## Re : [EMHL] Fermat point problem

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• Dear Barry and your student, You wrote ... I have the following proof. AA =BB =CC =d. 2d^2=aa+bb+cc+4Dsqrt3,where D=area(ABC).
Message 1 of 1 , Oct 1, 2003
You wrote
>The first Fermat point F is the intersection of the lines AA', BB',
>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*| .
I have the following proof.
AA'=BB'=CC'=d.
2d^2=aa+bb+cc+4Dsqrt3,where D=area(ABC).
AA*=(R1^2-O1A^2)/AA'=(R1^2-O1A^2)/d,and
analogously BB*==(R1^2-O1B^2)/d,
CC*==(R1^2-O1C^2)/d and we have
AA*+BB*+CC*=(1/d)[3R1^2-(O1A^2+O1B^2+O1C^2)].
Thus we need to prove that
3R1^2-(O1A^2+O1B^2+O1C^2)]=d^2 (1).
From Leibnitz theorem we have
O1A^2+O1B^2+O1C^2=3O1G^2+(1/3)(aa+bb+cc) (2).
From (1) and (2) it follows that we need to prove that
3R1^2-3O1G^2=d^2+(1/3)(aa+bb+cc).
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
a1^2+b1^2+c1^2=3d^2+a^2+b^2+c^2 =m(5).
From triangles FB'C',FC'A',FA'B' we get
a1^2=B'F^2+C'F^2+B'F*C'F,
b1^2=C'F^2+A'F^2+C'F*A'F,
c1^2=A'F^2+B'F^2+A'F*B'F,
Now we have to prove that
2(A'F^2+B'F^2+C'F^2)+B'F*C'F+C'F*A'F+A'F*B'F=m (6).
Since A'F=d-AF,B'F=d-BF,C'F=d-CF we have to prove that
2(AF^2+BF^2+CF^2)+BF*CF+CF*AF+AF*BF=aa+bb+cc (7).
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.
Best regards
Sincerely

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