## [EMHL] Re: ORTHOLINE

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• Dear Antreas, I checked when Point D is coinciding with X(13) then points 00, 11, 22, 33 are concyclic too (Cabri-proof). About the orthologic property between
Message 1 of 13 , Mar 24 7:31 AM
Dear Antreas,

I checked when Point D is coinciding with X(13) then points 00, 11, 22, 33 are concyclic too (Cabri-proof).

About the orthologic property between 2 triangles, it is even much more beautiful!

Denote A=P0, B=P1, C=P2, I=P3.
The points P0, P1, P2, P3 form a Complete Quadrangle.
Let L0 be the line through 00,01,02,03.
Let L1 be the line through 10,11,12,13.
Let L2 be the line through 20,21,22,23.
Let L3 be the line through 30,31,32,33.
The lines L0, L1, L2, L3 form a Complete Quadrilateral.
Now the Complete Quadrangle P0.P1.P2.P3 and the Complete Quadrilateral L0.L1.L2.L3 are orthologicically related (complete new notion to me!).
The four perpendiculars of P0,P1,P2,P3 on resp. L0,L1,L2,L3 are concurrent!
The six perpendiculars of S01, S02, S03, S12, S13, S23 on resp. P0.P1, P0.P2, P0.P3, P1.P2, P1.P3, P2.P3 form a network like the internal and external angle bisectors of a triangle producing an incenter and 3 excenters.

Last but not least, I forgot to tell that the 4 NPC's in my former message have a common point, which is the Euler-Poncelet Point (QA-P2)of the Quadrangle ABCI.

Best regards,

Chris van Tienhoven
www.chrisvantienhoven.nl

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> On Sun, Mar 24, 2013 at 2:50 PM, Chris Van Tienhoven
> <van10hoven@...>wrote:
>
> > **
> >
> >
> > Dear Antreas,
> >
> > When you replace point I by a random point D, then we have a random
> > Quadrangle ABCD (system of points without restrictions).
> > The 4 orthopoles of Component Triangles ABC, BCD, CDA, DAB are collinear.
> > Then this is also true when D coincides with I (Incenter X(1)) and when L
> > = some EulerLine.
> >
> > In a random Quadrangle points 00, 11, 22, 33 are not concyclic.
> > However when D coincides with I, then 00, 11, 22, 33 are concyclic indeed
> > (Cabri-proof).
> >
>
>
> Dear Chris
>
> Since Incenter I should not be the only one point with
> that property, I am wondering where are lying the other points (locus).
> Maybe it is a quite complicated locus. Or maybe Neuberg cubic is part of
> the locus.
>
> If Neuberg cubic indeed is, then Fermat points have the same property.
>
> So if you have time and want, you may check them (Fermat points instead of
> I)
>
> Antreas
>
>
>
> >
> > Further 00 lies on NPC of ABC.
> > 11, 22, 33 lie on NPC's ABD, BCD, CAD.
> >
> > Best regards,
> >
> > Chris van Tienhoven
> > www.chrisvantienhoven.nl
> >
> > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> > >
> > > Let ABC be a triangle and 0,1,2,3 the concurrent Euler lines
> > > of ABC, BCI, CAI, ABI, resp.
> > >
> > > Denote:
> > >
> > > 00 = the orthopole of 0 wrt ABC
> > >
> > > 01 = the orthopole of 0 wrt IBC
> > >
> > > 02 = the orthopole of 0 wrt ICA
> > >
> > > 03 = the orthopole of 0 wrt IAB
> > >
> > > ----
> > >
> > > 10 = the orthopole of 1 wrt ABC
> > >
> > > 11 = the orthopole of 1 wrt IBC
> > >
> > > 12 = the orthopole of 1 wrt ICA
> > >
> > > 13 = the orthopole of 1 wrt IAB
> > >
> > > ----
> > >
> > > 20 = the orthopole of 2 wrt ABC
> > >
> > > 21 = the orthopole of 2 wrt IBC
> > >
> > > 22 = the orthopole of 2 wrt ICA
> > >
> > > 23 = the orthopole of 2 wrt IAB
> > >
> > > ----
> > >
> > > 30 = the orthopole of 3 wrt ABC
> > >
> > > 31 = the orthopole of 3 wrt IBC
> > >
> > > 32 = the orthopole of 3 wrt ICA
> > >
> > > 33 = the orthopole of 3 wrt IAB
> > >
> > > In short:
> >
> > >
> > > Denote:
> > >
> > > Triangles ABC, IBC, ICA, IAB = (0), (1), (2), (3), resp.
> > >
> > > Euler lines of ABC, IBC, ICA, IAB = 0,1,2,3, resp.
> > >
> > > Orthopole of x [x in {0,1,2,3}] wrt (y)
> > > [(y) in {(0), (1), (2), (3)}] = xy
> > >
> > > How are organized the 16 points xy ?
> > >
> > > In four lines:
> > > (01 02 03 04), (11 12 13 14), (21 22 23 24) (31 32 33 34) (??)
> > >
> > > Are there points concyclic or on a conic in general?
> > >
> > > Antreas
> > >
> >
> > _
> >
>
>
> [Non-text portions of this message have been removed]
>
• Dear Chris Very nice!! Now, the problem is what points, other than I, have the concyclicy and orthology properties. If, for concyclicity, is necessary to be
Message 2 of 13 , Mar 24 10:06 AM
Dear Chris

