AoPS Academy
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
, prove there exists positive integer 
and positive real number 
, such that 
, for any 
, 
, 
holds.
be a acute triangle with 
. 
are the orthocenter and excenter. Let 
be a point on the same side of 
as 
, such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be a point on the same side of 
as , such that 
is equilateral. Let 
be the midpoint of 
. Prove that 
are collinear.
representation of 
?
has side length 
. Let the midpoint of 
be 
. What is the area of the overlapping region between the circle centered at with radius 
and the circle centered at with radius ? (You may express your answer using inverse trigonometry functions of noncommon values.)
occurs when 
for the function 
. room configurations:
Configuration : a one-room double, which is a square with side length of 
, Configuration 
: a two-room double, which is two connected rooms, each of them squares with a side length of 
.) to be exactly m
larger than the total area of a two-room double. Find the number of possible pairs of side lengths 
, where 
, 
, such that 
.
islands. One day, people decide to create island-states as follows. Each island randomly chooses one of the other five islands and builds a bridge between the two islands (it is possible for two bridges to be built between islands and if each island chooses the other). Then, all islands connected by bridges together form an island-state. What is the expected number of island-states Ur is divided into?
and 
be the roots of the polynomial 
. Compute 
.
, 
for prime 
and 
for 
. Find the smallest 
such that 
is a perfect power of 
.
be a triangle whose -excircle, -excircle, and 
-excircle have radii 
, 
, and 
, respectively. If 
and the perimeter of is 
, what is the area of ?
of functions 
satisfying:
,
,
.
can be written as 
where 
are distinct primes, compute 
.
and that the first (leftmost) two digits of 
are 10. Compute the number of integers with 
such that 
starts with either the digit 
or 
(in base 
).
be the circumcenter of . Let be the midpoint of , and let and 
be the feet of the altitudes from and , respectively, onto the opposite sides. 
intersects at . The line passing through and perpendicular to intersects the circumcircle of at 
(on the major arc ) and 
, and intersects at . Point lies on the line 
such that 
is perpendicular to 
. Given that 
and 
, compute 
.
be the isosceles triangle with side lengths 
. Arpit and Katherine simultaneously choose points and 
within this triangle, and compute 
, the squared distance between the two points. Suppose that Arpit chooses a random point within . Katherine plays the (possibly randomized) strategy which given Arpit’s strategy minimizes the expected value of . Compute this value. with sides, let 
denote the regular polygon with 
sides such that the vertices of are the midpoints of every other side of . Let 
denote the polygon that results after applying f a total of k times. The area of 
where is a pentagon of side length , can be expressed as 
for some positive integers 
where 
is not divisible by the square of any prime and does not share any positive divisors with 
and 
. Find 
.
. This function can be expressed in the form 
for sequences of integers 
, 
. Determine 
., let 
be the centroid and let the circumcenters of 
, 
, and 
be 
, and , respectively. The line passing through 
and the midpoint of intersects 
at 
. If the radius of circle is 
, the radius of circle 
is , and 
, what is the length of 
?
be a polynomial with nonnegative integer coefficients. If 
and 
, what is the remainder when 
is divided by 
are nonzero real numbers, find the minimum value of the expression![\[
P = \left| \frac{b + c + d}{a} \right| + \left| \frac{c + d + a}{b} \right| + \left| \frac{d + a + b}{c} \right| + \left| \frac{a + b + c}{d} \right|.
\]](http://latex.artofproblemsolving.com/e/3/e/e3e63937c4badb9bcb47e30778f2165e2df5ae46.png)
be non-negative real numbers such that 
and![\[
\frac{25a + 36b + 49c}{5a + 6b + 7c} + \frac{25b + 36c + 49a}{5b + 6c + 7a} + \frac{25c + 36a + 49b}{5c + 6a + 7b} = 18.
\]](http://latex.artofproblemsolving.com/a/8/a/a8a56fe414b713138e1adc48cdcedcc23200843c.png)
Prove that exactly two of the numbers are equal to 0.
be a triangle such that 
. Points 
lie on segments 
, respectively, such that lines 
and 
are the angle bisectors of triangle 
. Find the measure of angle 
.
, 
, 
, 
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.
and 
Prove that 
Prove that
Let 
Prove that
.
