I : Constructions " & Compass stature of has basic - - PowerPoint PPT Presentation

SLIDE 1

Applications

I

:

Geometric

" Ruler

&

Compass

"

Constructions

SLIDE 2

Motivation

:

Euclidean

Geometry

has

basic

stature of

definitions

, axioms , and prepositions

Axioms

:

①

given two

points

P

and

Q,

can draw segment Fa

① given

segment

POT ,

can

extend arbitrarily

in either dintin

③ givin

points P and

Q

,

can draw

a

circle with

center

P

passing though

Oi

④ all right

angles

are

congruent

why this

is

① parallel

postulate

(

geometry generated

by

ruler & compass

SLIDE 3

What

can

we

create

with tuna

tools ?

(Pnp 2) Ginn

segment

ED

and

a point

c

,

can crate

segment

CT

so that

ATgEmfm )

"

transport

a given

length

to

a given

point

"

④up 3) Givin

AT

and

ED

with

ATT

> CI

Thu

we can

find

E

• u

ATB

so

that

AT

= CI Ti

code for :

can construct

with ruler &

compass

SLIDE 4

Chop 9)

Bisect

a

given

angle

(Prep

12)

"

Drop

a perpendicular

"

to

live l

from

P

I

"

Despite

being

able

to

prove

lots of

new 113

. There

were

a few

" natural questions

" That

ancients

couldn't solve

SLIDE 5

Then

include

① Can

you bisect

a given

angle

?

⑦

"

Duplicating

a

cube

"

:

Gain

a cube of

volume

1,

create

a

cube

• f

volume

2 ( ie ,

can

we

create

a length

• f

size

352 given

a

segment

• f

unit

length ) ⑦

"

squaring the circle

"

:

Given

a

circle

• f

Area A,

create

a sgvun

et

area

A ?

( ie

, given leyth r

,

can

we

create

length

rift ?)

SLIDE 6

④

what

regular

n

• you

,

are

constructible ?

Eg

.

Prep I

says you

can

create equilateral

triangles lie ,

a

regular 3

• gon)

Then questions

were

unresolved for

• 2000

years

.

we'll

address ④ -③

Lin

The negate) today

.

SLIDE 7

To

resolve

these

questions

, we'll take

a

more

algebraic

view.

Petn (constructible

numbers)

A

number

aEIR

is

conductible if

,

using

a

ruler, compass

,

and

unit length ,

we

can

create

a segment at leythkl

.

Question

:

what

leyths

are

constructible ?

we'll

see

That

conductible

numbs

• n

a field,

Sime

1

is

c.uslmtible

,

we

got

Q

is subset of

the

constructible

.

SLIDE 8

Tim The

censtntible

numbers

term

a

field .

Pf

we'll

show

That it

xp

are

ceastuchbte

, Then

⑨ art p

④

a - p

clap

Cd) Xp

are

constantidle

.

T

THO

For

Cal , given

A, B

, C, D

with

T eo

g.to

• @

D

D

A

SLIDE 9

By

Axum 2

, extend

AT

Prop 2

: find

E

so that

Fe = ED

Axiom 3 : draw

a

curate

centred @ D , radius

BI

• r

Efi

. . .

• poof

j

• i

,

• I
• D

,

c

%

A 1

Then

AI

has

length

at P.

SLIDE 10

HIP

still

given

Ares . c. D

with

T

• ne
• D

c

&

A

Prep 3

says

we

can

find

some

E

between AIB

with FEZ ED

~

Then FB

if

• §

has length

Noooo

D

c

• E

D

L

• P

.

A

SLIDE 11

coats

him

A.B. c, D

with

T

• ne
• D

c

%

A

First,

dreg perpendicular Thgwh A

' .

"Mark off

"

unit

length

. .

Connect

E IB

fi

" Mark off

"

B length

E•¥t

paw

lion thigh F

¥9

G

goop

;

parallel

to

l

• ,

Similar Ingles

says

AI xp

.

"

At

SLIDE 12

(d) Mp

another

similar

Ingle

argument

.

Comdt Th

text.

DNA

what

can

we

say about this

fold of constantine

numbers?

Important

Fact : if

a

> O

is

constricting then

so

is

Ta

.

Cor

:

constructible

numbers

are

infinite

degree

• ur Q.

PI Consider

GQ E

(THE

( Vz ) E ⑥ ( %) E

SLIDE 13

To

delve

deeper

into

coastlines,

we

take

a new

point

• f

view .

Def's

( constructible point)

A-point

lay ) EIR'

it

constructible

from Cachao

) .k¥1

,

keyed

• ld

;d

if

• ne
• f

the Hlaing

holds

l " (x, y)

⇐ LIB) n LLC, D)

,

when

UAB) t LHP)

" line connecting

A-& B

"

SLIDE 14

C"

(x.y)

E

LIA , B)

n c(c,D)_

circle with

center c and

radios CDT

(3) (x.g)

E

CLA , B)

n CCC , D) we say

(x.g)

is

constructible iff

Hy) :(go)

• r 11,0)
• r if fun

is

a

sequence

• f

points

1901.4.01, R,

. . , Pk

so that

Pk

'

• ( x.y)

and each

Pi

is constructible relative to

some

subset at { ego) , 4,03, Pa.

