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.
be real numbers such that 
for all 
.
.
and 
be touch points of the incenter of 
at 
and 
, respectively. Let 
and 
be the circumcenter of triangles 
and 
, respectively. Show that 
and 
concurrent.
of one variable that fullfill the following for all real numbers 
and 
:
.
be a nonempty set of positive integers such that:
if 
then 
. for any prime 
, there exists 
such that 
. is finite.
be the unit circle in the complex plane. Let 
be the map given by 
. We define 
and 
for 
. The smallest positive integer 
such that 
is called the period of 
. Determine the total number of points in 
of period 
.
)
, we know that lines 
and 
intersect at 
, lines and 
intersect at , and lines 
and 
intersect at 
. Suppose that the circumcircle of 
intersects line 
at 
and 
, and the circumcircle of 
intersects line at 
and 
, where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
, then 
.
which satisfy the following equality for all 
![\[f(x)f(y)-f(x-1)-f(y+1)=f(xy)+2x-2y-4.\]](http://latex.artofproblemsolving.com/2/0/a/20ae7491750bac4c94deaed32d5b9aa66b26491f.png)
Proposed by Dániel Dobák, Budapest
such that 
for all 
and 
and coefficients in the set 
. Prove that the probability for this polynomial to be special is between 
and 
, where a polynomial 
is called special if for every 
in the sequence 
there are infinitely many numbers relatively prime with 
.
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let be the midpoint of the side . Let be a point on the side and a point on the side such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle .
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
, then the bigger number must be greater than 
. Thus, the smaller numbers do not exceed . Now, we have exactly smaller numbers which means that the smaller numbers are precisely the numbers 
. Now, note that since all the bigger numbers are 
, any number 
can have anything paired up with it. Also, is definitely paired with since there is no number large enough for it except . Now, we take cases on the number paired with 
(call it 
) and the number paired with 
(call it 
). Suppose 
. Then, 
. Suppose 
. Then, 
. Thus, we have a total of possible pairs for 
, for each of which we can arrange the remaining 
numbers in 
ways, which gives the answer as 
.
to 
must be the greatest in their pair. We will pair numbers from to 
. Every 
has to be paired with a number 
so there are 
options for , but since we have already paired 
numbers, there are 
options for left. This means our answer is 
must pair with an element in 
, as otherwise we'd have something in 
pairing up with another number in the same set, and that can't possibly happen due to size restrictions. pairs with ; pairs with 
, 
, or excluding ; pairs with 
, 
, , , or excluding the numbers or pairs with; and so on. There was option for 's number, 
options for 's, 
for 's, and proceed with this logic to get that the answer is 
.
to do not have a double 
, so we need to pair them with a corresponding number from to . must be paired with . has two choices 
, has three choices (
but one of them is already taken by ) has four choices, has three choices (
but 
have already been paired and other too.) has two choices, and has one.
choices.
must be paired with , has options, 
, has options, 
and the remaining option from the last. continuing, has options. But when we get to , the first few options are 
and have already been chosen so you must subtract , so it has 
options, similarly 
have already been chose for so we subtract , 
options and is the last remaining option. So in total 
. Prove that
,
has been picked already. Continuing in this fashion, there are 
.
and one element from 
.
,
,
.
We claim that all pairs satisfying the initial conditions must have one element in 
and one element in 
[independent of casework as each case for 12, 13 are both the same]. Continuing with the same logic, we obtain the answer as 
in that order to get our answer![\[\boxed{\begin{cases} (k!)^2 & \quad \text{for } n=2k \\ (k!)^2 \cdot (k+1) & \quad \text{for } n=2k+1 \end{cases}}\]](http://latex.artofproblemsolving.com/9/c/7/9c733bf90b2342b3486698a1b433940520f6562d.png)
Then, let us choose the number that is paired with 
then the number that is paired with 
and so on.
(it must be between 
and 
inclusive, but one of them will be taken), then there are 
(it must be between 
and 
ways to choose the number paired with 
Then, it is easy to see that
so there are 
ways to choose the numbers paired with 
Hence, the total number of ways to form the pairs is
giving an answer of
