## 22023Fwd: [EGML] An image made from the midpoints of a triangle-derived hexagon

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• Dec 8, 2013

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Date: Sun, Dec 8, 2013 at 12:23 PM
Subject: RE: [EGML] An image made from the midpoints of a triangle-derived hexagon
To: Anopolis@yahoogroups.com

Dear Seichii:

>> we denote by Ba and Ca the intersections of the line parallel to BC with AB and BC respectively….

What parallel line?

I assume you didn’t mean “the line parallel through P” because, in such case I get comcurrence occurs always at:

Q = ((b*q+c*r)*(p^2*a^2+c*b*q*r)+2*b*c*q*r*p*a)/a ::  for P=p:q:r  (trilinears)

and, for P=X(1)

Q(X(1))= ((b+c)*a^2+2*a*b*c+b*c*(b+c))/a ::

also non-ETC with ETC-search: 2.698384066352449657972..

Regards

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de seiichi kirikami
Enviado el: Domingo, 08 de Diciembre de 2013 01:54 a.m.
Para: Anopolis@yahoogroups.com
Asunto: [EGML] An image made from the midpoints of a triangle-derived hexagon

Dear friends,

Given a triangle ABC and a point P, we denote by Ba and Ca the intersections of the line parallel to BC with AB and BC respectively. The other intersections Cb, Ab and Ac, Bc are defined cyclically. The lines through the midpoints of AcAb and BcCb, BaBc and CaAc, CbCa and AbBa concur in a point Q.

If P={p, q, r}, then Q={(2p+q+r)(p^2+2p(q+r)+qr,  ,  }.

Example: If P=X(2), Q=X(2). If P=X(1), Q={(2a+b+c)(a^2+2a(b+c)+bc),  ,  }. The latter for a=6, b=9 and c=13 gives kx=1.93330058.. (non ETC).

I confirmed that this image was not a quadrangle point.

Best regards,

Seiichi.

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