HYACINTHOS 25395

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of I.

Denote:

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

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

A*B*C* = the orthic thriangle of A'B'C'

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

2nd point of intersection?

 

[César Lozada]:

Z2 = Inverse of X(104) in the incircle

= q*p^3*(4*p^2-4*p*q+4*q^2-3)+( 4*p^4-4*p^2+p*q+1)*(1-q^2) : : (trilinears), where p=sin(A/2), q=cos((B-C)/2)

= (b+c)*a^8-2*(b^2+c^2)*a^7-(2* b-c)*(b-2*c)*(b+c)*a^6+2*(3*b^ 2+4*b*c+3*c^2)*(b-c)^2*a^5-7*( b^2-c^2)*(b-c)*b*c*a^4-2*(3*b^ 4+3*c^4+2*b*c*(b^2+c^2))*(b-c) ^2*a^3+(b^2-c^2)*(b-c)*(2*b^4+ 2*c^4+3*b*c*(b-c)^2)*a^2+2*(b^ 2-c^2)^2*(b^4+c^4-2*b*c*(b-c)^ 2)*a-(b^2-c^2)^2*(b-c)^2*(b^3+ c^3) : : (trilinears)

= On lines: {1,104}, {7,151}, {56,11713}, {57,102}, {65,1359}, {117,226}, {124,1210}, {354,1361}, {518,3040}, {928,11028}, {942,2818}, {974,2779}, {1845,5902}, {2807,3664}, {3042,3812}, {3340,10696}, {3586,10732}, {3738,10015}, {3911,6711}, {4654,10709}, {5722,10747}, {9579,10726}

= midpoint of X(65) and X(1364)

= reflection of X(3042) in X(3812)

= incircle-inverse-of-X(104)

= [ 1.864597096611183, 1.90808711222661, 1.459097821160796 ]

 

César Lozada