Very nice!!

Now, the problem is what points, other than I, have the concyclicy and
orthology
properties.

If, for concyclicity, is necessary to be the orthopoles on the NPCs of
ABC, IBC, ICA, IAB
(and this happens when the concurrent lines pass through the circumcenters
of the
triangles, since the locus of the orthopoles of lines moving around O, is
the NPC)
then we can choose a variable point P and take (instead of the Euler lines
of the triangles)
the lines PO, POa, POb, POc, where O,Oa,Ob,Oc are the circumcenters of ABC,
IBC, ICA, IAB, resp.
and ask for which points P the orthopoles of PO, POa, POb, POc wrt ABC,
IBC, ICA, IAB
are concyclic. A point of the locus is Schiffler point S, on the Euler line.
And naturally, one expects that part of the locus is the Euler line!

Conjecture:
Let P be a point on the Euler line of ABC, and O, Oa,Ob,Oc the circumcenters
of ABC, IBC, ICA, IAB.
The orthopoles of PO, POa, POb, POc wrt ABC, IBC, ICA, IAB, resp.
are concyclic.

Question: Is it true for any point Q (instead of I) on the Neuberg cubic?

GENERAL CONJECTURE:

Let Q be a point on the Neuberg cubic, and P a point on the Euler line of
ABC.
Let O, Oa, Ob, Oc be the circumcenters of ABC, QBC, QCA, QAB.
The orthopoles of PO, POa, POb, POc wrt ABC, QBC, QCA, QAB, resp
are concyclic.

Antreas

On Sun, Mar 24, 2013 at 4:31 PM, Chris Van Tienhoven
<van10hoven@...>wrote:

