Ill-known/original/interesting investigations/applications of inversion (the geometric transform)












10












$begingroup$


Inversion transform with center $C$ and power $k^2$ is defined by:



$$tag{1}J_{C,k}:M leftrightarrow M' iff vec{CM'}=frac{k^2}{||vec{CM}||^2} vec{CM} $$



The double arrow is there to mean that it is an "involutive" transform: $M'$ is the image of $M$ iff $M$ is the image of $M'$.



This transform, presented for the first time in a systematic way by Plücker, (1834) is detailed in many books/sites.



See Appendix 1 for a comment about inversion properties displayed on Fig. 1.



enter image description here



Fig. 1. : Inversion with center $C$ and power 1. Shapes in black are exchanged by inversion with shapes in red. The blue circle, called "circle of inversion", is the locus of invariant points.



But there are ill-known features that I would like to gather.



Moreover, I would like to enlarge this "quest" to interesting (not fully standard) questions/applications of inversion.



For the initialization of this "collection", I propose 3 themes and a compendium of interesting facts:



1) "Champagne glass inversion" : Fig. 2 shows how inversion with center $C$ and radius of inversion $k$ (we find back the large and the small fish, inverted images one of the other...) can be realized by a 3 steps $PGamma P^{-1}$ operation with respect to a paraboloid $Pi$, where:




  • $P$ is the vertical projection from $Pi$ onto the horizontal plane, and


  • $Gamma$ is the conical projection from $Pi$ to $Pi$ with center $C_0(a,b,a^2+b^2-k^2)$ where $(a,b)$ are the coordinates of $C$.



This non-classical way to describe inversion deserves an explanation that we have placed in Appendix 2.



enter image description here



Fig. 2. Inversion by a conical projection $Pi to Pi$.



2) "Bubble inversion" This is a cousin representation of the first one with a sphere instead of a parabola (see Fig. 3). It necessitates to use 2 steps with the two stereographic projections $S_N$ and $S_S$ from the North and South pole resp. with respect to the equatorial plane. Take a look at Fig. 3.



Consider a certain figure, say the circle in red. It is transformed by $S_N$ into the circle on the unit sphere materialized by black little diamonds ; this circle is transformed in turn by $S_S$ into the blue circle on the equatorial plane... which is the inversive image of the initial red circle. Briefly said :



$$tag{$star$}I=S_S circ S_N$$



This shouldn't come as a surprize because stereographic projections are themselves 3D inversions.



See (composition of stereographic projections is inversion through the ball - a geometric way) for a proof.



enter image description here



Fig. 3. (planar) inversion realized by combining two stereographic projections (formula ($star$)).



3) "How inversion transform is connected with linear algebra" (in fact connected with item 1) ; this issue looks paradoxical because inversion is definitely not a linear transform. In fact, there exists a $4 times 4$ linear representation of the co-called conformal geometry including another category of non-linear transforms, translations. Reference: the very good book "Riemannian Geometry" by S. Gallot, D. Hullin, J. Lafontaine, 2nd edition 1993. Universitext, Springer, pages 175-176. Here is how this correspondence is done (Explanations will be found in this book or in the very interesting development by @MvG cited in 4a):



$$text{If} J text{is the basic inversion (center 0, power 1):}$$
$$[J]:=begin{pmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{pmatrix}.$$
$$text{If} H_r text{is the homothety with ratio} r text{and center} 0: $$
$$[H_r]:=begin{pmatrix}1&0&0&0\0&r&0&0\0&0&r&0\0&0&0&r^2end{pmatrix}.$$



$$text{If} T_V text{is the translation by vector} V=binom{a}{b} :$$ $$[T_V]:=begin{pmatrix}1&0&0&0\2a&1&0&0\2b&0&1&0\a^2+b^2&a&b&1end{pmatrix}.$$



All these transformations preserve the following quadratic form:



$$q(x_0;x_1,x_2,x_3):=x_1^2+x_2^2-x_0x_3$$



where $x_0$ should be considered as a homogeneous coordinate (i.e., is there to take into account the projective dimension). The signature of $q$ being $(+++-)$, we are dealing with elements of the group classically denoted $O(3,1)$.



This correspondence with 2D transforms can be extended in a straightforward way to higher dimensions (for $n$D transforms, the correspondence is with $(n+2) times (n+2)$ matrices).



4) Complex representation with an application :



As recalled by @José Carlos Santos, inversion with the unit circle $S^1={zinmathbb{C},|,|z|=1$} as invariant circle is the map



$$I : begin{cases} begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&frac{1}{overline z}end{array}end{cases}$$



Let us prove the following result :



Any (2D) rotation can be obtained as a certain combination of translations and inversions.



Let us consider the following identity (as given in https://mathoverflow.net/q/19965) :



$$e^{i theta} + frac{1}{-e^{-i theta} + frac{1}{e^{i theta} + frac{1}{z}}} = - e^{2 i theta} z. tag{%}$$



If we conjugate both sides of (%), we get :



$$e^{-i theta} + frac{1}{overline{-e^{-i theta} + frac{1}{overline{e^{-i theta} + frac{1}{overline{z}}}}}} = - e^{-2 i theta}overline{z}. tag{%%}$$



(The LHS of (%%) can be written in a symbolic way as : $T_{e^{-i theta}}circ I circ T_{-e^{-i theta}} circ I circ T_{e^{-i theta}} circ I$).



