A Midpoint on the Euler Line

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• Let ABC be a triangle and A B C the medial triangle. Denote: (Oab) = the circle with diameter A B (ie the circle centered on the midpoint Oab of A B, with
Message 1 of 6 , Mar 30, 2013
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Let ABC be a triangle and A'B'C' the medial triangle.

Denote:

(Oab) = the circle with diameter A'B
(ie the circle centered on the midpoint Oab of A'B,
(Oac) = the circle with diameter A'C

(Obc) = the circle with diameter B'C
(Oba) = the circle with diameter B'A

(Oca) = the circle with diameter C'A
(Ocb) = the circle with diameter C'B

X = the radical center of the circles (Oab), (Obc), (Oca)

Y = the radical center of the circles (Oac), (Oba), (Ocb)

Which line is the line (XY) (ie wwhose point is trilinear polar)
and which point is the midpoint of the line segment XY ?

Antreas
• The line is (4 a^4 - 11 a^2 b^2 - 3 b^4 - 11 a^2 c^2 + 24 b^2 c^2 - 3 c^4) x + (-3 a^4 - 11 a^2 b^2 + 4 b^4 + 24 a^2 c^2 - 11 b^2 c^2 - 3 c^4) y + (-3 a^4 + 24
Message 2 of 6 , Mar 31, 2013
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The line is

(4 a^4 - 11 a^2 b^2 - 3 b^4 - 11 a^2 c^2 + 24 b^2 c^2 -
3 c^4) x + (-3 a^4 - 11 a^2 b^2 + 4 b^4 + 24 a^2 c^2 -
11 b^2 c^2 - 3 c^4) y + (-3 a^4 + 24 a^2 b^2 - 3 b^4 -
11 a^2 c^2 - 11 b^2 c^2 + 4 c^4) z = 0

Its trilinear pole is not in ETC. The midpoint of XY is X3526.

If we take A'B'C' as the pedal triangle an arbitrary point P instead of O, it seems that there is an only point inside the triangle such that P is exactly the midpoint of XY.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle and A'B'C' the medial triangle.
>
> Denote:
>
> (Oab) = the circle with diameter A'B
> (ie the circle centered on the midpoint Oab of A'B,
> (Oac) = the circle with diameter A'C
>
> (Obc) = the circle with diameter B'C
> (Oba) = the circle with diameter B'A
>
> (Oca) = the circle with diameter C'A
> (Ocb) = the circle with diameter C'B
>
> X = the radical center of the circles (Oab), (Obc), (Oca)
>
> Y = the radical center of the circles (Oac), (Oba), (Ocb)
>
> Which line is the line (XY) (ie wwhose point is trilinear polar)
> and which point is the midpoint of the line segment XY ?
>
>
> Antreas
>
• Dear Antreas and Francisco-Javier if M is the midpoint of XY (on the Euler line as you ve notices), we have OM = 6.MG The line XY is parallel to the Simson
Message 3 of 6 , Mar 31, 2013
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Dear Antreas and Francisco-Javier
if M is the midpoint of XY (on the Euler line as you've notices), we have OM = 6.MG
The line XY is parallel to the Simson line of the Parry point X_111
Kind regards. Jean-Pierre

