## Re: [EMHL] Polar of H

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• ... Dear Alex! This fact follows from next theorem. Let two triangles ABC and A B C are perspective (AA , BB and CC concur in point P) and orthologic (the
Message 1 of 13 , Nov 3, 2003
>Dear friends!
>The polar of orthocenter is perpendicular to Euler line.
>May be it is well known fact, but short and syntnetic proof of it? Is it
>well known also?
>
Dear Alex!
This fact follows from next theorem. Let two triangles ABC and A'B'C' are
perspective (AA', BB' and CC' concur in point P) and orthologic (the
perpendiculars from A', B', C' to respectively BC, CA, AB concur in point
Q). Then PQ is perpendicular to perspective axis. IF A'B'C' is the
orthotriangle of ABC we obtain your fact. There are many another interesting
corollary from this theorem. Also it has a 3-dimensial analogous.

Sincerely Alexey
• Dear Alex and Alexey ... it? Is it ... A B C are ... (the ... in point ... interesting ... Just a little remark : in the present case, we have P = Q = H.
Message 2 of 13 , Nov 3, 2003
Dear Alex and Alexey

> >The polar of orthocenter is perpendicular to Euler line.
> >May be it is well known fact, but short and syntnetic proof of
it? Is it
> >well known also?
> >
> Dear Alex!
> This fact follows from next theorem. Let two triangles ABC and
A'B'C' are
> perspective (AA', BB' and CC' concur in point P) and orthologic
(the
> perpendiculars from A', B', C' to respectively BC, CA, AB concur
in point
> Q). Then PQ is perpendicular to perspective axis. IF A'B'C' is the
> orthotriangle of ABC we obtain your fact. There are many another
interesting
> corollary from this theorem. Also it has a 3-dimensial analogous.

Just a little remark : in the present case, we have P = Q = H. Thus,
you have to add to your theorem the fact that the perpendicular from
A,B,C to B'C', C'A',A'B' concur at a point Q' on the line PQ.
In the present case, Q' = O and we are done.
Of course, the proof of the theorem above is much more difficult
than a direct proof of the fact that the orthic axis is
perpendicular to the Euler line.
Another remark : if a point moves on a circumconic going through G,
his trilinear polar has a fixed direction. Thus, the trilinear polar
of M is perpendicular to the Euler line iff M lies on Kiepert
hyperbola.
Friendly. Jean-Pierre
• Dear Alex and Darij ... In general, if Q = (u:v:w) is a fixed point and P = (x:y:z) is a variable point, then the locus of P such that the trilinear polar of P
Message 3 of 13 , Nov 4, 2003
Dear Alex and Darij

[Alex]:
> >> The polar of orthocenter is perpendicular to Euler line.
>

In general, if Q = (u:v:w) is a fixed point and
P = (x:y:z) is a variable point, then the locus of P such
that the trilinear polar of P is perpendicular to QP line,
is the cubic with trilinear equation (if my calculations
were correct):

x(y^2(u - wcosB) - z^2(u - vcosC)) + cyclic = 0

If Q = O, then the locus is Darboux cubic, which contains H.

[Darij]:

> Yes, if you mean the trilinear polar. In fact, the trilinear
> polar of the orthocenter of a triangle ABC is the radical
> axis of the circumcircle and the nine-point circle of ABC.

In general, which is the locus of P such that the
trilinear polar of P is the radical axis of the circumcircle
and the pedal (or cevian) circle of P?

Antreas
• Let ABC be a triangle, P a pointt, and P* its isogonal conjugate. Which is the locus of P such that the trilinear polars of P and P* are (1) perpendicular (2)
Message 4 of 13 , Nov 4, 2003
Let ABC be a triangle, P a pointt, and P* its isogonal
conjugate.

Which is the locus of P such that the trilinear polars
of P and P* are (1) perpendicular (2) parallel?

Antreas
--
• Dear antreas ... As the trilinear polar of P has a fixed direction when P moves on a circumconic going through G, the trilinear polar of P and P* are parallel
Message 5 of 13 , Nov 4, 2003
Dear antreas
> Which is the locus of P such that the trilinear polars
> of P and P* are (1) perpendicular (2) parallel?

As the trilinear polar of P has a fixed direction when P moves on a
circumconic going through G, the trilinear polar of P and P* are
parallel iff the line PP* goes through K. Hence, the locus (2) is
the Grebe cubic - self-isogonal with pivot K - = K102 in Bernard's
site.
The other one is K107.
Friendly. Jean-Pierre
• Dear Antreas, ... I m not sure your computation is correct. The barycentric equation I have is : [ (SB u + a^2w)y - (SC u + a^2v)z] yz + cyclic = 0 It is a
Message 6 of 13 , Nov 4, 2003
Dear Antreas,

> [APH] In general, if Q = (u:v:w) is a fixed point and
> P = (x:y:z) is a variable point, then the locus of P such
> that the trilinear polar of P is perpendicular to QP line,
> is the cubic with trilinear equation (if my calculations
> were correct):
>
>      x(y^2(u - wcosB) - z^2(u - vcosC)) + cyclic = 0
>
> If Q = O, then the locus is Darboux cubic, which contains H.

