Straight-Edge and Compass: Constructing the Heptadecagonbaronmath.com/assets/heptadecagon.pdf · 2019. 10. 30. · Background The Construction. Straightedge-And-Compass Construction

BackgroundThe Construction.

Straight-Edge and Compass: Constructing theHeptadecagon

Daniel Baron

Western Washington University

May 15, 2013

Daniel Baron Construction of the Heptadecagon


Outline

1 BackgroundStraightedge-And-Compass ConstructionHistory

2 The Construction.Basic Tools and Techniques.Constructing Polygons.How To Draw a Heptadecagon



Straightedge-And-Compass ConstructionHistory

Outline






What Is Straightedge-And-Compass Construction?

It is a method for drawing “perfect” lengths and shapes withsimple tools:

An infinite, unruled straightedge.A compass which can draw a perfect circle of any radius.

For example, here is a construction of an equilateral triangle:




























































Applications.Why bother?

We can use geometry to prove that a construction is exactlycorrect. This is significant in many fields. In particular,consider. . .

Drafting, both by hand and with CAD software.The science of weights and measures.

And most importantly, it’s fun to do!




Applications.Why bother?

We can use geometry to prove that a construction is exactlycorrect. This is significant in many fields. In particular,consider. . .

Drafting, both by hand and with CAD software.The science of weights and measures.

And most importantly, it’s fun to do!




Outline






The Ancients.

The ancient Greeks were very skilled with compassconstructions. Euclid detailed a great many, including . . .

Division of line segments into equal parts.Bisection of angles.Construction of parallel and perpendicular lines.Construction of regular 3-, 4-, 5-, and 15-gons.

But very little progress was made after Euclid. In particular, nonew polygon constructions were found until . . .




The Ancients.







The Ancients.







The 18th Century and Beyond.

In 1796, Gauss discovered that the regular heptadecagon isconstructible by ruler and compass.

Gauss further proved that the regular n-gon is constructible ifand only if n is of the form

n = 2r p1p2 . . . ps,

where r ≥ 0 and each pi is a distinct Fermat prime;i.e., a prime of the form pi = 22k

+ 1.The known Fermat primes are 3, 5, 17, 257, and 65537.





In 1796, Gauss discovered that the regular heptadecagon isconstructible by ruler and compass.Gauss further proved that the regular n-gon is constructible ifand only if n is of the form

n = 2r p1p2 . . . ps,


+ 1.

The known Fermat primes are 3, 5, 17, 257, and 65537.





In 1796, Gauss discovered that the regular heptadecagon isconstructible by ruler and compass.Gauss further proved that the regular n-gon is constructible ifand only if n is of the form

n = 2r p1p2 . . . ps,


+ 1.The known Fermat primes are 3, 5, 17, 257, and 65537.



Basic Tools and Techniques.Constructing Polygons.How To Draw a Heptadecagon

Outline






Bisection.

We can use compasses to bisect a given line segment:

Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




Bisection.


Or an angle:




A Time-Saving Tip.

It is usually only necessary to draw short arcs of your circles:

But perhaps it is not as visually pleasing.




Perpendicular lines.

Given a line and a point, we can construct a perpendicular line:







































Outline






Polygons and Angles.

Suppose we wish to construct the regular pentagon. It turns outthat this is equivalent to constructing the angle 2π/5.

2π5






2π5






2π5






2π5






2π5






2π5






2π5






2π5






2π5






2π5






2π5






2π5






2π5




Primitive Roots of Primes.

In fact, if n is prime, we can draw the n-gon given any angle2kπ

n , with k an integer and 0 < k < n. We can find each of the nvertices of the polygon by copying 2kπ

n around a circle as we didon the previous slide with 2π

5 .

We can understand this in group theory terms, noting that anynonzero element k of the group Zn of integers mod n generatesthe whole group, so the sets

{0, k ,2k , . . . , (n − 1)k} = {0,1,2, . . . ,n − 1}

are equal under mod n arithmetic.





