Geometry of the Triangle

Home , Fermat point, Napoleon points, Spieker center

Paul Yiu

Summer 2007

Department of Mathematics Florida Atlantic University

Version 7.0707

July 2007

Contents

1 Preliminaries 1 1.1 Coordinatization of points on a line ...... 1 1.2 Centers of similitude of two circles ...... 2 1.3 Tangent circles ...... 3 1.4 Harmonic division ...... 4 1.5 Homothety ...... 5 1.6 The power of a point with respect to a circle ...... 6 1.7 Directed angles ...... 6 1.8 Menelaus and Ceva theorems ...... 7 1.8.1 Menelaus and Ceva Theorems ...... 7 1.8.2 Desargues Theorem ...... 9

2 The circumcircle and the incircle 11 2.1 The circumcircle and the law of sines ...... 11 2.2 The incircle and the Gergonne point ...... 12 2.3 The Heron formula ...... 15 2.4 The excircles and the Nagel point ...... 17 2.5 Euler’s formula and Steiner’s porism ...... 20 2.5.1 Euler’s formula ...... 20 2.5.2 Steiner’s porism ...... 21

3 The Euler line 23 3.1 The medial and antimedial triangles ...... 23 3.1.1 The medial triangle, the nine-point center, and the Spieker point . . 23 3.1.2 The antimedial triangle and the orthocenter ...... 24 3.1.3 The Euler line ...... 25 3.2 The nine-point circle ...... 28 3.2.1 The Euler triangle as a midway triangle ...... 28 3.2.2 The orthic triangle as a pedal triangle ...... 29 3.2.3 The nine-point circle ...... 30

4 The OI-line 33 4.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle . 33 4.1.1 Construction of mixtilinear incircles ...... 34 4.2 The reﬂection of I in O ...... 36 iv CONTENTS

4.2.1 The circumcircle of the excentral triangle ...... 36 4.3 The homothetic center of the intouch and excentral triangles ...... 38 4.4 The line OI as the Euler line of the intouch triangle ...... 38 4.4.1 The orthocenter of the intouch triangle ...... 39 4.4.2 The centroids of the excentral and intouch triangles ...... 40 4.4.3 The point which divides OI in the ratio R + r : −2r ...... 40

5 Homogeneous Barycentric Coordinates 43 5.1 Barycentric coordinates with reference to a triangle ...... 43 5.1.1 Homogeneous barycentric coordinates ...... 43 5.1.2 The centroid ...... 43 5.1.3 The incenter ...... 44 5.1.4 The Gergonne point ...... 44 5.1.5 Cevian triangle ...... 45 5.1.6 The Nagel point and the extouch triangle ...... 46 5.1.7 The orthocenter and the orthic triangle ...... 47 5.1.8 The circumcenter ...... 47 5.1.9 The excenters ...... 48 5.1.10 The barycenter of the perimeter ...... 48 5.1.11 The nine-point center ...... 49 5.1.12 The centers of similitudes T± ...... 50 5.2 The area formula ...... 53

6 Straight lines 55 6.1 Equations of straight lines ...... 55 6.1.1 Two-point form ...... 55 6.1.2 Intersection of two lines ...... 56 6.2 Perspectivity ...... 58 6.3 Trilinear pole and polar ...... 60 6.4 Anticevian triangles ...... 62 6.4.1 Construction of anticevian triangle from trilinear polar and polar properties ...... 62 6.4.2 Another construction of anticevian triangles ...... 63 6.5 The cevian nest theorem ...... 66 6.5.1 G/P ...... 67 6.5.2 H/P ...... 68 6.5.3 ...... 68 6.6 Conway’s formula ...... 74 6.6.1 Conway’s Notation ...... 74 6.6.2 Conway’s formula ...... 75

7 Kiepert perspectors 81 7.1 Jacobi’s Theorem ...... 81 7.2 The Kiepert triangle K(θ) ...... 84 7.3 The Kiepert perspector ...... 84 CONTENTS v

7.4 Iterated Kiepert triangles ...... 92 7.4.1 Some interesting properties of iterated Kiepert triangles ...... 92 7.4.2 Lemoine’s problem ...... 93 7.5 Perspectivity with the superior and excentral triangles ...... 93

8 Parallel and perpendicular lines 95 8.1 Infinite points and parallel lines ...... 95 8.1.1 The infinite point of a line ...... 95 8.1.2 Infinite point as vector ...... 96 8.2 Perpendicular lines ...... 98 8.3 Triangles bounded by lines parallel to the sidelines ...... 102 8.3.1 The Grebe symmedian point ...... 103 8.3.2 Gossard ...... 103 8.4 Distance formula ...... 107 8.5 Pedal of a point on a line ...... 107 8.6 Pedal and reflection triangles ...... 108 8.6.1 Pedal triangle ...... 108 8.6.2 Examples ...... 108 8.6.3 Reflection triangle ...... 110 8.7 The Brocardians ...... 111 8.7.1 Equal-parallelians point ...... 111 8.7.2 The Brocardians ...... 114 8.8 The orthic triangle ...... 117 8.8.1 The centroid of the orthic triangle ...... 117 8.8.2 The orthocenter of the orthic triangle ...... 117 8.8.3 Further notes ...... 118 8.9 The tangential triangle ...... 119 8.9.1 The Gob homothetic center of the tangential and orthic triangles . . 119 8.9.2 The centroid of the tangential triangle ...... 120 8.9.3 The circumcenter of the tangential triangle ...... 120 8.10 The intangents triangle ...... 122 8.11 The triangle of reflections ...... 123 8.12 The Evans perspector ...... 125

9 Some Basic Constructions 127 9.1 Isotomic conjugates ...... 127 9.1.1 The Gergonne and Nagel points ...... 128 9.1.2 The isotomic conjugate of the orthocenter ...... 129 9.1.3 Yff-Brocard points ...... 130 9.2 Isogonal conjugates ...... 132 9.2.1 Reﬂections and isogonal conjugates ...... 132 9.2.2 The pedal circle ...... 133 9.2.3 Coordinates of isogonal conjugate ...... 134 9.3 Examples of isogonal conjugates ...... 135 9.3.1 The circumcenter and orthocenter ...... 135 vi CONTENTS

9.3.2 The symmedian point and the centroid ...... 136 9.3.3 The Gergonne point and the insimilicenter T+ ...... 137 9.3.4 The Nagel point and the exsimilicenter T− ...... 140 9.3.5 Isogonal conjugates of the Kiepert perspectors ...... 142 9.4 Isogonal conjugate of an inﬁnite point ...... 143

10 The circumcircle 147 10.1 Simson lines and lines of reflections ...... 147 10.2 Simson line and line of reflections: computations ...... 152 10.2.1 The direction of Simson line ...... 152 10.2.2 The line of reflections ...... 153 10.3 Circumcevian triangle ...... 157 10.3.1 The circumcevian triangle of H ...... 157 10.4 Antipedal triangles ...... 157 10.4.1 The circum-tangential triangle ...... 159

11 Circles 161 11.1 Equation of a circle ...... 161 11.1.1 The power of a point with respect to a circle ...... 162 11.1.2 The incircle and the excircles ...... 163 11.1.3 Circle with given center and radius ...... 164 11.1.4 Circle with a given diameter ...... 164 11.2 The Feuerbach theorem ...... 165 11.2.1 Intersection of the incircle and the nine-point circle ...... 165 11.2.2 Condition for tangency of a line and the incircle ...... 165 11.3 Circles tangent to two sidelines ...... 167 11.4 The Brocard points and the Brocard circle ...... 168 11.4.1 The third Brocard point ...... 171 11.4.2 The Brocard circle ...... 171 11.5 The Taylor circle ...... 173 11.5.1 The Taylor circle of the excentral triangle ...... 175 11.6 The Dou circle ...... 176 11.6.1 August 17, 2002: Edward Brisse ...... 177 11.7 The Adams circles ...... 178

12 Four homothetic triangles with collinear homothetic centers 179 12.1 The tangential triangle ...... 179 12.2 The intangent triangle ...... 180 12.3 The extangent triangle ...... 180 12.4 The line of centers of similitude ...... 184 12.5 ...... 185 Chapter 1

Preliminaries

1.1 Coordinatization of points on a line

Let B and C be two ﬁxed points on a line L. Every point X on L can be coordinatized in one of several ways: BX (1) the ratio of division t = XC , (2) the absolute barycentric coordinates: an expression of X as a convex combination of B and C: X =(1− t)B + tC, which expresses for an arbitrary point P outside the line L, the vector PX as a combination of the vectors PB and PC. (3) the homogeneous barycentric coordinates: the proportion XC : BX, which are masses at B and C so that the resulting system (of two particles) has balance point at X.

B X C 2 Preliminaries

1.2 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 : TI = R : r. The absolute barycentric coordinates of P with respect to OI are R · I + r · O T = . R + r

Y T T O I Y X

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 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 1.3 Tangent circles 3

1.3 Tangent circles

If two circles are tangent to each other, the line joining their centers passes through the point of tangency, which is a center of similitude of the circles.

T O T I O I 4 Preliminaries

1.4 Harmonic division

Two points X and Y are said to divide two other points B and C harmonically if BX BY = − . XC YC They are harmonic conjugates of each other with respect to the segment BC.

Examples 1. For two given circles, the two centers of similitude divide the centers harmonically.

2. Given triangle ABC, let the internal bisector of angle A intersect BC at X. The harmonic conjugate of X in BC is the intersection of BC with the external bisector of angle A.

B X C X 3. Let A and B be distinct points. If M is the midpoint of the segment AB,itisnot possible to find a finite point N on the line AB so that M, N divide A, B harmon- AN − AM − − ically. This is because NB = MB = 1, requiring AN = NB = BN, and AB = BN −AN =0, a contradiction. We shall agree to say that if M and N divide A, B harmonically, then N is the infinite point of the line AB.

Exercise 1. If X, Y divide B, C harmonically, then B, C divide X, Y harmonically.

2. Given a point X on the line BC, construct its harmonic associate with respect to the segment BC. Distinguish between two cases when X divides BC internally and externally. 1

3. The centers A and B of two circles A(a) and B(b) are at a distance d apart. The line AB intersect the circles at A and B respectively, so that A, B are between A, B.

1Make use of the notion of centers of similitude of two circles. 1.5 Homothety 5

A B A B

4. Given two ﬁxed points B and C and a positive constant k =1 , the locus of the points P for which |BP| : |CP| = k is a circle.

1.5 Homothety

Given a point T and a nonzero constant k, the similarity transformation h(T,k) which carries a point X to the point X on the line TX satisfying TX : TX = k :1is called the homothety with center T and ratio k. Explicitly,

h(T,k)(P )=(1− k)T + kP.

Any two circles are homothetic. Let P and Q be the internal and external centers of h r h − r similitude of two circles O(R) and I(r). Both the homotheties (Q, R ) and (P, R ) transform the circle O(R) into I(r). Theorem 1.1. If the sidelines of two triangles are pairwise parallel, the lines joining the corresponding vertices are concurrent. Proof. Suppose triangles ABC and XY Z are such that the lines BC and YZare parallel, as are CA and ZX, AB and XY . The two triangles are similar. Let t be the (signed) ratio of similarity. If the lines BY and CZ do not intersect, there are two possibilities. (i) XY Z is a translation of ABC. (ii) ABC and XY Z are oppositely congruent at the common midpoint of AX, BY , and CZ. PY PZ If the lines BY and CZ intersect at P , then PB = PC = t. If the lines PX and AC XZ PZ intersect at A , then AC = PC = t. This shows that A C = AC (as lengths of directed segments), and A = A. Thus, AX, BY , and CZ concur at P . In this case, X, Y , Z are the images of A, B, C under the homothety h(P, t). We call P the homothetic center of the triangles. It divides corresponding points of triangles ABC and XY Z in the ratio 1:−t. 6 Preliminaries

1.6 The power of a point with respect to a circle

The power of a point P with respect to a circle C = O(R) is the quantity

C(P ):=OP 2 − R2.

This is positive, zero, or negative according as P is outside, on, or inside the circle C.Ifit is positive, it is the square of the length of a tangent from P to the circle.

Theorem 1.2 (Intersecting chords). If a line L through P intersects a circle C at two points X and Y , the product PX · PY (of signed lengths) is equal to the power of P with respect to the circle.

X Y P

1.7 Directed angles

A reference triangle ABC 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 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 1.3. (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. 1.8 Menelaus and Ceva theorems 7

1.8 Menelaus and Ceva theorems

1.8.1 Menelaus and Ceva Theorems Consider a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively. Theorem 1.4 (Menelaus). The points X, Y , Z are collinear if and only if BX CY AZ · · = −1. (1.1) XC YA ZB

X B C

Proof. (⇒) Construct a parallel to the line XY Z through B, to intersect the line AC at Y .

Z Y

X B C

It is clear that BX CY AZ Y Y CY AY · · = · · XC YA ZB YC YA YY Y Y CY AY = · · YY YC YA =(−1)(−1)(−1) = − 1.

(⇐) 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. 8 Preliminaries

Theorem 1.5 (Ceva). The lines AX, BY , CZ are concurrent if and only if BX CY AZ · · =+1. (1.2) XC YA ZB

Y P

B X C

Proof. (⇒) Applying 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 BZ and CZ intersect at P , and AP intersects BC at X , then BX CY AZ · · =+1. XC YA ZB

BX BX Comparsion 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 intersect at P . 1.8 Menelaus and Ceva theorems 9

1.8.2 Desargues Theorem As a simple illustration of the use of the Menelaus and Ceva theorems, we prove the following Desargues Theorem.

Proposition 1.6. 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), 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. 10 Preliminaries Chapter 2

The circumcircle and the incircle

For a generic triangle ABC, we shall denote by (i) a, b, c the sidelines BC, CA, AB respectively, and (ii) a, b, c their lengths.

B D C

2.1 The circumcircle and the law of sines

The circumcircle of triangle ABC is the unique circle passing through the three vertices A, B, C. Its center, the circumcenter O, is the intersection of the perpendicular bisectors of the three sides. The directed angle ∠(OB, OC)=2α. The circumradius R is given by the law of sines: a b c 2R = = = . sin α sin β sin γ 12 The circumcircle and the incircle

2.2 The incircle and the Gergonne point

The incircle is tangent to each of the three sides BC, CA, AB (without extension). Its center, the incenter I, is the intersection of the bisectors of the three angles. The inradius r is related to the area ∆ by 1 ∆= (a + b + c)r. 2

Z Ge I

B X C

If the incircle is tangent to the sides BC at X, CA at Y , and AB at Z, then

b + c − a c + a − b a + b − c AY = AZ = ,BZ= BX = ,CX= CY = . 2 2 2

1 These expressions are usually simpliﬁed by introducing the semiperimeter s = 2 (a+b+c):

AY = AZ = s − a, BZ = BX = s − b, CX = CY = s − c.

∆ Also, r = s . It follows easily from the Ceva theorem that AX, BY , CZ are concurrent. The point of concurrency Ge is called the Gergonne point of triangle ABC. Triangle XY Z is called the intouch triangle of ABC. Clearly,

β + γ γ + α α + β X = ,Y= ,Z= . 2 2 2

It is always acute angled, and

α β γ YZ=2r cos ,ZX=2r cos ,XY=2r cos . 2 2 2 2.2 The incircle and the Gergonne point 13

Y X

C Z A

Exercise ∠ a ≡ α 1. (i) (IA, ) 2 + γ mod π, ∠ ≡− β γ ≡ π α (ii) (IB,IC) 2 + 2 2 + 2 mod π.

2. Given three points A, B, C not on the same line, construct three circles, with centers at A, B, C, mutually tangent to each other externally.

3. 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. 1

Ge I

B C

4. Two circles are orthogonal to each other if their tangents at an intersection are perpendicular to each other. Given three points A, B, C not on a line, construct three circles with these as centers and orthogonal to each other. √ (4R+r)2+s2 1 This is called the Adams circle. It is concentric with the incircle, and has radius 4R+r · r. 14 The circumcircle and the incircle

(1) Construct the tangents from A to the circle B(b), and the circle tangent to these two lines and to A(a) internally. (2) Construct the tangents from B to the circle A(a), and the circle tangent to these two lines and to B(b) internally. (3) The two circles in (1) and (2) are congruent.

5. Given a point Z on a line segment AB, construct a right-angled triangle ABC whose incircle touches the hypotenuse AB at Z. 2

6. Let ABC be a triangle with incenter I. (1a) Construct a tangent to the incircle at the point diametrically opposite to its point of contact with the side BC. Let this tangent intersect CA at Y1 and AB at Z1.

(1b) Same in part (a), for the side CA, and let the tangent intersect AB at Z2 and BC at X2.

(1c) Same in part (a), for the side AB, and let the tangent intersect BC at X3 and CA at Y3.

(2) Note that AY3 = AZ2. Construct the circle tangent to AC and AB at Y3 and Z2. How does this circle intersect the circumcircle of triangle ABC?

7. The incircle of ABC touches the sides BC, CA, AB at D, E, F respectively. X is a point inside ABC such that the incircle of XBC touches BC at D also, and touches CX and XB at Y and Z respectively. (1) The four points E, F , Z, Y are concyclic. 3 (2) What is the locus of the center of the circle EFZY ? 4

2P. Yiu, G. Leversha, and T. Seimiya, Problem 2415 and solution, Crux Math. 25 (1999) 110; 26 (2000) 62 – 64. 3International Mathematical Olympiad 1996. 4IMO 1996. 2.3 The Heron formula 15

2.3 The Heron formula

The area of triangle ABC is given by ∆= s(s − a)(s − b)(s − c).

This formula can be easily derived from a computation of the inradius r and the radius of one of the tritangent circles of the triangle. Consider the excircle Ia(ra) whose center is the intersection of the bisector of angle A and the external bisectors of angles B and C, tangent to the sidelines BC, CA, AB at Aa, Ba, and Ca respectively.

s − a

Y r Z − I s c

C B X Aa s − b

Ba s − c

ra Ca

From the similarity of triangles AIY and AIaBa, r s − a = . ra s

From the similarity of triangles IY C and CBaIa, s − c r = a . r s − b From these two equations,

(s − a)(s − b)(s − c) s(s − b)(s − c) r2 = ,r2 = . s a s − a Similarly, s(s − c)(s − a) s(s − a)(s − b) r2 = ,r2 = . b s − b c s − c 16 The circumcircle and the incircle

Exercise 1. Make use of the fact that A B B C C A tan tan +tan tan +tan tan =1 2 2 2 2 2 2 to prove that r2s =(s − a)(s − b)(s − c).

2. Show that 16 2 =2b2c2 +2c2a2 +2a2b2 − a4 − b4 − c4.

abc 3. R = 4∆ . ∆ 4. ra = s−a . 5. Let AA be the bisector of angle A of triangle ABC. Show that the incenter I and the excenter Ia divide AA harmonically.

X Y

Z I

A C B X X

6. Suppose the A-excircle touches BC at X . Show that the antipode X of X on the incircle lies on the segment AX . 2.4 The excircles and the Nagel point 17

2.4 The excircles and the Nagel point

Let X , Y , Z be the points of tangency of the excircles (Ia), (Ib), (Ic) with the corresponding sides of triangle ABC. The lines AX , BY , CZ are concurrent. The common point Na is called the Nagel point of triangle ABC.

Z Y Z Na Y I

C B X X

Ia 18 The circumcircle and the incircle

Exercise 1. Let D, E, F be the midpoints of the sides BC, CA, AB, and let the incircle touch these sides at X, Y , Z respectively. The lines through X parallel to ID, through Y to IE and through Z to IF are concurrent. 5

Y F E Z I P

B X D C

2. Construct the tritangent circles of a triangle ABC.

Ic Z Y Mi Z Y T I

C B X X

Ia (1) Join each excenter to the midpoint of the corresponding side of ABC. These three lines intersect at a point Mi. (This is called the Mittenpunkt of the triangle). (2) Join each excenter to the point of tangency of the incircle with the corresponding side. These three lines are concurrent at another point T .

5 Crux Math. Problem 2250. The reﬂection of the Nagel point N a in the incenter. This is X145 of ETC. 2.4 The excircles and the Nagel point 19

(3) The lines AMi and AT are symmetric with respect to the bisector of angle A;so are the lines BMi, BT and CMi, CT (with respect to the bisectors of angles B and C).

3. Construct the excircles of a triangle ABC. (1) Let D, E, F be the midpoints of the sides BC, CA, AB. Construct the incenter 6 Sp of triangle DEF, and the tangents from S to each of the three excircles. (2) The 6 points of tangency are on a circle, which is orthogonal to each of the excircles.

Y Sp

C B X

6This is called the Spieker point of triangle ABC. 20 The circumcircle and the incircle

2.5 Euler’s formula and Steiner’s porism

2.5.1 Euler’s formula The distance between the circumcenter and the incenter of a triangle is given by

OI2 = R2 − 2Rr.

Let the bisector of angle A intersect the circumcircle at M. Construct the circle M(B) to intersect this bisector at a point I. This is the incenter since 1 1 1 ∠IBC = ∠IMC = ∠AMC = ∠ABC, 2 2 2 ∠ 1 ∠ and for the same reason ICB = 2 ACB. Note that A (1) IM = MB = MC =2R sin 2 , r (2) IA = sin A , and 2 (3) by the theorem of intersecting chords, OI2 − R2 = the power of I with respect to the circumcircle = IA · IM = −2Rr.

I O

B C

M 2.5 Euler’s formula and Steiner’s porism 21

2.5.2 Steiner’s porism Construct the circumcircle (O) and the incircle (I) of triangle ABC. Animate a point A on the circumcircle, and construct the tangents from A to the incircle (I). Extend these tangents to intersect the circumcircle again at B and C. The lines BC is always tangent to the incircle. This is the famous theorem on Steiner porism: if two given circles are the circumcircle and incircle of one triangle, then they are the circumcircle and incircle of a continuous family of poristic triangles.

