## RE: [EMHL] Re: Orthocenters

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• Dear Antreas and Paul, [APH]: Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Let Ha, Hb, Hc be the orthocenters of the triangles APbPc,
Message 1 of 6 , Feb 4, 2004
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Dear Antreas and Paul,

[APH]: Let ABC be a triangle P a point and PaPbPc the pedal
triangle of P.

Let Ha, Hb, Hc be the orthocenters of the triangles
APbPc, BPcPa, CPaPb, resp.

Which is the locus of P such that PaPbPc, HaHbHc
are perspective?

If the locus is the whole plane, then what is the locus
of the perspectors as P moves on the Euler line?

[PY]: It is indeed the whole plane. The two triangles are
oppositely congruent. If P = (u:v:w) in homogeneous barycentric
coordinates, then the homothetic center is the point

(a^2(b^2c^2u + c^2S_C v + b^2S_B w) : ... : ...).

For P = O, this is the nine-point center, and
for P = H, this is the center of the Taylor circle X(389).

***********
A little synthetic.
If Ma, Mb, Mc are the mid points of the sides of the pedal triangle
and Pg is the centroid of the pedal triangle then Ma is the mid point
of PHa and the mid point Q of PaHa is the complement of P in the
pedal triangle or Q is produced from P in the homothety (Pg, -1/2).
So PaHa, PbHb, PcHc are condurrent at Q and the triangles
pedal of P and HaHbHc are oppositely congruent.
For P = O then Pg = G and Q = X(5)
for P = H then Pg = X(51) and Q = X(389)

Best regards
• Let ABC be a triangle and A B C , A B C the pedal triangles of two isogonal conjugate points P,P*. Denote: Ha := The orthocenter of A B C Hb := The
Message 2 of 6 , Apr 1 1:28 AM
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Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles
of two isogonal conjugate points P,P*.

Denote:

Ha := The orthocenter of A'B"C"

Hb := The orthocenter of B'C"A"

Hc := The orthocenter of C'A"B"

Are the triangles ABC, HaHbHc perspective + parallelogic + orthologic
for all P's? (If no, then for which P's are they?)

Perspector, and parallelo/orthologic centers ?

Antreas
• Dear Antreas ... ABC and HaHbHc are indirectly similar; hence, they are parallelogic and orthologic (two of the centers are upon the circumcircle of ABC, the
Message 3 of 6 , Apr 1 2:56 AM
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Dear Antreas
> Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles
> of two isogonal conjugate points P,P*.
>
> Denote:
>
> Ha := The orthocenter of A'B"C"
>
> Hb := The orthocenter of B'C"A"
>
> Hc := The orthocenter of C'A"B"
>
> Are the triangles ABC, HaHbHc perspective + parallelogic + orthologic
> for all P's? (If no, then for which P's are they?)
>
> Perspector, and parallelo/orthologic centers ?

ABC and HaHbHc are indirectly similar; hence, they are parallelogic and orthologic (two of the centers are upon the circumcircle of ABC, the two other ones upon the circumcircle of HaHbHc)
I think that they are perspective only when P lies on the Neuberg cubic.
Friendly. Jean-Pierre
• Let ABC be a triangle, P = (x:y:z) a point and A B C the cevian triangle of P. Denote: Ab := BC / (perpendicular from A to BB ) Ac:= BC / (perpendicular
Message 4 of 6 , Jun 13, 2009
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Let ABC be a triangle, P = (x:y:z) a point and A'B'C' the
cevian triangle of P.

Denote:

Ab := BC /\ (perpendicular from A to BB')

Ac:= BC /\ (perpendicular from A to CC')

Similarly Bc, Ba and Ca, Cb.

Let A*, B*, C* be the orthocenters of AAbAc, BBcBa, CCaCb resp.

ABC, A*B*C* are orthologic by construction.

Which is the other than H orthologic center?

If P = I, then the orth. center lies on the circumcircle of ABC.

Which is the locus of P such that the other orthologic
center is on the circumcircle of ABC?

Antreas
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