Dear Antreas,

[APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC, ICA,
IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections of BC, CA, AB,
in La, Lb, Lc, resp.

The circumcenter Q of the triangle bounded by the lines (L1,L2,L3) is on the
line L.

True?

*** No.

Best regards
Sincerely
Paul

Dear Paul

This time there are too many points P on the Euler line..... !

Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in the
perpendicular
bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian triangle of P
wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

APH


On Thu, Feb 7, 2013 at 1:50 AM, Paul Yiu <yiu@...> wrote:

> **
>
>
> Dear Antreas,
>
> [APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC,
> ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections of
> BC, CA, AB, in La, Lb, Lc, resp.
>
>
> The circumcenter Q of the triangle bounded by the lines (L1,L2,L3) is on
> the line L.
>
> True?
>
> *** No.
>
> Best regards
> Sincerely
> Paul
>
>  _
>


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

Dear Antreas,

This is wonderful.

[APH]:  Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in
the
perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian
triangle of P
wrt A'B'C'. Which is the locus of P such that ABC, A"B"C" are perspective?

A ' = (a^2:-(b^2-c^2):b^2-c^2),

If P = (u:v:w), then
A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) :
-c^2(v+w)((b^2-c^2)u-a^2w))
etc.

A''B''C'' and ABC are perspective if and only if P lies on the Euler line.
The perspector Q also lies on the Euler line.

If OP P PH = t : 1-t, then
OQ : QH = (1+t) : -8t\cos A\cos B\cos C

Here are some examples:

  P       Q
------------
  G      X(25)
  O      H
  H      X(24)
  N      X(3518)
  L       O
  X(21) X (28)
  X(22)  G
  X(23)  X(468)
  X(186  X(403)

Best regards
Sincerely
Paul

The following paper has been published in Forum Geometricorum. It can be viewed
at

http://forumgeom.fau.edu/FG2013volume13/FG201305index.html

The editors
Forum Geometricorum
Martin Josefsson, Characterizations of trapezoids,
Forum Geometricorum, 13 (2013) 23--35.

Abstract. We review eight and prove an additional 13 necessary and sufficient
conditions for a convex quadrilateral to be a trapezoid. One aim for this paper
is to show that many of the known properties of trapezoids are in fact
characterizations.


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

The following paper has been published in Forum Geometricorum. It can be viewed
at

http://forumgeom.fau.edu/FG2013volume13/FG201304index.html

The editors
Forum Geometricorum
Martin Josefsson, Five proofs of an area characterization of rectangles,
Forum Geometricorum, 13 (2013) 17--21.

Abstract. There are a handful of well known characterizations of rectangles,
most of which concerns one or all four of the angles of the quadrilateral. One
example is that a parallelogram is a rectangle if and only if it has (at least)
one right angle. Here we shall prove that \emph{a convex quadrilateral with
consecutive sides a, b, c, d is a rectangle if and only if its area K satisfies
K = (1/2)Sqrt((a^2+c^2)(b^2+d^2)). We give five different proofs of this area
characterization.


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

Dear Paul

Note that the triangle A'B'C', where A',B',C' are the reflections of A,B,C
in the perpendicular bisectors,
is the triangle which is the antipodal triangle of the circumcevian
triangle of H.

So we can replace H with an other fixed point T on the Euler line.

That is:

Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the
circumcevian
triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the
antipodes of At, Bt,Ct).

Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

I think that the locus is the Euler Line.

We can furthermore generalize it by taking T as an arbitrary fixed point.
The locus will be some line, I guess.

Greetings

Antreas

On Thu, Feb 7, 2013 at 5:14 PM, Paul Yiu <yiu@...> wrote:

> **
>
>
> Dear Antreas,
>
> This is wonderful.
>
> [APH]: Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C
> in the
>
> perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian
> triangle of P
> wrt A'B'C'. Which is the locus of P such that ABC, A"B"C" are perspective?
>
> A ' = (a^2:-(b^2-c^2):b^2-c^2),
>
> If P = (u:v:w), then
> A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) :
> -c^2(v+w)((b^2-c^2)u-a^2w))
> etc.
>
> A''B''C'' and ABC are perspective if and only if P lies on the Euler line.
> The perspector Q also lies on the Euler line.
>
> If OP P PH = t : 1-t, then
> OQ : QH = (1+t) : -8t\cos A\cos B\cos C
>
> Here are some examples:
>
> P Q
> ------------
> G X(25)
> O H
> H X(24)
> N X(3518)
> L O
> X(21) X (28)
> X(22) G
> X(23) X(468)
> X(186 X(403)
>
> Best regards
> Sincerely
> Paul
>
>  _
>


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

Dear Antreas,

You are right.

