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.
, 
, 
, 
and 
. Extend 
to meet the circumcircle of triangle 
at 
. Prove that 
.
people at a party. Prove that there are two people such that, of the remaining 
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
.
such that![\[2^{n!}+1\mid 2^{m!}+19\]](http://latex.artofproblemsolving.com/8/7/e/87ede5616cffcb33aa187f5d8a13827470f741b4.png)
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these sequences, for at most how many of them does all the elements of the sequence give different remainders when divided by ?
with 
. Points 
and are on 
and 
respectively such that 
. Lines 
and 
meet at 
. 
is the incenter of 
and 
is the orthocenter of 
. 
and 
are circles with diameter 
and 
, respectively, intersecting each other at points 
and 
. Prove that and lie on 
.
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers such that all unit squares on the board can be made into all black after a finite number of moves.
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
, and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to 
inequalities, these given comparisons satisfy the following three constraints:
and 
, if the sums of elements of and are greater than those of subsets 
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
pairwise distinct positive integer numbers 
, find the maximum possible number of equal numbers among the fractions of the form![\[
\frac{a_i^2 + a_j^2}{a_i + a_j}
\]](http://latex.artofproblemsolving.com/3/5/3/3537ddec6d793d3831ba82a1e13b98d43b1c29bc.png)
students in the IMO team selection camp.
team selection tests were taken and the results have came out. Assume that no students have the same score on the same test.To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of these tests and starting from the first test, the person with the highest score between the remaining students will become a member of the team.The committee is having a session to choose the permutation. students have a chance of being a team member?
distinct divisors of a positive integer number form an arithmetic progression. Prove that the number has at least 
divisors.
for some 
. We may assume 
by replacing and 
by 
and 
respectively. Suppose 
for some 
, then 
we have 
, so 
. Since , 
, so for 
we have 
(
).
, plug in 
in) would imply 
, a contradiction. So 
and 
for some unique 
. Thus 
and 
are the only possible repetitions in the following 
divisors of : 
, which implies has at least 
divisors.
Prove that
,
in term of a function 
such that 
Prove that
