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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inspired by Balkan 2001
sqing   0
5 minutes ago
Source: Own
Let $a$, $b$, $c$ be positive real numbers with $abc \leq a+b+c$. Show that
$$ a^2+b^2 + kc(a+b) \geq 2\sqrt {2k-1}   abc $$Where $k>1. $
$$ a^2+b^2 + 5c(a+b) \geq 6abc $$
0 replies
1 viewing
sqing
5 minutes ago
0 replies
GHM and ABC are tangent to each other.
IndoMathXdZ   16
N 6 minutes ago by Ilikeminecraft
Source: DeuX Mathematics Olympiad 2020 Shortlist G5 (Level II Problem 3)
Given a triangle $ ABC$ with circumcenter $O$ and orthocenter $H$. Line $OH$ meets $AB, AC$ at $E,F$ respectively.
Define $S$ as the circumcenter of $ AEF$. The circumcircle of $ AEF$ meets the circumcircle of $ABC$ again at $J$, $J \not= A$. Line $OH$ meets circumcircle of $JSO$ again at $D$, $D \not= O$ and circumcircle of $JSO$ meets circumcircle of $ABC$ again at $K$, $K \not= J$. Define $M$ as the intersection of $JK$ and $OH$ and $DK$ meets circumcircle of $ABC$ at points $K,G$.

Prove that circumcircle of $GHM$ and circumcircle of $ABC$ are tangent to each other.

Proposed by 郝敏言, China
16 replies
IndoMathXdZ
Jul 12, 2020
Ilikeminecraft
6 minutes ago
Finding Max!
goldeneagle   3
N 15 minutes ago by Assassino9931
Source: Iran 3rd round 2013 - Algebra Exam - Problem 2
Real numbers $a_1 , a_2 , \dots, a_n$ add up to zero. Find the maximum of $a_1 x_1 + a_2 x_2 + \dots + a_n x_n$ in term of $a_i$'s, when $x_i$'s vary in real numbers such that $(x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_{n-1} - x_n)^2 \leq 1$.
(15 points)
3 replies
goldeneagle
Sep 11, 2013
Assassino9931
15 minutes ago
Number Theory Marathon!!!
starchan   436
N 20 minutes ago by zhihanpeng
Source: Possibly Mercury??
Number theory Marathon
Let us begin
P1
436 replies
starchan
May 28, 2020
zhihanpeng
20 minutes ago
An easy and classical inequality in 3 variables abc<=a+b+c
Valentin Vornicu   20
N 31 minutes ago by sqing
Source: Balkan MO 2001, problem 3
Let $a$, $b$, $c$ be positive real numbers with $abc \leq a+b+c$. Show that \[ a^2 + b^2 + c^2 \geq \sqrt 3 abc. \]
Cristinel Mortici, Romania
20 replies
Valentin Vornicu
Apr 24, 2006
sqing
31 minutes ago
Find smallest value of (x^2 + y^2 + z^2)/(xyz)
orl   8
N an hour ago by sqing
Source: CWMO 2001, Problem 4
Let $ x, y, z$ be real numbers such that $ x + y + z \geq xyz$. Find the smallest possible value of $ \frac {x^2 + y^2 + z^2}{xyz}$.
8 replies
1 viewing
orl
Dec 27, 2008
sqing
an hour ago
Painted cells
Titibuuu   0
an hour ago
A \( 2021 \times 2021 \) grid has \( k \) cells painted such that the following condition holds:
For every painted cell, at least one of its \emph{vertices} is also a vertex of \emph{another} painted cell.
Find the maximum possible value of \( k \).
0 replies
Titibuuu
an hour ago
0 replies
junior perpenicularity, 2 circles related
parmenides51   3
N an hour ago by LeYohan
Source: Greece Junior Math Olympiad 2024 p2
Consider an acute triangle $ABC$ and it's circumcircle $\omega$. With center $A$, we construct a circle $\gamma$ that intersects arc $AB$ of circle $\omega$ , that doesn't contain $C$, at point $D$ and arc $AC$ , that doesn't contain $B$, at point $E$. Suppose that the intersection point $K$ of lines $BE$ and $CD$ lies on circle $\gamma$. Prove that line $AK$ is perpendicular on line $BC$.
3 replies
parmenides51
Mar 2, 2024
LeYohan
an hour ago
Sequence
Titibuuu   0
an hour ago
Let \( a_1 = a \), and for all \( n \geq 1 \), define the sequence \( \{a_n\} \) by the recurrence
\[
a_{n+1} = a_n^2 + 1
\]Prove that there is no natural number \( n \) such that
\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]is a perfect square.
0 replies
Titibuuu
an hour ago
0 replies
Coolabra
Titibuuu   0
an hour ago
Let \( a, b, c \) be distinct real numbers such that
\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]Find the maximum possible value of \( a \).
0 replies
Titibuuu
an hour ago
0 replies
Inspired by Bet667
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
5 replies
sqing
Thursday at 1:03 PM
sqing
an hour ago
A tangent problem
hn111009   0
an hour ago
Source: Own
Let quadrilateral $ABCD$ with $P$ be the intersection of $AC$ and $BD.$ Let $\odot(APD)$ meet again $\odot(BPC)$ at $Q.$ Called $M$ be the midpoint of $BD.$ Assume that $\angle{DPQ}=\angle{CPM}.$ Prove that $AB$ is the tangent of $\odot(APD)$ and $BC$ is the tangent of $\odot(AQB).$
0 replies
hn111009
an hour ago
0 replies
3-var inequality
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
4 replies
sqing
May 7, 2025
sqing
an hour ago
Inspired by Kosovo 2010
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
2 replies
sqing
Yesterday at 3:56 AM
sqing
an hour ago
Wot n' Minimization
y-is-the-best-_   25
N May 2, 2025 by john0512
Source: IMO SL 2019 A3
Let $n \geqslant 3$ be a positive integer and let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2. Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of
\[
\left|1-\sum_{i \in X} a_{i}\right|
\]is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ with sum equal to 2 such that
\[
\sum_{i \in X} b_{i}=1.
\]
25 replies
y-is-the-best-_
Sep 23, 2020
john0512
May 2, 2025
Wot n' Minimization
G H J
Source: IMO SL 2019 A3
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y-is-the-best-_
16 posts
#1 • 3 Y
Y by centslordm, megarnie, Mango247
Let $n \geqslant 3$ be a positive integer and let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2. Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of
\[
\left|1-\sum_{i \in X} a_{i}\right|
\]is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ with sum equal to 2 such that
\[
\sum_{i \in X} b_{i}=1.
\]
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MarkBcc168
1595 posts
#2 • 2 Y
Y by centslordm, WTFrrrrrrrf
Comment

Solution
This post has been edited 1 time. Last edited by MarkBcc168, Sep 23, 2020, 6:42 AM
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dgrozev
2463 posts
#3 • 2 Y
Y by parmenides51, centslordm
This problem (with slightly different wording) had appeared here in this forum, as being given at some TST. I commented it in my blog without knowing it's from ISL 2019.
Z K Y
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rcorreaa
238 posts
#4 • 1 Y
Y by centslordm
Wow, I apparently have solved an A3. Hope it isn't a fakesolve! :P

