[/quote]
,
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
such that 
,
and 
lie on 
and 
respectively such that 
.
and 
are the reflections of 
lies on segment 
such that the circumcenter of triangle 
coincides with the circumcenter of triangle 
.
be a positive integer larger than 
.
of subsets of 
having the following properties:
distinct sets of 
that satisfy the equation 
?
and 
Find the value of 
be a triangle with a circle 
with center 
tangent to sides 
at 
respectively. Suppose the circle with diameter 
intersects the circumcircle of 
is the reflection of 
lie on 
are parallel to 
lie on the circumcircle of 
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be an acute triangle with 
and let 
be its orthocenter. Let 
such that 
and 
are on different sides of 
,
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
.
and 
square is filled with numbers 
.
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
intersects at point 
Prove that: 
are parallel.
,
be four points in the plane, with 
and 
.![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles 
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
Y by GoJensenOrGoHome, itslumi, Adventure10, cubres, NicoN9
For a positive integer 
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
should be a perfect square. Thus 
.
.![\[ \prod_{i = 2}^{n} \frac{i}{i - 1} = n \]](http://latex.artofproblemsolving.com/2/3/6/236859bfd4b55a1b1c1e8520fda5bca9170ce1c1.png)
initially, which is a perfect square. We wanted to reverse several fractions such that the final product is 
.![\[ \prod_{i = 2}^{\sqrt{n}} \frac{i}{i - 1} = \sqrt{n} \]](http://latex.artofproblemsolving.com/4/2/1/4217e7a703142d345d3f229b19996d811b6599c4.png)
is![\[ \left(\frac12\cdot \frac23\cdot \frac34\cdots \cdot \frac{k-1}{k} \right) \left( \frac{k+1}{k}\cdot \frac{k+2}{k+1}\cdots \cdot \frac{n}{n-1} \right) = \frac{n}{k^2}=1. \]](http://latex.artofproblemsolving.com/6/2/c/62c6bdd77184c5d5534488f124bf8a969b436145.png)
Suppose that ![\[ (2\cdot 1)(3\cdot 2)\cdots (n(n-1)) = (n-1)!^2n,\]](http://latex.artofproblemsolving.com/3/9/d/39d71adbc84873a602a9a22ac0ff9f67db4e5656.png)
which means 
,
fractions. If 
,
is bad. Whenever we flip a fraction 
,
,
for some positive integer 
.
fractions.
for distinct values of 
.![\[ \left( \frac{1}{2} \right)^2 \cdot \left( \frac{2}{3} \right)^2 \cdot \dots \cdot \left( \frac{\sqrt{n}-1}{\sqrt{n}} \right)^2 = \frac1{n}\]](http://latex.artofproblemsolving.com/2/4/7/2471572cbba208df93cc24511e84518bbe6632ba.png)
to 
.
.
in both the numerator and denominator. This implies that 
fractions. The product of these fractions will then be
![\[\frac{1}{2}\cdot\frac{2}{3}\dots\frac{\sqrt{n}-1}{\sqrt{n}}\cdot\frac{\sqrt{n}+1}{\sqrt{n}}\cdot\frac{\sqrt{n}+2}{\sqrt{n}+1}\dots\frac{n}{n-1}\]](http://latex.artofproblemsolving.com/d/3/2/d32d2e1a511041ebb1a124496522d57f9e9367bb.png)
And it is easy to see that this works.![\[\frac{2}{1}\cdot\frac{3}{2}\dots\frac{n}{n-1}=n\]](http://latex.artofproblemsolving.com/6/9/f/69fe2c2adae0749a09ce9644d85148ad2e9a5a16.png)
by the construction that works to get![\[\left(\frac{2}{1}\right)^{1+\epsilon_1}\cdot\left(\frac{3}{2}\right)^{1+\epsilon_2}\dots\left(\frac{n}{n-2}\right)^{1+\epsilon_n}\]](http://latex.artofproblemsolving.com/b/3/5/b35e94d6dab521618259f2f4ac9766a3b882ac94.png)
Where 
is either 
.
must be an even number, so 
![$$\left[\frac{1}{2} \cdot \frac{2}{3} \dotsm\frac{\sqrt{n}-1}{\sqrt{n}}\right] \times \left[\frac{\sqrt{n}+1}{\sqrt{n}} \dotsm \frac{n}{n-1}\right].$$](http://latex.artofproblemsolving.com/4/f/3/4f37f626ec7b4fc891b660eb1ca6d4a8fc6f11ac.png)
All of the first bracket simplifies to 
an the second to 
and 
has product 
and the denominator 
has product 
so 
Also note that for the value to be 
.
,
since otherwise for some prime, we have a number of those primes odd and thus we cannot group them into two groups having the same cardinality.
hence 
be the product of the fractions not flipped and 
the product of the fractions which are flipped. Then we have 
and 
,
.
,
.
.
is a perfect square and so is 
groups of 
fractions and reciprocate every group except the first. Here's an example for 
.
.
