Σάββατο, 19 Ιανουαρίου 2013

A MALFATTI - LIKE PROBLEM

Let ABC be a triangle. To draw three circles, each of which is tangent to the other two and to one side of ABC and to the circumcircle of ABC.

Perspective triangles.

Let (Ka), (Kb), (Kc) be the three circles.

(Ka) is tangent to the circumcircle at A', to the BC at A" and to the other two circles (Kb), (Kc) at C*, B*, resp.

(Kb) is tangent to the circumcircle at B', to the CA at B" and to the other two circles (Kc), (Ka) at A*, C*, resp.

(Kc) is tangent to the circumcircle at C', to the AB at C" and to the other two circles (Ka), (Kb) at B*, A*, resp.

The triangles A'B'C', A*B*C* are perspective (??)


Antreas P. Hatzipolakis, 19 Jan. 2013

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