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 real numbers satisfying 
and ![\[a_{i+1} - 2a_i + a_{i-1} = a^2_i\]](http://latex.artofproblemsolving.com/c/d/8/cd878e4ac230853182da75c54dcbf7be05b305bc.png)
for 
Prove that 
for 
![$ a,b\in [0,1] $](http://latex.artofproblemsolving.com/9/7/5/975edfc0cf5316d78283b313c6b5cfbdd7b92b6e.png)
. Prove that
Where 
Where 
, such that one of the 
letters 
can be placed in each cell of a table with rows and 
columns, and has the following property: For any two distinct rows in the table, there exists at most one column, such that the entries of these two rows in such a column are the same letter.
be an continuous function defined in ![$\text{[0,2015]},f(0)=f(2015)$](http://latex.artofproblemsolving.com/c/9/9/c99f63e726b504aea8c1a65ba2926155a873effa.png)
pairs of 
such that 
such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
be a triangle with 
. Let the incenter and incircle of triangle be 
and 
, respectively. Let 
be the point on line 
different from 
such that the line through parallel to 
is tangent to . Similarly, let 
be the point on line different from 
such that the line through parallel to 
is tangent to . Let 
intersect the circumcircle of triangle at 
. Let 
and 
be the midpoints of and , respectively.
.
with![\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]](http://latex.artofproblemsolving.com/a/b/f/abf10d7e679f6f04db556424a0a0eda26f371fd4.png)
for all 
.
be positive real numbers such that : 
. Prove that : 
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
of an isosceles triangle with base beyond point 
is taken point 
so that the angle 
is equal to the angle 
. Find the angle 
.
of an isosceles triangle with base beyond point is taken point so that the angle is equal to the angle . Find the angle .
be the midpoint of (so 
), and let 
intersect 
at 
.
be the midpoint of .
.
.
.
..
and 
, it is easy to see from here that 
.
. and as in the first proof, and let the perpendicular to at meet the extension of at 
.
, and since 
, it is easy to see that 
is a parallelogram.
, and by the problem statement, 
.
.
.
, so 
and 
. Prove that 
. Prove that
