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.![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
, prove there exists positive integer 
and positive real number 
, such that 
, for any 
, 
, 
holds.
be a acute triangle with 
. 
are the orthocenter and excenter. Let 
be a point on the same side of 
as 
, such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be the midpoint of 
. Prove that 
are collinear.
be real numbers satisfying 
and 
Prove that 
Let be real numbers satisfying 
and 
Prove that 
and a line 
in the plane.Let 
be reflections of 
across the line , respectively.Let 
be the midpoints of 
, respectively.Let 
be the reflections of across , respectively.Prove that the points lie on a line parallel to .
and 
are natural numbers such that 
is divisible by 
and 
is divisible by 
, then find the least possible value of 
.
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.
be a positive integer and let 
, 
,..., 
be positive integers from set 
such that every number from this set occurs exactly once. Is it possible that numbers , 
all have different remainders upon division by , if:
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
, 
, 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 an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
, 
, 
, 
are integers such that 
.
and 
of 
such that for each integer 
with 
, we have![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be a finite abelian group of order 
.
be arbitrary.
and pairwise distinct 
such that 
for 
.
and 
s.t 
and 
and there are exactly two i;j s.t 
and 
.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
Please post it here and now!
that add 
to 
and subtract from 
.
that add to and subtract from .
and 
of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
for some sequence ![$\{ a_i\}_{i\in [n]}$](http://latex.artofproblemsolving.com/f/6/e/f6e384c04b4aaf07513dacb66f8cadc0a597069e.png)
so that 
for all ![$i\in [n]$](http://latex.artofproblemsolving.com/e/e/5/ee51c5c5eb09111cc774a00e9970074ad9f9058c.png)
.![$\{ b_i\}_{i\in [n]}$](http://latex.artofproblemsolving.com/1/3/e/13e91e327fa3f2bc2d43edbe110540d350184040.png)
with sum divisible by that differ, in modulo , from in only two positions, say 
and 
.
and 
of so that 
for all .
for all that 
.
by, for each integer 
, define ![$i_{k+1}\in [n]$](http://latex.artofproblemsolving.com/c/e/5/ce54cff3726a2dae0e8cc97f7b358dd90622aeee.png)
to be the unique integer satisfy 
.
are the indices that 
with minimal 
, and then minimal 
.
. This means 
for all 
. for 
gives us
Plugging and use 
gives us 
.
, contradiction to the definition of 
.
. Let 
. Define the desired permutations and as follows:
and 
for the rest that 
.
is just to not use 
more than one time.
for all except when 
. and are permutations of , we've ![$\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$](http://latex.artofproblemsolving.com/4/0/c/40ce5067c15701668f9fc9a82369ce8ab1f4384c.png)
.
when too. This prove the validity of permutations we constructed.
for 
on the coordinate plane with coordinates modulo . We want to show that it is possible to choose the points plotted to have 
be any multiset of elements 
such that sum of the elements is 
. Note that the points plotted can be any points not in the same row or column.
points not in the same row or column such that the multiset 
of 
for all of the points 
can be any multiset of elements . Then, we can choose the unique point not in the same row or column as the points to obtain points for the original statement, as the sum of the coordinates of all points is necessarily 
. points in the following way. arbitrary points not in the same row or column. We will change the points in such a way that we can replace any element of by any residue . Suppose we change the element in corresponding to a point . Then, we will shift horizontally until the sum of the coordinates of is the desired value . Now, we might have two points in the same column. We will repeatedly perform the following operation. distinct rows taken up by the points. If the last point moved was and it is in the same column as , we will move on the line through with slope 
until occupies the previously empty row. to a point 
, then on the next step it moves to a point 
. We see that 
and 
both have slope . Now, note that and must be in the same row. Furthermore, after the operation moves to , the row containing must be empty, meaning that must be moved to that row. This implies that and are in the same row. Therefore, we see that 
is a cyclic isosceles trapezoid and in particular the perpendicular bisectors of and are the same, proving the claim. points to distinguish them. Now, there are finitely many states of the points, so eventually the operation must loop through a cycle of states. In this cycle, start at an arbitrary state. Now, since at least 
point must be moved and then return back to its original position, consider the point which returns to its original position the earliest. When was moved, the next point which was moved must have occupied the original row of . However, if was moved back to its original position, then the original row of must be vacant, so must have been moved. However, this implies that was moved back to its original position before by the claim, a contradiction, so we are done.
Prove that
Let 
Prove that
and 
Prove that 
.
.
