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• Dear All My Friends, Two points inverse each other in circumcircle will be mapped in one point, of course on the same line passing O. Denote the transform as
Message 1 of 2 , Aug 31, 2009
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Dear All My Friends,

Two points inverse each other in circumcircle will be mapped in one point, of course on the same line passing O. Denote the transform as Q=T(P), iX = inverse of X in circumcircle then: T(X) = T(iX)

Some example:

X(56) = T(X(1)) = T(X(36))
X(1995) = T(X(2)) = T(X(23))
X(24) = T(X(4)) = T(X(186))
X(1384) = T(X(6)) = T(X(187))

We can also create some points not in current ETC:
On line X(1)X(3):
T(X(35)) = T(X(484)) =
{ a^2*(a^2*(a - b - c) - a(b^2 + b*c + c^2) + (b + c)^3)/(a - b - c) : : }
Search value: +1.216112283210138

On line X(2)X(3):
T(X(21)) = T(X(1325)) =
{a*(a^3*(a - b - c) + a*(b^3 + a*b*c + c^3) - (b^2 - c^2)^2)/(b + c) : : }
Search value: +1.312427597052351

On line X(3)X(6):
T(X(39)) = T(X(2076)) =
{ a^2*(a^6 - b^6 - c^6 + 2*a^2*(a^2*(b^2 + c^2) + b^4 + c^4) + 5*a^2*b^2*c^2) : : }
Search value: +0.02937845634803689

Best regards,
Bui Quang Tuan

--- On Mon, 8/31/09, Quang Tuan Bui <bqtuan1962@...> wrote:

> From: Quang Tuan Bui <bqtuan1962@...>
> Subject: [EMHL] Orion Transform For Pedal Triangle
> To: hyacinthos@yahoogroups.com
> Date: Monday, August 31, 2009, 4:46 PM
> Dear All My Friends,
>
> Given triangle ABC, point P with barycentrics (p : q : r);
> A'B'C' is pedal triangle of P; P' is isogonal conjugate of P
> wrt A'B'C'; Pa, Pb, Pc are reflections of P' in B'C', C'A',
> A'B' respectively.
>
> Results:
>
> 1. ABC and PaPbPc are homothetic (proof is very easy) at Q
> with barycentrics:
> a^2*(p^2*b^2*c^2 + p*q*c^2*SC + p*r*b^2*SB -
> q*r*a^2*SA)  :  :
>
> 2. Denote gP as isogonal conjugate of P wrt ABC, O is
> circumcenter of ABC then:
>
> gP, P', Q are collinear
> O, P, Q are collinear
> PP' // OgP
>
> Best regards,
> Bui Quang Tuan
>
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