HYACINTHOS 25259

[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 AA', BB', CC', resp.

For P = I, the circumcircles of IMaA, IMbB, IMcC are coaxial. Second point of intersecton?

Which, other than I, points on the Neuberg cubic (so that AA', BB', CC' be concurrent) have that property?

[César Lozada]:

The locus for coaxial circles is: {sidelines of ABC} \/ {altitudes of ABC} \/ { the excentral-circum-septic q7 with barycentric equation:

CyclicSum[ y*z*((-(a^2+b^2-c^2)*c^4*y^5+( a^2-b^2+c^2)*b^4*z^5)*a^4+(-a^ 2+b^2+c^2)*(-(3*a^2-b^2+c^2)* c^2*y^3+(3*a^2+b^2-c^2)*b^2*z^ 3)*a^4*y*z-(b^2-c^2)*(3*a^4-6* a^2*b^2-6*a^2*c^2+3*b^4+2*b^2* c^2+3*c^4)*b^2*c^2*x^5-((2*a^ 6-6*a^4*b^2+6*a^2*b^4+5*a^2*b^ 2*c^2-6*a^2*c^4-2*b^6-3*b^4*c^ 2+b^2*c^4+4*c^6)*y+(-2*a^6+6* a^4*c^2+6*a^2*b^4-5*a^2*b^2*c^ 2-6*a^2*c^4-4*b^6-b^4*c^2+3*b^ 2*c^4+2*c^6)*z)*a^4*y^2*z^2+(- (a^8-4*a^6*c^2-a^4*b^2*c^2+6* a^4*c^4-4*a^2*b^6+7*a^2*b^4*c^ 2+a^2*b^2*c^4-4*a^2*c^6+3*b^8- 4*b^6*c^2+c^8)*c^2*y+(a^8-4*a^ 6*b^2+6*a^4*b^4-a^4*b^2*c^2-4* a^2*b^6+a^2*b^4*c^2+7*a^2*b^2* c^4-4*a^2*c^6+b^8-4*b^2*c^6+3* c^8)*b^2*z)*x^4+(b^2-c^2)*(a^ 6-3*a^4*b^2-3*a^4*c^2+3*a^2*b^ 4-2*a^2*b^2*c^2+3*a^2*c^4-b^6+ b^4*c^2+b^2*c^4-c^6)*a^2*x*y^ 2*z^2+(b^2-c^2)*(a^8-4*a^6*b^ 2-4*a^6*c^2+6*a^4*b^4-3*a^4*b^ 2*c^2+6*a^4*c^4-4*a^2*b^6+5*a^ 2*b^4*c^2+5*a^2*b^2*c^4-4*a^2* c^6+b^8+2*b^6*c^2-6*b^4*c^4+2* b^2*c^6+c^8)*x^3*y*z) ] = 0 }

 

q7 also pasess through the vertices of the reflection triangle and ETC’s: 1, 3, 4, 399

 

q7 shares with the Neuberg cubic these 13 points: vertices of triangles ABC, excentral and reflection and centers 1, 3, 4, 399. By numerical calculus I found exactly these 13 real intersections (plus 4 complex solutions) .

 

2^nd points of intersection Z(P):

Z(I) = X(5620)

 

Z(O) = X(5) = N

 

Z(X(399)) = reflection of X(550) in X(477)

= 8*S^4+2*(6*SA^2-2*(9*R^2+SW)* SA+3*R^2*(9*R^2+2*SW)-4*SW^2)* S^2-9*(27*R^2-8*SW)*(SB+SC)*R^ 2*SA: : (barycentrics)

= (31*cos(2*A)+3*cos(4*A)+55/2)* cos(B-C)+(-8*cos(A)+2*cos(3*A) )*cos(2*(B-C))+(-cos(2*A)+1)* cos(3*(B-C))-14*cos(3*A)-40* cos(A) : : (trilinears)

= 3*X(5)-4*X(3258) = 2*X(476)-3*X(549)

= On lines: {5,3258}, {30,146}, {476,549}, {477,550}, {3470,3627}

= reflection of X(550) in X(477)

= [ 14.170635474211770, 11.39095691812081, -10.785675911043150 ]

 

César Lozada