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Survey of Geometry

Paul Yiu

Department of MathematicsFlorida Atlantic University

Spring 2008

Last update: March 26

Contents

1 The circumcircle and the incircle 11.1 The law of cosines and its applications . . . . . . . . . . . . . . 11.2 The circumcircle and the law of sines . . . . . . . . . . . . . . . 41.3 Steiner - Lehmus Theorem . . . . . . . . . . . . . . . . . . . . 51.4 The incircle and excircles . . . . . . . . . . . . . . . . . . . . . 81.5 Heron’s formula for the area of a triangle . . . . . . . . . . . . .12

2 The shoemaker’s knife 172.1 The shoemaker’s knife . . . . . . . . . . . . . . . . . . . . . . . 172.2 Circles in the shoemaker’s knife . . . . . . . . . . . . . . . . . . 182.3 Archimedean circles in the shoemaker’s knife . . . . . . . . .. 21

3 Introduction to Triangle Geometry 253.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.1.1 Coordinatization of points on a line . . . . . . . . . . . . 253.1.2 Centers of similitude of two circles . . . . . . . . . . . . 263.1.3 Tangent circles . . . . . . . . . . . . . . . . . . . . . . . 263.1.4 Harmonic division . . . . . . . . . . . . . . . . . . . . . 273.1.5 Homothety . . . . . . . . . . . . . . . . . . . . . . . . . 293.1.6 The power of a point with respect to a circle . . . . . . . . 29

3.2 Menelaus and Ceva theorems . . . . . . . . . . . . . . . . . . . 303.2.1 Menelaus and Ceva Theorems . . . . . . . . . . . . . . . 303.2.2 Desargues Theorem . . . . . . . . . . . . . . . . . . . . 303.2.3 The incircle and the Gergonne point . . . . . . . . . . . . 323.2.4 The excircles and the Nagel point . . . . . . . . . . . . . 35

3.3 The nine-point circle . . . . . . . . . . . . . . . . . . . . . . . . 373.3.1 The Euler triangle as a midway triangle . . . . . . . . . . 373.3.2 The orthic triangle as a pedal triangle . . . . . . . . . . . 38

iv CONTENTS

3.3.3 The nine-point circle . . . . . . . . . . . . . . . . . . . . 403.4 TheOI-line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.4.1 The homothetic center of the intouch and excentral triangles 433.4.2 The centers of similitude of the circumcircle and the incircle 443.4.3 Reflection ofI in O . . . . . . . . . . . . . . . . . . . . 453.4.4 Orthocenter of intouch triangle . . . . . . . . . . . . . . . 463.4.5 Centroids of the excentral and intouch triangles . . . .. . 473.4.6 Mixtilinear incircles . . . . . . . . . . . . . . . . . . . . 51

3.5 Euler’s formula and Steiner’s porism . . . . . . . . . . . . . . . 523.5.1 Euler’s formula . . . . . . . . . . . . . . . . . . . . . . . 533.5.2 Steiner’s porism . . . . . . . . . . . . . . . . . . . . . . 53

4 Homogeneous Barycentric Coordinates 574.1 Barycentric coordinates . . . . . . . . . . . . . . . . . . . . . . 574.2 Cevian and traces . . . . . . . . . . . . . . . . . . . . . . . . . 574.3 Homogeneous barycentric coordinates of classical triangle centers 58

4.3.1 The centroid . . . . . . . . . . . . . . . . . . . . . . . . 584.3.2 The incenter . . . . . . . . . . . . . . . . . . . . . . . . 594.3.3 Gergonne point . . . . . . . . . . . . . . . . . . . . . . . 604.3.4 The Nagel point . . . . . . . . . . . . . . . . . . . . . . . 614.3.5 The orthocenter . . . . . . . . . . . . . . . . . . . . . . . 62

4.4 Area and barycentric coordinates . . . . . . . . . . . . . . . . . 634.4.1 Barycentric coordinates of the circumcenter . . . . . . .. 65

4.5 The inferior and superior triangles . . . . . . . . . . . . . . . . 664.5.1 The nine-point center . . . . . . . . . . . . . . . . . . . . 67

4.6 Isotomic conjugates . . . . . . . . . . . . . . . . . . . . . . . . 684.6.1 Congruent-parallelians point . . . . . . . . . . . . . . . . 69

4.7 Triangles bounded by lines parallel to the sidelines . . .. . . . . 704.7.1 The Grebe symmedian point . . . . . . . . . . . . . . . . 71

5 Straight lines 735.1 Equations of straight lines . . . . . . . . . . . . . . . . . . . . . 73

5.1.1 Two-point form . . . . . . . . . . . . . . . . . . . . . . . 735.1.2 Intersection of two lines . . . . . . . . . . . . . . . . . . 74

5.2 Perspectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.3 Infinite points and parallel lines . . . . . . . . . . . . . . . . . . 79

5.3.1 The infinite point of a line . . . . . . . . . . . . . . . . . 795.4 Trilinear pole and polar . . . . . . . . . . . . . . . . . . . . . . 82

CONTENTS v

5.5 Anticevian triangles . . . . . . . . . . . . . . . . . . . . . . . . 835.5.1 Construction of anticevian triangle from trilinear polar and

polar properties . . . . . . . . . . . . . . . . . . . . . . . 845.6 The cevian nest theorem . . . . . . . . . . . . . . . . . . . . . . 86

5.6.1 G/P . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Chapter 1

The circumcircle and the incircle

1.1 The law of cosines and its applications

Given a triangleABC, we denote bya, b, c the lengths of the sidesBC, CA, ABrespectively.

Theorem 1.1(The law of cosines).

c2 = a2 + b2 − 2ab cos C.

a

bc

A

B C a

bc

A

B CX

Theorem 1.2(Stewart). If X is a point on the sideBC (or its extension) such thatBX : XC = λ : µ, then

AX2 =λb2 + µc2

λ + µ− λµa2

(λ + µ)2.

2 The circumcircle and the incircle

Proof. Use the cosine formula to compute the cosines of the anglesAXB andAXC, and note thatcos ABC = − cos AXB.

Corollary 1.3 (Apollonius’ theorem). The lengthma of themedianAD is givenby

m2a =

1

4(2b2 + 2c2 − a2).

A

B C

FE

D

G

H

A

B CX

Y

Theorem 1.4(Angle bisector theorem). If AX bisects angleBAC, thenBX :XC = c : b.

Proof. Construct a line throughB parallel to the bisectorAX to intersect theextension ofCA at Y . Note that∠AY B = ∠CAX = ∠XAB = ∠ABY . ThismeansAY = AB. Clearly,BX : XC = Y A : AC = AB : AC = c : b.

Corollary 1.5. The lengthwa of the (internal) bisector of angleA is given by

w2a = bc

(

1 −(

a

b + c

)2)

.

Proof. Apply Stewart’s Theorem withλ = c andµ = b.

Remark.If we write s = 12(a + b + c), thenw2

a = 4bcs(s−a)(b+c)2

.

1.1 The law of cosines and its applications 3

Exercise 1.1

1. Show that the (4,5,6) triangle has one angle equal to twice ofanother.

2. If C = 2B, show thatc2 = (a + b)b.

3. Find a simple relation between the sum of the areas of the three squaresS1,S2, S3, and that of the squaresT1, T2, T3.

S3

S2

S1

T1

T2

T3

4. ABC is a triangle witha = 12, b + c = 18, andcos A = 738

. Calculate thelengths ofb andc, 1 and show that

a3 = b3 + c3.

5. The lengths of the sides of a triangle are 136, 170, and 174. Calculate thelengths of its medians.2

6. (a)mb = mc if and only if b = c.(b) m2

a + m2b + m2

c = 34(a2 + b2 + c2).

(c) If ma : mb : mc = a : b : c, show that the triangle is equilateral.

7. Supposemb : mc = c : b. Show that either

(i) b = c, or

1AMM E688, P.A. Piza. Here,b = 9 −√

5, andc = 9 +√

5.2Answers: 158, 131, 127.


(ii) the quadrilateralAEGF is cyclic.

Show that the triangle is equilateral if both (i) and (ii) hold. [Hint: Showthatb2m2

b − c2m2c = 1

4(c − b)(c + b)(b2 + c2 − 2a2)].

8. The lengths of the sides of a triangle are 84, 125, 169. Calculate the lengthsof its internal bisectors.3

1.2 The circumcircle and the law of sines

The perpendicular bisectors of the three sides of a triangleare concurrent at thecircumcenterof the triangle. This is the center of the circumcircle, the circlepassing through the three vertices of the triangle.

O

C

A

BD

EF O

C

A

BD

Theorem 1.6 (The law of sines). Let R denote the circumradius of a triangleABC with sidesa, b, c opposite to the anglesA, B, C respectively.

a

sin A=

b

sin B=

c

sin C= 2R.

Since the area of a triangle is given by△ = 12bc sin A, the circumradius can

be written as

R =abc

4△ .

3Answers:9757 , 26208253 , 12600

209 .

1.3 Steiner - Lehmus Theorem 5

Exercise 1.2

1. Where is the circumcenter of a right triangle? What is its circumradius?4

2. (Theorthocenter) Given triangleABC, construct lines throughA parallelto BC, throughB parallel toCA, and throughC parallel toAB. Thesethree lines bound a triangleA′B′C ′ (so thatA lies onB′C ′, B on C ′A′,andC on A′B′ respectively). Show that the altitudes of triangleABC arethe perpendicular bisectors of the sides ofA′B′C ′. Deduce that the threealtitudes of a triangle always intersect at a point (called its orthocenter).

3. Let H be the orthocenter of triangleABC. Show that

(i) A is the orthocenter of triangleHBC;

(ii) the trianglesHAB, HBC, HCA andABC have the same circumra-dius.

1.3 Steiner - Lehmus Theorem

Theorem 1.7(Steiner-Lehmus). A triangle is isosceles if it has two equal internalangle bisectors.

First proof. 5

Supposeβ < γ in triangleABC. We show that the bisectorBM is longer thanthe bisectorCN .

