HYACINTHOS 28938

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triange, A'B'C' the orthic triangle and L a line,

The parallels to L through A, B, C, intersect the circumcircle again at A", B", C", resp.

The cirumcircles of AA'A", BB'B", CC'C" are coaxial.

Radical trace ?

Special L = Euler line, Brocard axis, OI line.....  


 [Randy Hutson]:

A few more:

Napoleon axis -> X(3)X(18336)∩X(4)X(5965)

= 2 a^14 - 10 a^12 (b^2 + c^2) + a^10 (34 b^4 + 8 b^2 c^2 + 34 c^4) - a^8 (65 b^6 + 3 b^4 c^2 + 3 b^2 c^4 + 65 c^6) + a^6 (61 b^8 - 4 b^6 c^2 - 4 b^2 c^6 + 61 c^8) - a^4 (b^2 + c^2) (25 b^8 - 13 b^6 c^2 - 6 b^4 c^4 - 13 b^2 c^6 + 25 c^8) + a^2 (b^2 - c^2)^2 (3 b^8 + 32 b^6 c^2 - 22 b^4 c^4 + 32 b^2 c^6 + 3 c^8) - 5 b^2 c^2 (b^2 - c^2)^4 (b^2 + c^2) : :

= lies on these lines: {3,18336}, {4,5965}, {3545,18335}


IK -> X(3)X(667)∩X(4)X(518)

= a (a^8 (b^2 + c^2) - a^7 (3 b^3 + b^2 c + b c^2 + 3 c^3) + a^6 (b^4 - b^3 c + 6 b^2 c^2 - b c^3 + c^4) + a^5 (5 b^5 + 5 b^4 c - 4 b^3 c^2 - 4 b^2 c^3 + 5 b c^4 + 5 c^5) - a^4 (5 b^6 + b^5 c + b^4 c^2 + 6 b^3 c^3 + b^2 c^4 + b c^5 + 5 c^6) - a^3 (b^7 + 3 b^6 c + b^5 c^2 - 9 b^4 c^3 - 9 b^3 c^4 + b^2 c^5 + 3 b c^6 + c^7) + a^2 (b - c)^2 (3 b^6 + 7 b^5 c + 15 b^4 c^2 + 18 b^3 c^3 + 15 b^2 c^4 + 7 b c^5 + 3 c^6) - a (b - c)^2 (b^7 + 3 b^6 c + 7 b^5 c^2 + b^4 c^3 + b^3 c^4 + 7 b^2 c^5 + 3 b c^6 + c^7) + b c (b - c)^4 (b + c)^2 (b^2 + c^2)) : :

= lies on these lines: {3,667}, {4,518}, {6835,18343}, {15030,31849}, {31803,31851}


NK -> X(18348)

IH -> X(1)X(4)∩X(3)X(522)

= 2 a^10 - 2 a^9 (b + c) - a^8 (3 b^2 - 4 b c + 3 c^2) + a^7 (b + c) (5 b^2 - 6 b c + 5 c^2) - a^6 (3 b^4 + 5 b^3 c - 12 b^2 c^2 + 5 b c^3 + 3 c^4) - 3 a^5 (b - c)^4 (b + c) + a^4 (b - c)^2 (7 b^4 + 11 b^3 c + 4 b^2 c^2 + 11 b c^3 + 7 c^4) - a^3 (b - c)^2 (b + c) (b^4 + 8 b^3 c - 2 b^2 c^2 + 8 b c^3 + c^4) - a^2 (b - c)^4 (b + c)^2 (3 b^2 + b c + 3 c^2) + a (b - c)^4 (b + c)^5 - b c (b^2 - c^2)^4 : :

= lies on the circle O(1,3) and these lines: {1,4}, {2,18339}, {3,522}, {8,15633}, {102,1897}, {117,15252}, {124,952}, {2222,11491}, {2360,7452}, {2968,6711}, {3326,6284}, {5494,10265}, {5884,31849}, {6796,23981}, {11719,24030}

= midpoint of circumcircle intercepts of line X(1)X(4)
= X(102)-of-X(1)-Brocard-triangle


Best regards,
Randy Hutson.