PARALLEL NN-LINES

10. Let ABC be a triangle and P a point.

Denote:

Ab, Ac = the orthogonal projections of A on PB,PC, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc.

Similarly Nb1, Nb2 and Nc1, Nc2.

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel.

11. If P = I,

the lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

21. Let ABC be a triangle.

Denote:

Na1 = the NPC center of IBC

Na2 = the NPC center of Na1BC.

Similarly Nb1,Nb2, Nc1,Nc2

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

31. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Ab, Ac = the orthogonal projections of A on IaIc, IaIb, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

41. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Na1 = the NPC center of IaBC

Oa = the circumcenter of IaBC

Nao1 = The NPC center of OaBC.

Similarly Nb1, Nbo1, Nc1, Nco1.

The lines Na1Nao1, Nb1Nbo1, Nc1Nco1 are parallel to OI line of ABC.

Antreas P. Hatzipolakis, 1 October 2014

RADICAL CENTER - EULER LINE

Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P.

Denote:

Oab, Oac = the circumcenters of AA'B, AA'C, resp.

(O1) = the circumcircle of A'OabOac. Similarly (O2), (O3).

P* = the radical center of (O1),(O2), (O3)

For P = G, the G* is lyimg on the Euler line of ABC.

Locus: Which is the locus of P such that P* is lying 1. on the OP line 2. on the Euler Line of ABC?

Antreas P. Hatzipolakis, 1 September 2014

Peter Moses:

G* = X(140) = midpoint of ON

GENERALIZATION

Let P is a point on Euler line of triangle ABC and DEF is pedal triangle of P. Let Oab,Oac be the circumcenters of triangle DAB,DAC and (Oa) is circumcircle of triangle DOabOac. Similarly, we have circle (Ob),(Oc). Then radical center of (Oa),(Ob),(Oc) lies on Euler line, too.

Tran Quang Hung.

Response

HOMOTHETIC TRIANGLES and EULER LINES

Theorem 1.

Let ABC be an equilateral triangle and P a point. The Euler lines of the triangles PBC,PCA,PAB are concurent.Denote the point of concurrence with P*.

Reference:

APH, Hyacinthos #21592

Locus Problems:

As P moves on a line or on a circle [special case the incircle of ABC] which is the locus of P*?

Application:

Let ABC be a triangle and A',B',C' the apices of the equilateral triangles erected out/inwardly ABC, P a point and Pa, Pb, Pc the respective points of concurrences of the Euler lines wrt equil. triangles A'BC,B'CA,C'AB.

Which is the locus of P such that ABC, PaPbPc are perspective or orthologic?

Theorem 2 (generalization of Th. 1).

Let ABC, A'B'C' be two homothetic equilateral triangles. The Euler lines of the triangles AB'C', BC'A', CA'B' (and of the triangles A'BC, B'CA, C'AB) are concurrent.

Application to Morley Configuration with homothetic equilateral triangles:

1st Morley Triangle

2nd Morley Triangle

3rd Morley Triangle

Roussel Triangle

Theorem 3.

Let ABC, A'B'C' be two dilated triangles with scale factor 1. The Euler lines of AB'C', BC'A', CA'B' (and of the triangles A'BC, B'CA, C'AB) are concurrent.

Application:

Let ABC be a triangle and (O1),(O2),(O3) the reflections of the circumcircle (O) in BC,CA,AB, resp. Denote:

Ab = the second intersection of (O2) and the reflection of AN in the bisector of the angle HAC.

Ac = the second intersection of (O3) and the reflection of AN in the bisector of the angle HAB.

Similarly (cyclically) Bc, Ba and Ca, Cb

Oa, Ob, Oc = the circumcenters of the triangles ABcCb, BCaAc, CAbBa

The triangles ABC, OaObOc are dilated triangles with scalar facror 1. The Euler lines of AObOc, BOcOa, COaOb are concurrent and also the Euler lines of OaBC, ObCA, OcAB.

Note: The circumradii of the triangles ABcCb, BCaAc, CAbBa are equal.

Antreas P. Hatzipolakis, 20 April 2014

CIRCUMCENTERS ON THE LINES OH (Euler Line), OI

Let ABC be a triangle and

1. A'B'C' the pedal triangle of H (orthic triangle)

Denote:

(Oa) = the circumcircle of OBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in HA'.

(Ob) = the circumcircle of OCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in HB'.

(Oc) = the circumcircle of OAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in HC'.

The circumcenter of the triangle O'1O'2O'3 lies on the OH line (Euler line)

-----

2. A'B'C' the pedal triangle of I.

Denote:

(Oa) = the circumcircle of IBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in IA'.

