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 sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
such that 
, points 
and 
lie on 
and 
respectively such that 
. Points 
and 
are the reflections of and with respect to the midpoints of sides and , respectively. Point 
lies on segment 
such that the circumcenter of triangle 
coincides with the circumcenter of triangle . passes through the midpoint of segment 
.
be a positive integer larger than 
. Build an infinite set 
of subsets of 
having the following properties: distinct sets of have exactly one common element;
distinct sets of have void intersection.
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 again at 
is the reflection of over . Suppose points 
lie on such that 
are parallel to . Prove that: The intersection of 
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 be a point on the perpendicular bisector of 
such that 
and 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 w.r.t 
. Prove that 
and are collinear.
square is filled with numbers 
.The numbers inside four 
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points to the circumcircle of 
intersects at point 
Prove that: 
are parallel., 
, , 
be four points in the plane, with and on the same side of the line , such that 
and 
. Find the ratio![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](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 and are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles and at the point are perpendicular.)
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
such that 
does not imply 
, for example, 
.
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 = a_2 +1\\
a_2^2 + a_2 = a_3 +1\\
\hspace*{3.3em} \vdots \\
a_{n}^2 + a_n = a_1 +1\]](http://latex.artofproblemsolving.com/f/4/6/f466c4722151a2339ccfb9f7ce22f93c5a7ee2f4.png)
equations we get 
. (1)
, so we get
. (2)
for some 
. Then we get 
, 
, and so on. It is easy to see that 
is in fact a solution of the system.
. Using the AM-GM inequality, we get
,
. Put 
, then all the 
are 
.
)
, 
.
gives 
, 
, 
. This means 
or 
. Both are solutions.
and 
.
implies 
.
. holds if 
, so the remaining is 
.
, then 
.
,
and [2] 
are shown.
, [1] implies 
and [2] says 
also satisfies the supposition, so we can repeat to use [1] to show the chain:
, which also gives a contradiction.
.
. For the other side, 
,
, our system of equations is less of a system of equations than a system of definitions! Let and 
where 
is composed times. We seek all real solutions to 
.
, 
. For 
, 
. For 
, . Therefore, the only solutions may occur at 
. Indeed, these are solutions as 
. Therefore, the only solutions are 
and 
.
. We write each equation as 
for all integer ![$i \in [1,n-1]$](http://latex.artofproblemsolving.com/6/4/e/64ee1a7568a0cd9332a713ff47076a77c671007c.png)
and 
; multiplying all of them yields that 
. By the AM-GM inequality, we know that:![$$\sum_{i = 1}^n a_i^2 \ge n\sqrt[n]{\prod_{i = 1}^n a_i^2}=n$$](http://latex.artofproblemsolving.com/9/c/2/9c2058af6c713bf2f05892b7b1553ee736e63486.png)
Since equality clearly holds, 
. Plugging in 
into the first equation, we see that 
and it follows that 
. Therefore, we see that the only possible solutions are 
, which we can check do indeed work.
![\[ a_{k+1}+1 = a_k(a_k+1).\]](http://latex.artofproblemsolving.com/6/b/d/6bd253297aea1a36fa48ef60c6398bc74e310e5d.png)
Taking the cyclic product over all , we get 
. Taking the cyclic sum of the original sum, we get 
. By AM-GM, 
, and since equality is achieved, 
.
for some , then 
, and so on for all 
's, and similarly if then all are -1. So either all are 1 or all are -1.
and 
These work. Notice that if any one of the is equal to or 
the rest must be as well, so from now on, we will assume that no is equal to or 
. Furthermore, there can also be no 
as that would make 
(indices taken 
) equal to a contradiction.
and factoring, we get that 
. Taking the product of cyclic permutations, we get that 
and since 
we have that 
. However, by AM-GM, we have that 
and because equality holds, we get that the entire sequence is equal up to sign. From here, it is just a matter of checking cases!
. Notice that 
, but by IMO 2006/5 
has at most two fixed points which in this case are at and . Thus, either 
or 
for which it follows that all terms are equal.
. Notice that , but by IMO 2006/5 has at most two fixed points which in this case are at and . Thus, either or for which it follows that all terms are equal. has at most two integer fixed points.
and 
, both of which clearly work. Take indices modulo , so 
(when we say the sequence is periodic, this is by considering the infinite extension of the sequence in periodic manner). We now present two solutions to show that no other -tuples work. then all of them are as well, hence suppose none of them are . Add one to both sides of the equation to get 
. Since no equals , we have 
. On the other hand, summing gives 
. But by AM-GM we have
hence 
, which means all of them are either or , so we extract the given answer. 
, so there must exist some with 
. It is clear that if 
, 
, but this is absurd as it makes a periodic sequence strictly increasing, so we either have for all , for all , or there exists some such that 
. We will show that the third case is impossible. Indeed, suppose 
for 
. Since we have 
for all 
, we must have 
for all , hence ![$a \in (0,1/4]$](http://latex.artofproblemsolving.com/5/b/a/5ba8beb805920adb41f484955e03e796ef250159.png)
. Now we can calculate 
, and
We can check that 
for (factor it as 
), so 
is also negative but also strictly greater than 
. Thus, we have 
for all , but this is again impossible as 
is periodic. From this, we either have 
or 
—the desired solution. 
