Sorry, an error occurred while loading the content.

## 21330Re: Happy New Year and Nik's locus for 2013

Expand Messages
• Jan 1, 2013
Dear 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[3])v + (S_C+S Sqrt[3])w
: (S_B+S/Sqrt[3])v + (S_B+S Sqrt[3])w),
and similarly for B* and C*.

A*B*C* is perspective with ABC if and only if

Cyclic sum [(S_B+S/Sqrt[3])(S_C+S Sqrt[3]) uv^2
- (S_B+S Sqrt[3])(S_C+S/Sqrt[3]) uw^2]= 0.

Rewriting this equation as

Cyclic sum u/(S_A+S/Sqrt[3])[(S_C+S Sqrt[3])/(S_C+S/Sqrt[3]) v^2
- (S_B+S Sqrt[3])/(S_B+S/Sqrt[3])w^2] = 0 ,
we identify the locus of P as the isocubic K(Q,X(13)), where

Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)

has (6-9-13) search number 0.245203113787

Best regards
Sincerely
Paul
• Show all 14 messages in this topic