Advanced Euclidean Geometrymath.fau.edu/Yiu/AEG2013/2013AEG.pdf · 2013-08-02 · Advanced Euclidean Geometry Paul Yiu Summer 2013 Department of Mathematics Florida Atlantic University

Advanced Euclidean Geometry

Paul Yiu

Summer 2013

Department of MathematicsFlorida Atlantic University

a

bc

A

B C

August 2, 2013

Summer 2013

Contents

1 Some Basic Theorems 1011.1 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1011.2 Constructions of geometric mean . . . . . . . . . . . . . . . . . . . . . . . . . . 1041.3 The golden ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

1.3.1 The regular pentagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1061.4 Basic construction principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

1.4.1 Perpendicular bisector locus . . . . . . . . . . . . . . . . . . . . . . . . . 1081.4.2 Angle bisector locus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1091.4.3 Tangency of circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1101.4.4 Construction of tangents of a circle . . . . . . . . . . . . . . . . . . . . . 110

1.5 The intersecting chords theorem . . . . . . . . . . . . . . . . . . . . . . . . . .. 1121.6 Ptolemy’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

2 The laws of sines and cosines 1152.1 The law of sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1152.2 The orthocenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1162.3 The law of cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1172.4 The centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1202.5 The angle bisector theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

2.5.1 The lengths of the bisectors . . . . . . . . . . . . . . . . . . . . . . . . . 1212.6 The circle of Apollonius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

3 The tritangent circles 1253.1 The incircle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.2 Euler’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.3 Steiner’s porism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.4 The excircles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1303.5 Heron’s formula for the area of a triangle . . . . . . . . . . . . . . . . . . . .. . 131

4 The arbelos 1334.1 Archimedes’ twin circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

4.1.1 Harmonic mean and the equation1a+ 1

b= 1

t. . . . . . . . . . . . . . . . 134

4.1.2 Construction of the Archimedean twin circles . . . . . . . . . . . . . . . . 1344.2 The incircle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

4.2.1 Construction of the incircle of the arbelos . . . . . . . . . . . . . . . . . . 1364.2.2 Alternative constructions of the incircle . . . . . . . . . . . . . . . . . . . 138

4.3 Archimedean circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

iv CONTENTS

5 Menelaus’ theorem 2015.1 Menelaus’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2015.2 Centers of similitude of two circles . . . . . . . . . . . . . . . . . . . . . . . . . 203

5.2.1 Desargue’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2035.3 Ceva’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2045.4 Some triangle centers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

5.4.1 The centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2055.4.2 The incenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2055.4.3 The Gergonne point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2065.4.4 The Nagel point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

5.5 Isotomic conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

6 The Euler line and the nine-point circle 2116.1 The Euler line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

6.1.1 Inferior and superior triangles . . . . . . . . . . . . . . . . . . . . . . . . 2116.1.2 The orthocenter and the Euler line . . . . . . . . . . . . . . . . . . . . . . 212

6.2 The nine-point circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2136.3 Distances between triangle centers . . . . . . . . . . . . . . . . . . . . . . . . .215

6.3.1 Distance between the circumcenter and orthocenter . . . . . . . . . . . . .2156.3.2 Distance between circumcenter and tritangent centers . . . . . . . . . . .. 2166.3.3 Distance between orthocenter and tritangent centers . . . . . . . . . . .. . 217

6.4 Feuerbach’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 219

7 Isogonal conjugates 2217.1 Directed angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2217.2 Isogonal conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2227.3 The symmedian point and the centroid . . . . . . . . . . . . . . . . . . . . . . . 2237.4 Isogonal conjugates of the Gergonne and Nagel points . . . . . . . . .. . . . . . 225

7.4.1 The Gergonne point and the insimilicenterT+ . . . . . . . . . . . . . . . . 2257.4.2 The Nagel point and the exsimilicenterT− . . . . . . . . . . . . . . . . . 227

7.5 The Brocard points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2287.6 Kariya’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2307.7 Isogonal conjugate of an infinite point . . . . . . . . . . . . . . . . . . . . . .. 233

8 The excentral and intouch triangles 2358.1 The excentral triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

8.1.1 The circumcenter of the excentral triangle . . . . . . . . . . . . . . . . . . 2358.1.2 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 3019.1 Absolute and homogeneous barycentric coordinates . . . . . . . . . . . .. . . . 301

9.1.1 The centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3019.1.2 The incenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3029.1.3 The barycenter of the perimeter . . . . . . . . . . . . . . . . . . . . . . . 3039.1.4 The Gergonne point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

9.2 Cevian triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3059.2.1 The circumcenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306

CONTENTS v

9.2.2 The Nagel point and the extouch triangle . . . . . . . . . . . . . . . . . . 3079.2.3 The orthocenter and the orthic triangle . . . . . . . . . . . . . . . . . . . 308

9.3 Homotheties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3099.3.1 Superiors and inferiors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3099.3.2 Triangles bounded by lines parallel to the sidelines . . . . . . . . . . . . . 312

9.4 Infinite points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

10 Some applications of barycentric coordinates 31510.1 Construction of mixtilinear incircles . . . . . . . . . . . . . . . . . . . . . . . . 315

10.1.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle . 31510.1.2 Mixtilinear incircles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316

10.2 Isotomic and isogonal conjugates . . . . . . . . . . . . . . . . . . . . . . . . .. 31710.3 Isotomic conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31710.4 Equal-parallelians point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .319

11 Computation of barycentric coordinates 32111.1 The Feuerbach point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 32111.2 TheOI line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

11.2.1 The circumcenter of the excentral triangle . . . . . . . . . . . . . . . . . .32311.2.2 The centers of similitude of the circumcircle and the incircle . . . . . . . . 32311.2.3 The homothetic centerT of excentral and intouch triangles . . . . . . . . . 324

11.3 The excentral triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32511.3.1 The centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32511.3.2 The incenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325

12 Some interesting circles 32712.1 A fundamental principle on6 concyclic points . . . . . . . . . . . . . . . . . . . 327

12.1.1 The radical axis of two circles . . . . . . . . . . . . . . . . . . . . . . . . 32712.1.2 Test for6 concyclic points . . . . . . . . . . . . . . . . . . . . . . . . . . 328

12.2 The Taylor circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32912.3 Two Lemoine circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

12.3.1 The first Lemoine circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 33012.3.2 The second Lemoine circle . . . . . . . . . . . . . . . . . . . . . . . . . . 33112.3.3 Construction ofK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33112.3.4 The center of the first Lemoine circle . . . . . . . . . . . . . . . . . . . . 332

13 Straight line equations 40113.1 Area and barycentric coordinates . . . . . . . . . . . . . . . . . . . . . . .. . . 40113.2 Equations of straight lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .402

13.2.1 Two-point form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40213.2.2 Intersection of two lines . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

13.3 Perspective triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 40413.3.1 The Conway configuration . . . . . . . . . . . . . . . . . . . . . . . . . . 405

13.4 Perspectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40713.4.1 The Schiffler point: intersection of four Euler lines . . . . . . . . . . . .. 408

vi CONTENTS

14 Cevian nest theorem 40914.1 Trilinear pole and polar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409

14.1.1 Trilinear polar of a point . . . . . . . . . . . . . . . . . . . . . . . . . . . 40914.1.2 Tripole of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411

14.2 Anticevian triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41214.2.1 Construction of anticevian triangle . . . . . . . . . . . . . . . . . . . . . . 412

14.3 Cevian quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41514.3.1 The cevian nest theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 41514.3.2 Basic properties of cevian quotients . . . . . . . . . . . . . . . . . . . . . 419

15 Circle equations 42115.1 The power of a point with respect to a circle . . . . . . . . . . . . . . . . . .. . 42115.2 Circle equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42215.3 Points on the circumcircle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

15.3.1 X(101) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42315.3.2 X(100) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42315.3.3 The Steiner pointX(99) . . . . . . . . . . . . . . . . . . . . . . . . . . . 42415.3.4 The Euler reflection pointE = X(110) . . . . . . . . . . . . . . . . . . . 424

15.4 Circumcevian triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42515.5 The third Lemoine circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426

Chapter 1

Some Basic Theorems

1.1 The Pythagorean Theorem

Theorem 1.1(Pythagoras). The lengthsa ≤ b < c of the sides of a right triangle satisfythe relation

a2 + b2 = c2.

Proof.

a b

a

b

ab

a

b

4 1

23

4

1

2

3

a

a

b

b

Theorem 1.2 (Converse of Pythagoras’ theorem). If the lengths of the sides ofABCsatisfya2 + b2 = c2, then the triangle has a right angle atC.

ac

b

a

bC X

B Y

A Z

Proof. Consider a right triangleXY Z with ∠Z = 90, Y Z = a, andXZ = b. By thePythagorean theorem,XY 2 = Y Z2 + XZ2 = a2 + b2 = c2 = AB2. It follows thatXY = AB, andABC ≡ XY Z by the SSS test, and∠C = ∠Z = 90.

102 Some Basic Theorems

Example 1.1.Trigonometric ratios of30, 45, and60:

2

1

√3

60

1 1

1√2

45

θ sin θ cos θ tan θ

30 1

2

√32

√33

45

√22

√22 1

60

√32

12

√3

Example 1.2. For an arbitrary pointP on theminor arc BC of the circumcircle of anequilateral triangleABC, AP = BP + CP .

B C

A

P

Q

Proof. If Q is the point onAP such thatPQ = PC, then the isosceles triangleCPQ isequilateral since∠CPQ = ∠CBA = 60. Note that∠ACQ = ∠BCP . Thus,ACQ ≡BCP by the SAS congruence test. From this,AQ = BP , andAP = AQ + QP =BP + CP .

1.1 The Pythagorean Theorem 103

Example 1.3.Given a rectangleABCD, to construct pointsP onBC andQ onCD suchthat triangleAPQ is equilateral.

Let BCY andCDX are equilateral triangles inside the rectangleABCD. Extend thelinesAX andAY to intersectBC andCD respectively atP andQ. APQ is equilateral.

P

Q

Y

X

A B

CD

Proof. SupposeAB = 2a andBC = 2b. The distance ofX aboveAB = 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, AQ2 = (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), andPQ2 = CP 2 + CQ2 = 4((

√3a − b)2 + (

√3b − a)2) = 16(a2 +

b2 −√3ab). It follows thatAP = AQ = PQ, and triangleAPQ is equilateral.


1.2 Constructions of geometric mean

We present two ruler-and-compass constructions of the geometric means of two quanti-ties given as lengths of segments. These are based on Euclid’s proof of the Pythagoreantheorem.

Construct the altitude at the right angle to meetAB atP and the opposite sideZZ ′ ofthe squareABZZ ′ at Q. Note that the area of the rectangleAZQP is twice of the areaof triangleAZC. By rotating this triangle aboutA through a right angle, we obtain thecongruent triangleABY , whose area is half of the area of the square onAC. It follows thatthe area of rectangleAZQP is equal to the area of the square onAC. For the same reason,the area of rectangleBZ ′QP is equal to that of the square onBC. From these, the area ofthe square onAB is equal to the sum of the areas of the squares onBC andCA.

Q

PA B

C

Z Z′

Y

Y ′

X′

X

Construction.Given two segments of lengtha < b, mark three pointsP , A, B on a linesuch thatPA = a, PB = b, andA, B are on thesameside ofP . Describe a semicirclewith PB as diameter, and let the perpendicular throughA intersect the semicircle atQ.ThenPQ2 = PA · PB, so that the length ofPQ is the geometric mean ofa andb.

a

√ab

PA

B

Q

b

P

AB

Q

1.2 Constructions of geometric mean 105

Construction.Given two segments of lengtha, b, mark three pointsA, P , B on a line (PbetweenA andB) such thatAP = a, PB = b. Describe a semicircle withAB as diameter,and let the perpendicular throughP intersect the semicircle atQ. ThenPQ2 = PA · PB,so that the length ofPQ is the geometric mean ofa andb.

a b

√ab

AP

B

Q

A

PB

Q


1.3 The golden ratio

Given a segmentAB, a pointP in the segment is said to divide it in the golden ratio ifAP 2 = PB · AB. Equivalently,AP

PB=√5+12

. We shall denote this golden ratio byϕ. It isthe positive root of the quadratic equationx2 = x+ 1.

A BP BA

M

P

Q

Construction(Division of a segment in the golden ratio). Given a segmentAB,(1) draw a right triangleABM with BM perpendicular toAB and half in length,(2) mark a pointQ on the hypotenuseAM such thatMQ = MB,(3) mark a pointP on the segmentAB such thatAP = AQ.

ThenP dividesAB into the golden ratio.

SupposePB has unit length. The lengthϕ of AP satisfies

ϕ2 = ϕ+ 1.

This equation can be rearranged as

(

ϕ− 1

2

)2

=5

4.

Sinceϕ > 1, we have

ϕ =1

2

(√5 + 1

)

.

Note thatAP

AB=

ϕ

ϕ+ 1=

1

ϕ=

2√5 + 1

=

√5− 1

2.

This explains the construction above.

1.3.1 The regular pentagon

Consider a regular pentagonACBDE. It is clear that the five diagonals all have equallengths. Note that(1)∠ACB = 108,

1.3 The golden ratio 107

A

E D

BP

C

(2) triangleCAB is isosceles, and(3)∠CAB = ∠CBA = (180 − 108)÷ 2 = 36.

In fact, each diagonal makes a36 angle with one side, and a72 angle with another.It follows that

(4) trianglePBC is isosceles with∠PBC = ∠PCB = 36,(5)∠BPC = 180 − 2× 36 = 108, and(6) trianglesCAB andPBC are similar.

Note that triangleACP is also isosceles since(7)∠ACP = ∠APC = 72. This means thatAP = AC.

Now, from the similarity ofCAB andPBC, we haveAB : AC = BC : PB. In otherwordsAB · AP = AP · PB, orAP 2 = AB · PB. This means thatP dividesAB in thegolden ratio.

Construction.Given a segmentAB, we construct a regular pentagonACBDE with ABas a diagonal.(1) DivideAB in the golden ratio atP .(2) Construct the circlesA(P ) andP (B), and letC be an intersection of these two circles.(3) Construct the circlesA(AB) andB(C) to intersect at a pointD on the same side ofBC asA.(4) Construct the circlesA(P ) andD(P ) to intersect atE.

ThenACBDE is a regular pentagon withAB as a diagonal.


1.4 Basic construction principles

1.4.1 Perpendicular bisector locus

A variable pointP is equidistant from two fixed pointsA andB if and only ifP lies on theperpendicular bisector of the segmentAB.

The perpendicular bisectors of the three sides of a triangleare concurrent at thecircum-centerof the triangle. This is the center of the circumcircle, the circle passing through thethree vertices of the triangle.

O

A

B C

OB C

A

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 pointP is equidistant from two fixed linesℓ andℓ′ if and only if P lies on thebisector of one of the angles betweenℓ andℓ′.

The bisectors of the three angles of a triangle are concurrent at a point which is at equaldistances from the three sides. With this point as center, a circle can be constructed tangentto the sides of the triangle. This is theincircle of the triangle. The center is theincenter.

I

X

Y

Z

A

B C

Proof. Let the bisectors of anglesB andC intersect atI. Consider the pedals ofI on thethree sides. SinceI is on the bisector of angleB, IX = IZ. SinceI is also on the bisectorof angleC, IX = IY . It follows IX = IY = IZ, and the circle, centerI, constructedthroughX, also passes throughY andZ, and is tangent to the three sides of the triangle.


1.4.3 Tangency of circles

Two circles(O) and (O′) are tangent to each other if they are tangent to a lineℓ at thesame lineP , which is a common point of the circles. The tangency is internal or externalaccording as the circles are on the same or different sides ofthe common tangentℓ.

O IT O IT ′

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 circleO(A), thetangent to a circle atA is the perpendicular to the radiusOA atA.

A

O

A

O

B

PM

If P is a point outside a circle(O), there are two lines throughP tangent to the circle.Construct the circle withOP as diameter to intersect(O) at two points. These are thepoints of tangency.

The two tangents have equal lengths since the trianglesOAP andOBP are congruentby the RHS test.

1.4 Basic construction principles 111

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

x A B

K

T

Y

X

Let AB = a. Suppose the required circle has radiusr, andAT = x, whereT is thepoint of tangencyT with the center line.

(a+ x)2 + r2 = (a+ r)2,

x2 + r2 = (a− r)2.

From these, we have

x+a

2=

√3

2· a,

a

2+ x = 2r.

This means that ifM is the midpoint ofAB, thenMT =√32· a, which is the height of

the equilateral triangle onAB. In other words, ifC is an intersection of the two given cir-cles, thenCM andMT are two adjacent sides of a square. Furthermore, the side oppositetoCM is a diameter of the required circle!

x a

2

√3a

2

A B

K

T

Y

XC

D

M


1.5 The intersecting chords theorem

Theorem 1.3. Given a pointP and a circleO(r), if a line throughP intersects the circleat two pointsA andB, thenPA · PB = OP 2 − r2, independent of the line.

