If are given three Feurbach points, how can we
construct the fourth one?

aph

Let ABC be a triangle and HaHbHc the pedal triangle
of H (orthic triangle).
Let Ra, Rb, Rc be the reflections of A,B,C in H,
respectively.

The Euler Lines of ABC, HaRbRc, HbRcRa, HcRaRb
are concurrent.

Point?

APH

The point Q of concurrence, not currently on ETC, is on the Euler line,
satisfying HQ:QO = -2:7.

The locus of points satisfying this property (taking an arbitrary point instead
of H) is a septic

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle and HaHbHc the pedal triangle
> of H (orthic triangle).
> Let Ra, Rb, Rc be the reflections of A,B,C in H,
> respectively.
>
> The Euler Lines of ABC, HaRbRc, HbRcRa, HcRaRb
> are concurrent.
>
> Point?
>
> APH
>

[BG]
> given two points p:q:r and u:v:w in barycentrics, I wish to construct the
point
>
> c^2*q*u - b^2*r*u - a^2*r*v - b^2*r*v + c^2*r*v + a^2*q*w - b^2*q*w + c^2*q*w
:
>
> a^2*r*u + b^2*r*u - c^2*r*u - c^2*p*v + a^2*r*v + a^2*p*w - b^2*p*w -  c^2*p*w
:
>
> -a^2*q*u + b^2*q*u - c^2*q*u - a^2*p*v + b^2*p*v + c^2*p*v + b^2*p*w - 
a^2*q*w
>
> I already know one line passing through this point but cannot find another
one.
>
> Any idea ?
Dear Bernard,

Did you try splitting the barycentrics in terms?
Generally spoken a point P with barycentrics
P  (P1x+P2x+P3x+P4x : P1y+P2y+P3y+P4y : P1z+P2z+P3z+P4z)
can be split into:
P12 (P1x+P2x : P1y+P2y : P1z+P2z) and P34(P3x+P4x : P3y+P4y : P3z+P4z)
and also
P13 (P1x+P3x : P1y+P3y : P1z+P3z) and P24(P2x+P4x : P2y+P4y : P2z+P4z)
and also
P14 (P1x+P4x : P1y+P4y : P1z+P4z) and P23(P2x+P3x : P2y+P3y : P2z+P3z).
Now P = P12.P34 ^ P13.P24 ^ P14.P23.

When P12, P13, P14, P23, P24, P34 are difficult to construct a further split can
be done.
P1(P1x: P1y : P1z), P2(P2x : P2y : P2z), P3(P3x : P3y : P3z), P4(P4x : P4y :
P4z).
Now P12 is on line P1.P2,etc.

The point you mentioned has barycentrics:
(2 SB.q.w - 2 SC.r.v + c^2.q.u - b^2.r.u :
  2 SC.r.u - 2 SA.p.w + a^2.r.v - c^2.p.v :
  2 SA.p.v - 2 SB.q.u + b^2.p.w - a^2.q.w)
Because the barycentrics consist of 4 terms it can be split as described.

Best regards,

Chris van Tienhoven

Dear Chris,

> Did you try splitting the barycentrics in terms?
> Generally spoken a point P with barycentrics
> P (P1x+P2x+P3x+P4x : P1y+P2y+P3y+P4y : P1z+P2z+P3z+P4z)
> can be split into:
> P12 (P1x+P2x : P1y+P2y : P1z+P2z) and P34(P3x+P4x : P3y+P4y : P3z+P4z)
> and also
> P13 (P1x+P3x : P1y+P3y : P1z+P3z) and P24(P2x+P4x : P2y+P4y : P2z+P4z)
> and also
> P14 (P1x+P4x : P1y+P4y : P1z+P4z) and P23(P2x+P3x : P2y+P3y : P2z+P3z).
> Now P = P12.P34 ^ P13.P24 ^ P14.P23.
>
> When P12, P13, P14, P23, P24, P34 are difficult to construct a further split
can be done.
> P1(P1x: P1y : P1z), P2(P2x : P2y : P2z), P3(P3x : P3y : P3z), P4(P4x : P4y :
P4z).
> Now P12 is on line P1.P2,etc.
>
> The point you mentioned has barycentrics:
> (2 SB.q.w - 2 SC.r.v + c^2.q.u - b^2.r.u :
> 2 SC.r.u - 2 SA.p.w + a^2.r.v - c^2.p.v :
> 2 SA.p.v - 2 SB.q.u + b^2.p.w - a^2.q.w)
> Because the barycentrics consist of 4 terms it can be split as described.

