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Locus

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  • Antreas Hatzipolakis
    Let ABC be a triangle, P a point and (a),(Ob),(Oc) the circles with diameters PA,PB,PC, resp. The circles (Ob), (Oc) intersect in point A1, other than P, on BC
    Message 1 of 240 , Feb 28, 2013
      Let ABC be a triangle, P a point and (a),(Ob),(Oc) the circles
      with diameters PA,PB,PC, resp.

      The circles (Ob), (Oc) intersect in point A1, other than P, on BC
      Similarly B1,C1. The triangle A1B1C1 is the pedal triangle of P.

      Let A2B2C2 be the cevian triangle of P.

      The circles (Oa),(Ob), (Oc) intersect (O) at A3,B3,C3 [other than A,B,C, resp.]

      The lines AA3, BB3, CC3 [= readical axes of
      ((O),((Oa),((O),((Ob),((O),((Oc), resp.]
      bound a triangle A4B4C4.

      Loci:

      Which is the locus of P such that be perspective the triangles:

      1. ABC, A1B1C1 : it is Darboux cubic (since A1B1C1 is the pedal triangle of P)

      2. ABC, A4B4C4

      3. A1B1C1, A3B3C3

      4. A1B1C1, A4B4C4

      5. A2B2C2, A3B3C3,

      6. A2B2C2, A4B4C4

      7. A3B3C3, A4B4C4

      I think that in all cases the locus is Darboux cubic + possibly something else.

      APH
    • Antreas Hatzipolakis
      [APH]: Let ABC be a triangle. A line L passing through A intersects the circle with diameter AB again at Ab and the circle with diameter AC again at Ac.
      Message 240 of 240 , Feb 16

        [APH]:

         

        Let ABC be a triangle.

        A line L passing through A intersects the circle with diameter AB again at Ab and the circle with diameter AC again at Ac.

        [Equivalently: Let Ab, Ac be the orthogonal projections of B, C on L, resp.]
        Let A* be the intersection of BAc and CAb.

        Which is the locus of A* as L moves around A?


        [César Lozada]:


        Parametric trilinear equation:

         

        1/u(t) = a*((b^2+c^2-a^2)^2-4*b^2*c^2*c os(2*t)^2)/(2*S)

         

        1/v(t) = 2*(cos(2*t)*c-b)*S - c*sin(2*t)*(a^2+3*b^2-2*cos(2* t)*b*c-c^2)

         

        1/w(t) = 2*(cos(2*t)*b-c)*S + b*sin(2*t)*(a^2+3*c^2-2*cos(2* t)*b*c-b^2)

         

        Regards,

        César Lozada

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