If we take $theta=0$ in (%%), one gets :



$$1+ frac{1}{overline{-1 + frac{1}{overline{1 + frac{1}{overline{z}}}}}} = - overline{z}. tag{%%%}$$



(%%%) means that transformation $zmapsto-overline{z}$ can be obtained by combining only inversions and translation(s).



In a similar manner (%%) means that transformation $zmapsto -e^{-2itheta}overline{z}$, can obtained using as well inversions and translations.



Combining both transformations, we can thus transform $z$ into $e^{-2itheta}z$ for any theta, using translations and inversions only.



5) A compendium of interesting documents/issues




  • a) See the diaporama (http://www.cis.umac.mo/~fstitl/2013geometry/sterograph.pdf).


  • b) An in-depth analysis by MvG in (https://math.stackexchange.com/q/866403) related to conformal group analysed in 3).


  • c) A study of torii, Dupin cyclides, etc. in connection with 3D inversion : (inverting a cone to a torus).


  • d) An answer I made (https://math.stackexchange.com/q/2621344) explaining why, locally, in the vicinity of the circle of inversion, inversion behaves as a symmetry. This is also elegantly treated in ref. 3a).


  • e) An application to Voronoi centers (https://math.stackexchange.com/q/2378365).


  • f) An extension of a IMO problem solved by inversion/stereographic projection (https://pdfs.semanticscholar.org/3bcb/00d463fceb26ab103c9a3757ee85398eb187.pdf).




My (enlarged) question is : can you give examples (of your own) of interesting, possibly non-usual, uses of inversion ?




Appendix 1: Recall of properties of inversion as depicted by Fig. 1:




  • The image of a straight line is in general a circle that passes through the pole $C$, and, exceptionally, when the straight line passes through the origin, its image is the line itself (caution: in the latter case, the straight line is "globally" invariant, but its points, in general, aren't invariant).


  • The image $Gamma'$ of a circle $Gamma$ not passing through the origin is a circle of the same type (caution: the center of $Gamma'$ is not the image of the center of $Gamma$). The image of a circle passing through the origin is a straight line, as said before.


  • The image of the black fish made of line segments is the red fish made of circular arcs swimming in its blue round fishbowl.



Appendix 2: Explanation of inversion depicted in Fig. 2.



Let us begin by another definition of inversion.



$$tag{3}J_{C,k}:M to M' iff exists vec{U} text{(unit norm vector) s.t.} begin{cases}vec{CM}=lambda vec{U}\ vec{CM'}=lambda' vec{U}end{cases} text{and} lambdalambda'=k^2$$



Let $M(x,y)$ and $M'(x',y')$ exchanged by inversion $J_{C,k}$.



Let $M_0(x,y,x^2+y^2)$ and $M_0'(x',y',x'^2+y'^2)$ be their "lifted version" on paraboloid $Pi$.



Let $N(X,Y,Z) $ be any point of line $C_0M_0$. We can define an abscissa $mu$ on this line for point $N$ in the following way:



$$tag{$star$}vec{C_0N}=mu vec{C_0M_0} iff begin{cases}X=a+mu(x-a)\ Y=b+mu(y-b)\ Z=c+mu(x^2+y^2-c)end{cases}$$



with $$c:=a^2+b^2-k^2.$$



Please note that for $mu=1$, $N$ is in $M_0$. Let $lambda'$ be the value of $mu$ associated with $M'_0$.



Thus $N in Pi$ in two cases: when $mu=1$ ($N=M_0$) and when $mu=lambda'$ ($N=M'_0$). As



$$tag{3}N in Pi iff Z=X^2+Y^2,$$



plugging in (3) the expressions of $X,Y,Z$ in $(star)$, and replacing $c$ by its defining expression, we obtain the following quadratic equation with unknown $mu$:



$$mu^2[(x-a)^2+(y-b)^2] + mu[cdots] + k^2=0$$



Using the classical formula for the roots' product, we have



$$lambda' times 1 =dfrac{k^2}{(x-a)^2+(y-b)^2}=dfrac{k^2}{|vec{CM}|^2}$$



We find back here the definition of inversion given in (2) (recall that $vec{CM}$ hasn't been normalized).










share|cite|improve this question











$endgroup$

















    10












    $begingroup$


    Inversion transform with center $C$ and power $k^2$ is defined by:



    $$tag{1}J_{C,k}:M leftrightarrow M' iff vec{CM'}=frac{k^2}{||vec{CM}||^2} vec{CM} $$



    The double arrow is there to mean that it is an "involutive" transform: $M'$ is the image of $M$ iff $M$ is the image of $M'$.



    This transform, presented for the first time in a systematic way by Plücker, (1834) is detailed in many books/sites.



    See Appendix 1 for a comment about inversion properties displayed on Fig. 1.



    enter image description here



    Fig. 1. : Inversion with center $C$ and power 1. Shapes in black are exchanged by inversion with shapes in red. The blue circle, called "circle of inversion", is the locus of invariant points.



    But there are ill-known features that I would like to gather.



    Moreover, I would like to enlarge this "quest" to interesting (not fully standard) questions/applications of inversion.