In fact, if n is prime, we can draw the n-gon given any angle2kπ

n , with k an integer and 0 < k < n. We can find each of the nvertices of the polygon by copying 2kπ

n around a circle as we didon the previous slide with 2π

5 .We can understand this in group theory terms, noting that anynonzero element k of the group Zn of integers mod n generatesthe whole group, so the sets

{0, k ,2k , . . . , (n − 1)k} = {0,1,2, . . . ,n − 1}

are equal under mod n arithmetic.





Careful! This is only guaranteed if n is prime.In general, it works if gcd(k ,n) = 1.




Sines and Cosines.

Constructing the cosine of an angle (i.e., constructing a linesegment of that length) is just as good as constructing theangle itself:

cos 2π5cos 2π

5

Of course, the sine is also sufficient.




Sines and Cosines.


cos 2π5

cos 2π5





Sines and Cosines.


cos 2π5cos 2π

5





Sines and Cosines.


cos 2π5cos 2π

5





Sines and Cosines.


cos 2π5cos 2π

5





Sines and Cosines.


cos 2π5cos 2π

5





Sines and Cosines.


cos 2π5cos 2π

5





Sines and Cosines.


cos 2π5cos 2π

5





Sines and Cosines.


cos 2π5

cos 2π5





Sines and Cosines.


cos 2π5

cos 2π5





Sines and Cosines.


cos 2π5

cos 2π5





Sines and Cosines.


cos 2π5

cos 2π5





Sines and Cosines.


cos 2π5

cos 2π5





Outline






A Brief Acknowledgment.

The construction method I will demonstrate today is due toHerbert William Richmond, who published it in 1893. I haveonly seen a few techniques, but I find his to be very elegant andI am proud to share it with you today.




Getting Started.

As we have seen, constructing the 17-gon is equivalent toconstructing the length cos 2kπ

17 for any 0 < k < 17.In particular, today we will directly construct

cos 6π17

and cos 10π17 .

For convenience, denote α = 2π/17, so that we want cos 3αand cos 5α.Let’s see what they look like . . .




Getting Started.



cos 6π17

and cos 10π17 .

For convenience, denote α = 2π/17, so that we want cos 3αand cos 5α.

Let’s see what they look like . . .




Getting Started.



cos 6π17

and cos 10π17 .

For convenience, denote α = 2π/17, so that we want cos 3αand cos 5α.Let’s see what they look like . . .




The Big, Complicated Numbers.

We have cos 3α = 14096(

√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68

+√

17− 1)3 + 34096((

√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68

+√

17− 1)2 − 256)(√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68+√

17− 1).




The Big, Complicated Numbers.

And cos 5α = 11048576 (

√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68

+√

17− 1)5 + 5524288 ((

√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68

+√

17− 1)2 − 256)(√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68

+√

17− 1)3 + 51048576 ((

√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68

+√

17− 1)2 − 256)2(√−2√

17 + 34

+

√2√−2√

17 + 34(√

17− 1)− 16√

2√

17 + 34 + 12√

17 + 68+√

17− 1).




The Easy Way

Consider the acute angle φ such that tan 4φ = 4 (so φ isabout 18◦).

It can be shown that

tanφ = 2(cos 3α+ cos 5α)

andtan(φ− π

4) = 4 cos 3α cos 5α.

We will now construct these two lengths.




The Easy Way

Consider the acute angle φ such that tan 4φ = 4 (so φ isabout 18◦).It can be shown that

tanφ = 2(cos 3α+ cos 5α)

andtan(φ− π

4) = 4 cos 3α cos 5α.

We will now construct these two lengths.




First Steps.

OA

Start with a unit circle and arbi-trary diameter. The point A isyour 0th vertex.Next, find a perpendicularradius.

B

Bisect OB twice to find I at(0,1

4 ).IUsing your straightedge, drawthe line segment AI.




First Steps.

OA


B






First Steps.

OA


B






First Steps.

OA


B






First Steps.

OA


B


4 ).

IUsing your straightedge, drawthe line segment AI.




First Steps.

OA


B


4 ).





First Steps.

OA


B


4 ).





First Steps.

OA


B


4 ).





First Steps.

OA


B


4 ).





First Steps.

