EULER LINES - PARALLELOGIC TRIANGLES - I

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

Denote:

Ab = the orthogonal projection of A' on the parallel to BB' through A

Ac = the orthogonal projection of A' on the parallel to CC' through A

Similarly (cyclically) Bc, Ba and Ca,Cb.

Conjecture: ABC and The triangle bounded by the Euler lines of AAbAc, BBcBa, CCaCb are parallelogic.

Locus ?

Antreas P. Hatzipolakis. 7 September 2014

Generalization: Hyacinthos #22570

Geometrical Loci associated with the Euler Line

A SEQUENCE OF POINTS

hL Line:
Let ABC be a triangle, L a line passing through H, A'B'C' the pedal triangle of H (orthic triangle), A"B"C" the circumcevian triangle of H wrt A'B'C' (aka Euler triangle) and A1,B1,C1 the orthogonal projections of A,B,C, on L, resp. The lines A"A1, B"B1, C"C1 concur at point H11 on the pedal circle of H (NPC of ABC).
Let La,Lb,Lc be the parallels through A,B,C, to A"A1, B"B1, C"C1, resp. They are concurrent at point H12 on the circumcircle of ABC.
The line H11H12 passes through H. Let's call it hL

Concurrent Euler Lines:
Let A2,B2,C2 be the orthogonal projections of A,B,C on hL. The Euler lines of AA1A2, BB1B2, CC1C2 are concurrent at point P1.
Similarly we define hhL, hhhL...... h^nL, and we get a sequence of points Pn: P1 from L and hL, P2 from hL and hhL, etc

The parallels through A,B,C to the Euler lines of AA1A2, BB1B2, CC1C2 concur at point Q1.

Antreas P. Hatzipolakis, 16 Feb. 2013

A PONCELET POINT ON THE EULER LINE

Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of ABC,NBC,NCA,NAB, resp. [concurrent at pN]. The NPCs of the triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur at point ppN on the Euler Line of ABC.

Generalization:

Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs of ABC, PBC, PCA, PAB resp. [concurrent at pP]. If P is on the Euler line of ABC, then the NPCs of N1N2N3, PN1N2, PN2N3, PN3N1 concur at point ppP on the Euler Line of ABC.(??)

Antreas P. Hatzipolakis, 10 Febr. 2013

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pP is the center of the rectangular circumhyperbola through P.

ppP does not, in general, lie on the Euler line.

Some results:

P=X(1), the NPCs are concurrent, with center = non-ETC 1.121590125545969 (on lines 1,5 3,962).

P=X(2), ppP=non-ETC 1.690358502447462

P=X(3), ppP=X(140)

P=X(4), ppP=undefined

P=X(5), ppP=X(3628)

P=X(6), ppP=non ETC 0.780037257060191

P=X(7), ppP=non ETC 0.750876768572663

P=X(8), ppP=non ETC 2.966801160450799

P=X(9), ppP=non-ETC 0.972023454564163

P=X(10), ppP=non-ETC 2.238481946743318

P=X(20), ppP=non-ETC 6.363850996796102

P=X(21), ppP=non-ETC -1.717011738240629

P=X(22), ppP=non-ETC -4.036288926987237

Of these, only X(140) and X(3628) lie on the Euler line.Β The ppP for points P on the Euler line do not even lie on the same conic.

Locus?

Randy Hutson, Hyacinthos #21519

FOUR CONCURRENT LINES, CIRCUMCIRCLES

Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of ABC, IBC, ICA, IAB, resp. and S a point.

Denote:

L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.

1. For which S's the triangles ABC, O1O2O3 are perspective? (Locus)

2. The circumcenter of O1O2O3 is on the line L for all S's (??).

Special Case: L,La,Lb,Lc = Brocard axes

Antreas P. Hatzipolakis, 8 Febr. 2013

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Brocard Axes instead of Euler Lines case:

1. The triangles ABC, O1O2O3 are not perspective.

O1O2O3 appears to be perspective to

1/ excentral at

a (3 a^3+2 a^2 b+b^3+2 a^2 c+3 a b c+c^3):: = (r^2-3 s^2) X[1]+4 r (2 r+3 R) X[21] = (3 r^2-s^2) X[171] + 4 r R X[3753],

on lines {{1,21},{36,199},{171,3753},{210,5247},{511,3576},{740,4234},{976,4661},{978,1453},{986,4252},{1104,3742},{1125,1330},{1193,4881},{1247,2363},{1757,4134},{1961,5251},{2308,4511},{2938,4221},{3454,3624},{3880,5255}}

ETC search = 1.0647243333936774920

2/ intouch

3/ Hexyl,

4/ Yff

5/ cicrumcircleMidArc at X(58)

6/ firstCircumPerp

7/ second CircumPerp at X(58)

....

