- 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,

> [BG] Let P be a point on the circumcircle of triangle ABC with

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

> 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.

about I, angle +/-90°.

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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