2006 USAMO Problems/Problem 6
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.
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
