HYACINTHOS 25394

[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 triangle of A'B'C'

The circumcrcles of AA*Ma, BB*Mb, CC*Mc are coaxial.

If U,W are the points of intersections which is the midpoint of UW
[= intersection of the common radical axis and diacentric line] ?

 

 [César Lozada]:

They are coaxial.

 

Radical trace =

= ((b+c)*a^10-2*(b^2+4*b*c+c^2)* a^9-(b+c)*(3*b^2-14*b*c+3*c^2) *a^8+(8*b^4+8*c^4+b*c*(7*b^2+ 6*b*c+7*c^2))*a^7+(b+c)*(2*b^ 4+2*c^4-b*c*(23*b^2+3*b*c+23* c^2))*a^6-(12*b^6+12*c^6-b*c*( 7*b^2+13*b*c+7*c^2)*(b-c)^2)* a^5+(b+c)*(2*b^6+2*c^6+b*c*( 13*b^2-10*b*c+13*c^2)*(b+c)^2) *a^4+(8*b^6+8*c^6-(3*b^4+3*c^ 4+2*b*c*(17*b^2+25*b*c+17*c^2) )*b*c)*(b-c)^2*a^3-(b+c)*(3*b^ 8+3*c^8-(3*b^6+3*c^6-(15*b^4+ 15*c^4+b*c*(3*b^2-4*b*c+3*c^2) )*b*c)*b*c)*a^2-(b^2-c^2)^2*( 2*b^6+2*c^6-(13*b^4+13*c^4+b* c*(13*b^2-8*b*c+13*c^2))*b*c)* a+(b^3-c^3)*(b^2-c^2)^3*(b^2- 6*b*c+c^2))/(-a+b+c) : :

= on line {7,2475}

= [ 3.446187169266729, 0.33671613533652, 1.817005387012910 ]

 

César Lozada