2006 USAMO Problems/Problem 6
Contents
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>
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