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Hamilton-Cayley in Triangle Geometry

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  • Francois Rideau
    Dear all Sorry my chuff-chuffs are causing all this fuss overseas! As for me, I try it but I can t remember one single french soldier and though it was 62
    Message 1 of 1 , Mar 30, 2006
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      Dear all
      Sorry my chuff-chuffs are causing all this fuss overseas!
      As for me, I try it but I can't remember one single french soldier and
      though it was 62 years ago, I still have a vivid memory of tall and strong
      GIs giving me a lot of candies and incidentaly chasing the Nazis out off the
      landscape!

      To come back with Triangle Geometry, I give here a sketch of the proof of
      the problem:


      > As surprising as it is, even Hamilton Cayley theorem has numerous
      > applications in Triangle Geometry.
      > Here a problem in Triangle Geometry for Hyacinthos rookies proving what I
      > told:
      > In the real affine plane a triangle ABC is given.
      > We look at an inscribed triangle abc, point "a" on line BC, point "b" on
      > line CA, point "c" on line AB, such that signed areas S(A,B,C) and S(a,b,c)
      > are equal.
      > Let f be the affine map defined by f(A) = a; f(B) = b; f(C) = c.
      > 1° Prove that f is of order 3, that is: f^3 = id, has a unique fixed
      > point but no invariant lines.
      >


      Barycentrics of point "a" wrt ABC are: (0, 1-u , u)
      Barycentrics of point "b" wrt ABC are: (v, 0 , 1-v)
      Barycentrics of point "a" wrt ABC are: (1-w , w , 0)
      They are columns of a matrix M called matrix of the affine map f in the
      affine basis {A, B, C}.
      Let g be the associated linear map of the affine map f.
      Char(M) and Char(g) the characteristic polynomials of M and g
      Then we have:
      Char(M) = (X-1).Char(g)
      As consequence, we have:
      Trace(M) = 1 + Trace(g) and Det(M) = Det(g).
      But as matrix M as only 0 on its main diagonal, we have Trace(g) = -1
      and from S(A,B,C) = S(a,b,c), we get Det(g) = 1.
      So Char(g) = X^2 + X + 1 and Char(M) = X^3 - 1.
      From Hamilton-Cayley theorem, we get M^3 = I (3x3 identity matrix) and so
      f^3 = id.
      As 1 is not an eigenvalue of g, f has a single fixed point O and as Char(g)
      has no real roots(i.e g is not diagonalizable), f has no invariant lines.
      Note that f generates a cyclic group G(f) of order 3, hence O is the
      centroid of the orbit of any point under action of G(f).

      2° Given point "a" on line BC and point "b" on line CA, find an
      affineconstruction of point "c" on line AB such that:
      > S(A,B,C) = S(a,b,c).
      >

      I only give here an affine construction of point "c" without proof to let
      you folks thinking about it.
      I would be happy to read some proofs of yours!
      We complete parallelograms aCAa' and bCBb'.
      Line a'b cuts line BC on U and line b'a cuts line CA on V. Then line UV cuts
      line AB in the sought point "c".

      Friendly
      François
      PS
      I give me a 3-week break to visit cherries blossoming in Tokyo and look at
      some sungakus around here!


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