    For the initialization of this "collection", I propose 3 themes and a compendium of interesting facts:



    1) "Champagne glass inversion" : Fig. 2 shows how inversion with center $C$ and radius of inversion $k$ (we find back the large and the small fish, inverted images one of the other...) can be realized by a 3 steps $PGamma P^{-1}$ operation with respect to a paraboloid $Pi$, where:




    • $P$ is the vertical projection from $Pi$ onto the horizontal plane, and


    • $Gamma$ is the conical projection from $Pi$ to $Pi$ with center $C_0(a,b,a^2+b^2-k^2)$ where $(a,b)$ are the coordinates of $C$.



    This non-classical way to describe inversion deserves an explanation that we have placed in Appendix 2.



    enter image description here



    Fig. 2. Inversion by a conical projection $Pi to Pi$.



    2) "Bubble inversion" This is a cousin representation of the first one with a sphere instead of a parabola (see Fig. 3). It necessitates to use 2 steps with the two stereographic projections $S_N$ and $S_S$ from the North and South pole resp. with respect to the equatorial plane. Take a look at Fig. 3.



    Consider a certain figure, say the circle in red. It is transformed by $S_N$ into the circle on the unit sphere materialized by black little diamonds ; this circle is transformed in turn by $S_S$ into the blue circle on the equatorial plane... which is the inversive image of the initial red circle. Briefly said :



    $$tag{$star$}I=S_S circ S_N$$



    This shouldn't come as a surprize because stereographic projections are themselves 3D inversions.



    See (composition of stereographic projections is inversion through the ball - a geometric way) for a proof.



    enter image description here



    Fig. 3. (planar) inversion realized by combining two stereographic projections (formula ($star$)).



    3) "How inversion transform is connected with linear algebra" (in fact connected with item 1) ; this issue looks paradoxical because inversion is definitely not a linear transform. In fact, there exists a $4 times 4$ linear representation of the co-called conformal geometry including another category of non-linear transforms, translations. Reference: the very good book "Riemannian Geometry" by S. Gallot, D. Hullin, J. Lafontaine, 2nd edition 1993. Universitext, Springer, pages 175-176. Here is how this correspondence is done (Explanations will be found in this book or in the very interesting development by @MvG cited in 4a):



    $$text{If} J text{is the basic inversion (center 0, power 1):}$$
    $$[J]:=begin{pmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{pmatrix}.$$
    $$text{If} H_r text{is the homothety with ratio} r text{and center} 0: $$
    $$[H_r]:=begin{pmatrix}1&0&0&0\0&r&0&0\0&0&r&0\0&0&0&r^2end{pmatrix}.$$



    $$text{If} T_V text{is the translation by vector} V=binom{a}{b} :$$ $$[T_V]:=begin{pmatrix}1&0&0&0\2a&1&0&0\2b&0&1&0\a^2+b^2&a&b&1end{pmatrix}.$$



    All these transformations preserve the following quadratic form:



    $$q(x_0;x_1,x_2,x_3):=x_1^2+x_2^2-x_0x_3$$



    where $x_0$ should be considered as a homogeneous coordinate (i.e., is there to take into account the projective dimension). The signature of $q$ being $(+++-)$, we are dealing with elements of the group classically denoted $O(3,1)$.



    This correspondence with 2D transforms can be extended in a straightforward way to higher dimensions (for $n$D transforms, the correspondence is with $(n+2) times (n+2)$ matrices).



    4) Complex representation with an application :



    As recalled by @José Carlos Santos, inversion with the unit circle $S^1={zinmathbb{C},|,|z|=1$} as invariant circle is the map



    $$I : begin{cases} begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&frac{1}{overline z}end{array}end{cases}$$



    Let us prove the following result :



    Any (2D) rotation can be obtained as a certain combination of translations and inversions.



    Let us consider the following identity (as given in https://mathoverflow.net/q/19965) :



    $$e^{i theta} + frac{1}{-e^{-i theta} + frac{1}{e^{i theta} + frac{1}{z}}} = - e^{2 i theta} z. tag{%}$$



    If we conjugate both sides of (%), we get :



    $$e^{-i theta} + frac{1}{overline{-e^{-i theta} + frac{1}{overline{e^{-i theta} + frac{1}{overline{z}}}}}} = - e^{-2 i theta}overline{z}. tag{%%}$$



    (The LHS of (%%) can be written in a symbolic way as : $T_{e^{-i theta}}circ I circ T_{-e^{-i theta}} circ I circ T_{e^{-i theta}} circ I$).



    If we take $theta=0$ in (%%), one gets :



    $$1+ frac{1}{overline{-1 + frac{1}{overline{1 + frac{1}{overline{z}}}}}} = - overline{z}. tag{%%%}$$



    (%%%) means that transformation $zmapsto-overline{z}$ can be obtained by combining only inversions and translation(s).



    In a similar manner (%%) means that transformation $zmapsto -e^{-2itheta}overline{z}$, can obtained using as well inversions and translations.



    Combining both transformations, we can thus transform $z$ into $e^{-2itheta}z$ for any theta, using translations and inversions only.



    5) A compendium of interesting documents/issues




    • a) See the diaporama (http://www.cis.umac.mo/~fstitl/2013geometry/sterograph.pdf).


    • b) An in-depth analysis by MvG in (https://math.stackexchange.com/q/866403) related to conformal group analysed in 3).


    • c) A study of torii, Dupin cyclides, etc. in connection with 3D inversion : (inverting a cone to a torus).