Y C

Z I O

B X C

Exercise ≤ 1 1. r 2 R. When does equality hold?

2. Suppose OI = d. Show that there is a right-angled triangle whose sides are d, r and R − r. Which one of these is the hypotenuse?

3. Given a point I inside a circle O(R), construct a circle I(r) so that O(R) and I(r) are the circumcircle and incircle of a (family of poristic) triangle(s).

4. Given the circumcenter, incenter, and one vertex of a triangle, construct the triangle.

5. Construct an animation picture of a triangle whose circumcenter lies on the incircle. 7

6. What is the locus of the centroids of the poristic triangles with the same circumcircle and incircle of triangle ABC? How about the orthocenter?

7Hint: OI = r. 22 The circumcircle and the incircle

7. Let ABC be a poristic triangle with the same circumcircle and incircle of triangle ABC, and let the sides of BC, CA, AB touch the incircle at X, Y , Z. (i) What is the locus of the centroid of XY Z? (ii) What is the locus of the orthocenter of XY Z? (iii) What can you say about the Euler line of the triangle XY Z? Chapter 3

The Euler line

3.1 The medial and antimedial triangles

3.1.1 The medial triangle, the nine-point center, and the Spieker point The three medians of a triangle intersect at the centroid G, which divides each median in the ratio 2:1. The medial triangle DEF is the image of triangle ABC under the homothety − 1 1 h(G, 2 ). The circumcircle of the medial triangle has radius 2 R. Its center is the point − 1 1 N = h(G, 2 )(O). This is called the nine-point center of triangle ABC. It divides the segment OG in the ratio OG : GN =2:1. h − 1 The image of a point P under the homothety (G, 2 ) is called its inferior, which we denote by P−.

− 1 − P = 2 (3G P ).

For example, the incenter of the medial triangle is the inferior of the incenter I, namely, 1 − 2 2 (3G I). It is called the Spieker point Sp of triangle ABC.

1 The reason for this will be clear in §?. The nine-point center appears as X 5 in ETC. 2 The Spieker point appears as X10 in ETC. 24 The Euler line

F E

G O N Sp

B D C

3.1.2 The antimedial triangle and the orthocenter The antimedial triangle ABC is the triangle bounded by the parallels of the sidelines of ABC through the corresponding opposite vertices. 3 It is homothetic to ABC at the centroid G with ratio of homothety −2. Since the altitudes of triangle ABC are the perpendicular bisectors of the sides of triangle ABC, they intersect at the homothetic image of the circumcenter O. This point is called the orthocenter of triangle ABC, and is usually denoted by H. Note that OG : GH =1:2. The image of a point P under the homothety h(G, −2) is called its superior, and is denote by P +. Thus,

P + =3G − 2P .

The incenter of the antimedial triangle is the superior of I, which is the Nagel point Na.

3It is also called the anticomplementary triangle. 3.1 The medial and antimedial triangles 25

C A B

G N O H

B C

A 3.1.3 The Euler line The four points O, G, N and H are collinear. The line containing them is called the Euler line of triangle ABC. The Euler line is undeﬁned for the equilateral triangle, since these points coincide.

O G

B C

On the Euler line is also the deLongchamps point Lo, which is the reﬂection of H in O.

HN : NG : GO : OLo =3:1:2:6. 26 The Euler line

Exercise 1. A triangle is equilateral if and only if two of its circumcenter, centroid, and orthocenter coincide.

2. The circumcenter N of the medial triangle is the midpoint of OH.

3. (a) Show that the triangle HBC has circumradius R, and deduce that its circumcenter A is the reﬂection of O in the line BC. (b) Show that AA is the Euler line of triangle HBC. (c) Deduce that the Euler lines of triangles HBC, HCA, HAB intersect at a point on the Euler line of triangle ABC. What is this intersection?

B C

A 3.1 The medial and antimedial triangles 27

4. The Euler lines of triangles IBC, ICA, IAB also intersect at a point on the Euler line of triangle ABC. 4

5. Identify the point on the Euler line where it intersects the line joining the incenter to the Gergonne point.

Z I Ge O

B X C

4Problem 1018, Crux Mathematicorum. 28 The Euler line

3.2 The nine-point circle

The famous nine-point circle of a triangle is the common circumcircle of three associated triangles: the medial triangle, 5 the Euler triangle, and the orthic triangle.

3.2.1 The Euler triangle as a midway triangle Let P be a point in the plane of triangle ABC. The midpoints of the segments AP , BP, 1 CP form the midway triangle of P . It is the image of ABC under the homothety h(P, 2 ). The midway triangle of the orthocenter H is called the Euler triangle. The circumcenter of the midway triangle of P is the midpoint of OP. In particular, the circumcenter of the Euler triangle is the midpoint of OH, which is the same as N, the circumcenter of the medial triangle. (See Exercise ??.). The medial triangle and the Euler triangle have the same circumcircle.

B C

5See §3.1.1 3.2 The nine-point circle 29

3.2.2 The orthic triangle as a pedal triangle The pedals of a point are the intersections of the sidelines with the corresponding perpendiculars through P . They form the pedal triangle of P . The pedal triangle of the orthocenter H is called the orthic triangle of ABC.

Z P

B X C

The pedal X of the orthocenter H on the side BC is also the pedal of A on the same line, and can be regarded as the reﬂection of A in the line EF. It follows that

∠EXF = ∠EAF = ∠EDF, since AEDF is a parallelogram. From this, the point X lies on the circumcircle of the medial triangle DEF; similarly for the pedals Y and Z of H on the other two sides CA and AB. The orthic triangle degenerates if ABC contains a right angle.

The ob-incenter of the orthic triangle π − Since B, X, H, Z are concyclic, (ZX,XH)=(ZB,BH)= 2 α. Simiarly, (YX,XH)= − π − (YC,CH)= 2 α . It follows that XH bisects angle ZXY . Similarly, H lies on the bisectors of angles XY Z and YZX. If triangle ABC is acute, then H is the incenter of the orthic triangle XY Z. When ABC is obtuse, H is an excenter of the orthic triangle. We say that H is the ob-incenter of the orthic triangle. Remark. The orthocenter is the only point which is an incenter (or excenter) of its own cevian triangle. 30 The Euler line

Z H

B X C

Z Y A

B X C

3.2.3 The nine-point circle From §§3.2.1, 3.2.2 above, the medial triangle, the Euler triangle, and the orthic triangle all have the same circumcircle. This is called the nine-point circle of triangle ABC. Its center N, the midpoint of OH, is called the nine-point center of triangle ABC. 3.2 The nine-point circle 31

A Y

F E

O N

Z H

C B

B X D C

Exercise 1. Show that the incenter is the orthocenter of the excentral triangle.

2. Let P be a point on the circumcircle. What is the locus of the midpoint of HP? Why?

3. If the midpoints of AP , BP, CP are all on the nine-point circle, must P be the orthocenter of triangle ABC? 6

4. Let ABC be a triangle and P a point. The perpendiculars at P to PA, PB, PC intersect BC, CA, AB respectively at A, B, C. (1) A, B, C are collinear. 7 (2) The nine-point circles of the (right-angled) triangles PAA, PBB, PCC are concurrent at P and another point P . Equivalently, their centers are collinear. 8

6P. Yiu and J. Young, Problem 2437 and solution, Crux Math. 25 (1999) 173; 26 (2000) 192. 7B. Gibert, Hyacinthos 1158, 8/5/00. 8A.P. Hatzipolakis, Hyacinthos 3166, 6/27/01. The three midpoints of AA , BB, CC are collinear. The three nine-point circles intersect at P and its reﬂection in this line. 32 The Euler line

P C A B C

5. (Triangles with nine-point center on the circumcircle) Begin with a circle, center O and a point N on it, and construct a family of triangles with (O) as circumcircle and N as nine-point center. (1) Construct the nine-point circle, which has center N, and passes through the midpoint M of ON. (2) Animate a point D on the minor arc of the nine-point circle inside the circumcircle. (3) Construct the chord BC of the circumcircle with D as midpoint. (This is simply the perpendicular to OD at D). (4) Let X be the point on the nine-point circle antipodal to D. Complete the parallelogram ODXA (by translating the vector DO to X). The point A lies on the circumcircle and the triangle ABC has nine-point center N on the circumcircle. Here is a curious property of triangles constructed in this way: let A, B, C be the reﬂections of A, B, C in their own opposite sides. The reﬂection triangle ABC degenerates, i.e., the three points A, B, C are collinear. 9

9O. Bottema, Hoofdstukken uit de Elementaire Meetkunde, Chapter 16. Chapter 4

The OI-line

The OI-line of a triangle is the line joining the circumcenter and the incenter. We consider several interesting triangle centers on this line. Many of these are associated with the intouch and the excentral triangles.

4.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle

These are the centers of similitude of the circumcircle and the incircle. These are the points T+ and T− which divide the segment OI harmonically in the ratio of the circumradius and the inradius. 1 T+ = (r · O + R · I), R + r 1 T− = (−r · O + R · I). R − r

M A

I Z T− O T+

B X C

We give an interesting application of these centers of similitude. 34 The OI-line

4.1.1 Construction of mixtilinear incircles A mixtilinear incircle of triangle ABC is one that is tangent to two sides of the triangle and to the circumcircle internally. Denote by A the point of tangency of the mixtilinear incircle K(ρ) in angle A with the circumcircle. The center K clearly lies on the bisector of angle A, and AK : KI = ρ : −(ρ − r). In terms of barycentric coordinates, 1 K = (−(ρ − r)A + ρI) . r Also, since the circumcircle O(A) and the mixtilinear incircle K(A) touch each other at A,wehaveOK : KA = R − ρ : ρ, where R is the circumradius. From this, 1 K = (ρO +(R − ρ)A) . R

A A

O O I I T− K KA

B C B C

A A M Comparing these two equations, we obtain, by rearranging terms,

RI − rO R(ρ − r)A + r(R − ρ)A = . R − r ρ(R − r)

We note some interesting consequences of this formula. First of all, it gives the intersection of the lines joining AA and OI. Note that the point on the line OI represented by the left hand side is T−, the exsimilicenter of the circumcircle and the incircle. This leads to a simple construction of the mixtilinear incircle. Given a triangle ABC, extend AT− to intersect the circumcircle at A . The intersection of AI and A O is the center KA of the mixtilinear incircle in angle A. The other two mixtilinear incircles can be constructed similarly.

Exercise

1. Give an alterative construction of the A-mixtilinear incircle using the formula ρA = r cos2 A for its radius. 2 4.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle 35

2. Deﬁne the A-mixtilinear excircle as the one tangent to the lines AB, AC, and to the circumcircle externally. Show that the line joining A to the point of tangency with the circumcircle contains the insimilicenter T+. Construct the circle.

3. There are three circles each tangent internally to the circumcircle at a vertex, and externally to the incircle. It is known that the three lines joining the points of tangency of each circle with (O) and (I) pass through the insimilicenter T+ of similitude of (O) and (I). Construct these three circles. 1

T+ I O

B C

4. Let T+ be the insimilicenter of (O) and (I), with pedals Y and Z on CA and AB respectively. If Y and Z are the pedals of Y and Z on BC, calculate the length of Y Z. 2

Y Z T+ O I

B Z X Y C

5. Can any of the centers of similitude T± lie outside triangle ABC?

1A.P. Hatzipolakis and P. Yiu, Triads of circles, preprint. 2A.P. Hatzipolakis and P. Yiu, Pedal triangles and their shadows, Forum Geom., 1 (2001) 81 – 90. 36 The OI-line

4.2 The reﬂection of I in O

4.2.1 The circumcircle of the excentral triangle

The excentral triangle has vertices the excenters Ia, Ib, Ic. Its sides are the external bisectors of the angles of triangle ABC, so that the internal bisectors are its altitudes, and I its orthocenter. Thus, ABC is the orthic triangle of IaIbIc, and its circumcircle is the nine-point circle of the excentral triangle.

M A

O I

C B

Ia From this, we have the following interesting facts. (i) The midpoints of the segments IIa, IIb, and IIc are on the circumcircle of ABC. (ii) The midpoints of IbIc, IcIa, and IaIb are also on the circumcircle of ABC. (iii) In particular, the midpoints M of IIa and M of IbIc are antipodal on the circumcircle, and MM is the perpendicular bisector of BC.

Ic O I I

C B

Ia More surprising are the following facts about the circumcirle of the excentral triangle. 4.2 The reﬂection of I in O 37

(iv) The circumradius of the excentral circle is 2R. (v) Since the excentral triangle has nine-point center O and orthocenter I, its circumcenter is the reﬂection of I in O, i.e., I := 2O − I. Note that the line I Ia is perpendicular to BC. It therefore contains the point of tangency with the A-excircle. From this, we deduce two more interesting facts.

Proposition 4.1. The perpendiculars from three excenters to the corresponding sides concur at the reﬂection of the incenter in the circumcenter.

M A

I O I

X C Ac B D Aa Ab

Theorem 4.2. ra + rb + rc =4R + r. Proof. Consider the diameter MM of the circumcircle (of ABC) perpendicular to BC. (i) rb + rc =2· DM =2(R + OD). (ii) 2 · OD = r + I Aa = r +2R − ra. (iii) It follows that ra + rb + rc =4R + r. 38 The OI-line

4.3 The homothetic center of the intouch and excentral triangles

The intouch triangle and the excentral triangle are homothetic since their sides are perpendicular to the angle bisectors of ABC. Since these have circumradii r and 2R respectively, r − the ratio homothety is 2R . Their circumcenters being I and I =2O I, the homothetic center is the point T such that r r r I = h T, (2O − I)= 1 − T + (2O − I). 2R 2R 2R This gives (2R + r)I − 2r · O T = . 2R − r It is the point which divides OI externally in the ratio 2R + r : −2r. 3

Z T I

C B X

4.4 The line OI as the Euler line of the intouch triangle

The OI-line is the Euler line of the excentral triangle, since O and I are the nine-point center and orthocenter respectively. The corresponding sides of the intouch triangle and the excentral triangle are parallel, being perpendicular to the respective angle bisectors. Their Euler lines are parallel. Since the intouch triangle has circumcenter I, its Euler line is actually the line OI, which therefore contains its orthocenter.

3 The point T is X57 in ETC. 4.4 The line OI as the Euler line of the intouch triangle 39

4.4.1 The orthocenter of the intouch triangle h r The orthocenter of the excentral triangle being I, we apply the homothety T, 2R to ﬁnd the orthocenter of the intouch triangle: r r r h T, (I)= 1 − T + · I 2R 2R 2R r (2R + r)I − 2r · O r = 1 − + · I 2R 2R − r 2R (R + r)I − r · O = . R

This is the point which divides OI in the ratio R + r : −r. 4

Z H I O Ge Na

B X C

Exercise Not the right place! 1. Show that the orthocenter of the intouch triangle has coordinates a(b + c) b(c + a) c(a + b) : : . b + c − a c + a − b a + b − c

2. Show that this point lies on the line joining the Gergonne and Nagel points. 5

4 This is the point X65 of ETC. 2 5 a(b+c) s s−a = s−a − (s − a). 40 The OI-line

4.4.2 The centroids of the excentral and intouch triangles The centroid of the excentral triangle is the point which divides OI in the ratio −1:4. 6 h r From the homothety (T, 2R ), it is easy to see that the centroid of the intouch triangle is the point which divides OI in the ratio 3R + r : −r. 7

Y O Z T I

C B X

4.4.3 The point which divides OI in the ratio R + r : −2r This is the perspector of the excentral and the orthic triangle. 8

Exercise 1. Construct the external common tangent of each pair of excircles. These three external common tangents bound a triangle.

6 This is the point which divides the segment I I in to the ratio 1:2.ItisX165 of ETC. 7 The centroid of the intouch triangle is X354 of ETC. 8J.T. Groenman, Problem 1295, Crux Math. 16 (1990). 4.4 The line OI as the Euler line of the intouch triangle 41

(i) This triangle is perpsective with ABC. Identify the perspector. 9 (ii) Identify the incenter of this triangle. 10

2. Show that the distance between the circumcenter and the Nagel point is R − 2r. 11

3. Identify the midpoint of the Nagel point and the deLongchamps point as a point on the OI-line. 12

4. Show that the orthic triangle of the intouch triangle is homothetic to ABC. Identify the homothetic center. 13

Z O H I

B X C

5. Extend AB and AC by length a and join the two points by a line. Similarly deﬁne 14 the two other lines. The three lines bound a triangle with perspector X65

6. Let P be the point which divides OI in the ratio OP : PI = R :2r. There is a Rr circle, center P , radius R+2r , which is tangent to three congruent circles of the same radius, each tangent to two sides of the triangle. Construct these circles. 15

9Orthocenter of intouch triangle. 10Reflection of I in O. 11Feuerbach’s theorem. 12 Reflection of I in O. This is because the pedal triangle of X 40 is the cevian triangle of the Nagel point, and the reflections of the pedals of the Nagel point in the respective traces form the pedals of the de Longchamps point. 13T . 142/18/03. 15A. P. Hatzipolakis, Hyacinthos message 793, April 18, 2000. 42 The OI-line

B C

7. Let P be the centroid of the excentral triangle, with pedals X, Y , Z on the sides BC, CA, AB respectively. Show that 16 1 AY + AZ = BZ + BX = CX + CY = (a + b + c). 3

16The projections of O and I on the side BC are the midpoint D of BC, and the point of tangency D of a 1 the incircle with this side. Clearly, BD = 2 and BD = 2 (c + a − b). It follows that 4 4 1 1 BX = BD + DD = BD − BD = (3a + b − c). 3 3 3 6

1 1 Similarly, BZ = 6 (3c+b−a), and BX +BZ = 3 (a+b+c). A similar calculation shows that AY +AZ = 1 CX + CY = 3 (a + b + c). Chapter 5

Homogeneous Barycentric Coordinates

5.1 Barycentric coordinates with reference to a triangle

5.1.1 Homogeneous barycentric coordinates The notion of barycentric coordinates dates back to A. F. M¨obius (–). Given a reference triangle ABC, we put at the vertices A, B, C masses u, v, w respectively, and determine the balance point. The masses at B and C can be replaced by a single mass v + w at the v·B+w·C point X = v+w . Together with the mass at A, this can be replaced by a mass u + v + w at the point P which divides AX in the ratio AP : PX = v + w : u. This is the point with u·A+v·B+w·C 1 absolute barycentric coordinate u+v+w , provided u + v + w =0. We also say that the balance point P has homogeneous barycentric coordinates (u : v : w) with reference to ABC.

5.1.2 The centroid The midpoints of the sides are

B + C C + A A + B D = ,E= ,F= . 2 2 2 The centroid G divides each median in the ratio 2:1. Thus,

A +2D A + B + C G = = . 3 3 This is the absolute barycentric coordinate of G (with reference to ABC). Its homogeneous barycentric coordinates are simply

G =(1:1:1).

1A triple (u : v : w) with u + v + w =0does not represent any ﬁnite point on the plane. We shall say that it represents an inﬁnite point. See §?. 44 Homogeneous Barycentric Coordinates

A A

Y F E Z I G

B D C B X C

5.1.3 The incenter The bisector AX divides the side BC in the ratio BX : XC = c : b. This gives X = bB+cC ca b+c . Note that BX has length b+c . Now, in triangle ABX, the bisector BI divides AX ca in the ratio AI : IX = c : b+c = b + c : a. It follows that aA +(b + c)X aA + bB + cC I = = . a + b + c a + b + c The homogeneous barycentric coordinates of the incenter are I =(a : b : c).

5.1.4 The Gergonne point

We follow the same method to compute the coordinates of the Gergonne point Ge. Here, BX = s − b and XC = s − c, so that (s − b)b +(s − c)C X = . a The ratio AGrme : GeX, however, is not immediate obvious. It can nevertheless be found by applying the Menelaus theorem to triangle ABX with transversal CZ. Thus,

AGe XC BZ · · = −1. GeX CB ZA From this, AGe CB ZA −a s − a a(s − a) = − · = − · = . GeX XC BZ s − c s − b (s − b)(s − c) Therefore, (s − b)(s − c)A + a(s − a)X Ge = (s − b)(s − c)+a(s − a) (s − b)(s − c)A +(s − a)(s − c)B +(s − a)(s − b)C = . (s − b)(s − c)+a(s − a) 5.1 Barycentric coordinates with reference to a triangle 45

Z I Ge

B X C The homogeneous barycentric coordinates of the Gergonne point are

Ge =(s − b)(s − c):(s − c)(s − a):(s − a)(s − b) 1 1 1 = s−a : s−b : s−c .

5.1.5 Cevian triangle It is clear that the calculations in the preceding section applies in the general case. We summarize the results in the following useful alternative of the Ceva theorem. Theorem 5.1 (Ceva). Let X, Y , Z be points on the lines BC, CA, AB respectively. The lines AX, BY , CZ are collinear if and only if the given points have coordinates of the form X =(0:y : z), Y =(x :0:z), Z =(x : y :0), for some x, y, z. If this condition is satisﬁed, the common point of the lines AX, BY , CZ is P =(x : y : z). Remarks. (1) 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 triangle ABC). Sometimes, we shall adopt the more functional notation for the cevian triangle and its vertices:

cev(P ): AP =(0:y : z),BP =(x :0:z),CP =(x : y :0).