P

A

B

M

O

P

A

B

M

O

Proof. Let M be the midpoint ofAB. Note thatOM is perpendicular toAB. If P isoutside the circle, then

PA · PB = (PM +MA)(PM −BM)

= (PM +MA)(PM −MA)

= PM2 −MA2

= (OM2 + PM2)− (OM2 +MA2)

= OP 2 − r2.

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

The quantityOP 2− r2 is called thepower of P with respect to the circle. It is positive,zero, or negative according asP is outside, on, or inside the circle.

Corollary 1.4 (Intersecting chords theorem). If two chordsAB andCD of a circle inter-sect, extended if necessary, at a pointP , thenPA · PB = PC · PD.

In particular, if the tangent atT intersectsAB at P , thenPA · PB = PT 2.

P

A

B

O

C

D

T

P

A

B

O

C

D

The converse of the intersecting chords theorem is also true.

1.5 The intersecting chords theorem 113

Theorem 1.5. Given four pointsA, B, C, D, if the linesAB andCD intersect at a pointP such thatPA · PB = PC · PD (as signed products), thenA, B, C, D are concyclic.

In particular, if P is a point on a lineAB, andT is a point outside the lineAB suchthatPA · PB = PT 2, thenPT is tangent to the circle throughA, B, T .

Example 1.5.LetABC be a triangle withB = 2C. Thenb2 = c(c+ a).

A

B C

F

E

Construct a parallel throughC to the bisectorBE, to intersect the extension ofAB atF . Then

∠AFC = ∠ABE =1

2· ∠ABC = ∠ACB.

This means thatAC is tangent to the circle throughB, C, F . By the intersecting chordtheorem,AC2 = AB · AF , i.e., b2 = c(c+ a).


1.6 Ptolemy’s theorem

Theorem 1.6(Ptolemy). A convex quadrilateralABCD is cyclic if and only if

AB · CD + AD ·BC = AC ·BD.

AD

B C

P

dA

D

B C

P ′

Proof. (Necessity) Assume, without loss of generality, that∠BAD > ∠ABD. Choose apointP on the diagonalBD such that∠BAP = ∠CAD. TrianglesBAP andCAD aresimilar, since∠ABP = ∠ACD. It follows thatAB : AC = BP : CD, and

AB · CD = AC ·BP.

Now, trianglesABC andAPD are also similar, since∠BAC = ∠BAP + ∠PAC =∠DAC + ∠PAC = ∠PAD, and∠ACB = ∠ADP . It follows thatAC : BC = AD :PD, and

BC · AD = AC · PD.

Combining the two equations, we have

AB · CD + BC · AD = AC(BP + PD) = AC ·BD.

(Sufficiency). LetABCD be a quadrilateral satisfying (**). Locate a pointP ′ suchthat∠BAP ′ = ∠CAD and∠ABP ′ = ∠ACD. Then the trianglesABP andACD aresimilar. It follows that

AB : AP ′ : BP ′ = AC : AD : CD.

From this we conclude that(i) AB · CD = AC · BP ′, and(ii) trianglesABC andAP ′D are similar since∠BAC = ∠P ′AD andAB : 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 pointP ′ lies on diagonalBD. From this,∠ABD = ∠ABP ′ = ∠ACD, and the pointsA, 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). LetR denote the circumradius of a triangleABC.

2R =a

sinα=

b

sin β=

c

sin γ.

O

C

A

BD

EF O

C

A

BD

α

α

Since the area of a triangle is given by∆ = 12bc sinα, the circumradius can be written

as

R =abc

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 triangleA′B′C ′. NotethatABCB′ andACBC ′ are both parallelograms since each has two pairs of parallelsides.It follows thatB′A = BC = AC ′ andA is the midpoint ofB′C ′.

H

X

Y

Z

A

B C

A′

B′C′

Consider the altitudeAX of triangleABC. Seen in triangleA′B′C ′, this line is theperpendicular bisector ofB′C ′ since it is perpendicular toB′C ′ through its midpointA.Similarly, the altitudesBY andCZ of triangleABC are perpendicular bisectors ofC ′A′

andA′B′. As such, the three linesAX, BY , CZ concur at a pointH. This is called theorthocenter of triangleABC.

Proposition 2.2. The reflections of the orthocenter in the sidelines lie on thecircumcircle.

O

Oa

A

B C

Ha

H

Proof. It is enough to show that the reflectionHa of H in BC lies on the circumcircle.Consider also the reflectionOa of O in BC. SinceAH andOOa are parallel and have thesame length (2R cosα), AOOaH is a parallelogram. On the other hand,HOOaHa is aisosceles trapezoid. It follows thatOHa = HOa = AO, andHa lies on the circumcircle.

2.3 The law of cosines 117

2.3 The law of cosines

Given a triangleABC, we denote bya, b, c the lengths of the sidesBC, CA, AB respec-tively.

Theorem 2.3(The law of cosines).

c2 = a2 + b2 − 2ab cos γ.

a

bc

A

B CX a

b

c

A

B C X

Proof. LetAX be the altitude onBC.

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). LetX be a point on the sidelineBC of triangleABC.

a · AX2 = BX · b2 +XC · c2 − a ·BX ·XC.

Here, the lengths of the directed segments on the lineBC are signed. Equivalently, ifBX : XC = λ : µ, then

AX2 =λb2 + µc2

λ+ µ− λµa2

(λ+ µ)2.

a

bc

A

B CX

Proof. Use the cosine formula to compute the cosines of the anglesAXB andAXC, andnote thatcosAXC = − cosAXB.


Example 2.1. (Napoleon’s theorem). If similar isosceles trianglesXBC, Y CA andZAB(of base angleθ) are constructed externally on the sides of triangleABC, the lengthsof the segmentsY Z, ZX, XZ can be computed easily. For example, in triangleAY Z,AY = b

2sec θ, AZ = c

2sec θ and∠Y AZ = α + 2θ.

A

B C

Z

X

Yθ

θ

θ

θ θ

θ

By the law of cosines,

Y Z2 = AY 2 + AZ2 − 2AY · AZ · cosY AZ

=sec2 θ

4(b2 + c2 − 2bc cos(α + 2θ))

=sec2 θ

4(b2 + c2 − 2bc cosα cos 2θ + 2bc sinα sin 2θ)

=sec2 θ

4(b2 + c2 − (b2 + c2 − a2) cos 2θ + 4∆ sin 2θ)

=sec2 θ

4(a2 cos 2θ + (b2 + c2)(1− cos 2θ) + 4∆ sin 2θ).

Likewise, we have

ZX2 =sec2 θ

4(b2 cos 2θ + (c2 + a2)(1− cos 2θ) + 4∆ sin 2θ),

XY 2 =sec2 θ

4(c2 cos 2θ + (a2 + b2)(1− cos 2θ) + 4∆ sin 2θ).

It is easy to note thatY Z = ZX = XY if and only if cos 2θ = 12, i.e., θ = 30.

In this case, the pointsX, Y , Z are the centers of equilateral triangles erected externallyon BC, CA, AB respectively. The same conclusion holds if the equilateraltriangles areconstructed 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 centersare the vertices of an equilateraltriangle.

2.3 The law of cosines 119

A

B C

Z

X

Y

A

B C

Z′

X′

Y ′

Example 2.2. (Orthogonal circles) Given three pointsA, B, C that form an acute-angledtriangle, construct three circles with these points as centers that are mutually orthogonal toeach other.

A

B C

Y

X

Z

HF

D

E

Let BC = a, CA = b, andAB = c. If these circles have radiiRa, Rb, Rc respectively,then

R2b +R2

c = a2, R2c +R2

a = b2, R2a +R2

b = c2.

From these,

R2a =

1

2(b2 + c2 − a2), R2

b =1

2(c2 + a2 − b2), R2

c =1

2(a2 + b2 − c2).

These are all positive sinceABC is an acute triangle. Consider the perpendicular footE ofB onAC. Note thatAE = c cosA, so thatR2

a =12(b2+ c2−a2) = bc cosA = AC ·AE. It

follows if we extendBE to intersect atY the semicircle constructed externally on the sideAC as diameter, then,AY 2 = AC · AE = R2

a. Therefore we have the following simpleconstruction 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. LabeltheseX, Y , Z on the semicircles onBC, CA, AB respectively. These satisfyAY = AZ,BZ = BX, andCX = CY .(3) The circlesA(Y ), B(Z) andC(X) are mutually orthogonal to each other.


2.4 The centroid

Let E andF be the midpoints ofAC andAB respectively, andG the intersection of themediansBE andCF .

Construct the parallel throughC to BE, and extendAG to intersectBC atD, and thisparallel atH. .

A

B C

FE

D

G

H

By the converse of the midpoint theorem,G is the midpoint ofAH, andHC = 2 ·GEJoinBH. By the midpoint theorem,BH//CF . It follows thatBHCG is a parallel-

ogram. Therefore,D is the midpoint of (the diagonal)BC, andAD is also a median oftriangleABC. We have shown that the three medians of triangleABC intersect atG,which we call thecentroid of the triangle.

Furthermore,

AG =GH = 2GD,

BG =HC = 2GE,

CG =HB = 2GF.

The centroidG divides each median in the ratio2 : 1.

Theorem 2.6(Apollonius). If ma denotes the length of the median on the sideBC,

m2a =

1

4(2b2 + 2c2 − a2).

Example 2.3.Suppose the mediansBE andCF of triangleABC are perpendicular. Thismeans thatBG2 + CG2 = BC2, whereG is the centroid of the triangle. In terms of thelengths, we have4

9m2

b+49m2

c = a2; 4(m2b+m2

c) = 9a2; (2c2+2a2−b2)+(2a2+2b2−c2) =9a2; b2 + c2 = 5a2.

This relation is enough to describe, given pointsB andC, the locus ofA for which themediansBE andCF of triangleABC are perpendicular. Here, however, is a very easyconstruction: Fromb2 + c2 = 5a2, we havem2

a =14(2b2 +2c2 − a2) = 9

4a2; ma =

32a. The

locus ofA is the circle with center at the midpoint ofBC, and radius32·BC.

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 itsopposite side in the ratio of the remaining sides. IfAX andAX ′ respectively the internaland external bisectors of angleBAC, thenBX : XC = c : b andBX ′ : X ′C = c : −b.

c b

Z′

Z

A

B CX X′

Proof. Construct lines throughC parallel to the bisectorsAX andAX ′ to intersect the lineAB atZ andZ ′.

(1) Note that∠AZC = ∠BAX = ∠XAC = ∠ACZ. This meansAZ = AC. Clearly,BX : XC = BA : AZ = BA : AC = c : b.

(2) Similarly,AZ ′ = AC, andBX ′ : 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 angleA arerespectively

ta =2bc

b+ ccos

α

2and t′a =

2bc

|b− c| sinα

2.

c bta t′a

A

B CX X′

Proof. LetAX andAX ′ be the bisectors of angleA.(1) Consider the area of triangleABC as the sum of those of trianglesAXC andABX.

We have1

2ta(b+ c) sin

α

2=

1

2bc sinα.


From this,

ta =bc

b+ c· sinαsin α

2

=2bc

b+ c· cos α

2.

(2) Consider the area of triangle as the difference between those ofABX ′ andACX ′.

Remarks.(1) 2bcb+c

is the harmonic mean ofb andc. It can be constructed as follows. If theperpendicular toAX atX intersectsAC andAB atY andZ, thenAY = AZ = 2bc

b+c.

c b

ta

Y

Z

A

BC

X

(2) Applying Stewart’s Theorem withλ = c andµ = ±b, we also obtain the followingexpressions for the lengths of the angle bisectors:

t2a = bc

(

1−(

a

b+ c

)2)

,

t′2a = bc

(

(

a

b− c

)2

− 1

)

.

Example 2.4. (Steiner-Lehmus theorem). A triangle with two equal angle bisectors isisosceles. More precisely, if the bisectors of two angles ofa triangle have equal lengths,then the two angles are equal.

Proof. We show that ifa < b, thenta > tb. Note that froma < b we conclude(i) α < β andcos α

2> cos β

2;

(ii) bc > ac, b(c+ a) > a(b+ c); bb+c

> ac+a

; 2bcb+c

> 2cac+a

.From (i) and (ii), we have

ta =2bc

b+ c· cos α

2>

2ca

c+ a· cos β

2= tb.

The same reasoning shows thata > b ⇒ ta < tb. It follows that if ta = tb, thena = b.

2.6 The circle of Apollonius 123

2.6 The circle of Apollonius

Theorem 2.9.A andB are two fixed points. For a given positive numberk 6= 1, 1 thelocus of pointsP satisfyingAP : PB = k : 1 is the circle with diameterXY , whereXandY are points on the lineAB such thatAX : XB = k : 1 andAY : Y B = k : −1.

AB

XY

P

O B′

Proof. Sincek 6= 1, pointsX andY can be found on the lineAB satisfying the aboveconditions.

Consider a pointP not on the lineAB with AP : PB = k : 1. Note thatPX andPYare respectively the internal and external bisectors of angle APB. This means that angleXPY is a right angle, andP lies on the circle withXY as diameter.

Conversely, letP be a point on this circle. We show thatAP : BP = k : 1. LetB′ be a point on the lineAB such thatPX bisects angleAPB′. SincePA andPB areperpendicular to each other, the linePB is the external bisector of angleAPB′, and

AY

Y B′= − AX

XB′=

XA

XB′=

AY −XA

YX.

On the other hand,AY

Y B= −AX

XB=

XA

XB=

AY −XA

YX.

Comparison of the two expressions shows thatB′ coincides withB, andPX is the bisectorof angleAPB. It follows that PA

PB= AX

XB= k.

1If k = 1, the locus is clearly the perpendicular bisector of the segmentAB.


Chapter 3

The tritangent circles

3.1 The incircle

Let the incircle of triangleABC touch its sidesBC, CA, AB atX, Y , Z respectively.

s − b s − c

s − c

s − a

s − a

s − b

I

X

Y

Z

A

B C

If s denotes the semiperimeter of triangleABC, then

AY =AZ = s− a,

BZ =BX = s− b,

CX =CY = s− c.

The inradius of triangleABC is the radius of its incircle. It is given by

r =2∆

a+ b+ c=

∆

s.

126 The tritangent circles

Example 3.1. If triangleABC has a right angle atC, then the inradiusr = s− c.

s − a

s − b

s − c s − a

s − b

s − c

r

r

r

A

B

C

I

It follows that if d is the diameter of the incircle, thena+ b = c+ d.

Example 3.2.An equilateral triangle of side2a is partitioned symmetrically into a quadri-lateral, an isosceles triangle, and two other congruent triangles. If the inradii of the quadri-lateral and the isosceles triangle are equal, find this inradius. 1

A

B C

X

Y

θ

r

r

Suppose each side of the equilateral triangle has length 2, each of the congruent circleshas radiusr, and∠ACX = θ.

(i) From triangleAXC, r = 2cot 30+cot θ

.(ii) Note that∠BCY = 1

2(60 − 2θ) = 30 − θ. It follows thatr = tan(30 − θ) =

1cot(30−θ) =

cot θ−cot 30cot 30 cot θ+1

.

By puttingcot θ = x, we have 2√3+x

= x−√3√

3x+1; x2−3 = 2

√3x+2; x2−2

√3x−5 = 0,

andx =√3 + 2

√2. (The negative root is rejected). From this,r = 2√

3+x= 1√

3+√2=√

3−√2.

To construct the circles, it is enough to markY on the altitude throughA such thatAY =

√3− r =

√2. The construction is now evident.

1(√3−

√2)a.

3.1 The incircle 127

Example 3.3. (Conway’s circle). Given triangleABC, extend(i) CA andBA to Ya andZa such thatAYa = AZa = a,(ii) AB andCB toZb andXb such thatBZb = BXb = b,(iii) BC andAC toXc andYc such thatCXc = CYc = c.

The six pointsXb, Xc, Yc, Ya, Za, Zb are concyclic. The circle containing them hascenterI and radius

√r2 + s2.

a

bc

b

b

c

c

a

a

X

Y

Z

A

B CXcXb

Za

Zb

Yc

Ya

I

r

Proof. Let the incircle be tangent toBC atX. BX = s− b =⇒ XbX = b+(s− b) = s.From this,IX2

b = r2 + s2. Similarly, each of the remaining five points is at the samedistance fromI. They lie on a circle, centerI, radius

√r2 + s2.


3.2 Euler’s formula

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

OI

M

A

BC

Proof. ExtendAI to intersect the circumcircle atM . Since∠BAM = ∠MAC = A2, M

is the midpoint of the arcBC of the circumcircle. Note that

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

It follows that IM = BM . By the law of sines,BM = 2R sin A2. SinceAI = r

sin A2

,

AI · IM = 2Rr. This is equal to the power ofI with respect to the circumcircle, namely,= R2 −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 triangleABC with incircle (I) and circumcircleO), let A′ be an arbitrary pointon circumcircle. JoinA′ to I, to intersect the circumcircle again atM ′, and letA′Y ′, A′Z ′

be the tangents to the incircle. Construct a circle, centerM ′, throughI to intersect thecircumcircle atB′ andC ′.

