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Re: Triangle Constructions problems A, b, ;

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  • jpehrmfr
    Dear Nikolaos and Luis exactly as as Nikolas said, I started from the classical relations 1/ra+1/rb=1/r-1/rc=2/hc They give immediately the equations for r or
    Message 1 of 12 , Dec 7, 2010
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      Dear Nikolaos and Luis
      exactly as as Nikolas said, I started from the classical relations
      1/ra+1/rb=1/r-1/rc=2/hc
      They give immediately the equations for r or ra
      In the case R+r or R-r,you can start from the more difficult relation
      (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
      which leads to a quadratic equation for r or R when you know A,b and R+r or R-r
      Friendly. Jean-Pierre

      > >The equation giving r with L=r+R is a bit ...
      > I forgot L=R-r. I will be back to both of them
      > later, as well as to a few others with R.
      >
      > >... more complicated, but it exists.
      > Do you mean the equations you gave below are
      > already known?
      >
      >
      > >It remains to get nice geometric constructions.
      > I have no idea on how to deal with such data
      > geometrically.
      >
      > So let's go back to the algebraic solution.
      >
      > r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
      >
      > r^2+L.r-b.L.sin(A)/2=0 with L=rc-r
      >
      >
      > I have performed numerical tests with a=5, b=7 c=8
      > and the above equations are correct. I have now two
      > questions:
      >
      > 1) could you give me a hint on how to deduce them?
      >
      > I assume the others with ra, rb will follow the same way.
      >
      > 2) the equation with L=r+rc has 2 positive roots.
      > Is it possible to have data that give two solutions?
      > I mean two different triangles?
      >
      > Friendly,
      > Luis
      >
      >
      >
      > To: Hyacinthos@yahoogroups.com
      > From: jean-pierre.ehrmann@...
      > Date: Sat, 4 Dec 2010 16:57:20 +0000
      > Subject: [EMHL] Re: Triangle Constructions problems A, b, ;
      >
      >
      >
      >
      > Dear Luis
      >
      > > Consider the problems A, b, ....
      >
      > > r + R
      >
      > > r + r_c
      >
      > > r_c - r
      >
      > > r_a + r_b
      >
      > > r_a - r_b
      >
      > > r_b - r_a
      >
      > >
      >
      > > Do they have a R&C construction?
      >
      >
      >
      > The answer is yes for all the problems.
      >
      > In any case, you have a quadratic equation to get r or ra.
      >
      > For instance :
      >
      > r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
      >
      > r^2+L.r-b.l.sin(A)/2=0 with L=rc-r
      >
      > ra^2-ra.L+b.L.sin(A)/2=0 with L=ra+rb
      >
      > ra^2-(L+b.sin(A))ra+L.b.sin(A)/2=0 with L=ra-rb
      >
      > The equation giving r with with L=r+R is a bit more complicated, but it exists
      >
      > It remains to get nice geometric constructions.
      >
      > Friendly.
      >
      > Jean-Pierre
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      >
      > [Non-text portions of this message have been removed]
      >
    • Nikolaos Dergiades
      Dear Jean-Pierre Is it easy to explain how did you found this relation? (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0 Best regards Nikos Dergiades
      Message 2 of 12 , Dec 7, 2010
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        Dear Jean-Pierre
        Is it easy to explain
        how did you found this relation?
        (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