    • d) An answer I made (https://math.stackexchange.com/q/2621344) explaining why, locally, in the vicinity of the circle of inversion, inversion behaves as a symmetry. This is also elegantly treated in ref. 3a).


    • e) An application to Voronoi centers (https://math.stackexchange.com/q/2378365).


    • f) An extension of a IMO problem solved by inversion/stereographic projection (https://pdfs.semanticscholar.org/3bcb/00d463fceb26ab103c9a3757ee85398eb187.pdf).




    My (enlarged) question is : can you give examples (of your own) of interesting, possibly non-usual, uses of inversion ?




    Appendix 1: Recall of properties of inversion as depicted by Fig. 1:




    • The image of a straight line is in general a circle that passes through the pole $C$, and, exceptionally, when the straight line passes through the origin, its image is the line itself (caution: in the latter case, the straight line is "globally" invariant, but its points, in general, aren't invariant).


    • The image $Gamma'$ of a circle $Gamma$ not passing through the origin is a circle of the same type (caution: the center of $Gamma'$ is not the image of the center of $Gamma$). The image of a circle passing through the origin is a straight line, as said before.


    • The image of the black fish made of line segments is the red fish made of circular arcs swimming in its blue round fishbowl.



    Appendix 2: Explanation of inversion depicted in Fig. 2.



    Let us begin by another definition of inversion.



    $$tag{3}J_{C,k}:M to M' iff exists vec{U} text{(unit norm vector) s.t.} begin{cases}vec{CM}=lambda vec{U}\ vec{CM'}=lambda' vec{U}end{cases} text{and} lambdalambda'=k^2$$



    Let $M(x,y)$ and $M'(x',y')$ exchanged by inversion $J_{C,k}$.



    Let $M_0(x,y,x^2+y^2)$ and $M_0'(x',y',x'^2+y'^2)$ be their "lifted version" on paraboloid $Pi$.



    Let $N(X,Y,Z) $ be any point of line $C_0M_0$. We can define an abscissa $mu$ on this line for point $N$ in the following way:



    $$tag{$star$}vec{C_0N}=mu vec{C_0M_0} iff begin{cases}X=a+mu(x-a)\ Y=b+mu(y-b)\ Z=c+mu(x^2+y^2-c)end{cases}$$



    with $$c:=a^2+b^2-k^2.$$



    Please note that for $mu=1$, $N$ is in $M_0$. Let $lambda'$ be the value of $mu$ associated with $M'_0$.



    Thus $N in Pi$ in two cases: when $mu=1$ ($N=M_0$) and when $mu=lambda'$ ($N=M'_0$). As



    $$tag{3}N in Pi iff Z=X^2+Y^2,$$



    plugging in (3) the expressions of $X,Y,Z$ in $(star)$, and replacing $c$ by its defining expression, we obtain the following quadratic equation with unknown $mu$:



    $$mu^2[(x-a)^2+(y-b)^2] + mu[cdots] + k^2=0$$



    Using the classical formula for the roots' product, we have



    $$lambda' times 1 =dfrac{k^2}{(x-a)^2+(y-b)^2}=dfrac{k^2}{|vec{CM}|^2}$$



    We find back here the definition of inversion given in (2) (recall that $vec{CM}$ hasn't been normalized).










    share|cite|improve this question











    $endgroup$















      10












      10








      10


      2



      $begingroup$


      Inversion transform with center $C$ and power $k^2$ is defined by:



      $$tag{1}J_{C,k}:M leftrightarrow M' iff vec{CM'}=frac{k^2}{||vec{CM}||^2} vec{CM} $$



      The double arrow is there to mean that it is an "involutive" transform: $M'$ is the image of $M$ iff $M$ is the image of $M'$.



      This transform, presented for the first time in a systematic way by Plücker, (1834) is detailed in many books/sites.



      See Appendix 1 for a comment about inversion properties displayed on Fig. 1.



      enter image description here



      Fig. 1. : Inversion with center $C$ and power 1. Shapes in black are exchanged by inversion with shapes in red. The blue circle, called "circle of inversion", is the locus of invariant points.



      But there are ill-known features that I would like to gather.



      Moreover, I would like to enlarge this "quest" to interesting (not fully standard) questions/applications of inversion.



      For the initialization of this "collection", I propose 3 themes and a compendium of interesting facts:



      1) "Champagne glass inversion" : Fig. 2 shows how inversion with center $C$ and radius of inversion $k$ (we find back the large and the small fish, inverted images one of the other...) can be realized by a 3 steps $PGamma P^{-1}$ operation with respect to a paraboloid $Pi$, where:




      • $P$ is the vertical projection from $Pi$ onto the horizontal plane, and


      • $Gamma$ is the conical projection from $Pi$ to $Pi$ with center $C_0(a,b,a^2+b^2-k^2)$ where $(a,b)$ are the coordinates of $C$.



      This non-classical way to describe inversion deserves an explanation that we have placed in Appendix 2.



      enter image description here



      Fig. 2. Inversion by a conical projection $Pi to Pi$.



      2) "Bubble inversion" This is a cousin representation of the first one with a sphere instead of a parabola (see Fig. 3). It necessitates to use 2 steps with the two stereographic projections $S_N$ and $S_S$ from the North and South pole resp. with respect to the equatorial plane. Take a look at Fig. 3.



