ANOPOLIS 170

 

Dear Randy,

[RH]: If we call the centers Q13 and Q14 respectively, then:

Q13 = midpoint of X(2) and X(13)
...
Q14 = midpoint of X(2) and X(14)
...

*** This is wonderful. Since the two circles have the centroid G as a common point, there is a second intersection. This is the pedal of G
on the line joining the Fermat points, and is the fourth vertex of the rectangle with three vertices at G, the Kiepert center, and the Jerabek center. It has homogeneous barycentric coordinates

2 a^10 - 4 a^8 (b^2 + c^2) + 4 a^6 (2 b^4 - b^2 c^2 + 2 c^4)
+ a^4 (-b^2 - c^2) (7 b^4 - 10 b^2 c^2 + 7 c^4)
+ a^2 (-b^8 + 14 b^6 c^2 - 24 b^4 c^4 + 14 b^2 c^6 - c^8)
+ (b - c)^2 (b + c)^2 (b^2 - 2 c^2) (2 b^2 -c^2) (b^2 + c^2)
: ... : ...

with (6-9-13) search number 1.36378555272...

Best regards
Sincerely
Paul Yiu

 

 

***********************************************

Three related centers:

Q15 = midpoint of X(2) and X(15)
Q15 = {X(549),X(597)}-harmonic conjugate of Q16
Q15 trilinears: bc[4a(sin(A + pi/3)) + b(sin(B + pi/3)) + c(sin(C + pi/3))] : :
Q15 lies on lines 2,14 549,597.

Q16 = midpoint of X(2) and X(16)
Q16 = {X(549),X(597)}-harmonic conjugate of Q15
Q16 trilinears: bc[4a(sin(A - pi/3)) + b(sin(B - pi/3)) + c(sin(C - pi/3))] : :
Q16 lies on lines 2,13 549,597.

M2 = midpoint of Q15 and Q16
M2 = midpoint of X(2) and X(187)
M2 lies on lines 2,187 549,597.

Randy Hutson

 

--- In Anopolis@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
>
> Also, Q13 and Q14 are {X(5),X(597)}-harmonic conjugates.
>
> Randy
>
> --- In Anopolis@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
> >
> > Also, the ETC search values are:
> >
> > Q13: 1.352656972146667
> > Q14: 3.501400282251073
> > M: 2.427028627198870
> >
> > Randy
> >
> > --- In Anopolis@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
> > >
> > > Apparently Yahoo doesn't like the pi symbol. It translated it to 'Ï€' in the trilinears below.
> > >
> > > Randy
> > >
> > > --- In Anopolis@yahoogroups.com, Randy Hutson <rhutson2@> wrote:
> > > >
> > > > Antreas,
> > > >
> > > > If we call the centers Q13 and Q14 respectively, then:
> > > >
> > > > Q13 = midpoint of X(2) and X(13)
> > > > Q13 = center of (equilateral) triangle formed by nine-point centers of BCX(13), CAX(13), ABX(13)
> > > > Q13 = center of circle through X(2), X(13) and nine-point centers of BCX(13), CAX(13), ABX(13)
> > > > Q13 trilinears: bc[4a(csc(A + π/3)) + b(csc(B + π/3)) + c(csc(C + π/3))] : :
> > > > Q13 lies on lines 2,13 5,542.
> > > >
> > > > Q14 = midpoint of X(2) and X(14)
> > > > Q14 = center of (equilateral) triangle formed by nine-point centers of BCX(14), CAX(14), ABX(14)
> > > > Q14 = center of circle through X(2), X(14) and nine-point centers of BCX(14), CAX(14), ABX(14)
> > > > Q14 trilinears: bc[4a(csc(A - π/3)) + b(csc(B - π/3)) + c(csc(C - π/3))] : :
> > > > Q14 lies on lines 2,14 5,542.
> > > >
> > > > Let M be the midpoint of Q13 and Q14.  Then:
> > > > M = midpoint of X(2) and X(115)
> > > > M trilinears: bc[4(b^2 - c^2)^2 + (c^2 - a^2)^2 + (a^2 - b^2)^2] : :
> > > > M lies on lines 2,99 5,542.
> > > >
> > > > Best regards,
> > > > Randy
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > >________________________________
> > > > > From: Antreas Hatzipolakis <anopolis72@>
> > > > >To: anopolis@yahoogroups.com
> > > > >Cc: "Kimberling, Clark" <ck6@>
> > > > >Sent: Thursday, April 25, 2013 2:55 AM
> > > > >Subject: [EMHL] X(5453) + Hutson-Fermat Circles
> > > > >
> > > > >
> > > > >
> > > > > 
> > > > >In the discussion about the concyclic points I, Na,Nb,Nc 
> > > > >
> > > > >
> > > > >(see X(5453) and here the related messages to Hyacinthos:
> > > > >http://anthrakitis.blogspot.gr/2013/04/concyclic-points-locus.html%c3%82%c2%a0)
> > > > >
> > > > >
> > > > >
> > > > >Randy wrote about the related locus for ABC (instead of the
> > > > >cevian triangle of I).
> > > > >
> > > > >
> > > > >In the locus, lie the points X(13), X(14).
> > > > >
> > > > >
> > > > >So we have these two circles:
> > > > >
> > > > >
> > > > >Let ABC be a triangle, Q = X13 or X14, and Na, Nb, Nc the
> > > > >NPC centers of QBC, QCA, QAB.
> > > > >
> > > > >
> > > > >The points Q, Na, Nb, Nc are concyclic.
> > > > >
> > > > >
> > > > >Centers of the circles?
> > > > >
> > > > >
> > > > >Antreas