(2) The point P divides the segment AX in the ratio PX : AX = x : x + y + z. (3) It follows that the areas of the oriented triangles PBC and ABC are in the ratio ∆(PBC):∆(ABC)=x : x + y + z. This leads to the following interpretation of homogeneous barycentric coordinates: the homogeneous barycentric coordinates of a point P can be taken as the proportions of (signed) areas of oriented triangles:

P =∆(PBC):∆(PCA):∆(PAB). 46 Homogeneous Barycentric Coordinates

A A

BP CP

P P

B AP C B X C

5.1.6 The Nagel point and the extouch triangle

The A-excircle touches the side BC at a point X such that BX = s − c and XC = s − b. From this, the homogeneous barycentric coordinates of X are 0:s − b : s − c; similarly for the points of tangency Y and Z of the B- and C-excircles:

Ic Z Na Y

C − B s − c X s − b s b

s − c

X =(0 : s − b : s − c), Y =(s − a :0:s − c), Z =(s − a : s − b :0), 5.1 Barycentric coordinates with reference to a triangle 47

From these we conclude that AX , BY , and CZ concur. Their common point is called the Nagel point and has coordinates

Na =(s − a : s − b : s − c).

The triangle X Y Z is called the extouch triangle.

5.1.7 The orthocenter and the orthic triangle For the orthocenter H with traces X, Y , Z on BC, CA, AB respectively, we have BX = c cos B, XC = b cos C. This gives

cos B cos C BX : XC = c cos B : b cos C = : = S : S ; b c β γ similarly for the other two traces.

X =0:1 : 1 Sβ Sγ 1 1 Y = Sα :0:Sγ Z = 1 : 1 :0 Sα Sβ H = 1 : 1 : 1 Sα Sβ Sγ

A A

O Z H

B X C B C

The triangle XY Z is called the orthic triangle.

5.1.8 The circumcenter For the circumcenter, it is easier to compute the coordinates in terms of areas.

O =(a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2)). 48 Homogeneous Barycentric Coordinates

Proof. If R denotes the circumradius, the coordinates of the circumcenter O are 2 O = OBC : OCA : OAB 1 1 1 = R2 sin 2α : R2 sin 2β : R2 sin 2γ 2 2 2 =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).

5.1.9 The excenters

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

Proof.

Ia =∆(IaBC):∆(AIaC):∆(ABIa) 1 1 1 = − a · r : b · r : c · r 2 a 2 a 2 a = − a : b : c.

5.1.10 The barycenter of the perimeter Consider the barycenter (center of mass) of the perimeter of triangle ABC. The edges B+C BC, CA, AB can be replaced respectively by masses a, b, c at their midpoint D = 2 , C+A A+B E = 2 , and F = 2 . With reference to the medial triangle DEF, this has coordinates a : b : c. Since the sidelengths of triangle DEF are in the same proportions, this barycenter is the incenter of the medial triangle, also called the Spieker center Sp of ABC. The barycenter center of mass is therefore the point · B+C · C+A · A+B a · D + b · E + c · F a 2 + b 2 + c 2 (b + c)A +(c + a)B +(a + b)C Sp = = = . a + b + c a + b + c 2(a + b + c) In homogeneous barycentric coordinates,

Sp =(b + c : c + a : a + b) .

2 In ETC, the circumcenter appears as X3. 5.1 Barycentric coordinates with reference to a triangle 49

c b

s − c Aa s − b C B s − b

Ba s − c ra

ra Ca

ra Ia

F E

B D C

5.1.11 The nine-point center

O+H The nine point is the midpoint of the circumcenter and the orthocenter: N = 2 . In the expressions for the coordinates of O and H O =a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2), H =(c2 + a2 − b2)(a2 + b2 − c2):(a2 + b2 − c2)(b2 + c2 − a2) :(b2 + c2 − a2)(c2 + a2 − b2) both have coordinate sum 16 2. The a-coordinate of N can be taken as a2(b2 + c2 − a2)+(c2 + a2 − b2)(a2 + b2 − c2)=a2(b2 + c2) − (b2 − c2)2. This gives N =(a2(b2 + c2) − (b2 − c2)2 : b2(c2 + a2) − (c2 − a2)2 : c2(a2 + b2) − (a2 − b2)2). 50 Homogeneous Barycentric Coordinates

5.1.12 The centers of similitudes T± We compute the homongeneous barycentric coordinates of the insimilicenter from its abso- R·I+r·O lute barycentric coordinates T+ = R+r . It is convenient to perform calculations avoiding denominators of fractions. Begin with the known formulas

homogeneous barycentric coordinates coeﬃcient sum I = a : b : c 2s O = a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2)16 2 we equalize their sums and work with

16s 2 · I =(8 2a, 8 2b, 8 2c), 16s 2 · O =(sa2(b2 + c2 − a2),sb2(c2 + a2 − b2),sc2(a2 + b2 − c2)).

abc 2 Since R : r = 4 : s = sabc :4 , The a-component of the homogeneous barycentric coordinates of R · I + r · O can be taken as

sabc · 8 2a +4 2 · sa2(b2 + c2 − a2)+ =4s 2a2(2bc + b2 + c2 − a2) =4s 2a2((b + c)2 − a2) =4s 2a2(b + c + a)(b + c − a) =8s2 2 · a2(b + c − a).

The other two components have similar expressions obtained by cyclically permuting a, b, c. It is clear that 8s2∆2 is a factor common to the three components. Thus, in homogeneous barycentric coordinates,

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

Remark. An easy modiﬁcation of the above calculations gives the coordinates of T− = R·I−r·O · − · R−r . The a-component of the homogeneous barycentric coordinates of R I r O can be taken as

sabc · 8 2a − 4 2 · sa2(b2 + c2 − a2)+ =4s 2a2(2bc − b2 − c2 + a2) =4s 2a2(a2 − (b − c)2) =4s 2a2(a − b + c)(a + b − c).

This gives a2 b2 c2 T− = b+c−a : c+a−b : a+b−c .

Exercise

1. Find the coordinates of the excenter Ia by using the fact that AIa : AI = s : s − a. 5.1 Barycentric coordinates with reference to a triangle 51

2. Find the coordinates of P =(u : v : w) in its cevian triangle. 3

3. Show that the sum of the coordinates of O given above is 16 2.

4. Let XY Z be the intouch triangle of ABC. Show that the lines IX and YZintersect on the A-median. 4

5. Given triangle ABC, construct a circle tangent to AC at Y and AB at Z such that the line YZpasses through the centroid G. Show that YG: GZ = c : b. 5

Y G

B C

6. Establish the coordinates of the following points on the OI-line: a b c (a) T = b+c−a : c+a−b : a+b−c , (b) I =(a(a3 + a2(b+ c) −a(b+ c)2 −(b+ c)(b−c)2):···: ···), where the second and third coordinates are obtained from the ﬁrst by cyclically permuting a, b, c, (c) the reﬂection of O in I: 2I − O =(a(a3 − 2a2(b + c) − a(b2 − 4bc + c2)+2(b − c)2(b + c)) : ···: ···).

7. Find a point P for which the midpoints of the segments AP , BP, CP are equidistant from the sides BC, CA, AB respectively. The three perpendiculars are concurrent. Identify the point of concurrency. 6

3Let XYZ be the cevian triangle of P =(u : v : w). From 2(u, v, w)=(0,v,w)+(u, 0,w)+(u, v, 0), we have 2(u + v + w)P =(v + w)X +(w + u)Y +(u + v)Z. It follows that with reference to XYZ, P =(v + w : w + u : u + v). 4The intersection of IX and YZis the point (a : s − a : s − a). 5If AY = AZ = t, Y =(b − t :0:t) and Z =(c − t : t :0). If the line YZ contains I, then λ(b − t, 0,t)+µ(c − t, t, 0) = (1, 1, 1) for some λ and µ. Considering the last two coordinates, we take λ = µ =1. Thus, b·Y +c·Z =(b+c)G, and YG: GZ = c : b. Also, from (b+c−2t : t : t)=(1:1:1), b+c we have t = 3 . The line joining G to the center of the circle is perpendicular to BC. 6P =(3a − b − c :3b − c − a :3c − a − b) is the Nagel point of the superior triangle. The common distance is the diameter of the incircle. The three perpendiculars are concurrent at the reﬂection of O in I. 52 Homogeneous Barycentric Coordinates

8. Find a point P for which the midpoints of the segments AP , BP, CP are equidistant from the corresponding sides of the superior triangle. The three perpendiculars are concurrent. Identify the point of concurrency. 7

• Ma is the unique point on the segment BC such that

MaB + d(Ma,AB)=MaC + d(Ma,AC)

with d for distance; deﬁne Mb, Mc cyclically. Then MaMbMc and ABC are 2R ··· ··· 8 perspective at the point of the line IG with trilinear (1 + a : : ).

• Na is the unique point on the segment BC such that

NaB + d(Na,AC)=NaC + d(Na,AB)

with d for distance; deﬁne Nb, Nc cyclically. Then NaNbNc and ABC are 2R − ··· ··· 9 perspective at the point of the line IG with trilinear ( a 1: : ).

7P is the Nagel point. The common distance is the radius of the incircle. The three perpendiculars are concurrent at the midpoint of the orthocenter and the Nagel point. 8 wa va Let Ma =(0:v : w). We require v+w (1 + sin B)= v+w (1 + sin C). This means v : w =1+sinB : 1+sinC. Ma =(0:1+sinB :1+sinC). This gives M =(1+sinA :1+sinB :1+sinC)=(2R + a :2R + b :2R + c)=2R · 3G +2s · I.

This point divides IG in the ratio IM : MG =3R : s. (E. Lemoine, Suite de theoremes et de resultats concernant la geometrie du triangle, Congr`es de Paris1872(?). Check Hyacinthos mesage 8752). 9E. Lemoine, Suite de theoremes et de resultats concernant la geometrie du triangle, Congr`es de Paris1872(?). Check Hyacinthos mesage 8752. 5.2 The area formula 53

5.2 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 , x3 y3 z3 where 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 ∆. (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 The area of GIO is

111 1 abc · 3(a + b + c) · 16 2 a2(b2 + c2 − a2) b2(c2 + a2 − b2) c2(a2 + b2 − c2) 001 1 = a − bb− cc · 3(a + b + c) · 16 2 −(a − b)(a + b)(a2 + b2 − c2) −(b − c)(b + c)(b2 + c2 − a2) c2(a2 + b2 − c2) 001 (a − b)(b − c) = 11c · 3(a + b + c) · 16 2 −(a + b)(a2 + b2 − c2) −(b + c)(b2 + c2 − a2) c2(a2 + b2 − c2) (a − b)(b − c) 11 = · 3(a + b + c) · 16 2 −(a + b)(a2 + b2 − c2) −(b + c)(b2 + c2 − a2) (a − b)(b − c)(c − a)(a + b + c)2 = · 3(a + b + c)(a + b + c)(b + c − a)(c + a − b)(a + b − c) (a − b)(b − c)(c − a) = · 3(b + c − a)(c + a − b)(a + b − c) 54 Homogeneous Barycentric Coordinates Chapter 6

Straight lines

6.1 Equations of straight lines

6.1.1 Two-point form The area formula has an easy and extremely important consequence: the equation of the line joining two points with coordinates (x1 : y1 : z1) and (x2 : y2 : z2) is x1 y1 z1 x2 y2 z2 =0, xyz or (y1z2 − y2z1)x +(z1x2 − z2x1)y +(x1y2 − x2y1)z =0.

Examples (1) The equations of the sidelines BC, CA, AB are respectively x =0, y =0, z =0. (2) Given a point P =(u : v : w), the cevian line AP has equation wy − vz =0; similarly for the other two cevian lines BP and CP. These lines intersect corresponding sidelines at the traces of P :

AP =(0:v : w),BP =(u :0:w),CP =(u : v :0). (3) 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. (4) The equations of some important lines: 56 Straight lines 2 − 2 2 2 − 2 Euler line OH cyclic(b c )(b + c a )x =0 − − OI-line OI cyclic bc(b c)(b + c a)x =0 − − 2 Soddy line IGe cyclic(b c)(s a) x =0 2 2 2 − 2 Brocard axis OK cyclic b c (b c )x =0 van Aubel line HK cyclic Sαα(Sβ − Sγ )x =0

6.1.2 Intersection of two lines The intersection of the two lines

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

Proposition 6.1. 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

Examples (1) The intersection of the Euler line and the Soddy line is the point (c − a)(s − b)2 (a − b)(s − c)2 : ···: ··· (c2 − a2)(c2 + a2 − b2)(a2 − b2)(a2 + b2 − c2) − 2 − 2 − − (s b) (s c) ··· ··· =(c a)(a b) 2 2 − 2 2 2 − 2 : : (c + a)(c + a b )(a + b)(a + b c ) − 2 − − − (s b) a(b c) ··· ··· =(c a)(a b) 2 2 − 2 − 2 : : (c + a)(c + a b )(b c)(a + b + c) − 2 − − − (s b) a ··· ··· =(b c)(c a)(a b) 2 2 − 2 2 : : (c + a)(c + a b )(a + b + c) 2 1 (c + a − b) 4a = (b − c)(c − a)(a − b) 2 2 2 2 : ···: ··· 4 (c + a)(c + a − b )(a + b + c) 1 = (b − c)(c − a)(a − b)(−3a4 +2a2(b2 + c2)+(b2 − c2)2):···: ··· 4 2 Writing a = Sβ + Sγ etc., we have

− 3a4 +2a2(b2 + c2)+(b2 − c2)2 2 2 = − 3(Sβ + Sγ) +2(Sβ + Sγ)(2Sα + Sβ + Sγ)+(Sβ − Sγ )

=4(Sαβ + Sγα − Sβγ). 6.1 Equations of straight lines 57

This intersection has homogeneous barycentric coordinates

−Sβγ + Sγα + Sαβ : Sβγ − Sγα + Sαβ : Sβγ + Sγα − Sαβ.

This is the reﬂection of H in O, and is called the deLongchamps point Lo.

Exercise 1. Find the equation of the line joining the centroid to a given point P =(u : v : w). 1

2. Find the equations of the angle bisectors and determine from them the coordinates of the incenter and excenters.

3. Show that the Nagel point lies on (the perimeter of) the inferior triangle if and only if the centroid lies on (the perimeter of) the intouch triangle. 2

4. Let D, E, F be the midpoints of the sides of BC, CA, AB of triangle ABC, and X, Y , Z the midpoints of the altitudes from A, B, C respectively. Find the equations of the lines DX, EY , FZ, and show that they are concurrent. What are the coordinates of their intersection? 3

X F E

H Z B D C

5. 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. Find the equations of the lines DX, EY , FZ, and show that they are concurrent. What are the coordinates of their intersection? 4

1Equation: (v − w)x +(w − u)y +(u − v)z =0. 2In this case, the Nagel point lies on the incircle, which is tangent to a side of the inferior triangle. 3This intersection is the symmedian point K =(a2 : b2 : c2). 4The intersection is the point dividing the segment PGin the ratio 3:1. 58 Straight lines

F E B X P

Y P Z

B AP D C

6.2 Perspectivity

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 concurrent. The point of concurrency, ∧(X1Y1Z1,X2Y2Z2), is called the perspector. Along with the perspector, there is an axis of perspectivity,orthe perspectrix, which is the line joining containing

Y1Z2 ∩ Z1Y2,Z1X2 ∩ X1Z2,X1Y2 ∩ Y1X2.

We denote this line by L∧(X1Y1Z1,X2Y2Z2). We justify this in §below. If one of the triangles is the triangle of reference, it shall be omitted from the notation. Thus, ∧(XY Z)=∧(ABC, XY Z) and L∧(XY Z)=L∧(ABC, XY Z). Homothetic triangles are clearly prespective. If triangles T and T, their perspector is the homothetic center, which we shall denote by ∧0(T, T ). Proposition 6.2. 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 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 6.2 Perspectivity 59 and AB respectively at (−(u − U):0:w − W ) and (u − U : −(v − V ):0). These three points are collinear on the trilinear polar of (u − U : v − V : w − W ). The triangles XY Z and ABC are homothetic if the perspectrix is the line at inﬁnity.

Exercise

1. 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.

B C

Ac B C Ab

2. Given two points P1 and P2 with cevian triangles X1Y1Z1 and X2Y2Z2, the line joining X1Y1 ∩ X2Z2 and X1Z1 ∩ X2Y2 passes through A. Similarly for the other two lines. The three lines are concurrent. Their common point has coordinates

(u1u2(v1w2 + w1v2):v1v2(w1u2 + u1w2):w1w2(u1v2 + v1u2)). 60 Straight lines

6.3 Trilinear pole and polar

If the intersections of a line L with the side lines are

X =(0:v : −w),Y=(−u :0:w),Z=(u : −v :0), the equation of the line L is x y z + + =0. u v w We shall call the point P =(u : v : w) the trilinear pole (or simply tripole) of L, and the line L the trilinear polar (or simply tripolar) of P .

Construction of tripolar of a point

Given P with traces AP , BP , and CP on the side lines, let

X = BP CP ∩ BC, Y = CP AP ∩ CA, Z = AP BP ∩ AB.

These points X, Y , Z lie on the tripolar of P .

CP P

X B AP C

Z 6.3 Trilinear pole and polar 61

Construction of tripole of a line Given a line L intersecting BC, CA, AB at X, Y , Z respectively, let

A = BY ∩ CZ, B = CZ ∩ AX, C = AX ∩ BY.

The lines AA, BB and CC intersect at the tripole P of L.

C Y

A X B C

Exercise 1. The trilinear polar of P intersects the sidelines BC, CA, AB at X, Y , Z respectively. The centroid of the degenerate triangle XY Z is called the tripolar centroid of P .If P =(u : v : w), calculate the coordinates of its tripolar centroid. 5

2. The trilinear polar of P intersects the sidelines BC, CA, AB at X, Y , Z respectively. The centroid of the degenerate triangle XY Z is called the tripolar centroid of P .If P =(u : v : w), calculate the coordinates of its tripolar centroid. 6

5(u(v − w)(v + w − 2u):···: ···). 6(u(v − w)(v + w − 2u):···: ···). 62 Straight lines

6.4 Anticevian triangles

The vertices of the anticevian triangle of a point P =(u : v : w)

−1 cev (P ): Pa =(−u : v : w),Pb =(u : −v : w),Pc =(u : v : −w) are the harmonic conjugates of P with respect to the cevian segments AAP , BBP and CCP , i.e., AP : PAP = −APa : PaAP ; similarly for Pb and Pc. This is called the anticevian triangle of P since ABC is the cevian triangle P in PaPbPc. It is also convenient to regard P , Pa, Pb, Pc as a harmonic quadruple in the sense that any three of the points constitute the harmonic associates of the remaining point.

6.4.1 Construction of anticevian triangle from trilinear polar and polar properties If the trilinear polar LP of P intersects the sidelines BC, CA, AB at X , Y , Z respectively, then the anticevian triangle cev−1(P ) is simply the triangle bounded by the lines AX, BY , and CZ.

CP Pc P

X B AP C

Pb 6.4 Anticevian triangles 63

6.4.2 Another construction of anticevian triangles Here is an alternative construction of cev−1(P ). 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 OA 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. 64 Straight lines

Examples of anticevian triangles (1) The anticevian triangle of the centroid is the superior triangle, bounded by the lines through the vertices parallel to the opposite sides. (2) The anticevian triangle of the incenter is the excentral triangle whose vertices are the excenters. (3) The vertices of the tangential triangle being

2 2 2 2 2 2 2 2 2 Ka =(−a : b : c ),Kb =(a : −b : c ),Kc =(a : b : −c ), these clearly form are the anticevian triangle of a point with coordinates (a2 : b2 : c2), which we call the symmedian point K.

K O

B C

Ka 6.4 Anticevian triangles 65

(4) Here is an interesting property of the anticevian triangle of the circumcenter. Let the perpendiculars to AC and AB at A intersect BC at Ab and Ac respectively. We call AAbAc −1 an orthial triangle of ABC. The circumcenter of AAbAc is the vertex Oa of cev (O); similarly for the other two orthial triangles.

Ab Ac B C

Exercise 1. Find the area of the anticevian triangle of (u : v : w).

2. The anticevian triangles of O and K have equal areas.

3. The areas of the orthic triangle, the reference triangle, and the anticevian triangle of O are in geometric progression.

4. Find the coordinates of P =(u : v : w) in its anticevian triangle. 7

5. If P does not lie on the sidelines of triangle ABC and is the centroid of its own anticevian triangle, show that P is the centroid of triangle ABC.

7(v + w − u : w + u − v : u + v − w). 66 Straight lines

6.5 The cevian nest theorem

Theorem 6.3. For arbitrary points P and Q, the cevian triangle cev(P ) and the anticevian triangle cev−1(Q) are always perspective. If P =(u : v : w) and Q =(u : v : w), then

−1 u v w v w u w u v ∧(cev(P ), cev (Q)) = u − u + v + w : v − v + w + u : w − w + u + v .

Proof. 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 the point u v w v w u w u v u − + + : v − + + : w − + + u v w v w u w u v whose coordinates are invariant under the above cyclic permutations. This point therefore also lies on the lines YY and ZZ. Remark. The perspectrix of the two triangles is the line 1 u v w − + + x =0. cyclic u u v w

Corollary 6.4. If T is a cevian triangle of T and T is a cevian triangle of T, then T is a cevian triangle of T.

Proof. With reference to T2, the triangle T1 is anticevian.

Remark. Suppose T = cevT(P ) and T = cevT (Q).IfP =(u : v : w) with respect to T, and Q =(u : v : w) with respect to T, then, u v w ∧(T, T)= (v + w): (w + u): (u + v) u v w with respect to triangle T. The equation of the perspectrix L∧(T, T ) is 1 −v + w w + u u + v + + x =0. cyclic u u v w

These formulae, however, are quite difﬁcult to use, since they involve complicated changes of coordinates with respect to different triangles. 6.5 The cevian nest theorem 67

We shall simply write P/Q := (cev(P ), cev−1(Q)) and call it the cevian quotient of P by Q.

6.5.1 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 .