[APH]:  Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the
circumcevian triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C'
are the
antipodes of At, Bt,Ct).

Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

*** The locus is the line OT, and the perspector Q lies on OT as well.

If OP : PT = t: 1-t, then
OQ : QT = R^2(1+t) : - (R^2-OT^2) t.

Best regards
Sincerely
Paul

Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB, resp.
(concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB, resp.

Denote:

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.

1. The triangles ABC, O1O2O3 are perspective.

2. The circumcenter of O1O2O3 is on the line L.

True??

Figure
http://anthrakitis.blogspot.gr/2013/02/euler-lines-circumcircles.html

APH

Both are true.

The circumcenter of O1O2O3 divides the segment HO in the ratio
(4 r + 5 R)/R.


--- In Hyacinthos@yahoogroups.com, "Antreas"  wrote:
>
> Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB, resp.
(concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB, resp.
>
> Denote:
>
> Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
>
> Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
>
> Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
>
> O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.
>
> 1. The triangles ABC, O1O2O3 are perspective.
>
> 2. The circumcenter of O1O2O3 is on the line L.
>
> True??
>
> Figure
> http://anthrakitis.blogspot.gr/2013/02/euler-lines-circumcircles.html
>
> APH
>

Are the points in ETC?

aph

On Fri, Feb 8, 2013 at 10:19 AM, Francisco Javier
<garciacapitan@...>wrote:

> **
>
>
> Both are true.
>
> The circumcenter of O1O2O3 divides the segment HO in the ratio
> (4 r + 5 R)/R.
>
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
> >
> > Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB,
> resp. (concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB,
> resp.
> >
> > Denote:
> >
> > Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
> >
> > Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
> >
> > Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
> >
> > O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.
> >
> > 1. The triangles ABC, O1O2O3 are perspective.
> >
> > 2. The circumcenter of O1O2O3 is on the line L.
> >
> > True??
> >
> > Figure
> > http://anthrakitis.blogspot.gr/2013/02/euler-lines-circumcircles.html
> >
> > APH
> >
>
>
>


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

Peter J. C. Moses sent me complete answers for both, I have added
in the problem. See

http://anthrakitis.blogspot.gr/2013/02/euler-lines-circumcircles.html

Thanks !

Antreas

PS There are more things in the configuration, but later.... :-)


--- In Hyacinthos@yahoogroups.com, "Antreas"  wrote:
>
> Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB, resp.
(concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB, resp.
>
> Denote:
>
> Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
>
> Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
>
> Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
>
> O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.
>
> 1. The triangles ABC, O1O2O3 are perspective.
>
> 2. The circumcenter of O1O2O3 is on the line L.
>
> True??
>
> Figure
> http://anthrakitis.blogspot.gr/2013/02/euler-lines-circumcircles.html
>
> APH
>

GENERALIZATION

  Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of ABC, IBC, ICA, IAB,
resp. and S a point.

Denote:

L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.

1. For which S's the triangles ABC, O1O2O3 are perspective? (Locus)

2. The circumcenter of O1O2O3 is on the line L for all S's (??).

http://anthrakitis.blogspot.gr/2013/02/four-concurrent-lines-circumcircles.html


Special Caså:
L, La,Lb,Lc = Brocard Axes (they are concurrent)

APH

--- In Hyacinthos@yahoogroups.com, "Antreas"  wrote:
>
> Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB, resp.
(concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB, resp.
>
> Denote:
>
> Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
>
> Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
>
> Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
>
> O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.
>
> 1. The triangles ABC, O1O2O3 are perspective.
>
> 2. The circumcenter of O1O2O3 is on the line L.
>
> True??
>
> Figure
> http://anthrakitis.blogspot.gr/2013/02/euler-lines-circumcircles.html
>
> APH
>

Dear Antreas,

the locus is the conic

b^3 c x^2 - b c^3 x^2 - 2 a^3 c x y - a^2 b c x y + a b^2 c x y +
  2 b^3 c x y + 2 a c^3 x y - 2 b c^3 x y - a^3 c y^2 + a c^3 y^2 +
  2 a^3 b x z - 2 a b^3 x z + a^2 b c x z + 2 b^3 c x z -
  a b c^2 x z - 2 b c^3 x z + 2 a^3 b y z - 2 a b^3 y z - 2 a^3 c y z -
   a b^2 c y z + a b c^2 y z + 2 a c^3 y z + a^3 b z^2 - a b^3 z^2=0

Yes . The circumcenter of O1O2O3 is on the line L for all S's.

The three centers Oa, Ob, Oc are the mid points of the arcs
BC, CA, AB of the circumcircle of ABC.
The points O1, O2, O3 are the mid points of SOa, SOb, SOc.
If K is the mid point of SO then KO1 = OOa/2 = R/2.
Hence K is the center of the circle O1O2O3 with radious R/2.

Best regards
Nikos






> GENERALIZATION
>
>  Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
> ABC, IBC, ICA, IAB, resp. and S a point.
>
> Denote:
>
> L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
>
> Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
>
> Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
>
> Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
>
> O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
> resp.
>
> 1. For which S's the triangles ABC, O1O2O3 are perspective?
> (Locus)
>
> 2. The circumcenter of O1O2O3 is on the line L for all S's
> (??).
>

Dear Nikos,

We can replace the incenter I with any other fixed point and ask for the
loci,
but I guess they are complicated in that general case.

Probably we get interesting loci for points on the Neuberg cubic (like I)

How about H instead of I ?

I will ask about H in an other post.

APH


On Fri, Feb 8, 2013 at 10:32 PM, Nikolaos Dergiades <ndergiades@...>wrote:

> **
>
>
> Dear Antreas,
>
> the locus is the conic
>
> b^3 c x^2 - b c^3 x^2 - 2 a^3 c x y - a^2 b c x y + a b^2 c x y +
> 2 b^3 c x y + 2 a c^3 x y - 2 b c^3 x y - a^3 c y^2 + a c^3 y^2 +
> 2 a^3 b x z - 2 a b^3 x z + a^2 b c x z + 2 b^3 c x z -
> a b c^2 x z - 2 b c^3 x z + 2 a^3 b y z - 2 a b^3 y z - 2 a^3 c y z -
> a b^2 c y z + a b c^2 y z + 2 a c^3 y z + a^3 b z^2 - a b^3 z^2=0
>
> Yes . The circumcenter of O1O2O3 is on the line L for all S's.
>
> The three centers Oa, Ob, Oc are the mid points of the arcs
> BC, CA, AB of the circumcircle of ABC.
> The points O1, O2, O3 are the mid points of SOa, SOb, SOc.
> If K is the mid point of SO then KO1 = OOa/2 = R/2.
> Hence K is the center of the circle O1O2O3 with radious R/2.
>
> Best regards
> Nikos
>
> > GENERALIZATION
> >
> > Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
> > ABC, IBC, ICA, IAB, resp. and S a point.
> >
> > Denote:
> >
> > L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
> >
> > Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
> >
> > Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
> >
> > Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
> >
> > O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
> > resp.
> >
> > 1. For which S's the triangles ABC, O1O2O3 are perspective?
> > (Locus)
> >
> > 2. The circumcenter of O1O2O3 is on the line L for all S's
> > (??).
> >
>


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

Case of H:

Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
ABC, HBC, HCA, HAB, resp. and S a point.

Denote:

L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
resp.

1. For which S's the triangles ABC, O1O2O3 are perspective?
[they are homothetic, I think]
2. For which S's the circumcenter of O1O2O3 is on the line L ?

Interesting Cases:

1/ S is lying on the Euler Line of ABC (so L = SO = Euler line).
I think that for both 1,2 the answer is the Euler line!

2/ L, La, Lb, Lc = the Brocard axes of ABC, HBC, HCA, HAB, concurrent
at a point S.
The triangles ABC, O1O2O3 are perspective, but the circumcenter of O1O2O3 is not
on L.

APH

Dear Antreas,
Let P be an arbitrary point
Let ABC be a triangle, O, Oa, Ob, Oc the circumcenters of
ABC, PBC, PCA, PAB, resp. and S a point.
Denote:
L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
resp. and O' the circumcenter of O1O2O3.
Fact1 = the triangles ABC, O1,O2,O3 are perspective
Fact2 = O' lies on L

The locus of S for Fact1  is a conic and
the locus of S for Fact2 is a line.

If P = I = Ia = Ib = Ic Fact2 holds for every S

If P = H  then Fact1 holds for every S and Fact2 if S lies on
the Euler line.

ND


[APH]
> Case of H:
>
> Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
> ABC, HBC, HCA, HAB, resp. and S a point.
>
> Denote:
>
> L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
>
> Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
>
> Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
>
> Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
>
> O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
> resp.
>
> 1. For which S's the triangles ABC, O1O2O3 are perspective?
> [they are homothetic, I think]
> 2. For which S's the circumcenter of O1O2O3 is on the line L
> ?
>
> Interesting Cases:
>
> 1/ S is lying on the Euler Line of ABC (so L = SO = Euler
> line).
> I think that for both 1,2 the answer is the Euler line!
>
> 2/ L, La, Lb, Lc = the Brocard axes of ABC, HBC, HCA, HAB,
> concurrent
> at a point S.
> The triangles ABC, O1O2O3 are perspective, but the
> circumcenter of O1O2O3 is not on L.
>
> APH
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>

Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of ABC,NBC,NCA,NAB, resp.
[concurrent at pN]. The NPCs of the triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur
at point ppN on the Euler Line of ABC.

Coordinates of ppN?

Generalization:

Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs of ABC, PBC, PCA,
PAB resp. [concurrent at pP]. If P is on the Euler line of ABC, then the NPCs of
N1N2N3, PN1N2, PN2N3, PN3N1 concur at point ppP on the Euler Line of ABC.

True??

APH

Dear Antreas,

pP is the center of the rectangular circumhyperbola through P.  ppP does not,
in general, lie on the Euler line.

Some results:

P=X(1), the NPCs are concurrent, with center = non-ETC 1.121590125545969 (on
lines 1,5 3,962).
P=X(2), ppP=non-ETC 1.690358502447462
P=X(3), ppP=X(140)
P=X(4), ppP=undefined
P=X(5), ppP=X(3628)
P=X(6), ppP=non ETC 0.780037257060191
P=X(7), ppP=non ETC 0.750876768572663
P=X(8), ppP=non ETC 2.966801160450799
P=X(9), ppP=non-ETC 0.972023454564163
P=X(10), ppP=non-ETC 2.238481946743318
P=X(20), ppP=non-ETC 6.363850996796102
P=X(21), ppP=non-ETC -1.717011738240629
P=X(22), ppP=non-ETC -4.036288926987237

Of these, only X(140) and X(3628) lie on the Euler line.  The ppP for points P
on the Euler line do not even lie on the same conic.  Locus?

Best regards,

Randy




>________________________________
> From: Antreas <anopolis72@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Sunday, February 10, 2013 6:07 PM
>Subject: [EMHL] Poncelet points on the Euler line
>
>
> 
>Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of ABC,NBC,NCA,NAB, resp.
[concurrent at pN]. The NPCs of the triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur
at point ppN on the Euler Line of ABC.
>
>Coordinates of ppN?
>
>Generalization:
>
>Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs of ABC, PBC, PCA,
PAB resp. [concurrent at pP]. If P is on the Euler line of ABC, then the NPCs of
N1N2N3, PN1N2, PN2N3, PN3N1 concur at point ppP on the Euler Line of ABC.
>
>True??
>
>APH
>
>
>
>
>

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

Dear Randy

Thanks!!

