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14973Re: [EMHL] Four Circum Equilateral Triangles

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  • Alexey.A.Zaslavsky
    Mar 5, 2007
    • 0 Attachment
      Dear Quang!
      The point P is isogonally conjugated to the infinite point of Euler line, i.e the common point of the circumcircle and the Neuberg cubic distinct from A, B, C and two circular points.

      Sincerely Alexey

      I
      Dear All My Friends,
      > Given triangle ABC with circumcircle (O). There is and exactly one
      > equilateral triangle inscribed in (O) taken A as one vertex. We denote its
      > bottom line wrt A as La, intesection of La with sideline BC as A'. Similarly
      > define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one triangle A''B''C''.
      > Please prove following results:
      > 1. ABC and A''B''C'' are perspective with perspector P on circumcircle
      > (O).
      > 2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
      > 3. PQR are our fourth equilateral.
      > Thank you and best regards,
      >
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