[/quote]
,
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
have the property that 
is a perfect square for infinitely many positive integers 
?
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
.
are chosen on 
,
is the foot of the altitude from 
to 
.
are similarly defined, prove that lines 
and 
concurr on 
.
and let 
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
passing through 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
that is not isosceles at 
,
be its incircle, which is tangent to 
at 
,
and 
intersect the line passing through 
at 
and 
,
and perpendicular to 
intersect 
and 
at 
and 
,
are collinear and that 
are concurrent.
positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining 
numbers that are not hidden. What is the maximum number of the 
satisfying: 
.
min
is a prime number and 
are positive integers. Prove that there exist infinitely 
are divisible by 
if 
if 
primes, and let 
be the sum of the divisors of ![\[\left|\sum_{d|n}\frac{\mu(d)\sigma(d)}{d}\right| \geq \frac{1}{n}, \]](http://latex.artofproblemsolving.com/8/e/b/8eb88de7cd7ea8e2853dbc80cae24bdc6b77f9be.png)
be a sequance of natural integers. Proof that, any number in the sequance can be written only as 
for any 
and not necessarily distinct primes 
for 
.
are same in some case.
![\[2^{xy}f(xy-1)+2^{x+y+1}f(x)f(y)=4xy-2,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/d/1/d/d1de7196fd3ba60732c14fa2b156ec880b4829d0.png)
has no integer solutions
play a game with a stack of 
coins from the stack. The player who takes the last coin wins. Prove that there are infinitely many integers 
Y by Ankoganit, anantmudgal09, ATimo, rafayaashary1, ahmedAbd, laegolas, kk108, Tawan, Adventure10, Mango247, sami1618
In the country of Sugarland, there are 
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.
Is it possible that all students have a chance of being a team member?
Proposed by Morteza Saghafian
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.Is it possible that all
students have a chance of being a team member?Proposed by Morteza Saghafian
This post has been edited 3 times. Last edited by bgn, Apr 7, 2017, 6:42 PM
il all the tests. He has no chance for being a team member?
,
students. But don't have an idea about the maximum number.
.
This is in fact possible, with the following setup, where each column represents the top 6 scorers on each problem, and the contestants are numbered 
.
t
t
:
into the team first, you have to get 
from Test 
from Test 
before you start taking 
,
corresponds to the rank of student 
on problem 
.
can get on the team as if the problem they are ranked first on is chosen first;
;
;
can get on the team if the problems are chosen in the order 
;
can get on the team by symmetry (with 
replaced with 
and so on).
;
can get on the team if the problems are chosen in the order 
;
.
square to easily be able to eliminate top ranks; then, for each pair, we would be able to uplift three students who were ranked lower, which suffices for the 
satisfies
.
This is not necessary but it is sufficient in that every student that appears at the bottom of a pyramid (size 
It is not difficult to verify how this works.
the first row (rank) on which the 
appeared, this was done as I saw that if a person is on the 5th rank then for that guy to be picked up by there we must have a sort of upside-down pyramid, like
Here, if 
is to be picked from the bottom row, then we must have that pyramid shape above it (not necessarily in order, but you get it).
over all men(essentially extremal principle) and case bashed on its values (i.e., when it is 2, 3,... 6). Case of 2 was trivially false, but I tried 3 and saw that I could just put 3 people twice in row 2, and then if person 1 fills row 1 completely, then we can get up to 10 people. ( the reason to do this was that I wanted to put people in row three, i.e as 
![\[\begin{array}{|c|c|c|c|c|c|}
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline
A & A & A & A & A & A \\
\hline
B & B & B & B & B & B \\
\hline
C & C & C & C & C & C \\
\hline
D & D & D & D & D & D \\
\hline
E & E & F & F & G & G \\
\hline
H & I & J & K & L & M \\
\hline
\end{array}\]](http://latex.artofproblemsolving.com/c/8/2/c822faec25af888c2bf8e08367c205ee86117993.png)
