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Re: [EMHL] an angle question

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  • Ricardo Barroso Campos
    Amigo Ignacio: es BAC=96 gracias ... From: Ignacio Larrosa Cañestro To: Sent: Tuesday, October 15, 2002
    Message 1 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 2 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 3 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 4 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 5 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 6 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 7 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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