> **
>
>
> Dear Antreas,
>
> I checked when Point D is coinciding with X(13) then points 00, 11, 22, 33
> are concyclic too (Cabri-proof).
>
> About the orthologic property between 2 triangles, it is even much more
> beautiful!
>
> Denote A=P0, B=P1, C=P2, I=P3.
> The points P0, P1, P2, P3 form a Complete Quadrangle.
> Let L0 be the line through 00,01,02,03.
> Let L1 be the line through 10,11,12,13.
> Let L2 be the line through 20,21,22,23.
> Let L3 be the line through 30,31,32,33.
> The lines L0, L1, L2, L3 form a Complete Quadrilateral.
> L0.L1.L2.L3 are orthologicically related (complete new notion to me!).
> The four perpendiculars of P0,P1,P2,P3 on resp. L0,L1,L2,L3 are concurrent!
> The six perpendiculars of S01, S02, S03, S12, S13, S23 on resp. P0.P1,
> P0.P2, P0.P3, P1.P2, P1.P3, P2.P3 form a network like the internal and
> external angle bisectors of a triangle producing an incenter and 3
> excenters.
>
> Last but not least, I forgot to tell that the 4 NPC's in my former message
> have a common point, which is the Euler-Poncelet Point (QA-P2)of the
> See:
>
>
> Best regards,
>
> Chris van Tienhoven
> www.chrisvantienhoven.nl
>
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
> wrote:
> >
> > On Sun, Mar 24, 2013 at 2:50 PM, Chris Van Tienhoven
> > <van10hoven@...>wrote:
> >
> > > **
>
> > >
> > >
> > > Dear Antreas,
> > >
> > > When you replace point I by a random point D, then we have a random
> > > Quadrangle ABCD (system of points without restrictions).
> > > The 4 orthopoles of Component Triangles ABC, BCD, CDA, DAB are
> collinear.
> > > Then this is also true when D coincides with I (Incenter X(1)) and
> when L
> > > = some EulerLine.
> > >
> > > In a random Quadrangle points 00, 11, 22, 33 are not concyclic.
> > > However when D coincides with I, then 00, 11, 22, 33 are concyclic
> indeed
> > > (Cabri-proof).
> > >
> >
> >
> > Dear Chris
> >
> > Since Incenter I should not be the only one point with
> > that property, I am wondering where are lying the other points (locus).
> > Maybe it is a quite complicated locus. Or maybe Neuberg cubic is part of
> > the locus.
> >
> > If Neuberg cubic indeed is, then Fermat points have the same property.
> >
> > So if you have time and want, you may check them (Fermat points instead
> of
> > I)
> >
> > Antreas
> >
> >
> >
> > >
> > > Further 00 lies on NPC of ABC.
> > > 11, 22, 33 lie on NPC's ABD, BCD, CAD.
> > >
> > > Best regards,
> > >
> > > Chris van Tienhoven
> > > www.chrisvantienhoven.nl
> > >
> > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> > > >
> > > > Let ABC be a triangle and 0,1,2,3 the concurrent Euler lines
> > > > of ABC, BCI, CAI, ABI, resp.
> > > >
> > > > Denote:
> > > >
> > > > 00 = the orthopole of 0 wrt ABC
> > > >
> > > > 01 = the orthopole of 0 wrt IBC
> > > >
> > > > 02 = the orthopole of 0 wrt ICA
> > > >
> > > > 03 = the orthopole of 0 wrt IAB
> > > >
> > > > ----
> > > >
> > > > 10 = the orthopole of 1 wrt ABC
> > > >
> > > > 11 = the orthopole of 1 wrt IBC
> > > >
> > > > 12 = the orthopole of 1 wrt ICA
> > > >
> > > > 13 = the orthopole of 1 wrt IAB
> > > >
> > > > ----
> > > >
> > > > 20 = the orthopole of 2 wrt ABC
> > > >
> > > > 21 = the orthopole of 2 wrt IBC
> > > >
> > > > 22 = the orthopole of 2 wrt ICA
> > > >
> > > > 23 = the orthopole of 2 wrt IAB
> > > >
> > > > ----
> > > >
> > > > 30 = the orthopole of 3 wrt ABC
> > > >
> > > > 31 = the orthopole of 3 wrt IBC
> > > >
> > > > 32 = the orthopole of 3 wrt ICA
> > > >
> > > > 33 = the orthopole of 3 wrt IAB
> > > >
> > > > In short:
> > >
> > > >
> > > > Denote:
> > > >
> > > > Triangles ABC, IBC, ICA, IAB = (0), (1), (2), (3), resp.
> > > >
> > > > Euler lines of ABC, IBC, ICA, IAB = 0,1,2,3, resp.
> > > >
> > > > Orthopole of x [x in {0,1,2,3}] wrt (y)
> > > > [(y) in {(0), (1), (2), (3)}] = xy
> > > >
> > > > How are organized the 16 points xy ?
> > > >
> > > > In four lines:
> > > > (01 02 03 04), (11 12 13 14), (21 22 23 24) (31 32 33 34) (??)
> > > >
> > > > Are there points concyclic or on a conic in general?
> > > >
> > > > Antreas
> > > >
> > >
> > > _
> > >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>

--
http://anopolis72000.blogspot.com/

[Non-text portions of this message have been removed]
• [APH] ... If the Conjecture is true (as I think), then we have an interesting locus: As P moves on the Euler Line of ABC, all circles pass through the
Message 3 of 13 , Mar 29 6:08 PM
[APH]
> Conjecture:
> Let P be a point on the Euler line of ABC, and O, Oa,Ob,Oc
> the circumcenters of ABC, IBC, ICA, IAB.
> The orthopoles of PO [=Euler line of ABC], POa, POb, POc wrt
> ABC, IBC, ICA, IAB, resp.
> are concyclic.
>
> Question: Is it true for any point Q (instead of I) on the
> Neuberg cubic?

If the Conjecture is true (as I think), then we have an interesting
locus: As P moves on the Euler Line of ABC, all circles pass through
the orthopole of the Euler Line wrt ABC.

Which is the locus of their centers?

http://anthrakitis.blogspot.gr/2013/03/orthopolar-circles.html

APH
• Let ABC be a triangle, P, Q two points and O,Q1,Q2,Q3 the circumcenters of ABC, QBC, QCA, QAB, resp. Denote: P0 = the orthopole of PO wrt ABC P1 = the
Message 4 of 13 , Mar 30 6:34 AM
Let ABC be a triangle, P, Q two points and O,Q1,Q2,Q3 the circumcenters of ABC,
QBC, QCA, QAB, resp.

Denote:

P0 = the orthopole of PO wrt ABC

P1 = the orthopole of PQ1 wrt QBC

P2 = the orthopole of PQ2 wrt QCA

P3 = the orthopole of PQ3 wrt QAB

We have:

1. P0, P1, P2, P3 lie on the NPCs (N),(N1),(N2),(N3) of ABC,
QBC, QCA, QAB, resp. (since the respective lines pass through the circumcenters
of the respective triangles)

2. The NPCs of ABC, QBC, QCA, QAB concur at the Poncelet point Q*
of Q wrt ABC.

CONJECTURE:

The points P0, P1, P2, P3, Q* are concyclic.

Figure:

http://anthrakitis.blogspot.gr/2013/03/orthopolar-circles_30.html

Antreas

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> [APH]
> > Conjecture:
> > Let P be a point on the Euler line of ABC, and O, Oa,Ob,Oc
> > the circumcenters of ABC, IBC, ICA, IAB.
> > The orthopoles of PO [=Euler line of ABC], POa, POb, POc wrt
> > ABC, IBC, ICA, IAB, resp.
> > are concyclic.
> >
> > Question: Is it true for any point Q (instead of I) on the
> > Neuberg cubic?
>
> If the Conjecture is true (as I think), then we have an interesting
> locus: As P moves on the Euler Line of ABC, all circles pass through
> the orthopole of the Euler Line wrt ABC.
>
> Which is the locus of their centers?
>
> http://anthrakitis.blogspot.gr/2013/03/orthopolar-circles.html
>
> APH
>
• For quadrilateral: Let ABCD be a quadrilateral, A ,B ,C ,D the circumcenters of BCD, CDA, DAB, ABC, resp. and P a point. Denote: 1 = the orthopole of PA wrt
Message 5 of 13 , Mar 30 9:53 AM

Let ABCD be a quadrilateral, A',B',C',D' the circumcenters of BCD,
CDA, DAB, ABC, resp. and P a point.

Denote:

1 = the orthopole of PA' wrt BCD

2 = the orthopole of PB' wrt CDA

3 = the orthopole of PC' wrt DAB

4 = the orthopole of PD' wrt ABC

Conjecture:

The points 1,2,3,4 are concyclic. The circle passes through the Poncelet point of ABCD (=the point where the NPCs of BCD, CDA, DAB,
ABC concur)

Figure:

