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 real numbers, such that 
and 
. Determine the value of 
.
, determine the smallest integer 
, such that it is possible to place numbers 
around a circle so that the sum of every consecutive numbers takes one of at most values.
. 
amount of people stand on coordinates 
where 
. Every person got a water cup and two people are considered to be neighbour if the distance between them is 
. At the first minute, the person standing on coordinates 
got litres of water, and the other 
people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups. will not have more than 
litres of water.
be an integer. You have two 
boards. Each board contains the numbers to inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all 
and 
, the number that is in the 
row and 
column of the first board is different from the number that is in the row and column of the second board.
be a convex pentagon such that![\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]](http://latex.artofproblemsolving.com/d/3/d/d3d9a82f4318190298a8f008d417a364e03f1fca.png)
The diagonals 
and 
meet at 
. Prove that the line 
bisects the side 
.
, with 
, 
is the incenter, 
is the intersection of 
-excircle and 
. Point 
lies on the external angle bisector of 
such that and lieas on the same side of the line 
and 
. Point 
lies on such that 
. Circle 
is tangent to 
and 
at 
, circle 
is tangent to and 
at 
and both circles pass through the inside of triangle . if 
is the Midpoint od the arc , which does not contain , prove that lies on the radical axis of and .
be a sequence of positive real numbers, and 
be a positive integer, such that![\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]](http://latex.artofproblemsolving.com/f/1/2/f12c9c5e0f154549118a33a2f359329333f432a7.png)
and 
, such that![\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]](http://latex.artofproblemsolving.com/b/0/b/b0b2bd4507621302eeee27ae50ed9c0afd72c6ef.png)
on ![$[1,n]$](http://latex.artofproblemsolving.com/2/2/4/2240bb122bdadbe086c0a6291dcd5a3eb82ad5cf.png)
as a word with 
in some order, hence 
is a permutation on 5 symbols. We write 
for the wellness of namely the amount of pairs 
that appear in the same order in the word. Here 
.
for the number of 
with wellness . Hence we want to determine 
. hence 
. The symbol k adds 
to the wellness hence now we count the permutations on ![$[1,7]\setminus \{k\}$](http://latex.artofproblemsolving.com/2/a/3/2a3fb0361b0937dbfba193cae8ae422eb3328c31.png)
with wellness . Since we do not consider the symbol anymore, we may as well count the number of permutations on ![$[1,6]$](http://latex.artofproblemsolving.com/a/b/0/ab036e7da17bee704b3b7ee0ca5ea79f424465e5.png)
with wellness . This equals 
.
.. If the first symbol is 
then it contributes 
to the wellness hence 
. We find 
. The lower bound may be 
, in that case we start counting at 
. This reduces the problem to finding the numbers 
. Etc.
and then work your way upwards.
of 
by a unique 
-tuple of integers 
such that 
for all valid , where each 
represents the number of pairs of integers 
satisfying 
. Thus, we seek the number of -tuples such that for all valid and 
.
in the expansion of 
. But notice that 
is a root of 
. Hence, the sum of the coefficients of the terms of with degree divisible by is 
. Observe that the coefficients of 
and 
are trivially each , and the coefficients of 
and 
are equal due to symmetry. Thus, the coefficient of is simply 
.
. Prove that
.
Where 
,
,
.![\[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]](http://latex.artofproblemsolving.com/2/6/4/264d110e2a40c7f787c650aa8b8dc30af5d28263.png)
satisfying 
and 
.