[/quote]
,
be non-negative real numbers such that 
Prove that
be an acute triangle with orthocenter 
and circumcenter 
.
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
be the set of positive real numbers. Determine all functions 
such that ![\[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]](http://latex.artofproblemsolving.com/e/7/1/e71b6e0d0b858eb46b81149f1e6be8c41e13d301.png)
for every 
.
,
and 
be pairwise distinct nonnegative real numbers. Prove that![\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\]](http://latex.artofproblemsolving.com/2/0/9/209f9807481e7a1e97f023bf65a603cb462c62cb.png)
(Karl Czakler)
that can be written in the form![\[
n = \frac{a^2 - b^2}{b},
\]](http://latex.artofproblemsolving.com/f/9/0/f90a3ce2e1b51d6e1cb53a15cad6f36c2b30492a.png)
where 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
,
as 
,
as 
.
s
n
t
hats, numbered from 
hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
.![\[\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}.\]](http://latex.artofproblemsolving.com/3/e/4/3e4b336137b17e55bf172d45bd18fe6d5561a827.png)
be real numbers such that 
.![\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]](http://latex.artofproblemsolving.com/5/a/8/5a83eaa7470bcca6546599ff4e296e4efec69b16.png)
and determine all the cases when the equality occurs.
be the midpoint of side 
.
intersect 
at 
.
is tangent to 
,
.
Y by
Say that on the AMC 10, you do better on the A than the B, but you still qualify for AIME thru both. Then after your AIME, it turns out that you didn’t make JMO through the A+AIME index but you did pass the threshold for the B+AIME index.
does MAA consider your B+AIME index over the A+AIME index and consider you a JMO qualifier even tho Your A test score was higher?
does MAA consider your B+AIME index over the A+AIME index and consider you a JMO qualifier even tho Your A test score was higher?
