## Re: More 3D centers

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• Dear Jeff and Hauke [JB] ... locus ... [Hauke] ... line L ... straight ... is ... straight ... is ... is ... Except if t = 1, the locus of M (in the plane ABC)
Message 1 of 11 , Nov 29, 2005
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Dear Jeff and Hauke

[JB]
> Anyway,...can you elaborate some more on your results below? This
> looks interesting to me. Is there a simple way to see that the
locus
> exists only in a plane perpendicular to the plane of the triangle
> (and why always through X3) using your examples?
[Hauke]
> > > P is a point in space.
> > > Place P such that PA:PB:PC=a:b:c. The result is a straight
line L
> > which pierces ABC in #3, and the projection of L into ABC is the
> > Euler line.
> > >
> > > Place P such that PA:PB:PC=1/a:1/b:1/c. The result is a
straight
> > line L which pierces ABC in #3, and the projection of L into ABC
is
> > the Brocard axis.
> > >
> > > Place P such that PA:PB:PC=s-a:s-b:s-c. The result is a
straight
> > line L which pierces ABC in #3, and the projection of L into ABC
is
> > the OI line (#1-#3).
> > >
> > > Place P such that PA:PB:PC=1/(s-a):1/(s-b):1/(s-c). The result
is
> a
> > straight line L which pierces ABC in #3, and the projection of L
> into
> > ABC is the line through #3 and Mittenpunkt #9.

Except if t = 1, the locus of M (in the plane ABC) such as MA = t.MB
is a circle and the power of O wrt this circle is allways R^2.
If u,v,w are positive and distinct, the points M (in the plane ABC)
such as MA:MB:MC = u:v:w are the common points (if they are real) Q,
Q' of three coaxal circles and the radical axis QQ' goes through O.
Now the points P in the space such as PA:PB:PC = u:v:w will be the
common points of three coaxal spheres, ie the circle with diameter
QQ' in a plane perpendicular to the plane ABC.
Obviously, the projection of this circle upon the plane ABC is the
segment [QQ']; the line QQ' goes through O but O never lies in the
segment [QQ'].

To be more precise, if l = (ua+vb-wc)(-ua+vb+wc)(ua-vb+wc)
If l>0 then Q,Q' are real and distinct; thus the locus of P is the
circle above
If l=0 then Q=Q' and the locus of P is the line perpendicular at Q
to the plane ABC
If l<0, a point P such as PA:PB:PC = u:v:w cannot exist

Friendly. Jean-Pierre
• Dear friends, let P be a point on the circumcircle of triangle ABC with incenter I. denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB. I, Pa, Pb,
Message 2 of 11 , Dec 1, 2005
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Dear friends,

let P be a point on the circumcircle of triangle ABC with incenter I.

denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.

I, Pa, Pb, Pc are the vertices of a rectangle.

Find P such that the rectangle becomes a square.

Perspectivities ?...

Best regards

Bernard

[Non-text portions of this message have been removed]
• Obviously the incenters Pa, Pb, etc. are on the bisectors of angles of the quadrangle ABCP and it seems that the condition is that the four bisectors pass
Message 3 of 11 , Dec 2, 2005
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Obviously the incenters Pa, Pb, etc. are on the bisectors of angles
of the quadrangle ABCP and it seems that the condition is that the
four bisectors pass through a common point.
Best regards
Paris Pamfilos

-----Original Message-----
From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com]On
Behalf Of Bernard Gibert
Sent: Thursday, December 01, 2005 6:02 PM
To: Hyacinthos@yahoogroups.com
Subject: [EMHL] incenters & squares

Dear friends,

let P be a point on the circumcircle of triangle ABC with incenter I.

denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.

I, Pa, Pb, Pc are the vertices of a rectangle.

Find P such that the rectangle becomes a square.

Perspectivities ?...

Best regards

Bernard

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

--
No virus found in this incoming message.
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• Dear Bernard ... Suppose that P lies on the arc BC of the circumcircle opposite to A. The lines PaPb and PaPc go respectively through the excenters Ib and Ic.
Message 4 of 11 , Dec 3, 2005
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Dear Bernard
> let P be a point on the circumcircle of triangle ABC with incenter I.
>
> denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.
>
> I, Pa, Pb, Pc are the vertices of a rectangle.
>
> Find P such that the rectangle becomes a square.

Suppose that P lies on the arc BC of the circumcircle opposite to A.
The lines PaPb and PaPc go respectively through the excenters Ib and
Ic.
The rectangle will be a square if PaI is bisector of IbPaIc.
Hence the construction :
Pa lies on the circle (Ca) of diameter IbIc (center Wa) such as <IbPaI
= <IPaIc = 45°. P lies on the line WaPa.
Do the same for P lying on the arcs CA or AB of the circumcircle.
Friendly. Jean-Pierre
• ... I prefer to replace cannot exist by does not lie in real space ; I encountered many geometrical problems where e.g. a locus is imaginary except one
Message 5 of 11 , Dec 3, 2005
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--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <Jean-
Pierre.Ehrmann.70@n...> wrote:
>
> To be more precise, if l = (ua+vb-wc)(-ua+vb+wc)(ua-vb+wc)
> If l>0 then Q,Q' are real and distinct; thus the locus of P is the
> circle above
> If l=0 then Q=Q' and the locus of P is the line perpendicular at Q
> to the plane ABC
> If l<0, a point P such as PA:PB:PC = u:v:w cannot exist
>
I prefer to replace "cannot exist" by "does not lie in
real space"; I encountered many geometrical problems
where e.g. a locus is imaginary except one special real point.
Also, I don't shriek back from using negative or imaginary
distances inbetween if it gives a cute result :-)

THX for verifying it's a circle - where (projected back on ABC)
does its highest/lowest point lie?

Hauke
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