2006 USAMO Problems/Problem 6

Problem

(Zuming Feng, Zhonghao Ye) 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.

Solutions

Solution 1

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, $XAE\sim XBF$ 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, $YED\sim YFC$ 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 $XED\sim XFC$ 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.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2006 USAMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Problem
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2006 USAMO Problems/Problem 6

Contents

Problem

Solutions

Solution 1

See also