Geometry of the Triangle

Home , Feuerbach point, Nagel point, Tangential triangle

Paul Yiu

2016

Department of Mathematics Florida Atlantic University

b c

B a C

August 18, 2016

Contents

1 Some basic notions and fundamental theorems 101 1.1 Menelaus and Ceva theorems ...... 101 1.2 Harmonic conjugates ...... 103 1.3 Directed angles ...... 103 1.4 The power of a point with respect to a circle ...... 106 1.4.1 Inversion formulas ...... 107 1.5 The 6 concyclic points theorem ...... 108

2 Barycentric coordinates 113 2.1 Barycentric coordinates on a line ...... 113 2.1.1 Absolute barycentric coordinates with reference to a segment . . . 113 2.1.2 The circle of Apollonius ...... 114 2.1.3 The centers of similitude of two circles ...... 115 2.2 Absolute barycentric coordinates ...... 120 2.2.1 Homotheties ...... 122 2.2.2 Superior and inferior ...... 122 2.3 Homogeneous barycentric coordinates ...... 122 2.3.1 Euler line and the nine-point circle ...... 124 2.4 Barycentric coordinates as areal coordinates ...... 126 2.4.1 The circumcenter O ...... 126 2.4.2 The incenter and excenters ...... 127 2.5 The area formula ...... 128 2.5.1 Conway’s notation ...... 129 2.5.2 Conway’s formula ...... 130 2.6 Triangles bounded by lines parallel to the sidelines ...... 131 2.6.1 The symmedian point ...... 132 2.6.2 The ﬁrst Lemoine circle ...... 133

3 Straight lines 135 3.1 The two-point form ...... 135 3.1.1 Cevian and anticevian triangles of a point ...... 136 3.2 Infinite points and parallel lines ...... 138 3.2.1 The infinite point of a line ...... 138 3.2.2 Infinite point as vector ...... 141 3.3 Perpendicular lines ...... 145 iv CONTENTS

3.4 The distance formula ...... 148 3.4.1 The distance from a point to a line ...... 150

4 Cevian and anticevian triangles 203 4.1 Cevian triangles ...... 203 4.1.1 The orthic triangle ...... 205 4.1.2 The intouch triangle and the Gergonne point ...... 205 4.1.3 The Nagel point and the extouch triangle ...... 206 4.2 The trilinear polar ...... 208 4.3 Anticevian triangles ...... 209 4.3.1 The excentral triangle cev−1(I) ...... 211

5 Isotomic and isogonal conjugates 213 5.1 Isotomic conjugates ...... 213 5.1.1 Example: the Gergonne and Nagel points ...... 214 5.1.2 Example: isotomic conjugate of the orthocenter ...... 214 5.1.3 The equal-parallelians point ...... 215 5.1.4 Crelle-Yff points ...... 216 5.2 Isogonal conjugates ...... 218 5.2.1 The symmedian point and the centroid ...... 220 5.2.2 The tangential triangle cev−1(K) ...... 221 5.2.3 The Gergonne point and the insimilicenter T+ ...... 223 5.2.4 The Nagel point and the exsimilicenter T− ...... 226 5.2.5 The Brocard points ...... 229 5.3 Isogonal conjugate of an inﬁnite point ...... 233 5.3.1 Homogeneous barycentric equation of the circumcircle ...... 234 5.4 The isotomic conjugates of inﬁnite points ...... 236

6 Some basic constructions 237 6.1 Perspective triangles ...... 237 6.2 Jacobi’s Theorem ...... 242 6.2.1 The Kiepert perspectors ...... 246 6.3 Gossard’s theorem ...... 247 6.4 Cevian quotients ...... 251 6.4.1 ...... 254 6.5 The cevian quotient G/P ...... 255 6.6 The cevian quotient H/P ...... 256 6.7 Pedal and reflection triangles ...... 259 6.7.1 Reflections and isogonal conjugates ...... 259 6.7.2 The pedal circle ...... 259 6.7.3 Pedal triangle ...... 262 6.7.4 Examples ...... 262 6.7.5 Reflection triangle ...... 264 6.8 Barycentric product ...... 266 6.8.1 Barycentric square ...... 266 CONTENTS v

6.8.2 Barycentric square root ...... 269

7 Orthology 271 7.1 Triangle determined by orthology centers ...... 271 7.1.1 Examples ...... 273 7.2 Perspective orthologic triangles ...... 275 7.2.1 ...... 275

8 The circumcircle 303 8.1 The circumcircle ...... 303 8.1.1 Tangents to the circumcircle ...... 304 8.2 Simson lines ...... 306 8.3 Line of reflections ...... 310 8.4 Reflections of a line in the sidelines of T ...... 311 8.4.1 ...... 314 8.4.2 Perspectivity of reflection triangles ...... 316 8.5 Circumcevian triangles ...... 317 8.5.1 Circumcevian triangle ...... 319 8.5.2 The circumcevian triangle of H ...... 320 8.5.3 The circum-tangential triangle ...... 321 8.5.4 Circumcevian triangles congruent to the reference triangle . ....322 8.6 Triangle bounded by the reflections of a tangent to the circumcircle in the sidelines ...... 323

9 Circles 325 9.1 Generic circles ...... 325 9.2 Power of a point with respect to a circle ...... 328 9.2.1 Radical axis and radical center ...... 328 9.3 The tritangent circles ...... 330 9.3.1 The Feuerbach theorem ...... 331 9.3.2 Radical axes of the circumcircle with the excircles ...... 334 9.3.3 The radical center of the excircles ...... 336 9.3.4 The Spieker radical circle ...... 337 9.4 The Conway circle ...... 339 9.4.1 Sharp’s triad of circles ...... 341 9.5 The Taylor circle ...... 343 9.5.1 The Taylor circle of the excentral triangle ...... 345 9.6 Some triads of circles ...... 346 9.6.1 Circles with sides as diameters ...... 346 9.6.2 Circles with cevians as diameters ...... 347 9.6.3 Circles with centers at vertices and altitudes as radii ...... 348 9.6.4 Circles with altitudes as diameters ...... 349 9.6.5 Excursus: A construction problem ...... 350 9.6.6 Excursus: Radical center of a triad of circles ...... 352 9.6.7 Concurrency of three Euler lines ...... 353 vi CONTENTS

10 Tucker circles 401 10.1 The Taylor circle ...... 401 10.1.1 The pedals of pedals ...... 401 10.1.2 The Taylor circle ...... 402 10.1.3 The Taylor center ...... 402 10.1.4 The Taylor circle of the excentral triangle ...... 404 10.1.5 A triad of Taylor circles ...... 404 10.2 The Taylor circle ...... 405 10.2.1 ...... 406 10.3 Tucker circles ...... 408 10.3.1 The center of Tucker circle ...... 409 10.3.2 Construction of Tucker circle with given center ...... 411 10.3.3 Dao’s construction of the Tucker circles ...... 412 10.4 The Lemoine circles ...... 414 10.4.1 The ﬁrst Lemoine circle ...... 414 10.4.2 The second Lemoine circle ...... 416 10.4.3 Ehrmann’s third Lemoine circle ...... 417 10.4.4 Bui’s fourth Lemoine circle ...... 418 10.5 The Apollonius circle ...... 419 10.6 Tucker circles which are pedal circles ...... 421 10.6.1 The Gallatly circle ...... 422 10.7 Tucker circles which are cevian circumcircles ...... 423 10.7.1 Tucker circle congruent to the circumcircle ...... 424 10.8 Torres’ circle ...... 425 10.9 Tucker circles tangent to the tritangent circles ...... 426 10.9.1 The incircle ...... 426 10.9.2 The excircles ...... 427

11 Some special circles 431 11.1 The Dou circle ...... 431 11.2 The Adams circles ...... 433 11.3 Hagge circles ...... 434 11.3.1 ...... 434 11.4 The deLongchamps triad of circles (A(a),B(b),C(c)) ...... 435 11.5 The orthial circles ...... 437 11.6 Bui’s triad of circles ...... 438 11.7 Appendix: More triads of circles ...... 439 11.7.1 The triad {A(AH )} ...... 439 11.7.2 The triad of circles (AG(AH ),BG(BH ),CG(CH )) ...... 442 11.8 The triad of circles (AG(AH ),BG(BH ),CG(CH )) ...... 443 11.9 The Lucas circles ...... 444 CONTENTS vii

12 The triangle of reflections 447 12.1 The triangle of reflections T† ...... 447 12.2 Triangles with degenerate triangle of reflections ...... 449 12.3 Triads of concurent circles ...... 451 12.3.1 Musselman’s theorem ...... 451 12.3.2 The triad of circles AB†C†, BC†A†, CA†B† ...... 453 12.4 Triangle of reflections and cev−1(O) ...... 455 † 12.4.1 The Parry reflection point OE ...... 458 12.5 Triangle of reflections and the excenters ...... 459 12.5.1 The Evans perspector ...... 459 12.5.2 Triangle of reflections and the tangential triangle ...... 460

13 Circumconics 501 13.1 The perspector of a circumconic ...... 502 13.2 Circumconics as isotomic and isogonal conjugates of lines ...... 503 13.2.1 The fourth intersection of two circumconics ...... 504 13.3 Circumconic with a given center ...... 506 13.4 The center of a circumconic ...... 508 13.5 The Steiner circum-ellipse ...... 509 13.6 The Kiepert hyperbola ...... 511 13.6.1 The Kiepert center ...... 512 13.7 The Jerabek hyperbola ...... 514 13.8 The Feuerbach hyperbola ...... 516

14 The Steiner circumellipse 519 14.1 The Steiner circum-ellipse ...... 519 14.2 The Steiner point ...... 520 14.2.1 Bailey’s theorem on the Steiner point ...... 520 14.3 Construction of the Steiner point ...... 520 14.3.1 Steiner point and the nine-point center ...... 522

15 Circum-hyperbolas 523 15.1 Circum-hyperbolas with given inﬁnite points ...... 523 15.1.1 The circum-hyperbola with perspector St ...... 525 15.2 Circum-hyperbola with a prescribed asymptote ...... 527 15.2.1 The Euler asymptotic hyperbola ...... 528 15.2.2 The orthic asymptotic hyperbola ...... 529 15.2.3 The Lemoine asymptotic hyperbola ...... 530 15.3 Pencil of hyperbolas with parallel asymptotes ...... 531 u + v + w =0 ( : : ) 15.4 The circum-hyperbola x y z for an inﬁnite point u v w ...532 15.4.1 Perspective and orthologic triangles ...... 535 viii CONTENTS

16 Rectangular circum-hyperbolas 537 16.1 The center of a rectangular hyperbola ...... 538 16.2 Construction of asymptotes ...... 539 16.2.1 Antipodes on rectangular circum-hyperbola ...... 539 16.2.2 The Huygens hyperbola ...... 540 16.3 The rectangular circum-hyperbola through a given point ...... 541 16.3.1 The center ...... 541 16.3.2 The fourth intersection with the circumcircle ...... 541 16.3.3 The tangent at H and P ...... 541 16.3.4 The rectangular hyperbola through the Euler reflection point . . . . 543 16.4 Reflection conjugates as antipodal points on a rectangular circum-hyperbola545 16.5 Rectangular circum-hyperbola with a prescribed infinite point ...... 546 16.5.1 The Euler (rectangular) circum-hyperbola ...... 547 16.5.2 The circum-hyperbola with asymptotes parallel to the Brocard and Lemoine axes ...... 548 16.5.3 The circum-hyperbola with asymptotes parallel and perpendicular to the OI line ...... 549 Chapter 1

Some basic notions and fundamental theorems

1.1 Menelaus and Ceva theorems

Let B and C be two ﬁxed points on a line L. Every point X on L, apart BX from B and C, can be coordinatized by the ratio of division XC, where BX and XC are signed lengths. We begin with two classical theorems for the collinearity of three points and concurrency of three lines. Consider a triangle T with points X, Y , Z on the side lines BC, CA, AB respectively. Theorem (Menelaus). The points X, Y , Z are collinear if and only if BX CY AZ · · = −1. (1.1) XC YA ZB

A A

Y Y Z Z Y

X B C X B C

Figure 1.1: The Menelaus theorem

Proof. (⇒) Construct a parallel to the line XY Z through B, to intersect the line AC at Y . It is clear that BX CY AZ Y Y CY AY Y Y CY AY · · = · · = · · =(−1)(−1)(−1) = −1. XC YA ZB YC YA YY YY YC YA 102 Some basic notions and fundamental theorems

(⇐) If the lines YZand BC intersect at X, then BX CY AZ · · = −1. XC YA ZB BX = BX Comparsion with (1.1) gives XC XC . The points X and X divide BC in the same ratio. They are necessarily the same point. This means that X, Y , Z are collinear. Theorem (Ceva). The lines AX, BY , CZ are concurrent if and only if BX CY AZ · · =+1. (1.2) XC YA ZB A

Z P Y

B X C

Figure 1.2: The Ceva theorem

Proof. (⇒) Suppose that the lines are concurrent at a point P . Applying the Menelaus theorem to triangle AXC with transversal BPY ,wehave XB CY AP · · = −1. BC YA PX Likewise, for triangle ABX with transversal CPZ, XP AZ BC · · = −1. PA ZB CX Combining the two relations, with appropriate reversal of signs, we obtain (1.2). (⇐) If the lines BY and CZ intersect at P , and AP intersects BC at X, then BX CY AZ · · =+1. XC YA ZB BX = BX Comparison with (1.2) gives XC XC . The points X and X divide BC in the same ratio. They are necessarily the same point. This shows means that AX, BY , CZ are concurrent at P . 1.2 Harmonic conjugates 103

1.2 Harmonic conjugates

Given four points B, C, X, Y on a line, X and Y are said to divide B and C harmonically if BX BY = − . XC YC In this case, X and Y are harmonic conjugates of each other with respect to the segment BC. Construction. Given a point X on the line BC, to construct the harmonic conjugate of X with respect to the segment BC,

Z Y

B X C X

Figure 1.3: Harmonic conjugates

(1) take an arbitrary point A outside the line BC and construct the lines AB and AC; (2) take an arbitrary point P on the line AX and construct the lines BP and CP to intersect the lines CA and AB at Y and Z respectively; (3) construct the line YZto intersect BC at X. Then X and X divide B and C harmonically (see Figure 1.3). If M is the midpoint of a segment BC, it is not possible to find a finite point N on the line BC so that M, N divide B, C harmonically. This is BN = −BM = −1 = − = = because NC MC requires BN NC CN, and BC CN − BN =0, a contradiction. We shall agree to say that if M and N divide B, C harmonically, then N is the infinite point of the line BC.

1.3 Directed angles

A reference triangle T in a plane induces an orientation of the plane, with respect to which all angles are signed. For two given lines L and L, the 104 Some basic notions and fundamental theorems directed angle ∠(L, L) between them is the angle of rotation from L to L in the induced orientation of the plane. It takes values of modulo π. The following basic properties of directed angles make many geometric reasoning simple without the reference of a diagram. Theorem. (1) ∠(L, L)=−∠(L, L). (2) ∠(L1, L2)+∠(L2, L3)=∠(L1, L3) for any three lines L1, L2 and L3. (3) Four points P , Q, X, Y are concyclic if and only if ∠(PX,XQ)= ∠(PY,YQ). Remark. In calculations with directed angles, we shall slightly abuse notations by using the equality sign instead of the sign for congruence modulo π. It is understood that directed angles are deﬁned up to multiples of π.For example, we shall write β +γ = −α even though it should be more properly β + γ = π − α or β + γ ≡−α mod π. In terms of the angles of T are the directed angles between the sidelines: ∠(c, b)=α, ∠(a, c)=β, ∠(b, a)=γ. Let be a line with ∠(a,)=θ. Then ∠(b,)=γ + θ and ∠(c,)=−β + θ. The tangents to the circumcircle at the vertices of T are characterized by

∠(b, tA)=β, ∠(tA, c)=γ, ∠(c, tB)=γ, ∠(tB, a)=α, ∠(a, tC)=α, ∠(tC, b)=β.

Therefore, ∠(tC, tB)=∠(tC, a)+∠(a, tB)=−α − α = −2α. Similarly, ∠(tA, tC)=−2β and ∠(tB, tA)=−2γ. Theorem (Miquel). Let X, Y , Z be points on the sidelines BC, CA, AB of T respectively. The circles AY Z, BZX, CXY are concurrent. Proof. Let M be the intersection of the circles BZX and CXY . We prove that M also lies on the circle AY Z. This follows from ∠(YM,MZ)= ∠(YM,MX)+∠(XM, MZ) = ∠(YC,CX)+∠(XB, BZ) = ∠(AC, BC)+∠(BC, AB) = ∠(AC, AB) = ∠(YA,AZ). 1.3 Directed angles 105

M is called the Miquel point associated with X, Y , Z.IfX, Y , Z are the traces of P , we call M the Miquel associate of P .

P Miquel associate GO HH

Ge I

Exercise 1. If a, b, c are the sidelines of triangle T, then ∠(a, b)=−γ etc. 106 Some basic notions and fundamental theorems

1.4 The power of a point with respect to a circle

Theorem. Given a point P and a circle O(r), if a line through P intersects the circle at two points A and B, then PA· PB = OP 2 − r2, independent of the line.

O O P

B B P M M A A

Proof. Let M be the midpoint of AB. Note that OM is perpendicular to AB.IfP is outside the circle, then

PA· PB =(PM + MA)(PM − BM) =(PM + MA)(PM − MA) = PM2 − MA2 =(OM2 + PM2) − (OM2 + MA2) = OP 2 − r2.

The same calculation applies to the case when P is inside or on the circle, provided that the lengths of the directed segments are signed.

The quantity OP 2 −r2 is called the power of P with respect to the circle. It is positive, zero, or negative according as P is outside, on, or inside the circle.

Corollary (Intersecting chords theorem). If two chords AB and CD of a circle intersect, extended if necessary, at a point P , then PA· PB = PC · PD. In particular, if the tangent at T intersects AB at P , then PA · PB = PT2.

The converse of the intersecting chords theorem is also true. 1.4 The power of a point with respect to a circle 107

C C

D O O P

B B P D

A T A

Theorem. Given four points A, B, C, D, if the lines AB and CD intersect at a point P such that PA· PB = PC · PD (as signed products), then A, B, C, D are concyclic. In particular, if P is a point on a line AB, and T is a point outside the line AB such that PA·PB = PT2, then PT is tangent to the circle through A, B, T .

1.4.1 Inversion formulas Proposition. The inversive image of P in a circle O(ρ) is the point P −1 which divides OP in the ratio OP −1 : OP = ρ2 : OP 2. Proof. The inversive image is the point P −1 on the half line OP such that OP · OQ = ρ2. OP −1 : OP = OP · OP −1 : OP 2 = ρ2 : OP 2.

Proposition. The center of the inversive image of the circle Q(r) in the circle O(ρ) is the point Q which divides OQ in the ratio OQ : OQ = ρ2 : OQ2 − r2.

Proof. Let Q− and Q+ be the points of the diameter of (Q) through O. −−−→ OQ + r−→ OQ+ = OQ, OQ −−−→ OQ − r−→ OQ− = OQ; OQ −−−→ 2 2 2 −1 ρ −−−→ ρ OQ + r−→ ρ −→ OQ+ = 2 · OQ+ = 2 · OQ = OQ, OQ+ OQ+ OQ OQ(OQ + r) −−−→ 2 2 2 −1 ρ −−−→ ρ OQ − r−→ ρ −→ OQ− = 2 · OQ− = 2 · OQ = OQ. OQ− OQ− OQ OQ(OQ − r) 108 Some basic notions and fundamental theorems

−1 −1 The center of the image circle is the midpoint Q of Q+ and Q− . −−→ −−−→ −−−→ 1 −1 −1 OQ = OQ+ + OQ− 2 1 ρ2 ρ2 −→ = + OQ 2 OQ(OQ + r) OQ(OQ − r) 1 ρ2 2OQ −→ = · · OQ 2 OQ (OQ + r)(OQ − r) ρ2 −→ = OQ. OQ2 − r2

1.5 The 6 concyclic points theorem

The radical axis of two nonconcentric circles C1 and C2 is the locus of points of equal powers with respect to the circle. It is a straight line perpendicular to the line joining their centers. 1 Proposition. Given three circles with mutually distinct centers, the radical axes of the three pairs of circles are either concurrent or are parallel. Proof. If the centers of the circles are noncollinear, then two of the radical axes, being perpendiculars to two distinct lines with a common point, intersect at a point. This intersection has equal powers with respect to all three circles, and also lies on the third radical axis. If the three centers are collinear, then the three radical axes three parallel lines, which coincide if any two of them do. This is the case if and only if the three circles two points in common, or at mutually tangent at a point. In this case we say that the circles are coaxial. If the three circles have non-collinear centers, the unique point with equal powers with respect to the circles is called the radical center.

Example: The radical center of the excircles Consider the excircles of T. The excenter Ia is the intersection of the external bisectors of angles B and C; similarly for the excenters Ib and Ic.

1If the circles are concentric, there is no ﬁnite point with equal powers with respect to the circles. 1.5 The 6 concyclic points theorem 109

Gc Gb Sp

B Ga

Figure 1.4: Radical center of the excircles

Since the tangents to the B- and C-excircles from the midpoint Ga of BC b+c have equal lengths 2 , Ga is on the radical axis of the B- and C-excircle. This radical axis being a line perpendicular to IbIc, it is the bisector of angle GbGaGc, where Gb and Gc are the midpoints of CA and AB respectively; similarly for the other two radical axes. The radical center of the excircles is the incenter of the inferior triangle GaGbGc, also called the Spieker center Sp of T.

Theorem (The 6 concyclic points theorem). Let X1, X2 be points on the sideline BC, Y1, Y2 on CA, and Z1, Z2 on AB of triangle ABC.If

AY1·AY2 = AZ1·AZ2,BZ1·BZ2 = BX1·BX2,CX1·CX2 = CY1·CY2, then the six points X1, X2, Y1, Y2, Z1, Z2 are concyclic.

Proof. By the intersecting chords theorem, the points Y1, Y2, Z1, Z2 lie on a circle C1. Likewise, Z1, Z2, X1, X2 lie on a circle C2, and X1, X2, Y1, Y2 lie on a circle C3. If any two of these circles coincide, then all three circles coincide. If the circles are all distinct, then the sidelines of the triangle, being the three radical axes of the circles and nonparallel, should concur at a point, a contradiction. 110 Some basic notions and fundamental theorems

Example: The nine-point circle

Let Ga, Gb, Gc be the midpoints of the sides BC, CA, AB of triangle ABC, and Ha, Hb, Hc the pedals of A, B, C on their opposite sides. It is easy to see that 1 G · H = G · H = cos A b A b A c A c 2bc α, 1 G · H = G · H = cos B c B c B a B a 2ca β, 1 G · H = G · H = cos C a C a C b C b 2ab γ.

Therefore, the six points Ga, Gb, Gc, Ha, Hb, Hc are on a circle. This is called the nine-point circle of triangle ABC.

Example: The Conway circle

Given triangle ABC, extend (i) CA and BA to Ya and Za such that AYa = AZa = a, (ii) AB and CB to Zb and Xb such that BZb = BXb = b, (iii) BC and AC to Xc and Yc such that CXc = CYc = c.

a a

A s − a s − a Y r Z r s − c I s − b r X X b C c b B s − bX s − c c c

b Yc

Zb 1.5 The 6 concyclic points theorem 111

It is clear that

AYa · AYc = AZa · AZb = −a(b + c),

BZb · BZa = BXb · BXc = −b(c + a),

CXc · CXb = CYc · CYa = −c(a + b).

The six points Xb, Xc, Y√c, Ya, Za, Zb are concyclic. It is concentric with the incircle and has radius r2 + s2. 112 Some basic notions and fundamental theorems Chapter 2

Barycentric coordinates

2.1 Barycentric coordinates on a line

2.1.1 Absolute barycentric coordinates with reference to a segment Consider a line defined by two distinct points B and C. Every finite point BX q on the line is uniquely determined by the ratio of division XC. If this is p, = pB+qC + =0 then we write X p+q . Here p q . If X is the point on the line BC dividing the segment in the ratio BX : XC = t :1− t, we write X =(1− t)B + tC and call this the absolute barycentric coordinates of X with reference to BC. B+C Thus, the midpoint of the segment BC is 2 , and the trisection points are 2B+C B+2C 3 and 3 respectively. More generally, if BX : XC = q : p, then = pB+qC + =0 1 X p+q , provided p q . Example. If X and Y divide B and C harmonically, then B and C divide X and Y harmonically. Proof. Suppose BX : XC = q : p and BX : XC = −q : p. Then pB + qC pB − qC X = ,Y= . p + q p − q Solving for B and C in terms of X and Y ,wehave (p + q)X +(p − q)Y (p + q)X − (p − q)Y B = ,C= . 2p 2q 1Let B and C be distinct points. If p + q =0, then q : p =1:−1. There is no finite point on the line BC satisfying this condition. We shall say that the condition BX : XC =1:−1 defines the infinite point of the line. 114 Barycentric coordinates

Therefore, XB : BY = p − q : p + q and XC : CY = −(p − q):p + q. The points B and C divide X and Y harmonically.

2.1.2 The circle of Apollonius In a triangle, the two bisectors of an angle divide the opposite side harmonically. If X and X are points on the sideline BC of triangle T such that AX and AX are the internal and external bisectors of angle BAC, then BX c BX c = , = − . XC b XC b

B X C X Figure 2.1: Harmonic division by angle bisectors

Theorem. Given two ﬁxed points B, C, and a positive number k =1, 2 the locus of points P satisfying BP : PC = k :1is the circle with diameter XY , where X and Y are points on the line BC such that BX : XC = k :1 and BY : YC = k : −1.

C Y O B X C

Figure 2.2: Circle of Apollonius

2If k =1, the locus is clearly the perpendicular bisector of the segment AB. 2.1 Barycentric coordinates on a line 115

Proof. Since k =1, points X and Y can be found on the line BC satisfying the above conditions. Consider a point P not on the line BC with BP : PC = k :1. Note that PX and PY are respectively the internal and external bisectors of angle BPC. This means that angle XPY is a right angle, and P lies on the circle with XY as diameter. Conversely, let P be a point on this circle. We show that BP : CP = k : 1. Let C be a point on the line BC such that PX bisects angle BPC. Since PB and PC are perpendicular to each other, the line PC is the external bisector of angle BPC, and BY BX XB BY − XB = − = = . YC XC XC YX On the other hand, BY BX XB BY − XB = − = = . YC XC XC YX Comparison of the two expressions shows that C coincides with C, and PB = BX = PX is the bisector of angle BPC. It follows that PC XC k. Remark. If BC = d, and k =1, the radius of the Apollonius circle is k k2−1 d.

2.1.3 The centers of similitude of two circles Consider two circles O(R) and I(r), whose centers O and I are at a distance d apart. Animate a point X on O(R) and construct a ray through I oppositely parallel to the ray OX to intersect the circle I(r) at a point Y .You will ﬁnd that the line XY always intersects the line OI at the same point T . This we call the internal center of similitude, or simply the insimilicenter,of the two circles. It divides the segment OI in the ratio OT+ : T+I = R : r. The absolute barycentric coordinates of P with respect to OI are R · I + r · O T+ = . R + r If, on the other hand, we construct a ray through I directly parallel to the ray OX to intersect the circle I(r) at Y , the line XY always intersects OI at another point T−. This is the external center of similitude, or simply 116 Barycentric coordinates

T+ T−

O I Y X

the exsimilicenter, of the two circles. It divides the segment OI in the ratio OT− : T−I = R : −r, and has absolute barycentric coordinates R · I − r · O T− = . R − r

Example: Points with equal views of two circles

Given two disjoint circles (B) and (C), ﬁnd the locus of the point P such that the angle between the pair of tangents from P to (B) and that between the pair of tangents from P to (C) are equal.

B C

Figure 2.3: Circle of points of equal views of two circles

Let b and c be the radii of the circles. Suppose each of these angles is 2θ. b =sin = c : = : Then BP θ CP , and BP CP b c. From this, it is clear that the locus of P is the circle with the segment joining the centers of similitude of (B) and (C) as diameter. 2.1 Barycentric coordinates on a line 117

Example: Circles tangent to two given circles

P K

O1 O2

Given two circles O1(r1) and O2(r2), suppose there is a third circle K(ρ) tangent to O1(r1) internally at P and to O2(r2) externally at Q. Note that K divides O1P internally in the ratio O1K : KP = r1 − ρ : ρ, so that ρ · O1 +(r1 − ρ)P K = . r1 Similarly, the same point K divides O2Q externally in the O2K : KQ = r2 + ρ : −ρ, so that −ρ · O2 +(r2 + ρ)Q K = . r2 Eliminating K from these two equations, and rearranging, we obtain −r2(r1 − ρ)P + r1(r2 + ρ)Q r2 · O1 + r1 · O2 = . (r1 + r2)ρ r1 + r2 This equation shows that a point on the line PQis the same as a point on the line O1O2. This is the intersection of the lines PQand O1O2. Note that the point on the line O1O2 depends only on the two circles O1(r1) and O2(r2). It is indeed the insimilicenter T+ of the two circles, dividing O1O2 in the ratio O1T+ : T+O2 = r1 : r2. From the equation −r2(r1 − ρ)P + r1(r2 + ρ)Q = T+, (r1 + r2)ρ we note that each of P and Q determines the other, since the line PQpasses through T+. This leads to an easy construction of the point K as the intersection of the lines O1P and O2Q. From this, the circle K(ρ) can be constructed. 118 Barycentric coordinates

P K

O1 O2 O1 O2 T+ T+

Example: Desargues Theorem As a simple illustration of the use of the Menelaus and Ceva theorems, we prove the following Desargues Theorem. Proposition. Given three circles, the exsimilicenters of the three pairs of circles are collinear. Likewise, the three lines each joining the insimilicenter of a pair of circles to the center of the remaining circle are concurrent.

X A

Y C P Y Z X

Proof. We prove the second statement only. Given three circles A(r1), B(r2) and C(r3), the insimilicenters X of (B) and (C), Y of (C), (A), 2.1 Barycentric coordinates on a line 119 and Z of (A), (B) are the points which divide BC, CA, AB in the ratios

BX r2 CY r3 AZ r1 = , = , = . XC r3 YA r1 ZB r2 It is clear that the product of these three ratios is +1, and it follows from the Ceva theorem that AX, BY , CZ are concurrent. 120 Barycentric coordinates

2.2 Absolute barycentric coordinates

We consider a nondegenerate triangle T with vertices A, B, C as the reference triangle, and set up a coordinate system for points in the plane of the triangle. Theorem. Every finite point P of the plane can be expressed as P = uA + vB + wC for unique real numbers u, v, w satisfying u + v + w =1. Proof. This is clearly true if P is one of the vertices A, B, C. If P is a finite point other than the vertices, at most of one of the lines AP , BP, CP is parallel to its opposite sideline. We may assume AP intersecting BC at a finite point X.IfBX : XC = r : q, and AP : PX = s : t for q, r, s, t. Since q + r =0and s =0, we may rewrite AP : PX = q + r : p. = qB+rC = pA+(q+r)X Then, X q+r and P p+q+r . Substitution yields pA + qB + rC P = . p + q + r = p = q = r = + + With u p+q+r , v p+q+r , and w p+q+r ,wehaveP uA vB wC for u + v + w =1. Uniqueness. Suppose also P = uA + vB + wC for u + v + w =1.If u = u, then (v − v)B +(w − w)C uA + vB + wC = uA + vB + wC =⇒ A = , u − u and A is a point on the line BC, a contradiction. It follows that u = u, and similarly v = v, w = w. If P = uA + vB + wC with u + v + w =1, we shall say that P has absolute barycentric coordinates uA + vB + wC with reference to T.

Examples 1 (1) The centroid G = 3(A + B + C). (2) The incenter I (see Figure 2.4). The bisector AI intersects BC at Ia. I : I = : I = ca By the angle bisector theorem, B a aC c b, so that B a b+c.In triangle ABIa, BI is the bisector of angle ABIa, and ca AI : II = BA : BI = c : = b + c : a. a a b + c 2.2 Absolute barycentric coordinates 121

Therefore, aA +(b + c)I aA + bB + cC I = a = . a + b + c a + b + c

Ib Ic I

B Ia C

Figure 2.4: The incenter 122 Barycentric coordinates

2.2.1 Homotheties Let P be a given point, and k a real number. The homothety with center h( ) P and ratio k is the−→ transformation−−→ P, k which maps a point X to the point Y such that PY = k · PX. Equivalently, Y divides PX in the ratio PY : YX = k :1− k, and h(P, k)(X)=(1− k)P + kX.

P k Y 1 − k X

Figure 2.5: Y = h(P, k)(X)

2.2.2 Superior and inferior 1 The homotheties h(G, −2) and h G, −2 are called the superior and inferior operations respectively. Thus, sup(P ) and inf(P ) are the points dividing P and the centroid G according to the ratios P G : Gsup(P )= 1:2, P G : Ginf(P )= 2:1.

P G sup(P )

inf(P )

Figure 2.6: Superior and inferior In absolute barycentric coordinates, sup(P )= 3G − 2P, 1 inf( )= (3G − ) P 2 P . Remark. inf(P ) is the midpoint between P and sup(P ).

2.3 Homogeneous barycentric coordinates

If P = uA + vB + wC in absolute barycentric coordinates with reference to T, we say that P has homogeneous barycentric coordinates (u : v : w) 2.3 Homogeneous barycentric coordinates 123 or k(u : v : w) for any nonzero k. Homogeneous barycentric coordinates are often much simpler and easier to use than absolute coordinates. Thus, in homogeneous barycentric coordinates, G = (1:1:1), I =(a : b : c). Points on the line BC have homogeneous barycentric coordinates of the form (0 : v : w).IfBX : XC = q : p, then X =(0:p : q) in homogeneous barycentric coordinates. Likewise, those on CA are (u :0:w), and those on AB are (u : v :0)respectively. Proposition. If P =(u : v : w) in homogeneous barycentric coordinates, then sup(P )= (v + w − u : w + u − v : u + v − w), inf(P )= (v + w : w + u : u + v). Proof. In absolute barycentric coordinates, sup(P )= 3G − 2P 2(uA + vB + wC) =(A + B + C) − u + v + w ( + + )( + + ) 2( + + ) = u v w A B C − uA vB wC u + v + w u + v + w (v + w − u)A +(w + u − v)B +(u + v − w)C = . u + v + w Therefore, sup(P )=(v + w − u : w + u − v : u + v − w) in homogeneous barycentric coordinates. The case for inferior is similar.

Example: the inferior triangle

Consider the midpoints Ga, Gb, Gc of the sides BC, CA, AB respectively. Each of these is the inferior of the opposite vertex. The triangle GaGbGc is the image of T under the inferior operation; it is called the inferior triangle. The Spieker center Sp, being the incenter of the inferior triangle, is the inferior of the incenter I =(a : b : c). In homogeneous barycentric coordinates,

Sp =(b + c : c + a : a + b). 124 Barycentric coordinates

2.3.1 Euler line and the nine-point circle The superior triangle is the image of T under the superior operation h(G, −2). Its vertices are Ga := sup(A)=(−1:1:1), Gb := sup(B)=(1:−1:1), Gc := sup(C)=(1:1:−1).

Gc A Gb

G H

B Ga C

Ga Figure 2.7: The superior triangle and the Euler line

a a Since Ga is the common midpoint of BC and AG , ABG C is a parallelogram. Similarly, BCGbA and CAGcB are also parallelograms. It follows that Gb, A, and Gc are collinear, and GbGc is parallel to BC. Since A is the midpoint of GbGc, the A-altitude of T is the perpendicular bisector of GbGc; similarly for the B- and C-altitudes. Therefore, the three altitudes of T are concurrent at a point H which is the circumcenter of the superior triangle. This is the orthocenter H, which is the superior of the circumcenter O. In particular, H, G, and O are collinear. The line containing them is the Euler line of T. The circumcenter of the inferior triangle is the point 1 1 O + H N := inf(O)= (3G − O)= (2O + H − O)= 2 2 2 , the midpoint of O and H. The orthocenter of the superior triangle is the point

Lo := sup(H)=3G − 2H =(2O + H) − 2H =2O − H, 2.3 Homogeneous barycentric coordinates 125 the reﬂection of H in O, and is called the deLongchamps point Lo of T. Proposition. The circumcircles of the following three triangles are identi- cal: (a) the inferior triangle, (b) the orthic triangle, 1 (c) the image of T under the homothety h H, 2 .

Gb Gb

Hc N H

B Ha Ga C

Figure 2.8: The nine-point circle

Proof. The fact that the traces of the centroid and the orthocenter are concyclic has been established in §1.5. The circle containing them has N and 1 T radius 2R. Consider the circumcircle of the image of under the homoth- 1 1 ety h H, 2 . This clearly also radius 2R. Its center is the image of O under 1 the homothety, i.e., 2(H + O)=N. The common circumcircle of these three triangles is called the nine-point circle of T. 3 1 2 6

H N G O Lo

Figure 2.9: The Euler line 126 Barycentric coordinates

2.4 Barycentric coordinates as areal coordinates

Applying the Menelaus theorem to triangle ABPa with transversal CPPc, it is easy to see that P divides the segment APa in the ratio PPa : APa = u : u + v + w. It follows that the areas of the oriented triangles PBC and ABC are in the ratio Δ(PBC):Δ(ABC)=u : u + v + w. Similarly, Δ(PCA):Δ(ABC)=v : u + v + w and Δ(PAB):Δ(ABC)=w : u + v + w.

B C Figure 2.10:

This leads to the interpretation of the homogeneous barycentric coordinates of a point P as the proportions of (signed) areas of oriented triangles:

P =(Δ(PBC):Δ(PCA):Δ(PAB)).

2.4.1 The circumcenter O

B C 2.4 Barycentric coordinates as areal coordinates 127

If R denotes the circumradius, the coordinates of the circumcenter O are

O =ΔOBC :ΔOCA :ΔOAB 1 1 1 = 2 sin 2 : 2 sin 2 : 2 sin 2 2R α 2R β 2R γ =sinα cos α :sinβ cos β :sinγ cos γ b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 = a · : b · : c · 2bc 2ca 2ab = a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2).

2.4.2 The incenter and excenters The homogeneous barycentric coordinates of the incenter I can be easily computed as areal coordinates. Let r be the inradius of the triangle. I =(Δ(I ): Δ(I ): Δ(I )) BC CA AB 1 1 1 = : : 2ra 2rb 2rc =(a : b : c).

The A-excenter Ia is the center of the excircle tangent to BC and the extensions of AC and AB respectively. Let ra be the radius of the A-excircle. In homogeneous barycentric coordinates, Ia =(Δ(Ia ): Δ(Ia ): Δ(Ia )) BC CA AB 1 1 1 = − · : · : · 2ra a 2ra b 2ra c =(−a : b : c).