(Ob) = the circumcircle of ICA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in IB'.

(Oc) = the circumcircle of IAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in IC'.

The circumcenter of the triangle O'1O'2O'3 lies on the OI line

Generalizations (Loci):

Let ABC be a triangle, P,P* two isogonal conjugate points and A'B'C',A"B"C" the pedal triangles of P,P*.

Denote:

(Oa) = the circumcircle of PBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in PA'.

(O"1) = the reflection of (O1) in P*A"

(Ob) = the circumcircle of PCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in PB'.

(O"2) = the reflection of (O2) in P*B"

(Oc) = the circumcircle of PAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in PC'.

(O"3) = the reflection of (O3) in P*C"

R' = the circumcenter of O'O'2O'3

R" = the circumcenter of O"1O"2O"3

Which is the locus of P such that:

1. O, P, R'

2. O, P, R"

3. O, R', R"

4. P, P*, R'

5. P, R', R"

are collinear ?

The McCay cubic?

Antreas P. Hatzipolakis, 16 April 2014.

RADICAL CENTERS ON THE LINES OH (Euler line),OI

Let ABC be a triangle and

1. A'B'C' the pedal triangle of H (orthic triangle)

Denote:

(Oa) = the circumcircle of OBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in HA'.

(Ob) = the circumcircle of OCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in HB'.

(Oc) = the circumcircle of OAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in HC'.

The radical center of (O'1),(O'2),(O'3) lies on the OH line (Euler line)

-----

2. A'B'C' the pedal triangle of I.

Denote:

(Oa) = the circumcircle of IBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in IA'.

(Ob) = the circumcircle of ICA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in IB'.

(Oc) = the circumcircle of IAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in IC'.

The radical center of (O'1),(O'2),(O'3) lies on the OI line.

Generalizations (Loci):

Let ABC be a triangle, P,P* two isogonal conjugate points and A'B'C',A"B"C" the pedal triangles of P,P*.

Denote:

(Oa) = the circumcircle of PBC.

(O1) = the reflection of (Oa) in BC.

(O'1) = the reflection of (O1) in PA'.

(O"1) = the reflection of (O1) in P*A"

(Ob) = the circumcircle of PCA.

(O2) = the reflection of (Ob) in CA.

(O'2) = the reflection of (O2) in PB'.

(O"2) = the reflection of (O2) in P*B"

(Oc) = the circumcircle of PAB.

(O3) = the reflection of (Oc) in AB.

(O'3) = the reflection of (O3) in PC'.

(O"3) = the reflection of (O3) in P*C"

R' = the radical center of (O'1),(O'2),(O'3)

R" = the radical center of (O"1),(O"2),(O"3)

Which is the locus of P such that:

1. O, P, R'

2. O, P, R"

3. O, R', R"

4. P, P*, R'

5. P, R', R"

are collinear ?

The McCay cubic?

Antreas P. Hatzipolakis, 15 April 2014.

CONCURRENT CIRCLES -- EULER LINE

Theorem:

Let ABC be a triangle and A',B',C' three points. If the circumcircles of A'BC, B'CA, C'AB are concurrent, then also the circumcircles of AB'C', BC'A', CA'B' are concurrent.

Corollary:

Let ABC be a triangle and P a point. If A',B',C' are arbitrary points on the circumcircles of PBC,PCA,PAB, resp. then the circumcircles of AB'C', BC'A', CA'B' are concurrent.

Applications:

Let P,Q be two points and PaPbPc, Q1Q2Q3 the antipedal, pedal triangles of P,Q, resp. The orthogonal projections A',B',C' of Pa,Pb,Pc on PQ1,PQ2,PQ3, resp. lie on the circumcircles of PBC,PCA,PAB, resp.

The circumcircles of AB'C', BC'A', CA'B' concur at a point D.

1. For P = O, Q = H:

The point D lies on the Euler line of ABC

2. For P = Q = I:

The point D lies on the Euler line of ABC

3. For P = Q = H:

The Euler lines L,L1,L2,L3 of ABC, DBC,DCA,DAB are concurrent at a point D' on the Neuberg cubic.

The parallels to L1,L2,L3 through A,B,C, resp. are concurrent at a point D"

Coordinates of the points D's, D' and D"?

More pairs (P,Q) such that the circumcircles concur on the Euler line?

Antreas P. Hatzipolakis, 11 April 2014

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

It seems that the general case for D has truly appalling barycentrics.