PG/Pcoordinates IMi (a(s − a):b(s − b):c(s − c)) OK(a2 : b2 : c2) 2 2 2 KO(a Sα : b Sβ : c Sγ )

8 The point Mi = G/I is called the Mittenpunkt of triangle ABC. It is the symmedian point of the excentral triangle. The tangential triangle of the excentral triangle is homothetic to ABC at T .

Figure 6.1: The Mittenpunkt as the symemdian point of the excentral triangle

8 This appears as X9 in ETC. 68 Straight lines

6.5.2 H/P For P =(u : v : w),

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

Examples

H/G =(Sβ + Sγ − Sα : Sγ + Sα − Sβ : Sα + Sβ − Sγ) 3 2 2 2 2 H/I =(a(a + a (b + c) − a(b + c ) − (b + c)(b − c) ):···: ···) 2 2 2 H/K = a : b : c Sα Sβ Sγ H/O H/N Remarks. (1) H/G is the superior of H •. (2) H/I is a point on the OI -line, dividing OI in the ratio R + r : −2r. 9 2 2 2 (3) H/K = a : b : c is the homothetic center of the orthic and tangential trian- Sα Sβ Sγ gle. 10 (4) H/O is the orthocenter of the tangential triangle. 11 (5) H/N is the orthocenter of the orthic triangle. 12

Exercise 1. Show that H/G lies on the line GK and ﬁnd the ratio of division of the segment GK. 13

6.5.3 cev cev−1 a b c (3) 0( (Ge), (I)) = T = s−a : s−b : s−c . (4) The perspector of the intouch triangle and the tangential triangle is 14

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

9 This point appears as X46 in ETC. 10 This appears as X25 in ETC. It is a point on the Euler line. 11 This appears as X155 in ETC. 12 This appears as X52 in ETC. 134:−3. 14 This appears as X1486 in ETC. 6.5 The cevian nest theorem 69

O I H P

B C

Theorem 6.5. (1) P/P = P ; (2) If P/Q = M, then Q = P/M. 70 Straight lines

H O

B C

Figure 6.2: H/K

B C

Figure 6.3: Ge/K 6.5 The cevian nest theorem 71

CM BQ

CP BP

Q P M

B C AP

AQ AM

Figure 6.4: P/(P/Q)=Q

Exercise

1. Let X, Y , Z be the midpoints of the sides BP CP , CP AP , AP BP of the cevian triangle of P . Show that XY Z and ABC are perspective. 15

2. Given a point Q =(x : y : z), ﬁnd the point P such that cev(P ) and cev−1(Q) are homothetic. What is the homothetic center? 16

15 (u(v + w):v(w + u):w(u + v)) x + y + z =0 Perspector = , and perspectrix is the line u2 v2 w2 . 16 1 1 1 x y z P = y+z−x : z+x−y : x+y−z ; homothetic center = y+z−x : z+x−y : x+y−z . 72 Straight lines

Q P

3. Let P =(u : v : w) and Q =(u : v : w) be two given points. If

X = BP CP ∩ AAQ,Y= CP AP ∩ BBQ,Z= AP BP ∩ CCQ,

show that triangle XY Z and cev(P ) are perspective. 17

4. Let XY Z be the intouch triangle and 4 a line through the incenter I of triangle ABC. Construct the pedals X of A, Y of B, and Z of C on 4. 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.

17(uu(vw + wv):··· : ···); see J. H. Tummers, Points remarquables, associ´esa ` un triangle, Nieuw Archief voor Wiskunde IV 4 (1956) 132 – 139. O. Bottema, Une construction par rapporta ` un triangle, ibid., IV 5 (1957) 68 – 70, has subsequently shown that this is the pole of the line PQ with respect to the circumconic through P and Q. 6.5 The cevian nest theorem 73

CQ X' BQ B X P Q

CP Z Y' P Y Z'

B AP AQ C

5. Let XY Z be the intouch triangle and 4 a line through the incenter I of triangle ABC. Construct the pedals X of X, Y of Y , and Z of Z on 4. 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. 74 Straight lines

6.6 Conway’s formula

6.6.1 Conway’s Notation We shall make use of the following convenient notations introduced by John H. Conway. In- stead 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, b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 S = ,S= ,S= . α 2 β 2 γ 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 A + B + C = π, cot(A + B + C) is inﬁnite. Its denominator cot A · cot B +cotB · cot C +cotC · cot A − 1=0. 2 2 From this, Sαβ + Sβγ + Sγα = S (cot A · cot B +cotB · cot C +cotC · cot A)=S .

Examples We rewrite some of the coordinates obtained in the preceding section in terms of Conway’s notation.

2 2 2 O =(a Sα : b Sβ : c Sγ) =(Sα(Sβ + Sγ ):Sβ(Sγ + Sα):Sγ (Sα + Sβ)) H =(Sβγ : Sγα : Sαβ ) N =(2Sβγ + Sγα + Sαβ : Sβγ +2Sγα + Sαβ : Sβγ + Sγα +2Sαβ)

Exercise 1. Show that the point P dividing the segment OH in the ratio OP : PH = λ : µ has homogeneous barycentric coordinates

(2λSβγ + µSα(Sβ + Sγ ):2λSγα + µSβ(Sγ + Sα):2λSαβ + µSγ(Sα + Sβ)).

2. Establish the coordinates of the following points on the Euler line.

(a) N =(2Sβγ + Sγα + Sαβ :2Sγα + Sαβ + Sβγ :2Sαβ + Sβγ + Sγα);

(b) Lo =(−Sβγ + Sγα + Sαβ : −Sγα + Sαβ + Sβγ : −Sαβ + Sβγ + Sγα) 6.6 Conway’s formula 75

6.6.2 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 θ 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 ϕ reckoned positive if and only if it is away from the vertex A. The swing angles are chosen − π ≤ ≤ π in the range 2 θ, ϕ 2 . A

B θ ϕ C

Theorem 6.6 (Conway’s formula). In terms of the AB- and AC-swing angles θ and ϕ, the homogeneous barycentric coordinates of P are

2 (−a : Sγ + Sϕ : Sβ + Sθ).

Proof. The homogeneous barycentric coordinates of P are

area PBC :areaPCA:areaPAB a2 sin θ sin ϕ b · a sin θ c · a sin ϕ = − : · sin(ϕ + γ): · sin(θ + β) 2sin(θ + ϕ) 2sin(θ + ϕ) 2sin(θ + ϕ) ab sin(ϕ + γ) ca sin(θ + β) = − a2 : : sin ϕ sin θ = − a2 : ab cos γ + ab sin γ cot ϕ : ca cos β + ca sin β cot θ 2 = − a : Sγ + Sϕ : Sβ + Sθ. 76 Straight lines

Examples

(1) The Pythagorean conﬁguration. Consider the square BCXcXb erected externally on the side BC of triangle ABC. The swing angles of Xc with respect to the side BC are π π ∠CBX = , ∠BCX = . c 4 c 2 π π Since cot 4 =1and cot 2 =0,

2 Xc =(−a : Sγ : Sβ + S).

Similarly, 2 Xb =(−a : Sγ + S : Sβ). The coordinates of the squares erected on the other two sides can be similarly written down. They are

2 2 Yc =(Sγ : −b : Sα + S), Ya =(Sγ + S : −b : Sα); 2 2 Za =(Sβ + S : Sα : −c ), Zb =(Sβ : Sα + S : −c ).

A Yc

B C

Xb Xc 6.6 Conway’s formula 77

(2) The tangential triangle. The tangential triangle is the triangle bounded by the tangents to the circumcircle of triangle ABC at its vertices. Suppose the tangents at B and C intersect at A. The two swing angles of A are both equal to A. It follows that

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

Similarly, the other two vertices of the tangential triangle are B =(a2 : −b2 : c2) and C =(a2 : b2 : −c2).

B C

(3) The midpoint of the two arcs BC of the circumcircle of triangle ABC.

M A

C B

The midpoint M of the arc not containing the vertex A has coordinates α α −a2 : S + S · cot : S + S · cot . γ 2 β 2 78 Straight lines

Note that α α S + S · cot =S + bc sin A cot γ 2 γ 2 α =S +2bc cos2 γ 2 =Sγ + bc(1 + cos α) 1 1 = (a2 + b2 − c2)+ (2bc + b2 + c2 − a2) 2 2 =b(b + c). · α Similarly, Sβ + S cot 2 = c(b + c). It follows that M =(−a2 : b(b + c):c(b + c)). The midpoint M of the arc containing A is the antipode of M, i.e., M =2· OM. From this, M =(a2 : −b(b − c):c(b − c)).

Exercise

1. Find the midpoint of XbXc. 2. Find the coordinates of the centers of the squares erected externally on the sides BC, CA, AB of triangle ABC. Show that they form a triangle with perspector 1 1 1 : : . Sα + S Sβ + S Sγ + S

Ya Ya

Y Za Za A A Y Yc Yc

Z Z P Zb Zb

B C B C

Xb Xc Xb X Xc

3. Let X, Y , Z be the midpoints of XbXc, YcYa, ZaZb respectively. Show that XY Z is perspective with ABC by computing the coordinates of the perspector. 4. Find the vertices of the inscribed squares with a side along BC. 18

18 S Recall that this can be obtained from applying the homothety h(A, S+a2 ) to the square BCX1X2. 6.6 Conway’s formula 79

B C

Xb Xc 80 Straight lines Chapter 7

Kiepert perspectors

7.1 Jacobi’s Theorem

Theorem 7.1 (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

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) 82 Kiepert perspectors A Y D D

J E

B J C E

Figure 7.1: Jacobi’s Theorem

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. 1 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. The homogeneous barycentric coordinates of I1 can be easily written down: 2 I1 =(−a : S − S C : S − S B ). C 4 B 4 θ − 1+2 cos θ B Now, for arbitrary θ,wehavecot 2 cot 2θ = sin 2θ . Putting θ = 2 and S =2rs = 4Rrs abc 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 C − C − Similarly, SC S = ab 1+2cos 2 . It follows that 4 B C I1 = a : b 1+2cos : c 1+2cos . 2 2

1See § below. 7.1 Jacobi’s Theorem 83

I2 I3 I

B C

The coordinates of the other two centers I2 and I3 can be written down by cyclic permutations of these coordinates.

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. 2

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. 3 θ − 1+2 cos θ 3. Prove that cot 2 cot 2θ = sin 2θ .

2Floor van Lamoen. 3 Floor van Lamoen. X =(0:Sψ1 − Sψ2 : Sϕ1 − Sϕ2 ). 84 Kiepert perspectors

7.2 The Kiepert triangle K(θ)

More generally, consider similar isosceles triangles XBC, YCA, and ZAB with base angle θ = ∠XBC = ∠XCB = ∠YCA= ∠YAC = ∠ZAB = ∠ZBA. The vertices X, Y , Z form the Kiepert triangle K(θ). These have coordinates

2 2 X = − a : Sγ + Sθ : −b : Sβ + Sθ, 2 Y =Sγ + Sθ : −b : Sα + Sθ, 2 Z =Sβ + Sθ : Sα + Sθ : −c .

Proposition 7.2. The Kiepert triangle K(θ) has centroid G. 2 − 2 1−3tan2 θ 2 − 2 Proposition 7.3. bθ cθ = 4 (b c ).

7.3 The Kiepert perspector

The lines AX, BY , CZ are concurrent at a point 1 1 1 K(θ)= : : . Sα + Sθ Sβ + Sθ Sγ + Sθ

We call XY Z the Kiepert triangle and K(θ) the Kiepert perspector of parameter θ.

K(θ)

B C

Figure 7.2: The Kiepert perspector 7.3 The Kiepert perspector 85

Examples ± π 4 (1) The Fermat points F± = K 3 .

F− A X

Z F+

B C

Figure 7.3: The Fermat points

4 The positive Fermat point is also known as the ﬁrst isogonic center. It appears in ETC as the point X 13. 86 Kiepert perspectors ± π 5 (2) The The Napoleon points K 6 .

X N+ N−

B C

Figure 7.4: The Napoleon points

5 The positive Napoleon point appears in ETC as the point X 17; the negative Napoleon point appears in ETC as the point X18. 7.3 The Kiepert perspector 87

K ± π Theorem 7.4 (Napoleon). The Kiepert triangle (θ) is equilateral if and only if θ = 6 . Proof. Let K(θ)=XY Z. We calculate the length of the side YZ by applying the law of cosines to triangle AY Z in which b c AY = · cot θ, ∠YAZ =2θ + α, AZ = · cot θ. 2 2 Thus,

YZ2 =AY 2 + AZ2 − 2AY · AZ · cot YAZ cot2 θ = (b2 + c2 − 2bc cos(α +2θ)) 4 cot2 θ = (b2 + c2 − 2bc cos α cos 2θ +2bc sin α sin 2θ) 4 cot2 θ = (b2 + c2 − (b2 + c2 − a2)cos2θ +2S sin 2θ). 4

1 ± π This is symmetric in a, b, c provided cos 2θ = 2 . In that case, θ = 6 and 3 a2 + b2 + c2 √ YZ2 = + 3S . 4 2 88 Kiepert perspectors ± π (3) The Vecten points K 4 . The centers of the three squares erected externally on the sides of triangle ABC form a triangle perspective with ABC. The perspector is called the (positive) Vecten point. 6

B C

6 π This is K( 4 ), the positive Vecten point. It appears in ETC as X485. 7.3 The Kiepert perspector 89

(5) The Spieker center is a Kiepert perspector. To ﬁnd the Kiepert parameter, we write

b + c : c + a : a + b 1 1 1 = : : (c + a)(a + b) (a + b)(b + c) (b + c)(c + a) 1 1 1 = : : (a + b + c)2 − (b2 + c2 − a2) (a + b + c)2 − (c2 + a2 − b2) (a + b + c)2 − (a2 + b2 − c2) 1 1 1 = : : Sα − Sθ Sβ − Sθ Sγ − Sθ

− 1 2 2 − 2s2 − s for Sθ = 2 (a + b + c) =2s . This means that cot θ = S = r . Equivalently,

− r Sp = K( arctan s ) .

Sp I

B C

Figure 7.5: The Spieker center as a Kiepert perspector 90 Kiepert perspectors

Exercise 1. Show that every Kiepert triangle is perspective with the tangential triangle and ﬁnd the perspector.

2. Calculate the homogeneous barycentric coordinates of the midpoint of the segment 7 F+F−.

3. Inside triangle ABC, consider two congruent circles Iab(r1) and Iac(r1) tangent to each other (externally), both to the side BC, and to CA and AB respectively. Note that the centers Iab and Iac, together with their pedals on BC, form a rectangle of sides 2:1. This rectangle can be constructed as the image under the homothety 2r h(I, a ) of a similar rectangle erected externally on the side BC.

Figure

(a) Make use of these to construct the two circles. (b) Calculate the homogeneous barycentric coordinates of the point of tangency of the two circles. 8 (c) Similarly, there are two other pairs of congruent circles on the sides CA and AB. The points of tangency of the three pairs have a perspector 9 1 1 1 : : . bc + S ca + S ab + S

(d) Show that the pedals of the points of tangency on the respective side lines of ABC are the traces of 10 1 1 1 : : . bc + S + Sα ca + S + Sβ ab + S + Sγ

4. Let K(θ)=XY Z. Show that the lines joining the vertices of the K(θ) to their isogonal conjugates intersect at a point on the Euler line. Find the ratio of division of the segment GL by this point. 11

7 2 2 2 2 2 2 2 2 2 ((b − c ) :(c − a ) :(a − b ) ). This point appears as X115 in ETC. 8This divides ID (D = midpoint of BC) in the ratio 2r : a and has coordinates (a 2 : ab + S : ac + S). 9 This point appears in ETC as X1123. 10 This point appears in ETC as X1659. 11The line XX∗ has equation

2 2 2 −(Sβ − Sγ)(S +2SAθ − Sθθ)x +(Sγ + Sθ)(S − 2SBθ − Sθθ)y − (Sβ + Sθ)(S − 2SCθ − Sθθ)z =0.

This line always passes through the point

(−SBC + SCA + SAB − Sθθ : SBC − SCA + SAB − Sθθ : SBC + SCA − SAB − Sθθ), which is a point on the Euler line, dividing the segment GL in the ratio tan 2 θ : −3. 7.3 The Kiepert perspector 91

Appendix: Kiepert parameters

Center Kiepert Parameter centroid G 0 π Orthocenter H 2 − r Spieker center Sp tan θ = s Fermat points F± ±π/3 Napoleon points ±π/6 Vecten points ±π/4 1 1 1 − X76 = a2 : b2 : c2 ω 1 ··· ··· X83 = b2+c2 : : ω r2+4Rr+3R2−s2 X94 cot θ = 2rs 2rs X96 cot θ = r2+4Rr+2R2−s2 − π − Tarry point X98 θ = ( 2 ω) b+c ··· ··· 2R+r X226 = b+c−a : : cot θ = 2s 1 1 1 π X262 = 2 2 2 : 2 2 2 : 2 2 2 θ = − ω b c −a Sα c a −b Sβ a b −c Sγ 2 1 X275 = 2 : ···: ··· cot θ =cotα cot β cot γ Sα(S +Sβγ) b+c c+a a+b r2+2Rr−s2 X321 = a : b : c cot θ = 2rs 92 Kiepert perspectors

7.4 Iterated Kiepert triangles

We denote by K(θ, ϕ) the Kiepert triangle K(ϕ) of K(θ). Its vertices are the points

θ,ϕ 2 A = −(2S +(Sβ + Sγ )(Sθ + Sϕ)+2Sθϕ) 2 : S − Sθϕ + Sγ(Sθ + Sϕ) 2 : S − Sθϕ + Sβ(Sθ + Sϕ), θ,ϕ 2 B = S − Sθϕ + Sγ(Sθ + Sϕ) 2 : −(2S +(Sγ + Sα)(Sθ + Sϕ)+2Sθϕ) 2 : S − Sθϕ + Sα(Sθ + Sϕ), θ,ϕ 2 C = S − Sθϕ + Sβ(Sθ + Sϕ) 2 : S − Sθϕ + Sα(Sθ + Sϕ) 2 : −(2S +(Sα + Sβ)(Sθ + Sϕ)+2Sθϕ).

7.4.1 Some interesting properties of iterated Kiepert triangles (1). From the symmetry of these coordinates we infer that the triangles K(θ, ϕ) and K(ϕ, θ) coincide. (2) K(θ, ϕ) has perspector

1 1 1 2 : 2 : 2 . S − Sθϕ + Sα(Sθ + Sϕ) S − Sθϕ + Sβ(Sθ + Sϕ) S − Sθϕ + Sγ (Sθ + Sϕ)

(3) This point lies on the Kiepert hyperbola of ABC, since it is the perspector of K(θ), where 1 − cot θ cot ϕ cot θ = = − cot(θ + ϕ). cot θ +cotϕ We summarize this by writing

Kθ(ϕ)=K(−(θ + ϕ)).

K − 1 − (4) The triangle (θ, θ) is homothetic to ABC at G, with ratio of homothety 4 (1 3tan2 θ), as Floor has computed. Its vertices are

θ,−θ − 2 − 2 2 2 2 2 A = 2(S Sθ ): S + Sθ : S + Sθ , θ,−θ 2 2 − 2 − 2 2 2 B = S + Sθ : 2(S Sθ ): S + Sθ , θ,−θ 2 2 2 2 − 2 − 2 C = S + Sθ : S + Sθ : 2(S Sθ ).

(5) The iterated Kiepert triangle K(θ, θ) has vertices

θ,θ 2 2 2 2 2 2 A = −2(S +(Sβ + Sγ)Sθ + Sθ ): S − Sθ +2SγSθ : S − Sθ +2SβSθ, θ,θ 2 2 2 2 2 2 B = S − Sθ +2SγSθ : −2(S +(Sγ + Sα)Sθ + Sθ ): S − Sθ +2SαSθ, θ,θ 2 2 2 2 2 2 C = S − Sθ +2SβSθ : S − Sθ +2SαSθ : −2(S +(Sα + Sβ)Sθ + Sθ ). 7.5 Perspectivity with the superior and excentral triangles 93

7.4.2 Lemoine’s problem Property (4) above yields a very elegant solution of Lemoine’s problem: 12 given the vertices of the equilateral triangles externally on the sides of ABC , to construct the triangle. T K π This is the construction problem of ABC from := 3 . ABC is the medial triangle K − π T of the Kiepert triangle 3 of .

7.5 Perspectivity with the superior and excentral triangles

Theorem 7.5. The excentral triangle is perspective with every Kiepert triangle: the perspector is the point dividing I Mi in the ratio (4R + r)cosθ : −s · sin θ. 2 Proof. The line joining Ia =(−a : b : c) to X(θ)=(−a : Sγ + Sθ : Sβ + Sθ) is

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

Similarly, the lines IbY (θ) and IcZ(θ) have equations

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

−c((s − b)(s − c) − Sθ)x + c((s − c)(s − a) − Sθ)y +(a − b)(s(s − c) − Sθ)z =0. The intersection is the point

3 2 2 2 [a(a + a (b + c) − a(b + c) − (b + c)(b − c) )+4(s − a)Sθ].

This is a point on the line joining X40 (the reﬂection of the incenter in the circumcenter) to the Mittenpunkt Mi =(a(s − a):b(s − b):c(s − c)). Since the sum of the coordinates are −4S2 and 8r(4R + r). This is the point

2 −4S · I +8r(4R + r)S · Mi ∼−s · sin θ · I +(4R + r)cosθ · Mi.

Exercise 1. Show that every Kiepert triangle is perspective with the superior triangle and ﬁnd the perspector. 13 2. Find the Kiepert triangle K(θ) which is perspective with the excentral triangle at the Spieker center. 14 3. Find the Kiepert triangle K(θ) which is perspective with the excentral triangle at the orthocenter. 15

12 This problem was ﬁrst solved by Ludwig Kiepert. 13 (K(θ), cev−1(G)) = sup(K(θ)). 14 r θ =arctans . 15 2R+r θ =arctan s . 94 Kiepert perspectors Chapter 8

Parallel and perpendicular lines

8.1 Inﬁnite points and parallel lines

8.1.1 The infinite point of a line A line px + qy + rz =0contains the point (q − r : r − p : p − q), as is 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) Here are the infinite points of some common lines.

