HYACINTHOS 27760

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

Na,Nb,Nc = the NPC centers of NBC, NCA, NAB, resp.

A'B'C' = the pedal triangle of N wrt triangle NaNbNc

 

The reflections of NA', NB', NC' in BC,CA, AB, resp. bound triangle A*B*C*

ABC, A*B*C* are parallelogic.

Parallelogic centers?

[Peter Moses]:

Hi Antreas,

(ABC, A*B*C*):

(a^4-a^2 b^2+b^4-2 a^2 c^2-2 b^2 c^2+c^4) (a^4-2 a^2 b^2+b^4-a^2 c^2-2 b^2 c^2+c^4) (-a^8+2 a^6 b^2-2 a^2 b^6+b^8+2 a^6 c^2-a^4 b^2 c^2+a^2 b^4 c^2-2 b^6 c^2+a^2 b^2 c^4+2 b^4 c^4-2 a^2 c^6-2 b^2 c^6+c^8)::

5 X[1656] - 4 X[10615], 3 X[381] - 2 X[16337].

= lies on the cubics K025, K465, the sextic Q114 and the lines: {2,6150},{3,3432},{4,93},{5,252},{30,930},{265,14859},{381,16337},{571,2963},{1487,3850},{1510,11583},{1656,10615},{3447,5899},{3627,14141},....

= anticomplement X(6150).

= midpoint of X(3627) and X(14141).

= reflection of X(i) in X(j) for these {i,j}: {{3, 16336}, {1157, 5}, {14980, 18403}}.

= {X(4),X(3519)}-harmonic conjugate of X(18370).

= polar circle inverse of X(6152).

= (anticomplement of circumcirlce) inverse of X(12325).

= antigonal image of X(2070).

= barycentric product X(2070)X(11140).

= barycentric quotient X(i)/X(j) for these {i,j}: {{93, 9381}, {2070, 1994}, {9380, 49}}.

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(A*B*C*, ABC):

a^16-5 a^14 b^2+10 a^12 b^4-10 a^10 b^6+5 a^8 b^8-a^6 b^10-5 a^14 c^2+14 a^12 b^2 c^2-10 a^10 b^4 c^2-5 a^8 b^6 c^2+11 a^6 b^8 c^2-8 a^4 b^10 c^2+4 a^2 b^12 c^2-b^14 c^2+10 a^12 c^4-10 a^10 b^2 c^4-3 a^8 b^4 c^4-a^6 b^6 c^4+10 a^4 b^8 c^4-12 a^2 b^10 c^4+6 b^12 c^4-10 a^10 c^6-5 a^8 b^2 c^6-a^6 b^4 c^6-4 a^4 b^6 c^6+8 a^2 b^8 c^6-15 b^10 c^6+5 a^8 c^8+11 a^6 b^2 c^8+10 a^4 b^4 c^8+8 a^2 b^6 c^8+20 b^8 c^8-a^6 c^10-8 a^4 b^2 c^10-12 a^2 b^4 c^10-15 b^6 c^10+4 a^2 b^2 c^12+6 b^4 c^12-b^2 c^14::

= lies on the lines: {5,195},{30,11671},{252,15345},{382,15619},{523,2070},{539,16337},{930,6150},{1141,1154},....

= reflection of X(i) in X(j) for these {i,j}: {{930,6150},{14140,12026}}.

Best regards,
Peter Moses.