Advanced Euclidean Geometry

Paul Yiu

Summer 2013

Department of Mathematics Florida Atlantic University

b c

B a C

August 2, 2013

Summer 2013

Contents

1 Some Basic Theorems 101 1.1 ThePythagoreanTheorem...... 101 1.2 Constructions of geometric mean ...... 104 1.3 Thegoldenratio...... 106 1.3.1 Theregularpentagon...... 106 1.4 Basic construction principles ...... 108 1.4.1 Perpendicular bisector locus ...... 108 1.4.2 Angle bisector locus ...... 109 1.4.3 Tangencyofcircles...... 110 1.4.4 Construction of tangents of a circle ...... 110 1.5 The intersecting chords theorem ...... 112 1.6 Ptolemy’s theorem ...... 114

2 The laws of sines and cosines 115 2.1 Thelawofsines ...... 115 2.2 Theorthocenter ...... 116 2.3 Thelawofcosines...... 117 2.4 Thecentroid...... 120 2.5 The angle bisector theorem ...... 121 2.5.1 The lengths of the bisectors ...... 121 2.6 The circle of Apollonius ...... 123

3 The tritangent circles 125 3.1 Theincircle ...... 125 3.2 Euler’sformula ...... 128 3.3 Steiner’sporism ...... 129 3.4 Theexcircles...... 130 3.5 Heron’s formula for the area of a triangle ...... 131

4 The arbelos 133 4.1 Archimedes’ twin circles ...... 133 1 1 1 4.1.1 Harmonic mean and the equation a + b = t ...... 134 4.1.2 Construction of the Archimedean twin circles ...... 134 4.2 Theincircle ...... 135 4.2.1 Construction of the incircle of the arbelos ...... 136 4.2.2 Alternative constructions of the incircle ...... 138 4.3 Archimedean circles ...... 140 iv CONTENTS

5 Menelaus’ theorem 201 5.1 Menelaus’theorem ...... 201 5.2 Centers of similitude of two circles ...... 203 5.2.1 Desargue’stheorem...... 203 5.3 Ceva’stheorem...... 204 5.4 Some triangle centers ...... 205 5.4.1 Thecentroid ...... 205 5.4.2 Theincenter ...... 205 5.4.3 TheGergonnepoint ...... 206 5.4.4 TheNagelpoint...... 207 5.5 Isotomic conjugates ...... 208

6 The Euler line and the nine-point circle 211 6.1 TheEulerline ...... 211 6.1.1 Inferior and superior triangles ...... 211 6.1.2 The orthocenter and the Euler line ...... 212 6.2 The nine-point circle ...... 213 6.3 Distances between triangle centers ...... 215 6.3.1 Distance between the circumcenter and orthocenter ...... 215 6.3.2 Distance between circumcenter and tritangent centers ...... 216 6.3.3 Distance between orthocenter and tritangent centers ...... 217 6.4 Feuerbach’stheorem...... 219

7 Isogonal conjugates 221 7.1 Directedangles ...... 221 7.2 Isogonalconjugates ...... 222 7.3 The symmedian point and the centroid ...... 223 7.4 Isogonal conjugates of the Gergonne and Nagel points ...... 225 7.4.1 The Gergonne point and the insimilicenter T+ ...... 225 7.4.2 The Nagel point and the exsimilicenter T ...... 227 7.5 TheBrocardpoints ...... − 228 7.6 Kariya’stheorem...... 230 7.7 Isogonal conjugate of an inﬁnite point ...... 233

8 The excentral and intouch triangles 235 8.1 The excentral triangle ...... 235 8.1.1 The circumcenter of the excentral triangle ...... 235 8.1.2 Relation between circumradius and radii of tritangent circles ...... 237 8.2 The homothetic center of the intouch and excentral triangles ...... 238

9 Homogeneous Barycentric Coordinates 301 9.1 Absolute and homogeneous barycentric coordinates ...... 301 9.1.1 Thecentroid ...... 301 9.1.2 Theincenter ...... 302 9.1.3 The barycenter of the perimeter ...... 303 9.1.4 TheGergonnepoint ...... 304 9.2 Cevian triangle ...... 305 9.2.1 Thecircumcenter...... 306 CONTENTS v

9.2.2 The Nagel point and the extouch triangle ...... 307 9.2.3 The orthocenter and the orthic triangle ...... 308 9.3 Homotheties...... 309 9.3.1 Superiors and inferiors ...... 309 9.3.2 Triangles bounded by lines parallel to the sidelines ...... 312 9.4 Inﬁnitepoints ...... 314

10 Some applications of barycentric coordinates 315 10.1 Construction of mixtilinear incircles ...... 315 10.1.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle . 315 10.1.2 Mixtilinear incircles ...... 316 10.2 Isotomic and isogonal conjugates ...... 317 10.3 Isotomic conjugates ...... 317 10.4 Equal-parallelians point ...... 319

11 Computation of barycentric coordinates 321 11.1 TheFeuerbachpoint...... 321 11.2 The OI line ...... 323 11.2.1 The circumcenter of the excentral triangle ...... 323 11.2.2 The centers of similitude of the circumcircle and the incircle ...... 323 11.2.3 The homothetic center T of excentral and intouch triangles ...... 324 11.3 The excentral triangle ...... 325 11.3.1 Thecentroid ...... 325 11.3.2 Theincenter ...... 325

12 Some interesting circles 327 12.1 A fundamental principle on 6 concyclic points ...... 327 12.1.1 The radical axis of two circles ...... 327 12.1.2 Test for 6 concyclicpoints ...... 328 12.2 TheTaylorcircle...... 329 12.3 Two Lemoine circles ...... 330 12.3.1 The ﬁrst Lemoine circle ...... 330 12.3.2 The second Lemoine circle ...... 331 12.3.3 Construction of K ...... 331 12.3.4 The center of the ﬁrst Lemoine circle ...... 332

13 Straight line equations 401 13.1 Area and barycentric coordinates ...... 401 13.2 Equations of straight lines ...... 402 13.2.1 Two-pointform...... 402 13.2.2 Intersection of two lines ...... 403 13.3 Perspective triangles ...... 404 13.3.1 The Conway conﬁguration ...... 405 13.4 Perspectivity...... 407 13.4.1 The Schifﬂer point: intersection of four Euler lines ...... 408 vi CONTENTS

14 Cevian nest theorem 409 14.1 Trilinear pole and polar ...... 409 14.1.1 Trilinear polar of a point ...... 409 14.1.2 Tripole of a line ...... 411 14.2 Anticevian triangles ...... 412 14.2.1 Construction of anticevian triangle ...... 412 14.3 Cevian quotients ...... 415 14.3.1 Theceviannesttheorem ...... 415 14.3.2 Basic properties of cevian quotients ...... 419

15 Circle equations 421 15.1 The power of a point with respect to a circle ...... 421 15.2 Circle equation ...... 422 15.3 Points on the circumcircle ...... 423 15.3.1 X(101) ...... 423 15.3.2 X(100) ...... 423 15.3.3 The Steiner point X(99) ...... 424 15.3.4 The Euler reﬂection point E = X(110) ...... 424 15.4 Circumcevian triangle ...... 425 15.5 The third Lemoine circle ...... 426 Chapter 1

Some Basic Theorems

1.1 The Pythagorean Theorem

Theorem 1.1 (Pythagoras). The lengths a b

b a

3 2 2 a b b

4 b

a b 3 a a 4 1 1

a b

Theorem 1.2 (Converse of Pythagoras’ theorem). If the lengths of the sides of ABC satisfy a2 + b2 = c2, then the triangle has a right angle at C. △ B Y

c a a

C b A Z b X

Proof. Consider a right triangle XYZ with ∠Z = 90◦, YZ = a, and XZ = b. By the Pythagorean theorem, XY 2 = YZ2 + XZ2 = a2 + b2 = c2 = AB2. It follows that XY = AB, and ABC XYZ by the SSS test, and ∠C = ∠Z = 90◦. △ ≡△ 102 Some Basic Theorems

Example 1.1. Trigonometric ratios of 30◦, 45◦, and 60◦:

2 √2 1 √3

60◦ 45◦ 1 1

θ sin θ cos θ tan θ 30 1 √3 √3 ◦ 2 2 3 45 √2 √2 1 ◦ 2 2 60 √3 1 √3 ◦ 2 2

Example 1.2. For an arbitrary point P on the minor arc BC of the circumcircle of an equilateral triangle ABC, AP = BP + CP . A

B C

Proof. If Q is the point on AP such that PQ = PC, then the isosceles triangle CPQ is equilateral since ∠CPQ = ∠CBA = 60◦. Note that ∠ACQ = ∠BCP . Thus, ACQ BCP by the SAS congruence test. From this, AQ = BP , and AP = AQ△+ QP ≡= BP△ + CP . 1.1 The Pythagorean Theorem 103

Example 1.3. Given a rectangle ABCD, to construct points P on BC and Q on CD such that triangle AP Q is equilateral. Let BCY and CDX are equilateral triangles inside the rectangle ABCD. Extend the lines AX and AY to intersect BC and CD respectively at P and Q. AP Q is equilateral.

D Q C

P X

A B

Proof. Suppose AB = 2a and BC = 2b. The distance of X above AB = 2b √3a. By the Pythagorean theorem, AX2 = a2 +(2b √3a)2 = 4(a2 +b2 √3ab). X is the− midpoint of AP . Therefore, AP 2 = 16(a2 + b2 −√3ab). Similarly, AQ− 2 = (2AY )2 = 4AY 2 = 4 (b2 +(2a √3b)2) = 16(a2 +b2 √3ab−). Finally, CP =2b 2(2b √3a)=2(√3a b), CQ· = 2(√−3b a), and PQ2 = CP− 2 + CQ2 = 4((√3a b)−2 +(√−3b a)2) = 16(a−2 + b2 √3ab). It− follows that AP = AQ = PQ, and triangle−AP Q is equilateral.− − 104 Some Basic Theorems

1.2 Constructions of geometric mean

We present two ruler-and-compass constructions of the geometric means of two quantities given as lengths of segments. These are based on Euclid’s proof of the Pythagorean theorem. Construct the altitude at the right angle to meet AB at P and the opposite side ZZ′ of the square ABZZ′ at Q. Note that the area of the rectangle AZQP is twice of the area of triangle AZC. By rotating this triangle about A through a right angle, we obtain the congruent triangle ABY , whose area is half of the area of the square on AC. It follows that the area of rectangle AZQP is equal to the area of the square on AC. For the same reason, the area of rectangle BZ′QP is equal to that of the square on BC. From these, the area of the square on AB is equal to the sum of the areas of the squares on BC and CA.

X′

Y ′

X C

A B P

Z Q Z′ Construction. Given two segments of length a

Q Q

√ab

A A P B B a P b 1.2 Constructions of geometric mean 105

Construction. Given two segments of length a, b, mark three points A, P , B on a line (P between A and B) such that AP = a, PB = b. Describe a semicircle with AB as diameter, and let the perpendicular through P intersect the semicircle at Q. Then PQ2 = PA PB, so that the length of PQ is the geometric mean of a and b. ·

Q Q

√ab

P A B B a P b A 106 Some Basic Theorems

1.3 The golden ratio

Given a segment AB, a point P in the segment is said to divide it in the golden ratio if 2 AP √5+1 AP = PB AB. Equivalently, PB = 2 . We shall denote this golden ratio by ϕ. It is the positive root· of the quadratic equation x2 = x +1.

A P B A P B

Construction (Division of a segment in the golden ratio). Given a segment AB, (1) draw a right triangle ABM with BM perpendicular to AB and half in length, (2) mark a point Q on the hypotenuse AM such that MQ = MB, (3) mark a point P on the segment AB such that AP = AQ. Then P divides AB into the golden ratio. Suppose PB has unit length. The length ϕ of AP satisﬁes

ϕ2 = ϕ +1.

This equation can be rearranged as

1 2 5 ϕ = . − 2 4 Since ϕ> 1, we have 1 ϕ = √5+1 . 2 Note that AP ϕ 1 2 √5 1 = = = = − . AB ϕ +1 ϕ √5+1 2 This explains the construction above.

1.3.1 The regular pentagon Consider a regular pentagon ACBDE. It is clear that the ﬁve diagonals all have equal lengths. Note that (1) ∠ACB = 108◦, 1.3 The golden ratio 107 C

A B P

E D

(2) triangle CAB is isosceles, and (3) ∠CAB = ∠CBA = (180◦ 108◦) 2=36◦. − ÷ In fact, each diagonal makes a 36◦ angle with one side, and a 72◦ angle with another. It follows that (4) triangle PBC is isosceles with ∠PBC = ∠PCB = 36◦, (5) ∠BPC = 180◦ 2 36◦ = 108◦, and (6) triangles CAB and− PBC× are similar. Note that triangle ACP is also isosceles since (7) ∠ACP = ∠AP C = 72◦. This means that AP = AC. Now, from the similarity of CAB and PBC, we have AB : AC = BC : PB. In other words AB AP = AP PB, or AP 2 = AB PB. This means that P divides AB in the golden ratio.· · · Construction. Given a segment AB, we construct a regular pentagon ACBDE with AB as a diagonal. (1) Divide AB in the golden ratio at P . (2) Construct the circles A(P ) and P (B), and let C be an intersection of these two circles. (3) Construct the circles A(AB) and B(C) to intersect at a point D on the same side of BC as A. (4) Construct the circles A(P ) and D(P ) to intersect at E. Then ACBDE is a regular pentagon with AB as a diagonal. 108 Some Basic Theorems

1.4 Basic construction principles

1.4.1 Perpendicular bisector locus A variable point P is equidistant from two ﬁxed points A and B if and only if P lies on the perpendicular bisector of the segment AB. The perpendicular bisectors of the three sides of a triangle are concurrent at the circumcenter of the triangle. This is the center of the circumcircle, the circle passing through the three vertices of the triangle.

A A

B C O O

B C

The circumcenter of a right triangle is the midpoint of its hypotenuse. 1.4 Basic construction principles 109

1.4.2 Angle bisector locus

A variable point P is equidistant from two ﬁxed lines ℓ and ℓ′ if and only if P lies on the bisector of one of the angles between ℓ and ℓ′. The bisectors of the three angles of a triangle are concurrent at a point which is at equal distances from the three sides. With this point as center, a circle can be constructed tangent to the sides of the triangle. This is the incircle of the triangle. The center is the incenter. A

Z I

B X C

Proof. Let the bisectors of angles B and C intersect at I. Consider the pedals of I on the three sides. Since I is on the bisector of angle B, IX = IZ. Since I is also on the bisector of angle C, IX = IY . It follows IX = IY = IZ, and the circle, center I, constructed through X, also passes through Y and Z, and is tangent to the three sides of the triangle. 110 Some Basic Theorems

1.4.3 Tangency of circles

Two circles (O) and (O′) are tangent to each other if they are tangent to a line ℓ at the same line P , which is a common point of the circles. The tangency is internal or external according as the circles are on the same or different sides of the common tangent ℓ.

T ′ O T I O I

The line joining their centers passes through the point of tangency. The distance between their centers is the sum or difference of their radii, according as the tangency is external or internal.

1.4.4 Construction of tangents of a circle A tangent to a circle is a line which intersects the circle at only one point. Given a circle O(A), the tangent to a circle at A is the perpendicular to the radius OA at A.

O M P

If P is a point outside a circle (O), there are two lines through P tangent to the circle. Construct the circle with OP as diameter to intersect (O) at two points. These are the points of tangency. The two tangents have equal lengths since the triangles OAP and OBP are congruent by the RHS test. 1.4 Basic construction principles 111

Example 1.4. Given two congruent circles each with center on the other circle, to construct a circle tangent to the center line, and also to the given circles, one internally and the other externally.

K Y

T x A B

Let AB = a. Suppose the required circle has radius r, and AT = x, where T is the point of tangency T with the center line.

(a + x)2 + r2 =(a + r)2, x2 + r2 =(a r)2. − From these, we have

a √3 x + = a, 2 2 · a + x = 2r. 2

√3 This means that if M is the midpoint of AB, then MT = 2 a, which is the height of the equilateral triangle on AB. In other words, if C is an intersection· of the two given circles, then CM and MT are two adjacent sides of a square. Furthermore, the side opposite to CM is a diameter of the required circle!

