Introduction to the Geometry of the Triangle

Paul Yiu

Fall 2005

Department of Mathematics Florida Atlantic University

Version 5.0924

September 2005

Contents

1 The Circumcircle and the Incircle 1 1.1 Preliminaries ...... 1 1.1.1 Coordinatization of points on a line ...... 1 1.1.2 Centers of similitude of two circles ...... 2 1.1.3 Tangent circles ...... 3 1.1.4 Harmonic division ...... 4 1.1.5 Homothety ...... 5 1.1.6 The power of a point with respect to a circle ...... 6 1.2 Menelaus and Ceva theorems ...... 7 1.2.1 Menelaus and Ceva Theorems ...... 7 1.2.2 Desargues Theorem ...... 8 1.3 The circumcircle and the incircle of a triangle ...... 9 1.3.1 The circumcircle and the law of sines ...... 9 1.3.2 The incircle and the Gergonne point ...... 10 1.3.3 The Heron formula ...... 14 1.3.4 The excircles and the Nagel point ...... 16 1.4 The medial and antimedial triangles ...... 20 1.4.1 The medial triangle, the nine-point center, and the Spieker point . . 20 1.4.2 The antimedial triangle and the orthocenter ...... 21 1.4.3 The Euler line ...... 22 1.5 The nine-point circle ...... 25 1.5.1 The Euler triangle as a midway triangle ...... 25 1.5.2 The orthic triangle as a pedal triangle ...... 26 1.5.3 The nine-point circle ...... 27 1.6 The OI-line ...... 31 1.6.1 The homothetic center of the intouch and excentral triangles . . . . 31 1.6.2 The centers of similitude of the circumcircle and the incircle . . . . 32 1.6.3 Reﬂection of I in O ...... 33 1.6.4 Orthocenter of intouch triangle ...... 34 1.6.5 Centroids of the excentral and intouch triangles...... 35 1.6.6 Mixtilinear incircles ...... 39 1.7 Euler’s formula and Steiner’s porism ...... 41 1.7.1 Euler’s formula ...... 41 1.7.2 Steiner’s porism ...... 42 iv CONTENTS

2 Homogeneous Barycentric Coordinates 45 2.1 Barycentric coordinates with reference to a triangle ...... 45 2.1.1 Homogeneous barycentric coordinates ...... 45 2.1.2 Point of division of a segment ...... 49 2.1.3 The area formula ...... 51 2.2 Conway’s formula ...... 52 2.2.1 Conway’s Notation ...... 52 2.2.2 Conway’s formula ...... 53 2.3 Equations of Straight lines ...... 58 2.3.1 Two-point form ...... 58 2.3.2 Inﬁnite points and parallel lines ...... 59 2.3.3 Perpendicular lines ...... 61 2.3.4 Intercept form: trilinear pole and polar ...... 63 2.3.5 Intersection of two lines ...... 65 2.4 Cevian and anticevian triangles ...... 68 2.4.1 Cevian triangle ...... 68 2.4.2 Anticevian triangle ...... 73 2.4.3 Construction of anticevian triangle ...... 73 2.5 Perspectivity ...... 76 2.5.1 Perspector and perspectrix ...... 76 2.5.2 Triangles bounded by lines parallel to the sidelines ...... 77 2.5.3 The cevian nest theorem ...... 80 Chapter 1

The Circumcircle and the Incircle

1.1 Preliminaries

1.1.1 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 The Circumcircle and the Incircle

1.1.2 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 1.1 Preliminaries 3

1.1.3 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 The Circumcircle and the Incircle

1.1.4 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 = −1 = − = ically. This is because NB MB , 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. 1.1 Preliminaries 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.

1.1.5 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). 6 The Circumcircle and the Incircle

1.1.6 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 (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

T 1.2 Menelaus and Ceva theorems 7

1.2 Menelaus and Ceva theorems

1.2.1 Menelaus and Ceva Theorems Consider a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively.

Theorem 1.1 (Menelaus). The points X, Y , Z are collinear if and only if BX CY AZ · · = −1. XC YA ZB

X B C

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

Y P

B X C 8 The Circumcircle and the Incircle

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

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 ratio is +1, and it follows from the Ceva theorem that AX, BY , CZ are concurrent. 1.3 The circumcircle and the incircle of a triangle 9

1.3 The circumcircle and the incircle of a triangle

For a generic triangle ABC, we shall denote the lengths of the sides BC, CA, AB by a, b, c respectively.

B D C

1.3.1 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 circumradius R is given by the law of sines: a b c 2R = = = . sin A sin B sin C 10 The Circumcircle and the Incircle

1.3.2 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 S =(a + b + c)r.

Z G e 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 2 ,BZ BX 2 ,CX CY 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, + + + = B C = C A = A B X 2 ,Y 2 ,Z 2 . It is always acute angled, and

=2 cos A =2 cos B =2 cos C YZ r 2 ,ZX r 2 ,XY r 2 . 1.3 The circumcircle and the incircle of a triangle 11

Exercise 1. 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.

