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Re: [EMHL] (A,e_b,r_c)

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  • Nikolaos Dergiades
    Dear Luis, ... I think that it is not possible to construct 2 different such triangles. Let AB
    Message 1 of 6 , Jan 16, 2009
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      Dear Luis,

      > e_b := B external bisector
      >
      > One can construct (Euclides) 2 triangles given
      > (A,e_b,r_a). But how about (A,e_b,r_c)?
      > I wasn't able to construct more than one.
      > Are there data where one can construct 2
      > triangles?

      I think that it is not possible to construct 2
      different such triangles.
      Let AB < BC
      If D is the point where the B external bisector meets the
      line AC then the excenter Ic lies on BD and in triangle
      ABD we know the side BD = e_b, the opposite angle BAD = 180-A
      and the internal bisector AIc = r_c/cos(A/2).
      We know how to construct this unique triangle BAD
      and then ABC.
      Let AB > AC.
      Now the new construction is symmetric to the previous
      relative to the perpencicular bisector of AC and hence it
      leads to the same triangle ABC as the previous construction.
      Hence we have only one solution provided that e_b > 2r_c.
      I hope that these thoughts are correct.

      Best regards
      Nikos Dergiades









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    • Luís Lopes
      Dear Hyacinthists, Nikos, Thank you for this analysis. ... Do you confirm the letters on these inequalities? My reasoning was to start with A acute (20
      Message 2 of 6 , Jan 16, 2009
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        Dear Hyacinthists, Nikos,

        Thank you for this analysis.

        >Let AB < BC. Let AB > AC.
        Do you confirm the letters on these inequalities?

        My reasoning was to start with A acute (20<A<40),
        and a fix Ic on the A external bisector. Then moving
        C on the ray (A,C) (further away from A to closer)
        one makes B range from an obtuse to an acute angle.
        One does that keeping in mind that CIc is the internal
        bisector of <ACB.

        In the process your point D moves from DAC to ACD
        and in this transition I hoped to construct two different
        triangles. One with obtuse B and the other with acute B.

        Best regards,
        Luis

        PS. I keep the reply to this message in its entirety this
        time only.

        http://www.microsoft.com/windows/windowslive/

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      • Nikolaos Dergiades
        Dear Luis, ... Sorry. The correct is. ... Best regards Nikos Dergiades ___________________________________________________________ Χρησιμοποιείτε
        Message 3 of 6 , Jan 16, 2009
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          Dear Luis,

          > >Let AB < BC. Let AB > AC.
          > Do you confirm the letters on these inequalities?

          Sorry.
          The correct is.
          >Let AB < BC. Let AB > BC.

          Best regards
          Nikos Dergiades




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        • Luís Lopes
          Dear Hyacinthists, Niko, Francisco Capitán, Francisco Capitán was able to find two triangles with A=45^ circ, e_b=6, r_c=1. The values of {a,b,c} with this
          Message 4 of 6 , Jan 16, 2009
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            Dear Hyacinthists, Niko, Francisco Capitán,

            Francisco Capitán was able to find two triangles
            with A=45^\circ, e_b=6, r_c=1.
            The values of {a,b,c} with this data are {5.2818109246, 6.5540237833, 2.10063998345} and
            {0.94717120728,1.28858190882,1.16983782628}Algebraically, by eliminating a and b one gets p(c)
            in fonction of the data:
            c^8 (1 - 2 cosA^2 + cosA^4) +
            c^6 (-2 eb^2 + 4 cosA^2 eb^2 - 2 cosA^4 eb^2 + 4 cosA rc^2 - 4 cosA^2 rc^2 - 4 cosA^3 rc^2 + 4 cosA^4 rc^2) + c^4 (eb^4 - 2 cosA^2 eb^4 + cosA^4 eb^4 + 2 eb^2 rc^2 - 4 cosA eb^2 rc^2 - 2 cosA^2 eb^2 rc^2 + 4 cosA^3 eb^2 rc^2 + 4 rc^4 - 8 cosA rc^4 + 4 cosA^2 rc^4) + c^2 (-2 eb^4 rc^2 + 2 cosA^2 eb^4 rc^2 - 4 eb^2 rc^4 + 4 cosA eb^2 rc^4) + eb^4 rc^4Maybe a good problem for Crux, AMM etc would be to
            determine conditions to have 0,1,2 solutions.

            Best regards,
            Luis
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          • Nikolaos Dergiades
            Dear Luis, Francisco has given the stimulus to find the following result. In case BA
            Message 5 of 6 , Jan 19, 2009
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              Dear Luis,
              Francisco has given the stimulus to find
              the following result.
              In case BA < BC
              If 2.r_c < e_b.sin(A/2) and (cosA - 1)r_c < e_b.cosA
              then a solution is given by

              f = atan((e_b/r_c).cot(A/2)) (atan is inverse function of tan)
              B = 2.asin(cos(A/2).cot(f/2)) - A (asin inverse of sin)
              c = -e_b.cos(A + B/2)/sinA
              b = c.sinB/sin(A + B)
              a = b.sinA/sinB

              In case BA > BC then A must be acute
              if (1 - cosA).r_c < e_b.cosA then
              a second solution is given by

              f = atan((e_b/r_c).cot(A/2)) (as previously)
              B = 2.asin(cos(A/2).tan(f/2)) - A (different from previous)
              c = e_b.cos(A + B/2)/sinA (differ in sign)
              b = c.sinB/sin(A + B)
              a = b.sinA/sinB

              In some inequalities I did n't used the equality sign
              in order to get sure results due to computer
              approximation by truncation.

              Hence if A = pi/2 we have never the second solution.
              If A = pi/2, e_b < 2.srt(2).r_c we have 0 solutions
              If A = pi/2, e_b > 2.srt(2).r_c we have 1 solution.

              Best regards
              Nikos Dergiades

              > Dear Hyacinthists, Niko, Francisco Capitán,
              >
              > Francisco Capitán was able to find two triangles
              > with A=45^\circ, e_b=6, r_c=1.
              > The values of {a,b,c} with this data are {5.2818109246,
              > 6.5540237833, 2.10063998345} and
              > {0.94717120728,1.28858190882,1.16983782628}Algebraically,
              > by eliminating a and b one gets p(c)
              > in fonction of the data:
              > c^8 (1 - 2 cosA^2 + cosA^4) +
              > c^6 (-2 eb^2 + 4 cosA^2 eb^2 - 2 cosA^4 eb^2 + 4 cosA rc^2
              > - 4 cosA^2 rc^2 - 4 cosA^3 rc^2 + 4 cosA^4 rc^2) + c^4
              > (eb^4 - 2 cosA^2 eb^4 + cosA^4 eb^4 + 2 eb^2 rc^2 - 4
              > cosA eb^2 rc^2 - 2 cosA^2 eb^2 rc^2 + 4 cosA^3 eb^2 rc^2 +
              > 4 rc^4 - 8 cosA rc^4 + 4 cosA^2 rc^4) + c^2 (-2 eb^4 rc^2
              > + 2 cosA^2 eb^4 rc^2 - 4 eb^2 rc^4 + 4 cosA eb^2 rc^4)
              > + eb^4 rc^4Maybe a good problem for Crux, AMM etc would be
              > to
              > determine conditions to have 0,1,2 solutions.
              >
              > Best regards,
              > Luis




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