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.
such that for any 
, 
such that 
is a perfect square.
be real positive numbers such that: 
. Prove that: 
and 
and 
are real numbers for which the following inequality is satisfied :![\[\left|\sum_{i=1}^{n} b_i\cos(ka_i)\right|<\frac{1}{k}\]](http://latex.artofproblemsolving.com/5/e/4/5e4b3bdc2d5b71e42ba3e6339845794625080eb5.png)
. Prove that 
.
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
and 
be positive real numbers such that 
. Prove that
When equality holds?
and 
lie on the sides 
and 
of the square 
respectively, and 
. The circle through 
intersects 
and 
again at 
and 
, respectively. Prove that 
, 
, 
and 
are the points of tangency of incircle with the center of 
to , 
and 
respectively. Let be the foot of the perpendicular from to 
. is on 
such that 
. If 
is the orthocenter of 
, prove that 
bisects 
.
, let 
be a variable point on the line 
such that 
lies between 
and and the incircles of the triangles 
and 
intersect at two distinct points 
and 
Prove that the line 
passes through a point independent of .
with real coefficients such that for all reals such that 
we have the following relations![\[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]](http://latex.artofproblemsolving.com/4/4/9/4491a53f9c6f96c18e42eb984b8e83566140bfdb.png)
on the whiteboard. In each move, she selects two distinct numbers, 
and 
, erases them, and replaces them with 
. She repeats this process until only one number, 
, remains. What are all the possible values of ?
, and the sum of all numbers decreases by 
as well. This value is equal to 
, and it will always be equal to 
. We add to get the sum of all numbers when there is number, and the only possible value of is 
.
