## New centers?

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• Dear friends, I am new in using trilinears. As an exercise in learning trilinears I invented following construction. Let P = x : y : z, A B C be P-cevian
Message 1 of 2 , Jul 27 1:43 PM
Dear friends,
I am new in using trilinears. As an exercise in learning trilinears I
invented following construction.
Let P = x : y : z, A'B'C' be P-cevian tringle of ABC. Let LA be the radical
axis of circles (AC'C) and (AB'B). Similarly we define lines LB and LC. It's
not difficult to prove that LA, LB and LC concur in P' = a/(by+cz):
b/(ax+cz): c/(ax+by)= sinA/(ysinB+zsinC): ...
We have mapping P -> P'. By this mapping, for example, X(1)-> X(58),
X(2)->X(6), X(3)-> X(54), X(4)-> X(4), X(6)-> X(251), X(8)-> X(1), X(20)->
X(3), X(63)-> X(284), X(144)-> X(55), X(192)-> X(31), X(193)-> X(25),
X(329)-> X(9),...
What are the following centers:
X{7}-> a^2 - (b - c)^2 : b^2 - (c - a)^2 : c^2 - (a - b)^2 = sinA/[tan(B/2)
+ tan(C/2)] : ...
X(10) -> a/(2a + b + c) : b/(2b + c+ a) : c/(2c + a + c) ?
We can obtain more cenetrs (or new construction for known centers).
Best regards,
Tatiana
• Dear Tatiana, [TE]: Let P = x : y : z, A B C be P-cevian tringle of ABC. Let LA be the radical axis of circles (AC C) and (AB B). Similarly we define lines
Message 2 of 2 , Jul 27 5:38 PM
Dear Tatiana,

[TE]: Let P = x : y : z, A'B'C' be P-cevian tringle of ABC. Let LA
be the radical axis of circles (AC'C) and (AB'B). Similarly we define
lines LB and LC. It's not difficult to prove that LA, LB and LC
concur in
P' = a/(by+cz): b/(ax+cz): c/(ax+by)
= sinA/(ysinB+zsinC): ...
We have mapping P -> P'.

*** This is the isogonal conjugate of the inferior of P. The inferor
of P is the point dividing PG externally in the ratio 3:-1.

By this mapping, for example, X(1)-> X(58),
> X(2)->X(6), X(3)-> X(54), X(4)-> X(4), X(6)-> X(251), X(8)-> X(1), X
(20)->
> X(3), X(63)-> X(284), X(144)-> X(55), X(192)-> X(31), X(193)-> X
(25),
> X(329)-> X(9),...
> What are the following centers:

> X{7}-> a^2 - (b - c)^2 : b^2 - (c - a)^2 : c^2 - (a - b)^2
= sinA/[tan(B/2) + tan(C/2)] : ...

***This is X(57). It is a point on the line OI, and is the
intersection of the the three lines each joining an excenter to the
point of tangency of the incenter with the corresponding side.

> X(10) -> a/(2a + b + c) : b/(2b + c+ a) : c/(2c + a + c) ?
> We can obtain more cenetrs (or new construction for known centers).

***This point is not in ETC.

Best regards,
Sincerely,
Paul
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