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• ## Circumcenter Problem (was: Re: A concyclic problem)

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• [APH] ... Variation (easier?): Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), P a point, and A ,B , and C the reflections of P in
Message 1 of 21 , Apr 13, 2010
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[APH]
> Let ABC be a triangle, HaHbHc the pedal triangle of H
> (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> resp.
> Which is the locus of P such that the circumcenter of
> A2B2C2 is lying on the Euler Line of ABC?
>
> It is the Euler Line + ?????

Variation (easier?):

Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), P a point, and A',B', and C' the reflections
of P in Ha, Hb, Hc, resp.
Which is the locus of P such that the circumcenter of
A'B'C' is lying on the Euler Line of ABC?

Is it Euler Line +???

APH
• Thear Antreas, ... This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc Equation: -a^4 c^2 x^3 y^2 + 2 a^2 b^2 c^2 x^3 y^2 - b^4 c^2 x^3 y^2 + a^2
Message 1 of 21 , Apr 13, 2010
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Thear Antreas,

> [APH]
> > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> > resp.
> > Which is the locus of P such that the circumcenter of
> > A2B2C2 is lying on the Euler Line of ABC?
> >
> > It is the Euler Line + ?????

This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc

Equation:

-a^4 c^2 x^3 y^2 + 2 a^2 b^2 c^2 x^3 y^2 - b^4 c^2 x^3 y^2 +
a^2 c^4 x^3 y^2 - b^2 c^4 x^3 y^2 - a^4 c^2 x^2 y^3 +
2 a^2 b^2 c^2 x^2 y^3 - b^4 c^2 x^2 y^3 - a^2 c^4 x^2 y^3 +
b^2 c^4 x^2 y^3 + a^6 x^3 y z - 2 a^4 b^2 x^3 y z +
a^2 b^4 x^3 y z - 2 a^4 c^2 x^3 y z + 4 a^2 b^2 c^2 x^3 y z -
2 b^4 c^2 x^3 y z + a^2 c^4 x^3 y z - 2 b^2 c^4 x^3 y z +
a^6 x^2 y^2 z - a^4 b^2 x^2 y^2 z - a^2 b^4 x^2 y^2 z +
b^6 x^2 y^2 z - 2 a^4 c^2 x^2 y^2 z + 4 a^2 b^2 c^2 x^2 y^2 z -
2 b^4 c^2 x^2 y^2 z + c^6 x^2 y^2 z + a^4 b^2 x y^3 z -
2 a^2 b^4 x y^3 z + b^6 x y^3 z - 2 a^4 c^2 x y^3 z +
4 a^2 b^2 c^2 x y^3 z - 2 b^4 c^2 x y^3 z - 2 a^2 c^4 x y^3 z +
b^2 c^4 x y^3 z - a^4 b^2 x^3 z^2 + a^2 b^4 x^3 z^2 +
2 a^2 b^2 c^2 x^3 z^2 - b^4 c^2 x^3 z^2 - b^2 c^4 x^3 z^2 +
a^6 x^2 y z^2 - 2 a^4 b^2 x^2 y z^2 + b^6 x^2 y z^2 -
a^4 c^2 x^2 y z^2 + 4 a^2 b^2 c^2 x^2 y z^2 - a^2 c^4 x^2 y z^2 -
2 b^2 c^4 x^2 y z^2 + c^6 x^2 y z^2 + a^6 x y^2 z^2 -
2 a^2 b^4 x y^2 z^2 + b^6 x y^2 z^2 + 4 a^2 b^2 c^2 x y^2 z^2 -
b^4 c^2 x y^2 z^2 - 2 a^2 c^4 x y^2 z^2 - b^2 c^4 x y^2 z^2 +
c^6 x y^2 z^2 + a^4 b^2 y^3 z^2 - a^2 b^4 y^3 z^2 -
a^4 c^2 y^3 z^2 + 2 a^2 b^2 c^2 y^3 z^2 - a^2 c^4 y^3 z^2 -
a^4 b^2 x^2 z^3 - a^2 b^4 x^2 z^3 + 2 a^2 b^2 c^2 x^2 z^3 +
b^4 c^2 x^2 z^3 - b^2 c^4 x^2 z^3 - 2 a^4 b^2 x y z^3 -
2 a^2 b^4 x y z^3 + a^4 c^2 x y z^3 + 4 a^2 b^2 c^2 x y z^3 +
b^4 c^2 x y z^3 - 2 a^2 c^4 x y z^3 - 2 b^2 c^4 x y z^3 +
c^6 x y z^3 - a^4 b^2 y^2 z^3 - a^2 b^4 y^2 z^3 + a^4 c^2 y^2 z^3 +
2 a^2 b^2 c^2 y^2 z^3 - a^2 c^4 y^2 z^3

