HYACINTHOS 28896

[Tran Quang Hung]:

I call Kosnita line the line conneting NPC center Χ(5) and Kosnita point Χ(54) of a triangle.

Let ABC be a triangle with excenters Ia, Ib, Ic.Then refections L'a, L'b, L'c  of Kosnita lines La, Lb, Lc of triangles IaBC, IbCA, and IcAB in lines BC, CA, and AB respectively are concurrent.


[Angel Montesdeoca]:


***  L'a, L'b, L'c  concur in X(110) = focus of Kiepert parabola.

*** Aditional information:

=== L'a meets the circumcircle again at D.
L'b meets the circumcircle again at E.
L'c meets the circumcircle again at F.
Then DEF is the circumcevian triangle of X(80)=reflection of incenter in Feuerbach point.

===  The triangle AaBbCc formed by the  lines La, Lb, Lc  is perspective to ABC at X(1290) = reflection of X(110) in line X(1)X(3).


=== If Aa=Lb /\ Lc,  Ab=L'a /\ Lc,  Ac=L'a /\ Lb, then the line AAa, BAb, CAc concur in a point Wa,    and define Wb and Wc cyclically.  WaWbWc is perspective to ABC at X(1290) and  WaWbWc is perspective to AaBbCc at X(1290).
Then the three perspectrix of ABC and  WaWbWc,  ABC and AaBaBc, AaBbBc and  WaWbWc concur in W  with first barycentric coordinate:

a (a+b) (a+c) ((b-c)^6 (b+c)^7+(b^2-c^2)^4 (b^4-b^3 c+8 b^2 c^2-b c^3+c^4) a-(b-c)^2 (b+c)^3 (4 b^6-2 b^5 c+11 b^4 c^2-14 b^3 c^3+11 b^2 c^4-2 b c^5+4 c^6) a^2-(b^2-c^2)^2 (4 b^6-7 b^5 c-3 b^4 c^2+7 b^3 c^3-3 b^2 c^4-7 b c^5+4 c^6) a^3+(5 b^9-3 b^8 c+15 b^7 c^2+31 b^6 c^3-18 b^5 c^4-18 b^4 c^5+31 b^3 c^6+15 b^2 c^7-3 b c^8+5 c^9) a^4+(5 b^8-18 b^7 c-43 b^6 c^2-16 b^5 c^3+14 b^4 c^4-16 b^3 c^5-43 b^2 c^6-18 b c^7+5 c^8) a^5+b c (12 b^5+3 b^4 c-19 b^3 c^2-19 b^2 c^3+3 b c^4+12 c^5) a^6+b c (22 b^4+41 b^3 c+47 b^2 c^2+41 b c^3+22 c^4) a^7+(-5 b^5-13 b^4 c-17 b^3 c^2-17 b^2 c^3-13 b c^4-5 c^5) a^8+(-5 b^4-13 b^3 c-17 b^2 c^2-13 b c^3-5 c^4) a^9+2 (2 b^3+3 b^2 c+3 b c^2+2 c^3) a^10+(4 b^2+3 b c+4 c^2) a^11+(-b-c) a^12-a^13)


W lies on lines X(10)X(21).
 
 (6 - 9 - 13) - search numbers  of W: (3.49717725619609, 4.40513014363584, -1.02312281270017).
 
 
 Angel Montesdeoca