are studied in §§10 – 12.

CONTENTS 1. Introduction 175 2. Preliminaries 176 2.1. Intersection of a circle with the circumcircle 176 2.2. Construction of circle equation 177 2.3. Some common triangle center functions 177 3. The first Lester circle 178 4. Gibert’s generalization of the first Lester circle 182 5. 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 proof of Theorem 29. 176 P. Yiu

6. Equations of circles 184 6.1. The circle F+F−G 184 6.2. The circle F+F−H 185 6.3. The first Lester circle 186 7. The Brocard axis and the Brocard circle 186 7.1. The Brocard axis 186 7.2. The Brocard circle 186 7.3. The isodynamic points 187 8. The excentral triangle 189 8.1. Change of coordinates 189 9. The first Evans circle 190 9.1. The Evans perspector W 190 9.2. Perspectivity of the excentral triangle and Kiepert triangles 191 9.3. The first Evans circle 194 10. The Parry circle and the Parry point 196 10.1. The center of the Parry circle 198 11. The generalized Parry circles 198 12. Circles containing the Parry point 199 12.1. The circle F+F−G 199 12.2. The circle GOK 200 13. Some special circles 201 13.1. The circle HOK 201 13.2. The circle NOK 202 14. The second Evans circle 203 15. The second Lester circle 204 References 209

2. Preliminaries We refer to [15] for the standard notations of triangle geometry. Given a triangle ABC, with sidelengths a, b, c, the circumcircle is represented in homogeneous barycentric coordinates by the equation a2yz + b2zx + c2xy =0. The equation of a general circle C is of the form a2yz + 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)=0is the radical axis of the circle C and the circumcircle.

2.1. Intersection of a circle with the circumcircle . The intersections of the circle C 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 can be easily identified. We say that a triangle center function f(a, b, c) represents an infinite 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 which L(x, y, z)=F (a, b, c) · b2c2 · f(a, b, c) · g(a, b, c)x, (3) cyclic where F (a, b, c) is symmetric in a, b, c, and f(a, b, c), g(a, b, c) are triangle center functions representing infinite points, then the circle intersects the circumcircle at the points a2 b2 c2 Q := : : f f(a, b, c) f(b, c, a) f(c, a, b) and a2 b2 c2 Q := : : . g g(a, b, c) g(b, c, a) g(c, a, b)

Proof. The line L(x, y, z)=0clearly contains the point Qf , which by (2) is the isogonal conjugate of an infinite point, and so lies on the circumcircle. It is therefore a common point of the circumcircle and C. The same reasoning applies to the point Qg.  For an application, see Remark after Proposition 11. 2.2. Construction of circle equation. Suppose we know the equation of a circle through two points Q1 and Q2, in the form of (1), and the equation of the line Q1Q2, in the form L (x, y, z)=0. To determine the equation of the circle through Q1, Q2 and a third point Q =(x0,y0,z0) not on the line Q1Q2, it is enough to find t such that 2 2 2 a y0z0 + b z0x0 + c x0y0 +(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)) = 0 represents the circle Q1Q2Q. For an application of this method, see §6.3 (11) and Proposition 11. 2.3. Some common triangle center functions. We list some frequently occurring homogeneous functions associated with the coordinates of triangle centers or co- efficients in equations of lines and circles. An asterisk indicates that the function represents an infinite point. 178 P. Yiu

Quartic forms 2 2 2 2 2 2 f4,1 := a (b + c ) − (b − c ) 2 2 2 4 4 f4,2 := a (b + c ) − (b + c ) 4 4 2 2 4 f4,3 := a − (b − b c + c ) 2 2 2 2 2 2 f4,4 := (b + c − a ) − b c 4 2 2 2 2 2 2 * f4,5 := 2a − a (b + c ) − (b − c ) 4 2 2 2 2 2 2 f4,6 := 2a − 3a (b + c )+(b − c ) 4 2 2 2 4 2 2 4 * f4,7 := 2a − 2a (b + c ) − (b − 4b c + c )

Sextic forms 6 4 2 2 2 4 2 2 4 2 2 2 2 2 f6,1 := a − 3a (b + c )+a (3b − b c +3c ) − (b + c )(b − c ) 6 4 2 2 2 4 4 2 2 2 2 2 * f6,2 := 2a − 2a (b + c )+a (b + c ) − (b + c )(b − c ) 6 4 2 2 2 4 4 2 2 3 * f6,3 := 2a − 6a (b + c )+9a (b + c ) − (b + c )

