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

(standardized)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
Let <math> \displaystyle ABCD </math> be a quadrilateral, and let <math> \displaystyle  E </math> and <math> \displaystyle F </math> be points on sides <math> \displaystyle AD </math> and <math> \displaystyle BC </math>, respectively, such that <math> \displaystyle AE/ED = BF/FC </math>.  Ray <math> \displaystyle FE </math> meets rays <math> \displaystyle BA </math> and <math> \displaystyle CD </math> at <math> \displaystyle S </math> and <math> \displaystyle T </math> respectively. Prove that the circumcircles of triangles  <math> \displaystyle SAE, SBF, TCF, </math> and <math> \displaystyle TDE </math> pass through a common point.
+
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.
  
 
== Solution ==
 
== Solution ==
 +
 +
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>.
 +
 +
<math>BXF=BSF=AXE</math> because <math>BSF</math> tends both arcs <math>AE</math> and <math>BF</math>.
 +
<math>BFX=XSB=XEA</math> because <math>XSB</math> tends both arcs <math>XA</math> and <math>XB</math>.
 +
Thus, <math>XAE~XBF</math> by AA similarity, and <math>X</math> is the center of spiral similarity for <math>A,E,B,</math> and <math>F</math>.
 +
<math>FYC=FTC=EYD</math> because <math>FTC</math> tends both arcs <math>ED</math> and <math>FC</math>.
 +
<math>FCY=FTY=EDY</math> because <math>FTY</math> tends both arcs <math>YF</math> and <math>YE</math>.
 +
Thus, <math>YED~YFC</math> by AA similarity, and <math>Y</math> is the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>.
 +
 +
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.
  
 
{{solution}}
 
{{solution}}

Revision as of 02:05, 28 March 2009

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.

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

Invalid username
Login to AoPS