Loading ...
Sorry, an error occurred while loading the content.

9147Re: [EMHL] Re: excircle and circumcircle

Expand Messages
  • Darij Grinberg
    Jan 28, 2004
    • 0 Attachment
      Dear Alexey,

      In Hyacinthos message #9144, you wrote:

      >> I think that I is in HaHb.

      Exactly. My apologies for the typo.

      In Hyacinthos message #9145, you wrote:

      >> The equivalence of 2 and 5 is a particular case
      >> of next fact. Let given a triangle ABC and a
      >> point P. A1, B1 are the common point of AP and
      >> BC, BP and AC. Q - a point in A1B1, Q' is
      >> isogonally conjugated to Q, A2, B2 are the
      >> common points of AQ' and BC, BQ' and AC. Then
      >> the point P' isogonally conjugated to P is in
      >> A2B2. This can be easily proved by trilinear
      >> coordinates.

      Yes, and this was exactly the same generalization
      I found in 2002 when I solved the problem!

      Darij Grinberg
    • Show all 3 messages in this topic