- Dec 8, 2013---------- Forwarded message ----------

From:**César Lozada**

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.comDear 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

César Lozada

**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 hexagonDear 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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