2. The circumcenter of O1O2O3 is on the line L.

Yes, at

a^2 (2 a^5-3 a^3 b^2+a^2 b^3+a b^4-b^5+2 a^3 b c-a^2 b^2 c-2 a b^3 c+b^4 c-3 a^3 c^2-a^2 b c^2-2 b^3 c^2+a^2 c^3-2 a b c^3-2 b^2 c^3+a c^4+b c^4-c^5)::

on lines {{3,6},{140,3454},{540,549},{631,1330},{758,1385},{1046,3576},{1125,2792},{1511,2842},{3794,4189}}.

also P = (4r^2-SW) X[3]-SW X[6]

ETC Search = 3.7671475697791725975

Peter J. C. Moses

ETC X5429

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Generalization

1. The locus is the conic:

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

2, Yes . The circumcenter of O1O2O3 is on the line L for all S's.

The three centers Oa, Ob, Oc are the mid points of the arcs BC, CA, AB of the circumcircle of ABC. The points O1, O2, O3 are the mid points of SOa, SOb, SOc. If K is the mid point of SO then KO1 = OOa/2 = R/2. Hence K is the center of the circle O1O2O3 with radious R/2.

Nikos Dergiades

EULER LINES, CIRCUMCIRCLES

Let ABC be a triangle, L,La,Lb,Lc the Euler lines of ABC, IBC, ICA, IAB, resp. (concurrent at S) and Oa,Ob,Oc the circumcenters of IBC, ICA, IAB, resp.

Denote:

Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb, resp.

1. The triangles ABC, O1O2O3 are perspective.

2. The circumcenter of O1O2O3 is on the line L.

Antreas P. Hatzipolakis, 8 Febr. 2013

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1. The triangles ABC, O1O2O3 are perspective.

Yes at

Q = a (2 a^3-2 a^2 b-2 a b^2+2 b^3-a^2 c-a b c-b^2 c-2 a c^2-2 b c^2+c^3) (2 a^3-a^2 b-2 a b^2+b^3-2 a^2 c-a b c-2 b^2 c-2 a c^2-b c^2+2 c^3)::

ETC Search = 2.1298183373048648210

Q is on lines {{21,4867},{79,2646},{80,3584},{758,2320},{1389,3746}} and on the Feuerbach hyperbola.

Q = 6 (r+R) X[21] +(2 r-R) X[4867]

Q = R X[79] + 4 (2 r+R) X[2646]

The isogonal of Q, gQ, is on lines {{1,3},{2,4867},{8,3841},{21,4084},{79,3671},{80,226},{81,759},{100,3919},{191,4018},{515,3982},{519,5249},{758,3219},{956,3894},{958,3901},{993,4880},{1001,3899},{1100,5341},{1203,3924},{1411,2003},{2802,3957},{3585,3649},{3636,5330},{3868,5258},{3869,5259},{3874,5288},{3881,4861},{3918,4420},{4067,5260}};

gQ = (4 r + 5 R)X[1] - 2 r X[3]

gQ = (2 r + 3 R) X[21] + 2 R X[4084]

ETC Search = 1.2648627122986571860

O1O2O3 is also perspective to various other triangles too; for example:

1. Excentral at

a (3 a^3-a^2 b-3 a b^2+b^3-a^2 c-3 a b c-3 b^2 c-3 a c^2-3 b c^2+c^3) :: = X[1] + 2 X[21]

on lines {{1,21},{30,1699},{35,3753},{36,3742},{100,3968},{210,5251},{214,5284},{442,3586},{484,3919},{1125,2475},{1420,3649},{1698,1837},{2320,3065},{2646,5259},{3158,3679},{3219,4525},{3336,4189},{3337,5267},{3616,4299},{3636,3648},{3683,4867},{3746,3880},{3956,5260},{4316,5249},{4539,5302},{4677,4933}}. ETC search = 2.6815474100289784017