Similarly, the other two excenters are

Ib =(a : −b : c) and Ic =(a : b : −c). 128 Barycentric coordinates

2.5 The area formula

Let Pi, i =1, 2, 3, be points given in absolute barycentric coordinates

Pi = xiA + yiB + ziC, with xi + yi + zi =1. The oriented area of triangle P1P2P3 is x1 y1 z1 x2 y2 z2 Δ, x3 y3 z3 where Δ is the area of ABC. If the vertices are given in homogeneous coordinates, this area is given by x1 y1 z1 x2 y2 z2 x3 y3 z3 Δ. (x1 + y1 + z1)(x2 + y2 + z2)(x3 + y3 + z3) In particular, Since the area of a degenerate triangle whose vertices are collinear is zero, we have the following useful formula. Example. The area of triangle GIO is 111 1 · Δ 2 abc 3(a + b + c) · 16Δ 2 2 2 2 2 2 2 2 2 2 2 2 a (b + c − a ) b (c + a − b ) c (a + b − c ) 001 1 = − − · Δ 2 a bbcc 3(a + b + c) · 16Δ 2 2 2 2 2 2 2 2 2 2 −(a − b)(a + b)(a + b − c ) −(b − c)(b + c)(b + c − a ) c (a + b − c ) 001 ( − )( − ) = a b b c 11 · Δ 2 c 3(a + b + c) · 16Δ 2 2 2 2 2 2 2 2 2 2 −(a + b)(a + b − c ) −(b + c)(b + c − a ) c (a + b − c ) (a − b)(b − c) 11 = 2 2 2 2 2 2 · Δ 3(a + b + c) · 16Δ2 −(a + b)(a + b − c ) −(b + c)(b + c − a ) ( − )( − )( − )( + + )2 = a b b c c a a b c · Δ 3(a + b + c)(a + b + c)(b + c − a)(c + a − b)(a + b − c) ( − )( − )( − ) = a b b c c a · Δ 3(b + c − a)(c + a − b)(a + b − c) 2.5 The area formula 129

2.5.1 Conway’s notation We shall make use of the following convenient notations introduced by John H. Conway. Instead of for the area of triangle ABC, we shall ﬁnd it more convenient to use S := 2 .

For a real number θ, denote S · cot θ by Sθ. In particular, 2 + 2 − 2 2 + 2 − 2 2 + 2 − 2 = b c a = c a b = a b c Sα 2 ,Sβ 2 ,Sγ 2 .

For arbitrary θ and ϕ, we shall simply write Sθϕ for Sθ · Sϕ. We shall mainly make use of the following relations.

2 2 2 (1) a = Sβ + Sγ, b = Sγ + Sα, c = Sα + Sβ. a2+b2+c2 (2) Sα + Sβ + Sγ = 2 . 2 (3) Sβγ + Sγα + Sαβ = S .

Proof. (1) and (2) are clear. For (3), since α + β + γ = π, cot(α + β + γ) is inﬁnite. Its denominator

cot α · cot β +cotβ · cot γ +cotγ · cot α − 1=0.

2 From this, Sαβ +Sβγ +Sγα = S (cot α·cot β+cotβ·cot γ+cotγ·cot α)= S2.

Example: The circumcenter and orthocenter

In Conway’s notations,

2 2 2 O =(a Sα : b Sβ : c Sγ)

=(SγSα + SαSβ : SαSβ + SβSγ : SβSγ + SγSα)

= inf(SβSγ : SγSα : SαSβ).

Since O = inf(H), it follows that 1 1 1 H =(SβSγ : SγSα : SαSβ)= : : . Sα Sβ Sγ 130 Barycentric coordinates

2.5.2 Conway’s formula The position of a point can be speciﬁed by its “compass bearings” from two vertices of the reference triangle. Given triangle ABC and a point P , the AB- swing angle of P is the oriented angle CBP, of magnitude ϕ reckoned positive if and only if it is away from the vertex A. Similarly, the AC-swing angle is the oriented angle BCP, of magnitude ψ reckoned positive if and only if it is away from the vertex A. The swing angles are chosen in the π π range − 2 ≤ ϕ, ψ ≤ 2 . The point P is uniquely determined by ϕ and ψ in this range. We shall denote this by X(ϕ, ψ). Theorem (Conway’s formula). In homogeneous barycentric coordinates,

2 X(ϕ, ψ)=(−a : Sγ + Sψ : Sβ + Sϕ).

B ϕ ψ C

X(ϕ, ψ) Figure 2.11: The point X(ϕ, ψ)

Proof. Let X = X(ϕ, ψ). Its homogeneous barycentric coordinates are

Δ(XBC):Δ(XCA):Δ(XAB) a2 sin ϕ sin ψ b · a sin ϕ c · a sin ψ = − : · sin(ψ + γ): · sin(ϕ + β) 2sin(ϕ + ψ) 2sin(ϕ + ψ) 2sin(ϕ + ψ) ab sin(ψ + γ) ca sin(ϕ + β) = −a2 : : sin ψ sin ϕ = −a2 : ab cos γ + ab sin γ cot ψ : ca cos β + ca sin β cot ϕ 2 = −a : Sγ + Sψ : Sβ + Sϕ. 2.6 Triangles bounded by lines parallel to the sidelines 131

2.6 Triangles bounded by lines parallel to the sidelines

Theorem (Homothetic center theorem). If parallel lines XbXc, YcYa, ZaZb to the sides BC, CA, AB of triangle ABC are constructed such that

AB : BXc = AC : CXb =1 : t1,

BC : CYa = BA : AYc =1 : t2,

CA : AZb = CB : BZa =1 : t3, these lines bound a triangle A∗B∗C∗ homothetic to ABC with homothety ratio 1+t1 + t2 + t3. The homothetic center is a point P with homogeneous barycentric coordinates t1 : t2 : t3.

A∗ A∗

Zb Zb Yc Yc

A A

P P

Za Ya Za Ya B C B C

∗ ∗ ∗ ∗ B Xc Xb C B Xc Xb C

Proof. Let P be the intersection of B∗B and C∗C. Since ∗ ∗ ∗ ∗ B C = B Xc + XcXb + XbC = t3a +(1+t1)a + t2a =(1+t1 + t2 + t3)a, we we have ∗ ∗ PB : PB = PC : PC =1:1+t1 + t2 + t3. A similar calculation shows that AA∗ and BB∗ intersect at the same point P . This shows that A∗B∗C∗ is the image of ABC under the homothety h(P, 1+t1 + t2 + t3). Now we compare areas. Note that Δ( )=BXc · BZa · Δ( )= Δ( ) (1) BZaXc AB CB ABC t1t3 ABC , 132 Barycentric coordinates

Δ(PBC) PB CB 1 1 1 (2) ∗ = ∗ · = · = . Δ(BZaB ) BB BZa t1+t2+t3 t3 t3(t1+t2+t3)

Since Δ(BZ B∗)=Δ(BZ X ),wehaveΔ(PBC)= t1 ·Δ(ABC). a a c t1+t2+t3 Similarly, Δ(PCA)= t2 · Δ(ABC) and Δ(PAB)= t3 · t1+t2+t3 t1+t2+t3 Δ(ABC). It follows that

Δ(PBC):Δ(PCA):Δ(PAB)=t1 : t2 : t3.

Corollary. Two triangles with corresponding sidelines parallel are homothetic.

2.6.1 The symmedian point

Consider the square erected externally on the side BC of triangle ABC, The line containing the outer edge of the square is the image of BC under S + 2 2 a a a a the homothety h(A, 1+t1), where 1+t1 = S =1+ , i.e., t1 = . a S S Similarly, if we erect squares externally on the other two sides, the outer edges of these squares are on the lines which are the images of CA, AB h( 1+ ) h( 1+ ) = b2 = c2 under the homotheties B, t2 and C, t3 with t2 S and t3 S .

A∗

Ca A Bc

K Cb B C

∗ ∗ B Ab Ac C Figure 2.12: The Grebe triangle and the symmedian point 2.6 Triangles bounded by lines parallel to the sidelines 133

The triangle bounded by the lines containing these outer edges is called the Grebe triangle of ABC. It is homothetic to ABC at a2 b2 c2 K = : : =(a2 : b2 : c2), S S S the symmedian point, and the ratio of homothety is S + a2 + b2 + c2 1+(t1 + t2 + t3)= . S

2.6.2 The ﬁrst Lemoine circle Given a point P =(u : v : w), the parallel to BC through P intersects AC at Ya and AB at Za. We call the segment ZaYa the A-parallelian through P . Likewise, the B- and C-parallelians are the segments XbZb and YcXc. We locate the point(s) P for which the six parallelian endpoints lie on a circle. Note that v + w v + w AY = · b and AZ = · c. a u + v + w a u + v + w Also, w v AY = · b and AZ = · c. c u + v + w b u + v + w

K Za Ya

B Xc Xb C

2 2 Therefore, AYa · AYc = AZa · AZb if and only if wb = vc ,orv : w = 2 2 2 2 b : c . Similarly, BZb · BZa = BXb · BXc if and only if w : u = c : a , 134 Barycentric coordinates and 2 2 CXc · CXb = CYc · CYa if and only if u : v = a : b . It follows that the six parallelian endpoints are concyclic if and only if u : v : w = a2 : b2 : c2, i.e., P is the symmedian point K. The circle containing them is called the ﬁrst Lemoine circle. Remark. The center of this circle is the midpoint of OK.

Exercise 1. Show that for the ﬁrst Lemoine circle, 2 2 2 BXc : XcXb : XbC = c : a : b , 2 2 2 CYa : YaYc : YcA = a : b : c , 2 2 2 AZb : ZbZa : ZaB = b : c : a .

2. Show that YcZb = ZaXc = XbYa. Chapter 3

Straight lines

3.1 The two-point form

A straight line in the plane of T is represented by a homogeneous linear equation in the homogeneous barycentric coordinates of points the line. Given two points P1 =(x1 : y1 : z1) and P2 =(x2 : y2 : z2), a point P =(x : y : z) lies on the line P1P2 if and only if x1 y1 z1 x2 y2 z2 =0, xyz or (y1z2 − y2z1)x +(z1x2 − z2x1)y +(x1y2 − x2y1)z =0. 3 Proof. As vectors in R , (x1,y1,z1), (x2,y2,z1), and (x, y, z) are linearly dependent. Proposition. (a) The intersection of the two lines

p1x + q1y + r1z =0, p2x + q2y + r2z =0 is the point (q1r2 − q2r1 : r1p2 − r2p1 : p1q2 − p2q1).

(b) Three lines pix + qiy + riz =0, i =1, 2, 3, are concurrent if and only if p1 q1 r1 p2 q2 r2 =0. p3 q3 r3 136 Straight lines

Examples. (1) The equations of the sidelines BC, CA, AB are respectively x =0, y =0, z =0. (2) The equation of the line joining the centroid and the incenter is 111 abc =0, xyz or (b − c)x +(c − a)y +(a − b)z =0. (3) The equations of some important lines: OH ( 2 − 2)( 2 + 2 − 2) =0 Euler line cyclic b c b c a x OI ( − )( + − ) =0 OI-line cyclic bc b c b c a x OK 2 2( 2 − 2) =0 Brocard axis cyclic b c b c x 2 Soddy line IGe cyclic(b − c)(b + c − a) x =0

3.1.1 Cevian and anticevian triangles of a point Let P =(u : v : w) be a given point. (1) The equations of the lines AP , BP, CP are y z AP :+− =0, v w x z BP : − + =0, u w x y CP :+− =0. u v (2) These lines intersect the sidelines at the points

Pa := BC ∩ AP =(0:v : w),

Pb := CA ∩ BP =(u :0:w),

Pc := AB ∩ CP =(u : v :0).

The triangle PaPbPc is called the cevian triangle of P , and is denoted by cev(P ). 3.1 The two-point form 137

(3) The equations of the lines PbPc, PcPa, PaPb are

x y z P P : − + + =0, b c u v w x y z P P :+− + =0, c a u v w x y z P P :++ − =0. a b u v w

(4) These lines intersect the sidelines at := ∩ =(0: : − ) Pa BC APa v w , := ∩ =(− :0: ) Pb CA BPb u w , := ∩ =( : − :0) Pc AB CPc u v . (5) The three points Pa, Pb, Pc on the sidelines are collinear. The line containing them is x y z L : + + =0. P u v w This line is called the trilinear polar of P . (6) The equations of the lines APa, BPb, CPc are y z AP :++ =0, a v w x z BP : + =0, b u w x y CP : + =0. c u v

(7) The lines APa, BPb, CPc bound a triangle with vertices a := ∩ =(− : : ) P BPb CPc u v w , b := ∩ =( : − : ) P CPc APa u v w , c := ∩ =( : : − ) P APa BPb u v w . The triangle P aP bP c is called the anticevian triangle of P , and is denoted by cev−1(P ). 138 Straight lines

3.2 Inﬁnite points and parallel lines

3.2.1 The inﬁnite point of a line

A line px + qy + rz =0contains the point (q − r : r − p : p − q),asis easily verified. Since the sum of its coordinates is zero, this is not a finite point. We call it an infinite point. Thus, an infinite point (x : y : z) is one which satisfies the equation x + y + z =0, which we regard as defining the line at infinity L∞. Now, it is easy to see that unless p : q : r =1:1:1,a line px + qy + rz =0has a unique infinite point as given above. Therefore, the infinite point of a line determines its “direction”. Two lines are parallel if and only if they have the same infinite point. It follows that the line through (u : v : w) parallel to px + qy + rz =0has equation xyz uvw =0. q − rr− pp− q

Examples

(1) The inﬁnite points of the sidelines of ABC are as follows.

Line Equation Inﬁnite point a x =0 (0:1:−1) b y =0 (−1:0:1) c z =0 (1:−1:0)

(2) If P =(u : v : w), the line joining the centroid G to P has equation (v − w)x +(w − u)y +(u − v)z =0. It has inﬁnite point (v + w − 2u : w + u − 2v : u + v − w).

(3) The Euler line, being the line joining G to H =(Sβγ : Sγα : Sαβ) has inﬁnite point

E∞ := (Sγα + Sαβ − 2Sβγ :+Sαβ + Sβγ − 2Sγα :2SβγSγα − 2Sαβ).

This is called the Euler inﬁnity point. 3.2 Inﬁnite points and parallel lines 139

Parametrization of a line The ﬁnite points of the line ux + vy + wz =0can be parametrized as

((v − w)(vw + t): (w − u)(wu + t): (u − v)(uv + t)).

Exercise 1. Find the equations of the lines through P =(u : v : w) parallel to the sidelines. 2. (a) Find the infinite point of the bisector of angle A. 1 (b) Find the infinite point of the external bisector of angle A. 2 3. Find the infinite point of the trilinear polar of (u : v : w). 4. Let D, E, F be the midpoints of the sides BC, CA, AB of triangle ABC. For a point P with traces AP , BP , CP , let X, Y , Z be the midpoints of BP CP , CP AP , AP BP respectively. (a) Find P such that the lines DX, EY , FZ are parallel to the internal bisectors of angles A, B, C respectively. 3 (b) Explain why the lines DX, EY , FZ are concurrent and identify the point of concurrency. 4

1(−(b + c):: c). 2(b − c : −b¯: c). 3The Nagel point. 4Spieker center. 140 Straight lines

CP X F E

I P Y Z

B D AP C

5. Let P be a given point with cevian triangle XY Z. Consider the triangle XY Z bounded by the parallels to YZ, ZX, XY through A, B, C respectively. Show that XY Z is an anticevian triangle, and ﬁnd the homothetic center with cev(P ). 5

Z P

B X C

5If P =(u : v : w), this is the anticevian triangle of (u(v + w):v(w + u):w(u + v)). The homothetic center is (u2(v + w):v2(w + u):w2(u + v)). 3.2 Inﬁnite points and parallel lines 141

3.2.2 Inﬁnite point as vector

The infinite point of a line through two given points can be computed through a calculation of absolute barycentric coordinates. If P =(u : v : w) and Q =(u : v : w) are finite points, the infinite point of the line PQ can be computed from

u u v v w w Q−P = − , − , − u + v + w u + v + w u + v + w u + v + w u + v + w u + v + w

−→ which we regard as the vector PQ. From this, we obtain a simpler ex- pression for the homogeneous barycentric coordinates of the inﬁnite point, namely

(u(v + w) − u(v + w):v(w + u) − v(w + u):w(u + v) − w(u + v)).

The altitudes

The inﬁnite point of the A-altitude is given by the vector 1 1 (0,Sγ,Sβ) − (1, 0, 0) = (−(Sβ + Sγ),Sγ,Sβ). Sβ + Sγ Sβ + Sγ From this, we obtain the homogeneous barycentric coordinates of the inﬁ- nite points of the altitudes.

Line Inﬁnite point A − altitude −(Sβ + Sγ): Sγ : Sβ B − altitude Sγ : −(Sγ + Sα): Sα C − altitude Sβ : Sα : −(Sα + Sβ) It follows that the equation of the perpendicular bisector of BC is xyz 011 =0, −(Sβ + Sγ) Sγ Sβ or

(Sβ − Sγ)x − (Sβ + Sγ)(y − z)=0. 142 Straight lines

Since the intouch triangle is homothetic to the excentral triangle, its altitudes are parallel to the angle bisectors of triangle ABC. Thus, the altitude through X has inﬁnite point (b + c, −b, −c); it is the line 0 a + b − cc+ a − b b + c −b −c = −((b−c)(b+c−a)x−(b+c)(c+a−b)y+(b+c)(a+b−c) xy z Similarly, the other two altitudes are (c + a)(b + c − a)x +(c − a)(c + a − b)y − (c + a)(a + b − c)z =0, −(a + b)(b + c − a)x +(a + b)(c + a − b)y +(a − b)(a + b − c)z =0. The orthocenter of the intouch triangle is the intersection of these two lines, namely, ( + − ) :(+ − ) :( + − ) b c a x c a b y a b c z − −( + ) + −( + ) + − = c a c a : − c a c a : c aca a + ba− b −(a + b) a − b −(a + b) a + b =2a(b + c):2b(c + a):2c(a + b).

Therefore, the orthocenter of the intouch triangle is the point with homogeneous barycentric coordinates a(b + c) b(c + a) c(a + b) : : . b + c − a c + a − b a + b − c

Exercise 1. Let XY Z be the intouch triangle of triangle ABC. Show that the pedal (orthogonal projection) of X on YZis the point b + c b c X = : : . b + c − a c + a − b a + b − c Similarly deﬁne Y and Z. The triangle XY Z is homothetic to ABC at T . 2. Compute the coordinates of the inﬁnite point of the OI-line. 6

6(a(a2(b + c) − 2abc − (b + c)(b − c)2):···: ···). 3.2 Inﬁnite points and parallel lines 143

Exercise 1. Given triangle ABC, extend, if necessary, (i) AC and AB to Ya and Za such that AYa = AZa = a, (ii) BA and BC to Zb and Xb such that BZb = BXb = b, (iii) CB and CA to Xc and Yc such that CXc = CYc = c.

B Xc C Xb

7 (a) Find the coordinates of the points Xb, Xc, Yc, Ya, Za, Zb.

(b) Find the equations of the three lines YcZb, ZaXc, XbYa, and show that they are concurrent. 8

(d) Find the equations of the lines YaZa, ZbXb, XcYc, and show that 10 they are perpendicular to YcZb, ZaXc, XbYa respectively. (e) Let X = BC ∩ YaZa, Y = CA ∩ ZbXb and Z = AB ∩ XcYc. Show that XY Z is perspective with ABC and ﬁnd the perspector. 11

7 Xb =(0:a − b : b), Xc =(0:c : a − c) etc. 8 YcZb :(b − c)x + by − cz =0etc. These are parallel to the angle bisectors. They are concurrent at the Nagel point. 9 1 1 1 X =(0:c : b) etc. The triangle has perspector a : b : c . 10 YaZa : ax +(a − c)y +(a − b)z =0etc. These are parallel to the external angle bisectors. 11 1 1 1 X =(0:a − b : c − a) etc., and X Y Z is perspective with ABC at b−c : c−a : a−b . 144 Straight lines

(f) Let X = YcZb ∩ YaZa, Y = ZaXc ∩ ZbXb and Z = XbYa ∩ XcYc. Show that X Y Z is perspective with ABC and ﬁnd the perspector. 12

12X =(−a(b + c)+(b2 + c2): b(c + a − b): c(a + b − c)) etc. The triangle has perspector Mı =(a(b + c − a): b(c + a − b): c(a + b − c)). 3.3 Perpendicular lines 145

3.3 Perpendicular lines

Given a line L : px + qy + rz =0, we determine the infinite point of lines perpendicular to it. The line L intersects the side lines CA and AB at the points Y =(−r :0:f) and Z =(q : −p :0). To find the perpendicular from A to L, we first find the equations of the perpendiculars from Y to AB and from Z to CA. These are − 2 − 2 Sβ Sα c Sγ b Sα −r 0 f =0 and q −p 0 =0 xy z xyz These are 2 Sαfx+(c h − Sβf)y + Sαrz =0, 2 Sαfx+ Sαqy +(b g − Sγp)z =0.

Z Y X

B C

These two perpendiculars intersect at the orthocenter of triangle AY Z, which is the point 2 2 X =(∗∗∗: Sαp(Sαr − b q + Sγp):Sαp(Sαq + Sβp − c r) ∼ (∗∗∗: Sγ(p − q) − Sα(q − r):Sα(q − r) − Sβ(r − p)). The perpendicular from A to L is the line AX, which has equation 10 0 ∗∗∗ Sγ(p − q) − Sα(q − r) −Sα(q − r)+Sβ(r − p) =0, xy z 146 Straight lines or

−(Sα(q − r) − Sβ(r − p))y +(Sγ(p − q) − Sα(q − r))z =0.

This has inﬁnite point

(Sβ(r − p) − Sγ(p − q):Sγ(p − q) − Sα(q − r):Sα(q − r) − Sβ(p − q)).

Note that the infinite point of L is (q − r : r − p : p − q). We summarize this in the following theorem. Theorem. If a line L has infinite point (f : g : h) (satisfying f +g+h =0), the lines perpendicular to L have infinite point (f : g : h), where

f =Sβg − Sγh, g =Sγh − Sαf, h =Sαf − Sβg.

Equivalently, two lines with inﬁnite points (f : g : h) and (f : g : h) are perpendicular to each other if and only if

Sαff + Sβgg + Sγhh =0.

Corollary. The perpendicular from the point (u : v : w) to the line with inﬁnite point (f : g : h) is the line uvw f g h =0. xyz

Orthogonal inﬁnite points

Line Inﬁnite point Orthogonal inﬁnite po Euler line E∞ := (Sγα + Sαβ − 2Sβγ : ···: ···)(Sβ − Sγ : ···: ··· x + y + z =0 ( ( − ):···: ···)(( − ) − ( − ) u v w u v w SBv w u SCw u v OI (a(b − c):b(c − a):c(a 3.3 Perpendicular lines 147

Exercise 1. Given triangle ABC, construct a circle tangent to AC at Y and AB at Z such that the line YZpasses through the centroid G. (i) Show that YG: GZ = c : b, 13 (ii) Calculate the coordinates of the center P of the circle. 14 (iii) Show that the line GP is perpendicular to BC. A

G Y

Z P B C

2. Given triangle ABC with orthic triangle DEF, let the parallels through H to BC, CA, AB intersect EF, FD, DE respectively at X, Y , Z. The points X, Y , Z are collinear on a line perpendicular to the Euler line. 15 3. Given triangle ABC with orthic triangle DEF, let the parallels through H to EF, FD, DE intersect BC, CA, AB respectively at X, Y , Z. The points X, Y , Z are collinear on a line perpendicular to the Euler line. 16

13If AY = AZ = t, Y =(b − t :0:t) and Z =(c − t : t :0). If the line YZ contains G, then λ 1−λ 1 b b+c b (b − t, 0,t)+ c (c − t, t, 0) = 3 (1, 1, 1) for some λ. From this, we have λ = b+c and t = 3 . Thus, YG: GZ = c : b. 14(−3a2 +(b + c)2 :2b(b + c):2c(b + c)). 15Except for the perpendicularity, this is true if H and the orthic triangle are replaced by a generic point and its cevian triangle. If P =(u : v : w), the line is the trilinear polar of u : v : w . v+w−u w+u−v u+v−w 1 1 1 1 1 1 This line has inﬁnite point (u(v − w):v(w − u):w(u − v)) = v − w : w − u : u − v . In the case of the Euler line, this is (Sβ − Sγ : Sγ − Sα : Sα − Sβ). 16Except for the perpendicularity, this is true if H and the orthic triangle are replaced by a generic point and its cevian triangle. If P =(u : v : w), the line is the trilinear polar of u : v : w . This line v+w w+u u+v 1 1 1 1 1 1 has inﬁnite point (u(v − w):v(w − u):w(u − v)) = v − w : w − u : u − v . In the case of the Euler line, this is (Sβ − Sγ : Sγ − Sα : Sα − Sβ). 148 Straight lines

E Y O X F H

B D C

O F H

X B D C

3.4 The distance formula

Let P = uA + vB + wC and Q = uA + vB + wC be given in absolute barycentric coordinates. The distance between them is given by

2 2 2 2 PQ = Sα(u − u ) + Sβ(v − v ) + Sγ(w − w ) .

Proof. Through P and Q draw lines parallel to AB and AC respectively, intersecting at X. The barycentric coordinates of X can be determined in two ways: X = P + k(A − B)=Q + h(A − C) for some h and k. It follows that X =(u + k)A +(v − k)B + wC =(u + h)A + vB +(w − h)C. 3.4 The distance formula 149

B C

From these we have

h = −(w − w),k= v − v, and X =(1− w − v)A + vB + wC.

Applying the law of cosines to triangle XPQ (in which ∠PXQ = α,we have

PQ2 =(hb)2 +(kc)2 − 2(hb)(kc)cosα 2 2 2 2 = h b + k c − 2hkSα =( − )2( + )+( − )2( + )+2( − )( − ) w w Sγ Sα v v Sα Sβ v v w w Sα 2 2 2 2 = (w − w ) +(v − v ) +2(v − v )(w − w ) Sα +(v − v ) Sβ +(w − w ) Sγ 2 2 2 =((w − w )+(v − v )) Sα +(v − v ) Sβ +(w − w ) Sγ 2 2 2 = Sα(u − u ) + Sβ(v − v ) + Sγ(w − w ) .

In homogeneous barycentric coordinates, if P =(x : y : z) and Q = (u : v : w) are ﬁnite points, then the square distance between them is given by 2 = 1 (( + ) − ( + ))2 PQ (u+v+w)2(x+y+z)2 cyclic Sα v w x u y z 150 Straight lines

Exercise 1. Show that the distances from P =(x : y : z) to the vertices of triangle ABC are given by c2y2 +2S yz + b2z2 AP 2 = α , (x + y + z)2 a2z2 +2S zx + c2x2 BP2 = β , (x + y + z)2 b2x2 +2S xy + a2y2 CP2 = γ . (x + y + z)2

2. Show that the square distance between P =(x : y : z) and Q =(u : v : w) can be written as ⎛ ⎞ 1 2 2 +2 + 2 2 2 + 2 + 2 2 = ⎝ c v Sαvw b w ⎠−a yz b zx c xy PQ + + ( + + )2 x ( + + )2 . x y z cyclic u v w x y z

3.4.1 The distance from a point to a line Consider a line L with equation px+qy+rz =0. We shall assume p, q, r not all equal so that L is not the line at infinity x + y + z =0. For convenience, we write its infinite point as (f : g : h)=(q − r : r − p : p − q). The orthogonal infinite point being

(f ,g,h)=(Sβg − Sγh, Sγh − Sαp, Sαp − Sβq), we seek a quantity k such that A + k(f ,g,h) lies on the line L. Thus,

p(1 + kf)+q · kg + r · kh =0, and p k = − . pf + qg + rh The (signed) distance from A to L is k times the length of the vector (f ,g,h). Similar calculations give the signed distances from B and C, and we have the following simple characterization of the barycentric equation of a straight line. 3.4 The distance formula 201

Proposition. The signed distances from the vertices A, B, C to the line px + qy + rz =0are in the ratio p : q : r. Note that

− (pf + qg + rh)

= −p(Sβ(r − p) − Sγ(p − q) − q(Sγ(p − q) − Sα(q − r)) − r(Sα(q − r) − Sβ(r − p)) 2 2 2 = Sα(q − r) + Sβ(r − p) + Sγ(p − q) , which is the square distance between the two finite points (q : r : p) and (r : p : q), assuming p + q + r =1. 17 This is zero if and only if p = q = r, in which case L is the line at infinity. Proposition. The signed distance from a finite point (x : y : z) to the line px + qy + rz =0is px + qy + rz S · . + + 2 2 2 x y z Sα(q − r) + Sβ(r − p) + Sγ(p − q)

17This assumption is valid if p + q + r =0.Ifp + q + r =0, use instead (q + s : r + s : p + s) and (r + s : p + s : q + s) for some nonzero s. 202 Straight lines Chapter 4

Cevian and anticevian triangles

4.1 Cevian triangles

We begin with a convenient reformulation of the Ceva theorem. Theorem (Ceva). For points X on BC, Y on CA, and Z on AB, the lines AX, BY , CZ are concurrent if and only if their homogeneous barycentric coordinates are X =0:v : w, Y = u :0:w, Z = u : v :0, for some u, v, w. If this condition is satisﬁed, the point of concurrency is P =(u : v : w). A

Pb Pc P

B Pa C Figure 4.1: The traces and cevian triangle of P The points X, Y , Z are called the traces of P . We also say that XY Z is the cevian triangle of P (with reference to T). Sometimes, we shall adopt the more functional notation for the cevian triangle and its vertices:

cev(P ): Pa =(0:v : w),Pb =(u :0:w),Pc =(u : v :0). 204 Cevian and anticevian triangles

Examples. P cev(P ) G inferior triangle I incentral triangle H orthic triangle 4.1 Cevian triangles 205

4.1.1 The orthic triangle The orthic triangle is the cevian triangle of the orthocenter H. Its vertices are

Ha =(0:Sγ : Sβ), Hb =(Sγ :0:Sα), Hc =(Sβ : Sα :0).

Hc Hb A H c H

B Ha C B Ha C Figure 4.2: The orthic triangle and its ob-incicle

If triangle T is acute, then H is the incenter of the orthic triangle XY Z. If the triangle is obtuse, then H is an excenter of the orthic triangle. We say that H is the ob-incenter of the orthic triangle.

4.1.2 The intouch triangle and the Gergonne point The incircle touches the sides of T at the pedals of the incenter I. These deﬁne the Gergonne point Ge and the intouch triangle: I =0:1 : 1 [a] c+a−b a+b−c, I = 1 :0:1 [b] b+c−a a+b−c, I = 1 : 1 :0 [c] b+c−a c+a−b , G = 1 : 1 : 1 e b+c−a c+a−b a+b−c. 206 Cevian and anticevian triangles

s − c s − a

I[b]

I [c] G e I s − c s − b

B s − b I[a] s − c C

Figure 4.3: The Gergonne point and the intouch triangle

4.1.3 The Nagel point and the extouch triangle Ia Ia = − The A-excircle touches the side BC at a point [a] such that B [a] s c and Ia = − [a]C s b. From this, we have the homogeneous barycentric coordinates Ia Ib Ic of [a]; similarly, for the points of tangency [b] and [c] of the B- and C- excircles:

Ic Ic [c] Ib Na [b]

C B Ia s − b s − c [a] s − b

s − c

Ia Figure 4.4: The Nagel point and the extouch triangle

Ia =0:+ − : + − [a] c a b a b c, Ib = + − :0:+ − [b] b c a a b c, Ic = + − : + − :0 [c] b c a c a b , Na = b + c − a : c + a − b : a + b − c. 4.1 Cevian triangles 207

The cevian triangle of the Nagel point is called the extouch triangle. Its centroid is the point

G[cev(Na)] = (a(b+c)(b+c−a): b(c+a)(c+a−b): c(a+b)(a+b−c)). 208 Cevian and anticevian triangles

4.2 The trilinear polar

Consider the cevian triangle cev(P )=PaPbPc for P =(u : v : w). The line joining Pb and Pc has equation x y z − + + =0. u v w It intersects the sideline a at X =(0:−v : w). Similarly, the lines PcPa and PaPb intersect b and c at Y =(u :0:−w) and Z =(−u : v :0).

Pc P

X B Pa C

Figure 4.5:

The three points X, Y , Z are collinear. The line containing them is called the trilinear polar of P : L : x + y + z =0 P u v w . Examples.

P Trilinear polar barycentric equation G line at inﬁnity x + y + z =0 H orthic axis Sαx + Sβy + Sγz =0 Ge Gergonne axis (s − a)x +(s − b)y +(s − c)z =0

Remark. The orthic axis is perpendicular to the Euler line. 4.3 Anticevian triangles 209

4.3 Anticevian triangles

A triangle XY Z is said to be an anticevian triangle of P with respect to triangle ABC if ABC is a cevian triangle of XY Z (with the same perspector P ). Suppose P has coordinates (x : y : z) with respect to XY Z. We want to ﬁnd its coordinates, and those of X, Y , Z, with respect to ABC. Note that in absolute barycentric coordinates, yY + zZ xX + zZ xX + yY A = ,B= ,C= . y + z x + z x + y It is easier to write X, Y , Z in terms of A, B, C: −(y + z)A +(z + x)B +(x + y)C X = , 2x (y + z)A − (z + x)B +(x + y)C Y = , 2y (y + z)A +(z + x)B − (x + y)C Z = . 2z From these we also obtain the coordinates of P : xX + yZ + zZ (y + z)A +(z + x)B +(x + y)C P = = . x + y + z 2(x + y + z)

We relabel X, Y , Z by P a, P b, P c and call triangle cev−1(P ):=P aP bP c the anticevian triangle of P . If we write the coordinates of P with respect to ABC as (u : v : w), then the coordinates of P a, P b, P c with respect to ABC as as follows.

P a = −u : v : w P b = u : −v : w P c = u : v : −w P = u : v : w. 210 Cevian and anticevian triangles

Proposition (Construction of anticevian triangle). Let the trilinear polar LP intersect the sidelines a, b, c respectively at X, Y , Z. The triangle bounded by the lines AX, BY , CZ is the anticevian triangle of P .

P P c c P

X B Pa C

P a

P b

Figure 4.6:

Proof. Let cev−1(P )=P aP bP c be the anticevian triangle of P . The line b =( : − : ) c =( : : − ) v + w =0 joining P u v w and P u v w has equation y z . It clearly contains the vertex A and X =(0:v : −w). Therefore, P bP c and AX are the same line. Similarly, P cP a and P aP b are the same lines as BY and CZ respectively. 4.3 Anticevian triangles 211

The superior triangle The anticevian triangle of the centroid is the superior triangle.

4.3.1 The excentral triangle cev−1(I)

Consider cev(I)=IaIbIc. The harmonic conjugates of Ia, Ib, Ic on the I I I I I I respective sidelines are the points a, b, c for which A a, B b, and C c are the external bisectors of the T. Their pairwise intersections are the excenters of T. These are the points

Ia =(−a : b : c), Ib =(a : −b : c), Ic =(a : b : −c).

They form the excentral triangle cev−1(I). Its altitudes are the internal bisectors of the angles of T. From this we deduce some basic facts about the excentral triangle:

Element in excentral triangle Element in T orthocenter I orthic triangle T nine-point circle circumcircle O(R) circumradius 2R circumcenter O†(I) = reﬂection of I in O Euler line OI

Since the pedal triangle of I is the intouch triangle I[a]I[b]I[c], the pedal O†(I) Ia Ib Ic triangle of is the extouch triangle [a] [b] [c] whose vertices are the points of tangency of the excircles and the corresponding sides of T). It follows that O†(I) is the point of concurrency of the perpendiculars from the excenters to the sidelines of T. In homogeneous coordinates,

O[cev−1(I)] = O†(I)=(a(a3+a2(b+c)−a(b+c)2−(b+c)(b−c)2):···: ···). 212 Cevian and anticevian triangles

Ic O†(I) O I

C B Ia I[a] [a]

Ia Figure 4.7: The excentral triangle and its circumcircle Chapter 5

Isotomic and isogonal conjugates

5.1 Isotomic conjugates

Two points P and Q (not on any of the sidelines of the reference triangle) are said to be isotomic conjugates if their respective traces on the sidelines are symmetric with respect to the endpoints of the corresponding sides. Thus,

BPa = QaC, CPb = QbA, APc = QcB.

We shall denote the isotomic conjugate of P by P •. A

Qc Pb

Qb Pc Q P

B Pa Qa C =( : : ) =(0: : ) : = : If P u v w , then Pa v w , BPa PaC w v, : = : = : =(0: : )= 0: 1 : 1 BQa QaC PaC BPa v w, Qa w v v w . Sim- = 1 :0: 1 = 1 : 1 : ilarly, Qb u w and Qc u v . We conclude that the isotomic conjugate of P is the point 1 1 1 P • = : : =(vw : wu : uv). u v w 214 Isotomic and isogonal conjugates

5.1.1 Example: the Gergonne and Nagel points

1 1 1 Ge = : : , b + c − a c + a − b a + b − c Na =(b + c − a : c + a − b : a + b − c).

Z Y Z I Y Na Ge

B X X C

Figure 5.1: The Gergonne and Nagel points

5.1.2 Example: isotomic conjugate of the orthocenter The isotomic conjugate of the orthocenter is the point

• H =(Sα : Sβ : Sγ).

Its traces are the pedals of the deLongchamps point Lo, the reﬂection of H in O. 5.1 Isotomic conjugates 215

Z L Y o O Y • Z H H

B X X C Figure 5.2: The orthocenter and its isotomic conjugate

5.1.3 The equal-parallelians point

We want to ﬁnd the point P for which the three parallelian segments ZaYa, = v+w · XbZb and YcXc have equal lengths. Note that ZaYa u+v+w a etc. Equality of the three lengths follows if and only if (v + w)a =(w + u)b =(u + v)c. Equivalently, 1 1 1 v + w : w + u : u + v = : : , a b c or inf(P )=I•, P = sup(I•). This is called the equal-parallelians point of triangle ABC. It has coordinates

(ca + ab − bc : ab + bc − ca : bc + ca − ab).

P Za Ya G I•

B Xc Xb C

2abc Remark. The common length of the equal parallelians is bc+ca+ab. 216 Isotomic and isogonal conjugates

5.1.4 Crelle-Yff points

Consider a point P satisfying BPa = CPb = APc = t. By Ceva’s theorem,

t t t · · =1. a − t b − t c − t

The solutions of this equation are the roots of the polynomial

f(t):=2t3 − (a + b + c)t2 +(ab + bc + ca)t − abc.