        Best regards
        Nikos Dergiades


        > Dear Nikolaos and Luis
        > exactly as as Nikolas said, I started from the classical
        > relations
        > 1/ra+1/rb=1/r-1/rc=2/hc
        > They give immediately the equations for r or ra
        > In the case R+r or R-r,you can start from the more
        > difficult relation
        > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
        > which leads to a quadratic equation for r or R when you
        > know A,b and R+r or R-r
        > Friendly. Jean-Pierre
        >
        > > >The equation giving r with L=r+R is a bit ...
        > > I forgot L=R-r. I will be back to both of them
        > > later, as well as to a few others with R.
        > >
        > > >... more complicated, but it exists.
        > > Do you mean the equations you gave below are
        > > already known?
        > >
        > >
        > > >It remains to get nice geometric constructions.
        > > I have no idea on how to deal with such data
        > > geometrically.
        > >
        > > So let's go back to the algebraic solution.
        > >
        > > r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
        > >
        > > r^2+L.r-b.L.sin(A)/2=0 with L=rc-r
        > >
        > >
        > > I have performed numerical tests with a=5, b=7 c=8
        > > and the above equations are correct. I have now two
        > > questions:
        > >
        > > 1) could you give me a hint on how to deduce them?
        > >
        > > I assume the others with ra, rb will follow the same
        > way.
        > >
        > > 2) the equation with L=r+rc has 2 positive roots.
        > > Is it possible to have data that give two solutions?
        > > I mean two different triangles?
        > >
        > > Friendly,
        > > Luis
        > >
        > >
        > >
        > > To: Hyacinthos@yahoogroups.com
        > > From: jean-pierre.ehrmann@...
        > > Date: Sat, 4 Dec 2010 16:57:20 +0000
        > > Subject: [EMHL] Re: Triangle Constructions problems A,
        > b, ;
        > >
        > >
        > >
        > >
        > > Dear Luis
        > >
        > > > Consider the problems A, b, ....
        > >
        > > > r + R
        > >
        > > > r + r_c
        > >
        > > > r_c - r
        > >
        > > > r_a + r_b
        > >
        > > > r_a - r_b
        > >
        > > > r_b - r_a
        > >
        > > >
        > >
        > > > Do they have a R&C construction?
        > >
        > > 
        > >
        > > The answer is yes for all the problems.
        > >
        > > In any case, you have a quadratic equation to get r or
        > ra.
        > >
        > > For instance :
        > >
        > > r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
        > >
        > > r^2+L.r-b.l.sin(A)/2=0 with L=rc-r
        > >
        > > ra^2-ra.L+b.L.sin(A)/2=0 with L=ra+rb
        > >
        > > ra^2-(L+b.sin(A))ra+L.b.sin(A)/2=0 with L=ra-rb
        > >
        > > The equation giving r with with L=r+R is a bit more
        > complicated, but it exists
        > >
        > > It remains to get nice geometric constructions.
        > >
        > > Friendly.
        > >
        > > Jean-Pierre
        > >
        > >
        > >
        > >
        > >
        > >     
        > >     
        > >
        > >     
        > >     
        > >
        > >
        > >
        > >
        > >
        > >
        > >       
        >         
        >           
        >  
        > >
        > > [Non-text portions of this message have been removed]
        > >
        >
        >
        >
        >
        > ------------------------------------
        >
        > Yahoo! Groups Links
        >
        >
        >     Hyacinthos-fullfeatured@yahoogroups.com
        >
        >
        >
      • Antreas
        [ND] ... Dear Nikos, There are two different questions about this relation. 1. How JPE has constructed it ? 2. Once we have it, how to prove it? Let s try to
        Message 3 of 12 , Dec 7, 2010
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          [ND]
          > Dear Jean-Pierre
          > Is it easy to explain
          > how did you found this relation?
          > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

          Dear Nikos,

          There are two different questions about this
          relation.
          1. How JPE has constructed it ?
          2. Once we have it, how to prove it?

          Let's try to answer 2:
          that is, to prove that:

          (1-cos A)((hc-2r)R-b^2/2) = r(r-hc)

          We have:

          First member:

          1-cosA = 2sin^2(A/2)

          hc-2r = 2RsinAsinB - 8Rsin(A/2)sin(B/2)sin(C/2) =

          = 8Rsin(A/2)cos(A/2)sin(B/2)cos(B/2) - 8Rsin(A/2)sin(B/2)sin(C/2)=

          = 8Rsin(A/2)sin(B/2)[cos(A/2)cos(B/2) - sin(C/2)] =

          = 8Rsin^2(A/2)sin^2(B/2)

          (hc-2r)R - b^2/2 = 8R^2sin^2(A/2)sin^2(B/2) -
          - 8R^2sin^2(B/2)cos^2(B/2)

          = 8R^2sin^2(B/2)[sin^2(A/2) - cos^2(B/2)]

