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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