AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a non-equilateral with integer sides. Let 
and 
be respectively the mid-point of 
and 
; let 
be the centroid of 
. Suppose, 
are concyclic. Find the least possible perimeter of .
rows and 
columns, with 
. At least squares of this grid are shaded such that for all 
, the collection of shaded squares in any 
rows of this grid together occupy at least columns. For example, the following grid is not a valid shading because the shaded cells in three rows -- the 
st, 
rd, and 
th rows -- only occupy two columns, namely columns 
and . shaded square from each of the rows in this grid such that no two chosen squares lie in the same column.
such that![\[
x^2 + y^2 + z^2 = 2.
\]](http://latex.artofproblemsolving.com/a/8/d/a8db4ff2ec57b9e926f2fc4f4eede3d2d4b3c556.png)
Prove that![\[
x + y + z \leq xyz + 2.
\]](http://latex.artofproblemsolving.com/f/7/5/f75ae1a7eb4676271c869021a9457348f9893569.png)
, find the smallest real number 
such that no matter how each integer point in the coordinate plane 
is colored with one of different colors, there always exist three non-collinear points 
of the same color such that the area of 
is at most .
has the property for every positive integer 
, its 
term is greater than the mean of the first 
terms, and the sum of its first terms is a multiple of . Let 
be the number of such sequences satisfying 
. Compute the remainder when is divided by 
.
has side length . Point 
lies on segment 
such that line 
is perpendicular to line 
. Given that 
, the area of can be expressed as 
, where and are relatively prime positive integers. Compute 
.
and 
satisfy 
, 
, and 
. If the number of possible ordered pairs 
is equal to 
, compute the remainder when is divided by . be a triangle. Point 
lies on side , point 
lies on side 
, and point 
lies on side 
such that 
, 
, and 
. Let 
be the foot of the altitude from 
to . Given that 
, 
, and ![$[AQPR] = \tfrac{3}{7} \cdot [ABC]$](http://latex.artofproblemsolving.com/e/f/1/ef1f89d7ee2b45faac7265d8e2777ba3918ae16e.png)
, the value of 
can be expressed as , where and are relatively prime positive integers. Compute .
and tells Bob that the three resulting remainders are 
, 
, and , in some order. For how many values of is it possible for Bob to uniquely determine ?
satisfying the system of equations 
The value of 
can be expressed as 
, where 
and 
are positive integers such that is square-free. Compute 
. be the set of all monotonically increasing six-term sequences whose terms are all integers between 
and 
inclusive. We say a sequence 
in is symmetric if for every integer 
, the number of terms of 
that are at least 
is 
. The probability that a randomly chosen element of is symmetric is 
, where and are relatively prime positive integers. Compute ., let 
denote the value of the binary number obtained by reading the binary representation of from right to left. Find the smallest positive integer such that the equation 
has at least ten positive integer solutions . be a quadratic polynomial with a positive leading coefficient. There exists a positive real number 
such that 
and 
for 
. Compute 
.
such that 
and 
is a multiple of 
.
pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a final configuration. For each , show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of .
be pairwise distinct positive integers such that 
is an integer. Prove that 
is not a prime number.
be a convex hexagon such that 
and ![\[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1. \]](http://latex.artofproblemsolving.com/9/9/b/99be4560967e7521ee7160d01ffa634d0922f37f.png)
Prove that ![\[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1. \]](http://latex.artofproblemsolving.com/7/e/2/7e2a05831899978b85421d282ec38281cba280c9.png)
such that 
.![\[
a^3 + b^3 + c^3 \geq a\sqrt{b+c} + b\sqrt{c+a} + c\sqrt{a+b}
\]](http://latex.artofproblemsolving.com/7/9/e/79e23cd5fdf9630a62070feb0f2c38ae1d11163d.png)
be 
positive integers such that the 
products![\[a_1 a_2 a_3 \cdots a_n, b_1 a_2 a_3 \cdots a_n, b_1 b_2 a_3 \cdots a_n, \dots, b_1 b_2 b_3 \cdots b_n\]](http://latex.artofproblemsolving.com/0/2/c/02cbff15c4d0118156a9d24be09360facb264d5a.png)
form a strictly increasing arithmetic progression in that order. Determine the smallest possible integer that could be the common difference of such an arithmetic progression.
be a polynomial with integer coefficients such that 
and 
for some integer . Prove that there exists no integer such that 
.
. Let 
such that 
for 
. This yields:
an irrational number. Thus, no such 
exists under these conditions.
an irrational number. Thus, no such exists under these conditions.
just because 
In fact, this clearly wrong when 
and 
isn't irrational when 
, yet such 
and exist as alexheinis pointed out. The rest of your solution is neat, its just this one step that makes an assumption. In fact, I believe the statement is false for all 
except constant and quadratic polynomials. Here is a proof: has degree 
, then take
where 
has degree 
with integer coefficients. Then 
and 
(This type of construction is common when you have one polynomial behaving like a smaller degree polynomial, such as 
, for a set of 
.) For the quadratic case, take 
. Then and 
give, respectively,
and
The first equation tells us that 
from which we get
and
We can do similar analysis for the second equation:
and
and produce a nonzero, integral value for 
(nonzero since we are looking for quadratics), and hence no such quadratics exist.
is an integer for integers and 
so 
or 
so 
so 
works.
where 
such that: 
Prove that :
and 
.