OA


B


4 ).





First Steps.

OA


B


4 ).I

Using your straightedge, drawthe line segment AI.




First Steps.

OA


B






First Steps.

OA


B






The Angle 4φ.

OA

B

I14

1

Recall that φ is acute and thattan 4φ = 4. Since

tan ∠OIA = OA/OI= 4,

we have already constructedthe angle 4φ.




The Angle φ and Its Tangent.

OA

B

I

Next, bisect ∠OIA twice to con-struct φ.

E

There it is, φ = ∠OIE .14 And in fact,

tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E

There it is, φ = ∠OIE .

14 And in fact,

tanφ = tan∠OIE

= OE/OI

= 4 OE .





OA

B

I


E


tanφ = tan∠OIE

= OE/OI

= 4 OE .




The Angle φ− π4 .

OA

B

I

E

To construct φ − π4 , begin

by finding a perpendicular linesegment to EI through I.

Next, bisect that right angle . . .F

Next, bisect that right angle . . .And you have it.Further, notice that

tan(φ− π

4) = 4OF .





OA

B

I

E





tan(φ− π

4) = 4OF .





OA

B

I

E





tan(φ− π

4) = 4OF .





OA

B

I

E





tan(φ− π

4) = 4OF .





OA

B

I

E





tan(φ− π

4) = 4OF .





OA

B

I

E





tan(φ− π

4) = 4OF .





OA

B

I

E


by finding a perpendicular linesegment to EI through I.Next, bisect that right angle . . .

F


tan(φ− π

4) = 4OF .





OA

B

I

E



F


tan(φ− π

4) = 4OF .





OA

B

I

E



F


tan(φ− π

4) = 4OF .





OA

B

I

E



F


tan(φ− π

4) = 4OF .





OA

B

I

E



F


tan(φ− π

4) = 4OF .





OA

B

I

E



Next, bisect that right angle . . .

F

Next, bisect that right angle . . .And you have it.

Further, notice that

tan(φ− π

4) = 4OF .





OA

B

I

E



Next, bisect that right angle . . .

F


tan(φ− π

4) = 4OF .




Nearly There.

OA

B

I

EF

Now we use what we knowabout the relationships amongcos 3α, cos 5α, tanφ, andtan(φ − π

4 ) to construct the twolengths we want.Draw a circle with diameter AF .

K

This circle intersects OB at thepoint K . Now draw a circlethrough K with center E . . .




Nearly There.

OA

B

I

EF



K





Nearly There.

OA

B

I

EF



K





Nearly There.

OA

B

I

EF



K





Nearly There.

OA

B

I

EF



K





Nearly There.

OA

B

I

EF



K





Nearly There.

OA

B

I

EF



K





Nearly There.

OA

B

I

EF



K





Drumroll . . .

OA

B

I

EF

K

N3N5

This last circle intersects ouroriginal diameter at N3 and N5.I now claim that

ON3 = cos 3α

and

ON5 = − cos 5α.

There is only one thing left todo!

P3




Drumroll . . .

OA

B

I

EF

K

N3N5


ON3 = cos 3α

and

ON5 = − cos 5α.


P3




Drumroll . . .

OA

B

I

EF

K

N3N5


ON3 = cos 3α

and

ON5 = − cos 5α.


P3




Drumroll . . .

OA

B

I

EF

K

N3N5


ON3 = cos 3α

and

ON5 = − cos 5α.


P3




Drumroll . . .

OA

B

I

EF

K

N3N5


ON3 = cos 3α

and

ON5 = − cos 5α.


P3




Drumroll . . .

OA

B

I

EF

K

N3N5


ON3 = cos 3α

and

ON5 = − cos 5α.


P3




Drumroll . . .

OA

B

I

EF

K

N3N5


ON3 = cos 3α

and

ON5 = − cos 5α.


P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish.

A

P3




The Big Finish




The Big Finish




The Big Finish


Documents

Straight-Edge and Compass: Constructing the Heptadecagonbaronmath.com/assets/heptadecagon.pdf · 2019. 10. 30. · Background The Construction. Straightedge-And-Compass Construction