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19Re: Lemoine point

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  • Barry Wolk
    Dec 27, 1999
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      > Date: Thu, 23 Dec 1999 09:23:47 +0100
      > From: Floor van Lamoen <f.v.lamoen@...
      >
      > > From: Clark Kimberling <ck6@...>
      > >
      > > However, Honsberger doesn't mention (directly) a certain interesting
      > > property of the Lemoine point. For any point P, let A'B'C' denote the
      > > pedal triangle of P (i.e., A' is the point in which the line through P
      > > perpendicular to line BC meets line BC). Let S(P) be the vector sum
      > > PA'+PB'+PC'. Then S(P) is the zero vector if P is the Lemoine point.
      >
      > Isn't this property exactly equivalent to the fact that the
      > Lemoine/symmedian point K is the centroid of its pedal triangle? This
      > property of K is mentioned in O. Bottema's ``Hoofdstukken uit de
      > Elementaire Meetkunde''.

      Of course that is an equivalent property.

      >
      [snip]
      > > I conjecture that the converse is true: that if P is a "point" (i.e.,
      > > f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2
      > > (barycentric coordinates of the Lemoine point).

      > This is not immediatly apparent to me.

      It wasn't apparent to me either. However, the calculations are not
      difficult, and this converse is true.

      Write vec(X,Y) = vector from X to Y.
      In homogenous barycentrics, P=(x:y:z).
      In normalized barycentrics, if P=(x,y,z) with x+y+z=1, then
      vec(O,P) = x vec(O,A) + y vec(O,B) + z vec(O,C)
      for any choice of origin O.

      Now the B-pedal point of P(x,y,z) is P_B = (x+y SC/bb , 0 , z+y SA/bb),
      and this is normalized, since SA+SC=bb.
      Then vec(P,P_B) = vec(O,P_B) - vec(O,P)
      = y SC/bb vec(O,A) - y vec(O,B) + y SA/cc vec(O,C)
      with similar formulas for vec(P,P_A) and vec(P,P_C)

      So the equation S(P)=0 becomes
      (-x + y SC/bb + z SB/cc) vec(O,A)
      + (x SC/aa - y + z SA/cc) vec(O,B)
      + (x SB/aa + y SA/bb - z) vec(O,C) = 0

      Well, one solution is obviously (x,y,z)=const*(aa,bb,cc). Any others?
      Choose the origin O to be the circumcenter. Then the three basis
      vectors are dependent, and we have the dependency relation
      aaSA vec(O,A) + bbSB vec(O,B) + ccSC vec(O,C) = 0
      This is the only dependency relation between these three vectors.

      So any other solution {x,y,z} would have to satisfy
      (-x + y SC/bb + z SB/cc) = k aaSA
      (x SC/aa - y + z SA/cc) = k bbSB
      (x SB/aa + y SA/bb - z) = k ccSC
      for some non-zero constant k. However, adding these 3 equations
      shows k=0, so there are no other solutions

      I admit the calculations weren't as easy as I expected,
      but this converse is true.
      --
      Barry Wolk <wolkb@...>
      Winnipeg Manitoba Canada
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