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More homocentric circles..... (Re: [EMHL] Reflecting circles on the altitudes)

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  • Antreas
    Let ABC be a triangle and A B C the circumcevian triangle of I. The circle (I,IA ) intersects again the circumcircle at A , the (I,IB ) at B and the (I,IC )
    Message 1 of 5 , Feb 22, 2013
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      Let ABC be a triangle and A'B'C' the circumcevian triangle of I.

      The circle (I,IA') intersects again the circumcircle at A",
      the (I,IB') at B" and the (I,IC') at C".

      1. ABC, A"B"C" are perspective on OI line

      2. Antipedal triangle of I (excentral triangle), A"B"C" are perspective.

      Orthic triangle Version:

      Let ABC be a triangle, HaHbHc the orthic triangle and
      A'B'C' the circumcevian triangle of H wrt NPC.

      The circle (H,HA') intersects again the NPC at A",
      the (H,HB') at B" and the (H, HC') at C".

      a. HaHbHc, A"B"C" are perspective (on OI line of orthic = Euler line of ABC)

      b. ABC, A"B"C" are perspective.

      Perspectors?

      APH
    • Angel
      Dear Antreas, 1. ABC, A B C are perspective on OI line. Perspector: X(36) 2. Antipedal triangle of I (excentral triangle),A B C are perspective. The first
      Message 2 of 5 , Feb 22, 2013
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        Dear Antreas,

        1. ABC, A"B"C" are perspective on OI line. Perspector: X(36)


        2. Antipedal triangle of I (excentral triangle),A"B"C" are perspective. The first barycentric coordinate of the perspector is:

        a(3a^6 - 4(b + c)a^5 + (8b*c - 3(b^2 + c^2))a^4 +
        (10(b^3 + c^3) - 4b*c(b + c))a^3 +
        (9b^2c^2 - 10b*c(b^2 + c^2) - b^4 - c^4)a^2 +
        (8b*c(b^3 + c^3) - 6(b^5 + c^5))a +
        b^6 + 2b^5c - b^4c^2 - 4b^3c^3 - b^2c^4 + 2b*c^5 + c^6)


        a. HaHbHc, A"B"C" are perspective (on OI line of orthic=Euler line of ABC). Perspector: Infinity Point of the Euler line



        b. ABC,A"B"C" are perspective.

        The first barycentric coordinate of the perspector is (S, SA, SB, SC usual Conway notation):

        SA/( (S^2-3SA^2) (SA(c^2-b^2)^2 + a^2(SB*SC-SA^2)) )

        This point is the barycentric product X(265)X(2986) or X(69)X(1989))X(2986)

        Best regards,
        Angel Montesdeoca



        --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
        >
        > Let ABC be a triangle and A'B'C' the circumcevian triangle of I.
        >
        > The circle (I,IA') intersects again the circumcircle at A",
        > the (I,IB') at B" and the (I,IC') at C".
        >
        > 1. ABC, A"B"C" are perspective on OI line
        >
        > 2. Antipedal triangle of I (excentral triangle), A"B"C" are perspective.
        >
        > Orthic triangle Version:
        >
        > Let ABC be a triangle, HaHbHc the orthic triangle and
        > A'B'C' the circumcevian triangle of H wrt NPC.
        >
        > The circle (H,HA') intersects again the NPC at A",
        > the (H,HB') at B" and the (H, HC') at C".
        >
        > a. HaHbHc, A"B"C" are perspective (on OI line of orthic = Euler line of ABC)
        >
        > b. ABC, A"B"C" are perspective.
        >
        > Perspectors?
        >
        > APH
        >
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