AoPS Academy
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, 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.
chessboard into 
non-overlapping rectangles subject to the following conditions:
is the number of white squares in the 
-th rectangle, then 
. for which such a decomposition is possible. For this value of , determine all possible sequences 
.
denote the determinant of a square matrix 
. Let ![$g:\left[0, \frac{\pi}{2}\right] \rightarrow R$](http://latex.artofproblemsolving.com/5/3/4/534fcbbc20e62de2dfa2579cd0642663c9fc3380.png)
be the function defined by
where 
.
be a quadratic polynomial whose roots are the maximum and minimum values of the function 
and 
. Then, which of the following is/are TRUE?
is the incentre and point 
is defined such that 
and 
is the perpendicular bisector of 
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let be the point on line 
such that 
. Let 
and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
that 
is a prime.
sastifying 
such that 
with 
. Denote the orthocenter of 
as , the intersection of 
and as 
. Compute the length of .
be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
.
such that 
be a set of Natural numbers from 
to 
. (
) distinct elements from this set.
be a triangle and be a point on segment . The circumscribed circle of the triangle 
cuts 
again at and the circumference circle of the triangle 
cuts again at . Prove that 
if and only if 
is the interior bisector of 
.
be a triangle and be a point on segment . The circumscribed circle of the triangle cuts again at and the circumference circle of the triangle cuts again at . Prove that if and only if is the interior bisector of .
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
is the bisector of 
,
,
,
,
,
,
We're done: