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19646Re: A point related to a quadrilateral

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  • Chris Van Tienhoven
    Jan 2, 2011
      Dear Jean-Pierre,

      > [JP]
      > > > Consider a quadrilateral A_1,A_2,A_3,A_4
      > > > For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
      > > > Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
      > > > Is there a special name for this point M? Do you know some references?
      > [Chris]
      > > I do not know a special name for point M.
      > > However I noticed this. Maybe you know it already.
      > > Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
      > > Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
      > > T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same orientation as A_1,A_2,A_3,A_4.
      > [JP]
      > In fact, if O_1' is the circumcenter of O_2O_3O_4,...,
      > the same homothecy maps A_i to O_i' and B_i to O_i
      > The point M is characterized by the angular relations
      > <A_iMA_j = <A_iA_kA_j +<A_iA_lA_j (oriented angles modulo Pi) where (i,j,k,l) is any permutation of (1,2,3,4)
      > I would be very surprised if there were no references about this configuration

      Thank you for your comments.
      They helped me to search further.
      1. The angular relations can be described some more specific as:
      Aimj = Aikj + Ailj,
      where (i,j,k,l) is any permutation of (1,2,3,4) and
      Aimj = Angle Ai.M.Aj
      Aikl = Angle Ai.Ak.Aj when A_k on same side of line Ai.Aj as M
      Pi - Angle Ai.Ak.Aj when Ak on same side of line Ai.Aj as M
      Ailj = Angle Ai.Al.Aj when Al on same side of line Ai.Aj as M
      Pi - Angle Ai.Al.Aj when Al on other side of line Ai.Aj as M

      2. Conics M.A1.A2.A3.A4 and M.O1*.O2*.O3*.O4* (conics are ellipses when A1.A2.A3.A4 is convex) are mutually tangent in M.
      Conics M.B1.B2.B3.B4 and M.O1.O2.O3.O4 (conics are ellipses when A1.A2.A3.A4 is convex) are mutually tangent in M.

      3. Let Ma and Mo* be the Centers of Conics M.A1.A2.A3.A4 and M.O1*.O2*.O3*.O4*.
      Let Mb and Mo be the Centers of Conics M.B1.B2.B3.B4 and M.O1.O2.O3.O4.
      Now M, Ma, Mo* are collinear and M, Mb, Mo are also collinear.
      Further M.Mo* / M.Ma = M.Mo / M.Mb.

      4. General rules:
      Let T(A1,A2,A3,A4) = Transform A1,A2,A3,A4 --> O1,O2,O3,O4.
      As already stated T^2 produces a homothetic quadrilateral O1*.O2*.O3*.O4*.
      Now it comes out that the corresponding circumscribed conics (after T^2-transform) are tangent im M.
      Also it appears that there is a fixed similitude-ratio M.Mo*/M.Ma for transform T^2.
      I have the feeling just like you there must be more (known) about this figure.
      Best regards,

      Chris van Tienhoven
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