M

NL

A

B C

4The midpoint of the hypotenuse; half of the length of the hypotenuse.5Gilbert - McDonnell, American Mathematical Monthly, vol. 70 (1963) 79 – 80.


Choose a pointL on BM such that∠NCL = 12β. ThenB, N , L, C are

concyclic since∠NBL = ∠NCL. Note that

∠NBC = β <1

2(β + γ) = ∠LCB,

and both are acute angles. Since smaller chords of a circle subtend smaller acuteangles, we haveCN < BL. It follows thatCN < BM .

Second proof.6

Suppose the bisectorsBM andCN in triangleABC are equal. We shall showthat β = γ. If not, assumeβ < γ. Compare the trianglesCBM andBCN .These have two pairs of equal sides with included angles∠CBM = 1

2β < 1

2γ =

∠BCN , both of which are acute. Their opposite sides therefore satisfy the rela-tion CM < BN .

MN

GA

B C

Complete the parallelogramBMGN , and consider the triangleCNG. This isisosceles sinceCN = BM = NG. Note that

∠CGN =1

2β + ∠CGM,

∠GCN =1

2γ + ∠GCM.

Sinceβ < γ, we conclude that∠CGM > ∠GCM . From this,CM > GM =BN . This contradicts the relationCM < BN obtained above.

6M. Descube, 1880.

1.3 Steiner - Lehmus Theorem 7

Third proof. 7

Let BB′, CC ′ be the respective internal angle-bisectors of anglesB, C in triangleABC, and leta, b, c denote the side-lengthsBC, CA, AB respectively. In thediagram below,β = 2θ andγ = 2ϕ.

V

vu

U

A

B C

C′

B′

θϕ

AssumeBB′ = CC ′. We show thatβ < γ (and henceb < c) lead to acontradiction.

(1) bu− c

v= u+U

u− v+V

v= U

u− V

v= a

c− a

b< 0 by the angle bisector theorem.

From this,bu

< cv.

(2) On the other hand, from

b

u÷ c

v=

b

c· v

u=

sin β

sin γ· v

u=

2 cos θ sin θ

2 cos ϕ sin ϕ· v

u=

cos θ

cos ϕ· sin θ

u· v

sin ϕ

=cos θ

cos ϕ· sin α

BB′· CC ′

sin α=

cos θ

cos ϕ> 1,

we havebu

> cv.

(1) and (2) clearly contradict each other. Likewise, a contradiction results fromβ > γ. We must conclude thatβ = γ.

Exercise 1.3

1. The internal bisectors of anglesB andC intersect the circumcircle of△ABCatB′ andC ′.

(i) Show that ifB = C, thenBB′ = CC ′.

(ii) If BB′ = CC ′, does it follow thatB = C? 8

2. TriangleABC has two equal external bisectorsw′

b andw′c which are also

equal to the length ofBC. Calculate the angles of the triangle.

7M. Hajja, 2008,Forum Geom., to appear.8(ii) No. BB′ = CC′ if and only if B = C or A = π

3 .


w′

b

w′c

aB

C

A

E

F

1.4 The incircle and excircles

The internal angle bisectors of a triangle are concurrent atthe incenterof thetriangle. This is the center of theincircle, the circle tangent to the three sides ofthe triangle.

If the incircle touches the sidesBC, CA andAB respectively atX, Y , andZ,

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

Z

X

Y

I

C

A

B

Denote byr the inradius of the triangleABC.

r =2△

a + b + c=

△s

.

1.4 The incircle and excircles 9

The internal bisector of each angle and theexternalbisectors of the remainingtwo angles are concurrent at anexcenterof the triangle. Anexcirclecan be con-structed with this as center, tangent to the lines containing the three sides of thetriangle.

Z

X

Y

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 re-

spectively. Since

△ = −△IaBC + △IaCA + △IaAB,

we have

△ =1

2ra(−a + b + c) = ra(s − a),

from whichra = △

s−a.

Exercise 1.4

1. Show that the inradius of a right triangle (of hypotenusec) is s − c.


2. A square of sidea is partitioned into 4 congruent right triangles and a smallsquare, all with equal inradiir. Calculater.

3. The incenter of a right triangle is equidistant from the midpoint of the hy-potenuse and the vertex of the right angle. Show that the triangle contains a30◦ angle.

B A

C

I

4. The incircle of triangleABC touches the sidesAC andAB at Y andZrespectively. SupposeBY = CZ. Show that the triangle is isosceles.

5. A line parallel to hypotenuseAB of a right triangleABC passes throughthe incenterI. The segments included betweenI and the sidesAC andBChave lengths 3 and 4. Calculate the area of the triangle.

B A

C

I

1.4 The incircle and excircles 11

6. If the incenter is equidistant from the three excenters, show that the triangleis equilateral.

7. The circleBIC intersects the sidesAC, AB atE andF respectively. ShowthatEF is tangent to the incircle of△ABC. 9

Z

X

YI

C

A

B

E

F

8. (a) The triangle is isosceles and the three small circles have equal radii.Suppose the large circle has radiusR. Find the radius of the small circles.10

(b) The large circle has radiusR. The four small circles have equal radii.Calculate this common radius.11

9Hint: Show thatIF bisects angleAFE.10Let θ be the semi-vertical angle of the isosceles triangle. The inradius of the triangle is

2R sin θ cos2 θ1+sin θ

= 2R sin θ(1 − sin θ). If this is equal toR2 (1 − sin θ), thensin θ = 1

4 . Fromthi,s the inradius is38R.

11Let θ be the smaller acute angle of one of the right triangles. The inradius of the right triangleis 2R sin θ cos θ

1+sin θ+cos θ. If this is equal toR

2 (1 − sin θ), then5 sin θ − cos θ = 1. From this,sin θ = 513 ,

and the radius is413R.


9. An equilateral triangle of side2a is partitioned symmetrically into a quadri-lateral, an isosceles triangle, and two other congruent triangles. If the inradiiof the quadrilateral and the isosceles triangle are equal, find this inradius.12

1.5 Heron’s formula for the area of a triangle

Consider a triangleABC with area△. Denote byr the inradius, andra the radiusof theexcircleon the sideBC of triangleABC. It is convenient to introduce thesemiperimeters = 1

2(a + b + c).

Ia

YY ′

I

A

B

C

ra r

12(√

3 −√

2)a.

1.5 Heron’s formula for the area of a triangle 13

• △ = rs.

• From the similarity of trianglesAIZ andAI ′Z ′,

r

ra

=s − a

s.

• From the similarity of trianglesCIY andI ′CY ′,

r · ra = (s − b)(s − c).

• From these,

r =

√

(s − a)(s − b)(s − c)

s,

△ =√

s(s − a)(s − b)(s − c).

This latter is the famous Heron formula.

Exercise 1.5

1. Thealtitudesa triangle are 12, 15 and 20. What is the area of the triangle ?13

2. Find the inradius and the exradii of the (13,14,15) triangle.

3. If one of the ex-radii of a triangle is equal to its semiperimeter, then thetriangle contains a right angle.

4. 1ra

+ 1rb

+ 1rc

= 1r.

5. rarbrc = rs2.

13△ = 150. The lengths of the sides are 25, 20 and 15.


6. Show that

(i) ra + rb + rc = −s3+(ab+bc+ca)s△

;

(ii) (s − a)(s − b)(s − c) = −s3 + (ab + bc + ca)s − abc.

Deduce that

ra + rb + rc = 4R + r.

7. The length of each side of the square is6a, and the radius of each of the topand bottom circles isa. Calculate the radii of the other two circles.14

8. (a) ABCD is a square of unit side.P is a point onBC so that the incircleof triangleABP and the circle tangent to the linesAP , PC andCD haveequal radii. Show that the length ofBP satisfies the equation

2x3 − 2x2 + 2x − 1 = 0.

A B

P

CD

A B

Q

CD

y

x

14a and 34a.

1.5 Heron’s formula for the area of a triangle 15

(b) ABCD is a square of unit side.Q is a point onBC so that the incircle oftriangleABQ and the circle tangent toAQ, QC, CD touch each other at apoint onAQ. Show that the radiix andy of the circles satisfy the equations

y =x(3 − 6x + 2x2)

1 − 2x2,

√x +

√y = 1.

Deduce thatx is the root of

4x3 − 12x2 + 8x − 1 = 0.


Chapter 2

The shoemaker’s knife

2.1 The shoemaker’s knife

Let P be a point on a segmentAB. The region bounded by the three semicircles(on the same side ofAB) with diametersAB, AP andPB is called a shoemaker’sknife. Suppose the smaller semicircles have radiia andb respectively. LetQ bethe intersection of the largest semicircle with the perpendicular throughP to AB.This perpendicular is an internal common tangent of the smaller semicircles.

A BOO1 O2P A BOO1 O2P

Q

U

V

H

K

R

Exercise

1. Show that the area of the shoemaker’s knife isπab.

2. Let UV be the external common tangent of the smaller semicircles, and Rthe intersection ofPQ andUV . Show that

(i) UV = PQ;

(ii) UR = PR = V R = QR. Hence, withR as center, a circle can bedrawn passing throughP , Q, U , V .

18 The shoemaker’s knife

3. Show that the circle throughU , P , Q, V has the same area as the shoe-maker’s knife.

2.2 Circles in the shoemaker’s knife

Theorem 2.1(Archimedes). The two circles each tangent toCP , the largest semi-circle AB 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 heightofthe center of this 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.

Theorem 2.2(Archimedes). The circle tangent to each of the three semicircleshas radius given by

ρ =ab(a + b)

a2 + ab + b2.

2.2 Circles in the shoemaker’s knife 19

A BOO1 O2P

C

X

Y

Proof. Let ∠COO2 = θ. By the cosine formula, we have

(a + ρ)2 = (a + b − ρ)2 + b2 + 2b(a + b − ρ) cos θ,(b + ρ)2 = (a + b − ρ)2 + a2 − 2a(a + b − ρ) cos θ.

Eliminatingθ, we have

a(a + ρ)2 + b(b + ρ)2 = (a + b)(a + b − ρ)2 + ab2 + ba2.

