## RE: [EMHL] Re: d

Expand Messages
• Dear Jean-Pierre, Thank you very much for pointing out my mistake. Yesterday, after sending out the message, I did feel something not quite right,, but did not
Message 1 of 9 , Apr 5, 2002
Dear Jean-Pierre,

Thank you very much for pointing out my mistake. Yesterday, after sending out
the message, I did feel something not quite right,, but did not have the time
to go through my calculations once more.

[JPE]: Let ABC be a triangle, and A',B',C' points on the negative
sides of BC,CA,AB, resp. such that the triangles A'BC, B'CA,
C'AB are isosceles with all the legs equal to a length d.

ie

BA' = A'C = CB' = B'A = AC' = C'B := d

For which d's the triangles ABC, A'B'C' are perspective?

[PY]:
For EVERY d >= a/2, b/2, c/2, A'B'C' and ABC are perspective at
the point whose homogeneous barycentric coordinates are
...

[JPE]: Are you sure?
The first time that Antreas asked this question, I've answered that d
...

[APH]: Let's see how Paul might have worked out the problem
(in my words, of course! :-)

ABC, A'B'C' are perspective <==>

cotB + cot(A'BC)
---------------- * cyclic = 1 [for non-signed angles]
cotC + cot(A'CB)

Now, cot(A'BC) = (a/2) / A'A*,
where A* = orth. proj. of A' on BC.

A
/\
/ \
C' / \ B'
/ \
/ \
/ \
/ \
B------A*------C

A'

We have A'A* = sqrt(d^2 - (a/2)^2) = sqrt(4d^2 - a^2) / 2

==> cot(A'BC) = a / sqrt(4d^2 - a^2)

It seems that Paul MISREAD it as:

cot(A'BC) = b / sqrt(4d^2 - b^2)

*******
This was exactly my mistake. I did not even realize that I was not working
with isosceles triangles! Forgive me.

Best regards
Sincerely
Paul
• ... Let s construct it: 1st step: Let ABC be a triangle, and la, lb, lc three parallels to BC,CA,AB with common distance -d (ie they are on the negative sides
Message 2 of 9 , Apr 7, 2002
On Friday, April 5, 2002, at 02:14 AM, Antreas P. Hatzipolakis wrote:

> The barycentric parspector in this false case is:
>
> 1
> (-------------------------------- ::)
> cotA + [a / sqrt(4d^2 - a^2)]
>

Let's construct it:

1st step:
Let ABC be a triangle, and la, lb, lc three parallels to BC,CA,AB
with common distance -d (ie they are on the negative sides of
BC,CA,AB).

A
/\
/ \
/ \
/ \
/ \
/ \
/ \
B--------|-----C
|
|d
--------------------------------|--------------------- la

2nd step:
Let B*, C* be the reflections of B, C in the midpoints Mb,Mc of
AC, AB, resp.

A
C* /\
/ \ B*
/ \
Mc Mb
/ \
/ \
/ \
B--------------C

3rd step:
The circle (B,BC*) intersects BC at Ca (on the pos. side of AB)
The circle (C,CB*) inetrsects BC at Ba (on the pos. side of AC)

A
C* /\
/ \ B*
/ \
/ \
/ \
/ \
/ \
Ba--B--------------C--Ca

4th step:
The perp. bisector of BCa intersects la at A2
The perp. bisector of CBa intersects la at A3

A
C* /\
/ \ B*
/ \
/ \
/ \
/ \
/ \
Ba--B--------------C--Ca

A'
--------------------------A3---A2--------------------- la

5th step:
A' := BA2 /\ CA3. Similarly B',C'.

The triangles ABC, A'B'C' are perspective with the
perspector above.

Antreas
• ... I append below the last file I posted to ANOPOLIS. http://groups.yahoo.com/group/Anopolis/message/49 Please send corrections/additions/comments. Antreas
Message 3 of 9 , Apr 8, 2002
On Friday, April 5, 2002, at 01:49 AM, Antreas P. Hatzipolakis wrote:

> [JPE]:
>
>> The first time that Antreas asked this question,
>
> It is a problem for me to remember all the problems
> I have posted to Hyacinthos !
>
> This was the reason I created ANOPOLIS list, to archive/index
> all those problems, but unfortunately I have no much free time
> to work for it.
>

I append below the last file I posted to ANOPOLIS.

http://groups.yahoo.com/group/Anopolis/message/49

Antreas

------------------------------------------------------

From: "Antreas P. Hatzipolakis" <xpolakis@...>
Date: Mon Apr 08, 2002 03:42:51 AM US/Pacific
To: Anopolis@yahoogroups.com
Subject: [ANOPOLIS] CONCURRENT SIMSON LINES

Notation:
Let P be a point on the circumcircle of triangle ABC.
P(ABC) = The Simson line of P with respect to ABC.