Since f (t)=6t2 − 2(a + b + c)t +(ab + bc + ca) > 0, f(t) is increasing in t. It has a unique positive root μ

A A

Z Y

• Pc Pb P P

B Pa C B X C Figure 5.3: The Crelle-Yff points

From this, we have

BPa : PaC = μ : a − μ 3 1 3 1 =(μ ) 3 :((a − μ) ) 3 1 3 1 =((a − μ)(b − μ)(c − μ)) 3 :((a − μ) ) 3 1 2 =((b − μ)(c − μ)) 3 :(a − μ) 3 1 1 b − μ 3 a − μ 3 = : , a − μ c − μ and analogous expressions for CPb : PbA and APc : PcB. Hence, the 5.1 Isotomic conjugates 217 coordinates of P : 1 1 =0:a−μ 3 : b−μ 3 Pa c−μ a−μ , 1 1 = c−μ 3 :0:b−μ 3 Pb b−μ a−μ , 1 1 = c−μ 3 : a−μ 3 :0; Pc b−μ c−μ 1 1 1 = c−μ 3 : a−μ 3 : b−μ 3 P b−μ c−μ a−μ . The isotomic conjugate 1 1 1 b − μ 3 c − μ 3 a − μ 3 P • = : : c − μ a − μ b − μ has traces X, Y , Z that satisfy XC = Y A = ZB = μ. These points are called the Crelle-Yff points. 1 They were brieﬂy considered by A. L. Crelle. 2

1P. Yff, An analogue of the Brocard points, Amer. Math. Monthly, 70 (1963) 495 – 501. 2A. L. Crelle, 1815. 218 Isotomic and isogonal conjugates

5.2 Isogonal conjugates

Two cevian lines and through A are isogonal in angle A if ∠(c,)= ∠(, b). Lemma. Let and be isogonal lines through A intersecting the sideline BC at X and X respectively. If X =(0: y : z) in homogeneous barycen- = 0:b2 : c2 tric coordinates, then X y z . A

θθ

B X X C

Figure 5.4:

Proof. Let ∠(c,AX)=∠(AX, b)=θ. By Conway’s formula, 2 X =(0:Sα − Sθ : −c ), 2 X =(0:−b : Sα − Sθ). Therefore, if X =(0:y : z) and X =(0:y : z), then − − 2 2 2 2 y · y = Sα Sθ · b = b =⇒ : = b : c 2 2 y z . z z −c Sα − Sθ c y z

An application of this lemma shows that for a point P =(x : y : z), the isogonal lines of AP , BP, CP (in the respective angles) are concurrent at a point P ∗ which we call the isogonal conjugate of P .IfP =(x : y : z), and if the isogonal lines of AP , BP, CP intersecting the sidelines BC, CA, AB at X, Y , Z respectively, then =0:b2 : c2 X y z , = a2 :0:c2 Y x z , = a2 : b2 :0; Z x y ∗ = a2 : b2 : c2 P x y z . 5.2 Isogonal conjugates 219

Examples. (1) The incenter is the isogonal conjugate of itself: I∗ = I;so is each of the excenters. (2) The circumcenter and the orthocenter H are isogonal conjugates, since at each vertex, the altitude and the circumradius are isogonal in the corresponding angle:

∠( H)=π − = ∠( O ) AB, A 2 β A ,AC etc.

A Hb

Hc O H

B C

Figure 5.5:

Since OOa = AH, AOOaH is a parallelogram, and HOa = AO. This means that the circle of reflections of O is congruent to the circumcircle. Therefore, the circle of reflections of H is the circumcircle, and the reflections of H lie on the circumcircle. 220 Isotomic and isogonal conjugates

5.2.1 The symmedian point and the centroid Consider triangle ABC together with its tangential triangle ABC, the triangle bounded by the tangents of the circumcircle at the vertices. A simple application of the Ceva theorem shows that the lines AA, BB, CC are concurrent at a point K. 3 We call this point K the symmedian point of triangle ABC for the reason below.

B D C

Since A is equidistant from B and C, we construct the circle A(B)= A(C) and extend the sides AB and AC to meet this circle again at Z and Y respectively. Note that

∠(AY,AB)=π − 2(π − α − γ)=π − 2β, and similarly, ∠(AC,AZ)=π − 2γ. Since ∠(AB,AC)=π − 2α,we have

∠(AY,AZ)= ∠(AY,AB)+∠(AB,AC)+∠(AC,AZ) =(π − 2β)+(π − 2α)+(π − 2γ) = π ≡ 0modπ.

3If triangle ABC is acute, this is the Gergonne point of ABC. 5.2 Isogonal conjugates 221

This shows that Y , A and Z are collinear, so that (i) AA is a median of triangle AY Z, (ii) AY Z and ABC are similar. It follows that AA is the isogonal line of the A-median. We say that it is a symmedian of triangle ABC. Similarly, the BB and CC are the symmedians isogonal to B- and C-medians. The lines AA, BB, CC therefore intersect at the isogonal conjugate of the centroid G. This we call the symmedian point K of triangle ABC.

5.2.2 The tangential triangle cev−1(K) I I The circle with diameter a a is the A-Apollonian circle (containing points whose distances from B and C are in the ratio c : b). The center of the I I =(0: 2 : − 2) 4 circle is the midpoint of a a, namely, X b c . We claim that AX is tangent to the circumcircle at A. For this, it is enough to show that ∠(AX, c)=∠(b, a).

∠(AX , c)= ∠(AX ,AIa)+∠(AIa, c)

= ∠(AIa, a)+∠(b,AIa) = ∠(b, a). I I I I Similarly, if Y and Z are the midpoints of b b and c c respectively, the lines BY and CZ are tangents to the circumcircle. The lines AX, BY , CZ bound the tangential triangle of T. The harmonic conjugates of X in BC, Y in CA, Z in AB are the traces of a point K with coordinates (a2 : b2 : c2). This is called the symmedian point K of T.

X =0:b2 : c2, Y = a2 :0:c2, Z = a2 : b2 :0, K = a2 : b2 : c2. The tangential triangle is therefore the anticevian triangle cev−1(K) and 4 1 1 (0,b,c) (0,b,−c) Proof: 2 (Ia + Ia)= 2 b+c + b−c ≈ (0, (b − c)b +(b + c)b, (b − c)c +(b + c)(−c)) = (0, 2b2, −2c2) ≈ (b2, −c2). 222 Isotomic and isogonal conjugates has vertices Ka =(−a2 : b2 : c2), Kb =(a2 : −b2 : c2), Kc =(a2 : b2 : −c2).

O A

A Kc K

Kc B C K O a B C K

Ka Kb

Figure 5.6: The tangential triangle

The circumcircle of T is the incircle of the tangential triangle, provided that T is acute. Note that the tangential triangle degenerates when T contains a right angle. When T is obtuse, the circumcircle of T is no longer the incircle, but the excircle on the opposite side of the obtuse angle. We describe this situation by saying that the circumcircle of T is the ob-incircle of triangle T. Remark. If the A-Apollonian circle intersects the circumcircle again at X, then AX is the A-symmedian. It follows that the symmedian point K has equal powers with respect to the three Apollonian circles (which is the negative power of K in the circumcircle). On the other hand, the circumcenter O also has equal powers (R2) with respect to the Apollonian circles. There- fore, the three Apollonian circles are coaxial, with two real common points on the line OK. These are called the isodynamic points J±. These are the points satisfying 1 1 1 AJ : BJ : CJ = : : ε ε ε a b c for ε = ±1. 5.2 Isogonal conjugates 223

5.2.3 The Gergonne point and the insimilicenter T+ Theorem. The isogonal conjugate of the Gergonne point is the insimilicenter of the circumcircle and the incircle: ∗ T+ = Ge.

E F E

F T+ I O Ge

B D C

Proof. Consider the intouch triangle DEF of triangle ABC. (1) If D is the reﬂection of D in the bisector AI, then (i) D is a point on the incircle, and (ii) the lines AD and AD are isogonal with respect to A.

A A

E E F E F F G I e I P

D D B D C B D C

(2) Likewise, E and F are the reﬂections of E and F in the bisectors BI and CI respectively, then 224 Isotomic and isogonal conjugates

(i) these are points on the incircle, (ii) the lines BE and CF are isogonals of BE and CF with respect to angles B and C. Therefore, the lines AD, BE, and CF concur at the isogonal conjugate of the Gergonne point. A

E F E

F I

D B D C (3) In fact, EF is parallel to BC. This follows from

(ID, IE)= (ID, IE)+(IE, IE) =(ID, IE)+2(IE, IB) =(ID, IE) + 2((IE, AC)+(AC, IB)) π =(ID, IE)+2(AC, IB)since(IE,AC)= 2 =( − )+2 + B π C C 2 = B + C = −A (mod π)

Similarly,

(ID, IF)= (ID, IF)+(IF, IF) =(ID, IF)+2(IF, IC) =(ID, IF) + 2((IF, AB)+(AB, IC)) π =(ID, IF)+2(AB, IC)since(IF,AB)= 2 = −( − ) − 2 + C π B B 2 = −(B + C)=A (mod π) 5.2 Isogonal conjugates 225

(4) Similarly, F D and DE are parallel to CA and AB respectively. It follows that DEF is homothetic to ABC. The ratio of homothety is r : R. Therefore, the center of homothety is the point which divides OI in the ratio R : r. This is the insimilicenter of (O) and (I). 226 Isotomic and isogonal conjugates

5.2.4 The Nagel point and the exsimilicenter T−

Theorem. The isogonal conjugate of the Nagel point is the exsimilicenter of the circumcircle and the incircle:

∗ T− = Na .

Ic D1 Cc

O Bb I Q Na

E1 F1

B Aa C

Proof. (1) Consider the A-cevian of the Nagel point, which joins the vertex A to the point of tangency Aa of the excircle with BC. This contains the antipode D1 on the incircle of the point of tangency D with BC. Therefore, the reflection of AAa in the bisector AI contains the reflection D1 of D1. D1 is clearly on the incircle. Indeed, it is the antipode of D , the reflection of D in AI. 5.2 Isogonal conjugates 227

D1 D1

D C B D Aa

(2) Likewise, if we consider the isogonal lines of the BBb and CCc in the respective angle bisectors, these contains E1 and F1 which are the antipodes of E and F on the incircle. The lines AD1, BE1, and CF1 concur at the isogonal conjugate of the Nagel point. Clearly, D1E1F1 and D E F are oppositely congruent at the O. Since D E F is homothetic to ABC, so are D1E1F1 and ABC, with ratio of homothety −r : R. The center of homothety is the point which divides OI in the ratio R : −r. This is the exsimilicenter of (O) and (I).

Exercise 1. The triangles D1E1F1 is perspective with the intouch triangle DEF. The perspector is the orthocenter of DEF. (Hint: Show that DD1 is parallel to the bisector IA). 228 Isotomic and isogonal conjugates

D1 E

F I

E1 F1

B D C 2. Let DEF be the intouch triangle of ABC. The circumcircles of AID, BIE, CIF intersect the incircle again at X, Y , Z respectively. (a) Prove that the lines AX, BY , CZ are concurrent and identify the point of concurrence. 5 (b) Show that DX, EY , CZ are also concurrent, and identify the point of concurrence. 6 A

F I ?

B D C

5Problem 1864, Math. Mag., 84 (2011) 64. Solution: Since IX and ID are equal chords in the circle AID, ∠DAI = ∠XAI. The lines AX and AD are isogonal with respect to angle A. Similarly, BY and BE are isogonal, so are CZ and C. The lines AX, BY , CZ concur at the isogonal conjugate of Ge, which is T+. 6The centroid of the intouch triangle. 5.2 Isogonal conjugates 229

5.2.5 The Brocard points

Given triangle ABC, there is an interior point Ω→ satisfying

∠(AΩ→, b)=∠(BΩ→, c)=∠(CΩ→, a). It can be constructed as the common point of the three circles: CAAB: through B, tangent to CA at A, CBBC: through C, tangent to AB at B, and CCCA: through A, tangent to BC at C.

A ω

Ω← Ω→ ω ω B C ω ω B C

The center of the circle CAAB, for example, is the intersection of the perpendicular bisector of AB and the perpendicular to CA at A. This circle c has radius sin α; similarly for the other two circles. If we denote the common angle by ω, then c a b AΩ→ = · sin ω, BΩ→ = · sin ω, CΩ→ = · sin ω. sin α sin β sin γ From these we easily obtain in homogeneous barycentric coordinates Ω =(ΔΩ :ΔΩ :ΔΩ ) → →BC →CA →AB 1 1 1 = · b · sin2 : · c · sin2 : · a · sin2 2a sin ω 2b sin ω 2c sin ω γ α β 1 1 1 = : : . c2 a2 b2

Similarly, there is another interior point Ω← satisfying

∠(c,AΩ←)=∠(a,BΩ←)=∠(b,CΩ←). 230 Isotomic and isogonal conjugates

This is the common point of the three circles: CABB: through A, tangent to BC at B, CBCC: through B, tangent to CA at C, and CCAA: through C, tangent to AB at A.

In homogeneous barycentric coordinates, 1 1 1 Ω← = : : . b2 c2 a2

From their coordinates, it is easy to see that Ω← and Ω→ are isogonal conjugates. It follows that

∠(c,AΩ←)= ∠(AΩ→, b)=ω, ∠(a,BΩ←)= ∠(BΩ→, c)=ω, ∠(b,CΩ←)= ∠(CΩ→, a)=ω.

The points Ω← and Ω→ are called the Brocard points, and ω the Brocard angle of triangle ABC.

ω ω

Ω→

Ω←

ω ω ω ω B C

Proposition. (a) csc2 ω = csc2 α + csc2 β + csc2 γ. (b) cot ω =cotα +cotβ +cotγ.

Proof. (a) Since the area of triangle ABC is the sum of the areas of CAΩ→, 5.2 Isogonal conjugates 231

ABΩ→, and BCΩ→,wehave =(· Ω + · Ω + · Ω )sin S b A → c B → a C → ω = bc + ca + ab sin2 sin sin sin ω α β γ 1 1 1 = S + + sin2 ω. sin2 α sin2 β sin2 γ From this we obtain

csc2 ω = csc2 α +csc2 β + csc2 γ.

(b) Now, cot2 ω = csc2 ω − 1 = csc2 α + csc2 β + csc2 γ − 1 =cot2 α +cot2 β +cot2 γ +2 =cot2 α +cot2 β +cot2 γ +2(cotα cot β +cotβ +cotγ +cotγ cot α) =(cotα +cotβ +cotγ)2.

Since ω is an acute angle, we have

cot ω =cotα +cotβ +cotγ.

a2+b2+c2 Corollary. Sω = Sα + Sβ + Sγ = 2 . 232 Isotomic and isogonal conjugates

Exercise 1. Let XY Z be the intouch triangle and a line through the incenter I of triangle ABC. Construct the pedals X of A, Y of B, and Z of C on . The lines XX, YY, ZZ concur at a point P . The locus of the isogonal conjugate of P with respect to XY Z is the nine-point circle of XY Z. 2. Let XY Z be the intouch triangle and a line through the incenter I of triangle ABC. Construct the pedals X of X, Y of Y , and Z of Z on . The lines AX, BY , CZ concur at a point P . The locus of the isogonal conjugate of P with respect to ABC is a circle. 5.3 Isogonal conjugate of an inﬁnite point 233

5.3 Isogonal conjugate of an inﬁnite point

Proposition. Given a triangle ABC and a line , let a, b, c be the parallels to through A, B, C respectively, and a, b, c their reﬂections in the angle bisectors AI, BI, CI respectively. The lines a, b, c intersect at a point on the circumcircle of triangle ABC.

O P b

a B C

a b c c

Proof. Let P be the intersection of b and c. ( )=( ) BP, PC b,c =( )+( )+( ) b,IB IB, IC IC, c =(IB, b)+(IB, IC)+(c,IC) =(IB, )+(IB, IC)+(, IC) =2( ) IB, IC =2 π + A 2 2 =(BA, AC)(modπ).

Therefore, b and c intersect at a point on the circumcircle of triangle ABC. Similarly, a and b intersect at a point P on the circumcircle. Clearly, P and P are the same point since they are both on the reﬂection of b in the bisector IB. Therefore, the three reﬂections a, b, and c intersect at the same point on the circumcircle. 234 Isotomic and isogonal conjugates

Proposition. The isogonal conjugates of the inﬁnite points of two perpendicular lines are antipodal points on the circumcircle.

I O P

B C

Proof. If P and Q are the isogonal conjugates of the inﬁnite points of two perpendicular lines and through A, then AP and AQ are the reﬂections of and in the bisector AI.

( )=( )+( )=−( )−( )=−( )=π AP, AQ AP, IA IA, AQ , IA IA, , 2 . Therefore, P and Q are antipodal points.

5.3.1 Homogeneous barycentric equation of the circumcircle The interesting fact of Proposition 5.3 leads to the very simple equation of the circumcircle. Theorem. A point with homogeneous barycentric coordinates (x : y : z) lies on the circumcircle of triangle ABC if and only if a2yz + b2zx + c2xy =0. = a2 : b2 : c2 ( : : ) Corollary. If P f g h for an inﬁnite point f g h , its antipodal point on the circumcircle is the point a2 b2 c2 : : . Sβg − Sγh Sγh − Sαa Sαf − Sβg 5.3 Isogonal conjugate of an inﬁnite point 235

Exercise 1. Find the locus of the isotomic conjugates of points on the circumcircle. 7 2. Let P and Q be antipodal points on the circumcircle. The lines PQ• and QP • joining each of these points to the isotomic conjugate of the other intersect orthogonally on the circumcircle.

P •

Q O

B C

Q•

3. A transversal cuts the sidelines BC, CA, AB of triangle ABC at X, Y , Z respectively. The parallels to through A, B, C intersect the circumcircle at X, Y , Z respectively. Show that XX, YY, ZZ intersect on the circumcircle. 8

2 2 2 7The line a x + b y + c z =0perpendicular to the Euler line. 2 2 2 8 a b c If : ux + vy + wz =0, then the intersection is u(v−w) : v(w−u) : w(u−v) on the circumcircle, the isogonal conjugate of the inﬁnite point of the isotomic line of . 236 Isotomic and isogonal conjugates

5.4 The isotomic conjugates of inﬁnite points

The isotomic conjugate of an infinite point cannot be an infinite point. If P =(u : v : w) is an infinite point, then so is Q =(v − w : w − u : u − v). The isotomic conjugates of these two infinite points are symmetric with respect to G. Proof.

Q• =((w − u)(u − v), (u − v)(v − w), (v − w)(w − u)) =(−u2 + u(v + w) − vw, ··· , ···) =(2u(v + w) − vw, ··· , ···) =(2(vw + wu + uv) − 3vw, ··· , ···) =2(vw + wu + uv)(1, 1, 1) − 3(vw, wu, uv) =3(vw + wu + uv)(2G − P •). is the symmetric of (vw : wu : uv) in G =(1:1:1). The Steiner circum-ellipse and the circumcircle intersect at the Steiner point 1 1 1 St := : : . b2 − c2 c2 − a2 a2 − b2 A line through the Steiner point intersects the circumcircle and the Steiner circum-ellipse again at the isogonal and isotomic conjugates of the same in- ﬁnite point. The line joining the isogonal and isotomic conjugates of a point P (u : v : w) has equation

(b2 − c2)ux +(c2 − a2)vy +(a2 − b2)wz =0.

This contains the Steiner point if and only if u + v + w =0, i.e., P is an inﬁnite point. Chapter 6

Some basic constructions

6.1 Perspective triangles

Many interesting points and lines in triangle geometry arise from the perspectivity of triangles. We say that two triangles X1Y1Z1 and X2Y2Z2 are perspective, X1Y1Z1 X2Y2Z2, if the lines X1X2, Y1Y2, Z1Z2 are concur- ) rent. The point of concurrency, (X1Y1Z1,X2Y2Z2 , is called the perspector. If one of the triangle is T, we shall simply write (XY Z) for (T,XYZ). Along with the perspector, there is an axis of perspectivity, or the perspectrix, which is the line joining containing

Y1Z2 ∩ Z1Y2,Z1X2 ∩ X1Z2,X1Y2 ∩ Y1X2.

We denote this line by L∧(X1Y1Z1,X2Y2Z2). Homothetic triangles are clearly prespective. If triangles X1Y1Z1 and X2Y2Z2 are homothetic, their perspector is the homothetic center, which we denote by 0(X1Y1Z1,X2Y2Z2). Proposition. A triangle with vertices

X = U : v : w, Y = u : V : w, Z = u : v : W, for some U, V , W , is perspective to ABC at (XY Z)=(u : v : w). The perspectrix is the line x y z + + =0. u − U v − V w − W 238 Some basic constructions

Proof. The line AX has equation wy − vz =0. It intersects the sideline BC at the point (0 : v : w). Similarly, BY intersects CA at (u :0:w) and CZ intersects AB at (u : v :0). These three are the traces of the point (u : v : w). The line YZhas equation −(vw−VW)x+u(w−W )y+u(v−V )z =0. It intersects the sideline BC at (0 : v − V : −(w − W )). Similarly, the lines ZX and XY intersect CA and AB respectively at (−(u − U):0:w − W ) and (u − U : −(v − V ):0). It is easy to see that these three points are collinear on the line x y z + + =0. u − U v − V w − W 6.1 Perspective triangles 239

Examples. The Conway conﬁguration Given triangle ABC, extend (i) CA and BA to Ya and Za such that AYa = AZa = a, (ii) AB and CB to Zb and Xb such that BZb = BXb = b, (iii) BC and AC to Xc and Yc such that CXc = CYc = c.

a a A

Z Y

? X X b C c b B c

X c

b Yc

Zb Figure 6.1: The Conway conﬁguration

These points have coordinates

Ya =(a + b :0:−a),Za =(c + a : −a :0); Zb =(−b : b + c :0),Xb =(0:a + b : −b); Xc =(0:−c : c + a),Yc =(−c :0:b + c).

From the coordinates of Yc and Zb, we determine easily the coordinates of X = BYc ∩ CZb:

Yc = −c :0:b + c = −bc :0:b(b + c) Zb = −b : b + c :0=−bc : c(b + c): 0 X ==−bc : c(b + c):b(b + c)

Similarly, the coordinates of Y = CZa ∩ AXc, and Z = AXb ∩ BYa can be determined. The following table shows that the perspector of triangles ABC and XY Z is the point with homogeneous barycentric coordinates 1 : 1 : 1 a b c . 240 Some basic constructions

= − : ( + ): ( + )= −1 : 1 : 1 X bc c b c b b c b+c b c = ( + ): − : ( + )= 1 : −1 : 1 Y c c a ca a c a a c+a c = ( + ): ( + ): − = 1 : 1 : −1 Z b a b a a b ab a b a+b ?= = 1 : 1 : 1 a b c Remark. The points X, Y , Z, together with the vertices of T, lie on an ellipse with center G (the Steiner circum-ellipse).

Exercise

Given triangle ABC, extend the sides AC to Ba and AB to Ca such that CBa = BCa = a. Similarly deﬁne Cb, Ab, Ac, and Bc. Calculate the coordinates of the intersections A of BBa and CCa, B of CCb and AAb, C of AAc, BBc. Show that AA , BB and CC are concurrent by identifying their common point. 1 Cb

B C

Ac B C Ab

Exercise

Let P =(u : v : w) and cev(P )=PaPbPc its cevian triangle. Prove that (i) the inferior triangle of cev(P ) is perspective with ABC, (ii) if X, Y , Z are the midpoints of APa, BPb, CPc respectively, then XY Z is perspective with the inferior triangle of T. 2

1Spieker center. 2Answer: in each case, the perspector is (u(v + w):v(w + u):w(u + v)). 6.1 Perspective triangles 241

x + y + z =0 The perspectrix is the line u v w , the trilinear polar of P . 242 Some basic constructions

6.2 Jacobi’s Theorem

Theorem (Jacobi). Suppose X, Y , Z are points with swing angles ∠CAY = ∠BAZ = α, ∠ABZ = ∠CBX = β, ∠BCX = ∠ACY = γ. The lines AX, BY , CZ are concurrent at the point 1 1 1 : : . Sα + Sα Sβ + Sβ Sγ + Sγ A Y D D

J E

B J C E

Figure 6.2: Jacobi’s Theorem

Proof. 2 X = − a : Sγ + Sγ : Sβ + Sβ −a2 1 1 = : : , (Sβ + Sβ)(Sγ + Sγ) Sβ + Sβ Sγ + Sγ 1 −b2 1 Y = : : , Sα + Sα (Sγ + Sγ)(Sα + Sα) Sγ + Sγ 1 1 −c2 Z = : : . Sα + Sα Sβ + Sβ (Sα + Sα)(Sβ + Sβ) 6.2 Jacobi’s Theorem 243 244 Some basic constructions

Examples

(1) The Morley center. Let X, Y , Z be points such that AY , AZ trisect angle A, BZ, BX trisect angle B, and CX, CY trisect angle C. The famous Morley’s theorem states that XY Z is an equilateral triangle. 3 Here, we note that XY Z is perspective with ABC. The perspector is the point 1 1 1 a b c o = : : = : : M A B C . S − S A S − S B S − S C cos cos cos α 3 β 3 γ 3 3 3 3

(2) Consider triangle ABC with incenter I. Let I1, I2, I3 be the incenters of triangles IBC, ICA, IAB respectively.

I3 I

B C

The homogeneous barycentric coordinates of I1 can be easily written down: 2 I1 =(−a : S − S C : S − S B ). C 4 B 4 cot θ − cot 2 = 1+2 cos θ = B Now, for arbitrary θ,wehave 2 θ sin 2θ . Putting θ 2 =2 = 4Rrs = abc and S rs 2R 2R ,wehave B B abc 1+2cos 2 B SB−S B = S cot B − cot = − · = −ca 1+2cos . 4 4 2R sin B 2

3See § below. 6.2 Jacobi’s Theorem 245 C Similarly, S − S C = −ab 1+2cos . It follows that C 4 2 = : 1+2cosC : 1+2cosB I1 a b 2 c 2 .

The coordinates of the other two centers I2 and I3 can be written down by cyclic permutations of these coordinates. From these coordinates, we readily see that triangle I1I2I3 is perspective with ABC at a : b : c A B C . 1+2cos 2 1+2cos 2 1+2cos 2

Exercise 1. Let X, Y , Z be respectively the pedals of X on BC, Y on CA, and Z on AB. Show that XY Z is a cevian triangle. 4

2. For i =1, 2, let XiYiZi be the triangle formed with given angles θi, ϕi and ψi. Show that the intersections

X = X1X2 ∩ BC, Y = Y1Y2 ∩ CA, Z = Z1Z2 ∩ AB

form a cevian triangle. 5 cot θ −cot 2 = 1+2 cos θ cot 2 +cot π − θ = 1+2 sin θ 3. Prove (a) 2 θ sin 2θ ; (b) θ 4 2 sin 2θ .

4. Let Ia, Ib, Ic be the excenters of triangle ABC. Compute the coordinates of the incenter I1 of triangle IaBC, and show that if I2 and I3 are similarly deﬁned, then triangle I1I2I3 is perspective with ABC at a : b : c A B C . 1+2sin 2 1+2sin 2 1+2sin 2

4Floor van Lamoen. 5 =(0: − : − ) Floor van Lamoen. X Sψ1 Sψ2 Sϕ1 Sϕ2 . 246 Some basic constructions

6.2.1 The Kiepert perspectors The Kiepert triangle K(θ) is perspective with T at the Kiepert perspector K(θ):

Y (θ)

Z(θ)

K(θ)

B C

X(θ)

Figure 6.3: The Kiepert perspector

X(θ)=∗∗∗∗∗ : 1 : 1 , Sβ+Sθ Sγ +Sθ Y (θ)= 1 : ∗∗∗∗∗ : 1 , Sα+Sθ Sγ +Sθ Z(θ)= 1 : 1 : ∗∗∗∗∗; Sα+Sθ Sβ+Sθ K(θ)= 1 : 1 : 1 . Sα+Sθ Sβ+Sθ Sγ +Sθ

Kiepert perspector θ homogeneous barycentric coordinates centroid G 0(1:1:1) orthocenter H π 1 : 1 : 1 2 Sα Sβ Sγ π √ 1 √ 1 √ 1 Fermat points ± 3 : : 3Sα±S 3Sβ±S 3Sγ ±S π 1√ 1√ 1√ Napoleon points ± 6 : : Sα± 3S Sβ± 3S Sγ ±3S Vecten points ±π 1 : 1 : 1 4 Sα±S Sβ±S Sγ ±S 6.3 Gossard’s theorem 247

6.3 Gossard’s theorem

Proposition. Let L : ux + vy + wz =0be a line intersecting the sidelines at ﬁnite points: BC at X, AC at Y , and AB at Z. The following statements are equivalent: (a) L is parallel to the Euler line of triangle ABC. (b) The Euler line of triangle AY Z is parallel to the sideline BC. (c) The Euler line of triangle BZX is parallel to the sideline CA. (d) The Euler line of triangle CXY is parallel to the sideline AB. Proof. It is enough to prove the equivalence of (a) and (b). These intercepts are the points X =(0:w : −v),Y=(−w :0:u),Z=(v : −u :0). The centroid and orthocenter of triangle AY Z are the points 2 Ga =(u − 2u(v + w)+3vw : u(u − w): u(u − v)),

Ha =(Sβγ(u − v)(u − w) − Sγαw(u − v) − Sαβv(u − w)

: Sαu(Sγ(u − v) − Sα(v − w): Sαu(Sα(v − w) − Sβ(w − u))). The line containing, the Euler line of triangle AY Z, has equation

2 2 (Sααu(v − w)(2u − v − w) − Sγαu(u − v) + Sαβu(u − w) )x 2 2 +(−Sαα(v − w)(u +3vw − 2wu − 2uv)+Sβγ(u − v) (u − w) 2 2 2 − Sγαw(u − v) − Sαβ(u − w)(u − v +3vw − 2wu − uv))y 2 2 +(−Sαα(v − w)(u +3vw − 2wu − 2uv) − Sβγ(u − v)(u − w) 2 2 2 + Sγα(u − v)(u − w +3vw − wu − 2uv)+Sαβv(u − w) )z =0. This is parallel to the sideline BC if and only if it contains the inﬁnite point (0 : −1:−1). This condition reduces to

−(u−v)(u−w)((Sαβ+Sγα−2Sβγ)u+(Sβγ+Sαβ−2Sγα)v+(Sγα+Sβγ−2Sαβ)w)=0. Since u, v, w are distinct, we must have

(Sαβ + Sγα − 2Sβγ)u +(Sβγ + Sαβ − 2Sγα)v +(Sγα + Sβγ − 2Sαβ)w =0; equivalently, the line ux + vy + wz =0containing the inﬁnite point of the Euler line

E∞ =(Sαβ + Sγα − 2Sβγ : Sβγ + Sαβ − 2Sγα : Sγα + Sβγ − 2Sαβ). 248 Some basic constructions

Therefore the Euler line of AY Z is parallel to BC if and only if L is parallel to the Euler line of triangle ABC. From this we deduce the following theorem. Theorem (Gossard). Suppose the Euler line of triangle ABC intersects the side lines BC, CA, AB at X, Y , Z respectively. The Euler lines of the triangles AY Z, BZX and CXY bound a triangle homothetic to ABC at a point on the Euler line.

A Hc

Ga Ha

Z Gc Gb

X B C Hb

Let L be parallel to the Euler line through a point (X, Y, Z). We write p =: Sαβ +Sγα −2Sβγ,q=: Sβγ +Sαβ −2Sγα,r=: Sγα +Sβγ −2Sαβ. The Euler line of AY Z is the line

Sα(Sβ − Sγ)(rY − qZ)x − (qrX − Sγ(Sα − Sβ)rY + Sβ(Sγ − Sα)qZ)(y + z)=0

This line is the image of the sideline BC under the homothety h(A, 1+ta), where −qrX + S (S − S )rY − S (S − S )qZ t = γ α β β γ α . a qr(X + Y + Z) 6.3 Gossard’s theorem 249

Similarly, the Euler lines of triangles BZX and CXY are images of CA and AB under the homotheties h(B,tb) and h(C, tc), where tb and tc are obtained from ta above by cyclic permutations of parameters. By the homothetic center theorem, the homethetic center is the point

(ta,tb,tc)

=(−pqrX + Sγ(Sα − Sβ)rpY − Sβ(Sγ − Sα)pqZ,

− Sγ(Sα − Sβ)qrX − pqrY + Sα(Sβ − Sγ)pqZ,

Sβ(Sγ − Sα)qrX − Sα(Sβ − Sγ)rpY − pqrZ) = − pqr(X, Y, Z)

+(Sγ(Sα − Sβ)rpY − Sβ(Sγ − Sα)pqZ,

− Sγ(Sα − Sβ)qrX + Sα(Sβ − Sγ)pqZ,

Sβ(Sγ − Sα)qrX − Sα(Sβ − Sγ)rpY)

Proof. The intercepts of the Euler line

Sα(Sβ − Sγ)x + Sβ(Sγ − Sα)y + Sγ(Sα − Sβ)z =0 are the points

X =(0 : −Sγ(Sα − Sβ):Sβ(Sγ − Sα)),

Y =(Sγ(Sα − Sβ):0:−Sα(Sβ − Sγ)),

Z =(−Sβ(Sγ − Sα):Sα(Sβ − Sγ):0), 250 Some basic constructions

In Conway’s notation, these are :( − )2 2 ( + − 2 ):··· SC SA SB SBC SAB SAC and S + S − 2S ···: BC AB AC : ··· . SB(SC − SA) 6.4 Cevian quotients 251

6.4 Cevian quotients

Theorem (The cevian nest theorem). For arbitrary points P =(u : v : w) and Q =(u : v : w), the cevian triangle cev(P ) and the anticevian triangle cev−1(Q) are always perspective. (a) The perspector is the point u v w v w u w u v (cev(P ), cev−1(Q)) = u − + + : v − + + : w − + + , u v w v w u w u v

−1 (b) The perspectrix is the line L∧(cev(P ), cev (Q)) with equation 1 u v w − + + x =0. cyclic u u v w

M Q Z P

B X C

Figure 6.4:

Proof. (a) Let cev(P )=XY Z and cev−1(Q)=XY Z. Since X =(0: v : w) and X =(−u : v : w), the line XX has equation 1 w v 1 1 − x − · y + · z =0. u w v v w The equations of YY and ZZ can be easily written down by cyclic permutations of (u, v, w), (u,v,w) and (x, y, z). It is easy to check that the line XX contains the point u v w v w u w u v u − + + : v − + + : w − + + u v w v w u w u v 252 Some basic constructions whose coordinates are invariant under the above cyclic permutations. This point therefore also lies on the lines YY and ZZ. (b) The lines YZand Y Z have equations −x + y + z =0 u v w , y + z =0 v w . They intersect at the point U =(u(wv − vw): vwv : −vww). Similarly, the lines pairs ZX, ZX and XY , XY have intersections V =(−wuu : v(uw − wu): wuw) and W =(uvu : −uvv : w(vu − uv)). The three points U , V , W lie on the line with equation given above. We shall simply write P/Q := (cev(P ), cev−1(Q)) and call it the cevian quotient of P by Q. Clearly, P/P = P . Proposition. P/Q = M if and only if Q = P/M.

Proof. Let P =(u : v : w), Q =(u : v : w), and M =(x : y : z).We have x u u v w = − + + , u u u v w y v u v w = − + , v v u v w z w u v w = + − . w w u v w From these, x y z u v w u v w − + + = − + + − , u v w u v w u v w x y z u v w u v w − + = − + + + − , u v w u v w u v w x y z u v w u v w + − = − − + − + , u v w u v w u v w 6.4 Cevian quotients 253

Z M Q Z P

B X C

X X

Y and x −x + y + z : y x − y + z : z x + y − z u u v w v u v w w u v w u v w = : : . u v w It follows that x y z x y z x y z u : v : w = x − + + : y − + : z + − . u v w u v w u v w 254 Some basic constructions

6.4.1

Consider the cevian triangle cev(P )=PaPbPc of P =(u : v : w), and the anticevian triangle cev−1(Q)=QaQbQc of Q =(x : y : z). c b a c b a Now, let X = PbQ ∩ PcQ , Y = PcQ ∩ PaQ , Z = PaQ ∩ PbQ . x y z x y z x y z X = x + + : y + − : z − + u v w u v w u v w etc. The triangle XY Z is perspective with (i) ABC at x y z : : , −x + y + z x − y + z x + y − z u v w u v w u v w (ii) cev(P ) at Q, and cev−1( ) x2 : y2 : z2 (iii) Q at u v w . Example. If Q = I =(a : b : c), XY Z is perspective with (i) ABC at (I/P )∗, (ii) cev(P ) at I, and (iii) the excentral triangle cev−1(I) at P ∗. 6.5 The cevian quotient G/P 255

6.5 The cevian quotient G/P

If P =(u : v : w),

G/P =(u(−u + v + w):v(−v + w + u):w(−w + u + v)).

Some common examples of G/P .

P G/P coordinates IMi (a(b + c − a):b(c + a − b):c(a + b − c)) OK(a2 : b2 : c2) 2 2 2 KO(a Sα : b Sβ : c Sγ) Proposition. The cevian quotient G/P is the isotomic conjugate of P in the inferior triangle.

a Proof. Let GaP and P Ga intersect GbGc at X and X respectively. Note that X is the trace of the cevian quotient G/P on GbGc. From (u, v, w)=2u(1, 0, 0) + (−u, v, w),wehave a (i) APa : PaP =2u :(v + w − u); (ii) GaP : PX =2u : v + w − u, since P has coordinates (v + w − u : w + u − v : u + v − w) in the inferior triangle. a Therefore, AX//P Ga. From this, the triangles AGcX and GaGbX are congruent. This means that the traces of P and G/P on GbGc, namely, X and X , are isotomic points on GbGc. Analogous results hold for the traces of the other sides of the inferior triangle. Therefore, P and G/P are isotomic conjugates in the inferior triangle. 256 Some basic constructions

6.6 The cevian quotient H/P

Proposition. Let AH BH CH be the orthic triangle, and X the reﬂection of P in a, then the intersection of the lines AH X and AP is the harmonic conjugate Pa of P in AAP : AP AP a = − . PaAP PAP

A A

P P

B AH AP C B AH AP C

X X

Pa Pa

A Proof. Let A be the reﬂection of A in BC. Applying Menelaus’ theorem to triangle AP AA with transversal AH XPa,wehave APa · AP X · A AH = −1 . PaAP XA AH A This gives AP XA PA AP a = − = − = − , PaAP AP X AP P PAP showing that Pa and P divide AAP harmonically. Corollary. The anticevian triangle and the reﬂection triangle of P are perspective with the orthic triangle at H/P .

−1 1. Construction of anticevian triangle cev (P ): Pa = AH X ∩ AP etc.

2. If P =(u : v : w), then Pa =(−u : v : w). 6.6 The cevian quotient H/P 257

3. Let Ma be the midpoint of the altitude AAH . The line MaPa intersects a at the pedal of P on a. 4. The reﬂection triangle of P , the anticevian triangle of P , and the orthic triangle are pairwise perspective at the same point (H/P). 5. The pedal of P on a has coordinates

u(Sβ + Sγ,Sγ,Sβ)+(Sβ + Sγ)(−u, v, w)

=(0,uSγ + v(Sβ + Sγ),uSβ + w(Sβ + Sγ))

=(0,vSβ +(u + v)Sγ, (w + u)Sβ + wSγ).

Note coordinate sum =(Sβ + Sγ)(u + v + w). 6. The reﬂection of P in a:

X ∼ 2(0,uSγ + v(Sβ + Sγ),uSβ + w(Sβ + Sγ)) − (Sβ + Sγ)(u, v, w)

∼ (−(Sβ + Sγ)u, 2uSγ + v(Sβ + Sγ), 2uSβ + w(Sβ + Sγ)).

Note coordinate sum =(Sβ + Sγ)(u + v + w). 258 Some basic constructions

For P =(u : v : w),

H/P =(u(−Sαu + Sβv + Sγw):v(−Sβv + Sγw + Sαu):w(−Sγw + Sαu + Sβv)).