Firstly, observe that the condition of $
\left|1-\sum_{i \in X} a_{i}\right|
= \delta$ being minimal is equivalent to $$
\frac{\left|\sum_{i \in X^C} a_i -\sum_{i \in X} a_{i}\right|}{2} $$be minimal. Then, assume WLOG that $\sum_{i \in X^C} a_i > \sum_{i \in X} a_{i}$ (if the equality ocurred, we would be done). Also, for any strictly increasing sequence of $n$ positive real numbers $(c_1,c_2,...,c_n)$, let $$\Delta_{(c_1,c_2,...,c_n)}= \left|\sum_{i \in X^C} c_i -\sum_{i \in X} c_{i}\right|$$We want to prove that there exists a strictly increasing sequence of $n$ positive real numbers $(b_1,b_2,...,b_n)$ such that $$\Delta_{(b_1,b_2,...,b_n)}=0$$
Now, observe that since $\frac{\Delta_{(a_1,a_2,...,a_n)}}{2}$ is minimal, for all $a_y \in X^C$ and $a_x \in X$ (they exist, since $\Delta_{(a_1,a_2,...,a_n)}>0$, we have that $$\frac{\Delta_{(a_1,a_2,...,a_n)}}{2} < a_y-a_x \qquad(\star)$$otherwise we could place $a_y$ in $X$ and $a_x$ in $X^C$ and $\frac{\Delta_{(a_1,a_2,...,a_n)}}{2}$ would decrease and would be greater than $0$, by assumption, and we would have a better choice for $X$, a contradiction!

Then, take $x_0 \in X, y_0 \in X^C$ with $a_{x_0} < a_{y_0}$ such that $a_{y_0}-a_{x_0}=M$ is minimal. From $(\star)$, $ \delta= \frac{\Delta_{(a_1,a_2,...,a_n)}}{2} < M$. Thus, if we choose $b_i=a_i+\frac{\delta}{|X|}$ for all $i \in X$ and $b_i=a_i-\frac{\delta}{|X^C|}$, the order of the terms in $b_i \in X$ is the same of $a_i \in X$, as well as the order of the terms in $b_i \in X^C$ is the same of $a_i \in X^C$. Furthermore, $b_y-b_x < M$, from $(\star)$, so we don't have issues with an index $i \in X$ sweeping with and index $j \in X^C$ in the new choice of sequence (the sequence keeps strictly increasing). Hence, this new sequence satisfies $\Delta_{(b_1,b_2,...,b_n)}=0$, $\sum_{i=1}^n b_i = \sum_{i=1}^n a_i = 2$ and $\sum_{i \in X} b_{i}=1$ by construction.

Therefore, we are done.

$\blacksquare$
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Tintarn
9042 posts
#6 • 2 Y
Y by centslordm, sabkx
We solve the problem by characterizing all sets $X$ which can occur.
To this end, write $Y=X^c$ for the complement and $X>Y$ ("$X$ majorizes $Y$") if for every $k$, the set $X$ contains at least as many indices larger than $k$ as does $Y$. Similarly, define $Y>X$. (Note that we do not at all need to have either $X>Y$ or $Y>X$.)
Now if $X>Y$ it is clear that $\sum_{i \in X} b_i>1$. Similarly, if $Y>X$ it is clear that $\sum_{i \in X} b_i<1$.
So we better prove that such sets $X$ cannot occur.
Indeed, let us prove the following two facts:
1) If $X>Y$, then $X$ can not be a minimizing set as in the problem statement.
2) If we do not have $X>Y$, then we can find $b_i$ such that $\sum_{i \in X} b_i<1$.
Before we prove 1) and 2), let us see how this finishes the problem.
From 1) we get that the set $X$ from the problem must have neither $X>Y$ nor $Y>X$. Then from 2) we get a sequence $b_i$ such that $\sum_{i  \in X} b_i<1$ and a sequence $c_i$ such that $\sum_{i \in X} c_i>1$. Taking a convex linear combination (which will still be strictly increasing and have total sum equal to $2$) we find a sequence with sum over $X$ exactly $1$.
Finally, let us see how to prove 1) and 2):
Proof of 1): Indeed, write $X=\{x_1,\dots,x_k\}$ and $Y=\{y_1,\dots,y_m\}$ with the elements in decreasing order. So $x_1 >y_1, x_2 > y_2$ etc. So $\sum_{i \in Y} a_i<1<\sum_{i \in Y} x_i$. But if $k \ge 2$ and $m \ge 1$ the set $M=\{y_1,x_2,\dots,x_k\}$ will have a sum strictly between the two sums, hence $X$ can not be a minimizing set. So we must have $X=\{n\}$ or $X=\{1,2,\dots,n\}$ both of which are easily seen to be only possibly minimizing if $n \le 2$ (so this is where our proof uses $n \ge 3$). Done.
Proof of 2): So for some $k$ the set $X$ contains less than half of the indices from $\{k+1,\dots,n\}$. Choose $c_i=0$ for $i \le k$ and $c_i=\frac{2}{n-k}$ for $i>k$. This is an increasing sequence with $\sum_{i \in X} c_i<1$. Now just perturb $c_i$ to a sequence $b_i$ by a sufficiently small $\varepsilon$ to make it strictly increasing and still have $\sum_{i \in X} b_i<1$. Done.
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jj_ca888
2726 posts
#7 • 7 Y
Y by mijail, samrocksnature, centslordm, sabkx, Mango247, Mango247, Mango247
Ok. I realize that this problem isn't particularly hard; it's just really annoying.

Split $a_1, \ldots , a_n$ into $X = \{x_1, \ldots , x_k\}$ and $Y = \{y_1, \ldots , y_{\ell}\}$ where $k + \ell = n$. Denote the sum of all $x_i$ as $1 + \epsilon$ and the sum of all $y_i$ as $1 - \epsilon$. If $\epsilon$ is ever $0$, then we are done. So WLOG $\epsilon > 0$.

Note that $\epsilon \leq x_i - y_j$ for all $i, j$ (where $x_i > y_j$ at least). If otherwise, then we may send $y_j$ to $X$ and $x_i$ to $Y$ and obtain a smaller $\epsilon '$ since\[1 - \epsilon < \sum Y' = (1 - \epsilon) + (x_i - y_j) \leq 1.\]In fact, we may further assume $\epsilon < x_i - y_j$ since if equality occurs, then after the swap, $\epsilon ' = 0$.

Similarly, with this swapping argument, we obtain that all $x_i$ are greater than $\epsilon$.

Now we construct a sequence $b_1, \ldots , b_n$ that satisfies the requirements. Let $S$ be the set of indices $i$ for which $a_i$ is a $x$ term and $T$ be the set of indices $j$ for which $a_j$ is a $y$ term. Clearly $|S| = k$ and $|T| = \ell$. I claim that the following sequence, defined by\begin{align*}b_i = a_i - \tfrac{\epsilon}{k} \text{  }\forall i \in S\\b_i = a_i + \tfrac{\epsilon}{\ell} \text{  }\forall i \in T\end{align*}works. Indeed, check that $\sum b_i = \sum a_i + {\epsilon} - {\epsilon} = 2$. and $\sum_{i \in S} b_i = (1 + \epsilon) - \epsilon = 1$.

It remains to check that this sequence is strictly increasing. If $b_m, b_{m+1}$ are either both in $S$ or both in $T$, then it clearly follows that $b_m < b_{m+1}$ since $a_m < a_{m+1}$. If $b_m \in S$ and $b_{m+1} \in T$, then also $b_{m+1} - b_m = (a_{m+1} - a_m) + \tfrac{\epsilon}{k} + \tfrac{\epsilon}{\ell} > 0$.

