Curves and Sectioning Angles - LSU Math · 2021. 3. 11. · 2 Nicomedes (280-210 BC) also used a marked straight edge along with a conchoid The successful trisections: 1 Apolonius

IntroductionBasic Constructions

The HyperbolaThe Problem of Trisection

Curves and Sectioning Angles

Nicholas Molbert, Julie Fink, and Tia Burden

July 6, 2012

Nicholas Molbert, Julie Fink, and Tia Burden Curves and Sectioning Angles



Table of Contents

1 IntroductionHistoryNotation

2 Basic ConstructionsPropositions and Pictures

3 The HyperbolaGeneral DefinitionsBridging the Gap from General Definition to ΓThe Curve Γ

4 The Problem of TrisectionConstruction of TrisectionProof of TrisectionInvestigation of Further Curves




HistoryNotation

History of the Problem of Trisection

Plato paves the way for planar constructions

Three Greek problems of antiquity

The groundbreaking discoveries in classical geometry:1 Hippocrates (460-380 BC) labels points and lines2 Hippias (460-399 BC) and the quadratix3 Menaechmus (380-320 BC) and conic sections




HistoryNotation








HistoryNotation




The groundbreaking discoveries in classical geometry:

1 Hippocrates (460-380 BC) labels points and lines2 Hippias (460-399 BC) and the quadratix3 Menaechmus (380-320 BC) and conic sections




HistoryNotation




The groundbreaking discoveries in classical geometry:1 Hippocrates (460-380 BC) labels points and lines

2 Hippias (460-399 BC) and the quadratix3 Menaechmus (380-320 BC) and conic sections




HistoryNotation




The groundbreaking discoveries in classical geometry:1 Hippocrates (460-380 BC) labels points and lines2 Hippias (460-399 BC) and the quadratix

3 Menaechmus (380-320 BC) and conic sections




HistoryNotation








HistoryNotation


Trisections with extra tools:

1 Archimedes (287-217 BC) used a marked straight edge2 Nicomedes (280-210 BC) also used a marked straight edge

along with a conchoid

The successful trisections:1 Apolonius (250-175 BC) used conic sections2 Pappus (early fourth century) used a hyperbola3 Descartes (1596-1650) used the curve y = x2

The link of Francios Viete (1540-1603)

Wantzel’s proof of impossibility

From Greek constructions to abstract algebra




HistoryNotation


Trisections with extra tools:1 Archimedes (287-217 BC) used a marked straight edge

2 Nicomedes (280-210 BC) also used a marked straight edgealong with a conchoid








HistoryNotation


Trisections with extra tools:1 Archimedes (287-217 BC) used a marked straight edge2 Nicomedes (280-210 BC) also used a marked straight edge









HistoryNotation




The successful trisections:

1 Apolonius (250-175 BC) used conic sections2 Pappus (early fourth century) used a hyperbola3 Descartes (1596-1650) used the curve y = x2







HistoryNotation




The successful trisections:1 Apolonius (250-175 BC) used conic sections

2 Pappus (early fourth century) used a hyperbola3 Descartes (1596-1650) used the curve y = x2







HistoryNotation




The successful trisections:1 Apolonius (250-175 BC) used conic sections2 Pappus (early fourth century) used a hyperbola

3 Descartes (1596-1650) used the curve y = x2







HistoryNotation











HistoryNotation











HistoryNotation











HistoryNotation











HistoryNotation

Notation

←→XY denotes the line through points X and Y .

XY denotes the segment from point X to point Y .

C (X ,Y ) denotes a circle with center X and radius XY .

XY denotes the magnitude of segment XY

∠XYZ denotes the measure or name of an angle, dependingon the context.




HistoryNotation

Notation









HistoryNotation

Notation









HistoryNotation

Notation









HistoryNotation

Notation









Propositions and Pictures

Propositions

Here are the propositions used in our final trisectionconstruction along with pictures demonstrating construction

procedures:

Rusty Compass Theorem

Copying an Angle

Bisecting an Angle

Parallel Postulate

Raising a Perpendicular





Rusty Compass Theorem

Given points A, B, and C , we wish to construct a circle centeredat point A with radius equal to BC .









Copying an Angle

Given ∠ABC and a line l containing a point D, we can find E on land a point F such that ∠ABC = ∠EDF .









Bisecting an Angle

Given ∠ABC , there is a point D such that ∠ABD ∼= ∠DBC .









Dropping a Perpendicular

Given a line l and a point p not on l , we can construct a line l ′

which is perpendicular to l and passes through p.









Parallel Postulate (Playfair)

Given a line l and P not on l , we can construct l ′ through P andparallel to l .









Raising a Perpendicular

Given a point p on a line l , you can construct l ′ through point pperpendicular to line l .








