[/quote]
,
be a positive integer, and let 
denote the sum of the positive divisors of 
.
smallest positive integer relatively prime to 
-smooth if all of its prime factors are 
.
denote the number of 
.
be the least quadratic non-residue mod 
.
is a non-negative arithmetic function and
is convergent for some 
,
,
such that any graph with 
vertices and at least 
and 
,
be vertices. Define a directed graph 
as follows: for any integer point 
,
,
.
,![\( G[k, l] \)](http://latex.artofproblemsolving.com/6/3/a/63a5df1124231155f3240a3e21e24d293d0e8db3.png)
denote the subgraph of ![\([k, l]\)](http://latex.artofproblemsolving.com/2/9/f/29f08f190960a99d062fde7f2128bcb635657f36.png)
.
such that for any integer 
,![\( G[r, r + \alpha - 1] \)](http://latex.artofproblemsolving.com/5/1/a/51afcceb73daae97f8fc61456bf043b73b2689d7.png)
of ![\( G[r, r + \alpha] \)](http://latex.artofproblemsolving.com/8/8/7/887e55f5636b842e7b64c9c03982b0374cdf1c9e.png)
(i.e., how many connected components and what each component is like).
of a graph be partitioned into 
,
.
.
.
-
.
where no more than one circle is disjoint.
be a median in an acute-angled triangle 
.
is chosen on the line through 
tangent to the circumcircle of 
so that 
.
and 
.
lies on the line 
.
be a cyclic quadrilateral for which 
.
and 
externally on the sides 
and 
respectively.
meets 
at 
,
meets 
at 
and 
denotes circumcenter of 
and 
meets again on 
.
.
that does not contain 
and let 
be the midpoint of 
be the 
-
be the reflection of 
are concyclic.
be a convex pentagon with 
and 
For each point 
The least possible value of 
can be expressed as 
where 
and 
be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
.
Y by MathTwo, rdj5933mile5, dft, NewAlbionAcademy, Adventure10, and 5 other users
Hello everyone! My friend and I created a mock AMC 10, and it is almost ready. Naturally, we would need some proofreaders for the test, as we might have made some errors. Also, we are planning to also make a mock AMC 12, with some harder problems (in particular, #25 and #23, as well as some others; I will tell you in detail when you become a helper) in addition to around 12 of the same problems on the AMC 10. We would like some helpers to help us finish that. If you are interested, post on this thread (so we can keep an active list). (Note: If you help with the test, you will not be allowed to participate.)
However, if you wish to take the test, please choose only ONE contest (10 or 12) to take as there will be problems that are replicas. We will decide testing dates a little later when we are done with the test.
So, Happy New Year everyone, and thanks in advance for participating!
Copy this list.
EDIT: From now on, don't pm me saying which test you want to take. Say that when you submit solutions (e.g. put your first sentence saying "This is for the AMC 12 test" or something like that).
EDIT2: There is no longer any limit on the number of helpers. However, in order to be a helper, you can not be a participant, as I mentioned before.
However, if you wish to take the test, please choose only ONE contest (10 or 12) to take as there will be problems that are replicas. We will decide testing dates a little later when we are done with the test.
So, Happy New Year everyone, and thanks in advance for participating!
Copy this list.
EDIT: From now on, don't pm me saying which test you want to take. Say that when you submit solutions (e.g. put your first sentence saying "This is for the AMC 12 test" or something like that).
EDIT2: There is no longer any limit on the number of helpers. However, in order to be a helper, you can not be a participant, as I mentioned before.
This post has been edited 3 times. Last edited by Lord.of.AMC, Jan 9, 2012, 10:38 PM
has roots 
,
has roots 
.
has at most 2 roots, so we should be able to pair up the numbers 
in each pair. Given that 
,
must be equal to 
.
and 
and 
.
,
or 
.
.
and 
.
.
has 3 roots but is a quartic, one root must be repeated. We know that 
so the roots of 
are the roots of 
and 
are roots of 
and 
are roots of 
,
,
.
,
.
is a minimum, the other two roots of 
so 
.
.
whose remainder is 
,
for any nonzero 
.
,
gives 
.
.
and 
then 
is one third the length of side 
.
is 3 times the area of 
,
.
and 
to 
.
falls exactly 
of the way between 
and 
.
falls exactly 
and 
.
of the area 
,
.
.
.
,
and 
.
with 3 non-9 digits and 23 9's, so that gives 26 digits.
if 
.
is similar to 
,
,
and 
and so 
.
.
,
because 
is the circumcenter, we have 
.
with a ratio of 
.
.
is right, we know that 
is half of 
is similar to triangle 
with ratio 
is also half of 
by AA similarity, so to find the ratio we compare two corresponding sides. In this case we look at 
and 
as a combination of 
and 
.
to knock out the 
.
