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 positive integers to themselves st:
holds for any 
?
in primes 
.
angle 
is greater than
. Let 
is midpoint of 
. 
and 
are the feet of the altitude from and respectively. 
and 
are midpoint of 
and 
respectively. If 
intersect the line through 
parallel to in 
, prove that 
.
, and 
be positive real numbers, such that 
. Prove the inequality![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
be an acute triangle with orthocentre 
and circumcircle 
. Let be the point on . The line through , parallel to , intersects 
and 
at 
, respectively. Let 
intersect the circumcircle of triangle 
again at 
. Let the tangent to at intersect 
at . Prove that lines 
and 
intersect on .
numbers on circle, and no one number is divided by other. In same time for all numbers we make next operation:
are two neighbors (
is left neighbor) , then we write between 
number 
and erase 
we can receive ?
numbers 
then after operation we receive circle with numbers 
.
. Prove that![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.
. Let be a subset of such that no sum of three distinct elements of is divisible by 
. What is the maximum value of 
.
, we can include elements that are congruent to 
and 
, that is, numbers of the form 
and 
. is 40.
Prove that
Subject to the constraints:
?
(mod 5) are compatible with each other in terms of the constraints. In fact, this is the only residue duo that satisfies the constraints given for A. Thus, it maximizes the amount of elements.
elements in 