      Consider a certain figure, say the circle in red. It is transformed by $S_N$ into the circle on the unit sphere materialized by black little diamonds ; this circle is transformed in turn by $S_S$ into the blue circle on the equatorial plane... which is the inversive image of the initial red circle. Briefly said :



      $$tag{$star$}I=S_S circ S_N$$



      This shouldn't come as a surprize because stereographic projections are themselves 3D inversions.



      See (composition of stereographic projections is inversion through the ball - a geometric way) for a proof.



      enter image description here



      Fig. 3. (planar) inversion realized by combining two stereographic projections (formula ($star$)).



      3) "How inversion transform is connected with linear algebra" (in fact connected with item 1) ; this issue looks paradoxical because inversion is definitely not a linear transform. In fact, there exists a $4 times 4$ linear representation of the co-called conformal geometry including another category of non-linear transforms, translations. Reference: the very good book "Riemannian Geometry" by S. Gallot, D. Hullin, J. Lafontaine, 2nd edition 1993. Universitext, Springer, pages 175-176. Here is how this correspondence is done (Explanations will be found in this book or in the very interesting development by @MvG cited in 4a):



      $$text{If} J text{is the basic inversion (center 0, power 1):}$$
      $$[J]:=begin{pmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{pmatrix}.$$
      $$text{If} H_r text{is the homothety with ratio} r text{and center} 0: $$
      $$[H_r]:=begin{pmatrix}1&0&0&0\0&r&0&0\0&0&r&0\0&0&0&r^2end{pmatrix}.$$



      $$text{If} T_V text{is the translation by vector} V=binom{a}{b} :$$ $$[T_V]:=begin{pmatrix}1&0&0&0\2a&1&0&0\2b&0&1&0\a^2+b^2&a&b&1end{pmatrix}.$$



      All these transformations preserve the following quadratic form:



      $$q(x_0;x_1,x_2,x_3):=x_1^2+x_2^2-x_0x_3$$



      where $x_0$ should be considered as a homogeneous coordinate (i.e., is there to take into account the projective dimension). The signature of $q$ being $(+++-)$, we are dealing with elements of the group classically denoted $O(3,1)$.



      This correspondence with 2D transforms can be extended in a straightforward way to higher dimensions (for $n$D transforms, the correspondence is with $(n+2) times (n+2)$ matrices).



      4) Complex representation with an application :



      As recalled by @José Carlos Santos, inversion with the unit circle $S^1={zinmathbb{C},|,|z|=1$} as invariant circle is the map



      $$I : begin{cases} begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&frac{1}{overline z}end{array}end{cases}$$



      Let us prove the following result :



      Any (2D) rotation can be obtained as a certain combination of translations and inversions.



      Let us consider the following identity (as given in https://mathoverflow.net/q/19965) :



      $$e^{i theta} + frac{1}{-e^{-i theta} + frac{1}{e^{i theta} + frac{1}{z}}} = - e^{2 i theta} z. tag{%}$$



      If we conjugate both sides of (%), we get :



      $$e^{-i theta} + frac{1}{overline{-e^{-i theta} + frac{1}{overline{e^{-i theta} + frac{1}{overline{z}}}}}} = - e^{-2 i theta}overline{z}. tag{%%}$$



      (The LHS of (%%) can be written in a symbolic way as : $T_{e^{-i theta}}circ I circ T_{-e^{-i theta}} circ I circ T_{e^{-i theta}} circ I$).



      If we take $theta=0$ in (%%), one gets :



      $$1+ frac{1}{overline{-1 + frac{1}{overline{1 + frac{1}{overline{z}}}}}} = - overline{z}. tag{%%%}$$



      (%%%) means that transformation $zmapsto-overline{z}$ can be obtained by combining only inversions and translation(s).



      In a similar manner (%%) means that transformation $zmapsto -e^{-2itheta}overline{z}$, can obtained using as well inversions and translations.



      Combining both transformations, we can thus transform $z$ into $e^{-2itheta}z$ for any theta, using translations and inversions only.



      5) A compendium of interesting documents/issues




      • a) See the diaporama (http://www.cis.umac.mo/~fstitl/2013geometry/sterograph.pdf).


      • b) An in-depth analysis by MvG in (https://math.stackexchange.com/q/866403) related to conformal group analysed in 3).


      • c) A study of torii, Dupin cyclides, etc. in connection with 3D inversion : (inverting a cone to a torus).


      • d) An answer I made (https://math.stackexchange.com/q/2621344) explaining why, locally, in the vicinity of the circle of inversion, inversion behaves as a symmetry. This is also elegantly treated in ref. 3a).


      • e) An application to Voronoi centers (https://math.stackexchange.com/q/2378365).


      • f) An extension of a IMO problem solved by inversion/stereographic projection (https://pdfs.semanticscholar.org/3bcb/00d463fceb26ab103c9a3757ee85398eb187.pdf).




      My (enlarged) question is : can you give examples (of your own) of interesting, possibly non-usual, uses of inversion ?




      Appendix 1: Recall of properties of inversion as depicted by Fig. 1:




      • The image of a straight line is in general a circle that passes through the pole $C$, and, exceptionally, when the straight line passes through the origin, its image is the line itself (caution: in the latter case, the straight line is "globally" invariant, but its points, in general, aren't invariant).


