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## Re: Simson lines

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• Dear Alexei ... prove next interesting fact. ... P_A the point on the line PA such that IA is the bisector (internal or external) of angle PIP_A. The points
Message 1 of 15 , Jul 1, 2008
Dear Alexei
> The polar transformation and the properties of Simson lines allow to
prove next interesting fact.
> Let I be the incenter of ABC. For an arbitrary point P we denote as
P_A the point on the line PA such that IA is the bisector (internal or
external) of angle PIP_A. The points P_B, P_C are defined similarly.
> 1. If P_A, P_B, P_C are collinear then this line touches the incircle
of ABC.
> 2. The locus of points P with this property is the line.
> Can anybody prove this without using the polar transformation?

Here is a better presentation of a generalization :
Starting with a point J in the plane of ABC, the perpendicular lines
through J to JA, JB, JC intersect BC, CA, AB respectively at A', B', C'.
It is well known that A',B',C' are on a same line L(J).
For any point M,
Ma is the point of the line AM such as <MJA = <AJMa
Mb is the point of the line BM such as <MJB = <BJMb
Mc is the point of the line CM such as <MJC = <CJMc
Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
inscribed conic with focus J
This can be proved with projective tools using the fact that M->Ma is
the harmonic homology with center A and axis the line JA'
* The harmonic homology (I don't know if this English name is correct)
with center O, axis L maps M to his harmonic conjugate wrt O and OM
inter L

I think that, if we consider three harmonic homologies with centers
A,B,C, the locus of M with three colinear images is generally a cubic,
even when the three axis concur. In the particular case of three axis
La,Lb,Lc going through a point J and such as there exists an involution
swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in three
lines :
the fixed lines of the involution and the line through La inter BC, Lb
inter CA, Lc inter AB
Friendly. Jean-Pierre
• Dear Alxei ... lines ... B , C . ... is ... correct) ... cubic, ... axis ... involution ... three ... Lb ... Now, if we consider a line L with trilinear pole P
Message 2 of 15 , Jul 1, 2008
Dear Alxei
I wrote :
> Here is a better presentation of a generalization :
> Starting with a point J in the plane of ABC, the perpendicular
lines
> through J to JA, JB, JC intersect BC, CA, AB respectively at A',
B', C'.
> It is well known that A',B',C' are on a same line L(J).
> For any point M,
> Ma is the point of the line AM such as <MJA = <AJMa
> Mb is the point of the line BM such as <MJB = <BJMb
> Mc is the point of the line CM such as <MJC = <CJMc
> Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
> In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
> inscribed conic with focus J
> This can be proved with projective tools using the fact that M->Ma
is
> the harmonic homology with center A and axis the line JA'
> * The harmonic homology (I don't know if this English name is
correct)
> with center O, axis L maps M to his harmonic conjugate wrt O and OM
> inter L
>
>
> I think that, if we consider three harmonic homologies with centers
> A,B,C, the locus of M with three colinear images is generally a
cubic,
> even when the three axis concur. In the particular case of three
axis
> La,Lb,Lc going through a point J and such as there exists an
involution
> swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in
three
> lines :
> the fixed lines of the involution and the line through La inter BC,
Lb
> inter CA, Lc inter AB

Now, if we consider a line L with trilinear pole P and intersecting
the sidelines of ABC at A',B',C', we get :
Suppose that M is a point of L; the harmonic homology with center A,
axis JA' maps M to Ma,.., similarly Mb,Mc
Then
Ma, Mb, Mc are collinear
the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the inscribed conic
with perspector the cevian product J*P.
Looking at that, I've realized something which should be interesting
for Bernard :
For any point J, the cevian product J*o(J), where o(J) is the
orthocorrespondant of J, is the perspector of the inscribed conic
with focus J.
Friendly. Jean-Pierre
• Dear colleagues! Is next fact known? Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is
Message 3 of 15 , Jun 6, 2012
Dear colleagues!
Is next fact known?
Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.

Sincerely Alexey

[Non-text portions of this message have been removed]
• Yes, it is a particular case of S-triangles. Two triangles ABC and A B C inscribed in the same circle are S-triangles iff arc(AA )+arc(BB )+arc(CC )=0 (mod
Message 4 of 15 , Jun 6, 2012
Yes, it is a particular case of S-triangles.

