**A. 438.** On the circumcircle of the equilateral triangle *ABC*, choose an arbitrary point *P* on arc *AB*. Mark the points *Q* and *R* on arc *BC* and the points *S* and *T* on arc *CA* for which *AP*=*BQ*=*CR*=*CS*=*AT* holds. Let lines *BQ* and *PR* meet at *U* and let lines *AS* and *PT* meet at *V*. Finally, let lines *AU* and *BV* meet at *W*. Show that each two of lines *AU*, *BV *and *PW* enclose an angle of 60^{o}.

(5 points)

**Deadline expired on 17 December 2007.**

**Solution (sketch).** Triangles *APV* and *PBA* are similar since *VAP*=*SAP*=*APB*=*PBQ*=*PBU*=120^{o} and *APV*=*PBA*.

Let be the rotated homothety which transforms triangle *APV* into triangle *PBA*, i.e. (*V*)=*A*, (*A*)=*P* and (*P*)=*B*. The angle between vectors *AP* and *PB* is 60^{o}, so the angle of rotation is 60^{o}.

The image of line *PB* is *BQ*, because (*PB*) passes through (*P*)=*B* and its direction matches the direction of *BP* (since *PBQ*=120^{o}). It can be obtained similarly that (*AB*)=*PR*. Then

(*B*)=(*PB**AB*)=(*PB*)(*AB*)=*BQ**PR*=*U*.

Therefore transformation moves *V* to *A*, *A* to *P*, *P* to *B*, and *B* to *U*.

Let *W* be the homothety center. Then *AOV*=*POA*=*BOP*=*UOB*=60^{o}. So *O*=*W* and the lines *VW*, *AW*, *PW*, *BW* and *UW* enclose angles of 60^{o}.