Circle Concurrency and Spiral Similarity

Circle Concurrency and Spiral Similarity: What is this about?
A Mathematical Droodle

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This problem hails from the 46^th 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 BC = AD and BE = DF are not necessary. The circumcircles of triangles PQR will have another common point provided

(1)	BE/BC = DF/AD.

Note that (1) is equivalent to

(1')	CE/BC = AF/AD.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

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,

AR/CR	= AR/AF · CE/CR · AD/BC
	= sin(AFR)/sin(ARF) · sin(CRE)/sin(CER) · AD/BC
	= sin(DFQ)/sin(CRE) · sin(CRE)/sin(BEQ) · AD/BC
	= sin(DFQ)/sin(BEQ) · AD/BC
	= sin(DFQ)/sin(BEQ) · DF/BE
	= [sin(DFQ)·DF] / [sin(BEQ)·BE]
	= [sin(DQF)·DQ] / [sin(BQE)·BQ]
	= DQ / BQ.

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 ∠APD = ∠ROQ so 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 46^th 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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