Now, if $b_m \in T$ and $b_{m+1} \in S$, then\[b_{m+1} - b_m = (x_i - y_j) - (\tfrac{\epsilon}{k} + \tfrac{\epsilon}{\ell}) \geq (x_i - y_j) - \epsilon > 0\]by our first note, along with using the fact that $\tfrac{1}{k} + \tfrac{1}{\ell} \leq \tfrac{1}{2} + \tfrac{1}{2n - 4} \leq 1$ for $n \geq 3$ unless $k$ or $\ell$ is $1$.

If $k = 1$, then the only $x$ term has value $1 + \epsilon > 1$ which is bad since then the next largest term is a $y$-term and has value $< 1$, contradicting minimality of $\epsilon$.

If $\ell = 1$, then the only $y$ term has value $1 - \epsilon$. The sum of the $n - 1$ remaning terms is $1 + \epsilon$. The sum of the smallest $n - 2$ terms is $> (n-2)\epsilon$, so the largest $x$ term is $< 1 + (3 - n)\epsilon = 1$. So if $x_{max} >$ the only $y$ term, we get a contradiction since their difference is $< \epsilon$. Thus, the $y$ term is the largest $a_i$, meaning that $b_m \in T$ and $b_{m+1} \in S$ is impossible.

So indeed, this sequence $b_i$ works. $\blacksquare$
This post has been edited 12 times. Last edited by jj_ca888, Oct 18, 2020, 3:42 PM
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alexiaslexia
110 posts
#8 • 3 Y
Y by centslordm, Kingsbane2139, sabkx
"No way an A3 could be THAT hard ..." $\Rightarrow$ me during the test, when I finally did it for 2.5x the average time other TST takers needed. Set-theory oriented solution in a completely different flavor: (first idea akin to the third solution from the official Shortlist).
For notational purposes let $[n]$ be the set $\{ 1,2,\ldots,n\}$.
Denote $a_{i+1} - a_i = d_i$, and $a_1$ be $d_0$. Note that the objective is equivalent to finding a sequence $\{b_i, 1 \leq i \leq n\}$ so that
\[ \sum_{i\in X} b_i = \sum_{j \in [n]-X} b_j \]and also that the problem statement is equivalent to $X$ satisfying
\[ \left| \sum_{i\in X} a_i - \sum_{j \in [n]-X} a_j \right| \]being the lowest possible among all set $X \subseteq [n]$, with the real numbrers $a_i$ fixed.
$\textcolor{green}{\textbf{\text{Main idea.}}}$ Bash for the win! We directly compute the terms
\[ S_d = \sum_{j\in Y} a_j - \sum_{i \in X} a_i = \sum_{i = 0}^{n-1} c_id_i \]in terms of $d_i$ (here, we let $Y = [n] - X$ for simplicity). Without Loss of Generality we can assume $X$ to be the set with the smaller sum.
\[
\begin{tabular}{cccccccccccc}
$a_i|\,\, i\in X$=&0&&$a_1$&&&&$a_3$\\
$a_j|\,\, j\in Y$=&&&&&$a_2$&&&&$a_4$&&$a_5$ \\
&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep \\
&&$d_0$&&$d_1$&&$d_2$&&$d_3$&&$d_4$ \\[3pt]
&&&&&&&&&&\text{Y-block} 
\end{tabular}
\]For example, when $X = \{1,3\}$ and $Y = \{2,4,5\}$, $S_d$ (the letters $S_d$ represents the sum of "differences" $d_i$) = $1 \cdot d_0+ 2 \cdot d_1 + 1 \cdot d_2 + 2 \cdot d_3 + 1 \cdot d_4$.
Note that if not all coefficients $c_i$ is of "unified" sign, e.g. there exists a positive and negative coefficient, we can $\textbf{easily}$ set the $d_i$ so that
\[ \sum_{i; c_i > 0} |c_i|d_i = \sum_{j; c_j<0} |c_j|d_j \](one way to do it is to set all $d_i$ in the $LHS$ to be $1$, all $d_j$ except one on the $RHS$ to be $\frac{0.001}{n^2}$ and adjust the one not yet filled accordingly.) In these cases, setting $b_i$ equal to the new $d_0+\cdots + d_{i-1}$ fulfills the objective.
So let all $c_i$ be positive or $0$, as these are the exceptional cases where we would like to investigate further (and ultimately, prove that these cases do not exist by the statement given in the problem.) Before we continue, we present a simple, but central observation.
$\textcolor{blue}{\textbf{\text{Observation.}}}$ Let $i \in X$ and $i+1 \in Y$. If $X' = X \cup \{i+1\} - \{i\}$, then $S_d' = S_d - 2d_i$.
$\textcolor{blue}{\textbf{\text{Proof.}}}$ Obvious [Just note that $S_d' = S_d - (a_{i+1} - a_{i}) + (a_{i} - a_{i+1})].$
$\rule{25 cm}{3pt}$
We have two different finishes from here.
$\textcolor{red}{\textbf{\text{Finish 1: Polished, after redoing the problem.}}}$ Divide the indices into $\textit{blocks}$ of $X$ and $Y$. For example, in the set $X = \{1,4,6\}$ and $Y = \{2,3,5,7\}$, set the $Y-$blocks to be $\{2,3\}, \{5\}$ and $\{7\}$. Define an $X-Y$ transitional index to be the last index in any $X$- block, and say the $X-Y$ transitional index of an $X-$block to be its final index. (this is another way of categorizing the indices in which we can make a switch in the above $\textcolor{blue}{\textbf{\text{Observation.}}}$). We claim the following:
$\textbf{Claim 1.}$ Each $X-Y$ transitional index $i$ has coefficient $c_i$ at least $1$.
$\textbf{Proof 1.}$ Consider the coefficient $c_{i+1}$ of $d_{i+1}$. Then, we can be sure that $c_{i} = c_{i+1} +1 \geq 0+1 = 1$. Here we use the observation that $c_i$ is the difference between "the amount of indices in $Y$" and "the amount of indices in $X$" greater than $i$.

$\textbf{Claim 2.}$ There are at most one $X-Y$ transitional index.
$\textbf{Proof 2.}$ If $i$ and $j$ are $X-Y$ transitional indices, then
\[ S_d(X) \geq c_id_i + c_jd_j \geq 1 \cdot d_i + 1 \cdot d_j \]Consider $X'(i)$ and $X'(j)$ after switching $i$ with $i+1$, or $j$ with $j+1$ respectively. Then, one of $S_d(X'(i)) \geq d_i - d_j$ and $S_d(X'(j)) \geq d_j-d_i$ is at least $0$, but less than $S_d$.
$\rule{25cm}{0.5pt}$
These two claims greatly delimit the possible configurations of $X$, as we can just let $X$ to be an interval $[i,j] \in [n]$ or a collection of two intervals $[a,b], [c,d]$ where $0<a<b<c<d = n$. However, the second case can also be easily ruled out as $X$ does not contain the largest number $a_n$.
On the first case, if $|X| > n-j$; i.e. there are more terms in $X$ rather than the ending $Y-$block, then the coefficient of $d_{i-1}$ is negative. Otherwise we can be sure that the $X-$ block is kind of small.
First subcase: if the ending $Y-$ block's size is $2$ or greater: then we know that $d_j$ is at least 2. Then consider $X'$ obtaining from switching $j$ with $j+1$.
Second subcase: ending $Y-$ block's size is $1$. Then, the size of the $X-$ block is also 1, and a little experimentation yields that choosing $X = \{n\}$ contradicts the $\textbf{original}$ problem statement. (While the value of $a_n$ alone can certainly be BIG, it cannot be BIG enough to overrun the initial difference)
(Note: configurations with no $X-Y$ transitional indices may be dealt with in the same way as post $\#2$ does it)