Y X

C Z A

2. 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. 2

I Ge

B C

3. 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. (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. √ (4R+r)2+s2 2 This is called the Adams circle. It is concentric with the incircle, and has radius 4R+r · r. 12 The Circumcircle and the Incircle

(3) The two circles in (1) and (2) are congruent.

4. Given a point Z on a line segment AB, construct a right-angled triangle ABC whose incircle touches the hypotenuse AB at Z. 3

5. 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?

6. 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. 4 (2) What is the locus of the center of the circle EFZY ? 5

3P. Yiu, G. Leversha, and T. Seimiya, Problem 2415 and solution, Crux Math. 25 (1999) 110; 26 (2000) 62 Ð 64. 4International Mathematical Olympiad 1996. 5IMO 1996. 1.3 The circumcircle and the incircle of a triangle 13

7. 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.

Y G

B C 14 The Circumcircle and the Incircle

1.3.3 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.If the incircle I(r) and this excircle are tangent to the line AC at Y and Y respectively, then (1) from the similarity of triangles AIZ and AIaZ , − r = s a; ra s

(2) from the similarity of triangles BIZ and IaBZ ,

r · ra =(s − b)(s − c).

ra I

Y s − b C s − c Y s − a A

It follows that (s − a)(s − b)(s − c) r = . s From this we obtain the famous Heron formula for the area of a triangle: ∆=rs = s(s − a)(s − b)(s − c). 1.3 The circumcircle and the incircle of a triangle 15

Exercise 1. Show that 162 =2b2c2 +2c2a2 +2a2b2 − a4 − b4 − c4.

= abc 2. R 4∆ . = ∆ 3. ra s−a . 4. 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

5. Suppose the A-excircle touches BC at X . Show that the antipode X of X on the incircle lies on the segment AX . 16 The Circumcircle and the Incircle

1.3.4 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 1.3 The circumcircle and the incircle of a triangle 17

Exercise 1. Construct the tritangent circles of a triangle ABC. (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 .

(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).

Z Y

Z Mi Y T I

C B X X

Ia 18 The Circumcircle and the Incircle

2. 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.

Ic Z

Y Sp

C B X

6This is called the Spieker point of triangle ABC. 1.3 The circumcircle and the incircle of a triangle 19

3. 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. 7

Y F E

Z I P

B X D C

7 Crux Math. Problem 2250. The reﬂection of the Nagel point N a in the incenter. This is X145 of ETC. 20 The Circumcircle and the Incircle

1.4 The medial and antimedial triangles

1.4.1 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 )( ) 8 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.

F E

G O N Sp

B D C

h( − 1 ) The image of a point P under the homothety G, 2 is called its inferior, which we denote by P−.

− = 1 (3 − ) P 2 G P .

For example, the incenter of the medial triangle is the inferior of the incenter I, namely, 1 (3 − ) 9 2 G I . It is called the Spieker point Sp of triangle ABC.

8 The reason for this will be clear in §?. The nine-point center appears as X 5 in ETC. 9 The Spieker point appears as X10 in ETC. 1.4 The medial and antimedial triangles 21

1.4.2 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. 10 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.

C A B

G N O H

B C

A 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.

10It is also called the anticomplementary triangle. 22 The Circumcircle and the Incircle

1.4.3 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. 1.4 The medial and antimedial triangles 23

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 A 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? A

B C

A 24 The Circumcircle and the Incircle

4. The Euler lines of triangles IBC, ICA, IAB also intersect at a point on the Euler line of triangle ABC. 11

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

11Problem 1018, Crux Mathematicorum. 1.5 The nine-point circle 25

1.5 The nine-point circle

1.5.1 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

B C 26 The Circumcircle and the Incircle

1.5.2 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. 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.

Z P

B X C 1.5 The nine-point circle 27

1.5.3 The nine-point circle From §§1.5.1, 1.5.2 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.

A Y

F E

O N

Z H

C B

B X D C 28 The Circumcircle and the Incircle

Exercise 1. Show that (i) the incenter is the orthocenter of the excentral triangle (ii) the circumcircle is the nine-point circle of the excentral triangle, (iii) the circumcenter of the excentral triangle is the reﬂection of I in O.

0 I I

C B

Ia 1.5 The nine-point circle 29

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? 12

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. 13 (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. 14

A' B C

12P. Yiu and J. Young, Problem 2437 and solution, Crux Math. 25 (1999) 173; 26 (2000) 192. 13B. Gibert, Hyacinthos 1158, 8/5/00. 14A.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. 30 The Circumcircle and the Incircle

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. 15

15O. Bottema, Hoofdstukken uit de Elementaire Meetkunde, Chapter 16. 1.6 The OI-line 31

1.6 The OI-line

1.6.1 The homothetic center of the intouch and excentral triangles The OI-line of a triangle is the line joining the circumcenter and the incenter. We consider several interesting triangle centers on this line, which arise from the homothety of the intouch and excentral triangles. These triangles are homothetic since their corresponding sides are parallel, being perpendicular to the same angle bisector of the reference triangle.

