HYACINTHOS 29206

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and P a point.

Denote:

A'B'C' = the reflection triangle 
(ie A', B', C' = the reflections of A, B, C in BC, CA, AB, resp.)

Pa, Pb, Pc = the isogonal conjugates of P wrt triangles A'BC, AB'C, ABC', resp.

Which is the locus of P such that ABC, PaPbPc are cyclologic?
N lies on the locus.


[César Lozada]:

 

Locus = The entire plane.

 

For P=x:y:z (barys),

Qa = A -> Pa = (-(-a^2+b^2+c^2)*x^2+(a^2-b^2)*x*y+(a^2-c^2)*x*z+a^2*y*z)*(((a^2-b^2+c^2)^2-c^2*a^2)*y^2+((b^2-c^2)^2-a^2*b^2)*y*z+((b^2-a^2)^2-b^2*c^2)*y*x-b^2*(a^2-b^2+c^2)*z*x)*(((a^2+b^2-c^2)^2-a^2*b^2)*z^2+((c^2-a^2)^2-b^2*c^2)*z*x+((-b^2+c^2)^2-c^2*a^2)*z*y-c^2*(a^2+b^2-c^2)*x*y)

 

Sorry, could not find the general expression for Qp = (Pa->A)

 

If P lies on the circumcircle of ABC then Qa(P) = X(265)

Qa(P) = Isogonal(CircumcircleInverse(AntigonalConjugate(P)))

 

Some pairs (P, Qa(P)): (3, 5962), (5, 11584), (6, 316), (13, 11601), (14, 11600), (30, 4), (54, 19552), (65, 5080), (66, 5523), (67, 671), (68, 16172), (69, 5203), (79, 19658), (80, 80)

 

Some others:

 

Qa( X(1) ) = REFLECTION OF X(36) IN X(13141)

= (a^2-b^2+b*c-c^2)*(a^4-c*a^3-(2*b^2-c^2)*a^2+(b^2-c^2)*c*a+(b^2-c^2)^2)*(a^4-b*a^3+(b^2-2*c^2)*a^2-(b^2-c^2)*b*a+(b^2-c^2)^2) : : (barys)

= lies on the cubic K529 and these lines: {1, 15529}, {36, 13141}, {355, 1836}, {14008, 33140}

= reflection of X(36) in X(13141)

= antigonal conjugate of X(36)

= [ -0.7305456255537472, 4.2965173312319450, 1.0033273874639940 ]

 

Qp( X(1) ) = TRILINEAR POLE OF THE LINE X(6)X(563)

= a^2*(a-b+c)*(a^6-a^5*c+2*a^3*c^3-(2*b^2+c^2)*a^4+(b^4+2*b^2*c^2-c^4)*a^2+(b^4-c^4)*a*c+(b^2-c^2)^2*c^2)*(a-c)*(a+b-c)*(a^6-a^5*b+2*a^3*b^3-(b^2+2*c^2)*a^4-(b^4-2*b^2*c^2-c^4)*a^2-(b^4-c^4)*a*b+(b^2-c^2)^2*b^2)*(a-b) : :

=  lies on the circumcircle and this line: {925,4575}

= trilinear pole of the line X(6)X(563)

= [ 10.6096975777740200, 2.1702119918036450, -2.7585734713907660 ]

 

Qa(X(2)) = ISOGONAL CONJUGATE OF X(11643)

= (2*a^2-b^2-c^2)*(2*a^4-(6*b^2-c^2)*a^2+2*(b^2-c^2)*(2*b^2-c^2))*(2*a^4+(b^2-6*c^2)*a^2+2*(b^2-c^2)*(b^2-2*c^2)) : : (barys)

= lies on the cubic K300 and this line: {381, 576}

= isogonal conjugate of X(11643)

= antigonal conjugate of X(187)  

= [ -2.4800124099807950, 1.8484220203078930, 3.5056088109162060 ]

 

Qp( X(2) ) : not interesting

 

César Lozada