2. Intouch at

a (a+b-c) (a-b+c) (2 a^4-2 a^3 b-2 a^2 b^2+2 a b^3-2 a^3 c-2 a^2 b c+b^3 c-2 a^2 c^2+2 b^2 c^2+2 a c^3+b c^3) :: R (r+4 R) X[7] - r (4 r+7 R) X[21]

on lines {{7,21},{11,30},{12,5251},{100,5172},{191,1420},{392,3647},{758,1319},{1317,2078},{1411,1758},{2771,5126},{3651,5204},{4189,5221}}. ETC search -1.0652570342486130228

2. The circumcenter of O1O2O3 is on the line L.

Yes, at

P = a (2 a^6-2 a^5 b-4 a^4 b^2+4 a^3 b^3+2 a^2 b^4-2 a b^5-2 a^5 c+a^2 b^3 c+2 a b^4 c-b^5 c-4 a^4 c^2+4 a^2 b^2 c^2+4 a^3 c^3+a^2 b c^3+2 b^3 c^3+2 a^2 c^4+2 a b c^4-2 a c^5-b c^5)::

on lines {{2,3},{36,3649},{191,3576},{214,960},{758,1385},{952,5258},{1837,5010},{3579,3754},{3650,5303}}

also

P = 3 R X[2] + (4 r+3 R) X[3]

P = (2 r - R) X[36] + R X[3649]

P = X[191] + 3 X[3576]

P = (2 r +R) X[3579] + 2 R X[3754]

P = R X[3650] - (6 r+R) X[5303]

ETC Search = 4.9797648772556482798.

Peter J. C. Moses, 8 Febr. 2013

ETC X5424, X5426, X5427, X5428

EULER LINE [Generalization]

Let ABC be a triangle, T a fixed point, AtBtCt the circumcevian triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the antipodes of At, Bt,Ct).

Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

Antreas P. Hatzipolakis, 7 Feb. 2013

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

The locus is the line OT, and the perspector Q lies on OT as well.

If OP : PT = t: 1-t, then

OQ : QT = R^2(1+t) : - (R^2-OT^2) t.

Paul Yiu, Hyacinthos #21597

HOMOTHETIC EQUILATERAL TRIANGLES

Let A'B'C', A"B"C" be two homothetic equilateral triangles.

Conjecture: The Euler lines of the triangles A'B"C", B'C"A", C'A"B" are concurrent, and also the Euler lines of the triangles A"B'C', B"C'A', C"A'B', if no one of the 6 triangles is degenerated.

A. P. Hatzipolakis, Hyacithos #21357

Concurrent Euler Lines

Let ABC be a triangle and AaBbCc the orthic triangle.

The circle (Aa,AaA) intersects AB,AC at Ab,Ac, resp. (other than A)
The circle (Bb,BbB) intersects BC,BA at Bc,Ba, resp. (other than B)
The circle (Cc,CcC) intersects CA,CB at Ca,Cb, resp. (other than C)

The Euler lines of the triangles:

1. AAbAc, BBcBa, CCaCb

2. AaAbAc, BbBcBa, CcCaCb
[=perp. bisectors of AbAc, BcBa, CaCb, resp.]

are concurrent.

Points ?

APH, 21 February 2012

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

1) (-s1*s2^2+2*s1^2*s3+s2*s3)/(2*s1*s3)
2) (3*s1*s3-s2^2)/(2*s3)
It is the orhocenter of the orthic triangle.

Where I use Moley'trick and s1=a+b+c, s2=ab+bc+ca and s3=abc

Etienne Rousee, Hyacinthos #20850

Euler Lines Locus

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

Denote:

Ab := the orthogonal projection of A' on BB'
Ac: = the orthogonal projection of A' on CC'

L1 := the Euler Line of A'AbAc.
Similarly the lines L2,L3.

Which is the locus of P such that the L1,L2,L3 are concurrent?

ΑΠΧ, Hyacinthos #20600

The locus is the circumcircle and the quintic CYCLIC SUM {x^3*S_A*[(cy)^2-(bz)^2]}=0
Nikos Dergiades, Hyacinthos #20601

This is the Euler-Morley quintic Q003
Bernard Gibert, Hyacinthos #20603

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For P = O, the lines bound a triangle which seems to be parallelogic
with the Orthic triangle.
Parallelogic Centers? (The one lies on the NPC)

ΑΠΧ, Hyacinthos #20600

For P=O, the parallelogic center on NPC is X128.

