## REGULAR POLYGON AND CONCURRENT CENTRAL LINES

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• Let A1A2A3A4A5A6 be a regular hexagon and P a point. Name the triangles PA1A2, PA2A3, etc as 1,2,3,4,5,6. The Euler lines of the even triangles (ie 2,4,6) are
Message 1 of 9 , Feb 20, 2013
Let A1A2A3A4A5A6 be a regular hexagon and P a point.
Name the triangles PA1A2, PA2A3, etc as 1,2,3,4,5,6.

The Euler lines of the even triangles (ie 2,4,6)
are concurrent at a point Q and of the odd triangles
(ie 1,3,5) at a different point R.

In General:

Let A1A2A3....An, be a regular n-gon, with n =3k,
and P a point. Name the triangles PA1A2, PA2A3, etc
as 1,2,3,....n.

Whose triangles the Euler lines are concurrent?

APH
• One more: Let ABC be an equilateral triangle, P a point, A;B C the cevian triangle of P, and A ,B ,C the reflections of A,B,C in A ,B ,C , resp. The Euler
Message 2 of 9 , Feb 21, 2013
One more:

Let ABC be an equilateral triangle, P a point, A;B'C' the
cevian triangle of P, and A",B",C" the reflections of
A,B,C in A',B',C', resp.

The Euler lines of AB"C", BC"A", CA"B"
are concurrent.

APH

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Another problem:
>
> Let ABC, A'B'C' be two arbitrary equilateral triangles.
>
> Denote:
>
> Ab, Ac = the reflections of A in BB', CC', resp.
> Bc, Ba = the reflections of B in CC', AA', resp.
> Ca, Cb = the reflections of C in AA', BB', resp.
>
> The Euler lines of AAbAc, BBcBa, CCaCb are concurrent.
>
> Corollary:
>
> Let ABC be an equilateral triangle and P a point.
>
> Denote:
>
> Ab, Ac = the reflections of A in BP, CP, resp.
> Bc, Ba = the reflections of B in CP, AP, resp.
> Ca, Cb = the reflections of C in AP, BP, resp.
>
> The Euler lines of AAbAc, BBcBa, CCaCb are concurrent.
>
> Proofs?
>
> APH
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
> >
> > Let ABC be an equilateral triangle and P a point.
> >
> > Which same central lines of the triangles PBC, PCA, PAB
> > are concurrent for all P's ?
> >
> > The OH (Euler) lines? The OI lines? ......
> >
> > APH
> >
>
• ... One more: Let ABC be an equilateral triangle, P a point, A;B C the cevian triangle of P, and A ,B ,C the reflections of A,B,C in A ,B ,C , resp. The
Message 3 of 9 , Apr 17, 2014

---In Hyacinthos@yahoogroups.com, <anopolis72@...> wrote :

One more:

Let ABC be an equilateral triangle, P a point, A;B'C' the
cevian triangle of P, and A",B",C" the reflections of
A,B,C in A',B',C', resp.

The Euler lines of AB"C", BC"A", CA"B"
are concurrent.

APH

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

In general, they are not.

Which is the locus of P such that the Euler lines of AB"C", BC"A", CA"B" or
the Euler lines of A"BC, B"CA, C"AB are concurrent?

APH
• ... , ... Answer: Central lines Ln (OH,OK,OI...) as they are described in :
Message 4 of 9 , Jun 6 8:33 AM

--- In Hyacinthos #21592, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be an equilateral triangle and P a point.
>
> Which same central lines of the triangles PBC, PCA, PAB
> are concurrent for all P's ?
>
> The OH (Euler) lines? The OI lines? ......
>
> APH

Central lines Ln (OH,OK,OI...) as they are described in :
Concurrency of Four Euler Lines by Pau Yiu et all,
Forum Geometricorum Volume 1 (2001) 59–68

We can take as Euler line of an equilateral triangle any line passing through its center. So, the Euler lines of ABC, PBC, PCA are concurrent on the Euler line of PAB. (See the paper above)

APH

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GENERALIZATIONS:

Let A1A2A3A4A5A6 be a regular hexagon and P a point.
The Euler lines of PA2A3, PA4A5, PA6A1 are concurrent.
In general:
Let A1A2A3.... A3n be a regular 3n-gon and P a point.
The Euler lines of PAnAn+1, PA2nA2n+1, PA3nA1 are concurrent.

Let ABC be an equilatelar triangle and A'B'C' the medial triangle
Three circles of equal radius centered at A', B', C' intersect
BC, CA, AB at (A1, A2), (B1, B2), (C1, C2 resp. [order doesn't play role].
Let P be a point. The Euler lines of PA1A2, PB1B2, PC1C2 are concurrent.

APH
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