- Dear Jeff and Hauke

[JB]> Anyway,...can you elaborate some more on your results below? This

locus

> looks interesting to me. Is there a simple way to see that the

> exists only in a plane perpendicular to the plane of the triangle

[Hauke]

> (and why always through X3) using your examples?

> > > P is a point in space.

line L

> > > Place P such that PA:PB:PC=a:b:c. The result is a straight

> > which pierces ABC in #3, and the projection of L into ABC is the

straight

> > Euler line.

> > >

> > > Place P such that PA:PB:PC=1/a:1/b:1/c. The result is a

> > line L which pierces ABC in #3, and the projection of L into ABC

is

> > the Brocard axis.

straight

> > >

> > > Place P such that PA:PB:PC=s-a:s-b:s-c. The result is a

> > line L which pierces ABC in #3, and the projection of L into ABC

is

> > the OI line (#1-#3).

is

> > >

> > > Place P such that PA:PB:PC=1/(s-a):1/(s-b):1/(s-c). The result

> a

Except if t = 1, the locus of M (in the plane ABC) such as MA = t.MB

> > straight line L which pierces ABC in #3, and the projection of L

> into

> > ABC is the line through #3 and Mittenpunkt #9.

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, 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 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]

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Version: 7.1.362 / Virus Database: 267.13.10/190 - Release Date: 1/12/2005 - Dear Bernard
> let P be a point on the circumcircle of triangle ABC with incenter I.

Suppose that P lies on the arc BC of the circumcircle opposite to A.

>

> 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.

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 - --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <Jean-

Pierre.Ehrmann.70@n...> wrote:>

I prefer to replace "cannot exist" by "does not lie in

> 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

>

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