Re : [EMHL] Fermat point problem
- Dear Barry and your student,
>The first Fermat point F is the intersection of the lines AA', BB',I have the following proof.
>CC', where the triangles A'BC, B'CA, and C'AB are equilateral.
>Let d be the circle through A'B'C'. Line AFA' meets d at A' and A*,
>and B* and C* on d are defined cyclically. Show that
> |AA'| = |AA*| + |BB*| + |CC*| .
CC*==(R1^2-O1C^2)/d and we have
Thus we need to prove that
From Leibnitz theorem we have
From (1) and (2) it follows that we need to prove that
Since centroid G1 of triangle A'B'C' is equal to centroid G of
triangle ABC then we have that O1G^2=R1^2-(1/9)(a1^2+b1^2+c1^2) (3).
Thus we have to prove that
From triangles FB'C',FC'A',FA'B' we get
Now we have to prove that
Since A'F=d-AF,B'F=d-BF,C'F=d-CF we have to prove that
Since AF=(1/2dsqrt3)[4D+(bb+cc-aa)sqrt3] and analogously
BF,CF and 2d^2=a^2+b^2+c^2+4Dsqrt3 after some
calculation we confirm that (7) is true.
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