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[EMHL] Re: metric relation in triangle

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  • Francisco Javier
    I started from the formulas for R^2, ra^2 and s in terms on a, b, c, namely R^2 = ((a^2 b^2 c^2)/((a + b - c) (a - b + c) (-a + b + c) (a + b + c))), ra^2 =
    Message 1 of 7 , Oct 30, 2009
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      I started from the formulas for R^2, ra^2 and s in terms on a, b, c, namely

      R^2 = ((a^2 b^2 c^2)/((a + b - c) (a - b + c) (-a + b + c) (a + b + c))),

      ra^2 = -(((a + b - c) (a - b + c) (a + b + c))/(4 (a - b - c)))

      s = (a+b+c)/2,

      Then I made use of Mathematica to eliminate a,b,c, with the command

      Eliminate[{R^2 ==..., ra^2==...,s==(a+b+c)/2,T==ab+bc+ca},{a,b,c}]

      Best regards,

      Francisco Javier.


      --- In Hyacinthos@yahoogroups.com, Kafka Catalin <kafka_mate@...> wrote:
      >
      > Dear Francisco,
      > Thank you very much for information you sent me! When you are available, could you describe me the steps I must follow in order to determine it myself too?!
      >
      > Best regards,
      > Catalin Barbu
      > --- On Fri, 10/30/09, Francisco Javier <garciacapitan@...> wrote:
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      > From: Francisco Javier <garciacapitan@...>
      > Subject: [EMHL] Re: metric relation in triangle
      > To: Hyacinthos@yahoogroups.com
      > Date: Friday, October 30, 2009, 12:47 PM
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      > Dear Catalin,
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      > I get the following formula:
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      > ab + bc + ca =
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      > (s^6 + (4 R ra + 3 ra^2) s^4 + (-16 R^2 ra^2 + 3 ra^4) s^2 -4 R ra^5 + ra^6)/(ra^2 + s^2)^2.
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      > Best regards,
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      > Francisco Javier.
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      > --- In Hyacinthos@yahoogro ups.com, Kafka Catalin <kafka_mate@ ...> wrote:
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      > > Dear Hyacinthists,
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      > > Let a,b,c are sides of triangle ABC, s - semiperimeter, R - circumradius and r_a a excircle radius. Find the sum ab+bc+ca in function of s,R,r_a.
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      > > I kow this relation: ab+bc+ca=s^2 + r^2 +4Rr, where r is incircle radius. Exist a relation ab+bc+ca = F(s,R,r_a)?
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      > > Best regards,Catalin Barbu
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    • wolkbarry
      ... The following can all be expressed in terms of R, s, r_a tan(A/2) = r / (s-a) = r_a / s sin(A) = 2 sin(A/2) cos(A/2) = 2 t / (t^2+1), where t=tan(A/2) a =
      Message 2 of 7 , Nov 2, 2009
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        Kafka Catalin <kafka_mate@...> wrote:

        > Dear Hyacinthists,
        > Let a,b,c are sides of triangle ABC, s - semiperimeter, R - circumradius and r_a a excircle radius. Find the sum ab+bc+ca in function of s,R,r_a.  
        > I kow this relation: ab+bc+ca=s^2 + r^2 +4Rr, where r is incircle radius. Exist a relation               ab+bc+ca = F(s,R,r_a)?
        > Best regards,Catalin Barbu

        The following can all be expressed in terms of R, s, r_a

        tan(A/2) = r / (s-a) = r_a / s
        sin(A) = 2 sin(A/2) cos(A/2) = 2 t / (t^2+1), where t=tan(A/2)
        a = 2 R sin(A)
        r = (s-a) r_a / s

        Then ab+bc+ca = a (2 s - a) + 4 R r s / a
        --
        Barry Wolk
      • Kafka Catalin
        Dear Barry, I want obtain  a relation ab+bc+ca in function only s, R and r_a . Sir Francisco Javier send me a relation    ab + bc + ca = (s^6 + (4 R ra +
        Message 3 of 7 , Nov 4, 2009
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          Dear Barry, I want obtain  a relation ab+bc+ca in function only "s, R and r_a". Sir Francisco Javier send me a relation   
          ab + bc + ca =
          (s^6 + (4 R ra + 3 ra^2) s^4 + (-16 R^2 ra^2 + 3 ra^4) s^2 -4 R ra^5 + ra^6)/(ra^2 + s^2)^2 (with Mathematica program), i need a proof a this relation.Best regards, Catalin Barbu
          --- On Tue, 11/3/09, wolkbarry <wolkbarry@...> wrote:

          From: wolkbarry <wolkbarry@...>
          Subject: [EMHL] Re: metric relation in triangle
          To: Hyacinthos@yahoogroups.com
          Date: Tuesday, November 3, 2009, 2:59 AM













           





          Kafka Catalin <kafka_mate@ ...> wrote:



          > Dear Hyacinthists,

          > Let a,b,c are sides of triangle ABC, s - semiperimeter, R - circumradius and r_a a excircle radius. Find the sum ab+bc+ca in function of s,R,r_a.  

          > I kow this relation: ab+bc+ca=s^2 + r^2 +4Rr, where r is incircle radius. Exist a relation               ab+bc+ca = F(s,R,r_a)?

          > Best regards,Catalin Barbu



          The following can all be expressed in terms of R, s, r_a



          tan(A/2) = r / (s-a) = r_a / s

          sin(A) = 2 sin(A/2) cos(A/2) = 2 t / (t^2+1), where t=tan(A/2)

          a = 2 R sin(A)

          r = (s-a) r_a / s



          Then ab+bc+ca = a (2 s - a) + 4 R r s / a

          --

          Barry Wolk




































          [Non-text portions of this message have been removed]
        • Nikolaos Dergiades
          Dear Catalin, ... Barry sent you the following: If t = tan(A/2) = ra/s then a = 2.R.sinA = 4.R.t/(1 + t^2) = 4.R.s.ra/(s^2 + ra^2) (1) bc = (abc)/a = 2.R.S/a
          Message 4 of 7 , Nov 4, 2009
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            Dear Catalin,

            > Dear Barry, I want obtain  a relation ab+bc+ca in
            > function only "s, R and r_a". Sir Francisco Javier send me a
            > relation   
            > ab + bc + ca =
            > (s^6 + (4 R ra + 3 ra^2) s^4 + (-16 R^2 ra^2 + 3 ra^4) s^2
            > -4 R ra^5 + ra^6)/(ra^2 + s^2)^2 (with Mathematica program),
            > i need a proof a this relation.

            Barry sent you the following:
            If t = tan(A/2) = ra/s then
            a = 2.R.sinA = 4.R.t/(1 + t^2) = 4.R.s.ra/(s^2 + ra^2) (1)
            bc = (abc)/a = 2.R.S/a = 4.R.(s-a)ra/a
            Hence
            ab + ac + bc = a(b+c) + bc = a(2s-a) + 4.R.(s-a)ra/a
            or substituting a from (1) we get Francisco's formula.
            Hence Barry sent you the proof of Francisco's formula.
            Best regards
            Nikos Dergiades






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