21330Re: Happy New Year and Nik's locus for 2013
- Jan 1, 2013Dear Nik and Antreas,
Happy New Year to you and all friends of Hyacinthos.
ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2 are equilateral triangles and A* is the mid point of A1, A2 in the
opposite side of BC ralative to A, and similarly define B*, C*
what is the locus of P such that ABC, A*B*C* are perspective?
If A_0BC, B_0CA, C_0AB are the equilateral triangles constructed externally on the sides BC, CA, AB respectively, your A* is the midpoint of A'A_0. If P = (u:v:W) in homogeneous barycentric coordinates, then
A* = (-(S_B+S_C)(v+w) : (S_C+S/Sqrt)v + (S_C+S Sqrt)w
: (S_B+S/Sqrt)v + (S_B+S Sqrt)w),
and similarly for B* and C*.
A*B*C* is perspective with ABC if and only if
Cyclic sum [(S_B+S/Sqrt)(S_C+S Sqrt) uv^2
- (S_B+S Sqrt)(S_C+S/Sqrt) uw^2]= 0.
Rewriting this equation as
Cyclic sum u/(S_A+S/Sqrt)[(S_C+S Sqrt)/(S_C+S/Sqrt) v^2
- (S_B+S Sqrt)/(S_B+S/Sqrt)w^2] = 0 ,
we identify the locus of P as the isocubic K(Q,X(13)), where
Q = ((S_A+S Sqrt)/(S_A+S/Sqrt) : ... : ...)
has (6-9-13) search number 0.245203113787
- << Previous post in topic Next post in topic >>