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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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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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== Solutions ==
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== Solution ==
 
  
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=== Solution 1 ===
 
Let the intersection of the circumcircles of <math>SAE</math> and <math>SBF</math> be <math>X</math>, and let the intersection of the circumcircles of <math>TCF</math> and <math>TDE</math> be <math>Y</math>.
 
Let the intersection of the circumcircles of <math>SAE</math> and <math>SBF</math> be <math>X</math>, and let the intersection of the circumcircles of <math>TCF</math> and <math>TDE</math> be <math>Y</math>.
  
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From the similarity, we have that <math>XE/XF=AE/BF</math>. But we are given <math>ED/AE=CF/BF</math>, so multiplying the 2 equations together gets us <math>ED/FC=XE/XF</math>. <math>DEX,CFX</math> are the supplements of <math>AEX, BFX</math>, which are congruent, so <math>DEX=CFX</math>, and so <math>XED~XFC</math> by SAS similarity, and so <math>X</math> is also the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>. Thus, <math>X</math> and <math>Y</math> are the same point, which all the circumcircles pass through, and so the statement is true.
 
From the similarity, we have that <math>XE/XF=AE/BF</math>. But we are given <math>ED/AE=CF/BF</math>, so multiplying the 2 equations together gets us <math>ED/FC=XE/XF</math>. <math>DEX,CFX</math> are the supplements of <math>AEX, BFX</math>, which are congruent, so <math>DEX=CFX</math>, and so <math>XED~XFC</math> by SAS similarity, and so <math>X</math> is also the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>. Thus, <math>X</math> and <math>Y</math> are the same point, which all the circumcircles pass through, and so the statement is true.
  
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{{alternate solutions}}
  
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== See Also ==
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== See also ==
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* <url>viewtopic.php?t=84559 Discussion on AoPS/MathLinks</url>
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* [[2006 USAMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=490691#p490691 Discussion on AoPS/MathLinks]
 
  
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{{USAMO newbox|year=2006|num-b=5|after=Last Problem}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:34, 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, $XAE~XBF$ 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.

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)
Preceded by
Problem 5 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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