Euler line OH (2Sβγ − Sγα − Sαβ : −Sβγ +2Sγα − Sαβ : −Sβγ − Sγα +2Sαβ) OI−line OI a(a2(b + c) − 2abc − (b + c)(b − c)2):···: ··· Soddy line IGe (−2a + b + c : a − 2b + c : a + b − 2c) Brocard axis OK a2(a2(b2 + c2) − (b4 + c4)) : ···: ··· van Aubel line HK 2a6 − a4(b2 + c2) − (b2 + c2)(b2 − c2)2 : ···: ··· 96 Parallel and perpendicular lines

Exercise 1. Find the equations of the lines through P =(u : v : w) parallel to the sidelines.

2. 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. (b) Explain why the lines DX, EY , FZ are concurrent and identify the point of concurrency.

CP X F E BP

I P Y Z

B D AP C

8.1.2 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 PQcan 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 expression 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)). 8.1 Inﬁnite points and parallel lines 97

Examples (1) 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 inﬁnite 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) (2) Therefore, the perpendicular bisector of BC has equation xyz 011 =0, −(Sβ + Sγ) Sγ Sβ or (Sβ − Sγ )x − (Sβ + Sγ)(y − z)=0.

Exercise 1. Let P =(u : v : w) be a point. Consider the triangle bounded by lines parallel to the sides of cevian triangle of P through the vertices of ABC. This is the anticevian triangle of (u(v + w):v(w + u):w(u + v)). It is homothetic to the cevian triangle of P . The homothetic center is the point

(u2(v + w):v2(w + u):w2(u + v)).

2. (a) 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 PGhas this property. 1

3. Given triangle ABC, construct a circle tangent to AC at Y and AB at Z such that the line YZpasses through the centroid G. It is known that YG: GZ = c : b. Show that the line joining G to the center of the circle is perpendicular to BC.

12/14/04. 98 Parallel and perpendicular lines

Y G

B C

8.2 Perpendicular lines

Given a line L : fx+ gy + hz =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 =(−h :0:f) and Z =(g : −f :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 −h 0 f =0 and g −f 0 =0 xy z xyz

These are

2 Sαfx+(c h − Sβf)y + Sαhz =0, 2 Sαfx+ Sαgy +(b g − Sγf)z =0.

These two perpendiculars intersect at the orthocenter of triangle AY Z, which is the point

2 2 X =(∗∗∗: Sαf(Sαh − b g + Sγf):Sαf(Sαg + Sβf − c h) ∼ (∗∗∗: Sγ (f − g) − Sα(g − h):Sα(g − h) − Sβ(h − f)).

The perpendicular from A to L is the line AX , which has equation 10 0 ∗∗∗ Sγ (f − g) − Sα(g − h) −Sα(g − h)+Sβ(h − f) =0, xy z or

−(Sα(g − h) − Sβ(h − f))y +(Sγ(f − g) − Sα(g − h))z =0. 8.2 Perpendicular lines 99

Z Y X

B C

This has inﬁnite point

(Sβ(h − f) − Sγ(f − g):Sγ (f − g) − Sα(g − h):Sα(g − h) − Sβ(f − g)).

Note that the inﬁnite point of L is (g − h : h − f : f − g). We summarize this in the following theorem.

Theorem 8.1. 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 =Sβ(h − f) − Sγ (f − g), g =Sγ(f − g) − Sα(g − h), h =Sα(g − h) − Sβ(h − f).

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

Sαff + Sβgg + Sγ hh =0.

Corollary 8.2. The perpendicular from the point (u : v : w) to the line fx+ gy + hz =0 is the line uvw f g h =0. xyz 100 Parallel and perpendicular lines

Exercise 1. 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. Find the equations of the lines DX, EY , FZ, and show that they are concurrent. What are the coordinates of their intersection? 2

F E B X P

Y P Z

B AP D C

2. Let D, E, F be the midpoints of the sides of BC, CA, AB of triangle ABC, and X, Y , Z the midpoints of the altitudes from A, B, C respectively. Find the equations of the lines DX, EY , FZ, and show that they are concurrent. What are the coordinates of their intersection? 3

X F E

H Z

B D C

2The intersection is the point dividing the segment PGin the ratio 3:1. 3This intersection is the symmedian point K =(a2 : b2 : c2). 8.2 Perpendicular lines 101

3. Find the point P whose pedals on the altitudes are equidistant from the vertices. 4 How about the point whose pedals on the bisectors are equidistant from the vertices? Answer: X361.

4This is the Nagel point. The three pedals are on the circle with diameter NH, the Fuhrmann circle. 102 Parallel and perpendicular lines

8.3 Triangles bounded by lines parallel to the sidelines

Theorem 8.3. 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 has homogeneous barycentric coordinates

[PBC]:[PCA]:[PAB]=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 = t3a +(1+t1)a + t2a =(1+t1 + t2 + t3)a, 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 the homothety of the triangle, with ratio 1+t1 + t2 + t3. Now we compare areas. Note that (1) [BZaXc]:[ABC]=t3 · t1. ∗ 1 · 3 (2) [PBC]:[BZaB ]= t1+t2+t3 t .

∗ t1 · Since [BZaB ]=[BZaXc],wehave[PBC]= 1+t1+t2+t3 [ABC]. t2 · t3 · Similarly, [PCA]= 1+t1+t2+t3 [ABC] and [PAB]= 1+t1+t2+t3 [ABC]. It follows that [PBC]:PCA]:[PAB]=t1 : t2 : t3. 8.3 Triangles bounded by lines parallel to the sidelines 103

8.3.1 The Grebe 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 the homothety h(A, 1+t1), where S + 2 2 a a a a 1+t1 = S =1+ , i.e., t1 = . Similarly, if we erect squares externally on the other a S S two sides, the outer edges of these squares are on the lines which are the images of CA, b2 c2 h 2 h 3 2 3 AB under the homotheties (B,1+t ) and (C, 1+t ) with t = S and t = S . A∗

Ca A Bc

K Cb B C

∗ ∗ B Ab Ac C The triangle bounded by the lines containing these outer edges is called the Grebe a2 b2 c2 2 2 2 triangle of ABC. It is homothetic to ABC at S : S : S =(a : b : c ), and the ratio of homothety is S + a2 + b2 + c2 1+(t1 + t2 + t3)= . S This homothetic center is called the Grebe symmedian point

K =(a2 : b2 : c2).

Remark. Note that the homothetic center remains unchanged if we replaced t1, t2, t3 by kt1, kt2, kt3 for the same nonzero k. This means that if similar rectangles are constructed on the sides of triangle ABC, the lines containing their outer edges always bound a triangle with homothetic center K.

8.3.2 Gossard Theorem 8.4 (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 oppositely congruent to ABC at a point on the Euler line.

Proof. The intercepts of the Euler line

Sα(Sβ − Sγ )x + Sβ(Sγ − Sα)y + Sγ(Sα − Sβ)z =0 104 Parallel and perpendicular lines

A Hc

Ga Ha

Z Gc Gb

X B C Hb

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),

Let P = f : g : h. The P −Euler line has equation (g − h)x +(h − f)y +(f − g)z =0. This intersects the sides at the points X =0:g − f : h − f, Y = f − g :0:h − g, Z = f − h : g − h :0. The triangle AY Z has two P −altitudes g(g − h)x +(f 2 − fg − g2 + hf)y +(f − g)gz =0, h(g − h)x +(h − f)hy +(h2 + hf − f 2 − fg)z =0. These two lines intersect at the point f(f 2 − g2 − h2 +3gh − fg − hf):g(h − f)(g − h):h(f − g)(g − h), 8.3 Triangles bounded by lines parallel to the sidelines 105 which is the P −orthocenter of triangle AY Z. The centroid of this triangle is the point

3(f 2 − g2 − h2 +3gh − fg − hf):(g − h)(h + f − 2g):−(g − h)(f + g − 2h).

The P −Euler line of this triangle is therefore the line

(g − h)2x − (g2 + h2 − f 2 − 3gh + hf + fg)(y + z)=0.

This is a line parallel to BC. Similarly, the P −Euler lines of the triangles BZX and CXY are respectively

(h − f)2y − (h2 + f 2 − g2 − 3hf + fg + gh)(z + x)=0, and (f − g)2z − (f 2 + g2 − h2 − 3fg + gh + hf)(x + y)=0. These two latter lines intersect at

A = −(g − h)2(g + h − 2f) :(h + f − 2g)(h2 + f 2 − g2 − 3hf + fg + gh) :(f + g − 2h)(f 2 + g2 − h2 − 3fg + gh + hf).

Note that A depends only on the direction of the P −Euler line. The coordinates of A, in Conway’s notation, are

− 2 − 2 − SA(SB SC ) (SAB + SAC 2SBC) − 2 2 − 2 − :(SBC + SAB 2SCA)(SAB + SBC SAC + SABC(SA 3SB + SC )) − 2 2 − 2 − :(SAC + SBC 2SAB)(SAC + SBC SAB + SABC(SA + SB 3SC )).

Similarly, we write down the coordinates of B and C. It is clear that AA, BB, CC intersect at the point

(g + h − 2f)(g2 + h2 − f 2 − 3gh + hf + fg) :(h + f − 2g)(h2 + f 2 − g2 − 3hf + fg + gh) :(f + g − 2h)(f 2 + g2 − h2 − 3fg + gh + hf).

The centroid of the triangle ABC is the point

(g + h − 2f)((f − g)2 +(h − f)2 − 2(g − h)2) :(h + f − 2g)((g − h)2 +(f − g)2 − 2(h − f)2) :(f + g − 2h)((h − f)2 +(g − h)2 − 2(f − g)2).

Since the midpoint of the two centroids is the perspector, the two triangles are indeed congruent. Since the perspector lies on the P −Euler line GP , the P −Euler lines of the two triangles ABC and ABC coincide. 106 Parallel and perpendicular lines

The perspector (homothetic center) of the P −Gossard triangle and the medial triangle is [(g − h)2(g + h − 2f)].

The homothetic center of the P −Gossard triangle and the anticomplementary triangle is g + h − 2f [ ]. g − h In the classical case, these are the points

[(b2 − c2)2(b2 + c2 − a2)2(2a4 − a2(b2 + c2) − (b2 − c2)2)] and 2a4 − a2(b2 + c2) − (b2 − c2)2 [ ]. (b2 − c2)(b2 + c2 − a2) In Conway’s notation, these are − 2 2 − ··· :(SC SA) SB(SBC + SAB 2SAC): and S + S − 2S ···: BC AB AC : ··· . SB(SC − SA) The perspector in the case of the Euler line is the point ··· :(c4 + a4 − 2b4 − 2c2a2 + a2b2 + b2c2) (−b8 + b6(c2 + a2)+b4(2c4 − 5c2a2 +2a4) −3b2(c2 − a2)2(c2 + a2)+(c2 − a2)2(c4 +3c2a2 + a4)) : ··· .

In Conway’s notation, this is ··· − 2 2 − 2 − ··· :(SBC + SAB 2SCA)(SAB + SBC SAC + SABC (SA 3SB + SC )) : .

Conway has pointed out that a parallel translation of this line results in a parallel translation of the “Gossard triangle” 8.4 Distance formula 107

8.4 Distance formula

Proposition 8.5. The square length of a vector uA + vB + wC, u + v + w =0, is given by

2 2 2 Sαu + Sβv + Sγ w .

Examples (1) The square length of the vector (q − r, r − p, p − q) parallel to the line px + qy + rz =0 is 2 2 2 Sα(q − r) + Sβ(r − p) + Sγ(p − q) . (2) The square length of the vector

(Sβ(r − p) − Sγ (p − q),Sγ(p − q) − Sα(q − r),Sα(q − r) − Sβ(r − p)) orthogonal to the line px + qy + rz =0is

2 2 2 2 (Sα(q − r) + Sβ(r − p) + Sγ (p − q) )S .

8.5 Pedal of a point on a line

Given a point P =(u : v : w) and a line L : px + qy + rz =0, we determine the pedal ( orthogonal projection) of P on L. This can be determined from the pedals of the vertices A, B, C on the same line. Now, the pedal of A on L is the point

2 2 2 (Sα(q − r) + Sβ(r − p) + Sγ (p − q) )(1, 0, 0)

+p(Sβ(r − p) − Sγ(p − q),Sγ(p − q) − Sα(q − r),Sα(q − r) − Sβ(r − p)).

Similarly, we have the pedals of B and C. It follows that for P =(u, v, w) with u + v + w =1, the pedal on L is the point

2 2 2 (Sα(q − r) + Sβ(r − p) + Sγ (p − q) )(u, v, w)

+(pu + qv + rw)(Sβ(r − p) − Sγ (p − q),Sγ(p − q) − Sα(q − r),Sα(q − r) − Sβ(r − p)).

The square distance from P to the line L is (pu + qv + rw)2S2 2 2 2 2 . (u + v + w) (Sα(q − r) + Sβ(r − p) + Sγ (p − q) )

Exercise 1. Find the point P whose pedals on the angle bisectors are equidistant from the vertices. 5

5 s−a s−b s−c a − a + b + c : ···: ···. 108 Parallel and perpendicular lines

8.6 Pedal and reﬂection triangles

8.6.1 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 inﬁnite point AH − A =(0: 2 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 Sγ u + 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

8.6.2 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 ? 6

Figure

(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.

6 P =(Sα : Sβ : Sγ ) is the isotomic conjugate of the orthocenter. It appears in ETC as the point X 69. 8.6 Pedal and reﬂection triangles 109

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 7

2 2 2 2 2 − 2 2 ··· ··· (a (Sγ Sα + 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. 8

2. Let X be the pedal of A on the side BC of triangle ABC. Complete the squares 9 AXXbAb and AXXcAc with Xb and Xc on the line BC.

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

11 (b) Calculate the coordinates of A = BAc ∩ CAb. (c) Similarly deﬁne B and C. Triangle ABC is perspective with ABC. What is the perspector? 12 (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. 13

7 This point appears in ETC as X1078. Conway calls this point the logarithm of the de Longchamps point. 8These are all on the Euler line. See G. Leversha, Problem 2358 and solution, Crux Mathematicorum,24 (1998) 303; 25 (1999) 371 –372. 9A.P. Hatzipolakis, Hyacinthos, message 3370, 8/7/01. 10 2 2 Ab =(a : −S : S) and Ac =(a : S : −S). 11A =(a2 : S : S). 12The centroid. 13 1 1 1 ( : : ) X485 Sα+S Sβ +S Sγ +S . This is called the ﬁrst Vecten point; it appears as in ETC. 110 Parallel and perpendicular lines

8.6.3 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 8.6. 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. 8.7 The Brocardians 111

8.7 The Brocardians

8.7.1 Equal-parallelians point Given triangle ABC, we want to construct a point P the three lines through which parallel to the sides cut out equal intercepts. Let P = xA + yB + zC in absolute barycentric coordinates. The parallel to BC cuts out an intercept of length (1 − x)a. It follows that the three intercepts parallel to the sides are equal if and only if 1 1 1 1 − x :1− y :1− z = : : . a b c The right hand side clearly gives the homogeneous barycentric coordinates of I •, the isotomic conjugate of the incenter I. 14 This is a point we can easily construct. Now, translating into absolute barycentric coordinates: 1 1 I• = [(1 − x)A +(1− y)B +(1− z)C]= (3G − P ). 2 2 we obtain P =3G−2I • = h(G, −2)(I•), the superior of I •, and can be easily constructed. The point P is called the equal-parallelians point of triangle ABC. 15

P G • Za I Ya I

B Xc Xb C The equal-parallelians point can be constructed as the isotomic conjugate (with respect to the superior triangle) of incenter of the same triangle.

14 The isotomic conjugate of the incenter appears in ETC as the point X 75. 15 It appears in ETC as the point X192. 112 Parallel and perpendicular lines

Exercise

1. Given a point P , the parallel to BC through P intersects AC at Ya and AB at Za. Likewise, the parallel to CA through P intersects BA and BC at Zb and Xb respectively, the parallel to AB through P intersects CB and CA at Xc and Yc respectively. Here are the coordinates

Xb =(0:v : w + u),Xc =(0:u + v : w), Yc =(u + v :0:w),Ya =(u :0:v + w), Za =(u : v + w :0),Zb =(w + u : v :0).

(a) Let X1 = BYa ∩CZa, Y1 = CZa ∩AXa, Z1 = AXa ∩BYa. Show that X1Y1Z1 is perspective with ABC and ﬁnd the perspector. 16

(b) Show that the triangle X2Y2Z2 bounded by the lines YcZb, ZaXc and XbYa with perspector P . 17

A Y2

Zb Z2

P Za Ya

B Xc Xb C

(c) Let X3 = BYc ∩ CZb, Y3 = CZa ∩ AXc, Z3 = AXb ∩ BYa. Show that X3Y3Z3 is perspective with ABC and ﬁnd the perspector. 18

(d) Let X4 = ZaBYc ∩ YaZb, Y4 = XbZa ∩ ZbXc, Z4 = YcXb ∩ XcYa. Show that 19 X4Y4Z4 is perspective with ABC and ﬁnd the perspector.

2. Calculate the homogeneous barycentric coordinates of the equal-parallelians point and the length of the equal parallelians. 20

16G 2 17 P =(u : v : w) − u : v : w If , the vertices are v+w . 18 u v w v+w : w+u : u+v . 19(u2 : v2 : w2). 20 2abc (ca + ab − bc : ab + bc − ca : bc + ca − ab). The common length of the equal parallelians is ab+bc+ca . 8.7 The Brocardians 113

3. Given a point P , the parallel to BC through P intersects AC at Ya and AB at Za. Likewise, the parallel to CA through P intersects BA and BC at Zb and Xb respectively, the parallel to AB through P intersects CB and CA at Xc and Yc respectively. Find all possible positions of P such that the segments XcXb, YaYc, ZbZa all have equal lengths. 21 A

Zb Yc

P Za Ya

B Xc Xb C 4. Let ABC be the midway triangle of a point P , and

Ya = B C ∩ CA, Za = B C ∩ AB, Zb = C A ∩ AB, Xb = C A ∩ BC, Xc = A B ∩ BC, Yc = A B ∩ CA.

(a) Determine P for which the three segments XbXc, YaYc and ZaZb have equal lengths. 22 A

Yc A

B Za Ya C

B Xc Xb C 23 (b) Determine P for which YaZa = ZbXb = XcYc.

21The isotomic conjugate of I. 22A.P. Hatzipolakis, Hyacinthos, message 3190, 7/13/01. P =(3bc−ca−ab :3ca−ab−bc :3ab−bc−ca). • This point appears in ETC as X1278. It is the reﬂection of the equal-parallelians point in I . In this case, the 2abc common length of the segment is ab+bc+ca , as in the equal-parallelians case. 23This is the point which divides I •G in the ratio I •P : PG =6:−5. It is the reﬂection of I • in the 5 equal-parallelians point. This common length is 4 of the equal parallelians. 114 Parallel and perpendicular lines

A Yc

Zb A

Za Ya B C

B Xc Xb C 8.7.2 The Brocardians The Brocardians of a point P =(u : v : w) are the points 1 1 1 1 1 1 P→ = : : and P← = : : . w u v v w u

We call these the forward and the backward Brocardians respectively.

Construction of Brocardian points 8.7 The Brocardians 115

Examples

(1) The Brocard points Ω→ and Ω← are the Brocardians of the symmedian point K. (2) The Brocardians of the incenter are called the Jerabek points: 1 1 1 1 1 1 I→ = : : and I← = : : . c a b b c a

The oriented parallels through I→ to BC, CA, AB intersect the sides CA, BC, AB at Y , Z, X such that I→Y = I→Z = I→X. Likewise, the parallels through I← to BC, CA, AB intersect the sides AB, BC, CA at Z , X , Y such that I←Z = I←X = I←Y . These 1 1 1 1 6 segments have length 4 satisfying * = a + b + c , one half of the length of the equal − 1 1 1 ··· ··· parallelians drawn through ( a + b + c : : ). A

Z Y'

I -> Y I Z' I<-

B X X' C

Figure 8.1: The Jerabek points 116 Parallel and perpendicular lines

(3) If forward parallels are drawn through the forward Brocardian point of the (positive) Fermat point F+, and intersect the sides CA, AB, BC at X, Y , Z respectively, then the triangle XY Z is equilateral. 24

Z Y'

F+-> F Z' + F+<-

B X X' C

Figure 8.2: Equilateral triangle formed by parallelians

This is the same for the backward parallels through the backward Brocardian point of F+.

24S. Bier, Equilateral triangles formed by oriented parallelians, Forum Geometricorum, 1 (2001) 25 – 32. 8.8 The orthic triangle 117

8.8 The orthic triangle

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

AH =(0:Sγ : Sβ),BH =(Sγ :0:Sα),CH =(Sβ : Sα :0).

CH BH A CH H

B AH C B AH C

If triangle ABC 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.

8.8.1 The centroid of the orthic triangle

The centroid of the orthic triangle has absolute barycentric coordinate 1 (0,Sγ,Sβ) (Sγ, 0,Sα) (Sβ,Sα, 0) Gorthic = + + 3 a2 b2 c2 (a2(b2S + c2S ),b2(c2S + a2S ),c2(a2S + b2S )) = β γ γ α α β 3a2b2c2 (a2(S2 + S ),b2(S2 + S ),c2(S2 + S )) = βγ γα αβ . 3a2b2c2

In homogeneous coordinates, this is the point 25

2 2 2 2 2 2 (a (S + Sβγ),b(S + Sγα),c(S + Sαβ)).

