947Re: [EMHL] Where is the incenter?
- Jun 2, 2000
>It seems vectors can do the jog without too much trouble (but I confess I
> 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.
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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