$\rule{25 cm}{3pt}$
$\textcolor{red}{\textbf{\text{Finish 2: Original 3 hour solution "Y-block maniac".}}}$ We refer to $Y-X$ transitional indices to be the last index in a (particular) $Y-$ block. We claim the following:
$\textbf{Claim 3.}$ The length of any $Y-$ block aside from the first one is exactly $1$.
$\textbf{Proof 3.}$ Let a $Y-$ block has length $l \geq 2$. Consider the transitional index of that block $t$. Now, using a similar principle to $\textbf{Claim 1}$ above we get $c_{t-l} = c_t+l \geq 0+l = l \geq 2$. Letting $X'$ with switching indices $t-l$ and $t-l+1$ and using the initial $\textcolor{blue}{\textbf{\text{Observation}}}$, a contradiction takes place.
$\rule{25cm}{0.5pt}$
This claim also delimits the possibilities of $X-$ blocks preceding each $Y-$ block, as each block must have length at least $1$. It is illogical for any $X-$ block to have length $2$ or more: as the leftmost coefficient of said block will inevitably be negative.
If we have $2$ or more $X-$ blocks, we now pick two which are $\{i\}$ and $\{j\}$. Let the index with the smaller "difference" be $i$ (i.e. $d_i \leq d_j$). Then there is an easy switch of index $i$ and $i+1$ which yields a contradiction.
The case where the only $X-$ block available is of length $1$ is handled similarly to $\textcolor{red}{\textbf{\text{Finish 1}}}$.
Motivation: This MUST hold; proceed onwards despite unyielding cases!
This post has been edited 6 times. Last edited by alexiaslexia, Nov 12, 2020, 6:28 AM
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TheUltimate123
1740 posts
#9 • 4 Y
Y by centslordm, sabkx, channing421, eulerleonhardfan
Solved with nukelauncher.

We will show there exists an increasing sequence \((b_1,\ldots,b_n)\) satisfying \[\sum_{i\in X}b_i=\sum_{i\notin X}b_i.\]This clearly suffices by scaling.

Assume without loss of generality that \(\sum_{i\in X}a_i<1\), since otherwise we replace \(X\) with \(\{1,\ldots,n\}\setminus X\). Then if \(n\in X\), we win automatically by letting \(b_i=a_i\) for \(i\le n-1\) and varying \(b_n\ge a_n\). Henceforth \(n\notin X\).

Claim: We have \[\sum_{i\in X}a_{i+1}>2-\sum_{i\in X}a_i.\]
Proof. Since \(\sum_{i\in X}a_{i+1}\) is greater than \(\sum_{i\in X}a_i\) but farther from 1 than \(\sum_{i\in X}a_i\), we have \(\sum_{i\in X}a_{i+1}\ge2-\sum_{i\in X}a_i\).

Suppose that equality holds. Evidently there is some \(i\in X\) with \(i+1\notin X\); then replacing \(i\) with \(i+1\) in \(X\) yields a sum closer to 1 than \(\sum_{i\in X}a_i\), contradiction. \(\blacksquare\)

Consider the sequence \((c_1,\ldots,c_n)\) defined by \(c_i=a_{i+1}\) for \(i\in X\) and \(c_i=a_i\) otherwise.

Claim: We have \[\sum_{i\in X}c_i>\sum_{i\notin X}c_i.\]
Proof. By the first claim, \[\sum_{i\in X}c_i=\sum_{i\in X}a_{i+1}>2-\sum_{i\in X}a_i=\sum_{i\notin X}a_i=\sum_{i\notin X}c_i,\]proving the claim. \(\blacksquare\)

Finally, we have \[\sum_{i\in X}a_i<\sum_{i\notin X}a_i\quad\text{and}\quad\sum_{i\in X}c_i>\sum_{i\notin X}c_i.\]We are free to vary \(a_i\le b_i<c_i\) for \(i\in X\), so evidently there is an increasing sequence \((b_1,\ldots,b_n)\) satisfying \[\sum_{i\in X}b_i=\sum_{i\notin X}b_i,\]and we are done.
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goodar2006
1347 posts
#10 • 2 Y
Y by centslordm, Acclab
Is this solution correct? :huh:

Note that since the $a_{i}$'s sum to $2$, set $\bar{X} = \{1, 2, \ldots, n\} - X$ also satisfies the problem's minimization condition. Therefore, WLOG we can assume that $\sum_{i \in X} a_{i} = 1 - \epsilon$, where $\epsilon > 0$. We consider two cases:

Case 1: There is an index $j$, $1 \le j \le n -1$, such that $a_{j} \in X$ and $a_{j + 1} \in \bar{X}$. In this case, we have $a_{j + 1} - a_{j} \ge 2\epsilon$. Otherwise, adding $j + 1$ to $X$ and removing $j$ from it yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{j} = a_{j} + \epsilon$, $b_{j + 1} = a_{j + 1} - \epsilon$, and $b_{k} = a_{k}$ for every other $1 \le k \le n$.

Case 2: $j > k$ for every $j \in X$ and $k \in \bar{X}$. Note that in this case $a_{1} \in \bar{X} \ge 2\epsilon$. Otherwise, adding $1$ to $X$ yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{n} = a_{n} + \epsilon$, $b_{1} = a_{1} - \epsilon$, and $b_{k} = a_{k}$ for every other $2 \le k \le n - 1$.
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Afo
1002 posts
#12 • 1 Y
Y by centslordm
Solution.

Let $|X|=p$. Let $X_1,...,X_p$ be the elements of $X$ in increasing order. WLOG
$$a_{X_1}+...+a_{X_p}=1+s$$where $s>0$. The main idea is to show that $(a_{X_1}-s)+a_{X_2}+..+a_{X_p}=1$ works. We finish the problem by proving three things.

1. $a_{X_1}-s$ is positive.
2. We need to show that the order is still preserved: $a_{X_1}-s>a_{X_1-1}$. Note that if $X_1=1$, we're done. We assume that $X_1 \ge 2$.
3. To make the sum $2$, we show that there is an $i \notin X$, such that $a_i+s<a_{i+1}$ (Again, to preserve the order).

Claim 1. $a_{X_1}-s$ is positive
Proof.
Because of the minimality condition,
$$a_{X_2}+..+a_{X_p}=s+1-a_{X_1} \leq 1-s$$$$2s\leq a_{X_1} \implies a_{X_1}-s>0$$Claim 2. $a_{X_1}-s>a_{X_1-1}$
Proof.
$$a_{X_2}+..+a_{X_p}+a_{X_1-1}=s+1-a_{X_1}+a_{X_1-1}\leq 1-s$$$$2s\leq a_{X_1}-a_{X_1-1} \implies a_{X_1}-s>a_{X_1-1}$$Claim 3. Point 3, above..
Proof.
let $a_i$ be the largest value such that $i \notin X$. If $a_i=a_n$, then just add $a_i$ by $s$. Otherwise, $a_{i+1} \in X$. Similar to Claim 2, we have that $a_{i+1}-a_i \ge 2s \implies a_i+s < a_{i+1}$.
This post has been edited 3 times. Last edited by Afo, Mar 23, 2021, 1:24 AM
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Mano
46 posts
#13 • 1 Y
Y by centslordm
In my opinion not that hard for an A3, but the amount of intimidating notation makes solvng the problem and writing up the solution considerably harder.
solution
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pad
1671 posts
#14 • 1 Y
Y by centslordm
Let $\sum_{i\in X} a_i = 1-\varepsilon$. WLOG $\varepsilon>0$, since the complement sums to $1+\varepsilon$.

