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947Re: [EMHL] Where is the incenter?

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  • Lambrou Michael
    Jun 2, 2000
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      > So I hope that one of the hyacinths (Paul perhaps?) can help by
      > finding the expression for the squared distance from the incenter
      > to the midpoint of GH as a function of a,b,c.
      It seems vectors can do the jog without too much trouble (but I confess I
      did not do the dirty work to the last line):
      square of IM = inner product of ( (g+h)/2-i) by itself,
      where g=(a+b+c)/3 etc all known. Save work by taking a=0 (where of course
      I have changed notation to vectors, easily going back to lengths)

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