**A. 377.** The inscribed circle of the triangle *ABC* touches the side *AB* at *C*_{1}, the side *BC *at *A*_{1}, and the side *CA* at *B*_{1}. It is known that the line segments *AA*_{1}, *BB*_{1} and *CC*_{1} pass through a common point. Let *N* denote that point. Draw the three circles that pass through *N* and touches two of the sides. Prove that the six points of tangency are concyclic.

(5 points)

**Deadline expired on 17 October 2005.**

**Solution.** Use the notations of the Figure.

Circle *B*_{3}*C*_{2}*N* can be obtained by scaling the inscribed circle from vertex *A*. Triangle *NB*_{3}*C*_{2} is the image of triangle *A*_{1}*B*_{1}*C*_{1}, so the sides of these triangles are paralel, respectively. By similar scalings from points *B* and *C* we obtain that lines *B*_{2}*N* and *NC*_{3} are paralel to *B*_{1}*C*_{1}. This implies that line *B*_{2}*C*_{3} passes through point *N* and also this line is paralel to *B*_{1}*C*_{1}.

The perpendicular bisector of *B*_{1}*C*_{1} is the angle bisector *AK*. Since line segments *B*_{2}*C*_{3} and *B*_{3}*C*_{2} are paralel to *B*_{1}*C*_{1}, *AK* is also their perpendicular bisector.

Similarly, the line segments *A*_{2}*B*_{3} and *A*_{3}*B*_{2} share the common perpendicular bisector *CK* and segments *A*_{2}*C*_{3} and *A*_{3}*C*_{2} share *BK*.

All these three perpendicular bisectors pass through the incenter *K*, so *KB*_{2}=*KC*_{3}=*KA*_{2}=*KB*_{3}=*KC*_{2}=*KA*_{3} and points *A*_{2},*A*_{3},*B*_{2},*B*_{3},*C*_{2},*C*_{3} lie on a certain circle of center *K*.