AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be reals such that 
and 
Prove that
be a cyclic quadrilateral. The point 
is the reflection of 
the intersection of 
and 
, the point 
is the reflection of midpoint of 
. Also let 
be the reflection of 
midpoint of 
. Show that 
concyclic.
. Prove that: 
be a triangle, and let 
be a point on side . A line through intersects side 
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle again at point 
distinct from point . The lines 
and 
intersect again at 
and 
respectively.
such that 
for all real number 
and 
be a scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let intersect 
and 
at and respectively and intersect 
and 
at and respectively .
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
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 be the intersection point of lines 
. We call an "Interior Point" if there are points 
on 
and 
on 
such that is between and . Prove that there are at least 
Interior points.(
)
.
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
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 
. Prove that the points 
and lie on the same line.
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
a point is taken. Construct circles 
and 
of diameters 
and respectively. A line that passes through cuts in the points and , and the circles 
and 
respectively at points 
and 
. Prove that 
.
a point is taken. Construct circles and of diameters and respectively. A line that passes through cuts in the points and , and the circles and respectively at points and . Prove that .
and 
be the points where 
and 
intersect respectively. Then 
and 
, so 
and since is the diameter of we have 
(symmetry), and therefore 
(symmetry again)
and 
then prove 
be the center of 
the midpoint of 
The quadrilateral 
is a trapezoid, 
are the midpoints
which means 
Hence, 
and finally 
and 
.
,
and
We get
which implies the problem.