Octic forms 8 6 2 2 4 2 2 2 2 2 4 2 2 4 f8,1 := a − 2a (b + c )+a b c + a (b + c )(2b − b c +2c ) −(b8 − 2b6c2 +6b4c4 − 2b2c6 + c8) 8 6 2 2 4 4 2 2 4 * f8,2 := 2a − 2a (b + c ) − a (3b − 8b c +3c ) +4a2(b2 + c2)(b2 − c2)2 − (b2 − c2)2(b4 +4b2c2 + c4) 8 6 2 2 4 4 2 2 4 * f8,3 := 2a − 5a (b + c )+a (3b +8b c +3c ) +a2(b2 + c2)(b4 − 5b2c2 + c4) − (b2 − c2)4 8 6 2 2 4 4 2 2 4 f8,4 := 3a − 8a (b + c )+a (8b +7b c +8c ) −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 F−

F+ O N

H

B C

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 points intersects the Euler line at the midpoint of the orthocenter H and the centroid G. The circles of Lester, Evans, Parry, and their generalizations 179

Clearing denominators in the homogeneous barycentric coordinates of the Fermat point 1 1 1 F+ = √ , √ , √ , 3SA + S 3SB + S 3SC + S we rewrite it in the form √ 2 2 2 F+ =(3SBC +S , 3SCA+S , 3SAB +S )+ 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 2 2 2 M =(3SBC + S , 3SCA + S , 3SAB + S ) 2 =3(SBC,SCA,SAB)+S (1, 1, 1).

Note that M is the midpoint of the segment HG, where H =(SBC,SCA,SAB) is the 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 √ √ 2 MF+ : F+K =2 3S(SA + SB + SC ):6S =(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.

F− A

K O F+ G M H

B C

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. 2 (B) The circle F+F−G is tangent to the Euler line at G, i.e., MF+ ·MF− = MG . 2 (C) The circle F+F−H is tangent to the Euler line at H, i.e., MF+·MF− = MH . (D) The Fermat points are inverse in the orthocentroidal circle.

2d d d 2d H M N G O

Figure 3. The Euler line

Proof. Since M is the midpoint of HG, the statements (B), (C), (D) are clearly equivalent. On the other hand, putting OH =6d,wehave 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, it is enough to prove (D). We do this by a routine calculation. Theorem 5. The Fermat points are inverse in the orthocentroidal circle.

A F−

O

F+ G

M H

B C

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

Proof. The equation of the orthocentroidal circle is 2 2 2 3(a yz + b zx + c xy) − 2(x + y + z)(SAx + SBy + SC z)=0, The circles of Lester, Evans, Parry, and their generalizations 181 equivalently, 1 2 2 2 −2(SAx +SBy +SCz )+((SB +SC)yz+(SC +SA)zx+(SA +SB)xy)=0. This is represented by the matrix ⎛ ⎞ −4SA SA + SB SA + SC ⎝ ⎠ M = SA + 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 2 2 2 X = 3SBC + S 3SCA + S 3SAB + S , √ Y = 3S SB + SC SC + SA SA + SB . With these, we have t t 2 2 2 2 XMX = Y MY =6S (SA(SB − SC ) + SB(SC − SA) + SC (SA − SB) ), and t t t t F+MF− =(X + Y )M(X − Y ) = XMX − Y MY =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 through O, N, and F± the first Lester circle.

F− A

Ki Z1 Z0

F+ O G N M H

B C

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]) is the point M =(SA(SB + SC )+4SBC : SB (SC + SA)+4SCA : SC (SA + SB )+4SAB ) on the Euler 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 midpoint of 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 equations of 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.

Y1

F A −

Y0

Z1 Z0 Ki

F + O G N M H

B C

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 the Kiepert 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)=0and L0(x, y, z)=0represent the lines F+F− and GY0 respectively. We may assume the circle given by 2 2 2 2 2 2 k0((b − c )yz +(c − a )zx +(a − b )xy) − L(x, y, z) · L0(x, y, z)=0 for an appropriately chosen constant k0. The circles of Lester, Evans, Parry, and their generalizations 183

Replacing G by H and Y0 by another point Y1, the intersection of the Kiepert hyperbola with the perpendicular to the Euler line at H, we write the equation of the circle F+F−H in the form 2 2 2 2 2 2 k1((b − c )yz +(c − a )zx +(a − b )xy) − L(x, y, z) · L1(x, y, z)=0, where L1(x, y, z)=0is the equation of the line HY1. The midpoints of the two chords GY0 and HY1 are the centers of the two circles F+F−G and F+F−H. The line joining them is therefore the perpendicular bisector of 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)=0 for some real number t. Let kt := tk0 +(1− t)k1 correspondingly. Then the equation 2 2 2 2 2 2 kt((b − c )yz +(c − a )zx +(a − b )xy) − L(x, y, z) · Lt(x, y, z)=0 represents a circle Ct through the Fermat points and the intersections of the line Lt(x, y, z)=0and the Kiepert hyperbola. The perpendicular bisector of F+F− is the diameter of the family of parallel lines Lt(x, y, z)=0. Therefore the center of the 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 segment HG, then the circle intersects the Euler line at two points dividing the segment HG harmonically, say in the ratio τ :1− τ for τ<0 or τ>1. In this case, the line 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 HG in the ratio 3:∓1, the diameter of the first Lester circle perpendicular to the Euler line intersects the latter at the point L dividing HG in the ratio 9:−1. This is the midpoint 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 3

