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Re: [EMHL] Kn & Lester circle

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  • Bernard Gibert
    Dear Larry, your error leads to another point on the Lester circle, so no regret ! ... that is correct and the corresponding point on the Lester circle is
    Message 1 of 7 , Jun 1 12:55 AM
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      Dear Larry,

      your "error" leads to another point on the Lester circle, so no regret !
      >
      > Again I am sorry for another error. The triangles should have been
      > T(5,13,14), T(5,17,18), T(6,14,17), and T(6,13,18).

      that is correct and the corresponding point on the Lester circle is the
      reflection of X6 about the line through the centers of circumcircles of
      T(6,14,17), and T(6,13,18).

      1st bary :

      5*a^14 - 21*a^12*b^2 + 41*a^10*b^4 - 38*a^8*b^6 - 7*a^6*b^8 + 49*a^4*b^10 -
      39*a^2*b^12 + 10*b^14 - 21*a^12*c^2 +
      44*a^10*b^2*c^2 - 44*a^8*b^4*c^2 + 80*a^6*b^6*c^2 - 146*a^4*b^8*c^2 +
      155*a^2*b^10*c^2 - 50*b^12*c^2 + 41*a^10*c^4 -
      44*a^8*b^2*c^4 - 53*a^6*b^4*c^4 + 88*a^4*b^6*c^4 - 251*a^2*b^8*c^4 +
      90*b^10*c^4 - 38*a^8*c^6 + 80*a^6*b^2*c^6 +
      88*a^4*b^4*c^6 + 270*a^2*b^6*c^6 - 50*b^8*c^6 - 7*a^6*c^8 -
      146*a^4*b^2*c^8 - 251*a^2*b^4*c^8 - 50*b^6*c^8 +
      49*a^4*c^10 + 155*a^2*b^2*c^10 + 90*b^4*c^10 - 39*a^2*c^12 - 50*b^2*c^12 +
      10*c^14

      index : 3.22223

      The other point in message #5626 lies on the circles (6,13,17), (6,14,18)
      and (5,6,E221) where E221 is the point on the Euler line such that
      OE221=6/5 OG (vectors).

      The line through those two points meets the Euler line at X381 = GH
      midpoint.

      The two points are different from point E380 of message #5622.
      >
      > Are you able to receive GSP files?

      Unfortunately not. I use Cabri for my drawings.

      Best regards

      Bernard
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