[/quote]
,
and 
are the incenter and the 
-
,
be the midpoint of arc 
on the circumcircle, and 
be the foot of the
.
intersects the circumcircle again at 
. For
on the circumcircle of triangle 
intersect 
. Prove
are concyclic.
squares, we cover the area with several S-Tetrominos (=Z-Tetrominos) along the square so that in every square, there are two or fewer tiles covering that (tiles can be overlap). Find the maximum possible number of squares covered by at least one tile.
satisfies
for any non-negative integers 
and 
.
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
,
and orthocenter 
.
is constant.
be the positive real number. Prove that
,
,
grid. Alice goes first, and they take turns drawing a black point from the coordinate set![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](http://latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
.
or 
unit squares depicted in the drawing below.
Y by Adventure10, Mango247, and 2 other users
Now a senior in high school, I have participated in competitive math for 6 years. After reading forum posts and discussions about the similarity between #12 on the AIME II 2014 and a question on one of this year's AMC Advantage tests, I am extremely disappointed in the MAA. I will no longer promote the AMC competitions in the state of Hawaii, nor ever recommend this discipline to future generations. I have no malice toward AMC Advantage users, bur rather toward the greed and deception of the MAA. You can attribute my feelings to a shortcoming of USAMO (9 on AIME 1, so close!), and while this may be partially true, you cannot hold the MAA innocent for this incident. Nearly verbatim problems that occur within such a short time period? Please. What happened to the supposedly innovative, thought-provoking questions that the AMC tests are commended for? From their website, "the AMC program identifies, recognizes, and rewards excellence in mathematics through a series of national contests." They do not reward excellence; they reward expenses toward their corrupt AMC Advantage. I fear for the future of competitive math. I have been deceived for 6 years.
((((( AMC must be really evil to come up with a program that can do something like that. It doesn't even matter that the AMC's motivated countless numbers of kids to deeply study mathematics, that one aime problem is ridiculous!!!! I had a feeling they've been plotting something like this for six years...
,
random people, you would find maybe 
business majors and 
,
for relatively prime positive integers m and n. Find m+n.
.
?
such pairs; this is a quadratic function, so if we add people gradually, the number of pairs will increase and increase by more each time! Once you hit 
people, which is the answer to the problem, there are 
pairs, which is a lot of possible matches!
of the numbers 
,
,
,
is not a permutation of the numbers 
.
such that for each 
,