Z T I

C B X

Ia 2R =2 − The ratio of homothetic is clearly r . Their circumcenters are I O I and I. The homothetic center is the point T such that 2R 2R 2R 2O − I = h T, (I)= 1 − T + · I. r r r

This gives (2R + r)I − 2r · O T = . 2R − r It is the point which divides OI externally in the ratio 2R + r : −2r. 16

16 The point T is X57 in ETC. 32 The Circumcircle and the Incircle

1.6.2 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. 17

1 T+ = (r · O + R · I), R + r 1 T− = (−r · O + R · I). R − r

Z I O T− T+

B X C

17 T+ and T− are respectively X55 and X56 in ETC. 1.6 The OI-line 33

1.6.3 Reﬂection of I in O

The reﬂection I of I in O is the circumcenter of the excentral triangle. It is the intersections of the perpendiculars from the excenters to the sidelines. 18 The midpoint M of the arc BC is also the midpoint of IIa. Since IM and I Ia are parallel, I Ia =2R. Similarly, I Ib = Pc =2R. This shows that the excentral triangle has circumcenter I =2O − I and circumradius 2R. Since I is the orthocenter of IaIbIc, its follows that O, the midpoint of I and I , is the nine-point center of the excentral triangle. In other words, the circumcircle of triangle ABC is the nine-point circle of the excentral triangle. Apart from A, the second intersection of IbIc with the circumcircle of ABC is the 19 midpoint M of IbIc. From this we deduce the following interesting formula:

ra + rb + rc =4R + r.

M A

O I I

Xc B D Xb

18 I appears as X40 in ETC. 19 Proof. ra + rb + rc =2MD+(I Ia − I X )=2R +2OD + I Ia − I X =4R + IX =4R + r. 34 The Circumcircle and the Incircle

1.6.4 Orthocenter of 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. This is the point which divides OI in the ratio R + r : −r. 20 It also lies on the line joining the Gergonne and Nagel points.

Z H I O G e Na

B X C

20 This is the point X65 in ETC. 1.6 The OI-line 35

1.6.5 Centroids of the excentral and intouch triangles The centroid of the excentral triangle is the point which divides OI in the ratio −1:4. 21 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. 22

Y O Z T I

C B X

Exercises 1. Can any of the centers of similitude of (O) and (I) lie outside triangle ABC?

2. Show that the distance between the circumcenter and the Nagel point is R − 2r. 23

3. Identify the midpoint of the Nagel point and the deLongchamps point as a point on the OI-line. 24

21 This is the point which divides the segment I I in to the ratio 1:2.ItisX165 of ETC. 22 The centroid of the intouch triangle is X354 of ETC. 23Feuerbach’s theorem. 24 Reﬂection of I in O. This is because the pedal triangle of X 40 is the cevian triangle of the Nagel 36 The Circumcircle and the Incircle

4. Construct the orthic triangle of the intouch triangle. This is homothetic to ABC. Identify the homothetic center. 25

Z O H I

B X C

5. Construct the external common tangent of each pair of excircles. These three external common tangents bound a triangle. (i) This triangle is perpsective with ABC. Identify the perspector. 26 (ii) Identify the incenter of this triangle. 27

6. Extend AB and AC by length a and join the two points by a line. Similarly deﬁne 28 the two other lines. The three lines bound a triangle with perspector X65

7. Let H be the orthocenter of the intouch triangle XY Z, and X , Y , Z its pedals on the sides YZ, ZX, XY respectively. Identify the common point of the three lines AX , BY , CZ as a point on the OI-line. Homothetic center of intouch and excentral triangles.

8. 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. 29 point, and the reﬂections of the pedals of the Nagel point in the respective traces form the pedals of the de Longchamps point. 25T . 26Orthocenter of intouch triangle. 27Reﬂection of I in O. 282/18/03. 29A. P. Hatzipolakis, Hyacinthos message 793, April 18, 2000. 1.6 The OI-line 37

9. 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 internal center T+ of similitude of (O) and (I). Construct these three circles. 30

T+ O I

B C

30A.P. Hatzipolakis and P. Yiu, Triads of circles, preprint. 38 The Circumcircle and the Incircle

10. 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. 31

Z T+ O I

B Z X Y C

11. Let P be the centroid of the excentral triangle, with pedals X, Y , Z on the sides BC, CA, AB respectively. Show that 32 1 + = + = + = ( + + ) AY AZ BZ BX CX CY 3 a b c .

31A.P. Hatzipolakis and P. Yiu, Pedal triangles and their shadows, Forum Geom., 1 (2001) 81 Ð 90. 32The 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). 1.6 The OI-line 39

1.6.6 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

B C

A 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 external center of similitude of the circumcircle and the incircle. 40 The Circumcircle and the Incircle

This leads to a simple construction of the mixtilinear incircle. Given a triangle ABC, let P be the external center of similitude of the circumcircle (O) and incircle (I). Extend AP 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.

