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## Re: [EMHL] AltiEulerian Triangles

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• Dear Antreas, Interesting question generates interesting result: The rectangular circumhyperbola passing through A, B, C, H and P* has center at concurrent
Message 1 of 5 , Aug 1, 2006
Dear Antreas,
Interesting question generates interesting result:
The rectangular circumhyperbola passing through A, B, C, H and P* has center at concurrent point Q.
Here H is orthocenter of ABC, P* is isogonal conjugate of P wrt ABC and P* is also orthocenter of triangle PaPbPc.
Best regards,
Bui Quang Tuan

"Antreas P. Hatzipolakis" <xpolakis@...> wrote: Dear Tuan

[BQT]
> Let ABC be a triangle, HaHbHc its Euler triangle, P = (x:y:z) with
>respect >ABC, a point, and Pa, Pb, Pc, the corresponding points with
>respect AHbHc, >BHcHa, CHaHb, resp.
> Now I try to connect Pa, Pb, Pc with the vertices of Euler
>triangle Ea, Eb, >Ec, the circumcenters of AHbHc, BHcHa, CHaHb.
> The result is very nice: with any P point = (x:y:z), three lines
>PaEa, PbEb, >PcEc are always concurrent at one point, say Q, on the
>nine point circle. The >locus is whole plane.

and I am wondering the point of concurrence
whose r. c/hyperbola the center is.

APH
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