8.8.2 The orthocenter of the orthic triangle

O+H 26 Since the orthic triangle has circumcenter N = 2 , its orthocenter is the point

25 The centroid of the orthic triangle appears as X 51 in ETC. 26 The orthic center of the orthic triangle appears as X 52 in ETC. 118 Parallel and perpendicular lines

Horthic =3Gorthic − (O + H) (a2(S2 + S ),b2(S2 + S ),c2(S2 + S )) = βγ γα αβ a2b2c2 2 2 2 − (S + Sβγ,S + Sγα,S + Sαβ) 2S2 (a2(S2 + S )(2S2 − b2c2)), ··· , ···) = βγ 2a2b2c2S2 (a2(S2 + S )(S2 − S ), ··· , ···) = βγ αα . 2a2b2c2S2 In homogeneous barycentric coordinates, this is

2 2 2 2 2 2 2 2 2 Horthic =(a (S + Sβγ)(S − Sαα):b (S + Sγα)(S − Sββ):c (S + Sαβ)(S − Sγγ)).

8.8.3 Further notes The superior triangle of the orthic triangle has the same nine point circle as ABC. Its incircle touches the nine-point circle at the Jerabek point. See 000517. 8.9 The tangential triangle 119

8.9 The tangential triangle

The tangential triangle is bounded by the tangents to the circumcircle at the vertices of triangle ABC. The directed angles between these tangents are (2α, 2β, 2γ). 27 Its vertices are the points A =(−a2 : b2 : c2),B =(a2 : −b2 : c2),C =(a2 : b2 : −c2). It is therefore the anticevian triangle of the symmedian point K.

O A

A C

C B C

B C A

A B

The circumcircle of ABC is the incircle of the tangential triangle, provided that ABC is acute. Note that the tangential triangle degenerates when ABC contains a right angle. When ABC is obtuse, the circumcircle of ABC 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 ABC is the ob-incircle of triangle ABC.

8.9.1 The Gob homothetic center of the tangential and orthic triangles The tangential triangle is homothetic to the orthic triangle XY Z since their corresponding sides are perpendicular to the radii of the circumcircle through the vertices. Their homothetic center is the Gob perspector 28 a2 b2 c2 H/K = : : . Sα Sβ Sγ Note that the ob-incenters of these triangles are respectively the circumcenter O and the orthocenter H (of triangle ABC), it follows that H/K lies on the Euler line OH.

27 (tB, tC )=(tB,BC)+(BC,tC )=α + α =2α. 28 This is the triangle center X25 in ETC. 120 Parallel and perpendicular lines

Writing the coordinates of H/K as

2 2 2 2 2 2 2 2 (a Sβγ,b Sγα,c Sαβ)=−Sβγ(−a ,b ,c )+b c (0,Sγ,Sβ), we have 2 2 2 2 2 2 (a Sβγ + b Sγα + c Sαβ)(H/K)=−2Sαβγ · A + a b c · X

2Sαβγ This shows that the homothetic ratio is a2b2c2 .

8.9.2 The centroid of the tangential triangle

The centroid of the tangential triangle has absolute barycentric coordinate

(−a2,b2,c2) (a2, −b2,c2) (a2,b2, −c2) + + 2Sα 2Sβ 2Sγ (a2(−S + S + S ), ··· , ···) = βγ γα αβ . 2Sαβγ

In homogeneous barycentric coordinates, this is the point 29

2 2 2 a (−Sβγ + Sγα + Sαβ):b (Sβγ − Sγα + Sαβ ):c (Sβγ + Sγα − Sαβ).

Exercise

1. Show that the A-median of the tangential triangle intersects the line OA on BC. Deduce from this that the centroid of the tangential triangle is O/K.

2. Show that the image of the tangential triangle under a homothety at O is perspective to the reference triangle. 30

8.9.3 The circumcenter of the tangential triangle

Since the Gob homothetic center lies on the Euler line, and the circumcenter of the orthic triangle is the nine-point center N, the circumcenter of the tangential triangle also lies on the Euler line. Let this be the point O. From the homothety, we have

2 2 2 2 2 2 (a Sβγ + b Sγα + c Sαβ)(H/K)=−2Sαβγ · O + a b c · N.

29 The centroid of the tangential triangle appears as S 154 in ETC. 30For the image of the tangential triangle under h(O, t), this perspector is the isogonal conjugate of the point P on the Euler line which divides OH in the ratio OP : PH =1:2t. See K. L. Nguyen, A synthetic proof of Goormaghtigh’s generalization of Musselman’s theorem, Forum Geom. 5 (2005) 17–20. 8.9 The tangential triangle 121

From this, we have

2 2 2 a b c (2S + S + S , ··· , ···) − (a2S , ··· , ···) 4(Sβγ+Sγα+Sαβ) βγ γα αβ βγ O = 2Sαβγ (a2b2c2(2S + S + S ) − 4(S + S + S ) · a2S , ··· , ···) = βγ γα αβ βγ γα αβ βγ 8Sαβγ (Sβγ + Sγα + Sαβ) (a2b2c2(4S + S + S ) − 4(S + S + S ) · a2S − 2a2b2c2S , ··· , ···) = βγ γα αβ βγ γα αβ βγ βγ 8Sαβγ(Sβγ + Sγα + Sαβ ) ((a2S + b2S + c2S +6S )a2S − 2a2b2c2S , ··· , ···) = αα ββ γγ αβγ A βγ 8Sαβγ(Sβγ + Sγα + Sαβ) (a2S + b2S + c2S +6S ) · O − a2b2c2 · H = αα ββ γγ αβγ 4Sαβγ (a2b2c2 +4S ) · O − a2b2c2 · H = αβγ . 4Sαβγ

This is the point dividing OH in the ratio 31

2 2 2 2 2 2 OO : OH = −a b c : a b c +4Sαβγ .

31 The circumcenter of the tangential triangle appears as X 26 in ETC. 122 Parallel and perpendicular lines

8.10 The intangents triangle

The reﬂection of the sideline a in the internal bisector of angle A gives the A-intangent, the common internal tangent of the incircle and the A-excircle. It is the line

bcx +(b − c)(cy − bz)=0.

Its points of tangency are

((b − c)2(s − a):b2(s − b):c2(s − c)) on the incircle and (−(b − c)2s : b2(s − c):c2(s − b)) on the A-excircle. Similarly, we deﬁne the B- and C-intangents. The three intangents bound the intangents triangle with vertices

(a2(s − a):b(c − a)(s − b):−c(a − b)(s − c)), (−a(b − c)(s − a):b2(s − b):c(a − b)(s − c)), (a(b − c)(s − a):−b(c − a)(s − b):c2(s − c)). This is (oppositely) homothetic to the tangential triangle. Their (ob-)incircles are I and O respectively. The homothetic center is the insimilicenter T+.

I T+

B C 8.11 The triangle of reﬂections 123

8.11 The triangle of reﬂections

The triangle of reﬂections has vertices

A =(−a2 : a2 + b2 − c2 : c2 + a2 − b2), B =(a2 + b2 − c2 : −b2 : b2 + c2 − a2), C =(c2 + a2 − b2 : b2 + c−a2 : −c2) which are the reflections of A, B, C in their opposite sides. Proposition 8.7. The triangle of reflections is the image of the reflection triangle of N under the homothety h(O, 2).

C O N

B C

Exercise 1. Show that the triangle of reﬂections is the image of the pedal triangle of N under the homothety h(G, 4).

2 2S 2S 2 2 a2 = S γ − β + S () + S () α b2 c2 β γ 124 Parallel and perpendicular lines

a2b2c2 +8S2 · S = α b2c2 a2b2c2 +8S2 · S b2 = β c2a2 a2b2c2 +8S2 · S c2 = γ a2b2

• a is longer than, equal to, or shorter than a according as A is an acute, right, or obtuse angle.

◦ 2 2 2 • Suppose c = c, so that C =90 . We write c = a + b , Sγ =0, and obtain 8a2b2 8a2b2 a2 = a2 + ,b2 = b2 + . a2 + b2 a2 + b2 Here, a2 − b2 = a2 − b2. It is clearly impossible to have a = b and b = a. Actually, in this case, the area of the reflection triangle is always 3 times that of ABC, with the same orientation. • Therefore, if the two triangles are congruent, we have either ABC ≡ BCA or ABC ≡ CAB. In the first case, we have 2 2 2 8S · Sα = b c (Sα − Sβ), 2 2 2 8S · Sβ = c a (Sβ − Sγ ), 2 2 2 8S · Sγ = a b (Sγ − Sα). From these, 2 2 2 2 8S (a Sα + b Sβ + c Sγ)=0, a contradiction; similarly for the second case. Therefore, the reflection triangle is never congruent to ABC, in any order. • Bottema has studied the case when the reflection triangle degenerates into a line. 5R2 Note: cot ω = 2S . • Suppose they have the same oriented area. This is equivalent to 2 2 2 a b c +4Sαβγ =0. 2 2 2 2 Since Sαβγ = S · Sω − a b c , this is the same as 3a2b2c2 S = ω 4S2 3R2 This is equivalent to cot ω = S ,or s2 =(R + r)(3R + r). 8.12 The Evans perspector 125

• The two triangles have equal (oriented) areas if and only if

2 2 2 a b c +2Sαβγ =0.

2R2 This is equivalent to cot ω = S ,or

s2 =2R2 +4Rr + r2.

The area of triangle ABC is

3(Sααβ + Sαββ + Sααγ + Sαγγ + Sββγ + Sβγγ)+14Sαβγ a2b2c2

• ABC degenerates into a line if and only if

3(Sβ + Sγ )Sαα +(3Sββ +14Sβγ +3Sγγ)Sα +3(Sβ + Sγ)Sβγ =0.

Proposition 8.8. The triangle of reﬂections of ABC is degenerate if and only if the nine- point center of ABC lies on its circumcircle.

Exercise 1. If ABC is the triangle of reﬂections of ABC, then the circumcircles of ABC, ABC, and ABC intersect at the orthocenter H.

8.12 The Evans perspector

Theorem 8.9 (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. 126 Parallel and perpendicular lines

This is called the Evans perspector Ev. Note that the reﬂection of the incenter in the circumcenter is the point

I =2O − I =(a(a3 + a2(b + c) − a(b + c)2 − (b + c)(b − c)2):···: ···)

The sum of the coordinates is −4S2. is the point

−4S2(2O − I)+4RSs · I ∼−S(2O − I)+Rs · I ∼−2r(2O − I)+R · I.

−4S2(2O − I)+4RSs · I ∼−S(2O − I)+Rs · I ∼−2r(2O − I)+R · I = − 4r · O +(R +2r)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. 32

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.

32 The perspector is X80. Chapter 9

Some Basic Constructions

9.1 Isotomic conjugates

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

BAP = AQC, CBP = BQA, ACP = CQB.

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

CQ BP

BQ CP Q P

B AP AQ C If P =(x : y : z), then 1 1 1 P • =( : : )=(yz : zx : xy). x y z 128 Some Basic Constructions

9.1.1 The Gergonne and Nagel points

1 1 1 e a − − − G = s−a : s−b : s−c ,N=(s a : s : s c).

Z Y

Y Z I Na Ge

B X X C

Ia 9.1 Isotomic conjugates 129

9.1.2 The isotomic conjugate of the orthocenter The isotomic conjugate of the orthocenter is the point

H• =(b2 + c2 − a2 : c2 + a2 − b2 : a2 + b2 − c2).

Its traces are the pedals of the deLongchamps point Lo, the reﬂection of H in O.

Y Lo

O Y

H•

Z H

B X X C

Exercise 1. Show that H • is the perspector of the triangle of reﬂections of the centroid G in the sidelines of the medial triangle. 130 Some Basic Constructions

Y G H•

B X C

9.1.3 Yff-Brocard points Consider a point P =(u : v : w) with traces X, Y , Z satisfying BX = CY = AZ = µ. This means that w u v a = b = c = µ. v + w w + u u + v Elimination of u, v, w leads to 0 −µa− µ 3 0= b − µ 0 −µ =(a − µ)(b − µ)(c − µ) − µ . −µc− µ 0 Indeed, µ is the unique positive root of the cubic polynomial

(a − t)(b − t)(c − t) − t3.

This gives the point 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 9.1 Isotomic conjugates 131

A A

Z Y

Z Y P •

B X C B X C has traces X, Y , Z that satisfy

CX = AY = BZ = µ.

These points are called the Yff-Brocard points. 1 They were brieﬂy considered by A. L. Crelle. 2

Exercise 1. Justify that the polynomial f(t):=(a − t)(b − t)(c − t) − t3 has a unique positive root. 3

2. Is it possible to ﬁnd collinear points X on BC, Y on CA, and Z on AB with BX − CX = CY − AY = AZ − BZ? 4

1P. Yff, An analogue of the Brocard points, Amer. Math. Monthly, 70 (1963) 495 – 501. 2A. L. Crelle, 1815. 3f(0) = abc > 0 and f(t) is a decreasing function since f (t)=−(ab + bc + ca)+2(a + b + c)t − 6t2 has discriminant 4(a2 + b2 + c2 − 4ab − 4bc − 4ca)=−8(3r2 +12Rr + s2) < 0 and is always negative. 4No, if BX − CX == CY − AY = AZ − BZ = t, then (a + t)(b + t)(c + t)=t3; (a + b + c)t2 + (ab + bc + ca)t + abc =0. The quadratic has no real roots. 132 Some Basic Constructions

9.2 Isogonal conjugates

9.2.1 Reﬂections and isogonal conjugates Consider a point P with reﬂections P a, P b, P c 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.

P b

P c

B C Figure 9.1:

Clearly, the triangles AQP b and AQP c are congruent, so that Q is equidistant from P b and P c. For the same reason, any point on a line isogonal to BP is equidistant from P c and P a. It follows that the intersection P ∗ of two lines isogonal to AP and BP is equidistant from the three reﬂections P a, P b, P c. 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 .

P b

P c

P ∗

B C

P a

Figure 9.2: 9.2 Isogonal conjugates 133

9.2.2 The pedal circle Clearly, (P ∗)∗ = P . Moreover, the circles of reﬂections of P and P ∗ are congruent, since, ∗ ∗ a ∗ ∗ a in Figure 9.3, the trapezoid PP Pa P being isosceles, 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 ∗.

A Qb

P b

P c

c Q P M Q = P ∗

B C

P a

Figure 9.3:

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 of P . 134 Some Basic Constructions

9.2.3 Coordinates of isogonal conjugate Let P =(x : y : z) be a point other than the vertices of triangle ABC. Its isogonal conjugate is the point P ∗ =(a2yz : b2zx : c2xy). (9.1)

Proof. We show that the reﬂection of the cevian AP in the bisector of angle A intersects b2 c2 the line BC at the point X =(0: y : z ).

θθ

B X X C

Let X be the A-trace of P , with ∠BAP = θ. This is the point X =(0:y : z)= 2 2 (0 : Sα − Sθ : −c ) in Conway’s notation. It follows that Sα − Sθ : −c = y : z.If the reﬂection of AX (in the bisector of angle A) intersects BC at X ,wehaveX =(0: − 2 − − 2 2 2 − 2 2 b2 c2 b : Sα Sθ)=(0: b c : c (Sα Sθ)) = (0 : b z : c y)=(0: y : z ). Similarly, the reﬂections of the cevians BP and CP in the respective angle bisectors intersect CA at a2 c2 a2 b2 Y =(x :0: z ) and AB at Z =(x : y :0). These points X , Y , Z are the traces of the point P ∗ whose coordinates are given in (9.1) above. 9.3 Examples of isogonal conjugates 135

9.3 Examples of isogonal conjugates

9.3.1 The circumcenter and orthocenter The circumcenter O and the orthocenter H are isogonal conjugates, since π ∠(AO, AC)= − β = −∠(AH, AB), 2 and similarly ∠(BO,BA)=−∠(BH,BC), ∠(CO,CB)=−∠(CH,CA). 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.

b A H

O Hc

B C

Figure 9.4: 136 Some Basic Constructions

9.3.2 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. 5 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,wehave

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

This shows that Y , A and Z are collinear, so that (i) AA is a median of triangle AY Z,

5If triangle ABC is acute, this is the Gergonne point of ABC. 9.3 Examples of isogonal conjugates 137

(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 of triangle ABC.

9.3.3 The Gergonne point and the insimilicenter T+ Theorem 9.1. 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 138 Some Basic Constructions

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 Ge I 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 (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 Ger- gonne 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 B =(π − C)+2 C + 2 =B + C = −A (mod π) 9.3 Examples of isogonal conjugates 139

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 C = − (π − B) − 2 B + 2 = − (B + C)=A (mod π)

(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). 140 Some Basic Constructions

9.3.4 The Nagel point and the exsimilicenter T− Theorem 9.2. 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. (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. 9.3 Examples of isogonal conjugates 141

1 D D1

D C B D Aa

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). A

D1 E

F I

E1 F1

B D C 142 Some Basic Constructions

9.3.5 Isogonal conjugates of the Kiepert perspectors The isogonal conjugate of the Kiepert perspector K(θ) is the point

∗ 2 2 2 K (θ)=(a (Sα + Sθ):b (Sβ + Sθ):c (Sγ + Sθ)). As θ varies, K∗(θ) traverses the line containing the circumcenter O and the symmedian point K, which we call the Brocard axis. More precisely, the point K ∗(θ) divides OK in the ratio cot ω cot θ :1=(a2 + b2 + c2)cosθ :4 sin θ. Proof.

2 2 2 (a (Sα + Sθ),b (Sβ + Sθ),c (Sγ + Sθ)) 2 2 2 2 2 2 =(a Sα,b Sβ,c Sγ )+Sθ(a ,b ,c ) 2 2 2 2 =2S · O +(a + b + c )Sθ · K =2S2(O +cotω cot θ · K).

Examples (1) The isogonal conjugate of the Spieker center. This is the perspector of the reﬂections of Gergonne point in intouch triangle. 6

6From ED’s ﬁle. 9.4 Isogonal conjugate of an inﬁnite point 143

9.4 Isogonal conjugate of an inﬁnite point

Theorem 9.3. Given a triangle ABC and a line 4, let 4a, 4b, 4c be the parallels to 4 through A, B, C respectively, and 4a, 4b, 4c their reﬂections in the angle bisectors AI, BI, CI respectively. The lines 4a, 4b, 4c intersect at a point on the circumcircle of triangle ABC.

A A

*b I P *b I O P *a B C B C

* *a *b *c *c * *b *c *c

Proof. (1) Let P be the intersection of 4b and 4c.

(BP, PC)=(4b,4c) =(4b,IB)+(IB, IC)+(IC, 4c)

=(IB, 4b)+(IB, IC)+(4c,IC) =(IB, 4)+(IB, IC)+(4, IC) =2(IB, IC) π A =2 + 2 2 =(BA, AC)(modπ).

Therefore, 4b and 4c intersect at a point on the circumcircle of triangle ABC. (2) Similarly, 4a and 4b intersect at a point P on the circumcircle. Clearly, P and P are the same point since they are both on the reflection of 4b in the bisector IB. (3) Therefore, the three reflections 4a, 4b, and 4c intersect at the same point on the circumcircle. This point is the isogonal conjugate of the infinite point of 4.

Theorem 9.4. A point with homogeneous barycentric coordinates (x : y : z) lies on the circumcircle of triangle ABC if and only if

a2yz + b2zx + c2xy =0. 144 Some Basic Constructions

Proposition 9.5. The isogonal conjugates of the inﬁnite points of two perpendicular lines are antipodal points on the circumcircle.

* Q

* I O P

B C

Proof. If P and Q are the isogonal conjugates of the inﬁnite points of two perpendicular lines 4 and 4 through A, then AP and AQ are the reﬂections of 4 and 4 in the bisector AI.

π (AP, AQ)=(AP, IA)+(IA, AQ)=−(4, IA) − (IA, 4)=−(4, 4)= . 2 Therefore, P and Q are antipodal points.

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 •

P Q O

B C

Q• 7The line a2x + b2y + c2z =0perpendicular to the Euler line. 9.4 Isogonal conjugate of an inﬁnite point 145

L x y z 3. Let be the line u + v + w =0, intersecting the side lines BC, CA, AB of triangle ABC at U, V , W respectively.

(a) Find the equation of the perpendiculars to BC, CA, AB at U, V , W respectively. 8 (b) Find the coordinates of the vertices of the triangle bounded by these three perpendiculars. 9 (c) Show that this triangle is perspective with ABC at a point P on the circumcircle. 10 (d) Show that the Simson line of the point P is parallel to L.

H Z C B U V

P W

X Y

8 2 2 (SBv + SCw)x + a wy + a vz =0, etc. 9 2 2 2 2 2 2 (−S u + SABuv + SBCvw + SCAwu : b (c uv − SAuw − S + Bvw):c (b uw − SAuv − SC vw). 2 10 a P = 2 : ···: ··· −a vw+SB uv+SC uw . 146 Some Basic Constructions Chapter 10

The circumcircle

10.1 Simson lines and lines of reﬂections

Theorem 10.1. The pedals of a point on the sidelines of triangle are collinear if and only if the point lies on the circumcircle of the triangle.

Z Z P A A P

B X C B X C

Proof. Let X, Y , Z be the pedals of a point P on the lines BC, CA, AB respectively. ± π Since (AY, Y P )=(AZ, ZP )= 2 , the points A, Y , P , Z are concyclic, and (PY, YZ)=(PA, AZ)=(PA, AB). Similarly, P , Y , C, X are concyclic, and (PY, YX)=(PC, CX)=(PC, BC). It follows that (YZ,YX)=(YZ,PY)+(PY, YX)=(AB, P A)+(PC, BC)= (BA, AP)+(PC, CB). X, Y , Z are collinear if and only if (YZ,YX)=0, i.e., (BA, AP)=(BC, CP). This is the case if and only if P lies on the circumcircle of triangle ABC. If P is a point on the circumcircle of triangle ABC, the line containing its pedals on the three sides is called the Simson line of P (with respect to the triangle). 148 The circumcircle

Corollary 10.2. The reﬂections of a point in the sidelines of triangle are collinear if and only if the point lies on the circumcircle of the triangle.