C X

√3a 2 K Y

x a T A 2 M B

D 112 Some Basic Theorems

1.5 The intersecting chords theorem

Theorem 1.3. Given a point P and a circle O(r), if a line through P intersects the circle at two points A and B, then PA PB = OP 2 r2, independent of the line. · −

O O P

B B P M M A A

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

PA PB =(PM + MA)(PM BM) · − =(PM + MA)(PM MA) − = PM 2 MA2 − =(OM 2 + PM 2) (OM 2 + MA2) − = OP 2 r2. − The same calculation applies to the case when P is inside or on the circle, provided that the lengths of the directed segments are signed. The quantity OP 2 r2 is called the power of P with respect to the circle. It is positive, zero, or negative according− as P is outside, on, or inside the circle. Corollary 1.4 (Intersecting chords theorem). If two chords AB and CD of a circle intersect, extended if necessary, at a point P , then PA PB = PC PD. In particular, if the tangent at T intersects AB·at P , then PA· PB = PT 2. ·

C C

D O O P

B B P D

A T A

The converse of the intersecting chords theorem is also true. 1.5 The intersecting chords theorem 113

Theorem 1.5. Given four points A, B, C, D, if the lines AB and CD intersect at a point P such that PA PB = PC PD (as signed products), then A, B, C, D are concyclic. In particular,· if P is a point· on a line AB, and T is a point outside the line AB such that PA PB = PT 2, then PT is tangent to the circle through A, B, T . · Example 1.5. Let ABC be a triangle with B =2C. Then b2 = c(c + a).

B C

Construct a parallel through C to the bisector BE, to intersect the extension of AB at F . Then 1 ∠AF C = ∠ABE = ∠ABC = ∠ACB. 2 · This means that AC is tangent to the circle through B, C, F . By the intersecting chord theorem, AC2 = AB AF , i.e., b2 = c(c + a). · 114 Some Basic Theorems

1.6 Ptolemy’s theorem

Theorem 1.6 (Ptolemy). A convex quadrilateral ABCD is cyclic if and only if AB CD + AD BC = AC BD. · · ·

D d D A A

P ′ P

B C B C

Proof. (Necessity) Assume, without loss of generality, that ∠BAD > ∠ABD. Choose a point P on the diagonal BD such that ∠BAP = ∠CAD. Triangles BAP and CAD are similar, since ∠ABP = ∠ACD. It follows that AB : AC = BP : CD, and AB CD = AC BP. · · Now, triangles ABC and AP D are also similar, since ∠BAC = ∠BAP + ∠P AC = ∠DAC + ∠P AC = ∠P AD, and ∠ACB = ∠ADP . It follows that AC : BC = AD : PD, and BC AD = AC PD. · · Combining the two equations, we have AB CD + BC AD = AC(BP + PD)= AC BD. · · · (Sufﬁciency). Let ABCD be a quadrilateral satisfying (**). Locate a point P ′ such that ∠BAP ′ = ∠CAD and ∠ABP ′ = ∠ACD. Then the triangles ABP and ACD are similar. It follows that

AB : AP ′ : BP ′ = AC : AD : CD. From this we conclude that (i) AB CD = AC BP ′, and · · (ii) triangles ABC and AP ′D are similar since ∠BAC = ∠P ′AD and AB : AC = AP ′ : AD. Consequently, AC : BC = AD : P ′D, and

AD BC = AC P ′D. · · Combining the two equations,

AC(BP ′ + P ′D)= AB CD + AD BC = AC BD. · · · It follows that BP ′ + P ′D = BD, and the point P ′ lies on diagonal BD. From this, ∠ABD = ∠ABP ′ = ∠ACD, and the points A, B, C, D are concyclic. Chapter 2

The laws of sines and cosines

2.1 The law of sines

Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle ABC. a b c 2R = = = . sin α sin β sin γ

A A

F O E O

B C B C D D

1 Since the area of a triangle is given by ∆ = 2 bc sin α, the circumradius can be written as abc R = . 4∆ 116 The laws of sines and cosines

2.2 The orthocenter

Why are the three altitudes of a triangle concurrent? Let ABC be a given triangle. Through each vertex of the triangle we construct a line parallel to its opposite side. These three parallel lines bound a larger triangle A′B′C′. Note that ABCB′ and ACBC′ are both parallelograms since each has two pairs of parallel sides. It follows that B′A = BC = AC′ and A is the midpoint of B′C′.

C′ A B′

Z H

B X C

A′

Consider the altitude AX of triangle ABC. Seen in triangle A′B′C′, this line is the perpendicular bisector of B′C′ since it is perpendicular to B′C′ through its midpoint A. Similarly, the altitudes BY and CZ of triangle ABC are perpendicular bisectors of C′A′ and A′B′. As such, the three lines AX, BY , CZ concur at a point H. This is called the orthocenter of triangle ABC. Proposition 2.2. The reﬂections of the orthocenter in the sidelines lie on the circumcircle.

O H

B C

Proof. It is enough to show that the reﬂection Ha of H in BC lies on the circumcircle. Consider also the reﬂection Oa of O in BC. Since AH and OOa are parallel and have the same length (2R cos α), AOOaH is a parallelogram. On the other hand, HOOaHa is a isosceles trapezoid. It follows that OHa = HOa = AO, and Ha lies on the circumcircle. 2.3 The law of cosines 117

2.3 The law of cosines

Given a triangle ABC, we denote by a, b, c the lengths of the sides BC, CA, AB respectively. Theorem 2.3 (The law of cosines). c2 = a2 + b2 2ab cos γ. −

A A

b c c b

B X a C B a C X

Proof. Let AX be the altitude on BC. c2 = BX2 + AX2 =(a b cos γ)2 +(b sin γ)2 − = a2 2ab cos γ + b2(cos2 γ + sin2 γ) − = a2 + b2 2ab cos γ. −

Theorem 2.4 (Stewart). Let X be a point on the sideline BC of triangle ABC. a AX2 = BX b2 + XC c2 a BX XC. · · · − · · Here, the lengths of the directed segments on the line BC are signed. Equivalently, if BX : XC = λ : µ, then λb2 + µc2 λµa2 AX2 = . λ + µ − (λ + µ)2

b c

B X a C Proof. Use the cosine formula to compute the cosines of the angles AXB and AXC, and note that cos AXC = cos AXB. − 118 The laws of sines and cosines

Example 2.1. (Napoleon’s theorem). If similar isosceles triangles XBC, YCA and ZAB (of base angle θ) are constructed externally on the sides of triangle ABC, the lengths of the segments YZ, ZX, XZ can be computed easily. For example, in triangle AY Z, b c ∠ AY = 2 sec θ, AZ = 2 sec θ and Y AZ = α +2θ.

θ Y

θ θ

B θ θ C

By the law of cosines,

YZ2 = AY 2 + AZ2 2AY AZ cos Y AZ − · · sec2 θ = (b2 + c2 2bc cos(α +2θ)) 4 − sec2 θ = (b2 + c2 2bc cos α cos2θ +2bc sin α sin2θ) 4 − sec2 θ = (b2 + c2 (b2 + c2 a2)cos2θ + 4∆sin2θ) 4 − − sec2 θ = (a2 cos2θ +(b2 + c2)(1 cos2θ)+4∆sin2θ). 4 − Likewise, we have

sec2 θ ZX2 = (b2 cos2θ +(c2 + a2)(1 cos2θ)+4∆sin2θ), 4 − sec2 θ XY 2 = (c2 cos2θ +(a2 + b2)(1 cos2θ)+4∆sin2θ). 4 −

1 It is easy to note that YZ = ZX = XY if and only if cos2θ = 2 , i.e., θ = 30◦. In this case, the points X, Y , Z are the centers of equilateral triangles erected externally on BC, CA, AB respectively. The same conclusion holds if the equilateral triangles are constructed internally on the sides. This is the famous Napoleon theorem.

Theorem 2.5 (Napoleon). If equilateral triangles are constructed on the sides of a triangle, either all externally or all internally, then their centers are the vertices of an equilateral triangle. 2.3 The law of cosines 119

A A

Y Z Z′

X′ B C Y ′ X B C

Example 2.2. (Orthogonal circles) Given three points A, B, C that form an acute-angled triangle, construct three circles with these points as centers that are mutually orthogonal to each other.

Y A

Z F H

B D C

Let BC = a, CA = b, and AB = c. If these circles have radii Ra, Rb, Rc respectively, then 2 2 2 2 2 2 2 2 2 Rb + Rc = a , Rc + Ra = b , Ra + Rb = c . From these, 1 1 1 R2 = (b2 + c2 a2), R2 = (c2 + a2 b2), R2 = (a2 + b2 c2). a 2 − b 2 − c 2 − These are all positive since ABC is an acute triangle. Consider the perpendicular foot E of 2 1 2 2 2 B on AC. Note that AE = c cos A, so that Ra = 2 (b + c a )= bc cos A = AC AE. It follows if we extend BE to intersect at Y the semicircle constructed− externally on· the side 2 2 AC as diameter, then, AY = AC AE = Ra. Therefore we have the following simple construction of these circles. · (1) With each side as diameter, construct a semicircle externally of the triangle. (2) Extend the altitudes of the triangle to intersect the semicircles on the same side. Label these X, Y , Z on the semicircles on BC, CA, AB respectively. These satisfy AY = AZ, BZ = BX, and CX = CY . (3) The circles A(Y ), B(Z) and C(X) are mutually orthogonal to each other. 120 The laws of sines and cosines

2.4 The centroid

Let E and F be the midpoints of AC and AB respectively, and G the intersection of the medians BE and CF . Construct the parallel through C to BE, and extend AG to intersect BC at D, and this parallel at H. .

E F G

B D C

By the converse of the midpoint theorem, G is the midpoint of AH, and HC =2 GE Join BH. By the midpoint theorem, BH//CF . It follows that BHCG is a parallel-· ogram. Therefore, D is the midpoint of (the diagonal) BC, and AD is also a median of triangle ABC. We have shown that the three medians of triangle ABC intersect at G, which we call the centroid of the triangle. Furthermore,

AG =GH =2GD, BG =HC =2GE, CG =HB =2GF.

The centroid G divides each median in the ratio 2:1.

Theorem 2.6 (Apollonius). If ma denotes the length of the median on the side BC, 1 m2 = (2b2 +2c2 a2). a 4 − Example 2.3. Suppose the medians BE and CF of triangle ABC are perpendicular. This means that BG2 + CG2 = BC2, where G is the centroid of the triangle. In terms of the 4 2 4 2 2 2 2 2 2 2 2 2 2 2 lengths, we have 9 mb + 9 mc = a ; 4(mb +mc )=9a ; (2c +2a b )+(2a +2b c )= 9a2; b2 + c2 =5a2. − − This relation is enough to describe, given points B and C, the locus of A for which the medians BE and CF of triangle ABC are perpendicular. Here, however, is a very easy 2 2 2 2 1 2 2 2 9 2 3 construction: From b + c =5a , we have ma = 4 (2b +2c a )= 4 a ; ma = 2 a. The locus of A is the circle with center at the midpoint of BC, and− radius 3 BC. 2 · 2.5 The angle bisector theorem 121

2.5 The angle bisector theorem

Theorem 2.7 (Angle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If AX and AX′ respectively the internal and external bisectors of angle BAC, then BX : XC = c : b and BX′ : X′C = c : b. − Z

Z′ c b

B X C X′

Proof. Construct lines through C parallel to the bisectors AX and AX′ to intersect the line AB at Z and Z′. (1) Note that ∠AZC = ∠BAX = ∠XAC = ∠ACZ. This means AZ = AC. Clearly, BX : XC = BA : AZ = BA : AC = c : b. (2) Similarly, AZ′ = AC, and BX′ : X′C = BA : AZ′ = BA : AC = c : b. − − 2.5.1 The lengths of the bisectors Proposition 2.8. (a) The lengths of the internal and external bisectors of angle A are respectively 2bc α 2bc α t = cos and t′ = sin . a b + c 2 a b c 2 | − |

b c ta t′a

B X C X′

Proof. Let AX and AX′ be the bisectors of angle A. (1) Consider the area of triangle ABC as the sum of those of triangles AXC and ABX. We have 1 α 1 t (b + c)sin = bc sin α. 2 a 2 2 122 The laws of sines and cosines

From this, bc sin α 2bc α ta = α = cos . b + c · sin 2 b + c · 2

(2) Consider the area of triangle as the difference between those of ABX′ and ACX′.

2bc Remarks. (1) b+c is the harmonic mean of b and c. It can be constructed as follows. If the 2bc perpendicular to AX at X intersects AC and AB at Y and Z, then AY = AZ = b+c . A

ta c b Z

C B X

(2) Applying Stewart’s Theorem with λ = c and µ = b, we also obtain the following expressions for the lengths of the angle bisectors: ±

a 2 t2 = bc 1 , a − b + c ! a 2 t′2 = bc 1 . a b c − − ! Example 2.4. (Steiner-Lehmus theorem). A triangle with two equal angle bisectors is isosceles. More precisely, if the bisectors of two angles of a triangle have equal lengths, then the two angles are equal.

Proof. We show that if atb. Note that from a cos 2 ; b a 2bc 2ca (ii) bc > ac, b(c + a) >a(b + c); b+c > c+a ; b+c > c+a . From (i) and (ii), we have 2bc α 2ca β ta = cos > cos = tb. b + c · 2 c + a · 2

The same reasoning shows that a>b ta

2.6 The circle of Apollonius

Theorem 2.9. A and B are two ﬁxed points. For a given positive number k = 1, 1 the locus of points P satisfying AP : PB = k : 1 is the circle with diameter XY 6, where X and Y are points on the line AB such that AX : XB = k :1 and AY : YB = k : 1. −

B Y X O A B′

Proof. Since k = 1, points X and Y can be found on the line AB satisfying the above conditions. 6 Consider a point P not on the line AB with AP : PB = k :1. Note that PX and PY are respectively the internal and external bisectors of angle AP B. This means that angle XPY is a right angle, and P lies on the circle with XY as diameter. Conversely, let P be a point on this circle. We show that AP : BP = k : 1. Let B′ be a point on the line AB such that PX bisects angle AP B′. Since PA and PB are perpendicular to each other, the line PB is the external bisector of angle AP B′, and AY AX XA AY XA = = = − . YB′ −XB′ XB′ YX On the other hand, AY AX XA AY XA = = = − . YB −XB XB YX

Comparison of the two expressions shows that B′ coincides with B, and PX is the bisector PA AX of angle AP B. It follows that PB = XB = k.

1If k = 1, the locus is clearly the perpendicular bisector of the segment AB. 124 The laws of sines and cosines Chapter 3

The tritangent circles

3.1 The incircle

Let the incircle of triangle ABC touch its sides BC, CA, AB at X, Y , Z respectively.

s a − s a − Y

Z I s c − s b −

B s b X s c C − −

If s denotes the semiperimeter of triangle ABC, then

AY =AZ = s a, − BZ =BX = s b, − CX =CY = s c. − The inradius of triangle ABC is the radius of its incircle. It is given by 2∆ ∆ r = = . a + b + c s 126 The tritangent circles

Example 3.1. If triangle ABC has a right angle at C, then the inradius r = s c. − B

s b −

I r s a r − s c r −

C s c s a A − − It follows that if d is the diameter of the incircle, then a + b = c + d.

Example 3.2. An equilateral triangle of side 2a is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. If the inradii of the quadrilateral and the isosceles triangle are equal, ﬁnd this inradius. 1

r X

Y r

B C

Suppose each side of the equilateral triangle has length 2, each of the congruent circles has radius r, and ∠ACX = θ. (i) From triangle AXC, r = 2 . cot 30◦+cot θ ∠ 1 (ii) Note that BCY = 2 (60◦ 2θ) = 30◦ θ. It follows that r = tan(30◦ θ) = 1 cot θ cot 30◦ − − − cot(30 θ) = cot 30−cot θ+1 . ◦− ◦ 2 x √3 2 2 By putting cot θ = x, we have = − ; x 3=2√3x+2; x 2√3x 5=0, √3+x √3x+1 − − − and x = √3+2√2. (The negative root is rejected). From this, r = 2 = 1 = √3+x √3+√2 √3 √2. To− construct the circles, it is enough to mark Y on the altitude through A such that AY = √3 r = √2. The construction is now evident. −

1(√3 √2)a. − 3.1 The incircle 127

Example 3.3. (Conway’s circle). Given triangle ABC, extend (i) CA and BA to Ya and Za such that AYa = AZa = a, (ii) AB and CB to Zb and Xb such that BZb = BXb = b, (iii) BC and AC to Xc and Yc such that CXc = CYc = c. The six points Xb, Xc, Yc, Ya, Za, Zb are concyclic. The circle containing them has center I and radius √r2 + s2.

Y b c Z I r X X b C c b B X a c

b Yc

Proof. Let the incircle be tangent to BC at X. BX = s b = XbX = b +(s b)= s. 2 2 2 − ⇒ − From this, IXb = r + s . Similarly, each of the remaining ﬁve points is at the same distance from I. They lie on a circle, center I, radius √r2 + s2. 128 The tritangent circles

3.2 Euler’s formula

Theorem 3.1 (Euler’s formula). OI2 = R2 2Rr −

I O

C B

∠ ∠ A Proof. Extend AI to intersect the circumcircle at M. Since BAM = MAC = 2 , M is the midpoint of the arc BC of the circumcircle. Note that

∠MBI = ∠MBC + ∠CBI = ∠MAC + ∠CBI = ∠BAI + ∠IBA = ∠BIM.

A r It follows that IM = BM. By the law of sines, BM = 2R sin 2 . Since AI = A , sin 2 AI IM = 2Rr. This is equal to the power of I with respect to the circumcircle, namely, = R·2 OI2. Therefore, OI2 = R2 2Rr. − − Corollary 3.2. R 2r; equality holds if and only if the triangle is equilateral. ≥ 3.3 Steiner’s porism 129

3.3 Steiner’s porism

Given a triangle ABC with incircle (I) and circumcircle O), let A′ be an arbitrary point on circumcircle. Join A′ to I, to intersect the circumcircle again at M ′, and let A′Y ′, A′Z′ be the tangents to the incircle. Construct a circle, center M ′, through I to intersect the circumcircle at B′ and C′.

A′ A

Z′

O B′ I Y ′

B C

M ′

C′

Denote by θ between A′I and each tangent. r It is known that A′I IM ′ = 2Rr (power of incenter in (O)). Since A′I = , we · sin θ have IM ′ = 2R sin θ. Therefore, M ′B′ = M ′C′ = 2R sin θ, and by the law of sines, ∠B′A′M = ∠C′A′M = θ. It follows that A′B′ and A′C′ are tangent to the incircle (I). Since M ′ is the midpoint of the arc B′C′, the circle M ′(B′) passes through the incenter of triangle A′B′C′. This means that I is the incenter, and (I) the incircle of triangle A′B′C′. 130 The tritangent circles

3.4 The excircles

The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. An excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle.

X B C Y

ra Z ra

Ia The exradii of a triangle with sides a, b, c are given by ∆ ∆ ∆ r = , r = , r = . a s a b s b c s c − − − 1 1 1 The areas of the triangles IaBC, IaCA, and IaAB are 2 ara, 2 bra, and 2 cra respectively. Since ∆= ∆IaBC +∆IaCA +∆IaAB, − we have 1 ∆= ra( a + b + c)= ra(s a), 2 − − ∆ from which ra = s a . − 3.5 Heron’s formula for the area of a triangle 131

3.5 Heron’s formula for the area of a triangle

Consider a triangle ABC with area ∆. Denote by r the inradius, and ra the radius of the excircle on the side BC of triangle ABC. It is convenient to introduce the semiperimeter 1 s = 2 (a + b + c).

ra r

C A Y ′ Y

(1) From the similarity of triangles AIY and AI′Y ′, r s a = − . ra s

(2) From the similarity of triangles CIY and I′CY ′,

r ra =(s b)(s c). · − − (3) From these,

(s a)(s b)(s c) r = − − − , r s s(s b)(s c) r = − − . a s a r − Theorem 3.3 (Heron’s formula).

∆= s(s a)(s b)(s c). − − − Proof. ∆= rs. p Proposition 3.4.

α (s b)(s c) α s(s a) α (s b)(s c) tan = − − , cos = − , sin = − − . 2 s(s a) 2 bc 2 bc s − r r 132 The tritangent circles Chapter 4

The arbelos

4.1 Archimedes’ twin circles

Theorem 4.1 (Archimedes). The two circles each tangent to CP , the largest semicircle AB and one of the smaller semicircles have equal radii t, given by ab t = . a + b

A O1 O P O2 B A O1 O P O2 B

Proof. Consider the circle tangent to the semicircles O(a + b), O1(a), and the line PQ. Denote by t the radius of this circle. Calculating in two ways the height of the center of this circle above the line AB, we have

(a + b t)2 (a b t)2 =(a + t)2 (a t)2. − − − − − − From this, ab t = . a + b

The symmetry of this expression in a and b means that the circle tangent to O(a+b), O2(b), and PQ has the same radius t. 134 The arbelos

1 1 1 4.1.1 Harmonic mean and the equation a + b = t 2ab The harmonic mean of two quantities a and b is a+b . In a trapezoid of parallel sides a and b, the parallel through the intersection of the diagonals intercepts a segment whose length is 1 1 1 the harmonic mean of a and b. We shall write this harmonic mean as 2t, so that a + b = t .

D a C

b a

A b B

Here is another construction of t, making use of the formula for the length of an angle bisector in a triangle. If BC = a, AC = b, then the angle bisector CZ has length 2ab C C t = cos =2t cos . c a + b 2 2 The length t can therefore be constructed by completing the rhombus CXZY (by con- structing the perpendicular bisector of CZ to intersect BC at X and AC at Y ). In particular, if the triangle contains a right angle, this trapezoid is a square.

X t Y b M

t t t

A Z B a

4.1.2 Construction of the Archimedean twin circles

Construct the circle P (C3) to intersect the diameter AB at P1 and P2 so that P1 is on AP and P2 is on PB. The center C1 respectively C2 is the intersection of the circle O1(P2) respectively O2(P1) and the perpendicular to AB at P1 respectively P2.

C1 Q1 C2

A O1 P1 O P PO2 2 B 4.2 The incircle 135

4.2 The incircle

Theorem 4.2 (Archimedes). The incircle of the arbelos has radius

r1r2(r1 + r2) ρ = 2 2 . r1 + r1r2 + r2

X Y

A O1 O C O2 B

Proof. Let ∠COO3 = θ. By the law of cosines, we have

(r + ρ)2 =(r + r ρ)2 + r2 +2r (r + r ρ)cos θ, 1 1 2 − 2 2 1 2 − (r + ρ)2 =(r + r ρ)2 + r2 2r (r + r ρ)cos θ. 2 1 2 − 1 − 1 1 2 − Eliminating θ, we have

r (r + ρ)2 + r (r + ρ)2 =(r + r )(r + r ρ)2 + r r2 + r r2. 1 1 2 2 1 2 1 2 − 1 2 2 1 The coefﬁcients of ρ2 on both sides are clearly the same. This is a linear equation in ρ:

r3 + r3 + 2(r2 + r2)ρ =(r + r )3 + r r (r + r ) 2(r + r )2ρ, 1 2 1 2 1 2 1 2 1 2 − 1 2 from which

4(r2 + r r + r2)ρ =(r + r )3 + r r (r + r ) (r3 + r3)=4r r (r + r ), 1 1 2 2 1 2 1 2 1 2 − 1 2 1 2 1 2 and ρ is as above. 136 The arbelos

4.2.1 Construction of the incircle of the arbelos In [?], Bankoff published the following remarkable theorem which gives a construction of the incircle of the arbelos of the incircle, much simpler than the one we designed before from Archimedes’ proof. The simplicity of the construction is due to the existence of a circle congruent to Archimedes’ twin circles.

Theorem 4.3 (Bankoff). The points of tangency of the incircle of the arbelos with the semicircles (AC) and (CB), together with C, are the points of tangency of the incircle (W3) of triangle O1O2O3 with the sides of the triangle. This circle (W3) is congruent to Archimedes’ twin circles (W1) and (W2).

Q R W3

A O1 O C O2 B

Proof. Since O1Q = O1C, O2C = O2R, and O3R = O3Q, the points C, Q, R are the points of tangency of the incircle of triangle O1O2O3 with its sides. The semi-perimeter of the triangle is

3 2 r1r2(r1 + r2) (r1 + r2) (r1 + r2) ρ s = r1 + r2 + ρ = r1 + r2 + 2 2 = 2 2 = . r1 + r2r2 + r2 r1 + r2r2 + r2 r1r2 The inradius of the triangle is the square root of

2 2 r1r2ρ r1r2 2 = 2 = t . s (r1 + r2)

It follows that this inradius is t. The incircle of triangle O1O2O3 is congruent to Archimedes’ twin circles. 4.2 The incircle 137

Construction. Let M and N be the midpoints of the semicircles (AC) and (CB) respectively. Construct (1) the lines O1N and O2M to intersect at W3, (2) the circle with center W3, passing through C to intersect the semicircle (AC) at Q and (CB) at R, (3) the lines O1Q and O2R to intersect at O3. The circle with center O3 passing through Q touches the semicircle (CB) at R and also the semicircle (AB).

O M 3

N Q R

A O1 O C O2 B 138 The arbelos

4.2.2 Alternative constructions of the incircle Theorem 4.4 (Bankoff). Let P be the intersection (apart from C) of the circumcircles of the squares on AC and CB. Let Q be the intersection (apart from C) of the circumcircle of the square on CB and the semicircle (AC), and R that of the circumcircle of the square on AB and the semicircle (CB). The points P , Q, R are the points of tangency of the incircle of the arbelos with the semicircles.

L P

M O3

Q R

A O1 O C O2 B

Proposition 4.5. The intersection S of the lines AN and BM also lies on the incircle of the arbelos, and the line CS intersects (AB) at P .

L P

M O3

Q R S

A O1 O C O2 B 4.2 The incircle 139

Q R

A B O1 O C O2

L′

Construction. Let L′ be the “lowest point” of the circle (AB). Construct (1) the line L′C to intersect the semicircle (AB) at P , (2) the circle, center L′, through A and B, to intersect the semicircles (AC) and (CB) at Q and R.

Proposition 4.6. Let X be the midpoint of the side of the square on AC opposite to AC, and Y that of the side of the square on CB opposite to CB. The center O3 of the incircle of the arbelos is the intersection of the lines AY and BX.

L P Y

O M 3

N Q R

A O1 O C O2 B 140 The arbelos

4.3 Archimedean circles

We shall call a circle Archimedean if it is congruent to Archimedes’ twin circle, i.e., with radius t = r1r2 , and has further remarkable geometric properties. r1+r2

1. (van Lamoen) The circle (W3) is tangent internally to the midway semicircle (O1O2) at a point on the segment MN. 1 M D

M1 M ′ M2

A O1 O O′ C O2 B

2. (van Lamoem) The circle tangent to AB at O and to the midway semicircle is Archimedean. 2 M

A O1 O O′ C O2 B

3. (Schoch) Let MN intersect CD and OL at Q and K respectively. The smallest circles through Q and K tangent to the semicircle (AB) are Archimedean.

L D

W20 M W21

N K Q

A O1 O C O2 B

1van Lamoen, June 10, 1999. 2van Lamoen, June 10, 1999. 4.3 Archimedean circles 141

4. (a) The circle tangent to (AB) and to the common tangent of (AC) and (CB) is Archimedean. (b) The smallest circle through C tangent to AB is Archimedean.

W11

A O1 O C O2 B

5. Let EF be the common tangent of the semicircles (AC) and (CB). The smallest circles through E and F tangent to CD are Archimedean.

E W9

W10 F

A O1 O C O2 B

6. (Schoch) Let X and Y be the intersections of the semicircle (AB) with the circles through C, with centers A and B respectively. The smallest circle through X and Y tangent to CD are Archimedean.

D X W13 W14 Y

A B O1 O C O2Y ′ 142 The arbelos

7. (van Lamoen) Let Y and Z be the intersections of the midway semicircle with the semicircles (AC) and (CB). The circles with centers Y and Z, each tangent to the line CD, are Archimedean. D

Y Z

A O1 O C X2 O2 B

8. (Schoch) (a) The circle tangent to the semicircle (AB) and the circular arcs, with centers A and B respectively, each passing through C, is Archimedean. (b) The circle with center on the Schoch line and tangent to both semicircles (AC) and (CB) is Archimedean.

W15 W16

A O1 O C O2 B

9. (Woo) Let α be a positive real number. Consider the two circular arcs, each passing through C and with centers ( αr , 0) and (αr , 0) respectively. The circle with − 1 2 center Uα on the Schoch line tangent to both of these arcs is Archimedean.

The Woo circle (Uα) which is tangent externally to the semicircle (AB) touches it at D (the intersection with the common tangent of (AC) and (CB)).

W28

W15

A O1 O C O2 B 4.3 Archimedean circles 143

10. (Power) Consider an arbelos with inner semicircles C1 and C2 of radii a and b, and outer semicircle C of radius a + b. It is known the Archimedean circles have radius ab C C t = a+b . Let Q1 and Q2 be the “highest” points of 1 and 2 respectively.

A circle tangent to (O) internally and to OQ1 at Q1 (or OQ2 at Q2) is Archimedean.

r C1

C2 Q1

C′ 1 a + b r Q2 a − C2′

A O1 b O P O2 B

11. (van Lamoen) D

M1 U1 M M2

A O1 O C O2 B

12. (Bui)

D D

T1 M M T2

A O1 O O′ C O2 B A O1 O C O2 B Chapter 5

Menelaus’ theorem

5.1 Menelaus’ theorem

Theorem 5.1 (Menelaus). Given a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively, the points X, Y , Z are collinear if and only if BX CY AZ = 1. XC · YA · ZB −

Z W

X B C

Proof. (= ) Let W be the point on AC such that BW//XY . Then, ⇒ BX WY AZ AY = , and = . XC YC ZB YW It follows that BX CY AZ WY CY AY CY AY WY = = = 1. XC · YA · ZB YC · YA · YW YC · YA · YW −

( =) Suppose the line joining X and Z intersects AC at Y ′. From above, ⇐ BX CY AZ BX CY AZ ′ = 1= . XC · Y ′A · ZB − XC · YA · ZB It follows that CY CY ′ = . Y ′A YA The points Y ′ and Y divide the segment CA in the same ratio. These must be the same point, and X, Y , Z are collinear. 202 Menelaus’ theorem

Example 5.1. The external angle bisectors of a triangle intersect their opposite sides at three collinear points.

Y ′ A

c b C

X′ B a

Z′

Proof. If the external bisectors are AX′, BY ′, CZ′ with X′, Y ′, Z′ on BC, CA, AB respectively, then BX c CY a AZ b ′ = , ′ = , ′ = . X′C −b Y ′A − c Z′B −a

BX′ CY ′ AZ′ It follows that = 1 and the points X′, Y ′, Z′ are collinear. X′C · Y ′A · Z′B − 5.2 Centers of similitude of two circles 203

5.2 Centers of similitude of two circles

Given two circles O(R) and I(r), whose centers O and I are at a distance d apart, we 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 . The line joining X and Y intersects the line OI of centers at a point T which satisﬁes

OT : IT = OX : IY = R : r.

This point T is independent of the choice of X. It is called the internal center of similitude, or simply the insimilicenter, of the two circles.

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. − 5.2.1 Desargue’s theorem Given three circles with centers A, B, C and distinct radii, show that the exsimilicenters of the three pairs of circles are collinear.

A B Z

X 204 Menelaus’ theorem

5.3 Ceva’s theorem

Theorem 5.2 (Ceva). Given a triangle ABC with points X, Y , Z on the side lines BC, CA, AB respectively, the lines AX, BY , CZ are concurrent if and only if BX CY AZ = +1. XC · YA · ZB

B X C Proof. (= ) Suppose the lines AX, BY , CZ intersect at a point P . Consider the line BPY cutting⇒ the sides of triangle CAX. By Menelaus’ theorem, CY AP XB CY PA BX = 1, or = +1. YA · PX · BC − YA · XP · BC Also, consider the line CPZ cutting the sides of triangle ABX. By Menelaus’ theorem again, AZ BC XP AZ BC XP = 1, or = +1. ZB · CX · PA − ZB · XC · PA Multiplying the two equations together, we have

CY AZ BX = +1. YA · ZB · XC ( =) Exercise. ⇐ 5.4 Some triangle centers 205

5.4 Some triangle centers

5.4.1 The centroid If D, E, F are the midpoints of the sides BC, CA, AB of triangle ABC, then clearly AF BD CE =1. FB · DC · EA The medians AD, BE, CF are therefore concurrent. Their intersection is the centroid G of the triangle. Consider triangle ADC with transversal BGE. By Menelaus’ theorem, AG DB CE AG 1 1 1= = − . − GD · BC · EA GD · 2 · 1 It follows that AG : GD =2:1. The centroid of a triangle divides each median in the ratio 2:1.

