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Re: [EMHL] Feuerbach

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  • jpehrmfr
    Dear Darij, I wrote ... Of course, this result is slightly different if ABCP is cyclic: F is the midpoint of PH; NNaNbNc lie on the circle (F,R/2) and the
    Message 1 of 3 , Apr 30, 2003
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      Dear Darij, I wrote
      > For any point P, let F = common point of the NPC of ABC, PBC, PCA,
      > PAB = center of the rectangular circumhyperbola going through P
      > Let N, Na, Nb, Nc be the NP-centers of ABC, BCP, CAP, ABP.
      > Q = isogonal conjugate of P, QaQbQc= pedal triangle of Q.
      > Then the figures QQaQbQc and NNaNbNc are indirectly similar.
      > Moreover, if Q' = inverse of Q in the circumcircle = isogonal
      > conjugate of the reflection of P wrt F, NaNbNc and the pedal
      > triangle of Q' are homothetic; the ratio is OQ^2/(OQ^2-R^2).

      Of course, this result is slightly different if ABCP is cyclic:
      F is the midpoint of PH; NNaNbNc lie on the circle (F,R/2) and the
      triangle NaNbNc is translated from the medial triangle by
      NF = 1/2 OP.
      Friendly. Jean-Pierre
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