OI

A

B C

A′

B′

C′

M ′

Z′

Y ′

Denote byθ betweenA′I and each tangent.It is known thatA′I · IM ′ = 2Rr (power of incenter in(O)). SinceA′I = r

sin θ, we

haveIM ′ = 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 thatA′B′ andA′C ′ are tangent to the incircle(I).SinceM ′ is the midpoint of the arcB′C ′, the circleM ′(B′) passes through the incenter oftriangleA′B′C ′. This means thatI is the incenter, and(I) the incircle of triangleA′B′C ′.


3.4 The excircles

The internal bisector of each angle and theexternalbisectors of the remaining two anglesare concurrent at anexcenterof the triangle. Anexcirclecan be constructed with this ascenter, tangent to the lines containing the three sides of the triangle.

Z

X

Y

ra

ra

ra

Ic

Ib

Ia

C

A

B

The exradii of a triangle with sidesa, b, c are given by

ra =∆

s− a, rb =

∆

s− b, rc =

∆

s− c.

The areas of the trianglesIaBC, IaCA, andIaAB are 12ara, 1

2bra, and 1

2cra respectively.

Since∆ = −∆IaBC +∆IaCA+∆IaAB,

we have

∆ =1

2ra(−a+ b+ c) = ra(s− a),

from whichra = ∆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 triangleABC with area∆. Denote byr the inradius, andra the radius of theexcircleon the sideBC of triangleABC. It is convenient to introduce thesemiperimeters = 1

2(a+ b+ c).

Ia

YY ′

I

A

B

C

ra r

(1) From the similarity of trianglesAIY andAI ′Y ′,

r

ra=

s− a

s.

(2) From the similarity of trianglesCIY andI ′CY ′,

r · ra = (s− b)(s− c).

(3) From these,

r =

√

(s− a)(s− b)(s− c)

s,

ra =

√

s(s− b)(s− c)

s− a.

Theorem 3.3(Heron’s formula).

∆ =√

s(s− a)(s− b)(s− c).

Proof. ∆ = rs.

Proposition 3.4.

tanα

2=

√

(s− b)(s− c)

s(s− a), cos

α

2=

√

s(s− a)

bc, sin

α

2=

√

(s− b)(s− c)

bc.


Chapter 4

The arbelos

4.1 Archimedes’ twin circles

Theorem 4.1(Archimedes). The two circles each tangent toCP , the largest semicircleAB and one of the smaller semicircles have equal radiit, given by

t =ab

a+ b.

A BOO1 O2P A BOO1 O2P

Q

Proof. Consider the circle tangent to the semicirclesO(a + b), O1(a), and the linePQ.Denote byt the radius of this circle. Calculating in two ways the height of the center ofthis circle above the lineAB, we have

(a+ b− t)2 − (a− b− t)2 = (a+ t)2 − (a− t)2.

From this,

t =ab

a+ b.

The symmetry of this expression ina andb means that the circle tangent toO(a+b),O2(b),andPQ has the same radiust.

134 The arbelos

4.1.1 Harmonic mean and the equation1a+ 1

b= 1

t

The harmonic mean of two quantitiesa andb is 2aba+b

. In a trapezoid of parallel sidesa andb,the parallel through the intersection of the diagonals intercepts a segment whose length isthe harmonic mean ofa andb. We shall write this harmonic mean as2t, so that1

a+ 1

b= 1

t.

b

a

A B

CD

t

ab

Here is another construction oft, making use of the formula for the length of an anglebisector in a triangle. IfBC = a, AC = b, then the angle bisectorCZ has length

tc =2ab

a+ bcos

C

2= 2t cos

C

2.

The lengtht can therefore be constructed by completing the rhombusCXZY (by con-structing the perpendicular bisector ofCZ to intersectBC atX andAC atY ). In particu-lar, if the triangle contains a right angle, this trapezoid is a square.

A B

C

M

Z

X

Y

tt

a

bt

t

4.1.2 Construction of the Archimedean twin circles

Construct the circleP (C3) to intersect the diameterAB atP1 andP2 so thatP1 is onAPandP2 is onPB. The centerC1 respectivelyC2 is the intersection of the circleO1(P2)respectivelyO2(P1) and the perpendicular toAB atP1 respectivelyP2.

A BOO1 O2P

C3

Q1

Q2

P2P1

C1

C2

Q


4.2 The incircle

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

ρ =r1r2(r1 + r2)

r21 + r1r2 + r22.

A BOO1 O2C

O3

X

Y

θ

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

(r1 + ρ)2 =(r1 + r2 − ρ)2 + r22 + 2r2(r1 + r2 − ρ) cos θ,

(r2 + ρ)2 =(r1 + r2 − ρ)2 + r21 − 2r1(r1 + r2 − ρ) cos θ.

Eliminatingθ, we have

r1(r1 + ρ)2 + r2(r2 + ρ)2 = (r1 + r2)(r1 + r2 − ρ)2 + r1r22 + r2r

21.

The coefficients ofρ2 on both sides are clearly the same. This is a linear equation in ρ:

r31 + r32 + 2(r21 + r22)ρ = (r1 + r2)3 + r1r2(r1 + r2)− 2(r1 + r2)

2ρ,

from which

4(r21 + r1r2 + r22)ρ = (r1 + r2)3 + r1r2(r1 + r2)− (r31 + r32) = 4r1r2(r1 + r2),

andρ is as above.

136 The arbelos

4.2.1 Construction of the incircle of the arbelos

In [?], Bankoff published the following remarkable theorem whichgives a construction ofthe incircle of the arbelos of the incircle, much simpler than the one we designed beforefrom Archimedes’ proof. The simplicity of the constructionis due to the existence of acircle congruent to Archimedes’ twin circles.

Theorem 4.3 (Bankoff). The points of tangency of the incircle of the arbelos with thesemicircles(AC) and (CB), together withC, are the points of tangency of the incircle(W3) of triangleO1O2O3 with the sides of the triangle. This circle(W3) is congruent toArchimedes’ twin circles(W1) and(W2).

A BOO1 O2C

O3

P

Q

R

W3

Proof. SinceO1Q = O1C, O2C = O2R, andO3R = O3Q, the pointsC, Q, R are thepoints of tangency of the incircle of triangleO1O2O3 with its sides. The semi-perimeter ofthe triangle is

s = r1 + r2 + ρ = r1 + r2 +r1r2(r1 + r2)

r21 + r2r2 + r22=

(r1 + r2)3

r21 + r2r2 + r22=

(r1 + r2)2ρ

r1r2.

The inradius of the triangle is thesquare rootof

r1r2ρ

s=

r21r22

(r1 + r2)2= t2.

It follows that this inradius ist. The incircle of triangleO1O2O3 is congruent to Archimedes’twin circles.


Construction.Let M andN be the midpoints of the semicircles(AC) and(CB) respec-tively. Construct(1) the linesO1N andO2M to intersect atW3,(2) the circle with centerW3, passing throughC to intersect the semicircle(AC) atQ and(CB) atR,(3) the linesO1Q andO2R to intersect atO3.

The circle with centerO3 passing throughQ touches the semicircle(CB) atR and alsothe semicircle(AB).

A BOO1 O2C

O3

P

Q R

W3

M

N

138 The arbelos

4.2.2 Alternative constructions of the incircle

Theorem 4.4(Bankoff). LetP be the intersection (apart fromC) of the circumcircles ofthe squares onAC andCB. LetQ be the intersection (apart fromC) of the circumcircle ofthe square onCB and the semicircle(AC), andR that of the circumcircle of the square onAB and the semicircle(CB). The pointsP , Q, R are the points of tangency of the incircleof the arbelos with the semicircles.

A BOO1 O2C

O3

P

Q R

M

N

L

Proposition 4.5. The intersectionS of the linesAN andBM also lies on the incircle ofthe arbelos, and the lineCS intersects(AB) at P .

A BOO1 O2C

O3

P

Q R

M

N

L

S


A BCOO1 O2

O3

P

Q R

O

L′

Construction.LetL′ be the “lowest point” of the circle(AB). Construct(1) the lineL′C to intersect the semicircle(AB) atP ,(2) the circle, centerL′, throughA andB, to intersect the semicircles(AC) and(CB) atQ andR.

Proposition 4.6. LetX be the midpoint of the side of the square onAC opposite toAC,andY that of the side of the square onCB opposite toCB. The centerO3 of the incircleof the arbelos is the intersection of the linesAY andBX.

A BOO1 O2C

O3

P

QR

M

N

L

X

Y

140 The arbelos

4.3 Archimedean circles

We shall call a circle Archimedean if it is congruent to Archimedes’ twin circle,i.e., withradiust = r1r2

r1+r2, and has further remarkable geometric properties.

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

A O1 OO′C O2 B

M2

M1

M

M ′

D

W3

2. (van Lamoem) The circle tangent toAB at O and to the midway semicircle isArchimedean.2

A O1 OO′C O2 B

M2

M1

M

W?

3. (Schoch) LetMN intersectCD andOL at Q andK respectively. The smallestcircles throughQ andK tangent to the semicircle(AB) are Archimedean.

A O1 O C O2 B

N

M

L D

W20W21

QK

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) isArchimedean.(b) The smallest circle throughC tangent toAB is Archimedean.

BCA

D

OO1 O2

F

E

W4

W11

5. Let EF be the common tangent of the semicircles(AC) and(CB). The smallestcircles throughE andF tangent toCD are Archimedean.

BCA

D

OO1 O2

F

E W9

W10

6. (Schoch) LetX andY be the intersections of the semicircle(AB) with the circlesthroughC, with centersA andB respectively. The smallest circle throughX andYtangent toCD are Archimedean.

A BOO1 O2C

D

X

W14W13

Y

Y ′

142 The arbelos

7. (van Lamoen) LetY andZ be the intersections of the midway semicircle with thesemicircles(AC) and(CB). The circles with centersY andZ, each tangent to thelineCD, are Archimedean.

A BOO1 O2C

D

X2

Y

Z

8. (Schoch) (a) The circle tangent to the semicircle(AB) and the circular arcs, withcentersA andB respectively, each passing throughC, is Archimedean.

(b) The circle with center on the Schoch line and tangent to both semicircles(AC)and(CB) is Archimedean.

A BOO1 O2C

W15

W16

9. (Woo) Letα be a positive real number. Consider the two circular arcs, each passingthroughC and with centers(−αr1, 0) and (αr2, 0) respectively. The circle withcenterUα 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 atD (the intersection with the common tangent of(AC) and(CB)).

A BOO1 O2C

W15

W28

D

4.3 Archimedean circles 143

10. (Power) Consider an arbelos with inner semicirclesC1 andC2 of radii a andb, andouter semicircleC of radiusa + b. It is known the Archimedean circles have radiust = ab

a+b. LetQ1 andQ2 be the “highest” points ofC1 andC2 respectively.

A circle tangent to(O) internally and toOQ1 atQ1 (orOQ2 atQ2) is Archimedean.

C1

C2

C′1

C′2

A BOO1 O2P

Q1

Q2a

b

r

a + b − r

11. (van Lamoen)

U1

U2

D

M

A BOO1 O2C

M1

M2

12. (Bui)

T1

T2

D

M

O′A BOO1 O2C

T1

T2

D

M

A BOO1 O2C

Chapter 5

Menelaus’ theorem

5.1 Menelaus’ theorem

Theorem 5.1(Menelaus). Given a triangleABC with pointsX, Y , Z on the side linesBC, CA, AB respectively, the pointsX, Y , Z are collinear if and only if

BX

XC· CY

Y A· AZZB

= −1.

A

B CX

Y

Z

W

Proof. (=⇒) LetW be the point onAC such thatBW//XY . Then,

BX

XC=

WY

Y C, and

AZ

ZB=

AY

YW.

It follows thatBX

XC· CY

Y A· AZZB

=WY

Y C· CY

Y A· AY

YW=

CY

Y C· AYY A

· WY

YW= −1.

(⇐=) Suppose the line joiningX andZ intersectsAC atY ′. From above,

BX

XC· CY ′

Y ′A· AZZB

= −1 =BX

XC· CY

Y A· AZZB

.

It follows thatCY ′

Y ′A=

CY

Y A.

The pointsY ′ andY divide the segmentCA in the same ratio. These must be the samepoint, andX, Y , Z are collinear.

202 Menelaus’ theorem

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

a

bc

B

C

A

Y ′

Z′

X′

Proof. If the external bisectors areAX ′, BY ′, CZ ′ with X ′, Y ′, Z ′ on BC, CA, ABrespectively, then

BX ′

X ′C= −c

b,

CY ′

Y ′A= −a

c,

AZ ′

Z ′B= − b

a.

It follows that BX′

X′C· CY ′

Y ′A· AZ′

Z′B= −1 and the pointsX ′, Y ′, Z ′ are collinear.

5.2 Centers of similitude of two circles 203

5.2 Centers of similitude of two circles

Given two circlesO(R) and I(r), whose centersO and I are at a distanced apart, weanimate a pointX on O(R) and construct a ray throughI oppositely parallel to the rayOX to intersect the circleI(r) at a pointY . The line joiningX andY intersects the lineOI of centers at a pointT which satisfies

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

This pointT is independent of the choice ofX. It is called theinternal center of similitude,or simply the insimilicenter, of the two circles.

O I

XY ′

Y

T T ′

If, on the other hand, we construct a ray throughI directly parallel to the rayOX tointersect the circleI(r) atY ′, the lineXY ′ always intersectsOI at another pointT ′. This istheexternal center of similitude, or simply the exsimilicenter, of the two circles. It dividesthe segmentOI in the ratioOT ′ : T ′I = R : −r.

5.2.1 Desargue’s theorem

Given three circles with centersA, B, C and distinct radii, show that the exsimilicenters ofthe three pairs of circles are collinear.

A

B

C

X

Y

Z


5.3 Ceva’s theorem

Theorem 5.2(Ceva). Given a triangleABC with pointsX, Y , Z on the side linesBC,CA, AB respectively, the linesAX, BY , CZ are concurrent if and only if

BX

XC· CY

Y A· AZZB

= +1.

P

X

Y

Z

A

B C

Proof. (=⇒) Suppose the linesAX, BY , CZ intersect at a pointP . Consider the lineBPY cutting the sides of triangleCAX. By Menelaus’ theorem,

CY

Y A· APPX

· XB

BC= −1, or

CY

Y A· PA

XP· BX

BC= +1.

Also, consider the lineCPZ cutting the sides of triangleABX. By Menelaus’ theoremagain,

AZ

ZB· BC

CX· XP

PA= −1, or

AZ

ZB· BC

XC· XP

PA= +1.

Multiplying the two equations together, we have

CY

Y A· AZZB

· BX

XC= +1.

(⇐=) 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 sidesBC, CA, AB of triangleABC, then clearly

AF

FB· BD

DC· CE

EA= 1.

The mediansAD, BE, CF are therefore concurrent. Their intersection is thecentroidGof the triangle.

Consider triangleADC with transversalBGE. By Menelaus’ theorem,

−1 =AG

GD· DB

BC· CE

EA=

AG

GD· −1

2· 11.

It follows thatAG : GD = 2 : 1. The centroid of a triangle divides each median in theratio 2:1.

CB

A

G

F

D

E

CB

A

I

Z

X

Y

5.4.2 The incenter

LetX, Y , Z be points onBC, CA, AB such thatAX, BY , CZ bisect anglesBAC, CBAandACB respectively. Then

AZ

ZB=

b

a,

BX

XC=

c

b,

CY

Y A=

a

c.

It follows thatAZ

ZB· BX

XC· CY

Y A=

b

a· cb· ac= +1,

andAX, BY , CZ are concurrent. Their intersection is theincenterof the triangle.Applying Menelaus’ theorem to triangleABX with transversalCIZ, we have

−1 =AI

IX· XC

CB· BZ

ZA=

AI

IX· −b

b+ c· ab

=⇒ AI

IX=

b+ c

a.


5.4.3 The Gergonne point

Let the incircle of triangleABC be tangent to the sidesBC atX, CA atY , andAB atZrespectively. SinceAY = AZ = s− a, BZ = BX = s− b, andCX = CY = s− c, wehave

BX

XC· CY

Y A· AZZB

=s− b

s− c· s− c

s− a· s− a

s− b= 1.

By Ceva’s theorem, the linesAX, BY , CZ are concurrent. The intersection is called theGergonne pointGe of the triangle.

s − b s − c

s − c

s − a

s − a

s − b

CB

A

IGe

Z

X

Y

Lemma 5.3. The Gergonne pointGe divides the cevianAX in the ratio

AGe

GeX=

a(s− a)

(s− b)(s− c).

Proof. Applying Menelaus’ theorem to triangleABX with transversalCGeZ, we have

−1 =AGe

GeX· XC

CB· BZ

ZA=

AGe

GeX· −(s− c)

a· s− b

s− a=⇒ AGe

GeX=

a(s− a)

(s− b)(s− c).

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, thelinesAX ′, BY ′, CZ ′ are concurrent by Ceva’s theorem:

BX ′

X ′C· CY ′

Y ′A· AZ

′

Z ′B=

s− c

s− b· s− c

s− a· s− a

s− b= 1.

The point of concurrency is the Nagel pointNa.

A

B C

Ic

Ia

Ib

Z′

X′

Y ′

s− c s− b

s− a

s− cs− b

s− a Na

Lemma 5.4. If the A-excircle of triangleABC touchesBC at X ′, then the Nagel pointdivides the cevianAX ′ in the ratio

ANa

NaX ′=

a

s− a.

