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
meet and at and , respectively.
meet the line at .
![$2[BXY]=2[AXY]+2[CXY]$](http://latex.artofproblemsolving.com/d/7/e/d7e0f55c1e0c22a4093cec10bd4e2e9324f1a3f2.png)
![$1=\frac{[AXY]}{[BXY]}+ \frac{[CXY]}{[BXY]}$](http://latex.artofproblemsolving.com/a/2/4/a24df023af46e527258ca349c87a1242b897f962.png)
and
and 
, and we have to show this equals 
, or 
.
, which completes the proof.
be the midpoint of 
and 
the orthogonal projections of 
onto line 
Segment 
becomes the median of the trapezoid 
thus 
But from 
we get
be the circumradius of 
Using that the altitudes and circumdiameters issuing from the same vertices are isogonals, we obtain
yields
intersect 
at 
respectively.
meet and at and , respectively. meet the line at . and and , and we have to show this equals , or ., which completes the proof.
meet the line at 
and 
then prove 
.
lie on the same side of 
.