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A Geometric Inequality

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  • insight_love@hotmail.com
    Hi, all. My conjecture is the following : Let P be in the interior of triangle ABC, and let the lines AP, BP, CP intersect the sides BC, CA, AB in L, M, N,
    Message 1 of 4 , Jan 26, 2001
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      Hi, all.

      My conjecture is the following :

      Let P be in the interior of triangle ABC, and let the lines AP, BP,
      CP intersect the sides BC, CA, AB in L, M, N, respectively.
      Then, 3/2 >= MN/BC + NL/CA + LM/AB ??

      Is this already known result?
      Is this true?

      Yours sincerely,
      Hojoo Lee
    • Bernard Gibert
      Dear Hojoo, ... This is obviously false : take (BC,CA,AB)=(13,9,6) and P on the median AA . Let f=MN/BC + NL/CA + LM/AB When P- A, f- 5sqrt(65)/36 = 1.1...
      Message 2 of 4 , Jan 27, 2001
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        Dear Hojoo,

        you wrote :

        > My conjecture is the following :
        >
        > Let P be in the interior of triangle ABC, and let the lines AP, BP,
        > CP intersect the sides BC, CA, AB in L, M, N, respectively.
        > Then, 3/2 >= MN/BC + NL/CA + LM/AB ??
        >
        > Is this already known result?
        > Is this true?
        >
        This is obviously false :
        take (BC,CA,AB)=(13,9,6) and P on the median AA'.

        Let f=MN/BC + NL/CA + LM/AB

        When P->A, f->5sqrt(65)/36 = 1.1...
        When P->A', f->101/36 = 2.8...

        Now, my question : since f is bounded inside the triangle, what are its
        limit sup and limit inf ?

        Best regards

        Bernard
      • asin@ath.forthnet.gr
        ... If the triangle ABC is acute-angled and P is its orthocenter, then the inequality is true: MN/BC = cosA, NL/CA=cosB,LM/AB=cosC and cosA + cosB + cosC
        Message 3 of 4 , Jan 27, 2001
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          --- In Hyacinthos@y..., insight_love@h... wrote:
          > Hi, all.
          >
          > My conjecture is the following :
          >
          > Let P be in the interior of triangle ABC, and let the lines AP, BP,
          > CP intersect the sides BC, CA, AB in L, M, N, respectively.
          > Then, 3/2 >= MN/BC + NL/CA + LM/AB ??
          >
          > Is this already known result?
          > Is this true?
          >
          > Yours sincerely,
          > Hojoo Lee

          If the triangle ABC is acute-angled and P is its orthocenter, then
          the inequality is true:
          MN/BC = cosA, NL/CA=cosB,LM/AB=cosC and
          cosA + cosB + cosC <=3/2 (a well-known inequality)

          Achilleas Sinefakopoulos
        • mnas@fct.unl.pt
          If we take A=(0,0); B=(0,1) C=(n,1) L (in segment BC) close enough to B. M=(m2,m2) (in segment AC) with m1=1 It is easy to see( taking n big enough) that the
          Message 4 of 4 , Feb 3, 2001
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            If we take A=(0,0); B=(0,1) C=(n,1)
            L (in segment BC) close enough to B.
            M=(m2,m2) (in segment AC) with m1=1
            It is easy to see( taking n big enough) that the function
            MN/BC + NL/CA + LM/AB
            take values as small as we want.

            manuel


            > > Let P be in the interior of triangle ABC, and let the lines AP,
            BP,
            > > CP intersect the sides BC, CA, AB in L, M, N, respectively.
            > > Then, 3/2 >= MN/BC + NL/CA + LM/AB ??
            > >
            > > Is this already known result?
            > > Is this true?
            > >
            > This is obviously false :
            > take (BC,CA,AB)=(13,9,6) and P on the median AA'.
            >
            > Let f=MN/BC + NL/CA + LM/AB
            >
            > When P->A, f->5sqrt(65)/36 = 1.1...
            > When P->A', f->101/36 = 2.8...
            >
            > Now, my question : since f is bounded inside the triangle, what are
            its
            > limit sup and limit inf ?
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