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,

> 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

>