HYACINTHOS 26373

[Le Viet An]:

Let ABC be a triangle and HaHbHc the orthic triangle.

The line IH intersects HbHc, HcHa, HaHb at Ka,Kb,Kc, resp.

The perpendiculars to IA, IB, IC  from Ka, Kb, Kc, resp. bound a triangle A'B'C'.

Then the circumcircle of the triangle A'B'C' touches the NPC of ABC.

Which is

1. the center of the circle ?

2. the touchpoint ?


[Peter Moses]:


Hi Antreas,


1).
(a+b-c) (a-b+c) (a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^12 b-2 a^11 b^2-6 a^10 b^3+7 a^9 b^4+4 a^8 b^5-8 a^7 b^6+4 a^6 b^7+2 a^5 b^8-6 a^4 b^9+2 a^3 b^10+2 a^2 b^11-a b^12+2 a^12 c-8 a^11 b c+4 a^10 b^2 c+16 a^9 b^3 c-19 a^8 b^4 c-2 a^7 b^5 c+14 a^6 b^6 c-10 a^5 b^7 c+2 a^4 b^8 c+2 a^3 b^9 c-2 a^2 b^10 c+2 a b^11 c-b^12 c-2 a^11 c^2+4 a^10 b c^2-2 a^9 b^2 c^2-a^8 b^3 c^2+8 a^7 b^4 c^2-11 a^6 b^5 c^2-2 a^5 b^6 c^2+9 a^4 b^7 c^2-2 a^3 b^8 c^2-a^2 b^9 c^2-6 a^10 c^3+16 a^9 b c^3-a^8 b^2 c^3-28 a^7 b^3 c^3+17 a^6 b^4 c^3+10 a^5 b^5 c^3-11 a^4 b^6 c^3+8 a^3 b^7 c^3-3 a^2 b^8 c^3-6 a b^9 c^3+4 b^10 c^3+7 a^9 c^4-19 a^8 b c^4+8 a^7 b^2 c^4+17 a^6 b^3 c^4-16 a^5 b^4 c^4+6 a^4 b^5 c^4-11 a^2 b^7 c^4+9 a b^8 c^4-b^9 c^4+4 a^8 c^5-2 a^7 b c^5-11 a^6 b^2 c^5+10 a^5 b^3 c^5+6 a^4 b^4 c^5-20 a^3 b^5 c^5+15 a^2 b^6 c^5+4 a b^7 c^5-6 b^8 c^5-8 a^7 c^6+14 a^6 b c^6-2 a^5 b^2 c^6-11 a^4 b^3 c^6+15 a^2 b^5 c^6-16 a b^6 c^6+4 b^7 c^6+4 a^6 c^7-10 a^5 b c^7+9 a^4 b^2 c^7+8 a^3 b^3 c^7-11 a^2 b^4 c^7+4 a b^5 c^7+4 b^6 c^7+2 a^5 c^8+2 a^4 b c^8-2 a^3 b^2 c^8-3 a^2 b^3 c^8+9 a b^4 c^8-6 b^5 c^8-6 a^4 c^9+2 a^3 b c^9-a^2 b^2 c^9-6 a b^3 c^9-b^4 c^9+2 a^3 c^10-2 a^2 b c^10+4 b^3 c^10+2 a^2 c^11+2 a b c^11-a c^12-b c^12)::
on lines {{4, 109}, {65, 7649}}.

2).
(a-b-c) (b-c)^2 (a^2+b^2-c^2) (a^2-b^2+b c-c^2) (a^2-b^2+c^2) (a^6-a^5 b-a^4 b^2+2 a^3 b^3-a^2 b^4-a b^5+b^6-a^5 c+a^4 b c+a^3 b^2 c-a^2 b^3 c-a^4 c^2+a^3 b c^2+a b^3 c^2-b^4 c^2+2 a^3 c^3-a^2 b c^3+a b^2 c^3-a^2 c^4-b^2 c^4-a c^5+c^6)::
on lines {{4, 2222}, {11, 7649}, {117, 1737}, {119, 1877}, {650, 5190}, {867, 10017}, {5089, 5513}}.
On the NP circle.
inverse of X(5190) in the Stevanovic circle.
inverse of X(2222) in the polar circle.


Best regards,
Peter Moses.