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.
from the set of non-negative integers to itself such that 
, whenever 
, are non-negative integers satisfying 
.
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
. Assume that the inequality is strict for some 
.
for every 
or 
for every .
be an integer. Ion draws a regular 
-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the -gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
, 
, and 
, let 
denote the greatest common divisor of 
.
is an integer.
. Prove that:
. Prove that:![$(a+b+c)^4(ab+bc+ca) - 9\sum{\frac{a}{c}} \ge 54[(a+b)(b+c)(c+a)+abc-1]$](http://latex.artofproblemsolving.com/d/3/c/d3cf439c79cb84ca23083e51b6190ab5118f34e8.png)
.
and the domain 
passing through the origin touch 
.
be a mobile point on the parabola . Let denote 
the distances between the point and the lines respectively. Find the coordinate of the point giving the minimum value of 
on a coordinate plane such that for all points 
over the domain 
, the inequality 
holds.
is defined recursively by ![\[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]](http://latex.artofproblemsolving.com/f/4/b/f4b8f95e49de7274cccd8ee498e0b496f31ddfea.png)
Show that 
for all 
.
be real numbers such that 
. Determine the minimum value of 
and determine all values of 
such that the minimum value is achived.
must be in the interior of 
. In other case as is convex, we can take a line passing through and not intersecting . So let be our point in the interior of quadrilateral , which we will denote with 
. Let 
be a line passing through and intersecting 
and 
in points 
and 
respectively. Let 
and 
be such points on that 
and 
. Let lines 
and 
intersect side in 
,
respectively. By 
we will denote area of triangle 
and the same thing for quadrilaterals... As lines 
,
and 
are lines dividing into two parts of equal area, we obtain equations ![\[ S(KQX)+S(XPL)=S(QMX)+S(NPX) \]](http://latex.artofproblemsolving.com/1/4/e/14e11850bd3579356df0fff5d47457b2060135b8.png)
but as 
we have 
which means that 
.Now let 
be a line passing through and parallel to this line intersects with lines 
and in points 
and 
respectively. As we have we have also 
,but there is also equality , and so on, if 
it would mean that 
which is impossible. We obtained that 
, so 
.
which means that is a parallelogram! 
Now , if is a parallelogram for point we can take intersection point of it's diagonals 
such that 
coprime, 
and 
.
,such that 
P
.
and 
,
and 
. Prove that
.
satisfying
Prove