THEOREM (Lemoine, 1869):
Let ABCD be a cyclic quadrilateral.
The lines A(BCD), B(CDA), C(DAB), D(ABC) concur at a point S.
The four Nine Point Circles of the triangles BCD, CDA, DAB, ABC
concur also at the same point S.
[Reference: F.G.-M.: Exercices de geometrie, Theor. 387-II, 388-II.]

Also:
A,B,C,D are four points on a circle. Prove that the Simson lines
of A,B,C, and D with respect to the triangles BCD,CDA,DAB, and
ABC respectively, are concurrent, and make the same angles with each
other as the radii to the points A,B,C and D.
(Mathematical Questions and Solutions from the Educational Times
23 (1913) 57, #17196, by Krishna Prasad De)

THEOREM
[earliest reference?]:
If the Simson Lines of three points D,E,F with respect to a triangle
ABC be concurrent at S, then the Simson lines of A,B,C with respect
to DEF are also concurrent at the same point S. Also determine the
position of S with respect to the triangles ABC and DEF.
[The point S is the midpoint of HH', where H,H' are
the orthocenters of ABC, DEF.]
(Mathematical Questions and Solutions from the Educational Times
20 (1911) 109 #16819 by S. Narayanan)

Corrollary:
If HaHbHc, MaMbMc be the orthic, medial triangles resp.
of ABC, then the six Simson lines:

Ha(MaMbMc), Hb(MaMbMc), Hc(MaMbMc)
Ma(HaHbHc), Mb(HaHbHc), Mc(HaHbHc)

are concurrent on the midpoint M of the line segment ON
(where N is the Nine Point Circle Center).
[Antreas P. Hatzipolakis - Paul Yiu, 2 Apr 2002]

LOCUS PROBLEM:
Let ABC be a triange, P a point and XYZ the circumcevian triangle
of P. Which is the locus of P such that ABC, DEF are Simson-Wallace
logic?
("Simson-Wallace logic" triangles = the triangles with the
property : the six Simson Lines:
X(ABC), Y(ABC), Z(ABC), A(XYZ), B(XYZ), C(XYZ)
are concurrent)
[Floor van Lamoen, Hyacinthos, msg #3870, 21 Sep 2001)

The Simson lines sX, sY, sZ are concurrent iff one of the points
X,Y,Z is the isogonal conjugate of the point at infinity of the
line passing through the two others.
Now, if PaPbPc is the circumcevian triangle of P, the Simson
lines wrt ABC of Pa,Pb,Pc concur iff P lies on the cubic I call
Kjp with equation :

a^2 x (c^2y^2 + b^2z^2) + cyclic = 0

[Bernard Gibert, Hyacinthos, msg #3871, 21 Sep 2001]

What is the locus of point P for which the Simson lines of
the vertices of the circumcevian triangle are concurrent?
[Paul Yiu, Hyacinthos, msg #5114, 2 Apr 2002]

This is the circumcubic that Bernard kindly named Kjp.
This cubic is the locus of M such as :
- the sum of directed line angles (BC,AM) + (CA, BM) + (AB, CM) = 0
- M and M* (isogonal conjugate of M) are conjugate wrt the
circumcircle
- the pedal circle of M is orthogonal to the NPC
This cubic is self-isogonal (non pivotal), the asymptots are parallel
to the Morley sidelines and intersect at G, the cubic goes through
the Apollonian centers, the vertices of the circumtangential
triangle,...
The barycentric equation is
(y+z)(x/a)^2 + (z+x)(y/b)^2 + (x+y)(z/c)^2 = 0
You can find this in the coming paper for FG "Special isocubics in
the triangle plane" (Authors : Bernard + JPE)
[Jean-Pierre Ehrmann, Hyacinthos, msg #5117, 3 Apr 2002]

LOCUS PROBLEM:
Let ABC be a triangle and PaPbPc, P'aP'bP'c the pedal triangles
of the isogonal conjugate points P, P', resp.
Which is the locus of P such that the six Simson lines
Pa(P'aP'bP'c), Pb(P'aP'bP'c), Pc(P'aP'bP'c)
P'a(PaPbPc), P'b(PaPbPc), P'c(PaPbPc)
are concurrent?
[Antreas P. Hatzipolakis, 5 Apr 2002]

The locus is the Mac Cay [McCay] cubic.
If you consider:
1) the triangle T with side lines
Pa(P'aP'bP'c), Pb(P'aP'bP'c), Pc(P'aP'bP'c)
2) the triangle T' with side lines
P'a(PaPbPc), P'b(PaPbPc), P'c(PaPbPc)
then the 6 vertices are allways cyclic on the circle with
center the midpoint of H(P)H(P') (H(P) = orthocenter of PaPbPc)
This explains that the 6 Simson lines will concur iff P lies on the
Mac Cay cubic.
[Jean-Pierre Ehrmann, 5 and 6 Apr 2002]

LOCUS PROBLEM:
Let ABC be a triangle and PaPbPc, P'aP'bP'c the cevian triangles
of the cycloconjugate points P,P' resp.
Which is the locus of P such that the six Simson lines

Pa(P'aP'bP'c), Pb(P'aP'bP'c), Pc(P'aP'bP'c)
P'a(PaPbPc), P'b(PaPbPc), P'c(PaPbPc)

are concurrent?
[Antreas P. Hatzipolakis, 5 Apr 2002]

The locus is a sextic through G, H, Gergonne...
This sextic is the isogonal conjugate of the cubic
with equation in barycentrics:

((b^2-c^2)x+a^2(y-z)) x^2/a^4 + cyclic = 0
[Jean-Pierre Ehrmann, 5 and 6 Apr 2002]

THEOREM:
Let HaHbHc, MaMbMc be the orthic, medial triangles resp.
of ABC. The three Simson lines:

Ma(HaMbMc), Mb(MaHbMc), Mc(MaMbHc)

are concurrent on the center of the Taylor Circle of ABC.
[Antreas P. Hatzipolakis - Paul Yiu, 2 Apr 2002]

LOCUS PROBLEM:
Let ABC be a triangle and PaPbPc, P'aP'bP'c the pedal triangles
of the isogonal conjugate points P, P', resp.
Which is the locus of P such that:
The three Simson lines

Pa(P'aPbPc), Pb(PaP'bPc), Pc(PaPbP'c)

are concurrent?
[Antreas P. Hatzipolakis, 5 Apr 2002]

The locus is the circular quartic (in barycentrics)
(b^2.SB.z - c^2.SC.y) x^3/a^2 + cyclic = 0
It goes through I,O,K,the isodynamics, ...
[Jean-Pierre Ehrmann, 5 Apr 2002]

LOCUS PROBLEM:
Let ABC be a triangle and PaPbPc, P'aP'bP'c the cevian triangles
of the cycloconjugate points P,P' resp.
Which is the locus of P such that the three Simson lines

Pa(P'aPbPc), Pb(PaP'bPc), Pc(PaPbP'c)

are concurrent?
[Antreas P. Hatzipolakis, 5 Apr 2002]

I think the locus is a 15 degree curve.
[Jean-Pierre Ehrmann, 5 Apr 2002]

Antreas P. Hatzipolakis
8 Apr 2002

------------------------------------------------------
• ... on the center of the Taylor circle of ABC. ... XYZ ... on the midpoint of ON. ... Sorry for then confusion! APH
Message 4 of 9 , Apr 8, 2002
On Monday, April 8, 2002, at 03:45 AM, Antreas P. Hatzipolakis wrote:

> If HaHbHc, MaMbMc be the orthic, medial triangles resp.
> of ABC, then the six Simson lines:
>
> Ha(MaMbMc), Hb(MaMbMc), Hc(MaMbMc)
> Ma(HaHbHc), Mb(HaHbHc), Mc(HaHbHc)
>
> are concurrent on the midpoint M of the line segment ON

on the center of the Taylor circle of ABC.

> (where N is the Nine Point Circle Center).
> [Antreas P. Hatzipolakis - Paul Yiu, 2 Apr 2002]
>
> LOCUS PROBLEM:
> Let ABC be a triange, P a point and XYZ the circumcevian triangle
> of P. Which is the locus of P such that ABC, DEF are Simson-Wallace
XYZ
> logic?
> ("Simson-Wallace logic" triangles = the triangles with the
> property : the six Simson Lines:
> X(ABC), Y(ABC), Z(ABC), A(XYZ), B(XYZ), C(XYZ)
> are concurrent)
> [Floor van Lamoen, Hyacinthos, msg #3870, 21 Sep 2001)
>

> THEOREM:
> Let HaHbHc, MaMbMc be the orthic, medial triangles resp.
> of ABC. The three Simson lines:
>
> Ma(HaMbMc), Mb(MaHbMc), Mc(MaMbHc)
>
> are concurrent on the center of the Taylor Circle of ABC.

on the midpoint of ON.

> [Antreas P. Hatzipolakis - Paul Yiu, 2 Apr 2002]
>
>

Sorry for then confusion!

APH
Your message has been successfully submitted and would be delivered to recipients shortly.