Difference between revisions of "2006 USAMO Problems/Problem 6"
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== Problem == | == Problem == | ||
− | (''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 | + | (''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</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through a common point. |
== Solutions == | == Solutions == |
Revision as of 03:38, 6 August 2014
Contents
[hide]Problem
(Zuming Feng, Zhonghao Ye) 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.
Solutions
Solution 1
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.
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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