## Πέμπτη, 7 Φεβρουαρίου 2013

### 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 +(2 r-R) X

Q = R X + 4 (2 r+R) X

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 - 2 r X

gQ = (2 r + 3 R) X + 2 R X

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 + 2 X

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 - r (4 r+7 R) X

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 + (4 r+3 R) X

P = (2 r - R) X + R X

P = X + 3 X

P = (2 r +R) X + 2 R X

P = R X - (6 r+R) X

ETC Search = 4.9797648772556482798.

Peter J. C. Moses, 8 Febr. 2013