,
,
,
,
,
,
over and our product becomes ![\[\underbrace{\frac{1}{a-1}}_{\text{first } a-2\text{ terms}}\cdot\underbrace{\frac{1}{a+1}}_{\text{terms }a+1\text{ to }a^2-1}\cdot\frac{a-1}{a}\cdot\frac{a+1}{a}\cdot\frac{a^2}{a^2-1}=1\]](http://latex.artofproblemsolving.com/a/e/9/ae9e06bb37e39a92944e9c0f59c0b6fc8b96cc0e.png)
![\[\frac{1}{2}*\frac{2}{3}*\cdots{}*\frac{k-1}{k}=\frac{1}{k}\]](http://latex.artofproblemsolving.com/8/d/7/8d7b98d599001044af590ba0af55f9aa718f84f5.png)
and![\[\frac{k+1}{k}*\frac{k+2}{k+1}*\cdots{}*\frac{k^2}{k^2-1}=k\]](http://latex.artofproblemsolving.com/2/4/4/244cb4144bea2434fc89e9158e6a5d7aec6bdf4f.png)
which we claim to be ![\[x=\frac{1}{\sqrt{n}}.\]](http://latex.artofproblemsolving.com/9/6/a/96a5531d5f1dc37ee45a528f5213b868ef560007.png)
We also know that 
.![\[\frac21, \frac32, \cdots, \frac{a^2}{a^2-1},\]](http://latex.artofproblemsolving.com/c/b/e/cbe57c2292a17355eec7ff3ea8a8fbf821829e2d.png)
and we want a product of the reciprocal of some of these to be 
.
,![\[\frac{1}{2}\cdot \frac{2}{3}\cdot \dots \frac{a-1}{a}\]](http://latex.artofproblemsolving.com/2/0/d/20d7a5bf10c1b6b3245eb07da3670bcb1c2817fd.png)
![\[\frac{1}{a},\]](http://latex.artofproblemsolving.com/9/1/1/911adf4e5042e448721f34035540ccdd5eb2e29e.png)
which means that 
.
Let 
be a subset of 
such that all the numbers of the form 
are reciprocated.
.![\[\frac{n}{\left(\frac{t_1}{t_1-1}\cdot\frac{t_2}{t_2-1}\cdot\dots\cdot\frac{t_k}{t_k-1}\right)^2}=1\]](http://latex.artofproblemsolving.com/c/6/2/c62b3134e90ce0ff0bde358510c800ef3a5477df.png)
implying that we require![\[\frac{t_1}{t_1-1}\cdot\frac{t_2}{t_2-1}\cdot\dots\cdot\frac{t_k}{t_k-1} =\sqrt{n}.\]](http://latex.artofproblemsolving.com/8/a/8/8a8fe0daf964ba643018f72dbb24642151bc6d22.png)
Since the LHS product is rational, we require the RHS to be rational as well. But 
Let ![\[\frac{k^2}{k^2-1}\cdot\frac{k+1}{k}\cdot\frac{k-1}{k-2}\cdot\frac{k-2}{k-3}\dots\frac{3}{2}\cdot\frac{2}{1}=k.\]](http://latex.artofproblemsolving.com/2/6/a/26a4000da8494c0cb732a230dd8dcb450c0c8280.png)
So the answer is 
.
,
,
.
did work, and then I wondered if the composite numbers worked, so I tried 
,
,
.
Notice the product of the reciprocated fractions is 
we multiply by 
so each prime factor gets changed by a multiple of 
hence 
![\[ \left(\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdots \cdot \frac{k-1}{k} \right) \left( \frac{k+1}{k}\cdot \frac{k+2}{k+1}\cdots \cdot \frac{n}{n-1} \right). \]](http://latex.artofproblemsolving.com/6/9/2/6920428a850470fcda59431e7f3e817a95e2fc2c.png)
![\[n!(n-1)!=n((n-1)!)^2\]](http://latex.artofproblemsolving.com/7/4/d/74d724e8c64e2d772edb677359308fdc7e5367b0.png)
is a square, so 
and win. 
must be a perfect square, so 
where 
is an odd number it is impossible to divide it by two (such that 
of the numerator is equal to the denominator - so that the final product is 1). Now, notice that when 
for odd 
Thus, if ![\[\frac{1}{2}\cdot \frac{2}{3} \cdots \frac{k-1}{k}\cdot \frac{k+1}{k} \cdot \frac{k+2}{k+1} \cdots \frac{k^2}{k^2-1}\]](http://latex.artofproblemsolving.com/2/b/2/2b2977d9d34270b3deb7cd4f70bdd90330f1bfe0.png)
,
.
.
.
,
the product of all the numerators and denominators of the fractions must be a perfect square. Note that the product of the numerators is 
Therefore, since this quantity must be a perfect square, we know that 
as desired, concluding the proof. 
I claim 
work. We use induction. 
case is obvious. If the result is true for 
,
look at the final 
terms because the first 
, done.
work.
Dividing equations yields
since 
terms also multiply to 
Reciprocate either the first 
which should obviously be distinct. Then we have that 
Thus 
Proof of necessity:
Assume you reciprocate some of the fractions among the fractions to get 1 .
as 
.
,
where 
and 
but we know that 
.
.
as desired.
be the ``prime factorization" of the product of (possibly reciprocated) fractions, where each 
is a not necessarily positive integer. Clearly, the parity of each 
,
.
.
where 
,
.
we must have 
and thus 
and keep the rest the same. Then we have 
are valid. Let 
Thus, 
Now, I claim that if 
will just multiply the current value by the square number 
Thus, since 
