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## 22166Fwd: [EGML] Re: Envelopes

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• Apr 21, 2014
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---------- Forwarded message ----------
Date: Mon, Apr 21, 2014 at 4:36 PM
Subject: RE: [EGML] Re: Envelopes
To: Anopolis@yahoogroups.com

Please forget that A’B’C’ may be the medial triangle of ABC in the last answer. In such case,  the reflected lines concur on a point on the NPC of ABC.

Apologizes

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de César Lozada
Enviado el: Lunes, 21 de Abril de 2014 08:36 a.m.
Para: Anopolis@yahoogroups.com
Asunto: RE: [EGML] Re: Envelopes

Special cases:

If A’B’C’ is the medial or the antimedial triangle of ABC, the fixed point on Euler lines has trilinear coordinates:

(2*f(6)+6*f(4)+14*f(2)+7)*g(1)

-2*(2*f(5)+2*f(3)+5*f(1))*g(2)

+(2*f(2)+2*f(4)+3)*g(3)

-2*(f(3)+f(1))*g(4)

-(f(7)+f(5)+6*f(3)+6*f(1)) : :

Where f(n)=cos(n*A) and g(n)=cos(n*(B-C))

This point is on line  (1141,3484) and has ETC-6-9-13 numbers:

(4.923838044604385591130, 4.248841971282335390013, -1.573382134182338747412)

Best regards

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de Antreas Hatzipolakis
Enviado el: Lunes, 21 de Abril de 2014 05:52 a.m.
Para: anopolis@yahoogroups.com; Hyacinthos
Asunto: [EGML] Re: Envelopes

Conjecture:

Let A'B'C' be the antimedial triangle of ABC. The reflections of the parallel lines L1,L2,L3 trough A,B,C, resp. in the sidelines of A'B'C' (instead of ABC) bound a triangle whose the Euler line passes through a fixed point.

In general:

Let ABC, A'B'C' be two homothetic triangles. Let L1,L2,L3 be three parallel lines through A,B,C, resp. The reflections of L1,L2,L3 in the sidelines B'C',C'A',A'B', resp. of A'B'C' bound a triangle whose the Euler line passes through a fixed point.

On Sun, Apr 20, 2014 at 10:59 PM, Antreas Hatzipolakis <anopolis72@...> wrote:

On Sun, Apr 20, 2014 at 8:13 PM, Antreas Hatzipolakis <anopolis72@...> wrote:

The problem is:
Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and let A'B'C' be the triangle bounded by the reflections of L1,L2,L3, in BC,CA,AB, resp.
As the lines L1,L2,L3 move around A,B,C, being parallel, which is the envelope of the Euler line of A'B'C'?
Furthermore, which is the envelope of the circumcircle of A'B'C'?

APH

The problem was posed by the friend Dao Thanh Oai, in my wording:

Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and let A'B'C' be the triangle bounded by the reflections of L1,L2,L3, in BC,CA,AB, resp.

As the lines L1,L2,L3 move around A,B,C, the Euler line of the triangle
A1B1C1 passes through a fixed point.

At first glance I thought that if it was true, then the fixed point should be

the Parry Point X(399), the point the reflections in BC,CA,AB of the parallels

to Euler line through A,B,C, resp. concur at.

But I was not right!!! If the lines L1,L2,L3 are parallels to Euler line,

then the triangle A1B1C1 is degenerated in Parry point, and its Euler line

is ANY line passing through Parry point!

The fixed point has barycentrics (computed by Francisco Javier García Capitán):

The point is (f(a,b,c):f(b,c,a):f(c,a,b)) where f(a,b,c) is
a^22 - 8 a^20 b^2 + 28 a^18 b^4 - 56 a^16 b^6 + 70 a^14 b^8 -
56 a^12 b^10 + 28 a^10 b^12 - 8 a^8 b^14 + a^6 b^16 - 8 a^20 c^2 +
42 a^18 b^2 c^2 - 92 a^16 b^4 c^2 + 106 a^14 b^6 c^2 -
62 a^12 b^8 c^2 + 7 a^10 b^10 c^2 + 13 a^8 b^12 c^2 -
8 a^6 b^14 c^2 + 4 a^4 b^16 c^2 - 3 a^2 b^18 c^2 + b^20 c^2 +
28 a^18 c^4 - 92 a^16 b^2 c^4 + 113 a^14 b^4 c^4 - 62 a^12 b^6 c^4 +
17 a^10 b^8 c^4 - 9 a^8 b^10 c^4 + 5 a^6 b^12 c^4 - 6 a^4 b^14 c^4 +
13 a^2 b^16 c^4 - 7 b^18 c^4 - 56 a^16 c^6 + 106 a^14 b^2 c^6 -
62 a^12 b^4 c^6 + 4 a^10 b^6 c^6 + 4 a^8 b^8 c^6 + 8 a^6 b^10 c^6 -
6 a^4 b^12 c^6 - 18 a^2 b^14 c^6 + 20 b^16 c^6 + 70 a^14 c^8 -
62 a^12 b^2 c^8 + 17 a^10 b^4 c^8 + 4 a^8 b^6 c^8 - 12 a^6 b^8 c^8 +
8 a^4 b^10 c^8 + 3 a^2 b^12 c^8 - 28 b^14 c^8 - 56 a^12 c^10 +
7 a^10 b^2 c^10 - 9 a^8 b^4 c^10 + 8 a^6 b^6 c^10 + 8 a^4 b^8 c^10 +
10 a^2 b^10 c^10 + 14 b^12 c^10 + 28 a^10 c^12 + 13 a^8 b^2 c^12 +
5 a^6 b^4 c^12 - 6 a^4 b^6 c^12 + 3 a^2 b^8 c^12 + 14 b^10 c^12 -
8 a^8 c^14 - 8 a^6 b^2 c^14 - 6 a^4 b^4 c^14 - 18 a^2 b^6 c^14 -
28 b^8 c^14 + a^6 c^16 + 4 a^4 b^2 c^16 + 13 a^2 b^4 c^16 +
20 b^6 c^16 - 3 a^2 b^2 c^18 - 7 b^4 c^18 + b^2 c^20