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 
. Show that 
. When will equality hold?
which satisfy the condition![\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]](http://latex.artofproblemsolving.com/4/b/d/4bddce124cf0da82cb8bb6c6ae2325239354f27c.png)
Find all possible values of 
.
be a positive integer and 
be a positive integer coprime to 
. Let 
, and for 
, define
Find, in terms of and 
, the greatest positive integer 
for which there exists an index 
such that 
is divisible by 
.
table are initially white. Alice and Bob play a game. First Alice paints of the fields in red. Then Bob chooses 
rows and columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of such that Alice can win the game no matter how Bob plays.
be an interior point of the acute triangle 
with 
so that 
The point 
on the segment 
satisfies 
the point 
on the segment 
satisfies 
and the point 
on the line satisfies 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
, define 
as the product of the digits of (in decimal). Find all positive integers such that 
.
be a function satisfying the following conditions:
, if 
and 
, then 
;
is surjective.
. be a triangle with 
and 
. Let and be the midpoints of 
and 
respectively. The point is inside of 
such that 
is equilateral. Let 
and 
. Prove that 
.
. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size 
, the blue tiles are of size and the black tiles are of size 
. Let 
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
can be written in the form
where 
for some non-negative 
such that
for every 
.
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 
.
is divided into 
equal right triangles and the square 
, as shown in the figure. Find the area of the square if 
.
,
and we want 
.
and multiplying by 
gives 
.