Proof. Applying Menelaus’ theorem to triangleACX ′ with transversalBNaY′, we have

−1 =ANa

NaX ′· X

′B

BC· CY ′

Y ′A=

ANa

NaX ′· −(s− c)

a· s− a

s− c=⇒ ANa

NaX ′=

a

s− a.


5.5 Isotomic conjugates

Given pointsX on BC, Y on CA, andZ on AB, we consider their reflections in themidpoints of the respective sides. These are the pointsX ′ onBC, Y ′ onCA andZ ′ onABsatisfying

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 ifAX ′, BY ′, CZ ′ are concurrent.

P

Z

X

Y

A

B C

Z′

X′

Y ′

P •

Proof.(

BX

XC· CY

Y A· AZZB

)(

BX ′

X ′C· CY ′

Y ′A· AZ

′

Z ′B

)

=

(

BX

XC· BX ′

X ′C

)(

CY

Y A· CY ′

Y ′A

)(

AZ

ZB· AZ

′

Z ′B

)

= 1.

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)

I

X

Y

Z

X′

A

B C

Ia

Ib

Ic

Y ′

Z′

GeNa

Example 5.3. (The isotomic conjugate of the orthocenter) LetH• denote the isotomicconjugate of the orthocenterH. Its traces are the pedals of the reflection ofH in O. Thislatter point is the deLongchamps pointLo.

O

H•

Z′

X′

Y ′

A

B C

H

X

Y

Z

Lo


Example 5.4. (Yff-Brocard points) Consider a pointP = (u : v : w) with tracesX, Y , ZsatisfyingBX = CY = AZ = µ. This means that

w

v + wa =

u

w + ub =

v

u+ vc = µ.

Elimination ofu, v, w leads to

0 =

∣

∣

∣

∣

∣

∣

0 −µ a− µb− µ 0 −µ−µ c− µ 0

∣

∣

∣

∣

∣

∣

= (a− µ)(b− µ)(c− µ)− µ3.

Indeed,µ is the unique positive root of the cubic polynomial

(a− t)(b− t)(c− t)− t3.

This gives the point

P =

(

(

c− µ

b− µ

)13

:

(

a− µ

c− µ

)13

:

(

b− µ

a− µ

)13

)

.

X

YZ

A

B C

P

X′

Y ′Z′

A

B C

P •

The isotomic conjugate

P • =

(

(

b− µ

c− µ

)13

:

(

c− µ

a− µ

)13

:

(

a− µ

b− µ

)13

)

has tracesX ′, Y ′, Z ′ that satisfy

CX ′ = AY ′ = BZ ′ = µ.

These points are called theYff-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

G

D

EF

A

B C

G

A′

B′C′ A

B C

The inferior triangleof ABC is the triangleDEF whose vertices are the midpoints ofthe sidesBC, CA, AB.

The two triangles share the same centroidG, and are homothetic atG with ratio−1 : 2.The superior triangleof ABC is the triangleA′B′C ′ bounded by the parallels of the

sides through the opposite vertices.The two triangles also share the same centroidG, and are homothetic atG 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 analtitude of triangleABC 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 theorthocenter of the given triangle.

OG

H

A′

B′C′ A

B C

The circumcenter, centroid, and orthocenter of a triangle are collinear. This is becausethe orthocenter, being the circumcenter of the superior triangle, is the image of the circum-center under the homothetyh(G,−2). The line containing them is called theEuler line ofthe 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 whosecenter 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.

N O

H

D

EF

G

F ′

D′

E′

X

Y

Z

A

B C

Proof. (1) LetN be the circumcenter of the inferior triangleDEF . SinceDEF andABCare homothetic atG in the ratio1 : 2, N , G, O are collinear, andNG : GO = 1 : 2. SinceHG : GO = 2 : 1, the four are collinear, and

HN : NG : GO = 3 : 1 : 2,

andN is the midpoint ofOH.(2) Let X be the pedal ofH on BC. SinceN is the midpoint ofOH, the pedal of

N is the midpoint ofDX. Therefore,N lies on the perpendicular bisector ofDX, andNX = ND. Similarly,NE = NY , andNF = NZ for the pedals ofH onCA andABrespectively. This means that the circumcircle ofDEF also containsX, Y , Z.

(3) LetD′, E ′, F ′ be the midpoints ofAH, BH, CH respectively. The triangleD′E ′F ′

is homothetic toABC atH in the ratio1 : 2. Denote byN ′ its circumcenter. The pointsN ′,G, O are collinear, andN ′G : GO = 1 : 2. It follows thatN ′ = N , and the circumcircle ofDEF also containsD′, E ′, F ′.

This circle is called thenine-point circle of triangleABC. Its centerN is called thenine-point center. Its radius is half of the circumradius ofABC.


Theorem 6.2. Let Oa, Ob, Oc be the reflections of the circumcenterO in the sidelinesBC,CA, AB respectively.(1) The circle throughOa, Ob, Oc is congruent to the circumcircle and has center at theorthocenterH.(2) The reflections ofH in the sidelines lie on the circumcircle.

O

H

A

B C

Ha

Oa

D

Proof. (1) If D is the midpoint ofBC, OOa = 2OD = AH. This means thatAHOaO isa parallelogram, andHOa = AO. Similarly,HOb = BO andHOc = CO for the othertwo reflections. Therefore, andH is the center of the circle throughOa, Ob, Oc, and thecircle is congruent to the circumcircle.

(2) If Ha is the reflection ofH in BC, thenHHaOAO is a trapezoid symmetric in theline BC. Therefore,OHa = HOa = OA. This means thatHa lies on the circumcircle; sodoHb andHc.

6.3 Distances between triangle centers 215

6.3 Distances between triangle centers

6.3.1 Distance between the circumcenter and orthocenter

N

O

H

D

EF

X

Y

Z

A

B C

Proposition 6.3.OH2 = R2(1− 8 cosα cos β cos γ).

Proof. In triangleAOH, AO = R, AH = 2R cosα, and∠OAH = |β− γ|. By the law ofcosines,

OH2 = R2(1 + 4 cos2 α− 4 cosα cos(β − γ))

= R2(1− 4 cosα(cos(β + γ) + cos(β − γ))

= R2(1− 8 cosα cos β cos γ).


6.3.2 Distance between circumcenter and tritangent centers

Lemma 6.4. If the bisector of angleA intersects the circumcircle atM , thenM is thecenter of the circle throughB, I, C, andIa.

Proof. (1) SinceM is the midpoint of the arcBC, ∠MBC = ∠MCB = ∠MAB. There-fore,

∠MBI = ∠MBC + ∠CBI = ∠MAB + ∠IBA = ∠MIB,

andMB = MI. Similarly,MC = MI.(2) On the other hand, since∠IBIa andICIa are both right angles, the four pointsB,

I, C, IaM are concyclic, with center at the midpoint ofIIA. This is the pointM .

Theorem 6.5(Euler). (a)OI2 = R2 − 2Rr.(b)OI2a = R2 + 2Rra.

ra

Y ′

OI

Y

Ia

M

A

BC

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

R2 −OI2 = AI · IM = AI ·MB =r

sin α2

· 2R · sin α

2= 2Rr.

(b) Consider the power ofIa in the circumcircle.Note thatIaA = ra

sin α2

. Also, IaM = MB = 2R sin α2.

OI2a = R2 + IaA · IaM= R2 +

rasin α

2

· 2R sinα

2

= R2 + 2Rra.

6.3 Distances between triangle centers 217

6.3.3 Distance between orthocenter and tritangent centers

Proposition 6.6.

HI2 = 2r2 − 4R2 cosα cos β cos γ,

HI2a = 2r2a − 4R2 cosα cos β cos γ.

H

X

I

A

B C

Proof. In triangleAIH, we haveAH = 2R cosα, AI = 4R sin β

2sin γ

2and∠HAI =

|β−γ|2

. By the law of cosines,

HI2 = AH2 +AI2 − 2AI ·AH · cos β − γ

2

= 4R2

(

cos2 α+ 4 sin2β

2sin2

γ

2− 4 cosα sin

β

2sin

γ

2cos

β − γ

2

)

= 4R2

(

cos2 α+ 4 sin2β

2sin2

γ

2− 4 cosα sin

β

2sin

γ

2cos

β

2cos

γ

2− 4 cosα sin2

β

2sin2

γ

2

)

= 4R2

(

cos2 α+ 4 sin2β

2sin2

γ

2− cosα sinβ sin γ − 4

(

1− 2 sin2α

2

)

sin2β

2sin2

γ

2

)

= 4R2

(

cosα(cosα− sinβ sin γ) + 8 sin2α

2sin2

β

2sin2

γ

2

)

= 4R2

(

− cosα cosβ cos γ + 8 sin2α

2sin2

β

2sin2

γ

2

)

= 2r2 − 4R2 cosα cosβ cos γ.


(2) In triangleAHIa, AIa = 4R cos β

2cos γ

2.

IH

Aa

Ba

Ca

A

BC

Ia

By the law of cosines, we have

HI2a = AH2 +AI2a − 2AIa ·AH · cos β − γ

2

= 4R2

(

cos2 α+ 4 cos2β

2cos2

γ

2− 4 cosα cos

β

2cos

γ

2cos

β − γ

2

)

= 4R2

(

cos2 α+ 4 cos2β

2cos2

γ

2− 4 cosα cos2

β

2cos2

γ

2− 4 cosα cos

β

2cos

γ

2sin

β

2sin

γ

2

)

= 4R2

(

cos2 α+ 4 cos2β

2cos2

γ

2− 4

(

1− 2 sin2α

2

)

cos2β

2cos2

γ

2− cosα sinβ sin γ

)

= 4R2

(

cosα(cosα− sinβ sin γ) + 8 sin2α

2cos2

β

2cos2

γ

2

)

= 4R2

(

− cosα cosβ cos γ + 8 sin2α

2cos2

β

2cos2

γ

2

)

= 2r2a − 4R2 cosα cosβ 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 incircleandexternally to each of the excircles.

HN

I

Ia

O

Proof. (1) SinceN is the midpoint ofOH, IN is a median of triangleIOH. By Apollo-nius’ theorem,

NI2 =1

2(IH2 +OI2)− 1

4OH2

=1

4R2 −Rr + r2

=

(

R

2− r

)2

.

Therefore,NI is the difference between the radii of the nine-point circleand the incir-cle. This shows that the two circles are tangent to each otherinternally.

(2) Similarly, in triangleIaOH,

NI2a =1

2(HI2a +OI2a)−

1

4OH2

=1

4R2 +Rra + r2a

=

(

R

2+ ra

)2

.

This shows that the distance between the centers of the nine-point and an excircle is thesum of their radii. The two circles are tangent externally.

The point of tangencyFe of the incircle and the nine-point circle is called theFeuerbachpoint.


A

B C

Ia

Ib

Ic

Aa

Ba

Ca

Ab

Bb

Cb

Ac

Bc

Cc

Fc

Fa

FbFe

I

N

Chapter 7

Isogonal conjugates

7.1 Directed angles

A reference triangleABC in a plane induces anorientationof the plane, with respect towhich all angles aresigned. For two given linesL andL′, the directed angle∠(L, L

′)between them is the angle of rotation fromL to L

′ in the induced orientation of the plane.It takes values of moduloπ. The following basic properties of directed angles make manygeometric 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 linesL1, L2 andL3.(3) Four pointsP , Q, X, Y are concyclic if and only if∠(PX,XQ) = ∠(PY, Y Q).

Remark.In calculations with directed angles, we shall slightly abuse notations by using theequality sign instead of the sign for congruence moduloπ. It is understood that directedangles are defined up to multiples ofπ. For example, we shall writeβ + γ = −α eventhough it should be more properlyβ + γ = π − α or β + γ ≡ −α mod π.

Exercise

1. If a, b, c are the sidelines of triangleABC, then∠(a, b) = −γ etc.

222 Isogonal conjugates

7.2 Isogonal conjugates

LetP be a given point. Consider the reflections of the ceviansAP , BP , CP in the respec-tive bisectors of anglesA, B, C, i.e., By Ceva’s theorem, these reflections are concurrent.Their intersection is theisogonal conjugateof P .

LetP andQ be isogonal conjugates,AP andAQ intersectingBC atX andX ′ respec-tively. Then

BX

XC· BX ′

X ′C=

c2

b2.

Example 7.1. The incenter is the isogonal conjugate of itself. The same istrue for theexcenters.

Example 7.2. (The circumcenter and orthocenter) For a given triangle with circumcenterO, the lineOA and the altitude throughA are isogonal lines, similarly for the circumradiiand altitudes throughB andC. Since the circumradii are concurrent atO, the altitudes alsoare concurrent. Their intersection is theorthocenterH, which is the isogonal conjugate ofO.

O

H

X

Y

Z

A

B CD

EF

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 thesymmedians. The isogonal conjugate ofthe centroidG is called thesymmedian pointK of the triangle.

A

B C

GK

Consider triangleABC together with itstangential triangleA′B′C ′, the triangle boundedby the tangents of the circumcircle at the vertices.

O

A

B C

A′

B′

C′

Z

Y

D

K

SinceA′ is equidistant fromB andC, we construct the circleA′(B) = A′(C) andextend the sidesAB andAC to meet this circle again atZ andY respectively. Note that

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


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 thatY , A′ andZ ′ are collinear, so that(i) AA′ is a median of triangleAY Z,(ii) AY Z andABC are similar.It follows thatAA′ is the isogonal line of theA-median,i.e., a symmedian. Similarly, theBB′ andCC ′ are the symmedians isogonal toB- andC-medians. The linesAA′, 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 insimilicenterT+

Consider the intouch triangleDEF of triangleABC.

(1) If D′ is the reflection ofD in the bisectorAI, then(i) D′ is a point on the incircle, and(ii) the linesAD andAD′ are isogonal with respect toA.

I

D

E

F

A

B C

D′

D

I

D

E

F

A

B C

D′

E′F ′

Ge P

(2) Likewise,E ′ andF ′ are the reflections ofE andF in the bisectorsBI andCIrespectively, then(i) these are points on the incircle,(ii) the linesBE ′ andCF ′ are isogonals ofBE andCF with respect to anglesB andC.

Therefore, the linesAD′, BE ′, andCF ′ concur at the isogonal conjugate of the Ger-gonne point.

(3) In fact,E ′F ′ is parallel toBC.

I

D

E

F

A

B C

D′

E′F ′

This follows from


(ID, IE′) =(ID, IE) + (IE, IE′)

=(ID, IE) + 2(IE, IB)

=(ID, IE) + 2((IE, AC) + (AC, IB))

=(ID, IE) + 2(AC, IB) since (IE,AC) =π

2

=(π − γ) + 2

(

γ +β

2

)

=β + γ = −α (mod π);

(ID, IF ′) =(ID, IF ) + (IF, IF ′)

=(ID, IF ) + 2(IF, IC)

=(ID, IF ) + 2((IF, AB) + (AB, IC))

=(ID, IF ) + 2(AB, IC) since (IF,AB) =π

2

=− (π − β)− 2(

β +γ

2

)

=− (β + γ) = α (mod π)

SinceE ′ andF ′ are on the incircle, andID ⊥ BC, it follows thatE ′F ′ is parallel toBC.

(4) Similarly,F ′D′ andD′E ′ are parallel toCA andAB respectively. It follows thatD′E ′F ′ is homothetic toABC.

The ratio of homothety isr : R. Therefore, the center of homothety is the pointT+

which dividesOI in the ratioR : r. This is theinternal center of similitude, or simply theinsimilicenterof (O) and(I).

I

D

E

FO

A

B C

D′

E′F ′

Ge

T+

7.4 Isogonal conjugates of the Gergonne and Nagel points 227

7.4.2 The Nagel point and the exsimilicenterT−

The isogonal conjugate of the Nagel point is the pointT− which dividesOI in the ratioOT− : T−I = R : −r. This is theexternal center of similitude(or exsimilicenter) of thecircumcircle and the incircle.

Aa

Bb

Cc

IO

A

B C

NaT−

Ia

Ib

IcD′

1

E′1

F ′1


7.5 The Brocard points

Analogous to the Crelles points, we may ask if there are concurrent lines through the ver-tices making equal angles with the sidelines. More precisely, given triangleABC, doesthere exist a pointP satisfying

∠BAP = ∠CBP = ∠ACP = ω.

It turns out that is one such unique configuration.

A

B C

P

ω

ωω

Note that ifP is a point satisfying∠BAP = ∠CBP , then the circle throughP , A, Bis tangent toBC atB. This circle is unique and can be constructed as follows. Itscenter isthe intersection of the perpendicular bisector ofAB and the perpendicular toBC atB.

A

B C

P

ω

ω

Likewise, if ∠CBP = ∠ACP , then the circle throughP , B, C is tangent toCA atC. It follows thatP is the intersection of these two circles. With thisP , the circlePCA istangent toAB atA.

By Ceva’s theorem, the angleω satisfies the equation

sin3 ω = sin(β − ω) sin(α− ω) sin(γ − ω).

