Problem A. 438. (November 2007)
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 pont)
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)=(PBAB)=(PB)(AB)=BQPR=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}.
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