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• ## Re: [EMHL] Points on the MacBeath ellipse

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• Dear Antreas and Milorad [MS] ... [APH] ... (339) ... The vertices of the focal axis are the common points of the Euler line and the NP-circle; they are
Message 1 of 14 , Nov 11, 2004
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[MS]
> >Are some other easy constructible points on the
> >MacBeath ellipse are known?
[APH]
> "The only Kimberling center lying on the MacBeath inconic is X
(339)
> (Weisstein, Oct. 16, 2004)."
>
> http://mathworld.wolfram.com/MacBeathInconic.html

The vertices of the focal axis are the common points of the Euler
line and the NP-circle; they are X(1312) and X(1313) in ETC.
Friendly. Jean-Pierre
• Dear Antreas and Jean-Pierre, [MS] ... [APH] ... http://mathworld.wolfram.com/MacBeathInconic.html Does anyone know some geometric property or geometric
Message 1 of 14 , Nov 11, 2004
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Dear Antreas and Jean-Pierre,

[MS]
>Are some other easy constructible points on the
>MacBeath ellipse are known?

[APH]
>"The only Kimberling center lying on the MacBeath inconic is X(339)
>(Weisstein, Oct. 16, 2004)."

http://mathworld.wolfram.com/MacBeathInconic.html

Does anyone know some geometric property or geometric
caracterization of point X(339)(beside that it is the isotomic
conjugate of X(225))?

[JPE]
>The vertices of the focal axis are the common points of the Euler
>line and the NP-circle; they are X(1312) and X(1313) in ETC.

Thanks to both of you.
It could be of interest to find some other triangle center on
the MacBeath ellipse.
If it is too hard then to find some other point as the point of
tangency of some perpendicular bisector of PH, for some
triangle center P on the circumcircle,with the MacBeath
ellipse.
I think that it could be easier way then to use some
Droz-Farny line.
But,the open problem could be,for which points P on the
circumcircle,by the help of perpendicular bisector of PH
we get the triangle center on the MacBeath ellipse?

Friendly

[Non-text portions of this message have been removed]
• Dear Antreas and Jean-Pierre, [MS] ... [APH] ... http://mathworld.wolfram.com/MacBeathInconic.html The point X(339) is the point of tangency of the MacBeath
Message 1 of 14 , Nov 12, 2004
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Dear Antreas and Jean-Pierre,

[MS]
>Are some other easy constructible points on the
>MacBeath ellipse are known?

[APH]
>"The only Kimberling center lying on the MacBeath inconic is X(339)
>(Weisstein, Oct. 16, 2004)."

http://mathworld.wolfram.com/MacBeathInconic.html

The point X(339) is the point of tangency of the MacBeath ellipse
and of perpendicular bisector of HTa where Ta=X(98)-Tarry point.
The midpoint of HTa is the center of Kiepert hyperbola,or the
midpoint of F1F2,where F1 and F2 are Fermat points.
Antipode of Tarry point on the circumcircle is Steiner point St.
If we take perpendicular bisector of HSt then we have the
following triangle center on the MacBeath ellipse.

X=(a cosA)/u^2, y=(b cosB)/v^2, z=(c cosC)/w^2,
where
u=(4R^2-a^2)m^2-(a^2)(b^2-c^2)^2,
v=(4R^2-b^2)m^2-(b^2)(c^2-a^2)^2,
w=(4R^2-c^2)m^2-(c^2)(a^2-b^2)^2,
where
m^2=a^4+b^4+c^4-(ab)^2-(bc)^2-(ca)^2.

I hope that my calculations are good.

This new triangle center could be of interest for Clark and for Edward.

Best regards

[Non-text portions of this message have been removed]
• Dear Milorad and Jean-Pierre I am wondering what geometrical properties [locus, envelope] has the dual of the MacBeath inconic ie the MacBeath circumconic [=
Message 1 of 14 , Nov 12, 2004
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I am wondering what geometrical properties [locus, envelope] has
the "dual" of the MacBeath inconic ie the MacBeath circumconic
[= the conic centered at N, and passing through A,B,C.]

Antreas
--
• Dear Antreas and everyone, [APH] ... My limited understanding of a dual comes from here http://www.math.fau.edu/yiu/GeometryNotes020402.ps page 120. By my
Message 1 of 14 , Nov 12, 2004
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Dear Antreas and everyone,

[APH]
>I am wondering what geometrical properties [locus, envelope] has
>the "dual" of the MacBeath inconic ie the MacBeath circumconic
>[= the conic centered at N, and passing through A,B,C.]

