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