O I

T−

B C

A M 1.7 Euler’s formula and Steiner’s porism 41

1.7 Euler’s formula and Steiner’s porism

1.7.1 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 2 IMC 2 AMC 2 ABC, ∠ = 1 ∠ and for the same reason ICB 2 ACB. Note that = = =2 sin A (1) IM MB MC R 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 42 The Circumcircle and the Incircle

1.7.2 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

A 1.7 Euler’s formula and Steiner’s porism 43

Exercises ≤ 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. 33

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?

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?

33Hint: OI = r. 44 The Circumcircle and the Incircle Chapter 2

Homogeneous Barycentric Coordinates

2.1 Barycentric coordinates with reference to a triangle

2.1.1 Homogeneous barycentric coordinates The notion of barycentric coordinates dates back to A. F. M¬obius (Ð). 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 + + =0 1 absolute barycentric coordinate u+v+w , provided u v w . We also say that the balance point P has homogeneous barycentric coordinates (u : v : w) with reference to ABC. These masses can be taken in the proportions of the areas of triangle PBC, PCA and PAB. Allowing the point P to be outside the triangle, we use signed areas of oriented triangles, i.e., u : v : w =∆(PBC):∆(PCA):∆(PAB).

B C

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 §?. 46 Homogeneous Barycentric Coordinates

Examples (1) The centroid G has homogeneous barycentric coordinates (1:1:1). The areas of the triangles GBC, GCA, and GAB are equal. 2

B C

(2) The incenter I has homogeneous barycentric coordinates (a : b : c).Ifr denotes the 1 1 1 3 inradius, the areas of triangles IBC, ICA and IAB are respectively 2 ra, 2 rb, and 2 rc.

B C

2 The centroid appears as X2 in ETC. 3 The incenter appears as X1 in ETC. 2.1 Barycentric coordinates with reference to a triangle 47

(3) The circumcenter

O =(a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2)).

Proof. If R denotes the circumradius, the coordinates of the circumcenter O are 4

OBC : OCA : OAB 1 1 1 = 2 sin 2 : 2 sin 2 : 2 sin 2 2R A 2R B 2R C =sinA cos A :sinB cos B :sinC cos C

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).

B C

Exercise 1. Show that the sum of the coordinates of O given above is 162.

4 In ETC, the circumcenter appears as X3. 48 Homogeneous Barycentric Coordinates

(4) Points on the line BC have coordinates of the form (0 : y : z). In fact, if P is a point on BC, in terms of signed lengths of segments, it has (homogeneous) barycentric coordinates (0 : PC : BP). Likewise, points on CA and AB have coordinates of the forms (x :0:z) and (x : y :0)respectively. (5) Consider the center of mass of the perimeter of triangle ABC. The edges BC, CA, B+C C+A A+B AB can be replaced respectively by masses a, b, c at their midpoint 2 , 2 , and 2 . The center of mass is therefore the point

B+C C+A A+B a · + b · + c · (b + c)A +(c + a)B +(a + b)C 3G − I 2 2 2 = = , a + b + c 2(a + b + c) 2 = A+B+C since G 3 . This is the inferior of the incenter I, and is the incenter of the medial triangle, the Spieker center Sp. In homogeneous barycentric coordinates,

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

F E

B D C 2.1 Barycentric coordinates with reference to a triangle 49

2.1.2 Point of division of a segment If P and Q are given in absolute barycentric coordinates, the point X which divides PQ µ · P + λ · Q in the ratio PX : XQ = λ : µ has absolute barycentric coordinates .It p + q is, however, convenient to perform calculations avoiding denominators of fractions. We therefore adapt this formula in the following way: if P =(u : v : w) and Q =(u : v : w) are the homogeneous barycentric coordinates satisfying u + v + w = u + v + w, the point X dividing PQin the ratio PX : XQ = p : q has homogeneous barycentric coordinates (µu + λu : µv + λv : µw + λw).

(1) Consider the A-excenter Ia. This divides the segment AI in the ratio AIa : IaI = s : −a. With I =(a : b : c) and A =(a + b + c :0:0)with equal coordinate sums, we have

Ia =(−a(a + b + c)+sa : sb : sc) =(−2as + as : bs : cs) =(−a : b : c).

Similarly, the other two excenters are Ib =(a : −b : c) and Ic =(a : b : −c). (2) Insimilicenter T+ of the circumcircle and the incircle We compute the homongeneous barycentric coordinates of T , the insimilicenter (internal center of similitude) of the circumcircle and the incircle, from its absolute barycentric R·I+r·O : = abc : ∆ = :4∆2 =( : : ) coordinates R+r . Note that R r 4∆ s sabc . Since I a b c with coordinates sum 2s, and O =(a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2)) has coordinates sum 16∆2 and we equalize their sums and work with O =(sa2(b2 + c2 − a2),sb2(c2 + a2 − b2),sc2(a2 + b2 − c2)), I =(8∆2a, 8∆2b, 8∆2c). The a-component of its homogeneous barycentric coordinates can be taken as 4∆2 · sa2(b2 + c2 − a2)+sabc · 8∆2a. The simpliﬁcation turns out to be quite easy: 4∆2 · sa2(b2 + c2 − a2)+sabc · 8∆2a =4s∆2a2(b2 + c2 − a2 +2bc) =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)). 50 Homogeneous Barycentric Coordinates

