## 22025Fwd: [EGML] An image made from the midpoints of a triangle-derived hexagon

Expand Messages
• Dec 9, 2013
• 0 Attachment

---------- Forwarded message ----------
Date: Mon, Dec 9, 2013 at 5:37 PM
Subject: RE: [EGML] An image made from the midpoints of a triangle-derived hexagon
To: Anopolis@yahoogroups.com

Dear Seichii:

I see clearly why we didn’t coincide: I misread your construction and called Ab what you called Ba, and so on.

Let’s assume your names for intersections of parallel lines with the sides of ABC. In fact, we have two different points where parallel lines concur:

Yours:  Q = line(midpoint( Ab,Ac),midpoint(Bc,Cb)) /\ line(midpoint( Ba,Bc),midpoint(Ac,Ca)) /\ line(midpoint( Ca,Cb),midpoint(Ab,Ba))

Mine:  Q’ = line(midpoint( Ba,Ca),midpoint(Bc,Cb)) /\ line(midpoint( Ab,Cb),midpoint(Ac,Ca)) /\ line(midpoint( Ac,Bc),midpoint(Ab,Ba))

So, for P=[p : q : r ] (barycentrics)

Q  = [ (2*p+q+r)(p^2+2*p*(q+r)+q*r : :  ]

Q’ = [ (r+q)*(p^2+r*q)+2*p*q*r :  : ]

If P’ = isotomic conjugate of P, then

1)       Q and Q’ are on line ( P, P’ )

2)       Q’ = midpoint(P,P’)

3)       IF P lies on line at infinity, then Q=Q’=P

ETC-pairs (P,Q):

(2,2), (8,3696), (20,1350), (144,5223), (145,3243), (148,523), (192,3993), (193,1351), (4440,514)

ETC-pairs (P,Q’):

(2,2),  (4,1352), (7,2550), (8,2550), (13,623), (14,624),

(69,1352), (75,30002), (99,523), (190,514), (264,30003), (290,511), (298,623),

(299,624), (648,525), (664,522), (666,918), (668,513), (670,512), (671,524),

(886,888), (889,891), (892,690), (903,519), (1121,527), (1494,30), (2481,518),

(2966,2799), (3225,698), (3226,726), (3227,536), (3228,538), (4555,900), (4562,812),

(4569,3900), (4577,826), (4586,824), (4597,4777)

Regards

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de seiichi kirikami
Enviado el: Lunes, 09 de Diciembre de 2013 02:53 a.m.
Para: Anopolis@yahoogroups.com
Asunto: Re: [EGML] An image made from the midpoints of a triangle-derived hexagon [1 Attachment]

Dear Cesar,

I mean the lines through P( example, X(1)) parallel to BC, CA and AB. See the attached picture. Originally I made Cabri picture of a=6, b=9 and c=13 and measured QR, which coincided with the computed value. Q is the concurrent point and R is the pedal point of Q on BC.

(In the previous email, my P and Q are barycentric.)

Best regards,

Seiichi.

On Sun, Dec 8, 2013 at 7:23 PM, César Lozada <cesar_e_lozada@...> wrote:

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.

• Show all 3 messages in this topic