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sides in arithmetic progression

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  • Luís Lopes
    Dear Hyacinthists, A problem from a local list. The sides of a general triangle with standard points are in AP with common difference d. Find dist(G,I)
    Message 1 of 5 , Jul 15, 2010
      Dear Hyacinthists,



      A problem from a local list.



      The sides of a general triangle with standard points <I,G>

      are in AP with common difference d.



      Find dist(G,I) as a function of d.



      Best regards,
      Luis

      _________________________________________________________________
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      [Non-text portions of this message have been removed]
    • Nikolaos Dergiades
      Dear Luis, Let b = a + d, c = b + d, or a + c = 2b then s = (a + b + c) / 2 = 3b/2, s - b = b/2 and since s.r = (s - b) rb we get that rb = 3r where r, rb
      Message 2 of 5 , Jul 15, 2010
        Dear Luis,
        Let b = a + d, c = b + d, or a + c = 2b
        then s = (a + b + c) / 2 = 3b/2, s - b = b/2 and since
        s.r = (s - b) rb we get that rb = 3r
        where r, rb are the radi of incircle and B_excircle.

        The line IG has equation in barycentrics
        (b - c)x + (c - a)y + (a - b)z = 0 or
        -x + 2y - z = 0 and meets the line BC at D (0 : 1 : 2)
        which means that BD/DC = 2 and the line IG is
        parallel to AC. If M is the midpoint of AC and the
        line BI meets AC at J then
        CJ = ab/(a + c) = a/2
        MJ = b/2 - a/2 = d/2 and
        GI = 2MJ/3 = d/3.

        The Nagel point Na is known that lies on line IG
        and if BNa meets AC at K then
        AK = s - c = 3b/2 - c and
        KM = AM - AK = b/2 - (3b/2 - c) = c - b = d
        Hence NaG = 2KM/3 = 2d/3

        Best regards
        Nikos Dergiades

        > Dear Hyacinthists,
        >
        >
        >
        > A problem from a local list.
        >
        >
        >
        > The sides of a general triangle with standard points
        > <I,G>
        >
        > are in AP with common difference d.
        >
        >
        >
        > Find dist(G,I) as a function of d.
        >
        >
        >
        > Best regards,
        > Luis
        >      
        >         
        >           
      • Quang Tuan Bui
        Dear Luis and Nikos, We can solve the problem by construction triangle ABC (b = a+d = c-d) as following: Choose b = AC as one segment. X as any point on
        Message 3 of 5 , Jul 15, 2010
          Dear Luis and Nikos,

          We can solve the problem by construction triangle ABC (b = a+d = c-d) as following:

          Choose b = AC as one segment.
          X as any point on segment AC and CX = d.
          X1 = reflection of X in C
          X2 = reflection of X in midpoint of AC
          B = intersection of two circles centered at C passing X2 and centered at A passing X1
          Y1 = midpoint of BX1
          Y2 = midpoint of BX2
          A1 = midpoint of BC
          C1 = midpoint of AB
          I = intesection of AY1 and CY2
          G = intesection of AA1 and CC1

          From this construction: GI//AC and
          GI = 2/3*A1Y1 = 2/3*1/2*CX1 = 1/3*CX1 = d/3

          Best regards,
          Bui Quang Tuan

          --- On Thu, 7/15/10, Luís Lopes <qed_texte@...> wrote:

          > From: Luís Lopes <qed_texte@...>
          > Subject: [EMHL] sides in arithmetic progression
          > To: hyacinthos@yahoogroups.com
          > Date: Thursday, July 15, 2010, 10:39 PM
          >
          > Dear Hyacinthists,
          >
          >
          >
          > A problem from a local list.
          >
          >
          >
          > The sides of a general triangle with standard points
          > <I,G>
          >
          > are in AP with common difference d.
          >
          >
          >
          > Find dist(G,I) as a function of d.
          >
          >
          >
          > Best regards,
          > Luis
          >      
        • ndergiades
          Dear Tuan, very good! Another proof with vectors: We have GA + GB + GC = 0 a.IA + b.IB + c.IC = 0 or (a + b + c).IG + a.GA + b.GB + c.GC = 0 or 3b.IG + (a -
          Message 4 of 5 , Jul 15, 2010
            Dear Tuan,
            very good!
            Another proof with vectors:
            We have
            GA + GB + GC = 0
            a.IA + b.IB + c.IC = 0 or
            (a + b + c).IG + a.GA + b.GB + c.GC = 0 or
            3b.IG + (a - b)GA + (c - b)GC = 0 or
            3b.IG = d.(GA - GC) = d.CA
            and hence |IG| = |CA|.d/3b = d/3.
            Best regards
            Nikos Dergiades


            > Dear Luis and Nikos,
            >
            > We can solve the problem by construction triangle ABC (b = a+d = c-d) as following:
            >
            > Choose b = AC as one segment.
            > X as any point on segment AC and CX = d.
            > X1 = reflection of X in C
            > X2 = reflection of X in midpoint of AC
            > B = intersection of two circles centered at C passing X2 and centered at A passing X1
            > Y1 = midpoint of BX1
            > Y2 = midpoint of BX2
            > A1 = midpoint of BC
            > C1 = midpoint of AB
            > I = intesection of AY1 and CY2
            > G = intesection of AA1 and CC1
            >
            > From this construction: GI//AC and
            > GI = 2/3*A1Y1 = 2/3*1/2*CX1 = 1/3*CX1 = d/3
            >
            > Best regards,
            > Bui Quang Tuan
            >
            > --- On Thu, 7/15/10, Luís Lopes <qed_texte@...> wrote:
            >
            > > From: Luís Lopes <qed_texte@...>
            > > Subject: [EMHL] sides in arithmetic progression
            > > To: hyacinthos@yahoogroups.com
            > > Date: Thursday, July 15, 2010, 10:39 PM
            > >
            > > Dear Hyacinthists,
            > >
            > >
            > >
            > > A problem from a local list.
            > >
            > >
            > >
            > > The sides of a general triangle with standard points
            > > <I,G>
            > >
            > > are in AP with common difference d.
            > >
            > >
            > >
            > > Find dist(G,I) as a function of d.
            > >
            > >
            > >
            > > Best regards,
            > > Luis
            > >      
            >
          • Antreas
            Sorry. there were several typos in my solution. I repost it. ... Dear Luis Lemma 1: In every triangle: GI^2 = (bc+ca+ab)/3 - (a^2+b^2+c^2)/9 - 4Rr Lemma 2: If
            Message 5 of 5 , Jul 16, 2010
              Sorry. there were several typos in my solution.
              I repost it.

              [Luis Lopes]:

              > A problem from a local list.
              > The sides of a general triangle with standard points <I,G>
              > are in AP with common difference d.
              > Find dist(G,I) as a function of d.

              Dear Luis

              Lemma 1:

              In every triangle:

              GI^2 = (bc+ca+ab)/3 - (a^2+b^2+c^2)/9 - 4Rr

              Lemma 2:

              If 2b = a + c [: b = a+d, c = a+2d ] ==> ac = 6Rr


              L1 /\ L2 ==> 9GI^2 = b^2 - ac = (a+d)^2 - a(a+2d) = d^2

              ==> GI = d/3

              APH
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