. We have the equations 
for each . for any , then:
so by induction 
which is indeed a solution. Otherwise, suppose that 
for all . for 
gives:
so
Multiplying these equations gives:
so
![$$1=\sqrt{\frac1n\sum_{i=1}^na_i^2}\ge\sqrt[n]{\prod_{i=1}^na_i}=1,$$](http://latex.artofproblemsolving.com/e/3/4/e34f700b649029ee61a3a5ead9a6be30a10a54ee.png)
so equality holds and all of the are equal. Rearranging the first equation gives that 
, and since 
we have 
which is indeed a solution.
then 
instead of 
, this is false—take 
, so 
. In fact, if 
, we have 
for any , then everything becomes . We will add in this case later, so assume that 
for now. to get![\[a_1^2+a_2^2+\dots+a_n^2=n.\]](http://latex.artofproblemsolving.com/b/3/1/b311b22a2c45e14c1026387a0a4b48d7a7f7a028.png)
![\[\frac{a_{k+1}+1}{a_k+1}=a_k\]](http://latex.artofproblemsolving.com/8/2/8/8285e18cc9a2901cf825874f9585000dbab4c535.png)
and multiply over all to get![\[a_1a_2\dots a_n=1.\]](http://latex.artofproblemsolving.com/e/1/e/e1eef015505ba92611f38fd7da9c8c16cd9e622e.png)
![\[\sqrt{\frac{a_1^2+a_2^2+\dots+a_n^2}{n}}=\sqrt[n]{a_1a_2\dots a_n}.\]](http://latex.artofproblemsolving.com/9/d/6/9d62493490287cb02a5fa42e6f4b994d0b1b7ef8.png)
This implies QM-GM for the variables 
and only has the equality case of everything being equal. goes to 
, then 
so 
. We can manually check that for each of these cases, doesn't go to either of or . Thus the sequences are indeed constant.'s or all 's, so the answer is 
.
. is .
and so on, so 
Case 2: No is .
But 
, so 
Now AM-GM gives us 
Since equality holds, 
. But no is , so all are , yielding 
.
and 
both clearly work, we're done. 
![\[a_{k+1} = a_{k}^2 + a_{k} - 1\]](http://latex.artofproblemsolving.com/a/a/8/aa8a94649796d8e5b86e39a66d9bcc3ae489c764.png)
We can substitute this value of 
into the problem condition to obtain![\[a_{k+2} = a_{k+1}^2 + a_{k+1} - 1 = {(a_{k}^2 + a_{k} - 1)}^2 + (a_{k}^2 + a_{k} - 1) -1 = a_{k}^4 + 2a_{k}^3 - a_{k} - 1\]](http://latex.artofproblemsolving.com/c/0/6/c0686a6f87fad471a15c1b9ffb9981e679c1a328.png)
QED
, then 
, in other words 
will get closer to 
.
as 
for some positive 
. Substituting this value into the equation for gives us![\[a_{k+2} = {(-1+\delta)}^4 + 2{(-1+\delta)}^3 - (-1+\delta) -1 = \delta^4 - 2\delta^3 + \delta - 1\]](http://latex.artofproblemsolving.com/7/4/3/7437924caf638e58b7cc959ce1e50104b30058f0.png)
Our goal is to show that![\[\delta > |-1 - (\delta^4 - 2\delta^3 + \delta - 1)|\]](http://latex.artofproblemsolving.com/5/a/0/5a048ee00ca71b3ba47b5004a1b019558c26cf8b.png)
Simplifying, we have![\[|-1 - (\delta^4 - 2\delta^3 + \delta - 1)| = |\delta^4 -2\delta^3 + \delta| = \delta \cdot |(\delta-1)(\delta^2-\delta-1)|\]](http://latex.artofproblemsolving.com/7/2/3/723d7732e116e831f9794dd8f5e718a6bc98cc1a.png)
Thus it suffices to show![\[|(\delta-1)(\delta^2-\delta-1)| < 1\]](http://latex.artofproblemsolving.com/e/1/a/e1a69a58411f8ef73eb4e97f68d2654ba40ea893.png)
Which can be bashed. QED
, then , in other words will get closer to .
like before, however this time for negative delta (I don't want to reexpand the equation lol). Again it suffices to show![\[-\delta > |\delta(\delta-1)(\delta^2-\delta-1)|\]](http://latex.artofproblemsolving.com/8/a/2/8a24b293463c93954df56786584c607d33cf926b.png)
However notice that![\[-\delta > |\delta(\delta-1)(\delta^2-\delta-1)| \iff 1 > |(\delta-1)(\delta^2-\delta-1)|\]](http://latex.artofproblemsolving.com/7/0/7/70788fc6a14fcd793987f4e2b7c29e602e1d5665.png)
for 
. Which is true since 
is a root of 
. QED
, then 
, then![\[a_{k+1} > a_{k}\]](http://latex.artofproblemsolving.com/d/0/1/d01b095f746ae8805c27b965fc9d4ff9d3466f2f.png)
Also if there is an 
, then 
denote the -th composition of 
, notice that![\[a_{1} = f^{n}(a_{1})\]](http://latex.artofproblemsolving.com/f/a/e/fae05c32e8faf4c68dd876872348a8c6520a813c.png)
Claims 1 until 5 imply that 
. Thus the only possible sequences are 
for all 
or 
for all .