Examples

(1) H/G =(Sβ + Sγ − Sα : Sγ + Sα − Sβ : Sα + Sβ − Sγ) is the superior of H•. (2) H/I is a point on the OI-line, dividing OI in the ratio R + r : −2r. 6

H/I =(a(a3 + a2(b + c) − a(b2 + c2) − (b + c)(b − c)2):···: ···). 2 2 2 (3) H/K = a : b : c is the homothetic center of the orthic and tan- Sα Sβ Sγ gential triangle. 7 It is a point on the Euler line. (4) H/O is the orthocenter of the tangential triangle. 8 (5) H/N is the orthocenter of the orthic triangle. 9 A

B AH ⊥ AP C AP

† AP

6 This point appears as X46 in ETC. 7 This appears as X25 in ETC. 8 This appears as X155 in ETC. 9 This appears as X52 in ETC. 6.7 Pedal and reﬂection triangles 259

6.7 Pedal and reﬂection triangles

6.7.1 Reﬂections and isogonal conjugates

a b c Consider a point P with reﬂections P† , P† , P† in the sidelines BC, CA, AB. Let Q be a point on the line isogonal to AP with respect to angle A, i.e., the lines AQ and AP are symmetric with respect to the bisector of angle BAC.

b P† c P†

B C Figure 6.5:

b c Clearly, the triangles AQP† and AQP† are congruent, so that Q is equidis- b c tant from P† and P† . For the same reason, any point on a line isogonal to c a ∗ BP is equidistant from P† and P† . It follows that the intersection P of two lines isogonal to AP and BP is equidistant from the three reﬂections a b c ∗ P† , P† , P† . Furthermore, P is on a line isogonal to CP. For this reason, we call P ∗ the isogonal conjugate of P . It is the center of the circle of reﬂections of P .

6.7.2 The pedal circle

Clearly, P ∗ = P . Moreover, the circles of reﬂections of P and P ∗ are ∗ ∗ a congruent, since, in Figure 6.7, the trapezoid PP Pa P† being isosceles, ∗ = ∗ a ∗ PPa P P† . It follows that the pedals of P and P on the sidelines all lie on the same circle with center the midpoint of PP∗. We call this the common pedal circle of P and P ∗. 260 Some basic constructions

b P† c P†

P ∗

B C

a P†

Figure 6.6:

A Qb

b P†

c P†

c Q P M Q = P ∗

B C

a P†

Figure 6.7:

Exercise

1. The perpendiculars from the vertices of ABC to the corresponding sides of the pedal triangle of a point P concur at the isogonal conjugate 6.7 Pedal and reﬂection triangles 261

of P . 2. Given a point P with isogonal conjugate P ∗, let X, Y , Z be the pedals of P on the sidelines BC, CA, AB of triangle ABC. If the circle ∗ ∗ X(P ) intersects BC at X1, X2, Y (P ) intersects CA at Y1, Y2, and ∗ Z(P ) intersects AB at Z1, Z2, then the six points X1, X2, Y1, Y2, Z1, Z2 are on a circle center P . 262 Some basic constructions

6.7.3 Pedal triangle The pedals of a point P =(u : v : w) are the intersections of the side lines with the corresponding perpendiculars through P . The A−altitude has 2 inﬁnite point AH −A =(0:Sγ : Sβ)−(Sβ +Sγ :0:0)=(−a : Sγ : Sβ). The perpendicular through P to BC is the line − 2 a Sγ Sβ uvw =0, xyz or 2 2 −(Sβv − Sγw)x +(Sβu + a w)y − (Sγu + a v)z =0.

Figure

This intersects BC at the point

2 2 A[P ] =(0:Sγu + a v : Sβu + a w).

Similarly the coordinates of the pedals on CA and AB can be written down. The triangle A[P ]B[P ]C[P ] is called the pedal triangle of triangle ABC: ⎛ ⎞ ⎛ ⎞ 2 2 A[P ] 0 Sγu + a vSβu + a w ⎝ ⎠ ⎝ 2 2 ⎠ B[P ] = Sγv + b u 0 Sαv + b w 2 2 C[P ] Sβw + c uSαw + c v 0

6.7.4 Examples (1) The pedal triangle of the circumcenter is clearly the medial triangle. (2) The pedal triangle of the orthocenter is called the orthic triangle. Its vertices are clearly the traces of H, namely, the points (0 : Sγ : Sβ), (Sγ :0:Sα), and (Sβ : Sα :0). (3) Let L be the deLongchamps point, i.e., the reﬂection of the orthocenter H in the circumcenter O. Show that the pedal triangle of L is the cevian triangle of some point P . What are the coordinates of P ? 10

Figure

10 P =(Sα : Sβ : Sγ ) is the isotomic conjugate of the orthocenter. It appears in ETC as the point X69. 6.7 Pedal and reﬂection triangles 263

(4) Let L be the de Longchamps point again, with homogeneous barycentric coordinates

(Sγα + Sαβ − Sβγ : Sαβ + Sβγ − Sγα : Sβγ + Sγα − Sαβ).

Find the equations of the perpendiculars to the side lines at the corresponding traces of L. Show that these are concurrent, and ﬁnd the coordinates of the intersection. The perpendicular to BC at AL =(0:Sαβ +Sβγ−Sγα : Sβγ+Sγα−Sαβ) is the line −( + ) Sβ Sγ Sγ Sβ 0 Sαβ + Sβγ − Sγα Sβγ + Sγα − Sαβ =0. xy z

This is

2 2 2 S (Sβ − Sγ)x − a (Sβγ + Sγα − Sαβ)y + a (Sβγ − Sγα + Sαβ)z =0.

Similarly, we write down the equations of the perpendiculars at the other two traces. The three perpendiculars intersect at the point 11 ( 2( 2 2 + 2 2 − 2 2):···: ···) a SγSα SαSβ SβSγ .

Figure

Exercise 1. Let D, E, F be the midpoints of the sides BC, CA, AB, and A, B, C the pedals of A, B, C on their opposite sides. Show that X = EC ∩ FB, Y = FA ∩ DC, and Z = DB ∩ EC are collinear. 12

2. Let X be the pedal of A on the side BC of triangle ABC. Complete 13 the squares AXXbAb and AXXcAc with Xb and Xc on the line BC.

14 (a) Calculate the coordinates of Ab and Ac.

11 This point appears in ETC as X1078. Conway calls this point the logarithm of the de Longchamps point. 12These are all on the Euler line. See G. Leversha, Problem 2358 and solution, Crux Mathematicorum,24 (1998) 303; 25 (1999) 371 –372. 13A.P. Hatzipolakis, Hyacinthos, message 3370, 8/7/01. 14 2 2 Ab =(a : −S : S) and Ac =(a : S : −S). 264 Some basic constructions

15 (b) Calculate the coordinates of A = BAc ∩ CAb. (c) Similarly deﬁne B and C. Triangle ABC is perspective with ABC. What is the perspector? 16 (d) Let A be the pedal of A on the side BC. Similarly deﬁne B and C. Show that ABC is perspective with ABC by calculating the coordinates of the perspector. 17

6.7.5 Reflection triangle The reflection triangle of P =(u : v : w) have vertices ⎛ ⎞ ⎛ ⎞ 2 2 2 A −a u 2Sγu + a v 2Sβu + a w ⎝ ⎠ ⎝ 2 2 2 ⎠ B = 2Sγv + b u −b v 2Sαv + b w . 2 2 2 C 2Sβw + c u 2Sαw + c v −c w The construction of harmonic conjugates in §?? shows that the line containing the harmonic conjugate of P in AAP and the reflection of P in a and passe through AH . Therefore, Proposition. The anticevian and reflection triangles of P are perspective at H/P.

Examples (1) Since the reﬂection triangle of I is homothetic to the excentral triangle, with ratio 2r :2R = r : R, the homothetic center is the point dividing II in the ratio −r : R i.e., R · I − r · I (R + r)I − 2r · O = . R − r R − r This shows that H/I is the point dividing OI in the ratio R + r : −2r.

15A =(a2 : S : S). 16The centroid. 17 1 1 1 ( : : ) 485 Sα+S Sβ +S Sγ +S . This is called the ﬁrst Vecten point; it appears as X in ETC. 6.7 Pedal and reﬂection triangles 265

Exercise 1. Given a point P , construct a circle with center P whose reﬂections in the sidelines of triangle ABC are concurrent. 18 2. The perspector of the intouch triangle and the tangential triangle is the point 19

2 3 2 2 2 2 Ge/K =(a (a − a (b + c)+a(b + c ) − (b + c)(b − c) ):···: ···).

3. Let XY Z be the circumcevian triangle of P , and XY Z be that of P ∗. The lines XX, YY, ZZ bound a triangle homothetic to ABC. What is the homothetic center? 20

18The circumcircle of the reﬂection triangle of P ∗. 19 This appears as X1486 in ETC. 2 20 a =( : : ) 2 2 2 : ···: ··· If P u v w , the homothetic center is u(Sα(v+w) +Sβ v +Sγ w ) . 266 Some basic constructions

6.8 Barycentric product

Let X1, X2 be two points on the line BC, distinct from the vertices B, C, with homogeneous coordinates (0 : y1 : z1) and (0 : y2 : z2).Fori =1, 2, complete parallelograms AKiXiHi with Ki on AB and Hi on AC. The coordinates of the points Hi, Ki are

H1 =(y1 :0:z1),K1 =(z1 : y1 :0); H2 =(y2 :0:z2),K2 =(z2 : y2 :0).

H2 K1

H1 K2

B X1 X X2 C From these, we easily ﬁnd that

BK1 ∩ CH2 =(y1z2 : y1y2 : z1z2), BK2 ∩ CH1 =(y2z1 : y1y2 : z1z2). Both of these have A-trace

X =(0:y1y2 : z1z2). The line joining them passes through A. This simple observation leads to the notion of the barycentric product. Given two points P1 =(x1 : y2 : z1) and P2 =(x2 : y2 : z1), the above construction (applied to the traces on each side line) gives the traces of a point P1 · P2 =(x1x2 : y1y2 : z1z2).

6.8.1 Barycentric square In particular, the barycentric square of a point P =(x : y : z) is the point

P · P =(x2 : y2 : z2) 6.8 Barycentric product 267

Z2 Z Y1 Y Y Z1 2 P · P 1 2 P2 P1

B X1 X X2 C and can be constructed as follows. (i) Complete the parallelograms AYaAP Za, BZbBP Xb and CXcCP Yc. (ii) Let X = BYa ∩ CZa, Y = CZb ∩ AXb, and Z = AXc ∩ BYc. Then XY Z is perspective with ABC at the barycentric square of P .

Zb BP

Za X CP Y Yc P M Z

C B Xc AP Xb

Exercise

1. Find the equation of the circle through B and C, tangent (internally) to incircle. Show that the point of tangency has coordinates a2 (s − c)2 (s − b)2 : : . s − a s − b s − c 268 Some basic constructions

Construct this circle by making use of the barycentric cube of the Ger- gonne point. 2. Three parallel lines with infinite point (u : v : w) through the vertices of a triangle ABC intersect the sidelines BC, CA, AB at X, Y , Z respectively. Show that the centroid of the cevian triangle XY Z has coordinates (u3 : v3 : w3). 3. A circle is tangent to the side BC of triangle ABC at the A−trace of a point P =(u : v : w) and internally to the circumcircle at A. Show that the line AA passes through the point (au : bv : vw). Make use of this to construct the three circles each tangent internally to the circumcircle and to the side lines at the traces of P . 4. Two circles each passing through the incenter I are tangent to BC at B and C respectively. A circle (Ja) is tangent externally to each of these, and to BC at X. Similarly define Y and Z. Show that XY Z is perspective with ABC, and find the perspector. 21

5. Let P1 =(f1 : g1 : h1) and P2 =(f2 : g2 : h2) be two given points. Denote by Xi, Yi, Zi the traces of these points on the sides of the reference triangle ABC.

(a) Find the coordinates of the intersections X+ = BY1 ∩ CZ2 and 22 X− = BY2 ∩ CZ1. 23 (b) Find the equation of the line X+X−. (c) Similarly deﬁne points Y+, Y−, Z+ and Z−. Show that the three lines X+X−, Y+Y−, and Z+Z− intersect at the point

(f1f2(g1h2 + h1g2):g1g2(h1f2 + f1h2):h1h2(f1g2 + g1f2)). 6. The barycentric cube of an inﬁnite point is the centroid of the cevian triangle of the point. It happens that the barycentric cube of the Euler line is again on the Euler line, and the Euler line is the only line through O with this property. For a given point P = G, only the line PG has this property. 24

21 a b c The barycentric square root of ( s−a : s−b : s−c ). See Hyacinthos, message 3394, 8/9/01. 22 X+ = f1f2 : f1g2 : h1f2; X− = f1f2 : g1f2 : f1h2. 23 2 2 (f1 g2h2 − f2 g1h1)x − f1f2(f1h2 − h1f2)y + f1f2(g1f2 − f1g2)z =0.. 242/14/04. 6.8 Barycentric product 269

6.8.2 Barycentric square root

Let P =(u : v : √w) be a point in the interior of triangle ABC, the barycentric square root P is the point Q in the interior such that Q2 = P . This can be constructed as follows. (1) Construct the circle with BC as diameter. (2) Construct the perpendicular to BC at the trace AP to intersect the circle at X. 25 Bisect angle BXC to intersect BC at X. (3) Similarly obtain Y on CA and Z on AB. √ The points X, Y , Z are the traces of the barycentric square root P .

A Y

Z B Z P Y

CP √ P P

AP B X C

2 2 2 2 Proof. BX : X C = BX : XC = BAP : AP C etc.

25It does not matter which of the two intersections is chosen. 270 Some basic constructions Chapter 7

Orthology

7.1 Triangle determined by orthology centers

Given triangles ABC and a ﬁnite point Q =(u : v : w), consider a triangle ABC for which the perpendiculars from A to BC, B to CA, C to AB are concurrent at Q. We shall say that ABC is orthologic to ABC with orthology center Q =⊥ (ABC,ABC). In absolute barycentric coordinates

2 A = Q + t1(−a ,Sγ,Sβ), 2 B = Q + t2(Sγ, −b ,Sα), 2 C = Q + t3(Sβ,Sα, −c )

for some t1, t2, t3. These three points are collinear, i.e., triangle A B C is degenerate, if and only if

t2t3 + t3t1 + t1t2 =0.

The inﬁnite point of BC is

2 2 t3(Sβ,Sα, −c ) − t2(Sγ, −b ,Sα) 2 2 =(t3Sβ − t2Sγ,t3Sα + t2b , −(t3c + t2Sα)).

Note that

2 2 Sα(t2 + t3)(t3Sβ − t2Sγ)+Sβ(−t3)(t3Sα + t2b )+Sγ(−t2)(−(t3c + t2Sα)) = Sα(t2 + t3)(t3Sβ − t2Sγ) − Sβt3(t3Sα + t2(Sγ + Sα)) + Sγt2(t3(Sα + Sβ)+t2Sα)) =0. 272 Orthology

It follows that the inﬁnite point orthogonal to B C is (t2 + t3, −t3, −t2), and the perpendicular from A to BC is the line xyz 0= 100 = t2y − t3z. t2 + t3 −t3 −t2

Similarly, the perpendiculars from B to CA and C to AB are the lines t3z − t1x =0and t1x − t2y =0. These three lines intersect at Q =(t2t3 : t3t1 : t1t2). This is a ﬁnite point if and only if ABC is nondegenerate. If we write Q =(u : v : w) in homogeneous barycentric coordinates with reference to ABC, then ( )= t t t t1,t2,t3 u, v , w for some t. It follows that t A = Q + (−a2,S ,S ), u γ β t B = Q + (S , −b2,S ), v γ α t C = Q + (S ,S , −c2). w β α From these, uA + vB + wC =(u + v + w)Q, and Q has the same barycentric coordinates with reference to ABC as does Q with reference to ABC. We summarize these results in the following theorem. Theorem. Let ABC and ABC be nondegenerate triangles. (a) ABC is orthologic to ABC if and only if ABC is orthologic to ABC. (b) If the triangles are orthologic, the orthology center ⊥ (ABC, ABC) has the same barycentric coordinates with reference to ABC as the orthology center ⊥ (ABC,ABC) with reference to ABC. 7.1 Triangle determined by orthology centers 273

7.1.1 Examples (1) The tangential triangle is orthologic to T; both orthology centers are O. (2) The excentral triangle is orthologic to T: −1 −1 ⊥(T, cev (I)) = I, ⊥(cev (I), T)=Na. (3) Every pedal triangle is orthologic to T: ⊥(Ped(P ), T)=P, ⊥(T, Ped(P )) = P ∗. (4) Every Kiepert triangle is orthologic to T: ⊥(K( ) T)=O ⊥(T K( )) = π − θ , , , θ K 2 θ . (5) The triangle of reﬂections T† has vertices † 2 A =(−a , 2Sγ, 2Sβ), † 2 B =(2Sγ, −b , 2Sα), † 2 C =(2Sβ, 2Sα, −c ). It is orthologic to T: ⊥(T†, T)=H, ⊥(T, T†)=N∗. (6) Let P be a point on a given circle C . Construct perpendiculars from P to the sidelines of T, to intersect C again at X, Y , Z respectively. The triangle XY Z is oppositely similar to T. It is clearly orthologic to T: ⊥(XY Z, T)=P . Therefore, ⊥(T,XYZ) is a point on the circumcircle of T. (7)Let A B C be the cevian triangle of P =(u : v : w), and Oa the circumcenter of triangle PB C . Similarly deﬁne Ob and Oc. The line ObOc is the perpendicular bisector of PA. Therefore, AP is perpendicular to ObOc. Similarly, BP ⊥ OcOa and CP ⊥ OaOb. It follows that triangle OaObOc is orthologic to ABC, with P =⊥(T,OaObOc).

P ⊥(OaObOc, T) GN HH

St L E X(6759) = EL ∩ HN∗ X(107) X(6523) = LX(107) ∩ HL∗ 274 Orthology

The perpendicular from Oa to BC is the line. The three perpendiculars intersect at the point Q. The barycentrics of Q in OaObOc are the same as those of P in ABC. Nikolaos Dergiades has given a wonderful proof. 7.2 Perspective orthologic triangles 275

7.2 Perspective orthologic triangles

Now t is nonzero if ABC is nondegenerate. If these two triangles are perspective, then the perspector is the second intersection of the line QQ with the rectangular circum-hyperbola through Q. This is S y − S z P = B C : ···: ··· , wy − vz corresponding to cyclic SAux(wy − vz) t = . cyclic SBC(wy − vz)

7.2.1 Consider the functions

2 pa(t)= S (v + w)t + u(Sα(v + w)(v + w )+Sβvv + Sγww ), 2 pb(t)= S (w + u)t + v(Sβ(w + u)(w + u )+Sγww + Sαuu ), 2 pc(t)= S (u + v)t + w(Sγ(u + v)(u + v )+Sαuu + Sβvv ); 2 qa(t)= −S t + Sα(v + w)(v + w )+Sβvv + Sγww , 2 qb(t)= −S t + Sβ(w + u)(w + u )+Sγww + Sαuu , 2 qc(t)= −S t + Sγ(u + v)(u + v )+Sαuu + Sβvv . The circles with diameters AQ, BQ, CQ have equations

2 2 2 u(u + v + w)(u + v + w )(a yz + b zx + c xy) − (x + y + z)(pax + u(qby + qcz)=0, 2 2 2 v(u + v + w)(u + v + w )(a yz + b zx + c xy) − (x + y + z)(pby + v(qax + qcz)=0, 2 2 2 w(u + v + w)(u + v + w )(a yz + b zx + c xy) − (x + y + z)(pcz + w(qax + qby)=0. The circle with diameter AQ intersects BC at two points given by

2 (u + v + w)(u + v + w ) · a yz − (y + z)(qby + qcz)=0; similarly for the other two circles. The six intercepts on the three sidelines, if real, lie on the circle

(u + v + w)(u + v + w)(a2yz + b2zx + c2xy)

− (x + y + z)(qax + qby + qcz)=0. 7.2 Perspective orthologic triangles 301

This has center the midpoint of PQ. The square radius of the circle is S2 QQ2 + t · . (u + v + w)(u + v + w)

S2 · − 2 This circle is imaginary for (u+v+w)(u+v+w) t< QQ . The circle with diameter AQ has equation

2 2 2 (u + v + w )(a yz + b zx + c xy) − (x + y + z)((Sαu + Sβ(w + u ))y +(Sαu + Sγ(u +

The line BC has equation

2 qax − (S t + Sαu (v + w)+Sβv(w + u ) − Sγww )y 2 − (S t + Sαu (v + w) − Sβvv + Sγw(u + v ))z =0.

Now, consider

(u + v + w)(u + v + w)(a2yz + b2zx + c2xy)

− (x + y + z)(qax + qby + qcz) 2 +(x + y + z)(qax − (S t + Sαu (v + w)+Sβv (w + u) − Sγww )y 2 − (S t + Sαu (v + w) − Sβvv + Sγw (u + v))z).

2 qb +(S t + Sαu (v + w)+Sβv(w + u ) − Sγww ) 2 = −S t + Sβ(w + u)(w + u )+Sγww + Sαuu 2 +(S t + Sαu (v + w)+Sβv(w + u ) − Sγww ) =(u + v + w)(Sαu + Sβ(w + u )); 2 qc +(S t + Sαu (v + w) − Sβvv + Sγw(u + v )) =(u + v + w)(Sαu + Sγ(u + v )).

Therefore, the circle of diameter AQ is in the pencil generated by C and the line BC. 302 Orthology Chapter 8

The circumcircle

8.1 The circumcircle

The equation of the circumcircle of T, a2yz + b2zx + c2xy =0, is derived from the fact that the circumcircle consists of the isogonal conjugate of infinite points. Here are a few triangle centers on the circumcircle with simple coordinates. • E = a2 : b2 : c2 The Euler reflection point b2−c2 c2−a2 a2−b2 may be regarded as the isogonal conjugate of the infinite point of a2x + b2y + 2 =0 x + y + z =0 c z , the isotomic line of the Lemoine axis a2 b2 c2 . • The isogonal conjugate of the infinite point of the Lemoine axis is the Steiner point 1 1 1 St = : : . b2 − c2 c2 − a2 a2 − b2 • The isogonal conjugate of the infinite point of the trilinear polar of the I incenter : ∗ a b c (L (I)∞) = : : . b − c c − a a − b • The isogonal conjugate of the infinite point of the line IG: 2 2 2 ∗ a b c (IG∞) = : : . b + c − 2a c + a − 2b a + b − 2c 304 The circumcircle

8.1.1 Tangents to the circumcircle Proposition. A line fx+ gy + hz =0is tangent to the circumcircle if and only if

a4f 2 + b4g2 + c4h2 − 2b2c2gh − 2c2a2hf − 2a2b2fg =0. (8.1)

If this condition is satisﬁed, the point of tangency is

P =(a2(−a2f + b2g + c2h):b2(a2f − b2g + c2h):c2(a2f + b2g − c2h)).

Proof. Clearly at most one of f, g, h can be zero. Assume g, h =0. Elimi- nating x from fx+ gy + hz =0and a2yz + b2zx + c2xy =0, we obtain

c2gy2 +(−a2f + b2g + c2h)yz + b2hz2 =0.

This is a quadratic in y, z with discriminant

(−a2f + b2g + c2h)2 − 4b2c2gh = a4f 2 + b4g2 + c4h2 − 2b2c2gh − 2c2a2hf − 2a2b2fg.

The line fx+ gx + hz =0is tangent to the circumcircle if and only if this discriminant vanishes. This is the condition given in (8.1). The polynomial in (8.1) can be rewritten in two different ways, showing that when the condition (8.1) is satisfed, (i) the point P lies on the line fx+ gy + hz =0

a4f 2 + b4g2 + c4h2 − 2b2c2gh − 2c2a2hf − 2a2b2fg = f · a2(−a2f + b2g + c2h)+g · b2(a2f − b2g + c2h) + h · c2(a2f + b2g − c2h);

(ii) the point P also lies on the circumcircle:

a4f 2 + b4g2 + c4h2 − 2b2c2gh − 2c2a2hf − 2a2b2fg = a2 · b2(a2f − b2g + c2h) · c2(a2f + b2g − c2h) + b2 · c2(a2f + b2g − c2h) · a2(−a2f + b2g + c2h) + c2 · a2(−a2f + b2g + c2h) · b2(a2f − b2g + c2h)=0.

The point P is therefore the point of tangency of the line and the circumcircle. 8.1 The circumcircle 305

Corollary. The trilinear polar of a point P is tangent to the circumcircle if and only if its isogonal conjugate P ∗ lies on the Steiner inellipse. If this condition is satisﬁed, the point of tangency is the cevian quotient P/K. ( : : ) a2 : b2 : c2 Corollary. Let u v w be an inﬁnite point so that u v w lies on the circumcircle. The tangent to the circumcircle at this point is the trilinear polar of the isogonal conjugate of (u2 : v2 : w2). a : b : c Example. The tangent to the circumcircle at the point b−c c−a a−b is the line (b − c)2x +(c − a)2y +(a − b)2z =0. 306 The circumcircle

8.2 Simson lines

We begin with a fundamental theorem. Theorem. The pedals of a point on the sidelines of T are collinear if and only if the point lies on the circumcircle of T.

Proof. Let P be a point with pedals P[a], P[b], P[c] on the sidelines of T.

∠(P[a]P[b],P[a]P[c])= ∠(P[a]P[b],PP[a])+∠(PP[a],P[a]P[c])

= ∠(P[b]P[a],P[a]P )+∠(PP[a],P[a]P[c])

= ∠(P[b]C, CP)+∠(PB, BP[c]) = ∠(AC, CP )+∠(PB, BA) = ∠(AC, CP ) − ∠(AB, BP ).

It follows from this that ∠(P[a]P[b],P[a]P[c])=0if and only if ∠(AB, BP )= ∠(AC, CP ). In other words, P[a], P[b], P[c] are collinear if and only if A, B, C, P are concyclic. For a point P on the circumcircle of T, we denote by s(P ) the line containing its pedals on the sidelines, and call this the Simson line of P . If P is a vertex of T, s(P ) is the altitude through the vertex. If P is the antipode of a vertex on the circumcircle, s(P ) is the sideline opposite to the vertex. Let P be the intersection of the circumcircle with the bisector of angle A. What is the Simson line of P ? The external bisector of Ga of the inferior triangle. The bisectors of the inferior triangle are Simson lines. What are the corresponding points on the circumcircle?

Proposition. Let the line PP[a] intersect the circumcircle of T again at Qa. The line AQa is parallel to s(P ). Proof.

∠(AQa,P[b]P[a])= ∠(AQa,QaP )+∠(QaP, P[b]P[a])

= ∠(AC, CP )+∠(PP[a],P[a]P[b])

= ∠(AC, CP )+∠(PC, CP[b]) = ∠(AC, CP )+∠(CP, AC) =0. 8.2 Simson lines 307

Therefore, AQa is parallel to s(P )=P[b]P[a].

Let P be a point on the circumcircle with Simson line s(P ). is perpendicular to the isogonal lines of AP , BP, CP. Therefore, s(P ) is perpendicular to the lines defining P on the circumcircle. Proposition. The Simson line s(P ) is orthogonal to the lines defining P as the isogonal conjugate of their (common) infinite point. Proof. Let Q be the reflection of P in the bisector of angle A.

∠(P[b]P[c],AQ)= ∠(P[b]P[c],AC)+∠(AC, AQ)

= ∠(P[c]P[b],P[b]A)+∠(AC, AQ)

= ∠(P[c]P, PA)+∠(AC, AQ) = π − ∠( )+∠( ) 2 AP, AB AC, AQ = π 2 .

Corollary. The Simson lines of antipodal points are orthogonal. = a2 : b2 : c2 ( : : ) If P u v w for an inﬁnite point u v w , then the Simson line s(P ) has inﬁnite point

(Sβv − Sγw : Sγw − Sαu : Sαu − Sβv). T † Let P be a point on the circumcircle of . Consider the reﬂection Pa of P in the sideline BC. H † s( ) Proposition. The line Pa is parallel to the Simson line P .

Proof. Let Ha be the reflection of H in BC. Since Ha lies on the circumcircle of T, the circle Ca := HBC is the reflection of the circumcircle in BC. The † † pedal P[a] is the midpoint of PPa . Let Xa be the reflection of Xa in BC. This is a point on Ca. H † H† Note that AP and Xa are reflections of aXa in two parallel lines, BC † and the diameter of C parallel to BC. The equality AP = HXa of vectors AH = PX† = X P† HP† = AX H † gives a a a. Therefore, a a. In particular, Pa is parallel to AXa, and to the Simson line s(P ) by Proposition ?. 308 The circumcircle

Corollary. The Simson line s(P ) bisects the segment P H. † †H Proof. Since P[a] is the midpoint of PPa , the parallel to Pa through P[a] H †H s( ) bisects the segment P . Since Pa is parallel to the Simson line P , the parallel through P[a] is indeed the Simson line. Proposition. The Simson lines of antipodal points intersect orthogonally at a point on the nine-point circle. Proof. Let P and P be antipodal points on the circumcircle. The Simson lines s(P ) and s(P ) contain respectively the midpoints of the segments HP and HP . These midpoints are antipodal on the nine-point circle of T. Since the Simson lines are orthogonal to each other, their intersection also is a point on the nine-point circle. Examples. The Simson line of the Euler reﬂection point E is the parallel to the Euler line through the midpoint of EH, which is

4 2 2 2 2 2 2 4 2 2 2 4 2 2 4 2 2 2 2 2 X113 =(−2a +a (b +c )+(b −c ) )(a (b +c )−2a (b −b c +c )+(b −c ) (b +c )

This is the line − SB SC =0 ( + ) − 2 x . cyclic SA SB SC SBC The other intersection with the nine-point circle is

2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 X3258 =(b − c ) ((b + c − a ) − b c )(2a − a (b + c ) − (b − c ) ).

This line contains X(n) for the following values of n:

30, 113, 1495, 1511, 1514, 1524, 1525, 1531, 1533, 1539, 1544, 1545, 1546, 1553, 1554,

The Simson line of the Steiner point St is − Sβ Sγ =0 − x . cyclic Sαα Sβγ This line contains (i) X(114): antipode of Kiepert center on the nine-point circle, (ii) X(325) (iii) X(511): parallel to the Brocard axis (iv) X(1513) on the Euler line, 8.2 Simson lines 309

(v) X(2679): center of the rectangular hyperbola with asymptotes parallel to the Brocard and Lemoine axes. The line of reﬂections:

2 2 2 a (b − c )Sαx =0. cyclic is the line HH• containing X(n) for the following values of n:

4, 69, 76, 264, 286, 311, 314, 315, 317, 340, 511, 877, 1232, 1234, 1235, 1236, 1330, 1352, 1531, 310 The circumcircle

8.3 Line of reﬂections

Since the reflections of P in the sidelines of T are images of the pedals of P under the homothety h(P, 2), we conclude that the reflections of P in the sidelines are collinear if and only if P lies on the circumcircle of T. The line of reflections of P is the image of the Simson line s(P ) under the same homothety. Theorem. The line of reflections of a point on the circumcircle contains the orthocenter H. H † H † H † s( ) Proof. The lines Pa , Pb , Pc are all parallel to the Simson line P .It H † † † follows that the four points , Pa , Pb , Pc are collinear. = a2 : b2 : c2 ( ; : ) Corollary. If P u v w for an infinite point u v w , the line of reflections of P is Sαux + Sβvy + Sγwz =0. Proof. This clearly contains the orthocenter H = 1 : 1 : 1 and has the Sα Sβ Sγ same infinite point as s(P ). Example. (The Euler reflection point E) Since the equation of the Euler line is 2 2 2 2 2 2 Sα(b − c )x + Sβ(c − a )y + Sγ(a − b )z =0, by choosing the infinite point (u : v : w)=(b2 − c2 : c2 − a2 : a2 − b2),we obtain the point 2 2 2 E := a : b : c b2 − c2 c2 − a2 a2 − b2 on the circumcircle, whose reflections in the sidelines lie on the Euler line. This is called the Euler reflection point. 8.4 Reflections of a line in the sidelines of T 311

O H

B C

Figure 8.1: Not the right illustration

8.4 Reﬂections of a line in the sidelines of T

Let L be a given line. Consider its reﬂections La, Lb, Lc in the sidelines a, b, c.

∠(Lb, Lc)= ∠(Lb, L )+∠(L , Lc) =2∠(b, L )+2∠(L , c) =2∠(b, c).

1 π Important note: ∠(b, c)=2∠(Lb, Lc)+ 2 .

Let T†(L ) be the triangle bounded by these three reﬂection lines. This is oppositely similar to the tangential triangle since ∠(tB, tC)=−2∠(b, c) for the tangents tB and tC to the circumcircle of T at the vertices of T.

Proposition. The triangle T†(L ) is perspective with ABC at a point on the circumcircle of T.

Proof. The vertex A is equidistant from L , Lb and Lc. Therefore it lies on the bisector of an angle between Lb and Lc. Similarly, each of B and C is on the bisector of an angle between two of the reﬂection lines. The triangle ABC is perspective with T†(L ), at a point Q which is the incenter or an excenter of T†(L ) . 312 The circumcircle

∠(BQ,QC)= ∠(BQ,La)+∠(La,QC) 1 1 = ∠(L L )+ ∠(L L ) 2 c, a 2 a, b = ∠(c a)+π + ∠(a b)+π , 2 , 2 = ∠(c, b) = ∠(BA,AC). Therefore the perspector Q lies on the circumcircle of T. Note that

∠(AQ, AB)= ∠(AQ, Lc)+∠(Lc, c) 1 = ∠(L L )+∠(L c) 2 b, c c, = ∠(b c)+π + ∠(c L ) , 2 , = ∠(b L )+π , 2 .

∠( L ⊥)= ∠(b L )+∠(L L ⊥)=∠(b L )+π AC, , , , 2 . Therefore, Q is the isogonal conjugate of the infinite point of L ⊥. When are the reflection lines concurrent? If the reflection lines of L are concurrent, the point of concurrency must be the perspector Q on the circumcircle. The reflections of Q in the sidelines are all on the line L . This shows that L must contain the orthocenter H. Conversely, if the line L contains the orthocenter H, then the lines of reflections are concurrent. T†(L ) L † Theorem. The triangle is degenerate, i.e., the reflection lines a , L † L † L b , c are concurrent, if and only if contains the orthocenter.

Proof. Construct the parallel to L through A to intersect the circumcircle at A, and the perpendiculars to BC from A and A, to intersect the circumcircle again at X and Q respectively. The line XQ is the reﬂection of L in BC. 8.4 Reﬂections of a line in the sidelines of T 313

A A

B C

X E Now let the parallel of L through B intersect the circumcircle at B.We show that BQ ⊥ CA:

(BQ, AC)= (BQ, BC)+(BC, AC) =(AQ, AC)+(AB,BB) =(AQ, AC)+(AB,AB)+(AB,BB) =(AQ, AC)+(AB,AB)+(AC, BC) =(AQ, AC)+(AX, AX)+(AC, BC) =(AQ, AC)+(AX, AQ)+(AC, BC) =(AX, BC) = π 2 . L † Therefore, the reﬂection line b contains the point Q, so does the reﬂec- L † tion line c from a similar calculation. 314 The circumcircle

The Euler reﬂection point

O H

B C

8.4.1 We give a simple algebraic proof of Theorem ?. A line L through H has equation

uSαx + vSβy + wSγz =0 for an inﬁnite point (u : v : w). The reﬂection line a†(L ) has equation

px + uSαx + vSβy + wSγz =0 for some p. Since this contains the point H† =(− 2 :( 2 + ) :( 2 + ) ) a a Sβγ S Sβγ Sγ S Sβγ Sβ , we have

2 2 −(p + uSα) · a Sβγ +(S + Sβγ)(vSβ · Sγ + wSγSβ)=0. 2 2 −(p + uSα) · a − (S + Sβγ)u =0. 2 2 p = −u(a Sα + SβSγ)=−u · S . a†(L ) L † Therefore, (or a ) is the line 2 2 2 −(S + Sβγ)ux + a Sβvy + a Sγwz =0. 8.4 Reﬂections of a line in the sidelines of T 315 a2 : b2 : c2 It is easy to see that this contains the point u v w (on the circumcircle). L † L † Similar calculations show that the reﬂection lines b and c contain the same point on the circumcircle. Example. If P =(u : v : w), the line HP has equation

Sα(Sβv − Sγw)x + Sβ(Sγw − Sαu)+Sγ(Sαu − Sβv)z =0.

The reﬂections of HP in the sidelines of T are concurrent at a2 b2 c2 : : . Sβv − Sγw Sγw − Sαu Sαu − Sβv on the circumcircle. † † † Pb Pb P b. 316 The circumcircle

8.4.2 Perspectivity of reﬂection triangles Theorem. The reﬂection triangles of P =(u : v : w) and Q =(x : y : z) are perspective if and only if the line PQcontains the orthocenter H. Proof. The condition of perspectivity is ⎛ ⎞

⎝ ⎠ SA(SBv − SCw)x Q =0, cyclic where ⎛

Q =(u+v+w)2(a2yz+b2zx+c2xy)−(x+y+z) ⎝ (c2v2 +(b2 + c2 − a2)vw + b2w cyclic

This deﬁnes the square of the distance between P and Q. If the line PQcontains H, the perspector is the intersection of the reﬂec- tions of the line PQin the sidelines. 8.5 Circumcevian triangles 317

8.5 Circumcevian triangles

Let XY Z be the circumcevian triangle of P , and XY Z be that of P ∗. The lines XX, YY, ZZ bound a triangle homothetic to ABC. What is the homothetic center? 1 Lemma. The vertices of the circumcevian triangle cevo(P ) are the isogonal conjugates of the inﬁnite points of the cevian lines of P ∗. More generally, for P =(u : v : w) and Q =(x : y : z), the lines joining the corresponding vertices of the circumcevian triangles of P and Q bound a triangle perspective with ABC at ⎛ ⎞ a2 ⎝ : ···: ···⎠ . b2 + c2 b2 + c2 − a2 · a2 v w y z u x For the Gergonne and Nagel points, this yields 2 a : ···: ··· (b + c − a)(3a2 +(b − c)2) with ETC (6-9-13)-search number 0.682731527705 ···

2 1 a =( : : ) 2 2 2 : ···: ··· If P u v w , the homothetic center is u(Sα(v+w) +Sβ v +Sγ w ) . 318 The circumcircle

Proposition. The circumcevian triangle of P =(u : v : w) is perspective with the tangential triangle at a4 b4 c4 a2 − + + : ···: ··· . u2 v2 w2

When are the lines joining the corresponding vertices of the circumcevian triangles of P and Q concurrent? This is the case if and only if one of the points lies on the circumcircle (trivial) or they are inverse in the circumcircle. Proposition. Let P =(u : v : w). The tangents of the circumcircle at the vertices of cevo(P ) bound a triangle with perspector a2 : ···: ··· . −a2vw + b2wu + c2uv This is the isogonal conjugate of the superior of the isogonal conjugate of P . 8.5 Circumcevian triangles 319

8.5.1 Circumcevian triangle Let P =(u : v : w). The lines AP , BP, CP intersect the circumcircle again at the vertices of the circumcevian triangle ocev(P ) of P : ⎛ ⎞ ⎛ ⎞ 2 (P ) −a vw : : A c2v+b2w v w ⎜ 2 ⎟ ocev( )=⎝ (P )⎠ = ⎝ : −b wu : ⎠ P B u a2w+c2u w . C(P ) : : −c2uv u v b2u+a2v The circumcevian triangle ocev(P ) is similar to the pedal triangle ped(P ) and the reﬂection triangle rﬂ(P ). Theorem. The circumcevian triangle ocev(P ) is perspective with the tangential triangle cev−1(K) at ∧(ocev( ) cev−1( )) 2(− a4 + b4 + c4 ): 2( a4 − b4 + c4 ): 2( a4 + b4 − c4 ) P , K a u2 v2 w2 b u2 v2 w2 c u2 v2 w2 .