But, for the cases you mention we have:

1) PQ = OH, D= X(186)

2) PQ = II, D = X(1325)

3) PQ = HH, D = X(1157)

Also PQ = OI, D = X(36)

Peter Moses, 14 April 2014

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

QUINTIC

Let ABC be a triangle, P a point and A1,B1,C1 are the NPC centers of PBC, PCA, PAB, resp.

Which is the locus of P such that the circumcenter O0 of A1B1C1 lies on the Euler line?

P : X1 = I, X4= H, X5 = N.......

For P = I we have O0 = N

Antreas P. Hatzipolakis, 15 April 2013

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

It is a circular quintic through I, H and N. The other three intersection points with Euler line are X30 (the infinite point) and the instersections with the circumcircle X1113 and X1114.

Equation:

a^6 c^2 x^3 y^2 - 3 a^4 b^2 c^2 x^3 y^2 + 3 a^2 b^4 c^2 x^3 y^2 - b^6 c^2 x^3 y^2 - a^4 c^4 x^3 y^2 + a^2 b^2 c^4 x^3 y^2 - a^2 c^6 x^3 y^2 + c^8 x^3 y^2 + a^6 c^2 x^2 y^3 - 3 a^4 b^2 c^2 x^2 y^3 + 3 a^2 b^4 c^2 x^2 y^3 - b^6 c^2 x^2 y^3 - a^2 b^2 c^4 x^2 y^3 + b^4 c^4 x^2 y^3 + b^2 c^6 x^2 y^3 - c^8 x^2 y^3 + 3 a^2 b^4 c^2 x^3 y z - 3 b^6 c^2 x^3 y z - 3 a^2 b^2 c^4 x^3 y z + 3 b^2 c^6 x^3 y z + 2 a^6 c^2 x^2 y^2 z - 2 b^6 c^2 x^2 y^2 z - 4 a^4 c^4 x^2 y^2 z + 4 b^4 c^4 x^2 y^2 z + 2 a^2 c^6 x^2 y^2 z - 2 b^2 c^6 x^2 y^2 z + 3 a^6 c^2 x y^3 z - 3 a^4 b^2 c^2 x y^3 z + 3 a^2 b^2 c^4 x y^3 z - 3 a^2 c^6 x y^3 z - a^6 b^2 x^3 z^2 + a^4 b^4 x^3 z^2 + a^2 b^6 x^3 z^2 - b^8 x^3 z^2 + 3 a^4 b^2 c^2 x^3 z^2 - a^2 b^4 c^2 x^3 z^2 - 3 a^2 b^2 c^4 x^3 z^2 + b^2 c^6 x^3 z^2 - 2 a^6 b^2 x^2 y z^2 + 4 a^4 b^4 x^2 y z^2 - 2 a^2 b^6 x^2 y z^2 + 2 b^6 c^2 x^2 y z^2 - 4 b^4 c^4 x^2 y z^2 + 2 b^2 c^6 x^2 y z^2 + 2 a^6 b^2 x y^2 z^2 - 4 a^4 b^4 x y^2 z^2 + 2 a^2 b^6 x y^2 z^2 - 2 a^6 c^2 x y^2 z^2 + 4 a^4 c^4 x y^2 z^2 - 2 a^2 c^6 x y^2 z^2 + a^8 y^3 z^2 - a^6 b^2 y^3 z^2 - a^4 b^4 y^3 z^2 + a^2 b^6 y^3 z^2 + a^4 b^2 c^2 y^3 z^2 - 3 a^2 b^4 c^2 y^3 z^2 + 3 a^2 b^2 c^4 y^3 z^2 - a^2 c^6 y^3 z^2 - a^6 b^2 x^2 z^3 + b^8 x^2 z^3 + 3 a^4 b^2 c^2 x^2 z^3 + a^2 b^4 c^2 x^2 z^3 - b^6 c^2 x^2 z^3 - 3 a^2 b^2 c^4 x^2 z^3 - b^4 c^4 x^2 z^3 + b^2 c^6 x^2 z^3 - 3 a^6 b^2 x y z^3 + 3 a^2 b^6 x y z^3 + 3 a^4 b^2 c^2 x y z^3 - 3 a^2 b^4 c^2 x y z^3 - a^8 y^2 z^3 + a^2 b^6 y^2 z^3 + a^6 c^2 y^2 z^3 - a^4 b^2 c^2 y^2 z^3 - 3 a^2 b^4 c^2 y^2 z^3 + a^4 c^4 y^2 z^3 + 3 a^2 b^2 c^4 y^2 z^3 - a^2 c^6 y^2 z^3 = 0

Francisco Javier, Hyacinthos #21962

SEQUENCE OF POINTS ON THE EULER LINE

Let ABC be a triangle.

