AoPS Academy
has property 
if :![\[
\begin{cases}
\text{the set } A \text{ has at least } 4 \text{ elements} \\
\text{the element } 1+1 \text{ is invertible}\\
x+x^4=x^2+x^3 \text{ holds for all } x \in A
\end{cases}
\]](http://latex.artofproblemsolving.com/b/6/9/b6993ee76eb3792e08d441f5d5c463247c392c4a.png)
a) Prove that if a ring has property , and 
are distinct elements, such that 
and 
are units, then 
is also a unit, but 
is not a unit..
be a positive integer, and 
be two complex numbers such that 
and 
, for any 
. The matrices 
satisfy the relation 
. Prove that 
and 
are invertible.
be a convex polygon with 
sides, 
. Any set of 
diagonals of that do not intersect in the interior of the polygon determine a triangulation of into 
triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of .
![$[n] = \{1, 2, . . . , n\}$](http://latex.artofproblemsolving.com/4/f/f/4ff58e6e89a225ffa8db0ab50038fdadd9d4e2bc.png)
is a sequence 
such that each element of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
appears precisely one time as a term of the sequence. For example, 
is a permutation of ![$[5]$](http://latex.artofproblemsolving.com/c/4/2/c4216c69e041fca0904e8c5366d4480d1b2ca8b8.png)
. Let 
be the number of permutations of for which 
is a perfect square for all 
. Find with proof the smallest such that is a multiple of 
.
is solvable over 
for every 
. Here 
, 
.
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
, the digits in the base-
representation of 
are all greater than . and be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
and 
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 , there exists 
such[/center]
is directly similar to either 
or 
.[/center]
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from to , and let be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points and . Prove that is the midpoint of 
. such that ![\[\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k\]](http://latex.artofproblemsolving.com/3/c/5/3c5984d0003c1ca2df2e68841dd9e93ed6c2d27c.png)
is an integer for every positive integer .
and be positive integers with 
. There are cupcakes of different flavors arranged around a circle and people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person , it is possible to partition the circle of cupcakes into groups of consecutive cupcakes so that the sum of 's scores of the cupcakes in each group is at least 
. Prove that it is possible to distribute the cupcakes to the people so that each person receives cupcakes of total score at least with respect to .
be collinear points (in order) and 
a point in plane. Consider the disc 
of center and radius 
, for some 
. Prove that ![$\mathcal{D}\cap [AC]$](http://latex.artofproblemsolving.com/0/5/6/0569d4130943000380a87b72be211965b5ff02ad.png)
is either the empty set or a segment of length at most 
. be a positive integer and ![$P(X)\in\mathbb{C}[X]$](http://latex.artofproblemsolving.com/9/9/e/99e8cde28b0b9bc9414425ebc30fa2dc0386fb37.png)
be a polynomial of degree . Prove that ![\[\sup_{x\in[0,1]}|P(x)|\le(2n+1)^{n+1}\int\limits_{0}^{1}|P(x)|\mathrm{d}x.\]](http://latex.artofproblemsolving.com/9/1/6/91647cd75888c5e9d291280c8ae3f199cc6d1bfd.png)
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .
from each of the integers at two neighboring vertices and adding 
to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.
. If you are sending solutions digitally through physical scan, please make sure your handwriting is eligible. Inability to discern hand-written solutions may warrant point deductions.
-point scale, and solutions will be graded in terms of both elegance and correctness.
and a contour 
in positive direction. If all roots of 
are inside 
and 
where 
, then find the value of 
.
then 
.The rest is Newton sum I guess.
.
,
.
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.
,
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 
,
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 center of 
.
be the intersection of 
and the line through 
,
.
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 
to attempt to identify 
.
or 
,
.
.
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.
between the set of people except Evan 
.
denotes the set of neighbors of 
people.
.
from the graph, noting that no person in 