Proof. The reﬂection triangle of a point P is the image of its pedal triangle under the homothety h(P, 2). Since the pedal triangle degenerates into a line if and only if P lies on the circumcircle, so does the reﬂection triangle.

Z Z Z P A P A Hb

Y O Hc Y H E Y B X C B C

X X

Proposition 10.3. The line of reﬂections of a point on the circumcircle of a triangle passes through the orthocenter.

Proof. Let Ha be the reflection of H in BC. It is known that Ha lies on the circumcircle. (X H, HP)=(X Ha,HaP )=(EHa,HaP ) where E is the point of concurrency of the reflections of HP in the sidelines. For the same reason, (Y H, HP)=(EHb,HbP ). Since E, Ha, Hb, P are all on the circumcircle, this directed angle is the same as (X H, HP). It follows that H, X , Y are collinear. For the same reason, the same line also contains Z . 10.1 Simson lines and lines of reflections 149

Theorem 10.4. The triangle of reﬂections is degenerate if and only if the nine-point center lies on the circumcircle.

N A

O 150 The circumcircle

Theorem 10.5. The reﬂections of the Euler line in the sidelines of a triangle are concurrent at a point on the circumcircle.

B C

Proof. If AA is parallel to the Euler line, and AX, AE are perpendicular to BC, then the line XE is the reﬂection of the Euler line in BC.

A A A A

O O

H H

B C B C

X X E E

Now let BB be parallel to the Euler line. We claim that BE ⊥ CA so that the 10.1 Simson lines and lines of reﬂections 151 reﬂection of the Euler line in CA contains the point E. For this, note that

(BE, AC)=(BE, BC)+(BC, AC) =(AE, AC)+(AB,BB) =(AE, AC)+(AB,AB)+(AB,BB) =(AE, AC)+(AB,AB)+(AC, BC) =(AE, AC)+(AX, AX)+(AC, BC) =(AE, AC)+(AX, AE)+(AC, BC) =(AX, BC) π = . 2 Similarly, the reﬂection of the Euler line in AB also contains the point E. The proof of the theorem applies to an arbitrary line through the orthocenter H.

Theorem 10.6. The reﬂections of a line 4 in the sidelines of a triangle are concurrent if and only if 4 contains the orthocenter. 152 The circumcircle

10.2 Simson line and line of reﬂections: computations

Let P be a point on the circumcircle of triangle ABC. The Simson line s(P ) is the line containing its pedals on the sidelines of the triangle. If we write the coordinates of P in the a2 b2 c2 2 2 2 form ( f : g : h )=(a gh : b hf : c fg) for an inﬁnite point (f : g : h), then 2 2 2 2 2 2 A[P ] =(0 : a Sγgh + a b hf : a Sβgh + a c fg)

∼(0 : h(−Sγ (h + f)+(Sγ + Sα)f):g(−Sβ(f + g)+(Sα + Sβ)f))

∼(0 : −h(Sγ h − Sαf):g(Sαf − Sβg)) =(0 : −hg : gh) g h ∼ 0:− : g h where, for convenience, we have written (f : g : h )=(Sβg − Sγ h : Sγ h − Sαf : Sαf − Sβg) for the inﬁnite point of lines in the direction perpendicular to (f : g : h). Similarly, f − h − f g B[P ] = f :0: h and C[P ] = f : g :0 . The equation of the Simson line is

f g h f x + g y + h z =0.

10.2.1 The direction of Simson line It is easy to determine the infinite point of the Simson line. Note that g h gh − hg − = g h gh (S h − S f )h − (S f − S g)g = γ α α β gh −S f (g + h)+S g2 + S h2 = α β γ gh S f 2 + S g2 + S h2 = α β γ gh S f 2 + S g2 + S h2 = α β γ · f . f gh Similarly, h f S f 2 + S g2 + S h2 − = α β γ · g, h f f gh f g S f 2 + S g2 + S h2 − = α β γ · h. f g f gh The Simson line s(P ) therefore has infinite point (f : g : h). It is perpendicular to the line defining P . It passes through, as we have noted, the midpoint between H and P , which lies on the nine-point circle. 10.2 Simson line and line of reflections: computations 153

Exercise 1. Animate a point P on the circumcircle of triangle ABC and trace its Simson line. 2. Let H be the orthocenter of triangle ABC, and P a point on the circumcircle. Show that the midpoint of HP lies on the Simson line s(P ) and on the nine-point circle of triangle ABC.

10.2.2 The line of reﬂections a2 b2 c2 Proposition 10.7. If P = f : g : h is on the circumcircle, the line of reﬂection is

Sαfx+ Sβgy + Sγhz =0. Proof. The reﬂections of P on the sidelines are the points 2 X =(−a gh :(Sαf + Sγ(g − h))h :(Sαf − Sβ(g − h))g), 2 Y =((Sβg − Sγ(h − f))h : −b hf :(Sβg + Sα(h − f))f), 2 Z =((Sγh + Sβ(f − g))g :(Sγh − Sα(f − g))f : −c fg).

It is easy to verify that each of these lies on the line Sαfx + Sβgy + Sγhz =0, which is therefore the line of reﬂections of P .

It is clear that the line Sαfx+ Sβgy+ Sγ hz =0contains the orthocenter H. Every line through H is of this form for an inﬁnite point (f : g : h).

Proposition 10.8. The reﬂections of the line Sαfx + Sβgy + Sγhz =0in the sidelines a2 b2 c2 intersect at the point f : g : h on the circumcircle.

Proof. The reﬂections of Sαfx+ Sβgy + Sγhz =0in the sidelines are

2 a (Sαfx+ Sβgy + Sγ hz) − 2(Sαβ + Sβγ + Sγα)x =0, 2 b (Sαfx+ Sβgy + Sγ hz) − 2(Sαβ + Sβγ + Sγα)y =0, 2 c (Sαfx+ Sβgy + Sγhz) − 2(Sαβ + Sβγ + Sγα)z =0. Each of these contains the point (a2gh : b2hf : c2fg) on the circumcircle.

Examples Intersection of reﬂections of lines in the sideline of the reference triangle.

line intersection of reﬂections a2 b2 C2 Euler line 2 2 : 2 2 : 2 2 b −c c −a a −b 2 2 2 HI a : b : c (b−c)(b+c−a) (c−a)(c+a−b) (a−b)(a+b−c) a2 b2 c2 HK 2 2 : 2 2 : 2 2 (b −c )Sα (c −a )Sβ (a −b )Sγ 154 The circumcircle

Exercise 1. Let L be a line through the orthocenter H. (1) Choose an arbitrary point Q on the line L and reﬂect it in the side lines BC, CA, AB to obtain the points X, Y , Z. (2) Construct the circumcircles of AY Z, BZX and CXY . These circles have a common point P , which happens to lie on the circumcircle. (3) Construct the reﬂections of P in the side lines of triangle ABC.

2. Animate a point Q on the circumcircle, together with its antipode Q. (1) The reﬂections X, Y , Z of Q on the side lines BC, CA, AB are collinear; so are those X , Y , Z of Q. (2) The lines XX, YY, ZZ intersect at a point P , which happens to be on the circumcircle. (3) Construct the reﬂections of P in the side lines of triangle ABC.

3. Animate a point P on the circumcircle, together with its antipodal point P . (1) Construct the line PP to intersect the side lines BC, CA, AB at X, Y , Z respectively. (2) Construct the circles with diameters AX, BY , CZ. These three circles have two common points. One of these is on the circumcircle. Label this point P ∗, and the other common point Q. (3) What is the locus of Q? (4) The line P ∗Q passes through the orthocenter H. As such, it is the line of reﬂection of a point on the circumcircle. What is this point?1 (5) Construct the Simson lines of P and P . They intersect at a point on the nine-point circle. What is this point?

4. Let P be a given point, and ABC the homothetic image of ABC under h(P, −1) (so that P is the common midpoint of AA, BB and CC). (1) The circles ABC, BCA and CAB intersect at a point Q on the circumcircle. (2) The circles ABC, BCA and CAB intersect at a point Q such that P is the midpoint of QQ. 2

1The reflection of P ∗ in the diameter PP. 2Musselman, Amer. Math. Monthly, 47 (1940) 354 – 361. If P =(u : v : w), the intersection of the three circles in (1) is the point 1 : ···: ··· b2(u + v − w)w − c2(w + u − v)v on the circumcircle. This is the isogonal conjugate of the infinite point of the line u(v + w − u) x =0. a2 cyclic 10.2 Simson line and line of reflections: computations 155

C' B' O

B C

5. Construct parallel lines La, Lb, and Lc through the D, E, F be the midpoints of the sides BC, CA, AB of triangle ABC. Reﬂect the lines BC in La, CA in Lb, and 3 AB in Lc. These three reﬂection lines intersect at a point on the nine-point circle.

6. Construct parallel lines La, Lb, and Lc through the pedals of the vertices A, B, C on their opposite sides. Reﬂect these lines in the respective side lines of triangle ABC. The three reﬂection lines intersect at a point on the nine-point circle. 4

3This was ﬁrst discovered in May, 1999 by a high school student, Adam Bliss, in Atlanta, Georgia. A proof can be found in F.M. van Lamoen, Morley related triangles on the nine-point circle, Amer. Math. Monthly, 107 (2000) 941 – 945. See also, B. Shawyer, A remarkable concurrence, Forum Geom., 1 (2001) 69 – 74. 4Ibid. 156 The circumcircle

A Q'

C' B'

B C

A' 10.3 Circumcevian triangle 157

10.3 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

10.3.1 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 triangle ocev(P ) is similar to the pedal triangle ped(P ) and the re- ﬂection triangle rﬂ(P ).

Theorem 10.9. The circumcevian triangle ocev(P ) is perspective with the tangential triangle cev−1(K) at a4 b4 c4 a4 b4 c4 a4 b4 c4 a2(− + + ):b2( − + ):c2( + − ) . u2 v2 w2 u2 v2 w2 u2 v2 w2

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 quadruple 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.

10.4 Antipedal triangles

The antipedal triangle ped−1(P ) of P is bounded by the perpendiculars of AP at A, BP at B, and CP at C.IfP =(u : v : w), the vertices of ped−1(P ) are     2 2 [ ] SC v+b u SB w+c u P −1 2 2 A SC u+a v SBu+a w  2 2   [P ] SC u+a v SAw+c v B =  2 −1 2  . SC v+b u SAv+b w [ ] 2 2 P SBu+a w SAv+b w C 2 2 −1 SB w+c u SAw+c v P antipedal triangle I excentral triangle O tangential triangle H superior triangle 158 The circumcircle

Proposition 10.10. The circumcevian triangle ocev(P ) and the antipedal triangle ped−1(Q) are perspective if and only if O, P , Q are collinear.

Corollary 10.11. The antipedal triangle and the circumcevian triangle of P are always perspective.

4 2 (Sα(a Sαα − a Sα(Sββ + Sγγ) − Sβγ(Sββ + Sγγ)) : ···: ···).

Exercise 1. The circumcevian triangle of P is always similar to the pedal triangle.

2. The circumcevian triangle of the incenter is perspective with the intouch triangle. What is the perspector? 5

3. If P is a point on the circumcircle, what is the antipedal triangle of P ? 6

4. The coordinates of P =(u : v : w) in its antipedal triangle are (u · AP 2 : ···: ···).

5. The isogonal conjugate of P in the antipedal triangle is the symmetric of P in the circumcenter.

Always perspective with the circumcevian triangle of O. 7

P Perspector I X1490 O X1498 H Lo X64 O X84 X40 The antipedal triangle and the circumcevian triangle of P are always perspective.

P Perspector I I O X1498 H Lo X64 O X84 X40 Remark. The circumcevian and the antipedal triangles of a point P are always perspective. The coordinates of the perspector, however, is extremely complicated. Here is the perspector for P = H:

5 T−. 6The antipode of P . 7C.K. Kimberling and I.G. MacDonald, Problem E3407 and solution, Amer. Math. Monthly, 97 (1990) 848; 99 (1992) 369. 10.4 Antipedal triangles 159

10.4.1 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 2 2 b c 2 2 v + w b c x + y + z =0, a2 v2 w2

They bound a triangle with perspector a2 P/K = : ···: ··· , b2wu + c2uv − a2vw

Examples

R (1) The circumtangential triangle of I is homothetic to ABC with ratio of homothety r . The homothetic center is T−, the exsimilicenter of (O) and (I) (2) The circumtangential triangle of K has perspector K. a2 b2 c2 8 (3) The circumtangential triangle of G has perspector Sα : Sα : Sγ . (4) For P = T , Q = T+.

8 This is X25 in ETC. 160 The circumcircle Chapter 11

Circles

11.1 Equation of a circle

The most basic circle in triangle geometry, the circumcircle, has a very simple equation, namely, a2yz + b2zx + c2xy =0. This basic fact leads to a simple description of the equations of other circles. 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−k)(u+v+w) where t = k , lies on the circumcircle. In other words, 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 =(a2yz + b2zx + c2xy)+t( a2(wy + vz))(x + y + z) 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.

The line fx+ gy + hz =0is the radical axis of C and the circumcircle. Explain. 162 Circles

Examples

(1) The nine-point circle. To ﬁnd the equation of the nine-point circle, we make use of 1 the fact that it is obtained from the circumcircle by applying the homothety h(G, 2 ).If P =(x : y : z) is a point on the nine-point circle, then the point

Q =3G − 2P =(x + y + z)(1:1:1)− 2(x : y : z)=(y + z − x : z + x − y : x + y − z) is on the circumcircle. From the equation of the circumcircle, we obtain 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 a2yz + b2zx + c2xy + (x + y + z)(S x + S y + S z)=0. 2 α β γ

(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.

11.1.1 The 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 11.1. Let f, g, h be the powers of A, B, C with respect to a circle C. (1) The equation of the circle is

a2yz + b2zx + c2xy − (x + y + z)(fx+ gy + hz)=0.

(2) The center of the circle is the point

2 2 2 (a Sα+Sβ(h−f)−Sγ (f−g):b Sβ+Sγ(f−g)−Sα(g−h):c Sgamma+Sα(g−h)−Sβ(h−f).

(3) The radius ρ of the circle is given by

a2b2c2 − 2(a2S f + b2S g + c2S h)+S (g − h)2 + S (h − f)2 + S (f − g)2 ρ2 = α β γ α β γ . 4S2 11.1 Equation of a circle 163

11.1.2 The incircle and the excircles It follows that the equation of the incircle is a2yz + b2zx + c2xy − (x + y + z)((s − a)2x +(s − b)2y +(s − c)2z)=0. The radical axis with the circumcircle is the line (s − a)2x +(s − b)2y +(s − c)2z =0. The same method gives the equations of the excircles:

a2yz + b2zx + c2xy − (x + y + z)(s2x +(s − c)2y +(s − b)2z)=0, a2yz + b2zx + c2xy − (x + y + z)((s − c)2x + s2y +(s − a)2z)=0, a2yz + b2zx + c2xy − (x + y + z)((s − b)2x +(s − a)2y + s2z)=0.

Exercise 1. Show that a circle passing through A has equation of the form a2yz + bzx + 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. Verify that the point ((b2 − c2)2 :(c2 − a2)2 :(a2 − b2)2), lies on the nine-point circle. 1 4. Show that the equation of the circumcircle of the excentral triangle is a2yz + b2zx + c2xy +(x + y + z)(bcx + cay + abz)=0.

5. Verify that the circle 2(b + c)(c + a)(a + b)(a2yz + b2zx + c2xy) − (x + y + z)(fx+ gy + hz)=0, where f =bc((a + b + c)(b2 + c2 − a2)+abc), g =ca((a + b + c)(c2 + a2 − a2)+abc), h =ab((a + b + c)(a2 + b2 − c2)+abc), passes through the traces of the incenter.

1This is the midpoint between the Fermat points. 164 Circles

6. Show that the Nagel point of triangle ABC lies on its incircle if and only if one of s its sides has length 2 . Make use of this to design an animation picture showing a triangle with its Nagel point on the incircle.

7. (a) Show that the centroid of triangle ABC lies on the incircle if and only if 5(a2 + b2 + c2)=6(ab + bc + ca). (b) Let ABC be an equilateral triangle with center O, and C the circle, center O, radius half that of the incirle of ABC. Show that the distances from an arbitrary point P on C to the sidelines of ABC are the lengths of the sides of a triangle whose centroid is on the incircle.

11.1.3 Circle with given center and radius The circle with (u : v : w) and radius ρ has equation 2 2 2 2 − 2 2 2 2 2 2 − c v +(b + c a )vw + b w − 2 a yz + b zx + c xy (x + y + z) 2 ρ x =0. cyclic (u + v + w)

11.1.4 Circle with a given diameter

Let P =(u1 : v1 : w1) and Q =(u2 : v2 : w2). The circle with diameter PQhas equation

2 2 2 (u1 + v1 + w1)(u2 + v2 + w2)(a yz + b zx + c xy) − (x + y + z)(fx+ gy + hz), where

f =Sα(v1 + w1)(v2 + w2)+Sβv1v2 + Sγw1w2,

g =Sαu1u2 + Sβ(w1 + u1)(w2 + u2)+Sγw1w2,

h =Sαu1u2 + Sβv1v2 + Sγ (u1 + v1)(u2 + v2).

Examples (1) The orthocentroidal circle has diameter HG: 2 (a2yz + b2zx + c2xy) − (x + y + z)(S x + S y + S z)=0. 3 α β γ (2) The Brocard circle has diameter OK: x + y + z (a2yz + b2zx + c2xy) − (b2c2x + c2a2y + a2b2z)=0. a2 + b2 + c2 11.2 The Feuerbach theorem 165

11.2 The Feuerbach theorem

11.2.1 Intersection of the incircle and the nine-point circle We consider how the incircle and the nine-point circle intersect. The intersections of the two circles can be found by solving their equations simultaneously:

a2yz + b2zx + c2xy − (x + y + z)((s − a)2x +(s − b)2y +(s − c)2z)=0, 1 a2yz + b2zx + c2xy − (x + y + z)(S x + S y + S z)=0. 2 α β γ Note that 1 1 1 1 (s−a)2 − S = ((b+c−a)2 −(b2 +c2 −a2)) = (a2 −a(b+c)+bc)= (a−b)(a−c). 2 α 4 2 2 Subtracting the two equations we obtain the equation of the radical axis of the two circles:

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

We rewrite this as x y z + + =0. b − c c − a a − b

11.2.2 Condition for tangency of a line and the incircle Proposition 11.2. A line px + qy + rz =0is tangent to the incircle if and only if

s − a s − b s − c + + =0. p q r

Theorem 11.3 (Feuerbach). (1) The nine-point circle and the incircle are tangent internally to each other at the point

2 2 2 Fe =((b − c) (s − a):(c − a) (s − b):(a − b) (s − c)), the common tangent being the line x y z + + =0. b − c c − a a − b (2) The nine-point circle is also tangent to each of the excircles externally. The points of tangency form a triangle perspective with ABC at the point (b + c)2 (c + a)2 (a + b)2 F = : : . e s − a s − b s − c

Figure 166 Circles

x y z Proof. (1) The fact that the line b−c + c−a + a−b =0is tangent to the incircle follows from §??. The point of the tangency is the pedal of the incenter on this line. This has coordinates

((s − a)(b − c)2 :(s − b)(c − a)2 :(s − c)(a − b)2).

(2) The points of the tangency of the nine-point circle with the excircles are

(−(b − c)2s :(c + a)2(s − c):(a + b)2(s − b)), ((b + c)2(s − c):−(c − a)2s :(a + b)2(s − a)), ((b + c)2(s − b):(c + a)2(s − a):−(a − b)2s).

Exercise 1. Show that Fe and Fe divide I and N harmonically.

2. Find the equations of the common tangent of the nine-point circle and the excircles. 2

3. Apart from the common external tangent, the nine-point circle and the A-circle have another pair of common internal tangent, intersecting at their excenter of similitude A. Similarly deﬁne B and C. The triangle ABC is perspective with ABC. What is the perspector? 3

4. Let 4 be a diameter of the circumcircle of triangle ABC. Animate a point P on 4 and construct its pedal circle, the circle through the pedals of P on the side lines. The pedal circle always passes through a ﬁxed point on the nine-point circle. What is this ﬁxed point if the diameter passes through the incenter?

2 x y z Tangent to the A-excircle: b−c + c+a − a+b =0. 3The Feuerbach point. 11.3 Circles tangent to two sidelines 167

11.3 Circles tangent to two sidelines

Lemma 11.4. 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).

Exercise 1. Given a point P , there are two circles through P tangent to the half-lines AB and AC. Identify the second intersection X of the two circles. Similarly, there are two circles through P and Y tangent to both half-lines BC, BA, two circles through P and Z tangent to both half-lines CA, CB. Show that triangle XY Z is perspective with ABC and identify the perspector. 168 Circles

11.4 The Brocard points and the Brocard circle

There is a unique point Ω→ such that

∠ABΩ→ = ∠BCΩ→ = ∠CAΩ→. 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.

