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 a trapezium inscribed in a circle 
with diameter 
. Let 
be the intersection point of the diagonals 
and 
. The circle with center 
and radius 
meets at the points 
and 
(where is on the same side of as 
). The line perpendicular to at intersects 
at 
. Prove that 
is perpendicular to 
.
![\[ x + y^2 = p^m \]](http://latex.artofproblemsolving.com/e/c/f/ecf553036f22327dd457dc06611b6bc1b0ef072c.png)
![\[ x^2 + y = p^n \]](http://latex.artofproblemsolving.com/5/b/d/5bd672934a395aabd6865ca4b50d8da75c0fe07e.png)
For 
positive integers and 
being a prime.
and 
. Prove that
Equality holds when 
or 
, , and 
lie on a circle 
such that 
. The tangent line to at point intersects the tangent line to at and the line 
at and . The circle and the circumcircle of triangle 
intersect again at . Prove that 
. and . Prove that
Equality holds when or 
intersect its circumcircle at 
. The tangets to from 
determine a triangle 
, where 
is relative to and so on. Prove that lines 
are concurrent at a point 
and that belongs to the euler line of . be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
, and 
intersect 
at 
. Prove that lines 
and 
are either parallel or coincide.
. Prove that:
be reals such that 
Prove that 
Let be reals such that 
Prove that 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
touch at . intersect incircle at . Points 
on incircle are such points, that 
. 
are incenters of 
.
bisector of 
and 
.
and 
.
Where 
such that :
:
so 
similarly we get 
but 
is harmonic thus 
hence the claim follows and 
is perpendicular to the bisector of 
whence 
so the result required follows.