(P ) 2 2 2 Proof. The line joining A to Ka =(−a : b : c ) has equation (b4w2 − c4v2)x + a2b2w2y − c2a2v2z =0. (P ) [P ] Similarly, the lines B Pb and C Pc have equations −a2b2w2x +(c4u2 − a4w2)y + b2c2u2z =0, c2a2v2x − b2c2u2y +(a4v2 − b4u2)z =0. These lines are concurrent since a2((b4w2 − c4v2)x + a2b2w2y − c2a2v2z) +b2(−a2b2w2x +(c4u2 − a4w2)y + b2c2u2z) +c2(c2a2v2x − b2c2u2y +(a4v2 − b4u2)z) =0. These lines intersect at : : x y z ( 4 2 − 4 2) 2 2 2 − 2 2 2 2 2 2 − 2 2 2 ( 4 2 − 4 2) = c u a w b c u : − a b w b c u : a b w c u a w −b2c2u2 (a4v2 − b4u2) c2a2v2 (a4v2 − b4u2) c2a2v2 −b2c2u2 =(c4u2 − a4w2)(a4v2 − b4u2)+b2c2u2 · b2c2u2 : a2b2((a4v2 − b4u2)w2 + c4u2v2):a2c2(b4w2u2 − (c4u2 − a4w2)v2) =a4(−a4v2w2 + b4w2u2 + c4u2v2) : a2b2(a4v2w2 − b4w2u2 + c4u2v2):a2c2(a4v2w2 + b4w2u2 − c4u2v2). 320 The circumcircle

P ∧(ocev(P ), cev−1(K)) 2 4 4 4 2 4 4 4 2 4 4 4 G (a (b + c − a ):b (c + a − b ):c (a + b − c )) 2( 2− ) 2( 2− ) 2( 2− ) H a S Sαα : b S Sββ : c S Sγγ Sα Sβ Sγ

Note that this perspector is the same for the points in the harmonic quadru- ple of P . Indeed, if X, Y , Z are the second intersections of the circumcircle with the sides of the anticevian triangle of P , then the line XX passes through the point (−a2 : b2 : c2), and the circumcevian triangle of AP is XY Z.

8.5.2 The circumcevian triangle of H

The circumcevian triangle of H is homothetic to the tangential triangle, with ratio r : R. The homothetic center is The circumcevian triangles of P and Q are perspective if and only if one of them lies on the circumcircle or they are inversive. In the latter case, the two triangles are oppositely congruent about the line containing the two points.

Proposition. The lines joining corresponding vertices of the circumcevian triangles of (u : v : w) and (x : y : z) bound a triangle perspective with ABC at 2 a : ···: ··· ( 2 + 2 )( 2 + 2 ) − 4 ⎛ b wu c uv b zx c xy a vwxy⎞ a2 = ⎝ : ···: ···⎠ . b2 + c2 b2 + c2 − a2 · a2 v w y z u x 8.5 Circumcevian triangles 321

8.5.3 The circum-tangential triangle The circum-tangential triangle of P is bounded by the tangents to the circumcircle at the vertices of ocev(P ). These tangents are the lines 2 1 b2 c2 b2 c2 + x + y + z =0, a2 v w v2 w2 etc. They bound a triangle with perspector a2 : ···: ··· . −a2vw + b2wu + c2uv

Examples (1) The circumtangential triangle of I is homothetic to ABC with ratio of R ( ) homothety r . The homothetic center is T−, the exsimilicenter of O and (I) (2) The circumtangential triangle of K has perspectorK. 2 2 2 (3) The circumtangential triangle of G has perspector a : b : c . 2 Sα Sα Sγ (4) For P = T , Q = T+. Corollary. The circum-tangential triangle of (u : v : w) is perspective with ABC at a2 b2 c2 : : . b2wu + c2uv − a2vw c2uv + a2vw − b2wu a2vw + b2wu − c2uv

Exercise 1. Let DEF be the circumcevian triangle of G, then 3 AG BG CG + + =3. GD GE GF 2. The locus of P with circumcevian triangle DEF such that the sum of the ratios is 3 is the circle with diameter OG.

2 This is X25 in ETC. 3Sastry, Problem 1119, Math. Magazine, (Solution, 55:3 (1982) 180–182). This is true also for the circumcenter. See also Problem 1120. 322 The circumcircle

8.5.4 Circumcevian triangles congruent to the reference triangle The circumcevian triangle of P is congruent to ABC if and only if P is one of the following points on the Brocard circle and their inverse on the Lemoine axis: the circumcenter, the Brocard points, and the midpoints of the symmedian chords. Theorem. The circumcevian triangle of P is congruent to ABC if and only if P is one of the following points: (i) the circumcenter O, (ii) the Brocard points Ω→ and Ω←, (iii) their inversive images Ω→ and Ω← in the circumcircle, and (iv) the intersections Pa, Pb, Pc of the symmedians with the Brocard circle, (iv) the inversive image of Pa, Pb, Pc in the circumcircle.

Ω→

P Ω→ b

K O Pa Pc Ω←

B C

Ω← 8.6 Triangle bounded by the reﬂections of a tangent to the circumcircle in the sidelines 323 Ω→

Ω→ C K O

Ω←

B C

Ω←

B→ A

C→ B← C∗ B∗

Ω→

K C← O

Ω←

B C

A∗

A← A→ 8.6 Triangle bounded by the reﬂections of a tangent to the circumcircle in the sidelines

Given a point P on the circumcircle of triangle T, consider the triangle bounded by the reﬂections of the tangent at P in the sidelines. 324 The circumcircle

A Bb

Bc C c Pb

Ca K O Pa Pc

B C

Ac Ab Aa

(1) These reﬂections bound a triangle ABC which is perspective with ABC at a point Q on the circumcircle. (2) The circumcircles of ABC and ABC are tangent internally at a point T which is the intersection of the reﬂections of the line HP in the sidelines of ABC. (3) The circumcenter of ABC is the intersection of OT and EQ. Chapter 9

Circles

9.1 Generic circles

Proposition. A circle C is represented by an equation of the form a2yz + b2zx + c2xy − (x + y + z)(fx+ gy + hz)=0 in homogeneous barycentric coordinates. Proof. Every circle C is homothetic to the circumcircle by a homothety, say h(T,k), where T = uA + vB + wC (in absolute barycentric coordinate) is a center of similitude of C and the circumcircle. This means that if P (x : y : z) is a point on the circle C, then h(T,k)(P )=(1− k)T + kP ∼ (x + tu(x + y + z):y + tv(x + y + z):z + tw(x + y + z)), (1− )( + + ) where t = k u v w , lies on the circumcircle. In other words, k 0= a2(y + tv(x + y + z))(z + tw(x + y + z)) cyclic = a2(yz + t(wy + vz)(x + y + z)+t2vw(x + y + z)2) cyclic

+ t2(a2vw + b2wu + c2uv)(x + y + z)2. Note that the last two terms factor as the product of x + y + z and another linear form. It follows that every circle can be represented by an equation of the form a2yz + b2zx + c2xy − (x + y + z)(fx+ gy + hz)=0. 326 Circles

Example (1). The nine-point circle. Under the homothety h(G, −2), the nine-point circle is mapped onto the circumcircle. This means that if (x : y : z) is a point on the nine-point circle, then its superior (y + z − x : z + x − y : x + y − z) lies on the circumcircle. It follows that a2(z+x−y)(x+y−z)+b2(x+y−z)(y+z−x)+c2(y+z−x)(z+x−y)=0.

Simplifying this equation, we have 1 2 + 2 + 2 − ( + + )( + + )=0 a yz b zx c xy 2 x y z Sαx Sβy Sγz . Example (2). The circumcircle of the superior triangle. The equation of the circle can be obtained from that of the circumcircle by substituting (x : y : z) by (y + z : z + x : x + y). Thus,

a2yz + b2zx + c2xy +(x + y + z)(a2x + b2y + c2z)=0.

The center of the circle is the orthocenter H.

Example (3). Circumcircle of the excentral triangle. To ﬁnd the equation of the circle through the excenters, we solve the system of linear equations a2bc − ab2c − abc2 −fa+ gb + hc = = −abc, −a + b + c −a2bc + ab2c − abc2 fa− gb + hc = = −abc, a − b + c −a2bc − ab2c + abc2 fa+ gb − hc = = −abc. a + b − c From these,

fa = gb = hc = −abc =⇒ f = −bc, g = −ca, h = −ab.

It follows that the circumcircle of the excentral triangle is the circle

a2yz + b2zx + c2xy +(x + y + z)(bcx + cay + abz)=0. 9.1 Generic circles 327

Exercise 1. Show that a circle passing through A has equation of the form

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

for some q and r. 2. Show that a circle passing through B and C has equation of the form

a2yz + b2zx + c2xy − px(x + y + z)=0

for some p. 3. Show that the superior of the circle C is the circle ⎛ ⎞

(a2yz + b2zx + c2xy) − (x + y + z) ⎝ (2g +2h − a2)x⎠ =0. cyclic

4. Show that the inferior of the circle C is the circle 4( 2 + 2 + 2 ) a yz b zx⎛ c xy ⎞

−(x + y + z) ⎝ (b2 + c2 − a2 + g + h − f)x⎠ =0. cyclic 328 Circles

9.2 Power of a point with respect to a circle

Consider a circle C := O(ρ) and a point P . By the theorem on intersecting chords, for any line through P intersecting C at two points X and Y , the product PX · PY of signed lengths is constant. We call this product the power of P with respect to C. By considering the diameter through P ,we obtain |OP|2 − ρ2 for the power of a point P with respect to O(ρ). Proposition. With respect to the circle C : a2yz + b2zx + c2xy − (x + y + z)(fx+ gy + hz)=0, the powers of the vertices A, B, C of T are f, g, h respectively. More ( : : ) fx+gy+hz generally, the power of a ﬁnite point x y z is x+y+z . Proof. Let the circle intersect the sideline a at the points X =(0:v : w) and X =(0:v : w). These coordinates satisfy the equation a2yz − (y + z)(gy + hz)=0 =⇒ gy2 +(g + h − a2)yz + hz2 =0. v · v = h v + v = −g+h−a2 Note that w w g and w w g . The power of B with respect to the circle is wa wa ww BX · BX = · = · a2 v + w v + w (v + w)(v + w) 1 2 1 2 = · a = · a v +1 v +1 vv + v + v +1 w w ww w w 1 2 = 2 · a = g. h − g+h−a +1 g g Similarly, the powers of A and C are f and h.

9.2.1 Radical axis and radical center Corollary. (a) The radical axis of C : a2yz + b2zx + c2xy − (x + y + z)(fx+ gy + hz)=0 and the circumcircle is the line fx+ gy + hz =0. (b) The radical axis of the circles C and C : a2yz + b2zx + c2xy − (x + y + z)(f x + gy + hz)=0 9.2 Power of a point with respect to a circle 329 is the line (f − f )x +(g − g)y +(h − h)z =0. Proposition. The circle C is tangent to the circumcircle if and only if the line fx+ gy + hz =0is tangent to any one of them. Proposition. If, for i =1, 2, 3, the centers of the three circles

2 2 2 Ci : a yz + b zx + c xy − (x + y + z)(fix + giy + hiz)=0 are noncollinear, the radical center is the point given by

f1x + g1y + h1z = f2x + g2y + h2z = f3x + g3y + h3z.

Explicitly, 1 g1 h1 f1 1 h1 f1 g1 1 x : y : z = 1 g2 h2 : f2 1 h2 : f2 g2 1 . 1 g3 h3 f3 1 h3 f3 g3 1

Exercise 1. Find the radical center of the circumcircles of T, the superior triangle, and the excentral triangles. 1 2. Find the radical center of the circumcircles of T, the inferior triangle, and the excentral triangles. 2 3. Find the radical axis of the circumcircles of the superior and inferior triangles. 3 4. Find the radical axis of the circumcircles of the superior and excentral triangles. 4 5. Find the radical center of the circumcircles of the superior, inferior, and excentral triangles. 5

1X(1491) = (a(b3 − c3):b(c3 − a3):c(a3 − b3)). 2X(650) = (a(b − c)(b + c − a):b(c − a)(c + a − b):c(a − b)(a + b − c)). 3(3a2 + b2 + c2)x +(a2 +3b2 + c2)y +(a2 + b2 +3c2)z =0. 4(a2 − bc)x +(b2 − ca)y +(c2 − ab)z =0. 5((b − c)(a3 + a2(b + c)+a(3b2 +2bc +3c2)+(b + c)(b2 + c2)) : ···: ···). 330 Circles

9.3 The tritangent circles

By the tritangent circles of a triangle we mean the circles each tangent to the three sides of the triangle. These include the incircle and the three excircles.

Ib s − c

s − b A

s − a B C s − a s − b s − c

The powers of the vertices with respect to each of these circles can be easily written found, leading to the equations of the circles.

(I): a2yz + b2zx + c2xy − (x + y + z)((s − a)2x +(s − b)2y +(s − c)2z)=0; 2 2 2 2 2 2 (Ia): a yz + b zx + c xy − (x + y + z)(s x +(s − c) y +(s − b) z)=0, 2 2 2 2 2 2 (Ib): a yz + b zx + c xy − (x + y + z)((s − c) x + s y +(s − a) z)=0, 2 2 2 2 2 2 (Ic): a yz + b zx + c xy − (x + y + z)((s − b) x +(s − a) y + s z)=0. 9.3 The tritangent circles 331

9.3.1 The Feuerbach theorem The superior of the incircle is the image of the incircle under the homothety h(G, −2). Applying Proposition ??(b) with (f,g,h)=((s − a)2, (s − b)2, (s − c)2), we obtain the equation a2yz + b2zx + c2xy − (x + y + z)((b − c)2x +(c − a)2y +(a − b)2z)=0. This is tangent to the circumcircle at the point ∗ a b c (L (I)∞) = : : , b − c c − a a − b the isogonal conjugate of the inﬁnite point of the trilinear polar of the incenter. A

Y O Z G I Na N

B X C

∗ (L (I)∞)

Likewise, the superior of the A-excircle is the circle a2yz + b2zx + c2xy − (x + y + z)((b − c)2x +(c + a)2y +(a + b)2z)=0. This is tangent to the circumcircle at the point a2 b2 c2 a b c : : = : : . a(b − c) −b(c + a) c(a + b) b − c −(c + a) a + b Similarly, the superiors of the B-and C-excircles are also tangent to the circumcircle. From these, we deduce the famous Feuerbach theorem. Theorem (Feuerbach). The nine-point circle is tangent internally to the incircle at the Feuerbach point 2 2 2 Fe := ((b − c) (b + c − a): (c − a) (c + a − b): (a − b) (a + b − c)), 332 Circles and externally to the excircles respectively at 2 2 2 Fa =(−(b − c) (a + b + c): (c + a) (a + b − c): (a + b) (c + a − b)), 2 2 2 Fb =((b + c) (a + b − c): −(c − a) (a + b + c): (a + b) (b + c − a)), 2 2 2 Fc =((b + c) (c + a − b): (c + a) (b + c − a): −(a + b) (a + b + c)). Proof. The tangency follows from the fact the nine-point circle is the inferior of the circumcircle. The point of tangency is the Feuerbach point F := Inf a : b : c e − − − b c c a a b = b + c : c + a : a + b − − − − − − c a a b a b b c b c c a = (b − c)2(b + c − a):(c − a)2(c + a − b):(a − b)2(a + b − c) . On the other hand, the point of tangency with the A-excircle is the inferior of the point a : b : c b − c −(c + a) a + b on the circumcircle. This is Fa given above; similarly for Fb and Fc. Proposition. (a) The points of the tangency of the nine-point circle with the excircles form a triangle perspective with T at the outer Feuerbach point (b + c)2 (c + a)2 (a + b)2 : : . b + c − a c + a − b a + b − c (b) The common tangents of the nine-point circle with the excircles bound a triangle perspective with the inferior triangle at the Spieker center Sp. Proof. The common tangents of the circumcircle with the superiors of the excircles are the lines (b − c)2x+(c + a)2y+(a + b)2z =0, (b + c)2x+(c − a)2y+(a + b)2z =0, (b + c)2x+(c + a)2y+(a − b)2z =0. These bound a triangle with vertices (−(a2+bc):b(b+c):c(b+c)), (a(c+a):−(b2+ca):c(c+a)), (a(a+b):b(a+b) perspective with T at the incenter I. 9.3 The tritangent circles 333

Remark. The common tangents of the nine-point circle with the incircle and the excircles are the lines x + y + z =0; b − c c − a a − b x y z + − =0, b − c c + a a + b x y z − + + =0, b + c c − a a + b x y z − + =0. b + c c + a a − b 334 Circles

9.3.2 Radical axes of the circumcircle with the excircles Clearly the incircle and the circumcircle do not intersect at real points. From the equation of the incircle, the line (s − a)2x +(s − b)2y +(s − c)2z =0 is the radical axis with the circumcircle. It is the trilinear polar of the barycentric square of the Gergonne point. On the other hand, the radical axes of the circumcircle with the excircles are the lines s2x+(s − c)2y+(s − b)2z =0, (s − c)2x+ s2y+(s − a)2z =0, (s − b)2x+(s − a)2y+ s2z =0.

B C

Ia These lines bound a triangle with vertices (−(b + c)(a2 +(b + c)2): b(a2 + b2 − c2):c(c2 + a2 − b2)), (a(a2 + b2 − c2):−(c + a)(b2 +(c + a)2):c(b2 + c2 − a2)), (a(c2 + a2 − b2): b(b2 + c2 − a2):−(a + b)(c2 +(a + b)2) perspective with T at the Clawson point a b c Cw := : : . Sα Sβ Sγ 9.3 The tritangent circles 335

Exercise 1. The radical axis of the incircle and the A-excircle is the perpendicular to the bisector of angle A through the midpoint of BC. It is the image 1 of the external bisector of angle A under the homothety h G, −2 . 2. The same conclusion applies to the radical axes of the incircle with the B- and C-excircles. The triangle bounded by the three radical axes is therefore the image of the excentral triangle under the homothety 1 h G, −2 . 336 Circles

9.3.3 The radical center of the excircles

The radical axis of the B- and C-excircles contains the midpoint Ga of BC, and is perpendicular to the line joining the excenters Ib and Ic. This is parallel to the bisector of angle A. It is indeed the A-bisector of the inferior triangle cev(G). Similarly, the radical axis of the C- and A-excircles is the B-bisector. It follows that the radical center of the excircles is the incenter of the inferior triangle. This is the Spieker center Sp.

Sp I

B C

Ia 9.3 The tritangent circles 337

9.3.4 The Spieker radical circle The Spieker radical circle is the circle orthogonal to the three excircles. Its center is the radical center Sp. Its squared radius is the common power of Sp in the excircles. This is s2(b + c)+(s − c)2(c + a)2 +(s − b)(a + b) b + c + c + a + a + b 2( + )( + )+ 2( + )( + )+ 2( + )( + ) − a c a a b b a b b c c b c c a (b + c + c + a + a + b)2 a2(b + c)+b2(c + a)+c2(a + b)+abc = . 4(a + b + c) We compute the equation of the Spieker radical circle through its image under the homothety h(G, −2). This is a circle with center I. The square radius of the circle is 4 times the square radius of the radical circle. The power of A in this superior circle is 2( + )+ 2( + )+ 2( + )+ IA2 − a b c b c a c a b abc a + b + c ( + − ) 2( + )+ 2( + )+ 2( + )+ = bc b c a − a b c b c a c a b abc a + b + c a + b + c = −a(b + c).

Similarly, the powers of B and C with respect to the superior of the spieker radical circle are −b(c+a) and −c(a+b). This superior has barycentric equation a2yz + b2zx + c2xy +(x + y + z)(a(b + c)x + b(c + a)y + c(a + b)z)=0.

Replacing x, y, z by y + z − x, z + x − y, x + y − z respectively, we obtain the barycentric equation of the Spieker radical circle: a2yz+b2zx+c2xy+(x+y+z)((s−b)(s−c)x+(s−c)(s−a)y+(s−a)(s−b)z)=0.

Exercise 1. Find the radical axis of the Spieker radical circle and the A-excircle. 6

6(a(a + b + c)+2bc)x + b(a + b − c)y + c(c + a − b)z =0. 338 Circles

2. Show that the triangle bounded by the radical axes of the Spieker radical circle and the excircles bound a triangle perspective with ABC and ﬁnd the perspector. 7

Here is an interesting property of the Spieker radical circle. Theorem. The locus of P whose polars with respect to the excircles are concurrent is the Spieker radical circle. The point of concurrency is the antipode of P on the circle.

7 1 X(2051) = a3−a(b2−bc+c2)−bc(b+c) : ···: ··· . 9.4 The Conway circle 339

9.4 The Conway circle

If Ya, Za are points on the extensions of CA and BA such that AYa = AZa = a, and similarly deﬁne Zb, Xb, Xc, Yc, the six points lie on the Conway circle (see §??). The coordinates of these points are

Xb =(0:a + b : −b),Xc =(0:−c : c + a); Ya =(a + b :0:−a),Yc =(−c :0:b + c); Za =(c + a : −a :0),Zb =(−b : b + c :0). The power of A with respect to the circle is

AYa · AYc = AZa · AZb = −a(b + c). Similarly, the powers of B and C are −b(c + a) and −c(a + b). Therefore, the equation of the Conway circle is a2yz + b2zx + c2xy +(x + y + z)(a(b + c)x + b(c + a)y + c(a + b)z)=0.

a a

X X b C c b B c c

b Yc

Exercise 1. Find the radical axis of the Conway circle with the circumcicle of the excentral triangle. 8

8(a(b + c) − bc)x +(b(c + a) − ca)y +(c(a + b) − ab)z =0. 340 Circles

2. Find the radical axis of the Conway circle with the circumcircle of the superior triangle. 9 3. Find the radical center of the Conway circle and the circumcircles of the excentral and superior triangles. 10

4. Show that the circle AYaZa has equation a2yz + b2zx + c2xy +(x + y + z)(c(c + a)y + b(a + b)z)=0;

similarly for the circles BZbXb and CXcYc. The radical center of the three circles is the Schifﬂer point a(b + c − a) Sc = : ···: ··· . b + c

5. Show that the circle AYcZb has equation a2yz + b2zx + c2xy + bc(x + y + z)(y + z)=0;

similarly for the circles BZaXc and CXbYa. The centers of these circles are on the circumcircle of T. Find the radical center of the three circles. 11

9a(b + c − a)x + b(c + a − b)y + c(a + b − c)z =0. 10((b − c)(a2(b + c) − a(b2 − bc + c2)+bc(b + c)) : ···: ···). 11The Nagel point. 9.4 The Conway circle 341

9.4.1 Sharp’s triad of circles

We extend the Conway conﬁguration by considering the reﬂections Ya, Za of Ya, Za in the vertex A, and similarly for the other four points.

Xb =(0:a + b : −b), Xc =(0:−c : c + a), =(0: − : ) =(0: : − + ); Xb a b b , Xc c c a Ya =(a + b :0:−a), Yc =(−c :0:b + c), =(− + :0: ) =( :0: − ); Ya a b a , Yc c b c Za =(c + a : −a :0), Zb =(−b : b + c :0). =( − : :0) =( : − + :0) Za c a a , Zb b b c .

This results in a triad of circles C a through Xb, Xc, Yc, Ya, Za, Zb; C b through Xb, Xc, Yc , Ya, Za, Zb; C c through Xb, Xc, Yc , Ya, Za, Zb.

a a Zb A

Ya Xb

B c Xb b Xc C Xc c Za

b Yc

Zb 342 Circles

The equations of the circles are

2 2 2 Ca : a yz + b zx + c xy +(x + y + z)(−a(b + c)x + b(c − a)y + c(−a + b)z)= 0, 2 2 2 Cb : a yz + b zx + c xy +(x + y + z)(a(−b + c)x − b(c + a)y + c(a − b)z)=0, 2 2 2 Cc : a yz + b zx + c xy +(x + y + z)(a(b − c)x + b(−c + a)y − c(a + b)z)=0.

Exercise 1. Find the centers and the radii of the circles in the triad. 12 2. Find the radical center of the triad. 13 3. Find the radius of the radical circle. 14 4. Establish the barycentric equation of the Sharp radical circle, the radical circle of Sharp’s triad of circles: ( + + )( 2 + 2 + 2 ) a b c a⎛yz b zx c xy ⎞

− (x + y + z) ⎝ ((a + b + c)(2b +2c − a) − 10bc)x⎠ =0. cyclic

12 2 2 Cc has center Ia and radius ra +(s − a) ; similarly for the other two circles. 13 Na. 14 2abc The square of the radius is the (common) power of the radical center in the circles; it is a+b+c =4Rr. 9.5 The Taylor circle 343

9.5 The Taylor circle

Consider the orthic triangle HaHbHc, and the pedals of each of the points Ha, Hb, Hc on the two sides not containing it. Thus,

Line Pedal of Ha Pedal of Hb Pedal of Hc a Xb Xc b Ya Yc c Za Zb A

Zb Yc

Hb Ya

H c H

Za B C Xc Ha Xb

It is easy to write down the lengths of various segments. For example, 2 AYa = b sin γ sin γ = b sin γ, 2 AYc = b cos α cos α = b cos α; 2 AZa = c sin β sin β = c sin β, 2 AZb = c cos α cos α = c cos α.

Since b sin γ = c sin β, AYa · AYc = AZa · AZb. More precisely, S2 · S AY · AY = AZ · AZ = αα , a c a b a2b2c2 S2 · S BZ · BZ = BX · BX = ββ , b a b c a2b2c2 S2 · S CX · CX = CY · CY = γγ . c b c a a2b2c2 We conclude that the six pedals of pedals are concyclic. The circle containing them is called the Taylor circle, and has barycentric equation 2 2 2 2 2 2 2 a b c (a yz + b zx + c xy) − S (x + y + z)(Sααx + Sββy + Sγγz)=0. 344 Circles

Remark. The coordinates of the pedals of pedals are as follows.

Line Pedals of Ha Pedals of Hb Pedals of Hc 2 2 a Xb =(0:Sγγ : S ) Xc =(0:S : Sββ) 2 2 b Ya =(Sγγ :0:S ) Yc =(S :0:Sαα) 2 2 c Za =(Sββ : S :0) Zb =(S : Sαα :0)

Exercise = S2 = 1. Show that YaZa abc R .

2. Find the equations of the lines YaZa, ZbXb, XcYc, and show that they bound a triangle perspective to ABC at the symmedian point. 15

3. Show that the Euler lines of the triangles HaYaZa, XbHbZb, XcYcHc are concurrent and ﬁnd the coordinates of the intersection. 16

15 2 The line YaZa has equation −S x + Sββy + Sγγz =0 16J.-P. Ehrmann, Hyacinthos 3695. The common point of the Euler lines is 2 2 4 2 X973 =(a (S + Sβγ)((Sα + Sβ + Sγ )S + Sαβγ (2S − Sαα)) : ···: ···). 9.5 The Taylor circle 345

9.5.1 The Taylor circle of the excentral triangle Triangle T := ABC is the orthic triangle of its excentral triangle cev−1(I)= IaIbIc. The orthocenter is I since AIa is perpendicular to IbIc. The pedals of pedals have very simple coordinates.

Line Pedal of A Pedal of B Pedal of C b c I I Xb =(b + c : b : −c) Xc =(b + c : −b : c) c a I I Ya =(a : c + a : −c) Yc =(−a : c + a : c) a b I I Za =(a : −b : a + b) Zb =(−a : b : a + b)

For example, BZb is parallel to the bisector of angle C, and has inﬁnite point (a : b : −(a + b)). It is the line (a + b)x + az =0; it intersects the x + y =0 =(− : : + ) external bisector of angle C, namely, a b at Zb a b a b . The superior of these points are the point in the Conway conﬁguration:

Line superior of pedal of A superior of pedal of B superior of pedal of C a (−c :0:b + c)(−b : b + c :0) b (0 : −c : c + a)(c + a : −a :0) c (0 : a + b : −b)(a + b :0:−a)

Xc A Xb

Ya Za Sp I G

B C

Therefore, the Taylor circle of the excentral triangle is the inferior of the Conway circle, and is the same as the Spieker radical circle. 346 Circles

9.6 Some triads of circles

9.6.1 Circles with sides as diameters BC a Consider the circle with diameter . The radius of the circle is 2 . The A 2 − a 2 = 1(2 2+2 2− 2)− 1 2 = power of with respect to the circle is ma 2 4 b c a 4a b2+c2−a2 2 = Sα. From this we obtain the equation of the circle: 2 2 2 a yz + b zx + c xy − Sαx(x + y + z)=0.

Similarly, the circles with diameters CA and AB are

2 2 2 a yz + b zx + c xy − Sβy(x + y + z)=0, 2 2 2 a yz + b zx + c xy − Sγz(x + y + z)=0.

Consider the orthocenter H with the orthic triangle HaHbHc. The circle Ca contains the pedals Hb and Hc. Similarly, Cb contains Hc and Ha, and Cc contains Ha and Hb. It follows that AHa is the radical axis of Cb and Cc. Similarly, BHb is the radical axis of Cc and Ca, and CHc that of Ca and Cb. The radical center is the orthocenter H, and

HA · HHa = HB · HHb = HC · HHc. 9.6 Some triads of circles 347

9.6.2 Circles with cevians as diameters Let P =(u : v : w), with cevian triangle cev(P )=XY Z. Consider the circle with diameter AX. The circle intersects the sideline BC at X and the H B C Sβ · aw = Sβw Sγ v pedal a. The powers of and in this circle are a v+w v+w and v+w . This circle has equation

1 a2yz + b2zx + c2xy − (S wy + S vz)(x + y + z)=0. v + w β γ

Note that the power of H in this circle is HA · HHa. Since this is equal to HB · HHb and HC · HHc, the orthocenter H has equal powers with respect to the three circles. It is the radical center of the triad of circles. We rewrite this in a slightly different form as follows, along with the circles with diameters BY and CZ: 2 + 2 + 2 − vw Sβ + Sγ ( + + )= 0 a yz b zx c xy + y z x y z , v w v w 2 + 2 + 2 − wu Sγ + Sα ( + + )= 0 a yz b zx c xy + z x x y z , w u w u uv S S a2yz + b2zx + c2xy − α x + β y (x + y + z)= 0. u + v u v The radical center of the three circles is the point given by Sβ y + Sγ z Sγ z + Sα x Sα x + Sβ y v w = w u = u v . 1 + 1 1 + 1 1 + 1 v w w u u v From these, 1 1 1 Sαx = Sβy = Sγz =⇒ x : y : z = : : , Sα Sβ Sγ and the radical center is the orthocenter H.

Exercise The perpendiculars from H to the three cevians intersect the corresponding circle at two points. The six intersections with the three circles lie on a circle with center P . Find the equation of the circle. 17 17 2 2 2 (u + v + w)(a yz + b zx + c xy) − (x + y + z) cyclic Sα(v + w − u)x =0. 348 Circles

9.6.3 Circles with centers at vertices and altitudes as radii

Consider the circle Ca with center A and radius AHa. Since the length of 2 H S −S2 Sβ A a is a , the powers of A, B, C with respect to the circle are a2 , a , 2 Sγ and a , the equation of the circle is

−S2x + S2y + S2z a2yz + b2zx + c2xy − β γ (x + y + z)=0. a2

Similarly, we have the equations of the circles B(Hb) and C(Hc). The radical center of the three circles is the point given by −S2x + S2y + S2z S2x − S2y + S2z S2x + S2y − S2z β γ = α γ = α β . a2 b2 c2

Exercise 1. This radical center has coordinates

2 4 2 4 2 4 (a (S − Sαβγ · Sα):b (S − Sαβγ · Sβ):c (S − Sαβγ · Sγ)). 9.6 Some triads of circles 349

9.6.4 Circles with altitudes as diameters BC (0 : : ) B Consider a circle tangent to at the point v w . The powers of and C aw 2 av 2 in the circle are respectively v+w and v+w . If this circle also passes through A, then its equation is a2 a2yz + b2zx + c2xy − (w2y + v2z)(x + y + z)=0. (v + w)2 Similarly, we have the equations of the circles through the vertices and tangent to the opposite sides at the traces of P =(u : v : w). These are a2 a2yz + b2zx + c2xy − (w2y + v2z)(x + y + z)= 0, (v + w)2 b2 a2yz + b2zx + c2xy − (u2z + w2x)(x + y + z)= 0, (w + u)2 c2 a2yz + b2zx + c2xy − (v2x + u2y)(x + y + z)= 0. (u + v)2 The radical center of the three circles is the point given by a2(w2y + v2z) b2(u2z + w2x) c2(v2x + u2y) = = . (v + w)2 (w + u)2 (u + v)2

y z z x x y 2 + 2 2 + 2 2 + 2 v w = w u = u v . u2(v+w)2 v2(w+u)2 w2(u+v)2 a2 b2 c2 Therefore, x y z u2(v + w)2 v2(w + u)2 w2(u + v)2 inf : : = : : . u2 v2 w2 a2 b2 c2

u2(v + w)2 v2(w + u)2 w2(u + v)2 x : y : z = u2 − + + : ···: ··· . a2 b2 c2 350 Circles

9.6.5 Excursus: A construction problem Given a ﬁnite point P in the plane of triangle ABC, to construct three circles Ca, Cb, Cc, each passing through P and one vertex of triangle ABC, such that their second intersections X = Cb ∩ Cc, Y = Cc ∩ Ca, and Z = Ca ∩ Cb lie on AP , BP, CP respectively. Equivalently, construct three points X, Y , Z (distinct from P ) on the lines AP , BP, CP respectively, such that the circles (AY Z), (BZX) and (CXY ) are concurrent at P . Analysis. Let P =(u : v : w) and the circles be 2 2 2 Ca : a yz + b zx + c xy − (x + y + z)(q1y + r1z)=0, 2 2 2 Cb : a yz + b zx + c xy − (x + y + z)(p2x + r2z)=0, 2 2 2 Cc : a yz + b zx + c xy − (x + y + z)(p3x + q3y)=0, for undetermined coefﬁcients q1, r1, p2, r2, p3, q3. : y − z =0 Since we require the line AP v w to contain the point X,itis the radical axis of the circles Cb and Cc, we must have = : = 1 : 1 p2 p3 and q3 r2 v w . Similarly, = : = 1 : 1 q1 q3 and r1 p3 w u, and = : = 1 : 1 r1 r2 and p2 q2 u v . From these, we rewrite the equations of the circles as 2 2 2 y z Ca : a yz + b zx + c xy − k(x + y + z) + =0, v w 2 2 2 x z Cb : a yz + b zx + c xy − k(x + y + z) + =0, u w x y C : a2yz + b2zx + c2xy − k(x + y + z) + =0, b u v for some k. Since these circles contain the point P , we must have a2vw + b2wu + c2uv k = , 2(u + v + w) with a2vw + b2wu + c2uv =0, i.e., P not lying on the circumcircle. Consider the trilinear polar of P : x y z L : + + =0, u v w intersecting the sidelines at A, B, C respectively. The line AA has equa- y + z =0 C tion v w . It is the radical axis of a and the circumcircle. Therefore, 9.6 Some triads of circles 351

Ca is the circle through A, P , and the intersection of AA with the circumcircle; similarly for Cb and Cc. The points X, Y , Z are now the pairwise intersections of the circles Ca, Cb, Cc.

A A

O P Y

Z C B A

C C B

X B 352 Circles

9.6.6 Excursus: Radical center of a triad of circles Given a point P =(u : v : w), consider the triad of circles each through one vertex of T and tangent to opposite side at the trace of P . These are the circles a2(w2y + v2z) C : a2yz + b2zx + c2xy − (x + y + z)=0, a (v + w)2 b2(u2z + w2x) C : a2yz + b2zx + c2xy − (x + y + z)=0, b (w + u)2 c2(v2x + u2y) C : a2yz + b2zx + c2xy − (x + y + z)=0, c (u + v)2

We determine the radical center Q of the triad of circles. For example, if P = H, then Q = H. More generally, Q =(x : y : z) is given by a2(w2y + v2z) b2(u2z + w2x) c2(v2x + u2y) = (x + y + z)= . (v + w)2 (w + u)2 (u + v)2 Equivalently, 2 y z 2 z x 2 x y a 2 + 2 b 2 + 2 c 2 + 2 v w = w u = u v . 1 + 1 2 1 + 1 2 1 + 1 2 v w w u u v Therefore, 2 2 2 x y z 1 1 1 1 1 1 1 1 1 : : = sup + : + : + . u2 v2 w2 a2 v w b2 w u c2 u v From this, the coordinates of the radical center Q can be easily computed. Here are some examples.

PQ G − 1 + 1 + 1 : ···: ··· (194) a2 b2 c2 X I (a2(a(a + b + c) − bc):···: ···) X(595) HH G 1 : 1 : 1 (279) e (b+c−a)2 (c+a−b)2 (a+b−c)2 X 2 2 Na ((b + c − a) ((a + b + c) − 8bc):···: ···) X(6552) 9.6 Some triads of circles 353

9.6.7 Concurrency of three Euler lines

We have computed the equation of the line BZb: (a + b)x + az =0. Like- wise, CYc is the line (c + a)x + ay =0. It intersects the line BZb at the orthocenter of triangle IaBC. This is the point (−a : c + a : a + b). On the other hand, the centroid of triangle IaBC is the point (−a :2b + c − a : b +2c − a). With these we compute the equation of the Euler line of triangle IaBC, and similarly those of triangles IbCA and IcAB. These are the lines

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

These lines are concurrent at a point (x : y : z) given by ( − ) :( − ) :( − ) b c x c a y a bz −( + ) −( + ) = c a b : − bb : b c a c −(a + b) c −(a + b) cc = a : b : c.

Therefore, a b c x : y : z = : : . b − c c − a a − b

Proposition. The Euler lines of triangles IaBC, IbCA and IcAB are concurrent at the superior of the Feuerbach point (on the circumcircle). 354 Circles

We begin with the equation of the circle tangent to two sides of the reference triangle T. Lemma. The circle tangent to AC and AB with common tangent length t from A (measured along AC and AB)is

2 2 2 2 2 2 Ca(t): a yz +b zx+c xy −(x+y +z)(t x+(c−t) y +(b−t) z)=0.

t t

B C

Proof. If the two tangents from A have length t (measured along AC and AB), the points of tangency are

Y =(b − t :0:t),Z=(c − t : t :0).

