## Fwd: Reflections of (O) in cevians of I

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• Locus of P is circumcircle + McCay cubic. ... Francisco Javier García Capitán http://garciacapitan.99on.com El 2 de diciembre de 2011 23:14, Antreas
Message 1 of 5 , Dec 2, 2011
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Locus of P is circumcircle + McCay cubic.

---
Francisco Javier García Capitán
http://garciacapitan.99on.com

El 2 de diciembre de 2011 23:14, Antreas Hatzipolakis
<anopolis72@...>escribió:

http://anthrakitis.blogspot.com/2011/12/reflections-of-circumcircle-in-cevians.html

<http://anthrakitis.blogspot.com/2011/12/reflections-of-circumcircle-in-cevians.html>
>

http://anopolis72000.blogspot.com/

[Non-text portions of this message have been removed]
• Dear friends, 1) If two triangles ABC and A B C have the same circumcircle and are perspective at P (A B C is the circumcevian triangle of P relative to ABC)
Message 2 of 5 , Dec 3, 2011
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Dear friends,

1) If two triangles ABC and A'B'C' have the same circumcircle
and are perspective at P
(A'B'C' is the circumcevian triangle of P relative to ABC) and
La is the line of the centers of circles PBC', PB'C
Lb is the line of the centers of circles PCA', PC'A
Lc is the line of the centers of circles PAB', PA'B
then I think that the lines La, Lb, Lc are concurrent at a point Q.

2) The same I think holds for orthologic triangles with the restriction that
A, A' are not on the same arc BC
B, B' are not on the same arc CA
C, C' are not on the same arc AB
and the perpendiculars from A', B', C'
to the sides BC, CA, AB pass through P.

What are the barycentrics of Q in both cases?
Is there a point P such that the six circumcenters are concyclic?

Best regards
• Dear Nikolaos, In the first case, Q is midpoint of P and O. This can be easily derived from the fact that [circumcenter of PBC ] is image of [intersection of
Message 3 of 5 , Dec 3, 2011
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Dear Nikolaos,

In the first case, Q is midpoint of P and O.
This can be easily derived from the fact that [circumcenter of PBC'] is
image of [intersection of [line perpendicular to PB and through B] and
[line perpendicular to PC' and through C'] ] under h(P,1/2).

Regards,
Sung Hyun

[Non-text portions of this message have been removed]
• Dear Sung Hyun thank you. Francisco Garcia answered me privately that Q in first case is mid point of OP after finding the barycentrics of Q. Yes, you are
Message 4 of 5 , Dec 3, 2011
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Dear Sung Hyun
thank you.
Francisco Garcia answered me privately that Q
in first case is mid point of OP
after finding the barycentrics of Q.
Yes, you are right.
If Bc is circumcenter of PBC' and Cb is circumcenter of PCB'
then since triangles PBC', PB'C are inversally similar
the diameter line PBc in PBC' is altitute in PB'C.
Hence PBc perpendicular to CB' or parallel to OCb
and hence PBcOCb is a parallelogram.
Although simple I hadn't seen it before.

Best regards

> Dear Nikolaos,
>
> In the first case, Q is midpoint of P and O.
> This can be easily derived from the fact that [circumcenter
> of PBC'] is
> image of [intersection of [line perpendicular to PB and
> through B] and
> [line perpendicular to PC' and through C'] ] under
> h(P,1/2).
>
> Regards,
> Sung Hyun
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Hello, Sung and Nikolaos, In case 1), there exist two points P for which the six circumcenters are on a circle. These two points lie on the Euler line, its
Message 5 of 5 , Dec 3, 2011
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Hello, Sung and Nikolaos,

In case 1), there exist two points P for which the six circumcenters are on
a circle.
These two points lie on the Euler line, its midpoint is M=X468 and they are
harmonic conjugates with respect to G, H, hence the formula MP^2 = MG·MH
gives a construction.

The coordinates of the points are shown below, where R stands for the
square root of

(a - b - c) (a + b - c) (a - b + c) (a + b + c) (a^2 - b^2 - c^2) (a^2 +
b^2 - c^2) (a^2 - b^2 + c^2) (a^2 + b^2 + c^2)

Therefore there are no such points with the triangle is obtuse.

The coordinates of these points P in not symmetrical form are

{{2 a^2 (a^2 + b^2 - c^2) (a^2 - b^2 + c^2) (-b^4 - c^4 +
a^2 (b^2 + c^2)), -(a^2 + b^2 - c^2) (a^8 - a^6 (2 b^2 + c^2) +
a^4 (2 b^4 + b^2 c^2 - c^4) +
b^2 (b^6 - b^4 c^2 - b^2 c^4 + c^6 - R) +
a^2 (-2 b^6 + b^4 c^2 + c^6 + R)), -(a^2 - b^2 + c^2) (a^8 -
a^6 (b^2 + 2 c^2) + a^4 (-b^4 + b^2 c^2 + 2 c^4) +
c^2 (b^6 - b^4 c^2 - b^2 c^4 + c^6 - R) +
a^2 (b^6 + b^2 c^4 - 2 c^6 + R))}, {2 a^2 (a^2 + b^2 -
c^2) (a^2 - b^2 + c^2) (-b^4 - c^4 +
a^2 (b^2 + c^2)), -(a^2 + b^2 - c^2) (a^8 - a^6 (2 b^2 + c^2) +
a^4 (2 b^4 + b^2 c^2 - c^4) + a^2 (-2 b^6 + b^4 c^2 + c^6 - R) +
b^2 (b^6 - b^4 c^2 - b^2 c^4 + c^6 + R)), -(a^2 - b^2 +
c^2) (a^8 - a^6 (b^2 + 2 c^2) + a^4 (-b^4 + b^2 c^2 + 2 c^4) +
a^2 (b^6 + b^2 c^4 - 2 c^6 - R) +
c^2 (b^6 - b^4 c^2 - b^2 c^4 + c^6 + R))}}

> Dear Sung Hyun
> thank you.
> Francisco Garcia answered me privately that Q
> in first case is mid point of OP
> after finding the barycentrics of Q.
> Yes, you are right.
> If Bc is circumcenter of PBC' and Cb is circumcenter of PCB'
> then since triangles PBC', PB'C are inversally similar
> the diameter line PBc in PBC' is altitute in PB'C.
> Hence PBc perpendicular to CB' or parallel to OCb
> and hence PBcOCb is a parallelogram.
> Although simple I hadn't seen it before.
>
> Best regards
>
> > Dear Nikolaos,
> >
> > In the first case, Q is midpoint of P and O.
> > This can be easily derived from the fact that [circumcenter
> > of PBC'] is
> > image of [intersection of [line perpendicular to PB and
> > through B] and
> > [line perpendicular to PC' and through C'] ] under
> > h(P,1/2).
> >
> > Regards,
> > Sung Hyun
> >
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> > ------------------------------------
> >
> >
> >
> > Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
> >
>
>
> ------------------------------------
>
>
>
>
>

--
---
Francisco Javier García Capitán
http://garciacapitan.99on.com

[Non-text portions of this message have been removed]
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