          ===>

          (1-cos A)((hc-2r)R-b^2/2) = 16R^2sin^2(A/2)sin^2(B/2)*
          [sin^2(A/2) - cos^2(B/2)] <1>

          Second member:

          r(r-hc)= 4Rsin(A/2)sin(B/2)sin(C/2)[4Rsin(A/2)sin(B/2sin(C/2) - 2RsinAsinB] =

          4R^2sin(A/2)sin(B/2)sin(C/2)[4sin(A/2)sin(B/2)sin(C/2) -
          - 8sin(A/2)cos(A/2)sin(B/2)cos(B/2)]=
          = 16R^2sin^2(A/2)sin^2(B/2)sin(C/2)[sin(C/2) - 2cos(A/2)cos(B/2)] <2>

          So we have to prove that <1> = <2>

          or

          sin^2(A/2) - cos^2(B/2) - sin^2(C/2)= - 2cos(A/2)cos(B/2)sin(C/2)

          or

          sin^2(A/2)+sin^2(B/2)-sin^2(C/2)= 1 - 2cos(A/2)cos(B/2)sin(C/2)

          which is true.

          APH
        • jpehrmfr
          Dear Nikolaos and Antreas [ND] ... I m not very proud of my way to get this : As hc/r = (a+b+c)/c, we have c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c I ve tried
          Message 4 of 12 , Dec 8, 2010
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            Dear Nikolaos and Antreas
            [ND]
            > Is it easy to explain
            > how did you found this relation?
            > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

            I'm not very proud of my way to get this :
            As hc/r = (a+b+c)/c, we have
            c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c
            I've tried to combine these relations to get b and used Maple to do that. As Antreas said, the result is easy to prove but more difficult to find.
            The key was
            b(a+b+c)=a(a+b-c)+(b+c-a)(a+b) (1)
            Now b(a+b+c)=(a.c.b^2)/(2.r.R) because (a+b+c)R.r = a.b.c/2
            a(a+b-c) = a.c.(hc-2.r)/r
            As r/R/(1-cos A) = (b+c-a)/a, we have
            (b+c-a)(a+b) = a.c(hc-r)/R/(1-cos A)

            Thus (1) gives b^2/(2.r.R) = (hc-2.r)/r + (hc-r)/R/(1-cos A)
            which is the desired relation.
            Friendly. Jean-Pierre
            b^2/(2.r.R) = (hc-2.r)/r +
          • Nikolaos Dergiades
            Dear Jean-Pierre Very good. Thank you. Your key relation is (a+b+c).r = c.hc (1) Because from a = 2Rsin(A) = 2R.hc/b and from (1) we get c = (2Rhc + b^2)r /
            Message 5 of 12 , Dec 8, 2010
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              Dear Jean-Pierre
              Very good. Thank you.
              Your key relation is (a+b+c).r = c.hc (1)
              Because from a = 2Rsin(A) = 2R.hc/b and from (1)
              we get c = (2Rhc + b^2)r / (b(hc-r))
              and substituting a, c to your second relation
              R(1-cosA)(b+c-a) = ar (2)
              we get your final relation (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0.
              Instead of (2) we can substitute a, c to
              cosA = (bb + cc -aa)/(2bc).

              Best regards
              Nikos Dergiades

              >
              > Dear Nikolaos and Antreas
              > [ND]
              > > Is it easy to explain
              > > how did you found this relation?
              > > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
              >
              > I'm not very proud of my way to get this :
              > As hc/r = (a+b+c)/c, we have
              > c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c
              > I've tried to combine these relations to get b and used
              > Maple to do that. As Antreas said, the result is easy to
              > prove but more difficult to find.
              > The key was
              > b(a+b+c)=a(a+b-c)+(b+c-a)(a+b) (1)
              > Now b(a+b+c)=(a.c.b^2)/(2.r.R) because (a+b+c)R.r =
              > a.b.c/2
              > a(a+b-c) = a.c.(hc-2.r)/r
              > As r/R/(1-cos A) = (b+c-a)/a, we have
              > (b+c-a)(a+b) = a.c(hc-r)/R/(1-cos A)
              >
              > Thus (1) gives b^2/(2.r.R) = (hc-2.r)/r + (hc-r)/R/(1-cos
              > A)
              > which is the desired relation.
              > Friendly. Jean-Pierre
              > b^2/(2.r.R) = (hc-2.r)/r +
              >
              >
              >
              > ------------------------------------
              >
              > Yahoo! Groups Links
              >
              >
              >     Hyacinthos-fullfeatured@yahoogroups.com
              >
              >
              >
            • Luís Lopes
              Dear Hyacinthists, Nikos, Jean-Pierre, Thank you very much. Nice solutions. So TCs such as A,b,R+r_a(b)(c) can be solved in a similar way. For R+r_a one has
              Message 6 of 12 , Dec 14, 2010
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                Dear Hyacinthists, Nikos, Jean-Pierre,