H N G L O

Figure 7. The Euler line

Proposition 7. (a) Lines perpendicular to the Euler line have infinite point 2 2 2 2 2 2 X523 =(b − c ,c − a ,a − b ). 184 P. Yiu

(b) The diameter of the first Lester circle perpendicular to the Euler line is along the line f6,1(a, b, c)x =0. (4) cyclic

Proposition 8. (a) The equation of the line F+F− is 2 2 (b − c )f4,4(a, b, c)x =0. (5) cyclic

(b) The perpendicular bisector of F+F− is the line x y z + + =0. (6) b2 − c2 c2 − a2 a2 − b2 Proof. (a) The line F+F− contains the symmedian point K and the Kiepert center 2 2 2 2 2 2 2 2 2 Ki =((b − c ) , (c − a ) , (a − b ) ).

(b) The perpendicular bisector of F+F− is the perpendicular at Ki to the line KKi, which has infinite point 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 X690 =((b − c )(b + c − 2a ), (c − a )(c + a − 2b ), (a − b )(a + b − 2c )). 

Proposition 9. The center of the first Lester circle has homogeneous barycentric coordinates 2 2 2 2 2 2 ((b − c )f8,3(a, b, c):(c − a )f8,3(b, c, a):(a − b )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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Je =((b −c ) (b +c −a ), (c −a ) (c +a −b ), (a −b ) (a +b −c )), which is the center of the Jerabek hyperbola, the isogonal conjugate of the Euler line. It follows that Je is equidistant from the Fermat points. The points Ki and Je are 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 2 2 L(x, y, z)= (b − c )f4,4(a, b, c)x, (7) cyclic 2 2 2 L0(x, y, z)= (b + c − 2a )x (8) cyclic for the equation of the line F+F− (Proposition 8(a)) and the perpendicular to the Euler line at G. Now, we seek a quantity k0 such that the member 2 2 2 2 2 2 k0((b − c )yz +(c − a )zx +(a − b )xy) − L(x, y, z) · L0(x, y, z)=0 The circles of Lester, Evans, Parry, and their generalizations 185 of the pencil of conic through the four points F±, G, Y0 is a circle. For this,

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 k0 = −3(a (c − a )(a − b )+b (a − b )(b − c )+c (b − c )(c − a )), and the equation can be reorganized as

9(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy) ⎛ ⎞ ⎝ 2 2 2 2 2 ⎠ +(x + y + z) (b − c )(b + c − 2a )f4,4(a, b, c)x =0. (9) cyclic

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

2 2 2 2 2 2 Z0 := ((b − c )f4,7(a, b, c):(c − a )f4,7(b, c, a):(a − b )f4,7(c, a, b)). b2−c2 0 : ···: ··· The point Y has coordinates b2+c2−2a2 .

6.2. The circle F+F−H. With the line 2 2 2 4 2 2 2 2 2 2 L1(x, y, z)= (b + c − a )(2a − a (b + c ) − (b − c ) )x =0 cyclic perpendicular to the Euler line at H, we seek a number k1 such that 2 2 2 2 2 2 k1((b − c )yz +(c − a )zx +(a − b )xy) − L(x, y, z) · L1(x, y, z)=0 of the pencil of conic through the four points F±, H, Y1 is a circle. For this,

2 4 2 2 2 4 2 2 2 4 2 2 2 2 2 2 k1 = 16Δ (a (b + c − a )+b (c + a − b )+c (a + b − c ) − 3a b c ), and the equation can be reorganized as

48(b2 − c2)(c2 − a2)(a2 − b2)Δ2(a2yz + b2zx + c2xy) ⎛ ⎞ ⎝ 2 2 2 2 2 ⎠ − (x + y + z) (b − c )(b + c − a )f4,4(a, b, c)f4,5(a, b, c)x =0. cyclic (10)

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

2 2 2 2 2 2 Z1 := ((b − c )f8,2(a, b, c):(c − a )f8,2(b, c, a):(a − b )f8,2(c, a, b)).

