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.
, denote the sum of digits of in its decimal representation by 
. Find all polynomials 
with integer coefficients such that for any positive integer 
, the integer 
is positive and 
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)
such that 
for all 
?
be a monic polynomial with complex coefficients such that there exist a polynomial 
with complex coefficients for which ![\[P(x^2-2)=P(x)Q(x).\]](http://latex.artofproblemsolving.com/a/6/2/a62b06890b6bbcc5c9634f6de8c7ae5fe9f37828.png)
Determine all complex numbers that could be the root of .
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
. Find all functions 
such that for all 
,![\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]](http://latex.artofproblemsolving.com/f/9/3/f93b7526ff4338eea89a1c59bc009aa001ffc15d.png)
of integers that has the following property:
(possibly 
) then 
for every integer .
of nonzero integers such that the only admissible set containing both 
and 
is the set of all integers.
be the circumcenter and orthocenter of triangle , respectively. Let 
and 
intersect at point 
. If 
is an arbitrary point on, and 
and 
be the intersection of 
with 
and 
, respectively.
are concyclic.
with incenter . Line 
and 
intersects 
and 
at 
and 
, respectively. Let and 
be the midpoints of and . Line 
meets the external bisector of angle 
at 
and line 
meets the external bisector of angle 
at 
. Prove that 
are concylic.
, 
is parallel to one of the sides of a polygon. We have 
diagonals, and only 
sides, which means that there is a side which is parallel to at least 
diagonals, thus these digonals are pairwise parallel, which means that they cannot share a common end. So these diagonals give us 
different enpoints. Together with our chosen side, this gives us different endpionts (obvoiusly these diagonals cannot share a common endpoint with our side), which are all vertex of our polygon. Thus one of our diagonals must be a side of the polygon ( impossible because of assumption about convexity)
and 
Prove that
Prove that
be a convex polygon with an even number of corners. Prove that there exists a diagonal 
which is not parallel to any side of the polygon.
