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(A,e_b,r_c)

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  • Luís Lopes
    Dear Hyacinthists, 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
    Message 1 of 6 , Jan 15, 2009
      Dear Hyacinthists,

      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?
      Best regards,
      Luis

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

          [Non-text portions of this message have been removed]
        • Nikolaos Dergiades
          Dear Luis, ... Sorry. The correct is. ... Best regards Nikos Dergiades ___________________________________________________________ Χρησιμοποιείτε
          Message 4 of 6 , Jan 16, 2009
            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 5 of 6 , Jan 16, 2009
              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 6 of 6 , Jan 19, 2009
                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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