## Re: [EMHL] Re: incenters & squares

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• Dear Bernard and Jean-Pierre, [BG] Let P be a point on the circumcircle of triangle ABC with incenter I denote by Pa, Pb, Pc the incenters of triangles PBC,
Message 1 of 2 , Dec 3 9:12 AM
Dear Bernard and Jean-Pierre,

[BG] 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.

[JPE]: 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 I. 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 that angle IbPaI = angle IPaIc = 45°. P lies on the
line WaPa. Do the same for P lying on the arcs CA or AB of the
circumcircle.

*** Thanks for an excellent problem and a beautiful solution.
I tried to follow up to see if PaPbPc is the circumcevian triangle of
some point. Unfortunately it is not. Pa has homogeneous barycentric
coordinates

-a^2/( (b+c)(c+a-b)(a+b-c) + 2aS)
: b^2/((c+a-b)(b(a+b-c)+S))
: c^2/((a+b-c)(c(c+a-b)+S))

From this we get an alternative description of Pa: it is the isogonal
conjugate of the infinite point of the line joining the Nagel point to
the A-trace of the isotomic conjugate of the inner Soddy center.

Best regards
Sincerely
Paul
• Dear Jean-Pierre and Paul, ... I arrived to the same conclusions with the use of rotated circles about I, angle +/-90°. I have found that the triangle with
Message 2 of 2 , Dec 3 9:48 AM
Dear Jean-Pierre and Paul,

> [BG] 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.
>
> [JPE]: 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 I. 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 that angle IbPaI = angle IPaIc = 45°. P lies on the
> line WaPa. Do the same for P lying on the arcs CA or AB of the
> circumcircle.
>
> [PY] I tried to follow up to see if PaPbPc is the circumcevian
> triangle of
> some point. Unfortunately it is not. Pa has homogeneous barycentric
> coordinates
>
> -a^2/( (b+c)(c+a-b)(a+b-c) + 2aS)
> : b^2/((c+a-b)(b(a+b-c)+S))
> : c^2/((a+b-c)(c(c+a-b)+S))
>
> From this we get an alternative description of Pa: it is the isogonal
> conjugate of the infinite point of the line joining the Nagel point to
> the A-trace of the isotomic conjugate of the inner Soddy center.

I arrived to the same conclusions with the use of rotated circles

I have found that the triangle with vertices the 3 points Pa (or the
centers of the squares ) is similar to the pedal triangle of X1124,
hence a relation with the Paasche point and inscribed congruent circles.

I didn't find any interesting perspectivity yet.

Best regards

Bernard

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