Two triangles ABC and A'B'C' inscribed in the same circle are S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
The Simson lines of A',B',C', with respect to ABC and the Simson lines of A,B,C, with respect to A'B'C' are concurrent in the midpoint of [HH'] where H and H' are the orthocenters of ABC and A'B'C'.

The triangles CAB and CPQ are S-triangles iff PQ is parallel to AB.
So, the Simson lines of P,Q, and C are concurrent. But the Simson line of C is the altitude from C.

See
[1] Traian Lalescu, A Class of Remarkable Triangles,
Gazeta Matematica , vol XX, feb. 1915 ,p 213
(in Romanian)

[2] Trajan Lalesco, La geometrie du triangle ,Bucharest,
1937, Paris, Libraire Vuibert

Sincerely
Daniel Vacaretu

--- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
>
> Dear colleagues!
> Is next fact known?
> Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.
>
> Sincerely Alexey
>
> [Non-text portions of this message have been removed]
>
• Dear Daniel, Very good observation and proof. So we can generalize Alexey s problem as follows: If L is a line and S is the isogonal conjugate of its infinite
Message 5 of 15 , Jun 6, 2012
Dear Daniel,
Very good observation and proof.
So we can generalize Alexey's problem
as follows:
If L is a line and S is the isogonal conjugate
of its infinite point and the chord PQ of the
circumcircle of ABC is parallel to L then the
triangles ABC, SPQ are S-triangles.
Hence the Simson lines of P, Q relative to ABC
meet at M' midpoint of HH'. The locus of H' orthocenter
of SPQ is the perpendicular line from S to L.
Hence the locus of M' is the perpendicular line
from the midpoint M of HS to L.

If L is parallel to BC then S=A, M=H.
Best regards

> Από: Daniel <dvacaretu@...>
> Yes, it is a particular case of
> S-triangles.
>
> Two triangles ABC and A'B'C' inscribed in the same circle
> are S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
> The Simson lines of A',B',C', with respect to ABC and
> the Simson lines of A,B,C, with respect to A'B'C' are
> concurrent in the midpoint of [HH'] where H and H' are the
> orthocenters of ABC and A'B'C'.
>
> The triangles CAB and CPQ are S-triangles iff PQ is parallel
> to AB.
> So, the Simson lines of P,Q, and C are concurrent. But the
> Simson line of C is the altitude from C.

> "Alexey Zaslavsky" <zasl@...> wrote:
> >
> > Dear colleagues!
> > Is next fact known?
> > Let P, Q be two point on the circumcircle of triangle
> ABC. Then their Simson lines meet on the altitude from C iff
> PQ is parallel to AB.
• An easier proof would be to just recall that the Simson line bisects the segment determined by the orthocenter of the triangle and the point, thus if we let X,
Message 6 of 15 , Jun 7, 2012
An easier proof would be to just recall that the Simson line bisects the segment determined by the orthocenter of the triangle and the point, thus if we let X, Y be the midpoints of HP and HQ (where H is the orthocenter of ABC), the result follows at once by Desargues' theorem applied to triangles PXM, QYN, where M, N are the projections of P, Q on AB.

Best,
Cosmin

--- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
>
> Dear colleagues!
> Is next fact known?
> Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.
>
> Sincerely Alexey
>
> [Non-text portions of this message have been removed]
>
• ... A routine calculation using complex numbers as coordinates gives the following:   If A,B,C,D,E are on the unit circle, then the Simson lines of D and of
Message 7 of 15 , Jun 7, 2012
> From: Daniel <dvacaretu@...>
>
> Yes, it is a particular case of S-triangles.
>
> Two triangles ABC and A'B'C' inscribed in the same circle are
> S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
> The Simson lines of A',B',C', with respect to ABC and  the Simson
> lines of A,B,C, with respect to A'B'C' are concurrent in the
> midpoint of [HH'] where H and H' are the orthocenters of ABC and
> A'B'C'.

A routine calculation using complex numbers as coordinates gives
the following:   If A,B,C,D,E are on the unit circle, then the Simson
lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2

This formula is simpler than expected, even though it was known
in advance that it had to have some symmetry and scaling properties.
And the S-triangle result follows easily from this.
--
Barry Wolk
• Dear Daniel, Nikos, Cosmin and Barry! Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and
Message 8 of 15 , Jun 7, 2012
Dear Daniel, Nikos, Cosmin and Barry!
Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and its center coincides with the center of NPC. And what is the locus of point P such that the circumcevian triangle of P is S-triangle? It is clear that this is some cubic, but which?

