GeometryClass Examples (July 3)

Paul Yiu

Department of MathematicsFlorida Atlantic University

a

bc

A

B C

Summer 2014

27

Example C11(a): Fermat point.Given triangle ABC, construct externally similar isosceles triangles XBC,Y CA, and ZAB with base angles θ.

θ θ

A

B C

Z

X

Y

(a) Applying the law of cosines to triangle ABX ,

AX2 = AB2 +BX2 − 2 · AB · BX cosABX

= c2 +( a

2 cos θ

)2− ca

cos θcos(B + θ)

= c2 +a2

4 cos2 θ− ca(cosB cos θ − sinB sin θ)

cos θ

= c2 +a2

4 cos2 θ− ca cosB + ca sinB tan θ

= c2 +a2

4 cos2 θ− 1

2(c2 + a2 − b2) + 2Δ tan θ

=1

2(a2 + b2 + c2) +

1− 4 cos2 θ

4 cos2 θa2 + 2Δ tan θ.

(b) This expression is symmetric in a, b, c if and only if cos θ = 12 , i.e.,

θ = ±60◦.With θ = 60◦, we have

AX2 =1

2(a2 + b2 + c2 + 4

√3Δ).

This means that if equilateral triangles XBC, Y CA, and ZAB are con-structed externally of triangle ABC, the segments AX , BY , CZ have equallengths.

28

Example C11(b): The Fermat point.

Given triangle ABC, construct equilateral triangles XBC, Y CA, andZAB externally on the sides. Let the circumcircle of XBC intersect theline AX at F .

A

B C

F

Z

X

Y

(a) Prove that ∠BFX = ∠XFC = 60◦.(b) Prove that A, F , C, Y are concyclic. Similarly, A, F , B, Z are also

concyclic.(c) Prove that B, F , Y are collinear. Similarly, C, F , Z are also collinear.The point F is called the Fermat point of triangle ABC. It is the point of

concurrency of the lines AX , BY , CZ. It is also the common point of thecircumcircles of the equilateral triangles XBC, Y CA, ZAB.

29

Apollonius’ Theorem

Theorem. Given triangle ABC, let D be the midpoint of BC. The length ofthe median AD is given by

AB2 + AC2 = 2(AD2 +BD2).

A

B CD

Proof. Applying the law of cosines to triangles ABD and ACD, and notingthat cosADB = − cosADC, we have

AB2 = AD2 +BD2 − 2AD · BD · cosADB;

AC2 = AD2 + CD2 − 2AD · CD · cosADC,

AC2 = AD2 +BD2 + 2AD · BD · cosADB.

The result follows by adding the first and the third lines.

If ma denotes the length of the median on the side BC,

m2a =

1

4(2b2 + 2c2 − a2).

30

Example C12.

ABC is a triangle with a = 8, b = 9, c = 11. Two of its medians haverational lengths. What are these?

11

98

C

B A

F

D E

1

1

mb=172;mc=

132.

31

Example C13.The lengths of the sides of a triangle are 136, 170, and 174. Calculate thelengths of its medians.

1

4.1. Angle bisector theorem

Theorem (Angle bisector theorem). The bisectors of an angle of a triangledivide its opposite side in the ratio of the remaining sides. If AX and AX ′

respectively the internal and external bisectors of angle BAC, then BX :XC = c : b and BX ′ : X ′C = c : −b.

c b

Z′

Z

A

B CX X′

Proof. Construct lines through C parallel to the bisectors AX and AX ′ tointersect the line AB at Z and Z ′.

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

(2) Similarly, AZ ′ = AC, and BX ′ : X ′C = BA : AZ ′ = BA :−AC = c : −b.

2

Example D1.A square is inscribed in a right triangle with sides a and b. Show that eachside of the square has length

� =ab

a+ b.

b

a

�

�

A

B

C

3

Solution to Example D1.

A square is inscribed in a right triangle with sides a and b. Show that eachside of the square has length

� =ab

a+ b.

b

a

�

�

A

B

C

P

Solution. �BCP + �ACP = �ABC =⇒ 12at +

12bt = 1

2ab =⇒(a+ b)t = ab =⇒ t = ab

a+b .

4

Example D2.In triangle ABC, α = 120◦. AX is the bisector of angle A. Show that1t =

1b +

1c .

bc

t

A

B CX

5

Solution to Example D2.

In triangle ABC, α = 120◦. AX is the bisector of angle A. Show that1t =

1b +

1c .

bc

t

A

B CX

6

Example D3.In the diagram below, ABX , BCY , and CDZ are equilateral triangles.Suppose ∠XY Z = 120◦.Show that 1

b =1a +

1c .

Hint: Extend ZY to intersect AB at T . Show that TC = a.

a b cA B C D

X

Y

Z

7

Solution to Example D3.

In the diagram below, ABX , BCY , and CDZ are equilateral triangles.Suppose ∠XY Z = 120◦.Show that 1

b =1a +

1c .

a b cA B C D

X

Y

Z

T

Solution. Since ∠XY T = ∠XBT = 60◦, the points X , Y , B, T are con-cyclic. Therefore, ∠XTY = ∠XBY = 60◦ and triangle XY T is equilat-eral. By Ptolemy’s theorem, BX = BT + BY , and TC = BT + CB =BT + BY = BX = a. Now, triangle TCZ has a 120◦ angle at C withbisector CY = b. Therefore, 1

b =1a +

1c .

