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RE: [EMHL] T construction given

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  • Luís Lopes
    Dear Hyacinthists, Maybe this could help. But I wasn t able to follow it. Please notice that a simple translation to English is not enough. I wonder there
    Message 1 of 3 , Aug 1, 2008
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      Dear Hyacinthists,

      Maybe this could help. But I wasn't able to follow it.

      Please notice that a simple translation to English
      is not enough. I wonder there isn't a typo
      anywhere.

      Best regards,
      Luis

      Lösung. Macht man mittelst eines Kreises um A
      mit dem Radius = b auf der Verlängerung von c
      die Linie AG = b, und ebenso AE = c, wo E in a
      liegt, so ist das Hülfsdreieck BEG gegeben, da
      BE = p-q und, wie sich leicht ergiebt, der
      ist.

      _______________________________

      To: hyacinthos@yahoogroups.com
      From: qed_texte@...
      Date: Wed, 30 Jul 2008 13:03:32 +0000
      Subject: [EMHL] T construction given

      Dear Hyacinthists,

      Here I am again with another TC.

      In a triangle we have:

      B---Ha---B'-----Ma------------C

      where , and .
      Then CB'=p-q. Now how to construct the
      triangle?

      Best regards,
      Luis

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    • Luís Lopes
      Dear Hyacinthists, I repeat my text, because there are missed parts. Lösung. Macht man mittelst eines Kreises um A mit dem Radius = b auf der Verlängerung
      Message 2 of 3 , Aug 1, 2008
      • 0 Attachment
        Dear Hyacinthists,

        I repeat my text, because there are missed parts.

        Lösung. Macht man mittelst eines Kreises um A
        mit dem Radius = b auf der Verlängerung von c
        die Linie AG = b, und ebenso AE = c, wo E in a
        liegt, so ist das Hülfsdreieck BEG gegeben, da
        BE = p-q und, wie sich leicht ergiebt, der
        \measuredangle BEG = 1R + A/2 ist.

        In a triangle we have:

        B---Ha---B'-----Ma------------C

        where AB=AB'=c, BHa=B'Ha, CHa=p, and BHa=q.
        Then CB'=p-q.
        Now how to construct the triangle?

        Best regards,
        Luis

        _______________________________

        To: hyacinthos@yahoogroups.com
        From: qed_texte@...
        Date: Fri, 1 Aug 2008 21:10:53 +0000
        Subject: RE: [EMHL] T construction given

        Dear Hyacinthists,

        Maybe this could help. But I wasn't able to follow it.

        Please notice that a simple translation to English
        is not enough. I wonder there isn't a typo
        anywhere.

        Best regards,
        Luis

        Lösung. Macht man mittelst eines Kreises um A
        mit dem Radius = b auf der Verlängerung von c
        die Linie AG = b, und ebenso AE = c, wo E in a
        liegt, so ist das Hülfsdreieck BEG gegeben, da
        BE = p-q und, wie sich leicht ergiebt, der
        ist.

        _______________________________

        To: hyacinthos@yahoogroups.com
        From: qed_texte@...
        Date: Wed, 30 Jul 2008 13:03:32 +0000
        Subject: [EMHL] T construction given

        Dear Hyacinthists,

        Here I am again with another TC.

        In a triangle we have:

        B---Ha---B'-----Ma------------C

        where , and .
        Then CB'=p-q. Now how to construct the
        triangle?

        Best regards,
        Luis

        _________________________________________________________________
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