## 10187Re: [EMHL] Re: IMO 2004 Problem 1 revisited

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• Aug 1, 2004
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Dear Antreas and Bernard,

Thanks for another surprising perspector!

In Hyacinthos message #10184, Bernard Gibert
wrote:

>> > [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.]
>> > and SaSbSc the pedal triangle of I (aka
>> > intouch triangle of ABC).
>> >
>> > Are the triangles RaRbRc, SaSbSc perspective?
>>
>> They are indeed at : [...]

I think that the perspector of the triangles
RaRbRc and SaSbSc is the point X(847) of the
triangle SaSbSc (i. e. of the intouch triangle).

In fact, if we rewrite the problem from the
viewpoint of triangle SaSbSc, we get the
following:

Let ABC be a triangle. Further, let H' be the
orthocenter of the tangential triangle of
triangle ABC, and let D, E, F be the
orthogonal projections of the point H' on the
perpendicular bisectors of the segments BC,
CA, AB. Then show that the lines AD, BE, CF
are concurrent at the point X(847) of
triangle ABC.

In fact, by the definition of X(847), the
line AX(847) passes through the point
BAb /\ CAc, where the point Ab is the point
of intersection of the A-altitude of
triangle ABC with the perpendicular
bisector of the side CA, and the point Ac
is is the point of intersection of the
A-altitude with the perpendicular bisector
of the side AB. Thus, the lines
AD = AX(847), BAb and CAc are concurrent.

In fact, it turns out (conjecture!) that the
following four lines are concurrent: