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

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• 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
Message 1 of 9 , Feb 8, 2013
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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
> (??).
>
• 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 2 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 3 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 4 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
>
>
>
> ------------------------------------
>