[/quote]
,
such that 
,
,
be four points such that no three are collinear and 
.
,
,
be the orthocenters of 
,
,
,
,
,
are pairwise distinct and are concurrent. Show that the four points 
and 
be the circumcenter and orthocenter, respectively, of an acute scalene triangle 
intersects 
and 
at 
and 
respectively. Let 
denote the intersection of the circumcircles of triangles 
and 
other than 
and 
analogously by repeating this construction two more times. Prove that 
be a quadrilateral inscribed in a circle 
.
,
,
,
,
,
,
,
.
lies over 
.
table filled with natural numbers, we say it is a divisor table if:
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
be integers so that the integers 
are divisible by 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
and 
.
.
is circumcircle of an acute triangle 
and 
is incenter of triangle 
intersect 
and 
.
such that 
is parallel to 
.
.
.
and 
such that
Y by NO_SQUARES
A 
-sequence is a sequence of 
numbers 
each equal to either 
or 
. Determine the largest 
so that, for any -sequence, there exists an integer 
and indices 
so that 
for all 
, and 
-sequence is a sequence of 
numbers 
each equal to either 
or 
. Determine the largest 
so that, for any -sequence, there exists an integer 
and indices 
so that 
for all 
, and 
This post has been edited 7 times. Last edited by ike.chen, Aug 6, 2023, 10:00 PM
Reason: Typo
Reason: Typo
.
.
,
blocks with entries 
entries 
.
,
entries 
.
positive numbers in the initial sequence. We then choose all entries 
.
then one of the two adjacent 
buckets in a string of length 
,
,
.
ones. We will choose the 
as follows: divide the sequence into blocks of 
'
of the 
,
doesn't work, we will take the sequence 
,
.
.
.
.
(since we have an unit block).
,
.
,
in a gap(so say if we have 
we go through indices 
)
be the number of 
![\[C\le \text{max}\left(x-\frac{2022-x}{2},2022-x-\frac{x}{2}\right) = \text{max}\left(\frac{3x-2022}{2},\frac{4044-3x}{2}\right).\]](http://latex.artofproblemsolving.com/6/a/5/6a5f591b76bc00b336a54b13be6c0f53529a63fd.png)
This is the maximum of two numbers that sum up to 
,
,
is repeated 
is added to it. Notice the maximal sum we can get in each block of 
is 
is 
.
is 
,
.
numbers 
,
numbers 
In a block of 
numbers. Therefore, sum of 
So for every 
.
.
denote the number of 1s and -1s respectively. For each block of 
of them. Then the problem follows since![\[
c(1) - \sum_{\text{block of } x} \left\lfloor \frac x2 \right\rfloor \geq 1011 - \left\lfloor \frac{1011}{2} \right\rfloor = 506,
\]](http://latex.artofproblemsolving.com/c/f/e/cfea26dd28791bf66b1bd3133071a94e083c9bff.png)
as desired.
.
,
We will first prove a crucial lemma.
that satisfy the condition in the problems. Then, for any positive integer 
,
is in 
,
.
,
.
,
,
.
a contradiction. 
is maximized. Clearly 
,
gives the absolute value of the sum to be 
,
.
and 
as 
,
in the sequence of indices (i.e. making the new sequence of indices 
)
increase by 
over all 
,
which will make the absolute value of the sum bigger, contradicting the maximality of it. So we must have 
.
,
,
since 
since 
is 
and 
by the minimality of 
must be in 
.
is positive, so its absolute value is at least that of 
,
is maximal as well, so we may assume that 
,
and get a maximal 
,
.
for all 
.
,
,
are in 
'
has 
if 
if 
for some 
.
is at most 
,
,
of them. By the lemma, we have that for each odd 
,
,
,
.
,
of 
as desired.
.
for all 
,
so 
Now, note that 
is a valid sequence since 
,
for all 
,
Thus, we have that 
as desired.
so for any 
.
.
which would have at least as many 
,
for some 
for the same 
.
'
.
in order, where 
is the set of all 
.
,
,
'
odd numbers in 
,
so since all the 
'
Thus, we have that 
as desired.
works, so we're done. 
denote the sum in question; we seek the minimum of 
over all sequences 
.
,
terms; by symmetry, we can maximize the raw value of 
are among the chosen 
,
adjacent to but not among the chosen terms, we can include the terms 
whilst preserving conditions and incrementing 
.
,![\[S\le1-1+1+1-1+1+1-\dots-1+1+1=506,\]](http://latex.artofproblemsolving.com/a/4/e/a4edb6655d8673f0313b3ee5cb49f401abc79c1e.png)
as claimed.
denote the lengths of contiguous blocks of consecutive 
similarly. To maximize the magnitude of 
.
.![\[|S|_1=\sum_{i=1}^mr_i-\sum_{i=1}^n\left\lfloor\frac{b_i}{2}\right\rfloor\quad\text{and}\quad|S|_2=\sum_{i=1}^nb_i-\sum_{i=1}^m\left\lfloor\frac{r_i}{2}\right\rfloor\]](http://latex.artofproblemsolving.com/9/e/e/9eee5587035a47838e0dd5531df0dc0070f75850.png)
are both attainable. Summing, we have![\[|S|_1+|S|_2=\left(\sum_{i=1}^mr_i+\sum_{i=1}^nb_i\right)-\left(\sum_{i=1}^n\left\lfloor\frac{b_i}{2}\right\rfloor+\sum_{i=1}^m\left\lfloor\frac{r_i}{2}\right\rfloor\right)\ge2022-1011=1011,\]](http://latex.artofproblemsolving.com/5/1/2/51203d84ec9ce9aa552193e894e9702c98107368.png)
whence one of 
are 
,
By a straightforward check the exact bound is obtained for the sequence 
numbers respectively 
probably 
or 
equal to 
)
As a strategy lets pick all 
and in each block of 
instead of 
in the sigma notation?![\[\left| \sum_{i=1}^{k} a_{t_i} \right| \geqslant C.\]](http://latex.artofproblemsolving.com/7/7/b/77b6ddf8f9cbcab9c87ada489f8de339cf90f594.png)
in addition. In each 
we can get at most 
.
are the number of 
total then we have on a fixed sequence:
,
.
of the 
of the 
.