Sincerely Alexey

----- Original Message -----
From: Barry Wolk
To: Hyacinthos@yahoogroups.com
Sent: Thursday, June 07, 2012 11:33 PM
Subject: [EMHL] Re: Simson lines

> From: Daniel <dvacaretu@...>
>
> Yes, it is a particular case of S-triangles.
>
> Two triangles ABC and A'B'C' inscribed in the same circle are
> S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
> The Simson lines of A',B',C', with respect to ABC and the Simson
> lines of A,B,C, with respect to A'B'C' are concurrent in the
> midpoint of [HH'] where H and H' are the orthocenters of ABC and
> A'B'C'.

A routine calculation using complex numbers as coordinates gives
the following: If A,B,C,D,E are on the unit circle, then the Simson
lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2

This formula is simpler than expected, even though it was known
in advance that it had to have some symmetry and scaling properties.
And the S-triangle result follows easily from this.
--
Barry Wolk

[Non-text portions of this message have been removed]
• Dear Alexey, ... see K024 in http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html Best regards Bernard [Non-text portions of this message have been
Message 9 of 15 , Jun 8, 2012
Dear Alexey,

> Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and its center coincides with the center of NPC. And what is the locus of point P such that the circumcevian triangle of P is S-triangle? It is clear that this is some cubic, but which?

see K024 in

http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard! see K024 in http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html Thank you. Sincerely Alexey [Non-text
Message 10 of 15 , Jun 8, 2012
Dear Bernard!

see K024 in

http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html

Thank you.

Sincerely Alexey

[Non-text portions of this message have been removed]
• Let ABC be a triangle P a point and A B C , A B C the pedal, circumcevian triangles of P, resp. Denote: A*, B*, C* = the other (than A , B , C )
Message 11 of 15 , Jun 29
Let ABC be a triangle P a point and A'B'C', A"B"C" the pedal, circumcevian triangles of P, resp.

Denote:

A*, B*, C* = the other (than A", B", C")  intersections of A"A', B"B', C"C' and the circumcircle, resp.

S1, S2, S3 = the Simson lines of A*, B*, C*, resp.

AsBsCs = the triangle bounded by S1, S2, S3.

Which is the locus of P such that:

1. ABC, AsBsCs are parallelogic?
The entire plane?

2. S1, S2, S3 are concurrent?

APH
• [APH]: Let ABC be a triangle P a point and A B C , A B C the pedal, circumcevian triangles of P, resp. Denote: A*, B*, C* = the other (than A , B , C )
Message 12 of 15 , Jun 30
[APH]:

Let ABC be a triangle P a point and A'B'C', A"B"C" the pedal, circumcevian triangles of P, resp.

Denote:

A*, B*, C* = the other (than A", B", C")  intersections of A"A', B"B', C"C' and the circumcircle, resp.

S1, S2, S3 = the Simson lines of A*, B*, C*, resp.

AsBsCs = the triangle bounded by S1, S2, S3.

Which is the locus of P such that:

1. ABC, AsBsCs are parallelogic?
The entire plane?

2. S1, S2, S3 are concurrent?

2) Simson lines concur for P on: {sidelines} \/ { Line at infinity} \/ {circum-sixtic q6 through ETC’s:  4, 64, 84})

q6:  Σ[ v*w*((SB-SC)*SB*SC*a*v^2*w^2+( ((SB-SA)*S^2+3*SA^2*SB)*b*v-(( SC-SA)*S^2+3*SA^2*SC)*c*w)*u^ 3)] = 0 (trilinears)

ETC-pairs (P,Z2(P)=point of concurrence): (4, 4), (64, 5562), (84,10)

1)      Locus for parallelogic ABC and AsBsCs = the entire plane

The parallelogic center A->As is Za(P) = Isogonal(Reflection(Isogonal( P), O)),

i.e., the Isogonal-conjugate of the-reflection-in-O of the isogonal-conjugate-of-P.

The parallelogic center Zs(P)=As->A has a complicated expression. There are no ETC pairs (P,Zs(P)) for non-concurrent Simson lines.

Points Zs(P) seem not to be of interest: long coordinates and very few relations with existing 13600+ ETC centers.