HYACINTHOS 24880

 

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

A', B', C' = the reflections of P in BC, CA, AB, resp.

Ma, Mb, Mc = the midpoints of AP, BP, CP, resp

La, Lb, Lc = the Euler lines of A'MbMc, B'McMa,C'MaMb, resp.

A*B*C*  = the triangle bounded by La, Lb, Lc

Which is the locus of P such that:

1. La, Lb, Lc are concurrent?

I lies on the locus

2. ABC, A*B*C* are parallelogic?

N lies on the locus.

The parallelogic center (ABC,A*B*C*) lies on the circumcircle.

[César Lozada]:

1)      q7=Degree 7 excentral-circumcurve through ETC’s I, O, H and barycentrics eq:

CyclicSum[ y*z*((-(2*a^8-7*a^6*b^2-4*a^6* c^2+8*a^4*b^4+3*a^4*b^2*c^2-3* a^2*b^6+a^2*b^4*c^2-2*a^2*b^2* c^4+4*a^2*c^6+2*b^6*c^2-6*b^4* c^4+6*b^2*c^6-2*c^8)*c^2*y^3+( 2*a^8-4*a^6*b^2-7*a^6*c^2+3*a^ 4*b^2*c^2+8*a^4*c^4+4*a^2*b^6- 2*a^2*b^4*c^2+a^2*b^2*c^4-3*a^ 2*c^6-2*b^8+6*b^6*c^2-6*b^4*c^ 4+2*b^2*c^6)*b^2*z^3)*a^4*y*z- (b^2-c^2)*(-a^2+b^2+c^2)*(2*a^ 6-6*a^4*b^2-6*a^4*c^2+6*a^2*b^ 4+5*a^2*b^2*c^2+6*a^2*c^4-2*b^ 6+2*b^4*c^2+2*b^2*c^4-2*c^6)* b^2*c^2*x^5-(-(4*a^8+a^6*b^2- 14*a^6*c^2-19*a^4*b^4-10*a^4* b^2*c^2+18*a^4*c^4+18*a^2*b^6- 19*a^2*b^4*c^2+11*a^2*b^2*c^4- 10*a^2*c^6-4*b^8+10*b^6*c^2-6* b^4*c^4-2*b^2*c^6+2*c^8)*c^2* y+(4*a^8-14*a^6*b^2+a^6*c^2+ 18*a^4*b^4-10*a^4*b^2*c^2-19* a^4*c^4-10*a^2*b^6+11*a^2*b^4* c^2-19*a^2*b^2*c^4+18*a^2*c^6+ 2*b^8-2*b^6*c^2-6*b^4*c^4+10* b^2*c^6-4*c^8)*b^2*z)*a^4*y^2* z^2+((4*a^10-16*a^8*b^2-13*a^ 8*c^2+24*a^6*b^4+27*a^6*b^2*c^ 2+14*a^6*c^4-16*a^4*b^6-13*a^ 4*b^4*c^2-14*a^4*b^2*c^4-a^4* c^6+4*a^2*b^8+a^2*b^6*c^2-4*a^ 2*b^4*c^4+7*a^2*b^2*c^6-8*a^2* c^8-2*b^8*c^2+2*b^6*c^4+6*b^4* c^6-10*b^2*c^8+4*c^10)*y+(-4* a^10+13*a^8*b^2+16*a^8*c^2-14* a^6*b^4-27*a^6*b^2*c^2-24*a^6* c^4+a^4*b^6+14*a^4*b^4*c^2+13* a^4*b^2*c^4+16*a^4*c^6+8*a^2* b^8-7*a^2*b^6*c^2+4*a^2*b^4*c^ 4-a^2*b^2*c^6-4*a^2*c^8-4*b^ 10+10*b^8*c^2-6*b^6*c^4-2*b^4* c^6+2*b^2*c^8)*z)*b^2*c^2*x^4- (b^2-c^2)*(2*a^10-10*a^8*b^2- 10*a^8*c^2+20*a^6*b^4+9*a^6*b^ 2*c^2+20*a^6*c^4-20*a^4*b^6+ 22*a^4*b^4*c^2+22*a^4*b^2*c^4- 20*a^4*c^6+10*a^2*b^8-27*a^2* b^6*c^2+28*a^2*b^4*c^4-27*a^2* b^2*c^6+10*a^2*c^8-2*b^10+6*b^ 8*c^2-4*b^6*c^4-4*b^4*c^6+6*b^ 2*c^8-2*c^10)*a^2*x*y^2*z^2+( b^2-c^2)*(11*a^8-31*a^6*b^2- 31*a^6*c^2+25*a^4*b^4+21*a^4* b^2*c^2+25*a^4*c^4-a^2*b^6-2* a^2*b^4*c^2-2*a^2*b^2*c^4-a^2* c^6-4*b^8+16*b^6*c^2-24*b^4*c^ 4+16*b^2*c^6-4*c^8)*b^2*c^2*x^ 3*y*z) ] = 0

 

Points of concurrence Ze(P) of Euler lines:   Ze(O) = O,  Ze(H) = X(30)

Ze( I ) = (5*a^3-2*(b+c)*a^2-(5*b^2-9*b* c+5*c^2)*a+2*(b-c)^2*(b+c))*( 2*a-b-c) : : (barycentrics)

= midpoint of X(i) and X(j) for these {i,j}: {1,10031}, {2,7972}

= reflection of X(4669) in X(3035)
= On lines: {1,10031}, {2,7972}, {214,519}, {528,5542}, {547,551}, {2800,3655}, {2801,3898}, {2802,3241}, {3035,4669}, {3635,10609}

= [ 1.207012039184705, 0.03916848300952, 3.056465360200229 ]

 

2)      Locus = { circumcircle } \/ { Napoleon-Feuerbach cubic } \/ { same q7 than last one }

 

Parallelogic centers Za(P)=(A->ZA*) and Z*(P)=(A*->A):

For P on the circumcircle, Za(P)=reflection of P-in-N, i.e., the locus of Za(P) is the circle {H,R} and Z*(P) describes a deltoid.

 

Some centers for P on Napoleon-Feuerbach cubic:

Za( N ) = X(1141)

Z*(N) = (S^2*(R^2*(51*R^2-37*SW)-3*S^ 2+7*SW^2)-SW*(12*R^2*(R^2-SW)- S^2+3*SW^2)*SA+(R^2*(9*R^2-13* SW)-2*S^2+4*SW^2)*SA^2)*a : : (trilinears)

= On line: {5642,5943}

= [ 2.626040568795400, -0.38532744390082, 2.695410911317836 ]

 

Za( X(54) ) = X(265)

Z*(P) = 3*(R^2*(27*R^2-23*SW)+2*S^2+4* SW^2)*SA^2+(3*R^2*(27*R^2*SW+ 23*S^2-23*SW^2)-12*SW*(S^2-SW^ 2))*SA-(R^2*(126*R^2-89*SW)-4* S^2+14*SW^2)*S^2 : : (baryc.)

= [ -6.395577825254121, 10.07890632294154, -0.385234745396325 ]

 

Regards,

César Lozada