[/quote]
,
of relatively prime integers 
and 
such that 
is divisible by 
.
![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](http://latex.artofproblemsolving.com/4/d/2/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
,
,
,
are nonnegative real numbers such that 
.
be the set of all integers. Find all pairs of integers 
for which there exist functions 
and 
satisfying![\[ f(g(x))=x+a \quad\text{and}\quad g(f(x))=x+b \]](http://latex.artofproblemsolving.com/3/2/8/328146c3331a66cdc5cfc4a01b0609e91d278ec5.png)
for all integers 
.
satisfying 
.
such that 
be the sequence defined by 
and 
for each integer 
.
is a prime and 
is a positive integer. Prove that some term of the sequence 
.
is interesting if 2018 divides 
(the number of positive divisors of 
be integers and 
be a prime for which![\[ x^2-3xy+p^2y^2=12p \]](http://latex.artofproblemsolving.com/2/4/c/24cb51eb55144a9a1e0e489d115993496a052e79.png)
.
satisfying 
for all positive integers 
be an isosceles trapezoid with 
.
be the midpoint of 
.
and 
the circumcircles of the triangles 
and 
,
be the crossing point of the tangent to 
with the tangent to 
.
is tangent to 
,
denote the feet of the altitudes from 
,
.
and 
be the circles through 
,
other than 
such that![\[ f(x+f(y))=y+f(x+1),\]](http://latex.artofproblemsolving.com/b/a/3/ba35828951778c0d35f50b5b557796bd2cf9e338.png)
for all 
.
is defined by 
,
for 
.
of ordered pairs 
for which there is a finite set 
of positive integers such that 
,
.
,
,
with the following property: An ordered pair of nonnegative integers ![\[m < \alpha x+\beta y < M\]](http://latex.artofproblemsolving.com/a/9/b/a9bb35658173fdab821cc8180420e47d5437e0a2.png)
if and only if 
.
.
,![\[
f(xy)(f(x) - f(y)) = (x-y)f(x)f(y)
\]](http://latex.artofproblemsolving.com/0/9/b/09b8ed4e133ccf33b3f902fd9a9aa8e23c264341.png)
for all 
.
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 
,
,
because otherwise you could reflect 
.
,
.
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 
.
of the way between 
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 
.
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 
.