The triangle center b2 − c2 c2 − a2 a2 − b2 Y1 = : : f4,5(a, b, c) f4,5(b, c, a) f4,5(c, a, b) is X2394. 186 P. Yiu

6.3. The first Lester circle. Since the line joining the Fermat points has equation L(x, y, z)=0with L given by (7), every circle through the Fermat points is represented by 9(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy) ⎛ ⎞ ⎝ 2 2 2 2 2 ⎠ +(x + y + z) (b − c )(b + c − 2a + t)f4,4(a, b, c)x =0 (11) cyclic for an appropriate choice of t. The value of t for which this circle passes through the circumcenter is 2( 2 − 2)( 2 − 2)+ 2( 2 − 2)( 2 − 2)+ 2( 2 − 2)( 2 − 2) = a c a a b b a b b c c b c c a t 32Δ2 . The equation of the circle is 96Δ2(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy) ⎛ ⎞ ⎝ 2 2 ⎠ +(x + y + z) (b − c )f4,4(a, b, c)f6,1(a, b, c)x =0. cyclic

7. The Brocard axis and the Brocard circle 7.1. The Brocard axis . The isogonal conjugate of the Kiepert perspector K(θ) is the point ∗ 2 2 2 K (θ)=(a (SA + Sθ),b(SB + Sθ),c(SC + Sθ)), which lies on the line joining the circumcenter O and the symmedian point K. The line OK is called the Brocard axis. It is represented by the equation b2c2(b2 − c2)x =0. (12) cyclic

7.2. The Brocard circle. The Brocard circle is the circle with OK as diameter. It is represented by the equation 2 2 2 2 2 2 2 2 2 2 2 2 (a +b +c )(a yz+b zx+c xy)−(x+y+z)(b c x+c a y+a b z)=0. (13) It is clear from ∗ 2 2 2 2 2 2 K (θ)=(a SA,bSB,cSC )+Sθ(a ,b,c), ∗ 2 2 2 2 2 2 K (−θ)=(a SA,bSB,cSC ) − Sθ(a ,b,c) ∗ ∗ that K (θ) and K (−θ) divide O and K harmonically, and so are inverse in the ∗ π Brocard circle. The points K ± 3 are called the isodynamic points, and are more simply denoted by J±. Proposition 10. K∗(±θ) are inverse in the circumcircle if and only if they are the isodynamic points. The circles of Lester, Evans, Parry, and their generalizations 187

7.3. The isodynamic points. The isodynamic points J± are also the common points of 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 the bisectors of angle A on the sidelines BC. These are the points (b, ±c). The center of 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 axis x y z + + =0, (14) a2 b2 c2 which is the perpendicular bisector of the segment J+J−.

A

J+ O K

B C

J−

Figure 8. The Apollonian circles and the isodynamic points 188 P. Yiu

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

3(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy) ⎛ ⎞ 2 2 2 2 2 2 2 +(x + y + z) ⎝ b c (b − c )(b + c − 2a + t)x⎠ =0 (15) cyclic 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) a2(c2 − a2)(a2 − b2)(c2y − b2z)=0. cyclic

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

3(b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy) ⎛ ⎞ 2 2 2 2 2 2 2 +(x + y + z) ⎝ b c (b − c )(b + c − 2a )x⎠ =0. (16) cyclic

Now, the line containing the isodynamic points is the Brocard axis given by (12). It follows that every circle through J± is represented by (15) above for some choice of t (see §2.2). 

Remark. As is easily seen, equation (16) is satisfied by x = y = z =1, and so 2 2 2 2 2 represents the circle through J± and G. Since the factors b − c and b + c − 2a yield infinite points, applying Proposition 1, we conclude that this circle intersects a2 the circumcircle at the Euler reflection point E = 2 2 : ··· : ··· and the Parry b −c a2 : ··· : ··· point b2+c2−2a2 . This is the Parry circle we consider in §10 below.

Proposition 12. The circle through the isodynamic points and the orthocenter has equation

16Δ2 · (b2 − c2)(c2 − a2)(a2 − b2)(a2yz + b2zx + c2xy) ⎛ ⎞ ⎝ 2 2 2 2 2 2 2 ⎠ +(x + y + z) b c (b − c )(b + c − a )f4,1(a, b, c)x =0. cyclic

Its center is the point

2 2 2 2 2 2 Z3 := (a (b − c )(b + c − a )f4,1(a, b, c):···: ···). The circles of Lester, Evans, Parry, and their generalizations 189

8. The excentral triangle

The excentral triangle IaIbIc has as vertices the excenters of triangle ABC.It has circumradius 2R and circumcenter I, the reflection of I in O (see Figure 9). 1 1 1 Since the angles of the excentral triangle are 2 (B + C), 2 (C + A), and 2 (A + B), its sidelengths a = IbIc, b = IcIa, c = IaIb satisfy 2 : 2 : 2 =cos2 A :cos2 B :cos2 C a b c 2 2 2 = a(b + c − a):b(c + a − b):c(a + b − c).

Ib

A

Ic

O I I

C B

Ia 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 y z + + =0. a2 b2 c2 With reference to triangle ABC, the same line is represented by the equation a(b + c − a)(cy + bz) b(c + a − b)(az + cx) c(a + b − c)(bx + ay) + + =0, a(b + c − a) b(c + a − b) c(a + b − c) 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 a2 b2 c2 + + =0, x y z is represented by 1 1 1 + + =0 cy + bz az + cx bx + ay 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 of A in BC, B in CA, C in AB. The triangle A∗B∗C∗ is called the triangle of reflections of ABC. Larry Evans has discovered the perspectivity of the excentral triangle and A∗B∗C∗. Theorem 13. The excentral triangle and the triangle of reflections are perspective at a point which is the inverse image of the incenter in the circumcircle of the excentral triangle.