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

a3 + b3 + 2(a2 + b2)ρ = (a + b)3 + ab(a + b) − 2(a + b)2ρ,

from which

4(a2 + ab + b2)ρ = (a + b)3 + ab(a + b) − (a3 + b3) = 4ab(a + b),

andρ is as above.

Theorem 2.3(Leon Bankoff). If the incircleC(ρ) of the shoemaker’s knife touchesthe smaller semicircles atX andY , then the circle through the pointsP , X, Yhas the same radius as the Archimedean circles.

A BOO1 O2P

C

Z

X

Y


Proof. The circle throughP , X, Y is clearly the incircle of the triangleCO1O2,since

CX = CY = ρ, O1X = O1P = a, O2Y = O2P = b.

The semiperimeter of the triangleCO1O2 is

a + b + ρ = (a + b) +ab(a + b)

a2 + ab + b2=

(a + b)3

a2 + ab + b2.

The inradius of the triangle is given by

√

abρ

a + b + ρ=

√

ab · ab(a + b)

(a + b)3=

ab

a + b.

This is the same ast, the common radius of Archimedes’ twin circles.

Construction of incircle of shoemaker’s knife

Let Q1 andQ2 be the “highest” points of the semicirclesO1(a) andO2(b) re-spectively. The intersection ofO1Q2 andO2Q1 is a pointC3 “above” P , andC3P = ab

a+b= t. This gives a very easy construction of Bankoff’s circle in The-

orem 2.3 above. From this, we obtain the pointsX andY . The center of theincircle of the shoemaker’s knife is the intersectionC of the linesO1X andO2Y .The incircle of the shoemaker’s knife is the circleC(X). It touches the largestsemicircle of the shoemaker atZ, the intersection ofOC with this semicircle.

A BOO1 O2P

C

X

YC3

Q1

Q2

Z

Note thatC3(P ) is the Bankoff circle, which has the same radius as the Archimedeancircles.

2.3 Archimedean circles in the shoemaker’s knife 21

Exercise

1. Show that the area of triangleCO1O2 is ab(a+b)2

a2+ab+b2.

2. Show that the centerC of the incircle of the shoemaker’s knife is at a dis-tance2ρ from the lineAB.

3. Show that the area of the shoemaker’s knife to that of the heart (bounded bysemicirclesO1(a), O2(b) and thelowersemicircleO(a+ b)) is asρ to a+ b.

A BOO1 O2P

A BOO1 O2P

C

X Y

Z

Q1

Q2

4. Show that the points of contact of the incircleC(ρ) with the semicircles canbe located as follows:Y , Z are the intersections withQ1(A), andX, Z arethe intersections withQ2(B).

2.3 Archimedean circles in the shoemaker’s knife

Let UV be the external common tangent of the semicirclesO1(a) and O2(b),which extends to a chordHK of the semicircleO(a+b). LetC4 be the intersectionof O1V andO2U . Since

O1U = a, O2V = b, and O1P : PO2 = a : b,

C4P = aba+b

= t. This means that the circleC4(t) passes throughP and touchesthe common tangentHK of the semicircles atN .

Let M be the midpoint of the chordHK. SinceO and P are symmetric(isotomic conjugates) with respect toO1O2,

OM + PN = O1U + O2V = a + b.


A BOO1 O2P

N

U

V

H

K

C4

M

C5

it follows that (a + b) − QM = PN = 2t. From this, the circle tangent toHKand the minor arcHK of O(a + b) has radiust. This circle touches the minor arcat the pointQ.

Theorem 2.4(Thomas Schoch). The incircle of the curvilinear triangle boundedby the semicircleO(a + b) and the circlesA(2a) andB(2b) has radiust = ab

a+b.

A BOO1 O2P

S

Proof. Denote this circle byS(x). Note thatSO is a median of the triangleSO1O2. By Apollonius theorem,

(2a + x)2 + (2b + x)2 = 2[(a + b)2 + (a + b − x)2].

From this,

x =ab

a + b= t.

Exercise

1. The circles(C1) and (C ′1) are each tangent to the outer semicircle of the

shoemaker’s knife, and toOQ1 at Q1; similarly for the circles(C2) and(C ′

2). Show that they have equal radiit = aba+b

.

2.3 Archimedean circles in the shoemaker’s knife 23

C1

C2

C′

1

C′

2

A BOO1 O2P

Q1

Q2

2. We call the semicircle with diameterO1O2 the midwaysemicircle of theshoemaker’s knife.

Show that the circle tangent to the linePQ and with center at the intersec-tion of (O1) and the midway semicircle has radiust = ab

a+b.

C

C′

A BO PO1 O2

Q

3. Show that the radius of the circle tangent to the midway semicircle, theouter semicircle, and with center on the linePQ has radiust = ab

a+b.

C

A BO PO1 O2

Q


Chapter 3

Introduction to Triangle Geometry

3.1 Preliminaries

3.1.1 Coordinatization of points on a line

Let B andC be two fixed points on a lineL. Every pointX onL can be coordi-natized in one of several ways:

(1) the ratio of divisiont = BXXC

,(2) the absolute barycentric coordinates: an expression ofX as aconvexcom-

bination ofB andC:X = (1 − t)B + tC,

which expresses for an arbitrary pointP outside the lineL, the vectorPX as acombination of the vectorsPB andPC.

(3) the homogeneous barycentric coordinates: the proportion XC : BX,which are masses atB andC so that the resulting system (of two particles) hasbalance point atX.

P

B CX

26 Introduction to Triangle Geometry

3.1.2 Centers of similitude of two circles

Consider two circlesO(R) and I(r), whose centersO and I are at a distanced apart. Animate a pointX on O(R) and construct a ray throughI oppositelyparallel to the rayOX to intersect the circleI(r) at a pointY . You will find thatthe lineXY always intersects the lineOI at the same pointT . This we call theinternal center of similitude, or simply the insimilicenter, of the two circles. Itdivides the segmentOI in the ratioOT : TI = R : r. The absolute barycentriccoordinates ofP with respect toOI are

T =R · I + r · O

R + r.

O I

XY ′

Y

T T ′

If, on the other hand, we construct a ray throughI directly parallel to the rayOX to intersect the circleI(r) atY ′, the lineXY ′ always intersectsOI at anotherpointT ′. This is the external center of similitude, or simply the exsimilicenter, ofthe two circles. It divides the segmentOI in the ratioOT ′ : T ′I = R : −r, andhas absolute barycentric coordinates

T ′ =R · I − r · O

R − r.

3.1.3 Tangent circles

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

3.1 Preliminaries 27

O IT O IT ′

3.1.4 Harmonic division

Two pointsX andY are said to divide two other pointsB andC harmonically if

BX

XC= −BY

Y C.

They are harmonic conjugates of each other with respect to the segmentBC.

Examples

1. For two given circles, the two centers of similitude divide the centers har-monically.

2. Given triangleABC, let the internal bisector of angleA intersectBC atX. The harmonic conjugate ofX in BC is the intersection ofBC with theexternal bisector of angleA.

3. Let A andB be distinct points. IfM is the midpoint of the segmentAB,it is not possible to find afinite point N on the lineAB so thatM , Ndivide A, B harmonically. This is becauseAN

NB= −AM

MB= −1, requiring

AN = −NB = BN , andAB = BN − AN = 0, a contradiction. Weshall agree to say that ifM andN divideA, B harmonically, thenN is theinfinite pointof the lineAB.

Exercise

1. If X, Y divideB, C harmonically, thenB, C divideX, Y harmonically.


A

B CX X′

2. Given a pointX on the lineBC, construct its harmonic associate with re-spect to the segmentBC. Distinguish between two cases whenX dividesBC internally and externally.1

3. The centersA andB of two circlesA(a) andB(b) are at a distanced apart.The lineAB intersect the circles atA′ andB′ respectively, so thatA, B arebetweenA′, B′.

A′

A BB′

4. Given two fixed pointsB andC and a positive constantk 6= 1, the locus ofthe pointsP for which |BP | : |CP | = k is a circle.

1Make use of the notion of centers of similitude of two circles.

3.1 Preliminaries 29

3.1.5 Homothety

Given a pointT and a nonzero constantk, the similarity transformationh(T, k)which carries a pointX to the pointX ′ on the lineTX satisfyingTX ′ : TX =k : 1 is called thehomothetywith centerT and ratiok. Explicitly,

h(T, k)(P ) = (1 − k)T + kP.

Any two circles are homothetic. LetP andQ be the internal and externalcenters of similitude of two circlesO(R) andI(r). Both the homothetiesh(Q, r

R)

andh(P,− rR) transform the circleO(R) into I(r).

3.1.6 The power of a point with respect to a circle

Thepowerof a pointP with respect to a circleC = O(R) is the quantity

C(P ) := OP 2 − R2.

This is positive, zero, or negative according asP is outside, on, or inside the circleC. If it is positive, it is the square of the length of a tangent fromP to the circle.

Theorem (Intersecting chords)

If a line L throughP intersects a circleC at two pointsX andY , the productPX · PY (of signed lengths) is equal to the power ofP with respect to the circle.

P

T

T ′

X

Y

O


3.2 Menelaus and Ceva theorems

3.2.1 Menelaus and Ceva Theorems

Consider a triangleABC with pointsX, Y , Z on the side linesBC, CA, ABrespectively.

Theorem 3.1(Menelaus). The pointsX, Y , Z are collinear if and only if

BX

XC· CY

Y A· AZ

ZB= −1.

A

B CX

Y

Z

Theorem 3.2(Ceva). The linesAX, BY , CZ are concurrent if and only if

BX

XC· CY

Y A· AZ

ZB= +1.

3.2.2 Desargues Theorem

As a simple illustration of the use of the Menelaus and Ceva theorems, we provethe following rmDesargues Theorem: Given three circles, the exsimilicenters ofthe three pairs of circles are collinear. Likewise, the three lines each joining theinsimilicenter of a pair of circles to the center of the remaining circle are concur-rent.

We prove the second statement only. Given three circlesA(r1), B(r2) andC(r3), the insimilicentersX of (B) and(C), Y of (C), (A), andZ of (A), (B)are the points which divideBC, CA, AB in the ratios

BX

XC=

r2

r3,

CY

Y A=

r3

r1,

AZ

ZB=

r1

r2.