      • The image $Gamma'$ of a circle $Gamma$ not passing through the origin is a circle of the same type (caution: the center of $Gamma'$ is not the image of the center of $Gamma$). The image of a circle passing through the origin is a straight line, as said before.


      • The image of the black fish made of line segments is the red fish made of circular arcs swimming in its blue round fishbowl.



      Appendix 2: Explanation of inversion depicted in Fig. 2.



      Let us begin by another definition of inversion.



      $$tag{3}J_{C,k}:M to M' iff exists vec{U} text{(unit norm vector) s.t.} begin{cases}vec{CM}=lambda vec{U}\ vec{CM'}=lambda' vec{U}end{cases} text{and} lambdalambda'=k^2$$



      Let $M(x,y)$ and $M'(x',y')$ exchanged by inversion $J_{C,k}$.



      Let $M_0(x,y,x^2+y^2)$ and $M_0'(x',y',x'^2+y'^2)$ be their "lifted version" on paraboloid $Pi$.



      Let $N(X,Y,Z) $ be any point of line $C_0M_0$. We can define an abscissa $mu$ on this line for point $N$ in the following way:



      $$tag{$star$}vec{C_0N}=mu vec{C_0M_0} iff begin{cases}X=a+mu(x-a)\ Y=b+mu(y-b)\ Z=c+mu(x^2+y^2-c)end{cases}$$



      with $$c:=a^2+b^2-k^2.$$



      Please note that for $mu=1$, $N$ is in $M_0$. Let $lambda'$ be the value of $mu$ associated with $M'_0$.



      Thus $N in Pi$ in two cases: when $mu=1$ ($N=M_0$) and when $mu=lambda'$ ($N=M'_0$). As



      $$tag{3}N in Pi iff Z=X^2+Y^2,$$



      plugging in (3) the expressions of $X,Y,Z$ in $(star)$, and replacing $c$ by its defining expression, we obtain the following quadratic equation with unknown $mu$:



      $$mu^2[(x-a)^2+(y-b)^2] + mu[cdots] + k^2=0$$



      Using the classical formula for the roots' product, we have



      $$lambda' times 1 =dfrac{k^2}{(x-a)^2+(y-b)^2}=dfrac{k^2}{|vec{CM}|^2}$$



      We find back here the definition of inversion given in (2) (recall that $vec{CM}$ hasn't been normalized).










      share|cite|improve this question











      $endgroup$




      Inversion transform with center $C$ and power $k^2$ is defined by:



      $$tag{1}J_{C,k}:M leftrightarrow M' iff vec{CM'}=frac{k^2}{||vec{CM}||^2} vec{CM} $$



      The double arrow is there to mean that it is an "involutive" transform: $M'$ is the image of $M$ iff $M$ is the image of $M'$.



      This transform, presented for the first time in a systematic way by Plücker, (1834) is detailed in many books/sites.



      See Appendix 1 for a comment about inversion properties displayed on Fig. 1.



      enter image description here



      Fig. 1. : Inversion with center $C$ and power 1. Shapes in black are exchanged by inversion with shapes in red. The blue circle, called "circle of inversion", is the locus of invariant points.



      But there are ill-known features that I would like to gather.



      Moreover, I would like to enlarge this "quest" to interesting (not fully standard) questions/applications of inversion.



      For the initialization of this "collection", I propose 3 themes and a compendium of interesting facts:



      1) "Champagne glass inversion" : Fig. 2 shows how inversion with center $C$ and radius of inversion $k$ (we find back the large and the small fish, inverted images one of the other...) can be realized by a 3 steps $PGamma P^{-1}$ operation with respect to a paraboloid $Pi$, where:




      • $P$ is the vertical projection from $Pi$ onto the horizontal plane, and


      • $Gamma$ is the conical projection from $Pi$ to $Pi$ with center $C_0(a,b,a^2+b^2-k^2)$ where $(a,b)$ are the coordinates of $C$.



      This non-classical way to describe inversion deserves an explanation that we have placed in Appendix 2.



      enter image description here



      Fig. 2. Inversion by a conical projection $Pi to Pi$.



      2) "Bubble inversion" This is a cousin representation of the first one with a sphere instead of a parabola (see Fig. 3). It necessitates to use 2 steps with the two stereographic projections $S_N$ and $S_S$ from the North and South pole resp. with respect to the equatorial plane. Take a look at Fig. 3.



      Consider a certain figure, say the circle in red. It is transformed by $S_N$ into the circle on the unit sphere materialized by black little diamonds ; this circle is transformed in turn by $S_S$ into the blue circle on the equatorial plane... which is the inversive image of the initial red circle. Briefly said :



      $$tag{$star$}I=S_S circ S_N$$



      This shouldn't come as a surprize because stereographic projections are themselves 3D inversions.



      See (composition of stereographic projections is inversion through the ball - a geometric way) for a proof.



      enter image description here



      Fig. 3. (planar) inversion realized by combining two stereographic projections (formula ($star$)).



