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

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m (→‎Solution 1)
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<math>BXF=BSF=AXE</math> because <math>BSF</math> tends both arcs <math>AE</math> and <math>BF</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>.
 
<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>.
+
Thus, <math>XAE\sim 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>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>.
 
<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>.
+
Thus, <math>YED\sim 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.
+
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\sim 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.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 03:36, 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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