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Re: [EMHL] Another Problem (locus)

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  • xpolakis@otenet.gr
    ... Hmm..... Now, after Paul s solution, I recall this well-known exercise in Euclidean geometry; If the non-parallel sides of a trapezium meet at X, and its
    Message 1 of 1 , Oct 6, 2000
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      I wrote:

      >>Let P be a point on the plane of a triangle ABC, and A'B'C' its cevian
      >>triangle.
      >>From B', C' draw parallels to BC intersecting AB, AC at Ba, Ca, respectively.
      >>Let A* be the intersection of B'C'and BaCa. Similarly define B*,C*.
      >>
      >> A
      >> /\
      >> / \
      >> C' Ca
      >> / \
      >> / A* \
      >> Ba B'
      >> / P \
      >> / \
      >> B--------A'------C
      >>
      >>Which is the locus of P such that A*B*C*, ABC are in perspective?

      Paul Yiu wrote:

      >[...] the perspector is always the centroid.
      >The desmic system is
      >
      >
      >(u(v+w):vw:vw) & (::) & (::) & (u^2(v+w):v^2(w+u):w^2(u+v) \\
      >(uvw:v^2(w+u):w^2(u+v)) & (::) & (::) & (1:1:1)
      > Desmon & (u:v:w) \\
      > Harmon & (u(-\frac 1u+\frac 1v+ \frac 1w)::)

      Hmm.....

      Now, after Paul's solution, I recall this well-known exercise in Euclidean
      geometry;

      If the non-parallel sides of a trapezium meet at X, and its diagonals at Y,
      then XY bisects the parallel sides.

      Let's see it in the figure of the problem above:

      A
      /\
      / \
      C' Ca
      / \
      / A* \
      Ba ---------B'
      D

      Assume that AA* intersects BaB' at D.

      Then by Ceva theorem we get:

      DBa CaB' C'A
      --- x ---- x ----- = - 1
      DB' CaA C'Ba

      We have C'Ca // BaB', therefore D is the midpoint of B'Ba,
      and AA* bisects every parallel to it.
      And since BC is parallel to BaB', the AA* is the a-median of ABC etc

      Note that the points A',B',C' in the problem do not have to form
      a triangle in perspective with ABC (they can be arbitrarily chosen
      on the triangle sides)

      Antreas
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