## Re: [EMHL] Re: Euler Lines, Circumcircles (corrected)

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• Dear Nikos, We can replace the incenter I with any other fixed point and ask for the loci, but I guess they are complicated in that general case. Probably we
Message 1 of 9 , Feb 9, 2013
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Dear Nikos,

We can replace the incenter I with any other fixed point and ask for the
loci,
but I guess they are complicated in that general case.

Probably we get interesting loci for points on the Neuberg cubic (like I)

APH

> **
>
>
> Dear Antreas,
>
> 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
>
> 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.
>
> Best regards
> Nikos
>
> > GENERALIZATION
> >
> > 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
> > (??).
> >
>

[Non-text portions of this message have been removed]
• Case of H: Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of ABC, HBC, HCA, HAB, resp. and S a point. Denote: L, La, Lb ,Lc = the lines SO, SOa, SOb,
Message 2 of 9 , Feb 9, 2013
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Case of H:

Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
ABC, HBC, HCA, HAB, 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?
[they are homothetic, I think]
2. For which S's the circumcenter of O1O2O3 is on the line L ?

Interesting Cases:

1/ S is lying on the Euler Line of ABC (so L = SO = Euler line).
I think that for both 1,2 the answer is the Euler line!

2/ L, La, Lb, Lc = the Brocard axes of ABC, HBC, HCA, HAB, concurrent
at a point S.
The triangles ABC, O1O2O3 are perspective, but the circumcenter of O1O2O3 is not on L.

APH
• Dear Antreas, Let P be an arbitrary point Let ABC be a triangle, O, Oa, Ob, Oc the circumcenters of ABC, PBC, PCA, PAB, resp. and S a point. Denote: L, La, Lb
Message 3 of 9 , Feb 10, 2013
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Dear Antreas,
Let P be an arbitrary point
Let ABC be a triangle, O, Oa, Ob, Oc the circumcenters of
ABC, PBC, PCA, PAB, 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. and O' the circumcenter of O1O2O3.
Fact1 = the triangles ABC, O1,O2,O3 are perspective
Fact2 = O' lies on L

The locus of S for Fact1 is a conic and
the locus of S for Fact2 is a line.

If P = I = Ia = Ib = Ic Fact2 holds for every S

If P = H then Fact1 holds for every S and Fact2 if S lies on
the Euler line.

ND

[APH]
> Case of H:
>
> Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
> ABC, HBC, HCA, HAB, 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?
> [they are homothetic, I think]
> 2. For which S's the circumcenter of O1O2O3 is on the line L
> ?
>
> Interesting Cases:
>
> 1/ S is lying on the Euler Line of ABC (so L = SO = Euler
> line).
> I think that for both 1,2 the answer is the Euler line!
>
> 2/ L, La, Lb, Lc = the Brocard axes of ABC, HBC, HCA, HAB,
> concurrent
> at a point S.
> The triangles ABC, O1O2O3 are perspective, but the
> circumcenter of O1O2O3 is not on L.
>
> APH
>
>
>
> ------------------------------------
>