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2006 USAMO Problems/Problem 6

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Problem

Let ABCD be a quadrilateral, and let E and F be points on sides AD and BC, respectively, such that AE/ED = BF/FC. Ray FE meets rays BA and CD at S and T respectively. Prove that the circumcircles of triangles SAE, SBF, TCF, and TDE pass through a common point.

Solution

Let the intersection of the circumcircles of SAE and SBF be X, and let the intersection of the circumcircles of TCF and TDE be Y.

BXF=BSF=AXE because BSF tends both arcs AE and BF. BFX=XSB=XEA because XSB tends both arcs XA and XB. Thus, XAE~XBF by AA similarity, and X is the center of spiral similarity for A,E,B, and F. FYC=FTC=EYD because FTC tends both arcs ED and FC. FCY=FTY=EDY because FTY tends both arcs YF and YE. Thus, YED~YFC by AA similarity, and Y is the center of spiral similarity for E,D,F, and C.

From the similarity, we have that XE/XF=AE/BF. But we are given ED/AE=CF/BF, so multiplying the 2 equations together gets us ED/FC=XE/XF. DEX,CFX are the supplements of AEX, BFX, which are congruent, so DEX=CFX, and so XED~XFC by SAS similarity, and so X is also the center of spiral similarity for E,D,F, and C. Thus, X and Y are the same point, which all the circumcircles pass through, and so the statement is true.


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