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More Cubics

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  • xpolakis@otenet.gr
    Let ABC be a triangle and P a point on its plane. We construct [the how, and the number of solutions is another story !] three similar triangles PAbAc, PBaBc,
    Message 1 of 5 , Dec 2, 2000
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      Let ABC be a triangle and P a point on its plane.
      We construct [the how, and the number of solutions is another story !]
      three similar triangles PAbAc, PBaBc, PCaCb, whose the vertices Ai, Bi, Ci
      lie on the sides of ABC.

      A
      /\
      / \
      / \
      / \
      Ac Ma Ab
      / q f \
      / w \
      / P \
      Bc w w Cb
      / f q \
      / Mb Mc \
      / q f \
      B-----Ba--------Ca-------C


      Angles of the similar triangles PAbAc, PBaBc, PCaCb:
      P [= AcPAb = BcPBa = CaPCb] := omega [w; not be confused with
      Brocard's!]

      Ab [=PAbAc] = Bc = Ca := phi (f)
      Ac [=PAcAb] = Ba = Cb := theta (q)

      Let now Ma,Mb,Mc be the midpoints of AbAc, BcBa, CaCb.

      The locus of the points P such that MaMbMc be in perspective with ABC
      is a cubic [actually a diparametric family of cubics, since the
      equation has two parameters, namely f,q [the third, w, is expressd by the
      twos : w = Pi - (f+q)]).

      If the similar triangles are isosceles with f = q = (Pi - w)/2, then the
      cubic, according to my computations, is an isogonal one (actually a pencil
      of isogonal cubics) with equation:

      x(y^2 + z^2) (BC - sA) + [cyclically] = 0 (in trilinears)

      where A,B,C are shortcuts for sin(A+w) - sinA, B =..., C = ...., s = sinw.

      The pivot (sin(B+w) - sinB)(sin(C+w) - sinC) - sinw(sin(A+w) - sinA) ::)
      can be simplified, but I didn't make the calculations.

      The most interesting cases are of course for w = 60 d. or 90 d.

      If w = 60 d. then the three similar triangles are equilateral, and a
      natural question is: How about their centers K1,K2,K3 ? That is:
      Which is the locus of P such that K1K2K3 be in perspective with ABC?
      (I think that the locus is a sextic, but I don't know whether reduces or not)

      If w = 90 d. then the midpoints Ma,Mb,Mc are centers of three squares
      (Kenmotu configuration), whose the fourth vertices let be P1,P2,P3:

      A
      /\
      /P1\
      / \
      / \
      Ac Ma Ab
      / \
      / \
      / P \
      Bc Cb
      / \
      / Mb Mc \
      / \ P3
      P2 B-----Ba--------Ca-------C


      And now the question is about the P1,P2,P3:
      For which points P the triangle P1P2P3 is in perspective with ABC?
      (in general, we can also consider P1,P2,P3 as the forth vertices of
      three similar rhombi - FvL's configuration).
      (I haven't worked on this locus.)

      Antreas
    • xpolakis@otenet.gr
      ... ^ x(y^2 - z^2) (BC - sA) + [cyclically] = 0 (in trilinears) APH
      Message 2 of 5 , Dec 2, 2000
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        Correction of a typo:

        >x(y^2 + z^2) (BC - sA) + [cyclically] = 0 (in trilinears)
        ^
        x(y^2 - z^2) (BC - sA) + [cyclically] = 0 (in trilinears)

        APH
      • xpolakis@otenet.gr
        Let ABC be a triangle, and PaPbPc the pedal triangle of P. The circle (P, PPa) intersects the bisector of ang(PbPPc) at A , A [A near to A] Similary we
        Message 3 of 5 , Oct 19, 2001
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          Let ABC be a triangle, and PaPbPc the pedal triangle of P.

          The circle (P, PPa) intersects the bisector of ang(PbPPc) at A', A"
          [A' near to A]

          Similary we define the points B', C'; B", C".

          Which are the loci of P such that:

          1. ABC, A'B'C'

          2. ABC, A"B"C"

          are perspective?


          Let A1, A2 be the orth. proj. of A', A" (resp.) on BC.
          Similarly we define the points B1, C1; B2, C2.

          Which are the loci of P such that:

          3. ABC, A1B1C1

          4. ABC, A2B2C2

          are perspective?


          Antreas
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