It also follows that with the sameω, there is another triad of circles intersecting atanother pointQ such that

∠CAQ = ∠ABQ = ∠BCQ = ω.

7.5 The Brocard points 229

A

B C

P

ω

ωω

A

B C

Q

ω

ω

ω

The pointsP andQ are isogonal conjugates. They are called the Brocard points oftriangleABC.

A

B C

Q

ω

ω

ωP

ω

ωω


7.6 Kariya’s theorem

Given a triangleABC with incenterI, consider a pointX on the perpendicular fromI toBC, such thatIX = t. We regardt > 0 if X and the point of tangency of the incircle withthe sideBC are on the same side ofI.

Theorem 7.2(Kariya). LetI be the incenter of triangleABC. If pointsX, Y ,Z are chosenon the perpendiculars fromI toBC, CA, AB respectively such thatIX = IY = IZ, thenthe linesAX, BY , CZ are concurrent.

I

A

B C

X

Y

Z

Q

Proof. (1) We compute the length ofAX. Let the perpendicular fromA to BC and theparallel fromX to the same line intersect atX ′. In the right triangleAXX ′,

AX ′ =2∆

a− r + t =

2rs

a− r + t =

r(b+ c)

a+ t,

XX ′ = (s− b)− c cosB

=1

2(c+ a− b)− 1

2a(c2 + a2 − b2)

=a(c+ a− b)− (c2 + a2 − b2)

2a

=b2 − c2 − a(b− c)

2a=

(b− c)(b+ c− a)

2a=

(b− c)(s− a)

a.

Applying the Pythagorean theorem to the right triangleAXX ′, we have

7.6 Kariya’s theorem 231

I

A

B C

XX′

OI

A

B C

X

M ′

P

AX2 =

(

r(b+ c)

a+ t

)2

+

(

(b− c)(s− a)

a

)2

=r2(b+ c)2 + (b− c)2(s− a)2

a2+

2r(b+ c)t

a+ t2

=(s−a)(s−b)(s−c)(b+c)2

s+ (s− a)2(b− c)2

a2+

2r(b+ c)t+ at2

a

=(s− a)((s− b)(s− c)(b+ c)2 + s(s− a)(b− c)2)

a2s+

2r(b+ c)t+ at2

a

=(s− a) · a2bc

a2s+

2r(b+ c)t+ at2

a

=abc(s− a)

as+

2r(b+ c)t+ at2

a

=4Rr(s− a)

a+

2r(b+ c)t+ at2

a

=4Rr(s− a) + 2r(b+ c)t+ at2

a.

(2) LetM ′ be the midpoint of the arcBAC of the circumcircle, and letMX intersectOI atP . We shall prove that angleIAP = angleIAX.

First of all,

AI2 =(s− a)2

cos2 A2

= (s− a)2 · bc

s(s− a)=

bc(s− a)

s=

4Rr(s− a)

a.

For later use, we also establish

AI2 + 2Rr =4Rr(s− a)

a+ 2Rr =

2Rr(2s− 2a) + 2Rr · aa

=2Rr(b+ c)

a.

SinceIP : PO = t : R, IP = tR+t

· OI. Recall thatOI2 = R2 − 2Rr. Applying the law


of cosines to triangleAIP , we have

AP 2 = AI2 + IP 2 − 2 · AI · IP cosAIP

= AI2 +

(

t2

(R + t)2

)

(R2 − 2Rr)− t

R + t(AI2 +OI2 −R2)

= AI2 +

(

t2

(R + t)2

)

(R2 − 2Rr)− t

R + t(AI2 − 2Rr)

=R

R + t· AI2 + 2R2rt

(R + t)2+

R2t2

(R + t)2

=R(R + t)AI2 + 2R2rt+R2t2

(R + t)2

=R2 · AI2 +R(AI2 + 2Rr)t+R2t2

(R + t)2

=R2 · 4Rr(s−a)

a+ 2R2r(b+c)

at+R2t2

(R + t)2

=R2(4Rr(s− a) + 2r(b+ c)t+ at2)

a(R + t)2.

Note thatAP 2 = R2

(R+t)2· AX2. This means thatAP = R

R+t· AX.

(3) Let AI intersectPX at X ′′ and the circumcircle again atM , the antipode ofM ′.Note thatMM ′

M ′O= −2 andOI

IP= −R+t

t.

OI

A

B C

X

M ′

P

M

X′′

Applying Menelaus’ theorem to triangleM ′OP and transversalIX ′′M , we have

−1 =PX ′′

X ′′M ′ ·M ′M

MO· OI

IP=⇒ PX ′′

X ′′M ′ =−t

2(R + t)=⇒ PX ′′

PM ′ =−t

2R + t.

Now XPPM ′ =

tR

. Therefore,XPPX′′ =

2R+t−R , and

XX ′′

X ′′P=

R + t

R=

AX

AP.

7.7 Isogonal conjugate of an infinite point 233

This shows thatAX ′′ bisects angleXAP . SinceAX ′′ is the bisector of angleA, the linesAP andAX are isogonal with respect to angleA.

(4) Likewise, if pointsY andZ are chosen on the perpendiculars fromI to CA andAB such thatIY = IZ = t = IX, then with the same pointP onOI, the linesBP andBY are isogonal with respect to angleB, andCP , CZ isogonal with respect to angleC.Therefore the three linesAX, BY , CZ intersect at the isogonal conjugate ofP (whichdividesOI in the ratioOP : PI = R : t).

7.7 Isogonal conjugate of an infinite point

Proposition 7.3. Given a triangleABC and a lineℓ, let ℓa, ℓb, ℓc be the parallels toℓthroughA, B, C respectively, andℓ′a, ℓ

′b, ℓ′c their reflections in the angle bisectorsAI, BI,

CI respectively. The linesℓ′a, ℓ′b, ℓ′c intersect at a point on the circumcircle of triangle

ABC.

I

O

A

B C

P

ℓ ℓa ℓb ℓc

ℓ′a

ℓ′b

ℓ′c

Proof. Let P be the intersection ofℓ′b andℓ′c.

(BP, PC) =(ℓ′b, ℓ′c)

=(ℓ′b, IB) + (IB, IC) + (IC, ℓ′c)

=(IB, ℓb) + (IB, IC) + (ℓc, IC)

=(IB, ℓ) + (IB, IC) + (ℓ, IC)

=2(IB, IC)

=2

(

π

2+

A

2

)

=(BA, AC) (mod π).

Therefore,ℓ′b andℓ′c intersect at a point on the circumcircle of triangleABC.Similarly, ℓ′a andℓ′b intersect at a pointP ′ on the circumcircle. Clearly,P andP ′ are

the same point since they are both on the reflection ofℓb in the bisectorIB. Therefore, thethree reflectionsℓ′a, ℓ

′b, andℓ′c intersect at the same point on the circumcircle.


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

I O

A

B C

P

Q

ℓ′

ℓ

Proof. If P andQ are the isogonal conjugates of the infinite points of two perpendicularlinesℓ andℓ′ throughA, thenAP andAQ are the reflections ofℓ andℓ′ in the bisectorAI.

(AP, AQ) = (AP, IA) + (IA, AQ) = −(ℓ, IA)− (IA, ℓ′) = −(ℓ, ℓ′) =π

2.

Therefore,P andQ 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 ofABC has vertices the excentersIA, Ib, IC .SinceAIa ⊥ IbIc, BIb ⊥ IcIa, andCIc ⊥ IaIb,AIa, BIb, CIc are the altitudes, andI is the orthocenter of the excentral triangle.It follows that the circumcircle of triangleABC is the nine-point circle of the excentraltriangle.

O

I

Ia

Ib

Ic

I′

A

BC

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

236 The excentral and intouch triangles

Let the bisectorAIa intersect the circumcircle atM . M is the midpoint of the arcBCof the circumcircle (on the opposite side ofA). Therefore,OM is perpendicular toBC.

X′

O

I

Ia

Ib

Ic

I′

M

A

BC

The circumcenterI ′ is the reflection ofI in O. Join I ′ to Ia. SinceIO = OI ′ andIM = MIa, I ′Ia is parallel toOM (and I ′Ia = 2 · OM = 2R). Therefore,I ′Ia isperpendicular toBC, and it passes through the point of tangency ofX ′ of BC with theA-excircle.

Corollary. The perpendiculars from the excenters to the corresponding sides of a triangleare 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.

Aa AbAc

M ′

O

I

X

Ia

Ib

Ic

I′

M

A

BC

D

Proof. (1) Consider the four pointsIb, B, C, Ic. They are concyclic since bothIbBIc andIbCIc are right angles. The center of the circle must be the midpoint M ′ of IbIc, whichmust also lie on the perpendicular bisector ofBC, which is the lineDM ′. Therefore,M ′

is the antipode ofM 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 thatra + rb + rc = 4R + r.

238 The excentral and intouch triangles

8.2 The homothetic center of the intouch and excentraltriangles

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

Proposition 8.2. The excentral and the intouch triangles are homothetic at a point T di-vidingOI in the ratioOT : TI = 2R + r : −2r.

I

X

Y

ZI′

Ia

Ib

Ic

A

BC

TO

Proof. (1) The segmentsIbIc andY Z are parallel, since they are both perpendicular to thebisector of angleA. Similarly, IcIa//ZX andIaIb//XY . Therefore the two triangles arehomothetic.

(2) Since the excentral triangle has circumradius2R and the intouch triangle has cir-cumradiusr, the homothetic ratio is2R : r.

(3) The homothetic center is the point dividing the segmentI ′I externally in the ratioI ′T : TI = 2R : −r.

(4) SinceI ′ is the reflection ofI in O, the homothetic centerT dividesOI 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 referencetriangleABC, we put at the verticesA, B, C massesu, v, w respectively, and determinethe balance point. The masses atB andC can be replaced by a single massv + w at thepointX = v·B+w·C

v+w. Together with the mass atA, this can be replaced by a massu+ v+w

at the pointP which dividesAX in the ratioAP : PX = v +w : u. This is the point withabsolute barycentric coordinateu·A+v·B+w·C

u+v+w, providedu + v + w 6= 0. 1 We also say that

the balance pointP hashomogeneous barycentric coordinates(u : v : w) with referencetoABC.

9.1.1 The centroid

The midpoints of the sides are

D =B + C

2, E =

C + A

2, F =

A+ B

2.

The centroidG divides each median in the ratio2 : 1. Thus,

G =A+ 2D

3=

A+ B + C

3.

This is the absolute barycentric coordinate ofG (with reference toABC). Its homogeneousbarycentric 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 saythat it represents aninfinite point. See§9.4.

302 Homogeneous Barycentric Coordinates

G

D

EF

A

B C

I

A

B CX

Y

Z

9.1.2 The incenter

The bisectorAX divides the sideBC in the ratioBX : XC = c : b. This givesX =bB+cCb+c

. Note thatBX has length cab+c

. Now, in triangleABX, the bisectorBI dividesAXin the ratioAI : IX = c : ca

b+c= b+ c : a. It follows that

I =aA+ (b+ c)X

a+ b+ c=

aA+ bB + cC

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 triangleABC. The edgesBC, CA, AB can be replaced respectively by massesa, b, c at their midpointD = B+C

2,

E = C+A2

, andF = A+B2

. With reference to the medial triangleDEF , this has coordinatesa : b : c. Since the sidelengths of triangleDEF are in the same proportions, this barycenteris the incenter of the medial triangle, also called the Spieker centerSp of ABC.

Sp

D

EF

A

B C

The center of mass of the perimeter is therefore the point

Sp =a ·D + b · E + c · F

a+ b+ c

=a · B+C

2+ b · C+A

2+ c · A+B

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) .


9.1.4 The Gergonne point

We follow the same method to compute the coordinates of the Gergonne pointGe. Here,BX = s− b andXC = s− c, so that

X =(s− b)b+ (s− c)C

a.

I

X

Y

Z

A

B C

Ge

The ratioAGe : GeX, however, is not immediate obvious. It can nevertheless be foundby applying the Menelaus theorem to triangleABX with transversalCZ. Thus,

AGe

GeX· XC

CB· BZ

ZA= −1.

From this,AGe

GeX= −CB

XC· ZABZ

= − −a

s− c· s− a

s− b=

a(s− a)

(s− b)(s− c).

Therefore,

Ge =(s− b)(s− c)A+ a(s− a)X

(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

s−a : 1s−b :

1s−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. Wesummarize the results in the following useful alternative of the Ceva theorem.

Theorem 9.1(Ceva). LetX, Y , Z be points on the linesBC, CA, AB respectively. ThelinesAX, BY , CZ are collinear if and only if the given points have coordinates of theform

X = (0 : y : z),Y = (x : 0 : z),Z = (x : y : 0),

for somex, y, z. If this condition is satisfied, the common point of the linesAX, BY , CZis P = (x : y : z).

A

B C

CP

AP

BP

P P

A

B CX

Remarks.(1) The pointsX, Y , Z are called the traces ofP . We also say thatXY Z isthe cevian triangle ofP (with reference to triangleABC). Sometimes, we shall adopt themore 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 pointP divides the segmentAX in the ratioPX : AX = x : x+ y + z.(3) It follows that the areas of the oriented trianglesPBC andABC are in the ratio

∆(PBC) : ∆(ABC) = x : x + y + z. This leads to the following interpretation ofhomogeneous barycentric coordinates: the homogeneous barycentric coordinates of a pointP can be taken as the proportions of (signed) areas of orientedtriangles:

P = ∆(PBC) : ∆(PCA) : ∆(PAB).


9.2.1 The circumcenter

Consider the circumcenterO of triangleABC.

O

A

B C

Since∠BOC = 2A, the area of triangleOBC is

1

2·OB ·OC · sinBOC =

1

2R2 sin 2A.

Similarly, the areas of trianglesOCA andOAB are respectively12R2 sin 2B and1

2R2 sin 2C.

It follows that the circumcenterO has homogeneous barycentric coordinates

∆OBC : ∆OCA : ∆OAB

=1

2R2 sin 2A :

1

2R2 sin 2B :

1

2R2 sin 2C

= sin 2A : sin 2B : sin 2C

= a cosA : b cosB : c cosC

= a · b2 + c2 − a2

2bc: b · c

2 + a2 − b2

2ca: c · a

2 + b2 − c2

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

TheA-excircle touches the sideBC at a pointX ′ such thatBX ′ = s− c andX ′C = s− b.From this, the homogeneous barycentric coordinates ofX ′ are0 : s − b : s − c; similarlyfor the points of tangencyY ′ andZ of theB- andC-excircles:

s− c s− b

s− c

s− b

Ic

Ib

Na Y ′

Z′

Ia

X′

A

B

C

X ′ =(0 : s− b : s− c),

Y ′ =(s− a : 0 : s− c),

Z ′ =(s− a : s− b : 0),

From these we conclude thatAX ′, BY ′, andCZ ′ concur. Their common point is calledthe Nagel point and has coordinates

Na = (s− a : s− b : s− c).

The triangleX ′Y ′Z ′ is called the extouch triangle.


9.2.3 The orthocenter and the orthic triangle

For the orthocenterH with tracesX, Y , Z onBC, CA, AB respectively, we haveBX =c cosB, XC = b cosC. This gives

BX : XC = c cosB : b cosC =cosB

b:cosC

c;

similarly for the other two traces.

X = 0 : bcosβ

: ccos γ

Y = acosα

: 0 : ccos γ

Z = acosα

: bcosβ

: 0

H = acosα

: bcosβ

: ccos γ

A

B C

H

X

Y

Z

O

A

B C

The triangleXY Z is called the orthic triangle.

9.3 Homotheties 309

9.3 Homotheties

Let P be a given point, andk a real number. Thehomothetywith centerP and ratiok isthe transformationh(P, k) which maps a pointX to the pointY such that

−→PY = k · −−→PX.

Equivalently,Y dividesPX in the ratioPY : Y X = k : 1− k, and

h(P, k)(X) = (1− k)P + kX.

P XYk 1− k

9.3.1 Superiors and inferiors

The homothetiesh(G,−2) andh(

G,−12

)

are called thesuperiorand inferior operationsrespectively. Thus,sup(P ) andinf(P ) are the points dividingP and the centroidG accord-ing 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

= (A+ B + C)− 2(uA+ vB + wC)

u+ v + w

=(u+ v + w)(A+B + C)

u+ v + w− 2(uA+ vB + wC)

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 barycentriccoordinates.

The case for inferior is similar.


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

(2) The inferior and superior of the Gergonne point

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

are

inf(Ge) = ((s− c)(s− a) + (s− a)(s− b) : (s− a)(s− b) + (s− b)(s− c)

: (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(Ge) = ((s− c)(s− a) + (s− a)(s− b)− (s− b)(s− c)

: (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 ofO andH, is the inferior ofO.

H

N

G O

Lo

From the homogeneous barycentric ofO, we obtain

N = b2(c2 + a2 − b2) + c2(a2 + b2 − c2) : · · · : · · ·= a2(b2 + c2)− (b2 − c2)2 : · · · : · · · .

Similarly, the deLongchamps point, being the reflection ofH in O, is the superior ofH. It has coordinates

Lo = (−3a4 + 2a2(b2 + c2) + (b2 − c2)2 : −3b4 + 2b2(c2 + a2) + (c2 − a2)2

: −3c4 + 2c2(a2 + b2) + (a2 − b2)2).

