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

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== Problem == | == Problem == | ||

− | Let <math> | + | 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 be a quadrilateral, and let and be points on sides and , respectively, such that . Ray meets rays and at and respectively. Prove that the circumcircles of triangles and pass through a common point.

## Solution

Let the intersection of the circumcircles of and be , and let the intersection of the circumcircles of and be .

because tends both arcs and . because tends both arcs and . Thus, by AA similarity, and is the center of spiral similarity for and . because tends both arcs and . because tends both arcs and . Thus, by AA similarity, and is the center of spiral similarity for and .

From the similarity, we have that . But we are given , so multiplying the 2 equations together gets us . are the supplements of , which are congruent, so , and so by SAS similarity, and so is also the center of spiral similarity for and . Thus, and 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.*