HYACINTHOS 29158

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and P a variable point, Q a fixed point and QaQbQc the pedal triangle of Q.

Denote:

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

Which is the locus of P such that the perpendiculars from Qa, Qb, Qc to NbNc, NcNa, NbNa, resp. are concurrent?

For Q =

1. O
2. H
3. G 
4. N
5. I

 

[César Lozada]:

 

 

Let Q=x:y:z (barys)

 

·         If Q=O then the asked locus of P is the entire plane and the point of concurrence of the given lines is:

Z(P) = Complement-of-the-AntigonalConjugate-of-P

 

ETC-pairs (P,Z(P)) : (1,214), (2,2482), (3,1511), (5,6592), (6,6593), (7,10427), (8,1145), (9,6594), (13, 619), (14,618), (20,3184), (23,187), (54,11597), (64,11598), (66,15116), (67,141), (69,5181), (76,5976), (80,10), (83,8290), (186,12095), (265,5), (316,858), (621,33498), (622,33500), (671,2), (895,6), (1138,31378), (1156,9), (1177,206), (1263,140), (1320,1), (1325,36), (1916,39), (2070,6150), (2071,12096) , (3065,3647), (3254,142), (3407,10291), (3484,6760), (3557,2029), (3558,2028) and more.

 

 

·         If Q<>O the locus of P for such concurrence is :

LocusP = {Linf} ∪ {circumcircle} ∪ { K(Q): a rectangular circum-hyperbola described below} 

∪ { q4=∑ [y*z*(a^4*y*z+(a^4-2*(b^2+c^2)*a^2+b^4+c^4)*x^2)] =0, a circum-quartic through X(4)}

 

K(Q) has

Center: Oq = with center Oq = (SB*b^2*z-SC*c^2*y)*((b^2-c^2)*x+(y-z)*a^2 : :

Perspector: Tq = a^2*(c^2*SC*y-b^2*SB*z) : :

 

Oq lies on the NPC circle of ABC.

 

If Q lies on a line through O and having  tripole Q*, then

Oq = Inverse-in-POLAR CIRCLE-of-the-PolarConjugate-of-the-Isotomic-of-Q*

= Complement(Isotomic(Anticomplement(Isogonal(Q*))))

= Complement(Isotomic(Reflection(G, Reflection(Isogonal(Q*), G))))

= Complement(Isotomic(Reflection(H, Reflection(Isogonal(Q*), N)))), et als.

 

Tq = isogonal-conjugate-of-Q*.

 

Also, the given lines are concurrent for Q in the infinity.

 

Which is the locus of P such that the perpendiculars from Qa, Qb, Qc to NbNc, NcNa, NbNa, resp. are concurrent?

For Q =

1. O: The entire plane
2. H : Jerabek hyperbola
3. G : Jerabek hyperbola
4. N : Jerabek hyperbola
5. I : Feuerbach hyperbola

Some more and related centers:

6. K=X(6) : Kiepert hyperbola

7: Ge=X(7) : Conic {A, B, C, X(4), X(55)}, center X(15607), perspector T7 given below. 

8: Ng=X(8) : Conic {A, B, C, X(4), X(56)}, center X(3259), perspector X(3310). 

 

T7 = ORTHIC AXIS INTERSECT OF X(657)X(663)

= a^2*(b-c) *(-a+b+c)^2*((b+c)*a^2+2*b*c*a-(b^2-c^2)*(b-c))  : : (barys)

= lies on these lines: {230, 231}, {657, 663}, {665, 4017}, {905, 31605}, {4130, 4990}, {4171, 6608}, {25084, 27417}

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (657, 17412, 3063), (3709, 10581, 657), (6129, 6586, 647)

= [ -13.9679619824305000, 15.6274977990710400, -0.7316207717122605 ]

 

Z( X(15) ) = COMPLEMENT OF X(11600)

= (SB+SC)*(SA+sqrt(3)*S)*(sqrt(3)* (SB+SC)+2*S)*(3*SA^2-2*sqrt(3)*S*SA+S^2) : : (barys)

= lies on the cubic K900 and these lines: {2, 11600}, {15, 2380}, {16, 323}, {140, 619}, {532, 18803}, {618, 3479}, {624, 33496}, {630, 31376}, {3643, 22649}, {6107, 6669}, {6671, 15609}

= midpoint of X(15) and X(10409)

= reflection of X(15609) in X(6671)

= complement of X(11600)

= [ 2.6097172278494070, 0.8452888967171004, 1.8509796020958900 ]

 

Z( X(16) ) = COMPLEMENT OF X(11601)

= (SB+SC)*(SA-sqrt(3)*S)*(sqrt(3)* (SB+SC)-2*S)*(3*SA^2+2*sqrt(3)*S*SA+S^2) : : (barys)

= lies on the cubic K900 and these lines: {2, 11601}, {15, 323}, {16, 2381}, {140, 618}, {533, 18804}, {619, 3480}, {623, 33497}, {629, 31376}, {3642, 22648}, {6106, 6670}, {6672, 15610}, {8456, 11145}

= midpoint of X(16) and X(10410)

= reflection of X(15610) in X(6672)

= complement of X(11601)

= [ 3.6960425369520690, 2.7649249154691660, 0.0206199079893870 ]

 

César Lozada