I think it would be interesting to study this special case:

Let p(pN) be the point where concur the NPCs of N1N2N3, pNN1N2,
pNN2N3, pNN3N1 (ie we replace N with pN).

Is it interesting ? ie ls it lying on some interesting curves, lines?

In my figure it lies on the circle centered at N with radius NppN

APH

[Randy Hutson]
> pP is the center of the rectangular circumhyperbola through P.
> ppP does not, in general, lie on the Euler line.
>
> Some results:
>
> P=X(1), the NPCs are concurrent, with center = non-ETC 1.121590125545969 (on
lines 1,5 3,962).
> P=X(2), ppP=non-ETC 1.690358502447462
> P=X(3), ppP=X(140)
> P=X(4), ppP=undefined
> P=X(5), ppP=X(3628)
> P=X(6), ppP=non ETC 0.780037257060191
> P=X(7), ppP=non ETC 0.750876768572663
> P=X(8), ppP=non ETC 2.966801160450799
> P=X(9), ppP=non-ETC 0.972023454564163
> P=X(10), ppP=non-ETC 2.238481946743318
> P=X(20), ppP=non-ETC 6.363850996796102
> P=X(21), ppP=non-ETC -1.717011738240629
> P=X(22), ppP=non-ETC -4.036288926987237
>
> Of these, only X(140) and X(3628) lie on the Euler line. 
> The ppP for points P on the Euler line do not even lie on the
> same conic. Locus?
[APH]
> >Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of
> >ABC,NBC,NCA,NAB, resp. [concurrent at pN]. The NPCs of the
> >triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur at point ppN
> >on the Euler Line of ABC.
> >
> >Coordinates of ppN?
> >
> >Generalization:
> >
> >Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs
> >of ABC, PBC, PCA, PAB resp. [concurrent at pP]. If P is on
> >the Euler line of ABC, then the NPCs of N1N2N3, PN1N2, PN2N3,
> >PN3N1 concur at point ppP on the Euler Line of ABC.
> >
> >True??
> >
> >APH

Dear Hyacinthists,
an article concerning “A variant of IMO Beijing China 1990’’ has been put
on my website.


http://perso.orange.fr/jl.ayme    vol. 10

Sincerely
Jean-Louis

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

Let ABC be a triangle.

1. Let (N1),(N2),(N3) be the reflections of the NPC (N) in the
sidelines BC,CA,AB, resp. and A'B'C' the triangle
bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.

Are the triangles ABC, A'B'C' are perspective ?


2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the reflections
of the NPC (N) in the altitudes HHa,HHb,HHc,
resp. and and A'B'C' the triangle bounded by the radical axes of
((O),(N1)), ((O),(N2)), ((O),(N3)), resp.

Are the triangles HaHbHc, A'B'C' are perspective ?

Perspectors?

APH

Dear Antreas

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis  wrote:
>
> Let ABC be a triangle.
>
> 1. Let (N1),(N2),(N3) be the reflections of the NPC (N) in the
> sidelines BC,CA,AB, resp. and A'B'C' the triangle
> bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
> Are the triangles ABC, A'B'C' are perspective ?


****  Yes. Perspector X(54)

> 2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the reflections
> of the NPC (N) in the altitudes HHa,HHb,HHc,
> resp. and and A'B'C' the triangle bounded by the radical axes of
> ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
> Are the triangles HaHbHc, A'B'C' are perspective ?
>

**** Yes. Perspector: Baricentric product of  X(4) X(30) X(30).



Angel M.

Dear Antreas,

1. The perspector is X54.

2. The perspector is the product X4 and X3163.

Best regards
Nikos


> Let ABC be a triangle.
>
> 1. Let (N1),(N2),(N3) be the reflections of the NPC (N) in
> the
> sidelines BC,CA,AB, resp. and A'B'C' the triangle
> bounded by the radical axes of ((O),(N1)), ((O),(N2)),
> ((O),(N3)), resp.
>
> Are the triangles ABC, A'B'C' are perspective ?
>
>
> 2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the
> reflections
> of the NPC (N) in the altitudes HHa,HHb,HHc,
> resp. and and A'B'C' the triangle bounded by the radical
> axes of
> ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
> Are the triangles HaHbHc, A'B'C' are perspective ?
>
> Perspectors?
>
> APH
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>