• - -' Pi . .)

SLIDE 15

kegthem If

la ,.ae) , Ibi, but , Cayce),@ ,.dz)

are

in

Fh

for

Sonu

⑥ E F E IR

( say

: F'

( a.aybi.bz,c. end..dz))

Then

ay

lay)

constructible

relation to

AB, c, and D

lives in

some

quadratic

extension of

F

.

( ie ,

exists

ZEF

so

that

x.yet ( re ))

Ef

(sketch)

case

1

:

hey )

is

intersection

• f

LL AB)

and

LCGD)

.

Note

ABEFZ , Then The gratin for

LCA ,B)

SLIDE 16

takes The

form

rxtsytt

=D

for

some

Bs,TEF .

Likewise

LLC , D)

has

Sean q'

a

fxtsytt

• O ferris

,Iet

them the ilntoseltian

can

be

solved for

, and

solution

is

in

F .

If

Hy)

t

LLA.B) n Ccc , D)

, then

we get

UAB)

has

an equation

rxtsytt

• O

fer

r,Stef

and

CCC,D)

has

an

equation ( x-g)

't (y

• EY
• p

' for qc.pe#

Solve this

system t find

EEF with y,xE FIRE) .

SLIDE 17

(3)

same idea

if

Cx, y)

E

CCA.is/ncC4D)

TBT

Hew

is this

useful ?

Tim

IX.g) is

coustmtibte if

x.you

austmtibk numbers .

Cos

a

is

constable

implies [ ① Ca)

: @ 3=-2

"

for

some

n EIN

.

HI

2

is

constable

⇒

Cao)

is

constructible

.

SLIDE 18

Hence

we

hail

a sequence

{ 10,9, 4,0) , Hi,y .),

. " , lxk-i.gr) ,ko))

so that

each

term

in the

sequence

is

cealtrufibk

relative

to

some

preceding

terms

in the squirm

.

Hence

(Xi, y ;)

is

constable

relative

to

some

s . but of

199,407 , IX. , Yi) ;

. . , ( Xi . . , you, ) .

Sinn

(Xinyi) is

canstmtihk

relink

to

point

when

coordmh

,

an in

④ ( KH . ,

.

• > Xia , Yin)

,

we

know

[ ④ ( x ,,yy

. . ., Xia, Tia , Xi, yi)

: ⑥(Xi,Yy

.

Xii,Yi -D) c.2

.

SLIDE 19

Have

us

'

g th degree

ferula

along th tower ⑧ E

④ ( Xi

,y

.)

E ④ ( kyyyxayr) E

• - EQ (14,9 , ,

. -yxk.ik.in)

we

get

( ⑥ ( x ,,y . ,

. . .,Xk . "Yak)

: Q ) = 21

.

By the

degree

formula

again

we get

[ Ak )

:

Q]

• 2

"

for

some

net

.

DA

SLIDE 20

Cos

Not

all

angles

are

triseotnbk .

If

Prep

I

says

we

can

create

a

unit

length guilty Ingle

. IInternal

angles

are

60

.)

so

trisecting this

angle

is equivalent to couslnetibil.ly

af

cos 120

.)

.

However the

triple

angle

fermin

says

feeµf¥

cos 1307=4 cos' la)

• 3cos

On

So

c.ska) salish

O

.

SLIDE 21

So

we

get

[ Ql cos 200)

: Q .] =3

t 2

"

.

Hence

cos 1207

isn 't

constrictble .

Cos you

can't duplicate

a

cube.

PI

duplication of

cube

is equivalent

to aeustuetig Vz .

But

V2

has

irra (Vz)

• x3
• 2

, so

[atra) :

=3 t 2

"

• Dappy

SLIDE 22

Cos you

can't

square

a

circle .

PI

Equitant

te

: if you

have

'r

constructible

,

then

rift

is

constructible

lie

, ft is coustutibk) .

If

so , then [④ ( fu )

: Q)

2

"

for some

n

. This

fries

fit

to

be

algebraic ,

and

so

it

is

algebraic

. But

*

is

transcendental ,

so

→←

.

DH

SLIDE 23

If

you

are

keen

• n

ioastuetihilitg of regular

n

• gens ,

consult pg

137 . you'll find

th ( Gauss) If

p

is

an

• dd pain, then you

can

constrict

a

regular

p

• goniff

p

.

. get * i

• Fermat

for

sewn

t > 0

.

primes

C.g.,

2%1

=

2

'

th

• 3
• r I't 1=5,

22't 1=17 ,