Lemma 1: Suppose $i \in X$ and $j \not \in X$, and $i<j$. Then $a_j-a_i \ge 2\varepsilon$.

Proof: We know $a_i<a_j$ since the sequence is increasing. Suppose instead that $0<a_j-a_i< 2\varepsilon$. Then add $j$ to $X$ and remove $i$, so the new sum is $1-\varepsilon+a_j-a_i < 1+\varepsilon$. Now this new set is closer to $1$ than the original $X$ was, contradicting minimality. $\blacksquare$

Lemma 2: Suppose $j\not \in X$. Then $a_j \ge 2\varepsilon$.

Proof: If $a_j < 2\varepsilon$, then adding $j$ to $X$ makes the new sum between in the range $(1-\varepsilon, 1+\varepsilon)$, contradicting minimality. $\blacksquare$

Suppose there exists an index $i$ for which $a_i \in X$ and $a_{i+1} \not \in X$. By Lemma 1, $a_{i+1} - a_i \ge 2\varepsilon$. Create a new sequence where we increase $a_i$ by $\varepsilon$ and decrease $a_{i+1}$ by $\varepsilon$. This change preserves the strictly increasing property and all positive property, so we have a new sequence in which the $X$-indexed elements sum to $(1-\varepsilon)+\varepsilon=1$, finishing. (Technically, the change could make $a_i$ and $a_{i+1}$ equal, so increase $a_i$ by $\delta \ll \varepsilon$ less than $\varepsilon$ and increase some other $X$-indexed element by $\delta$, where $\delta$ is small enough so that this preserves strictly increasing.)

Else, we must have $X=\{k+1,\ldots,n\}$ for some $k\ge 1$. This guarantees that $a_1\not \in X$ and $a_n\in X$, so simply remove $\varepsilon$ from $a_1$ and add $\varepsilon$ onto $a_n$. This new sequence sums to $(1-\varepsilon)+\varepsilon=1$. It preserves all positive since $a_1 \ge 2\varepsilon>\varepsilon$ by Lemma 2, and it obviously preserves the strictly increasing property.

Remarks
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cmsgr8er
434 posts
#15 • 1 Y
Y by wzs26843545602
Note that either $S = \sum_{i\in X} a_i = 1 \pm \varepsilon,$ but since the sum is two let us WLOG consider $1-\varepsilon.$ Suppose for the sake of contradiction that no such sequence of $b_i$ exists.

Claim: For some integer $1\le k\le n,$ $a_1, a_2,\dots, a_k\not\in X$ and $a_{k+1}, \dots, a_n \in X.$

Proof. Suppose not. Consider $X$ such that $T = \sum_{i\in X} i$ is maximal. Then there must be some $i,j$ with $i<j$ and $a_j\not\in X$ and $a_i\in X.$ Note that $a_j-a_i\ne 2\varepsilon$ otherwise we could replace $i$ with $j$ in $X,$ contradicting maximality of $T$.

Moreover, if $a_j-a_i<2\varepsilon$ then replacing $i$ with $j$ in $X$ contradicts minimality of $S.$ As a result, $a_j-a_i > 2\varepsilon,$ so setting $b_i = a_i + \varepsilon$ and $b_j = a_j-\varepsilon$ and $b_k = a_k$ for $k\ne i,j$ gives a valid sequence of $b,$ contradiction. $\blacksquare$

Claim: For all $i\not\in X,$ $a_i \ge 2\varepsilon.$

Proof. If otherwise then add $i$ in $X,$ contradicting minimality. $\blacksquare$

Finally set,
\begin{align*}
b_i = a_i - \frac{\varepsilon}{k}  &\qquad 1\le i\le k  \\
b_i = a_i + \frac{\varepsilon}{n-k}  &\qquad k+1\le i\le n,
\end{align*}Whereby for $1\le i\le k,$ $b_i = a_i-\varepsilon/k \ge \varepsilon(2-1/k) > 0,$ giving a valid sequence for $b.$
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uvuvwevwevwe
12 posts
#16
Y by
I think I have quite a simple sketch of the solution but I have not worked it out on paper, so it might be a fake-solve.

Sketch:
We wish to prove the contrapositive of the statement. i.e. If the desired $\{ b_i \}$ does not exist then the first absolute sum is not minimised.
Lemma: The set $X$ satisfies the second condition iff $X$ neither majorizes $Y$ or be majorized by $Z$, where
$$Y=\{n, n-2, n-4, \dots ,2 \}, Z=\{n-1, n-3, n-5, \dots 1\} $$$$Y=\{n, n-2, n-4, \dots 1\}, Z=\{ n-1, n-3, n-5, \dots 2\}$$for even and odd $n$ respectively. Note that if $|A| > |B|$ then $B$ cannot majorize $A$, but $A$ can majorize $B$.

If $X$ majorizes $Y$, then
\[
\sum_{i \in X} a_{i} > \sum_{i \in Y} a_{i} > 1 \implies \left|1-\sum_{i \in X} a_{i}\right| > \left|1-\sum_{i \in Y} a_{i}\right|
\]If $X$ is majorized by $Z$, then
\[
\sum_{i \in X} a_{i} < \sum_{i \in Z} a_{i} < 1 \implies \left|1-\sum_{i \in X} a_{i}\right| > \left|1-\sum_{i \in Z} a_{i}\right|
\]which will imply that the first sum is not minimised anyways. Contradiction.
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CANBANKAN
1301 posts
#17 • 1 Y
Y by sabkx
The idea is to characterize all possible $X$. I am minimizing $$|\sum_{i\in [n]\backslash X} a_i - \sum_{i\in X} a_i|$$
If $X=\{x_1,\cdots,x_t\}$ and there exists $z_j\in [n]\backslash X$ such that $z_j>x_j$, then I should swap $(z_i,x_i)$ to reduce that absolute value. There is a similar condition for $[n]\backslash X$ (*)

Now, our goal is to construct $c_1<c_2<\cdots<c_n, d_1<\cdots<d_n$ such that

$$\sum\limits_{i\in [n]\backslash X} c_i - \sum\limits_{i\in X} c_i  < 0 < \sum\limits_{i\in [n]\backslash X} d_i - \sum\limits_{i\in X} d_i $$
Then let $-A=\sum\limits_{i\in [n]\backslash X} c_i - \sum\limits_{i\in X} c_i$, $B=\sum\limits_{i\in [n]\backslash X} d_i - \sum\limits_{i\in X} d_i$, then setting $$b_1:b_2:\cdots:b_n = Bc_1+Ad_1 : Bc_2+Ad_2 : \cdots : Bc_n+Ad_n$$works

WLOG $n\in [n]\backslash X$, then set $d_j=j$ for all $1\le j\le n-1$ and $d_n=10^{10^n}$

It remains to construct $c_i$. Let $X=\{x_1,\cdots,x_t\}, [n]\backslash X=\{y_1,\cdots,y_{n-t}\}$, and consider the smallest integer $j$ such that $x_{t-j}>y_{n-t-j}$. It must exist by (*), then consider the following construction: for all $k>x_{t-j}, c_k=10^{10^n}+k\epsilon^2 10^{-10^n}$

$x_{t-j}=10^{10^n}, y_{n-t-j}=10$ and for all $k<y_{n-t-j}$, $c_k=k\epsilon^2 10^{-10^n}$

Check this works.
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Number1048576
91 posts
#18
Y by
solution
This post has been edited 1 time. Last edited by Number1048576, Jun 25, 2022, 2:51 PM
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awesomeming327.
1717 posts
#19
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Let the elements of $X$ be $c_1$, $c_2$, $c_3$, $\dots$, $c_i$. Without loss of generality, we can assume that \[a_{c_1}+a_{c_2}+a_{c_3}+\dots+a_{c_i}=1+d > 1\]If $d=0$ we are already done by making the sequence $a$ the same as the sequence $b$.