W

A Y

X I O C B Y I

A∗ I∗

Ia

Figure 10. The Evans perspector W The circles of Lester, Evans, Parry, and their generalizations 191

∗ ∗ ∗ Proof. We show that the lines IaA , IbB , and IcC intersect the line OI at the same point. Let X be the intersection of the lines AA∗ and OI (see Figure 10). If ∗ ∗ ha is the A-altitude of triangle ABC, and the parallel from I to AA meets IaA at I∗, then since ∗ ∗ II = II = IaI = YY = a ∗ , 2ha AA IaA AY s ∗ = 2aha =4 WI = I Ia = 2R we have II s r and WI II∗ 4r . Therefore, W divides II in the ratio IW : WI = R : −2r. Since this ratio is a symmetric function of the sidelengths, we conclude that the ∗ ∗ = R · same point W lies on the lines IbB and IcC . Moreover, since I W R−2r I I, by the famous Euler formula OI2 = R(R − 2r),wehave R 4R 4R IW · II = · II2 = · OI2 = · R(R − 2r)=(2R)2. R − 2r R − 2r R − 2r 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 barycentric coordinates, 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 isogonal with respect to angle Ia the excentral triangle (see Figure 11).

Ib θ A

Ic X

B θ C

X

Ia

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 IbIc are antiparallel. In this similarity X and X are homologous points. Hence, the lines IaX and IaX are isogonal in the excentral triangle.  We shall denote Kiepert perspectors with reference to the excentral triangle by Ke(−). Theorem 15. The excentral triangle and the Kiepert triangle K(θ) are perspective at 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 Lemma 14 (see Figure 11). (i) IaX , IbY , IcZ concur at the Kiepert perspector Ke(−θ) of the excentral triangle. (ii) Since IaX and IaX are isogonal with respect to Ia, and similarly for the pairs IbY , IbY and IcZ and IcZ , the lines IaX, IbY , IcZ concur at the isogonal conjugate of Ke(−θ) in the excentral triangle. 

Ib

A A

Ic

C V−

B C

B

Ia V K − π Figure 12. Evans’ perspector − of 3 and excentral triangle

We denote the perspector in Theorem 15 by V (θ), and call this a generalized π π Evans perspector. In particular, V 3 and V − 3 are the isodynamic points of the excentral triangle, and are simply denoted by V+ and V− respectively (see The circles of Lester, Evans, Parry, and their generalizations 193

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 (b − c)(b2 + c2 − a2)x =0. (19) cyclic

Proof. The line V+V− is the Brocard axis of the excentral triangle, with equation 2 − 2 b c · =0 2 x cyclic a with reference to the excentral triangle (see §7.1). Replacing these by parameter with reference to triangle ABC,wehave 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- ∗ ∗ ∗ π ∗ flections A B C if and only if θ = ± 3 . The perspector is K (−θ), the isogonal conjugate of K(−θ). π This means that for ε = ±1, the Fermat triangle K ε · 3 and the triangle of reflections are perspective at the isodynamic point J−ε (see Figure 13 for the case ε = −1).

B∗ A A

C∗ J+

C

B C

B

A∗ K − π A∗B∗C∗ J Figure 13. 3 and perspective at + 194 P. Yiu

9.3. The first Evans circle. Since V± are the isodynamic points of the excentral triangle, they are inverse in the circumcircle of the excentral triangle. Since W and I are also inverse in the same circle, we conclude that V+, V−, I, and W are concyclic (see [1, Theorem 519]). We call this the first Evans circle. A stronger result holds in view of Proposition 10. Theorem 18. The four points V (θ), V (−θ), I, W are concyclic if and only if π θ = ± 3 .

Ib

I

B∗ A V+

Ic I C B W A∗ Ia

C∗

V−

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, from = IW = R2 the degenerate triangle II W we have OM 2 OI . This shows that the The circles of Lester, Evans, Parry, and their generalizations 195 midpoint of IW is the inverse of I in the circumcircle. Therefore, the perpendicular bisector of IW is the polar of I in the circumcircle. This is the line ⎛ ⎞ ⎛ ⎞ 2 2 0 c b x abc⎝c2 0 a2⎠ ⎝y⎠ =0, b2 a2 0 z 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, the perpendicular 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 excentral triangle, its Lemoine axis has equation

x y z + + =0. a2 b2 c2 Changing coordinates, we have, with reference to ABC, the same line represented by the equation

a(b + c − a)(cy + bz) b(c + a − b)(az + cx) c(a + b − c)(bx + ay) + + =0, a(b + c − a) b(c + a − b) c(a + b − c) which simplifies into (21). 

