HYACINTHOS 29006

[Angel Montesdeoca]:

 

Dear all,

Let  DEF  be the intouch triangle and A'B'C' be the complement of the excentral triangle.
Let Ba and Ca be the points of intersection of the line EF with the interior bisectors of angle B and angle C, respectively.
Let Oa be the circumcenter of ABaCa, and define Ob and Oc cyclically.
Then OaObOc and A'B'C' are perspective with perspector X(10). 

Also,  X(10) is the orthologic center of OaObOc to A'B'C'.  

The orthologic center of A'B'C' to OaObOc is 

U = X(2)X(2124)∩X(10)X(971)

= (b+c-a)(a^5 (b+c)-3 a^4 (b-c)^2+2 a^3 (b-c)^2 (b+c)+2 a^2 (b-c)^2 (b^2+6 b c+c^2)-a (b-c)^2 (3 b^3+13 b^2 c+13 b c^2+3 c^3)+(b-c)^6) : :   

= lies on these lines: {2,2124}, {10,971}, {142,10004}, {279,19605}, {2391,3452}, {6706,20205}

Best regards,
Angel Montesdeoca