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Θέμα: [EMHL] Distance from centroid G and incenter I

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  • Nikolaos Dergiades
    Dear friends, after many efforts, I found a good proof for the inequality IA+IB+IC
    Message 1 of 3 , Dec 1, 2006
      Dear friends,
      after many efforts, I found a good proof for the
      inequality IA+IB+IC <= GA+GB+GC (1)
      based on an interesting property of X(177).

      Let A'B'C' be the cevian triangle of the incenter I,
      let A1B1C1 be the intouch triangle of ABC and
      let A2B2C2 be the triangle with vertices on the
      incircle and the segments IA, IB, IC that are the
      mid-arc points of triangle A'B'C'.
      The line A1A2 is a internal bisector of triangle
      and is perpendicular to B2C2.
      Hence A1A2, B1B2, C1C2 pass through the
      incenter Q of A'B'C' that is the orthocenter of
      This point Q is the point X(177) in ETC.
      If c < b and M1 is the mid-point of BC then
      BA1 = s-b < BA' = ac/(b+c) < BM1 = a/2.
      Hence the point Q is on the other side of the centroid
      relative to the line AA'.
      The lines AA', BB', CC' devide the triangle ABC in 6
      Tba = IBA', Tcb = ICB', Tac = IAC'
      Tca = IA'C, Tab= IB'A, Tbc = IC'B
      If the points G, I coincide then (1) is equality.
      If G is on a bisector of ABC then Q is on the same
      and the angle GIQ = pi.
      If e.g. G is inside triangle Tbc then Q is inside
      triangle Tcb
      and since angle BIC is obtuse then also angle GIQ
      is also obtuse and hence always
      the dot product of vectors IG, IQ
      is negative or zero i.e
      IG.IQ <= 0 (2)

      Let i, j, k be the unit vectors along IA, IB, IC
      and r be the inradius of ABC.
      The points Q, I are the orthocenter and circumcenter
      of A2B2C2
      and for vectors Sylvester's theorem gives
      IA2 + IB2 + IC2 = IQ or
      r(i + j + k) = IQ or
      i + j + k = (1/r)IQ

      We have
      IA + IB + IC = IA.i + IB.j + IC.k
      = (IG +GA).i + (IG + GB).j + (IG + GC).k
      = IG.(i + j + k) + GA.i + GB.j + GC.k
      = (1/r)IG.IQ + GA.i + GB.j + GC.k
      <= GA.i + GB.j + GC.k due to (2)
      and hence
      IA + IB + IC <= | GA.i + GB.j + GC.k |
      <= | GA.i | + | GB.j | + | GC.k |
      <= GA + GB + GC.

      Best regards
      Nikos Dergiades

      > > ABC with centroid G and incenter I, then
      > > AG+BG+CG-AI-BI-CI=0 or >
      > > 0. Is that true?

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