HYACINTHOS 29224

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of I.

Denote:

D =  the Feuerbach point X(11)

A", B", C" = the reflections of A', B', C' in OI line.

D' = the orthogonal projection of D on the OI line.
D" = the reflection of D in the OI line

1. The NPCs of DD'A", DD'B", DD'C" are coaxial.

2. The NPCs of D'D"A', D'D"B', D'D"C' are coaxial.

2nd (other than the midpoints of DD', D'D") intersections?  

 

[Peter Moses]:

 

Hi Antreas,

 

1. X(12736)  

2.  MIDPOINT OF X(65) AND X(3025)  

= a*(a^8*b - 3*a^7*b^2 - 2*a^6*b^3 + 9*a^5*b^4 - 9*a^3*b^6 + 2*a^2*b^7 + 3*a*b^8 - b^9 + a^8*c - 2*a^7*b*c + 8*a^6*b^2*c - 8*a^5*b^3*c - 16*a^4*b^4*c + 22*a^3*b^5*c + 4*a^2*b^6*c - 12*a*b^7*c + 3*b^8*c - 3*a^7*c^2 + 8*a^6*b*c^2 - 10*a^5*b^2*c^2 + 17*a^4*b^3*c^2 - 25*a^2*b^5*c^2 + 13*a*b^6*c^2 - 2*a^6*c^3 - 8*a^5*b*c^3 + 17*a^4*b^2*c^3 - 26*a^3*b^3*c^3 + 19*a^2*b^4*c^3 + 12*a*b^5*c^3 - 8*b^6*c^3 + 9*a^5*c^4 - 16*a^4*b*c^4 + 19*a^2*b^3*c^4 - 32*a*b^4*c^4 + 6*b^5*c^4 + 22*a^3*b*c^5 - 25*a^2*b^2*c^5 + 12*a*b^3*c^5 + 6*b^4*c^5 - 9*a^3*c^6 + 4*a^2*b*c^6 + 13*a*b^2*c^6 - 8*b^3*c^6 + 2*a^2*c^7 - 12*a*b*c^7 + 3*a*c^8 + 3*b*c^8 - c^9) : :

= 3 X[354] - X[13756],5 X[18398] - X[23153]

= lies on these lines: {1, 901}, {57, 953}, {65, 3025}, {226, 31841}, {354, 13756}, {513, 12736}, {517, 5083}, {942, 24201}, {1210, 3259}, {2800, 14115}, {13411, 22102}, {18398, 23153}

= midpoint of X(65) and X(3025)
= reflection of X(24201) in X(942)
= reflection of X(12736) in the OI line
= incircle inverse of X(2718)

Best regards,
Peter Moses.