[/quote]
,
,
,
cut three common chords concurrent at 
.
,
which are internally tangent to all 
.
be nonnegative real numbers such that 
.
Prove that
Proposed by Phan Ngoc Chau
be a triangle, and let 
be a point on side 
.
at 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle 
distinct from point 
.
and 
intersect 
and 
respectively.
grid has had the central unit square removed. For which positive integers 
is it possible to cut the remaining area into 
and 
rectangles?
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
and 
be positive real numbers such that 
.![\[
\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)}
\geq \frac{3}{4}.
\]](http://latex.artofproblemsolving.com/5/6/5/56568d27a296f1c900fabbb1042549cd2e552e29.png)
![\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1. \]](http://latex.artofproblemsolving.com/b/2/f/b2ff456632d4dd3b9ed93e4cba437bdd34d54cd8.png)
be a parallelogram of center 
.
,
and 
such that 
.
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.
can get into the team with the order 
.
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
through 
.
is admitted.
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)
