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.
has incentre 
and incircle touching 
at 
. Let 
be the antipode of 
in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
for which:
for all 
.
particles on a circle situated at the vertices of a regular 
-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before.
be the smallest number of collisions after which all particles return to their original positions. Find .
![$Q\in \{-1,1,0\}[x]$](http://latex.artofproblemsolving.com/b/4/0/b404c8d6415aaf3d4442cc5381f582c1f8f8b996.png)
with 
.![$\exists P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/2/e/a2e67dd74df9971ca90d1a5d81c1c8e1f3982276.png)
palindrom with 
?
for which 
, where 
denote the integer part.
such that 
is divisible by 
, 
is divisible by 
, 
is divisible by 
.
such that if squares of a 
chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.
such that there exists a set 
of 
integers satisfying the following condition: for each 
, there exist 
such that 
is divisible by 
.
be the Fermat point of a 
. Prove that the Euler line of the triangles 
, 
, 
are concurrent and the point of concurrence is 
, the centroid of .
be a family of functions from 
. It is known that for all 
, there exists 
such that for all 
, the following equation holds:![\[
y^2 \cdot f\left(\frac{g(x)}{y}\right) = h(xy)
\]](http://latex.artofproblemsolving.com/f/7/1/f7196164e0639695529b06788ea7cb1e50eddbe7.png)
and all 
, the following identity is satisfied:![\[
f\left(\frac{x}{f(x)}\right) = 1.
\]](http://latex.artofproblemsolving.com/1/5/b/15b363bf1754a4edd72423c3d0ea631173897beb.png)
, 
, 
, , where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
and 
are given. Let be the midpoint of the segment joining their centers, , 
be arbitrary points on , respectively such that 
. Find the locus of the midpoints of segments 
.
of points in the plane is balanced if, for any two different points and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points in such that 
., there exists a balanced set consisting of points. for which there exists a balanced centre-free set consisting of points.
and
there exist 
such that 
because of Piegonhole principle.
. Therefore, 
.
,
such that 
. So 
.
but I can't prove.
, if doesn't exist some consecutive terms in the sequence with their sum equal to 
,then all of members in 
and 
can take 
values and 
is a permutation of 
.
because all members of 
are and cannot contain numbers 30 and 2007-30=1977 in it. (Use the symmetry!)
, then there is a sequence of consecutive numbers that add up to 30.
and 
Prove that
.
?