      3) "How inversion transform is connected with linear algebra" (in fact connected with item 1) ; this issue looks paradoxical because inversion is definitely not a linear transform. In fact, there exists a $4 times 4$ linear representation of the co-called conformal geometry including another category of non-linear transforms, translations. Reference: the very good book "Riemannian Geometry" by S. Gallot, D. Hullin, J. Lafontaine, 2nd edition 1993. Universitext, Springer, pages 175-176. Here is how this correspondence is done (Explanations will be found in this book or in the very interesting development by @MvG cited in 4a):



      $$text{If} J text{is the basic inversion (center 0, power 1):}$$
      $$[J]:=begin{pmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{pmatrix}.$$
      $$text{If} H_r text{is the homothety with ratio} r text{and center} 0: $$
      $$[H_r]:=begin{pmatrix}1&0&0&0\0&r&0&0\0&0&r&0\0&0&0&r^2end{pmatrix}.$$



      $$text{If} T_V text{is the translation by vector} V=binom{a}{b} :$$ $$[T_V]:=begin{pmatrix}1&0&0&0\2a&1&0&0\2b&0&1&0\a^2+b^2&a&b&1end{pmatrix}.$$



      All these transformations preserve the following quadratic form:



      $$q(x_0;x_1,x_2,x_3):=x_1^2+x_2^2-x_0x_3$$



      where $x_0$ should be considered as a homogeneous coordinate (i.e., is there to take into account the projective dimension). The signature of $q$ being $(+++-)$, we are dealing with elements of the group classically denoted $O(3,1)$.



      This correspondence with 2D transforms can be extended in a straightforward way to higher dimensions (for $n$D transforms, the correspondence is with $(n+2) times (n+2)$ matrices).



      4) Complex representation with an application :



      As recalled by @José Carlos Santos, inversion with the unit circle $S^1={zinmathbb{C},|,|z|=1$} as invariant circle is the map



      $$I : begin{cases} begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&frac{1}{overline z}end{array}end{cases}$$



      Let us prove the following result :



      Any (2D) rotation can be obtained as a certain combination of translations and inversions.



      Let us consider the following identity (as given in https://mathoverflow.net/q/19965) :



      $$e^{i theta} + frac{1}{-e^{-i theta} + frac{1}{e^{i theta} + frac{1}{z}}} = - e^{2 i theta} z. tag{%}$$



      If we conjugate both sides of (%), we get :



      $$e^{-i theta} + frac{1}{overline{-e^{-i theta} + frac{1}{overline{e^{-i theta} + frac{1}{overline{z}}}}}} = - e^{-2 i theta}overline{z}. tag{%%}$$



      (The LHS of (%%) can be written in a symbolic way as : $T_{e^{-i theta}}circ I circ T_{-e^{-i theta}} circ I circ T_{e^{-i theta}} circ I$).



      If we take $theta=0$ in (%%), one gets :



      $$1+ frac{1}{overline{-1 + frac{1}{overline{1 + frac{1}{overline{z}}}}}} = - overline{z}. tag{%%%}$$



      (%%%) means that transformation $zmapsto-overline{z}$ can be obtained by combining only inversions and translation(s).



      In a similar manner (%%) means that transformation $zmapsto -e^{-2itheta}overline{z}$, can obtained using as well inversions and translations.



      Combining both transformations, we can thus transform $z$ into $e^{-2itheta}z$ for any theta, using translations and inversions only.



      5) A compendium of interesting documents/issues




      • a) See the diaporama (http://www.cis.umac.mo/~fstitl/2013geometry/sterograph.pdf).


      • b) An in-depth analysis by MvG in (https://math.stackexchange.com/q/866403) related to conformal group analysed in 3).


      • c) A study of torii, Dupin cyclides, etc. in connection with 3D inversion : (inverting a cone to a torus).


      • d) An answer I made (https://math.stackexchange.com/q/2621344) explaining why, locally, in the vicinity of the circle of inversion, inversion behaves as a symmetry. This is also elegantly treated in ref. 3a).


      • e) An application to Voronoi centers (https://math.stackexchange.com/q/2378365).


      • f) An extension of a IMO problem solved by inversion/stereographic projection (https://pdfs.semanticscholar.org/3bcb/00d463fceb26ab103c9a3757ee85398eb187.pdf).




      My (enlarged) question is : can you give examples (of your own) of interesting, possibly non-usual, uses of inversion ?




      Appendix 1: Recall of properties of inversion as depicted by Fig. 1:




      • The image of a straight line is in general a circle that passes through the pole $C$, and, exceptionally, when the straight line passes through the origin, its image is the line itself (caution: in the latter case, the straight line is "globally" invariant, but its points, in general, aren't invariant).


      • The image $Gamma'$ of a circle $Gamma$ not passing through the origin is a circle of the same type (caution: the center of $Gamma'$ is not the image of the center of $Gamma$). The image of a circle passing through the origin is a straight line, as said before.


      • The image of the black fish made of line segments is the red fish made of circular arcs swimming in its blue round fishbowl.



      Appendix 2: Explanation of inversion depicted in Fig. 2.



      Let us begin by another definition of inversion.



      $$tag{3}J_{C,k}:M to M' iff exists vec{U} text{(unit norm vector) s.t.} begin{cases}vec{CM}=lambda vec{U}\ vec{CM'}=lambda' vec{U}end{cases} text{and} lambdalambda'=k^2$$



      Let $M(x,y)$ and $M'(x',y')$ exchanged by inversion $J_{C,k}$.



      Let $M_0(x,y,x^2+y^2)$ and $M_0'(x',y',x'^2+y'^2)$ be their "lifted version" on paraboloid $Pi$.