9.3 Homotheties 311

A

B C

Na

H

X

Y

Z

Example 9.2.The distance betweenO andNa is equal toR− 2r.

Proof. SinceO = sup(N) andNa = sup(I), ONa = 2 ·NI = 2(

R2− r)

= R− 2r.

Example 9.3. Pedals on altitudes equidistant from vertices We locate thepoint P whosepedals on the altitudes are equidistant from the vertices.

Let P = (u : v : w). We require

v + w

u+ v + w· 2∆a

=w + u

u+ v + w· 2∆

b=

u+ v

u+ v + w· 2∆

c=⇒ v + w

a=

w + u

b=

u+ v

c.

This means thatinf(P ) = I andP = 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)

(b+ c− a) + (c+ a− b) + (a+ b− c)· 2∆a

=4∆

a+ b+ c= 2r,

the diameter of the incircle. The three pedals lie on the circle with diameterNaH, calledthe Fuhrmann circle.


9.3.2 Triangles bounded by lines parallel to the sidelines

Theorem 9.3(Homothetic center theorem). If parallel linesXbXc, YcYa, ZaZb to the sidesBC, CA, AB of triangleABC 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 triangleA∗B∗C∗ homothetic toABC with homothety ratio1 + t1 +t2 + t3. The homothetic center is a pointP with homogeneous barycentric coordinatest1 : t2 : t3.

A

B C

A∗

B∗ C∗

P

Xc Xb

Ya

YcZb

Za

A

B C

A∗

B∗ C∗

P

Xc Xb

Ya

YcZb

Za

Proof. Let P be the intersection ofB∗B andC∗C. Since

B∗C∗ = B∗Xc +XcXb +XbC∗

= t3a+ (1 + t1)a+ t2a = (1 + t1 + t2 + t3)a,

we we havePB : PB∗ = PC : PC∗ = 1 : 1 + t1 + t2 + t3.

A similar calculation shows thatAA∗ andBB∗ intersect at the same pointP . This showsthatA∗B∗C∗ is the image ofABC under the homothetyh(P, 1 + t1 + t2 + t3).

Now we compare areas. Note that(1)∆(BZaXc) =

BXc

AB· BZa

CB·∆(ABC) = t1t3∆(ABC),

(2) ∆(PBC)∆(BZaB∗)

= PBBB∗ · CB

BZa= 1

t1+t2+t3· 1t3= 1

t3(t1+t2+t3).

Since∆(BZaB∗) = ∆(BZaXc), we have∆(PBC) = t1

t1+t2+t3·∆(ABC).

Similarly, ∆(PCA) = t2t1+t2+t3

· ∆(ABC) and∆(PAB) = t3t1+t2+t3

· ∆(ABC). Itfollows that

∆(PBC) : ∆(PCA) : ∆(PAB) = t1 : t2 : t3.

9.3 Homotheties 313

The Grebe symmedian point

Consider the square erected externally on the sideBC of triangleABC, The line containingthe outer edge of the square is the image ofBC under the homothetyh(A, 1 + t1), where

1 + t1 =2∆a

+a2∆a

= 1 + a2

2∆, i.e., t1 = a2

2∆. Similarly, if we erect squares externally on the

other two sides, the outer edges of these squares are on the lines which are the images ofCA, AB under the homothetiesh(B, 1 + t2) andh(C, 1 + t3) with t2 =

b2

2∆andt3 = c2

2∆.

Ab Ac

Bc

Ba

Ca

Cb

A

B C

A∗

B∗ C∗

K

The triangle bounded by the lines containing these outer edges is called theGrebe

triangle of ABC. It is homothetic toABC at(

a2

2∆: b2

2∆: c2

2∆

)

= (a2 : b2 : c2), and the

ratio of homothety is

1 + (t1 + t2 + t3) =2∆ + a2 + b2 + c2

2∆.

This homothetic center is called theGrebe symmedian point

K = (a2 : b2 : c2).

Remark.Note that the homothetic center remains unchanged if we replacedt1, t2, t3 bykt1, kt2, kt3 for the same nonzerok. This means that if similar rectangles are constructedon the sides of triangleABC, the lines containing their outer edges always bound a trianglewith homothetic centerK.


9.4 Infinite points

A point with zero sum of homogeneous coordinates is called aninfinite point. It can beexpressed as the difference of the absolute barycentric coordinates of two points (withequal nonzero sums of homogeneous coordinates). As such, aninfinite point defines thedirection of a family of parallel lines. All infinite points form the line at infinity: consistingof points(x : y : z) satisfyingx+y+z = 0. The infinite point of a line is defined uniquelyup to a nonzero scalar multiple.

Example 9.4. 1.The infinite points of the side linesBC, CA, AB are(0 : −1 : 1),(1 : 0 : −1), (−1 : 1 : 0) respectively.

2. The infinite point of theA−altitude has homogeneous coordinates

(0 : Sγ : Sβ)− a2(1 : 0 : 0) = (−a2 : Sγ : Sβ).

3. The infinite point of the Euler line is the point

O −H =(Sα(Sβ + Sγ), Sβ(Sγ + Sα), Sγ(Sα + Sβ)

2S2− (Sβγ, Sγα, Sαβ)

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 theOI-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 barycentriccoordinates

10.1 Construction of mixtilinear incircles

10.1.1 The insimilicenter and the exsimilicenter of the circumcircleand incircle

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

T+ =1

R + r(r ·O +R · I),

T− =1

R− r(−r ·O +R · I).

OI

A

B CX

Y

Z

T+

T−

M

M ′

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 triangleABC is one that is tangent to two sides of the triangleand to the circumcircle internally. Denote byA′ the point of tangency of the mixtilinearincircleK(ρ) in angleA with the circumcircle. The centerK clearly lies on the bisector ofangleA, andAK : KI = ρ : −(ρ− r). In terms of barycentric coordinates,

K =1

r(−(ρ− r)A+ ρI) .

Also, since the circumcircleO(A′) and the mixtilinear incircleK(A′) touch each other atA′, we haveOK : KA′ = R− ρ : ρ, whereR is the circumradius. From this,

K =1

R(ρO + (R− ρ)A′) .

I

O

K

A′

A

B C

O

KA

A′

A

B C

T−

M

I

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

RI − rO

R− r=

R(ρ− r)A+ r(R− ρ)A′

ρ(R− r).

We note some interesting consequences of this formula. First of all, it gives the intersectionof the lines joiningAA′ andOI. Note that the point on the lineOI represented by the lefthand side isT−, the exsimilicenter of the circumcircle and the incircle.

This leads to a simple construction of the mixtilinear incircle. Given a triangleABC,extendAT− to intersect the circumcircle atA′. The intersection ofAI andA′O is the centerKA of the mixtilinear incircle in angleA.

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 pointsP andQ (not on any of the side lines of the reference triangle) are said to be isotomic conjugatesif their respective traces are symmetric with respect to themidpoints of the correspondingsides. Thus,

BX = X ′C, CY = Y ′A, AZ = Z ′B.

AC

B

P P •X

Y

ZX′

Y ′

Z′

We shall denote theisotomic conjugateof P by P •. If P = (x : y : z), then

P • =

(

1

x:1

y:1

z

)

= (yz : zx : xy).


Example 10.1.(The Gergonne and Nagel points)

Ge =(

1s−a : 1

s−b :1

s−c)

, Na = (s− a : s− b : s− c).

I

X

Y

Z

X′

A

B C

Ia

Ib

Ic

Y ′

Z′

GeNa

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

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

Its traces are the pedals of the deLongchamps pointLo, the reflection ofH in O.

O

H•

Z′

X′

Y ′

A

B C

H

X

Y

Z

Lo

10.4 Equal-parallelians point 319

10.4 Equal-parallelians point

Given triangleABC, we want to construct a pointP the three lines through which parallelto the sides cut out equal intercepts. LetP = xA + yB + zC in absolute barycentriccoordinates. The parallel toBC cuts out an intercept of length(1− x)a. It follows that thethree intercepts parallel to the sides are equal if and only if

1− x : 1− y : 1− z =1

a:1

b:1

c.

The right hand side clearly gives the homogeneous barycentric coordinates ofI•, the iso-tomic conjugate of the incenterI. 1 This is a point we can easily construct. Now, translatinginto absolutebarycentric coordinates:

I• =1

2[(1− x)A+ (1− y)B + (1− z)C] =

1

2(3G− P ).

we obtainP = 3G − 2I•, and can be easily constructed as the point dividing the segmentI•G externally in the ratioI•P : PG = 3 : −2. The pointP is called the congruent-parallelians point of triangleABC.

AC

B

I•

P G

1The isotomic conjugate of the incenter appears inETC as the pointX75.


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 pointFe is the point of (internal) tangency of the incircle and thenine-point circle. It dividesNI in the ratioNFe : FeI = R

2: −r = R : −2r. Therefore,

Fe =R · I − 2r ·N

R− 2r

in absolute barycentric coordinates.A

B C

I O

H

Fe

N

From the homogeneous barycentric coordinates ofN ,

(a2(b2 + c2)− (b2 − c2)2, b2(c2 + a2)− (c2 − a2)2, c2(a2 + b2)− (a2 − b2)2) = 32∆2 ·N.

we have

2r ·N =r

16∆2

(

a2(b2 + c2)− (b2 − c2)2, · · · , · · ·)

=R

4sabc

(

a2(b2 + c2)− (b2 − c2)2, · · · , · · ·)

,

R · I =R

2s(a, b, c) =

R

4sabc· 2abc(a, b, c).

322 Computation of barycentric coordinates

Therefore,

Fe ∼ R · I − 2r ·N∼ 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 toNFe.(b)ONa = R− 2r.(c)NaH is parallel toOI.(d) The reflection ofH in I and the reflection ofNa in O are the same point.(e) IN andOSp intersect at the midpoint ofNaH.

A

B C

I O

H

Fe

NSp

Na

11.2 TheOI line 323

11.2 TheOI line

11.2.1 The circumcenter of the excentral triangle

Let I ′ be the reflection ofI in O. Show thatI ′ is also the midpoint ofNaLo.Let N ′ be the midpoint ofOLo. The trianglesOI ′N ′ andOIN are congruent.I ′N ′ is

parallel toIN and henceNaO. Furthermore,I ′N ′ = IN = 12NaO. It follows thatI ′ is the

midpoint ofNaLo.

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

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

16∆2(a2(b2 + c2 − a2), b2(c2 + a2 − b2), c2(a2 + b2 − c2)) +

R

2s(a, b, c)

∼ (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), · · · , · · · ).

T− =(

a2

s−a : b2

s−b :c2

s−c

)

.

Proof.

T− ∼ r ·O −R · I

=r

16∆2(a2(b2 + c2 − a2), b2(c2 + a2 − b2), c2(a2 + b2 − c2))− R

2s(a, b, c)

∼ (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), · · · , · · · ).


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 centerT of excentral and intouch triangles

The two triangles are homothetic since their correspondingsides are perpendicular to theangle bisectors of triangleABC. Denote byT the homothetic center. This is clearly theexsimilicenter of their circumcircles. It is therefore thepoint dividingI ′ andI in the ratioI ′T : IT = 2R : r. It follows thatOT : TI = 2R + r : −2r.

T

I

O

I′

T =(

as−a : b

s−b :c

s−c)

.

Proof.

T ∼ −2r ·O + (2R + r)I

=−2r

16r2s2(a2(b2 + c2 − a2), · · · , · · · ) + 2R + r

2s(a, b, c)

∼ −(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

2a(a+ b+ c)(c+ a− b)(a+ b− c), · · · , · · ·

)

∼(

a

b+ c− a, · · · , · · ·

)

.

11.3 The excentral triangle 325

Exercise

1. Show thatTI : IO = 2r : 2R− r.

2. Find the ratio of divisionT+T : TT−.

3. Show thatG, Ge, andT are collinear by findingp, 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) andq = a+ b+ c.

4. Given thatGGe : GeT = 2(2R− r) : 3r, show thatGe, I, Lo are collinear.Apply the converse of Menelaus’ theorem to triangleOGT with Ge on GT , I onTO, andLo onOG.

GGe

GeT· TIIO

· OLo

LoG=

2(2R− r)

3r· 2r

2R− r· −3

4= −1.

11.3 The excentral triangle

11.3.1 The centroid

The centroid of the excentral triangle is the point

Ia + Ib + Ic3

=1

3

(

(−a, b, c)

b+ c− a+

(a,−b, c)

c+ a− b+

(a, b,−c)

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

∼ cosA

2· (−a, b, c)

b+ c− a+ cos

B

2· (a,−b, c)

c+ a− b+ cos

C

2· (a, b,−c)

a+ b− c

∼ cos A2

2r cot A2

(−a, b, c) +cos B

2

2r cot B2

(a,−b, c) +cos C

2

2r cot C2

(a, b,−c)

∼ sinA

2(−a, b, c) + sin

B

2(a,−b, c) + sin

C

2(a, b,−c)

∼(

a

(

− sinA

2+ sin

B

2+ sin

C

2

)

, · · · , · · ·)

.


Chapter 12

Some interesting circles

12.1 A fundamental principle on6 concyclic points

12.1.1 The radical axis of two circles

Given two nonconcentric circlesC1 andC2. The locus of points of equal powers withrespect to the circle is a straight line perpendicular to theline joining their centers. In fact,if the circles are concentric, there is no finite point with equal powers with respect to thecircles. On the other hand, if the centers are distinct pointsA andB at a distanced apart,there is a unique pointP with distancesAP = x andPB = d− x such that

r21 − x2 = r22 − (d− x)2.

If this common value ism, then every pointQ on the perpendicular toAB atP has powerm− PQ2 with respect to each of the circles. This line is called the radical axis of the twocircles.

If the two circles intersect at two distinct points, then theradical axis is the line joiningthese common points. If the circles are tangent to each other, then the radical axis is thecommon tangent.

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

Proof. (1) If any two of the circles are concentric, there is no finitepoint with equal powerswith respect to the three circles.

(2) If the centers of the circles are distinct and noncollinear, then two of the radicalaxes, being perpendiculars to two distinct lines with a common point, intersect at a point.This intersection has equal powers with respect to all threecircles, and also lies on the thirdradical axis.

(3) If the three centers are distinct but collinear, then thethree radical axes three parallellines, which coincide if any two of them do. This is the case ifand only if the three circlestwo points in common, or at mutually tangent at a point. In this case we say that the circlesare coaxial.

328 Some interesting circles

12.1.2 Test for6 concyclic points

Proposition 12.2.LetX,X ′ be points on the sidelinea, Y , Y ′ onb, andZ,Z ′ onc. The sixpoints are on a circle if and only if the four points on each pair of sidelines are concyclic.

A

B

C

X XY

Y ′

Z

Z′

Proof. It is enough to prove the sufficiency part. LetCa be the circle through the four pointsY , Y ′, Z, Z ′ onb andc, andCb the one throughZ, Z ′, X, X (on c anda), andCc throughX, X, Y , Y ′ (ona and(b). We claim that these three circles are identical. If not, then theyare pairwise distinct. The three pairs among them have radical axesa (for Cb andCc), b (forCc andCa), andc (for Ca andCb) respectively. Now, the three radical axes of three distinctcircles either intersect at a common point (the radical center), or are parallel (when theircenters are on a line), or coincide (when, in additon, the three circles are coaxial). In nocase can the three radical axes form a triangle (with sidelinesa, b, c. This shows that thethree circles coincide.

12.2 The Taylor circle 329

12.2 The Taylor circle

Consider the orthic triangleXY Z, and the pedals of each of the pointsX, Y , Z on the twosides not containing it. Thus,

Sideline Pedals ofX Pedals ofY Pedals ofZa Xb Xc

b Ya Yc

c Za Zb

H

X

Y

Z

Xc

Yc

Ya

Za

Zb

Xb

A

B C

It is easy to write down the lengths of various segments. Fromthese we easily determinethe coordinates of these pedals. For example, fromYaC = b cos2 γ, we haveAYa =b− b cos2 γ = b sin2 γ. Note that

AYa · AYc =(b− b cos2 γ)(b cos2 α) = b2 cos2 α sin2 γ = 4R2 cos2 α sin2 β sin2 γ,

AZa · AZb =(c− c cos2 β)(c cos2 α) = c2 cos2 α sin2 β = 4R2 cos2 α sin2 β sin2 γ,

givingAYa·AYc = AZa·AZb =Sαα

4R2 = S2·Sαα

a2b2c2. Similarly,BXb·BXc = BZb·BZa =

S2·Sββ

a2b2c2

andCYc · CYa = CXc · CXb =S2·Sγγ

a2b2c2. By Proposition 12.2, the six pointsXb, Xc, Yc, Ya,

Za, Zb are concyclic. The circle containing them is called theTaylor circle.

Exercise

1. Calculate the length ofXbXc.

2. Find the equation of the lineXbXc. 1

3. The three linesXbXc, YcYa, ZaZb bound a triangle with perspectorK.

1−S2x+ SBBy + SCCz = 0.


12.3 Two Lemoine circles

12.3.1 The first Lemoine circle

Given triangleABC, how can one choose a pointK so that when parallel lines are con-structed through it to intersect each sideline at two points, the resulting six points are on acircle?

