10187Re: [EMHL] Re: IMO 2004 Problem 1 revisited
- Aug 1, 2004Dear Antreas and Bernard,
Thanks for another surprising perspector!
In Hyacinthos message #10184, Bernard Gibert
>> > [APH] Let Ra, Rb, Rc be the orthogonalI think that the perspector of the triangles
>> > 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 : [...]
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
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
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:
AD, BAb, CAc and X(5)X(6)X(68)X(155).
I would be glad to see a proof of this (BTW,
I know the synthetic proof of the
collinearity of the points X(5), X(6), X(68),
X(155), so you can define the line
X(5)X(6)X(68)X(155) by any two of them).
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