A A

Y F E Z G I

B D C B X C

5.4.2 The incenter Let X, Y , Z be points on BC, CA, AB such that AX, BY , CZ bisect angles BAC, CBA and ACB respectively. Then AZ b BX c CY a = , = , = . ZB a XC b YA c It follows that AZ BX CY b c a = = +1, ZB · XC · YA a · b · c and AX, BY , CZ are concurrent. Their intersection is the incenter of the triangle. Applying Menelaus’ theorem to triangle ABX with transversal CIZ, we have AI XC BZ AI b a AI b + c 1= = − = = . − IX · CB · ZA IX · b + c · b ⇒ IX a 206 Menelaus’ theorem

5.4.3 The Gergonne point Let the incircle of triangle ABC be tangent to the sides BC at X, CA at Y , and AB at Z respectively. Since AY = AZ = s a, BZ = BX = s b, and CX = CY = s c, we have − − − BX CY AZ s b s c s a = − − − =1. XC · YA · ZB s c · s a · s b − − − By Ceva’s theorem, the lines AX, BY , CZ are concurrent. The intersection is called the Gergonne point Ge of the triangle.

s a −

Z I s c Ge −

s b −

B s b X s c C − −

Lemma 5.3. The Gergonne point Ge divides the cevian AX in the ratio AG a(s a) e = − . G X (s b)(s c) e − −

Proof. Applying Menelaus’ theorem to triangle ABX with transversal CGeZ, we have AG XC BZ AG (s c) s b AG a(s a) 1= e = e − − − = e = − . − G X · CB · ZA G X · a · s a ⇒ G X (s b)(s c) e e − e − − 5.4 Some triangle centers 207

5.4.4 The Nagel point

If X′, Y ′, Z′ are the points of tangency of the excircles with the respective sidelines, the lines AX′, BY ′, CZ′ are concurrent by Ceva’s theorem:

BX′ CY ′ AZ′ s c s c s a = − − − =1. X C · Y A · Z B s b · s a · s b ′ ′ ′ − − − The point of concurrency is the Nagel point Na.

s b s c Ic − −

Z′

Y ′ Na s a s a − −

B s c X s b C − ′ −

Lemma 5.4. If the A-excircle of triangle ABC touches BC at X′, then the Nagel point divides the cevian AX′ in the ratio AN a a = . N X s a a ′ − Proof. Applying Menelaus’ theorem to triangle ACX′ with transversal BNaY ′, we have

AN X′B CY ′ AN (s c) s a AN a 1= a = a − − − = a = . − N X · BC · Y A N X · a · s c ⇒ N X s a a ′ ′ a ′ − a ′ − 208 Menelaus’ theorem

5.5 Isotomic conjugates

Given points X on BC, Y on CA, and Z on AB, we consider their reﬂections in the midpoints of the respective sides. These are the points X′ on BC, Y ′ on CA and Z′ on AB satisfying

BX′ = XC, BX = X′C; CY ′ = Y A, CY = Y ′A; AZ′ = ZB, AZ = Z′B.

Clearly, AX, BY , CZ are concurrent if and only if AX′, BY ′, CZ′ are concurrent. A

Z ′ Y

Y Z ′ P P •

B C X X′

Proof.

BX CY AZ BX CY AZ BX BX CY CY AZ AZ ′ ′ ′ = ′ ′ ′ = 1. XC · Y A · ZB X C · Y A · Z B XC · X C Y A · Y A ZB · Z B ′ ′ ′ ′ ′ ′ 

The points of concurrency of the two triads of lines are called isotomic conjugates. 5.5 Isotomic conjugates 209

Example 5.2. (The Gergonne and Nagel points)

Z′ Y

Z I Y ′ Na Ge

B C X X′

Example 5.3. (The isotomic conjugate of the orthocenter) Let H• denote the isotomic conjugate of the orthocenter H. Its traces are the pedals of the reﬂection of H in O. This latter point is the deLongchamps point Lo.

Z′

Y Lo

O Y ′

H• Z H

B C X X′ 210 Menelaus’ theorem

Example 5.4. (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 µ a µ − − 0= b µ 0 µ =(a µ)(b µ)(c µ) µ3. − − − − − − µ c 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 − − − !

A A

Z′ Y ′

Z Y P •

B C B C X X′ The isotomic conjugate

1 1 1 b µ 3 c µ 3 a µ 3 P • = − : − : − c µ a µ b µ − − − ! has traces X′, Y ′, Z′ that satisfy

CX′ = AY ′ = BZ′ = µ.

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

1P. Yff, An analogue of the Brocard points, Amer. Math. Monthly, 70 (1963) 495 – 501. 2A. L. Crelle, 1815. Chapter 6

The Euler line and the nine-point circle

6.1 The Euler line

6.1.1 Inferior and superior triangles

C′ A B′ A

B C F E G

A′ B D C

The inferior triangle of ABC is the triangle DEF whose vertices are the midpoints of the sides BC, CA, AB. The two triangles share the same centroid G, and are homothetic at G with ratio 1:2. − The superior triangle of ABC is the triangle A′B′C′ bounded by the parallels of the sides through the opposite vertices. The two triangles also share the same centroid G, and are homothetic at G with ratio 2: 1. − 212 The Euler line and the nine-point circle

6.1.2 The orthocenter and the Euler line The three altitudes of a triangle are concurrent. This is because the line containing an altitude of triangle ABC is the perpendicular bisector of a side of its superior triangle. The three lines therefore intersect at the circumcenter of the superior triangle. This is the orthocenter of the given triangle.

C′ A B′

O G H

B C

A′

The circumcenter, centroid, and orthocenter of a triangle are collinear. This is because the orthocenter, being the circumcenter of the superior triangle, is the image of the circumcenter under the homothety h(G, 2). The line containing them is called the Euler line of the reference triangle (provided it− is non-equilateral). The orthocenter of an acute (obtuse) triangle lies in the interior (exterior) of the triangle. The orthocenter of a right triangle is the right angle vertex. 6.2 The nine-point circle 213

6.2 The nine-point circle

Theorem 6.1. The following nine points associated with a triangle are on a circle whose center is the midpoint between the circumcenter and the orthocenter: (i) the midpoints of the three sides, (ii) the pedals (orthogonal projections) of the three vertices on their opposite sides, (iii) the midpoints between the orthocenter and the three vertices.

D′ Y F E G N O Z H

E′ F ′ B X D C

Proof. (1) Let N be the circumcenter of the inferior triangle DEF . Since DEF and ABC are homothetic at G in the ratio 1:2, N, G, O are collinear, and NG : GO =1:2. Since HG : GO =2:1, the four are collinear, and

HN : NG : GO =3:1:2, and N is the midpoint of OH. (2) Let X be the pedal of H on BC. Since N is the midpoint of OH, the pedal of N is the midpoint of DX. Therefore, N lies on the perpendicular bisector of DX, and NX = ND. Similarly, NE = NY , and NF = NZ for the pedals of H on CA and AB respectively. This means that the circumcircle of DEF also contains X, Y , Z. (3) Let D′, E′, F ′ be the midpoints of AH, BH, CH respectively. The triangle D′E′F ′ is homothetic to ABC at H in the ratio 1:2. Denote by N ′ its circumcenter. The points N ′, G, O are collinear, and N ′G : GO =1:2. It follows that N ′ = N, and the circumcircle of DEF also contains D′, E′, F ′. This circle is called the nine-point circle of triangle ABC. Its center N is called the nine-point center. Its radius is half of the circumradius of ABC. 214 The Euler line and the nine-point circle

Theorem 6.2. Let Oa, Ob, Oc be the reﬂections of the circumcenter O in the sidelines BC,CA, AB respectively. (1) The circle through Oa, Ob, Oc is congruent to the circumcircle and has center at the orthocenter H. (2) The reﬂections of H in the sidelines lie on the circumcircle.

B D C

Proof. (1) If D is the midpoint of BC, OOa =2OD = AH. This means that AHOaO is a parallelogram, and HOa = AO. Similarly, HOb = BO and HOc = CO for the other two reﬂections. Therefore, and H is the center of the circle through Oa, Ob, Oc, and the circle is congruent to the circumcircle. (2) If Ha is the reﬂection of H in BC, then HHaOAO is a trapezoid symmetric in the line BC. Therefore, OHa = HOa = OA. This means that Ha lies on the circumcircle; so do Hb and Hc. 6.3 Distances between triangle centers 215

6.3 Distances between triangle centers

6.3.1 Distance between the circumcenter and orthocenter

Y F E

O Z N H

B X D C

Proposition 6.3. OH2 = R2(1 8cos α cos β cos γ). − Proof. In triangle AOH, AO = R, AH =2R cos α, and ∠OAH = β γ . By the law of cosines, | − |

OH2 = R2(1+4cos2 α 4cos α cos(β γ)) − − = R2(1 4cos α(cos(β + γ) + cos(β γ)) − − = R2(1 8cos α cos β cos γ). − 216 The Euler line and the nine-point circle

6.3.2 Distance between circumcenter and tritangent centers Lemma 6.4. If the bisector of angle A intersects the circumcircle at M, then M is the center of the circle through B, I, C, and Ia. Proof. (1) Since M is the midpoint of the arc BC, ∠MBC = ∠MCB = ∠MAB. There- fore, ∠MBI = ∠MBC + ∠CBI = ∠MAB + ∠IBA = ∠MIB, and MB = MI. Similarly, MC = MI. (2) On the other hand, since ∠IBIa and ICIa are both right angles, the four points B, I, C, IaM are concyclic, with center at the midpoint of IIA. This is the point M. Theorem 6.5 (Euler). (a) OI2 = R2 2Rr. 2 2 − (b) OIa = R +2Rra.

Y O I

C B

Y ′ M

Proof. (a) Considering the power of I in the circumcircle, we have

2 2 r α R OI = AI IM = AI MB = α 2R sin =2Rr. − · · sin 2 · · 2

(b) Consider the power of Ia in the circumcircle. ra α Note that IaA = α . Also, IaM = MB =2R sin . sin 2 2

2 2 OI = R + IaA IaM a · 2 ra α = R + α 2R sin sin 2 · 2 2 = R +2Rra. 6.3 Distances between triangle centers 217

6.3.3 Distance between orthocenter and tritangent centers Proposition 6.6.

HI2 =2r2 4R2 cos α cos β cos γ, − HI2 =2r2 4R2 cos α cos β cos γ. a a −

I H

B X C β γ ∠ Proof. In triangle AIH, we have AH = 2R cos α, AI = 4R sin 2 sin 2 and HAI = β γ | − | 2 . By the law of cosines,

2 2 2 β γ HI = AH + AI 2AI AH cos − − · · 2 2 2 2 β 2 γ β γ β γ = 4R cos α + 4sin sin 4cos α sin sin cos − 2 2 − 2 2 2 2 β 2 γ β γ β γ 2 β 2 γ = 4R2 cos2 α + 4sin sin 4cos α sin sin cos cos 4cos α sin sin 2 2 − 2 2 2 2 − 2 2 2 β 2 γ 2 α 2 β 2 γ = 4R2 cos2 α + 4sin sin cos α sin β sin γ 4 1 2sin sin sin 2 2 − − − 2 2 2 2 α 2 β 2 γ = 4R2 cos α(cos α sin β sin γ)+8sin sin sin − 2 2 2 2 α 2 β 2 γ = 4R2 cos α cos β cos γ + 8sin sin sin − 2 2 2 = 2r2 4R2 cos α cos β cos γ. − 218 The Euler line and the nine-point circle

β γ (2) In triangle AHIa, AIa =4R cos 2 cos 2 .

I H

C B Aa

Ia By the law of cosines, we have

2 2 2 β γ HIa = AH + AIa 2AIa AH cos − − · · 2 2 2 2 β 2 γ β γ β γ = 4R cos α + 4cos cos 4cos α cos cos cos − 2 2 − 2 2 2 β γ β γ β γ β γ = 4R2 cos2 α + 4cos2 cos2 4cos α cos2 cos2 4cos α cos cos sin sin 2 2 − 2 2 − 2 2 2 2 β γ 2 α β γ = 4R2 cos2 α + 4cos2 cos2 4 1 2sin cos2 cos2 cos α sin β sin γ 2 2 − − 2 2 2 − 2 α β γ = 4R2 cos α(cos α sin β sin γ)+8sin cos2 cos2 − 2 2 2 2 α β γ = 4R2 cos α cos β cos γ + 8sin cos2 cos2 − 2 2 2 2 2 = 2ra 4R cos α cos β cos γ. − 6.4 Feuerbach’s theorem 219

6.4 Feuerbach’s theorem

Theorem 6.7 (Feuerbach). The nine-point circle is tangent internally to the incircle and externally to each of the excircles.

H O N

Proof. (1) Since N is the midpoint of OH, IN is a median of triangle IOH. By Apollo- nius’ theorem, 1 1 NI2 = (IH2 + OI2) OH2 2 − 4 1 = R2 Rr + r2 4 − R 2 = r . 2 − Therefore, NI is the difference between the radii of the nine-point circle and the incircle. This shows that the two circles are tangent to each other internally. (2) Similarly, in triangle IaOH,

1 1 NI2 = (HI2 + OI2) OH2 a 2 a a − 4 1 = R2 + Rr + r2 4 a a R 2 = + r . 2 a This shows that the distance between the centers of the nine-point and an excircle is the sum of their radii. The two circles are tangent externally.

The point of tangency Fe of the incircle and the nine-point circle is called the Feuerbach point. 220 The Euler line and the nine-point circle

Cc Fe Fb

Fc Bb I

Ac B Aa C Ab Fa

Ia Chapter 7

Isogonal conjugates

7.1 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 7.1. (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. − − ≡− Exercise 1. If a, b, c are the sidelines of triangle ABC, then ∠(a, b)= γ etc. − 222 Isogonal conjugates

7.2 Isogonal conjugates

Let P be a given point. Consider the reﬂections of the cevians AP , BP , CP in the respective bisectors of angles A, B, C, i.e., By Ceva’s theorem, these reﬂections are concurrent. Their intersection is the isogonal conjugate of P . Let P and Q be isogonal conjugates, AP and AQ intersecting BC at X and X′ respectively. Then 2 BX BX′ c = 2 . XC · X′C b Example 7.1. The incenter is the isogonal conjugate of itself. The same is true for the excenters.

Example 7.2. (The circumcenter and orthocenter) For a given triangle with circumcenter O, the line OA and the altitude through A are isogonal lines, similarly for the circumradii and altitudes through B and C. Since the circumradii are concurrent at O, the altitudes also are concurrent. Their intersection is the orthocenter H, which is the isogonal conjugate of O.

Y F E O

Z H

B X D C 7.3 The symmedian point and the centroid 223

7.3 The symmedian point and the centroid

The isogonal lines of the medians are called the symmedians. The isogonal conjugate of the centroid G is called the symmedian point K of the triangle.

G K

B C

Consider triangle ABC together with its tangential triangle A′B′C′, the triangle bounded by the tangents of the circumcircle at the vertices.

C′

B′

B D C

A′ Z

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

∠(A′Y,A′B′)= π 2(π α γ)= π 2β, − − − − 224 Isogonal conjugates

and similarly, ∠(A′C′,A′Z′)= π 2γ. Since ∠(A′B′,A′C′)= π 2α, we have − −

∠(A′Y,A′Z)=∠(A′Y,A′B′)+ ∠(A′B′,A′C′)+ ∠(A′C′,A′Z) =(π 2β)+(π 2α)+(π 2γ) − − − =π 0 mod π. ≡

This shows that Y , A′ and Z′ are collinear, so that (i) AA′ is a median of triangle AY Z, (ii) AY Z and ABC are similar. It follows that AA′ is the isogonal line of the A-median, i.e.,a symmedian. 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. 7.4 Isogonal conjugates of the Gergonne and Nagel points 225

7.4 Isogonal conjugates of the Gergonne and Nagel points

7.4.1 The Gergonne point and the insimilicenter T+

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 I Ge 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.

(3) In fact, E′F ′ is parallel to BC.

E F ′ E′

F I

D′ B D C

This follows from 226 Isogonal conjugates

(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 β =(π γ) + 2 γ + − 2 =β + γ = α (mod π); − (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 γ = (π β) 2 β + − − − 2 = (β + γ)= α (modπ) −

Since E′ and F ′ are on the incircle, and ID BC, it follows that E′F ′ is parallel to BC. ⊥ (4) Similarly, F ′D′ and D′E′ are parallel to CA and AB respectively. It follows that D′E′F ′ is homothetic to ABC. The ratio of homothety is r : R. Therefore, the center of homothety is the point T+ which divides OI in the ratio R : r. This is the internal center of similitude, or simply the insimilicenter of (O) and (I).

E F ′ E′

T+ F I O Ge

D′ B D C 7.4 Isogonal conjugates of the Gergonne and Nagel points 227

7.4.2 The Nagel point and the exsimilicenter T − The isogonal conjugate of the Nagel point is the point T which divides OI in the ratio − OT : T I = R : r. This is the external center of similitude (or exsimilicenter) of the circumcircle− − and the− incircle.

Ic D1′ Cc

O B I b T N − a

E1′ F1′

B Aa C

Ia 228 Isogonal conjugates

7.5 The Brocard points

Analogous to the Crelles points, we may ask if there are concurrent lines through the vertices making equal angles with the sidelines. More precisely, given triangle ABC, does there exist a point P satisfying ∠BAP = ∠CBP = ∠ACP = ω. It turns out that is one such unique conﬁguration.

P ω ω B C Note that if P is a point satisfying ∠BAP = ∠CBP , then the circle through P , A, B is tangent to BC at B. This circle is unique and can be constructed as follows. Its center is the intersection of the perpendicular bisector of AB and the perpendicular to BC at B.