.
and the minimum is 
.![\[M+(-m)=\left \lceil \frac{a}{2}\right \rceil + \left \lceil \frac{b}{2}\right \rceil\]](http://latex.artofproblemsolving.com/b/0/c/b0c4a1ab52e28530605566f241d3e6381d38ed29.png)
which is equal to 
.
Here 
which on summing would give 
where 
are the sequence of 
are the sequences of 
,![\[ \sum \left\lfloor x_i/2 \right\rfloor \le \left\lfloor (x_1+x_2+\dots)/2 \right\rfloor \le 505, \]](http://latex.artofproblemsolving.com/b/a/d/badff2a6857ed7f674c8e55113ee4bccc04a9f71.png)
so adding in all of the positive terms yields that 
![\[\underbrace{+1, \, +1, \, \ldots, \, +1}_{1011}, \, \underbrace{-1, \, -1, \, \ldots, \, -1}_{1011}.\]](http://latex.artofproblemsolving.com/1/6/5/165b995f3f164decf633b4aeced6b2bc9c2edf1f.png)
Then our subsequence 
;
.
,
,
Clearly it is optimal for Sally to begin on a ``block" of red buckets as if she begins on blue, she may just shift one ``block" back. Thus say she attempts to maximize 
.
,
,
and blocks of blue buckets 
,
,
.
to end at.
.
Then we guarantee at least,
and hence we are done. 
for subsequence 
to be the sum of the lengths of consecutive subsequences of 
in subsequence 
is the number of 
,
.
,
.
satisfies the conditions. Observe that both 
and 
are equal to 
followed by 
negative ones. Then, in each consecutive group of negative ones we can eliminate at least half of them so there are at most 
negative ones left. Hence, the sum is at least 
,
of them. So there exists a sequence with at least 
.
Now for the lower bound, suppose the number of 
through 
and similarly the number of 
.
and thus,
.
.
.
and 
pick 
and if 
pick 
.
.
.
so by repeating the same algorithm from the start (without forgetting about the last number) we can choose every positive number and skip at least 
negative numbers so the sum will be at least 
.
with 
.
of these 
.
of these 
of 
of them and hence 
,
.
,
we have we can guarantee a sum at least 
é 
o 
e 
o 
-
e 
alternadamente entre os termos de 
começando com 
.
e, para cada 
Como queríamos mostrar.
?
,
of the 
is at least 
.
,
In particular, consider the sequence
Where the first term is 
somehow (if we can find 
,
)
,
.
WLOG.
.
.
.![$[a_1, a_{t_k}]$](http://latex.artofproblemsolving.com/2/d/8/2d83cb5e1158d65c0ff5d94068275011262f7565.png)
;
,
.
But note that 
,
.
.
(the leftmost group of plusses aren't countable in the range), so 
yet again as desired.
Consider the following algorithm 
with 
.
is positive, let 
,
.
.
groups of length 
respectively.
while the other one gives us 
. Therefore, the average value of 
Since the average of 
,
and 
respectively. WLOG assume 
.
.
Here there are 
Similar work holds if we let the number of 
ones in our sequence, so that there are 
negative ones. As before, we choose our sequence 
where 
In a run of 
of them, requiring us to choose 
of them. Therefore, when we choose the sequence which satisfies this, we get our sum is 
Since the sum is an integer, it is at least 
showing sufficiency.
as claimed.
Consider all the 
)
It is easy to check that the maximum possible sum in this case is 506.
There are 
chunks with an anomaly at the end. Say we want to maximise the number of 
more 
are, we can always achieve
for some 
We do so according to the following algorithm,
![\[ a_i = \begin{cases}1 & i = 1 \\ -1 & i = 2022 \\ -1 & i \equiv 2, 3 \pmod 4 \text{ and } i \neq 2022 \\ 1&i \equiv 0, 1 \pmod 4 \text{ and } i \neq 1.\end{cases}\]](http://latex.artofproblemsolving.com/a/6/5/a65bbd9b4d8aecea0e9ca6c7ae820a1a09ee6994.png)
This configuration is symmetric because 
for all 
so without loss of generality, it is sufficient to prove that 
for all possible configurations of the 
for each 
,
so we must take a 
,
If we choose to take 
so that we get 
so our maximum would be 
Otherwise, our maximum would be 
Assume without loss of generality that the number of 
is at least 
(If instead the number of 
apply a similar strategy as described below but with signs flipped.)
Then "fill in the gaps" where needed. If, in 
there is a run of 
instances of 
instances of 
of them. Now, 
by our without-loss-of-generality, so we can take ![\[ \sum\limits_{i=1}^{k} a_{t_i} \ge 1011 \cdot 1 + 505 \cdot -1 = 506.\]](http://latex.artofproblemsolving.com/3/8/3/3831e1e3163fe53396d37cf72637d66daeacc76a.png)
