## 21333Re: Happy New Year and Nik's locus for 2013

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• Jan 1, 2013
Dear Nik and Francisco Javier,

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?

PY: If A_0BC, B_0CA, C_0AB are the equilateral triangles
constructed externally on the sides BC, CA, AB respectively,
... A*B*C* is perspective with ABC if and only if P [lies on]
the isocubic K(Q,X(13)), where
Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
has (6-9-13) search number 0.245203113787â¦

***

I think we can go a little bit further and get a Euclidean constructible example. The point $Q$ can be regarded as the isogonal conjugate of the barycentric product of $X(15)$ (isodynamic point) and $X(17)$ (Napoleon point):
$$Q = (X(15)*X(17))^*.$$
It is an interior point of triangle $ABC$ if all angles are less than $120^\circ$.

In this case, the barycentric square root $\sqrt Q$ is a point on the locus. With $P = \sqrt Q$, the perspector is the barycentric square root of
$$(1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...) = (1/(S_A+S \sqrt 3)^2 : ... : ...)* ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...) = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,$$
and is the barycentric product $X(17)*P$.

Best regards
Sincerely
Paul
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