11784Re: incenters & squares
- Dec 3, 2005Dear 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
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.
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