The other center is

{-4 a^22 + 24 a^20 b^2 - 62 a^18 b^4 + 94 a^16 b^6 - 99 a^14 b^8 +
77 a^12 b^10 - 35 a^10 b^12 - 3 a^8 b^14 + 11 a^6 b^16 - a^4 b^18 -
3 a^2 b^20 + b^22 + 24 a^20 c^2 - 108 a^18 b^2 c^2 +
194 a^16 b^4 c^2 - 172 a^14 b^6 c^2 + 67 a^12 b^8 c^2 -
2 a^10 b^10 c^2 + 13 a^8 b^12 c^2 - 24 a^6 b^14 c^2 -
3 a^4 b^16 c^2 + 18 a^2 b^18 c^2 - 7 b^20 c^2 - 62 a^18 c^4 +
194 a^16 b^2 c^4 - 230 a^14 b^4 c^4 + 120 a^12 b^6 c^4 -
7 a^10 b^8 c^4 - 29 a^8 b^10 c^4 + 20 a^6 b^12 c^4 +
14 a^4 b^14 c^4 - 41 a^2 b^16 c^4 + 21 b^18 c^4 + 94 a^16 c^6 -
172 a^14 b^2 c^6 + 120 a^12 b^4 c^6 - 56 a^10 b^6 c^6 +
19 a^8 b^8 c^6 + 8 a^6 b^10 c^6 - 14 a^4 b^12 c^6 +
36 a^2 b^14 c^6 - 35 b^16 c^6 - 99 a^14 c^8 + 67 a^12 b^2 c^8 -
7 a^10 b^4 c^8 + 19 a^8 b^6 c^8 - 30 a^6 b^8 c^8 + 4 a^4 b^10 c^8 +
12 a^2 b^12 c^8 + 34 b^14 c^8 + 77 a^12 c^10 - 2 a^10 b^2 c^10 -
29 a^8 b^4 c^10 + 8 a^6 b^6 c^10 + 4 a^4 b^8 c^10 -
44 a^2 b^10 c^10 - 14 b^12 c^10 - 35 a^10 c^12 + 13 a^8 b^2 c^12 +
20 a^6 b^4 c^12 - 14 a^4 b^6 c^12 + 12 a^2 b^8 c^12 -
14 b^10 c^12 - 3 a^8 c^14 - 24 a^6 b^2 c^14 + 14 a^4 b^4 c^14 +
36 a^2 b^6 c^14 + 34 b^8 c^14 + 11 a^6 c^16 - 3 a^4 b^2 c^16 -
41 a^2 b^4 c^16 - 35 b^6 c^16 - a^4 c^18 + 18 a^2 b^2 c^18 +
21 b^4 c^18 - 3 a^2 c^20 - 7 b^2 c^20 + c^22,
a^22 - 3 a^20 b^2 - a^18 b^4 + 11 a^16 b^6 - 3 a^14 b^8 -
35 a^12 b^10 + 77 a^10 b^12 - 99 a^8 b^14 + 94 a^6 b^16 -
62 a^4 b^18 + 24 a^2 b^20 - 4 b^22 - 7 a^20 c^2 + 18 a^18 b^2 c^2 -
3 a^16 b^4 c^2 - 24 a^14 b^6 c^2 + 13 a^12 b^8 c^2 -
2 a^10 b^10 c^2 + 67 a^8 b^12 c^2 - 172 a^6 b^14 c^2 +
194 a^4 b^16 c^2 - 108 a^2 b^18 c^2 + 24 b^20 c^2 + 21 a^18 c^4 -
41 a^16 b^2 c^4 + 14 a^14 b^4 c^4 + 20 a^12 b^6 c^4 -
29 a^10 b^8 c^4 - 7 a^8 b^10 c^4 + 120 a^6 b^12 c^4 -
230 a^4 b^14 c^4 + 194 a^2 b^16 c^4 - 62 b^18 c^4 - 35 a^16 c^6 +
36 a^14 b^2 c^6 - 14 a^12 b^4 c^6 + 8 a^10 b^6 c^6 +
19 a^8 b^8 c^6 - 56 a^6 b^10 c^6 + 120 a^4 b^12 c^6 -
