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.
such that 
is a perfect square.
as the set of all numbers of the form 
for some nonnegative 
and 
. Find (with proof) all pairs 
such that 
and 
.
, Elbek writes 
. Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
- Greatest common divisor of numbers 
and 
board, there is initially a tree of height 
. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
. Determine the largest 
such that the gardener can ensure there are eventually majestic trees on the board, no matter how the lumberjack plays.
such that the equation
be a positive integer and 
be a polynomial with integer coefficients such that 
.
is divisible by 
.
such that for every positive real 
, there exists an integer 
satisfying![\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]](http://latex.artofproblemsolving.com/4/4/4/4443cf7166cc65e67e2e504a826c4ac5f91aab2b.png)
be a positive integer and let 
be a strictly increasing sequence of positive real numbers with sum equal to 2. Let 
be a subset of 
such that the value of![\[
\left|1-\sum_{i \in X} a_{i}\right|
\]](http://latex.artofproblemsolving.com/2/5/1/2515e074de1782f38453e1fce8ab7c2ec7cfe4c0.png)
is minimised. Prove that there exists a strictly increasing sequence of positive real numbers 
with sum equal to 2 such that![\[
\sum_{i \in X} b_{i}=1.
\]](http://latex.artofproblemsolving.com/3/7/4/37410820fc4c546f11d57212cfce53a7ec4941e3.png)
be a scalene triangle with circumcircle 
, and suppose the incircle of touches 
at 
. The angle bisector of 
meets and at 
and 
. The circumcircle of 
intersects the 
-excircle at 
, 
, and at 
. Prove that line 
passes through either or .
be positive real numbers such that : 
. Prove that : 
![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
be relatively prime nonconstant polynomials. Show that there can be at most three real numbers 
such that 
is the square of a polynomial.
with 
is called historic if 
. Show that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.
with is called historic if . Show that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.
be the smallest integer that has not been put in a set. If 
has not been put in a set make the set 
and otherwise make the set 
. Suppose that this fails at some point and we can make neither of those sets. This would mean that 
was already in a set. It cannot be the least element of a set because otherwise we would have made a set with as the least element instead when we first put in a set. If is the middle element of a set then the least element has to be 
since is not in a set yet, but this would be the same contradiction as being the least element. This means that is the largest element of its set, so the least element would be 
. But then we would have made our middle element , a contradiction. Hence, this algorithm works and we are done.
so obviously we don't want to end up in a situation such that all the points except one in the set of 
are covered at all.So we need to devise an algorithm such that we don't get into that position.
if possible, if not then use 
.
be the -th turn.We say that the algorithm has failed on the 
-th turn. Where it failed must have been at 
or 
, since 
hasn't been covered at all.Furthermore 
must have been the largest members of their respected set.So by backtracking we have that:
But notice how then 
would have been covered by our algorithm.Thus a contradiction is taking place here., we also get a contradiction.
th step, we construct a new historic set 
with as the smallest unclogged element, as 
if it's unclogged and 
otherwise, and 
. I claim that it will not terminate. and are clogged. Note that must be a 
element by size, which means that either or 
is clogged as an element. The former is obviously impossible, and in the latter case, must have been clogged in the first place.
. Consider the following algorithm: consider a set . Let 
be the smallest nonnegative integer not in , then let be 
whenever possible, and let be 
if not possible, and we call an operator.
Note that if 
is in the set, then there must've been an operator that is at least . However, that raises a contradiction because if a number larger than is used as an operator, then by then must have been in already, but now that we are using as an operator, must not have been in .
is unavailable, then either 
was an operator before or 
was an operator before. If 
is unavailable, then 
was an operator before. Since is currently being used as an operator, the historic set associated with is 
, but the set with 
is only used when is unavailable! Obviously this cannot be the case, so we're done.
set if 
and 
. Similarly, call a historic set a 
set if 
and 
. with an set, then repeatedly tries to cover the lowest uncovered number with the same type of set until the of the new set has to be more than one greater than the of the previous set, then it switches to the other type of set. If at one point, the lowest uncovered number is greater than any covered number, restart by covering the lowest uncovered number with an set.
sets are made, there will be a gap between the largest and the smallest . When we start to use the sets, note how the previously largest and the previously largest are now one-upped by the and in the new set. This continues until a new gap is formed between the largest and the smallest , and the algorithm is continued. If at any point the gap doesn't exist anymore, we can simply restart the algorithm. Now, we note that the and cannot possibly clash( is supposed to always be 
above which increases every time), and cannot clash because it one-ups the previous largest . QED.
(which we will call a 1776 set) or 
(a 2001 set). be the smallest number not yet covered. If none of the numbers in the 1776 set have been covered already, then we use a 1776 set, otherwise we use a 2001 set. element can't have already been covered.
element has been covered, then that means that on some previous step, we either used the 1776 set on or the 2001 set on . However, this is impossible as 
. has already been covered, then either we used the 1776 set on 
, or we used any set on 
. Clearly the first one is impossible as 
. Now we consider the second case. If we used the 1776 set on , then it would imply that we would have already covered , a contradiction. If we used the 2001 set, then that means that some element in the 1776 set was already covered. If was already covered, then this implies that was already covered, contradiction. Otherwise, this implies that was already covered, another contradiction.
that hasn't already been covered. be the lowest if and that isn't covered. be 
.
. is already covered. It can't be the 1st number in its triplet, since it's higher than and the triplets are added in strictly increasing order based on the 1st number. For the same reason, it can't be the 2nd number in its triplet. So, it is the third in its triplet. The first must be 1776. However, then, because has not yet been covered, when this triplet was created, would've been made the 2nd number, a contradiction. So, the 2nd step must be possible. is already covered, then it must be the 3rd in its triplet. However, then would be the first, a contradiction.
and 
with any positive integers 
with 
Note that historic sets would be of the form 
or 
We use the greedy algorithm, where we operate on the smallest nonnegative integer 
not yet covered. If 
is not yet covered, we use the former set, otherwise we use the latter.
is the smallest number not covered, but 
are already covered. Assume that this is the first occurence of such a situation. Observe that 
cannot be covered as it is larger than any 
covered previously. must not have been the smallest numbers in their sets as they are both larger than 
so 
must have been covered with 
while 
was covered along with 
For this the latter cannot occur as 
so 
must occur. But this overlaps with any of the first two historic sets, contradiction. Hence the algorithm works. QED
, such that for all 
:![\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]](http://latex.artofproblemsolving.com/1/4/1/1418fad9fd00b31c38bfbc5657e2a322d66ef450.png)
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