Exercise 1. Establish the coordinates of the following points on the OI-line. = a2 : b2 : c2 (a) T− b+c−a c+a−b a+b−c , = a : b : c (b) T b+c−a c+a−b a+b−c , (c) 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. 2.1 Barycentric coordinates with reference to a triangle 51

2.1.3 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) · 162 a2(b2 + c2 − a2) b2(c2 + a2 − b2) c2(a2 + b2 − c2)

001 1 = a − bb− cc · 3(a + b + c) · 162 −(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) · 162 −(a + b)(a2 + b2 − c2) −(b + c)(b2 + c2 − a2) c2(a2 + b2 − c2)

(a − b)(b − c) 11 = · 3(a + b + c) · 162 −(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) 52 Homogeneous Barycentric Coordinates

2.2 Conway’s formula

2.2.1 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, 2 + 2 − 2 2 + 2 − 2 2 + 2 − 2 = b c a = c a b = a b c SA 2 ,SB 2 ,SC 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 = SB + SC , b = SC + SA, c = SA + SB. + + = a2+b2+c2 (2) SA SB SC 2 . 2 (3) SBC + SCA + SAB = 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, SAB + SBC + SCA = S (cot A · cot B +cotB · cot C +cotC · cot A)=S .

Example The circumcenter is the point 2 2 2 O =(a SA : b SB : c SC )=(SA(SB + SC ):SB(SC + SA):SC (SA + SB)). 2 Note that in this form, the coordinate sum is 2(SAB + SBC + SCA)=2S .

Exercise 1. Make use of the fact that the orthocenter H divides the segment OG externally in the ratio OH : HG =3:−2 to show that in homogeneous barycentric coordinates, 1 1 1 H =(SBC : SCA : SAB)= : : . SA SB SC

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

(2λSBC + µSA(SB + SC ):2λSCA + µSB(SC + SA):2λSAB + µSC(SA + SB)).

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

(a) N =(2SBC + SCA + SAB :2SCA + SAB + SBC :2SAB + SBC + SCA);

(b) Lo =(−SBC + SCA + SAB : −SCA + SAB + SBC : −SAB + SBC + SCA) 2.2 Conway’s formula 53

2.2.2 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 2.1 (Conway’s formula). In terms of the AB- and AC-swing angles θ and ϕ, the homogeneous barycentric coordinates of P are

2 (−a : SC + Sϕ : SB + 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 + C): · sin(θ + B) 2sin(θ + ϕ) 2sin(θ + ϕ) 2sin(θ + ϕ) ab sin(ϕ + C) ca sin(θ + B) = − a2 : : sin ϕ sin θ = − a2 : ab cos C + ab sin C cot ϕ : ca cos B + ca sin B cot θ 2 = − a : SC + Sϕ : SB + Sθ. 54 Homogeneous Barycentric Coordinates

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 CBXc 4 , BCXc 2 . cot π =1 cot π =0 Since 4 and 2 ,

2 Xc =(−a : SC : SB + S).

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

2 2 Yc =(SC : −b : SA + S), Ya =(SC + S : −b : SA); 2 2 Za =(SB + S : SA : −c ), Zb =(SB : SA + S : −c ).

A Yc

B C

Xb Xc 2.2 Conway’s formula 55

(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 : SC + SA : SB + SA)=(−a : b : c ).

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

B C

A (3) The midpoint of the two arcs BC of the circumcircle of triangle ABC. The midpoint M of the arc not containing the vertex A has coordinates − 2 : + · cot A : + · cot A a SC S 2 SB S 2 .

Note that

+ · cot A = + sin cot A SC S 2 SC bc A 2 = +2 cos2 A SC bc 2 =SC + bc(1 + cos A) 1 1 = ( 2 + 2 − 2)+ (2 + 2 + 2 − 2) 2 a b c 2 bc b c a =b(b + c). 56 Homogeneous Barycentric Coordinates

+ · cot A = ( + ) Similarly, SB S 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, dividing OM in the ratio OM : M M = −1:2. It is, however, also the midpoint of the of segment Ib and Ic. Therefore, it has coordinates (a + b − c)(a, −b, c)+(c + a − b)(a, b, −c)=(2a2, −2b(b − c), 2c(b − c)). In homogeneous form, this is (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 : : . SA + S SB + S SC + S

A Y Yc

P Zb

B C

Xb 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. 5. 2.2 Conway’s formula 57

A Yc

B C

Xb X Xc

B C

Xb Xc

5 S Recall that this can be obtained from applying the homothety h(A, S+a2 ) to the square BCX1X2 58 Homogeneous Barycentric Coordinates

