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 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 
.
+1 w
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.
be a positive integer that is not divisible by 
. Furthermore, let 
be a positive integer. Prove that none of the numbers of the form 
is a square number., 
, 
, 
lie on a semicircle 
in this order with diameter 
, and is not parallel to . Points and lie on segments 
and 
respectively such that 
and 
. A circle 
passes through and is tangent to , and intersects again at 
. Prove that the lines 
, and 
are concurrent.
if the following conditions are satisfied: 
, 
, 
.
and 
denote two sequences of integers defined as follows:
Prove that, except for the "1", there is no term which occurs in both sequences.
and the same for 
then,
![\[x_n=X(-1)^n+Y2^n.\]](http://latex.artofproblemsolving.com/b/6/4/b64103ce6b36840afb72e204ce9d8bde7c3ae59e.png)
Using 
and 
yields ![\[x_n=\frac{2^n+(-1)^{n+1}}{3}.\]](http://latex.artofproblemsolving.com/0/e/9/0e91ecb368a2adce3368cc089356ac41c89c7f0a.png)
Similarly, solving for the second recurrence yields ![\[y_n=2\cdot 3^{n-1}+(-1)^n.\]](http://latex.artofproblemsolving.com/b/e/9/be9151fca1e0c704e75c410c3d6963099b9170e1.png)
So if 
then 
or 
.
or 
, then 
is the only solution, corresponding the the fact that the term is in both sequences. If 
, then 
. But we have 
, so ![\[2\cdot 3^n+3(-1)^n+(-1)^m\equiv 5(-1)^n+(-1)^m\pmod 8\]](http://latex.artofproblemsolving.com/c/9/6/c965e92b534b4e7746ecd90274be64ceaa3d0122.png)
which cannot be 
. Hence there are no solutions for , and the only integer in both sequences is indeed . 
cannot be equal to 
. 
.
. Now suppose this is true for some . Since 
, and 
, so adding these two gives 
, which means will always be greater than for all . 
,
,![\[ f(x)f(yf(x))=f(x+y). \]](http://latex.artofproblemsolving.com/5/6/0/560049dd70944cafa0423e5f1229d215f78215a0.png)
and the same for 
then we would have 
.
,
and 
,
,
.
.
,
,
or 
,
.
.
.
if 
,
.
where 
and 
are roots of the equation 
.
and 
we get 
.
.
we get 
.
if 
, so we get 
.
we get 
which is not true hence , sequence 
and 
shares no common numbers other than 