[/quote]
,
be a function such that for every regular 
-
we have 
.
for all reals 
.
be a triangle 
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
and 
.
passes through the midpoint of 
.
& 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
and puts 
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
.
,
.
and 
to be the remainder when 
is divided by 
.
and 
is odd for 
.
or an equivalent formulation. To prove this construction works, we need to show that
is always an integer between 
and 
inclusive.
is divided by 
.
,
since it telescopes.
,![\[\frac{r_{i+1} n - r_i}{2^i} > \frac{r_{i+1} n}{2^i} - 1 \geq \frac{n}{2^i} - 1 \geq \frac{n}{2^k} - 1.\]](http://latex.artofproblemsolving.com/4/2/0/420d26b6099ca264846885e7eef23b2c9c7b2dd0.png)
Thus, each digit is lower bounded by 
,
fails.
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
,
such that 
has no nonreal roots. We can also assume 
case is trivial as the constant term must be nonzero, so fix 
.
.
.
which has the 
for all integers 
.
be the coefficient of the 
term of 
for all 
,
for which 
.
and 
.
and 
.
is the coefficient of the 
term in 
and 
.
,
.
,
)
has 
distinct real roots as well, along with two consecutive nonleading coefficients equal to zero, but one degree lower, by the Power Rule. By doing this repeatedly, we eventually end up with a polynomial where the two consecutive coefficients have degree 
,
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
distinct from 
and 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
,
to 
,
to 
.
to be the set of points strictly outside the circle with diameter 
.
is always acute. Suppose two roads 
and 
cross. Then, 
is a convex quadrilateral. Since 
is not acute but there does not exist an 
such that 
are connected by some path. This is trivially true for 
,
are also connected.
is obtuse, we have 
.
,
,
is not mentioned. Do not reward points if the induction argument is incomplete or incorrect.
be the orthocenter of acute triangle 
be the foot of the altitude from 
intersects line 
.
be the center of 
.
be the intersection of 
and the line through 
,
.
be the foot of the altitude from 
and 
,
is a midline and thus 
.
is a midline of 
,
and thus 
.
(by definition), by either HL congruence or the Pythagorean theorem, we must have 
.
.
and 
.
or 
,
.
.![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](http://latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer 
is divisible by 
,
be a prime power dividing 
.
.
,
if 
and 
otherwise, so for each term, either both the numerator are divisible by 
,
,
,
.
has an inverse modulo 
.![\[\binom ni \equiv \pm 1 \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor} \pmod{p^a}.\]](http://latex.artofproblemsolving.com/0/1/c/01c00beb6efc07ed028fd0273f262b8bfb965d33.png)
modulo 
.
,![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \equiv p \equiv 0 \pmod p.\]](http://latex.artofproblemsolving.com/2/e/c/2ececb6afa8a10e116056cdc57a6c948298d0c29.png)
where 
,
satisfies the induction hypothesis for the prime 
.![\[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (\pm 1)^k \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \equiv p\sum_{i=0}^d \binom di^k \pmod {p^a}\]](http://latex.artofproblemsolving.com/6/c/9/6c9a4b6aac62b1befef1511451c9fa30993d2cf7.png)
By the induction hypothesis, 
divides the inner binomial sum, so since we are multiplying it by 
.
.
is incorrect, reward up to 
points total.
and 
.
between the set of people except Evan 
.
denotes the set of neighbors of 
people.
.
from the graph, noting that no person in 
.
-
do not necessarily imply that the answer is not an integer. If the answer is an integer 
.
Y by fermat007, El_Ectric, distortedwalrus, Hydroxide, mathgenius64, ssilwa, DiscipulusBonus, ahaanomegas, steve123456, dantx5, vlchen888, yugrey, baijiangchen, AkshajK, countyguy, Amir Hossein, r31415, Adventure10, parmenides51, aidan0626, and 9 other users
The ASIA TEAM is a mock AIME written by members of AAST (explanation of title: AAST AIME rearranges to ASIA TEAM). Specifically, these include myself, fermat007, yugrey, and USAMOREAPER as the primary contributors. The test is nearly completed and we hope to release it very soon, well in time for forum-goers to take it before the actual AIME occurs.
