Circle Concurrency and Spiral Similarity: What is this about?
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Copyright © 1996-2012 Alexander Bogomolny
This problem hails from the 46th International Mathematical Olympiad, Mérida, Mexico, July 13-14, 2005. The problems with solutions have been published in Mathematics Magazine. The whole collection of problems from the 2005 USAMO, Team Selection Test, and IMO have been also published by MAA.
| Let ABCD be a given convex quadrilateral with sides BC and AD equal in length and not parallel. Let points E and F lie on sides BC and AD, respectively, such that BE = DF. Lines AC and BD meet at P, lines BD and EF meet at Q, and lines EF and AC meet at R. Consider all the triangles PQR as E and F vary. Show that the circumcircles of these triangles have a common point other than P. |
The applet presents a strengthened variant of the problem. The two conditions
| (1) | BE/BC = DF/AD. |
Note that (1) is equivalent to
| (1') | CE/BC = AF/AD. |
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Well, the other point common to all circumcircles PQR as E and F vary under (1), is the center of the unique spiral similarity that maps AC to DB. When BC||AD, P serves as the center of that transformation, EF is incident with P so that the three points P, Q, R, coincide and all the circles PQR degenerate to a point -- P.
Using law of sines applied several times,
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So it follows that the spiral similarity that maps A to D and B to C also maps R to Q. Assuming O denotes the center of this transformation, this means that
| ∠ROQ + ∠RPQ = 180° |
which makes quadrilateral OQPR cyclic. This is true for all possible positions of E and F satisfying (1) and the problem is solved.
References
- The 46th IMO, Mathematics Magazine, v 79, n 3, June 2006, pp 233-236.
- Z. Feng et al (ed.), USA and IMO Olympiads 2005, MAA, 2006
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Copyright © 1996-2012 Alexander Bogomolny
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