My limited understanding of a dual comes from here
http://www.math.fau.edu/yiu/GeometryNotes020402.ps page 120.
By my reckoning, I think the dual is the circumconic,
a^2 SA y z + cyclic = 0, centered on X(6) passing through
110, 287, 648, 651, 677, 895, 1331, 1332, 1797, 1813, 1814, 1815

If it is of interest, here are a few centers, other than X(399,
1312, 1313), that I think are on the MacBeath conic,
a^4 SA^2 x^2 - 2 b^2 c^2 SB SC y z + Cyclic = 0.

rfl X(339) in X(5)
a^2 SB SC (SB SC - SA^2)^2
on lines {3,112},{4,147},{25,110},{114,132}

rfl X(339) in Euler line.
a^2 SA (b^2 - c^2)^2 (S^2 + SA^2 - 4 SB SC)^2

(b - c)^2 (b + c - a)^2 SA
on lines {3, 8}, {11, 123}, {116, 122}

(b - c)^2 SB SC
on lines {4, 145}, {25,105},{124,136}

b^2 c^2 (b^2 - c^2)^2 SB SC
on lines {4,94},{25,98},{115,135},{125,136}

a^2 (b^2 - c^2) SB SC (SB - SC)
on lines {4,147},{25,111},{127,136}

b^2 c^2 SA (SB - SC)^2
on lines {3,76},{115,127}

a^2 SA^3 (SB - SC)^2
on lines {3,74},{0,122,125},{0,127,136}

b^2 c^2 (b - c)^2 SB SC
on line {4, 150}

a^2 SA (a^2 (SB SC - SA^2) + SA (SB - SC)^2)^2
on lines {3,74},{113,131}

b^2 c^2 SB SC (S^2 - 3 SB SC)^2
on lines {3,107},{113,133}

SA (a b c (b + c) - 2 S^2 + a (b + c) (b c - 2 SA) - 2 SB SC)^2
on lines {3,100},{117,131}

note that many of these are intersections of {Focus, NP},{NP,NP}

In general, a point on the conic has 3 friends, its reflection in X
(5) and their reflection in the Euler line.

Also, I find here
http://www.genealogy.ams.org/html/id.phtml?id=24339
an entry for Alexander Murray MacBeath. The same one?

Best regards,
Peter.
• Dear Peter, and all other friends. I have found the following result: For any point P(x:y:z) on the circumcircle, point Q((a cosA)/m^2:(b cosB)/n^2:(c
Message 1 of 14 , Nov 13, 2004
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Dear Peter,
and all other friends.

I have found the following result:

For any point P(x:y:z) on the circumcircle,
point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
is on the MacBeath ellipse for
m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).

This could be useful as input for good
computer program to find a lot of new points
on the MacBeath ellipse.

Best regards

[Non-text portions of this message have been removed]
• Dear Milorad, [MS] ... What a beautiful result !! a^2 SA / (a^2 SA^2 x - SB y (b^2 SB + 2 SA SC) - SC z (c^2 SC + 2 SA SB))^2 X(339) is your transform of X(98)
Message 1 of 14 , Nov 15, 2004
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[MS]
>I have found the following result:

>For any point P(x:y:z) on the circumcircle,
>point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
>is on the MacBeath ellipse for
>m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
>n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
>p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).

>This could be useful as input for good
>computer program to find a lot of new points
>on the MacBeath ellipse.

What a beautiful result !!

a^2 SA / (a^2 SA^2 x - SB y (b^2 SB + 2 SA SC) -
SC z (c^2 SC + 2 SA SB))^2

X(339) is your transform of X(98)

Best regards
Peter.
• Dear Milorad and Peter you wrote, ... I would like to add 2 other pairs X(1113) transforms to X(1312) X(1114) transforms to X(1313) I couldn t find other
Message 1 of 14 , Nov 15, 2004
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you wrote,
> [MS]
> >I have found the following result:
>
> >For any point P(x:y:z) on the circumcircle,
> >point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
> >is on the MacBeath ellipse for
> >m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
> >n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
> >p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).
>
> >This could be useful as input for good
> >computer program to find a lot of new points
> >on the MacBeath ellipse.
>
> What a beautiful result !!
>
> a^2 SA / (a^2 SA^2 x - SB y (b^2 SB + 2 SA SC) -
> SC z (c^2 SC + 2 SA SB))^2
>
> X(339) is your transform of X(98)