Proposition 21. The center of the first Evans circle is the point a(b − c) b(c − a) c(a − b) : : . b + c c + a 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 also the 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) ⎝ bc(b − c)(c + a)(a + b)(b + c − a)x⎠ =0. (22) cyclic 196 P. Yiu

X23

J−

X351

X352 A E

K

J + G X187 X111 X353 B C O

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 of interesting triangle centers, including the isodynamic points and the centroid. We shall define the Parry circle as the circle through these three points, and seek to explain the incidence of the other points. First of all, since the isodynamic points are inverse in each of the circumcircle and the Brocard circle, the Parry circle is orthogonal to each of these circles. In particular, it contains the inverse of G in the circumcircle. This is the triangle center

2 4 4 2 2 4 2 4 4 2 2 4 2 4 4 2 2 4 X23 =(a (a −b +b c −c ),b(b −c +c a −a ),c(c −a +a b −b )). (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 a2 b2 c2 E = , , (24) b2 − c2 c2 − a2 a2 − b2 The circles of Lester, Evans, Parry, and their generalizations 197 and the point a2 b2 c2 P = , , (25) b2 + c2 − 2a2 c2 + a2 − 2b2 a2 + b2 − 2c2 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 2 2 (b − c )f4,2(a, b, c)x =0. (26) cyclic 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 the Brocard circle.

F− A

O

K G

F+

B M C

E

H

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 at its midpoint, the center of the Brocard circle. Since the circle through E, G, J± is orthogonal to the Brocard circle, E and G are inverse to each other with respect to this 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, 2 4 2 2 2 4 2 2 4 X352 =(a (a − 4a (b + c )+(b +5b c + c )), ··· , ···),

(ii) the second intersection with the line joining X23 to K, namely, 2 4 2 2 2 4 2 2 4 X353 =(a (4a − 4a (b + c ) − (2b + b c +2c )), ··· , ···), which is the inverse of the Parry point P in the Brocard circle, and also the inverse of X352 in the circumcircle. 10.1. The center of the Parry circle. Proposition 25. The perpendicular bisector of the segment GE is the line x y z + + =0. (27) b2 + c2 − 2a2 c2 + a2 − 2b2 a2 + b2 − 2c2 Proof. The midpoint of EG is the point 2 2 2 2 2 2 2 2 2 Z4 := ((b +c −2a )f4,5(a, b, c):(c +a −2b )f4,5(b, c, a):(a +b −2c )f4,5(c, a, b)). By Lemma 23 and Proposition 8(b), the perpendicular bisector of EG has infinite point X690. The line through Z4 with this infinite point is the perpendicular bisector of 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 and the points K∗(±θ) on the Brocard axis. Since K∗(θ) and K∗(−θ) are inverse in the Brocard circle (see §7.2), the generalized Parry circle CP(θ) is orthogonal to the Brocard circle, and must also contain the Euler reflection point E. Its equation is 3(b2 − c2)(c2 − a2)(a2 − b2)(16Δ2 sin2 θ − (a2 + b2 + c2)2 cos2 θ)(a2yz + b2zx + c2xy) 2 2 2 2 2 2 +(x + y + z) b c (b − c )(f6,2(a, b, c)sin θ + f6,3(a, b, c)cos θ)x =0. cyclic The second intersection with the circumcircle is the point 2 ( )= a ··· ··· Q θ 2 2 , , . f6,2(a, b, c)sin θ + f6,3(a, b, c)cos θ π The Parry point P is Q(θ) for θ = 3 . Here are two more examples. π (i) With θ = 2 , we have the circle GEO tangent to the Brocard axis and with center 2 2 2 2 2 2 Z5 := (a (b − c )(b + c − 2a )f4,4(a, b, c):···: ···). The circles of Lester, Evans, Parry, and their generalizations 199

It intersects the circumcircle again at the point a2 b2 c2 , , . (28) f6,2(a, b, c) f6,2(b, c, a) f6,2(c, a, b)

This is the triangle center X842. (ii) With θ =0, we have the circle GEK tangent to the Brocard axis and with center 2 2 2 2 2 2 2 2 2 2 2 2 Z6 := (a (b − c )(b + c − 2a )((a + b + c ) − 9b c ): ··· : ···). It intersects the circumcircle again at the point a2 b2 c2 Z7 := : : . (29) f6,3(a, b, c) f6,3(b, c, a) f6,3(c, a, b)

A

O K

G X842

Z6 B C

Z5

E

Z7

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 computed in §6.1, and is given by (9). Applying Proposition 1, we see that the circle F+F−G contains the Parry point and the point 1 1 1 = : : Q 2 2 2 2 2 2 . (b − c )f4,4(a, b, c) (c − a )f4,4(b, c, a) (a − b )f4,4(c, a, b) 200 P. Yiu

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

P

F− A

O

J+

K G

X476 F+ B C

E

J−

X23

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) ⎝ b2c2((b2 − c2)t +1)x⎠ =0 cyclic (30) for suitably chosen t. = − a4+b4+c4−b2c2−c2a2−a2b2 With t 3(b2−c2)(c2−a2)(a2−b2) , and clearing denominators, we obtain the equation of the circle GOK.