General DefinitionsBridging the Gap from General Definition to ΓThe Curve Γ

Important Features of Hyperbolas

ImportantFeatures are:

1 focus

2 directrix

3 vertex

4 asymptotes

5 transverseaxis

6 conjugateaxis





Definition of Hyperbolas

The locus of all points Pin the plane thedifference of whosedistances r1 = F1P andr2 = F2P from two fixedpoints, called foci, is aconstant k , withk = r2 − r1.





Definition of Hyperbolas

The locus of all pointsfor which the ratio ofdistances from one focusto a line (the directrix) isa constant e (theeccentricity), with e > 1.

These loci createtwo distinctbranches of thecurve.





From Definitions to Derivation

So how do we tie all of this together?

Derive a curve that satisfies the specific conditions of theproblem.

Use this curve to trisect an arbitrary acute angle.

Interpret the features of the curve as it fits into the trisectionpicture.
































Process of Deriving Γ





Solving for w

Values from Derivation Triangle

β = x + w y = r sin θ x = r cos θ s = r sin θsin 2θ

w2 = s2 − y2

w2 =r2 sin 2θ

sin 22θ− r2 sin 2θ

w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)

=r(cos 2θ − sin 2θ)

2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)

= r sin θ√

csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for w



w2 = s2 − y2

w2 =r2 sin 2θ


w =

√r2 sin2 θ

(1

sin2 2θ− 1

)= r sin θ

√csc 22θ − 1

= r sin θ√

cot 22θ

= r sin θ

(cos 2θ

2 sin θ cos θ

)


2 cos θ

w =r cos θ

2− r sin 2θ

2 cos θ





Solving for β

β = r cos θ +r cos θ

2− r sin2 θ

2 cos θ

2r2 cos2 θ + r2 cos2 θ − r2 sin2 θ = 2βr cos θ

(x − β

3

)2− y2

3=β2

9

(x − β3 )2(

β3

)2− y2(

β√3

)2= 1





Solving for β


2− r sin2 θ

2 cos θ


(x − β

3

)2− y2

3=β2

9

(x − β3 )2(

β3

)2− y2(

β√3

)2= 1





Solving for β


2− r sin2 θ

2 cos θ


(x − β

3

)2− y2

3=β2

9

(x − β3 )2(

β3

)2− y2(

β√3

)2= 1





Solving for β


2− r sin2 θ

2 cos θ


(x − β

3

)2− y2

3=β2

9

(x − β3 )2(

β3

)2− y2(

β√3

)2= 1





Picture of Γ




Construction of TrisectionProof of TrisectionInvestigation of Further Curves

Construction of the Trisection

Trisection Construction

Given AB (with A = (0, 0) and B = (1, 0) on the Cartesian plane)and an angle θ, we can construct the trisection of an arbitraryangle using our derived hyperbola, Γ, along with basicconstructions previously outlined.





Step 1

Construct the hyperbola Γ with right branch’s focus at point B.





Step 2

Using the Rusty Compass Theorem, construct point D at ( 12 , 0).





Step 3

Construct line l perpendicular to AB at point D by dropping aperpendicular. Note that line l is the perpendicular bisector of AB.





Step 4

Bisect angle θ to obtain θ2 .





Step 5

From the axioms of basic planar constructions, we are able toconstruct ray l ′ with right endpoint on l at an angle θ

2 from lmeasured anti-clockwise.





Step 6

If line l ′ does not contain point A, construct l ′′ ‖ l ′ through pointA.





Step 7

If line l ′ does contain point A, obtain point O such that∠AOD = θ

2 .





Step 8

Construct AO.





Step 9

Reflect ∠AOD about line l such that it creates ∠DOB by usingthe construction to copy an angle.





Step 10

Construct OB. Note that AO = OB, so 4AOB is isosceles.





Step 11

Construct C (O,A). Note that C (O,A) contains both points A andB because OA and OB are radii. Also, ∠AOB = θ.





Step 12

Obtain point P from the intersection of_AB and Γ.





Step 13

Construct OP, AP, and PB. Note that 4APB, obtained inprevious step, is the triangle such that 2∠PAB = ∠PBA.





Theorem

If ∠PBA = 2∠PAB, then ∠AOP = 2∠POB. Hence, ∠POBtrisects ∠AOB.

Proof

We know that 12ψ = φ and

2φ = 12α. Solving for φ in

both equations andequating them, we arrive at12ψ = 1

4α. So, 2ψ = α orψ = 1

2α. Notice thatα + ψ = ∠AOB, so∠AOB = 3ψ orψ = 1

3∠AOB. Therefore, ψtrisects ∠AOB.





Geometric Interpretations of Features of Γ





Investigation of Further Curves

Can we modify this triangle to section angles into however manyparts we want just as we have trisected an angle using this triangle?





Investigation of Further Curves

Yes, we can by using this triangle!


Documents

Curves and Sectioning Angles - LSU Math · 2021. 3. 11. · 2 Nicomedes (280-210 BC) also used a marked straight edge along with a conchoid The successful trisections: 1 Apolonius