>
> Variation (easier?):
>
> Let ABC be a triangle, HaHbHc the pedal triangle of H
> (orthic triangle), P a point, and A',B', and C' the reflections
> of P in Ha, Hb, Hc, resp.
> Which is the locus of P such that the circumcenter of
> A'B'C' is lying on the Euler Line of ABC?
>
> Is it Euler Line +???
>

This is Euler line + nothing

> APH
>
• Dear Linh Nguyen Van and Nikos! If triangle ABC is regular we obtain next interesting result. Given regular triangle ABC and point P. A_1B_1C_1 is the
Message 1 of 21 , Apr 13, 2010
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Dear Linh Nguyen Van and Nikos!

If triangle ABC is regular we obtain next interesting result.
Given regular triangle ABC and point P. A_1B_1C_1 is the circucmevian
triangle of P, points A_2, B_2, C_2 are the reflections of A_1, B_1, C_1 in
the midpoints of BC, CA, AB
Then the circumcircle of A_2B_2C_2 pass through the center O of ABC and its
center is the reflection of P in O.

Proof. Points O and A_2 lie on the circle with center in p[oint A_0 opposite
to A. Thus the bisector of segment OA_2 is also the bisectrix of angle
OA_0A_2 which is parallel to AP.
It is clear that this line passes through point Q symmetric to P wrt O.
Similarly the bisectors of OB_2 and OC_2 pass through Q.

Sincerely
Alexey
• Another problem with circumcenters: Let ABC be a triangle and X,Y,Z three collinear points and XaXbXc, YaYbYc the pedal triangles of X,Y resp. Let A1,B1,C1 be
Message 1 of 21 , Apr 13, 2010
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Another problem with circumcenters:

Let ABC be a triangle and X,Y,Z three collinear points and
XaXbXc, YaYbYc the pedal triangles of X,Y resp.

Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.

The circumcircle of A2B2C2 passes through Z and has center
on the line XYZ.

True??

Antreas
• Yes, it is true, If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = -2 - 1/t. Francisco Javier.
Message 1 of 21 , Apr 13, 2010
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Yes, it is true,

If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = -2 - 1/t.

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
>
> Another problem with circumcenters:
>
> Let ABC be a triangle and X,Y,Z three collinear points and
> XaXbXc, YaYbYc the pedal triangles of X,Y resp.
>
> Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
> and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.
>
> The circumcircle of A2B2C2 passes through Z and has center
> on the line XYZ.
>
> True??
>
> Antreas
>
• ... Is it listed in Bernard s list of 5ics? APH [Non-text portions of this message have been removed]
Message 1 of 21 , Apr 14, 2010
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>
>
> > [APH]
> > > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> > > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> > > resp.
> > > Which is the locus of P such that the circumcenter of
> > > A2B2C2 is lying on the Euler Line of ABC?
> > >
> > > It is the Euler Line + ?????
>
> [Francisco]

> This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc
>
> Equation:
>

Is it listed in Bernard's list of 5ics?

APH

[Non-text portions of this message have been removed]
• [APH] ... Another variation: Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), Q a fixed point on the Euler line, and P a point. Let
Message 1 of 21 , Apr 14, 2010
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[APH]
> > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > (orthic triangle), P a point, and A',B', and C' the reflections
> > of P in Ha, Hb, Hc, resp.
> > Which is the locus of P such that the circumcenter of
> > A'B'C' is lying on the Euler Line of ABC?
> >
> > Is it Euler Line +???

[Francisco]:
> This is Euler line + nothing

Another variation:

Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), Q a fixed point on the Euler line, and
P a point.

Let A1,B1,C1 be the reflections of Q in Ha,Hb,Hc, resp.
and A2,B2,C2 the reflections of A1,B1,C1 in P, resp.

Which is the locus of P such that the circumcenter
of A2B2C2 is on the Euler line of ABC?

Is it the Euler Line ??

Perspectivity:

Are the triangles HaHbHc, A2B2C2 perspective?

In general:

Let X,Y,Z be three collinear points, XaXbXc the
pedal triangle of X, and A1,B1,C1 the reflections
of Y in Xa,Xb,Xc resp. and A2,B2,C2 the reflections
of A1,B1,C1 in Z, resp.

Are the triangles A2B2C2, XaXbXc perspective
(with perspector on the line XYZ)?

APH
• Dear friends of Hyacinthos: The synthetics solutions of this problem can be found at here:
Message 1 of 21 , May 2, 2010
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Dear friends of Hyacinthos:
The synthetics solutions of this problem can be found at here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&p=1863882&sid=ce1cf3b9df63a1477cc4699f020559dc#p1863882

Best regard,
Linh Nguyen Van
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