HYACINTHOS 25420

[Antreas P. Hatzipolakis]:

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

Denote:

AaAbAc, BaBbBc, CaCbCc = the orthic triangles of AB'C', BC'A', CA'B', resp.
[Other definition of the points:

Denote: Aa = the orthogonal projection of A on B'C', Ba = the orthogonal projection of A' on AB, Ca = the orthogonal projection of A' on AC and similarly the others]

Oa, Ob, Oc = the circumcenters of AaBaCa, AbBbCb, AcBcCc, resp.

ABC, OaObOc are orthologic.

Conjecture:

Denote:

Pa, Pb, Pc = same points on the Euler lines of AaBaCa, AbBbCb, AcBcCc, resp.

ABC, PaPbPc are orthologic.

If true, which are the loci of the orthologic centers as Pa,Pb,Pc move on the Euler lines of the respective triangles being same?

 

[César Lozada]:

 

 

> ABC, OaObOc are orthologic

 

Orthologic centers: X(6145) and X(6146)

 

> Conjecture:

 

Proved. If t =OaPa/OaHa=ObPb/ObHb=OcPc/ OcHc, orthologic centers are:

 

Za = (ABC->PaPbPc) =

= 1/(4*(a^4-(b^2+c^2)^2+b^2*c^2) *S^2*t+3*a^8-6*(b^2+c^2)*a^6+ 3*b^2*c^2*a^4+6*(b^6+c^6)*a^2- 3*(b^6-c^6)*(b^2-c^2)) : : (barycentrics)

= 1/((t-3)*(2*cos(2*A)+1)*cos(B- C)+t*(4*cos(A)-cos(3*A))+3* cos(3*A)) : :  (trilinears)

= on Jerabek hyperbola = isogonal(Euler line)

= isogonal conjugate of Z* such that OZ*/OH = (t-3)*R^2/((3*R^2-2*SW)*t-3*( 7*R^2-2*SW))

 

 

Zp = (PaPbPc->ABC) =

= -4*S^2*(2*a^6-(b^2+c^2)*a^4-( b^4-c^4)*(b^2-c^2))*t+3*(-a^2+ b^2+c^2)*(2*a^8-3*(b^2+c^2)*a^ 6+(b^2-c^2)^2*a^4-(b^4-c^4)*( b^2-c^2)*a^2+(b^2-c^2)^4) : : (barycentrics)

= (-cos(A)*(-2+cos(2*(B-C)))+(- 1+cos(2*A))*cos(B-C)-cos(3*A)) *t+3*cos(A)*cos(2*(B-C))+3* cos(3*A)-3*cos(B-C)*(1+cos(2* A)) : : (trilinears)

= on line {4,6}

= Zp such that HZp/HK = -(t-3)*SW/(6*R^2)   (K=X(6))

 

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For P=H

 

Za(H) = X(265)

 

Zp(H) = X(4)X(6)/\X(5)X(49)

= 2*a^10-5*(b^2+c^2)*a^8+2*(2*b^ 4+3*b^2*c^2+2*c^4)*a^6-2*(b^4- c^4)*(b^2-c^2)*a^4+2*(b^2-c^2) ^2*(b^4-b^2*c^2+c^4)*a^2-(b^4- c^4)*(b^2-c^2)^3 : : (barycentrics)

= (cos(2*A)+2)*cos(B-C)-cos(A)*( cos(2*(B-C))+1)-cos(3*A) : : (trilinears)

= (3*R^2-SW)*X(4)+SW*X(6) = 2*(5*R^2-SW)*X(5)-(7*R^2-2*SW) *X(49)

= On lines: {3,3580}, {4,6}, {5,49}, {30,568}, {51,7576}, {68,7503}, {125,11430}, {141,7550}, {184,403}, {185,1986}, {235,1614}, {378,1899}, {381,11402}, {389,6240}, {436,6761}, {468,11464}, {511,11660}, {539,5891}, {546,11423}, {550,3581}, {569,9927}, {578,1594}, {1593,11457}, {1885,6241}, {1994,3153}, {3448,7527}, {3542,9707}, {3564,11459}, {3567,3575}, {3628,11704}, {5446,11750}, {5562,5965}, {5876,11264}, {6193,6816}, {6756,9781}, {7507,11426}, {9545,9820}, {9818,11442}, {9833,10594}, {10018,10182}, {10127,11451}, {10282,10619}, {10295,11438}, {10297,11422}

= reflection of X(i) in X(j) for these (i,j): (5890,11245), (7576,51)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (4,6776,11456), (265,567,5)

= [ 0.103591190306201, 0.41990272012237, 3.302151280142953 ]

 

-----------------------------

For P=N

 

Za(N) = 1/(5*a^8-10*(b^2+c^2)*a^6+9*b^ 2*c^2*a^4+2*(b^2+c^2)*(5*b^4- 7*b^2*c^2+5*c^4)*a^2-5*(b^6-c^ 6)*(b^2-c^2)) : : (barycentrics)

= 1/((10*cos(2*A)+5)*cos(B-C)-4* cos(A)-5*cos(3*A)) : : (trilinears)

= on Jerabek hyperbola

= isogonal conjugate of {2,3}/\{323,11202}

= [ 1.339630572876544, 2.38992529506769, 1.367809782456033 ]

 

Zp(N) = 10*a^10-25*(b^2+c^2)*a^8+4*(5* b^4+6*b^2*c^2+5*c^4)*a^6-10*( b^4-c^4)*(b^2-c^2)*a^4+2*(b^2- c^2)^2*(5*b^4-2*b^2*c^2+5*c^4) *a^2-5*(b^4-c^4)*(b^2-c^2)^3 : : (barycentrics)

= (cos(2*A)+2)*cos(B-C)-cos(A)*( cos(2*(B-C))+1)-cos(3*A) : : (trilinears)

= (12*R^2-5*SW)*X(4)+5*SW*X(6)

= On lines: {4,6}, {30,11225}, {1899,11410}, {3628,5972}

= [ 1.548643840614386, 2.15766946727431, 1.432134616587836 ]

 

César Lozada