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Difference between revisions of "2006 USAMO Problems/Problem 6"

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== Problem ==
 
== Problem ==
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(''Zuming Feng, Zhonghao Ye'') Let <math>ABCD </math> be a quadrilateral, and let <math> E </math> and <math>F </math> be points on sides <math>AD </math> and <math>BC </math>, respectively, such that <math>AE/ED = BF/FC </math>.  Ray <math>FE </math> meets rays <math>BA </math> and <math>CD </math> at <math>S </math> and <math>T </math> respectively. Prove that the circumcircles of triangles <math>SAE, SBF, TCF, </math> and <math>TDE </math> pass through a common point.
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(''Zuming Feng, Zhonghao Ye'') Let <math>ABCD</math> be a quadrilateral, and let <math>E</math> and <math>F</math> be points on sides <math>AD</math> and <math>BC</math>, respectively, such that <math>AE/ED = BF/FC</math>.  Ray <math>FE</math> meets rays <math>BA</math> and <math>CD</math> at <math>S</math> and <math>T</math> respectively. Prove that the circumcircles of triangles <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through a common point.
  
 
== Solutions ==
 
== Solutions ==

Revision as of 04:38, 6 August 2014

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.

See also

  • <url>viewtopic.php?t=84559 Discussion on AoPS/MathLinks</url>
2006 USAMO (Problems • Resources)
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