Advanced Euclidean Geometry - Florida Atlantic Universitymath.fau.edu/yiu/AEG2016/160711.pdf · 2016. 7. 9. · Advanced Euclidean Geometry Paul Yiu Department of Mathematics Florida

Advanced Euclidean Geometry

Paul Yiu

Department of MathematicsFlorida Atlantic University

Summer 2016

July 11

Menelaus and Ceva Theorems

Menelaus’ theorem

Theorem 0.1 (Menelaus). Given a triangle ABC with points X , Y , Z onthe side lines BC, CA, AB respectively, the points X , Y , Z are collinear ifand only if

BX

XC· CY

Y A· AZZB

= −1.

A

B CX

Y

Z

W

Yiu: Advanced Euclidean Geometry 2016

1

Menelaus’ theorem

Proof. (=⇒)

A

B CX

Y

Z

W

Let W be the point on AC such that BW//XY . Then,

BX

XC=

WY

Y C, and

AZ

ZB=

AY

YW.

It follows that

BX

XC· CY

Y A· AZZB

=WY

Y C· CY

Y A· AYYW

=CY

Y C· AYY A

· WY

YW= −1.


2

Menelaus theorem

(⇐=) Suppose the line joining X and Z intersects AC at Y ′. From above,

BX

XC· CY ′

Y ′A· AZZB

= −1 =BX

XC· CY

Y A· AZZB

.

It follows thatCY ′

Y ′A=

CY

Y A.

The points Y ′ and Y divide the segment CA in the same ratio. These mustbe the same point, and X , Y , Z are collinear.


3

Example: The external bisectors

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 are AX ′, BY ′, CZ ′ with X ′, Y ′, Z ′ on BC,CA, AB respectively, 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 points X ′, Y ′, Z ′ are collinear.


4

Example

Given triangle ABC and pointsY on CA, Z on AB, and X on the extension of BC,such that X , Y , Z are collinear.If CX = x, BZ = z, and AY = y,and two of these lengths are given, calculate the remaining one.

c

ba

x

y

z AB

C

Z

X

Y


5

c

ba

x

y

z AB

C

Z

X

Y

(a, b, c) x y z

(3, 4, 5) 1 2 4

(3, 5, 6) 1 4

(3, 5, 7) 1 6

(4, 5, 6) 3 4

(4, 5, 7) 2 4

(5, 6, 7) 1 6

(6, 7, 8) 4 6


6

(2, 3, 4) triangle

ABC is a triangle with a = 2, b = 3, c = 4.A transversal intersects the sidelines at X , Y , Zsuch that AY = BZ = CX = t.Calculate t.

4

32

AB

C

Z

X

Y


7

(7, 12, 18) triangle

ABC is a triangle with a = 7, b = 12, c = 18.A transversal intersects the sidelines at X , Y , Zsuch that AZ = BX = CY = t.Calculate t.

18

127

AB

C

Z

Y

X


8

(9, 10, 12) triangle

ABC is a triangle with a = 9, b = 10, c = 12.A transversal intersects the sidelines at X , Y , Zsuch that BX = CY = AZ = t.Calculate t.

12

109

AB

C

Z

X

Y


9

Example

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

A

B

C

X

Y

Z


10

Line with equal intercepts on sidelines of a given triangle

Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internally so that CX = CY = AZ.

x

x

x

A

B C

Z

X

Y


11

Solution

Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internally so that CX = CY = AZ.

rr

x

I

A

B C

Z

X

Y

If CX = x, by Menelaus’ theorem, we requirea+ x

−x· x

b− x· x

c− x= −1.

From this,

x =bc

a+ b+ c.

Note that

x =bc sinA

2s sinA=

Δ

s· 1

sinA=

r

sinA.

This means that IZ is parallel to CA, and suggests the following simpleconstruction of the line.


12

Construction

r

r

x

I

A

B C

Z

X

Y

(1) Construct the incenter I of triangleABC.(2) Construct a line through I parallel to AC, to intersect AB at Z.(3) Construct a circle with center C, radius AZ, to intersect BC externallyat X and CA internally at Y .

Then X , Y , Z are collinear with CX = CY = AZ = rsinA .


13

Line with equal intercepts on sidelines of a given triangle

Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internallyso that CX = CY = BZ.

y

y

y

A

B C

Z

X

Y


14

Solution

Given triangle ABC,construct a line intersecting BC at X externally,CA at Y and AB at Z internally so that CX = CY = BZ.

y y

y

A

B C

Z

X

Y

Ac

Bc

Cc

Ic

If CX = CY = BZ = y, by Menelaus’ theorem, we require

a+ y

−y· y

b− y· c− y

y= −1.

