HYACINTHOS 28692

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

Fe = the Feuerbach point.

 

The NPCs of IFeNa, IFeNb, IFeNc are coaxial.

Second (other than the midpoint of IFe) intersection?

 

[Peter Moses]:

Hi Antreas,

= X(1)X(18341)∩X(11)X(11700)

2 a^10-3 a^9 b-5 a^8 b^2+9 a^7 b^3+2 a^6 b^4-9 a^5 b^5+4 a^4 b^6+3 a^3 b^7-4 a^2 b^8+b^10-3 a^9 c+14 a^8 b c-7 a^7 b^2 c-22 a^6 b^3 c+24 a^5 b^4 c+a^4 b^5 c-15 a^3 b^6 c+8 a^2 b^7 c+a b^8 c-b^9 c-5 a^8 c^2-7 a^7 b c^2+34 a^6 b^2 c^2-14 a^5 b^3 c^2-27 a^4 b^4 c^2+22 a^3 b^5 c^2+a^2 b^6 c^2-a b^7 c^2-3 b^8 c^2+9 a^7 c^3-22 a^6 b c^3-14 a^5 b^2 c^3+44 a^4 b^3 c^3-10 a^3 b^4 c^3-8 a^2 b^5 c^3-3 a b^6 c^3+4 b^7 c^3+2 a^6 c^4+24 a^5 b c^4-27 a^4 b^2 c^4-10 a^3 b^3 c^4+6 a^2 b^4 c^4+3 a b^5 c^4+2 b^6 c^4-9 a^5 c^5+a^4 b c^5+22 a^3 b^2 c^5-8 a^2 b^3 c^5+3 a b^4 c^5-6 b^5 c^5+4 a^4 c^6-15 a^3 b c^6+a^2 b^2 c^6-3 a b^3 c^6+2 b^4 c^6+3 a^3 c^7+8 a^2 b c^7-a b^2 c^7+4 b^3 c^7-4 a^2 c^8+a b c^8-3 b^2 c^8-b c^9+c^10 : : 

 

= lies on these lines: {1,18341}, {11,11700}, {109,16173}, {117,11715}, {942,1387}, {1125,3738}, {2802,6718}, {2817,6713}, {3576,10771}, {5450,11798}

Best regards,
Peter Moses.