Now, we will show that we can decrease a variable in $a_X$ by $s$, increase a variable not in $a_X$ by $s$, and maintain the relation $0<a_1<a_2<\dots < a_{n-1}<a_n$. Let us decrease $a_{c_j}$ by $d$, then we will show that $a_{c_j} - d > a_{c_j-1}$. Clearly, \[a_{c_1}+\dots+a_{c_{j-1}}+a_{c_j-1}+a_{c_{j+1}}+\dots+c_i\ge 1-d\]due to minimality. Now, we have $a_{c_j}-a_{c_j-1}\ge 2d$ which means we can subtract $d$ from $a_{c_j}$ without getting into any trouble. The same applies to adding $d$ to a non-$a_X$.
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asdf334
7585 posts
#20
Y by
Idea is somewhat similar to 2022 USAJMO 2.

Essentially, if there exist sequences $(b_1,b_2,\dots,b_n)$ and $(c_1,c_2,\dots,c_n)$ where the second condition has the first sum less and the second sum greater, some sort of smoothing argument that changes $b_i$ to $c_i$ will give a correct value. Then $X$ must be majorized by its complement (WLOG) and we can find some better subset than $X$ by replacing an element of $X$ with the corresponding element in $X^C$. Done.

Example
This post has been edited 1 time. Last edited by asdf334, Mar 11, 2023, 2:15 AM
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thdnder
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Hopefully, it isn't fakesolve.

WLOG, we can assume $\sum_{i \in X} a_i < 1$. Let $\varepsilon = 1 - \sum_{i \in X} a_i$.Let $i_1 < i_2 < \dots < i_k$ be the elements of $X$. We split the problem into two cases:

Case 1: $n \in X$.

If there exists $s$ such that $i_{s + 1} > i_s + 1$, then $i_s + 1 \not\in X$, which means that $\sum_{j \in X \backslash \{i_s\} \cup \{i_s + 1\}} a_j \ge 1 + \varepsilon$, thus $a_{i_s + 1} \ge a_{i_s} + 2\varepsilon$, so replacing $a_n \to a_n + \varepsilon$ and $a_{i_s + 1} \to a_{i_s + 1} - \varepsilon$ works.

If there doesn't exists such $s$, then $X = \{n, n - 1, n - 2, \dots, n - k + 1 \}$. Then we can easily scale it.

Case 2: $n \not\in X$.

Let $m = i_k$. Then considering $X \backslash \{m\} \cup \{m+1\}$, we get $a_{m+1} - a_m \ge 2\varepsilon$. If $a_{m+1} > a_m + 2\varepsilon$, then replacing $a_{m+1} \to a_{m+1} - \varepsilon$ and $a_m \to a_m + \varepsilon$ works. Thus we may assume $a_{m+1} - a_m = 2\varepsilon$.

If there exists $k \neq j \in X$, then replace $a_{m+1} \to a_{m+1} - \varepsilon$ and $a_m \to a_m + \varepsilon_1$ and $a_j \to a_j + \varepsilon - \varepsilon_1$ and take $\varepsilon_1$ such that $a_{j+1} > a_j + \varepsilon - \varepsilon_1$.

Otherwise, since $n \ge 3$, there exists $j$ different from $k, k + 1$ and $j \not\in X$. Then replace $a_m \to a_m + \epsilon$ and $a_{m+1} \to a_{m+1} - \varepsilon_1$ and $a_j \to a_j - (\varepsilon - \varepsilon_1)$ and take $\varepsilon_1$ such that $\varepsilon - \varepsilon_1$ is close to $0$. Then it works perfectly.

Hence, in all cases, we can scale the sequence $(a_i)_{1 \le i \le n}$, so we're done. $\blacksquare$
This post has been edited 1 time. Last edited by thdnder, Aug 10, 2024, 4:32 PM
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vsamc
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Solution
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asdf334
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Obviously redefine $X$ to be a subset of $\{a_1,\dots,a_n\}$. Let $s=\displaystyle{\sum_{i\in X}i}<1$ and let $Y$ be the complement of $X$.
Suppose there exist $y\in Y$ and $x\in X$ where $y>x$. Then consider set $X'$ formed by removing $x$ and adding $y$. Clearly
\[s-x+y\ge 2-s\implies y-x\ge 2-2s.\]Now if there exist $y\in Y$ and $x\in X$ where $y>x$ at all, then there must exist $i$ such that $a_{i+1}\in Y$ and $a_i\in X$. Consider the sequence $b$ where
  • If $j\not\in \{i,i+1\}$ then $b_j:=a_j$.
  • We have $b_i:=a_i+(1-s)$ and $b_{i+1}:=a_{i+1}-(1-s)$.
This works provided that $a_{i+1}-a_i\ne 2-2s$. In this case set $b_i=a_i+(1-s)-\epsilon$ instead and increase some other element in $X$ by $\epsilon$.
Now suppose that there do not exist $y\in Y$ and $x\in X$ such that $y>x$. In this case divide all values by $2-s$ so that the sum of values in $Y$ is $1$. Now increase the largest value in $X$ by
\[1-\frac{s}{2-s}=\frac{2-2s}{2-s}\]and we are done. $\blacksquare$
This post has been edited 1 time. Last edited by asdf334, Apr 27, 2024, 12:35 AM
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HamstPan38825
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Giving explicit constructions is too hard :(

We will characterize all sets $X$ that could possibly satisfy the given condition, then show that such a sequence $\{b_i\}$ exists for all such sets $X$. We say that a set $X$ dominates a set $Y$ if we can assign a distinct $f(y) \in X$ for each $y \in Y$ such that $f(y) > y$ for all $y$.

Claim. [First Part] For any sequence $\{a_i\}$, the given value can only be minimized when neither $X$ nor $X^c$ dominate each other.

Proof. Assume otherwise, and suppose $X^c$ dominates $X$. Let $\sum_{i \in X} a_i = 1-\varepsilon$ for $\varepsilon > 0$.

If $|X| = 1$, assume for the sake of contradiction that $i < n$. It follows that $1-\varepsilon = a_i < a_n \leq 1+\varepsilon$, so $a_n$ is closer to $1$ than $a_i$, contradiction.

