CONCYCLIC CIRCUMCENTERS - 2

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

Denote:

Ab,Ac = the reflections of A' in CC',BB', resp. (lying on AC, AB, resp.)

Similarly Bc,Ba = the reflections of B' in AA',CC', resp. and Ca,Cb = the reflections of C' in BB',AA', resp.

A1B1C1 = the antipedal triangle of I wrt triangle A'AbAc

Similarly A2B2C2 = the antipedal triangle of I wrt triangle B'BcBa and A3B3C3 = the antipedal triangle of I wrt triangle C'CaCb,

Oa,Ob,Oc = the circumcenters of the triangles A1B1C1, A2B2C2, A3B3C3, resp.

The circumcenter O of ABC and the circumcenters Oa,Ob,Oc of A1B1C1,A2B2C2,A3B3C3, resp. are concyclic.

Which point is the center X of the circle?

Antreas P. Hatzipolakis, 4 June 2014

X is X(5495)

Peter Moses 5 June 2014

CONCYCLIC POINTS. LOCUS

A CIRCLE:

Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3 the NPC centers of IB'C', IC'A', IA'B', resp.

The points I, N1,N2,N3 are concyclic.

ETC X(5453)

Center of the circle?

LOCUS:

Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1, N2, N3 the NPC centers of PB'C', PC'A', PA'B', resp.

Which is the locus of P such that the points P,N1,N2,N3 are concyclic ?

Antreas P. Hatzipolakis, 17 April 2013, Hyacinthos #21970

**********************************************************

The center X of the circle has trilinears:

2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : :

ETC search: 2.387773069046934.., 2.38593313995937.., 0.886815506990847..

X = Midpoint of X(I),X(J) for these (I,J): (1,500)

X lies on line X(I),X(J) for these (I,J):

(1,30), (3,81), (5,581), (21,323), (58,5428), (140,3216), (186,2906), (386,549), (511,1385), (550,991), (1154,2646), (2771,3743)

Cιsar Lozada Hyacinthos #21972

**********************************************************

A related locus:

Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle intercepts of line X(5)X(523), and the point Qi (of ABC) such that I = Qi of the cevian triangle of I.

Randy Hutson, Hyacinthos #21977

This is the tricircular sextic: Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z) where S = twice the area of ABC. (tricircular = the circular points are triple points of the curve).

Francisco Javier Hyacinthos #21981

.... the point Qi on this curve is the point which is the incenter of its anticevian triangle, and has ETC search value 1.999434154060428. Coordinates? Its isogonal conjugate Qi* is the point which is the nine-point center of its pedal triangle (ETC search 1.142779079509848). The line QiQi* passes through X(5).

Randy Hutson, Hyacinthos #21982

Does this locus contain any ETC centers or bicentric pairs besides X(13), X(14), and PU(5)? Are the circumcircle intercepts of line X(5)X(523) triangle centers or another bicentric pair?

Randy Hutson, Hyacinthos #21983

ORTHOPOLAR CIRCLES [orthic triangle]

Let ABC be a triangle and A'B'C' the orthic triangle.

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.

Let L be a line passing through H.

Denote:

0 = the orthopole of L wrt A'B'C'

1 = the orthopole of L wrt A'AbAc

2 = the orthopole of L wrt B'BcBa

3 = the orthopole of L wrt C'CaCb

The points 0,1,2,3 are concyclic.

Special Case: L = Euler line of ABC.

L passes through the common circumcenter of the triangles A'AbAc, B'BcBa, C'CaCb [= the H of ABC] and the circumcenter of A'B'C' [=the N of ABC]. The points 0,1,2,3 coincide with the Poncelet point U of H wrt A'B'C'. (The U is the point of concurrence of 7 NPCs: The NPCs of A'B'C', HB'C', HC'A', HA'B', A'AbAc, B'BcBa, C'CaCb.)

Problem:

Which is the locus of the centers of the circles 0123 as L moves around H?

Antreas P. Hatzipolakis, 6 April 2013

ORTHOPOLAR CIRCLES

Let ABCD be a quadrilateral [quadragon], 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)

The circle (1,2,3,4) is a line when ABCD is cyclic (ie A' = B' = C'= D' = O ==> PO = PA' = PB' = PC' := L, a line psiing through the circumcenter of the cyclic ABCD)

Antreas P. Hatzipolakis, 30 March 2013

ORTHOPOLAR CIRCLES [triangle]

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.

Antreas P. Hatzipolakis, 30 March 2013.

Concyclic Os

Let ABC be a triangle and P a point.

For which P's the circumcenters O,Oa,Ob,Oc of ABC, PBC, PCA, PAB, resp. are concyclic?

APH, 28 February 2012