Τρίτη 29 Οκτωβρίου 2019

HYACINTHOS 29052

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, PaPbPc the pedal triangle of I and D a poiint.

Denote:

A', B', C' = the midpoints of AI, BI, CI, resp.

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

Ab, Ac = the reflections of A' in DNb, DNc, resp.
Bc, Ba = the reflections of B' in DNc, DNa, resp.
Ca, Cb = the reflections of C' in DNa, DNb, resp.

La, Lb, Lc = the perpendicular bisectors of AbAc, BcBa, CaCb, resp.
(concurrent at D = the common circumcenter of A'AbAc, B'BcBa, C'CaCb)

The reflections of La, Lb, Lc in IPa, IPb, IPc, resp. are concurrent for:

D = Feuerbach point (the point of concurrence, as we have seen in Hyacinthos 29040, is X(13756))
D = N. Point of concurrence?
 
 
[Peter Moses]:

Hi Antreas,

a*(2*a^9 - 5*a^8*b - a^7*b^2 + 12*a^6*b^3 - 9*a^5*b^4 - 6*a^4*b^5 + 13*a^3*b^6 - 4*a^2*b^7 - 5*a*b^8 + 3*b^9 - 5*a^8*c + 20*a^7*b*c - 19*a^6*b^2*c - 19*a^5*b^3*c + 46*a^4*b^4*c - 22*a^3*b^5*c - 15*a^2*b^6*c + 21*a*b^7*c - 7*b^8*c - a^7*c^2 - 19*a^6*b*c^2 + 56*a^5*b^2*c^2 - 36*a^4*b^3*c^2 - 34*a^3*b^4*c^2 + 57*a^2*b^5*c^2 - 21*a*b^6*c^2 - 2*b^7*c^2 + 12*a^6*c^3 - 19*a^5*b*c^3 - 36*a^4*b^2*c^3 + 84*a^3*b^3*c^3 - 38*a^2*b^4*c^3 - 21*a*b^5*c^3 + 18*b^6*c^3 - 9*a^5*c^4 + 46*a^4*b*c^4 - 34*a^3*b^2*c^4 - 38*a^2*b^3*c^4 + 52*a*b^4*c^4 - 12*b^5*c^4 - 6*a^4*c^5 - 22*a^3*b*c^5 + 57*a^2*b^2*c^5 - 21*a*b^3*c^5 - 12*b^4*c^5 + 13*a^3*c^6 - 15*a^2*b*c^6 - 21*a*b^2*c^6 + 18*b^3*c^6 - 4*a^2*c^7 + 21*a*b*c^7 - 2*b^2*c^7 - 5*a*c^8 - 7*b*c^8 + 3*c^9) : :
= lies on this line: {1, 23279}.

Best regards,
Peter Moses.

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