3(b2 − c2)(c2 − a2)(a2 − b2)(a2 + b2 + c2)(a2yz + b2zx + c2xy) ⎛ ⎞ +(x + y + z) ⎝ b2c2(b2 − c2)(b2 + c2 − 2a2)2x⎠ =0. cyclic

2 To 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. The circles of Lester, Evans, Parry, and their generalizations 201

X691 A

P

K G O H

B C

X112

X842 E

Figure 19. The circles GOK and HOK

This circle GOK contains the Parry point P and the point a2 Q = : ··· : ··· , (b2 − c2)(b2 + c2 − 2a2)

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

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 a4(c2 − a2)(a2 − b2)+b4(a2 − b2)(b2 − c2)+c4(b2 − c2)(c2 − a2) t = − 16Δ2(b2 − c2)(c2 − a2)(a2 − b2)

3 2 2 This may be checked by computing the infinite point of the line EQ as X512 =(a (b − 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) ⎛ ⎞ ⎝ 2 2 2 2 2 2 2 ⎠ +(x + y + z) b c (b − c )(b + c − a )f6,2(a, b, c)x =0. cyclic

Therefore, the circle HOK intersects the circumcircle at a2 b2 c2 , , , (b2 − c2)(b2 + c2 − a2) (c2 − a2)(c2 + a2 − b2) (a2 − b2)(a2 + b2 − c2) which is the triangle center X112, and X842 given by (28).

Remarks. (1) The circle HOK has center

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

(2) X112 is the second intersection of the circumcircle with the line joining X74 with the symmedian point. (3) X842 is the second intersection of the circumcircle with the parallel to OK through E. It is also the antipode of X691, which is the reflection of E in the Brocard axis. (4) The radical axis with the circumcircle intersects the Euler line at X186 and the 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 both O, N and K, Ki are inverse in the orthocentroidal circle.

A F−

Ki

K O F+ N M H

B C

Figure 20. The circle NOK The circles of Lester, Evans, Parry, and their generalizations 203

This circle has equation 32(a2 − b2)(b2 − c2)(c2 − a2)(a2 + b2 + c2)Δ2(a2yz + b2zx + c2xy) 2 2 2 2 + b c (b − c )f8,4(a, b, c)x =0. cyclic Its center is the point 2 2 2 Z9 := (a (b − c )f8,1(a, b, c):··· : ···).

14. The second Evans circle π Evans also conjectured that the perspectors V± = V (± 3 ) and X74, X399 are concyclic. Recall that a2 b2 c2 X74 = : : f4,5(a, b, c) f4,5(b, c, a) f4,5(c, a, b) is the antipode on the circumcircle of the Euler reflection point E and X339, the Parry reflection point, is the reflection of O in E. We confirm Evans’ conjecture indirectly, by first finding the circle through V± and the point 2 2 2 X101 =(a (c − a)(a − b): b (a − b)(b − c): c (b − c)(c − a)) on the circumcircle. Making use of the equation (22) of the first Evans circle, and the 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) ⎝ (bc(b − c)(c + a)(a + b)(b + c − a)+t(b − c)(b2 + c2 − a2))x⎠ =0, cyclic represents a circle through the point X101. For this, we require abc(a2(b + c − a)+b2(c + a − b)+c2(a + b − c)+abc) t = − , (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) ⎛ ⎞ ⎝ 2 2 ⎠ − (x + y + z) b c (b − c)f4,5(a, b, c)x =0. cyclic

It is clear that this circle does also contain the point X74. The center of the circle is the point 2 2 2 2 2 2 2 Z10 =(a (b − c)((b + c − a ) − b c ), ··· , ···). Now, the perpendicular bisector of the segment OE is the line x =0 2 , a f4,4(a, b, c) 204 P. Yiu which clearly contains the center of the circle. Therefore, the circle also contains the point which is the reflection of X74 in the midpoint of OE. This is the same as the reflection of O in E, the Parry reflection point X399.

