HYACINTHOS 25098

[Tran Quang Hung.]

Let ABC be a triangle with the first Fermat point F.

 

A'B'C' is cevian triangle of F.

 

Fa,Fb,Fc are the first Fermat points of AB'C',BC'A',CA'B'.

 

Then the circles (FFaA),(FFbB),(FFcC) are coaxal, and their radical axis is parallel to Euler line of ABC.

 

[Peter Moses]:

Hi Antreas,

 

The coaxal circles meet at X(13) and ...

(a^2-a b+b^2-c^2) (a^2+a b+b^2-c^2) (a^2-b^2-a c+c^2) (a^2-b^2+a c+c^2) (3 (3 a^8-8 a^6 b^2+6 a^4 b^4-b^8-8 a^6 c^2+11 a^4 b^2 c^2-5 a^2 b^4 c^2+2 b^6 c^2+6 a^4 c^4-5 a^2 b^2 c^4-2 b^4 c^4+2 b^2 c^6-c^8)-2 Sqrt[3] (a^6-a^4 b^2-a^2 b^4+b^6-a^4 c^2+3 a^2 b^2 c^2-b^4 c^2-a^2 c^4-b^2 c^4+c^6) S)::

on lines {{13,15},{14,476},{23,6104},{ 187,1989},{477,5995},{531, 11078},{8737,10295}}.

Reflection of X(13) in X(11537).

On K061a, K262a.

 

The line of circle centers passes through X{395,523,6138,9195,9200, 11537}.

The radical axis is parallel to Euler through X{13,523}.

 

Best regards

Peter Moses