HYACINTHOS 26145

[Antreas P. Hatzipolakis]:

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

Denote:

(Oa) = the reflection of the circle (P, PA') in BC
(Ob) = the reflection of the circle (P, PB') in CA
(Oc) = the reflection of the circle (P, PC') in AB.

T = the radical center of (Oa), (Ob), (Oc)

Which is the locus of P such that T lies on the Euler line?
O, H lie on the locus.

 

Note:
T is the point on PP* such that P*T / P*P = 1/4, where P* = the isogonal conjugate of P.

 

[César Lozada]:

For P=u:v:w (trilinears) , T(P)=4*a*u*v*w+u^2*(b*w+c*v)+ 3*v*w*(b*v+c*w) : :

T(P)=P for P in the infinity

T(P)=isogonal(P), for P on the circumcircle

otherwise

> T is the point on PP* such that P*T / P*P = 1/4, where P* = the isogonal conjugate of P

(confirmed)

 

ETC-pairs (P,T(P)): (1,1), (3,546), (4,140), (6,3589), (15,11542), (16,11543), (80,3911), (3062,9)

 

 

> Which is the locus of P such that T lies on the Euler line?

This circum-cubic:

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

 

through ETCs X(3), X(4), X(30), X(74), X(10620) and vertices of the antipedal triangle of X(110).

 

T( X(10620)) = 8*S^4+(918*R^4-360*R^2*SW+12* SA^2-12*SW*SA+32*SW^2)*S^2+( 63*R^2-16*SW)*(27*R^2-4*SW)*( SA-SW)*SA : : (barycentrics)

= 3*X(548)-2*X(7471)

= on line {2,3}

= [ 28.810281409421600, 27.86475923671581, -28.947375640167310 ]

 

 

César Lozada