                Thank you very much. Nice solutions.

                So TCs such as A,b,R+r_a(b)(c) can be solved in a similar way.
                For R+r_a one has

                (1-cos A)((h_c-2r_a)R+b^2/2)-r_a(h_c-r_a)=0.

                TC given A, a, r_a+r_b+r_c has an easy construction.
                How about A, b, r_a+r_b+r_c ?

                As always, thank you very much for your help.

                Luis


                > To: Hyacinthos@yahoogroups.com
                > From: ndergiades@...
                > Date: Wed, 8 Dec 2010 21:34:09 +0000
                > Subject: Re: [EMHL] Re: Triangle Constructions problems A, b, ;
                >
                > Dear Jean-Pierre
                > Very good. Thank you.
                > Your key relation is (a+b+c).r = c.hc (1)
                > Because from a = 2Rsin(A) = 2R.hc/b and from (1)
                > we get c = (2Rhc + b^2)r / (b(hc-r))
                > and substituting a, c to your second relation
                > R(1-cosA)(b+c-a) = ar (2)
                > we get your final relation (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0.
                > Instead of (2) we can substitute a, c to
                > cosA = (bb + cc -aa)/(2bc).
                >
                > Best regards
                > Nikos Dergiades
                >
                > >
                > > Dear Nikolaos and Antreas
                > > [ND]
                > > > Is it easy to explain
                > > > how did you found this relation?
                > > > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
                > >
                > > I'm not very proud of my way to get this :
                > > As hc/r = (a+b+c)/c, we have
                > > c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c
                > > I've tried to combine these relations to get b and used
                > > Maple to do that. As Antreas said, the result is easy to
                > > prove but more difficult to find.
                > > The key was
                > > b(a+b+c)=a(a+b-c)+(b+c-a)(a+b) (1)
                > > Now b(a+b+c)=(a.c.b^2)/(2.r.R) because (a+b+c)R.r =
                > > a.b.c/2
                > > a(a+b-c) = a.c.(hc-2.r)/r
                > > As r/R/(1-cos A) = (b+c-a)/a, we have
                > > (b+c-a)(a+b) = a.c(hc-r)/R/(1-cos A)
                > >
                > > Thus (1) gives b^2/(2.r.R) = (hc-2.r)/r + (hc-r)/R/(1-cos
                > > A)
                > > which is the desired relation.
                > > Friendly. Jean-Pierre
                > > b^2/(2.r.R) = (hc-2.r)/r +
                > >
                > >
                > >
                > > ------------------------------------
                > >
                > > Yahoo! Groups Links
                > >
                > >
                > > Hyacinthos-fullfeatured@yahoogroups.com
                > >
                > >
                > >
                >
                >
                >
                >
                > ------------------------------------
                >
                > Yahoo! Groups Links
                >
                >
                >


                [Non-text portions of this message have been removed]
              • Nikolaos Dergiades
                Dear Luis, ... We know that r_a+r_b+r_c = 4R + r. Hence we have a discussed case. Best regards Nikos Dergiades
                Message 7 of 12 , Dec 14, 2010
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                  Dear Luis,

                  > How about A, b, r_a+r_b+r_c ?

                  We know that r_a+r_b+r_c = 4R + r.
                  Hence we have a discussed case.
                  Best regards
                  Nikos Dergiades
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