172 a^2 b^14 c^6 + 94 b^16 c^6 + 34 a^14 c^8 + 12 a^12 b^2 c^8 +
4 a^10 b^4 c^8 - 30 a^8 b^6 c^8 + 19 a^6 b^8 c^8 - 7 a^4 b^10 c^8 +
67 a^2 b^12 c^8 - 99 b^14 c^8 - 14 a^12 c^10 - 44 a^10 b^2 c^10 +
4 a^8 b^4 c^10 + 8 a^6 b^6 c^10 - 29 a^4 b^8 c^10 -
2 a^2 b^10 c^10 + 77 b^12 c^10 - 14 a^10 c^12 + 12 a^8 b^2 c^12 -
14 a^6 b^4 c^12 + 20 a^4 b^6 c^12 + 13 a^2 b^8 c^12 -
35 b^10 c^12 + 34 a^8 c^14 + 36 a^6 b^2 c^14 + 14 a^4 b^4 c^14 -
24 a^2 b^6 c^14 - 3 b^8 c^14 - 35 a^6 c^16 - 41 a^4 b^2 c^16 -
3 a^2 b^4 c^16 + 11 b^6 c^16 + 21 a^4 c^18 + 18 a^2 b^2 c^18 -
b^4 c^18 - 7 a^2 c^20 - 3 b^2 c^20 + c^22,
a^22 - 7 a^20 b^2 + 21 a^18 b^4 - 35 a^16 b^6 + 34 a^14 b^8 -
14 a^12 b^10 - 14 a^10 b^12 + 34 a^8 b^14 - 35 a^6 b^16 +
21 a^4 b^18 - 7 a^2 b^20 + b^22 - 3 a^20 c^2 + 18 a^18 b^2 c^2 -
41 a^16 b^4 c^2 + 36 a^14 b^6 c^2 + 12 a^12 b^8 c^2 -
44 a^10 b^10 c^2 + 12 a^8 b^12 c^2 + 36 a^6 b^14 c^2 -
41 a^4 b^16 c^2 + 18 a^2 b^18 c^2 - 3 b^20 c^2 - a^18 c^4 -
3 a^16 b^2 c^4 + 14 a^14 b^4 c^4 - 14 a^12 b^6 c^4 +
4 a^10 b^8 c^4 + 4 a^8 b^10 c^4 - 14 a^6 b^12 c^4 +
14 a^4 b^14 c^4 - 3 a^2 b^16 c^4 - b^18 c^4 + 11 a^16 c^6 -
24 a^14 b^2 c^6 + 20 a^12 b^4 c^6 + 8 a^10 b^6 c^6 -
30 a^8 b^8 c^6 + 8 a^6 b^10 c^6 + 20 a^4 b^12 c^6 -
24 a^2 b^14 c^6 + 11 b^16 c^6 - 3 a^14 c^8 + 13 a^12 b^2 c^8 -
29 a^10 b^4 c^8 + 19 a^8 b^6 c^8 + 19 a^6 b^8 c^8 -
29 a^4 b^10 c^8 + 13 a^2 b^12 c^8 - 3 b^14 c^8 - 35 a^12 c^10 -
2 a^10 b^2 c^10 - 7 a^8 b^4 c^10 - 56 a^6 b^6 c^10 -
7 a^4 b^8 c^10 - 2 a^2 b^10 c^10 - 35 b^12 c^10 + 77 a^10 c^12 +
67 a^8 b^2 c^12 + 120 a^6 b^4 c^12 + 120 a^4 b^6 c^12 +
67 a^2 b^8 c^12 + 77 b^10 c^12 - 99 a^8 c^14 - 172 a^6 b^2 c^14 -
230 a^4 b^4 c^14 - 172 a^2 b^6 c^14 - 99 b^8 c^14 + 94 a^6 c^16 +
194 a^4 b^2 c^16 + 194 a^2 b^4 c^16 + 94 b^6 c^16 - 62 a^4 c^18 -
108 a^2 b^2 c^18 - 62 b^4 c^18 + 24 a^2 c^20 + 24 b^2 c^20 - 4 c^22}

Francisco Javier García Capitán
28 December 2011