## Re: A Midpoint on the Euler Line

Expand Messages
• 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 1 of 6 , Mar 31, 2013
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 2 of 6 , Mar 31, 2013
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
> >
>
Your message has been successfully submitted and would be delivered to recipients shortly.