HYACINTHOS 28437

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle and P a point.

 

Denote:

 

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

 

D = the Poncelet point of ABCP

 

La, Lb, Lc  =; the reflections of DP in AP, BP, CP, resp.

 

Which is the locus of P such that the parallels to La, Lb, Lc through A, B, C, resp. are concurrent?

 

I, N lie on the locus.

 

APH

 

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[Ercole Suppa]

 

  

*** locus of point P such that the parallels to La, Lb, Lc through A, B, C, resp. are concurrent is

     {Linf} U {q5=excentral-circumquintic = Q038 (of B. Gibert catalogue)} 

 

q5: ∑ [2 b^2 (b-c) c^2 (b+c) (a^2-b^2-c^2) x^3 y z+(a-b) (a+b) c^2 (a^2+b^2-c^2)^2 x^2 y^2 z+a^2 (a^2-b^2+c^2) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) y^3 z^2-a^2 (a^2+b^2-c^2) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) y^2 z^3] = 0

 

q5: pass through points X(i) for these i: {1, 4, 5, 80, 1113, 1114, 1263, 2009, 2010}

 

 

*** pairs (P ∈ q5, Q(P)) : {1,953},{80,4},{1113,3},{1114,3},{1263,143}

 

 

*** some points Q=Q(P) :

 

-- Q(X(4)) = (0:0:0) undefined

 

-- P1=Q(X(5)) = X(4)X(14980) ∩ X(1154)X(14072)

 

= -a^2 (a^4+b^4-b^2 c^2+c^4-2 a^2 (b^2+c^2)) (-(b^2-c^2)^2+a^2 (b^2+c^2)) (a^6-2 a^5 c+2 a c^3 (b^2-c^2)-a^4 (b^2+c^2)+(b^2-c^2)^2 (b^2+c^2)+2 a^3 c (b^2+2 c^2)-a^2 (b^4+b^2 c^2+c^4)) (a^6-2 a^5 b-a^4 (b^2+c^2)+(b^2-c^2)^2 (b^2+c^2)+2 a^3 b (2 b^2+c^2)-2 a (b^5-b^3 c^2)-a^2 (b^4+b^2 c^2+c^4)) (a^6+2 a^5 b-a^4 (b^2+c^2)+(b^2-c^2)^2 (b^2+c^2)-2 a^3 b (2 b^2+c^2)+2 a (b^5-b^3 c^2)-a^2 (b^4+b^2 c^2+c^4)) (a^6+2 a^5 c-a^4 (b^2+c^2)+(b^2-c^2)^2 (b^2+c^2)-2 a^3 c (b^2+2 c^2)-a^2 (b^4+b^2 c^2+c^4)+a (-2 b^2 c^3+2 c^5)) : : (barys)

 

= lies on these lines : {4,14980},{1154,14072},{5663,15907}

 

= (6-8-13) search numbers [1.96202118971729647, 6.27098399161003823, -1.60633421523055343]

 

 

Best regards

Ercole Suppa