3.2 Menelaus and Ceva theorems 31

P

X

Y

Z

A

B C

A

B

C

X

Y

Z

P

X′

Y ′

Z′

It is clear that the product of these three ratio is+1, and it follows from the Cevatheorem thatAX, BY , CZ are concurrent.


3.2.3 The incircle and the Gergonne point

The incircle is tangent to each of the three sidesBC, CA, AB (without extension).Its center, the incenterI, is the intersection of the bisectors of the three angles. Theinradiusr is related to the area∆ by

S = (a + b + c)r.

I

A

B C

Ge

X

Y

Z

If the incircle is tangent to the sidesBC atX, CA atY , andAB atZ, then

AY = AZ =b + c − a

2, BZ = BX =

c + a − b

2, CX = CY =

a + b − c

2.

These expressions are usually simplified by introducing thesemiperimeters =12(a + b + c):

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

Also, r = ∆s.

It follows easily from the Ceva theorem thatAX, BY , CZ are concurrent.The point of concurrencyGe is called the Gergonne point of triangleABC.

TriangleXY Z is called the intouch triangle ofABC. Clearly,

X =B + C

2, Y =

C + A

2, Z =

A + B

2.


It is always acute angled, and

Y Z = 2r cosA

2, ZX = 2r cos

B

2, XY = 2r cos

C

2.

Exercise

1. Given three pointsA, B, C not on the same line, construct three circles,with centers atA, B, C, mutually tangent to each other externally.

C A

B

Z

Y

X

2. Construct the three circles each passing through the Gergonne point andtangent to two sides of triangleABC. The 6 points of tangency lie on acircle. 2

3. Two circles are orthogonal to each other if their tangents atan intersectionare perpendicular to each other. Given three pointsA, B, C not on a line,construct three circles with these as centers and orthogonal to each other.

(1) Construct the tangents fromA′ to the circleB(b), and the circle tangentto these two lines and toA(a) internally.

(2) Construct the tangents fromB′ to the circleA(a), and the circle tangentto these two lines and toB(b) internally.

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

2This is called the Adams circle. It is concentric with the incircle, and has radius√

(4R+r)2+s2

4R+r·

r.


I

A

B C

Ge

4. Given a pointZ on a line segmentAB, construct a right-angled triangleABC whose incircle touches the hypotenuseAB atZ. 3

5. Let ABC be a triangle with incenterI.

(1a) Construct a tangent to the incircle at the point diametrically oppositeto its point of contact with the sideBC. Let this tangent intersectCA atY1

andAB atZ1.

(1b) Same in part (a), for the sideCA, and let the tangent intersectAB atZ2 andBC atX2.

(1c) Same in part (a), for the sideAB, and let the tangent intersectBC atX3 andCA atY3.

(2) Note thatAY3 = AZ2. Construct the circle tangent toAC andAB atY3

andZ2. How does this circle intersect the circumcircle of triangleABC?

6. The incircle of△ABC touches the sidesBC, CA, AB atD, E, F respec-tively. X is a point inside△ABC such that the incircle of△XBC touchesBC atD also, and touchesCX andXB atY andZ respectively.

3P. Yiu, G. Leversha, and T. Seimiya, Problem 2415 and solution, Crux Math. 25 (1999) 110;26 (2000) 62 – 64.


(1) The four pointsE, F , Z, Y are concyclic.4

(2) What is the locus of the center of the circleEFZY ? 5

7. Given triangleABC, construct a circle tangent toAC at Y andAB at Zsuch that the lineY Z passes through the centroidG. Show thatY G : GZ =c : b.

A

B C

GY

Z

3.2.4 The excircles and the Nagel point

Let X ′, Y ′, Z ′ be the points of tangency of the excircles(Ia), (Ib), (Ic) with thecorresponding sides of triangleABC. The linesAX ′, BY ′, CZ ′ are concurrent.The common pointNa is called the Nagel point of triangleABC.

Exercise

1. Construct the tritangent circles of a triangleABC.

(1) Join each excenter to the midpoint of the corresponding side of ABC.These three lines intersect at a pointMi. (This is called the Mittenpunkt ofthe triangle).

4International Mathematical Olympiad 1996.5IMO 1996.


Ic

Ib

Na Y ′

Z′

Ia

X′

I

A

B

C

X

Y

Z

(2) Join each excenter to the point of tangency of the incircle with the cor-responding side. These three lines are concurrent at another pointT .

(3) The linesAMi andAT are symmetric with respect to the bisector of an-gleA; so are the linesBMi, BT andCMi, CT (with respect to the bisectorsof anglesB andC).

2. Construct the excircles of a triangleABC.

(1) LetD, E, F be the midpoints of the sidesBC, CA, AB. Construct theincenterSp of triangleDEF , 6 and the tangents fromS to each of the threeexcircles.

(2) The 6 points of tangency are on a circle, which is orthogonal to each ofthe excircles.

6This is called the Spieker point of triangleABC.

3.3 The nine-point circle 37

Ic

Ib

Y ′

Z′

Ia

X′

I

A

B

C

X

Y

Z Mi

T

3. Let D, E, F be the midpoints of the sidesBC, CA, AB, and let the incircletouch these sides atX, Y , Z respectively. The lines throughX parallel toID, throughY to IE and throughZ to IF are concurrent.7

3.3 The nine-point circle

3.3.1 The Euler triangle as a midway triangle

Let P be a point in the plane of triangleABC. The midpoints of the segmentsAP , BP , CP form the midway triangle ofP . It is the image ofABC under thehomothetyh(P, 1

2). The midway triangle of the orthocenterH is called the Euler

7Crux Math. Problem 2250. The reflection of the Nagel pointNa in the incenter. This isX145

of ETC.


Ia

Ic

Ib

SpY ′

Z′

X′

A

B

C

triangle. The circumcenter of the midway triangle ofP is the midpoint ofOP . Inparticular, the circumcenter of the Euler triangle is the midpoint ofOH, which isthe same asN , the circumcenter of the medial triangle. (See Exercise??.). Themedial triangle and the Euler triangle have the same circumcircle.

3.3.2 The orthic triangle as a pedal triangle

The pedals of a point are the intersections of the sidelines with the correspondingperpendiculars throughP . They form the pedal triangle ofP . The pedal triangleof the orthocenterH is called the orthic triangle ofABC.

The pedalX of the orthocenterH on the sideBC is also the pedal ofA on thesame line, and can be regarded as the reflection ofA in the lineEF . It followsthat

∠EXF = ∠EAF = ∠EDF,


D

EF

I

A

B CX

Y

Z

P

O

A

B C

A′

B′ C′

P

O′

sinceAEDF is a parallelogram. From this, the pointX lies on the circumcircleof the medial triangleDEF ; similarly for the pedalsY andZ of H on the othertwo sidesCA andAB.


A

B CX

Y

ZP

3.3.3 The nine-point circle

From§§3.3.1, 3.3.2 above, the medial triangle, the Euler triangle, and the orthictriangle all have the same circumcircle. This is called the nine-point circle oftriangleABC. Its centerN , the midpoint ofOH, is called the nine-point centerof triangleABC.

Exercise

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

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

3. If the midpoints ofAP , BP , CP are all on the nine-point circle, mustP bethe orthocenter of triangleABC? 8

8P. Yiu and J. Young, Problem 2437 and solution,Crux Math.25 (1999) 173; 26 (2000) 192.


ON

D

EF

A

B C

H

X

Y

Z

A′

B′ C′

4. Let ABC be a triangle andP a point. The perpendiculars atP to PA, PB,PC intersectBC, CA, AB respectively atA′, B′, C ′.

(1) A′, B′, C ′ are collinear.9

(2) The nine-point circles of the (right-angled) trianglesPAA′, PBB′,PCC ′ are concurrent atP and another pointP ′. Equivalently, their cen-ters are collinear.10

5. (Triangles with nine-point center on the circumcircle)

Begin with a circle, centerO and a pointN on it, and construct a family oftriangles with(O) as circumcircle andN as nine-point center.

(1) Construct the nine-point circle, which has centerN , and passes through

9B. Gibert, Hyacinthos 1158, 8/5/00.10A.P. Hatzipolakis, Hyacinthos 3166, 6/27/01. The three midpoints ofAA′, BB′, CC′ are

collinear. The three nine-point circles intersect atP and its reflection in this line.


0

Ic

Ib

Ia

I

A

B

C

I′

A

BC

P

B′

C′

A′

P ′

3.4 TheOI-line 43

the midpointM of ON .

(2) Animate a pointD on the minor arc of the nine-point circleinsidethecircumcircle.

(3) Construct the chordBC of the circumcircle withD as midpoint. (Thisis simply the perpendicular toOD atD).

(4) LetX be the point on the nine-point circle antipodal toD. Complete theparallelogramODXA (by translating the vectorDO to X).

The pointA lies on the circumcircle and the triangleABC has nine-pointcenterN on the circumcircle.

Here is a curious property of triangles constructed in this way: letA′, B′,C ′ be the reflections ofA, B, C in their own opposite sides. The reflectiontriangleA′B′C ′ degenerates,i.e., the three pointsA′, B′, C ′ are collinear.11

3.4 TheOI-line

3.4.1 The homothetic center of the intouch and excentral tri-angles

The OI-line of a triangle is the line joining the circumcenter and the incenter.We consider several interesting triangle centers on this line, which arise from thehomothety of the intouch and excentral triangles. These triangles are homotheticsince their corresponding sides are parallel, being perpendicular to the same anglebisector of the reference triangle.

The ratio of homothetic is clearly2Rr

. Their circumcenters areI ′ = 2O − IandI. The homothetic center is the pointT such that

2O − I = h

(

T,2R

r

)

(I) =

(

1 − 2R

r

)

T +2R

r· I.

This gives

T =(2R + r)I − 2r · O

2R − r.

It is the point which dividesOI externally in the ratio2R + r : −2r. 12

11O. Bottema,Hoofdstukken uit de Elementaire Meetkunde, Chapter 16.12The pointT is X57 in ETC.


