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

Where is the incenter ?

Expand Messages
  • Paul Yiu
    Dear John, The squared distance between the incenter I and the center of the orthocentroidal circle (midpoint of GH) is, in terms of a, b, c: the cyclic sum
    Message 1 of 1 , Jun 2, 2000
    • 0 Attachment
      Dear John,

      The squared distance between the incenter I and the center of the
      orthocentroidal circle (midpoint of GH) is, in terms of a, b, c:
      the cyclic sum

      Sigma [4a^6 - 6a^5(b+c) - 4a^4(b^2+c^2) + 12b^3c^3 + 21a^4bc
      - 15a^3bc(b+c) + 39a^2b^2c^2]

      divided by 36SS, S being twice the area.

      I shall try to understand the rest of your message.

      Best regards
      Sincerely
      Paul
    Your message has been successfully submitted and would be delivered to recipients shortly.