ω B C

Likewise, if ∠CBP = ∠ACP , then the circle through P , B, C is tangent to CA at C. It follows that P is the intersection of these two circles. With this P , the circle PCA is tangent to AB at A. By Ceva’s theorem, the angle ω satisﬁes the equation sin3 ω = sin(β ω)sin(α ω)sin(γ ω). − − − It also follows that with the same ω, there is another triad of circles intersecting at another point Q such that ∠CAQ = ∠ABQ = ∠BCQ = ω. 7.5 The Brocard points 229

A A

ω ω

Q ω

P ω ω ω B C B C

The points P and Q are isogonal conjugates. They are called the Brocard points of triangle ABC.

ω ω

ω P ω ω ω B C 230 Isogonal conjugates

7.6 Kariya’s theorem

Given a triangle ABC with incenter I, consider a point X on the perpendicular from I to BC, such that IX = t. We regard t> 0 if X and the point of tangency of the incircle with the side BC are on the same side of I.

Theorem 7.2 (Kariya). Let I be the incenter of triangle ABC. If points X, Y , Z are chosen on the perpendiculars from I to BC, CA, AB respectively such that IX = IY = IZ, then the lines AX, BY , CZ are concurrent.

I Q

B C

Proof. (1) We compute the length of AX. Let the perpendicular from A to BC and the parallel from X to the same line intersect at X′. In the right triangle AXX′,

2∆ 2rs r(b + c) AX′ = r + t = r + t = + t, a − a − a XX′ = (s b) c cos B − − 1 1 = (c + a b) (c2 + a2 b2) 2 − − 2a − a(c + a b) (c2 + a2 b2) = − − − 2a b2 c2 a(b c) (b c)(b + c a) (b c)(s a) = − − − = − − = − − . 2a 2a a

Applying the Pythagorean theorem to the right triangle AXX′, we have 7.6 Kariya’s theorem 231

M ′ A A

O I I P

B C B C

X′ X X

r(b + c) 2 (b c)(s a) 2 AX2 = + t + − − a a r2(b + c)2 +(b c)2(s a)2 2r(b + c)t = − − + + t2 a2 a 2 (s a)(s b)(s c)(b+c) 2 2 − − − +(s a) (b c) 2r(b + c)t + at2 = s − − + a2 a (s a)((s b)(s c)(b + c)2 + s(s a)(b c)2) 2r(b + c)t + at2 = − − − − − + a2s a (s a) a2bc 2r(b + c)t + at2 = − · + a2s a abc(s a) 2r(b + c)t + at2 = − + as a 4Rr(s a) 2r(b + c)t + at2 = − + a a 4Rr(s a)+2r(b + c)t + at2 = − . a

(2) Let M ′ be the midpoint of the arc BAC of the circumcircle, and let MX intersect OI at P . We shall prove that angle IAP = angle IAX. First of all,

(s a)2 bc bc(s a) 4Rr(s a) AI2 = − =(s a)2 = − = − . cos2 A − · s(s a) s a 2 − For later use, we also establish

4Rr(s a) 2Rr(2s 2a)+2Rr a 2Rr(b + c) AI2 +2Rr = − +2Rr = − · = . a a a

Since IP : PO = t : R, IP = t OI. Recall that OI2 = R2 2Rr. Applying the law R+t · − 232 Isogonal conjugates of cosines to triangle AIP , we have AP 2 = AI2 + IP 2 2 AI IP cos AIP − · · t2 t = AI2 + (R2 2Rr) (AI2 + OI2 R2) (R + t)2 − − R + t − t2 t = AI2 + (R2 2Rr) (AI2 2Rr) (R + t)2 − − R + t − R 2R2rt R2t2 = AI2 + + R + t · (R + t)2 (R + t)2 R(R + t)AI2 +2R2rt + R2t2 = (R + t)2 R2 AI2 + R(AI2 +2Rr)t + R2t2 = · (R + t)2 2 2 4Rr(s a) 2R r(b+c) 2 2 R − + t + R t = · a a (R + t)2 R2(4Rr(s a)+2r(b + c)t + at2) = − . a(R + t)2

2 Note that AP 2 = R AX2. This means that AP = R AX. (R+t)2 · R+t · (3) Let AI intersect PX at X′′ and the circumcircle again at M, the antipode of M ′. Note that MM ′ = 2 and OI = R+t . M ′O IP t − − M ′

O I P

X′′

B C

X M Applying Menelaus’ theorem to triangle M ′OP and transversal IX′′M, we have

PX′′ M ′M OI PX′′ t PX′′ t 1= = = − = = − . − X′′M ′ · MO · IP ⇒ X′′M ′ 2(R + t) ⇒ PM ′ 2R + t XP t XP 2R+t Now PM = R . Therefore, PX = R , and ′ ′′ − XX R + t AX ′′ = = . X′′P R AP 7.7 Isogonal conjugate of an inﬁnite point 233

This shows that AX′′ bisects angle XAP . Since AX′′ is the bisector of angle A, the lines AP and AX are isogonal with respect to angle A. (4) Likewise, if points Y and Z are chosen on the perpendiculars from I to CA and AB such that IY = IZ = t = IX, then with the same point P on OI, the lines BP and BY are isogonal with respect to angle B, and CP , CZ isogonal with respect to angle C. Therefore the three lines AX, BY , CZ intersect at the isogonal conjugate of P (which divides OI in the ratio OP : PI = R : t).

7.7 Isogonal conjugate of an inﬁnite point

Proposition 7.3. Given a triangle ABC and a line ℓ, let ℓa, ℓb, ℓc be the parallels to ℓ through A, B, C respectively, and ℓa′ , ℓb′ , ℓc′ their reﬂections in the angle bisectors AI, BI, CI respectively. The lines ℓa′ , ℓb′ , ℓc′ intersect at a point on the circumcircle of triangle ABC.

ℓ O P b′

ℓa′ B C

ℓ ℓa ℓb ℓc ℓc′

Proof. Let P be the intersection of ℓb′ and ℓc′ .

(BP, PC) =(ℓb′ , ℓc′ )

=(ℓb′ , IB)+(IB, IC)+(IC, ℓc′ )

=(IB, ℓb)+(IB, IC)+(ℓc, IC) =(IB, ℓ)+(IB, IC)+(ℓ, IC) =2(IB, IC) π A =2 + 2 2 =(BA, AC) (mod π).

Therefore, ℓb′ and ℓc′ intersect at a point on the circumcircle of triangle ABC. Similarly, ℓa′ and ℓb′ intersect at a point P ′ on the circumcircle. Clearly, P and P ′ are the same point since they are both on the reﬂection of ℓb in the bisector IB. Therefore, the three reﬂections ℓa′ , ℓb′ , and ℓc′ intersect at the same point on the circumcircle. 234 Isogonal conjugates

Proposition 7.4. The isogonal conjugates of the inﬁnite points of two perpendicular lines are antipodal points on the circumcircle.

ℓ′ Q

ℓ I O P

B C

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

π (AP, AQ)=(AP, IA)+(IA, AQ)= (ℓ, IA) (IA, ℓ′)= (ℓ, ℓ′)= . − − − 2 Therefore, P and Q are antipodal points. Chapter 8

The excentral and intouch triangles

8.1 The excentral triangle

8.1.1 The circumcenter of the excentral triangle

The excentral triangle of ABC has vertices the excenters IA, Ib, IC . Since AIa IbIc, BIb IcIa, and CIc IaIb, ⊥ ⊥ ⊥ AIa, BIb, CIc are the altitudes, and I is the orthocenter of the excentral triangle. It follows that the circumcircle of triangle ABC is the nine-point circle of the excentral triangle.

O I′ I

C B

Ia Corollaries: (1) Each side of the excentral triangle intersects the circumcircle (O) at its own midpoint (apart from the vertices of triangle ABC). (2) The circumcenter of the excentral triangle is the reﬂection of I in O. 236 The excentral and intouch triangles

Let the bisector AIa intersect the circumcircle at M. M is the midpoint of the arc BC of the circumcircle (on the opposite side of A). Therefore, OM is perpendicular to BC.

O I′ I

C B X′

The circumcenter I′ is the reﬂection of I in O. Join I′ to Ia. Since IO = OI′ and IM = MIa, I′Ia is parallel to OM (and I′Ia = 2 OM = 2R). Therefore, I′Ia is · perpendicular to BC, and it passes through the point of tangency of X′ of BC with the A-excircle.

Corollary. The perpendiculars from the excenters to the corresponding sides of a triangle are concurrent. The point of concurrency is the circumcenter of the excentral triangle. 8.1 The excentral triangle 237

8.1.2 Relation between circumradius and radii of tritangent circles

Proposition 8.1. ra + rb + rc =4R + r.

M ′ A

I′ O I

X C Ac B D Aa Ab

Proof. (1) Consider the four points Ib, B, C, Ic. They are concyclic since both IbBIc and IbCIc are right angles. The center of the circle must be the midpoint M ′ of IbIc, which must also lie on the perpendicular bisector of BC, which is the line DM ′. Therefore, M ′ is the antipode of M on the circumcircle. (2) Therefore, rb + rc =2 DM ′ = 2(R + OD). · (3) On the other hand, 2 OD = r + I′Aa = r +2R ra. · − (4) It follows that ra + rb + rc =4R + r. 238 The excentral and intouch triangles

8.2 The homothetic center of the intouch and excentral triangles

The intouch triangle has vertices the points of the tangency of the incircle with the sidelines.

Proposition 8.2. The excentral and the intouch triangles are homothetic at a point T dividing OI in the ratio OT : TI =2R + r : 2r. −

Ic Y I Z ′ T O I

C B X

Proof. (1) The segments IbIc and YZ are parallel, since they are both perpendicular to the bisector of angle A. Similarly, IcIa//ZX and IaIb//XY . Therefore the two triangles are homothetic. (2) Since the excentral triangle has circumradius 2R and the intouch triangle has circumradius r, the homothetic ratio is 2R : r. (3) The homothetic center is the point dividing the segment I′I externally in the ratio I′T : TI =2R : r. − (4) Since I′ is the reﬂection of I in O, the homothetic center T divides OI in the ratio

OT : TI =2R + r : 2r. − Chapter 9

Homogeneous Barycentric Coordinates

9.1 Absolute and homogeneous barycentric coordinates

The notion of barycentric coordinates dates back to A. F. Mobius¨ (–). 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 6: v : w) with reference to ABC.

9.1.1 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 = 0 does not represent any ﬁnite point on the plane. We shall say that it represents an inﬁnite point. See 9.4. § 302 Homogeneous Barycentric Coordinates

A A

Y F E Z I G

B D C B X C

9.1.2 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). 9.1 Absolute and homogeneous barycentric coordinates 303

9.1.3 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. A

F E

B D C

The center of mass of the perimeter is therefore the point

a D + b E + c F S = · · · p a + b + c a B+C + b C+A + c A+B = · 2 · 2 · 2 a + b + c (b + c)A +(c + a)B +(a + b)C = . 2(a + b + c) In homogeneous barycentric coordinates,

Sp =(b + c : c + a : a + b) . 304 Homogeneous Barycentric Coordinates

9.1.4 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

Z I Ge

B X C The ratio AGe : GeX, however, is not immediate obvious. It can nevertheless be found by applying the Menelaus theorem to triangle ABX with transversal CZ. Thus, AG XC BZ e = 1. GeX · CB · ZA − From this, AG CB ZA a s a a(s a) e = = − − = − . G X −XC · BZ −s c · s b (s b)(s c) e − − − − Therefore,

(s b)(s c)A + a(s a)X G = − − − e (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) − − − 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 . − − − 9.2 Cevian triangle 305

9.2 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 9.1 (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).

A A

BP CP P P

B AP C B X C

Remarks. (1) The points X, Y , Z are called the traces of P . We also say that XYZ 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):∆(P AB). 306 Homogeneous Barycentric Coordinates

9.2.1 The circumcenter Consider the circumcenter O of triangle ABC.

B C

Since ∠BOC =2A, the area of triangle OBC is 1 1 OB OC sin BOC = R2 sin2A. 2 · · · 2

1 2 1 2 Similarly, the areas of triangles OCA and OAB are respectively 2 R sin2B and 2 R sin2C. It follows that the circumcenter O has homogeneous barycentric coordinates

∆OBC :∆OCA :∆OAB 1 1 1 = R2 sin2A : R2 sin2B : R2 sin2C 2 2 2 = sin2A : sin2B : sin2C = a cos A : b cos B : c 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). − − − 9.2 Cevian triangle 307

9.2.2 The Nagel point and the extouch triangle

The A-excircle touches the side BC at a point X′ such that BX′ = s c and X′C = 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:

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), − −

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

N =(s a : s b : s c). a − − −

The triangle X′Y ′Z′ is called the extouch triangle. 308 Homogeneous Barycentric Coordinates

9.2.3 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 = : ; b c similarly for the other two traces.

b c X = 0 : cos β : cos γ a c Y = cos α : 0 : cos γ a b Z = cos α : cos β : 0 a b c H = cos α : cos β : cos γ

A A

O Z H

B X C B C

The triangle XYZ is called the orthic triangle. 9.3 Homotheties 309

9.3 Homotheties

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

P k Y 1 k X −

9.3.1 Superiors and inferiors

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

PG : Gsup(P )= 1:2, PG : Ginf(P )= 2:1.

P G sup(P )

inf(P )

Proposition 9.2. If P =(u : v : w) in homogeneous barycentric coordinates, then

sup(P )= (v + w u : w + u v : u + v w), − − − inf(P )= (v + w : w + u : u + v).

Proof. In absolute barycentric coordinates,

sup(P )= 3G 2P − 2(uA + vB + wC) = (A + B + C) − u + v + w (u + v + w)(A + B + C) 2(uA + vB + wC) = u + v + w − u + v + w (v + w u)A +(w + u v)B +(u + v w)C = − − − . u + v + w

Therefore, sup(P )=(v + w u : w + u v : u + v w) in homogeneous barycentric coordinates. − − − The case for inferior is similar. 310 Homogeneous Barycentric Coordinates

Example 9.1. (1) The superior of the incenter is the Nagel point. The inferior of the incenter is the Spieker center, the barycenter of the perimeter of the triangle.

(2) The inferior and superior of the Gergonne point

G = ((s b)(s c): (s c)(s a): (s a)(s b)) e − − − − − − are

inf(G )= ((s c)(s a)+(s a)(s b): (s a)(s b)+(s b)(s c) e − − − − − − − − : (s b)(s c)+(s c)(s a)) − − − − = ((s a)(s c + s b): (s b)(s a + s c): (s c)(s b + s a)) − − − − − − − − − = (a(s a): b(s b): c(s c)), − − − and

sup(G )= ((s c)(s a)+(s a)(s b) (s b)(s c) e − − − − − − − : (s a)(s b)+(s b)(s c) (s c)(s a) − − − − − − − : (s b)(s c)+(s c)(s a) (s a)(s b)) − − − − − − − = (a(s a) (s b)(s c): b(s b) (s c)(s a) − − − − − − − − : c(s c) (s a)(s b)) − − − − = ( s2 +(a + b + c)s a2 bc : s2 +(a + b + c)s b2 ca − − − − − − : s2 +(a + b + c)s ab) − − = (s2 a2 bc : s2 b2 ca : s2 c2 ab) − − − − − − = ((a + b + c)2 4a2 4bc : (a + b + c)2 4b2 4ca − − − − : (a + b + c)2 4c2 4ab) − − = ( 3a2 +2a(b + c)+(b c)2 : 3b2 +2b(c + a)+(c a)2 − − − − : 3c2 +2c(a + b)+(a b)2). − − (2) The nine-point center, being the midpoint of O and H, is the inferior of O.

N Lo H G O

From the homogeneous barycentric of O, we obtain

N = b2(c2 + a2 b2)+ c2(a2 + b2 c2): : − − ··· ··· = a2(b2 + c2) (b2 c2)2 : : . − − ··· ··· Similarly, the deLongchamps point, being the reﬂection of H in O, is the superior of H. It has coordinates

L =( 3a4 +2a2(b2 + c2)+(b2 c2)2 : 3b4 +2b2(c2 + a2)+(c2 a2)2 o − − − − : 3c4 +2c2(a2 + b2)+(a2 b2)2). − − 9.3 Homotheties 311

X Na

H Z

B C

Example 9.2. The distance between O and N is equal to R 2r. a − Proof. Since O = sup(N) and N = sup(I), ON =2 NI =2 R r = R 2r. a a · 2 − − Example 9.3. Pedals on altitudes equidistant from vertices We locate the point P whose pedals on the altitudes are equidistant from the vertices. Let P =(u : v : w). We require

v + w 2∆ w + u 2∆ u + v 2∆ v + w w + u u + v = = = = = . u + v + w · a u + v + w · b u + v + w · c ⇒ a b c

This means that inf(P )= I and P = sup(I)= Na, the Nagel point. With P = Na, the common distance from the vertices to the pedals is (c + a b)+(a + b c) 2∆ 4∆ − − = =2r, (b + c a)+(c + a b)+(a + b c) · a a + b + c − − − the diameter of the incircle. The three pedals lie on the circle with diameter NaH, called the Fuhrmann circle. 312 Homogeneous Barycentric Coordinates

9.3.2 Triangles bounded by lines parallel to the sidelines

Theorem 9.3 (Homothetic center theorem). If parallel lines XbXc, YcYa, ZaZb to the sides BC, CA, AB of triangle ABC are constructed such that

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

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

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

A∗ A∗

Zb Zb Yc Yc

A A

P P

Za Ya Za Ya B C B C

B∗ Xc Xb C∗B∗ Xc Xb C∗

Proof. Let P be the intersection of B∗B and C∗C. Since

B∗C∗ = B∗Xc + XcXb + XbC∗

= t3a +(1+ t1)a + t2a = (1+ t1 + t2 + t3)a, we we have PB : PB∗ = PC : PC∗ =1:1+ t1 + t2 + t3.

