- 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 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:

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

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 - --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>

If ABCD is cyclic (PO = PA' = PB' = PC' := L , a line passing through

> 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 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:

>

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

>

> Antreas

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 = 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.

Explanation of above Quadrangle Points:

QA-P2 = (Euler-)Poncelet Point

QA-P3 = Gergonne Steiner Point

QA-P4 = Isogonal Center

QA-P12= Orthocenter Diagonal Triangle

More information at http://www.chrisvantienhoven.nl/quadrangle-objects/17-mathematics/4-quadrangle-objects.html

Best regards,

Chris van Tienhoven

www.chrisvantienhoven.nl

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:

>

> 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 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:

>

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

>

> 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

>