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Inversive Geometry Inversive Geometry Wojciech Wieczorek following: Harold S.M. Coxeter Geometry Revisited So far you know the following maps of the plane: Translation: Rotation: b Line symmetry: All of the above: 1. Send every point of the plane to some other point of the plane. 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. b b b O P P' I(P ) = P 0 PO · OP 0 = r2 I(O) is not defined. Q = I(Q) b b b b O P P' If you want to catch a lion: If you want to catch a lion: If you want to catch a lion: Geometric construction of the image. O b b P' Geometric construction of the image. U b O b b P' b T Geometric construction of the image. U b O b b b P P' b T Geometric construction of the image. U b O b b b P P' b T Geometric construction of the image. OP OU U b = OU OP 0 b b b O 0 P P' OP · OP = r2 b T Instrument for constructing images. U O P' P T Instrument for constructing images. U a b X b O P' P T Instrument for constructing images. U OP · OP 0 = a b (OX + XP ) · (OX − XP ) = 2 − 2 b X OX XP = O P' P (a2 − UX2) − (b2 − UX2) = a2 − b2 T What are the images of lines and circles? Start with the easiest answer: ! P = I(P ) b b O I(!) = ! Almost as easy: b b b l O ! I(l) = l Almost as easy: b b b l O ! I(l) = l Remember: point O goes nowhere. Most interesting case: circle α passing through O. 0 0 0 OX · OX = OP · OP b X 0 X b ◦ 90 OX OP b b b = 0 O P P' OP OX and \XOP = \X0OP 0 thus: ∆OP X ∼ ∆OX0P 0 More than I (α) = l: If OP is diameter of α l then: α OP ? l P P' O bb b b A Special cases: l I (α) is tangent to α. α P O b b α intersects !. l α Pb O b b Q What are the images of other circles? • As a tool we would like to describe circles using distance between points. • We would like something similar to the familiar distance formula: AC + CB ≥ AB. b C b b A B AC + CB ≥ AB Definition: For four distinct ordered points A, B, C and D, define their cross ratio fAB; CDg to be the number: AC·BD fAB; CDg = AD·BC Theorem: Cross ratio of four distinct ordered points A, B, C, D satisfies: fAD; BCg + fAB; DCg = 1 if and only if: b b D A B C D C either: b b b b or: b B b A Main idea of the proof: Any 3 (not collinear) points define a A circle. b b b B C Main idea of the proof: Any 3 (not collinear) points define a A b P circle. b b Put fourth point P on it and project b onto sides of the triangle ∆ABC. b b b B C Main idea of the proof: Any 3 (not collinear) points define a A b P circle. b b Put fourth point P on it and project b onto sides of the triangle ∆ABC. Red dots are on one line if and only if b b b P is on the circle. B C Theorem: 0 0 0 0 If points A, B, C, D are mapped onto A , B , C and D under an inversion, then their cross ratios are equal: fA0B0;C0D0g = fAB; CDg Corollary: Inversion maps circles onto circles or lines. Proof of Theorem: 0 B0 · 0 · 0 ) OA OB OA OA = OB OB = OB = OA0 0 0 ∆OAB and ∆OB A share the same angle at O. B 0 0 b Thus: ∆OAB ∼ ∆OB A . O A A0 AB A0B0 Thus: OA = OB0 and: A0B0·OA 0 0 OA·OB AB = OB0 = A B r2 and: f g AC·BD A0C0·B0D0 f 0 0 0 0g AB; CD = AD·BC = A0D0·B0C0 = A B ;C D Angle between two circles. θ b b α b b β b a Angle between two circles. θ b b α b b β b a Angle between two circles. θ b b α b b β b a θ Ad a point at infinity: I(O) = 1 Line with one point added is a circle! Theorem: Inversion sends circles into circles. Inversion preserves the angles between circles: \ fI(α);I(β)g = \ fα; βg Proof: I(α) b O α Proof: I(α) b O α parallel! Proof: β I(α) I(β) b O α Proof: β I(α) 0 b P b P I(β) b O α Thus \ fI(α);I(β)g = \ fα; βg when α and β pass through O. a I(a) I(α) α Take two arbitrary intersecting circles: a θ b \fα; βg = \fa; bg b α = \fI(a);I(b)g β = \fI(α);I(β)g Orthogonal circles. b b b 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. A b β α b b O b B Corollary: If β ? α then Iα (β) = β. I(β) ? I(α) = α A α b and β I(A) = A and I(B) = B So: I(β) passes b B through A and B. Take ! ? α and ! ? β α ! I(β) = β and I(α) = α so P b I(P ) = P 0 b P 0 β Similar to line symmetry: α ! b b b b P P 0 β b b b b b b r r+a = sin(180=n) b b b (Here: n = 8) r b b b a r b b b .
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