A similar calculation shows that AA∗ and BB∗ intersect at the same point P . This shows that A∗B∗C∗ is the image of ABC under the homothety h(P, 1+ t1 + t2 + t3). Now we compare areas. Note that BXc BZa (1) ∆(BZaXc)= ∆(ABC)= t t ∆(ABC), AB · CB · 1 3 (2) ∆(PBC) = PB CB = 1 1 = 1 . ∆(BZaB∗) BB∗ · BZa t1+t2+t3 · t3 t3(t1+t2+t3)

t1 Since ∆(BZaB∗)=∆(BZaXc), we have ∆(PBC)= ∆(ABC). t1+t2+t3 · Similarly, ∆(PCA) = t2 ∆(ABC) and ∆(P AB) = t3 ∆(ABC). It t1+t2+t3 t1+t2+t3 follows that · · ∆(PBC):∆(PCA):∆(P AB)= t1 : t2 : t3. 9.3 Homotheties 313

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 2∆ 2 2 a +a a a 1+ t1 = 2∆ = 1+ 2∆ , i.e., t1 = 2∆ . Similarly, if we erect squares externally on the a other two sides, the outer edges of these squares are on the lines which are the images of b2 c2 CA, AB under the homotheties h(B, 1+ t2) and h(C, 1+ t3) with t2 = 2∆ and t3 = 2∆ .

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 2∆ : 2∆ : 2∆ = (a : b : c ), and the ratio of homothety is 2∆ + a2 + b2 + c2 1+(t + t + t )= . 1 2 3 2∆ 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. 314 Homogeneous Barycentric Coordinates

9.4 Inﬁnite points

A point with zero sum of homogeneous coordinates is called an infinite point. It can be expressed as the difference of the absolute barycentric coordinates of two points (with equal nonzero sums of homogeneous coordinates). As such, an infinite point defines the direction of a family of parallel lines. All infinite points form the line at infinity: consisting of points (x : y : z) satisfying x+y +z =0. The infinite point of a line is defined uniquely up to a nonzero scalar multiple.

Example 9.4. 1. The infinite points of the side lines BC, CA, AB are (0 : 1 : 1), (1:0: 1), ( 1:1:0) respectively. − − − 2. The infinite point of the A altitude has homogeneous coordinates − 2 2 (0 : Sγ : Sβ) a (1:0:0)=( a : Sγ : Sβ). − − 3. The infinite point of the Euler line is the point

(S (S + S ),S (S + S ),S (S + S ) (S ,S ,S ) O H = α β γ β γ α γ α β βγ γα αβ − 2S2 − S2 (Sα(Sβ + Sγ) 2Sβγ,Sβ(Sγ + Sα) 2Sγα,Sγ(Sα + Sβ) 2Sαβ = − − − . 2S2 In homogeneous coordinates, this is

(Sα(Sβ + Sγ) 2Sβγ : Sβ(Sγ + Sα) 2Sγα : Sγ(Sα + Sβ) 2Sαβ). − − − 4. The inﬁnite point of the OI-line is

(ca(c a)(s b) ab(a b)(s c): : ) (a(a2(−b + c) − 2abc− (b +− c)(b −c)2):··· ···: ). ∼ − − − ··· ··· Chapter 10

Some applications of barycentric coordinates

10.1 Construction of mixtilinear incircles

10.1.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle

The centers of similatitude of two circles are the points dividing the centers in the ratio of their radii, either internally or externally. For the circumcircle and the incircle, these are

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

M ′ A

Z T I O − T+

B X C

We give an interesting application of these centers of similitude. 316 Some applications of barycentric coordinates

10.1.2 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′, we have OK : 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. 10.2 Isotomic and isogonal conjugates 317

10.2 Isotomic and isogonal conjugates 10.3 Isotomic conjugates

The Gergonne and Nagel points are examples of 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 are symmetric with respect to the midpoints of the corresponding sides. Thus, BX = X′C, CY = Y ′A, AZ = Z′B.

X′ Z

X Z′ P P •

C A Y Y ′

We shall denote the isotomic conjugate of P by P •. If P =(x : y : z), then

1 1 1 P • = : : =(yz : zx : xy). x y z 318 Some applications of barycentric coordinates

Example 10.1. (The Gergonne and Nagel points)

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

Z′ Y

Z I Y ′ Na Ge

B C X X′

Example 10.2. (The isotomic conjugate of the orthocenter) The isotomic conjugate of the orthocenter is the point

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

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

Z′ L Y o

O Y ′

H• Z H

B C X X′ 10.4 Equal-parallelians point 319

10.4 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. 1 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•, and can be easily constructed as the point dividing the segment − I•G externally in the ratio I•P : PG =3: 2. The point P is called the congruent- parallelians point of triangle ABC. − B

P G

I•

C A

1 The isotomic conjugate of the incenter appears in ETC as the point X75. 320 Some applications of barycentric coordinates Chapter 11

Computation of barycentric coordinates

11.1 The Feuerbach point

Proposition 11.1. The homogeneous barycentric coordinates of the Feuerbach point are

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

Proof. The Feuerbach point Fe is the point of (internal) tangency of the incircle and the nine-point circle. It divides NI in the ratio NF : F I = R : r = R : 2r. Therefore, e e 2 − − R I 2r N F = · − · e R 2r − in absolute barycentric coordinates. A

I O H N

B C From the homogeneous barycentric coordinates of N,

(a2(b2 + c2) (b2 c2)2,b2(c2 + a2) (c2 a2)2,c2(a2 + b2) (a2 b2)2) = 32∆2 N. − − − − − − · we have r 2r N = a2(b2 + c2) (b2 c2)2, , · 16∆2 − − ··· ··· R = a2(b2 + c2) (b2 c2)2, , , 4sabc − − ··· ··· R R R I = (a,b,c )= 2abc(a,b,c). · 2s 4sabc · 322 Computation of barycentric coordinates

Therefore,

F R I 2r N e ∼ · − · 2abc a (a2(b2 + c2) (b2 c2)2) ∼ · − − − = (a2(2bc b2 c2)+(b c)2(b + c)2, , ) − − − ··· ··· = ( a2(b c)2 +(b c)2(b + c)2, , ) − − − ··· ··· = (((b + c)2 a2)(b c)2, , ) − − ··· ··· = ((a + b + c)(b + c a)(b c)2, , ) − − ··· ··· ((b + c a)(b c)2, , ). ∼ − − ··· ···

Proposition 11.2. (a) ONa is parallel to NFe. (b) ON = R 2r. a − (c) NaH is parallel to OI. (d) The reﬂection of H in I and the reﬂection of Na in O are the same point. (e) IN and OSp intersect at the midpoint of NaH.

I O N Na Sp H

B C 11.2 The OI line 323

11.2 The OI line

11.2.1 The circumcenter of the excentral triangle

Let I′ be the reﬂection of I in O. Show that I′ is also the midpoint of NaLo. Let N ′ be the midpoint of OLo. The triangles OI′N ′ and OIN are congruent. I′N ′ is 1 parallel to IN and hence NaO. Furthermore, I′N ′ = IN = 2 NaO. It follows that I′ is the midpoint of NaLo.

3 2 2 2 I′ =(a(a + a (b + c) a(b + c) (b + c)(b c) ): : ). − − − ··· ··· 11.2.2 The centers of similitude of the circumcircle and the incircle

T =(a2(s a): b2(s b): c2(s c)). + − − − Proof.

T r O + R I + ∼ · · r R = (a2(b2 + c2 a2),b2(c2 + a2 b2),c2(a2 + b2 c2)) + (a,b,c) 16∆2 − − − 2s (a2(b2 + c2 a2),b2(c2 + a2 b2),c2(a2 + b2 c2))+8Rrs(a,b,c) ∼ − − − = (a2(b2 + c2 a2),b2(c2 + a2 b2),c2(a2 + b2 c2))+2abc(a,b,c) − − − = (a2(b2 +2bc + c2 a2), , ) − ··· ··· = (a2(a + b + c)(b + c a), , ) − ··· ··· (a2(b + c a), , ). ∼ − ··· ···

a2 b2 c2 T = s a : s b : s c . − − − − Proof.

T r O R I − ∼ · − · r R = (a2(b2 + c2 a2),b2(c2 + a2 b2),c2(a2 + b2 c2)) (a,b,c) 16∆2 − − − − 2s (a2(b2 + c2 a2),b2(c2 + a2 b2),c2(a2 + b2 c2)) 8Rrs(a,b,c) ∼ − − − − = (a2(b2 + c2 a2),b2(c2 + a2 b2),c2(a2 + b2 c2)) 2abc(a,b,c) − − − − = (a2(b2 2bc + c2 a2), , ) − − ··· ··· = (a2(b c + a)(b c a), , ) − − − ··· ··· (a2(c + a b)(a + b c), , ). ∼ − − ··· ··· 324 Computation of barycentric coordinates

Example 11.1. (a) G, T+, Fe are collinear. (b) H, T , Fe are collinear. −

Proof. (a) This follows from

(a2(b + c a),b2(c + a b),c2(a + b c)) − − − ((b + c a)(b c)2, (c + a b)(c a)2, (a + b c)(a b)2) − − − − − − − = ((b + c a)(a2 (b c)2), (c + a b)(b2 (c a)2), (a + b c)(c2 (a b)2) − − − − − − − − − = ((b + c a)(a b + c)(a + b c), (c + a b)(b c + a)(b + c a), − − − − − − (a + b c)(c a + b)(c + a b)) − − − =(b + c a)(c + a b)(a + b c)(1, 1, 1). − − −

11.2.3 The homothetic center T of excentral and intouch triangles

The two triangles are homothetic since their corresponding sides are perpendicular to the angle bisectors of triangle ABC. Denote by T the homothetic center. This is clearly the exsimilicenter of their circumcircles. It is therefore the point dividing I′ and I in the ratio I′T : IT =2R : r. It follows that OT : TI =2R + r : 2r. −

I I′ T O

a b c T = s a : s b : s c . − − − Proof.

T 2r O + (2R + r)I ∼ − · 2r 2R + r = − (a2(b2 + c2 a2), , )+ (a,b,c) 16r2s2 − ··· ··· 2s (a2(b2 + c2 a2), , )+(2R + r)(4rs)(a,b,c) ∼ − − ··· ··· = (a2(b2 + c2 a2), , )+(2abc + 4(s a)(s b)(s c))(a,b,c) − − ··· ··· − − − = (a( a(b2 + c2 a2)+(2abc + 4(s a)(s b)(s c))), , ) − − − − − ··· ··· 1 = a(a + b + c)(c + a b)(a + b c), , 2 − − ··· ··· a , , . ∼ b + c a ··· ··· − 11.3 The excentral triangle 325

Exercise 1. Show that TI : IO =2r :2R r. − 2. Find the ratio of division T+T : TT . −

3. Show that G, Ge, and T are collinear by ﬁnding p, q satisfying p(1, 1, 1) + q((c + a b)(a + b c), (a + b c)(b + c a), (b + c a)(c + a b)) − − − − − − = 2(a(c + a b)(a + b c),b(a + b c)(b + c a),c(b + c a)(c + a b)). − − − − − − Answer: p = (b + c a)(c + a b)(a + b c) and q = a + b + c. − − − − 4. Given that GG : G T = 2(2R r):3r, show that G , I, L are collinear. e e − e o Apply the converse of Menelaus’ theorem to triangle OGT with Ge on GT , I on TO, and Lo on OG. GG TI OL 2(2R r) 2r 3 e o = − − = 1. G T · IO · L G 3r · 2R r · 4 − e o − 11.3 The excentral triangle

11.3.1 The centroid The centroid of the excentral triangle is the point

Ia + Ib + Ic 1 ( a,b,c) (a, b,c) (a,b, c) = − + − + − 3 3 b + c a c + a b a + b c − − − (c + a b)(a + b c)( a,b,c)+(a + b c)(b + c a)(a, b,c) ∼ − − − − − − +(b + c a)(c + a b)(a,b, c) − − − (a( (s b)(s c)+(s c)(s a)+(s a)(s b)), , ) ∼ − − − − − − − ··· ··· (a(s2 2as bc + ca + ab), , ) ∼ − − ··· ··· (a(s2 a2 bc), , ) ∼ − − ··· ··· (a( 3a2 +2a(b + c)+(b c)2), , ). ∼ − − ··· ··· 11.3.2 The incenter

A ( a,b,c) B (a, b,c) C (a,b, c) cos − + cos − + cos − ∼ 2 · b + c a 2 · c + a b 2 · a + b c A − B − C − cos 2 cos 2 cos 2 A ( a,b,c)+ B (a, b,c)+ C (a,b, c) ∼ 2r cot 2 − 2r cot 2 − 2r cot 2 − A B C sin ( a,b,c)+sin (a, b,c)+sin (a,b, c) ∼ 2 − 2 − 2 − A B C a sin + sin + sin , , . ∼ − 2 2 2 ··· ··· 326 Computation of barycentric coordinates Chapter 12

Some interesting circles

12.1 A fundamental principle on 6 concyclic points

12.1.1 The radical axis of two circles

Given two nonconcentric circles C1 and C2. The locus of points of equal powers with respect to the circle is a straight line perpendicular to the line joining their centers. In fact, if the circles are concentric, there is no ﬁnite point with equal powers with respect to the circles. On the other hand, if the centers are distinct points A and B at a distance d apart, there is a unique point P with distances AP = x and PB = d x such that − r2 x2 = r2 (d x)2. 1 − 2 − − If this common value is m, then every point Q on the perpendicular to AB at P has power m PQ2 with respect to each of the circles. This line is called the radical axis of the two circles.− If the two circles intersect at two distinct points, then the radical axis is the line joining these common points. If the circles are tangent to each other, then the radical axis is the common tangent.

Theorem 12.1. Given three circles with distinct centers, the radical axes of the three pairs of circles are either concurrent or are parallel.

Proof. (1) If any two of the circles are concentric, there is no ﬁnite point with equal powers with respect to the three circles. (2) If the centers of the circles are distinct and noncollinear, then two of the radical axes, being perpendiculars to two distinct lines with a common point, intersect at a point. This intersection has equal powers with respect to all three circles, and also lies on the third radical axis. (3) If the three centers are distinct but collinear, then the three radical axes three parallel lines, which coincide if any two of them do. This is the case if and only if the three circles two points in common, or at mutually tangent at a point. In this case we say that the circles are coaxial. 328 Some interesting circles

12.1.2 Test for 6 concyclic points

Proposition 12.2. Let X, X′ be points on the sideline a, Y , Y ′ on b, and Z, Z′ on c. Thesix points are on a circle if and only if the four points on each pair of sidelines are concyclic.

Y ′ A

Z′ C B X X Y

Proof. It is enough to prove the sufﬁciency part. Let Ca be the circle through the four points Y , Y ′, Z, Z′ on b and c, and Cb the one through Z, Z′, X, X (on c and a), and Cc through X, X, Y , Y ′ (on a and (b). We claim that these three circles are identical. If not, then they are pairwise distinct. The three pairs among them have radical axes a (for Cb and Cc), b (for Cc and Ca), and c (for Ca and Cb) respectively. Now, the three radical axes of three distinct circles either intersect at a common point (the radical center), or are parallel (when their centers are on a line), or coincide (when, in additon, the three circles are coaxial). In no case can the three radical axes form a triangle (with sidelines a, b, c. This shows that the three circles coincide. 12.2 The Taylor circle 329

12.2 The Taylor circle

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

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

Zb Yc

Z H

B Xc X Xb C

It is easy to write down the lengths of various segments. From these we easily determine 2 the coordinates of these pedals. For example, from YaC = b cos γ, we have AYa = b b cos2 γ = b sin2 γ. Note that − 2 2 2 2 2 2 2 2 2 AYa AYc =(b b cos γ)(b cos α)= b cos α sin γ =4R cos α sin β sin γ, · − 2 2 2 2 2 2 2 2 2 AZa AZb =(c c cos β)(c cos α)= c cos α sin β =4R cos α sin β sin γ, · − 2 S S2 S S Sββ αα αα · giving AYa AYc = AZa AZb = 4R2 = a2·b2c2 . Similarly, BXb BXc = BZb BZa = a2b2c2 2 · · S Sγγ · · and CYc CYa = CXc CXb = 2· 2 2 . By Proposition 12.2, the six points Xb, Xc, Yc, Ya, · · a b c Za, Zb are concyclic. The circle containing them is called the Taylor circle.

