Κυριακή, 24 Μαρτίου 2013

ORTHOLINES

Let ABC be a triangle and 0,1,2,3 the concurrent (at S) 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 ?

1. Collinearity:

The four lines: (01 02 03 04), (11 12 13 14), (21 22 23 24) (31 32 33 34)

are concurrent. Proof ?

2. Concyclicity:

The points 00, 10, 20, 30 are concyclic (?).

Is it true, for other than I points, on the Neuberg cubic (Fermat points etc)?

3. Orthology:

The triangles ABC, Triangle A'B'C' bounded by the lines [(11 12 13 14), (21 22 23 24) (31 32 33 34)] are orthologic. The one orthologic center (ABC, A'B'C') is S. The other one S' of (A'B'C', ABC) ?

Antreas P. Hatzipolakis, 24 March 2013

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

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.

See also Hyacinthos message #21070.

Then this is also true when D coincides with I (Incenter X(1)) and when L = some Euler Line.

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).

Further 00 lies on NPC of ABC. 11, 22, 33 lie on NPC's ABD, BCD, CAD.

Chris van Tienhoven, Hyacinthos #21831

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

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.

See HERE:

Chris van Tienhoven, Hyacinthos #21834

CONJECTURES:

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.

APH, Hyacinthos #21835

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