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Re: [EMHL] an angle question

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  • Ignacio Larrosa Cañestro
    ... From: Ricardo Barroso Campos To: dick tahta ; Sent: Tuesday, October 15, 2002
    Message 1 of 10 , Oct 15, 2002
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      ----- Original Message -----
      From: "Ricardo Barroso Campos" <rbarroso@...>
      To: "dick tahta" <dick@...>; <Hyacinthos@yahoogroups.com>
      Sent: Tuesday, October 15, 2002 7:46 PM
      Subject: Re: [EMHL] an angle question


      Friend Dick, friends of Hyacintos:

      I was saying m ABC =12, not B/2 =12.

      It(He,She) would be B/2 = 6...

      I AGREE with Ioana, BAC=92

      ================
      Ricardo,

      You want to say <BAC = 96º, not 92º.

      Did you tray to inscribe a 30-gon or 60-gon in angle BAC?

      Saludos,

      Ignacio Larrosa Cañestro
      A Coruña (España)
      ilarrosa@...
    • Ricardo Barroso Campos
      Amigo Ignacio: es BAC=96 gracias ... From: Ignacio Larrosa Cañestro To: Sent: Tuesday, October 15, 2002
      Message 2 of 10 , Oct 15, 2002
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        Amigo Ignacio:

        es BAC=96

        gracias



        ----- Original Message -----
        From: "Ignacio Larrosa Cañestro" <ilarrosa@...>
        To: <Hyacinthos@yahoogroups.com>
        Sent: Tuesday, October 15, 2002 7:44 PM
        Subject: Re: [EMHL] an angle question


        ----- Original Message -----
        From: "Ricardo Barroso Campos" <rbarroso@...>
        To: "dick tahta" <dick@...>; <Hyacinthos@yahoogroups.com>
        Sent: Tuesday, October 15, 2002 7:46 PM
        Subject: Re: [EMHL] an angle question


        Friend Dick, friends of Hyacintos:

        I was saying m ABC =12, not B/2 =12.

        It(He,She) would be B/2 = 6...

        I AGREE with Ioana, BAC=92

        ================
        Ricardo,

        You want to say <BAC = 96º, not 92º.

        Did you tray to inscribe a 30-gon or 60-gon in angle BAC?

        Saludos,

        Ignacio Larrosa Cañestro
        A Coruña (España)
        ilarrosa@...








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      • Nikolaos Dergiades
        ... [AC]. ... m(ABC)=? ... ******* If I is the incenter of ABC and x = B/2 then B/2 + C/2 = 42 IE=IB*b/(a+c), ID = IC*c/(a+b) IC/IB = sin(B/2)/sin(C/2) and
        Message 3 of 10 , Oct 15, 2002
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          >ABC is a triangle. D is a point on [AB] and E is a point on
          [AC].
          >m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 ,
          m(ABC)=?
          >
          *******
          If I is the incenter of ABC and x = B/2
          then B/2 + C/2 = 42
          IE=IB*b/(a+c), ID = IC*c/(a+b)
          IC/IB = sin(B/2)/sin(C/2)
          and
          sin(24)/sin(18) = ID/IE and after trig. calculations
          = sin(84 - x)/sin(42 + x)
          (1)
          We'll prove that
          sin(24)/sin(18) = sin(78)/sin(48) or
          2sin12cos12/sin18 = cos12/sin48 or
          2sin12sin48 = sin18 or
          cos36 - cos60 = sin18 or
          4(cos18)^2 - 3 = 2sin18 or
          cos18*[4(cos18)^2 - 3] = 2sin18cos18 or
          cos(3*18) = sin(2*18) or
          cos54 = sin36 which is true.

          Hence from (1) we get
          sin(84 - 6)/sin(42 + 6) = sin(84 - x)/sin(42 + x)
          or
          (tan84-tan6)/(tan42+tan6)=(tan84-tanx)/(tan42+tanx)
          or tanx = tan6 or x = 6 or B = 12, C = 42, A = 96.

          Best regards
          Nikos Dergiades
        • dick tahta
          Ricardo: a thousand apologies - I misread your statement. Indeed the trigonometry does yield B/2 = 6. best wishes, Dick
          Message 4 of 10 , Oct 15, 2002
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            Ricardo:

            a thousand apologies - I misread your statement. Indeed the trigonometry
            does yield B/2 = 6.

            best wishes, Dick

            > From: "Ricardo Barroso Campos" <rbarroso@...>

            > I was saying m ABC =12, not B/2 =12.
            >
            > It(He,She) would be B/2 = 6...
          • Vijaya Prasad Nalluri
            Let r be the radius of the inscribed circle D.Taking advantage of the symmetry of the situation it can be shown by elementary trigonometry that w.r.t. D the
            Message 5 of 10 , Oct 16, 2002
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              Let r be the radius of the inscribed circle D.Taking advantage of the symmetry of the situation it can be shown by elementary trigonometry that w.r.t. D the areas of the escribed equilateral triangle,the inscribed hexagon,the inscribed equilateral triangle are respectively
              (3 r^2 sqrt3) ,(3 r^2 sqrt3)/2,(3 r^2 sqrt3)/4 which are obviously in G.P.
              Best Regards,
              Vijaya Prasad Nalluri



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            • Sharygin I. F.
              Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on straight GP, ED = EG. m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD),
              Message 6 of 10 , Oct 19, 2002
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                Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on
                straight GP, ED = EG.
                m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD), m(EDP)=36
                PD=ED=EG = EF, m(FED) = 72 - (60-36) = 48.
                C and B on straight FP, CD is the perpendicular to EP, EB is perpendicular
                FD.
                m(EDC) = m(PDC) = half m(EDP) = 18, m(DEB) = m(FEB) = 24,
                CE and BD intersect in A.
                So: ABC, BE and CD are bisectors, m(EDC) = 18, m( BED) =24.
                m(EBC) = m(FEB) - m(EFC) = 24 - 15 = 6, m(ABC) = 12.
                Best wishes, Igor Sharygin.
                m(ECD)= m(DCB), m(DBE)=m(EBC).


