Geometry Class Examples (July 3) Paul Yiu Department of Mathematics Florida Atlantic University a b c A B C Summer 2014
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.
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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
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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.
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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 .