## More Orth. Proj. of H (was: IMO 2004 Problem 1 revisited)

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• Let ABC be a triangle, P a point, A , B , C the orth. projections of H on AP, BP, CP, resp. and Ea, Eb, Ec the midpoints of AH, BH, CH (ie EaEbEc = the Euler
Message 1 of 11 , Jul 31, 2004
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Let ABC be a triangle, P a point, A', B', C' the orth.
projections of H on AP, BP, CP, resp. and Ea, Eb, Ec
the midpoints of AH, BH, CH (ie EaEbEc = the Euler
triangle of ABC)

Which is the locus of P such that A'B'C', EaEbEc
are perspective? (Locus of the perspectors?)

I think that I, N are points of the locus.

APH
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• Dear Antreas and Bernard, Thanks for another surprising perspector! In Hyacinthos message #10184, Bernard Gibert ... I think that the perspector of the
Message 2 of 11 , 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:

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).

Sincerely,
Darij Grinberg
• ... In my acute-angled triangle figure, the Euler lines of BCRa, CARb, ABRc are concurrent. Is it true ? APH --
Message 3 of 11 , Aug 1, 2004
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>[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.

Is it true ?

APH
--
• Dear Antreas, ... Good idea... but unfortunately, zooming in on my dynamic geometry sketch shows that they are not exactly concurrent... Friendly, Darij
Message 4 of 11 , Aug 1, 2004
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Dear Antreas,

In Hyacinthos message #10188, you wrote:

>> In my acute-angled triangle figure, the
>> Euler lines of BCRa, CARb, ABRc are concurrent.

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

Friendly,
Darij Grinberg
• Dear Antreas, [APH]: Let ABC be a triangle, P a point, A , B , C the orth. projections of H on AP, BP, CP, resp. and Ea, Eb, Ec the midpoints of AH, BH, CH
Message 5 of 11 , Aug 2, 2004
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Dear Antreas,

[APH]: Let ABC be a triangle, P a point, A', B', C' the orth.
projections of H on AP, BP, CP, resp. and Ea, Eb, Ec
the midpoints of AH, BH, CH (ie EaEbEc = the Euler
triangle of ABC)

Which is the locus of P such that A'B'C', EaEbEc
are perspective? (Locus of the perspectors?)

I think that I, N are points of the locus.

*** Yes, the locus the quintic

cyclic sum a^2(S_{CA}+S_{AB}+2S_{BC})v^2w^2(S_Bv - S_Cw)
+ cyclic uvw S_A(S_B-S_C)(b^2c^2u^2- a^2S_Avw) = 0.

It also contains H (trivially), X_{80}, and X_{1263}. Note that X(80)
and X(1263) are respectively the reflection conjugates of I and N
respectively.

I have verified that the quintic is indeed invariant reflection
conjugation. [The reflection conjugate of a point P is the common
point of the reflections of the circles PBC, PCA, PAB in the sidelines
BC, CA, AB respectively].

Best regards
Sincerely
Paul
• Dear Darij ... Hmmmm...... Apo idees .... allo tipota!! [free translation: I am full of ideas!!] Here is another one (idea, I mean): What point is (where is
Message 6 of 11 , 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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