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

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  • Nikolaos Dergiades
    A partial answer. It is easy to prove that IA^2+IB^2+IC^2 = GA^2+GB^2+GC^2+3IG^2 Hence IA^2+IB^2+IC^2 = GA^2+GB^2+GC^2 The inequality GA+GB+GC = IA+IB+IC
    Message 1 of 3 , Nov 30, 2006
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      A partial answer.
      It is easy to prove that
      IA^2+IB^2+IC^2 = GA^2+GB^2+GC^2+3IG^2
      Hence
      IA^2+IB^2+IC^2 >= GA^2+GB^2+GC^2

      The inequality
      GA+GB+GC >= IA+IB+IC
      from a computer test seems to be true
      but I can't give a proof.

      Best regards
      Nikos Dergiades





      > ABC with centroid G and incenter I, then
      > AG+BG+CG-AI-BI-CI=0 or >
      > 0. Is that true?
      > And how about AG^n+BG^n+CG^n-AI^n-BI^n-CI^n, for
      > n>1?
      > Thanks!







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    • Nikolaos Dergiades
      Dear friends, after many efforts, I found a good proof for the inequality IA+IB+IC
      Message 2 of 3 , Dec 1, 2006
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        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
        A'B'C'
        and is perpendicular to B2C2.
        Hence A1A2, B1B2, C1C2 pass through the
        incenter Q of A'B'C' that is the orthocenter of
        A2B2C2.
        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
        G
        relative to the line AA'.
        The lines AA', BB', CC' devide the triangle ABC in 6
        triangles
        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
        bisector
        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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