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.
:
and 
are in the coordinate plane such that 
. Let 
denote the locus of all points 
in the coordinate plane satisfying 
, and let 
be the midpoint of 
. Points 
and 
are on such that 
and 
. The value of 
can be expressed in the form 
, where 
and 
are relatively prime positive integers. Find 
.
be the positive number satisfied 
Find the minimum of 
) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
be positive integers such that no number divides some other number. If 
, prove that 
.
with the property that ![\[f(x-f(y))=f(f(x))-f(y)-1\]](http://latex.artofproblemsolving.com/f/2/5/f25cc1e8ae0be1fd02b347fd94be4fab88af1d46.png)
holds for all 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.. There are 
mutually non-attacking rooks placed on a grid 
. The grid is splitted into two connected parts, symmetric with respect to the center of the grid. What is the largest number of rooks that could lie in the same part?
such that the largest prime divisor of 
is equal to the largest prime divisor of 
.
, is circumscribed around the balls. Then equals
? is about 
the intersection of 
's angle bisector and . We'll prove 
.
, we know that 
must be an isosceles trapezoid (the two opposite angle bisectors are symmetric) where 
and 
is cyclic, then 
, so 
.
is the perpendicular bisector of 
, so 
must be isosceles. By 
, 
.
, so 
is the inner angle bisector of 
. We conclude that 
.
.
be real numbers such that 
.
is isosceles, so 
is a trapezoid.