10

Theorem. (a) The lengths of the internal and external bisectors of angle Aare respectively

ta =2bc

b+ ccos

α

2and t′a =

2bc

|b− c| sinα

2.

c bta t′a

A

B CX X′

Proof. Let AX and AX ′ be the bisectors of angle A.(1) Consider the area of triangle ABC as the sum of those of triangles

AXC and ABX . We have

1

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 of ABX ′

and ACX ′.

Remarks. (1) 2bcb+c is the harmonic mean of b and c. It can be constructed as

follows. If the perpendicular to AX at X intersects AC and AB at Y andZ, then AY = AZ = 2bc

b+c .

c b

ta

Y

Z

A

BC

X

12

Example D5.The lengths of the sides of a triangle are 84, 125, 169. Calculate the lengthsof its internal bisectors. 1

1Answers: 9757 , 26208

253 , 12600209 .

13

Example D6.ABC is a right triangle in which the bisector of the right angle, and themedian to the hypotenuse have lengths 24

√2 and 35 respectively. Calculate

the sidelengths of the triangle.

14

Example D7.Find an isosceles triangle for which the bisector of a base angle is the geo-metric mean of the two segments it divides on the opposite side. 2

C

A

B

YZ

2Answer: (a, b, c) = (1, 1 +√2, 1 +

√2).

15

Example D8.In �ABC, α = 60◦, and B < C.The bisector of ∠A intersects BC at X .If AX is a mean proportional between BX and CX ,find angle B.

A

B CX

16

Solution to Example D8.

In �ABC, α = 60◦, and B < C.The bisector of ∠A intersects BC at X .If AX is a mean proportional between BX and CX ,find angle B.

A

B CX

Solution. By the angle bisector theorem, BX = acb+c , XC = ab

b+c .

The length of the bisector AX is 2bcb+c cos

A2 =

√3bc

b+c since A = 60◦.If AX is a mean proportional of BX and CX , then

3b2c2

(b+ c)2=

a2bc

(b+ c)2.

From this, a2 = 3bc.By the law of cosines, a2 = b2 + c2 − bc.Therefore, b2 + c2 − bc = 3bc, b2 + c2 − 4bc = 0.Solving this equation, we obtain c = (2±√

3)b.Since b < c, we have c = (2 +

√3)b.

c− b

c+ b=

√3 + 1√3 + 3

=1√3.

This means tan C−B2

tan C+B2

= 1√3.

But C + B = 120◦ =⇒ tan C+B2 =

√3.

It follows that tan C−B2 = 1, and C − B = 90◦.

Hence, C = 105◦ and B = 15◦.

17

Example D.Suppose ABC is a triangle with AB �= AC, and let D, E, X , Y be pointson the line BC defined as follows: D is the midpoint of BC, E is the footof the perpendicular from A to BC, AX , AY bisect angle A. Prove thatAB · AC = DE ·XY .

A

BC

XY DE

18

The circle of Apollonius

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

AB

XY

P

O B′

Proof. Since k �= 1, points X and Y can be found on the line AB satisfyingthe above conditions.

Consider a point P not on the line AB with AP : PB = k : 1. Note thatPX and PY are respectively the internal and external bisectors of angleAPB. This means that angle XPY is a right angle, and P lies on the circlewith XY as diameter.

Conversely, let P be a point on this circle. We show that AP : BP = k :1. Let B′ be a point on the line AB such that PX bisects angle APB′. SincePA and PB are perpendicular to each other, the line PB is the externalbisector of angle APB′, 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 that B′ coincides with B, andPX is the bisector of angle APB. It follows that PA

PB = AXXB = k.

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

19

Example D9.If AB = d, and k �= 1, the radius of the Apollonius circle is k

k2−1d.

20

Example D10.Given two disjoint circles (A) and (B), find the locus of the point P suchthat the angle between the pair of tangents from P to (A) and that betweenthe pair of tangents from P to (B) are equal.

A B

P

21

Steiner-Lehmus theorem

Lemma. Let triangles ABC and XY Z be equiangular. If AB > XY , thenAC > XZ and BC > Y Z.

A B

C

X Y

Z

B′

C ′

Given: Equiangular triangles ABC and XY Z with AB > XY .To prove: AC > XZ and BC > Y Z.Construction: Let B′ be a point on AB such that AB′ = XY .Construct the parallel through B′ to BC, intersecting the line AC at C ′.Proof : C ′ must be between A and C,for otherwise, B′ and C ′ are on opposite sides of BC, and the lines BC,B′C ′ intersect.This contradicts their parallelism.Now, triangles AB′C ′ and XY Z are congruent. [ASA]Therefore, AC > AC ′ = XZ.

22

Theorem. A triangle is isosceles if and only if it has two equal internal anglebisectors.

Given: Triangle ABC with B < C and bisectors BE, CF .To prove: BE > CF .Construction: Point G on AB such that ∠GCF = ∠ABE.Join CG and let it intersect BF at H .

B C

A

EF

G

H

Proof : The triangles GBH and GCF are equiangular with GB > GC.Therefore, BH > CF . (Lemma above)Since BE > BH , BE > CF .