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

Ib

X74

I

O V+ A

Ic C B E

Ia

X399

V−

Figure 21. The second Evans circle

15. The second Lester circle

In Kimberling’s first list of triangle centers [8], the point X19, the homothetic center of the orthic triangle and the triangular hull of the three excircles, was called the crucial point. Kimberling explained that this name “derives from the name of the 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 The circles of Lester, Evans, Parry, and their generalizations 205 commented that this is “possibly the earliest record of this point”.4 a b c Cw = , , . (31) b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2

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

A

A

O

Cw C B C B

Figure 22. The Clawson point

Proof. The equations of the excircles are given in [15, §6.1.1]. The radical axes with circumcircle are the lines 2 2 2 La := s x +(s − c) y +(s − b) z =0, 2 2 2 Lb := (s − c) x + s y +(s − a) z =0, 2 2 2 Lc := (s − b) x +(s − a) y + s z =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 ABC are perspective at a point whose coordinates 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 the Fermat points, Lester has discovered another circle through the symmedian point, the Clawson point, the Feuerbach point and the homothetic center of the orthic and the intangents triangle. The intangents are the common separating tangents of the incircle and the excircles apart from the sidelines. These are the lines := +( − ) − ( − ) =0 La bcx b c cy b c bz , := −( − ) + +( − ) =0 Lb c a cx cay c a az , := ( − ) − ( − ) + =0 Lc a b bx a b ay abz . 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 a(b + c − a) b(c + a − b) c(a + b − c) To = , , . (32) b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2

A

To

C B

Figure 23. The intangents triangle

Theorem 29 (Lester). The symmedian point, the Feuerbach point, the Clawson point, and the homothetic center of the orthic and the intangent triangles are con- cyclic. The circles of Lester, Evans, Parry, and their generalizations 207

A

A

F To e K Cw

C B C

B

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 containing these four points. Alternatively, one may compute distances and invoke the inter- secting chords theorem. These proofs all involve polynomials of large degrees. We present here a proof given by Nikolaos Dergiades which invokes only polynomials of 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)=0 where D(u1,u2,u3) D(v1,v2,v3) D(w1,w2,w3) p = ,q= ,r= , s1s2s3D(1, 2, 3) s1s2s3D(1, 2, 3) s1s2s3D(1, 2, 3) with u1 v1 w1 s1 = u1+v1+w1,s2 = u2+v2+w2,s3 = u3+v3+w3,D(1, 2, 3) = u2 v2 w2 , u3 v3 w3 208 P. Yiu and 2 2 2 a v1w1 + b w1u1 + c u1v1 s1v1 s1w1 2 2 2 D(u1,u2,u3)= a v2w2 + b w2u2 + c u2v2 s2v2 s2w2 , 2 2 2 a v3w3 + b w3u3 + c u3v3 s3v3 s3w3 2 2 2 s1u1 a v1w1 + b w1u1 + c u1v1 s1w1 2 2 2 D(v1,v2,v3)= s2u2 a v2w2 + b w2u2 + c u2v2 s2w2 , 2 2 2 s3u3 a v3w3 + b w3u3 + c u3v3 s3w3 2 2 2 s1u1 s1v1 a v1w1 + b w1u1 + c u1v1 2 2 2 D(w1,w2,w3)= s2u2 s2v2 a v2w2 + b w2u2 + c u2v2 . 2 2 2 s3u3 s3v3 a v3w3 + b w3u3 + c u3v3 Proof. This follows from applying Cramer’s rule to the system of linear equations 2 2 2 a v1w1 + b w1u1 + c u1v1 − s1(pu1 + qv1 + rw1)= 0, 2 2 2 a v2w2 + b w2u2 + c u2v2 − s2(pu2 + qv2 + rw2)= 0, 2 2 2 a v3w3 + b w3u3 + c u3v3 − s3(pu3 + qv3 + rw3)= 0. 

Lemma 31. Four points Pi =(ui : vi : wi), i =1, 2, 3, 4, are concyclic if and only if D(u1,u2,u4) D(v1,v2,v4) D(w1,w2,w4) s4D(1, 2, 4) = = = . D(u1,u2,u3) D(v1,v2,v3) D(w1,w2,w3) 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)(px + qy + rz)= 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 and ( ) p = p q = q r = r p = p D u1,u2,u4 = only if , , . The condition is equivalent to 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 2 2 2 P1 = K =(a : b : c ), 2 2 2 P2 = Fe =((b − c) (b + c − a):(c − a) (c + a − b):(a − b) (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) (b + c − a)(c + a − b)(a + b − c) s4D(1, 2, 4) = = . D(u1,u2,u3) a + b + c s3D(1, 2, 3) The cyclic symmetry also shows that D(v1,v2,v4) D(w1,w2,w4) (b + c − a)(c + a − b)(a + b − c) = = . D(v1,v2,v3) D(w1,w2,w3) a + b + c The circles of Lester, Evans, Parry, and their generalizations 209

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 Lester circle, namely,

Z11 := (a(b − c)f5(a, b, c)f12(a, b, c): ···: ···), where 5 4 3 4 3 2 2 3 4 2 3 f5(a, b, c)= a − a (b + c)+2a bc − a(b +2b c − 2b c +2bc + c )+(b − c) (b + c) , 12 11 10 9 2 2 f12(a, b, c)= a − 2a (b + c)+9a bc + a (b + c)(2b − 13bc +2c ) − 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.

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Paul Yiu: Department of Mathematical Sciences, Florida Atlantic University, 777 Glades Road, Boca Raton, Florida 33431-0991, USA E-mail address: [email protected]