From this,y =

ca

a+ b− c=

rcsinB

,

where rc is the radius of the excircle on the sideAB.


15

Ceva’s theorem

Theorem 0.2 (Ceva). Given a triangleABC with pointsX , Y , Z on the sidelines BC, CA, AB respectively, the lines AX , BY , CZ are concurrent ifand only if

BX

XC· CY

Y A· AZZB

= +1.

P

X

Y

Z

A

B C


16

Proof

P

X

Y

Z

A

B C

Proof. (=⇒) Suppose the lines AX , BY , CZ intersect at a point P . Con-sider the line BPY cutting the sides of triangle CAX . By Menelaus’ theo-rem,

CY

Y A· APPX

· XB

BC= −1, or

CY

Y A· PA

XP· BX

BC= +1.

Also, consider the line CPZ cutting the sides of triangle ABX . ByMenelaus’ theorem again,

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.


17

Example

Given triangle ABC and points X on BC, Y on CA, Z on AB

such that the cevians AX , BY , CZ are concurrent.If CX = x, BZ = z, and AY = y,and two of these lengths are given, calculate the remaining one.

c

ba

z

y

x

AB

C

Z

X Y

(a, b, c) x y z

(3, 4, 6) 1 2 4

(3, 5, 6) 1 4

(3, 5, 7) 1 3

(4, 5, 6) 3 4

(4, 5, 7) 1 4

(5, 6, 7) 3 4

(6, 7, 9) 5 4


18

(3, 4, 6) triangle

ABC is a triangle with a = 3, b = 4, c = 6.X , Y , Z are points on BC, CA, AB respectivelysuch that BZ = AY = CX = t.If the cevians AX , BY , CZ are concurrent, calculate t.

6

43

AB

C

Z

X Y


19

Example

ABC is a right triangle.Show that the lines AX , BY , and CQ are concurrent.

Q

PA B

C

Z Z′

Y

Y ′

X′

X


20

Solution

ABC is a right triangle.Show that the lines AX , BY , and CQ are concurrent.

Q

Z0A B

C

Z Z′

Y

Y ′X′

X

X0Y0

Let AX intersect BC at X0,BY intersect CA at Y0, and CQ intersect AB at Z0.

AZ0

Z0B· BX0

X0C· CY0

Y0A=

AZ0 · AZZ0B · BZ ′ ·

BX

AC· BC

AY

=AC2

BC2· BC

AC· BC

AC= 1.

By Ceva’s theorem, the lines AX0, BY0, CZ0 are concurrent.Yiu: Advanced Euclidean Geometry 2016

21

The centroid

If D, E, F are the midpoints of the sides BC, CA, AB of triangle ABC,then clearly

AF

FB· BD

DC· CE

EA= 1.

The medians AD, BE, CF are therefore concurrent. Their intersection isthe centroid G of the triangle.

CB

A

GF

D

E

Consider the line BGE intersecting the sides of triangle ADC. By theMenelaus theorem,

−1 =AG

GD· DB

BC· CE

EA=

AG

GD· −1

2· 11.

It follows that AG : GD = 2 : 1.The centroid of a triangle divides each median in the ratio 2:1.


22

The incenter

Let X , Y , Z be points on BC, CA, AB such that AX , BY , CZ bisectangles BAC, CBA and ACB respectively. Then

AZ

ZB=

b

a,

BX

XC=

c

b,

CY

Y A=

a

c.

CB

A

IZ

X

Y

It follows that

AZ

ZB· BX

XC· CY

Y A=

b

a· cb· ac= +1,

and AX , BY , CZ are concurrent. Their intersection is the incenter of thetriangle.


23

Example

Given a point P , let the lines AP , BP , CP intersectBC, CA, AB respectively at X , Y , Z.Construct the circle through X , Y , Z,to intersect the lines BC, CA, AB again at X ′, Y ′, Z ′.Then the lines AX ′, BY ′, CZ ′ are concurrent.


24

Example

Suppose two cevians, each through a vertex of a triangle, trisect each other.Show that these are medians of the triangle.

A

B C

YZ

P


25

Solution

Suppose two cevians, each through a vertex of a triangle, trisect each other.Show that these are medians of the triangle.

A

B C

YZP

Given: Triangle ABC with cevians BY and CZ intersecting at Psuch that P trisects BY and CZ.To prove: BY and CZ are medians,i.e., Y and Z are midpoints of CA and AB respectively.

