AoPS Academy
![$P \in \mathbb Q[x,y]$](http://latex.artofproblemsolving.com/a/3/8/a384cbf38fd3a7f812b22baaeb0860e7f6eb26c0.png)
with 
and ![$\forall (a,b) \in \mathbb Z^2 \cap [0,d]^2, P(a,b) \in \mathbb Z$](http://latex.artofproblemsolving.com/a/6/5/a65ecb4bb2d9b5e6096f322c2de8bd11b945ed6a.png)
.
?
with 
.![$P(x,y) \in \mathbb Q[x,y]$](http://latex.artofproblemsolving.com/b/3/7/b37942b05292a493c8e350322adc122205bc778d.png)
?
be a quadrilateral inscribed in a circle 
, and let 
be a point on the extension of 
such that 
and 
are tangent to . The tangent at 
intersects at 
and the line 
at 
. Let 
be the second point of intersection between 
and . Prove that 
, , are collinear.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
be a parallelogram with 
. A circle tangent to sides 
, 
, and 
intersects diagonal 
at points and with 
, as shown. Suppose that 
, 
, and 
. Then the area of can be expressed in the form 
, where 
and 
are positive integers, and is not divisible by the square of any prime. Find 
.
and 
intersect at points 
and 
. Line 
is tangent to and at 
and , respectively, with line 
closer to point than to . Circle passes through and intersecting again at 
and intersecting again at 
. The three points , , 
are collinear, 
, 
, and 
. Find 
.
, we know that lines and 
intersect at , lines and 
intersect at 
, and lines 
and 
intersect at 
. Suppose that the circumcircle of 
intersects line 
at and , and the circumcircle of 
intersects line at and , where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
, then 
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
, points 
, 
, and 
are collinear if and only if 
.
, 
, 
, 
are integers such that 
.
and 
of 
such that for each integer 
with 
, we have![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be a finite abelian group of order 
.
be arbitrary.
and pairwise distinct 
such that 
for 
.
and 
s.t 
and 
and there are exactly two i;j s.t 
and 
.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
Please post it here and now!
that add 
to 
and subtract from 
.
that add to and subtract from .
and 
of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
for some sequence ![$\{ a_i\}_{i\in [n]}$](http://latex.artofproblemsolving.com/f/6/e/f6e384c04b4aaf07513dacb66f8cadc0a597069e.png)
so that 
for all ![$i\in [n]$](http://latex.artofproblemsolving.com/e/e/5/ee51c5c5eb09111cc774a00e9970074ad9f9058c.png)
.![$\{ b_i\}_{i\in [n]}$](http://latex.artofproblemsolving.com/1/3/e/13e91e327fa3f2bc2d43edbe110540d350184040.png)
with sum divisible by that differ, in modulo , from in only two positions, say 
and 
.
and 
of so that 
for all .
for all that 
.
by, for each integer 
, define ![$i_{k+1}\in [n]$](http://latex.artofproblemsolving.com/c/e/5/ce54cff3726a2dae0e8cc97f7b358dd90622aeee.png)
to be the unique integer satisfy 
.
are the indices that 
with minimal 
, and then minimal 
.
. This means 
for all 
. for 
gives us
Plugging and use 
gives us 
.
, contradiction to the definition of 
.
. Let 
. Define the desired permutations and as follows:
and 
for the rest that 
.
is just to not use 
more than one time.
for all except when 
. and are permutations of , we've ![$\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$](http://latex.artofproblemsolving.com/4/0/c/40ce5067c15701668f9fc9a82369ce8ab1f4384c.png)
.
when too. This prove the validity of permutations we constructed.
for 
on the coordinate plane with coordinates modulo . We want to show that it is possible to choose the points plotted to have 
be any multiset of elements 
such that sum of the elements is 
. Note that the points plotted can be any points not in the same row or column.
points not in the same row or column such that the multiset 
of 
for all of the points 
can be any multiset of elements . Then, we can choose the unique point not in the same row or column as the points to obtain points for the original statement, as the sum of the coordinates of all points is necessarily 
. points in the following way. arbitrary points not in the same row or column. We will change the points in such a way that we can replace any element of by any residue . Suppose we change the element in corresponding to a point . Then, we will shift horizontally until the sum of the coordinates of is the desired value . Now, we might have two points in the same column. We will repeatedly perform the following operation. distinct rows taken up by the points. If the last point moved was and it is in the same column as , we will move on the line through with slope 
until occupies the previously empty row. to a point 
, then on the next step it moves to a point 
. We see that 
and 
both have slope . Now, note that and must be in the same row. Furthermore, after the operation moves to , the row containing must be empty, meaning that must be moved to that row. This implies that and are in the same row. Therefore, we see that 
is a cyclic isosceles trapezoid and in particular the perpendicular bisectors of and are the same, proving the claim. points to distinguish them. Now, there are finitely many states of the points, so eventually the operation must loop through a cycle of states. In this cycle, start at an arbitrary state. Now, since at least 
point must be moved and then return back to its original position, consider the point which returns to its original position the earliest. When was moved, the next point which was moved must have occupied the original row of . However, if was moved back to its original position, then the original row of must be vacant, so must have been moved. However, this implies that was moved back to its original position before by the claim, a contradiction, so we are done.
.
