[/quote]
,
be reals such that 
and 
Prove that
be a cyclic quadrilateral. The point 
the intersection of 
and 
,
.
be the reflection of 
.
concyclic.
. Prove that: 
at 
at 
intersects the circumcircle 
distinct from point 
and 
intersect 
and 
respectively.
such that 
for all real number 
and 
and 
on the plane such that 
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
Let 
be an equiangular triangle with circumcircle 
. Let point 
and point 
so that 
. The circumcircle of triangle 
intersects the circle in the point 
. The halflines 
and 
intersect the line through 
and 
in the points 
and 
. Prove that the incenter of the triangle 
is independent of the choice of 
and 
.
(The angles in the problem statement are not directed. It is assumed that and are chosen in such a way that the halflines and indeed intersect the line through and .)
be an equiangular triangle with circumcircle 
. Let point 
and point 
so that 
. The circumcircle of triangle 
intersects the circle in the point 
. The halflines 
and 
intersect the line through 
and 
in the points 
and 
. Prove that the incenter of the triangle 
is independent of the choice of 
and 
.(The angles in the problem statement are not directed. It is assumed that
and are chosen in such a way that the halflines and indeed intersect the line through and .)
.
degrees and the angle condition, we deduce 
and 
.
is cyclic, and since the two chords 
and 
(by the rotation), it follows that 
is the bisector of 
.
,
.
,
