be a triangle with orthocenter 
and let 
be the second intersection of the circumcircle of triangle 
with the internal bisector of the angle 
.
be the circumcenter of triangle 
and 
the orthocenter of triangle 
.
is equal to the circumradius of triangle 
such that the following property holds: if we list the divisors of 
in increasing order as 
,![\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]](http://latex.artofproblemsolving.com/a/d/c/adc94af88b3579164aabba51bd021442133e364a.png)
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
be the incircle of a fixed equilateral triangle 
be a variable line that is tangent to 
and 
at points 
,
is chosen such that 
and 
.
be positive real numbers such that ![\(a+b+c=4\sqrt[3]{abc}\)](http://latex.artofproblemsolving.com/2/8/2/28214a1814762b71858db1543fae1a50212eaafa.png)
.![\[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]](http://latex.artofproblemsolving.com/7/0/4/7043a170ac5535465a5a1b4daccb1f1de3043280.png)
are multiples of 
or 
but not 
?
![\[222{,}222-22{,}222-2{,}222-222-22-2?\]](http://latex.artofproblemsolving.com/1/5/6/156c3662fe4c874a80e54887f76c302c13b9d14c.png)
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 
,
,
must be the vertex 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 
and 
,
is half of 
because triangle 
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 
.
