## 22480Re: A Circle

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• Jun 30, 2014

[APH]:
Sometime ago I posted this:

Let ABC be a triangle and La, Lb, Lc the Euler lines of IBC,ICA,IAB, resp.

Denote:

Ea = the reflection of point of La wrt triangle IBC
(ie the point of concurrence of La in the sidelines of IBC,
lying on the circumcircle of IBC)
Similarly Eb, Ec.
The points Ea,Eb,Ec are lying on the circle with diameter OI.

A natural question is this:
How about if we replace the Euler lines with Brocard axes (OK-lines)
(they are concurrent as well) and take the points by the generalized reflection theorem?

That is:

Let La, Lb, Lc be the Brocard axes of IBC,ICA,IAB resp.

The reflections of La in the sidelines of IBC bound a triangle
whose the incenter or an excenter, call it Ja, is lying on the
circumcircle of IBC.
Similarly, the reflections of Lb in the sidelines of ICA bound
a triangle, whose the incenter or an excenter Jb is lying on
the circumcircle of ICA and the reflections of Lc in the sidelines
of IAB bound a triangle whose the incenter or an excenter Jc
is lying on the circumcircle of IAB.

I think the points Ja,Jb,Jc and I are concyclic.

Which point is the center of the circle?

Antreas

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[Randy Hutson]:

Dear Antreas,

The circle as center X(4295) [corrected: X(4297)], segment X(1)X(20) as diameter, and also passes through X(3109).

Best regards,
Randy
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