HYACINTHOS 25422

[Antreas P. Hatzipolakis]:

• Let ABC be a triangle.

  Denote:

  A', B', C' = the reflections of I in BC, CA, AB, resp.

  Ma, Mb, Mc = the midpoints of AA', BB', CC, resp.

  M1, M2, M3 = the midpoints of AI, BI, CI, resp.

  The circumcircles of the trianges IMaM1, IMbM2, IMcM3 are coaxial.
  
  2nd intersection?
  
  

  Generalization:

  Let ABC be a triangle.

  Denote:

  A', B', C' = the reflections of I in BC, CA, AB, resp.

  A*, B*, C* = points on AI, BI, CI, resp. such that:

  A*A/A*I = B*B/B*I = C*C/C*I = t

  The circumcircles of IA*Ma, IB*Mb, IC*Mc are coaxial.

  Locus of the 2nd intersection as t varies?

   

  [César Lozada]:
  
  
  

  1)

  Q2 = a*(a^4-2*b*a^3+(b^2+2*b*c-2*c^ 2)*a^2+2*b*(b-c)^2*a-(b^2-c^2) *(2*b^2-2*b*c+c^2))*(2*a^4-2*( b+c)*a^3-(b^2-4*b*c+c^2)*a^2+ 2*(b^2-c^2)*(b-c)*a-(b^2-c^2)^ 2)*(a^4-2*c*a^3-(2*b^2-2*b*c- c^2)*a^2+2*c*(b-c)^2*a+(b^2-c^ 2)*(b^2-2*b*c+2*c^2))*S-2*(a+ b+c)*(a^3-(b+c)*a^2-(b^2-3*b* c+c^2)*a+(b^2-c^2)*(b-c))*(a^ 6-3*(b+c)*a^5-(b^2-11*b*c+c^2) *a^4+2*(b+c)*(3*b^2-7*b*c+3*c^ 2)*a^3-(b^2+10*b*c+c^2)*(b-c)^ 2*a^2-(b^2-c^2)*(b-c)*(3*b^2- 8*b*c+3*c^2)*a+(b^2-3*b*c+c^2) *(b^2-c^2)^2)*(2*a^4-2*(b+c)* a^3-(b^2-4*b*c+c^2)*a^2+2*(b^ 2-c^2)*(b-c)*a-(b^2-c^2)^2) : : (barycentrics)

  = on line {1,5}

  = [ -0.254694503731151, -1.74405697616215, 4.965639851741781 ]

   

  2) The 2^nd point of intersection has barycentric coordinates:

  Q2 = F(t, a, b, c)* G(t, a, b, c) : :

   

  Where

  F(t, a, b, c) = 2*a^4-(3*t-1)*(b+c)*a^3+(-b^2+ 6*t*b*c-2*b*c-c^2)*a^2+(b^2-c^ 2)*(b-c)*(3*t-1)*a-(b^2-c^2)^2

   

  G(t, a, b, c) =

  (a^4-(3*t-1)*c*a^3+(-2*b^2+3* t*b*c-b*c+c^2)*a^2+c*(b-c)^2*( 3*t-1)*a+(b^2-c^2)*(b^2-3*t*b* c+b*c+2*c^2))*(a^4-(3*t-1)*b* a^3+(b^2+3*t*b*c-b*c-2*c^2)*a^ 2+b*(b-c)^2*(3*t-1)*a+(b^2-c^ 2)*(-2*b^2+3*t*b*c-b*c-c^2))*( (1+2*t)*a-(t-1)*(b+c))*S-6*(a+ b+c)*(a^9-(3*t+1)*(b+c)*a^8+(- 2*b^2+3*t*b^2+9*t^2*b*c+18*t* b*c-7*b*c-2*c^2+3*t*c^2)*a^7+( b+c)*(9*t*b^2+2*b^2-27*t^2*b* c-9*t*b*c+2*c^2+9*t*c^2)*a^6+( -9*t*b^4-18*t*b^3*c+7*b^3*c-9* t^2*b^3*c+99*t^2*b^2*c^2-30*t* b^2*c^2+6*b^2*c^2+7*b*c^3-9*t^ 2*b*c^3-18*t*b*c^3-9*t*c^4)*a^ 5-(b+c)*(9*t*b^4-54*t^2*b^3*c+ 2*b^3*c-15*t*b^3*c+4*b^2*c^2- 12*t*b^2*c^2+126*t^2*b^2*c^2+ 2*b*c^3-15*t*b*c^3-54*t^2*b*c^ 3+9*t*c^4)*a^4+(b-c)^2*(9*t*b^ 4+2*b^4-9*t^2*b^3*c+11*b^3*c+ 14*b^2*c^2+18*t*b^2*c^2-90*t^ 2*b^2*c^2-9*t^2*b*c^3+11*b*c^ 3+9*t*c^4+2*c^4)*a^3+(b^2-c^2) *(b-c)*(3*t*b^4-2*b^4-27*t^2* b^3*c+3*t*b^3*c+72*t^2*b^2*c^ 2+4*b^2*c^2-24*t*b^2*c^2-27*t^ 2*b*c^3+3*t*b*c^3-2*c^4+3*t*c^ 4)*a^2-(b^2-c^2)^2*(3*t*b^4+b^ 4+7*b^3*c-18*t*b^3*c-9*t^2*b^ 3*c+27*t^2*b^2*c^2+6*t*b^2*c^ 2+7*b*c^3-9*t^2*b*c^3-18*t*b* c^3+c^4+3*t*c^4)*a+(b^2-c^2)^ 3*(b-c)*(b^2-3*t*b*c+c^2))

   

  F(t, a, b, c) is a degree-1 polynomial w/r to t and G(t, a, b, c) ) is a degree-3 polynomial w/r to t. No more factorization is possible. What is the locus?. A quartic?.

   

  César Lozada