Exercise

1. Calculate the length of XbXc.

1 2. Find the equation of the line XbXc.

3. The three lines XbXc, YcYa, ZaZb bound a triangle with perspector K.

1 2 S x + SBBy + SCC z = 0. − 330 Some interesting circles

12.3 Two Lemoine circles

12.3.1 The ﬁrst Lemoine circle Given triangle ABC, how can one choose a point K so that when parallel lines are constructed through it to intersect each sideline at two points, the resulting six points are on a circle?

Zb L K Za Ya

B Xc Xb C

Analysis. By the intersecting chords theorem, AYa AYc = AZa AZb. We have · · AY AY AZ AZ b2 a c = c2 a b · AC · AC · AB · AB AY AZ = b2 c = c2 b ⇒ · AC · AB BX CX X C = b2 c = c2 b = c2 b ⇒ · BC · CB · BC 2 BXc c = = 2 . ⇒ XbC b 2 BZa a Similarly, = 2 , and ZbA b 2 XcXb XcXb XcK BZa a = = = = 2 . XbC KYa KYc ZbA b 2 2 2 Therefore, BXc : XcXb : XbC = c : a : b . These proportions determine the points Xb and Xc on BC, and subsequently the other points: the parallels XcYc and XbZb (to c and b respectively) intersect at K, and the parallel through K to a determines the points Ya and Za. The point K is called the Lemoine symmedian point of triangle ABC, and the circle containing these six points is called the ﬁrst Lemoine circle. Denote by L the center of the ﬁrst Lemoine circle. Note that L lies on the perpendicular bisector of each of the parallel segments XcXb and ZaYa. This means that the line joining the midpoints of these segments is the common perpendicular bisector of the segments, and the trapezoid XcXbYaZa is symmetric; so are YaYcZbXb and ZbZaXcYc by the same reasoning. It follows that the segments YcZb, ZaXc, and XbYa have equal lengths. (Exercise: abc Show that this common length is a2+b2+c2 ). 12.3 Two Lemoine circles 331

12.3.2 The second Lemoine circle Now, if we construct the parallels of these segments through K to intersect the sidelines, we obtain another three equal segments with common midpoint K. The 6 endpoints, Xb′, abc Xc′ on a, Yc′, Ya′ on b, and Za′ , Zb′ on c, are on a circle center K and radius a2+b2+c2 . This circle is called the second Lemoine circle of triangle ABC.

Za′ Ya′

Yc′

Zb′

B C Xb′ Xc′

12.3.3 Construction of K

The length of AK is twice the median of triangle AYcZb on the side YcZb. If we denote by 2bcma ma etc the lengths of medians of triangle ABC, then AK = a2+b2+c2 . Similarly, BK = 2camb 2abmc a2+b2+c2 and CK = a2+b2+c2 . This allows us to determine the angles KAB etc and the radii AB AK BK of the circles KAB etc. The radius of the circle KAB, for example, is Rc = 4∆(· KAB· ) = abcmamb BK ∆ 2 2 2 . It follows that sin KAB = = . (a +b +c )∆(ABC) 2Rc bma

I K G

K G

B C B C

AC AG CG bmamc Consider also the circle GAC. This has radius Rb′ = 4∆(· GAC· ) = 3∆ . From this, sin GAC = CG = ∆ . This shows that ∠KAB = ∠GAC, and AK and the median AG 2Rb′ bma are isogonal lines with respect to the sides AB and AC. Similarly, BK and CK are the lines isogonal to the medians BG and CG respectively, and K is the symmedian point of triangle ABC. 332 Some interesting circles

12.3.4 The center of the ﬁrst Lemoine circle We show that the center L of the ﬁrst Lemoine circle is the midpoint between K and the circumcenter O. It is clear that this center is on the perpendicular bisector of XbXc. Let M be the midpoint of XbXc, and X the orthogonal projection of K on BC.

Zb O K Za Ya

B XcX M DXb C

We have ac2 1 a3 a(a2 +2c2) BM = + = , a2 + b2 + c2 2 · a2 + b2 + c2 2(a2 + b2 + c2) and

BX = BXc + KXc cos B ac2 a2c c2 + a2 b2 = + − a2 + b2 + c2 a2 + b2 + c2 · 2ca ac2 + a(c2 + a2 b2) = − 2(a2 + b2 + c2) a(a2 b2 +3c2) = − . 2(a2 + b2 + c2) It follows that a(a2 b2 +3c2) a(a2 + b2 + c2) a(a2 +2c2) BX + BD = − + = =2 BM. 2(a2 + b2 + c2) 2(a2 + b2 + c2) a2 + b2 + c2 · This means that M is the midpoint of XD, and the perpendicular to a at M contains the midpoint of OK. The same reasoning shows that the midpoint of OK also lies on the perpendiculars to b and c respectively at the midpoints of YcYa and ZaZb. It is therefore the center L of the ﬁrst Lemoine circle. (Exercise. Calculate the radius of the ﬁrst Lemoine circle). Chapter 13

Straight line equations

13.1 Area and barycentric coordinates

Theorem 13.1. If for i = 1, 2, 3, Pi = xi A + yi B + zi C (in absolute barycentric · · · coordinates) , then the area of the oriented triangle P1P2P3 is

x1 y1 z1 ∆P P P = x y z ∆ABC. 1 2 3 2 2 2 · x y z 3 3 3

Example 13.1. (Area of cevian triangle) Let P =(u : v : w) be a point with cevian triangle XYZ. The area of the cevian triangle XYZ is

0 v w 1 2uvw u 0 w ∆= ∆. (v + w)(w + u)(u + v) · (v + w)(w + u)(u + v) · u v 0

If P is an interior point (so that u, v, w are positive), then v + w 2√vw, w + u 2√wu, and u + v 2√uv. It follows that ≥ ≥ ≥ 2uvw 2uvw 1 = . (v + w)(w + u)(u + v) ≤ 2√vw 2√wu 2√uv 4 · · Equality holds if and only if u = v = w, i.e., P =(1:1:1)= G, the centroid. The centroid is the interior point with largest cevian triangle. 402 Straight line equations

13.2 Equations of straight lines

13.2.1 Two-point form

The area formula has an easy and extremely important consequence: three points Pi = (ui,vi,wi) are collinear if and only if

u1 v1 w1 u v w =0. 2 2 2 u v w 3 3 3

Consequently, the equation of the line joining two points with coordinates (x1 : y1 : z1) and (x2 : y2 : z2) is

x1 y1 z1 x y z =0, 2 2 2 x y z

(y z y z )x +(z x z x )y +(x y x y )z =0. 1 2 − 2 1 1 2 − 2 1 1 2 − 2 1

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

1 1 1 a b c =0,

x y z

or (b c)x +(c a)y +(a b)z =0 . − − − (4) The equations of some important lines:

Euler line OH (b2 c2)(b2 + c2 a2)x =0 cyclic − − OI-line OI bc(b c)(b + c a)x =0 Pcyclic − − Soddy line IG (b c)(s a)2x =0 e Pcyclic − − Brocard axis OK b2c2(b2 c2)x =0 Pcyclic − van Aubel line HK Sαα(Sβ Sγ)x =0 Pcyclic − P 13.2 Equations of straight lines 403

13.2.2 Intersection of two lines The intersection of the two lines

p1x + q1y + r1z =0, p2x + q2y + r2z =0 is the point (q r q r : r p r p : p q p q ). 1 2 − 2 1 1 2 − 2 1 1 2 − 2 1

Proposition 13.2. Three lines pix + qiy + riz =0, i =1, 2, 3, are concurrent if and only if

p1 q1 r1 p q r =0. 2 2 2 p q r 3 3 3

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) ··· ··· − − − − (s b)2 (s c)2 =( c a)(a b) − − : : − − (c + a)(c2 + a2 b2) (a + b)(a2 + b2 c2) ··· ··· − − (s b)2 a(b c) =(c a)(a b) − − : : − − (c + a)(c2 + a2 b2) (b c)(a + b + c)2 ··· ··· − − (s b)2 a =(b c)(c a)( a b) − : : − − − (c + a)(c2 + a2 b2) (a + b + c)2 ··· ··· − 1 (c + a b)2 4a = (b c)(c a)(a b ) − : : 4 − − − (c + a)(c2 + a2 b2) (a + b + c) 2 ··· ··· − 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). − 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. 404 Straight line equations

13.3 Perspective triangles

Many interesting points and lines in triangle geometry arise from the perspectivity of triangles. We say that two triangles X1Y1Z1 and X2Y2Z2 are perspective, X1Y1Z1 ⊼ X2Y2Z2, if the lines X1X2, Y1Y2, Z1Z2 are concurrent. The point of concurrency, (X1Y1Z1,X2Y2Z2), is called the perspector. Along with the perspector, there is an axis of∧ perspectivity, or the perspectrix, which is the line joining containing

Y Z Z Y ,Z X X Z , X Y Y X . 1 2 ∩ 1 2 1 2 ∩ 1 2 1 2 ∩ 1 2

We denote this line by L (X1Y1Z1,X2Y2Z2). ∧ Homothetic triangles are clearly prespective. If triangles T and T′, their perspector is the homothetic center, which we shall denote by (T, T′). ∧0 Proposition 13.3. 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 (XYZ)=(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 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). It is easy to see that these three points− are− collinear on the− line − − − x y z + + =0. u U v V w W − − − 13.3 Perspective triangles 405

The excentral triangle The excentral triangle is perspective with ABC; the perspector is the incenter I:

Ia = a : b : c − Ib = a : b : c − Ic = a : b : c − I = a : b : c

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

Z Y

? X X b C c b B c

X c

b Yc

Zb These points have coordinates

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

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

Yc = c : 0 : b + c = bc : 0 : b(b + c) − − Zb = b : b + c : 0 = bc : c(b + c): 0 − − X = = bc : c(b + c) : b(b + c) − 406 Straight line equations

Similarly, the coordinates of Y = CZa AXc, and Z = AXb BYa can be determined. The following table shows that the perspector∩ of triangles ABC and∩ XYZ is the point with 1 1 1 homogeneous barycentric coordinates a : b : c . 1 1 1 X = bc : c(b + c) : b(b + c) = b−+c : b : c − 1 1 1 Y = c(c + a) : ca : a(c + a) = a : c−+a : c − 1 1 1 Z = b(a + b) : a(a + b) : ab = a : b : a−+b − 1 1 1 ?= = a : b : c 13.4 Perspectivity 407

13.4 Perspectivity

Y Z Z Y ,Z X X Z , X Y Y X . 1 2 ∩ 1 2 1 2 ∩ 1 2 1 2 ∩ 1 2

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, (XYZ)= (ABC, XY Z) and L (XYZ)= 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 (T, T′). ∧0 Proposition 13.4. 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 (XYZ)=(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 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 XYZ and ABC are homothetic if the perspectrix is the line at inﬁnity. 408 Straight line equations

13.4.1 The Schiffler point: intersection of four Euler lines Theorem 13.5. Let I be the incenter of triangle ABC. The Euler lines of the triangles IBC, ICA, IAB are concurrent at a point on the Euler line of ABC, namely, the Schiffler point a(b + c a) b(c + a b) c(a + b c) S = − : − : − . c b + c c + a a + b Proof. Let I be the incenter of triangle ABC. We first compute the equation of the Euler line of the triangle IBC. The centroid of triangle IBC is the point (a : a+2b+c : a+b+2c). The circumcenter 2 of triangle is the midpoint of IIa. This is the point ( a : b(b + c): c(b + c)). From these we obtain the equation of the Euler line: −

x y z 0= a a +2b + c a + b +2c =(b c)(b + c)x + a(c + a)y a(a + b)z. − − a2 b(b + c) c(b + c) −

The equations of the Euler lines of the triangles ICA and IAB can be obtained by cyclic permutations of a,b,c and x,y,z. Thus the three Euler lines are

(b c)(b + c)x + a(c + a)y a(a + b)z = 0, − b(b + c)x + (c a)(c + a)y −+ b(a + b)z = 0, −c(b + c)x − c(c + a)y + (a b)(a + b)z = 0. − − Computing the intersection of the latter two lines, we have the point

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

=( c + a)(a b)+ bc : b(c + a b): c(b (c a)) − − − − = a(b + c a): b(c + a b): c(a + b c). − − − It is easy to verify that this point also lies on the Euler line of IBC given by the ﬁrst equation:

(b c)a(b + c a)+ ab(c + a b) ac(a + b c) − − − − − = a((b c)(b + c a)+ b(c + a b) c(a + b c)) − − − − − = a(b2 c2 ab + ca + bc + ab b2 ca bc + c2) − − − − − =0.

It is routine to verify that this point also lies on the Euler line of ABC, with equation

(b2 c2)(b2 + c2 a2)x +(c2 a2)(c2 + a2 b2)y +(a2 b2)(a2 + b2 c2)z =0. − − − − − − This shows that the four Euler lines are concurrent. Chapter 14

Cevian nest theorem

14.1 Trilinear pole and polar

14.1.1 Trilinear polar of a point

Given a point P with traces AP , BP , and CP on the sidelines of triangle ABC, let

X = BP CP BC, Y = CP AP CA, Z = AP BP AB. ∩ ∩ ∩ These points X, Y , Z lie on a line called the trilinear polar (or simply tripolar) of P .

C P P

X B AP C

Z 410 Cevian nest theorem

If P = (u : v : w), then BP = (u :0: w) and CP = (u : v : 0). The line BP CP has equation x y z + + =0. −u v w It intersects the sideline BC at the point X =(0: v : w). − Similarly, AP =(0: v : w) and the points Y , Z are

Y =( u :0: w),Z =(u : v : 0). − − The line containing the three points X, Y , Z is x y z + + =0. u v w This is the tripolar of P . 14.1 Trilinear pole and polar 411

14.1.2 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′ are concurrent. The point of concurrency is the tripole P of L.

C′ Y

A ′ X B C

B′

Clearly P is the tripole of L if and only if L is the tripolar of P . 412 Cevian nest theorem

14.2 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.

14.2.1 Construction of anticevian triangle

If the trilinear polar LP of P intersects the sidelines BC, CA, AB at X′, Y ′, Z′ respec- 1 tively, then the anticevian triangle cev− (P ) is simply the triangle bounded by the lines AX′, BY ′, and CZ′.

CP Pc P

X B AP C

Pb 14.2 Anticevian triangles 413

Another construction of anticevian triangle

1 Here is an alternative construction of cev− (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, we have AP A X A A a P ′ H = 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. 414 Cevian nest theorem

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 A′ =( a : b : c ), B′ =(a : b : c ), C′ =(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. (4) The anticevian triangle of the circumcenter. Here is an interesting property of 1 cev− (O). Let the perpendiculars to AC and AB at A intersect BC at Ab and Ac respectively. We call AAbAc an orthial triangle of ABC. The circumcenter of AAbAc is the 1 vertex Oa of cev− (O); similarly for the other two orthial triangles. (See ??). §

Ab Ac B C

Oa 14.3 Cevian quotients 415

14.3 Cevian quotients

14.3.1 The cevian nest theorem Theorem 14.1. For arbitrary points P and Q, the cevian triangle cev(P ) and the antice- 1 vian triangle cev− (Q) are always perspective. If P = (u : v : w) and Q = (u′ : v′ : w′), then (i) the perspector is the point

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

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

Y ′

Z′

M Q Z P

B X C

X′

1 Proof. (i) Let cev(P ) = XYZ and cev− (Q) = X′Y ′Z′. Since X = (0 : v : w) and X′ =( u′ : v′ : w′), the line XX′ has equation − 1 w v 1 1 ′ ′ x y + z =0. u w − v − v · w · ′ The equations of YY ′ and ZZ′ can be easily written down by cyclic permutations of (u,v,w), (u′,v′,w′) and (x,y,z). It is easy to check that the line XX′ contains the point

u′ v′ w′ v′ w′ u′ w′ u′ v′ u′ + + : v′ + + : w′ + + − u v w − v w u − w u v 416 Cevian nest theorem whose coordinates are invariant under the above cyclic permutations. This point therefore also lies on the lines YY ′ and ZZ′.