Ω→

ω B C

Consider the circle CAAB. Since the circle passes through A and B, its equation is of the form a2yz + b2zx + c2xy − rz(x + y + z)=0 for some constant r. Since it is tangent to CA at A, when we set y =0, the equation should reduce to z2 =0. This means that r = b2 and the circle is 2 2 2 2 CAAB : a yz + b zx + c xy − b z(x + y + z)=0. Similarly, we consider the analogous circles 2 2 2 2 CBBC : a yz + b zx + c xy − c x(x + y + z)=0. and 2 2 2 2 CCCA : a yz + b zx + c xy − a y(x + y + z)=0. 11.4 The Brocard points and the Brocard circle 169

These three circles intersect at the forward Brocard point 1 1 1 → Ω = c2 : a2 : b2 .

This common angle is an acute angle ω given by

a2 + b2 + c2 cot ω = . 2S

It is called the Brocard angle of ABC. This follows by writing the coordinates of Ω→ in Conway’s notation

Analogously, there is a unique point Ω← such that

∠BAΩ← = ∠CBΩ←BA = ∠ACΩ←.

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.

Ω←

ω ω B C 170 Circles

The equations of these circles are

2 2 2 2 CABB : a yz + b zx + c xy − a z(x + y + z)=0,

2 2 2 2 CBCC : a yz + b zx + c xy − b x(x + y + z)=0,

2 2 2 2 CABB : a yz + b zx + c xy − c y(x + y + z)=0. These three circles intersect at the backward Brocard point 1 1 1 Ω← = b2 : c2 : a2 .

The common angle is the same ω given above. It follows that the two points Ω→ and Ω← are isogonal conjugates.

ω ω

Ω→

Ω←

ω ω ω ω B C

Exercise

4 1. The midpoint of the segment Ω→Ω← is the Brocard midpoint

(a2(b2 + c2):b2(c2 + a2):c2(a2 + b2)).

Show that this is a point on the line OK.

4 The Brocard midpoint appears in ETC as the point X 39. 11.4 The Brocard points and the Brocard circle 171

2. Let XY Z be the pedal triangle of Ω→ and X Y Z be that of Ω←.

(a) Find the coordinates of these pedals. (b) Show that YZ is parallel to BC. (c) The triangle bounded by the three lines YZ, ZX and XY is homothetic to triangle ABC. What is the homothetic center? 5 (d) The triangles ZXY and Y ZX are congruent.

11.4.1 The third Brocard point

Consider the Brocard points Ω→ and Ω←. The lines BΩ← and CΩ→ intersect at X(−ω). Similarly, we have Y (−ω)=CΩ← ∩ AΩ→, and Z(−ω)=AΩ← ∩ BΩ→. Clearly, the triangle X( − ω)Y( − ω)Z(−ω) is perspective to ABC at the point − 1 1 1 ∼ 1 1 1 K( ω)= : : 2 : 2 : 2 , Sα − Sω Sβ − Sω Sγ − Sω a b c which is the isotomic conjugate of the symmedian point. 6

11.4.2 The Brocard circle The Brocard circle is the circle through the three points X(−ω), Y (−ω), and Z(−ω).It has equation a2b2c2 x y z a2yz + b2zx + c2xy − (x + y + z) + + =0. a2 + b2 + c2 a2 b2 c2

Show that this circle also contains the two Brocard points Ω→ and Ω←, the circumcenter, and the symmedian point.

5The symmedian point. 6 This is also known as the third Brocard point. It appears as the point X 76 in ETC. 172 Circles

Z Z

: Z X

. Z Z :! Y Z Z B Z C

Figure 11.1: The third Brocard point 11.5 The Taylor circle 173

11.5 The Taylor circle

Consider the orthic triangle XY Z with

X =(0:Sγ : Sβ),Y=(Sβ :0:Sα),Z=(Sβ : Sα :0). The pedals of the vertices of the orthic triangle on the sidelines of triangle ABC are

Line Pedals of X Pedals of Y Pedals of Z 2 2 a Ya =(0:SCC : S ) Za =(0:S : SBB) 2 2 b Xb =(SCC :0:S ) Zb =(S :0:SAA) 2 2 c Xc =(SBB : S :0) Yc =(S : SAA :0)

Yc Zb

Z H

B Za X Ya C

These six points are on a circle, with equation 1 a2yz + b2zx + c2xy − (x + y + z)(S x + S y + S z)=0. 4R2 αα ββ γγ This is called the Taylor circle. Its center is called the Taylor center and has coordinates

2 4 2 4 2 4 (a (S − Sαβγ Sα):b (S − SαβγSβ):c (S − Sαβγ Sγ )). This is a point on the Brocard axis. Its coordinates can be written as

2 2 2 (a (SA + Sθ):b (SB + Sθ):c (SC + Sθ)) 174 Circles for −S4 Sθ = . SASBSC In other words, tan θ = − tan A tan B tan C.

The radius of the Taylor circle is

R sin ω . sin(θ + ω)

The orthocenter of triangle AZbYc is the point

4 2 2 7 Ha =(S − Sαβγ Sα : b Sααβ : c Sααγ ).

Similarly, let Hb and Hc be the orthocenters of triangles BXcZa and CYaXb respectively. 2 2 Note that triangle HaHbHc is perspective to ABC at the circumcenter O =(a SA : b SB : 2 c SC ).

Proposition 11.5. The triangle HaHbHc is oppositely congruent to the orthic triangle at the Taylor center.

Proof. 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 : SC : SB)+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 SC + a b SAAB : b c S SB + a c SAAC) 2 4 2 2 2 2 2 2 2 2 =(a (S − SAABC ):b (c S SC + a SAAB):c (b S SB + a SAAC))

Now

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

2 4 Similarly, the third coordinate is c (S − SCCAB). The midpoint of XHa is the Tay- lor center; similarly for YHb and ZHc. The triangles HaHbHc and XY Z are therefore oppositely congruent at the Taylor center.

7The sum of the coordinates is b2c2S2. 11.5 The Taylor circle 175

11.5.1 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 triangle. This shows that the radius of the Taylor circle 1 2 2 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. 8

3. Show that the Euler lines of the triangles XXbXc, YYcYa, ZZaZb are concurrent and ﬁnd the coordinates of the intersection. 9

8 2 The line XbXc has equation −S x + SBBy + SCCz =0 9 J.-P. Ehrmann, Hyacinthos 3695. The common point of the Euler lines is X 973. 176 Circles

11.6 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 veriﬁcation 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 )

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). Its distance uS from BC is (u+v+w)a , and its pedal on BC is the point

2 2 (0 : SC u + 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 − SC w) a2(u + v + w)2 11.6 The Dou circle 177

11.6.1 August 17, 2002: Edward Brisse Theorem 11.6 (Brisse). The Dou circle is orthogonal to the circle through the centroid, X111, and the anticomplement of X110.

10 [This circle intersects the Euler line at X858]. The center of this circle is the point

[(b2 − c2)(a2(b2 + c2)+(b4 − 4b2c2 + c4)], which is the superior of the point 11

[(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)

10Identiﬁcation number 2.33589872509 ···. 11Identiﬁcation number 2.77610382619 ···. 178 Circles

11.7 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. 12

Ge I

B C

the Adams circle. It has radius (4R + r)2 + s2 · r. 4R + r Lemma 11.7. 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 Gergonne 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 12 This is called the Adams circle. It is concentric with the incircle, and has radius 4R+r · r. Chapter 12

Four homothetic triangles with collinear homothetic centers

Given a triangle ABC, there is a quadruple of homothetic triangles with four centers of similitude lying on a line. These four triangles are (1) the orthic triangle, (2) the tangential triangle, bounded by the tangents to the circumcircle at the vertices, (3) the intangent triangle, bounded by the “fourth” common tangents of the incircle and each excircle, (4) the extangent triangle, bounded by the “fourth” common tangents of each pair of excircles.

12.1 The tangential triangle

The tangential triangle is bounded by the tangents to the circumcircle at the vertices. Each of these tangents is the reﬂection of a side of the superior triangle in the corresponding angle bisector. The tangential triangle is the anticevian triangle of the symmedian point K. It is clear that the reﬂection of each side of the orthic triangle about the corresponding angle bisector is parallel to the side of the triangle. It follows that the sides of the tangential triangle and the orthic triangle are parallel. The two triangles are homothetic. Their homothetic center is the Gob perspector 1

a2 b2 c2 H/K = : : . Sα Sβ Sγ

Note that the incenters of these triangles are respectively the circumcenter O and the orthocenter H (of triangle ABC), it follows that H/K lies on the Euler line OH.

1 This is the triangle center X25 in ETC. 180 Four homothetic triangles with collinear homothetic centers

12.2 The intangent triangle

The side lines of the intangent triangle are the reﬂections of BC, CA, AB respectively in the lines IoIa, IoIb, IoIc. They are parallel to the sides of the tangential triangle. The intangent triangle is oppositely homothetic to the tangential triangle.

B C

Since the circles O(R) and I(r) are the ob-incircles of the tangential and the intangent triangles respectively, the homothetic center is the insimilicenter T+.

12.3 The extangent triangle

The sides of the extangent triangle are the reﬂections of the same lines BC, CA, AB respectively in the lines IbIc, IcIa, and IaIb. Since IbIc ⊥ IoIa etc., these are parallel to the corresponding sides of the intangent triangles. The two triangles are homothetic. Proposition 12.1. The ob-incircle of the extangent triangle is the circle I (2R + r), where I =2O − I is the reﬂection of I in O. Corollary 12.2. 0(Tint, Text)=T+. 12.3 The extangent triangle 181

C A

B C

Figure 12.1:

The equations of the fourth extangents are

bcx + c(b + c)y + b(b + c)z =0, c(c + a)x + cay + a(c + a)z =0, b(a + b)x + a(a + b)y + abz =0.

The vertices of the extangent triangle are

2 −a s : b(c + a)sc : c(a + b)sb, 2 a(b + c)sc : −b sb : c(a + b)sa, 2 a(b + c)sb : b(c + a)sa : −c sc. 182 Four homothetic triangles with collinear homothetic centers

This has perspector T a(b + c) b(c + a) c(a + b) ( ext)= − : − : − . 0 b + c a c + a b a + b c

Since the reﬂections of the sides of the orthic triangle about the respective bisectors are parallel to the sides of the reference triangle ABC, these three triangles are homothetic. A line L : px + qy + rz =0is tangent to the incircle if and only if s s s a + b + c =0. p q r By extraversion, the condition of tangency of L with the excircles are

− − s sc sb a excircle : p + q + r =0; − sc − s sa b excircle : p q + r =0; − sb sa − s c excircle : p + q r =0. It follows that the “fourth” common tangent of the incircle and a−excircle is the line with 1 1 1 : : =s2 − s2 : −(ss + s ):ss + s p q r b c c ab b ca =a(b − c):ab : −ac =b − c : b : −c.

The side lines of the intangent triangle are therefore

x y − z b−c + b c =0, − x y z a + c−a + c =0, x − y z a b + a−b =0. Its vertices are the points

2 a sa : b(c − a)sb : −c(a − b)sc, 2 −a(b − c)sa : b sb : c(a − b)sc, 2 a(b − c)sa : −b(c − a)sb : c sc. The fourth common tangent of the b− and c−excircles is the line

bcx + c(b + c)y + b(b + c)z =0, c(c + a)x + cay + a(c + a)z =0, b(a + b)x + a(a + b)y + abz =0.

with 1 1 1 : : = s2 − s2 :(ss + s ):(ss + s )=b + c : b : c. p q r a c ab b ca 12.3 The extangent triangle 183

The lines joining the vertices of the intangent triangle to the corresponding vertices of the extangent triangle are respectively bc(b − c)x + a2cy − a2bz =0, −b2cx + ca(c − a)y + ab2z =0, bc2x − c2ay + ab(a − b)z =0. Each of these lines passes through the corresponding vertex of the tangential triangle. It follows that the tangential, the intangent, and the extangent triangle have a common center of similitude at a2(b + c − a):b2(c + a − b):c2(a + b − c), which is the pro-Nagel point T+, the internal center of similitude of the circumcircle and the incircle. In general, for four mutually homothetic triangles, there are 6 centers of similitude. Those of any three of the triangles are collinear. In the present case, however, there are only 4 centers of similitude, and they are collinear.

orthic tangential intangent extangent orthic ∗ X25 X33 X19 tangential ∗ T+ T+ intangent ∗ T+ extangent ∗ The line joining the vertices of the intangent triangle to the corresponding vertices of the orthic triangle are

(b − c)Sαx + aSβy − aSγz =, 0,

−bSαx +(c − a)Sβy + bSγ z =0,

cSαx − cSβy +(a − b)Sγ z =0. From these, the homothetic center of the intangent and the orthic triangle is the point as bs cs a : b : c . Sα Sβ Sγ

This is the point X33 in [ETC]. The lines joining the vertices of the extangent triangle to the corresponding vertices of the orthic triangle are

(b − c)Sαx − aSβy + aSγ z =0, bSαx +(c − a)Sβy − bSγ z =0, −cSαx + cSβy +(a − b)Sγ z =0. The homothetic center of the extangent triangle and the orthic triangle is the Clawson point a b c X19 = : : . Sα Sβ Sγ 184 Four homothetic triangles with collinear homothetic centers

12.4 The line of centers of similitude

Sine the centroid, the incenter and the Nagel point are collinear, (the line containing them being (b − c)x +(c − a)y +(a − b)z =0), the three points X19 (Clawson), X25 (Gob), and X33 are on the line (b − c)S (c − a)S (a − b)S α x + β y + γ z =0. a b c

This also contains the insimilicenter T+. 12.5 185

12.5

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.

a Ab =(0:−b : a + b) Ac =(0:c + a : −c) b Ba =(−a :0:a + b) Bc =(b + c :0:−c) c Ca =(−a : c + a :0) Cb =(b + c : −b :0)

Ac B C Ab

Ca 186 Four homothetic triangles with collinear homothetic centers

1. Let A = BBa ∩ CCa, B = CCb ∩ AAb and C = AAc ∩ BBc. Cb

B C

Ac B C Ab

Ca A simple calculation shows that AA , BB, CC intersect at the Spieker center Sp:

Ba = −a :0:a + b Ca = −a : c + a :0 A = −a : c + a : a + b B = b + c : −b : a + b C = b + c : c + a : −c b + c : c + a : a + b In fact, Sp is the common midpoint of the segments AA , BB , CC , and A B C is oppositely congruent to ABC.

Therefore, the lines BBc and CCb are parallel. They are indeed parallel to the bisector of angle A: 1 1 a + b + c B − B = (b + c, 0, −c) − (0, 1, 0) = (b + c, −b, −c)= (A − I). c b b b

Similarly, CCa and AAc are parallel to the bisector of angle B; and AAb, BBa are parallel to the bisector of angle C. 2. A B C is perspective to the excentral triangle IaIbIc and the extouch triangle X Y Z, both at O := 2 · O − I, the reﬂection O of the incenter in the circumcenter, which is also the circumcenter of the excentral triangle. From 2s · A =(−a, b, c)+(0,c+ a − b, a + b − c)=2(s − a)Ia +2a · X ,wehave s · A =(s − a)Ia + a · X . Now, it is known that IaX = ra and IaO =2R. Therefore, 2R · X = ra · I +(2R − ra)Ia. 12.5 187

3. The line BaCa is parallel to the side BI CI of the cevian triangle of I:

1 1 B − C = (−a, 0,a+ b) − (−a, c + a, 0) a a b c 1 = (c(−a, 0,a+ b) − b(−a, c + a, 0)) bc 1 = (a(b − c), −b(c + a),c(a + b)) bc (c + a)(a + b) (a, 0,c) (a, b, 0) = − bc c + a a + b (c + a)(a + b) = (B − C ). bc I I

Therefore, the triangle bounded by BaCa, CbAb, and AcBc is homothetic to cev(I).

4. Let XY Z be the triangle bounded by the lines BaCa, CbAb, AcBc. Show that XY Z is perspective with ABC and ﬁnd the perspector. 2

B C Ab

Z Ba

Y Ca

The equation of the line BaCa is x y z + + =0. a c + a a + b

Similarly, CbAb and AcBc are the lines 2 a(b+c) b(c+a) c(a+b) b+c−a : c+a−b : a+b−c . This is the orthocenter of the intouch triangle. See §4.4.1. 188 Four homothetic triangles with collinear homothetic centers

x y z + + =0, b + c b a + b x y z + + =0. b + c c + a c These two lines intersect at 1 1 1 1 1 1 b a+b b+c a+b b+c b X = 1 1 : − 1 1 : 1 1 c+a c b+c c b+c c+a 1 1 1 1 1 1 1 1 = − : − : − bc (c + a)(a + b) b + c a + b c b + c c + a b =a(b + c)(a + b + c):−b(c + a)(a + b − c):−c(a + b)(c + a − b) a(b + c)(a + b + c) b(c + a) c(a + b) = : − : − . (c + a − b)(a + b − c) c + a − b a + b − c

5. The homothetic center of XY Z and cev(I) is

2abc · X +2(c + a)(a + b)(b + c)AI =(a(b + c)(a + b + c), −b(c + a)(a + b − c), −c(a + b)(c + a − b)) +2(c + a)(a + b)(0,b,c) =(a + b + c) · (a(b + c),b(c + a),c(a + b)).

• Therefore, ∧0(XY Z, cev(I)) = (a(b + c):b(c + a):c(a + b)) is the inferior of I , − (b+c)(c+a)(a+b) and the homothetic ratio is abc .

6. The segment YZand BaCa have a common midpoint. The midpoint of BaCa is

Ma = c(−a, 0,a+ b)+b(−a, c + a, 0) = (−a(b + c),b(c + a),c(a + b)).

Similarly for CbAb and AcBc. These midpoints form the anticevian triangle of (a(b+ c):b(c + a):c(a + b)). The points Y and Z are Y =(−a(b + c)(a + b − c):b(c + a)(a + b + c):−c(a + b)(b + c − a)), Z =(−a(b + c)(c + a − b):−b(c + a)(b + c − a):c(a + b)(a + b + c)).

Note that the coefﬁcient sum in each case is 2abc. It follows that the midpoint of YZ is the point (−a(b + c)(a + b − c),b(c + a)(a + b + c), −c(a + b)(b + c − a)) +(−a(b + c)(c + a − b), −b(c + a)(b + c − a),c(a + b)(a + b + c)) =(−2a2(b + c), 2ab(c + a), 2ac(a + b)) =2a(−a(b + c),b(c + a),c(a + b)).

This is the same point Ma above. Similarly, we have Mb and Mc. 12.5 189

7. The triangle MaMbMc is

(a) perspective with ABC at (a(b + c):b(c + a):c(a + b)), 3 4 (b) perspective with A B C at (a(b + c)Sα : b(c + a)Sβ : c(a + b)Sγ ), (c) is the inferior triangle of XY Z; their common centroid is the point 5

a(−a(b + c),b(c + a),c(a + b)) +b(a(b + c), −b(c + a),c(a + b)) +c(a(b + c),b(c + a), −c(a + b)) =(a(b + c)(b + c − a),b(c + a)(c + a − b),c(a + b)(a + b − c)),

which coincides with the centroid of the extouch triangle.

8. The homothetic center 6

∧0(cev(I),MaMbMc)=I/(a(b + c):b(c + a):c(a + b)) =(a2(b + c):b2(c + a):c2(a + b)).

9. The circumcenter of triangle XY Z is the point which divides IO in the ratio 3:−1.

Proof. The equation of the circle ABaCa is

a2yz + b2zx + c2xy + a(x + y + z)(cy + bz)=0.

Therefore, the radical axis with the circumcircle is the external bisector of angle A, the line cy + bz =0. This shows that the point (a2 : −b(b − c):c(b − c)).

2(b2 + bc + c2 − a2)(a2, −b(b − c),c(b − c)) +(c + a − b)(a + b − c)(−a(b + c),b(c + a),c(a + b)) = − (b + c − a)(c + a − b)(a + b − c)(a, b, c) +(−a2(c + a − b)(a + b − c),b(c + a − b)(a + b − c)(b +2c),c(c + a − b)(a + b − c)(2b + c)) +2(b2 + bc + c2 − a2)(a2, −b(b − c),c(b − c)) = − (b + c − a)(c + a − b)(a + b − c)(a, b, c) +3(a2(b2 + c2 − a2),b2(c2 + a2 − b2),c2(a2 + b2 − c2))

Try to ﬁnd p, q, h, k such that

3 This appears as X37 in ETC. 4 This appears as X72 in ETC. 5 This appears as X210 in ETC. 6 This appears as X42 in ETC. 190 Four homothetic triangles with collinear homothetic centers

p(a2, −b(b − c),c(b − c)) + q(−a(b + c),b(c + a),c(a + b)) =h(a, b, c)+k(a2(b2 + c2 − a2),b2(c2 + a2 − b2),c2(a2 + b2 − c2)).

h + a(b2 + c2 − a2)k − ap +(b + c)q =0, h + b(c2 + a2 − b2)k +(b − c)p − (c + a)q =0, h + c(a2 + b2 − c2)k − (b − c)p − (a + b)q =0.

This has solution h = −(b + c − a)(c + a − b)(a + b − c),k=3; p =2(b2 + bc + c2 − a2),q=(c + a − b)(a + b − c).

This means that the intersection is the point

−(b + c − a)(c + a − b)(a + b − c) · 2s · I +3· 4S2 · O = −4S2 · I +12S2 · O =4S2(3 · O − I).

This is the point which divides IO in the ratio 3:−1. It is the midpoint of O and I.

10. The circumcenter of the cevian triangle of I: From

abc(3 · O − I)+2(c + a)(a + b)(b + c)O =(a + b + c)(ab + bc + ca)(3 · G − I•),

we have

4(c + a)(a + b)(b + c)O =2(a + b + c)(ab + bc + ca)(3 · G − I•) −3abc(2 · O − I)+abc · I =3abc · I +2(a + b + c)(ab + bc + ca)(3 · G − I•)+abc · I =3((a + b + c)(ab + bc + ca)G − abc · O) +(3abc · I − (a + b + c)(ab + bc + ca)I•)