## envelope (point)

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• Let ABC be a point, L a line and A ,B ,C the orth. projections of A,B,C in L, resp. Let P be a point on L and Pa,Pb,Pc its reflections in AA ,BB ,CC , resp.
Message 1 of 3 , Feb 27, 2013
Let ABC be a point, L a line and A',B',C' the orth. projections
of A,B,C in L, resp.
Let P be a point on L and Pa,Pb,Pc its reflections in AA',BB',CC', resp.
The perpendiculars from Pa,Pb,Pc to BC, CA, AB concur at a point dP.
As P moves on L, which is the envelope of the lines PdP ?

APH
• Dear Antreas If the equation (barycentric) of L is px + qy + rz = 0, when P moves on the line L the lines PdP through the point L of first coordinate: L =(
Message 2 of 3 , Feb 27, 2013
Dear Antreas

If the equation (barycentric) of L is px + qy + rz = 0,
when P moves on the line L the lines PdP
through the point L' of first coordinate:

L'=( a^6p(2p-q-r)+
a^4(b^2(p(-5q+r)+2q(q+r))+c^2(p(q-5r)+2r(q+r)))-
a^2(b^2-c^2)(b^2(2p^2+4q*r-p(5q+r))+c^2(-2p^2-4q*r+ p(q+5r)))+
(b^2-c^2)^2(b^2(p-2q)-c^2(p-2r))(q-r) : ... : ... )

L=Central line associated with a triangle center X then L'= orthojoin of X

(The orthojoin of a point X other than X(6) is defined in ETC,
notes just before X(1512))

----------------------------------

Why not on ETC the ORTHOJOIN OF X(12)?

With (6-9-13)-search number -0.44020430340184767186343281...
and first coordinate:

(4a^8+a^6(-5b^2+2b*c-5c^2)
-4a^5b*c(b+c)-a^4(b^2-c^2)^2
+4a^3b*c(b-c)^2(b+c)+a^2(b-c)^4(b+c)^2
+(b^2-c^2)^4)
(a^6(b^2+c^2)-3a^4(b^4+c^4)
-2a^3b^2c^2(b+c)+a^2(3b^6-b^4c^2+4b^3c^3-b^2c^4+3c^6)+
2a*b^2(b-c)^2c^2(b+c)-(b^4-c^4)^2).

Angel Montesdeoca

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> Let ABC be a point, L a line and A',B',C' the orth. projections
> of A,B,C in L, resp.
> Let P be a point on L and Pa,Pb,Pc its reflections in AA',BB',CC', resp.
> The perpendiculars from Pa,Pb,Pc to BC, CA, AB concur at a point dP.
> As P moves on L, which is the envelope of the lines PdP ?
>
> APH
>
• I thought it was the orthopole of L. So I was wrong Francois ... [Non-text portions of this message have been removed]
Message 3 of 3 , Feb 28, 2013
I thought it was the orthopole of L.
So I was wrong
Francois

On Thu, Feb 28, 2013 at 1:42 AM, Angel <amontes1949@...> wrote:

> **
>
>
> Dear Antreas
>
> If the equation (barycentric) of L is px + qy + rz = 0,
> when P moves on the line L the lines PdP
> through the point L' of first coordinate:
>
> L'=( a^6p(2p-q-r)+
> a^4(b^2(p(-5q+r)+2q(q+r))+c^2(p(q-5r)+2r(q+r)))-
> a^2(b^2-c^2)(b^2(2p^2+4q*r-p(5q+r))+c^2(-2p^2-4q*r+ p(q+5r)))+
> (b^2-c^2)^2(b^2(p-2q)-c^2(p-2r))(q-r) : ... : ... )
>
> L=Central line associated with a triangle center X then L'= orthojoin of X
>
> (The orthojoin of a point X other than X(6) is defined in ETC,
> notes just before X(1512))
>
> ----------------------------------
>
> Why not on ETC the ORTHOJOIN OF X(12)?
>
> With (6-9-13)-search number -0.44020430340184767186343281...
> and first coordinate:
>
> (4a^8+a^6(-5b^2+2b*c-5c^2)
> -4a^5b*c(b+c)-a^4(b^2-c^2)^2
> +4a^3b*c(b-c)^2(b+c)+a^2(b-c)^4(b+c)^2
> +(b^2-c^2)^4)
> (a^6(b^2+c^2)-3a^4(b^4+c^4)
> -2a^3b^2c^2(b+c)+a^2(3b^6-b^4c^2+4b^3c^3-b^2c^4+3c^6)+
> 2a*b^2(b-c)^2c^2(b+c)-(b^4-c^4)^2).
>
> Angel Montesdeoca
>
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
> >
> > Let ABC be a point, L a line and A',B',C' the orth. projections
> > of A,B,C in L, resp.
> > Let P be a point on L and Pa,Pb,Pc its reflections in AA',BB',CC', resp.
> > The perpendiculars from Pa,Pb,Pc to BC, CA, AB concur at a point dP.
> > As P moves on L, which is the envelope of the lines PdP ?
> >
> > APH
> >
>
>
>

[Non-text portions of this message have been removed]
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