USAMO 2008 Problem #2
by KingSmasher3, Apr 16, 2013, 11:37 PM
Let 
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
_______
Clearly,
and 
intersect at 
the circumcenter of 
Let 
and 
Then 
Furthermore, 
Hence 
is cyclic. Let the tangents to the circumcircle of 
meet at 
We know that 
also lies on the circumcircle of 
Now,
is a symmedian of 
so it follows that 
and 
Let 
be the intersection of with the circumcircle of 
We know that 
so 
Then 
Since 
and both lie on the same circle, 
In other words, 
and are collinear.
Since
is right, 
is right as well. Thus, 
is cyclic as desired.
Main Point(s): First step is to notice that is cyclic. This leads to the fact that is right, which leads to trying to prove that 
is a line. Symmedians should come to mind.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle._______
Clearly,
and 
intersect at 
the circumcenter of 
Let 
and 
Then 
Furthermore, 
Hence 
is cyclic. Let the tangents to the circumcircle of 
meet at 
We know that 
also lies on the circumcircle of 
Now,
is a symmedian of 
so it follows that 
and 
Let 
be the intersection of with the circumcircle of 
We know that 
so 
Then 
Since 
and both lie on the same circle, 
In other words, 
and are collinear.Since
is right, 
is right as well. Thus, 
is cyclic as desired.Main Point(s): First step is to notice that
is cyclic. This leads to the fact that is right, which leads to trying to prove that 
is a line. Symmedians should come to mind.
be a given point inside quadrilateral 
.
and 
are located within 
if and only if 
.
then obviously we are done. If not, then let 
be the intersection of 
and 
Without loss of generality let 
and 
than 
and 
Now note that from the given angles, 
and the isogonal conjugate of 
This implies that both 
over the angle bisector of 
In other words, 
are collinear. However, this is a contradiction if either 
so we are done.
March 2014
October 2013