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> The line is
>
> (4 a^4 - 11 a^2 b^2 - 3 b^4 - 11 a^2 c^2 + 24 b^2 c^2 -
> 3 c^4) x + (-3 a^4 - 11 a^2 b^2 + 4 b^4 + 24 a^2 c^2 -
> 11 b^2 c^2 - 3 c^4) y + (-3 a^4 + 24 a^2 b^2 - 3 b^4 -
> 11 a^2 c^2 - 11 b^2 c^2 + 4 c^4) z = 0
>
> Its trilinear pole is not in ETC. The midpoint of XY is X3526.
>
> If we take A'B'C' as the pedal triangle an arbitrary point P instead of O, it seems that there is an only point inside the triangle such that P is exactly the midpoint of XY.
>
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > Let ABC be a triangle and A'B'C' the medial triangle.
> >
> > Denote:
> >
> > (Oab) = the circle with diameter A'B
> > (ie the circle centered on the midpoint Oab of A'B,
> > (Oac) = the circle with diameter A'C
> >
> > (Obc) = the circle with diameter B'C
> > (Oba) = the circle with diameter B'A
> >
> > (Oca) = the circle with diameter C'A
> > (Ocb) = the circle with diameter C'B
> >
> > X = the radical center of the circles (Oab), (Obc), (Oca)
> >
> > Y = the radical center of the circles (Oac), (Oba), (Ocb)
> >
> > Which line is the line (XY) (ie wwhose point is trilinear polar)
> > and which point is the midpoint of the line segment XY ?
> >
> >
> > Antreas
> >
>
• Dear Francisco and Jean-Pierre, Generalization: Define the points Oab and Oac on BC, Obc and Oba on CA and Oca and Ocb on AB such that: BOab = -COac = t*|BC|
Message 4 of 6 , Mar 31, 2013
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Dear Francisco and Jean-Pierre,

Generalization:

Define the points Oab and Oac on BC, Obc and Oba on CA and
Oca and Ocb on AB such that:

BOab = -COac = t*|BC|

CObc = -AOba = t*|CA|

AOca = -BOcb = t*|AB|

Let X be the radical center of the circles:
(Oab, OabB), (Obc, ObcC), (Oca, OcaA) and Y
of the circles: (Oac, OacC), (Oba, ObaA), (Ocb, OcbB)

The locus of the midpoint M of the line segment XY,
as the real number t varies, is the Euler line.

Antreas

--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...> wrote:
>
> Dear Antreas and Francisco-Javier
> if M is the midpoint of XY (on the Euler line as you've notices), we have OM = 6.MG
> The line XY is parallel to the Simson line of the Parry point X_111
> Kind regards. Jean-Pierre
>
> --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
> >
> > The line is
> >
> > (4 a^4 - 11 a^2 b^2 - 3 b^4 - 11 a^2 c^2 + 24 b^2 c^2 -
> > 3 c^4) x + (-3 a^4 - 11 a^2 b^2 + 4 b^4 + 24 a^2 c^2 -
> > 11 b^2 c^2 - 3 c^4) y + (-3 a^4 + 24 a^2 b^2 - 3 b^4 -
> > 11 a^2 c^2 - 11 b^2 c^2 + 4 c^4) z = 0
> >
> > Its trilinear pole is not in ETC. The midpoint of XY is X3526.
> >
> > If we take A'B'C' as the pedal triangle an arbitrary point P instead of O, it seems that there is an only point inside the triangle such that P is exactly the midpoint of XY.
> >
> >
> > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> > >
> > > Let ABC be a triangle and A'B'C' the medial triangle.
> > >
> > > Denote:
> > >
> > > (Oab) = the circle with diameter A'B
> > > (ie the circle centered on the midpoint Oab of A'B,
> > > with radius OabB=OabA')
> > > (Oac) = the circle with diameter A'C
> > >
> > > (Obc) = the circle with diameter B'C
> > > (Oba) = the circle with diameter B'A
> > >
> > > (Oca) = the circle with diameter C'A
> > > (Ocb) = the circle with diameter C'B
> > >
> > > X = the radical center of the circles (Oab), (Obc), (Oca)
> > >
> > > Y = the radical center of the circles (Oac), (Oba), (Ocb)
> > >
> > > Which line is the line (XY) (ie wwhose point is trilinear polar)
> > > and which point is the midpoint of the line segment XY ?
> > >
> > >
> > > Antreas
• Another midpoint: Let ABC be a triangle, P a point and A B C the cevian triangle of P. Denote: Ma, Mb, Mc = the midpoints of AA , BB , CC resp. X = the
Message 5 of 6 , Mar 31, 2013
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Another midpoint:

Let ABC be a triangle, P a point and A'B'C' the cevian
triangle of P.

Denote:

Ma, Mb, Mc = the midpoints of AA', BB', CC' resp.

X = the radical center of (Ma, MaB), (Mb, MbC), (Mc, McA)

Y = the radical center of (Ma, MaC), (Mb, MbA), (Mc, McB).

