**A. 564.** Let *k* be the incircle in the triangle *ABC*, which is tangent to the sides *AB*, *BC*, *CA* at the points *C*_{0}, *A*_{0} and *B*_{0}, respectively. The angle bisector starting at *A* meets *k* at *A*_{1} and *A*_{2}, the angle bisector starting at *B *meets *k* at *B*_{1} and *B*_{2}; *AA*_{1}<*AA*_{2} and *BB*_{1}<*BB*_{2}. The circle *k*_{1}*k* is tangent externally to the side *CA* at *B*_{0} and it is tangent to the line *AB*. The circle *k*_{2}*k* is tangent externally to the side *BC* at *A*_{0}, and it is tangent to the line *AB*. The circle *k*_{3} is tangent to *k* at *A*_{1}, and it is tangent to *k*_{1} at point *P*. The circle *k*_{4} is tangent to *k* at *B*_{1}, and it is tangent to *k*_{2} at point *Q*. Prove that the radical axis between the circles *A*_{1}*A*_{2}*P* and *B*_{1}*B*_{2}*Q* is the angle bisector starting at *C*.

(5 points)

**A. 565.** The positive integers are coloured with a finite number of colours. A function *f *from the set of positive integers to itself has the following two properties:

(*a*) if *x**y*, then *f*(*x*)*f*(*y*); and

(*b*) if *x*, *y*, and *z* are (not necessarily distinct) positive integers of the same colour and *x*+*y*=*z*, then *f*(*x*)+*f*(*y*)=*f*(*z*).

Does it follow that the function is bounded from above?

(Based on *Romanian Master in Mathematics,* problem 2012/3)

(5 points)

**B. 4458.** Lines *a* and *b* are given lines in the plane, *A* is a given point not lying on *a*, and *B* is a given point not lying on *b*. If point *O* does not coincide with *A* or *B*, then there exists a transformation composed of a rotation about *O *and a central enlargement from *O*, such that *A* is mapped to *B*. What is the locus of those points in the plane for which this transformation maps the intersection of lines *OA* and *a* to the intersection of lines *OB* and *b*?

Suggested by *A. Hraskó,* Budapest

(4 points)

**B. 4460.** Regular triangles *ABD*, *BCE* and *CAF* are drawn over the sides of a triangle *ABC*, on the outside. Let *G* and *H* denote the midpoints of the line segments *BD* and *BE*, respectively, and let the midpoint of triangle *CAF* be *I*. Prove that the lines *AH*, *CG* and *BI* are concurrent.

Suggested by *Sz. Miklós,* Herceghalom

(5 points)