Τετάρτη 23 Οκτωβρίου 2019

HYACINTHOS 25550

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

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

R1 = the radical axis of the circles (Nb, NbA'), (Nc, NcA')
R2 = the radical axis of the circles (Nc, NcB'), (Na, NaB')
R3 = the radical axis of the circles (Na, NaC'), (Nb, NbC')

A*B*C* = the triangle bounded by R1, R2, R3

Which is the locus of P such that:

1. R1, R2, R3 are concurrent?
I lies on the locus
 
2. ABC, A*B*C* are parallelogic?
The entire plane?
If yes, which are the loci of the parallelogic centers if P lies on the Euler line?


[César Lozada]:

1)       

{Infinity} \/ {circumcircle} \/ {McCay cubic} \/ {circum-quartic CyclicSum(y*z*(2*(S^2-SA^2)*x^ 2-(SB+SC)^2*y*z))=0, through H}

ETC pairs (P, point of concurrence): (1, 1317), (3, 1511)

 

2)      The entire plane. For P=u:v:w ( trilinears)

Z(P) = PC(A->A*) = Antigonal-conjugate-of-P

Z*(P) = PC(A*->A) =

(2*a*b*c*(-2*a^2*c^2-4*b^2*c^ 2+2*b^4-2*a^2*b^2+2*c^4+3*a^4) *w^2*v^2*u^2+a*(3*c^6-a^2*b^2* c^2-3*b^2*c^4+b^4*c^2+a^6+a^4* c^2-b^6-5*a^2*c^4)*v*u^2*w^3+ a*(-a^2*b^2*c^2-c^6-3*b^4*c^2- 5*a^2*b^4+3*b^6+b^2*c^4+a^6+a^ 4*b^2)*v^3*u^2*w+(a^2-c^2)*b* u*w*v^4*(a^2-b^2+c^2)^2+c*(-c^ 6+a^4*b^2+a^6+2*a^2*b^4-2*b^6+ 3*b^4*c^2-2*a^2*b^2*c^2)*w^2* v^3*u+(a^2-b^2)*c*u*v*w^4*(a^ 2+b^2-c^2)^2+b*(-b^6-2*c^6+a^ 4*c^2+2*a^2*c^4+a^6+3*b^2*c^4- 2*a^2*b^2*c^2)*w^3*v^2*u+b*a^ 2*(-b^2*c^2+5*b^4+8*c^4+5*a^4- 7*a^2*c^2-10*a^2*b^2)*w*v^2*u^ 3+2*a^3*(a^4-2*a^2*b^2-2*a^2* c^2+b^4+c^4)*w*v*u^4+c*a^2*(- 7*a^2*b^2-b^2*c^2+8*b^4-10*a^ 2*c^2+5*a^4+5*c^4)*w^2*v*u^3+ b*a^2*(-2*a^2*c^2-a^2*b^2+3*b^ 2*c^2+c^4+a^4)*w^3*u^3+c*a^2*( -2*a^2*b^2+a^4-a^2*c^2+b^4+3* b^2*c^2)*v^3*u^3+2*a^3*u^4*b* v^2*c^3+2*a^3*b^3*u^4*c*w^2+a* b*c*(a^2+b^2-c^2)*(a^2-b^2+c^ 2)*v^2*w^4-a*b*c*(a^2+b^2-c^2) *(b^2+c^2-a^2)*u^2*w^4-(b^2-c^ 2)^2*a*v^3*w^3*(b^2+c^2-a^2)+ a*b*c*(a^2+b^2-c^2)*(a^2-b^2+ c^2)*w^2*v^4-a*b*c*(a^2-b^2+c^ 2)*(b^2+c^2-a^2)*u^2*v^4)/a : :

 

ETC pairs (P, Z*(P)): (1, 1317), (3, 1511) (A*=B*=C*, according to 1)

 

Examples:

Z(G) = X(671)


Z*(G)= a*((b^2+c^2)*a^2-b^4-c^4)*((a^ 2+b^2+c^2)^2-9*b^2*c^2) : : (trilinears)

= On lines: {2,2854}, {114,325}, {183,9775}, {526,9185}, {1995,9145}, {2871,7998} , {5640,11163}, {5663,6054}, {5968,9155}, {9770,11002}, {9872,11580}, {10748,11185}

= [ 2.522021817587868, -1.25334028403642, 3.344351531969041 ]

 

Z(K) = X(67)

 

Z*(K) = 2*a^10+3*(b^2+c^2)*a^8-(7*b^4+ 4*b^2*c^2+7*c^4)*a^6-(b^2+c^2) *(3*b^4-11*b^2*c^2+3*c^4)*a^4+ (b^2-c^2)^2*(5*b^4+13*b^2*c^2+ 5*c^4)*a^2-6*(b^4-c^4)*(b^2-c^ 2)*b^2*c^2 : : (barycentrics)

= ((15*cos(A)-16*cos(3*A)+7*cos( 5*A))*cos(B-C)+(-3*cos(2*A)-5* cos(4*A)-4)*cos(2*(B-C))-6* cos(A)*cos(3*(B-C))+6*cos(4*A) -cos(6*A)+18*cos(2*A)-11)*csc( A) : : (trilinears)

= On line {543,3629}

= [ -1.504682679728710, 1.26735321785658, 3.457735029420001 ]

 

César Lozada

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