HYACINTHOS 25112

[Antreas P. Hatzipolakis]:

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

Denote:

Oa, Ob, Oc = the circumcenters of PB'C', PC'A', PA'B', resp.

Ab, Ac = the orthogonal projections of Oa on POb, POc, resp.

Bc, Ba = the orthogonal projections of Ob on POc, POa, resp.

Ca, Cb = the orthogonal pojections of Oc on POa, POb, resp.

La, Lb, Lc = the Euler lines of OaAbAc, ObBcBa, OcCaCb, resp.

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

The triangles ABC, A*B*C* are parallelogic.

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

[Peter Moses]:

Hi Antreas,

 

If P = {p,q,r}, then the parallelogic center (ABC, A*B*C*) lies on the circumcircle of ABC at
1/(b^2 (a^2-b^2) c^2 p^2 q+c^2 (a^2-c^2) (a^2-b^2+c^2) p q^2+b^2 c^2 (a^2-c^2) p^2 r+c^2 (a^4-a^2 b^2-b^4+2 b^2 c^2-c^4) q^2 r+b^2 (a^2-b^2) (a^2+b^2-c^2) p r^2+b^2 (a^4-b^4-a^2 c^2+2 b^2 c^2-c^4) q r^2)::

Examples:

Points on K003 –> X(74).

P=X(2) -> X(6325).

P=X(13) -> {{110,5463},{111,5916},{112,91 12},{530,691},{542,9202},...}.

P=X(14) –> {{110,5464},{111,5917},{112,91 13},{531,691},{542,9203},...}.

 

P = X(6) :

 

 

a^2/(2 a^6 b^2-2 a^4 b^4+2 a^6 c^2+2 a^4 b^2 c^2-a^2 b^4 c^2-2 b^6 c^2-2 a^4 c^4-a^2 b^2 c^4+4 b^4 c^4-2 b^2 c^6)::

on lines {{476,8705},{526,6325},{5663,6 236}}

 

It’s isogonal at infinity is on lines {{23,183},[30,511},{141,5112}, {468,10163},{476,6325},{477,62 36},{566,858},{1316,6322},{515 9,10173},{5189,7774},{6232,679 5},{7426,9829},{10989,11163}}.

 

Best regards,

Peter Moses.

 

[César Lozada]:

La, Lb, Lc concur for P on the Euler line of ABC or P in its sidelines.

 

For P=u:v:w (trilinears)

 

Z(P) =OC(ABC, A*B*C*) =

a/(a*u*(a^2-b^2)*(b*u*v*a+2*w^ 2*SC)+a*u*(a^2-c^2)*(c*u*w*a+2 *v^2*SB)-b*(S^2+SB^2-4*SB*SC)* v*w^2-c*(S^2+SC^2-4*SB*SC)*v^2 *w) : :

 

ETC pairs  (P,Za(P)): (1,74),(2,6325), (3,74), (4,74), (5, {110,10203}/\{933,6143}), (6, {476,8705}/\{526,6325} ), (523,476), (733,1291) , (1113,1113), (3864,5966), (7250,721)

 

César Lozada