HYACINTHOS 25543

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the pedal triangle of H (orthic triangle).

Denote:

A* := (The Reflection BC in BB') /\ (Reflection (BC in CC')
B* := (The Reflection CA in CC') /\ (Reflection (CA in AA')
C* := (The Reflection AB in AA') /\ (Reflection (AB in BB')

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

(Note: They are cyclologic for any point P, not just for H)

H* := The cyclologic center (A*B*C*, ABC) (ie the point of concurrence of the circumcircles of A*BC, B*CA, C*AB)
H** = The cyclologic center (ABC, A*B*C*) (ie the point of concurrence of the circumcircles of AB*C*, BC*A*, CA*B*)

Oa, Ob, Oc := The circumcenters of A*BC, B*CA, C*AB, resp.

Note:

The triangles ABC, OaObOc are circumcyclologic triangles.

(ie the circumcircles of OaBC, ObCA, OcAB, OaObOc are concurrent and the circumcircles of AObOc, BOcOa, COaOb, ABC are concurrent)

La := The Reflection of H*Oa in B'C'
Lb := The Reflection of H*Ob in C'A'
Lc := The Reflection of H*Oc in A'B'

1. Which is the point H* ?

2. Which point is the point H**?

3. La, Lb, Lc are concurrent. Point of concurrence?
4. The parallels to La, Lb, Lc through A, B, C, resp. are concurrent at a point on the circumcircle. Point?

 

[César Lozada]:

 

1)      H* = X(265)

2)      H**=X(10152)

3)      Z3 = cos(A)*((80*cos(A)+30*cos(3*A) +2*cos(5*A))*cos(B-C)+(-24* cos(2*A)-2*cos(4*A)-23)*cos(2* (B-C))+10*cos(A)*cos(3*(B-C))- cos(4*(B-C))-8*cos(4*A)-36* cos(2*A)-27) : : (trilinears)

= On lines: {3,476}, {30,1986}, {131,2072}, {523,7723}, {1368,11749}

= [ -2.750215411376614, -2.30151713799492, 6.503352690385452 ]

 

4)      X(477)

 

More:

>The triangles ABC, OaObOc are circumcyclologic triangles.

 

Centers:

(Oa->A) = X(5964)

 

(A->Oa) = a/(b^2-c^2)/((b^2+c^2)*a^4-(2* b^4-b^2*c^2+2*c^4)*a^2+( b^4-c^4)*(b^2-c^2)) : : (trilinears)

= on the circumcircle and line {74,11250}

= [ 4.106329136640178, -1.45560299885301, 2.753160802664071 ]

 

 

César Lozada