YaZa

Zb

Xb

Yc

Xc

L

A

B C

K

Analysis. By the intersecting chords theorem,AYa · AYc = AZa · AZb. We have

b2 · AYa

AC· AYc

AC= c2 · AZa

AB· AZb

AB

=⇒ b2 · AYc

AC= c2 · AZb

AB

=⇒ b2 · BXc

BC= c2 · CXb

CB= c2 · XbC

BC

=⇒ BXc

XbC=

c2

b2.

Similarly, BZa

ZbA= a2

b2, and

XcXb

XbC=

XcXb

KYa

=XcK

KYc

=BZa

ZbA=

a2

b2.

Therefore,BXc : XcXb : XbC = c2 : a2 : b2.These proportions determine the pointsXb andXc onBC, and subsequently the other

points: the parallelsXcYc andXbZb (to c andb respectively) intersect atK, and the par-allel throughK to a determines the pointsYa andZa. The pointK is called theLemoinesymmedian pointof triangleABC, and the circle containing these six points is called thefirst Lemoine circle.

Denote byL the center of the first Lemoine circle. Note thatL lies on the perpendicularbisector of each of the parallel segmentsXcXb andZaYa. This means that the line joiningthe midpoints of these segments is the common perpendicularbisector of the segments, andthe trapezoidXcXbYaZa is symmetric; so areYaYcZbXb andZbZaXcYc by the same rea-soning. It follows that the segmentsYcZb, ZaXc, andXbYa have equal lengths. (Exercise:Show that this common length is abc

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 throughK to intersect the sidelines,we obtain another three equal segments with common midpointK. The6 endpoints,X ′b,X ′c on a, Y ′c , Y ′a on b, andZ ′a, Z

′b on c, are on a circle centerK and radius abc

a2+b2+c2. This

circle is called thesecond Lemoine circleof triangleABC.

A

B CX′b X′

c

Y ′c

Y ′a

Z′a

Z′b

K

12.3.3 Construction ofK

The length ofAK is twice the median of triangleAYcZb on the sideYcZb. If we denote byma etc the lengths of medians of triangleABC, thenAK = 2bcma

a2+b2+c2. Similarly, BK =

2camb

a2+b2+c2andCK = 2abmc

a2+b2+c2. This allows us to determine the anglesKAB etc and the radii

of the circlesKAB etc. The radius of the circleKAB, for example, isRc =AB·AK·BK4∆(KAB)

=abcmamb

(a2+b2+c2)∆(ABC). It follows thatsinKAB = BK

2Rc= ∆

bma.

A

B C

KG

I

A

B C

GK

Consider also the circleGAC. This has radiusR′b = AC·AG·CG4∆(GAC)

= bmamc

3∆. From this,

sinGAC = CG2R′

b

= ∆bma

. This shows that∠KAB = ∠GAC, andAK and the medianAGare isogonallines with respect to the sidesAB andAC. Similarly, BK andCK are thelines isogonal to the mediansBG andCG respectively, andK is thesymmedian pointoftriangleABC.


12.3.4 The center of the first Lemoine circle

We show that the centerL of the first Lemoine circle is the midpoint betweenK and thecircumcenterO. It is clear that this center is on the perpendicular bisector of XbXc. LetMbe the midpoint ofXbXc, andX the orthogonal projection ofK onBC.

YaZa

Zb

Xb

Yc

Xc

A

B C

K

DX M

O

We have

BM =ac2

a2 + b2 + c2+

1

2· a3

a2 + b2 + c2=

a(a2 + 2c2)

2(a2 + b2 + c2),

and

BX = BXc +KXc cosB

=ac2

a2 + b2 + c2+

a2c

a2 + b2 + c2· c

2 + a2 − b2

2ca

=ac2 + a(c2 + a2 − b2)

2(a2 + b2 + c2)

=a(a2 − b2 + 3c2)

2(a2 + b2 + c2).

It follows that

BX + BD =a(a2 − b2 + 3c2)

2(a2 + b2 + c2)+

a(a2 + b2 + c2)

2(a2 + b2 + c2)=

a(a2 + 2c2)

a2 + b2 + c2= 2 ·BM.

This means thatM is the midpoint ofXD, and the perpendicular toa atM contains themidpoint ofOK. The same reasoning shows that the midpoint ofOK also lies on theperpendiculars tob andc respectively at the midpoints ofYcYa andZaZb. It is therefore thecenterL of the first Lemoine circle. (Exercise. Calculate the radius of the first Lemoinecircle).

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 barycentriccoordinates) , then the area of the oriented triangleP1P2P3 is

∆P1P2P3 =

∣

∣

∣

∣

∣

∣

x1 y1 z1x2 y2 z2x3 y3 z3

∣

∣

∣

∣

∣

∣

·∆ABC.

Example 13.1.(Area of cevian triangle) LetP = (u : v : w) be a point with cevian triangleXY Z. The area of the cevian triangleXY Z is

1

(v + w)(w + u)(u+ v)

∣

∣

∣

∣

∣

∣

0 v wu 0 wu v 0

∣

∣

∣

∣

∣

∣

·∆ =2uvw

(v + w)(w + u)(u+ v)·∆.

If P is an interior point (so thatu, v, w are positive), thenv + w ≥ 2√vw, w + u ≥

2√wu, andu+ v ≥ 2

√uv. It follows that

2uvw

(v + w)(w + u)(u+ v)≤ 2uvw

2√vw · 2√wu · 2√uv

=1

4.

Equality holds if and only ifu = v = w, i.e., P = (1 : 1 : 1) = G, the centroid. Thecentroid 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 pointsPi =(ui, vi, wi) are collinear if and only if

∣

∣

∣

∣

∣

∣

u1 v1 w1

u2 v2 w2

u3 v3 w3

∣

∣

∣

∣

∣

∣

= 0.

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

∣

∣

∣

∣

∣

∣

x1 y1 z1x2 y2 z2x y z

∣

∣

∣

∣

∣

∣

= 0,

or

(y1z2 − y2z1)x+ (z1x2 − z2x1)y + (x1y2 − x2y1)z = 0.

Examples

(1) The equations of the sidelinesBC, CA, AB are respectivelyx = 0, y = 0, z = 0.(2) Given a pointP = (u : v : w), the cevian lineAP has equationwy − vz = 0;

similarly for the other two cevian linesBP andCP . These lines intersect correspondingsidelines at the traces ofP :

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 1a b cx y z

∣

∣

∣

∣

∣

∣

= 0,

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

Euler line OH∑

cyclic(b2 − c2)(b2 + c2 − a2)x = 0

OI-line OI∑

cyclic bc(b− c)(b+ c− a)x = 0

Soddy line IGe

∑

cyclic(b− c)(s− a)2x = 0

Brocard axis OK∑

cyclic b2c2(b2 − c2)x = 0

van Aubel line HK∑

cyclic Sαα(Sβ − Sγ)x = 0

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(q1r2 − q2r1 : r1p2 − r2p1 : p1q2 − p2q1).

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

∣

∣

∣

∣

∣

p1 q1 r1p2 q2 r2p3 q3 r3

∣

∣

∣

∣

∣

∣

= 0.

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)

∣

∣

∣

∣

: · · · : · · ·

=(c− a)(a− b)

∣

∣

∣

∣

(s− b)2 (s− c)2

(c+ a)(c2 + a2 − b2) (a+ b)(a2 + b2 − c2)

∣

∣

∣

∣

: · · · : · · ·

=(c− a)(a− b)

∣

∣

∣

∣

(s− b)2 a(b− c)(c+ a)(c2 + a2 − b2) (b− c)(a+ b+ c)2

∣

∣

∣

∣

: · · · : · · ·

=(b− c)(c− a)(a− b)

∣

∣

∣

∣

(s− b)2 a(c+ a)(c2 + a2 − b2) (a+ b+ c)2

∣

∣

∣

∣

: · · · : · · ·

=1

4(b− c)(c− a)(a− b)

∣

∣

∣

∣

(c+ a− b)2 4a(c+ a)(c2 + a2 − b2) (a+ b+ c)2

∣

∣

∣

∣

: · · · : · · ·

=1

4(b− c)(c− a)(a− b)(−3a4 + 2a2(b2 + c2) + (b2 − c2)2) : · · · : · · ·

Writing a2 = Sβ + Sγ etc., we have

− 3a4 + 2a2(b2 + c2) + (b2 − c2)2

=− 3(Sβ + Sγ)2 + 2(Sβ + Sγ)(2Sα + Sβ + Sγ) + (Sβ − Sγ)

2

=4(Sαβ + Sγα − Sβγ).

This intersection has homogeneous barycentric coordinates

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

This is the reflection ofH in O, and is called the deLongchamps pointLo.


13.3 Perspective triangles

Many interesting points and lines in triangle geometry arise from theperspectivityof trian-gles. We say that two trianglesX1Y1Z1 andX2Y2Z2 are perspective,X1Y1Z1 ⊼X2Y2Z2, ifthe linesX1X2, Y1Y2,Z1Z2 are concurrent. The point of concurrency,∧(X1Y1Z1, X2Y2Z2),is called theperspector. Along with the perspector, there is anaxis of perspectivity, or theperspectrix, which is the line joining containing

Y1Z2 ∩ Z1Y2, Z1X2 ∩X1Z2, X1Y2 ∩ Y1X2.

We denote this line byL∧(X1Y1Z1, X2Y2Z2).Homothetic triangles are clearly prespective. If trianglesT andT′, their perspector is

the homothetic center, which we shall denote by∧0(T,T′).

Proposition 13.3.A triangle with vertices

X = U : v : w,Y = u : V : w,Z = u : v : W,

for someU , V , W , is perspective toABC at ∧(XY Z) = (u : v : w). The perspectrix isthe line

x

u− U+

y

v − V+

z

w −W= 0.

Proof. The lineAX has equationwy − vz = 0. It intersects the sidelineBC at the point(0 : v : w). Similarly,BY intersectsCA at (u : 0 : w) andCZ intersectsAB at (u : v : 0).These three are the traces of the point(u : v : w).

The lineY Z has equation−(vw−VW )x+u(w−W )y+u(v−V )z = 0. It intersectsthe sidelineBC at (0 : v−V : −(w−W )). Similarly, the linesZX andXY intersectCAandAB respectively at(−(u− U) : 0 : w −W ) and(u− U : −(v − V ) : 0). It is easy tosee that these three points are collinear on the line

x

u− U+

y

v − V+

z

w −W= 0.

13.3 Perspective triangles 405

The excentral triangle

The excentral triangle is perspective withABC; the perspector is the incenterI:

Ia = −a : b : cIb = a : −b : cIc = a : b : −cI = a : b : c

13.3.1 The Conway configuration

Given triangleABC, extend(i) CA andBA to Ya andZa such thatAYa = AZa = a,(ii) AB andCB toZb andXb such thatBZb = BXb = b,(iii) BC andAC toXc andYc such thatCXc = CYc = c.

b

b

c

c

a

a

A

B CXcXb

Za

Zb

Yc

Ya

?

X

YZ

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 ofYc andZb, we determine easily the coordinates ofX = BYc∩CZb:

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


Similarly, the coordinates ofY = CZa ∩ AXc, andZ = AXb ∩ BYa can be determined.The following table shows that the perspector of trianglesABC andXY Z is the point withhomogeneous barycentric coordinates

(

1a: 1

b: 1

c

)

.

X = −bc : c(b+ c) : b(b+ c) = −1b+c

: 1b

: 1c

Y = c(c+ a) : −ca : a(c+ a) = 1a

: −1c+a

: 1c

Z = b(a+ b) : a(a+ b) : −ab = 1a

: 1b

: −1a+b

? = = 1a

: 1b

: 1c

13.4 Perspectivity 407

13.4 Perspectivity

Many interesting points and lines in triangle geometry arise from theperspectivityof trian-gles. We say that two trianglesX1Y1Z1 andX2Y2Z2 are perspective,X1Y1Z1 ⊼X2Y2Z2, ifthe linesX1X2, Y1Y2,Z1Z2 are concurrent. The point of concurrency,∧(X1Y1Z1, X2Y2Z2),is called theperspector. Along with the perspector, there is anaxis of perspectivity, or theperspectrix, which is the line joining containing

Y1Z2 ∩ Z1Y2, Z1X2 ∩X1Z2, X1Y2 ∩ Y1X2.

We denote this line byL∧(X1Y1Z1, X2Y2Z2). We justify this in§below.If one of the triangles is the triangle of reference, it shallbe omitted from the notation.

Thus,∧(XY Z) = ∧(ABC,XY Z) andL∧(XY Z) = L∧(ABC,XY Z).Homothetic triangles are clearly prespective. If trianglesT andT′, their perspector is

the homothetic center, which we shall denote by∧0(T,T′).

Proposition 13.4.A triangle with vertices

X = U : v : w,Y = u : V : w,Z = u : v : W,

for someU , V , W , is perspective toABC at ∧(XY Z) = (u : v : w). The perspectrix isthe line

x

u− U+

y

v − V+

z

w −W= 0.

Proof. The lineAX has equationwy − vz = 0. It intersects the sidelineBC at the point(0 : v : w). Similarly,BY intersectsCA at (u : 0 : w) andCZ intersectsAB at (u : v : 0).These three are the traces of the point(u : v : w).

The lineY Z has equation−(vw−VW )x+u(w−W )y+u(v−V )z = 0. It intersectsthe sidelineBC at (0 : v−V : −(w−W )). Similarly, the linesZX andXY intersectCAandAB respectively at(−(u− U) : 0 : w −W ) and(u− U : −(v − V ) : 0). These threepoints are collinear on the trilinear polar of(u− U : v − V : w −W ).

The trianglesXY Z andABC are homothetic if the perspectrix is the line at infinity.


13.4.1 The Schiffler point: intersection of four Euler lines

Theorem 13.5.Let I be the incenter of triangleABC. The Euler lines of the trianglesIBC, ICA, IAB are concurrent at a point on the Euler line ofABC, namely, the Schifflerpoint

Sc =

(

a(b+ c− a)

b+ c:b(c+ a− b)

c+ a:c(a+ b− c)

a+ b

)

.

Proof. Let I be the incenter of triangleABC.We first compute the equation of the Euler line of the triangleIBC.The centroid of triangleIBC is the point(a : a+2b+c : a+b+2c). The circumcenter

of triangle is the midpoint ofIIa. This is the point(−a2 : b(b+ c) : c(b+ c)). From thesewe obtain the equation of the Euler line:

0 =

∣

∣

∣

∣

∣

∣

x y za a+ 2b+ c a+ b+ 2c

−a2 b(b+ c) c(b+ c)

∣

∣

∣

∣

∣

∣

= (b− c)(b+ c)x+ a(c+ a)y − a(a+ b)z.

The equations of the Euler lines of the trianglesICA and IAB can be obtained bycyclic permutations ofa, b, c andx, 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−c a− b

∣

∣

∣

∣

: −∣

∣

∣

∣

−b bc a− b

∣

∣

∣

∣

:

∣

∣

∣

∣

−b c− ac −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 lineof IBC given by the firstequation:

(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 ofABC, 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 pointP with tracesAP , BP , andCP on the sidelines of triangleABC, let

X = BPCP ∩ BC, Y = CPAP ∩ CA, Z = APBP ∩ AB.

These pointsX, Y , Z lie on a line called thetrilinear polar (or simply tripolar) ofP .

A

B CX

Z

Y

P

AP

BP

CP

410 Cevian nest theorem

If P = (u : v : w), thenBP = (u : 0 : w) andCP = (u : v : 0). The lineBPCP hasequation

−x

u+

y

v+

z

w= 0.

It intersects the sidelineBC at the pointX = (0 : v : −w).Similarly,AP = (0 : v : w) and the pointsY , Z are

Y = (−u : 0 : w), Z = (u : −v : 0).

The line containing the three pointsX, Y , Z is

x

u+

y

v+

z

w= 0.

This is the tripolar ofP .

14.1 Trilinear pole and polar 411

14.1.2 Tripole of a line

Given a lineL intersectingBC, CA, AB atX, Y , Z respectively, let

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

The linesAA′, BB′ andCC ′ are concurrent. The point of concurrency is thetripole P ofL.

A

B C

X

Y

Z

P

A′

B′

C′

ClearlyP is the tripole ofL if and only ifL is the tripolar ofP .


14.2 Anticevian triangles

The vertices of theanticevian triangleof a pointP = (u : v : w)

cev−1(P ) : Pa = (−u : v : w), Pb = (u : −v : w), Pc = (u : v : −w)

are the harmonic conjugates ofP with respect to the cevian segmentsAAP , BBP andCCP , i.e.,

AP : PAP = −APa : PaAP ;

similarly forPb andPc. This is called the anticevian triangle ofP sinceABC is the ceviantriangleP in PaPbPc. It is also convenient to regardP , Pa, Pb, Pc as a harmonic quadruplein the sense that any three of the points constitute the harmonic associates of the remainingpoint.

14.2.1 Construction of anticevian triangle

If the trilinear polarLP of P intersects the sidelinesBC, CA, AB at X ′, Y ′, Z ′ respec-tively, then the anticevian trianglecev−1(P ) is simply the triangle bounded by the linesAX ′, BY ′, andCZ ′.