2.3 Equations of Straight lines

2.3.1 Two-point form

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 line AP has equation wy − vz =0. It intersects the sideline BC at the point AP =(0:v : w), which we call the trace of P on BC. The traces of P on the other two sidelines are

BP =(u :0:w) and 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:

2 2( 2 − 2) =0 Euler line OH cyclic b c b c x ( − )( + − ) =0 OI-line OI cyclic bc b c b c a x ( − )( − )2 =0 Soddy line IGe cyclic b c s a x 2 2( 2 − 2) =0 Brocard axis OK cyclic b c b c x van Aubel line HK cyclic SAA(SB − SC )x =0 2.3 Equations of Straight lines 59

2.3.2 Infinite points and parallel lines A line px + qy + rz =0contains the point (q − r : r − p : p − q), as is easily verified. This is clearly an infinite point, and it is the only infinite point of satisfying px + qy + rz =0. We shall call it the infinite point of the line px + qy + rz =0. All infinite points have their coordinates (x : y : z) satisfy the equation x + y + z = 0. We say that the infinite points constitute the line at infinity L∞ which has equation x + y + z =0. 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

Exercise 1. If P =(u : v : w) and Q =(u : v : w) are finite points, the infinite point of the line PQis the point u − u : v − v : w − w u + v + w u + v + w u + v + w u + v + w u + v + w u + v + w −→ corresponding to the vector PQ. 2. (a) Find the infinite points of the sidelines BC, CA, AB respectively. (b) Find the equations of the lines through P =(u : v : w) parallel to the sidelines. 3. Establish the coordinates of the infinite points the following lines: (a) the line IG: (b + c − 2a : c + a − 2b : a + b − 2c);

(b) the Euler line:

(−2SBC + SCA + SAB : −2SCA + SAB + SBC : −2SAB + SBC + SCA);

(a(a2(b + c) − 2abc − (b + c)(b − c)2):···: ···)

in which the second and third coordinates are obtained from the ﬁrst by cyclic permutations of a, b, c; (d) the Brocard axis OK:

(a2(a2(b2+c2)−(b4+c4)) : b2(b2(c2+a2)−(c4+a4)) : c2(c2(a2+b2)−(a4+b4)). 60 Homogeneous Barycentric Coordinates

4. Let D, E, F be the midpoints of the sides BC, CA, AB of triangle ABC. For a point P with traces AP , BP , CP , let X, Y , Z be the midpoints of BP CP , CP AP , AP BP respectively. (a) Find P such that the lines DX, EY , FZ are parallel to the internal bisectors of angles A, B, C respectively. (b) Explain why the lines DX, EY , FZ are concurrent and identify the point of concurrency.

CP X F E

I P Y Z

B D AP C 2.3 Equations of Straight lines 61

2.3.3 Perpendicular lines We begin with a simple observation. The altitudes of triangle ABC have inﬁnite points

2 2 2 (−a : SC : SB), (SC : −b : SA), (SB : SA : −c ) respectively. These are useful in finding equation of lines perpendicular to the sidelines of triangle ABC. 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 SB SA −c SC −b SA

−h 0 f =0 and g −f 0 =0 xy z xyz

These are

2 SAfx+(c h − SBf)y + SAhz =0, 2 SAfx+ SAgy +(b g − SC f)z =0.

X' Y

B C

Figure 2.1: Perpendicular to a given line

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

2 2 X =(∗∗∗: SAf(SAh − b g + SC f):SAf(SAg + SBf − c h) ∼ (∗∗∗: SC (f − g) − SA(g − h):SA(g − h) − SB(h − f)). 62 Homogeneous Barycentric Coordinates

The perpendicular from A to L is the line AX , which has equation

10 0

∗∗∗ SC (f − g) − SA(g − h) −SA(g − h)+SB(h − f) =0, xy z or −(SA(g − h) − SB(h − f))y +(SC (f − g) − SA(g − h))z =0. This has inﬁnite point

(SB(h − f) − SC (f − g):SC (f − g) − SA(g − h):SA(g − h) − SB(f − g)).

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

Theorem 2.2. 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 =SB(h − f) − SC (f − g), g =SC (f − g) − SA(g − h), h =SA(g − h) − SB(h − f).

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

SAff + SBgg + SC hh =0.

Corollary 2.3. The perpendicular from the point (u : v : w) to the line fx+ gy + hz =0 is the line uvw f g h =0. xyz 2.3 Equations of Straight lines 63

2.3.4 Intercept form: 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

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

C X B AP

Figure 2.2: Construction of tripolar 64 Homogeneous Barycentric Coordinates

Construction of tripole 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'

B C X

Figure 2.3: Construction of tripole 2.3 Equations of Straight lines 65

2.3.5 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). The inﬁnite point of a line L can be regarded as the intersection of L with the line at inﬁnity L∞ : x + y + z =0.