(ii) The lines YZ and Y ′Z′ have equations

x y z u + v + w = 0, − y z v′ + w′ = 0. They intersect at the point

U ′ =(u(wv′ vw′): vwv′ : vww′). − −

Similarly, the lines pairs ZX, Z′X′ and XY , X′Y ′ have intersections

V ′ =( wuu′ : v(uw′ wu′): wuw′) − − and W ′ =(uvu′ : uvv′ : w(vu′ uv′)). − − The three points U ′, V ′, W ′ lie on the line with equation given above.

Corollary 14.2. 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.

T T T T Remark. Suppose ′ = cev (P ) and ′′ = cev ′ (Q). If P = (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. u − u′ v′ w′ cyclicX These formulae, however, are quite difﬁcult to use, since they involve complicated changes of coordinates with respect to different triangles. We shall simply write

1 P/Q := (cev(P ), cev− (Q)) and call it the cevian quotient of P by QV. 14.3 Cevian quotients 417

The cevian quotients of the centroid G/P If P =(u : v : w),

G/P =(u( u + v + w): v( v + w + u): w( w + u + v)). − − − Some common examples of G/P .

P G/P coordinates IM (a(s a): b(s b): c(s c)) i − − − O K (a2 : b2 : c2) 2 2 2 K O (a Sα : b Sβ : c Sγ)

1 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 . (i) We compute the symmedian point of the excentral triangle. Note that

a2bc ab2c abc2 I I2 = , I I2 = , I I2 = . b c (s b)(s c) c a (s c)(s a) a b (s a)(s b) − − − − − − For the homogeneous barycentric coordinates of the symmedian point of the excentral triangle IaIbIc, we have

2 2 2 IbI Ia + IcI Ib + IaI Ic c · a · b · a2bc ( a,b,c) ab2c (a, b,c) abc2 (a,b, c) = − + − + − (s b)(s c) · 2(s a) (s c)(s a) · 2(s b) (s a)(s b) · 2(s c) − − − − − − − − − abc = (a( a,b,c)+ b(a, b,c)+ c(a,b, c)) 2(s a)(s b)(s c) − − − − − − abc = (a( a + b + c),b(a b + c),c(a + b c)) . 2(s a)(s b)(s c) − − − − − −

From this, we obtain the Mittenpunkt Mi. (ii) Since the excentral triangle is homothetic to the intouch triangle at T , their tangential triangles are homothetic at the same triangle center. Using the cevian nest theorem, we have 1 a b c T = 0(cev(Ge), cev− (I)) = s a : s b : s c . − − − V

1 This appears as X9 in ETC. 418 Cevian nest theorem

The cevian quotients of the orthocenter H/P For P =(u : v : w),

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

Examples

(1) H/G =(Sβ + Sγ Sα : Sγ + Sα Sβ : Sα + Sβ Sγ) is the superior of H•. (2) H/I is a point− on the OI-line, dividing− OI in the− ratio R + r : 2r. 2 − H/I =(a(a3 + a2(b + c) a(b2 + c2) (b + c)(b c)2): : ). − − − ··· ··· 2 2 2 (3) H/K = a : b : c is the homothetic center of the orthic and tangential trian- Sα Sβ Sγ gle. 3 It is a point on the Euler line. (4) H/O is the orthocenter of the tangential triangle. 4 (5) H/N is the orthocenter of the orthic triangle. 5

The cevian quotient Ge/K This is the perspector of the intouch triangle and the tangential triangle 6

G /K =(a2(a3 a2(b + c)+ a(b2 + c2) (b + c)(b c)2): : ). e − − − ··· ···

2 This point appears as X46 in ETC. 3 This appears as X25 in ETC. 4 This appears as X155 in ETC. 5 This appears as X52 in ETC. 6 This appears as X1486 in ETC. 14.3 Cevian quotients 419

14.3.2 Basic properties of cevian quotients Proposition 14.3. (1) P/P = P . (2) If P/Q = M, then Q = P/M.

Y ′

Z′

Z M Q ′′ Z P

B X C

X′′ X′

Y ′′

Proof. (2) Let P =(u : v : w), Q =(u′ : v′ : w′), and M =(x : y : z). We have x u u v w = ′ ′ + ′ + ′ , u u − u v w y v u v w = ′ ′ ′ + ′ , v v u − v w z w u v w = ′ ′ + ′ ′ . w w u v − w From these, x y z u v w u v w + + = ′ ′ + ′ ′ + ′ ′ , −u v w u − v w u v − w x y z u v w u v w + = ′ + ′ + ′ ′ + ′ ′ , u − v w − u v w u v − w x y z u v w u v w + = ′ ′ + ′ ′ ′ + ′ , u v − w − u − v w u − v w 420 Cevian nest theorem and x x y z y x y z z x y z + + : + : + u −u v w v u − v w w u v − w u v w = ′ : ′ : ′ . u v w It follows that x y z x y z x y z u′ : v′ : w′ = x + + : y + : z + . −u v w u − v w u v − w Chapter 15

Circle equations

15.1 The power of a point with respect to a circle

The power of P with respect to a circle Q(ρ) is the quantity P(P ) := PQ2 ρ2. A point P is in, on, or outside the circle according as P is negative, zero, or positive.− Proposition 15.1. Let P = xA + yB + zC in absolute barycentric coordinates. P(P )= xP(A)+ yP(B)+ zP(C) (a2yz + b2zx + c2xy). − A

B X C yB+zC Proof. Let X be the trace of P on BC. In absolute coordinates, X = y+z , so that P = xA +(y + z)X. Applying Stewart’s theorem in succession to triangles QAX, QBC and ABC, we have

PQ2 = xAQ2 +(y + z)XQ2 x(y + z)AX2 − y z yz = xAQ2 +(y + z) BQ2 + CQ2 BC2 x(y + z)AX2 y + z y + z − (y + z)2 − yz = xAQ2 + yBQ2 + zCQ2 BC2 x(y + z)AX2 − y + z − yz = xAQ2 + yBQ2 + zCQ2 BC2 − y + z z y yz x(y + z) AC2 + AB2 BC2 − y + z y + z − (y + z)2 · (1 x)yz = xAQ2 + yBQ2 + zCQ2 − BC2 zx AC2 xy AB2 − y + z · − · − · = xAQ2 + yBQ2 + zCQ2 (a2yz + b2zx + c2xy). − 422 Circle equations

From this it follows that

P(P )= PQ2 ρ2 − = x(AQ2 ρ2)+ y(BQ2 ρ2)+ z(CQ2 ρ2) (a2yz + b2zx + c2xy) − − − − = xP(A)+ yP(B)+ zP(C) a2yz b2zx c2xy. − − −

15.2 Circle equation

Using homogeneous barycentric coordinates for P and writing

f := P(A), g := P(B), h := P(C), for the powers of A, B, C with respect to a circle Q(ρ), we have,

fx + gy + hz a2yz + b2zx + c2xy P(P )= x + y + z − (x + y + z)2 (a2yz + b2zx + c2xy) (x + y + z)(fx + gy + hz) = − . − (x + y + z)2

Therefore, the equation of the circle is

(a2yz + b2zx + c2xy) (x + y + z)(fx + gy + hz)=0. − Example 15.1. (1) The equation of the circumcircle is a2yz + b2zx + c2xy = 0 since f = g = h =0. (2) For the incircle, we have

f =(s a)2, g =(s b)2, h =(s c)2. − − − The equation of the incircle is

a2yz + b2zx + c2xy (x + y + z)((s a)2x +(s b)2y +(s c)2z)=0. − − − − (3) Similarly, the A-excircle has equation

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

b b Sα 1 (4) For the nine-point circle, we have f = 2 c cos A = 2 b = 2 Sα. Similarly, 1 1 · · g = 2 Sβ and h = 2 Sγ. Therefore, the equation of the nine-point circle is

2 2 2 2(a yz + b zx + c xy) (x + y + z)(Sαx + Sβy + Sγz)=0. − 15.3 Points on the circumcircle 423

Exercise 1. Find the equation of the Conway circle.

2. Find the equations of the circles (i) CBBC passing through B and C, and tangent to BC at B, (ii) CBCC passing through B and C, and tangent to BC at C.

3. Compute the coordinates of the Brocard points: (i) Ω as the intersection of the circles CBBC , CCCA, and CAAB, → (ii) Ω as the intersection of the circles CBCC , CCAA, and CABB. ← 4. Find the equation of the circle with diameter BC.

15.3 Points on the circumcircle

The equation of the circumcircle can be written in the form

a2 b2 c2 + + =0. x y z

This shows that the circumcircle consists of the isogonal conjugates of inﬁnite points.

15.3.1 X(101) The point a2 b2 c2 X(101) = : : b c c a a b − − − is clearly on the circumcircle.

15.3.2 X(100) The point a b c X(100) = : : b c c a a b − − − is clearly on the circumcircle. It is the isogonal conjugate of the inﬁnite point

(a(b c): b(c a): c(a b)) − − − x y z (on the trilinear polar of the incenter, namely, the line a + b + c =0). Its inferior is a point on the nine-point circle. To ﬁnd this, we rewrite

X(100) = (a(c a)(a b): b(a b)(b c): c(b c)(c a)). − − − − − − 424 Circle equations

From this,

inf(X(100)) = (b(a b)(b c)+ c(b c)(c a): : ) − − − − ··· ··· = ((b c)(b(a b)+ c(c a)) : : ) − − − ··· ··· = ((b c)2(b + c a): : ), − − ··· ··· the Feuerbach point!

1. The distance from X(100) to the Nagel point is the diameter of the incircle.

2. X(100) is the intersection of the Euler lines of the triangles IaBC, IbCA, IcAB.

15.3.3 The Steiner point X(99) The Steiner point 1 1 1 X(99) = : : b2 c2 c2 a2 a2 b2 − − − The inferior of the Steiner point is

X(115) = ((b2 c2)2 : (c2 a2)2 : (a2 b2)2). − − − This is the midpoint between the Fermat points.

15.3.4 The Euler reﬂection point E = X(110) Theorem 15.2. The reﬂections of the Euler line in the sidelines of triangle ABC are concurrent at a2 b2 c2 E = : : b2 c2 c2 a2 a2 b2 − − − on the circumcircle.

Proof. The Euler line

Sα(Sβ Sγ)x + Sβ(Sγ Sα)y + Sγ(Sα Sβ)z =0 − − − intersects the sideline BC at

X =(0: Sγ(Sα Sβ): Sβ(Sγ Sα). − − − We ﬁnd the reﬂection of H in BC as follows. From the relation

(Sβγ + Sγα + Sαβ)(0,Sγ,Sβ)+ Sβγ(Sβ + Sγ, Sγ, Sβ) − − = (Sβ + Sγ)(Sβγ,Sγα,Sαβ), we have Sβγ H = X + (Sβ + Sγ, Sγ, Sβ). (Sβ + Sγ)(Sβγ + Sγα + Sαβ) − − 15.4 Circumcevian triangle 425

Therefore, its reﬂection in BC is the point

Sβγ X′ = X (Sβ + Sγ, Sγ, Sβ). − (Sβ + Sγ)(Sβγ + Sγα + Sαβ) − − In homogeneous barycentric coordinates, this is

X′ = (Sβγ + Sγα + Sαβ)(0,Sγ,Sβ) Sβγ(Sβ + Sγ, Sγ, Sβ) − − − = ( Sβγ(Sβ + Sγ),Sγ(2Sβγ + Sγα + Sαβ),Sβ(2Sβγ + Sγα + Sαβ)) −

The inferior of the Euler reﬂection is the point

X(125) = ((b2 c2)2(b2 + c2 a2):(c2 a2)2(c2 + a2 b2):(a2 b2)2(a2 + b2 c2)). − − − − − − This is the intersection of the Euler lines of the triangles AY Z, BZX, CXY , where XYZ is the orthic triangle.

15.4 Circumcevian triangle

Let P = (u : v : w). The lines AP , BP , CP intersect the circumcircle again at X, Y , Z. The triangle XYZ is called the circumcevian triangle of P . Since X lies on the line AP , X =(x : v : w) for some x. This point lies on the circumcircle if and only if

a2vw + b2xw + c2xv =0.

a2vw This gives x = b−2w+c2v . Therefore,

X =( a2vw :(b2w + c2v)v :(b2w + c2v)w). − Similarly,

Y = ((c2u+a2w)u : b2wu :(c2u+a2w)w),Z = ((a2v+b2u)u :(a2v+b2u)v : c2uv). − − Proposition 15.3. The circumcevian triangle of P = (u : v : w) is perspective with the tangential triangle at

a4 b4 c4 a2 + + : : . −u2 v2 w2 ··· ··· Proof. The vertices of the tangential triangle are ( a2,b2,c2), (a2, b2,c2), (a2,b2, c2). The line joining ( a2,b2,c2) to X is − − − − x y z a2 b2 c2 =0. − a2vw (b2w + c2v)v (b2w + c2v)w −

426 Circle equations

This is (b4w2 c4v2)x + a2b2w2y a2c2v2z =0. − − Similarly, the lines joining (a2, b2,c2) to Y and (a2,b2, c2) to Z are − − a2b2w2x +(c4u2 a4w2)y + b2c2u2z = 0, − − a2b2v2x b2c2u2y +(b4v2 b4u2)z = 0. − − These three lines concur at a point with coordinates given above.

Example 15.2. 1. G: X(22) = (a2( a4 + b4 + c4): : ). − ··· ··· a2(a4+b4+c4 2a2b2 2a2c2) − − 2. H: X(24) = b2+c2 a2 : : . − ··· ··· These two points are on the Euler line, and are the centers of similitude of the circumcircle and incircle of the tangential triangle.

15.5 The third Lemoine circle

Given a point P =(u : v : w), it is easy to ﬁnd the equation of the circle Ca through P , B, C. Since P(B)= P(C)=0, the equation of the circle is

2 2 2 Ca : a yz + b zx + c xy (x + y + z) fx =0 − · for some f. Since the circle passes through P =(u : v : w), we must have

a2vw + b2wu + c2uv f = . u(u + v + w)

This circle Ca intersects the lines AC and AB each again at another point. To ﬁnd the 2 intersection with AC, we put y =0 in the equation of (Ca) and obtain b zx fx(x+z)=0, 2 − x((b f)z fx))=0. Therefore, apart from C = (0, 0, 1), the circle Ca intersects AC at − − 2 2 2 2 2 2 2 2 2 Ba =(b f :0: f)=(b u + b uv a vw c uv :0: a vw + b wu + c uv). − − −

Similarly, the circle Ca intersects AB again at

2 2 2 2 2 2 2 2 2 Ca =(c f : f :0)=(c u + c wu a vw b wu : a vw + b wu + c uv : 0). − − − Similarly, with

a2vw + b2wu + c2uv a2vw + b2wu + c2uv g = and h = , v(u + v + w) w(u + v + w) the circles Cb through P , C, A and Cc through P , A, B intersect the sidelines again at

2 2 2 2 Cb =(g : c g : 0),Ab =(0: a g : g),Ac =(0: h : a h), Bc =(h :0: b h). − − − − 15.5 The third Lemoine circle 427

Note the lengths of the segments:

f f b2 h b2 h ABa = b = , ABc = − b = − , b2 · b b2 · b and f f c2 g c2 g ACa = c = , ACb = − c = − . c2 · c c2 · c The four points Ba, Bc, Ca, Cb are concyclic if and only if

f(b2 h) f(c2 g) b2 h v2 ABa ABc = ACa ACb = − = − = = = . · · ⇒ b2 c2 ⇒ c2 g w2

c2 w Likewise, the four points Cb, Ca, Ab, Ac are concyclic if and only if a2 = u , and the a2 u four points Ac, Ab, Bc, Ba are concyclic if and only if b2 = v . By the principle of 6 concyclic points, the six points Ab, Ac, Bc, Ba, Ca, Cb are concyclic if and only if u : v : w = a2 : b2 : c2, namely, P = (u : v : w)=(a2 : b2 : c2), the symmedian point. The circle C containing these 6 points is the third Lemoine circle. For this choice of P ,

3b2c2 3c2a2 3a2b2 f = , g = , h = . a2 + b2 + c2 a2 + b2 + c2 a2 + b2 + c2 With respect to the circle C containing these 6 points, we have

f(b2 h) 3b2c2 b2(b2 + c2 2a2) 3b2c2(b2 + c2 2a2) P(A)= − = · − = − . b2 b2(a2 + b2 + c2)2 (a2 + b2 + c2)2 Similarly,

3c2a2(c2 + a2 2b2) 3a2b2(a2 + b2 2c2) P(B)= − , P(C)= − . (a2 + b2 + c2)2 (a2 + b2 + c2)2 From these, we obtain the equation of the third Lemoine circle:

(a2 + b2 + c2)2(a2yz + b2zx + c2xy) 3(x + y + z) b2c2(b2 + c2 2a2)x + c2a2(c2 + a2 2b2)y + a2b2(a2 + b2 2c2)z =0. − − − −