Ic

Ib

Ia

I

A

B

C

X

Y

Z T

3.4.2 The centers of similitude of the circumcircle and the in-circle

These are the pointsT+ andT− which divide the segmentOI harmonically in theratio of the circumradius and the inradius.13

T+ =1

R + r(r · O + R · I),

T− =1

R − r(−r · O + R · I).

13T+ andT− are respectivelyX55 andX56 in ETC.

3.4 TheOI-line 45

OI

A

B CX

Y

Z

T+T−

M

M ′

3.4.3 Reflection ofI in O

The reflectionI ′ of I in O is the circumcenter of the excentral triangle. It is theintersections of the perpendiculars from the excenters to the sidelines.14

The midpointM of the arcBC is also the midpoint ofIIa. SinceIM andI ′Ia are parallel,I ′Ia = 2R. Similarly, I ′Ib = Pc = 2R. This shows that thermexcentral triangle has circumcenterI ′ = 2O − I and circumradius2R. SinceI is the orthocenter ofIaIbIc, its follows thatO, the midpoint ofI andI ′, is thenine-point center of the excentral triangle. In other words, the circumcircle oftriangleABC is the nine-point circle of the excentral triangle. Apart from A, thesecond intersection ofIbIc with the circumcircle ofABC is the midpointM ′ ofIbIc. From this we deduce the following interesting formula:15

ra + rb + rc = 4R + r.

14I ′ appears asX40 in ETC.15Proof. ra+rb+rc = 2MD+(I ′Ia−I ′X ′) = 2R+2OD+I ′Ia−I ′X ′ = 4R+IX = 4R+r.


O

Ic

Ib

XbXc

Ia

I

A

B

C

I′

M

M ′

D

3.4.4 Orthocenter of intouch triangle

The OI-line is the Euler line of the excentral triangle, sinceO and I are thenine-point center and orthocenter respectively. The corresponding sides of thermintouch triangle and the rmexcentral triangle are parallel, being perpendicularto the respective angle bisectors. Their Euler lines are parallel. Since the intouchtriangle has circumcenterI, its Euler line is actually the lineOI, which thereforecontains its orthocenter. This is the point which dividesOI in the ratioR + r :−r. 16 It also lies on the line joining the Gergonne and Nagel points.

16This is the pointX65 in ETC.

3.4 TheOI-line 47

OI

A

B CX

Y

Z

Ge Na

H′

3.4.5 Centroids of the excentral and intouch triangles

The centroid of the excentral triangle is the point which dividesOI in the ratio−1 : 4. 17 From the homothetyh(T, r

2R), it is easy to see that the centroid of the

intouch triangle is the point which dividesOI in the ratio3R + r : −r. 18

Exercises

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

17This is the point which divides the segmentI ′I in to the ratio1 : 2. It is X165 of ETC.18The centroid of the intouch triangle isX354 of ETC.


O

Ic

Ib

Ia

I

A

B

C

X

Y

ZT

2. Show that the distance between the circumcenter and the Nagel point isR − 2r. 19

3. Identify the midpoint of the Nagel point and the deLongchamps point as apoint on theOI-line. 20

4. Construct the orthic triangle of the intouch triangle. Thisis homothetic to

19Feuerbach’s theorem.20Reflection ofI in O. This is because the pedal triangle ofX40 is the cevian triangle of the

Nagel point, and the reflections of the pedals of the Nagel point in the respective traces form thepedals of the de Longchamps point.

3.4 TheOI-line 49

ABC. Identify the homothetic center.21

O

I

A

B CX

Y

ZH′

5. Construct the external common tangent of each pair of excircles. Thesethree external common tangents bound a triangle.

(i) This triangle is perpsective withABC. Identify the perspector.22

(ii) Identify the incenter of this triangle.23

6. ExtendAB andAC by lengtha and join the two points by a line. Similarlydefine the two other lines. The three lines bound a triangle with perspectorX65

24

7. Let H ′ be the orthocenter of the intouch triangleXY Z, andX ′, Y ′, Z ′ itspedals on the sidesY Z, ZX, XY respectively. Identify the common pointof the three linesAX ′, BY ′, CZ ′ as a point on theOI-line. Homotheticcenter of intouch and excentral triangles.

21T .22Orthocenter of intouch triangle.23Reflection ofI in O.242/18/03.


8. Let P be the point which dividesOI in the ratioOP : PI = R : 2r. Thereis a circle, centerP , radius Rr

R+2r, which is tangent to three congruent circles

of the same radius, each tangent to two sides of the triangle.Construct thesecircles.25

9. There are three circles each tangent internally to the circumcircle at a vertex,and externally to the incircle. It is known that the three lines joining thepoints of tangency of each circle with(O) and(I) pass through the internalcenterT+ of similitude of(O) and(I). Construct these three circles.26

OI

A

B C

T+

10. LetT+ be the insimilicenter of(O) and(I), with pedalsY andZ onCA andAB respectively. IfY ′ andZ ′ are the pedals ofY andZ on BC, calculatethe length ofY ′Z ′. 27

11. Let P be the centroid of the excentral triangle, with pedalsX, Y , Z on the

25A. P. Hatzipolakis, Hyacinthos message 793, April 18, 2000.26A.P. Hatzipolakis and P. Yiu, Triads of circles, preprint.27A.P. Hatzipolakis and P. Yiu, Pedal triangles and their shadows,Forum Geom., 1 (2001) 81 –

90.

3.4 TheOI-line 51

O

I

A

B C

T+

X

Y

Z

Y ′Z′

sidesBC, CA, AB respectively. Show that28

AY + AZ = BZ + BX = CX + CY =1

3(a + b + c).

3.4.6 Mixtilinear incircles

A mixtilinear incircle of triangleABC is one that is tangent to two sides of thetriangle and to the circumcircle internally. Denote byA′ the point of tangencyof the mixtilinear incircleK(ρ) in angleA with the circumcircle. The centerKclearly lies on the bisector of angleA, andAK : KI = ρ : −(ρ − r). In terms ofbarycentric coordinates,

K =1

r(−(ρ − r)A + ρI) .

28The projections ofO and I on the sideBC are the midpointD of BC, and the point oftangencyD′ of the incircle with this side. Clearly,BD = a

2 andBD′ = 12 (c + a − b). It follows

that

BX = BD′ +4

3D′D =

4

3BD − 1

3BD′ =

1

6(3a + b − c).

Similarly,BZ = 16 (3c + b− a), andBX + BZ = 1

3 (a + b + c). A similar calculation shows thatAY + AZ = CX + CY = 1

3 (a + b + c).


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

K =1

R(ρO + (R − ρ)A′) .

I

O

K

A′

A

B C

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 theintersection of the lines joiningAA′ andOI. Note that the point on the lineOIrepresented by the left hand side isT−, the external center of similitude of thecircumcircle and the incircle.

This leads to a simple construction of the mixtilinear incircle. Given a triangleABC, letP be the external center of similitude of the circumcircle(O) and incir-cle (I). ExtendAP 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.

3.5 Euler’s formula and Steiner’s porism

3.5 Euler’s formula and Steiner’s porism 53

O

KA

A′

A

B C

T−

M

I

3.5.1 Euler’s formula

The distance between the circumcenter and the incenter of a triangle is given by

OI2 = R2 − 2Rr.

Let the bisector of angleA intersect the circumcircle atM . Construct thecircleM(B) to intersect this bisector at a pointI. This is the incenter since

∠IBC =1

2∠IMC =

1

2∠AMC =

1

2∠ABC,

and for the same reason∠ICB = 12∠ACB. Note that

(1) IM = MB = MC = 2R sin A2,

(2) IA = r

sin A2

, and

(3) by the theorem of intersecting chords,OI2−R2 = thepowerof I with respectto the circumcircle =IA · IM = −2Rr.

3.5.2 Steiner’s porism

Construct the circumcircle(O) and the incircle(I) of triangleABC. Animate apoint A′ on the circumcircle, and construct the tangents fromA′ to the incircle(I). Extend these tangents to intersect the circumcircle againat B′ andC ′. ThelinesB′C ′ is always tangent to the incircle. This is the famous theoremon Steiner


OI

A

B C

M

porism: if two given circles are the circumcircle and incircle of one triangle, thenthey are the circumcircle and incircle of a continuous family of poristic triangles.

Exercises

1. r ≤ 12R. When does equality hold?

2. SupposeOI = d. Show that there is a right-angled triangle whose sides ared, r andR − r. Which one of these is the hypotenuse?

3. Given a pointI inside a circleO(R), construct a circleI(r) so thatO(R)and I(r) are the circumcircle and incircle of a (family of poristic) trian-gle(s).

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

5. Construct an animation picture of a triangle whose circumcenter lies on theincircle.29

29Hint: OI = r.

3.5 Euler’s formula and Steiner’s porism 55

OI

A

B CX

Y

Z

B′

A′

C′

6. What is the locus of the centroids of the poristic triangles with the samecircumcircle and incircle of triangleABC? How about the orthocenter?

7. Let A′B′C ′ be a poristic triangle with the same circumcircle and incircle oftriangleABC, and let the sides ofB′C ′, C ′A′, A′B′ touch the incircle atX,Y , Z.

(i) What is the locus of the centroid ofXY Z?

(ii) What is the locus of the orthocenter ofXY Z?

(iii) What can you say about the Euler line of the triangleXY Z?


Chapter 4

Homogeneous BarycentricCoordinates

4.1 Barycentric coordinates

The notion of barycentric coordinates dates back to Mobius. In a given triangleABC, every pointP is coordinatized by a triple of numbers(x : y : z) in sucha way that the system of massesx at A, y at B, andz at C will have itsbalancepoint atP . A massy atB and a massz atC will balance at the pointX = yB+zC

y+z

on the lineBC. A massx atA and a massy + z atX will balance at the point

P =xA + (y + z)X

x + (y + z)=

xA + yB + zC

x + y + z.

We say that with reference to triangleABC, the pointP has

(i) absolutebarycentric coordinatexA + yB + zC

x + y + zand

(ii) homogeneousbarycentric coordinates(x : y : z).

