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 monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
.
such that 
sastifying 
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, 
)
be an acute triangle with 
, and let 
be the projection from onto 
. Let be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
such that the relation ![\[|z^{2n}+z^nw^n+w^{2n} | = 2^{2n}+2^n+1 \]](http://latex.artofproblemsolving.com/1/2/2/1229a609403904449c447225094b3d97e063efd4.png)
holds for all positive integers .
be a parallelogram with 
. Inside the parallelogram there is a point such that the triangle 
is equilateral and 
. Let K be the midpoint of side . Calculate the measure of the angle 
.
be a parallelogram with . Inside the parallelogram there is a point such that the triangle is equilateral and . Let K be the midpoint of side . Calculate the measure of the angle .
to be the symmetric point to wrt to 
Then triangle 
is equilateral, triangle 
is isosceles with base angle 
and triangle 
is isosceles and right. Therefore, 
Let 
intersect at 
But then 
and therefore triangle 
is isosceles and right; thus 
But triangles 
and 
are congruent because 
and 
Therefore 
From the above we get 
and therefore 
is the midpoint of and thus 
Finally, 
because 
is a rhombus.
we get 
where 
Similarly from triangle 
we get 
where 
From 
we get 
and substituting it in 
we get 
and therefore 
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 
.
.