The circle Ca(t) intersects the line b : y =0at (x :0:z) satisfying b2zx − (x + z)(t2x +(b − t)2z)=0 =⇒ (tx − (b − t)z)2 =0.

Therefore, x : z = b − t : t. The circle is tangent to the sideline b at Y given above. A similar calculation shows that it is tangent to c at Z. Chapter 10

Tucker circles

10.1 The Taylor circle

10.1.1 The pedals of pedals

Consider the orthic triangle XY Z, and the pedals of each of the points X, Y , Z on the two sides not containing it. Thus,

Line Pedals of X Pedals of Y Pedals of Z a Xb Xc b Ya Yc c Za Zb

Zb Yc

Z H

B Xc X Xb C 402 Tucker circles

It is easy to write down the lengths of various segments. From these we easily determine the coordinates of these pedals. For example, from 2 2 2 YaC = b cos γ,wehaveAYa = b − b cos γ = b sin γ. In homogeneous coordinates,

2 2 2 2 Ya =(cos γ :0:sin γ)=(cot γ :0:1)=(Sγγ :0:S ).

Similarly we obtain the coordinates of the remaining pedals.

Line Pedals of X Pedals of Y Pedals of Z 2 2 a Xb =(0:Sγγ : S ) Xc =(0:S : Sββ) 2 2 b Ya =(Sγγ :0:S ) Yc =(S :0:Sαα) 2 2 c Za =(Sββ : S :0) Zb =(S : Sαα :0)

10.1.2 The Taylor circle

Note that

2 2 2 2 2 2 2 2 2 AYa · AYc =(b − b cos γ)(b cos α)=b cos α sin γ =4R cos α sin β sin γ, 2 2 2 2 2 2 2 2 2 AZa · AZb =(c − c cos β)(c cos α)=c cos α sin β =4R cos α sin β sin γ,

2 · = · = Sαα = S ·Sαα · = giving AYa AYc AZa AZb 4 2 2 2 2 . Similarly, BXb BXc 2 R a b c 2 · = S ·Sββ · = · = S ·Sγγ BZb BZa a2b2c2 and CYc CYa CXc CXb a2b2c2 . By Proposition ??, the six points Xb, Xc, Yc, Ya, Za, Zb are concyclic. The circle containing them is called the Taylor circle with equation

2 2 2 2 2 2 2 a b c (a yz + b zx + c xy) − S (x + y + z)(Sααx + Sββy + Sγγz)=0.

10.1.3 The Taylor center

Proposition. The center of the Taylor circle 1 has homogeneous barycentric coordinates

2 4 2 4 2 4 (a (S − SαβγSα):b (S − SαβγSβ):c (S − SαβγSγ)).

1 The Taylor center appears as X389 in ETC. 10.1 The Taylor circle 403

Proof. We compute the center of the Taylor circle by Proposition ??. This is S2 a2S + (S (S − S ) − S (S − S )) : ···: ··· α a2b2c2 β γγ αα γ αα ββ S2 =a2S − · a2(S − S ):···: ··· α a2b2c2 αα βγ 2 2 2 2 2 =a (a b c Sα − S (Sαα − Sβγ)) : ···: ··· 2 2 2 2 =a ((S + Sαα)(S − Sβγ) − S (Sαα − Sβγ)) : ···: ··· 2 4 =a (S − Sαβγ · Sα):···: ··· .

This is a point on the Brocard axis, namely,

∗ 2 2 2 K(θ) =(a (Sα + Sθ):b (Sβ + Sθ):c (Sγ + Sθ)) for cot θ = − tan α tan β tan γ.

Exercise 1. Show that the centroid of the perimeter of the orthic triangle is the center of the Taylor circle. 2. (a) Let XY Z be the orthic triangle of ABC. Show that the orthocenter of the residual triangle AY Z is

4 2 2 Ha =(S − SαβγSα : b Sααβ : c Sααγ).

(b) Similarly deﬁne Hb and Hc. Show that triangle HaHbHc is oppositely congruent to the orthic triangle at the Taylor center. 2

2 The midpoint between Ha and X is

2 2 2 2 4 2 2 b c S (0 : Sγ : Sβ)+a (S − Sαβγ · Sα : b Sααβ : c Sααγ ) 2 4 2 2 2 2 2 2 2 2 =(a (S − Sαβγ · Sα):b (c S Sγ + a Sααβ):c (b S Sβ + a Sααγ )) 2 4 2 2 2 2 2 2 2 2 =(a (S − Sαβγ · Sα):b (c S Sγ + Sαβ(S − Sβγ)) : c (b S Sβ + Sαγ (S − Sβγ)) 2 4 2 2 2 2 2 2 =(a (S − Sαβγ · Sα):b (S (c Sγ + Sαβ) − Sαβγ · Sβ)) : c (S (b Sβ + Sαγ ) − Sαβγ · Sγ )) 2 4 2 4 2 4 =(a (S − Sαβγ · Sα):b (S − Sαβγ · Sβ)) : c (S − Sαβγ · Sγ )).

This is the Taylor center. Similarly the midpoints of YHb and ZHc are the same point. Therefore, the triangles are oppositely congruent at the Taylor center. 404 Tucker circles

10.1.4 The Taylor circle of the excentral triangle Triangle ABC is the orthic triangle of its excentral triangle. To ﬁnd the Taylor center of the excentral triangle, we ﬁnd the orthocenter of triangle IABC. It is the point (−a : c + a : a + b). The midpoint of this orthocenter and A is the point

(−a : c + a : a + b)+(a + b + c)(1 : 0 : 0) = (b + c : c + a : a + b), the Spieker center. The pedals themselves are very simple: (−a : b : a + b), (−a : c + a : c), (b + c : −b : c), (a : −b : a + b), (a : c + a : −c), (b + c : b : −c).

The midpoint between (a : −b : a + b) and (a : c + a : −c) is (2a : c + a − b : a + b − c), and the distance is s. This midpoint is precisely the point of tangency of the incircle with the corresponding side√ of the medial 1 2 2 triangle. This shows that the radius of the Taylor circle is 2 r + s . This proves that the Taylor circle of the excentral triangle is the Spieker radical circle of the excircles.

Exercise = S2 = 1. Show that XbXc abc R .

2. Find the equations of the lines XbXc, YcYa, ZaZb, and show that they bound a triangle perspective to ABC at the symmedian point. 3

3. Show that the Euler lines of the triangles XXbXc, YYcYa, ZZaZb are concurrent and ﬁnd the coordinates of the intersection. 4

10.1.5 A triad of Taylor circles Consider the Taylor circle of HBC. This passes through the points

3 2 The line XbXc has equation −S x + Sββy + Sγγz =0 4 J.-P. Ehrmann, Hyacinthos 3695. The common point of the Euler lines is X973. 10.2 The Taylor circle 405

10.2 The Taylor circle

The pedals of the vertices of the orthic triangle on the side lines are

2 Xb =(SCC :0:S ), 2 Xc =(SBB : S :0),

2 Yc =(S : SAA :0), 2 Ya =(0:SCC : S ),

2 Za =(0:S : SBB), 2 Zb =(S :0:SAA).

The orthocenter of triangle AZbYc is the point 4 2 2 5 Ha =(S − SAABC : b SAAB : c SAAC). These orthocenters form a triangle perspective with ABC at the circumcen- 2 2 2 ter O =(a Sα : b Sβ : c Sγ). It is more interesting to note that HaHbHc is homothetic to the orthic triangle. The midpoint of HaX is on the perpendicular bisector of ZbXb, and also of XcYc. It is necessarily the Taylor center. Similarly for the other two segments HbY and HcZ. The midpoint between Ha and X is 2 2 2 2 4 2 2 b c S (0 : Sγ : Sβ)+a (S − SAABC : b SAAB : c SAAC) 2 4 2 2 2 2 2 2 2 2 2 2 =(a (S − SAABC):b c S Sγ + a b SAAB : b c S Sβ + a c SAAC) 2 4 2 2 2 2 2 2 2 2 =(a (S − SAABC):b (c S Sγ + a SAAB):c (b S Sβ + a SAAC))

Now 2 2 2 2 2 2 c S Sγ + a SAAB =c S Sγ + a SAAB + SBBCA − SBBCA 2 2 =c S Sγ + SAABB + SAABC + SBBCA − SBBCA 2 2 =c S Sγ + SAB(SAB ++SAC + SBC) − SBBCA 2 2 2 =c S Sγ + SABS − SBBCA 4 =S − SBBCA

5The sum of the coordinates is b2c2S2. 406 Tucker circles

2 4 Similarly, the third coordinate is c (S −SCCAB). The triangles HaHbHc and XY Z therefore are homothetic at the triangle center 2 4 2 4 2 4 (a (S − SAABC):b (S − SBBCA):c (S − SCCAB)) with ratio of homothety −1. This homothetic center is the Taylor center. The Taylor center is a point on the Brocard axis. Its coordinates can be written as 2 2 2 (a (Sα + Sθ):b (Sβ + Sθ):c (Sγ + Sθ)) for −S4 Sθ = . SαSβSγ In other words, tan θ = − tan A tan B tan C. The radius of the Taylor circle is R sin ω sin(θ + ω)

10.2.1 Triangle ABC is the orthic triangle of its excentral triangle. To ﬁnd the Taylor center of the excentral triangle, we ﬁnd the orthocenter of triangle IαBC. It is the point (−a : c + a : a + b). The midpoint of this orthocenter and A is the point (−a : c + a : a + b)+(a + b + c)(1 : 0 : 0) = (b + c : c + a : a + b), the Spieker center. The pedals themselves are very simple: (−a : b : a + b), (−a : c + a : c), (b + c : −b : c), (a : −b : a + b), (a : c + a : −c), (b + c : b : −c). The midpoint between (a : −b : a + b) and (a : c + a : −c) is (2a : c + a − b : a + b − c), and the distance is s. This midpoint is precisely the point of tangency of the incircle with the corresponding side√ of the medial 1 2 2 triangle. This shows that the radius of the Taylor circle is 2 r + s . This proves that the Taylor circle of the excentral triangle is the radical circle of the excircles. 10.2 The Taylor circle 407

Exercise = S2 = (i) XbXc abc R . (ii) The line XbXc has equation 2 −S x + Sββy + Sγγz =0.

(iii) The three lines XbXc, YcYa, ZaZb bound a triangle with perspector K. (iv) The Euler lines of XXbXc, YYcYa, ZZaZb are concurrent. (Ehrmann, Hyacinthos 3695).

2 2 2 2 a (S + Sβγ)(S (S (Sα + Sβ + Sγ)+Sαβγ) − 2Sαβγ · Sαα) : ···: ···).

(v) The Euler lines of AXbXc, BYcYa, CZaZb are concurrent. These are the triangle centers X(973) and X(974). 408 Tucker circles

10.3 Tucker circles

Given a triangle, through an arbitrary point on any one sideline, a point on a second sideline is obtained by alternately constructing parallels and antiparallels to a third sideline. The process terminates in six steps resulting in three pairs of points on the sidelines which lie on a Tucker circle. If the ﬁrst point is Ab =(0:1− τ : τ) on a, then by constructing alternatively parallels and antiparallels to c, a, b, c, a, b, we obtain the 6 points with coordinates indicated in the ﬁgure below. A

Ba =(1− τ :0:τ)

2 2 2 Ca =(c − b τ : b τ :0)

2 2 2 2 2 2 Cb =(a τ,c − a τ,0) Bc =(a τ :0:c − a τ)

B 2 2 2 C Ab =(0:1− τ : τ) Ac =(0:b τ : c − b τ)

These coordinates can be put in a symmetric form if we rewrite the co- 2 ordinates of the ﬁrst point as Ab =(0:Sγ + t : c ) for some t. The above construction leads to the six points

2 2 Ab =(0:Sγ + t : c ),Ba =(Sγ + t :0:c ), 2 2 Ac =(0:b : Sβ + t),Ca =(Sβ + t : b :0), 2 2 Bc =(a :0:Sα + t),Cb =(a : Sα + t :0).

The Tucker circle through these six points has barycentric equation

C ( ): ( + + + )2( 2 + 2 + 2 ) Tucker t Sα Sβ Sγ ⎛ t a yz b zx ⎞c xy

⎝ 2 2 ⎠ − (x + y + z) b c (Sα + t)x =0. cyclic 10.3 Tucker circles 409

10.3.1 The center of Tucker circle

Proposition. The center of the Tucker circle Ct is the point ∗ 2 2 2 2 2 2 K (t):=(a (S + t · Sα): b (S + t · Sβ): c (S + t · Sγ)) on the Brocard axis.

Proof. The perpendicular bisectors of the segments AbAc, BcBa, CaCb are the lines

2 2 2 2 2 2 (b − c )(Sα + t)x +a (c + Sβ)y −a (b + Sγ)z =0, 2 2 2 2 2 2 −b (c + α)x+(c − a )(Sβ + t)y +b (a + Sγ)z =0, 2 2 2 2 2 2 c (b + Sα)x −c (a + Sβ)y+(a − b )(Sγ + t)z =0. These lines are concurrent at the point K∗(t) given above, which clearly lies on the Brocard axis OK. Remark. K∗(t) is the isogonal conjugate of the Kiepert perspector K(arctan t).

Kα Ca

T K

Cb Bc

Kβ Kγ

B Ab Ac C

Figure 10.1:

The radius of the circle is R sin ω , sin(φ + ω) where R is the circumradius of ABC. 410 Tucker circles

The square radius of the circle is 2 + 2 2 · 2 csc2 2 sin2 S t · 2 = R S φ = R ω 2 R 2 2 2 . (Sα + Sβ + Sγ + t) S (cot ω +cotφ) sin (ω + φ) 10.3 Tucker circles 411

10.3.2 Construction of Tucker circle with given center Let P be a point on the Brocard axis. To construct the Tucker circle with center P , draw parallels through P to OA, OB, OC to intersect the symmedians at X, Y , Z. The antiparallels through X, Y , Z give the Tucker hexagon. 412 Tucker circles

10.3.3 Dao’s construction of the Tucker circles Let A be the point dividing the A-altitude in the ratio AA : AHa = τ :1. The circle with diameter AA intersects AC and AB at

2 2 Ba =(S (1 − τ)+Sγγ :0:S τ), 2 2 Ca =(S (1 − τ)+Sββ : S τ :0).

The line BaCa is antiparallel to BC. Similarly,

2 2 Cb =(S τ : S (1 − τ)+Sαα :0), 2 2 Ab =(0:S (1 − τ)+Sγγ : S τ); 2 2 Ac =(0:S τ : S (1 − τ)+Sββ), 2 2 Bc =(S τ :0:S (1 − τ)+Sαα).

Note that BcCb is parallel to BC. Therefore the six points are concyclic on a Tucker circle. The equation of the circle is 2 2 2( 2 + 2 + 2 ) a b c a yz b⎛zx c xy ⎞

−S2τ(x + y + z) ⎝ (b2c2 − S2τ)x⎠ =0 cyclic with center

2 2 2 2 2 (a (a b c Sα − S τ(Sαα − Sβγ)) : ···: ···).

This divides OK in the ratio

2 2 2 2 2 τS (Sα + Sβ + Sγ):a b c − τS (Sα + Sβ + Sγ).

This is the Tucker circle C(t) with a2b2c2 − τS t = αβγ . τS2 The three circles have radical center S α : ···: ··· . b2c2 − τS2 10.3 Tucker circles 413

This is the isogonal conjugate of the point

2 2 2 2 (a Sβγ(b c − τS ):···: ···) on the Euler line which divides OH in the ratio

2 2 2 (1 − τ)a b c :2τSαβγ.

The locus of the radical center is the Jerabek hyperbola. 414 Tucker circles

10.4 The Lemoine circles

10.4.1 The ﬁrst Lemoine circle

We ﬁnd a point P the parallelians through which intersect the sidelines at 6 concyclic points. If the A-parallelian intersects b and c respectively at Ya and Za, then

v + w v + w AY = · b and AZ = · c. a u + v + w a u + v + w

A = w · A = v · A ·A = A ·A Also, Yc u+v+w b and Zb u+v+w c. Therefore, Ya Yc Za Zb if and only if

v + w w v + w v · b · · b = · c · · c. u + v + w u + v + w u + v + w u + v + w

v = w B · B = B · B This reduces to b2 c2 . Similarly, Zb Za Xb Xc if and only if w = u C · C = C · C u = v c2 a2 , and Xc Xb Yc Ya if and only if a2 b2 . It follows that the six points are on a circle if and only if u : v : w = a2 : b2 : c2, i.e., P is the symmedian point K. This gives the ﬁrst Lemoine circle.

K Za Ya

B C Xc Xb 10.4 The Lemoine circles 415

The powers of A, B, C with respect to the circle are b2c2(b2 + c2) AY · AZ = , a a (a2 + b2 + c2)2 c2a2(c2 + a2) BZ · BX = , b b (a2 + b2 + c2)2 a2b2(a2 + b2) CX · CY = . c c (a2 + b2 + c2)2 From these we obtain the equation of the ﬁrst Lemoine circle: ⎛ ⎞

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

Remark. The center of this circle is the midpoint of OK. We shall establish this by showing that the circle is a Tucker circle. Clearly, YcZb = ZaXc = XbYa.

Exercise 1. Show that for the ﬁrst Lemoine circle, 2 2 2 BXc : XcXb : XbC = c : a : b , 2 2 2 CYa : YaYc : YcA = a : b : c , 2 2 2 AZb : ZbZa : ZaB = b : c : a . 416 Tucker circles

10.4.2 The second Lemoine circle 10.4 The Lemoine circles 417

10.4.3 Ehrmann’s third Lemoine circle Construct the circles KBC, KCA, KAB and note the intersections of these circles with the sidelines. circle abc 2 2 2 2 2 2 2 2 KBC : Ba =(a + b − 2c :0:3c ),Ca =(c + a − 2b :3b :0); 2 2 2 2 2 2 2 2 KCA : Ab =(0:a + b − 2c :3c ),Cb =(3a : b + c − 2a :0); 2 2 2 2 2 2 2 2 KAB : Ac =(0:3b : c + a − 2b ),Bc =(3a :0:b + c − 2a ).

The six points are on a circle, ( 2 + 2 + 2)2( 2 + 2 + 2 ) a b c ⎛ a yz b zx c xy ⎞

−3(x + y + z) ⎝ b2c2(b2 + c2 − 2a2)x⎠ =0. cyclic

The center of the circle is the point

2 2 2 X576 =((Sα + Sβ + Sγ)a SA − 3a · S : ···: ···), dividing OK in the ratio 3:−1. 418 Tucker circles

10.4.4 Bui’s fourth Lemoine circle

Construct the three circles Ca, Cb, Cc each through the symmedian point K and tangent to the circumcircle at the vertices A, B, C respectively. The intersections of these circles with the sidelines are the points

circle ab c 2 2 2 2 2 2 2 2 Ca Ba =(3a :0:2b +2c − a ),Ca =(3a :2b +2c − a :0); 2 2 2 2 2 2 2 2 Cb Ab =(0:3b :2c +2a − b ),Cb =(2c +2a − b :3b :0); 2 2 2 2 2 2 2 2 Cc Ac =(0:2a +2b − c :3c ),Bc =(2a +2b − c :3c :0).

These six points are concyclic. The circle containing them is called Bui’s fourth Lemoine circle. It has barycentric equation 4( 2 + 2 + 2)2( 2 + 2 + 2 ) a b ⎛c a yz b zx c xy ⎞

−(x + y + z) ⎝ 3b2c2(2b2 +2c2 − a2)x⎠ =0, cyclic and center

2 2 2 X(575) = ((Sα + Sβ + Sγ)a SA +3a · S : ···: ···), which divides OK in the ratio 3:1. 10.5 The Apollonius circle 419

10.5 The Apollonius circle

00907, 0210003 The Apollonius circle is the circular hull of the excircles. It is the inversive image of the nine-point circle in the Spieker radical circle. S F S F S F The touch points are the (second) intersections of the lines p a, p b, p c with the excircles. These are the points =(− 2( ( + )+( 2 + 2))2 :42( + )2 ( − ): 4 2( + )2 ( − )) Fa a a b c b c b c a s s c c a b s s b , =(42( + )2 ( − ): − 2( ( + )+( 2 + 2))2 :42( + )2 ( − )) Fb a b c s s c b b c a c a c a b s s a , =(42( + )2 ( − ): 4 2( + )2 ( − ): − 2( ( + )+( 2 + 2))2) Fc a b c s s b b c a s s a c c a b a b .

Ab Fc

Fa Fc I N Bc Ba C A Fb

Ac Ca

Fb The equation of the Apollonius circle is

4abc(a2yz+b2zx+c2xy)+(a+b+c)(x+y+z) bc(a(a+b+c)+2bc)x =0. cyclic

The intersections with the sidelines are as follows.

Ab =(0:cs + ab : −cs),Ba =(cs + ab :0:−cs); Ac =(0:−bs : bs + ca),Ca =(bs + ca : −bs :0); Bc =(−as :0:as + bc),Cb =(−as : as + bc :0). 420 Tucker circles

Clearly, BcCb, CaAc, and AbBa are parallel to the sidelines. Note that −cs −bs AB · AC = · b b = · c c = AC · AB. a ab ca a

This means that B, C, Ca, Ba are concyclic, and BaCa is antiparallel to BC. Similarly, BaCa, CbAb, AcBc are antiparallel to CA and AB respectively. Therefore, the Apollonius circle is a Tucker circle CTucker (t). The parameter is indeed a2 + b2 + c2 2abc t = − + . 2 a + b + c 10.6 Tucker circles which are pedal circles 421

10.6 Tucker circles which are pedal circles

Consider a Tucker circle which is the common pedal triangle of an isogonal conjugate pair P and Q. Denote these pedals by P[a] and Q[a] etc. Suppose the two endpoints of one of the “antiparallel” sides of the Tucker hexagon are both pedals of P . P must lie on an altitude of T, and the endpoint of this altitude (a pedal of the orthocenter H) is a vertex of the hexagon. One of the two sides of the hexagon containing this pedal of H must be another “antiparallel” side of the hexagon, and is parallel to a side of the orthic triangle of T. This means that one side of the Tucker hexagon is a side of the orthic triangle. This is enough to determine the Tucker circle. There are three such circles. Their centers are the intersections of the Brocard axis with the perpendicular bisectors of the sides of the orthic triangle.

P Q Ab Ba Ca Ac Bc Cb BH ∩ CO BO ∩ CH Pa Pb = H[b] Qc = H[c] Qa Qb Pc CH ∩ AO CO ∩ AH Qa = H[a] Qb Qc Pa Pb Pc = H[c] AH ∩ BO AO ∩ BH Qa Pb Pc Pa = H[a] Qb = H[b] Qc

2 2 2 2 2 2 2 b c (a yz + b zx + c xy) − Sα(x + y + z)((Sαα + Sβγ)x + a Sβy + a Sγz)=0, 2 2 2 2 2 2 2 c a (a yz + b zx + c xy) − Sβ(x + y + z)(b Sαx +(Sββ + Sγα)y + b Sγz)=0, 2 2 2 2 2 2 2 a b (a yz + b zx + c xy) − Sγ(x + y + z)(c Sαx + c Sβ +(Sγγ + Sαβ)z)=0.

Centers:

2 2 2 2 2 2 2 Sβγ(a Sα,b Sβ,c Sγ)+S · Sα(a ,b ,c ), 2 2 2 2 2 2 2 Sγα(a Sα,b Sβ,c Sγ)+S · Sβ(a ,b ,c ), 2 2 2 2 2 2 2 Sαβ(a Sα,b Sβ,c Sγ)+S · Sγ(a ,b ,c ). 422 Tucker circles

10.6.1 The Gallatly circle

We may assume the segments P[b]Q[c], P[c]Q[a], P[a]Q[b] parallel to the sideline. Let P =(x : y : z).

a2yz + b2zx + c2xy − b2x(x + y + z)=0, a2yz + b2zx + c2xy − c2y(x + y + z)=0, a2yz + b2zx + c2xy − a2z(x + y + z)=0. Γ = 1 : 1 : 1 These have → b2 c2 a2 as their common point. The isogonal conju- Γ = 1 : 1 : 1 gate is ← c2 a2 b2 . The pedal circle is the Gallatly circle

( + + +3 2)2( 2 + 2 + 2 ) SAA SBB SCC S a yz⎛ b zx c xy ⎞

⎝ 2 2 ⎠ − (SA + SB + SC)(x + y + z) b c (SAA +2SAB +2SAC + SBC)x =0. cyclic

The radius of the circle is

S · = √ abc 2 2 R 2 2 2 2 2 2 . S +(SA + SB + SC) 2 a b + b c + c a π It is a Tucker circle with Kiepert parameter 2 − ω. The Gallatly circle is the smallest Tucker circle. 10.7 Tucker circles which are cevian circumcircles 423

10.7 Tucker circles which are cevian circumcircles

Consider a Tucker circle which is the cevian circumcircle of P (and its cyclocevian conjugate Q = P o). Suppose the two endpoints of one of the “parallel” sides of the Tucker hexagon are the traces of P . P must lie on a median of T, and the endpoint of this median is a vertex of the hexagon. One of the two sides of the hexagon containing this midpoint must be another “parallel” side of the hexagon, and is parallel to a side of T. This means that one side of the Tucker hexagon is a side of the inferior triangle of T. This is enough to determine the Tucker circle.

2 2 2 2 2 2 2 GbGc (−Sα + Sβ + Sγ)(a Sα,b Sβ,c Sγ)+S (a ,b ,c ), 2 2 2 2 2 2 2 GcGa (Sα − Sβ + Sγ)(a Sα,b Sβ,c Sγ)+S (a ,b ,c ), 2 2 2 2 2 2 2 GaGb (Sα + Sβ − Sγ)(a Sα,b Sβ,c Sγ)+S (a ,b ,c ). These are the circles

4a2(a2yz + b2zx + c2xy) − (x + y + z)(b2c2x + c2(2a2 − b2)y + b2(2a2 − c2)z)=0, 4b2(a2yz + b2zx + c2xy) − (x + y + z)(c2(2b2 − a2)x + c2a2y + a2(2b2 − c2)z)=0, 4c2(a2yz + b2zx + c2xy) − (x + y + z)(b2(2c2 − a2)x + a2(2c2 − b2)y + a2b2z)=0.

These circles are the cevian circumcircles of the following cyclocevian conjugate pairs: (2a2 − b2 : b2 :2a2 − b2), (2a2 − c2 :2a2 − c2 : c2); (2b2 − c2 :2b2 − c2 : c2), (a2 :2b2 − a2 :2b2 − a2); (a2 :2c2 − a2 :2c2 − a2), (2c2 − b2 : b2 :2c2 − b2). Remark. There is a Tucker circle which is a cevian circumcircle of P and P o the endpoints of whose “parallel” (and antiparallel) sides are not traces of the same point P or P o. 424 Tucker circles

10.7.1 Tucker circle congruent to the circumcircle is the one with (a2 + b2 + c2)c2 t = . a2b2 + b2c2 + c2a2 The Tucker hexagon has vertices

(0, −(c4 − a2b2),c2(a2 + b2 + c2)), (−(c4 − a2b2), 0,c2(a2 + b2 + c2)), (−(b4 − c (0,,b2(a2 + b2 + c2), −(b4 − c2a2)), (a2(a2 + b2 + c2), 0, −(a4 − b2c2)), (a2(a2 +

Its center is the point

X(3095) = (a2(a2(b4 +3b2c2 + c4) − (b6 + c6)) : ···: ···).

The Tucker circles are symmetric about the line joining the Brocard points: a4 − b2c2 b4 − c2a2 c4 − a2b2 x + y + z =0. a2 b2 c2

Envelope of the Tucker circles From the equation of the Tucker circles, we obtain the envelope:

4 2 2 2 4 2 2 2 4 2 2 2 Gt := a (b + c ) yz + b (c + a ) zx + c (a + b ) xy − (x + y + z)(b4c4x + c4a4y + a4b4z)=0.

This is the inscribed conic with perspector K and center X(39). This is called the Brocard ellipse. Its foci are the Brocard points. Now,

2 2 2 2 2 2 2 2 2 2 2 4a b Ft = −Gt +(b (c − 2a t)x + a (c − 2b t)y + a b (1 − 2t)z)

This line has inﬁnite point (a2(b2 − c2):b2(c2 − a2):c2(a2 − b2)), and is perpendicular to the Brocard axis. The Tucker circle and the Brocard ellipse are bitangent if they intersect at real points. 10.8 Torres’ circle 425

10.8 Torres’ circle

Let ABC be the triangle of reﬂections. Consider the pedals of A, B, C on the sidelines of T. These are the points

BC CA AB 2 2 2 2 A Ba =(SCC − S :0:2S ),Ca =(SBB − S :2S :0); 2 2 2 2 B Ab =(0:SCC − S :2S ),Cb =(2S : SAA − S :0); 2 2 2 2 C Ac =(0:2S : SBB − S ),Bc =(2S :0:SAA − S ).

The segments BcCb, CaAc, AbBa are parallel to BC, CA, AB respectively. The segments BaCa, CbAb, AcBc are antiparallel to BC, CA, AB respectively. Therefore, these six pedals deﬁne a Tucker circle CTucker (t). The parameter is 1 t = − S2(S + S + S )+S . 2S2 α β γ αβγ

Bc Cb

C A

X(52) H

K O Ba

A C Ac B b

The equation of the circle containing these six pedals is 2 2 2( 2 + 2 + 2 ) a b c a yz b zx c xy 2 2 2 2 − 2S (x + y + z) (Sαα − S )x +(Sββ − S )y +(Sγγ − S )z =0. The center of the circle is the orthocenter of the orthic triangle: 2 2 2 2 2 2 2 2 2 X(52) = (a (Sαα−S )(S +Sβγ):b (Sββ−S )(S +Sγα):c (Sγγ−S )(S +Sαβ)). 426 Tucker circles

10.9 Tucker circles tangent to the tritangent circles

10.9.1 The incircle The radical axis of the Tucker circle C (t) and the incircle is the line Proposition. The Tucker circle C (t) is tangent to the incircle if t is one of the following values: a4 + b4 + c4 − 2(ab + bc + ca)(a2 + b2 + c2 − ab − bc − ca) t0 = , 2(2bc +2ca +2ab − a2 − b2 − c2) a3 − a2(b + c)+a(b2 +4bc + c2) − (b + c)(b2 + c2) t = , a 2(b + c − a) and the two values tb, tc obtained by cyclic permutations of a, b, c.

With t = t0, the equation of the Tucker circle is 4 ( + + )2( 2 + 2 + 2 ) abc a b c a yz b zx c xy ⎛

− (2ab +2bc +2ca − a2 − b2 − c2)(x + y + z) ⎝ bc(b + c − a)(2bc + a(b + c − cyclic The point of tangency is a2(a(b + c) − (b2 + c2))2 b2(b(c + a) − (c2 + a2))2 c2(c(a + b) − (a X(1362) = : : b + c − a c + a − b a + b − The radius of the circle is −(a3b − 2a2b2 + ab3 + a3c − a2bc − ab2c + b3c − 2a2c2 − abc2 − 2b2c2 + ac3 + bc 2abc(2s) r((4R + r)2 + s2) = . 4s2 X(1362) can be constructed as the second intersection of the incircle with the line through Fe and X(354), the centroid of the intouch circle. For the value of ta, the point of tangency with the incircle is 2 2 2 2 2 2 2 2 Ta =(a (a(b+c)−(b +c )) : b (c−a) (b+c−a)(c+a−b): c (a−b) (b+c−a)(a+b−

Together with Tb and Tc, this forms a triangle perspective with ABC at X(3271) = (a2(b−c)2(b+c−a):b2(c−a)2(c+a−b):c2(a−b)2(a+b−c)), 10.9 Tucker circles tangent to the tritangent circles 427 and with perspectrix the line ( − )( + − ) ( − )( + − ) ( − )( + − ) b c b c a x + c a c a b y + a b a b c z =0 a b c which is the line OI. The circle corresponding to ta:

4abc(a2yz + b2zx + c2xy) − (b + c − a)(x + y + z)· (bc(2bc − a(b + c − a))x + ca(2ca − b(b + c − a))y + ab(2ab − c(b + c − a))z)=0.

It has radius a2(b + c) − a(b2 + bc + c2)+bc(b + c) · R 2abc ( − )3 + ( − )( − ) = s a s s b s c 4Δ ( − )2 + 2 = s a ra 4ra This Tucker circle is also tangent to the B- and C-excircles. It is the inversive image of the line BC in the Spieker radical circle.

10.9.2 The excircles (i) a3 + a2(b + c)+a(b2 +4bc + c2)+(b + c)(b2 + c2) t = − , 2(a + b + c)

Point of tangency: (−a2(a(b+c)+(b2 +c2))2 : b2(c+a)2(a+b+c)(a+ b − c):c2(a + b)2(a + b + c)(c + a − b)). With two circles tangent to the B- and C-excircles, these points of tangency are perspective with ABC at a2(b + c)2 b2(c + a)2 c2(a + b)2 X(181) = : : b + c − a c + a − b a + b − c x + y + z =0 and with perspectrix a b c , the trilinear polar of I. Note that X(181) is the Apollonius point of the excircles. This gives the Apollonius circle. 428 Tucker circles

(ii) With t = −(a3 + a2b + ab2 + b3 − a2c − 4abc − b2c + ac2 + bc2 − c3)/(2a +2b − 2c), we have the circle

4abc(a2yz + b2zx + c2xy) − (c + a − b)(x + y + z)· (bc(2bc − a(c + a − b))x + ca(2ca − b(c + a − b))y + ab(2ab − c(c + a − b))z)=0. with radius −(a2b − ab2 − a2c + abc − b2c − ac2 + bc2) s(s − c)(s − a)+(s − b)3 r2 + ·R = = b 2abc 4Δ The point of tangency is

(−a2(b−c)2(a+b+c)(c+a−b):b2(c+a)2(c+a−b)(a+b−c): c2(a2+b2+c(a−b))2).

(iii) With t = −(a3 − a2b + ab2 − b3 + a2c − 4abc + b2c + ac2 − bc2 + c3)/(2a − 2b +2c), we have the circle

4abc(a2yz + b2zx + c2xy) − (a + b − c)(x + y + z)· (bc(2bc − a(a + b − c))x + ca(2ca − b(a + b − c))y + ab(2ab − c(a + b − c))z)=0. with radius a2b + ab2 − a2c − abc − b2c + ac2 + bc2 s(s − a)(s − b)+(s − c)3 r2 +(s − ·R = = c 2abc 4Δ 4rc The point of tangency is

(−a2(b−c)2(a+b+c)(a+b−c):b2(c2+a2−b(c−a))2 : c2(a+b)2(c+a−b)(a+b−c)).

(iv) With t = −(a4 +2a3b +2a2b2 +2ab3 + b4 +2a3c +2a2bc − 2ab2c − 2b3c +2a2c2 − 2abc2 +2b2c2 +2ac3 − 2bc3 + c4)/(2(a2 + b2 + c2 − 2bc + 2ca +2ab)), we have the circle

4abc(b + c − a)2(a2yz + b2zx + c2xy)+(a2 + b2 + c2 − 2bc +2ca +2ab)(x + y + z (bc(a + b + c)(2bc + a(a + b + c))x + ca(a + b − c)(2ca + b(a + b − c))y + ab(c + a − b)(2ab + c(c + a − b))z)=0. with radius a3b +2a2b2 + ab3 + a3c + a2bc − ab2c − b3c +2a2c2 − abc2 +2b2c2 + ac3 − bc3 ·R = 2abc(b + c − a) 10.9 Tucker circles tangent to the tritangent circles 429

The point of tangency is

(−a2(c + a − b)(a + b − c)(b2 + c2 + a(b + c))2 :b2(a + b + c)(c + a − b)(c2 + a2 − b(c − a))2 :c2(a + b + c)(a + b − c)(a2 + b2 + c(a − b))2).

This is 2 2 2 u : v : w a + b + c −a − b + c −a + b − c for the inﬁnite point

(u : v : w)=(−a(b2+c2+a(b+c)) : b(c2+a2−b(c−a)) : c(a2+b2+c(a−b)).

The line through Ia =(−a : b : c) with this inﬁnite point intersects the sideline a at b c (0 : b(a + b):c(c + a)) = 0: : . c + a a + b The common length of the antiparallels is (a + b + c)2 − 4bc . 2(b + c − a) 430 Tucker circles Chapter 11

Some special circles

11.1 The Dou circle

Crux 1140: construction of a circle from which the chords cut out on the sidelines subtend right angles at their opposite vertices. In the published solution [Crux 13 (1987) 232–234], it was established that if P (ρ) is the circle, then 2 = 2 + 2 = 2 + 2 = 2 + 2 ρ PD ha PE hb PF hc where D, E, F are the vertices of the orthic triangle, and ha, hb, hc the altitudes. 2 + 2 = 2 + 2 The locus of point P such that PE hb PF hc is a line perpendicular to EF. This line contains the point (−a2 : b2 : c2), which is a vertex of the tangential triangle. It has equation S2 − S2 S2 − S2 (S − S )x + C y − B z =0. B C b2 c2 Since the tangential triangle is homothetic to the orthic triangle, this line is indeed an altitude of the tangential triangle. Nikolaos Dergiades [Hyacinthos 5815, 7/27/02] has given a simple ver- iﬁcation of the fact that the A-vertex of the tangential triangle lies on this line. It follows that the center of the circle we are seeking is the orthocenter of the tangential triangle. This is the point X155. This is a ﬁnite point if and only if ABC does not contain a right angle. The radius of the circle is the square root of 4 6 2 2 S (4R − R S + SASBSC) 2 . (SASBSC) 432 Some special circles

The equation of the circle is

2 2 2 2 2 2 2SABC(a yz+b zx+c xy)+(x+y+z)( SA(−a SAA+b SBB+c SCC)x)=0.

Suppose the center has homogeneous barycentric coordinates (u : v : w). uS Its distance from BC is (u+v+w)a, and its pedal on BC is the point 2 2 (0 : SCu + a v : SBu + a w).

The square distance from this pedal to A is S2 (S v − S w)2 + B C . a2 a2(u + v + w)2 The square radius of the circle is then 2 2 2 2 (u +(u + v + w) )S +(SBv − SCw) a2(u + v + w)2 Theorem (Brisse). The Dou circle is orthogonal to the circle through the centroid, X111, and the anticomplement of X110.

[This circle intersects the Euler line at X858]. The center of this circle is the point 1 [(b2 − c2)(a2(b2 + c2)+(b4 − 4b2c2 + c4)], which is the superior of the point 2

[(b2 − c2)(b2 + c2 − 2a2)(b2 + c2 − 3a2)].

