HYACINTHOS 6546

Darij Grinberg wrote:

> In message #6469, I wrote:
>
> >> 1. (Floor van Lamoen in message #4547)
> >>
> >> Let A'B'C' be the reflections of ABC through a
> >> point D, then the circumcircles of AB'C', A'BC'
> >> and A'B'C concur in a point.
> >>
> >> 3. (Reduced to elementary version of 2.)
> >>
> >> The point of concurrence lies on the circumcircle
> >> of triangle ABC.

[big snip]

> Well, it's time to find a general formula!
>
> Darij Grinberg

Here is your general formula, in barycentrics. If D=(x:y:z),
then the point of concurrence is

(a^2/(y'-z') : b^2/(z'-x') : c^2/(x'-y') )
where x' = x(y+z-x)/a^2, y' and z' similarly.

--
Barry Wolk