[/quote]
,
Prove that
and 
Prove that
'is a perfect square.
be a point on side 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle 
distinct from point 
.
and 
intersect 
and 
respectively.
such that 
for all real number 
and 
is an equilateral . Let 
and 
at 
and 
at 
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
be a positive integer such that 
.
.
be lines in the plane such that no two of them are parallel and no three of them are concurrent. Let 
.
on 
and 
on 
such that 
Interior points.
)
.
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
and 
then prove 
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
be non-negative real numbers such that 
Prove that
Y by
In triangle 
, points 
and 
on side 
are the feet of the median and angle bisector, respectively, drawn from vertex 
. Points 
and 
are the feet of perpendiculars drawn from point on sides 
and 
, respectively. Point 
lies on the median 
such that 
. Prove that points 
and are collinear.
, points 
and 
on side 
are the feet of the median and angle bisector, respectively, drawn from vertex 
. Points 
and 
are the feet of perpendiculars drawn from point on sides 
and 
, respectively. Point 
lies on the median 
such that 
. Prove that points 
and are collinear.
be the incenter and let the incircle meet 
at 
respectively. It's well-known that 
concur, so because the homothety at 
also sends 
and 
it follows that 
must be collinear as well and we're done. 