I would like to add 2 other pairs

X(1113) transforms to X(1312)
X(1114) transforms to X(1313)

I couldn't find other points in the ETC

Greetings from Bruges

Eric

PS

Sorry if this message appears twice
• Dear Milorad and Peter, ... Let M = u:v:w be a point and M2 = u^2:v^2:w^2 its barycentric square. for any M at infinity, the M2 isoconjugate of O lies on the
Message 1 of 14 , Nov 15, 2004
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> [MS]
> >I have found the following result:
>
> >For any point P(x:y:z) on the circumcircle,
> >point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
> >is on the MacBeath ellipse for
> >m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
> >n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
> >p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).
>
> [PM] This could be useful as input for good
> >computer program to find a lot of new points
> >on the MacBeath ellipse.

Let M = u:v:w be a point and M2 = u^2:v^2:w^2 its barycentric square.

for any M at infinity, the M2 isoconjugate of O lies on the MacBeath
conic.
this is the point u^2/(a^2SA) : : .
with M=X525, we find X339.

consequently, for any point N = p:q:r on the circumcircle, the X(25)
isoconjugate of M2 also lies on the MacBeath conic.
this is the point a^2/(p^2 SA) : : .
with N = X112, we find X339.

this gives a very simple way to find many simple points on the conic,
and in fact on any in-conic with given perspector or center.

I do not think this is very interesting...

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Peter,Eric,Bernard I could add a new result. For any point P which is on circumcircle and on ortho cubic of triangle ABC we have 1.Perpendicular bisector
Message 1 of 14 , Nov 15, 2004
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Dear Peter,Eric,Bernard

I could add a new result.
For any point P which is on circumcircle
and on ortho cubic of triangle ABC we have
1.Perpendicular bisector of PH is Wallace line
of P wrt triangle ABC.
2.This line is tangent to the MacBeath ellipse.

Are the coordinates of points P known?
Can we find the corresponding points on the
MacBeath ellipse ?

If the Wallace line of point P wrt triangle ABC
is orthogonal to line PH then P is also on the
ortho cubic of triangle ABC.

Best regards

[Non-text portions of this message have been removed]
• Dear friends, ... (339) ... see also at http://pages.infinit.net/spqrsncf/ngorecent.htm#L2I-11 but I think that this inconic is known in F.G-M fifth edition
Message 1 of 14 , Nov 16, 2004
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Dear friends,

> "The only Kimberling center lying on the MacBeath inconic is X
(339)
> (Weisstein, Oct. 16, 2004)."

http://pages.infinit.net/spqrsncf/ngorecent.htm#L2I-11

but I think that this inconic is known in F.G-M fifth edition
problem §130. In my Greek translation there is no reference
but in the French edition 5 or 8 there is the reference that
this conic is due to Paul Seret N.A. 1865 p. 428.

Best regards
• Dear Nikolaos and friends: In FGM problem 130, it is the ellipse of centers H and O tangent to sides of triangle one talks about Paul Serret 1865, to M.E.
Message 1 of 14 , Nov 18, 2004
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Dear Nikolaos and friends:

In FGM problem 130, it is the ellipse of centers H and O
tangent to sides of triangle one talks about Paul Serret
1865, to M.E. Lemoine 1858 and Picquet 1866.

University of Michigan Historical Math Collection

Exercices de géométrie, comprenant l'esposé des méthodes
géométriques et 2000 questions résolues par F. G.-M. ---
Frère Gabriel Marie, 1820-1891.
5. ed.: 3 p. L., [iii]-xxiv, 1302 p. diagrs. 22 cm.
Tours,
A. Mame et fils; [etc., etc.]
1912.

In :
http://www.personal.us.es/rbarroso/ellipsefocusOHtangent.htm

there is a link to the pdf of the FGM

Greetings

Ricardo
>
> Dear friends,
>
> > "The only Kimberling center lying on the MacBeath
> inconic is X
> (339)
> > (Weisstein, Oct. 16, 2004)."
>
> http://pages.infinit.net/spqrsncf/ngorecent.htm#L2I-11
>
> but I think that this inconic is known in F.G-M fifth
> edition
> problem §130. In my Greek translation there is no
> reference
> but in the French edition 5 or 8 there is the reference
> that
> this conic is due to Paul Seret N.A. 1865 p. 428.
>
> Best regards
>
>
>
>

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