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

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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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== See Also == | == See Also == |

## 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.