The midpoint of XY is the circumcenter O.

Now, denote:

Ka, Kb, Kc points on AA', BB', CC', resp. such that:

1. KaA / KaP = KbB / KbP = KcB / KcP = k

La, Lb, Lc on the same cevians such that:

2. LaA' / LaP = LbB' / LbP = LcC' / LcP = l

and Ma, Mb, Mc similarly such that:

3. MaA / MaA' = MbB / MbB' = McC / McC' = m

X1 = the radical center of (Ka, KaB), (Kb, MbC), (Kc, McA)

Y1 = the radical center of (Ka, KaC), (Kb, KbA), (Kc, KcB).

X2 = the radical center of (La, LaB), (Lb, LbC), (Lc, LcA)

Y2 = the radical center of (La, LaC), (Lb, LbA), (Lc, LcB).

X3 = the radical center of (Ma, MaB), (Mb, MbC), (Mc, McA)

Y3 = the radical center of (Ma, MaC), (Mb, MbA), (Mc, McB).

M1, M2, M3 = the midpoints of X1Y1, X2Y2, X3C3, resp.

Which are the loci of M1,M2,M3 as k,l,m vary?

Special Case:

P = G. Since GA / GA' = GB / GB' = GC / GC', the three
loci are identical: the Euler line of ABC.

By the way, G is the ONLY one point P with the property

PA / PA' = PB / PB' = PC / PC' ?

Antreas

[APH]:
> Let ABC be a triangle and A'B'C' the medial triangle.
>
> Denote:
>
> (Oab) = the circle with diameter A'B
> (ie the circle centered on the midpoint Oab of A'B,
> (Oac) = the circle with diameter A'C
>
> (Obc) = the circle with diameter B'C
> (Oba) = the circle with diameter B'A
>
> (Oca) = the circle with diameter C'A
> (Ocb) = the circle with diameter C'B
>
> X = the radical center of the circles (Oab), (Obc), (Oca)
>
> Y = the radical center of the circles (Oac), (Oba), (Ocb)
>
> Which line is the line (XY) (ie wwhose point is trilinear polar)
> and which point is the midpoint of the line segment XY ?
>
>
> Antreas
>
• There is, in fact, only one point inside ABC such that P is the midpoint of XY, and that point is P=X=Y=I. Randy
Message 6 of 6 , Mar 31, 2013
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There is, in fact, only one point inside ABC such that P is the 'midpoint' of XY, and that point is P=X=Y=I.

Randy

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> The line is
>
> (4 a^4 - 11 a^2 b^2 - 3 b^4 - 11 a^2 c^2 + 24 b^2 c^2 -
> 3 c^4) x + (-3 a^4 - 11 a^2 b^2 + 4 b^4 + 24 a^2 c^2 -
> 11 b^2 c^2 - 3 c^4) y + (-3 a^4 + 24 a^2 b^2 - 3 b^4 -
> 11 a^2 c^2 - 11 b^2 c^2 + 4 c^4) z = 0
>
> Its trilinear pole is not in ETC. The midpoint of XY is X3526.
>
> If we take A'B'C' as the pedal triangle an arbitrary point P instead of O, it seems that there is an only point inside the triangle such that P is exactly the midpoint of XY.
>
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > Let ABC be a triangle and A'B'C' the medial triangle.
> >
> > Denote:
> >
> > (Oab) = the circle with diameter A'B
> > (ie the circle centered on the midpoint Oab of A'B,
> > (Oac) = the circle with diameter A'C
> >
> > (Obc) = the circle with diameter B'C
> > (Oba) = the circle with diameter B'A
> >
> > (Oca) = the circle with diameter C'A
> > (Ocb) = the circle with diameter C'B
> >
> > X = the radical center of the circles (Oab), (Obc), (Oca)
> >
> > Y = the radical center of the circles (Oac), (Oba), (Ocb)
> >
> > Which line is the line (XY) (ie wwhose point is trilinear polar)
> > and which point is the midpoint of the line segment XY ?
> >
> >
> > Antreas
> >
>
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