The equation of this circle is x + y + z (S + S − 2S )(S2 − S2 ) a2yz + b2zx + c2xy + B C A A x =0. 3 (SC − SA)(SA − SB)

1Identiﬁcation number 2.33589872509 ···. 2Identiﬁcation number 2.77610382619 ···. 11.2 The Adams circles 433

11.2 The Adams circles

Construct the three circles each passing through the Gergonne point and tangent to two sides of triangle ABC. The 6 points of tangency lie on a circle. 3

I Ge

B C

the Adams circle. It has radius (4R + r)2 + s2 · r. 4R + r Lemma. Every circle tangent to AB and AC has equation of the form a2yz + b2zx + c2xy − (x + y + z)(t2x +(c − t)2y +(b − t)2z)=0. Proof. The circle touches AB at (c − t : t :0)and AC at (b − t :0:t). There are two values of t for which this circle passes through the Ger- gonne point. These are −a(b + c − a)2 −(b + c − a)(a2 + ab + ac − 2b2 − 2c2 +4bc) t = , a2 + b2 + c2 − 2ab − 2bc − 2ca a2 + b2 + c2 − 2ab − 2bc − 2ca The radius of the second circle is (4R + r)2 +9s2 · r. 4R + r √ (4R+r)2+s2 3 This is called the Adams circle. It is concentric with the incircle, and has radius 4R+r · r. 434 Some special circles

11.3 Hagge circles

Given a point P =(u : v : w) with circumcevian triangle ABC, let A, B, C be the reﬂections of A, B, C in BC, CA, AB respectively. The circle ABC is called the Hagge circle of P . It has center

2 2 2 2 2 2 (−(a vw + b wu + c uv)(S + SBC)+2a S vw : ···: ···), which is the symmetric of the isogonal conjugate P ∗ in the nine-point center. This circle passes through the orthocenter H. The reﬂections of A in the triangle HBC are on the circles ABC, AHC, and ABH respectively. If the line HP intersects the circles HBC, AHC, ABH at X, Y , Z respectively, these are the reﬂections of a point Q on the circumcircle.

11.3.1 Theorem (Luong). Let A, B, C be points on the circumcircle of ABC, and A, B, C be their reﬂections in the respective sides. The circle ABC passes through H if and only if ABC is the circumcevian triangle of a point P . † ∗ If this condition is satisﬁed, the center OP of the circle A B C is N (P ). 11.4 The deLongchamps triad of circles (A(a),B(b),C(c)) 435

11.4 The deLongchamps triad of circles (A(a),B(b),C(c))

Consider the triad of circles (A(a),B(b),C(c)). Let ABC be the superior triangle, and A, B, C the reﬂections of A, B, C respectively in the perpendicular bisectors of BC, CA, AB. These latter are the points A =(−a2 : b2 − c2 : c2 − b2), B =(a2 − c2 : −b2 : c2 − a2), C =(a2 − b2 : b2 − a2 : −c2).

A A B C

Lo C O

B C B

From the obvious incidence relations circle points A(a) B,C,B,C B(b) C,A,C,A C(c) A, B,A,B we obtain the radical axes of the circles: circle radical axis B(b) and C(c) AA C(c) and A(a) BB A(a) and B(b) CC 436 Some special circles

These radical axes are the altitudes of the superior triangle. It follows that the radical center of the triad is the orthocenter of the superior triangle, which is the deLongchamps point Lo. The equations of the circles are as follows. circle equation A(a): a2yz + b2zx + c2xy +(x + y + z)(a2x +(a2 − c2)y +(a2 − b2)z)=0 B(b) a2yz + b2zx + c2xy +(x + y + z)((b2 − c2)x + b2y +(b2 − a2)z)=0 C(c) a2yz + b2zx + c2xy +(x + y + z)((c2 − b2)x +(c2 − a2)y + c2z)=0 11.5 The orthial circles 437

11.5 The orthial circles

Consider the A-orthial triangle in §??. Its circumcircle is tangent to that of ABC at A. Its equation can be written in the form y z a2yz + b2zx + c2xy − k(x + y + z) + =0 b2 c2 for some k. Since it contains Xb =(0:Sγ +Sα : −Sα), we easily determine 2 2 2 k = −a b c Sα . From this, we have the triad of orthial circles: Sβγ

2 2 2 2 + 2 + 2 + a b c Sα ( + + ) y + z =0 a yz b zx c xy x y z 2 2 , Sβγ b c 2 2 2 2 + 2 + 2 + a b c Sβ ( + + ) y + z =0 a yz b zx c xy x y z 2 2 , Sγα b c 2 2 2 2 + 2 + 2 + a b c Sγ ( + + ) y + z =0 a yz b zx c xy x y z 2 2 . Sαβ b c The radical center (x : y : z) is given by 2 2 2 2 2 2 2 2 2 a b c Sα y + z = a b c Sβ z + x = a b c Sγ x + y 2 2 2 2 2 2 , Sβγ b c Sγα c a Sαβ a b 1 1 1 y + z : z + x : x + y = : : 2 2 2 2 2 2 2 2 2 . b c c a a b Sα Sβ Sγ From these,

x : y : z =a2(−S2S2 + S2S2 + S2S2):b2(S2S2 − S2S2 + S2S2):c2(S2S2 + S2S2 − S2S2) β γ γ α α β β γ γ α α β β γ γ α α β 4 4 4 4 4 4 = 2 − a + b + c : 2 a − b + c a ( 2 )2 ( 2 )2 ( 2 )2 b ( 2 )2 ( 2 )2 ( 2 )2 a Sα b Sβ c Sγ a Sα b Sβ c Sγ 4 4 4 : 2 a + b − c c 2 2 2 2 2 2 . (a Sα) (b Sβ) (c Sγ) This is the perspector of the tangential triangle and the circumcevian triangle of O. Corollary. The radical axis of the B- and C-orthial circles contains the antipode of A on the circumcircle, and the intersection of the tangents of the circumcircle at B and C. 438 Some special circles

11.6 Bui’s triad of circles

Let XY Z be the cevian triangle of O, and ABC the circumcevian triangle. The circle X(A) has equation

2 2 2 1 2 S (S + SBC)(a yz + b 2zx + c xy) 2 2 2 2 2 2 − a SA(x + y + z)(a b c x + c SCCy + b SBBz)=0.

This circle intersects BC at two points. Similarly for the other two circles. The six points lie on the conic ⎛ ⎞

2 2 2 ⎝ 2 2 ⎠ S a SA(S + SBC)yz − (x + y + z) b c SBBSCCx =0. cyclic cyclic

This conic has center O. It is homothetic to the circumconic with center N.

C B

X B C

This is true for points on the Euler line. 11.7 Appendix: More triads of circles 439

11.7 Appendix: More triads of circles

11.7.1 The triad {A(AH )} The equation of the A-circle:

2 2 2 2 2 −a (a yz + b zx + c xy)+(x + y + z)(−S x + SBBy + SCCz)=0. The radical center is therefore given by −S2x + S y + S z S x − S2y + S z S x + S y − S2z BB CC = AA CC = AA BB . a2 b2 c2 This is

2 4 2 4 2 4 x : y : z = a (S − SAABC):b (S − SBBCA : c (S − SCCAB), the center of the Taylor circle.

1. (a) Construct the circle tangent to the circumcircle internally at A and also to the side BC. (b) Find the coordinates of the point of tangency with the side BC. (c) Find the equation of the circle. 4 (d) Similarly, construct the two other circles, each tangent internally to the circumcircle at a vertex and also to the opposite side. (e) Find the coordinates of the radical center of the three circles. 5 2. Construct the three circles each tangent to the circumcircle externally at a vertex and also to the opposite side. Identify the radical center, which is a point on the circumcircle. 6 3. Let X, Y , Z be the traces of a point P on the side lines BC, CA, AB of triangle ABC. (a) Construct the three circles, each passing through a vertex of ABC and tangent to opposite side at the trace of P . (b) Find the equations of these three circles.

2 4 2 2 2 a 2 2 a yz + b zx + c xy − (b+c)2 (x + y + z)(c y + b z)=0. 5 2 2 (a (a + a(b + c) − bc):···: ···). This point appears as X595 in ETC. 2 2 2 6 a b c b−c : c−a : a−b . This point appears as X110 in ETC. 440 Some special circles

4. Find the equations of the three circles each through a vertex and the traces of the incenter and the Gergonne point on the opposite side. What is the radical center of the triad of circles? 7

5. Let P =(u : v : w). Find the equations of the three circles with the cevian segments AAP , BBP , CCP as diameters. What is the radical center of the triad ? 8

6. Given a point P . The perpendicular from P to BC intersects CA at Ya and AB at Za. Similarly deﬁne Zb, Xb, and Xc, Yc. Show that the circles AYaZA, BZbXb and CXcYc intersect at a point on the circumcircle of ABC. 9

Exercises

1. Consider triangle ABC with three circles A(Ra), B(Rb), and C(Rc). The circle B(Rb) intersects AB at Za+ =(Rb : c − Rb :0)and Za− =(−Rb : c + Rb :0). Similarly, C(Rc) intersects AC at Ya+ = 10 (Rc :0:b − Rc) and Ya− =(−Rc :0:b + Rc).

(a) Show that the centers of the circles AYa+Za+ and AYa−Za− are symmetric with respect to the circumcenter O. 11 (b) Find the equations of the circles AYa+Za+ and AYa−Za−. (c) Show that these two circles intersect at −a2 b −c Q = : : bRb − cRc Rb Rc on the circumcircle.

7The external center of similitude of the circumcircle and incircle. 8Floor van Lamoen, Hyacinthos, message 214, 1/24/00. 2 2 2 9 =( : : ) ( a : b : c ) If P u v w , this intersection is vSB −wSC wSC −uSA uSA−vSB ; it is the inﬁnite point of the line perpendicular to HP. A.P. Hatzipolakis and P. Yiu, Hyacinthos, messages 1213, 1214, 1215, 8/17/00. 10A.P. Hatzipolakis, Hyacinthos, message 3408, 8/10/01. 11 2 2 2 a yz + b zx + c xy − (x + y + z)(c · Rby + b · Rcz)=0for = ±1. 11.7 Appendix: More triads of circles 441

(d) Find the equations of the circles AYa+Za− and AYa−Za+ and show that they intersect at −a2 b c Q = : : bRb + cRc Rb Rc on the circumcircle. 12 (e) Show that the line QQ passes through the points (−a2 : b2 : c2) and 13 =( 2(− 2 2 + 2 2 + 2 2):···: ···) P a a Ra b Rb c Rc .

(f) If W is the radical center of the three circles A(Ra), B(Rb), and C(Rc), then P =(1− t)O + t · W for 2 2 2 2 = a b c t 2 2 + 2 2 + 2 2 . Raa SA Rb b SB Rcc SC

14 (g) Find P if Ra = a, Rb = b, and Rc = c. 15 (h) Find P if Ra = s − a, Rb = s − b, and Rc = s − c.

(i) If the three circles A(Ra), B(Rb), and C(Rc) intersect at W = (u : v : w), then

2 2 2 2 2 2 2 P =(a (b c u − a SAvw + b SBwu + c SCuv):···: ···).

(j) Find P if W is the incenter. 16 (k) If W =(u : v : w) is on the circumcircle, then P = Q = Q = W .

2. Given triangle ABC, construct a circle Ca tangent to AB at Za and AC at Ya such that YaZa passes through the centroid G. Similarly construct 17 the circles Cb and Cc. What is the radical center of the three circles?

12 2 2 2 a yz + b zx + c xy − (x + y + z)(c · Rby − b · Rcz)=0for = ±1. 13 2 2 2 2 2 2 2 QQ :(b Rb − c Rc )x + a (Rb y − Rc z)=0. 14 2 4 4 4 2 4 4 4 2 4 4 4 (a (b + c − a ):b (c + a − b ):c (a + b − c )). This point appears as X22 in ETC. 2 2 2 2 15 a (a −2a(b+c)+(b +c )) ( s−a : ···: ···). This point appear in ETC as X1617. 2 2 2 16 a b c ( s−a : s−b : s−c ). 17h(G, 2)(I). See Problem 2945, Crux Math., 30 (2004) 233. 442 Some special circles

11.7.2 The triad of circles (AG(AH ),BG(BH ),CG(CH ))

Consider the circle whose center is the midpoint AG of BC, passing through 1 | 2 − 2| the pedal AH on BC. This circle has radius 2a b c . The power of A with respect to this circle

(b2 − c2)2 a2 (b2 − c2)2 S = m2 − = S + − = S + BC . a 4a2 A 4 4a2 A a2 Those of B and C are each equal to

a2 (b2 − c2)2 S − = BC . 4 4a2 a2 The equation of this circle is therefore

S a2YZ+ b2ZX + c2XY S X + BC (X + Y + Z)= . A a2 X + Y + Z Similarly, we write down the equations of the other two circles, and ﬁnd the radical center of the three circles by solving

S S S S X+ BC (X+Y +Z)=S Y + CA(X+Y +Z)=S Z+ AB (X+Y +Z). A a2 B b2 C c2 This gives

: : = 2 ( 2 2 + 2 2 ): 2 ( 2 2 + 2 2 ): 2( 2 2 + 2 2 )) X Y Z a SA b SB c SC b SB c SC a SA c a SA b SB .

This is the point X185 in Kimberling’s list. It is the Nagel point of the orthic triangle! The radius of the orthogonal circle is

S abc cos A cos B cos C ABC = =4R cos A cos B cos C, abcS S the diameter of the incircle of the orthic triangle. This is the incircle of the antimedial triangle of the orthic triangle. Therefore it touches the nine-point circle. The point of tangency is the Jerabek point. 11.8 The triad of circles (AG(AH ),BG(BH ),CG(CH )) 443

11.8 The triad of circles (AG(AH),BG(BH),CG(CH))

Consider the circle whose center is the midpoint AG of BC, passing through 1 | 2 − 2| the pedal AH on BC. This circle has radius 2a b c . The power of A with respect to this circle (b2 − c2)2 a2 (b2 − c2)2 S = m2 − = S + − = S + BC . a 4a2 A 4 4a2 A a2 Those of B and C are each equal to a2 (b2 − c2)2 S − = BC . 4 4a2 a2 The equation of this circle is therefore S a2YZ+ b2ZX + c2XY S X + BC (X + Y + Z)= . A a2 X + Y + Z Similarly, we write down the equations of the other two circles, and ﬁnd the radical center of the three circles by solving

S S S S X+ BC (X+Y +Z)=S Y + CA(X+Y +Z)=S Z+ AB (X+Y +Z). A a2 B b2 C c2 This gives : : = 2 ( 2 2 + 2 2 ): 2 ( 2 2 + 2 2 ): 2( 2 2 + 2 2 )) X Y Z a SA b SB c SC b SB c SC a SA c a SA b SB .

This is the point X185 in Kimberling’s list. It is the Nagel point of the orthic triangle! The radius of the orthogonal circle is S abc cos A cos B cos C ABC = =4R cos A cos B cos C, abcS S the diameter of the incircle of the orthic triangle. This is the incircle of the antimedial triangle of the orthic triangle. Therefore it touches the nine-point circle. The point of tangency is the Jerabek point. 444 Some special circles

11.9 The Lucas circles

Consider the square AbAcAcAb inscribed in triangle ABC, with Ab, Ac on BC. Since this square can be obtained from the square erected externally ( S ) C on BC via the homothety h A, a2+S , the equation of the circle A through A, Ab and Ac can be easily written down: a2 C : a2yz + b2zx + c2xy − · (x + y + z)(c2y + b2z)=0. A a2 + S

Figure 11.1: Lucas circle

The center of the circle CA is the point 2 2 2 Oa =(a (SA +2S):b SB : c SC). 11.9 The Lucas circles 445

Figure 11.2: Radical center of Lucas circles

Figure 11.3: Radical center of Lucas circles 446 Some special circles

Likewise if we construct inscribed squares BcBaBaBc and CaCbCbCa on the other two sides, the corresponding Lucas circles are b2 C : a2yz + b2zx + c2xy − · (x + y + z)(c2x + a2z)=0, B b2 + S and c2 C : a2yz + b2zx + c2xy − · (x + y + z)(b2x + a2y)=0. C c2 + S The coordinates of the radical center satisfy the equations a2(c2y + b2z) b2(a2z + c2x) c2(b2x + a2y) = = . a2 + S b2 + S c2 + S Since this can be rewritten as y z z x x y + : + : + = a2 + S : b2 + S : c2 + S, b2 c2 c2 a2 a2 b2 it follows that x y z : : = b2 + c2 − a2 + S : c2 + a2 − b2 + S : a2 + b2 − c2 + S, a2 b2 c2 and the radical center is the point 2 2 2 (a (2SA + S):b (2SB + S):c (2SC + S)). This is K∗(arctan 2) on the Brocard axis. 18 The three Lucas circles are mutually tangent to each other, the points of tangency being 2 2 2 A =(a SA : b (SB + S):c (SC + S)), 2 2 2 B =(b (SA + S):b SB : c (SC + S)), 2 2 2 C =(a (SA + S):b (SB + S):c SC). These point of tangency form a triangle perspective with ABC at the point 19 ∗ π =( 2( + ): 2( + ): 2( + )) K 4 a SA S b SB S c SC S .

Exercise

1. Show that the Lucas circles CB and CC also touch the A-Apollonian circle at A.

18 This appears as X1151 in ETC. 19 This point appears in ETC as X371, and is called the Kenmotu point. Chapter 12

The triangle of reﬂections

12.1 The triangle of reﬂections T†

The triangle of reﬂections T† has vertices the reﬂections of A, B, C in their opposite sides: † := † † := † † := † A Aa,BBb ,CCc .

H N

B C Ga

† Oa N† † a Ha

A† −→ −→ −→ H =2OG = OO† OO† H Since A a a, a A is a parallelogram. From this simple fact, we deduce several interesting results. OO† H† H OH† = 1. Note that a a is a symmetric trapezoid. This means that a O† H = O H† T a A. Therefore, a lies on the circumcircle of ; similarly for H† H† H b and c. The reﬂection triangle and the circumcevian triangle of coincide. 448 The triangle of reﬂections

2. Since the midpoint of OH is the nine-point center N, this is also the O† midpoint of A a. From this we conclude that the reflection triangle of O is oppositely congruent to T at N. O † O† N† O † 3. Since A is the reflection of aA in BC, a is the midpoint of A . N† N† O † O † Similarly, b and c are the midpoints of B and C . From this we conclude that the triangle of reflections T† is the image of the reflection triangle of N under the homothety h(O, 2). B† A

C† N O H

B C

† Na

A† 4. The circumcenter of T† is the reﬂection of O in N∗.

Proof. The circumcenter of the reﬂection triangle of a point P is the isogonal conjugate P ∗. Since T† is the image of the reﬂection triangle of N under h(O, 2), the circumcenter of T† is the point h(O, 2)(N∗)=2N∗ − O.

5. AN∗ is perpendicular to B†C†.

Proof. B†C† is parallel to the a-side of the reﬂection triangle of N, which is orthogonal to the line AN∗.

Exercise 1. Show that the triangle of reﬂections is the image of the pedal triangle of N under the homothety h(G, 4). 12.2 Triangles with degenerate triangle of reﬂections 449

12.2 Triangles with degenerate triangle of reﬂections

Proposition. The triangle of reﬂections is degenerate if and only if the nine- point center lies on the circumcircle.

N A

Given a circle O(R) and a point N on its circumference, let H be the reﬂection of O in N. For an arbitrary point P on the minor arc of the circle R N( 2 ) inside O(R), let (i) A be the intersection of the segment HP with O(R), (ii) the perpendicular to HP at P intersect O(R) at B and C. Then triangle ABC has nine-point center N on its circumcircle O(R). It is clear that O(R) is the circumcircle of triangle ABC. Let M be the midpoint of BC so that OM is orthogonal to BC and parallel to PH. Thus, OMPH is a (self-intersecting) trapezoid, and the line joining the midpoints of PM and OH is parallel to PH. Since the midpoint of OH is N and PH is orthogonal to BC, we conclude that N lies on the perpendicular bisector of R R PM. Consequently, NM = NP = 2 , and M lies on the circle N( 2 ). This circle is the nine-point circle of triangle ABC, since it passes through the R pedal P of A on BC and through the midpoint M of BC and has radius 2 .

Exercise Suppose the nine-point center N of triangle ABC lies on the circumcircle. In this case, T† degenerates into a line L. 450 The triangle of reﬂections

1. If X, Y , Z are the centers of the circles BOC, COA, AOB, the lines AX, BY , CZ are all perpendicular to L. 2. The circles AOA†, BOB†, COC† are mutually tangent at O. The line joining their centers is the parallel to L through O. 3. The circles AB†C†, BC†A†, CA†B† pass through O. 12.3 Triads of concurent circles 451

12.3 Triads of concurent circles

12.3.1 Musselman’s theorem Theorem (Musselman). The circles AOA†, BOB†, and COC† have a second common point, the inversive image of N∗ in the circumcircle.

N O

C B

A∗ † Oa

A†

Proof. Let A∗, B∗, C∗ be the inverses of A†, B†, C† in the circumcircle. It is enough to show that the lines AA∗, BB∗, CC∗ are concurrent in N∗. 1 Since OA† · OA∗ = OA2, the line OA is tangent to the circle AA∗A† at A. ∠O ∗ = ∠ ∗ † = ∠O † = ∠O† † AA A A A A A aAA . It follows that ∠ ∗ = ∠ O + ∠O ∗ = ∠ † + ∠ † O† = ∠ O† CAA CA AA BAA A A a BA a. ∗ O† Therefore, AA and A a are isogonal lines with respect to AC and AB. O† N ∗ Since A a contains the nine-point center , the line AA contains its isogonal conjugate N∗. Similarly, the lines BB∗ and CC∗ also contain N∗. The three lines therefore are concurrent at N∗.

1The isogonal conjugate of the nine-point center, N∗, is usually called the Kosnita point. 452 The triangle of reﬂections

Corollary. Let AO, BO, CO be the centers of the circles OBC, OCA, OAB ∗ respectively. The lines AAO, BBO, CCO are concurrent at N . † Proof. The center AO is the inverse of Oa in the circumcircle. By symmetry † O † in BC, the point Oa lies on the circle A A . Therefore, the line AAO is the inverse of this circle, and it contains the point N∗ by Musselman’s theorem. 12.3 Triads of concurent circles 453

12.3.2 The triad of circles AB†C†, BC†A†, CA†B† Since the reﬂections of the orthocenter H in the sidelines of T lie on the circumcircle, the circumcircles of A†BC, AB†C, and ABC† intersect at the orthocenter H. By the dual triads of circles theorem, the circles AB†C†, A†BC†, A†B†C also have a common point. Theorem. The circles AB†C†, BC†A†, CA†B† are concurrent at the inverse of N∗ in the circumcircle.

B† Q

† C A

N∗

O B C

Proof. Let Q be the common point of the circles BOB† and COC† apart from O. We prove that Q lies on the circle AB†C†, by calculating direct angles, with equalities modulo π:

∠(QB†,QC†) = ∠(QB†,QO)+∠(QO,QC†) = ∠(BB†,BO)+∠(CO,CC†) = ∠(BB†,BC)+∠(BC, BO)+∠(CO,BC)+∠(BC, CC†) = ∠(BB†,CC†)+∠(CO,BO) = ∠(BH,CH)+2∠(CA,BA) =(−α)+2(−α) =(−α)+(−α)+(−α) = ∠(AB†,AC)+∠(AC, AB)+∠(AB, AC†) = ∠(AB†,AC†). 454 The triangle of reﬂections

Therefore, B†, Q, A, C† are concyclic. Equivalently, Q lies on the circle AB†C†. By Musselman’s theorem, Q is also lies on the circle AOA†, and is the inverse of N∗ in the circumcircle. The same reasoning shows that Q also lies on the circles BC†A† and CA†B†, and is a common point of the three circles.

Exercise 1. The centers of the circles A†BC, AB†C and ABC† form a triangle homothetic to ABC. Identify the center of homothety. 2 2. Let P be the center of the circle A†B†C†, and M the midpoint of OP. Identify the isogonal conjugate of M. 3

2 X5. 3The nine-point center. M = N∗. 12.4 Triangle of reﬂections and cev−1(O) 455

12.4 Triangle of reﬂections and cev−1(O)

Theorem. The triangle of reflections and the anticevian triangle of the circumcenter are perspective at the reflection of the circumcenter in the Euler reflection point. † −1 † ∧(T , cev (O)) = OE.

B† A

C† Y

H O

B C

A†

Proposition. If the line OA intersects BC at P and the reﬂection of the OQ = OA Euler line in BC in Q, then QP OP .

Proof. Applying Menelaus’ theorem to triangle AP D with transversals XO 456 The triangle of reﬂections

X B D P C

H†

O†

Q and XQ respectively, we have PX DH AO · · = − 1, (12.1) XD HA OP DX PQ AH† · · = − 1. (12.2) XP QA H†D PX · DX =1 = † Note that XD XP , and since DH H D,

DH AH† AH† AH† QA · = = = . HA H†D HA O†O OQ Multiplying the two equations (12.1) and (12.2), we have, after obvious cancellations, PQ AO · =1. OQ OP From this the result follows. Proposition. Let T be the reﬂection of O in the Euler reﬂection point E. The line A†T is parallel to H†E.

Proof. AH = OO† = H†A† = EY †. A†Y † is parallel to HE and intersects the extension of OE at T . 12.4 Triangle of reﬂections and cev−1(O) 457 Y A

B D C H† E

O†

A† Y † † AZ = − AP Proposition. The lines A T and OA intersect at Z such that ZP PO. Proof. Since A†T is parallel to H†E and OT =2· OE, OZ =2· OQ. AO = OQ Since OP PQ,

AO OQ AO +2· OQ AO +2· OQ AZ AZ = = = = = − . OP PQ OP +2· PQ 2 · OQ − OP PZ ZP This shows that Z and O divide OP harmonically, and Z is the A-vertex of the anticevian triangle of O. Theorem. The triangle of reﬂections is perspective to the anticevian triangle of the nine-point center. The line joining corresponding vertices are parallel to the Euler line. † −1 ∧(T , cev (N)) = E∞. 458 The triangle of reﬂections

† 12.4.1 The Parry reflection point OE † The point OE is also called the Parry reflection point. Theorem (C. F. Parry). 4 Construct parallel lines through the vertices of ABC, and reflect them in the corresponding sidelines. The three reflection lines are concurrent if and only if they are reflections of lines parallel to the † Euler line. In this case, the point of concurrency is OE. Proof. Consider a line ux + vy + wz =0. The parallel through A is the line (u − v)y +(u − w)z =0. Its reflection in BC is the line

(2a2u − (a2 + b2 − c2)v − (c2 + a2 − b2)w)x + a2(u − v)y + a2(u − w)z =0.

Similarly we find the equations of the reflections in CA, AB of the parallels through B and C respectively. These three reflections are concurrent if and only if (2a4 − a2(b2 + c2)+(b2 − c2)2)u (b2 + c2 − a2)(v − w)2 =0.

The second factor is always nonzero unless u = v = w. Thus, the line ux + vy + wz =0contains the point

(2a4 − a2(b2 + c2)+(b2 − c2)2 : ···: ···), which is the inﬁnite point of the Euler line. It is parallel to the Euler line.

Find a synthetic proof for the intersection to be the reﬂection of O in E.

4Amer. Math. Monthly, 105 (1998) 68; solution, ibid. 106 (1999) 779–780. 12.5 Triangle of reﬂections and the excenters 459

12.5 Triangle of reﬂections and the excenters

12.5.1 The Evans perspector Theorem (Evans). The triangle of reﬂections is perspective with the excentral triangle. The perspector is the point (a(a3 + a2(b + c) − a(b2 + bc + c2) − (b + c)(b − c)2):···: ···)

† Proof. The line joining A to the excenter Ia =(−a : b : c) has equation (b−c)(a+b+c)(b+c−a)x−a(c2−ca+a2−b2)y+a(a2−ab+b2−c2)z =0. This line intersects the OI line bc(b − c)(b + c − a)x + ca(c − a)(c + a − b)y + ab(a − b)(a + b − c)z =0 at W =(a(a3 + a2(b + c) − a(b2 + bc + c2) − (b + c)(b − c)2):···: ···). The coordinates are symmetric in a, b, c. Therefore the three lines are concurrent.

This is called the Evans perspector Ev. Note that the reﬂection of the incenter in the circumcenter is the point † 3 2 2 2 IO =2O − I =(a(a + a (b + c) − a(b + c) − (b + c)(b − c) ):···: ···) The sum of the coordinates is −4S2.

Proposition. Ev is the inverse of I in the circumcircle of the excentral triangle cev−1(I).

Exercise

1. Let X be the reflection of the excenter Ia in BC; similarly define Y and Z. Calculate the coordinates of X, Y , Z. Show that XY Z is perspective with ABC. 5 2. Let P be the perspector in the preceding exercise. Show that the cevian and the reflection triangles of P are perspective at the Evan perspector.

5 The perspector is X80. 460 The triangle of reﬂections

† † † Proposition. The circles A IbIc, IaB Ic, IaIbC have the Parry reflection point as a common point. † † † † † † Proposition. The circles IaB C , A IbC and A B Ic have a common point. Their centers are perspective with ABC at a point on the OI line. Proposition. The reflections of the excenters in the respective sides of T form a triangle perspective with T at I†. Proposition. The triangle of reflections is perspective with the Kiepert tri- π angle K(θ) if and only if θ = ± 3 .

12.5.2 Triangle of reﬂections and the tangential triangle Let XY Z be the tangential triangle. The circles XB†C†, A†YC†, A†B†Z have a common point, the Parry † reﬂection point OE. What is the common point of the triad of circles A†YZ, B†ZX, C†XY ?

A† =(−a2 : a2 + b2 − c2 : c2 + a2 − b2), B† =(a2 + b2 − c2 : −b2 : b2 + c2 − a2), C† =(c2 + a2 − b2 : b2 + c2 − a2 : −c2) which are The distance from A† to O is the same as that of the reﬂection of O in BC † † to A. More directly, in triangle OBA ,wehaveOB = R, BA = BA = c, † π π and ∠OBA = β + 2 − α = 2 − (α − β). This square distance is 2 + 2 − 2 cos π − ( − ) R c cR 2 α β = R2 + c2 − 2cR sin(α − β) = R2 + c2 − c(a cos β − b cos α) 1 1 = 2 + 2 − ( 2 + 2 − 2)+ ( 2 + 2 − 2) R c 2 c a b 2 b c a = R2 + b2 + c2 − a2.

Therefore the inverse is the point which divides OA† in the ratio

OA∗ : A∗A† = R2 :(R2 + b2 + c2 − a2) − R2 = R2 : b2 + c2 − a2 12.5 Triangle of reﬂections and the excenters 461

1 A∗ = (b2 + c2 − a2)O + R2 · A† . R2 + b2 + c2 − a2

† 2 2 2 2 2 2 2 2 A =(−a :2Sγ :2Sβ)=(−a : a + b − c : c + a − b ) with coordinate sum a2.

2S R2 α (a2S ,b2S ,c2S )+ (−a2, 2S , 2S ). 2S2 α β γ a2 γ β Since 2RS = abc, the homogeneous coordinates of A∗ can be taken as

2 2 2 2 2 2 4Sα(a Sα,b Sβ,c Sγ)+b c (−a , 2Sγ, 2Sβ) 2 2 2 2 2 2 =(∗∗∗:4b Sαβ +2b c Sγ :4c Sγα +2b c Sβ) 2 2 2 2 ∼ (∗∗∗: b (2Sαβ + c ):c (2Sγα + b Sβ)) ∼ (∗∗∗: 2( 2 + ): 2( 2 + )) b S Sαβ c S Sγα 2 2 ∼ ∗∗∗: b : c 2 2 . S + Sγα S + Sαβ It is clear that AA∗, BB∗, CC∗ intersect at the isogonal conjugate of the nine-point center. The inverse in the circumcircle is the common point of the circles AOA†, BOB†, COC†.

Proposition. The triangle of reﬂections is perspective to the anticevian triangle of P if and only if P lies on the (Napoleon) isogonal cubic with pivot N, the nine-point center.

P perspector IX484Evans perspector N Euler inﬁnity point O Parry reﬂection point ∗ X54 = N X1157 = inversive image of X54 in circumcircle X195 O

The excenters are also on the Napoleon cubic. Let W = X484 be the Evans perspectors. For each of the excenters, the anticevian triangle is also 462 The triangle of reﬂections

point anticevian triangle perspector ∗ Ia AIcIb Wa ∗ Ib IcBIa Wb ∗ Ic IbIaCWc Chapter 13

Circumconics

In this chapter we introduce some basic examples of circumconics as isogonal (and isotomic) conjugates of lines. We show how to determine the center of a circumconic, and prove that among the circumconics, the rectangular hyperbolas are characterized as those passing through the orthocenter of the reference triangle, and that their centers are on the nine- point circle. One of the most remarkable results in this chapter is that a circum-hyperbola is uniquely determined by an asymptote. Throughout this

chapter, we highlight specific examples and explicit geometric constructions. We give a large number of examples of circum-hyperbolas with two given infinite points, those with a specified asymptote, and the rectangular circum-hyperbola through a given point. We study a large number of con- crete examples. Apart from the classical ones, the Steiner circum-ellipse and the Kiepert, Jerabek, and Feuerbach hyperbolas, we also study the rectangular circum-hyperbolas through the Steiner point, the Euler reflection points etc. We also study some important triangle centers assoicated with them. For example, we prove important properties of the Steiner point, the Kiepert center, and the Schiffler point. We also study the focus of a circum- parabola with a specified direction for the axis. The last section on general conics actually belongs to another chapter, but is included here to give a flavor of the treatment of conics in the book. 502 Circumconics

13.1 The perspector of a circumconic

A homogeneous quadratic equation 2 2 2 a11x + a22y + a33z +2a23yz +2a13zx +2a12xy =0 in barycentric coordinates (x : y : z) with reference to T represents a conic. The conic passes through the vertices of T if and only if a11 = a22 = a33 = 0. In this case we call it a circumconic and write its equation in the form pyz + qzx + rxy =0. (13.1)

Z A Y

B C

Figure 13.1: The circumconic Cp(P ) with perspector P

The tangents to the conic at the vertices of T are the lines ry + qz =0,rx+ pz =0,qx+ py =0. They bound a triangle whose vertices are the points X =(−p : q : r),Y=(p : −q : r),Z=(p : q : −r), forming the anticevian triangle of P =(p : q : r). For this reason we call P the perspector of the circumconic, and denote the conic by Cp(P ).For example, the circumcircle a2yz + b2zx + c2xy =0 has perspector K =(a2 : b2 : c2), the symmedian point, since the tangential triangle is the anticevian triangle cev−1(K). By rewriting (13.1) in the form p q r + + =0, (13.2) x y z we note an interesting property of the perspector of a circumconic: the trilinear polar for every point on Cp(P ) passes through the perspector P . 13.2 Circumconics as isotomic and isogonal conjugates of lines 503

13.2 Circumconics as isotomic and isogonal conjugates of lines

The circumconic Cp(P ) is the image of the line px + qy + rz =0un- der isotomic conjugation. We shall simply say that Cp(P ) is the isotomic conjugate of the line. For example, the circumconic 1 1 1 Cp(G): + + =0 x y z is the isotomic conjugate of the line at inﬁnity L∞ : x + y + z =0. It is the Steiner circum-ellipse (see §13.5 below). This is an ellipse because it does not contain an inﬁnite point. 1 The circumcircle

2 2 2 Cp(K): a yz + b zx + c xy =0 is the isotomic conjugate of the line a2x + b2y + c2z =0. 2 More generally, • • Cp(P )=(L (P )) .

We shall, however, more often regard circumconics as isogonal conjugates of lines instead of isotomic conjugates. Thus, Cp(P ) is the isogonal conjugate of the line px qy rz + + =0. a2 b2 c2 For example, the circumcircle is the isogonal conjugate of the line at inﬁnity. More generally, ∗ ∗ Cp(P )=(L (P )) .

We deﬁne a few basic circumconics as isogonal conjugates of lines.

1 Proof: If (u : v : w) is an inﬁnite point on Cp(G), then u + v + w =0and uv + vw + wu =0. Eliminating w, we obtain u2 + uv + v2 =0, an impossibility for u, v ∈ R. 2The equation of the circumcircle is established in an earlier chapter. 504 Circumconics

Circumconic as isogonal conjugate of as L • OK GK Kiepert hyperbola Brocard axis b2−c2 =0 2 2( 2 − 2) =0 cyclic x cyclic b c b c x OH HH• Jerabek hyperbola Euler line a2(b2−c2)(b2+c2−a2) =0 ( 2 − 2)( 2 + 2 − 2) =0 cyclic x cyclic b c b c a x OI G N Feuerbach hyperbola line e a a(b−c)(b+c−a) =0 ( − )( + − ) =0 cyclic x cyclic bc b c b c a x G K GK GK• through and a2(b2−c2) =0 ( 2 − 2) =0 cyclic x cyclic b c x Each of these circumconics, being the isogonal conjugate of a line through the circumenter O, is a hyperbola. It contains two inﬁnite points which are the isogonal conjugates of the intersections of the line with the circumcircle (see §??). These hyperbolas can also be regarded as isotomic conjugates of lines indicated in the rightmost column in the table.

13.2.1 The fourth intersection of two circumconics

If we regard two circumconics as the isogonal conjugates of two lines L1 and L2, then, apart from the vertices of T, they intersect at the isogonal conjugate of L1 ∩ L2. We call this the fourth intersection of the two circumconics. ∗ ∗ ∗ L1 ∩ L2 = {A, B, C, (L1 ∩ L2) }. In particular, the fourth intersection of the circumconic L ∗ with the circumcircle is the isogonal conjugate of the inﬁnite point of L .

Circumconic Fourth intersection with circumcircle line S := 1 : 1 : 1 L (K) Steiner circumellipse Steiner point t b2−c2 c2−a2 a2−b2 1 Kiepert hyperbola Tarry point Ta := 4 4 2 2 2 : ···: ··· OK b +c −a (b +c ) 2 Jerabek hyperbola X(74) := a : ···: ··· OH SA(SB+SC )−2SBC a Feuerbach hyperbola X(104) := 2 2 : ···: ··· OI a (b+c)−2abc−(b+c)(b−c) G K P := a2 : ···: ··· GK through and Parry point a b2+c2−2a2 Analogous results hold if we treat the circumconics as isotomic conjugates of lines. 13.2 Circumconics as isotomic and isogonal conjugates of lines 505

Example. (Steiner and Tarry points) These are the intersections of the circumcircle with the Steiner circum-ellipse and the Kiepert hyperbola respec- L (K): x + y + z =0 tively. Since the Lemoine axis a2 b2 c2 is the polar of K with respect to the circumcircle, it is orthogonal to the Brocard axis OK. Therefore, the Steiner and the Tarry points are antipodal on the circumcircle (see Figure 13.2).

