HYACINTHOS 25401

[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.

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Notice that MaMbMc, A*B*C* are perspectve (at the orthoceter of A'B'C' = pedal triangle of I)

Now I construct another triangle A*B*C* perspective with MaMbMc:

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.

A*, B*, C* = the reflections of Ma, Mb, Mc in I, resp.

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

"Radical trace" (*) ? [ = midpoint of the line segment joining the points of the intersections of the circles]

 

(*) The term used by CL is a nice term, I don't know if it is an old one or a neologism in geometry (it is a term used in Algebra)

[César Lozada]:

 

 

T = (b+c)*a^8-2*(3*b^2+4*b*c+3*c^ 2)*a^7-(b+c)*(2*b^2-35*b*c+2* c^2)*a^6+2*(9*b^4+9*c^4-2*b*c* (4*b^2+15*b*c+4*c^2))*a^5-b*c* (b+c)*(77*b^2-137*b*c+77*c^2)* a^4-2*(9*b^6+9*c^6-2*(12*b^4+ 12*c^4+b*c*(9*b^2-14*b*c+9*c^ 2))*b*c)*a^3+(b+c)*(2*b^6+2*c^ 6+(41*b^4+41*c^4-b*c*(165*b^2- 229*b*c+165*c^2))*b*c)*a^2+6*( b^2-c^2)^2*(b^4+c^4-b*c*(4*b^ 2-3*b*c+4*c^2))*a+(b^2-c^2)^2* (b+c)^3*(3*b*c-b^2-c^2) : : (trilinears)

= on line {2475,2802}

= [ 0.758582699112975, 3.36208228083251, 0.962953964817422 ]

 

César Lozada

 

PD:

Radical trace: In Yahoo Hyacinthos web page,  search conversations containing “radical trace” (quotes included). You will find some posts using this term, the first one dated Jan 11, 2007