[/quote]
,
be the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
,
for all integers 
.
.
and 
Prove that
Let 
and 
Prove that
which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
.
in natural numbers m and n.
,
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
.
and 
intersect at 
.
of the two circles touches 
at 
respectively. Let 
be the symmetric points of 
intersects circles 
respectively which are distinct from 
and 
are equal.
,
.
intersects 
at 
.
cuts segment 
at 
respectively. If 
and 
,
equals 
.
are points on a circle 
,
,
.
at 
.
is on 
such that 
.
intersects 
.
is the reflection of 
wrt 
.
lies on the same circle.
in its exterio
so that 
,
,
.
and
such that 
and 
.
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
has a right angle at 
.
such that 
and 
lies between 
.
is chosen so that 
and 
.
is chosen so that 
and 
is the bisector of 
.
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
such that the number of 
which satisfy the following condition is minimal:
has periodic 
and 
are tangent to the circumcircle of 
be the reflection of 
be the reflection of 
.
is the circumcenter of 
,
is perpendicular to 
.
Y by megarnie, Adventure10
Let 
and 
be distinct positive integers. The following infinite process takes place on an initially empty board.
, operation 
will be performed only finitely many times.
Proposed by Serbia.
and 
be distinct positive integers. The following infinite process takes place on an initially empty board.- If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by and the other by .
- If no such pair exists, we write two times the number
.
, operation 
will be performed only finitely many times.Proposed by Serbia.
This post has been edited 4 times. Last edited by Tsukuyomi, Jul 17, 2019, 12:56 PM
such that we have to use operation 
infinitely many time. We called the final position of each time as 
(Final position before we use operation 2)
.
and 
.
zero, 
then we are done.
satisfy the Claim. The proof is bashing everything out. (The actual proof is too messy and I'm too lazy to show that.)
and 
.
and 
are on the board, and we add 
,
will appear on the board twice for all nonnegative 
,
.
then 
and 
will both appear twice by the induction hypothesis, and these will each "spawn" a copy of 
and 
,
will appear twice, and so we get at least one copy of 
.
.
where 
and 
.
,
appears on the board twice, by the induction hypothesis. This number will hence "spawn" another copy of 
.
(because 
,
will appear in the terminating state of the board if we are on a round whose binary representation has a 
in the 
,
.
(in binary, where there are 
,
nonnegative integers and 
.
,
,
denotes the real root of 
greater than 1.
be the multiset of integers currently written on the board. Define the quantity![\[S=\sum_{n\in\mathcal{B}}f(n). \]](http://latex.artofproblemsolving.com/d/1/f/d1f49d2fc3cf8e45c9f61b39069ce62bbd6112d8.png)
We claim that 
is preserved under type (i) operations. It suffices to show that 
,
,
.
and 
(by rearrangements of 
)
.![\[S<\sum_{x=0}^{\infty}\sum_{y=0}^{a-1}\varphi^{2y-x}=\left(\sum_{x=0}^{\infty}\varphi^{-x}\right)\left(\sum_{y=0}^{a-1}\varphi^{2y}\right), \]](http://latex.artofproblemsolving.com/5/0/e/50e3a80294f35ad2c3031177f6720036690ded25.png)
which is finite since the left factor is a decreasing geometric series and the right factor is finite.
![\[2f(n) = 2\varphi^{2y-x}\]](http://latex.artofproblemsolving.com/f/a/f/faf56742ab8395ff3af1a44d0ec6f273e7393f04.png)
![\[f(n+a)+f(n+b) = \varphi^{2y-x+2} + \varphi^{2y-x-1} = \varphi^{2y-x}(\varphi^2 + \varphi^{-1}) = 2\varphi^{2y-x+1}\]](http://latex.artofproblemsolving.com/3/3/f/33f6e2b0d2b40e8cdc58d6dd8d27beae22dd8cca.png)
and these two are not equal? I had a similar idea of firing points, with 
where 
denotes 
.
and 
when 
There is no reason to care whether it could be achieved by the condition that enforces the second rule on using operation. We could swap in any way we want.
that is the function which counts the number of times 
for 
iff 
for some constant 
.
with 
we needed operation (ii) an infinite number of times.
s
.
.
.
.
,
.
.
this multiset. But then if you apply the move to 
,
until we get to an "end position" where we must apply a (ii) move. If we apply an arbitrary sequence 
of type (i) moves on the same starting position, we will get to the same end position.
consists of replacing 
with 
then take the first move 
consisting of replacing 
(there must exist one, otherwise 
Note that 1) all moves are still valid since the 
'
Then apply the inductive hypothesis. 
for an integer 
.
times of operation (ii), I claim that this will go on forever. Let 
,
for 
,
,
things, then each of these numbers will get written at least twice, so we are done by the above paragraph. 
and define 
Inductively. If we have 
on the board, we can add 
to the board. Adding another 
on the board. Keep adding 
and 
will be on the board. Now, proceed to pair numbers arbitrarily until we reach all numbers being distinct. To show that we will use (ii) finitely often, it is sufficient to show there are finitely many 
'
,
,
.
are distinct. So, ![\[ k(b-a) \geq 2^k \]](http://latex.artofproblemsolving.com/e/5/a/e5a6b93b3b2211c7ab4d8708306b420c1c325621.png)
Which is not true for sufficiently large 
)
type 
)
type 
the base case 
is trivial. Notice that due to our assumption sequence 
must end at some point. Now suppose that 
replaces 
in particular the initial configuration of numbers contain two copies of 
so some move 
has to make replacing 
.
lead to the same final configuration as 
It suffices to apply inductive step for configuration appeared after 
of moves starting from an empty board, which contains 
zeroes will lead to the same final configuartion.
for 
with some sequence of moves type 
,
with 
nonnegative integers, then 
is an integer, 
,
.
,
consecutive integers that are used during some round. Let 
be an integer such that on or before round 
,
.
and 
,
to have infinite occurences of 
.
between them.
.
.
points, and apply operation (i) any number of times. Then using the coordinates it is easy to construct the set in the claim via a sequence of right moves and up moves, so by the claim we are done.
to the first quadrant by a series of knight moves, which was used in a number of other solutions.
and 
.
in that order such that 
has at least 
,
can be written in this form. Thus if we make 
and at least 
for some large enough 
.
,
,
The configuration of the problem clearly prompts us to consider an alternate formulation of the problem. The following infinite process takes place on an initially empty coordinate grid.
,
represents a number at 
,
,
,
and 
,
,
,
and 
.
with 
.
and 
.
we assign a frequency of numbers 
,![\[(x,y),(x,y)\to (x+1,y),(x,y+1).\]](http://latex.artofproblemsolving.com/0/7/0/0702c2cf5500ae88799c8c7154d17a62becdcdb5.png)
If 
,![\[(x,y),(x,y)\to (x+1,y),(x+b,0).\]](http://latex.artofproblemsolving.com/6/5/0/650558d2b3cadce0c018bf05d0e7df9a0c1c9d9c.png)
Operation 
.
operations of 
are all 
for 
may only be incremented through the step![\[(r-1,0),(r-1,0)\to (r,0).\]](http://latex.artofproblemsolving.com/3/3/7/3370b57cf36a1d905df22be3dc57fc33dc9dc919.png)
Evidently we begin with a value of 
at 
times at 
.
are all 
t
consists of lattice points 
.
.
times. We claim that operation 
such that 
,
,
are linear combinations of 
and at least one copy of 
,