A Ta

(GK∞)•

K O G H

B C

Figure 13.2: The Steiner circum-ellipse and the Kiepert hyperbola

The Kiepert hyperbola intersects the Steiner circum-ellipse at 1 1 1 (GK∞)• = : : . b2 + c2 − 2a2 c2 + a2 − 2b2 a2 + b2 − 2c2 506 Circumconics

13.3 Circumconic with a given center

Given a (finite) point Q =(u : v : w), we compute the equation of the circumconic which has Q as center. We denote this conic by Co(Q).It clearly contains the reflections of A, B, C in Q, namely, the points Q†(A)= (−(v + w − u): 2v :2w), Q†(B)= (2u : −(w + u − v): 2w), Q†(C)= (2u :2v : −(u + v − w)). The isotomic conjugates of these points are collinear. The line containing them is u(v + w − u)x + v(w + u − v)y + w(u + v − w)z =0, as is easily verified. Therefore, Co(Q) is the conic u(v + w − u) v(w + u − v) w(u + v − w) + + =0. x y z Note that the perspector is the cevian quotient G/Q. Therefore, we have established Co(Q)=Cp(G/Q). (13.3)

Proposition. For Q =(u : v : w), the fourth intersection of Co(Q) with (a) the circumcircle is the point 1 (L ((G/Q)∗)∞)∗ = : ···: ··· , b2w(u + v − w) − c2v(w + u − v) C ( ) u : v : w (b) the circumconic p Q is v−w w−u u−v . Example. (The superior of the Feuerbach point) The circum-ellipse Co(I) intersects the circumcircle and Cp(I) at the same point a b c : : . b − c c − a a − b The inferior of this point has coordinates b + c : c + a : a + b c − a a − b a − b b − c b − c c − a =(b(a − b)(b − c)+c(b − c)(c − a):c(b − c)(c − a)+a(c − a)(a − b) : a(c − a)(a − b)+b(a − b)(b − c)) =((b − c)2(b + c − a):(c − a)2(c + a − b):(a − b)2(a + b − c)). 13.3 Circumconic with a given center 507

This is the Feuerbach point Fe, the point of tangency of the nine-point circle and the incircle. Therefore, Co(I) intersects the circumcircle at sup(Fe). −1 Since Co(O) is the nine-point circle of the superior triangle cev (G), this point is the point of tangency with the incircle of the superior triangle, which has center Na (see Figure 13.3).

Gb Gc A

† I†(C) I (B)

I Na G

B C

I†(A)

sup(Fe)

Figure 13.3: Co(I) and Cp(I) intersect at sup(Fe) on the circumcircle

Remark. If Q =(u : v : w) is an inﬁnite point, i.e., u + v + w =0, the points Q†(A), Q†(B), Q†(C) all coincide with Q. Nevertheless, we may still speak of the circumconic Co(Q) with equation

u2 v2 w2 + + =0. x y z

Since this conic has a unique inﬁnite point, namely, Q =(u : v : w),itis a parabola. 3 We shall continue to speak of Q as the center of the parabola, though more accurately it is the inﬁnite point of the axis of the parabola. Circum-parabolas are discussed in more details in §??.

3Proof: Let (x : y : z) be an inﬁnite point on the conic. With x = −(y +z),wehave0=u2yz+x(v2z + w2y)=u2yz − (y + z)(v2z + w2y)=−w2y2 +(u2 − v2 − w2)yz − v2z2 = −(w2y2 − 2vwyz + v2z2)= −(wy − vz)2. Therefore, y : z = v : w and x : y : z = −(v + w):v : w = u : v : w. 508 Circumconics

13.4 The center of a circumconic

Since P = G/Q if and only if Q = G/P , 4 we conclude from (13.3) that the center of the circumconic Cp(P ) is the cevian quotient G/P . Figure 13.4 illustrates Cp(P )=Co(G/P ).

P b

P c Q†(B) Q†(C)

Gc G Q b

B Ga C

Q†(A)

P a

Figure 13.4: Cp(P )=Co(Q) with Q = G/P p + q + r =0 More explicitly, the center of the circumconic x y z is the point (p(q + r − p): q(r + p − q): r(p + q − r)).

Here are the centers of some common circumconics.

Circumconic Perspector P Center G/P Steiner circumellipse G G circumcircle K O 2 2 2 2 2 Kiepert hyperbola (b − c : ···: ···) Ki := ((b − c ) : ···: ···) 2 2 2 2 2 Jerabek hyperbola ((b − c )SA : ···: ···) Je := ((b − c ) SA : ···: ···) 2 Feuerbach hyperbola (a(b − c)(b + c − a):···: ···) Fe := ((b − c) (b + c − a):···: ···) through G and K (a2(b2 − c2):···: ···) (a4(b2 − c2)2 : ···: ···)

4A basic theorem on cevian quotients established in an earlier chapter. 13.5 The Steiner circum-ellipse 509

13.5 The Steiner circum-ellipse

The Steiner circum-ellipse Cp(G)=Co(G) consists of isotomic conjugates of inﬁnite points. It intersects the circumcircle at the Steiner point 1 1 1 St = : : . b2 − c2 c2 − a2 a2 − b2

A line through St intersects the Steiner circum-ellipse and the circumcircle each at one other point. These are the isotomic and isogonal conjugates of the same inﬁnite point P . Figure 13.5 shows three parallel lines through A, B, C, intersecting their opposite sides at X, Y , Z respectively. (1) The isotomic lines AX, BY , CZ intersect at P • on the Steiner circum- ellipse. (2) The isogonal lines, which are the reﬂections of AX, BY , CZ in the respective bisectors of angles A, B, C, intersect at P ∗ on the circumcircle. • ∗ (3) The line through P and P passes through the Steiner point St. In fact, for arbitrary P =(u : v : w), the line containing P • and P ∗ has equation (b2 − c2)ux +(c2 − a2)vy +(a2 − b2)wz =0.

If u + v + w =0, then this line clearly contains the Steiner point St.

A P ∗

P • Y O Y G

X B C X

Figure 13.5: The Steiner circum-ellipse and the Steiner point

Example. (Antipodes on the Steiner circum-ellipse) If (u : v : w) is an inﬁnite point, Q• is a point on the Steiner circum-ellipse. The antipode of 510 Circumconics

Q• is the isotomic conjugate of (v − w : w − u : u − v), the inﬁnite point of the line ux + vy + wz =0. The antipode of St on the Steiner circum-ellipse is the point (GK∞)• on the Kiepert hyperbola (see Example 13.2.1). 13.6 The Kiepert hyperbola 511

13.6 The Kiepert hyperbola

The Kiepert hyperbola 2 − 2 2 − 2 2 − 2 H (K): b c + c a + a b =0 x y z is the isogonal conjugate of the Brocard axis OK. Its center is the point

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

(HK∞)∗ A

K O H G

B C

X(112)

Figure 13.6: The Kiepert hyperbola

The tangent at H is the line HK. The reﬂections of HK in the sidelines are concurrent at 2 2 2 X(112) = a : b : c 2 2 2 2 2 2 (b − c )SA (c − a )SB (a − b )SC which also lies on the circle OHK. 512 Circumconics

13.6.1 The Kiepert center Proposition. The Kiepert center is the inferior of the Steiner point. Proof. It is enough to determine the superior of the Kiepert center. 2 2 2 2 2 2 2 2 2 sup(Ki)= ((c − a ) +(a − b ) − (b − c ) : ···: ···) =((c2 − a2)2 +(a2 − c2)(a2 + c2 − 2b2):···: ···) =((c2 − a2)(c2 − a2 + a2 + c2 − 2b2):···: ···) =((2 − 2)( 2 − 2):···: ···) c a a b 1 = : ···: ··· b2 − c2 = St.

Proposition. The Kiepert center is the point of concurrency of the Brocard axes of the residuals of the orthic triangle.

Ta A

Oa Ki Hb

K O Hc G H

C B Ha

Figure 13.7: Brocard axis of residual of orthic triangle

Proof. Since HbHc is antiparallel to BC, the Brocard axis of the residual triangle AHbHc is the parallel to ATa through the midpoint of AH, the circumcenter Oa. (The symmedian point is the intersection of this line with 13.6 The Kiepert hyperbola 513 the median AG). This parallel clearly passes through the midpoint of TaH, which is the Kiepert center Ki (Corollary 16.1). The same reasoning applies to the other two residuals, showing that the three Brocard axes are concurrent at Ki. 514 Circumconics

13.7 The Jerabek hyperbola

The Jerabek hyperbola 2( 2 − 2) 2( 2 − 2) 2( 2 − 2) H (O): a b c SA + b c a SB + c a b SC =0 x y z is the isogonal conjugate of the Euler line. It clearly contains H, O, K = G∗, ∗ • N , and H =(SA : SB : SC). Its center is the point 2 2 2 2 2 2 2 2 2 Je := ((b − c ) SA :(c − a ) SB :(a − b ) SC) called the Jerabek center. This is also the point of concurrency of the Euler lines of the “residuals” of the orthic triangle cev(H) (see Figure 13.9).

X(1294) A

X(74)

G[cev(H)]

H O N G

B C

X(107)

Figure 13.8: The Jerabek hyperbola

The tangent at H is the line

2( 2 − 2) 3 =0 a b c SAx , cyclic 13.7 The Jerabek hyperbola 515 which contains the centroid of the orthic triangle. The reﬂections of this tangent in the sidelines are concurrent at 1 1 1 X(107) = : : 2 2 2 2 2 2 . (b − c )SAA (c − a )SBB (a − b )SCC

Proposition. (a) The Jerabek center Je is the inferior of the Euler reﬂection point E. (b) The Jerabek center is the point of concurrency of the Euler lines of the residuals of the orthic triangle.

(E∞)∗ A

X(265)

Je Hb

N∗ • Hc H K O H X(3426)

B Ha C

Figure 13.9: The Jerabek hyperbola and three Euler lines 516 Circumconics

13.8 The Feuerbach hyperbola

The Feuerbach hyperbola H (I): ( − )( + − ) ( − )( + − ) ( − )( + − ) a b c b c a + b c a c a b + c a b a b c =0 x y z is the isogonal conjugate of the line OI. Its center is the Feuerbach point Fe, the point of tangency of the incircle and the nine-point circle. It intersects the Euler line at the Schifﬂer point a(b + c − a) b(c + a − b) c(a + b − c) Sc := : : . b + c c + a a + b

(OI∞)∗ A

X(1836) ∗ O SC I S N c H

B C

X(108)

Figure 13.10: The Schifﬂer point on the Feuerbach hyperbola and the Euler line Note that the OI line is tangent to the Feuerbach hyperbola at I. The tangent at H is the line

a(b − c)(b + c − a)SAAx =0. cyclic 13.8 The Feuerbach hyperbola 517

∗ It intersects the OI line at Sc . The reﬂections of this tangent in the sidelines are concurrent at a b c X(108) = : : . (b − c)(b + c − a)SA (c − a)(c + a − b)SB (a − b)(a + b − c)SC 518 Circumconics Chapter 14

The Steiner circumellipse

14.1 The Steiner circum-ellipse

The Steiner circum-ellipse is the isotomic conjugate of the inﬁnite line. It contains X(n) for the following values of n:

99, 190, 290, 648, 664, 666, 668, 670, 671, 886, 889, 892, 903, 1121, 1494, 2481, 2966, 3225, 3226, 3227, 3228, 4555, 4562, 4569, 4577, 4586, 4597,...

(190) = 1 1. X b−c: Yff parabolic point, perspector of inscribed parabola with focus X(101).

1 2. X648 = 2 2 (b −c )Sα

• Trilinear pole of the Euler line.

• Orthocorrespondent of X107 and X125. • “Third” orthoassociate of Steiner point. See 7/29/04.

= 1 3. X671 b2+c2−2a2

1 • Reﬂection conjugate of centroid X2. • Reﬂection of centroid in Kiepert center.

1Also in ED’s ﬁle. 520 The Steiner circumellipse

14.2 The Steiner point

The Steiner point St is the fourth intersection of the circumcircle and the Steiner ellipse yz + zx + xy =0. It has coordinates 1 1 1 : : . b2 − c2 c2 − a2 a2 − b2

14.2.1 Bailey’s theorem on the Steiner point Theorem (Bailey). Let P be a point with cevian triangle XY Z, and the line L joining P ∗ and P •.IfL intersects the sidelines of T at X, Y , Z respectively, then the circles AXX, BY Y , CZZ are concurrent at the Steiner point St.

Proof. Let P =(u : v : w). The line joining P ∗ and P • is

(b2 − c2)ux +(c2 − a2)vy +(a2 − b2)wz =0.

It intersects the sidelines of T at

X =(0:−(a2−b2)w :(c2−a2)v),Y =((a2−b2)w :0:−(b2−c2)u),Z =(−(c

The circle AXX has equation

(v + w)((c2 − a2)v − (a2 − b2)w)(a2yz + b2zx + c2xy) − a2vw(x + y + z)((c2 − a2)y − (a2 − b2)z)=0.

This clearly contains the Steiner point; so do the other two circles BY Y and CZZ.

14.3 Construction of the Steiner point

Let C be a circumconic with center Q. Denote by A, B, C the symmetrics of A, B, C in Q. Then the fourth intersection of C with the circumcircle is the common point of the circles ABC, BCA and CAB.

Let P =(u : v : w) (in barycentrics), with pedal triangle A[P ]B[P ]C[P ]. Let A, B, C be the reflections of A, B, C in the respective sides of the 14.3 Construction of the Steiner point 521 pedal triangle. Then ABC is perspective with ABC at the isogonal conjugate of P ∗. (This is well known). Now the circles A[P ]B C , B[P ]C A and C[P ]A B concur at a point. This is a corollary of the following. Let P =(u : v : w) (in barycentrics), with antipedal triangle A[P ]B[P ]C[P ]. Let A, B, C be the reflections of A[P ], B[P ], C[P ] in the lines BC, CA, AB respectively. Now the circles ABC, BCA and CAB concur at a point. The first component of the barycentric coordinates of this common point can be taken 2 as a F1F2G where

2 2 4 • F1 =(S − SBB)vw +(S − SAA)wu − c uv,

2 4 2 • F2 =(S − SCC)vw − b wu +(S − SAA)uv, and

6 2 2 4 2 2 4 2 2 2 • G = −a SAv w + b SBCw u + c SBCu v +2SAABCu vw + 2 2 2 2 2 2 2 2 a (b S − 2SACC)uvw + a (c S − 2SABB)uv w.

Gb Gc A

O I G Na

B C

sup(Fe)

Figure 14.1: 522 The Steiner circumellipse

14.3.1 Steiner point and the nine-point center

Let X be the circumcenter of the AA+A−, where A± are the apices of the Fermat triangles. Similarly deﬁne Y and Z. Darij Grinberg has noted that XY Z has the Steiner point as perspector. Simply, X =(0:SA − SB : SC − SA). The circle has equation 2 2 2 2 a a yz + b zx + c xy = (x + y + z)((SA − SB)y +(SC − SA)z). SB − SC The radical center of the three circles is the point given by

(S − S )y +(S − S )z A B C A = ···= ··· . SB−SC 2 a 2 x (SB−SC ) This means that : ···: ··· is the superior of 2 : ···: ··· , SB−SC a and the radical center is the nine-point center. Chapter 15

Circum-hyperbolas

15.1 Circum-hyperbolas with given inﬁnite points

We begin with the simple fact that a hyperbola has two asymptotes. A circum-hyperbola is uniquely determined by its two inﬁnite points (giving the directions of its two asympotes). If these two inﬁnite points have homogeneous barycentric coordinates (u1 : v1 : w1) and (u2 : v2 : w2), then the equation of the hyperbola is simply u1u2 v1v2 w1w2 + + =0. (15.1) x y z From this we make the following simple conclusions.

1. The perspector of the circumhyperbola is the point P =(u1u2 : v1v2 : w1w2). 2. The center of the hyperbola is the cevian quotient G/P :

G/P =(u1u2(−u1u2 + v1v2 + w1w2):···: ···) =(u1u2(v1w2 + v2w1):···: ···).

3. The asymptote with inﬁnite point (u1 : v1 : w1) is the line xyz 0= u1 v1 w1 ( + ) ( + ) ( + ) u1u2 v1w2 v2w1 v1v2 w1u2 w2u1 w1w2 u1v2 u2v1 v1 w1 = x + ··· v1v2(w1u2 + w2u1) w1w2(u1v2 + u2v1)

= v1w1(w2(u1v2 + u2v1) − v2(w1u2 + w2u1))x + ··· = u2v1w1(v1w2 − w1v2)x + v2w1u1(w1u2 − u1w2)y + w2u1v1(u1v2 − v1u2)z. 524 Circum-hyperbolas

Now, since

v1w2−w1v2 = −(w1+u1)w2+w1(w2+u2)=w1u2−u1w2 = u1v2−v1u2,

the equation of the asymptote can be simpliﬁed as

u2v1w1x + v2w1u1y + w2u1v1z =0

or u2 v2 w2 x + y + z =0. u1 v1 w1

4. Similarly the asymptote with inﬁnite point (u2 : v2 : w2) is the line

u1 v1 w1 x + y + z =0. u2 v2 w2

Proposition. The asymptotes of the circum-hyperbola with inﬁnite points (u1 : v1 : w1) and (u2 : v2 : w2) are the isotomic lines

u1 v1 w1 x + y + z =0, u2 v2 w2 u2 v2 w2 x + y + z =0. u1 v1 w1 15.1 Circum-hyperbolas with given inﬁnite points 525

15.1.1 The circum-hyperbola with perspector St The circumconic yz zx xy Cp(St): + + =0. b2 − c2 c2 − a2 a2 − b2 has center 4 − 2( 2 + 2) − ( 4 − 3 2 2 + 4) G S = a a b c b b c c : ···: ··· / t 2 − 2 b c (c2 − a2)(a2 − b2)+(b2 − c2)2 = : ···: ··· . b2 − c2

A X(892)

G/St

• Ki

C B

Figure 15.1: The circumconic Cp(St) It contains the two inﬁnite points (a2 − b2 : b2 − c2 : c2 − a2) and (c2 − a2 : a2 − b2 : b2 − c2). Therefore, the asymptotes are parallel to the lines c2x + a2y + b2z =0 and b2x + c2y + a2z =0. This hyperbola also contains • 1 1 1 Ki = : : . (b2 − c2)2 (c2 − a2)2 (a2 − b2)2 526 Circum-hyperbolas

It intersects (i) the circumcircle at 1 : ···: ··· , (b2 − c2)(a2(b2 + c2) − 2b2c2) (ii) the Steiner circum-ellipse at 1 X(892) = : ···: ··· , (b2 − c2)(b2 + c2 − 2a2) • the antipode of Ki (see Figure 15.1). 15.2 Circum-hyperbola with a prescribed asymptote 527

15.2 Circum-hyperbola with a prescribed asymptote

Since the two asymptotes of a circum-hyperbola are isotomic lines, the hyperbola is uniquely determined by an asymptote. If it has one asymptote + + =0 x + y + z =0 fx gy hz , then the other asymptote is the line f g h . Their intersection is the center of the hyperbola:

Q =(f(g2 − g2): g(h2 − f 2): h(f 2 − g2)).

The perspector is the cevian quotient

G/Q =(f(g − h)2 : g(h − f)2 : h(f − g)2).

The equation of the hyperbola is therefore f(g − h)2 g(h − f)2 h(f − g)2 + + =0. x y z 528 Circum-hyperbolas

15.2.1 The Euler asymptotic hyperbola The Euler asymptote hyperbola is the circum-hyperbola with the Euler line as an asymptote: S (S − S )(S (S + S ) − 2S )2 A B C A B C BC =0. cyclic x Its center is the point 2 X1650 =(SAA(SB − SC) (SA(SB + SC) − 2SBC): ···: ···).

A X(265) X(1294)

Je X(1650)

O H

B C

Figure 15.2: The Euler asymptotic hyperbola

This hyperbola also contains the following triangle centers: 1 • X1494 = : ···: ··· , intersection with the Steiner SA(SB+SC )−2SBC circum-ellipse, 1 • X1294 = : ···: ··· , intersection with cir- SAA(SBB−SBC+SCC)−SBBSCC cumcircle, SA • X265 = 2 : ···: ··· , which is the reﬂection conjugate of the 3SAA−S circumcenter O. 1

1 If Oa, Ob, Oc are the reﬂections of O in a, b, c respectively, the three circles OaBC, ObCA, OcAB intersect at the reﬂection conjugate of O. 15.2 Circum-hyperbola with a prescribed asymptote 529

15.2.2 The orthic asymptotic hyperbola The orthic asymptote hyperbola is the circum-hyperbola with the orthic axis SAx + SBy + SCz =0as an asymptote: S (S − S )2 A B C =0. cyclic x Its center is the point

X647 =(SA(SBB − SCC): SB(SCC − SAA): SC(SAA − SBB)).

K G

B C

This hyperbola also contains the following triangle centers: 1 • X2966 = : ···: ··· , intersection with the Steiner ((SB−SC )(SAA−SBC) circum-ellipse, 1 • X935 = : ···: ··· , intersection with SA(SB−SC )(4SBC−(SC +SA)(SA+SB)) circumcircle, SA(SB−SC ) • X879 = : ···: ··· , which is the intersection with the SAA−SBC Jerabek hyperbola. 2 SB−SC • X2394 = : ···: ··· , which is the intersection of the SA(SB+SC )−2SBC parallels to the orthic axis through the orthocenter and to the isotomic line through the centroid.

2Also of the parallels to the orthic axis through the symmedian point and to the isotomic line through the circumcenter. 530 Circum-hyperbolas

15.2.3 The Lemoine asymptotic hyperbola The Lemoine aymptotic hyperbola has the Lemonie axis

x + y + z =0 a2 b2 c2 and its isotomic line a2x + b2y + c2z =0 for asymptotes. It is the circumconic 2( 2 − 2)2 2( 2 − 2)2 2( 2 − 2)2 a b c + b c a + c a b =0 x y z with center

X(3005) = (a2(b4 − c4):b2(c4 − a4):c2(a4 − b4)), and intersects the circumcircle at the reﬂection of E in the Brocard axis, namely, a2 b2 c2 X(691) = : : . (b2 − c2)(b2 + c2 − 2a2) (c2 − a2)(c2 + a2 − 2b2) (a2 − b2)(a2 + b2 − 2c2) 15.3 Pencil of hyperbolas with parallel asymptotes 531

15.3 Pencil of hyperbolas with parallel asymptotes

p + q + r =0 Suppose a circum-hyperbola x y z has a prescribed infinite point (u : v : w), i.e., an asymptote with prescribed direction. Then the other p : q : r p + q + r =0 infinite point is u v w . It follows that u v w , and the perspector lies on the trilinear polar of the given infinite point (u : v : w). A typical point on the trilinear polar of (u : v : w) has coordinates (u(v − w + tu):v(w − u + tv):w(u − v + tw)). The hyperbola is one in the pencil 3 u(v − w) v(w − u) w(u − v) u2 v2 w2 + + + t + + =0. x y z x y z The center of the hyperbola, being the cevian quotient G/(p : q : r), lies on the conic x(y + z − x) y(z + x − y) z(x + y − z) + + =0. u v w This is the (bicevian) conic which intersects the sidelines at the midpoints of the sides and the traces of (u : v : w). This is a parabola.

2 2 2 3 u v w The circumconic x + y + z =0has only one inﬁnite point; it is a parabola. 532 Circum-hyperbolas u + v + w =0 15.4 The circum-hyperbola x y z for an inﬁnite point (u : v : w)

( : : ) u + v + w =0 If u v w is an infinite point, the circumconic x y z contains the centroid and is a hyperbola. The center of the hyperbola is the point (u2 : v2 : w2) on the Steiner inellipse. Let ((v − w)(u + t):(w − u)(v + t):(u − v)(w + t)) be an infinite point of the hyperbola. Then u v w + + =0. (v − w)(u + t) (w − u)(v + t) (u − v)(w + t) u + v + w =0 ( : : ) 15.4 The circum-hyperbola x y z for an infinite point u v w 533

Exercise 1. The circumconic through the Brocard points is

(a4 − b2c2)yz +(b4 − c2a2)zx +(c4 − a2b2)xy =0.

It contains (i) X99 the Steiner point, = 1 (ii) X83 b2+c2 , = a4−b2c2 (iii) X880 a2(b2−c2). 2. Given a point P =(u : v : w), consider the circumconic C through P and its isotomic conjugate. For every point Q on C with cevian triangle XY Z, let X, Y , Z be the midpoints of AX, BY , CZ respectively. XY Z is perspective with cev(P ). = 1 : ···: ··· ( + : + : + ) If Q tu+vw , then this perspector is t u t v t w . = =1 ( + − ) = −1 For P the Gergonne point with u / b c a , with t a+b+c, Q = H, the orthocenter. The perspector is X57. If we take P to be the Nagel point, and Q = H, the perspector is the incenter I. The locus of the perspector is the line GP .

3. X(1648) and X(1649)

Let P =(u : v : w) be an inﬁnite point. Consider the point

Q =(u2(v − w):v2(w − u):w2(u − v)).

(a) Q is the intersection of the three lines: (i) ux + vy + wz =0, u + v + w =0 (ii) x y z , x + y + z =0 (iii) u(v−w) v(w−u) w(u−v) . (b) The cevian quotient of Q is the point Q =(u(v−w)2 : v(w−u)2 : w(u − v)2). (c) The line QQ is the trilinear polar of (u(v − w):v(w − u): w(u − v)). 534 Circum-hyperbolas

Therefore, for the circumconic with perspector Q: u2(v − w) v2(w − u) w2(u − v) + + =0, x y z (a) the center is Q, (b) the inﬁnite points are P =(u : v : w) and (u(v − w):v(w − u): w(u − v)), (c) the asymptotes are the isotomic lines (v − w)x +(w − u)y +(u − v)z =0and x + y + z =0 v−w w−u u−v . These are deﬁned as the tripolar centroids of X(523) and X(524) respectively. They have barycentric coordinates

2 2 2 2 2 2 X1648 =((b − c ) (2a − b − c ):···: ···), 2 2 2 2 2 2 X1649 =((b − c )(2a − b − c ) : ···: ···).

G/X1648 = X1649. The hyperbola with perspector X1648 have asymptotes which are the = 1 : ···: ··· =(2 2− tripolars of the points X671 2a2−b2−c2 and X524 a b2 − c2 : ···: ···).

According to ETC, X1648 can be constructed as the intersection of GK and the line joining the Kiepert and Jerabek centers.

The other hyperbola with perspector X1649 has asymptotes which are the trilinear polars of X99 and X523.

X2398 and X2400 are isotomic conjugates. Their trilinear polars are the asymptotes of the hyperbola with center X1566 and perspector X676. These two hyperbolas have a common point

2 2 2 2 2 X690 =((b − c )(2a − b − c ):···: ···). u + v + w =0 ( : : ) 15.4 The circum-hyperbola x y z for an inﬁnite point u v w 535

15.4.1 Perspective and orthologic triangles Theorem. If T and a triangle XY Z is both perspective and orthologic to T, then ⊥(T,XYZ)is the second intersection of the line joining Q :=⊥ (XY Z, T) to P := (XY Z, T) and the rectangular circum-hyperbola H (P ) through the perspector.

Q H Q

Y Z

B C

Figure 15.3: Perspectivity and orthology of two triangles Proof. Let (XY Z,T)=P =(u : v : w) and ⊥(XY Z, T)=Q =(u : v : w). The point X must lie on the line AP and the perpendicular from Q to BC. on the line AQ; similarly for Y and Z. These are the points X =(u + v + w )(SBv − SCw)(1, 0, 0) + (SBv − SCw )(−(v + w),v,w), Y =(u + v + w )(SCw − SAu)(0, 1, 0) + (SCw − SAu )(u, −(w + u),w), Z =(u + v + w )(SAu − SBv)(0, 0, 1) + (SAu − SBv )(u, v, −(u + v)). The perpendiculars from A, B, C to YZ, ZX, XY respectively are concurrent at S v − S w S w − S u S u − S v Q = B C : C A : A B . vw − wv wu − uw uv − vu 536 Circum-hyperbolas

Note that this point lies on the rectangular circum-hyperbola u(S v − S w) H (P ): B C =0, cyclic x since u(vw − wv)+v(wu − uw)+w(uv − vu)=0. Therefore this second orthology center Q is the intersection of the line PQ with the circum- hyperbola H (P ). Chapter 16

Rectangular circum-hyperbolas

A hyperbola is rectangular if its asymptotes are perpendicular to each other. Theorem. A circumconic is a rectangular hyperbola if and only if it contains the orthocenter.

Proof. Two lines with inﬁnite points (u1 : v1 : w1) and (u2 : v2 : w2) are perpendicular to each other if and only if 1

SAu1u2 + SBv1v2 + SCw1w2 =0. (16.1)

Thus, the orthocenter H lies on the hyperbola (15.1). Conversely, if a circumconic contains H, its isogonal conjugate is a line containing the circumcenter O, an interior point of the circumcircle. Therefore, the conic is a hyperbola, and we may assume its equation in the form (15.1). Now the condition (16.1) guarantees that the two asymptotes are perpendicular to each other. The Kiepert, Jerabek, and Feuerbach hyperbolas are all rectangular. On the other hand, the circumconic through G and K is a hyperbola (since these are interior points of the cirumcircle), albeit not rectangular (since the line GK does not contain O). Nor is the hyperbola Cp(St) in Example ??.

1This condition of perpendicularity is established in an earlier chapter. 538 Rectangular circum-hyperbolas

16.1 The center of a rectangular hyperbola

Theorem. The center of a rectangular hyperbola is a point on the nine-point circle.

Proof. From the equations of the asymptotes in (??) and (??), if P1 =(u1 : v1 : w1) and P2 =(u2 : v2 : w2) are orthogonal, then these are the Simson ∗ ∗ 2 lines of the points P1 and P2 on the circumcircle. Furthermore, these points are antipodal. Their Simson lines are orthogonal and intersect on the nine-point circle. Corollary. The fourth intersection of a rectangular circum-hyperbola with the circumcircle is the antipode of the orthocenter H. Proof. Since the circumcircle is the image of the nine-point circle under the homothety h(H, 2), the antipode of H must be a point on the circumcircle. It is the fourth intersection of the rectangular circum-hyperbola and the circumcircle. The coordinates of the fourth intersections of the basic circumconics can be found in §13.2.1

2This is a nontrivial result. It follows from the equations of Simson lines established in the chapter on the circumcircle. 16.2 Construction of asymptotes 539

16.2 Construction of asymptotes

Given the center W of a rectangular hyperbola, and the tangent at a point P , the asymptotes can be easily constructed. Construct (1) the circle P (W ) to intersect the tangent at two points Q and Q, (2) the lines WQand WQ. These are the asymptotes. 3 Example. The case of the Feuerbach hyperbola is strikingly easy. Since the OI is the tangent at I, the lines joining the Feuerbach point Fe to the intersections of the incircle with OI are the asymptotes.

O I Q N Q H

B C

Figure 16.1: Construction of asymptotes of the Feuerbach hyperbola

16.2.1 Antipodes on rectangular circum-hyperbola Theorem. Two points on a rectangular circum-hyperbola are antipodal if and only if their isogonal conjugates are inverse with respect to the circumcircle.

2 3In Cartesian coordinates, let P be the point t, c on the rectangular hyperbola xy = c2. The tangent t 1 c2 2 2c2 at P is the line 2 t x + ty = c . This tangent intersects the axes (asymptotes) at (2t, 0) and 0, t . These two points are clearly on the circle P (O). 540 Rectangular circum-hyperbolas

16.2.2 The Huygens hyperbola The Huygens hyperbola (Alperin chose this name for convenience only) is the isogonal transform of the tangent to the Jerabek hyperbola at O.Itis tangent to the Euler line at H.

SBC(SB − SC)yz =0. cyclic

It contains X1826 (on line joining H to Mittenspunkt), X225 (on HI), • X264 (isotomic conjugate of O,onHH ), X393 (square of orthocenter, on HK), X1093, X93, X847. It intersects the circumcircle at X1300.

X1300

X393 X1093 B C

Figure 16.2: The Huygens hyperbola 16.3 The rectangular circum-hyperbola through a given point 541

16.3 The rectangular circum-hyperbola through a given point

Let P =(u : v : w) be a given point other than the vertices and orthocenter H of T. The rectangular circum-hyperbola through P is the isogonal conjugate of the line OP ∗. It is the hyperbola u(S v − S w) H (P ): B C =0. cyclic x Apart from the orthocenter H, it intersects the Euler line at the point u(S v − S w) v(S w − S u) w(S u − S v) B C : C A : A B . SB − SC SC − SA SA − SB

16.3.1 The center The center of H (P ) is the cevian quotient

W (P ):= G/(u(SBv − SCw):v(SCw − SAu):w(SAu − SBv))

=(u(SBv − SCw)(−u(SBv − SCw)+v(SCw − SAu)+w(SAu − SBv) : ···: ···) 2 2 =(u(SBv − SCw)(b (u + v)w − c (w + u)v):···: ···) on the nine-point circle.

16.3.2 The fourth intersection with the circumcircle The rectangular circum-hyperbola H (P ) intersects the circumcircle at the point 1 ((O ∗)∞)∗ = : ···: ··· P 2 2 . c v(SCw − SAu) − b w(SAu − SBv)

16.3.3 The tangent at H and P The tangent to H (P ) at the orthocenter is the line

SAAu(SBv − SCw)x =0. cyclic 542 Rectangular circum-hyperbolas

Proposition. The following three points on the circumcircle coincide and have coordinates 2 2 2 a : b : c : SAu(SBv − SCw) SBv(SCw − SAu) SCw(SAu − SBv) (1) the antipode of the isogonal conjugate of the inﬁnite point of the tangent at H, (2) the point of concurrency of the reﬂections of the tangent in the sideline, (3) the second intersection of the circumcircle of the line joining its intersection with H (P ) and the orthocenter H. 16.3 The rectangular circum-hyperbola through a given point 543

16.3.4 The rectangular hyperbola through the Euler reﬂection point

2 2 2 2 a (−a SAA + b SBB + c SCC − 2SABC)yz =0. cyclic

Center of conic X113. Intersection with the Euler line a2((−a2S + b2S + c2S − 2S )) AA BB CC ABC : ···: ··· . b2 − c2 Tangent at H:

2 2 2 2 a SAA(−a SAA + b SBB + c SCC − 2SABC)x =0. cyclic

B C

Figure 16.3: The rectangular circumhyperbola through the Euler reﬂection point 544 Rectangular circum-hyperbolas

Exercise 1. Consider the rectangular circum-hyperbola H(P ) through a given point P . Denote by Q the fourth intersection with the circumcircle. The reﬂections of the tangent to H(P ) at H in the sidelines intersect on the circumcircle at its second intersection with the line HQ. Jerabek hyperbola X(107) Kiepert hyperbola X112 16.4 Reﬂection conjugates as antipodal points on a rectangular circum-hyperbola545

16.4 Reﬂection conjugates as antipodal points on a rectangular circum-hyperbola

=( : : ) † † † Let Q u v w be a finite point with reflection triangle QaQbQc. These are the points † =(− 2 :2 + 2 :2 + 2 ) Qa a u SCu a v SBv a w , † =(2 + 2 : − 2 :2 + 2 ) Qb SCv b u b v SAv b w , † =(2 + 2 :2 + 2 : − 2 ) Qc SBw c u SAw c v c w . The reflection conjugate of Q is the common point Q† of the circles † † † QaBC, QbCA, QcAB. Theorem. (a) Q† is the antipode of Q on the rectangular circum-hyperbola H (Q). † (b) Q lies on the circumconic Co(inf(Q)). † † † † † † The three circles AQbQc, BQcQa,CQaQb are concurrent at the the fourth intersection of Co(Q) and the circumcircle.

(inf(Q)) ∩ (O) O Q G H (Q) ∩ (O) W (Q) inf(Q)

H (Q) ∩ Co(inf(Q)) B C

Figure 16.4: Intersections of H (Q) and Co(inf(Q)) and the circumcircle 546 Rectangular circum-hyperbolas

16.5 Rectangular circum-hyperbola with a prescribed in- ﬁnite point

If a rectangular circum-hyperbola has an inﬁnite point (u : v : w), the other inﬁnite point is (SBv − SCw : SCw − SAu : SAu − SBv). The hyperbola is u(S v − S w) B C =0. cyclic x

The asymptote with the given inﬁnite point is the line S v − S w S w − S u S u − S v B C x + C A y + A B z =0. u v w 16.5 Rectangular circum-hyperbola with a prescribed inﬁnite point 547

16.5.1 The Euler (rectangular) circum-hyperbola The Euler (rectangular) circum-hyperbola has asymptotes parallel and perpendicular to the Euler line. The two inﬁnite points are 4

E∞ =(SA(SB + SC) − 2SBC : SB(SC + SA) − 2SCA : SC(SA + SB) − 2SAB),

(L∗(H))∞ =(SB − SC : SC − SA : SA − SB).

It follows that the Euler hyperbola has equation (S − S )(S (S + S ) − 2S ) B C A B C BC =0. cyclic x

X476

O G

B C

Figure 16.5: The Euler hyperbola Its center is the point

2 2 X3258 =((SB − SC) (SA(SB + SC) − 2SBC)(S − 3SAA):···: ···).

This is the reﬂection of the Jerabek center in the Euler line.

4 The second infinite point is the infinite point of the orthic axis SAx + SBy + SC z =0. Its orthogonal infinite point is (SA(SB − SC ) − SB(SC − SA):···: ···)=(SA(SB + SC ) − 2SBC : ···: ···)=E∞. 548 Rectangular circum-hyperbolas

16.5.2 The circum-hyperbola with asymptotes parallel to the Brocard and Lemoine axes This is the hyperbola

4 a (SB−SC )(SAA−SBC) =0 cyclic x with center

2 2 X(2679) = (a (SB−SC) (SAA−SBC)(SA(SA+SB+SC)−(SBB+SBC+SCC)) : ···: · ·

It contains the point X(32) = (a4 : b4 : c4) on the Brocard axis, which is the isogonal conjugate of the Kiepert perspector K(−ω). 16.5 Rectangular circum-hyperbola with a prescribed inﬁnite point 549

16.5.3 The circum-hyperbola with asymptotes parallel and perpendicular to the OI line This is the hyperbola a2(b−c)(a2(b+c)−2abc−(b+c)(b−c)2) =0 cyclic x with center

X(3259) = ((b+c−2a)(b−c)2(a2(b+c)−2abc−(b+c)(b−c)2):···: ···).

It contains the exsimilicenter T− of the circumcircle and incircle.

G O X(4226) H N

W(St)

C B

Figure 16.6: The rectangular circum-hyperbola through the Steiner poinrt

S = 1 : 1 : 1 Example. Since the Steiner point t b2−c2 c2−a2 a2−b2 is the antipode of the Tarry point (the fourth intersection with the Kiepert hyperbola) on the circumcircle, the center of the rectangular circum-hyperbola 2a4 − a2(b2 + c2)+(b2 − c2)2 H (St): =0 cyclic x 550 Rectangular circum-hyperbolas is the antipode of the Kiepert center on the nine-point circle. This is the inferior of the Tarry point: 2 2 2 4 4 4 2 2 2 2 2 2 W(St)=inf(Ta)= (a (b + c ) − (b + c ))(2a − a (b + c )+(b − c ) ):···: ··

The hyperbola intersects the Euler line again at 2a4 − a2(b2 + c2)+(b2 − c2)2 X(4226) = : ···: ··· . b2 − c2

More generally, let P1 and P2 be antipodes on the circumcircle. The centers Q1 = W (P1) and Q2 = W (P2) of the hyperbolas H (P1) and H (P2) are antipodes on the nine-point circles, since these are the midpoints of the segments HP1 and HP2. Furthermore, Q1 = inf(P2) and Q2 = inf(P1).