That's what I did for the line I've found which corresponds to your P1.P2 but
I'm still stuck !

Thanks

Bernard

[Non-text portions of this message have been removed]

[APH]
> If are given three Feuerbach points, how can we
> construct the fourth one?

Let ABC be a triangle. Find a point F on the circumcircle
of ABC such that: if A,B,C, are the ex-Feuerbach points
of some triangle, then F is the in-Feuerbach point of that
triangle.

Which are the coordinates of F (with respect ABC)?

Antreas

[APH]
> > Let ABC be a triangle and HaHbHc the pedal triangle
> > of H (orthic triangle).
> > Let Ra, Rb, Rc be the reflections of A,B,C in H,
> > respectively.
> >
> > The Euler Lines of ABC, HaRbRc, HbRcRa, HcRaRb
> > are concurrent.
> >
> > Point?

[Francisco]
> The point Q of concurrence, not currently on ETC, is on the
> Euler line, satisfying HQ:QO = -2:7.
>
> The locus of points satisfying this property (taking an
> arbitrary point instead of H) is a septic

I think we can parametrize it (probably we have
already seen it here before?):

Let Ra, Rb, Rc be the homothetic images of A,B,and C
under the homothety (H, t).

The Euler lines of HaRbRc, HbRcRa, HcRaRb are concurrent.

The locus of the point of concurrence, as t varies,
is the Euler Line of ABC.

If A',B',C' are the midpoints of AH,BH,CH, resp. then we can take
Ra, Rb, Rc as homothetic images of A',B',C' with homothety (H,t).

We have that A',B',C' are on the circumcircle of HaHbHc (= NPC of
ABC) and A,B,C are the excenters of the triangle HaHbHc (with
incenter H).

So we have the variation:

Let ABC be a triangle and A'B'C' the circumcevian triangle of I.

Let Ra, Rb, Rc be the homothetic images of A',B',C', resp.
with homothety (I,t).

The Euler lines of the triangles ARbRc, BRcRa, CRaRb are concurrent.

The locus of the point of concurrence, as t varies, is the
Euler line of the Excentral triangle of ABC.

APH

--- In Hyacinthos@yahoogroups.com, "Philippe" <chephip@...> wrote:
>
>
> *) the idea I have is related with conjugate directions.
> you can project the conjugate directions of one conic section
> on a circle, same for the other conic section.
> To find two common conjugate directions, that is the fixed points
> of the composition of these homographies on the circle.
> To be studied further...
>
> Best Regards.
> Philippe.
>

Dear Philippe, I followed your idea and found the common conjugate directions:
so we have the common autopolar triangle of the two conics, whose sides are the
line at infinity and the two conjugate diameters. But I can't go further and
need help.
In "Traitι des sections coniques" Chasles says that from the common autopolar
triangle we can construct (by ruler and compass) the common chords of the conics
(or the degenerate conics of the pencil generated by the two concentric conics):
can someone explicate how this can be done ?
Another question: if we know one vertex of the common autopolar triangle of two
conics, how could we construct the other two vertices? [I think that this should
be possible by ruler and compass]
Thanks in advance to all giving help.

