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