Denote:

A1, B1, C1 = The NPC centers of NBC, NCA, NAB, resp.

A2, B2 ,C2 = The NPC centers of A1BC, B1CA, C1AB, resp.

A3, B3, C3 = The NPC centers of A2BC, B2CA, C2AB

An, Bn, Cn = The NPC centers of A_n-1BC, B_n-1CA, C_n-1AB.

On = the circumcenter of the triangle AnBnCn.

The points O1, O2,......, On, ..... lie on the Euler Line of ABC.

The point O1 is now in ETC: X5501

Coordinates of On? Ratio of OnO / OnN ?

Antreas P. Hatzipolakis, 14 April 2013

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

I calculated the ratios NO1:O1O and NO2:O2O. The expression for the first one is quite long, and that for the second one is enormous:

If p stands for the semiperimeter, we have

NO1:O1O = (2 p^4 - 12 p^2 r^2 + 2 r^4 - 16 p^2 r R + 16 r^3 R - 8 p^2 R^2 + 40 r^2 R^2 + 32 r R^3 + 9 R^4)/(2 p^4 + 20 p^2 r^2 + 2 r^4 - 16 p^2 r R + 16 r^3 R - 16 p^2 R^2 + 48 r^2 R^2 + 64 r R^3 + 23 R^4)

Francisco Javier, Hyacinthos #21952

The conjecture is false !!

From Cesar Lozada (24 June 2016)

Dear Antreas,

Sorry. O1 does lie on Euler line but O2 does not. Algebraically confirmed.

I didn´t try O3 because calculus are awful

Regards,

César Lozada

INCENTER AND NPCs

Let ABC be a triangle.

Denote:

N1,N2,N3 = The NPC centers of IBC, ICA, IAB, resp.

N11, N22, N33 = The NPC centers of N1BC, N2CA, N3AB, resp.

1. The lines N1N11, N2N22, N3N33 are parallel to Euler Line of ABC.

2. The Circumcenter of N1N2N3 is N, the NPC center of ABC.

Synthetic Proofs?

Antreas P. Hatzipolakis, 13 April 2013

ORTHOLOGIC, EULER LINE

Let ABC be a triangle, A'B'C', A"B"C" the medial, orthic triangles, resp. and A*,B*,C* points on AA",BB",CC", resp. such that: A*A/A*A" = B*B/B*B" = C*C/C*C" = t.

Denote:

Ab = (Parallel to BC through A*) /\ AC

Ac = (Parallel to BC through A*) /\ AB

Bc = (Parallel to CA through B*) /\ BA

Ba = (Parallel to CA through B*) /\ BC

Ca = (Parallel to AB through C*) /\ CB

Cb = (Parallel to AB through C*) /\ CA

1. Oa, Ob, Oc = the circumcenters of A'AbAc, B'BcBa, C'CaCb, resp.

The triangles ABC, OaObOc are orthologic

The locus of the orthologic center (OaObOc, ABC), as t varies, is the Euler line of ABC.

Which is the locus of the other orthologic center (ABC, OaObOc)?

2. Na, Nb, Nc = the NPCs centers of A'AbAc, B'BcBa, C'CaCb, resp.

The triangles ABC, NaNbNc are orthologic

The locus of the orthologic center (NaNbNc, ABC), as t varies, is the Euler line of ABC.

Which is the locus of the other orthologic center (ABC, NaNbNc)?

Antreas P. Hatzipolakis, 6 April 2013

CEVIANS, RADICAL CENTERS, EULER LINE

Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and Ma, Mb, Mc points on AA',BB',CC', resp.

Denote:

X = the radical center of the ciecles (Ma, MaB), (Mb, MbC), (Mc, McA)

Y = the radical center of the circles (Ma, MaC), (Mb, MbA), (Mc, McB)

M = the midpoint of the line segmant XY

1.

Let Ma, Mb, Mc be points such that: MaA / MaA' = MbB / MbB' = McC / McC' = t

Which is the locus of M as t varies?

For P = G, the locus is the Euler line.

For t = -1 (ie Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.)

==> M is the circumcenter O of ABC for all P's.

2. Let Ma, Mb, Mc be points such that: MaA / MaP = MbB / MbP = McC / McP = t

Which is the locus of M as t varies?

For P = O ==> M = the NPC center N

For P = G ==> The locus is the Euler Line.

3. Let Ma, Mb, Mc be points such that: MaP / MaA' = MbP / MbB' = McP / McC' = t

Which is the locus of M as t varies?

For P = G ==> the locus is the Euler line.

Antreas P. Hatzipolakis, 31 March 2013