[/quote]
,
kawaii if it satisfies 
is the number of integers not more than 
with 
and 
.
and 
intersect at 
,
and 
intersect at 
.
be the midpoints of sides 
,
and 
be points on segment 
and 
,
is the angle bisector of 
and 
is the angle bisector of 
.
is parallel to 
if and only if 
divides 
(i dont remember exactly if it was 8032038 or some other number, so correct me on this)
(i posted this at 12:29) lol
?
for which, for all 
,![\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]](http://latex.artofproblemsolving.com/9/0/c/90c180110402e1a32b70edb2b0a03a28727457d1.png)
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
be a 
makes a fixed angle with the tangent to 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
on the spiral, the polar of 
such that every positive integer appears exactly once and the product of the first 
power ?
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be the set of black cells at the end of the game. Ali and Shayan respectively aim to minimize and maximise the perimeter of the shape 
Y by Adventure10, Mango247, and 19 other users
For the 2012 IMO team from the USA the AMC is adopting a new team selection process. The details are enumerated on the AMC website in the Frequently Asked Questions, MOSP #5
http://amc.maa.org/faq/faq.shtml#mosp
and it is repeated at the end of the announcement about the 2011 IMO team
http://amc.maa.org/imo/2011imo.shtml
Steve Dunbar
MAA Director. American Mathematics Competitions
http://amc.maa.org/faq/faq.shtml#mosp
and it is repeated at the end of the announcement about the 2011 IMO team
http://amc.maa.org/imo/2011imo.shtml
Steve Dunbar
MAA Director. American Mathematics Competitions
student per year. (Yes, that is basically the number of people who qualify for both Blue/Black MOP and RSI junior year. In fact, I don't think any of RSI 2011 made Blue/Black; in 2010 there was 1, in 2009 there was 1, and in 2008 there was 1.) I'm not sure why everyone feels the need to protect this one poor soul--someone who, if he/she could've made the IMO under the previous system, would likely be chosen by Zuming's special selection to be in the team. (And look: 1 person affected by the new policy, and 1 person specially selected by the IMO coach...!)
