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.
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 
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.
. If 
and 
, show that the area of 
is not less than the area of 
, and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to 
inequalities, these given comparisons satisfy the following three constraints:
and 
, if the sums of elements of and are greater than those of subsets 
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than ) of the line 
and the circle . Prove that 
must hold.
the length of the line segment .)
such that 
(where 
), and a functional equation 
that satisfies 
for all , has a finite number of solutions. Find such .
is the function obtained by composing 
times, that is, 
)
with 
. Suppose that the inequality 
takes place for all choiches of the 
signs. Show that there exists a line 
through such that all points 
are all on one side of .
with . Suppose that the inequality takes place for all choiches of the signs. Show that there exists a line through such that all points are all on one side of .
. If we let 
be a line perpendicular to 
at , the claim is that all the vectors lie on the same side of as 
. Simply notice that by the given condition the following holds for all 
. ![\[\big|\overrightarrow{OX}\big|\ge\big|\overrightarrow{OX}-2\overrightarrow{OA_i}\big|\]](http://latex.artofproblemsolving.com/9/2/8/9289f3a56c8ef343774b1521fc4c1c7efdb52a40.png)
Therefore, if we suppose that 
is not on the same side of as and let 
be the point such that 
we get ![\[\angle XOB_i>90^\circ\implies XB_i>OX\]](http://latex.artofproblemsolving.com/4/f/d/4fd511aa35161befbb8a1b30e56facce8b1ad26c.png)
Which is a contradiction to the above.
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 
.
.
