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 function such that
for all 
. Determine all such functions 
.
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 .
be real numbers such that 
Prove that 
Where 
. Find all 
such that for all 
, 
, 
and 
be points on the same circumference with 
. Let 
and 
be the projections of onto 
and 
, respectively. Prove that 
cuts the segment 
into equal parts.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
such that for any 
:
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
that satisfy 
and![\[
f(f(n)) = f(f(n+2)+2) = n
\]](http://latex.artofproblemsolving.com/4/d/8/4d8d7420e51a7529996473badd8a4e9976c2f01a.png)
for all integers 
.
such that, for all 
, 
, 
, and![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
. Determine all positive integers 
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
with exactly 9 elements that is subset of 
. Prove that there exist two subsets and that satisfy the following: and are non-empty subsets from , and are equal, and
is an empty subset.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
with 
and 
. Lines 
and 
intersect at 
, and lines and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to if and only if divides into two triangles with equal area.
and 
, Since, 
at 
and 
at 
After some angle chasing, 
Also, 
is Rhombus, Hence, 
and similarly, 
, Hence, 
is equilateral
be a triangle with 
and 
,
be the reflection of 
be the reflection of 
;
be the reflection of 
is an equilateral triangle.
and 
.
due to the property of reflection.
and 
,
.
.
and its reflection over 
.
(so that the new triangle is called 
)
,
.
is an equilateral triangle, and thus 
and 
.
,
.
is an equilateral triangle.
be the intersection of 
be the intersection of 
and 
then 
.
is isosceles 
.
be the center.
where 
is the intersection between the circle and 
.
.