Giovanni Artico

Dear Yakub and Hyacintists,
an article concerning "A new point on Euler line" can be seeing on:
http://perso.orange.fr/jl.ayme   vol.5. (First synthetic proof)
Sincerely
Jean-Louis


--- In Hyacinthos@yahoogroups.com, "jeanlouisayme" <jeanlouisayme@...> wrote:
>
> Dear Yakub,
> if I am not wrong, I found a complete synthetic proof of your problem
> based on Desargues (the theorem of the two triangles), Terquem (the
cyclocevian conjugate)and Pascal's theorem.
> Sincerely
> Jean-Louis
>
> --- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@> wrote:
> >
> > Dear friends, let H be the orthocenter (intersection point of altitudes AH1,
BH2, CH3), and and G be the centroid (intersection point of medians AM1, BM2,
CM3). Draw the parallel to AC through H, which intersects BA and BC at C1 and
A2, respectively. Analogously, draw the parallel through H to BA (and to BC), to
find the points A1 and B2 (and B1 and C2). Extend the lines C1B2, A1C2, and B1A2
to form a new triangle XYZ (X is opposite to A, Y to B, Z to C). Denote
midpoints of C1B2, A1C2,  and B1A2 by T1, T2 and T3. It is known that H1, H2,
H3, M1, M2, M3, T1, T2 and T3 are concyclic (Euler circle). Let this circle
intersect C1B2, A1C2, and B1A2 again at the points S1, S2 and S3. I find that
the lines XS1, YS2, and ZS3 intersect at a new point which is collinear with H,
G, Oe (center of Euler cirle) (Euler line).  Is this point a new triangle
center? Is it in the Encyclopedia of Kimberling?
> > Yakub.
> >
>

Dear Hyacinthians,  

The reflections of the vertices of triangle ABC in their opposite sides are the
points
A'=(-a^2:a^2+b^2-c^2:c^2+a^2-b^2),   B'=(a^2+b^2-c^2:-b^2:b^2+c^2-a^2),
C'=(c^2+a^2-b^2:a^2+b^2-c^2-c^2).

The triangle A'B'C' is called the TRIANGLE OF REFLECTIONS.


Points in the Encyclopedia of Kimberling (ETC):


The triangle center X(3060) of ABC is triangle center X(2) of A'B'C'.

The triangle center X(195) of ABC is triangle center X(3) of A'B'C'.


Is there another triangle center X(i) of ABC which is a triangle center X(j) of
A'B'C'?

Best regards

Angel






[Non-text portions of this message have been removed]

Let ABC be a triangle, P a point and
PaPbPc the pedal triangle of P.

Let A' be the other than P intersection of
the circles (Pb, PbP) and (Pc, PcP)
and similarly B', C'.

Which is the locus of P such that:

1. ABC, A'B'C' are perspective

2. PaPbPc, A'B'C' are perspective

3. ABC, A'B'C' are orthologic.

The triangles PaPbPc, A'B'C' are orthologic.
One orthologic center is P. The other one?

Antreas

zeroprof wrote:
>
> Dear Philippe, I followed your idea and found the common conjugate
> directions: so we have the common autopolar triangle of the two
> conics, whose sides are the line at infinity and the two conjugate
> diameters. But I can't go further and need help.
> In "Traitι des sections coniques" Chasles says that from the common
> autopolar triangle we can construct (by ruler and compass) the common
> chords of the conics (or the degenerate conics of the pencil
> generated by the two concentric conics): can someone explicate how
> this can be done ?

Dear Giovanni,

I finally ended my construction of the intersection of two conic
sections (Ca) and (Cb) sharing the same center O.
I summarize it here, I'll set it soon in action on my web site.
But it is rather complicated, as all the details have to be
explicitely constucted.
So if anybody has a simpler construction, please post it !

First of all I search for the common conjugate diameters
to both conic sections.
These are the fixed points in the homography of the pencil of
lines O* defined by the composition of the conjugation through
both conic sections.
That is any line (d) through O -> conjugate (d') with respect
to (Ca), then (d") conjugate of (d') wrt (Cb)
The homography (d) -> (d") has fixed points which are the searched
directions.
To contruct these fixed points, I project any three (d) lines and
the corresponding (d") lines on any circle going through O
The axis of the induced homography on the circle is constructed,
giving the fixed points as intersections with the circle.

So now we have the *directions* of the parallelogram made by the
intersection points of (Ca) and (Cb). Say Ou and Ov.
But this doesn't give at once the sides themselves !
You said this problem was solved by Chasles. I don't know how he did
However here is my method.

Consider any line (L) intersecting both conic sections in Ma, M'a
and Mb, M'b.
There is a large choice of lines but we require :
- not through O, not parallel to the directions Ou and Ov [1]
- one of MaM'a or MbM'b is entirely inside the other. [2]

The pencil of conic sections defined by (Ca),(Cb) defines a
Desargues involution on (L) : M -> M'
Because of the requirements [2] on (L), this involution has two
real fixed points I, J, and at finite distance because of [1]

We may construct them by projecting this onto any circle, and
construct the axis of the induced involution on the circle as usual.
Two points (Ma,M'a) and (Mb,M'b) suffice as it is an involution.

The involution is then defined by (M,M',I,J) = -1

Let Q the intersection of (L) with Ou, and let E and E' the
intersections with (L) of the degenerated conic section parallel to Ou.
Q is midpoint of EE' (for O is the midpoint of the searched base
points of the pencil)

So just to construct E and E' from QE^2 = QE'^2 = QI.QJ
So we get the degenerated conic section, that is the two common
chords parallel to Ou, as the parallels to Ou from E, E'.
Then the searched intersections with (Ca) or (Cb).

Best Regards.
Philippe

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle, P a point and
> PaPbPc the pedal triangle of P.
>
> Let A' be the other than P intersection of
> the circles (Pb, PbP) and (Pc, PcP)
> and similarly B', C'.
>
> Which is the locus of P such that:
>
> 1. ABC, A'B'C' are perspective
>
> 2. PaPbPc, A'B'C' are perspective
>
> 3. ABC, A'B'C' are orthologic.
>
> The triangles PaPbPc, A'B'C' are orthologic.
> One orthologic center is P. The other one?
>
> Antreas
>


A'B'C' is the reflection triangle of P(x:y:z) in its pedal triangle.

1. ABC, A'B'C' are perspective if and only if P lies on the orthocubic cubic
(Antreas P Hatzipolakis and Paul Yiu, Reflections in triangle geometry, Forum
Geometricorum, 9 (2009) 301--348. Proposition 26)

3. ABC, A'B'C' are orthologic if and only if P lies (infinity lines,
circumcircle, McCay Cubic).


The triangles PaPbPc, A'B'C' are orthologic.
  One orthologic center is P(x:y:z). The other one:
a^2(2b^2c^2x^2 + c^2(a^2+b^2-c^2)x y + b^2(a^2-b^2+c^2)x z +
(a^2b^2-b^4+a^2c^2+2b^2c^2-c^4)y z):  ...  : ....

Angel

--- In Hyacinthos@yahoogroups.com, "Philippe" <chephip@...> wrote:
>
>
> I finally ended my construction of the intersection of two conic
> sections (Ca) and (Cb) sharing the same center O.
> I summarize it here, I'll set it soon in action on my web site.
> But it is rather complicated, as all the details have to be
> explicitely constucted.
> So if anybody has a simpler construction, please post it !
>
Dear Philippe,
in the meantime I explored better Chasles' book and found an alternative
solution that seems interesting.
It solves this problem: given a vertex V of the common autopolar triangle,
construct the corresponding degenerate conic.
Construction: take a point P, different from V, and draw its polars wrt the two
conics; find the point Q where the two polars intersect. Repeat with another
point P' and find its corresponding Q'. The pairs of lines VP,VQ and VP',VQ'
define an involution; the two double elements of this involution define the
searched degenerate conic.

In our case, we already have a vertex of the autopolar triangle, the common
center of the conics, and following this procedure we can construct the
corresponding degenerate conic and successively its intersections with one of
the given conics. Alternatively, if we already found also the other vertices, we
could construct the other degenerate conics and reduce the problem to the
intersection of lines.

There is also an alternative solution that I had no time to explore: if one of
the conics is an ellipse, we could transform it into a circle by an orthogonal
projection; the other conic is also transformed into another concentric conic,
whose axes can be found by a known procedure; so the problem is reduced to the
intersection of a conic with a concentric circle, that is easy. But how could be
modified this procedure if the two conics are hyperbolae?

A problem remains open for me: if we know a vertex of the common autopolar
triangle, how to construct the other two?

Best regards

Giovanni

P.S. Hope you could present the various solutions in your very wonderful site

Dear Yakub and Jean-Louis,
--- In Hyacinthos@yahoogroups.com, "jeanlouisayme" <jeanlouisayme@...> wrote:
>
> Dear Yakub and Hyacintists,
> an article concerning "A new point on Euler line" can be seeing on:
> http://perso.orange.fr/jl.ayme   vol.5. (First synthetic proof)
> Sincerely
> Jean-Louis

One of the result of the former discussion, started here by the Yakub's message,
is that all the properties described in #18419 hold in fact for any other choice
of the starting point P, as soon as we replace the Euler circle by CV(P,G), the
cevian ellipse of G and P. A figure that recalls this discussion is
http://www.les-mathematiques.net/phorum/file.php?8,file=15257

I have read carefully the JLA paper. In this 20 pages document, only the special
case where P=H=X(4) is discussed, the core of the proof being that angles like
MaSaTa are inscribed into half a circle and therefore are right angles. This
discards any possibility of extension to the general case.

By the way, I have worked again on the topic.
(1) A side result is : triangle SaSbSc is perspective with ABC at a point Z =
p^2/(q+r-p):: that is the image of the anticomplement of P under the
isoconjugacy that fixes P.
(2) I have tried to apply a synthetic point of view to the general case. This
involves a tower of hexagons in order to masquerade all these determinants that
won the battle. I am not sure that the result is more clearly established, but
another sight is ever useful.
(3) the reason why the Steiner circumellipse was involved has been found : this
conic is a contour line of the ratio GY(P) / GP (these vectors are ever
collinear).
More details, figures and a geogebra worksheet are provided at
http://www.douillet.info/~douillet/triangle/

Best regards, Pierre.

zeroprof wrote:
> --- Philippe wrote:
>>
>> I finally ended my construction of the intersection of two conic
>> sections (Ca) and (Cb) sharing the same center O.
>> I'll set it soon in action on my web site.

Dear Hyacinthists,

done :
(http://mathafou.free.fr/pbw_en/pb440.html)


> in the meantime I explored better Chasles' book and found an
> alternative solution that seems interesting.
> It solves this problem: given a vertex V of the common autopolar
> triangle, construct the corresponding degenerate conic.
> Construction: take a point P, different from V, and draw its polars
> wrt the two conics; [...]

Nice, but this is not a much "simpler" method than mine, as it requires
intersection points of several lines with the conic section to construct
the polars, and this construction is not "cheap".

>  Alternatively, [...] we could construct the other degenerate conics
> and reduce the problem to the intersection of lines.

Yep, that's how my applet works, as most of the job is already done
in finding the first degenerate conic, the second one is then a child's
play.

>
> There is also an alternative solution that I had no time to explore:
> if one of the conics is an ellipse, we could transform it into a
> circle by an orthogonal projection; the other conic is also
> transformed into another concentric conic, [...]
> so the problem is reduced to the intersection of a
> conic with a concentric circle, that is easy. But how could be
> modified this procedure if the two conics are hyperbolae?
>

The problem of transforming any conic section into a circle is
"easy", through a projective homology for instance.
But the result is then generally not two concentric conic sections
From definition of center = pole of line at infinity, the two transformed
conic sections have then just the same pole for the transformed of
line at infinity in that homology.
The case you mention (orthogonal projection) of an affine transform
is that of a projective homology which holds the line at infinity, hence
the concentric property.
Hyperbola being defined as conic sections which intersect the line
at infinity, any affine transform can't change the kind of conic,  as it
holds the line at infinity, as a whole.
Ellipses are transformed into ellipses (a circle is a specific case of
ellipse) hyperbolas into hyperbolas and parabolas into parabolas.

> A problem remains open for me: if we know a vertex of the common
> autopolar triangle, how to construct the other two?

Still no idea in the general case, that is when the given vertex is not
the center of two concentric conics.
As the polar property is a projective one, it is held in the above mentioned
homology, hence we always can reduce the problem to one of the two
conic sections being a circle.
Then the properties (orthocentric system) of autopolar triangles wrt a
circle might help.

Best Regards.

Dear friends of Hyacinthos,
I found an interesting concyclic problem. Can anyone solve it synthetically?
Given triangle ABC with its circumcircle (O) and its orthocenter H. Let
H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on
Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2
be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that
H, A_2,B_2,C_2 are concyclic.
Best regard,
Linh Nguyen Van

Dear Linh Nguyen Van,

We have that the result is also true when P is on the circumcircle.
In this case A1=B1=C1=P and the circumcircle of A2B2C2 is image of the nine
point circle by a homothety with P as center and ratio 2. Is easy that H is on
this circumcircle.

Finally, we have a degenerate case when P is on an altitude, say AH. In this
case we have A2=H, thus H,A_2,B_2,C_2 are trivially concyclic.

--- In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...>
wrote:
>
> Dear friends of Hyacinthos,
> I found an interesting concyclic problem. Can anyone solve it synthetically?
> Given triangle ABC with its circumcircle (O) and its orthocenter H. Let
H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on
Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2
be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that
H, A_2,B_2,C_2 are concyclic.
> Best regard,
> Linh Nguyen Van
>

Dear Linh Nguyen Van

See Hagge circle construction in FG 2007 pp. 231-247

M.T.

--- In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...>
wrote:
>
> Dear friends of Hyacinthos,
> I found an interesting concyclic problem. Can anyone solve it synthetically?
> Given triangle ABC with its circumcircle (O) and its orthocenter H. Let
H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on
Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2
be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that
H, A_2,B_2,C_2 are concyclic.
> Best regard,
> Linh Nguyen Van
>

Dear Francisco,
Yes, two cases you mentioned are trivival. But my problem is stated that P lies
on Euler line.

Dear Armpist,
Please read the message above carefully! My problem is only similar to Hagge
circle, but it is not Hagge.

Best regard,
Linh Nguyen Van

Dear Linh Nguyen Van,

I haven't a synthetic solution at the moment, but perhaps is useful to know that
there are another (although trivial) solutions of the problem.

--- In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...>
wrote:
>
> Dear friends of Hyacinthos,
> I found an interesting concyclic problem. Can anyone solve it synthetically?
> Given triangle ABC with its circumcircle (O) and its orthocenter H. Let
H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on
Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2
be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that
H, A_2,B_2,C_2 are concyclic.
> Best regard,
> Linh Nguyen Van
>

[Linh Nguyen Van]
> Given triangle ABC with its circumcircle (O) and its orthocenter
> H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is
> an arbitrary point on Euler line. AP, BP, CP intersect (O) again
> at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1,
> B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2
> are concyclic.

As locus problem:

Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic
triangle), P a point, A1B1C1 the circumcevian triangle of P
and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc, resp.

Which is the locus of P such that the circumcircle of
A2B2C2 passes through H?

Questions:

1. Is the answer: "Circumcircle of ABC + Three Altitudes + Euler
Line of ABC" complete?

2. As P moves on the Euler line, where is moving the circumcenter
of A2B2C2?

Variation:

Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
Which is the locus of P such that the circumcircle of
A2B2C2 passes through H?

Antreas

Dear Antreas

> As locus problem:
>
> Let ABC be a triangle, HaHbHc the pedal triangle of H
> (orthic
> triangle), P a point, A1B1C1 the circumcevian triangle of P
>
> and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> resp.
>
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
> Questions:
>
>
> 2. As P moves on the Euler line, where is moving the
> circumcenter
> of A2B2C2?

It seems to be part of the Euler line.

Best regards
Nikos

__________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam);   Το Yahoo! Mail διαθέτει την καλύτερη
δυνατή προστασία κατά των ενοχλητικών μηνυμάτων
http://mail.yahoo.gr

We may want to know what is the locus of points P such that the orthocenter lies
on one of the sides of A2B2C2.

Answer: For example, B2, C2 and H are collinear if and only if P is on the
inverse with respect to the circumcircle of ABC of the circumcircle of BCH.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> [Linh Nguyen Van]
> > Given triangle ABC with its circumcircle (O) and its orthocenter
> > H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is
> > an arbitrary point on Euler line. AP, BP, CP intersect (O) again
> > at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1,
> > B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2
> > are concyclic.
>
> As locus problem:
>
> Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic
> triangle), P a point, A1B1C1 the circumcevian triangle of P
> and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc, resp.
>
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
> Questions:
>
> 1. Is the answer: "Circumcircle of ABC + Three Altitudes + Euler
> Line of ABC" complete?
>
> 2. As P moves on the Euler line, where is moving the circumcenter
> of A2B2C2?
>
> Variation:
>
> Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
> B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
> Antreas
>

Dear Linh Nguyen Van,

you wrote
> Given triangle ABC with its circumcircle (O) and
> its orthocenter H. Let HaHbHc be the orthic triangle of
> triangle ABC. P is an arbitrary point on Euler line.
> AP, BP, CP intersect (O) again at A1, B1, C1.
> Let A2, B2, C2 be the reflections of A1, B1, C1 wrt Ha, Hb, Hc,
> respectively. Prove that H, A2, B2, C2 are concyclic.

I tried to give a synthetic proof but did nothing.
I tried to give a proof with barycentrics but also did nothing
because the computations were very complicated.

Here is a proof with complex numbers.
Let the circumcircle of ABC be the unit circle and at the
points A, B, C, O be the complex numbers a, b, c, 0.
Since the conjugate(a) = 1/a ... we can do all we need.
Then it is known and easy to prove that at H is the number h=a+b+c.
At the point P on the Euler line is the number t*(a+b+c).
At the point A1 is the number a1=bc[a-t(a+b+c)]/[t(ab+bc+ca)-bc].
At the point Ha is the number ha=[a(a+b+c)-bc]/(2a).
The reflection of O in Ha is the point O1 at the number o1=2ha.
The point A2 is at the number a2 = 2ha - a1.
The triangle O1A2H is isosceles because
O1A2 = OA1 = R = OH1 = O1H  where H1 (on the circumcircle)
is the reflection of H in Ha.
The mid point M1 of A2H is at the point m1 = (a2+h)/2.
Finally the line O1M1 meets the Euler line at the point Q
that is equidistant from A2 and H.
This point Q is at the number q = m*(a+b+c)
where m = [(s+2abc)t-2abc]/[(s+abc)t-abc]
where s = (a+b)(b+c)(c+a).
Since this point Q is symmetric relative to a,b,c
is equidistant from B2, H  and from C2, H.
Hence the points A2, B2, C2, H are concyclic
with center Q on the Euler line.

If P = H then t = 1 and Q = H
If P = O then t = 0 and Q = X382 the reflection of O in H.

Best regards
Nikos Dergiades




__________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam);   Το Yahoo! Mail διαθέτει την καλύτερη
δυνατή προστασία κατά των ενοχλητικών μηνυμάτων
http://mail.yahoo.gr

Dear friends,
Francisco mentioned me that the points
A2, B2, C2, H are collinear if P is the inverse
of H in (O).
Yes this is true if Q lies at infinity
or the denominator of q is zero
that is (s+abc)t-abc = 0 or
t = abc / [(a+b)(b+c)(c+a)+abc]
and for this t the point P is the inverse
of H in (O).
Thank you Francisco.
Best regards
Nikos Dergiades


> Dear Linh Nguyen Van,
>
> you wrote
> > Given triangle ABC with its circumcircle (O) and
> > its orthocenter H. Let HaHbHc be the orthic triangle
> of
> > triangle ABC. P is an arbitrary point on Euler line.
> > AP, BP, CP intersect (O) again at A1, B1, C1.
> > Let A2, B2, C2 be the reflections of A1, B1, C1 wrt
> Ha, Hb, Hc,
> > respectively. Prove that H, A2, B2, C2 are concyclic.
>
> I tried to give a synthetic proof but did nothing.
> I tried to give a proof with barycentrics but also did
> nothing
> because the computations were very complicated.
>
> Here is a proof with complex numbers.
> Let the circumcircle of ABC be the unit circle and at the
> points A, B, C, O be the complex numbers a, b, c, 0.
> Since the conjugate(a) = 1/a ... we can do all we need.
> Then it is known and easy to prove that at H is the number
> h=a+b+c.
> At the point P on the Euler line is the number t*(a+b+c).
> At the point A1 is the number
> a1=bc[a-t(a+b+c)]/[t(ab+bc+ca)-bc].
> At the point Ha is the number ha=[a(a+b+c)-bc]/(2a).
> The reflection of O in Ha is the point O1 at the number
> o1=2ha.
> The point A2 is at the number a2 = 2ha - a1.
> The triangle O1A2H is isosceles because
> O1A2 = OA1 = R = OH1 = O1H  where H1 (on the
> circumcircle)
> is the reflection of H in Ha.
> The mid point M1 of A2H is at the point m1 = (a2+h)/2.
> Finally the line O1M1 meets the Euler line at the point Q
> that is equidistant from A2 and H.
> This point Q is at the number q = m*(a+b+c)
> where m = [(s+2abc)t-2abc]/[(s+abc)t-abc]
> where s = (a+b)(b+c)(c+a).
> Since this point Q is symmetric relative to a,b,c
> is equidistant from B2, H  and from C2, H.
> Hence the points A2, B2, C2, H are concyclic
> with center Q on the Euler line.
>
> If P = H then t = 1 and Q = H
> If P = O then t = 0 and Q = X382 the reflection of O in H.
>
> Best regards
> Nikos Dergiades
>
>
>  
>
> __________________________________________________
> Χρησιμοποιείτε Yahoo!;
> Βαρεθήκατε τα ενοχλητικά
> μηνύματα (spam);   Το Yahoo! Mail
> διαθέτει την καλύτερη δυνατή
> προστασία κατά των ενοχλητικών
> μηνυμάτων 
> http://mail.yahoo.gr
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

__________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam);   Το Yahoo! Mail διαθέτει την καλύτερη
δυνατή προστασία κατά των ενοχλητικών μηνυμάτων
http://mail.yahoo.gr

Thanks you dear Nikolaos for your solution. I am also finding the synthetic
solution but false.

>
>[Francisco Javier]
> Variation:
>
> Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
> B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
Dear Francisco,
For this variation you can see 'Hagge circle'. Every points P on the plane
satisfy this condition.

Dear Linh Nguyen Van!
   I haven't a synthetic proof, but using  complex numbers it is easy to prove
your assertion and also obtain that the circumcenter of A_2B_2C_2 lie on the
Euelr line.

   Sincerely                                   Alexey



   Dear friends of Hyacinthos,
   I found an interesting concyclic problem. Can anyone solve it synthetically?
   Given triangle ABC with its circumcircle (O) and its orthocenter H. Let
H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on
Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2
be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that
H, A_2,B_2,C_2 are concyclic.
   Best regard,
   Linh Nguyen Van



   .



[Non-text portions of this message have been removed]

[APH]
> > As locus problem:
> >
> > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > (orthic
> > triangle), P a point, A1B1C1 the circumcevian triangle of P
> >
> > and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> > resp.
> >
> > Which is the locus of P such that the circumcircle of
> > A2B2C2 passes through H?
> >
> > Questions:
> >
> >
> > 2. As P moves on the Euler line, where is moving the
> > circumcenter
> > of A2B2C2?

[ND]
> It seems to be part of the Euler line.

Dear Nikos

So if P lies on the Euler line, then the circumcenter
of A2B2C2 lies also on the Euler line.

Reversely: If the circumcenter of A2B2C2 is on the
Euler line, then where is lying P? That is:

Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), P a point, A1B1C1 the circumcevian triangle
of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
resp.
Which is the locus of P such that the circumcenter of
A2B2C2 is lying on the Euler Line of ABC?

It is the Euler Line + ?????

APH

[APH]
> Let ABC be a triangle, HaHbHc the pedal triangle of H
> (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> resp.
> Which is the locus of P such that the circumcenter of
> A2B2C2 is lying on the Euler Line of ABC?
>
> It is the Euler Line + ?????

Variation (easier?):

Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), P a point, and A',B', and C' the reflections
of P in Ha, Hb, Hc, resp.
Which is the locus of P such that the circumcenter of
A'B'C' is lying on the Euler Line of ABC?

Is it Euler Line +???

APH