Theorem

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

Exercise 1. Find the equation of the line joining the centroid to a given point P =(u : v : w). 6

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

3. 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. 7

4. Show the the Soddy line IGe contains the deLongchamps point.

6Equation: (v − w)x +(w − u)y +(u − v)z =0. 7(u(v − w)(v + w − 2u):···: ···). 66 Homogeneous Barycentric Coordinates

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? 8

F E B X P

Y P Z

B AP D C

6. 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? 9

X F E

H Z

B D C

8The intersection is the point dividing the segment PGin the ratio 3:1. 9This intersection is the symmedian point K =(a2 : b2 : c2). 2.3 Equations of Straight lines 67

7. 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

Ca 68 Homogeneous Barycentric Coordinates

2.4 Cevian and anticevian triangles

2.4.1 Cevian triangle The cevian triangle of a point P has vertices its traces on the sidelines:

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

BP CP

B AP C

Theorem 2.4 (Ceva). Three points X, Y , Z on BC, CA, AB respectively are the traces of a point if and only if they 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, XY Z = cev(P ) for P =(x : y : z). 2.4 Cevian and anticevian triangles 69

Examples (1) The cevian triangle of the centroid is the medial triangle. (2) The intouch triangle is the cevian triangle of the Gergonne point Ge. From the points of tangency of the incircle with the sidelines, we determine easily the coordinates of Ge. =(0:− : − )= 0: 1 : 1 X s c s b s−b s−c , =(− :0: − )= 1 :0: 1 Y s c s a s−a s−c , =(− : − :0) = 1 : 1 :0 Z s b s a s−a s−b .

Z I Ge

B X C

It follows that = 1 : 1 : 1 Ge s−a s−b s−c . 70 Homogeneous Barycentric Coordinates

(3) The Nagel point and the extouch triangle. From the points of tangency of the excircles with the corresponding sides we easily ﬁnd the coordinates of the Nagel point Na.

Ic Z

Na Y

C B X

Ia Since

X =(0 : s − b : s − c), Y =(s − a :0:s − c), Z =(s − a : s − b :0), it follows that Na =(s − a : s − b : s − c). 2.4 Cevian and anticevian triangles 71

(4) The orthocenter and the orthic triangle. The pedal of the orthocenter H on BC is the same as that of A on the same line. The trace of H on BC is the same as the pedal of A on the side BC. From BX = c cos B and XC = b cos C, cos B cos C BX : XC = c cos B : b cos C = : =cotB :cotC = SB : SC , b c we write down the coordinates of X, and similarly of Y and Z, easily: 1 1 X =(0:SC : SB)= 0: : , SB SC 1 1 Y =(SC :0:SA)= :0: , SA SC 1 1 Z =(SB : SA :0) = : :0 . SA SB It follows that = 1 : 1 : 1 H SA SB SC .

Z H

B X C 72 Homogeneous Barycentric Coordinates

Exercise 1. Find the area of the cevian triangle of (u : v : w). 10

2. Show that the intouch and the extouch triangles have equal areas.

3. Find the coordinates of P =(u : v : w) in its cevian triangle. 11

10 2uvw (v+w)(w+u)(u+v) ·. 11(v + w : w + u : u + v). 2.4 Cevian and anticevian triangles 73

2.4.2 Anticevian triangle 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.

2.4.3 Construction of anticevian triangle 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.

Z BP P Y B C X X' AP Y'

Figure 2.4: Anticevian triangle 74 Homogeneous Barycentric Coordinates

Examples of anticevian triangles (1) The anticevian triangle of the centroid is the antimedial 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 2.4 Cevian and anticevian triangles 75

(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. Find the coordinates of P =(u : v : w) in its anticevian triangle. 12

3. 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.

12(v + w − u : w + u − v : u + v − w). 76 Homogeneous Barycentric Coordinates

2.5 Perspectivity

2.5.1 Perspector and perspectrix 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). Two triangles T and T are homothetic if and only if their perspectrix is the line at inﬁnity. In this case, their perspector ∧(T, T) is called the homothetic center instead, and written as ∧0(T, T ).

Examples (1) ∧(cev(P )) = P , and ∧(cev−1(P )) = P . (2) For P = G, the centroid, the cevian and anticevian triangles are each homothetic to the reference triangle.

−1 −1 G = ∧0(cev(G)) = ∧0(cev (G)) = ∧0(cev(G), cev (G)).

Consider a triangle XY Z with perspector P =(u : v : w). The vertices X, Y , Z have coordinates of the form X = U : v : w, Y = u : V : w, Z = u : v : W, for some U, V , W . The perspectrix is the line x y z + + =0. u − U v − V w − W Proof. The line YZ has equation −(vw − VW)x + u(w − W )y + u(v − V )z =0.It intersects the sideline BC at (0 : v − V : −(w − W )). Similarly, the lines ZX and XY intersect CA and AB respectively at (−(u − U):0:w − W ) and (u − U : −(v − V ):0). 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. 2.5 Perspectivity 77

2.5.2 Triangles bounded by lines parallel to the sidelines

Proposition 2.5. Let 6a, 6b, 6c be the images of the side lines by the similarities h(A, 1+ta), h(B,1+tb), h(C, 1+tc) respectively. The three lines bound a triangle homothetic to ABC at the point (ta : tb : tc), the ratio of homothety being 1+ta + tb + tc.

