USAMO 2006 Problem #6
by KingSmasher3, Mar 19, 2013, 5:04 AM
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.
_______
Lemma 1 ("Mean Geometry Theorem"): Given directly similar triangles
and 
and points 
on 
respectively such that 
then 
is also directly similar to the other two.
Proof: If
then all three triangles are congruent and we are done. Otherwise, there exists a unique center of spiral similarity 
that takes to 
This implies that the three triangles 
are all similar by SAS. Hence, the three triangles 
are all similar as well, so is the center of spiral similarity that takes to so they are directly similar. 
Now, let
be the intersection of the circumcircles of 
and 
Since 
is cyclic, 
Moreover, 
is cyclic as well, so 
This implies that 
so 
by AA. From Lemma 1, we now have have 
so 
implying that 
is cyclic. It follows that 
so 
is cyclic as well, and we are done.
Main Point(s): When two points divide two line segments into equal ratios, consider MGT.
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._______
Lemma 1 ("Mean Geometry Theorem"): Given directly similar triangles
and 
and points 
on 
respectively such that 
then 
is also directly similar to the other two.Proof: If
then all three triangles are congruent and we are done. Otherwise, there exists a unique center of spiral similarity 
that takes to 
This implies that the three triangles 
are all similar by SAS. Hence, the three triangles 
are all similar as well, so is the center of spiral similarity that takes to so they are directly similar. 
Now, let
be the intersection of the circumcircles of 
and 
Since 
is cyclic, 
Moreover, 
is cyclic as well, so 
This implies that 
so 
by AA. From Lemma 1, we now have have 
so 
implying that 
is cyclic. It follows that 
so 
is cyclic as well, and we are done.Main Point(s): When two points divide two line segments into equal ratios, consider MGT.
This post has been edited 2 times. Last edited by KingSmasher3, Mar 25, 2013, 8:25 PM
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