Antreas

[APH]
> Let ABC be a triangle, P, Q two points and O,Q1,Q2,Q3 the circumcenters of ABC,
> QBC, QCA, QAB, resp.
>
> Denote:
>
> P0 = the orthopole of PO wrt ABC
>
> P1 = the orthopole of PQ1 wrt QBC
>
> P2 = the orthopole of PQ2 wrt QCA
>
> P3 = the orthopole of PQ3 wrt QAB
>
> We have:
>
> 1. P0, P1, P2, P3 lie on the NPCs (N),(N1),(N2),(N3) of ABC,
> QBC, QCA, QAB, resp. (since the respective lines pass through the circumcenters
> of the respective triangles)
>
> 2. The NPCs of ABC, QBC, QCA, QAB concur at the Poncelet point Q*
> of Q wrt ABC.
>
> CONJECTURE:
>
> The points P0, P1, P2, P3, Q* are concyclic.
>
> Figure:
>
> http://anthrakitis.blogspot.gr/2013/03/orthopolar-circles_30.html
>
>
> Antreas
• ... If ABCD is cyclic (PO = PA = PB = PC := L , a line passing through the circumcenter O of ABCD), the circle (1,2,3,4) is a line. APH
Message 6 of 13 , Mar 30 11:52 AM
--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
>
> Let ABCD be a quadrilateral, A',B',C',D' the circumcenters of BCD,
> CDA, DAB, ABC, resp. and P a point.
>
> Denote:
>
> 1 = the orthopole of PA' wrt BCD
>
> 2 = the orthopole of PB' wrt CDA
>
> 3 = the orthopole of PC' wrt DAB
>
> 4 = the orthopole of PD' wrt ABC
>
> Conjecture:
>
> The points 1,2,3,4 are concyclic. The circle passes through the Poncelet point of ABCD (=the point where the NPCs of BCD, CDA, DAB,
> ABC concur)
>
> Figure:
>
>
> Antreas

If ABCD is cyclic (PO = PA' = PB' = PC' := L , a line passing through
the circumcenter O of ABCD), the circle (1,2,3,4) is a line.

APH
• Dear Antreas, Very nice observation! Let s call the circle through points 1,2,3,4 the QA-Orthopole-circle. Then: when P = QA-P3 then the QA-Orthopole-circle =
Message 7 of 13 , Mar 30 11:55 AM
Dear Antreas,

Very nice observation!
Let's call the circle through points 1,2,3,4 the QA-Orthopole-circle.
Then:
when P = QA-P3 then the QA-Orthopole-circle = circle with diameter QA-P2.QA-P3,
when P = QA-P4 then the QA-Orthopole-circle = point QA-P2,
when P = QA-P12 then the QA-Orthopole-circle = circumcircle of the Diagonal Triangle.

QA-P2 = (Euler-)Poncelet Point
QA-P3 = Gergonne Steiner Point
QA-P4 = Isogonal Center
QA-P12= Orthocenter Diagonal Triangle

Best regards,

Chris van Tienhoven
www.chrisvantienhoven.nl

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
>
> Let ABCD be a quadrilateral, A',B',C',D' the circumcenters of BCD,
> CDA, DAB, ABC, resp. and P a point.
>
> Denote:
>
> 1 = the orthopole of PA' wrt BCD
>
> 2 = the orthopole of PB' wrt CDA
>
> 3 = the orthopole of PC' wrt DAB
>
> 4 = the orthopole of PD' wrt ABC
>
> Conjecture:
>
> The points 1,2,3,4 are concyclic. The circle passes through the Poncelet point of ABCD (=the point where the NPCs of BCD, CDA, DAB,
> ABC concur)
>
> Figure:
>
>
> Antreas
>
> [APH]
> > Let ABC be a triangle, P, Q two points and O,Q1,Q2,Q3 the circumcenters of ABC,
> > QBC, QCA, QAB, resp.
> >
> > Denote:
> >
> > P0 = the orthopole of PO wrt ABC
> >
> > P1 = the orthopole of PQ1 wrt QBC
> >
> > P2 = the orthopole of PQ2 wrt QCA
> >
> > P3 = the orthopole of PQ3 wrt QAB
> >
> > We have:
> >
> > 1. P0, P1, P2, P3 lie on the NPCs (N),(N1),(N2),(N3) of ABC,
> > QBC, QCA, QAB, resp. (since the respective lines pass through the circumcenters
> > of the respective triangles)
> >
> > 2. The NPCs of ABC, QBC, QCA, QAB concur at the Poncelet point Q*
> > of Q wrt ABC.
> >
> > CONJECTURE:
> >
> > The points P0, P1, P2, P3, Q* are concyclic.
> >
> > Figure:
> >
> > http://anthrakitis.blogspot.gr/2013/03/orthopolar-circles_30.html
> >
> >
> > Antreas
>
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