4.2 Cevian and traces

Because of the fundamental importance of the Ceva theorem intriangle geometry,we shall follow traditions and call the three lines joining apointP to the verticesof the reference triangleABC the ceviansof P . The intersectionsX, Y , Z ofthese cevians with the side lines are called thetracesof P . The coordinates of the

58 Homogeneous Barycentric Coordinates

traces can be very easily written down:

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

AC

B

P

X

Y

Z

Theorem 4.1(Ceva theorem). Three pointsX, Y , Z onBC, CA, AB respectivelyare the traces of a point if and only if they have coordinates of the form

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

for somex, y, z.

4.3 Homogeneous barycentric coordinates of classi-cal triangle centers

4.3.1 The centroid

The midpoint points of the sides have coordinates

X = (0 : 1 : 1),Y = (1 : 0 : 1),Z = (1 : 1 : 0).

The centroidG has coordinates(1 : 1 : 1). 1

1The centroid appears in Kimberling’sEncyclopedia of Triangle centers(ETC, available athttp://faculty.evansville.edu/ck6/encyclopedia/ETC.html) as the pointX2.

4.3 Homogeneous barycentric coordinates of classical triangle centers 59

AC

B

G

X

Y

Z

4.3.2 The incenter

AC

B

IX

Y

Z

The traces of the incenter have coordinates

X = (0 : b : c),Y = (a : 0 : c),Z = (a : b : 0).

The incenterI has coordinatesI = (a : b : c). 2

2The incenter appears inETC as the pointX1.


4.3.3 Gergonne point

s − b s − c

s − c

s − a

s − a

s − b

AC

B

IGe

X

Y

Z

The points of tangency of the incircle with the side lines are

X = 0 : s − c : s − b,Y = s − c : 0 : s − a,Z = s − b : s − a : 0.

These can be reorganized as

X = 0 : 1s−b

: 1s−c

,

Y = 1s−a

: 0 : 1s−c

,

Z = 1s−a

: 1s−b

: 0.

It follows thatAX, BY , CZ intersect at a point with coordinates(

1

s − a:

1

s − b:

1

s − c

)

.

This is called theGergonne pointGe of triangleABC. 3

3The Gergonne point appears inETC as the pointX7.

4.3 Homogeneous barycentric coordinates of classical triangle centers 61

4.3.4 The Nagel point

The points of tangency of the excircles with the corresponding sides have coordi-nates

X ′ = (0 : s − b : s − c),Y ′ = (s − a : 0 : s − c),Z ′ = (s − a : s − b : 0).

These are the traces of the point with coordinates

(s − a : s − b : s − c).

This is theNagel pointNa of triangleABC. 4

A

B C

Ia

Ib

Ic

Z′

X′

Y ′

s − c s − b

s − a

s − cs − b

s − a Na

4The Nagel point appears inETC as the pointX8.


4.3.5 The orthocenter

The traceX of the orthocenterH dividesBC in the ratio

BX : XC = c cosB : b cos C.

AC

B

HX

Y

Z

In homogeneous barycentric coordinates,

X = (b cos C : c cos B) =

(

0 :b

cos B:

c

cos C

)

.

Similarly, the other two traces are

Y =( a

cos A: 0 :

c

cos C

)

,

Z =

(

a

cos A:

b

cos B: 0

)

.

The orthocenter is the point5

H =

(

a

cos A:

b

cos B:

c

cos C

)

.

By the law of cosines, these coordinates can be rewritten as

H =

(

1

b2 + c2 − a2:

1

c2 + a2 − b2:

1

a2 + b2 − c2

)

.

5The orthocenter appears inETC as the pointX4.

4.4 Area and barycentric coordinates 63

4.4 Area and barycentric coordinates

Theorem 4.2. If in homogeneous barycentric coordinates with reference to trian-gleABC, P = (x : y : z), then

△PBC : △APC : △ABP = x : y : z.

Theorem 4.3.If for i = 1, 2, 3, Pi = x1 ·A+y1 ·B+zi ·C (in absolute barycentriccoordinates) , then the area of the oriented triangleP1P2P3 is

△P1P2P3 =

∣

∣

∣

∣

∣

∣

x1 y1 z1

x2 y2 z2

x3 y3 z3

∣

∣

∣

∣

∣

∣

· △ABC.

Theorem 4.4(Routh theorem). If X, Y , Z are points on the linesBC, CA, ABrespectively such that

BX : XC = λ : 1, CY : Y A = µ : 1, AZ : ZB = µ : 1,

then the cevian linesAX, BY , CZ bound a triangle with area

(λµν − 1)2

(µν + µ + 1)(νλ + ν + 1)(λµ + λ + 1)· △ABC.

λ 1

µ

1

ν

1

A

B CX

Z

Y

P Q

R

Proof. In homogeneous barycentric coordinates with reference to triangleABC,

X = (0 : 1 : λ), Y = (µ : 0 : 1), Z = (1 : ν : 0).

Those ofP , Q, R can be worked out easily:

P = BY ∩ CZ Q = CZ ∩ AX R = AX ∩ BY

Y = (µ : 0 : 1) Z = (1 : ν : 0) X = (0 : 1 : λ)Z = (1 : ν : 0) X = (0 : 1 : λ) Y = (µ : 0 : 1)

P = (µ : µν : 1) Q = (1 : ν : νλ) R = (λµ : 1 : λ)


This means that theabsolute barycentric coordinatesof P , Q, R are

P =1

µν + µ + 1(µA + µνB + C),

Q =1

νλ + ν + 1(A + νB + νλC),

R =1

λµ + λ + 1(λµA + B + λC).

From these,

Area(PQR) =

∣

∣

∣

∣

∣

∣

µ µν 11 ν νλ

λµ 1 λ

∣

∣

∣

∣

∣

∣

(µν + µ + 1)(νλ + ν + 1)(λµ + λ + 1)· △

=(λµν − 1)2

(µν + µ + 1)(νλ + ν + 1)(λµ + λ + 1)· △.

Example 4.1. Given a triangleABC, X, Y , Z are points on the side lines speci-fied by the ratios of divisions

BX : XC = 2 : 1, CY : Y A = 5 : 3, AZ : ZB = 3 : 2.

The linesAX, BY , CZ bound a trianglePQR. Suppose triangleABC has area△. Find the area of trianglePQR.

2 1

5

3

3

2

A

B CX

Z

Y

P Q

R

We make use ofhomogeneous barycentric coordinateswith respect toABC.

X = (0 : 1 : 2), Y = (5 : 0 : 3), Z = (2 : 3 : 0).

4.4 Area and barycentric coordinates 65

Those ofP , Q, R can be worked out easily:

P = BY ∩ CZ Q = CZ ∩ AX R = AX ∩ BY

Y = (5 : 0 : 3) Z = (2 : 3 : 0) X = (0 : 1 : 2)Z = (2 : 3 : 0) X = (0 : 1 : 2) Y = (5 : 0 : 3)

P = (10 : 15 : 6) Q = (2 : 3 : 6) R = (10 : 3 : 6)

This means that theabsolute barycentric coordinatesof X, Y , Z are

P =1

31(10A + 15B + 6C), Q =

1

11(2A + 3B + 6C), R =

1

19(10A + 3B + 6C).

The area of trianglePQR

=1

31 · 11 · 19

∣

∣

∣

∣

∣

∣

10 15 62 3 610 3 6

∣

∣

∣

∣

∣

∣

· △ =576

6479△.

4.4.1 Barycentric coordinates of the circumcenter

Consider the circumcenterO of triangleABC.

O

A

B C

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

1

2· OB · OC · sin BOC =

1

2R2 sin 2A.


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

12R2 sin 2C. It follows that the circumcenterO has homogeneous barycentric co-

ordinates

△OBC : △OCA : △OAB

=1

2R2 sin 2A :

1

2R2 sin 2B :

1

2R2 sin 2C

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

=a cos A : b cos B : c cos C

=a · b2 + c2 − a2

2bc: b · c2 + a2 − b2

2ca: c · a2 + b2 − c2

2ab=a2(b2 + c2 − a2) : b2(c2 + a2 − b2) : c2(a2 + b2 − c2).

4.5 The inferior and superior triangles

The triangle whose vertices are the midpoints of the sides oftriangleABC iscalled theinferior triangle ofABC. Its vertices are the points

A′ = (0 : 1 : 1), B′ = (1 : 0 : 1), C ′ = (1 : 1 : 0).

This triangle ishomothetictoABC at the centroidG, with ratio−12. Equivalently,

we say that it is the image ofABC under the homothetyh(G,−12). This means

that h(G,−12) is a one-to-one correspondence of points on the plane such that

P andP ′ have the same homogeneous barycentric coordinates with reference toABC andA′B′C ′ wheneverX, X ′ andG are collinear and

GP ′

GP= −1

2.

We callP ′ the inferior of P . More explicitly,P ′ dividesPG in the ratioPP ′ :P ′G = 3 : −1, so thatP ′ = 1

2(3G−P ). SupposeP has homogeneous barycentric

coordinates(x : y : z) with reference toABC. It is the pointP = xA+yBzC

x+y+z. 6

6The conversion from homogeneous barycentric coordinates to absolute barycentric coordi-nates is callednormalization.

4.5 The inferior and superior triangles 67

Thus,

P ′ =1

2(3G − P )

=1

2

(

A + B + C − xA + yBzC

x + y + z

)

=1

2

(

(x + y + z)(A + B + C) − (xA + yB + zC)

x + y + z

)

=(y + z)A + (z + x)B + (x + y)C

2(x + y + z).

It follows that the homogenous coordinates ofP ′ are(y + z : z + x : x + y).Thesuperiortriangle ofABC is its image under the homothetyh(G,−2). Its

vertices are the points

A′′ = (−1 : 1 : 1), B′′ = (1 : −1 : 1), C ′′ = (1 : 1 : −1).

4.5.1 The nine-point center

The nine-point circle is the circumcircle of the inferior triangle. Its centerN is theinferior of the circumcenter. FromN = 3G−O

2andO = 1

2(3G − H), we obtain

N = 12(O + H). The nine-point centerN is the midpoint ofOH. 7

Exercise

1. The Nagel pointNa lies on the line joining the incenter to the centroid; itdividesIG in the ratioINa : NaG = 3 : −2.

2. Find the coordinates of the circumcenterO by using the fact thatOG :GH = 1 : 2.

3. The superiorof P is the pointP ′′ on the lineGP for which GP ′′

GP= −2.