Google docs is the most convenient way for us to collect answer submissions. However, Google Docs has a minor flaw in that it depends upon users self-reporting their information. This, unfortunately, allows for misrepresenting your username (aka trolling). Therefore, we request that you sign up to take the exam by posting below in some (preferably a bolded "signing up" or similar) distinctive fashion. You may send a PM to myself if you prefer (but please, do not do both). I will then send you a passcode that you will put into the Gdoc when submitting answers (you can instead send me a distinctive password if you prefer). If you are absolutely dead-set against this, you can alternatively send your answers by Private Message to myself after the contest starts in a clear manner, understanding that the chance of your submission being lost or misinterpreted is far greater. You can do this at any time until official solutions are released, but sooner is better. All scores will be released publicly at the same time as official solutions, unless you specifically request your score to be released only privately.
The format of the contest is analogous to the actual AIME. That is, 15 problems to be solved in 3 hours, each with a nonnegative integer answer ranging from 000 to 999. We expect that the difficulty of the ASIA TEAM will be greater than that of the the actual AIME.
Good luck
Google docs is the most convenient way for us to collect answer submissions. However, Google Docs has a minor flaw in that it depends upon users self-reporting their information. This, unfortunately, allows for misrepresenting your username (aka trolling). Therefore, we request that you sign up to take the exam by posting below in some (preferably a bolded "signing up" or similar) distinctive fashion. You may send a PM to myself if you prefer (but please, do not do both). I will then send you a passcode that you will put into the Gdoc when submitting answers (you can instead send me a distinctive password if you prefer). If you are absolutely dead-set against this, you can alternatively send your answers by Private Message to myself after the contest starts in a clear manner, understanding that the chance of your submission being lost or misinterpreted is far greater. You can do this at any time until official solutions are released, but sooner is better. All scores will be released publicly at the same time as official solutions, unless you specifically request your score to be released only privately.
The format of the contest is analogous to the actual AIME. That is, 15 problems to be solved in 3 hours, each with a nonnegative integer answer ranging from 000 to 999. We expect that the difficulty of the ASIA TEAM will be greater than that of the the actual AIME.
Good luck
This post has been edited 9 times. Last edited by BOGTRO, Mar 3, 2013, 2:27 AM
).
.
for #9 darn). Then #6 and #7 were quite nice but #7 was better; #8 was okay, kind of straightforward and the amount of computation is not really necessary. I really really liked #10, really liked #13, #12 was not bad, #14 was pretty nice (and somewhat missplaced), and #15 was okay.
.
.
denote the number of integers between 
denote the number of integers between ![\[\phi(n) + 4\tau(n) = n.\]](http://latex.artofproblemsolving.com/6/2/f/62f501b9b920a7b4170eeb54ae638172775b2f06.png)
.
for some positive integer 
,![\[2^{a-1} + 4(a+1) = 2^a \implies 4(a+1) = 2^{a-1} \implies a+1 = 2^{a-3}.\]](http://latex.artofproblemsolving.com/9/5/1/951fe06df4b825ec9150497caca06a5e04b18a7f.png)
Solutions can only occur when 
.
and 
are not solutions and when 
,
outgrows 
.
,
,
,
are distinct odd primes, 
,
,
are positive integers, and 
.
.![\[n = \phi(n) + 4\tau(n) \leq \frac{n}{2} + 4\tau(n) \implies \frac{n}{\tau(n)} \leq 8.\]](http://latex.artofproblemsolving.com/2/8/7/287a556837c6246b3b8b5320927636cf6ac00dcc.png)
.![\[8 \geq \frac{n}{\tau(n)} = \frac{2^a}{a+1} \cdot \frac{p_1^{e_1}}{e_1 + 1} \cdot \frac{p_1^{e_2}}{e_2 + 1} \cdot ... \cdot \frac{p_k^{e_k}}{e_k + 1} = f(2, a) \cdot f(p_1, e_1) \cdot f(p_2, e_2) \cdot ... \cdot f(p_k, e_k). \quad(*)\]](http://latex.artofproblemsolving.com/c/9/7/c9774a20c19c9230343ce38e06aa28abf528ecc0.png)
Clearly, 
increases as 
.
,
,![\[8 \geq f(2, a) \cdot f(p_1, e_1) \cdot f(p_2, e_2) \cdot ... \cdot f(p_k, e_k) \geq f(2, 1) \cdot f(3, 1) \cdot f(5, 1) \cdot f(7, 1) = \frac{105}{8}.\]](http://latex.artofproblemsolving.com/3/e/e/3ee622729adc7a7ba27a394b101432e4f875663f.png)
Thus, 
,
.