Dear Antreas,

Some interesting results from (2.):

Triangle N1N2N3 is inversely similar to ABC, with similitude center X(195).
X(3) of N1N2N3 = X(4)
X(4) of N1N2N3 = X(5) of circumorthic triangle
X(5) of N1N2N3 = X(143)
X(6) of N1N2N3 = X(576)
X(30) of N1N2N3 = isogonal conjugate of X(477)
X(2070) of N1N2N3 = X(265)

The perspector of (2.) = isogonal conjugate wrt orthic triangle of X(1990)
= pole, wrt polar circle, of tangent to Steiner circumellipse at X(1494) (line
X(1494)X^-1(1304))
= polar conjugate of isotomic conjugate of X(3163)

Randy





>________________________________
> From: Antreas Hatzipolakis <anopolis72@...>
>To: Hyacinthos <Hyacinthos@yahoogroups.com>
>Sent: Monday, February 11, 2013 6:43 AM
>Subject: [EMHL] Perspective?
>
>
> 
>Let ABC be a triangle.
>
>1. Let (N1),(N2),(N3) be the reflections of the NPC (N) in the
>sidelines BC,CA,AB, resp. and A'B'C' the triangle
>bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
>Are the triangles ABC, A'B'C' are perspective ?
>
>2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the reflections
>of the NPC (N) in the altitudes HHa,HHb,HHc,
>resp. and and A'B'C' the triangle bounded by the radical axes of
>((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
>Are the triangles HaHbHc, A'B'C' are perspective ?
>
>Perspectors?
>
>APH
>
>
>
>

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

3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
the reflections of the NPC (N) in the perp. bisectors
OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
The triangles ABC, A'B'C' are perspective.

Perspector?

APH

On Mon, Feb 11, 2013 at 2:43 PM, Antreas Hatzipolakis
<anopolis72@...> wrote:
> Let ABC be a triangle.
>
> 1. Let (N1),(N2),(N3) be the reflections of the NPC (N) in the
> sidelines BC,CA,AB, resp. and A'B'C' the triangle
> bounded by the radical axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
> Are the triangles ABC, A'B'C' are perspective ?
>
>
> 2. Let HaHbHc be the orthic triangle, (N1),(N2),(N3) the reflections
> of the NPC (N) in the altitudes HHa,HHb,HHc,
> resp. and and A'B'C' the triangle bounded by the radical axes of
> ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
>
> Are the triangles HaHbHc, A'B'C' are perspective ?
>
> Perspectors?
>
> APH

Dear Antreas,

The triangles ABC, A'B'C' are perspective.

  Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)

Best regards.
Angel M.

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis  wrote:
>
> 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
> the reflections of the NPC (N) in the perp. bisectors
> OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
> axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
> The triangles ABC, A'B'C' are perspective.
>
> Perspector?
>
> APH
>

Another problem:

Let ABC, A'B'C' be two arbitrary equilateral triangles.

Denote:

Ab, Ac = the reflections of A in BB', CC', resp.
Bc, Ba = the reflections of B in CC', AA', resp.
Ca, Cb = the reflections of C in AA', BB', resp.

The Euler lines of AAbAc, BBcBa, CCaCb are concurrent.

Corollary:

Let ABC be an equilateral triangle and P a point.

Denote:

Ab, Ac = the reflections of A in BP, CP, resp.
Bc, Ba = the reflections of B in CP, AP, resp.
Ca, Cb = the reflections of C in AP, BP, resp.

The Euler lines of AAbAc, BBcBa, CCaCb are concurrent.

Proofs?

APH

--- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
>
> Let ABC be an equilateral triangle and P a point.
>
> Which same central lines of the triangles PBC, PCA, PAB
> are concurrent for all P's ?
>
> The OH (Euler) lines? The OI lines? ......
>
> APH
>

The following paper has been published in Forum Geometricorum. It can be viewed
at

http://forumgeom.fau.edu/FG2013volume13/FG201306index.html

The editors
Forum Geometricorum
Maria Calvo and Vicente Munoz, The most inaccessible point of a convex domain,
Forum Geometricorum, 13 (2013) 37--52.