Show that ifP = (x : y : z), thenP ′′ = (−x+y+z : x−y+z : x+y−z).

4. Find the coordinates of the superior of the incenter.

5. Find the coordinates of the inferior of the incenter. Show that it is the cen-troid of theperimeterof triangleABC.

7The nine-point center appears inETC as the pointX5.


6. The midway triangle of a pointP is the triangle whose vertices are themidpoints ofAP , BP , CP . If P = (x : y : z), find the coordinates of thecentroid and the incenter of the midway triangle with reference toABC.

7. Make use of the fact thatOH : HG = 3 : −2 to compute the homogeneousbarycentric coordinates of the circumcenterO.

8. Make use of the fact that the nine-point centerN divides the segmentOGin the ratioON : GN = 3 : −1 to show that its barycentric coordinates canbe written as8

N = (a cos(B − C) : b cos(C − A) : c cos(A − B)).

4.6 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 beisotomic conjugates if their respective traces are symmetric with respect to themidpoints of the corresponding sides. 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)

.8In ETC, the nine-point center appears as the pointX5.

4.6 Isotomic conjugates 69

4.6.1 Congruent-parallelians point

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

1 − x : 1 − y : 1 − z =1

a:1

b:

1

c.

The right hand side clearly gives the homogeneous barycentric coordinates ofI•,the isotomic conjugate of the incenterI. 9 This is a point we can easily construct.Now, translating intoabsolutebarycentric 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 thesegmentI•G externally in the ratioI•P : PG = 3 : −2. The pointP is called thecongruent-parallelians point of triangleABC. 10

AC

B

I•

P G

Exercises

1. Calculate the homogeneous barycentric coordinates of the congruent-parallelianpoint and the length of the equal parallelians.11

9The isotomic conjugate of the incenter appears inETC as the pointX75.10This point appears inETC as the pointX192.11(ca + ab− bc : ab + bc− ca : bc + ca− ab). The common length of the equal parallelians is2abc

ab+bc+ca.


2. Let A′B′C ′ be the midway triangle of a pointP . The lineB′C ′ intersectsCA at

Ba = B′C ′ ∩ CA, Ca = B′C ′ ∩ AB,Cb = C ′A′ ∩ AB, Ab = C ′A′ ∩ BC,Ac = A′B′ ∩ BC, Bc = A′B′ ∩ CA.

DetermineP for which the three segmentsBaCa, CbAb and AcBc haveequal lengths.

4.7 Triangles bounded by lines parallel to the side-lines

Theorem 4.5. If parallel linesXbXc, YcYa, ZaZb to the sidesBC, CA, AB oftriangleABC 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 has homogeneous barycentriccoordinates

[PBC] : [PCA] : [PAB] = t1 : t2 : t3.

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

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

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

A similar calculation shows thatAA∗ andBB∗ intersect at the same pointP . Thisshows the homothety of the triangle, with ratio1 + t1 + t2 + t3.

Now we compare areas. Note that(1) [BZaXc] : [ABC] = t3 · t1.(2) [PBC] : [BZaB

∗] = 1t1+t2+t3

· t3.

Since[BZaB∗] = [BZaXc], we have[PBC] = t1

1+t1+t2+t3· [ABC].

4.7 Triangles bounded by lines parallel to the sidelines 71

A

B C

A∗

B∗ C∗

P

Xc Xb

Ya

YcZb

Za

A

B C

A∗

B∗ C∗

P

Xc Xb

Ya

YcZb

Za

Similarly, [PCA] = t21+t1+t2+t3

· [ABC] and[PAB] = t31+t1+t2+t3

· [ABC]. Itfollows that

[PBC] : [PCA] : [PAB] = t1 : t2 : t3.

4.7.1 The Grebe symmedian point

Consider the square erected externally on the sideBC of triangleABC, The linecontaining the outer edge of the square is the image ofBC under the homothety

h(A, 1 + t1), where1 + t1 =Sa+aSa

= 1 + a2

S, i.e., t1 = a2

S. Similarly, if we erect

squares externally on the other two sides, the outer edges ofthese squares are onthe lines which are the images ofCA, AB under the homothetiesh(B, 1+ t2) andh(C, 1 + t3) with t2 = b2

Sandt3 = c2

S.

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

Grebe triangleof ABC. It is homothetic toABC at(

a2

S: b2

S: c2

S

)

= (a2 : b2 :

c2), and the ratio of homothety is

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

S.

This homothetic center is called theGrebe symmedian point

K = (a2 : b2 : c2).


Ab Ac

Bc

Ba

Ca

Cb

A

B C

A∗

B∗ C∗

K

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 areconstructed on the sides of triangleABC, the lines containing their outer edgesalways bound a triangle with homothetic centerK.

Chapter 5

Straight lines

5.1 Equations of straight lines

5.1.1 Two-point form

The area formula has an easy and extremely important consequence: the equationof the line joining two points with coordinates(x1 : y1 : z1) and(x2 : y2 : z2) is

∣

∣

∣

∣

∣

∣

x1 y1 z1

x2 y2 z2

x 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 intersectcorresponding sidelines at the traces ofP :

AP = (0 : v : w), BP = (u : 0 : w), CP = (u : v : 0).

74 Straight lines

(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

5.1.2 Intersection of two lines

The intersection of the two lines

p1x + q1y + r1z = 0,p2x + q2y + r2z = 0

is the point

(∣

∣

∣

∣

q1 r1

q2 r2

∣

∣

∣

∣

: −∣

∣

∣

∣

p1 r1

p2 r2

∣

∣

∣

∣

:

∣

∣

∣

∣

p1 q1

p2 q2

∣

∣

∣

∣

)

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

Proposition 5.1. Three linespix + qiy + riz = 0, i = 1, 2, 3, are concurrent ifand only if

∣

∣

∣

∣

∣

∣

p1 q1 r1

p2 q2 r2

p3 q3 r3

∣

∣

∣

∣

∣

∣

= 0.

Examples

(1) The intersection of the Euler line and the Soddy line is the point

5.1 Equations of straight lines 75

∣

∣

∣

∣

(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) : · · · : · · ·

This intersection has homogeneous barycentric coordinates

−3a4+2a2(b2+c2)+(b2−c2)2 : −3b4+2b2(c2+a2)+(c2−a2)2 : −3c4+2c2(a2+b2)+(a2−b2)2.

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

Exercise

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

2. Find the equations of the angle bisectors and determine fromthem the co-ordinates of the incenter and excenters.

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

4. LetD, E, F be the midpoints of the sides ofBC, CA, AB of triangleABC,andX, Y , Z the midpoints of the altitudes fromA, B, C respectively. Findthe equations of the linesDX, EY , FZ, and show that they are concurrent.What are the coordinates of their intersection?3

1Equation:(v − w)x + (w − u)y + (u − v)z = 0.2In this case, the Nagel point lies on the incircle, which is tangent to a side of the inferior

triangle.3(b2 − c2)x − a2(y − z) = 0 and two similar ones. The intersection is the symmedian point

K = (a2 : b2 : c2).

76 Straight lines

D

EF

H

A

B C

X

Y

Z

Sp

D

EF

A

B C

5. Show that each angle bisector of the inferior triangle bisects the perimeterof the reference triangle.

6. Given triangleABC, extend, if necessary,(i) AC andAB to Ya andZa such thatAYa = AZa = a,(ii) BA andBC to Zb andXb such thatBZb = BXb = b,(iii) CB andCA to Xc andYc such thatCXc = CYc = c.

Show that the linesYcZb, ZaXc, andXbYa are concurrent and find the coor-dinates of the common point.4

7. Let D, E, F be the midpoints of the sidesBC, CA, AB of triangleABC.For a pointP with tracesX, Y , Z, let X ′, Y ′, Z ′ be the midpoints ofY Z, ZX, XY respectively. Find the equations of the linesDX ′, EY ′,FZ ′, and show that they are concurrent. What are the coordinatesof theirintersection?5

5.2 Perspectivity

Many interesting points and lines in triangle geometry arise from theperspectiv-ity of triangles. We say that two trianglesX1Y1Z1 andX2Y2Z2 are perspective,X1Y1Z1 ⊼ X2Y2Z2, if the linesX1X2, Y1Y2, Z1Z2 are concurrent. The pointof concurrency,∧(X1Y1Z1, X2Y2Z2), is called theperspector. Along with the

4The Nagel point.5The intersection is the point dividing the segmentPG in the ratio3 : 1.

5.2 Perspectivity 77

A

B CXc Xb

Za

Zb

Yc

Ya

A

B C

P

D

EF

X

YZ

X′

Y ′

Z′

perspector, there is anaxis of perspectivity, or theperspectrix, which is the linejoining 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 per-

78 Straight lines

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

Proposition 5.2. 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). Theperspectrix is the line

x

u − U+

y

v − V+

z

w − W= 0.

Proof. The lineAX has equationwy − vz = 0. It intersects the sidelineBC atthe 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 − V W )x + u(w − W )y + u(v − V )z = 0.It intersects the sidelineBC at (0 : v − V : −(w − W )). Similarly, the linesZX andXY intersectCA andAB respectively at(−(u − U) : 0 : w − W ) and(u − U : −(v − V ) : 0). These three points are collinear on the trilinear polar of(u − U : v − V : w − W ).

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

Exercise

1. Given triangleABC, extend the sidesAC to Ba andAB to Ca such thatCBa = BCa = a. Similarly defineCb, Ab, Ac, andBc. Calculate thecoordinates of the intersectionsA′ of BBa andCCa, B′ of CCb andAAb, C ′

of AAc, BBc. Show thatAA′, BB′ andCC ′ are concurrent by identifyingtheir common point.

