HYACINTHOS 24747

[Antreas P. Hatzipolakis]:

 

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

Denote:

A", B", C" = the orthogonal projections of A', B', C' on the Euler line, resp.

La, Lb, Lc  = the reflections of BC, CA, AB in A'A", B'B", C'C" resp.

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

1. ABC, A*B*C* are parallelogic.

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

2. Which is the locus of the parallelogic center (A*B*C*, ABC) as P moves on the Euler line?

3. Which is the locus of P such that La, Lb, Lc are concurrent?

O lies on the locus

 

[César Lozada]:

Suppose P=u:v:w (trilinears). Then

1)      Parallelogic center (ABC,A*B*C*) = X(476)

Parallelogic center (A*B*C*,ABC) = Z*(P) =

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

 

Z*(O) = X(3258)

Z*(H) = (-2*cos(A)*cos(B-C)+1)*((2* cos(4*A)-3)*cos(B-C)+(-2*cos( A)-2*cos(3*A))*cos(2*(B-C))- cos(3*A)-cos(5*A)+5*cos(A)) : :

= on line {113,403}

= [ -2.473491099993588, -3.83676727166032, 7.438499254592410 ]

 

Z*(N) = (-cos(2*A)+5*cos(4*A)+cos(6*A) -13/2)*cos(B-C)+(5*cos(A)-cos( 3*A)-cos(5*A))*cos(2*(B-C)) +(cos(2*A)+cos(4*A)+1/2)*cos( 3*(B-C))-2*cos(5*A)+4*cos(A)- 4*cos(3*A) : :

= on lines: {511,3233}, {1154,10096}

= [ 0.741910787392766, -0.61380402264393, 3.723185364941206 ]

 

2)      For P such that OP=t*OH, OZ*(P)=t*X(3258)Z*(H)

 

3)      The line {O,X(526)} (no other ETC center on it), but for P=X(526) the lines La,Lb,Lc coincide.

 

César Lozada