Abstract. The inaccessibility of a point p in a bounded domain D subset R^n is
the minimum of the lengths of segments through p with boundary at bd D. The
points of maximum inaccessibility I_D are those where the inaccessibility
achieves its maximum. We prove that for strictly convex domains, I_D is either a
point or a segment, and that for a planar polygon I_D is in general a point. We
study the case of a triangle, showing that this point is not any of the
classical notable points.


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

Dear Antreas and Angel,

A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269 (coordinates?)

N1N2N3 is inversely similar to ABC, with similitude center X(3).
N1N2N3 is perspective to ABC at X(3519).
X(3) of N1N2N3 = X(3)
X(186) of N1N2N3 = X(1511)
X(2070) of N1N2N3 = X(110)
The Euler line of N1N2N3 is line X(3)X(74).
Line IO of N1N2N3 is line X(3)X(191).
The Brocard axis of N1N2N3 is line X(3)X(67).

Randy

--- In Hyacinthos@yahoogroups.com, "Angel"  wrote:
>
> Dear Antreas,
>
> The triangles ABC, A'B'C' are perspective.
>
>  Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)
>
> Best regards.
> Angel M.
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis  wrote:
> >
> > 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
> > the reflections of the NPC (N) in the perp. bisectors
> > OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
> > axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
> > The triangles ABC, A'B'C' are perspective.
> >
> > Perspector?
> >
> > APH
> >
>

Dear Randy,

A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269 (coordinates?)

Its coordinates are very complicated; the first coordinate of its center is:

a^20(b^2+c^2)-
2a^18(3b^4+7b^2c^2+3c^4)+
a^16(b^2+c^2)(13b^4+37b^2c^2+13c^4)-
2a^14(4b^8+39b^6c^2+68b^4c^4+39b^2c^6+4c^8)-
a^12(b^2+c^2)(14b^8-71b^6c^2-106b^4c^4-71b^2c^6+14c^8)+
2a^10(14b^12-3b^10c^2-64b^8c^4-74b^6c^6- 64b^4c^8-3b^2c^10+14c^12)-
a^8(b^2+c^2)(14b^12+27b^10c^2-110b^8c^4+ 66b^6c^6-   
110b^4c^8+27b^2c^10+14c^12)-
2a^6(b^2-c^2)^2(4b^12-27b^10c^2- 8b^8c^4-2b^6c^6-8b^4c^8-27b^2c^10+4c^12)+
a^4(b^2-c^2)^4(b^2+c^2)(13b^8-23b^6c^2+4b^4c^4- 23b^2c^6+13c^8) -
2a^2(b^2-c^2)^6(b^2+c^2)^2(3b^4-2b^2c^2+3c^4)+
(b^2-c^2)^8(b^2+c^2)^3


Best regards.
  Angel M.

--- In Hyacinthos@yahoogroups.com, "rhutson2"  wrote:
>
> Dear Antreas and Angel,
>
> A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269
(coordinates?)
>
> N1N2N3 is inversely similar to ABC, with similitude center X(3).
> N1N2N3 is perspective to ABC at X(3519).
> X(3) of N1N2N3 = X(3)
> X(186) of N1N2N3 = X(1511)
> X(2070) of N1N2N3 = X(110)
> The Euler line of N1N2N3 is line X(3)X(74).
> Line IO of N1N2N3 is line X(3)X(191).
> The Brocard axis of N1N2N3 is line X(3)X(67).
>
> Randy
>
> --- In Hyacinthos@yahoogroups.com, "Angel"  wrote:
> >
> > Dear Antreas,
> >
> > The triangles ABC, A'B'C' are perspective.
> >
> >  Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)
> >
> > Best regards.
> > Angel M.
> >
> > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis  wrote:
> > >
> > > 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)
> > > the reflections of the NPC (N) in the perp. bisectors
> > > OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical
> > > axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.
> > > The triangles ABC, A'B'C' are perspective.
> > >
> > > Perspector?
> > >
> > > APH
> > >
> >
>