[/quote]
,
is written on the board. A move consists of adding either 
or 
to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is 
,
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
be a triangle. Let 
be a circle passing through 
intersecting 
at 
and 
at 
.
be the intersection of 
and 
and 
be the intersections of 
and 
with the tangent to 
at 
be the second intersection of 
and 
and 
be a polynomial with integer coefficients such that 
,
be an integer. Define 
and 
for all integers 
.
.
such that 
coprime, 
and 
.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
.
and 
such that 
.
and 
,
.
,
,
be non-negative real numbers, no two of which are zero. Prove that :
that, for all 
,![\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]](http://latex.artofproblemsolving.com/4/8/b/48b9fc4af5a20de7fd9dfc45a74d5aa27e615e46.png)
squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of 
squares.
Y by Adventure10, Danielzh, Mango247, kiyoras_2001, and 1 other user
For any nonempty set 
of real numbers, let 
denote the sum of the elements of . Given a set 
of 
positive integers, consider the collection of all distinct sums as ranges over the nonempty subsets of . Prove that this collection of sums can be partitioned into classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.
of real numbers, let 
denote the sum of the elements of . Given a set 
of 
positive integers, consider the collection of all distinct sums as ranges over the nonempty subsets of . Prove that this collection of sums can be partitioned into classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.
.
,
.
and if 
is one of them we have:
for some 
denote the group of sums satisfying the above inequality. We claim that these groups have the desired property. to prove this, it is sufficient enough to prove that the following inequality implies that:
and since 
such that 
.
we have:
we are done.
)
then the collection of sums can be partitioned into 
term (not necessarily greater than 
)
classes. To do this, we analyze two cases: when 
and when 
everything works. (all subsets can fit into a single class, which can be split into n+1 classes in numerous ways)
.
through 
was already shown. If 
,
had n elements) already had the maximum ratio of 2, then there were only 
of them, so that we have room for a second new class to hold the sum of all the elements (if it's even possible to have all the classes with a ratio of 2). If all the 
elements of 
twice as the maximal sum of one class AND the minimal sum of the next class, even though you can find it in two different ways.
'
Let us pick an element, WLOG, 
The remaining elements, by the inductive hypothesis can be split into k classes with the desired condition. Now, we just place 
into the (k+1)st class.
,
because it doesn't give us a new subset sum.
be 
,
for all 
be the maximum element of each class, gives a construction that satisfies the problems conditions. For any subset 
of 
such that for suitable choice of 
,
is the sum of the elements of 
.
.
then we have that 
.
,
.
and consider partial sums 
,
,
.
,
,![$[a_i, a_1+\cdots+a_i]$](http://latex.artofproblemsolving.com/6/d/2/6d25f42dcf5c16657cbcb20950be044f79dda93c.png)
,
)![$[a_1+\cdots+a_{i-1}, a_1+\cdots+a_i]$](http://latex.artofproblemsolving.com/2/1/9/219e85b31a2f1bb52956f188087555494c56b583.png)
.
be some non-empty subset of 
such that 
.![$[a_1+\cdots+a_{j-1}, a_1+\cdots+a_j]$](http://latex.artofproblemsolving.com/4/d/a/4dab8e1942f21db2c64510cedb2fc364b10e2baf.png)
and ![$[a_j, a_1+\cdots+a_j]$](http://latex.artofproblemsolving.com/3/a/f/3af615c4c16a748bd17d45b0c7f96a6260ad38b7.png)
(the latter being the case since 
)
being easy.![\[ A = \left\{ a_1 > a_2 > \dots > a_n \right\}. \]](http://latex.artofproblemsolving.com/d/c/4/dc4dfe849f2fb48dec49316d8da67b47f169fbc4.png)
Fix any index ![\[ a_k > \frac{\sigma(A)}{2^k} \]](http://latex.artofproblemsolving.com/7/d/b/7dbee4387f2d949f01bbb68d4e8c801bdc917107.png)
(which must exist since 
)
and 
;
.
classes via induction hypothesis on 
.
.
of all 
.
,
,
.![$[G,G+x]$](http://latex.artofproblemsolving.com/5/f/a/5fa92991d419c55e02c488443291dc71daa44e3c.png)
can be held in a class.
,
smallest elements with 
,
smallest elements with 
,
,
be the elements of 
to be 
for 
.
.
and thus 
,
.
,
,
to 
to 
