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Re: Three Rectangles Around Inscribed Triangle

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  • Kostas Vittas
    Dear Tuan, Vijaya Prasad and all. The results as you mentioned in your configuration, is true in general case of a triangle A B C and three arbitrary
    Message 1 of 4 , Nov 7, 2006
      Dear Tuan, Vijaya Prasad and all.

      The results as you mentioned in your configuration, is true in
      general case of a triangle A'B'C' and three arbitrary rectangles
      B'C'CaBa, C'A'AbCb, A'B'BcAc, erected on sidesegments B'C', C'A',
      A'B' respectively, outwardly (or inwardly) to it. So, we have again
      the results as follows:

      1). – The lines A'A'', B'B'', C'C'', are concurrent at one point, so
      be it P, as the homothetic center of A'B'C', A''B''C''.

      2). – The midperpendiculars of the segments AbAc, BaBc, CaCb, are
      concurrent at one point, so be it Q.

      2). – The lines through A'', B'', C'', and perpendicular to AbAc,
      BaBc, CaCb respectively, are concurrent at one point, so be it R.

      3). – The lines through A', B', C', and perpendicular to AbAc, BaBc,
      CaCb respectively, are concurrent at one point, so be it R' (and so,
      QR = QR').

      4). – The points P, Q, R, R', are collinear.

      5). – We draw two lines through Ba, Ca and perpendicular to BaBc,
      CaCb respectively and we denote as A1, their intersection point.
      Similarly we define the points B1, C1. So, we can prove that the
      quadrilaterals B1C1AcAb, A1C1BcBa, A1B1CbCa, are rectangles and
      hence, the point Q, is the circumcenter of the triangle A1B1C1.

      But all the above results are well known. A very interesting work by
      Nikos Dergiades and Flour van Lamoen, has already been posted in
      journal Forumgeom. Please see at:

      http://forumgeom.fau.edu/FG2003volume3/FG200316index.html

      The proofs are given by barycentrics and it is a nice challenge, to
      try someone for the solutions by classical geometry.

      I have in mind some elementary proofs of the above results and with
      pleasure, I will present here next time.

      Best regards.
      Kostas Vittas.



      --- In Hyacinthos@yahoogroups.com, Quang Tuan Bui <bqtuan1962@...>
      wrote:
      >
      > Dear All My Friends,
      > Given triangle ABC with points A', B', C' on sidelines BC, CA, AB
      respectively. One line La passing through vertex A and parallel with
      B'C'; Ba, Ca are orthogonal projections of B', C' on the line La
      respectively. We have one rectangle B'C'CaBa. Similarly from vertices
      B and C we can construct two other lines Lb, Lc and two other
      rectangles: C'A'AbCb, A'B'BcAc. Three lines La, Lb, Lc bound one
      triangle A''B''C''. It is clear that A''B''C'' and A'B'C' are
      homothetic with homothetic center, say P.
      > Please prove that:
      > 1. Three perpendicular bisectors of three segments AbAc, BcBa,
      CaCb are concurrent at one point say Q.
      > 2. Three perpendicular lines from A'', B'', C'' to AbAc, BcBa,
      CaCb respectively are concurrent at one point say R.
      > 3. Three points P, Q, R are collinear.
      > 4. What we can say about the ratio PQ/PR?
      > 5. Are there somethings interesting when A'B'C' is Cevian
      triangle of ABC?
      > Best regards,
      > Bui Quang Tuan
      >
      >
      > ---------------------------------
      > Check out the all-new Yahoo! Mail - Fire up a more powerful email
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      >
      > [Non-text portions of this message have been removed]
      >
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