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 
such that 
where indices are modulo 
. Let 
. Consider a set 
such that each element of is of the form 
where 
and if we have any 
with 
, then 
should occur in 
elements of . The existence of such a set can be proven using Hall's Marriage Theorem by considering a set to have 
copies of each subset of 
of size 
and then noticing that there is a matching from subsets of size and those of size 
because they can be shown to satisfy Hall's condition and gluing the matchings would give rise to . be any set satisfying the given conditions. Then ![\[|K| = \frac{\sum_{A\in K}\dbinom{n}{|A|}\cdot \sum_{P}^{A\in P \in T} 1}{|T|} = \frac{\sum_{P\in T} \sum _{A\in P\cap K } \dbinom{n}{|A|}}{|T|}\le \frac{\sum_{P\in T} M}{|T|} = M\]](http://latex.artofproblemsolving.com/7/6/c/76c5b3219f67501f15dad96152bd971868f46469.png)
![\[K = \bigcup_{j=0}^{\left\lfloor\frac{n-k}{3}\right\rfloor} \{ S| S\subset\{1, 2, \ldots, n\}, |S|=k+3j\}\]](http://latex.artofproblemsolving.com/6/9/f/69fcfa2250ccef81f4311276a1030526d25f715f.png)
( is even) or 
( is odd) satisfy the requirements of the problem, the number of the subsets in this group is 
or 
one of which equals 
. that the maximum number of subsets is .
is obvious.
, and we will prove for .
subgroups, on of which contains all subsets which do not contain the element 
and the other subgroup contains the rest of the subsets. The first subgroup has at most 
by induction, the second subgroup has at most 
elements (we can consider subsets 
instead of if 
, then add to the number of that subsets 
for the subset 
) by induction. This gives the result in case is odd, still for even more tricks must be done.
be the maximum.
.
, from which we can easily finish by induction.
. Note for every 
, at most one of 
is in the family. For the remaining sets (which contain and exclude 
), we can pick at most more sets. Hence our conclusion follows.
subsets into the floor of 
chains of elements and then the last is either a singleton or a chain of , depending on the parity of . The induction step is more or less "clone and combine."
the answer is ? we can get subsets whereas you predicted to have at most 
. Here are the subsets, 
. It seems to have at most 
subsets instead. Sorry, in advance if I am wrong somewhere.
and 
Prove that
such that for any two of them like 
if 
then 
.
is the number of elements of the set 
_
is set of the elements that are in 
the solution of the problem for 
and then continue on knowing what sum of binomial coefficents to take(module 
)
be sets which satisfies the conditions of the problem.
.
and
,
and so 
.
，
.