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

15394Re:Cevian Trace Equal Area Points

Expand Messages
  • Nikolaos Dergiades
    Aug 1, 2007
      Dear Tuan,

      > I have found this point and this construction by
      > combine following small well known elementary locus
      > problem into triangle:
      > Two segments XY, X'Y' are on given two lines. To
      > find the locus of Z such that Area(ZXY) =
      > Area(ZX'Y').
      > If areas are not signed then the locus is two
      > lines passing intersection of two given lines.

      Yes the same method I also used.
      If A is the intersection of XY, X'Y' and
      M, M' are the mid points of XY', X'Y
      then the locus such that the signed areas
      are equal is the line parallel to MM' that
      passes through A.

      Best regards
      Nikos Dergiades




      ___________________________________________________________
      Χρησιμοποιείτε Yahoo!;
      Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
      διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
      μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
    • Show all 18 messages in this topic