EXCENTERS - CIRCUMCENTERS - CONICS

Let ABC be a triangle and P a point.

Denote:

Iab, Iac = the excenters of IBC respective to angles PBC, PCB.

Ibc, Iba = the excenters of ICA respective to angles PCA, PAC.

Ica, Icb = the excenters of IAB respective to angles PAB, PBA.

For P = O, the six excenters are concyclic, lying on the circumcircle (O).

Are they always lying on a conic? And for which points P the conic is circle?

Denote:

Oa, O'a = the circumcenters of PIbcIcb, PIbaIca, resp.

Ob, O'b = the circumcenters of PIcaIac, PIcbIab, resp.

Oc, O'c = the circumcenters of PIabIba, PIacIbc, resp.

The triangles OaObOc, O'aO'bO'c are perspective.

The line segments OaO'a, ObO'b, OcO'c are bisected by the perspector P' of the triangles, therefore the six points lie on a conic with center P'.

For which points P the conic is circle?

Antreas P. Hatzipolakis, 20 May 2013

ANOPOLIS CIRCLE

Let ABC be a triangle.

Denote:

(N1),(N2), (N3) = the NPCs of IBC, ICA, IAB, resp.

(12), (13) = the reflections of (N1) in BI, CI, resp.

(23), (21) = the reflections of (N2) in CI, AI, resp.

(31), (32) = the reflections of (N3) in AI, B1, resp.

The six centers 12,13,23,21,31,32 are concyclic.

The circle has diameter IO.

Perspectivity:

The circles (I), (21), (31) concur at a point A'.

The circles (I), (32), (12) concur at a point B'.

The circles (I), (13), (23) concur at a point C'.

The triangles ABC, A'B'C' are perspective (??)

Antreas P. Hatzipolakis, 19 May 2013

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Midpoint of OI = X(1385)

12 = {a (a^2-b^2+a c-2 b c-c^2),-a^3+a b^2+a^2 c+2 a b c-2 b^2 c+a c^2-3 b c^2-c^3,-c^2 (b+c)} 13 = {a (a^2+a b-b^2-2 b c-c^2),-b^2 (b+c),-a^3+a^2 b+a b^2-b^3+2 a b c-3 b^2 c+a c^2-2 b c^2}

ETC points on the Anopolis circle {1,3,1083,3109,3110}

The orthogonal projection of I on a line through O is also on the circle .. hence X(3109) & X(3110). So too X(1083) as it is the orthogonal projection of I on line X(3) X(667)

Also the orthogonal projection of O on a line through I is also on the circle

A non ETC point on the Anopolis circle

a^2 (a^4 b^2-a^2 b^4-2 a^4 b c+a b^4 c+a^4 c^2+a^2 b^2 c^2-b^3 c^3-a^2 c^4+a b c^4) ::

on lines {{1,667},{3,238},{35,1083},{41,813}}

A’ = reflection of midpoint of AI in 21 31 = {(a-b-c) (b-c)^2 (a+b-c) (a-b+c),-(a-b)^2 b^2 (a+b-c),-(a-c)^2 c^2 (a-b+c)}

A’B’C’ is perspective to ABC at X(59)

Also to the tangential triangle at:

a^2 (a^9-3 a^8 b+8 a^6 b^3-6 a^5 b^4-6 a^4 b^5+8 a^3 b^6-3 a b^8+b^9-3 a^8 c+12 a^7 b c-12 a^6 b^2 c-12 a^5 b^3 c+30 a^4 b^4 c-12 a^3 b^5 c-12 a^2 b^6 c+12 a b^7 c-3 b^8 c-12 a^6 b c^2+33 a^5 b^2 c^2-21 a^4 b^3 c^2-21 a^3 b^4 c^2+33 a^2 b^5 c^2-14 a b^6 c^2+2 b^7 c^2+8 a^6 c^3-12 a^5 b c^3-21 a^4 b^2 c^3+48 a^3 b^3 c^3-21 a^2 b^4 c^3-4 a b^5 c^3+2 b^6 c^3-6 a^5 c^4+30 a^4 b c^4-21 a^3 b^2 c^4-21 a^2 b^3 c^4+18 a b^4 c^4-2 b^5 c^4-6 a^4 c^5-12 a^3 b c^5+33 a^2 b^2 c^5-4 a b^3 c^5-2 b^4 c^5+8 a^3 c^6-12 a^2 b c^6-14 a b^2 c^6+2 b^3 c^6+12 a b c^7+2 b^2 c^7-3 a c^8-3 b c^8+c^9)::

= (2r - R) (r^2 + 6 r R + 8 R^2 - s^2) X[1486] – 4 r (r^2 + 5 r R + 4 R^2 - s^2) X[1618]

On line {1486, 1618}

Search = 0.89807482985690351940

Peter J. C. Moses, 19 May 2013

COLLINEARITY

Let ABC be a triangle and P, Q two points.

