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Re: [EMHL] From Atul Dixit: Kenmotu configuration

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  • xpolakis@otenet.gr
    ... Probably we have already discussed this, but let s try it anyway: C* A B* / / / A / Ab
    Message 1 of 1 , Jun 1, 2001
      Atul Dixit asked:

      >If we consider the Kenmotu configuration. Draw lines through the
      >vertices of the three squares, nearer to the respective vertices of
      >the triangle, parallel to the respective diagonals of the three
      >squares. Do these three lines form a triangle perspective to triangle
      >ABC?

      Probably we have already discussed this, but let's try it anyway:


      C* A B*
      /\
      / \
      / A' \
      / \
      Ab Ac
      / \
      / \
      Ba J Ca
      / \
      / \
      / B' C' \
      / \
      B--------Bc---Cb---------C






      A*


      Let JAbA'Ac, JBcB'Ba, JCaC'Cb be the three congruent squares.

      Let A*B*C* be the vertices of the triangle formed by the parallels.
      (B*C* // AbAc, etc)

      By some angle computations, which I hope they were correct, I found that
      the triangles A*BC, B*CA, C*AB are isosceles: (A*BC) = (A*CB) = A, etc.

      Now (by Kiepert triangles theory) we have:

      Triangles: A*B*C*, ABC perspective <==>

      cot(A*BC) + cotB cot(B*CA) + cotC cot(C*AB) + cotA
      ---------------- x ----------------- x ---------------- = 1
      cot(A*CB) + cotC cot(B*AC) + cotA cot(C*BA) + cotB

      or

      cotA + cotB cotB + cotC cotC + cotA
      ----------- x ------------ x ----------- = 1
      cotA + cotC cotB + cotA cotC + cotB

      True.

      Perspector: The perspector is the point:

      ((cotB + cotC) ::) = ((sinA / sinBsinC) ::) = (sin^2A ::) = K : Lemoine P.
      in Barycentrics.


      And a problem (generalization), by the way:

      Construct the three Kenmotu squares but so that their sides
      be proportional to given p, q, r. Special case: p:q:r = a:b:c
      (In the standard Kenmotu problem we have that p:q:r = 1:1:1)

      Antreas
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