InversiveGeometry WW

7/30/2019 InversiveGeometry WW

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Inversive Geometry

Wojciech Wieczorek

following:

Harold S.M. Coxeter

Geometry Revisited



So far you know the following maps of the plane:

Translation:



Rotation:



Line symmetry:



All of the above:

1. Send every point of the plane to some other point of theplane.

2. Preserve lengths, angles.

3. Send lines lines and circles circles.

4. Every map can be inverted.



One more map: dilation:

Still sends lines lines and circles circles.



New map: Inversion.

P P' O

I (P ) = P ′

P O · OP ′ = r2



I (O) is not defined.

P

Q = I(Q)

P' O



If you want to catch a lion:



Geometric construction of the image.

P'

O




T

U

P'

O




P

T

U

P'

O




P

T

U

P'

O

OP

OU =

OU

OP ′

OP ·OP ′

= r2



Instrument for constructing images.

P

T

U

P' O




P

T

U

P' O

X

a b




P

T

U

P' O

X

a b

OP · OP ′ =

(OX + XP ) · (OX −XP

OX 2 − XP 2 =

(a2 − U X 2)− (b2 − U X 2)

a2 − b2



What are the images of lines and circles?

Start with the easiest answer:

P = I (P )

O

ω

I (ω) = ω



Almost as easy:

l

Oω

I (l) = l



Almost as easy:

l

Oω

I (l) = l

Remember: point O goes nowhere.



Most interesting case: circle α passing through O.

X

X ′

90◦

P P' O

OX ·OX ′ = OP · O

OX

OP

= OP ′

OX ′

and

∠XOP = ∠X ′OP ′

thus:

∆OP X ∼ ∆OX ′P ′



More than I (α) = l:

A

P P'

O

α

lIf OP is diameter of α

then:

OP ⊥ l



Special cases:

P

O

α

l I (α) is tangent to α.



α intersects ω.

P

Q

O

αl



What are the images of other circles?

• As a tool we would like to describe circles using distancebetween points.

• We would like something similar to the familiar distance

formula: AC + CB ≥ AB.

A B

C



Definition:

For four distinct ordered points A, B, C and D, define their

cross ratio {AB, CD} to be the number:

{AB, CD} =AC ·BD

AD·BC



Theorem:

Cross ratio of four distinct ordered points A, B, C , D

satisfies:

{AD, BC }+ {AB, DC } = 1

if and only if:

either:

A B C D or:



Main idea of the proof:

A

B C

Any 3 (not collinear) points de

circle.




A

B C

P


circle.

Put fourth point P on it and pr

onto sides of the triangle ∆AB




A

B C

P


circle.

Put fourth point P on it and pr

onto sides of the triangle ∆AB

Red dots are on one line

if and only ifP is on the circle.



Theorem:

If points A, B, C , D are mapped onto A′, B′, C ′ and D′

under an inversion, then their cross ratios are equal:

{A′B′, C ′D′} = {AB, CD}

Corollary:

Inversion maps circles onto circles or lines.

P f f Th



Proof of Theorem:

O A A′

B

B′ OA ·OA′ = OB ·OB ′ =⇒ OA

OB= OB′

OA′

∆OAB and ∆OB′

A′

share

the same angle at O.

Thus: ∆OAB ∼ ∆OB ′A′.

Thus: ABOA = A

′

B′

OB′and:

AB = A′B′·OA

OB′= A′B′ OA·OB

r2and:

{AB, CD} = AC ·BDAD·BC = A

′

C ′

·B′

D′

A′D′·B′C ′ = {A′B′, C ′D′}



Angle between two circles.

θ

α

β

b

a




θ

α

β

b

a




θ

θ

α

β

b

a



Ad a point at infinity:

I (O) = ∞

Line with one point added is a circle!

Theorem:

Inversion sends circles into circles.

Inversion preserves the angles between circles:

∠ {I (α), I (β )} = ∠ {α, β }



Proof:

O

α

I (α)



Proof:

O

α

I (α)

parallel!



Proof:

O

α

β

I (α)

I (β )



Proof:

O

α

β

I (α)

I (β )

P

P ′

Thus

∠ {I (α), I (β )} = ∠ {α, β }

when α and β pass through O.



a

α

I (a)

I (α)



Take two arbitrary intersecting circles:

θ

α

β

b

a

∠{α, β } = ∠{a, b}

= ∠{I (a), I (b)}

= ∠{I (α), I (β )}



Orthogonal circles.



Given two points A and B on α there is unique circle β

such that: α ⊥ β and β is passing through A and B.



Corollary:

If β ⊥ α then I α (β ) = β .

A

B

αβ

I (β ) ⊥ I (α) = α

and

I (A) = A and I (B) = B

So: I (β ) passes

through A and B.

k d



Take ω ⊥ α and ω ⊥ β

P

P ′

ω

α

β

I (β ) = β and I (α) = α

so

I (P ) = P ′

Si il li



Similar to line symmetry:

P P ′

ω

α

β



r

r+a= sin(180/



a r

r

r+a( /

(Here: n = 8)

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