## RE: [EMHL] Locus

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• Dear Antreas, Let ABC be a triangle, P = (u:v:w) with cevian triangle A B C and Q=(x:y:z) with anticevian triangle QaQbQc. Denote A* = B Qc / C Qb, B* =
Message 1 of 240 , Feb 1, 2013
Dear Antreas,

Let ABC be a triangle, P = (u:v:w) with cevian triangle A'B'C' and Q=(x:y:z) with anticevian triangle QaQbQc.

Denote A* = B'Qc /\ C'Qb, B* = C'Qa /\ A'Qc, C* = A'Qb /\ B'Qa.

A*B*C* is perspective with

(1) ABC at ( x/(-x/u+y/v+z/w) : y/(x/u-y/v+z/w) : z/(x/u+y/v-z/w)),
(2) A'B'C' at Q,
(3) QaQbQc at (x^2/u : y^2/v: z^2/w).

Note: A'B'C' and QaQbQc are perspective at the cevian quotient
P/Q = ( x(-x/u+y/v+z/w) : y(x/u-y/v+z/w) : z(x/u+y/v-z/w)).

Best regards
Sincerely
Paul
• [APH]: Let ABC be a triangle. A line L passing through A intersects the circle with diameter AB again at Ab and the circle with diameter AC again at Ac.
Message 240 of 240 , Feb 16

[APH]:

Let ABC be a triangle.

A line L passing through A intersects the circle with diameter AB again at Ab and the circle with diameter AC again at Ac.

[Equivalently: Let Ab, Ac be the orthogonal projections of B, C on L, resp.]
Let A* be the intersection of BAc and CAb.

Which is the locus of A* as L moves around A?

Parametric trilinear equation:

1/u(t) = a*((b^2+c^2-a^2)^2-4*b^2*c^2*c os(2*t)^2)/(2*S)

1/v(t) = 2*(cos(2*t)*c-b)*S - c*sin(2*t)*(a^2+3*b^2-2*cos(2* t)*b*c-c^2)

1/w(t) = 2*(cos(2*t)*b-c)*S + b*sin(2*t)*(a^2+3*c^2-2*cos(2* t)*b*c-b^2)

Regards,