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## Re: [EMHL] Orion Transform For Pedal Triangle

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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
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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