Proof. (1) Since P is a trisection of BY and CQ,ZCCP = −3

p for p = 1 or 2,PYY B = −q

3 for q = 1 or 2.(2) Applying Menelaus’ theorem to triangle BPZ with transversal CY A,we have. . .

26

(3) Similarly, by applying Menelaus’ theorem to triangleCPY with transver-sal AZB, we conclude that Y is the midpoint of CA, and BY is also amedian.


27

Example

Let AX , BY , CZ be cevians of �ABC intersecting at a point P .(i) Show that if AX bisects angle A, and BX · CY = XC · BZ, then

�ABC is isosceles.(ii) Show if if AX , BY , CZ are bisectors, and BP · ZP = BZ · AP ,

then �ABC is a right triangle.


28

Example

ABC is an isosceles triangle with β = γ = 40◦. Y and Z are points on CA

and AB respectively such that BY bisects angle B and ∠ZCB = 30◦. LetP be the intersection of BY and CZ. Show that BP = AB.

A

B C

Y

Z

P


29

Example

Let ABC be a triangle with A = 40◦, B = 60◦. Let E and F be pointslying on the sides AC and AB respectively, such that ∠EBC = 40◦ and∠FCB = 70◦. Let BE and CF intersect at P . Show that AP is perpendic-ular to BC.

A

B C

EP

F


30

Example

Let A,B,C,D,E, F be six consecutive points on a circle. Show that thechords AD, BE, CF are concurrent if and only if AB · CD · EF = BC ·DE · FA.

A

B

C

D

E

F


31

Example

A1A2A3 · · ·A11A12 is a regular 12-gon. Show that the diagonals A1A5,A3A6, and A4A8 are concurrent.

A1

A2

A3

A4

A5

A6

A7

A8

A9

A10

A11

A12


32

Kiepert perspectors

If similar isosceles triangles A′BC, B′CA and C ′AB (of base angle θ)are constructed on the sides of triangle ABC,either all externally or all internally,the lines AA′, BB′, and CC ′ are concurrent.

A

B C

C′

A′

B′

P

α1 α2

θ

β1

β2

γ2

γ1

θ

θ

θ θ

θ


33

Proof

Proof. Applying the law of sines to triangles ABA′ and ACA′, we have

sinα1

sinα2=

sinα1

sin(β + θ)· sin(β + θ)

sin(γ + θ)· sin(γ + θ)

sinα2

=BA′

AA′ ·sin(β + θ)

sin(γ + θ)· AA

′

CA′

=sin(β + θ)

sin(γ + θ).

Likewise, sinβ1

sinβ2= sin(γ+θ)

sin(α+θ) and sin γ1sin γ2

= sin(α+θ)sin(β+θ) . From these

sinα1

sinα2· sin β1sin β2

· sin γ1sin γ2

= +1.

and the lines AA′, BB′, CC ′ are concurrent.

The point of intersection is called the Kiepert perspector K(θ). In par-ticular, it is called(1) a Fermat point if θ = ±60◦,(2) a Napoleon point if θ = ±30◦,(3) a Vecten point if θ = ±45◦.


34

Trigonmetric version of the Ceva Theorem

Let X be a point on the side BC of triangle ABC such that the directedangles ∠BAX = α1 and ∠XAC = α2. By the sine formula,

BX

XC=

BX/AX

XC/AX=

sinα1/ sin β

sinα2/ sin γ=

sin γ

sin β· sinα1

sinα2=

c

b· sinα1

sinα2.

X

A

B C

Z

Y

α1 α2

β1

β2

γ2

γ1

Likewise, if Y and Z be points on the lines CA, AB respectively, with∠CBY = β1, ∠Y BA = β2 and ∠ACZ = γ1, ∠ZCB = γ2, we have

CY

Y A=

a

c· sin β1sin β2

,AZ

ZB=

b

a· sin γ1sin γ2

.

These lead to the following trigonometric version of the Ceva theorem.Yiu: Advanced Euclidean Geometry 2016

35

Theorem 0.3 (Ceva). The lines AX , BY , CZ are concurrent if and only if

sinα1


· sin γ1sin γ2

= +1.

Proof.AZ

ZB· BX

XC· CY

Y B=

sinα1


· sin γ1sin γ2

.


36

Advanced Euclidean Geometry - Florida Atlantic Universitymath.fau.edu/yiu/AEG2016/160711.pdf · 2016. 7. 9. · Advanced Euclidean Geometry Paul Yiu Department of Mathematics Florida

Documents