                ----- Original Message -----
                From: RICARDO BARROSO CAMPOS <rbarroso@...>
                To: <yagcimustafa@...>; <Hyacinthos@yahoogroups.com>
                Sent: Tuesday, October 15, 2002 3:00 PM
                Subject: Re: [EMHL] an angle question


                > Dear Hyacinthos:
                >
                > ABC is a triangle. D is a point on [AB] and E is a point on [AC].
                > m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 , m(ABC)=?
                >
                > I'm a conjecture with Cabri:
                > m (ABC) =12
                >
                > geometric reasons? I do not know it
                >
                >
                > Ricardo
                >
                >
                >
                >
                >
                >
                >
                >
                >
                >
                >
                >
                > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
                >
              • Sharygin I. F.
                ... From: Sharygin I. F. To: ; Sent: Saturday, October 19, 2002 9:47 PM Subject: Re:
                Message 7 of 10 , Oct 19, 2002
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                  ----- Original Message -----
                  From: Sharygin I. F. <sharygin@...>
                  To: <Hyacinthos@yahoogroups.com>; <yagcimustafa@...>
                  Sent: Saturday, October 19, 2002 9:47 PM
                  Subject: Re: [EMHL] an angle question



                  Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on
                  straight GP, ED = EG.
                  m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD), m(EDP)=36
                  PD=ED=EG = EF, m(FED) = 72 - (60-36) = 48.
                  C and B on straight FP, CD is the perpendicular to EP, EB is perpendicular
                  FD.
                  m(EDC) = m(PDC) = half m(EDP) = 18, m(DEB) = m(FEB) = 24,
                  CE and BD intersect in A.
                  So: ABC, BE and CD are bisectors, m(EDC) = 18, m( BED) =24.
                  m(EBC) = m(EFC) - m(FEB) = 30 - 24 = 6, m(ABC) = 12.
                  Best wishes, Igor Sharygin.
                  .
                  >
                  >
                  > ----- Original Message -----
                  > From: RICARDO BARROSO CAMPOS <rbarroso@...>
                  > To: <yagcimustafa@...>; <Hyacinthos@yahoogroups.com>
                  > Sent: Tuesday, October 15, 2002 3:00 PM
                  > Subject: Re: [EMHL] an angle question
                  >
                  >
                  > > Dear Hyacinthos:
                  > >
                  > > ABC is a triangle. D is a point on [AB] and E is a point on [AC].
                  > > m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 , m(ABC)=?
                  > >
                  > > I'm a conjecture with Cabri:
                  > > m (ABC) =12
                  > >
                  > > geometric reasons? I do not know it
                  > >
                  > >
                  > > Ricardo
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > >
                  > > Your use of Yahoo! Groups is subject to
                  http://docs.yahoo.com/info/terms/
                  > >
                  >
                • RICARDO BARROSO CAMPOS
                  Friends of Hyacintos, Igor: It is a perfect euclidea construction.Congratulations My conjecture of 12 was correct. Greetings Ricardo Barroso Campoos University
                  Message 8 of 10 , Oct 19, 2002
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                    Friends of Hyacintos, Igor:
                    It is a perfect euclidea construction.Congratulations

                    My conjecture of 12 was correct.

                    Greetings
                    Ricardo Barroso Campoos
                    University of Seville Spain

                    http://www.pdipas.us.es/r/rbarroso/trianguloscabri
                    >>> sharygin@... 19/10/02 20:43 >>>

                    ----- Original Message -----
                    From: Sharygin I. F. <sharygin@...>
                    To: <Hyacinthos@yahoogroups.com>; <yagcimustafa@...>
                    Sent: Saturday, October 19, 2002 9:47 PM
                    Subject: Re: [EMHL] an angle question



                    Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on
                    straight GP, ED = EG.
                    m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD), m(EDP)=36
                    PD=ED=EG = EF, m(FED) = 72 - (60-36) = 48.
                    C and B on straight FP, CD is the perpendicular to EP, EB is perpendicular
                    FD.
                    m(EDC) = m(PDC) = half m(EDP) = 18, m(DEB) = m(FEB) = 24,
                    CE and BD intersect in A.
                    So: ABC, BE and CD are bisectors, m(EDC) = 18, m( BED) =24.
                    m(EBC) = m(EFC) - m(FEB) = 30 - 24 = 6, m(ABC) = 12.
                    Best wishes, Igor Sharygin.
                    .
                    >
                    >
                    > ----- Original Message -----
                    > From: RICARDO BARROSO CAMPOS <rbarroso@...>
                    > To: <yagcimustafa@...>; <Hyacinthos@yahoogroups.com>
                    > Sent: Tuesday, October 15, 2002 3:00 PM
                    > Subject: Re: [EMHL] an angle question
                    >
                    >
                    > > Dear Hyacinthos:
                    > >
                    > > ABC is a triangle. D is a point on [AB] and E is a point on [AC].
                    > > m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 , m(ABC)=?
                    > >
                    > > I'm a conjecture with Cabri:
                    > > m (ABC) =12
                    > >
                    > > geometric reasons? I do not know it
                    > >
                    > >
                    > > Ricardo
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > >
                    > > Your use of Yahoo! Groups is subject to
                    http://docs.yahoo.com/info/terms/
                    > >
                    >





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