A

B CX

Z

Y

P

AP

BP

CPPc

Pa

Pb

14.2 Anticevian triangles 413

Another construction of anticevian triangle

Here is an alternative construction ofcev−1(P ).LetAHBHCH be the orthic triangle, andX the reflection ofP in a, then the intersection

of the linesAHX andOA is the harmonic conjugatePa of P in AAP :

APa

PaAP

= − AP

PAP

.

A

B CAH

P

X

Pa

AP

A

B CAH

P

X

Pa

AP

A′

Proof. Let A′ be the reflection ofA in BC. Applying Menelaus’ theorem to triangleAPAA

′ with transversalAHXPa, we have

APa

PaAP

· APX

XA′· A′AH

AHA= −1.

This givesAPa

PaAP

= − XA′

APX= − PA

APP= − AP

PAP

,

showing thatPa andP divideAAP harmonically.


Examples of anticevian triangles

(1) The anticevian triangle of the centroid is the superior triangle, bounded by the linesthrough the vertices parallel to the opposite sides.

(2) The anticevian triangle of the incenter is the excentraltriangle whose vertices arethe excenters.

(3) The vertices of the tangential triangle being

A′ = (−a2 : b2 : c2), B′ = (a2 : −b2 : c2), C ′ = (a2 : b2 : −c2),

these clearly form are the anticevian triangle of a point with coordinates(a2 : b2 : c2),which we call thesymmedian pointK.

(4) The anticevian triangle of the circumcenter. Here is an interesting property ofcev−1(O). Let the perpendiculars toAC andAB at A intersectBC at Ab andAc re-spectively. We callAAbAc anorthial triangle of ABC. The circumcenter ofAAbAc is thevertexOa of cev−1(O); similarly for the other two orthial triangles. (See§??).

A

B C

Oa

Ob

Oc

Ab Ac

14.3 Cevian quotients 415

14.3 Cevian quotients

14.3.1 The cevian nest theorem

Theorem 14.1.For arbitrary pointsP andQ, the cevian trianglecev(P ) and the antice-vian trianglecev−1(Q) are always perspective. IfP = (u : v : w) andQ = (u′ : v′ : w′),then(i) the perspector is the point

∧(cev(P ), cev−1(Q)) =

(

u′(

−u′

u+

v′

v+

w′

w

)

: v′(

−v′

v+

w′

w+

u′

u

)

: w′(

−w′

w+

u′

u+

v′

v

))

,

(ii) the perspectrix is the lineL∧(cev(P ), cev−1(Q)) with equation

∑

cyclic

1

u

(

−u′

u+

v′

v+

w′

w

)

x = 0.

A

B C

Q

Y ′

Z′

X′

P

X

Y

ZM

Proof. (i) Let cev(P ) = XY Z andcev−1(Q) = X ′Y ′Z ′. SinceX = (0 : v : w) andX ′ = (−u′ : v′ : w′), the lineXX ′ has equation

1

u′

(

w′

w− v′

v

)

x− 1

v· y + 1

w· z = 0.

The equations ofY Y ′ andZZ ′ 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 lineXX ′ contains the point

(

u′(

−u′

u+

v′

v+

w′

w

)

: v′(

−v′

v+

w′

w+

u′

u

)

: w′(

−w′

w+

u′

u+

v′

v

))


whose coordinates are invariant under the above cyclic permutations. This point thereforealso lies on the linesY Y ′ andZZ ′.

(ii) The linesY Z andY ′Z ′ have equations

−xu

+ yv

+ zw

= 0,yv′

+ zw′

= 0.

They intersect at the point

U ′ = (u(wv′ − vw′) : vwv′ : −vww′).

Similarly, the lines pairsZX, Z ′X ′ andXY , X ′Y ′ have intersections

V ′ = (−wuu′ : v(uw′ − wu′) : wuw′)

andW ′ = (uvu′ : −uvv′ : w(vu′ − uv′)).

The three pointsU ′, V ′, W ′ lie on the line with equation given above.

Corollary 14.2. If T′ is a cevian triangle ofT andT′′ is a cevian triangle ofT′, thenT′′

is a cevian triangle ofT.

Proof. With reference toT2, the triangleT1 is anticevian.

Remark.SupposeT′ = cevT(P ) andT′′ = cevT′(Q). If P = (u : v : w) with respect toT, andQ = (u′ : v′ : w′) with respect toT′, then,

∧(T,T′′) =( u

u′(v + w) :

v

v′(w + u) :

w

w′(u+ v)

)

with respect to triangleT. The equation of the perspectrixL∧(T,T′′) is

∑

cyclic

1

u

(

−v + w

u′+

w + u

v′+

u+ v

w′

)

x = 0.

These formulae, however, are quite difficult to use, since they involve complicatedchanges of coordinates with respect to different triangles.

We shall simply write

P/Q :=∧

(cev(P ), cev−1(Q))

and call it thecevian quotientof P byQ.


The cevian quotients of the centroidG/P

If P = (u : v : w),

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

Some common examples ofG/P .

P G/P coordinatesI Mi (a(s− a) : b(s− b) : c(s− c))O K (a2 : b2 : c2)K O (a2Sα : b2Sβ : c2Sγ)

The pointMi = G/I is called theMittenpunktof triangleABC. 1 It is the symmedianpoint of the excentral triangle. The tangential triangle ofthe excentral triangle is homothetictoABC atT .

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

IbI2c =

a2bc

(s− b)(s− c), IcI

2a =

ab2c

(s− c)(s− a), IaI

2b =

abc2

(s− a)(s− b).

For the homogeneous barycentric coordinates of the symmedian point of the excentral tri-angleIaIbIc, we have

IbI2c · Ia + IcI

2a · Ib + IaI

2b · Ic

=a2bc

(s− b)(s− c)· (−a, b, c)

2(s− a)+

ab2c

(s− c)(s− a)· (a,−b, c)

2(s− b)+

abc2

(s− a)(s− b)· (a, b,−c)

2(s− c)

=abc

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

=abc

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

From this, we obtain the MittenpunktMi.(ii) Since the excentral triangle is homothetic to the intouch triangle atT , their tangen-

tial triangles are homothetic at the same triangle center. Using the cevian nest theorem, wehave

T =∧

0(cev(Ge), cev−1(I)) =

(

as−a : b

s−b :c

s−c)

.

1This appears asX9 in ETC.


The cevian quotients of the orthocenterH/P

ForP = (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 ofH•.(2)H/I is a point on theOI-line, dividingOI in the ratioR + r : −2r. 2

H/I = (a(a3 + a2(b+ c)− a(b2 + c2)− (b+ c)(b− c)2) : · · · : · · · ).

(3) H/K =(

a2

Sα: b2

Sβ: c2

Sγ

)

is the homothetic center of the orthic and tangential trian-

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 quotientGe/K

This is the perspector of the intouch triangle and the tangential triangle6

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

2This point appears asX46 in ETC.3This appears asX25 in ETC.4This appears asX155 in ETC.5This appears asX52 in ETC.6This appears asX1486 in ETC.


14.3.2 Basic properties of cevian quotients

Proposition 14.3. (1) P/P = P .(2) If P/Q = M , thenQ = P/M .

A

B C

Q

Y ′

Z′

X′

P

X

Y

ZMZ′′

Y ′′

X′′

Proof. (2) LetP = (u : v : w), Q = (u′ : v′ : w′), andM = (x : y : z). We have

x

u=

u′

u

(

−u′

u+

v′

v+

w′

w

)

,

y

v=

v′

v

(

u′

u− v′

v+

w′

w

)

,

z

w=

w′

w

(

u′

u+

v′

v− w′

w

)

.

From these,

−x

u+

y

v+

z

w=

(

u′

u− v′

v+

w′

w

)(

u′

u+

v′

v− w′

w

)

,

x

u− y

v+

z

w=

(

−u′

u+

v′

v+

w′

w

)(

u′

u+

v′

v− w′

w

)

,

x

u+

y

v− z

w=

(

−u′

u− v′

v+

w′

w

)(

u′

u− v′

v+

w′

w

)

,


and

x

u

(

−x

u+

y

v+

z

w

)

:y

v

(x

u− y

v+

z

w

)

:z

w

(x

u+

y

v− z

w

)

=u′

u:v′

v:w′

w.

It follows that

u′ : v′ : w′ = x(

−x

u+

y

v+

z

w

)

: y(x

u− y

v+

z

w

)

: z(x

u+

y

v− z

w

)

.

Chapter 15

Circle equations

15.1 The power of a point with respect to a circle

The power ofP with respect to a circleQ(ρ) is the quantityP(P ) := PQ2 − ρ2. A pointP is in, on, or outside the circle according asP is negative, zero, or positive.

Proposition 15.1.LetP = xA+ yB + zC in absolute barycentric coordinates.

P(P ) = xP(A) + yP(B) + zP(C)− (a2yz + b2zx+ c2xy).A

B C

Q

P

X

Proof. Let X be the trace ofP on BC. In absolute coordinates,X = yB+zC

y+z, so that

P = xA+ (y + z)X. Applying Stewart’s theorem in succession to trianglesQAX, QBCandABC, we have

PQ2 = xAQ2 + (y + z)XQ2 − x(y + z)AX2

= xAQ2 + (y + z)

(

y

y + zBQ2 +

z

y + zCQ2 − yz

(y + z)2BC2

)

− x(y + z)AX2

= xAQ2 + yBQ2 + zCQ2 − yz

y + zBC2 − x(y + z)AX2

= xAQ2 + yBQ2 + zCQ2 − yz

y + zBC2

− x(y + z)

(

z

y + zAC2 +

y

y + zAB2 − yz

(y + z)2·BC2

)

= xAQ2 + yBQ2 + zCQ2 − (1− x)yz

y + z·BC2 − zx ·AC2 − xy ·AB2

= 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 forP and writing

f := P(A), g := P(B), h := P(C),

for the powers ofA, B, C with respect to a circleQ(ρ), we have,

P(P ) =fx+ gy + hz

x+ y + z− a2yz + b2zx+ c2xy

(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 isa2yz + b2zx + c2xy = 0 sincef = 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, theA-excircle has equation

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

(4) For the nine-point circle, we havef = b2· c cosA = b

2· Sα

b= 1

2Sα. Similarly,

g = 12Sβ andh = 1

2Sγ. Therefore, the equation of the nine-point circle is

2(a2yz + b2zx+ c2xy)− (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 throughB andC, and tangent toBC atB,(ii) CBCC passing throughB andC, and tangent toBC atC.

3. Compute the coordinates of the Brocard points:(i) Ω→ as the intersection of the circlesCBBC , CCCA, andCAAB,(ii) Ω← as the intersection of the circlesCBCC , CCAA, andCABB.

4. Find the equation of the circle with diameterBC.

15.3 Points on the circumcircle

The equation of the circumcircle can be written in the form

a2

x+

b2

y+

c2

z= 0.

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

15.3.1 X(101)

The point

X(101) =

(

a2

b− c:

b2

c− a:

c2

a− b

)

is clearly on the circumcircle.

15.3.2 X(100)

The point

X(100) =

(

a

b− c:

b

c− a:

c

a− b

)

is clearly on the circumcircle. It is the isogonal conjugateof the infinite point

(a(b− c) : b(c− a) : c(a− b))

(on the trilinear polar of the incenter, namely, the linexa+ y

b+ z

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)).


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 fromX(100) to the Nagel point is the diameter of the incircle.

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

15.3.3 The Steiner pointX(99)

The Steiner point

X(99) =

(

1

b2 − c2:

1

c2 − a2:

1

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 pointE = X(110)

Theorem 15.2.The reflections of the Euler line in the sidelines of triangleABC are con-current at

E =

(

a2

b2 − c2:

b2

c2 − a2:

c2

a2 − b2

)

on the circumcircle.

Proof. The Euler line

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

intersects the sidelineBC at

X = (0 : Sγ(Sα − Sβ) : −Sβ(Sγ − Sα).

We find the reflection ofH 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

H = X +Sβγ

(Sβ + Sγ)(Sβγ + Sγα + Sαβ)(Sβ + Sγ,−Sγ ,−Sβ).

15.4 Circumcevian triangle 425

Therefore, its reflection inBC is the point

X ′ = X − Sβγ

(Sβ + Sγ)(Sβγ + Sγα + Sαβ)(Sβ + Sγ ,−Sγ ,−Sβ).

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 trianglesAY Z, BZX, CXY , whereXY Zis the orthic triangle.

15.4 Circumcevian triangle

Let P = (u : v : w). The linesAP , BP , CP intersect the circumcircle again atX, Y , Z.The triangleXY Z is called the circumcevian triangle ofP . SinceX lies on the lineAP ,X = (x : v : w) for somex. This point lies on the circumcircle if and only if

a2vw + b2xw + c2xv = 0.

This givesx = −a2vwb2w+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 ofP = (u : v : w) is perspective with thetangential triangle at

(

a2(

−a4

u2+

b4

v2+

c4

w2

)

: · · · : · · ·)

.

Proof. The vertices of the tangential triangle are(−a2, b2, c2), (a2,−b2, c2), (a2, b2,−c2).The line joining(−a2, b2, c2) toX is

∣

∣

∣

∣

∣

∣

x y z−a2 b2 c2

−a2vw (b2w + c2v)v (b2w + c2v)w

∣

∣

∣

∣

∣

∣

= 0.


This is(b4w2 − c4v2)x+ a2b2w2y − a2c2v2z = 0.

Similarly, the lines joining(a2,−b2, c2) to Y and(a2, b2,−c2) toZ 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) : · · · : · · · ).

2. H: X(24) =(

a2(a4+b4+c4−2a2b2−2a2c2)b2+c2−a2 : · · · : · · ·

)

.

These two points are on the Euler line, and are the centers of similitude of the circum-circle and incircle of the tangential triangle.

15.5 The third Lemoine circle

Given a pointP = (u : v : w), it is easy to find the equation of the circleCa throughP , B,C. SinceP(B) = P(C) = 0, the equation of the circle is

Ca : a2yz + b2zx+ c2xy − (x+ y + z) · fx = 0

for somef . Since the circle passes throughP = (u : v : w), we must have

f =a2vw + b2wu+ c2uv

u(u+ v + w).

This circleCa intersects the linesAC andAB each again at another point. To find theintersection withAC, we puty = 0 in the equation of(Ca) and obtainb2zx−fx(x+z) = 0,x((b2− f)z− fx)) = 0. Therefore, apart fromC = (0, 0, 1), the circleCa intersectsAC at

Ba = (b2 − f : 0 : f) = (b2u2 + b2uv − a2vw − c2uv : 0 : a2vw + b2wu+ c2uv).

Similarly, the circleCa intersectsAB again at

Ca = (c2 − f : f : 0) = (c2u2 + c2wu− a2vw − b2wu : a2vw + b2wu+ c2uv : 0).

Similarly, with

g =a2vw + b2wu+ c2uv

v(u+ v + w)and h =

a2vw + b2wu+ c2uv

w(u+ v + w),

the circlesCb throughP , C, A andCc throughP , A, B intersect the sidelines again at

Cb = (g : c2−g : 0), Ab = (0 : a2−g : g), Ac = (0 : h : a2−h), Bc = (h : 0 : b2−h).

15.5 The third Lemoine circle 427

Note the lengths of the segments:

ABa =f

b2· b = f

b, ABc =

b2 − h

b2· b = b2 − h

b,

and

ACa =f

c2· c = f

c, ACb =

c2 − g

c2· c = c2 − g

c.

The four pointsBa, Bc, Ca, Cb are concyclic if and only if

ABa · ABc = ACa · ACb =⇒ f(b2 − h)

b2=

f(c2 − g)

c2=⇒ b2

c2=

h

g=

v2

w2.

Likewise, the four pointsCb, Ca, Ab, Ac are concyclic if and only ifc2

a2= w

u, and the

four pointsAc, Ab, Bc, Ba are concyclic if and only ifa2

b2= u

v.

By the principle of6 concyclic points, the six pointsAb, Ac, Bc, Ba, Ca, Cb are con-cyclic if and only if

u : v : w = a2 : b2 : c2,

namely,P = (u : v : w) = (a2 : b2 : c2), the symmedian point. The circleC containingthese6 points is the third Lemoine circle.

For this choice ofP ,

f =3b2c2

a2 + b2 + c2, g =

3c2a2

a2 + b2 + c2, h =

3a2b2

a2 + b2 + c2.

With respect to the circleC containing these6 points, we have

P(A) =f(b2 − h)

b2=

3b2c2 · b2(b2 + c2 − 2a2)

b2(a2 + b2 + c2)2=

3b2c2(b2 + c2 − 2a2)

(a2 + b2 + c2)2.

Similarly,

P(B) =3c2a2(c2 + a2 − 2b2)

(a2 + b2 + c2)2, P(C) =

3a2b2(a2 + b2 − 2c2)

(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.

Documents

Advanced Euclidean Geometrymath.fau.edu/Yiu/AEG2013/2013AEG.pdf · 2013-08-02 · Advanced Euclidean Geometry Paul Yiu Summer 2013 Department of Mathematics Florida Atlantic University