Denote:

PQa = the isogonal conjugate of P wrt QBC

PQb = the isogonal conjugate of P wrt QCA

PQc = the isogonal conjugate of P wrt QAB

QPa = the isogonal conjugate of Q wrt PBC

QPb = the isogonal conjugate of Q wrt PCA

QPc = the isogonal conjugate of Q wrt PAB

Apq = PQaQPa /\ BC

Bpq = PQbQPb /\ CA

Cpq = PQcQPc /\ AB

Conjecture: The points Apq, Bpq, Cpq are collinear.

Let R be another point. We have the lines (if the conjecture is true):

ApqBpqCpq, AqrBqrCqr, ArpBrpCrp.

Are they concurrent? Do they bound a triangle in perspective with ABC?

Antreas P. Hatzipolakis, 18 May 2013

ORTHOCENTERS

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

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.

Ha, Hb, Hc = the orthocenters of the triangles A'AbAc, B'BcCa, C'CaCb, resp.

H'a, H'b, H'c = the reflections of Ha, Hb, Hc in AA', BB', CC', resp.

Conjecture 1.:

The triangles ABC, HaHbHc are perspective.

Conjecture 2.:

The points H'a, H'b, H'c are collinear

Locus of variable P such that

1. ABC, HaHbHc are perspective

2. H'a, H'b, H'c are collinear?

Antreas P. Hatzipolakis, 17 May 2013

CONCURRENT CIRCUMCIRCLES - CONCYCLIC POINTS

Let ABC be a triangle, Na,Nb,Nc the NPC centers of IBC, ICA, IAB, resp. and Oa, Ob, Oc the circumcenters of NaBC, NbCA, NcAB, resp.

1. The Euler line of NaNbNc is the line INF of ABC (O of NaNbNc = N of ABC, H of NaNbNc = I of ABC, F of ABC = ?? of NaNbNc).

2. The circumcircles (Oa),(Ob),(Oc) of NaBC, NbCA, NcAB, resp. concur at a point Q on the circumcircle of NaNbNc.

Which point is the Q wrt 1. ABC 2.NaNbNc ?

Antreas P. Hatzipolakis, 17 May 2013

RADICAL CENTER- EULER LINE

Let ABC be a triangle and A'B'C' the cevian triangle of P = H (orthic tr.)

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.

Na, Nb, Nc = The NPC centers of A'AbAc, B'BcBa, C'CaCb, resp.

O* = The circumcenter of NaNbNc. It is the NPC center of A'B'C'.

R* = the radical center of (Na), (Nb), (Nc)

The NPC circles (Na), (Nb), (Nc) concur on the NPC of A'B'C'. The point of concurrence, the R*, is the Poncelet point of H wrt A'B'C' and since the H of ABC is the I of A'B'C', the point is the Feuerbach point [1] of A'B'C' = the center of the Feuerbach hyperbola of A'B'C'.

The points R*, H, O* are collinear. The line is the INF-line of A'B'C'

Generalization:

Let P = point on the Euler line of ABC.

Conjecture:

The points R*, P, O* are collinear.

Locus:

P = a variable point.

Which is the locus of P such that R*,P,O* are collinear?

Euler Line + ???

Is the incenter I on the locus?

Note [1]:

The Feuerbach point of ABC:

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

Ab, Ac = the reflections of A in BB', CC', resp.

Bc, Ba = the reflectuons of B in CC', AA', resp.

Ca, Cb = the reflections of C in AA', BB', resp.

The NPCs of AAbAc, BBcBa, CCaCb (and ABC) concur at Feuerbach point of ABC.

Antreas P. Hatzipolakis, 16 May 2013

NPC center of the Cevian triangle of I

Let ABC be a triangle, A'B'C' the cevian triangle of I, I' the isogonal conjugate of I wrt A'B'C' and N' the NPC center of A'B'C'.

Conjecture 1:

The points I, N',I' are collinear.

Conjecture 2:

Let A"B"C" be the pedal triangle of I' wrt A'B'C'.

I' is the NPC center A"B"C".(Randy Hutson)

The line II' is the Euler line of A"B"C"

Antreas P. Hatzipolakis, 15 My 2013