Suppose that $|X| \geq 2$. It follows that for every index $i \in S$, the difference $a_{f(i)} - a_i$ is at least $2\varepsilon$, as otherwise, by replacing $i$ with $f(i)$ in $X$ we get a sum closer to $1$. Hence \[\sum_{j \in X^c} a_j - \sum_{i \in X} a_i \geq 2\varepsilon |X| > 2\varepsilon\]yields a contradiction. $\blacksquare$

Claim. [Second Part] If neither $X$ nor $X^c$ dominates the other, then we can find such a sequence $\{b_i\}$.

Proof. Let $|X| = r$. It suffices to construct a sequence $\{b_i\}$ such that $\sum_{i \in X} b_i = \sum_{j \in X^c} b_j$, as we can scale by a suitable amount. Thus, we will instead construct a sequence $(c_1, c_2, \dots, c_n)$ with $c_i > 0$ for each $i$ such that
\[\sum_{i=0}^{r-1} i \sum_{j=x_i}^{x_{i+1}} c_j = \sum_{i=0}^{r-1} (n-r-i) \sum_{j = y_i}^{y_{i+1}} c_j\]where $\{x_i\}$ is the increasing sequence of all indices in $X$ and $\{y_i\}$ is the increasing sequence of all indices in $X^c$. Then, taking $b_i = c_1+c_2+\cdots+c_i$ for each $i$ does the trick. (Here $c_i$ is just the difference sequence of $b_i$.)

Notice that every $c_i$ appears $f(i)$ times in the LHS sum, where $f(i)$ denotes the number of elements in $X$ at least $a_i$. This element is then counted $n-i+1 - f(i)$ times in the RHS sum, as it is counted $n-i+1$ times total. So it suffices to demonstrate such a sequence $\{c_i\}$ with $c_i > 0$ satisfying only the condition \[\sum_{i=1}^n c_i(2f(i) - n+i-1) = 0.\]There obviously exists such a $\{c_i\}$ as long as $2f(i) - n+i-1 > 0$ and $2f(j) - n+j-1 < 0$ for some $i, j$. Assume for the sake of contradiction that $2f(i) \leq n-i+1$ for all $i$; the other case is analogous, say by swapping $X$ with $X^c$. We will arrive at a contradiction by inductively pairing off every index $x_i \in X$ with a $y_j \in X^c$ with $x_i \leq y_j$.

For our base case, notice that for $i=n$, the inequality reads $2f(n) \leq 1$, i.e. $f(n) = 0$, so $n \not \in X$. So we can pair $x_r$ with $y_{n-r} = n$. For the inductive step, for $i = x_j$ the inequality reads $f\left(x_j\right) \leq \frac{n-x_j + 1}2$. In particular, there are at least $\frac{n-x_j + 1}2$ elements $y_i$ among $x_j + 1, \dots, n$. Because there are only at most $\frac{n-x_j-1}2$ elements $x_i$ among these which we have paired off with an equal number of $y_j$, there exists at least $\frac{n-x_j+1}2 - \frac{n-x_j-1}2 = 1$ unpaired $y_\ell$, which we can pair with our $x_j$, completing the induction.

Thus we can create a pairing $(x_i, y_j)$ for all $x_i \in X$, which contradicts the assumption. It follows that such a sequence $\{c_i\}$ must exist, completing the problem.
This post has been edited 1 time. Last edited by HamstPan38825, Jun 17, 2024, 7:11 PM
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Acclab
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Why do everyone's solution include majorization? I find the construction the same as @goodar2006, which I think is quite intuitive.

Call $X$ good if there is $a_n$ satisfying the constraint so $ | \sum_{i \in X} a_i - 1 | $ is minimised. We wish to modify $a_n$ while keeping its stricty monoticity and sum, so that $\sum_{i \in X} a_i=1$. Since $X$ is good if and only if its complement $\bar{X}$ is, we can assume without loss of generality that $\sum_{i \in X} a_i < 1$.

Heuristic


The rigorous way of writing the solution follows:
Pick $k$ so $k \in X$ but $k \notin X$, we claim $a_k + (1-s) < a_{k+1} - (1-s)$, where $s = \sum_{i \in X} a_i$. Indeed, let $X' = X + \{k+1\} - \{k\}$, then by minimality we have $|s - a_k + a_{k+1} - 1| \geq |s - 1| = 1-s$. This simplifies to $a_{k+1}-(1-s) \geq a_k+(1-s)$. So we can let $b_k = a_k+1-s$, $b_{k+1} = a_{k+1}-(1-s)$ and $b_m=a_m$ otherwise. This keeps the monoticity except when equality is a held, but in this case we can decrease $a_k$ by some $\epsilon$ and then add to any other $a_i$ with $i \in X$. This is unless $X$ only has one element (which is why $n = 2$ is excluded). If this is the case, we have two cases:

1. $X = \{1\}$. Then $2 - (a_3 + ...) = a_1 + a_2 > 2 - a_1 \implies a_3 \leq a_3+...<a_1$, a contradiction.

2. $X = \{m\}$ for some $m \geq 2$. If $a_m \leq \frac{1}{2}$, then $a_1 + a_m < 1$ which is closer to $1$ than $a_m$, contradiction. If $a_m > \frac{1}{2}$, then $a_1 + a_m > 2 - a_m \implies 2-(a_1+a_m) < a_m$. L.H.S. is the sum of all $a_i$ except $i=1,m$, so this it must be the case that $m = n$. In this case, we can still indeed construct $b_i$ so $b_n = 1$ while keeping $0<b_1 < b_2 < ... < b_n$.

If we cannot pick such $k$ describe above, then $X = \{m, m+1, ..., n\}$ for some $m$ (X = $\phi$ is not possible). By using the minimality on the set $X = X + \{ a_1 \}$ we have that $a_1 - (1-s) \geq 1-s > 0$, so we can subtract $1-s$ from $a_1$ and add to $a_n$, keeping the positivity, strict monoticity and giving the desired sum.
This post has been edited 1 time. Last edited by Acclab, Jul 14, 2024, 2:38 PM
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Acclab
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#28 • 1 Y
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goodar2006 wrote:
Is this solution correct? :huh:

Note that since the $a_{i}$'s sum to $2$, set $\bar{X} = \{1, 2, \ldots, n\} - X$ also satisfies the problem's minimization condition. Therefore, WLOG we can assume that $\sum_{i \in X} a_{i} = 1 - \epsilon$, where $\epsilon > 0$. We consider two cases:

Case 1: There is an index $j$, $1 \le j \le n -1$, such that $a_{j} \in X$ and $a_{j + 1} \in \bar{X}$. In this case, we have $a_{j + 1} - a_{j} \ge 2\epsilon$. Otherwise, adding $j + 1$ to $X$ and removing $j$ from it yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{j} = a_{j} + \epsilon$, $b_{j + 1} = a_{j + 1} - \epsilon$, and $b_{k} = a_{k}$ for every other $1 \le k \le n$.

Case 2: $j > k$ for every $j \in X$ and $k \in \bar{X}$. Note that in this case $a_{1} \in \bar{X} \ge 2\epsilon$. Otherwise, adding $1$ to $X$ yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{n} = a_{n} + \epsilon$, $b_{1} = a_{1} - \epsilon$, and $b_{k} = a_{k}$ for every other $2 \le k \le n - 1$.

