## Re: Area problem

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• Better I add a pic: http://pic-hoster.net/code/af0e139a00fff39403ed36b97bcabf36.gif So, we can immediately ask for several natural possible loci: 1. Green
Message 1 of 3 , Jun 1, 2010
http://pic-hoster.net/code/af0e139a00fff39403ed36b97bcabf36.gif

So, we can immediately ask for several natural possible loci:

1. Green areas equal (fiendish)
2. Yellow areas equal (trivial, incenter)
3. Arcs DE=FG=HI (middle difficulty, I give it a try)
4. Arcs EF=GH=ID (same as 2)
5. Triangle aress ADE=BFG=CHI (straight lines DE,FG,HI
have been omitted for clarity - middle difficulty)

Obviously, these can be used to combine two loci into
a point, but only 3+5 makes good sense. One solution
surely is the circumcenter, others might exist.

(Pic approximates case 3)

Hauke
• [superseding reply - why does Yahoo have no edit mode, as any lousy BBS can do?] ... Algebraically I get the Brocard axis as solution. Note that when
Message 2 of 3 , Jun 2, 2010
[superseding reply - why does Yahoo have no
edit mode, as any lousy BBS can do?]

> 3. Arcs DE=FG=HI (middle difficulty, I give it a try)

Algebraically I get the Brocard axis as solution.
Note that when DE=FG=HI=0, the circumcircle arises
naturally (trivial). Somehow I took the wrong surd
at Albuquerque, as I get quarter-circle arcs for
the cosine circle (at least P=symmedian and
r=abc/(a^2+b^2+c^2) results) which is of course nonsense:
http://mathworld.wolfram.com/KenmotuPoint.html
THIS are the quarter-circle arcs. <shrug>

Hauke
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