HYACINTHOS 28607

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and P, Q two isogonal conjugate points.

Denote:

 

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

 

QaQbQc = the pedal triangle of Q.

 

Which is the locus of P such that:

 

1. the reflections of PNa, PNb, PNc in NbNc, NcNa, NaNb, resp. are concurrent?

 

2. the parallels to PNa, PNb, PNc through Qa, Qb, Qc, resp. are concurrent?

 

 

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[Ercole Suppa]

 

1. The locus is very complicated (Mathematica gives no answer!)

 

2. The locus of point P such that the parallels to PNa, PNb, PNc through Qa, Qb, Qc, resp. are concurrent:

 

Γ= {Linf} U {sidelines of ABC} U {c = circumcircle of ABC} U {K005 Napoleon - Feuerbach cubic} 

 

-- K005:  ∑ a^2 (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) y^2 z-a^2 (a^4-a^2 b^2-2 a^2 c^2-b^2 c^2+c^4) y z^2 = 0   (barys)

 

-- K005 passes through X(i) for these i: 1,3,4,5,17,18,54,61,62,195,627,628,2120,2121,3336,3459,3460,3461,3462,3463,3467,3468,3469,3470,3471,3489,3490,6191,6192,7344,7345,8837,8839,8918,8919,8929,8930 

 

-- Let Q = be the point of concurrency. Pairs {P=X(i) ∈ K005, Q=X(j)} : {{1, 1317}, {3, 4}, {4,30}, {54, 10272}}

 

-- some points:

 

 

Q(X(5)) =  X(5)X(49) X(128)X(539)

 

= (a^4+b^4-b^2 c^2-a^2 (2 b^2+c^2)) (a^4-b^2 c^2+c^4-a^2 (b^2+2 c^2)) (2 a^8+(b^2-c^2)^4-4 a^6 (b^2+c^2)-2 a^2 (b^2-c^2)^2 (b^2+c^2)+3 a^4 (b^4+c^4)) (a^6-3 a^4 (b^2+c^2)-(b^2-c^2)^2 (b^2+c^2)+a^2 (3 b^4-b^2 c^2+3 c^4)) : : (barys)

 

= X[137]-2*X[12242], 3*X[195]+X[13512], X[930]+X[15801], X[6343]+X[20424]

 

= lies on these lines: {5,49}, {128,539}, {137,12242}, {195,13512}, {930,15801}, {1154,14071}, {1157,6592}, {1493,25150}, {5965,12060}, {6343,20424}, {14072,25044}, {14857,18400}, {15959,19468}

 

= reflection of X(137) in X(12242)

 

= (6-8-13) search numbers [-8.99419970726164326, 15.3643624789451281, -2.84503275401072039]

 

 

Best regards

Ercole Suppa