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## 10195Re: [EMHL] Re: IMO 2004 Problem 1 revisited

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• Aug 2, 2004
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Dear Darij

>[APH] Let Ra, Rb, Rc be the orthogonal
>projections of H on the cevians of I [ie
>on the angle bisectors of A,B,C, resp.]
>In my acute-angled triangle figure, the Euler lines
>of BCRa, CARb, ABRc are concurrent.

[DG]:
>Good idea... but unfortunately, zooming in on my
>dynamic geometry sketch shows that they are not
>exactly concurrent...

Hmmmm...... Apo idees .... allo tipota!!
[free translation: I am full of ideas!!]

Here is another one (idea, I mean):

What point is (where is lying on?) the Radical Center of the
circumcircles of the triangles bounded by the lines:
(AB,AC, parallel - to - BC - through - Ra)
(BC,BA, parallel - to - CA - through - Rb)
(CA,CB, parallel - to - AB - through - Rc) ?

Now, let Ta, Tb, Tc be three points on the angle bisectors
AI, BI, CI such that:

ITa / IRa = ITb / IRb = ITc / IRc = t

As t varies, which is the locus of the Radical Center of the
circumcircles of the triangles bounded by the lines:

(AB,AC, parallel - to - BC - through - Ta)
(BC,BA, parallel - to - CA - through - Tb)
(CA,CB, parallel - to - AB - through - TC) ?

Greetings

Antreas

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