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A simple occurrence of X(31) = (a^3 : b^3 : c^3)

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  • Paul Y Yiu
    Dear Antreas, [APH](19845): Given triangle ABC and its circumcircle (O). Construct the circle (Oa) which is tangent to [the segment] BC and tangent to (O) at
    Message 1 of 3 , Feb 15, 2011
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      Dear Antreas,


      [APH](19845): Given triangle ABC and its circumcircle (O). Construct the circle (Oa) which is tangent to [the segment] BC and tangent to (O) at (A). Let A' be the intersection of the tangents of (Oa) through B,C.

      *** Similarly define B' and C'.

      The triangle A'B'C' is perspective with ABC at X(31) = (a^3 : b^3 : c^3) in homogeneous barycentric coordinates.

      Best regards
      Sincerely
      Paul
    • Antreas
      Dear Paul We can construct triads of circles (Oa),(Ob),(Oc), each circle of whose is tangent to one (only) side of triangle, and construct similarly the
      Message 2 of 3 , Feb 15, 2011
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        Dear Paul

        We can construct triads of circles (Oa),(Ob),(Oc),
        each circle of whose is tangent to one (only) side of triangle,
        and construct similarly the intersections of tangens A',B',C'
        and ask if ABC and A'B'C' are perspective.

        Here is another example.

        Let HaHbHc be the orthic triangle, and DaDbDc the circumcevian
        tiangle of H. The circle (Da,DaHa) is tangent to BC at Ha.
        From B,C we draw tangents to the circle (Da,DaHa) intersecting
        at A'. Similarly B',C'.

        Are the triangles ABC, A'B'C' perspective?

        Antreas

        PS1: We can take the circles (A,AHa), (B,BHb), (C,CHc)
        (tangent to BC,CA,AB at Ha,Hb,Hc, resp.)

        PS2: A generalization is this:
        Let P be a point, PaPbPc the pedal (or cevian) triangle of P,
        (Oa) the circle centered on the circumcircle (O) and tangent
        the BC. Similarly (Ob),(Oc). All three circles are centered
        on either the positive or the negative sides of BC,CA,AB, resp.
        (so we have two cases).
        Let A' be the intersection of the tangents to (Oa) from B,C.
        Similarly B',C'. Which is the locus of P such that ABC, A'B'C'
        are perspective?

        APH



        --- In Hyacinthos@yahoogroups.com, Paul Y Yiu <yiu@...> wrote:
        >
        > Dear Antreas,
        >
        >
        > [APH](19845): Given triangle ABC and its circumcircle (O). Construct the circle (Oa) which is tangent to [the segment] BC and tangent to (O) at (A). Let A' be the intersection of the tangents of (Oa) through B,C.
        >
        > *** Similarly define B' and C'.
        >
        > The triangle A'B'C' is perspective with ABC at X(31) = (a^3 : b^3 : c^3) in homogeneous barycentric coordinates.
        >
        > Best regards
        > Sincerely
        > Paul
        >
      • xpolakis
        [APH] ... I think they are not. If A is the reflection of A in BC and A* is the intersection of AA and BC, and similarly B*,C*, then A*,B*,C* are collinear.
        Message 3 of 3 , Feb 16, 2011
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          [APH]
          > Let HaHbHc be the orthic triangle, and DaDbDc the circumcevian
          > tiangle of H. The circle (Da,DaHa) is tangent to BC at Ha.
          > From B,C we draw tangents to the circle (Da,DaHa) intersecting
          > at A'. Similarly B',C'.
          >
          > Are the triangles ABC, A'B'C' perspective?

          I think they are not.

          If A" is the reflection of A' in BC and A* is the
          intersection of AA" and BC, and similarly B*,C*,
          then A*,B*,C* are collinear.

          And since Da,Db,Dc are the reflections of H in Ha,Hb,Hc, resp.
          the problem in simpler form is:

          Let HaHbHc be the orthic triangle. The tangents from B,C to the
          circle (H,HHa) intersect at Aa. The lines AAa and BC intersect
          at A*. Similarly B*,C*.
          The points A*,B*,C* are collinear.

          [APH]
          > PS1: We can take the circles (A,AHa), (B,BHb), (C,CHc)
          > (tangent to BC,CA,AB at Ha,Hb,Hc, resp.)

          They are perspective at O

          Antreas
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