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The triangle center X(5020) as radical center

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  • Angel
    A simple property of X(5020): Let ABC be an triangle, DEF is the medial triangle. Let (Ca) be a circle through E and F that is tangent to the circumcircle at
    Message 1 of 2 , Jan 30, 2013
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      A simple property of X(5020):

      Let ABC be an triangle, DEF is the medial triangle. Let (Ca) be a circle through E and F that is tangent to the circumcircle at a point A' different of A. Defined similarly the circles (Cb) and (Cc), and the points B' and C'.

      The triangles ABC and A'B'C' are perspective and the perspector is X(25)
      The orthic triangle and A'B'C' are perspective and the perspector is X(2)

      The radical center of the circles (Ca), (Cb) and (Cc) is X(5020)= a^2(a^4-b^4-c^4+6b^2c^2): ... : ...


      A' = (a^4-(b^2-c^2)^2: -2b^2(a^2+b^2-c^2): -2c^2(a^2-b^2+c^2)).

      (Ca) = c^2x*y + b^2x*z + a^2y*z -
      (x+y+z)/(2(a^2+(b-c)^2)(a^2+(b+c)^2))
      (4a^2b^2c^2x + c^2(a^2-b^2+c^2)^2y + b^2(a^2+b^2-c^2)^2z).

      Angel M.
    • Randy Hutson
      Dear Angel, Here are a couple more properties of X(5020): X(5020) = homothetic center of medial triangle and 3rd antipedal triangle of X(3) X(5020) = trilinear
      Message 2 of 2 , Jan 30, 2013
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        Dear Angel,

        Here are a couple more properties of X(5020):

        X(5020) = homothetic center of medial triangle and 3rd antipedal triangle of X(3)
        X(5020) = trilinear pole of polar of X(3527) wrt 2nd Lemoine circle

        Best regards,
        Randy





        >________________________________
        > From: Angel <amontes1949@...>
        >To: Hyacinthos@yahoogroups.com
        >Sent: Wednesday, January 30, 2013 6:29 AM
        >Subject: [EMHL] The triangle center X(5020) as radical center
        >
        >

        >A simple property of X(5020):
        >
        >Let ABC be an triangle, DEF is the medial triangle. Let (Ca) be a circle through E and F that is tangent to the circumcircle at a point A' different of A. Defined similarly the circles (Cb) and (Cc), and the points B' and C'.
        >
        >The triangles ABC and A'B'C' are perspective and the perspector is X(25)
        >The orthic triangle and A'B'C' are perspective and the perspector is X(2)
        >
        >The radical center of the circles (Ca), (Cb) and (Cc) is X(5020)= a^2(a^4-b^4-c^4+6b^2c^2): ... : ...
        >
        >A' = (a^4-(b^2-c^2)^2: -2b^2(a^2+b^2-c^2): -2c^2(a^2-b^2+c^2)).
        >
        >(Ca) = c^2x*y + b^2x*z + a^2y*z -
        >(x+y+z)/(2(a^2+(b-c)^2)(a^2+(b+c)^2))
        >(4a^2b^2c^2x + c^2(a^2-b^2+c^2)^2y + b^2(a^2+b^2-c^2)^2z).
        >
        >Angel M.
        >
        >
        >
        >
        >

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