      Let $N(X,Y,Z) $ be any point of line $C_0M_0$. We can define an abscissa $mu$ on this line for point $N$ in the following way:



      $$tag{$star$}vec{C_0N}=mu vec{C_0M_0} iff begin{cases}X=a+mu(x-a)\ Y=b+mu(y-b)\ Z=c+mu(x^2+y^2-c)end{cases}$$



      with $$c:=a^2+b^2-k^2.$$



      Please note that for $mu=1$, $N$ is in $M_0$. Let $lambda'$ be the value of $mu$ associated with $M'_0$.



      Thus $N in Pi$ in two cases: when $mu=1$ ($N=M_0$) and when $mu=lambda'$ ($N=M'_0$). As



      $$tag{3}N in Pi iff Z=X^2+Y^2,$$



      plugging in (3) the expressions of $X,Y,Z$ in $(star)$, and replacing $c$ by its defining expression, we obtain the following quadratic equation with unknown $mu$:



      $$mu^2[(x-a)^2+(y-b)^2] + mu[cdots] + k^2=0$$



      Using the classical formula for the roots' product, we have



      $$lambda' times 1 =dfrac{k^2}{(x-a)^2+(y-b)^2}=dfrac{k^2}{|vec{CM}|^2}$$



      We find back here the definition of inversion given in (2) (recall that $vec{CM}$ hasn't been normalized).







      geometry linear-transformations circle big-list






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 9:18







      Jean Marie

















      asked Jan 29 '18 at 11:23









      Jean MarieJean Marie

      29.9k42051




      29.9k42051






















          4 Answers
          4






          active

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          3












          $begingroup$

          An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.




          Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
            $endgroup$
            – Jean Marie
            Jan 29 '18 at 21:13





















          3












          $begingroup$

          Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem.
          For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1={zinmathbb{C},|,|z|=1$} is simply the map$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&overline z^{,-1}.end{array}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
              $endgroup$
              – Jean Marie
              Jan 29 '18 at 11:55





















            2












            $begingroup$

            Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:



            enter image description here



            Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.



            The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.






            share|cite|improve this answer









            $endgroup$













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              4 Answers
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              4 Answers
              4






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              active

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              3












              $begingroup$

              An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.




              Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”







              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
                $endgroup$
                – Jean Marie
                Jan 29 '18 at 21:13


















              3












              $begingroup$

              An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.




              Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”







              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
                $endgroup$
                – Jean Marie
                Jan 29 '18 at 21:13
















              3












              3








              3





              $begingroup$

              An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.




              Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”







              share|cite|improve this answer









              $endgroup$



              An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.




              Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”








              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 29 '18 at 21:09









              Moishe CohenMoishe Cohen

              47.1k343108




              47.1k343108












              • $begingroup$
                Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
                $endgroup$
                – Jean Marie
                Jan 29 '18 at 21:13




















              • $begingroup$
                Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
                $endgroup$
                – Jean Marie
                Jan 29 '18 at 21:13


















              $begingroup$
              Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
              $endgroup$
              – Jean Marie
              Jan 29 '18 at 21:13






              $begingroup$
              Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote).
              $endgroup$
              – Jean Marie
              Jan 29 '18 at 21:13













              3












              $begingroup$

              Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem.
              For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem.
                For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem.
                  For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html






                  share|cite|improve this answer









                  $endgroup$



                  Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem.
                  For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 29 '18 at 21:18









                  Soumil AggarwalSoumil Aggarwal

                  578




                  578























                      2












                      $begingroup$

                      Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1={zinmathbb{C},|,|z|=1$} is simply the map$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&overline z^{,-1}.end{array}$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
                        $endgroup$
                        – Jean Marie
                        Jan 29 '18 at 11:55


















                      2












                      $begingroup$

                      Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1={zinmathbb{C},|,|z|=1$} is simply the map$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&overline z^{,-1}.end{array}$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
                        $endgroup$
                        – Jean Marie
                        Jan 29 '18 at 11:55
















                      2












                      2








                      2





                      $begingroup$

                      Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1={zinmathbb{C},|,|z|=1$} is simply the map$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&overline z^{,-1}.end{array}$$






                      share|cite|improve this answer









                      $endgroup$



                      Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1={zinmathbb{C},|,|z|=1$} is simply the map$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb{C}setminus{0}\z&mapsto&overline z^{,-1}.end{array}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 29 '18 at 11:27









                      José Carlos SantosJosé Carlos Santos

                      162k22128232




                      162k22128232












                      • $begingroup$
                        It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
                        $endgroup$
                        – Jean Marie
                        Jan 29 '18 at 11:55




















                      • $begingroup$
                        It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
                        $endgroup$
                        – Jean Marie
                        Jan 29 '18 at 11:55


















                      $begingroup$
                      It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
                      $endgroup$
                      – Jean Marie
                      Jan 29 '18 at 11:55






                      $begingroup$
                      It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition...
                      $endgroup$
                      – Jean Marie
                      Jan 29 '18 at 11:55













                      2












                      $begingroup$

                      Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:



                      enter image description here



                      Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.



                      The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:



                        enter image description here



                        Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.



                        The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:



                          enter image description here



                          Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.



                          The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.






                          share|cite|improve this answer









                          $endgroup$



                          Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:



                          enter image description here



                          Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.



                          The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 30 '18 at 0:34









                          dxivdxiv

                          57.9k648101




                          57.9k648101






























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