HYACINTHOS 25125

[Tran Quang Hung]:

Let ABC be a triangle with incenter I.

A'B'C' is pedal triangle of I.

A1 is reflection of I though B'C'.
A2 is reflection of A1 through BC.

A3 is reflection of I though BC.
A4 is reflection of A3 through B'C'.

Line A2A4 is da. Similarly we have line db,dc.

Then lines da,db,dc are concurrent.

Point of concurrence?

[César Lozada]:

 

 

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

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

= (2*R+5*r)*X(7)-5*r*X(631)

= On lines: {7,631}, {72,527}, {144,5177}, {390,11278}, {516,10950}, {1770,5851}, {3555,5856}, {4312,5220}, {5728,5762}, {5729,5735}, {5791,6172}

= [ -1.276582869521125, -0.69680160914437, 4.712257689555618 ]

 

César Lozada