2. Given two pointsP1 andP2 with cevian trianglesX1Y1Z1 andX2Y2Z2, theline joining X1Y1 ∩ X2Z2 andX1Z1 ∩ X2Y2 passes throughA. Similarlyfor the other two lines. The three lines are concurrent. Their common pointhas coordinates

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

5.3 Infinite points and parallel lines 79

B′

A′

C′

A

B CAc

Bc

Ba

Ca

Cb

Ab

5.3 Infinite points and parallel lines

5.3.1 The infinite point of a line

A line px + qy + rz = 0 contains the point(q − r : r − p : p − q), as is easilyverified. Since the sum of its coordinates is zero, this is nota finite point. Wecall it an infinite point. Thus, an infinite point(x : y : z) is one which satisfiesthe equationx + y + z = 0, which we regard as defining theline at infinityL∞.Now, it is easy to see that unlessp : q : r = 1 : 1 : 1, a linepx + qy + rz = 0has a unique infinite point as given above. Therefore, the infinite point of a linedetermines its “direction”.

Two lines areparallel if and only if they have the same infinite point. It followsthat the line through(u : v : w) parallel topx + qy + rz = 0 has equation

∣

∣

∣

∣

∣

∣

x y zu v w

q − r r − p p − q

∣

∣

∣

∣

∣

∣

= 0.

80 Straight lines

Examples

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

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

(2) The infinite point of the Soddy lineIGe has coordinates

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

Exercise

1. Find the equations of the lines throughP = (u : v : w) parallel to thesidelines.

2. Let D, E, F be the midpoints of the sidesBC, CA, AB of triangleABC.For a pointP with tracesAP , BP , CP , let X, Y , Z be the midpoints ofBPCP , CPAP , AP BP respectively.

(a) FindP such that the linesDX, EY , FZ are parallel to the internalbisectors of anglesA, B, C respectively.

(b) Explain why the linesDX, EY , FZ are concurrent and identify thepoint of concurrency.

3. Given a pointP in the plane of triangleABC, three parallelograms areconstructed with a common vertex atP and each with two sides parallel totwo sidelines of triangleABC, and a diagonal along the third side. Showthat the “fourth” vertices of these parallelograms form a triangle perspectivewith ABC.

4. Let P = (u : v : w) be a point. Consider the triangle bounded by linesparallel to the sides of cevian triangle ofP through the vertices ofABC.This is the anticevian triangle of(u(v + w) : v(w + u) : w(u + v)). It ishomothetic to the cevian triangle ofP . The homothetic center is the point

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

5.3 Infinite points and parallel lines 81

I P

AP

BP

CP

A

B CD

EF

X

YZ

A

B C

P

5. The barycentric cube of an infinite point is the centroid of the cevian triangleof the point. It happens that the barycentric cube of the Euler line is againon the Euler line, and the Euler line is the only line throughO with this

82 Straight lines

property. For a given pointP 6= G, only the linePG has this property.6

5.4 Trilinear pole and polar

If the intersections of a lineL with the side lines are

X = (0 : v : −w), Y = (−u : 0 : w), Z = (u : −v : 0),

the equation of the lineL is

x

u+

y

v+

z

w= 0.

We shall call the pointP = (u : v : w) thetrilinear pole (or simply tripole) ofL,and the lineL thetrilinear polar (or simply tripolar) ofP .

Construction of tripolar of a point

GivenP with tracesAP , BP , andCP on the side lines, let

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

These pointsX, Y , Z lie on the tripolar ofP .

Construction of 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 ′ intersect at the tripoleP of L.

Exercise

1. The trilinear polar ofP intersects the sidelinesBC, CA, AB at X, Y , Zrespectively. The centroid of the degenerate triangleXY Z is called thetripolar centroidof P . If P = (u : v : w), calculate the coordinates of itstripolar centroid.7

62/14/04.7(u(v − w)(v + w − 2u) : · · · : · · · ).

5.5 Anticevian triangles 83

A

B CX

Z

Y

P

AP

BP

CP

2. The trilinear polar ofP intersects the sidelinesBC, CA, AB at X, Y , Zrespectively. The centroid of the degenerate triangleXY Z is called thetripolar centroidof P . If P = (u : v : w), calculate the coordinates of itstripolar centroid.8

5.5 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

8(u(v − w)(v + w − 2u) : · · · : · · · ).

84 Straight lines

A

B C

X

Y

Z

P

A′

B′

C′

andCCP , i.e.,

AP : PAP = −APa : PaAP ;

similarly for Pb andPc. This is called the anticevian triangle ofP sinceABCis the cevian triangleP in PaPbPc. It is also convenient to regardP , Pa, Pb, Pc

as a harmonic quadruple in the sense that any three of the points constitute theharmonic associates of the remaining point.

5.5.1 Construction of anticevian triangle from trilinear polarand polar properties

If the trilinear polarLP of P intersects the sidelinesBC, CA, AB at X ′, Y ′, Z ′

respectively, then the anticevian trianglecev−1(P ) is simply the triangle bounded

by the linesAX ′, BY ′, andCZ ′.

Examples of anticevian triangles

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

5.5 Anticevian triangles 85

A

B CX

Z

Y

P

AP

BP

CPPc

Pa

Pb

(2) The anticevian triangle of the incenter is the excentraltriangle whose ver-tices are the excenters.

(3) The vertices of the tangential triangle being

Ka = (−a2 : b2 : c2), Kb = (a2 : −b2 : c2), Kc = (a2 : b2 : −c2),

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

Exercise

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

2. The anticevian triangles ofO andK have equal areas.

86 Straight lines

O

Ka

Kb

Kc

A

B C

K

3. The areas of the orthic triangle, the reference triangle, and the anticeviantriangle ofO are in geometric progression.

4. Find the coordinates ofP = (u : v : w) in its anticevian triangle.9

5. If P does not lie on the sidelines of triangleABC and is the centroid of itsown anticevian triangle, show thatP is the centroid of triangleABC.

5.6 The cevian nest theorem

Theorem 5.1. For arbitrary pointsP andQ, the cevian trianglecev(P ) and theanticevian trianglecev−1(Q) are always perspective. IfP = (u : v : w) andQ = (u′ : v′ : w′), then

∧(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

))

.

9(v + w − u : w + u − v : u + v − w).

5.6 The cevian nest theorem 87

Proof. Let cev(P ) = XY Z andcev−1(Q) = X ′Y ′Z ′. SinceX = (0 : v : w) and

X ′ = (−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 permutationsof (u, v, w), (u′, v′, w′) and(x, y, z). It is easy to check that the lineXX ′ 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 pointtherefore also lies on the linesY Y ′ andZZ ′.

Remark.The perspectrix of the two triangles is the line∑

cyclic

1

u

(

−u′

u+

v′

v+

w′

w

)

x = 0.

Corollary 5.2. 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 compli-cated changes of coordinates with respect to different triangles.

We shall simply write

P/Q :=∧

(cev(P ), cev−1(Q))

and call it thecevian quotientof P by Q.

Exercise. 1. Find the coordinates of the cevian quotientP/Q with referenceto the cevian triangle ofP . 10

10(

v+w−xvw+ywu+zuv

: w+uxvw−ywu+zuv

: u+vxvw+ywu−zuv

)

.

88 Straight lines

5.6.1 G/P

If P = (u : v : w),

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

Some common examples ofG/P .

P G/P coordinatesI Mi (a(s − a) : b(s − b) : c(s − c))O K (a2 : b2 : c2)K O (a2(b32 + c2 − a2) : b2(c2 + a2 − b2) : c2(a2 + b2 − c2))

The pointMi = G/I is called theMittenpunktof triangleABC. 11 It is thesymmedian point of the excentral triangle. The tangential triangle of the excentraltriangle is homothetic toABC atT .

(3)∧

0(cev(Ge), cev−1(I)) = T =

(

as−a

: bs−b

: cs−c

)

.(4) The perspector of the intouch triangle and the tangential triangle is12

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

Theorem5.3. (1) P/P = P ;(2) If P/Q = M , thenQ = P/M .

Exercise. 1. (a) Show that the line joining the excenterIa to the point of tan-gency of the incircle witha passes through the midpoint of theA-altitude.13

(b) Show that the three lines each joining the midpoint of an altitude on aside to the point of tangency of the incircle with that side are concurrent.What is the point of concurrency?14

2. Let X, Y , Z be the midpoints of the sidesBP CP , CP AP , AP BP of thecevian triangle ofP . Show thatXY Z andABC are perspective.15

3. Given a pointQ = (x : y : z), find the pointP such thatcev(P ) andcev

−1(Q) are homothetic. What is the homothetic center?16

11This appears asX9 in ETC.12This appears asX1486 in ETC.13a · M1 + (s − a)Ia = s · X .14T .15Perspector =(u(v +w) : v(w +u) : w(u+ v)), and perspectrix is the linex

u2 + y

v2 + zw2 = 0.

16P =(

1y+z−x

: 1z+x−y

: 1x+y−z

)

; homothetic center=(

xy+z−x

: y

z+x−y: z

x+y−z

)

.

5.6 The cevian nest theorem 89

4. Let P = (u : v : w) andQ = (u′ : v′ : w′) be two given points. If

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

show that triangleXY Z andcev(P ) are perspective.17

5. Let XY Z be the intouch triangle andℓ a line through the incenterI oftriangleABC. Construct the pedalsX ′ of A, Y ′ of B, andZ ′ of C on ℓ.The linesXX ′, Y Y ′, ZZ ′ concur at a pointP . The locus of the isogonalconjugate ofP with respect toXY Z is the nine-point circle ofXY Z.

6. Let XY Z be the intouch triangle andℓ a line through the incenterI oftriangleABC. Construct the pedalsX ′ of X, Y ′ of Y , andZ ′ of Z on ℓ.The linesAX ′, BY ′, CZ ′ concur at a pointP . The locus of the isogonalconjugate ofP with respect toABC is a circle.

17(uu′(vw′ + wv′) : · · · : · · · ); see J. H. Tummers, Points remarquables, associes a un triangle,Nieuw Archief voor WiskundeIV 4 (1956) 132 – 139. O. Bottema, Une construction par rapport aun triangle, ibid., IV 5 (1957) 68 – 70, has subsequently shown that this is the pole of the linePQ

with respect to the circumconic throughP andQ.

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