I think this is almost correct except when an equality case occurs (in your case 1, "closer to $1$ than $X$" is almost true, but it could be the case that they are both equidistant).
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maromex
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Let $S_n = \{1, 2, 3, \ldots n\}$. Then \[
\left|1-\sum_{i \in S_n \backslash X} a_{i}\right| = \left|1 - (2 - \sum_{i \in X} a_i)\right| = \left|1 - \sum_{i \in X} a_i \right|
\]Both $S_n \backslash X$ and $X$ achieve the minimum value. Also, $(1 - \sum\limits_{i \in S_n \backslash X} a_i) + (1 - \sum\limits_{i \in X} a_i) = 2 - \sum\limits_{i \in S} a_i = 0$, which implies at least one of these addends is nonnegative. We may assume without loss of generality that $1 - \sum\limits_{i \in X} a_i \ge 0 \quad (1)$. Also, let $\epsilon = 1 - \sum\limits_{i \in X} a_i$. The value of $\epsilon$ only depends on the values in sequence $\{a_n\}$.

We will proceed by induction. Let $k$ be the largest element of $X$.

-- Base case: Goal holds for $k = n$

Our approach is to scale all terms before $a_n$ by scalar $s_1$ and scale $a_n$ by scalar $s_2$ for the construction of $\{ b_n \}$. For the condition $\sum\limits_{i \in X} b_i = 1$ to hold, we must have \[ s_1(1 - \epsilon - a_n) + s_2a_n = 1 \tag{1.1} \]For the sum of terms to be 2, we must have \[ s_1(2 - a_n) + s_2a_n = 2 \tag{1.2} \]Take (1.2) - (1.1) to get $$s_1(1 + \epsilon) = 1$$Thus, $$s_1 = \frac{1}{1 + \epsilon}$$Solving for $s_2$ in (1.2) gives $$s_2 = \frac{1}{a_n}(2 - \frac{2-a_n}{1+ \epsilon}) = \frac{1}{a_n}(\frac{2\epsilon + a_n}{1 + \epsilon}) = s_1 + \frac{2\epsilon}{a_n(1 + \epsilon)} > s_1$$Thus $s_2a_n > s_1a_{n-1} > s_1a_{n-2} > s_1a_{n-3} > \ldots$ We construct $\{ b_n \}$ with $b_i = s_1a_i \quad\forall i < n$, and $b_n = s_2a_n$. This sequence satisfies all of the requirements by construction.

-- Induction case: If the goal holds for $k = m$, it also holds for $k = m-1$ for all $2 \le m \le n$

We assume the goal holds for $k = m$. For the case where $k = m-1$, observe that \[ \sum\limits_{i \in (X \backslash \{a_{m-1}\}) \cup \{a_m\}} a_i = a_m - a_{m-1} + \sum\limits_{i \in X} a_i \tag{2.1}\]Also, it is given that the minimum value of $\left|1-\sum\limits_{i \in T} a_{i}\right|$ is achieved when $T = X$. Thus, we must have \[ \left| 1 - \sum\limits_{i \in (X \backslash \{ a_{m-1} \}) \cup \{ a_m \}} a_i \right| \ge \left| 1 - \sum\limits_{i \in X} a_i\right| \]After application of (1) and (2.1), this is equivalent to \[ \left| 1 - a_m + a_{m-1} - \sum\limits_{i \in X} a_i  \right| \ge 1 - \sum\limits_{i \in X} a_i \tag{2.2} \]Observe that $a_m > a_{m-1}$ because $\{a_n \}$ is strictly increasing. This means $ 1 - a_m + a_{m-1} + \sum\limits_{i \in X} a_i < 1 - \sum\limits_{i \in X} a_i$. This shows that for (2.2) to hold true, $1 - a_m + a_{m-1} + \sum\limits_{i \in X} a_i$ cannot be positive. We may transform (2.2) as follows: \[ 1 - a_m + a_{m-1} - \sum\limits_{i \in X} a_i \le \sum\limits_{i \in X} a_i - 1\]\[ - a_m + a_{m-1} \le 2\sum\limits_{i \in X} a_i - 2\]\[ - a_m + a_{m-1} \le -2\epsilon\]\[ a_m \ge a_{m-1}  +2\epsilon \tag{2.3}\]Now, if equality holds, that implies \[ \left| 1 - \sum\limits_{i \in (X \backslash \{ a_{m-1} \}) \cup \{ a_m \}} a_i \right| = \left| 1 - \sum\limits_{i \in X} a_i\right| \]This means that the minimum value of $\left|1 - \sum\limits_{i \in T} a_i \right|$ is achieved by $T = (X \backslash \{ a_{m-1} \}) \cup \{ a_m \}$. Applying the inductive hypothesis on $(X \backslash \{ a_{m-1} \}) \cup \{ a_m \}$ finishes for the equality case.

If equality does not hold for (2.3), we have $a_m > a_{m-1} + 2\epsilon \quad (2.4)$. Then we may construct $\{b_n \}$ with $b_i = a_i \quad\forall i \notin \{ m, m-1 \}$, $b_{m-1} = a_{m-1} + \epsilon$, and $b_m = a_m - \epsilon$. The sum of terms is 2 by construction. The sequence is strictly increasing due to (2.4), and otherwise by construction. We may use the definition of $\epsilon$ to get $\sum\limits_{i \in X} a_i = 1 - \epsilon$. Observe that $\sum\limits_{i \in X} b_i = \epsilon + \sum\limits_{i \in X} a_i = 1$, as the only difference between the sums is the $+ \epsilon$ for $b_{m-1}$. Thus, $\{b_n \}$ satisfies all of the requirements.
This post has been edited 2 times. Last edited by maromex, Apr 4, 2025, 1:59 AM
Reason: fixed some syntax issues alright latex PLEASE WORK
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john0512
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#30
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OMG SO GOOD

We say that a subset $S$ of $[n]$ is dominating if, for all $i$, at least as many of the last $i$ numbers are in $S$ as not in $S$. Similarly, we say that $S$ is dominated if $[n]\backslash S$ is dominating.

Main Claim: There exists a strictly increasing sequence $b_i$ with sum $2$ and $\sum_{i\in S}b_i=1$ if and only if $S$ is neither dominating nor dominated.

If $S$ is dominating, then we can match $S$ to elements $[n]\backslash S$ such that in every pair the former element is larger, so it has sum greater than $1$, similarly if $S$ is dominated, it has sum less than $1$.

Consider if $S$ is neither. Assume the largest element does not belong in $S$ (otherwise flip it). Then, $\sum_{i\in S}b_i$ can be arbitrarily small by letting $a_n$ be close to $2$. However, since $S$ is not dominated, there is some $k$ where the last $k$ elements have a majority in $S$. By setting the first $n-k$ elements close to $0$ and the last $k$ close to $\frac{2}{k}$, we can obtain a sum greater than $1$. Thus by IVT a sum of $1$ is attainable.

Thus, it remains to show that the closest subset sum to $1$ cannot be achieved by a dominating subset. Consider a dominating matching
$$(x_1,y_1),(x_2,y_2),\dots,(x_k,y_k)$$with $x_i>y_i$ for all $i$ (pad the dominated subset with zeroes if it runs out).

Let $d_i=x_i-y_i>0$. Clearly $k\geq 2$ since a dominating subset must contain at least half of the elements. Thus, flipping a single $d_i$ to negative will clearly decrease $|\sum d_i|$, since all of the $d_i$ pointing in the same sign will maximize it. Thus, by swapping one of the pairs we obtain a closer to equal distribution, contradicting minimality. We are done.
This post has been edited 1 time. Last edited by john0512, May 2, 2025, 6:12 PM
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