P B C

B' Z Y C'

Figure 2.5: The Pythagorean conﬁguration and the symmedian point

Proof. The line 6a has equation x = −ta(x + y + z). Likewise, the lines 6b and 6c have equations y = −tb(x + y + z) and z = −tc(x + y + z) respectively. It follows that they bound a triangle with vertices

A =1+tb + tc : −tb : −tc, B = −ta :1+tc + ta : −tc, C = −ta : −tb :1+ta + tb.

It is clear that 0(A B C )=P =(ta : tb : tc). Since

(ta,tb,tc)=(1+ta + tb + tc)(1, 0, 0) − (1 + tb + tc, −tb, −tc), the homothetic center P divides the segment AA in the ratio

AP : PA = −1:1+ta + tb + tc. PA =1+ + + From this, the ratio of homothety is AP ta b tc. 78 Homogeneous Barycentric Coordinates

Examples (1) 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), S +a 2 2 1+ = a =1+a = a where t1 S S , i.e., t1 S . Similarly, if we erect squares externally on a the other two sides, the outer edges of these squares are on the lines which are the images h( 1+ ) h( 1+ ) = b2 = c2 of CA, AB under the homotheties B, t2 and C, t3 with t2 S and t3 S . The triangle bounded by the lines containing these outer edges is called the Grebe triangle a2 : b2 : c2 =( 2 : 2 : 2) of ABC. It is homothetic to ABC at S S S a b c , the symmedian point, and the ratio of homothety is

S + a2 + b2 + c2 1+(t1 + t2 + t3)= . S

K B C

Figure 2.6: The Pythagorean conﬁguration and the symmedian point

(2) 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. 2.5 Perspectivity 79

Exercise 1. Let P be a point. Consider the triangle bounded by lines parallel to the sides of cevian triangle of P through the vertices of ABC. Find the homothetic center 13

2. (Gossard’s Theorem) Suppose the Euler line of triangle ABC intersects the side lines BC, CA, AB at X, Y , Z respectively. The Euler lines of the triangles AY Z, BZX and CXY bound a triangle homothetic to ABC with ratio −1 and with homothetic center on the Euler line of ABC.

H c

C' H O a a B' Y O b G o Z O H

X B C

H b A'

O c

Figure 2.7: Gossard theorem

13(u(v + w):v(w + u):w(u + v)). 80 Homogeneous Barycentric Coordinates

2.5.3 The cevian nest theorem Theorem 2.6. 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 2.7. 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 =0 x . 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. 2.5 Perspectivity 81

We shall simply write

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

Examples (1) 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 SA : b SB : c SC )

14 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 2.8: The Mittenpunkt as the symemdian point of the excentral triangle

14 This appears as X9 in ETC. 82 Homogeneous Barycentric Coordinates

(2) If P =(u : v : w),

H/P =(u(−SAu + SBv + SC w):v(−SBv + SC w + SAu):w(−SC w + SAu + SBv)).

In particular, (i) H/I =(a(a3 + a2(b + c) − a(b2 + c2) − (b + c)(b − c)2):···: ···) divides is a point on the OI-line, dividing OI in the ratio −2r : R + r; 15

O I H P

B C

Ia = a2 : b2 : c2 (ii) H/K SA SB SC is the homothetic center of the orthic and tangential triangle; 16 (iii) H/O is the orthocenter of the tangential triangle; 17 (iv) H/N is the orthocenter of the orthic triangle. 18

15 This point appears as X46 in ETC. 16 This appears as X25 in ETC. It is a point on the Euler line. 17 This appears as X155 in ETC. 18 This appears as X52 in ETC. 2.5 Perspectivity 83

H O

B C

Figure 2.9: H/K (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 19

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

19 This appears as X1486 in ETC. 84 Homogeneous Barycentric Coordinates

B C

Figure 2.10: Ge/K

Theorem 2.8. (1) P/P = P ; (2) If P/Q = M, then Q = P/M. 2.5 Perspectivity 85

CM BQ

CP BP

Q P M

B C AP

AQ AM

Figure 2.11: 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. 20

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? 21

20 (u(v + w):v(w + u):w(u + v)) x + y + z =0 Perspector = , and perspectrix is the line u2 v2 w2 . 21 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 . 86 Homogeneous Barycentric Coordinates

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. 22

22(uu(vw + wv):··· : ···); see J. H. Tummers, Points remarquables, associ«esa ` un triangle, Nieuw Archief voor Wiskunde IV 4 (1956) 132 Ð 139. O. Bottema, Une construction par rapporta ` un triangle, ibid., IV 5 (1957) 68 Ð 70, has subsequently shown that this is the pole of the line PQ with respect to the circumconic through P and Q. 2.5 Perspectivity 87

CQ X' BQ B X P Q

CP Z Y' P Y Z'

B AP AQ C