.But we have that 
and defining the sequence 
gives us 
via AM-GM.This gives us all the results.
which clearly work.
is sufficient enough to imply the solution set 
, so assume otherwise. We can rearrange the relation 
(indices 
) to 
which summing over all telescopes to 
. Moreover, we also have 
which multiplying over all gives 
.![$\dfrac{|a_1|^2 + |a_2|^2 + \dots + |a_n|^2}{n} = \sqrt[n]{|a_1|^2 |a_2|^2 \dots |a_n|^2}$](http://latex.artofproblemsolving.com/9/9/b/99bfb74eddcdfee1ccaa49ce63f7ef2643431d29.png)
which is the equality case of AM-GM. Then 
easily gives 
as a solution set, done.
We also have that 
and so 
Thus, by AM-GM, we have that 
If 
then the rest are all 1. If 
the rest are -1.
so multiplying these yields 
Meanwhile 
But by the Power Mean Inequality ![$$\sqrt{\frac{\sum a_i^2}{n}} \geq \sqrt[n]{\prod a_i} \iff 1=1$$](http://latex.artofproblemsolving.com/3/1/8/318da8bfe58f9db28bea7a0f144732a88385ae47.png)
so equality in fact holds and all are equal. 
so the only solutions are 
. Summing the condition over gives 
. The condition can also be manipulated to 
, which gives 
by multiplying over . This gives the equality case of AM-GM on 
which implies 
. Note that if 
, then 
as well, and likewise, if , then 
. The result follows by induction.
. We aim to show that there are no other solutions.
into four regions , , , with y-coordinates 
, ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
, ![$[-1,0]$](http://latex.artofproblemsolving.com/3/1/d/31d4af7eb18014c3073c9fa861cdbfc4afb367a0.png)
, 
. Then we can see that the mapping 
maps![\[A \mapsto A, B \mapsto \text{down}, C \mapsto D, D \mapsto C.\]](http://latex.artofproblemsolving.com/3/b/6/3b6dca09ce9df82e3bd165685bfa379dd00e7b8b.png)
. Thus the only other solutions we could have are from![\[(x^2+x-1)^2+(x^2+x-1)-1=x \implies (x-1)(x+1)^3 = 0.\]](http://latex.artofproblemsolving.com/4/0/3/4032c5b7e0b5c599b4ad45be9899e9a1ce6cee03.png)
![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
(think about the dotted portion)
and plugging back see 
assuming 
for any 
.
it is easy to see that by AM-gm and keeping in mind the sum 
that all 
such that 
then automatically all 
,
,
,
and 
both work, and our casework shows that these are the only solutions.
Plugging this into the second line, then the third, etc gives 
which implies either the product of 
is 1, or 
.
as a solution. The product of 
is 1, but no value can be greater 1. This means if any 
,
.
as well, and the chain continues, so they're all 
.
.
)
.![$n=\sum_{i=1}^{n} {a_i}^2\geq n\sqrt[n]{\prod_{i=1}^{n} (a_i)^2}=n$](http://latex.artofproblemsolving.com/6/b/1/6b1408c1a40b80a2cf41b5fcd1d2ba9c55f0f937.png)
,
are equal. Note that the product of all of them is 1, so they must all be equal to 
and 
are the only solutions.
![\[ a_{1}^2 + a_{2}^2 + \dots + a_{n}^2 = n\]](http://latex.artofproblemsolving.com/a/0/0/a005763d2af9406b3f750c3d44282e09cbf9b758.png)
Also, note that![\[ a_{k+1} + 1 = a_k(a_k + 1) \]](http://latex.artofproblemsolving.com/5/9/8/598eab636db5facd5682dea34bb325a8ed73b1a4.png)
As such, ![\[ a_1a_2\dots a_n(a_i + 1) = (a_i + 1) \]](http://latex.artofproblemsolving.com/c/0/b/c0b7b19921ef343f6fc16b6cb444b2d0432ae70d.png)
If 
for some 
and no ![\[ \frac{a_1^2 + a_2^2 + \dots + a_n^2}{n} \ge \sqrt[n]{a_1^2a_2^2\dots a_n^2} = 1 \]](http://latex.artofproblemsolving.com/b/6/d/b6d172dbb1a1e6acb08b19bc11aa3549d264e1d7.png)
Equality can only hold if 
,