I'm not sure your computation is correct. The barycentric equation I
have is :

[ (SB u + a^2w)y - (SC u + a^2v)z] yz + cyclic = 0

It is a circum-cubic K passing through G and having its asymptotes
parallel to those of the Thomson cubic which is obtained with Q = O.

1. When Q lies at infinity, K decomposes into the line at infinity and
a circum-conic through G.

2. K is a pK iff Q lies on Darboux and then, the pole W lies on
Thomson, the pivot R on Lucas.
We find a family of pK having all their asymptotes parallel to the
Thomson cubic.

question : what are the mappings Q -> W, Q -> R ?

3. K is a nK0 iff Q lies on the circumcircle and then, the pole is Q
and the root is the tripole of the Simson line of Q (a point on the
Simson cubic).

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, ... It seems that Q - R is the classical Pinkernell transfo. (Same algebraïc formulas). Regards. Fred
Message 7 of 13 , Nov 4, 2003
Dear Bernard,
> 2. K is a pK iff Q lies on Darboux and then, the pole W lies on
> Thomson, the pivot R on Lucas.
> We find a family of pK having all their asymptotes parallel to the
> Thomson cubic.
>
> question : what are the mappings Q -> W, Q -> R ?

It seems that Q -> R is the classical Pinkernell transfo.
(Same algebraïc formulas).
Regards.
Fred
• Dear Jean-Pierre! ... Of course this fact must be used. It is known that if the perpendicular from A,B,C to B C , C A ,A B concur at a point Q then the
Message 8 of 13 , Nov 5, 2003
Dear Jean-Pierre!
>
>Just a little remark : in the present case, we have P = Q = H. Thus,
>you have to add to your theorem the fact that the perpendicular from
>A,B,C to B'C', C'A',A'B' concur at a point Q' on the line PQ.
>In the present case, Q' = O and we are done.
>
Of course this fact must be used. It is known that if the perpendicular from
A,B,C to B'C', C'A',A'B' concur at a point Q then the erpendicular from
A',B',C' to BC, CA, AB concur at a point Q'. So if two triangles are
perspective and orthologic the theorem prove that the perspective center and
two orthologic centers are collinear. Another interesting fact: if Q=Q' then
AA', BB' and CC' concur. Infortunately, my proof is very hard.

Sincerely Alexey
• Dear Fred, ... you re right ! W is the perspector of the pedal triangle of Q and the anticevian triangle of Q, R is the perspector of the pedal triangle of Q
Message 9 of 13 , Nov 5, 2003
Dear Fred,

> [BG]
> > 2. K is a pK iff Q lies on Darboux and then, the pole W lies on
> > Thomson, the pivot R on Lucas.
> > We find a family of pK having all their asymptotes parallel to the
> > Thomson cubic.
> >
> > question : what are the mappings Q -> W, Q -> R ?
> [FL]
> It seems that Q -> R is the classical Pinkernell transfo.
> (Same algebraïc formulas).

you're right !

W is the perspector of the pedal triangle of Q and the anticevian
triangle of Q,

R is the perspector of the pedal triangle of Q and ABC.

Thank you.

Best regards

Bernard

[Non-text portions of this message have been removed]
• ... Dear Jean-Pierre There are more similar problems on tripolars, and probably we have discussed some of them in the past. Two other ones: Which is the locus
Message 10 of 13 , Nov 5, 2003
[APH]:

>> Which is the locus of P such that the trilinear polars
>> of P and P* are (1) perpendicular (2) parallel?

[JPE]:

>As the trilinear polar of P has a fixed direction when P moves on a
>circumconic going through G, the trilinear polar of P and P* are
>parallel iff the line PP* goes through K. Hence, the locus (2) is
>the Grebe cubic - self-isogonal with pivot K - = K102 in Bernard's
>site.
>The other one is K107.

Dear Jean-Pierre

There are more similar problems on tripolars,
and probably we have discussed some of them
in the past.

Two other ones:

Which is the locus of P such that the tripolars of
P, P*, and the line PP* are concurrent?
[P* = isog. conj. of P]

Which is the locus of P such that the tripolars
of P, P*, and P# are concurrent?
[P# = isot. conj. of P]

Greetings

Antreas

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