[/quote]
,
be non-negative real numbers. Prove that![\[\sum_{i=1}^n \frac{\displaystyle a_i^2 \left( \sum_{j=1}^n a_j - a_i \right)^2}{\displaystyle \sum_{j=1}^n a_j^2 - a_i^2} \geq \left(\sum_{i=1}^n a_i\right)^2 - \sum_{i=1}^n a_i^2.\]](http://latex.artofproblemsolving.com/b/1/6/b16f7db4a2f1bb30df4f6534bc6fa18b0fd6b01b.png)
Assume all denominators are non-zero.
be non-negative real numbers. Prove that![\[\sum_{i=1}^n \frac{\displaystyle a_i^2 \left( \sum_{j=1}^n a_j - a_i \right)}{\displaystyle \sum_{j=1}^n a_j^2 - a_i^2} \geq \sum_{i=1}^n a_i.\]](http://latex.artofproblemsolving.com/5/c/2/5c2e78940e212cd290f76cc566b8c8018c8fea44.png)
Assume all denominators are non-zero.
,
for all positive 
and 
and 
be two triangles with a common orthocenter 
.
are concurrent
Prove that
Prove that
,
,
,
be reals, and let 
.![\[ f(5,5)f(5,6) = f(2,4)f(3,4) = f(1,4)f(4,4) \implies f(1,1)f(1,2)f(4,5)f(4,6) = f(2,3)f(3,3)f(1,5)f(1,6) \]](http://latex.artofproblemsolving.com/b/9/6/b9663a84c82661cb6a5e1cfdf3894c9943892a6f.png)
be an acute triangle and let 
be the midpoint of 
.
passing through 
and 
and 
at points 
and 
respectively. Let 
be the point such that 
is a parallelogram. Suppose that 
.
of positive integers for which the numbers 
and 
are both positive integers.
![\[\tau\big(\varphi(n)\big) \geq \varphi\big(\tau(n)\big)\]](http://latex.artofproblemsolving.com/1/e/f/1ef41ab06470c85fd23226c2b470b1811629b559.png)
with all integer n.
of integers is defined by![\[
-\frac{1}{2} < a_{n+1} - \frac{a_n^2}{a_{n-1}} \leq \frac{1}{2}
\]](http://latex.artofproblemsolving.com/5/2/f/52ffcd88e92af6bbff0073c8f1a536cfd4644a9a.png)
with 
,
,
is odd for all values of 
.
grid of unit squares. Two distinct unit squares are adjacent if they share a common side. Each unit square is to be coloured either black or white. Such a colouring is called 
if every unit square in the grid is adjacent to an even number of black unit squares. Determine the number of 
be positive integer such that ![\[ \text{LCM}(x+2, y+2)+\text{LCM}(x, y)=2\text{LCM}(x+1, y+1).\]](http://latex.artofproblemsolving.com/7/3/3/7338e4d6058ededa9b8cd9fcc945ae362db4bad4.png)
Prove that 
Y by pifinity, Adventure10, Mango247
For a positive integer 
define a sequence of zeros and ones to be balanced if it contains zeros and ones. Two balanced sequences 
and 
are neighbors if you can move one of the 
symbols of to another position to form . For instance, when 
, the balanced sequences 
and 
are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set 
of at most 
balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in .
define a sequence of zeros and ones to be balanced if it contains zeros and ones. Two balanced sequences 
and 
are neighbors if you can move one of the 
symbols of to another position to form . For instance, when 
, the balanced sequences 
and 
are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set 
of at most 
balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in .Attachments:
This post has been edited 1 time. Last edited by orl, Oct 25, 2004, 12:12 AM
let 
be the sum of the positions of the 
’
.
.
balanced sequences into 
classes according to the residue of 
,
be a class of minimum size. Then 
,
.
obtained from 
-
satisfies 
.
is the 
,
places and 
neighbors of 
for 
consecutive integers. In particular, one of these 
.
.
be the sequence of 
s
s
,
.
,![\[
S=\{s : f(s)\equiv m\pmod{n+1}\}
\]](http://latex.artofproblemsolving.com/6/f/2/6f22659a580760e482013dc7cc92d9bae7495815.png)
is sufficient (and since there are 
such sets and 
total sequences, at least one must have appropriate size).
be the sequence where 
,
in particular.
,
.
is a sequence of integers increasing by 
.
,
.
is a sequence of integers decreasing by 
?
-
,
,
adjacent to 
.![\[S_k = \{s : f(s)\equiv k\pmod{n+1}\}\]](http://latex.artofproblemsolving.com/5/b/3/5b3e9c914f8865975ad78c4236c79da2b4cb3674.png)
works. To finish, note that ![\[|S_0|+\cdots+|S_n|=\binom{2n}{n},\]](http://latex.artofproblemsolving.com/9/4/9/94932eb4da44b25ad00a8f7d309cd24a20c8f7c8.png)
so we can pick some 
.
![$f:\{\text{balanced sequence}\}\to [n+1],$](http://latex.artofproblemsolving.com/d/3/0/d300bbae04ec6ff20ef09f457c6ef3de6d9b9658.png)
such that for any balanced sequence 
let 
be the set of neighbors of ![$\text{Im}(f(N(\alpha)))=[n+1].$](http://latex.artofproblemsolving.com/c/f/1/cf19db370c7442c34d08e55672036922c822f5ca.png)
If so by Pigeonhole principle there exists ![$j\in [n+1]$](http://latex.artofproblemsolving.com/0/b/6/0b64283b96b9ab716efb25865c88a37484254ce5.png)
such that 
take the set 
and we are done.
Define 
Then
Because 
Therefore from 
to 
add(minus) 1 for ![$[n+1]\subseteq\{f(\alpha_1),\ldots ,f(\alpha_{2n})\},$](http://latex.artofproblemsolving.com/a/7/c/a7c4b83718d9a3e0d95beee31993de18f12208b1.png)
done