![\[8 \geq f(2, a) \cdot f(p_1, e_1) \geq f(2, a) \cdot f(3, 1) \implies \frac{16}{3} \geq f(2, a) \implies a \leq 5.\]](http://latex.artofproblemsolving.com/8/f/5/8f51c0a3b61bd1db7fd985c66e871b660ffbf827.png)
into the given equation implies ![\[2^{a-1} \cdot (p_1 - 1) + 4 \cdot 2(a+1) = 2^a \cdot p_1 \implies p_1 = \frac{8(a+1)}{2^{a-1}}-1.\]](http://latex.artofproblemsolving.com/a/3/8/a38ed8e1706fb10297173fd246b89a4c4a5737d4.png)
,
.
,
.
.
.
.
,
.
,
.
,![\[8 \geq f(2, a) \cdot f(p_1, e_1) \geq f(2, 1) \cdot f(5, 2) = \frac{25}{3},\]](http://latex.artofproblemsolving.com/4/b/d/4bd9fa61d923407d136883efb7d8f0729efce27a.png)
which implies 
.
into the given equation implies ![\[2^a \cdot 3^{e_1-1} + 4(a+1)(e_1 + 1) = 2^a \cdot 3^{e_1} \implies (a+1)(e_1 + 1) = 2^{a-1} \cdot 3^{e_1-1}\]](http://latex.artofproblemsolving.com/6/7/9/679c8c316b21e848cb9b3cc85eb8c4a32d152e22.png)
![\[\implies \frac{2^a}{a+1} \cdot \frac{3^{e_1}}{e_1+1} = 6 \implies f(2, a) \cdot f(3, e_1) = 6.\]](http://latex.artofproblemsolving.com/6/3/a/63a391c175ad7064b097d2e1aa127df51121d410.png)
If 
,![\[f(2, a) \cdot f(3, e_1) \geq f(2, 1) \cdot f(3, 3) = \frac{27}{4} > 6,\]](http://latex.artofproblemsolving.com/7/3/4/734ec2231e1f490222016d99e15ef8497f613908.png)
which is a contradiction. Thus, the only possible value of ![\[f(2, a) \cdot f(3, e_1) = 6 \implies f(2, a) = 2 \implies a = 3.\]](http://latex.artofproblemsolving.com/d/b/9/db9269f892a793cec5365d5e12b09dd21638145c.png)
This yields the solution 
.![\[8 \geq f(2, a) \cdot f(p_1, e_1) \cdot f(p_2, e_2) \geq f(2, 1) \cdot f(5, 1) \cdot f(7, 1) = \frac{35}{4},\]](http://latex.artofproblemsolving.com/2/8/9/2894db3b15dfe16a0c1a6e60158b329c4bc4f20c.png)
which is a contradiction. Hence, 
.
can be revised as follows: ![\[\frac{\phi(n)}{n} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{p_2 - 1}{p_2} < \frac{1}{3} \implies \phi(n) < \frac{n}{3}.\]](http://latex.artofproblemsolving.com/6/6/0/66027159be4a59fe0b16abe880054ba2957f6cf9.png)
Thus, ![\[n = \phi(n) + 4\tau(n) < \frac{n}{3} + 4\tau(n) \implies \frac{n}{\tau(n)} < 6,\]](http://latex.artofproblemsolving.com/9/7/e/97e4a325038476c888c5cd3be1c7f0224f3ce135.png)
which implies ![\[6 > f(2, a) \cdot f(3, e_1) \cdot f(p_2, e_2). \quad (**)\]](http://latex.artofproblemsolving.com/c/5/2/c52a946e88ba76ae2664fc8e6a853dd83d835528.png)
If 
,![\[6 > f(2, a) \cdot f(3, e_1) \cdot f(p_2, e_2) \geq f(2, 1) \cdot f(3, 2) \cdot f(5, 1) = \frac{15}{2},\]](http://latex.artofproblemsolving.com/b/0/0/b0077c966918629dc1cdc654b740ae73f3753f9a.png)
which implies 
.
,![\[6 > f(2, a) \cdot f(3, 1) \cdot f(p_2, e_2) \geq f(2, 1) \cdot f(3, 1) \cdot f(5, 2) = \frac{25}{2},\]](http://latex.artofproblemsolving.com/2/2/2/22242b6caf32b2b912ba2970721854c1e0dfca13.png)
which implies 
.
.![\[6 > f(2, a) \cdot f(3, 1) \cdot f(p_2, 1) \geq f(2, 3) \cdot f(3, 1) \cdot f(5, 1) = \frac{15}{2},\]](http://latex.artofproblemsolving.com/0/d/2/0d2086789e70e679fe9fb7df9b89f17cf4989d6b.png)
which implies 
.
.
,
.
.
,
.
,
,
,
.
,
.
can be expressed as 
.
.
,
or 
(depending on your orientation).
.
can be written as 
.
