## Re: [EMHL] Darboux and a line with 5 points GENERALIZED

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• It is known. I recalled it in an earlier message to Hyacinthos. You can find it in A Generalization of a Theorem of Euler , by D.Mitrea and M.Mitrea, American
Message 1 of 2 , Mar 5, 2003
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It is known. I recalled it in an earlier message to Hyacinthos. You can find
it in "A Generalization of a Theorem of Euler", by D.Mitrea and M.Mitrea,
American Mathematical Monthly January 1994.

I would like to find a construction of something stated in this article: how
to construct a triangle A"B"C" congruent to a given one ABC and perspective
and orthologic to another given triangle A'B'C'. I have managed to construct
the perspective part, but not yet the orthologic. If somebody can help ...

Quim Castellsaguer
Barcelona

En/Na "Darij Grinberg " ha escrit:

> In Hyacinthos #6609, I wrote:
>
> >> Dear friends,
> >>
> >> Most of us know the following theorem (which
> >> can be proven with trilinears):
> >>
> >> DARBOUX THEOREM
> >>
> >> If ABC is a triangle, P is a point with pedal
> >> triangle XYZ. Then the lines AX, BY, CZ
> >> concur if and only if the point P, the
> >> isogonal conjugate P' of P and the Longchamps
> >> point L of triangle ABC are collinear.
> >>
> >> But here is MY CONJECTURE:
> >>
> >> If the lines AX, BY, CZ concur, then the
> >> line through P, P' and L also contains the
> >> point of concurrence of the lines AX, BY, CZ.
>
> GENERALIZATION (conjecture):
>
> If two orthologic triangles are perspective,
> then the perspector is collinear with the
> two orthologic centers!
>
> NOTE: Two triangles are called orthologic if
> the perpendiculars from the vertices of the
> first triangle to the sides of the second
> triangle concur, and if the perpendiculars
> from the vertices of the first triangle to
> the sides of the second triangle concur. In
> fact, these two conditions are equivalent,
> and the two points of concurrences are
> called orthologic centers of the triangles.
>
> I am quite sure that this generalization is
> known (Johnson?). If not, it will be a great
> discovery:-)
>
> More is true:
>
> If two orthologic triangles are perspective,
> then the perspector is collinear with the
> two orthologic centers, and the join of the
> three points is orthogonal to their
> perspectrix!
>
> [BTW, this theorem reminds of III) in
> message #6228, but I don't think it is
> equivalent.]
>
> Darij Grinberg
>
> >>
> >> Evident COROLLARY of Darboux Theorem (evident
> >> if we know the basic theorems on pedal
> >> circles):
> >>
> >> The line through P, P' and L passes through
> >> the circumcenter of XYZ.
> >>
> >> Nontrivial EXAMPLE: If P is the Bevan point
> >> X(40), then we get the collinear points
> >> P = Bevan point X(40);
> >> P' = Hexyl point X(84);
> >> L = Longchamps point X(20);
> >> AX /\ BY /\ CZ = Nagel point X(8);
> >> circumcenter of XYZ = X(1158).
> >> True!
> >>
> >> Can somebody prove my conjecture?
> >>
> >> Darij Grinberg
>
>
>
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• I am currently overrun by new theorems and conjectures around pedal- cevian points. I would admire any (synthetic) proof of one of them. (1) If a triangle XYZ
Message 2 of 2 , Mar 6, 2003
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I am currently overrun by new theorems and conjectures around pedal-
cevian points. I would admire any (synthetic) proof of one of them.

(1) If a triangle XYZ is a pedal-cevian triangle inscribed in a
triangle ABC, and the perpendiculars from X to BC, etc. concur at P,
and the cevians AX, BY, CZ concur at Q, then P, Q and the isogonal
conjugate of P are collinear.

(2) In the same configuration, P, the isogonal conjugate of Q and the
circumcenter of ABC are collinear.

(3) In particular: If Q is the Longchamps point of ABC, then the
circumcenter of ABC is the midpoint between P and the isogonal
conjugate of Q.

(4) Back to the general case: If the perpendicular to BC at X meets
AB at K, and the perpendicular to AB at Z meets BC at K', and Tb is
the intersection of AK' and CK - i. e., K is some kind of
generalization of the b-Triplex point of triangle ABC -, then we can
easily prove (by Pappos) that K lies on the line PQ. What is much
more difficult: The intersection of BK with CA lies on the line
through P, the isogonal conjugate of Q and the circumcenter of ABC.

Darij Grinberg
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