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 a right triangle with a right angle in 
and a circumcenter 
. On the sides 
and 
, the points 
and 
lie in such a way that 
. Let 
and 
be the projection of and on 
, respectively. Prove that 
is half as long as .
be an isosceles triangle with 
and circumcircle 
. The line through 
perpendicular to is denoted by 
. Furthermore, let 
be any point on . The circle 
with center and radius 
intersects once more at point 
and the circumcircle once more at point 
. Prove that the points 
and lie on a straight line.
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
be the incentre of a non-equilateral triangle , 
be the 
-excentre, 
be the reflection of in , and 
be the reflection of line 
in 
. Define points 
, 
and line 
analogously. Let be the intersection point of and . lies on line 
where 
is the circumcentre of triangle . to the incircle of triangle meet the circumcircle at points 
and 
. Show that 
.
such that 
is a perfect square and 
is a prime number.
, 
, and 
,![\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]](http://latex.artofproblemsolving.com/9/f/a/9fa6cca20b3620fbb32ee782196e96f6fffa0063.png)
such that for any real value of 
we have:
ones are written on a blackboard. Grisha and Gleb are playing game, by taking turns; Grisha goes first. On one move Grisha squares some 
numbers on the board. On his move, Gleb picks a few (perhaps none) numbers on the board and increases each of them by 
. If in 
moves on the board a number divisible by 
appears, Gleb wins, otherwise Grisha wins. Which of the players has a winning strategy?
be a nonempty subset of the points in the Cartesian plane such that for each 
exactly one of 
or 
also belongs to . Prove that for each positive integer 
there is a line in the plane (possibly different lines for different ) which contains at least points of .
such that for all positive real numbers 
,![\[
f^2(x) \geq f(x+y)\left(f(x) + y\right).
\]](http://latex.artofproblemsolving.com/7/1/f/71fb0d51492ee4c3b2bf4049e8224882fec16020.png)
such that 
is a divisor of 
.
players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left. updates their vote, one at a time, according to the following procedure: At the time of the update, if is voting for 
and is voting for 
then updates their vote to 
(Note that 
and need not be distinct; if 
then 's vote does not change for this update.) Every player updates their vote exactly once in each round, in an order determined by the players (possibly different across different rounds). rounds. Prove that at this time, all players will unanimously vote for the same person.
, because last four digits would have to be divisible by 
, but in each of this cases numbers 
are not. Similarly, we can see that 
leads to contradiction. is coprime to . We know that 
is coprime with 
and 
for all , so function 
is bijective on integers relatively prime to 
(we count modulo ). To see that this function is bijective, we can also see that 
iff 
divides 
, but cannot divide neither 
, nor 
if 
is not divisible by , so we would have to have 
. The same argument works modulo 
. Therefore, we can have 
. Obviously, we can also have 
, so our answer is 
.
,
,![\[ f(x)f(yf(x))=f(x+y). \]](http://latex.artofproblemsolving.com/5/6/0/560049dd70944cafa0423e5f1229d215f78215a0.png)
such that for any positive integer 
there exists a positive integer 
are equal to 
