Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
distance of a point from incircle equals to a diameter of incircle
parmenides51   4
N 22 minutes ago by LeYohan
Source: 2019 Oral Moscow Geometry Olympiad grades 8-9 p1
In the triangle $ABC, I$ is the center of the inscribed circle, point $M$ lies on the side of $BC$, with $\angle BIM = 90^o$. Prove that the distance from point $M$ to line $AB$ is equal to the diameter of the circle inscribed in triangle $ABC$
4 replies
parmenides51
May 21, 2019
LeYohan
22 minutes ago
f(a + b) = f(a) + f(b) + f(c) + f(d) in N-{O}, with 2ab = c^2 + d^2
parmenides51   8
N 3 hours ago by TiagoCavalcante
Source: RMM Shortlist 2016 A1
Determine all functions $f$ from the set of non-negative integers to itself such that $f(a + b) = f(a) + f(b) + f(c) + f(d)$, whenever $a, b, c, d$, are non-negative integers satisfying $2ab = c^2 + d^2$.
8 replies
parmenides51
Jul 4, 2019
TiagoCavalcante
3 hours ago
Functional Inequality Implies Uniform Sign
peace09   33
N 3 hours ago by ezpotd
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
33 replies
peace09
Jul 17, 2024
ezpotd
3 hours ago
Labelling edges of Kn
oVlad   1
N 4 hours ago by TopGbulliedU
Source: Romania Junior TST 2025 Day 2 P3
Let $n\geqslant 3$ be an integer. Ion draws a regular $n$-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the $n$-gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
1 reply
oVlad
May 6, 2025
TopGbulliedU
4 hours ago
c^a + a = 2^b
Havu   8
N 4 hours ago by MathematicalArceus
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
8 replies
Havu
May 10, 2025
MathematicalArceus
4 hours ago
Concurrence of lines defined by intersections of circles
Lukaluce   1
N 4 hours ago by sarjinius
Source: 2025 Macedonian Balkan Math Olympiad TST Problem 2
Let $\triangle ABC$ be an acute-angled triangle and $A_1, B_1$, and $C_1$ be the feet of the altitudes from $A, B$, and $C$, respectively. On the rays $AA_1, BB_1$, and $CC_1$, we have points $A_2, B_2$, and $C_2$ respectively, lying outside of $\triangle ABC$, such that
\[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\]If the intersections of $B_1C_2$ and $B_2C_1$, $C_1A_2$ and $C_2A_1$, and $A_1B_2$ and $A_2B_1$ are $A', B'$, and $C'$ respectively, prove that $AA', BB'$, and $CC'$ have a common point.
1 reply
Lukaluce
Apr 14, 2025
sarjinius
4 hours ago
Factorial Divisibility
Aryan-23   45
N 4 hours ago by MathematicalArceus
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
$$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$
45 replies
Aryan-23
Jul 9, 2023
MathematicalArceus
4 hours ago
Multiple of multinomial coefficient is an integer
orl   14
N 4 hours ago by mickeymouse7133
Source: Romanian Master in Mathematics 2009, Problem 1
For $ a_i \in \mathbb{Z}^ +$, $ i = 1, \ldots, k$, and $ n = \sum^k_{i = 1} a_i$, let $ d = \gcd(a_1, \ldots, a_k)$ denote the greatest common divisor of $ a_1, \ldots, a_k$.
Prove that $ \frac {d} {n} \cdot \frac {n!}{\prod\limits^k_{i = 1} (a_i!)}$ is an integer.

Dan Schwarz, Romania
14 replies
orl
Mar 7, 2009
mickeymouse7133
4 hours ago
Functional Equation from IMO
prtoi   1
N 4 hours ago by KAME06
Source: IMO
Question: $f(2a)+2f(b)=f(f(a+b))$
Solve for f:Z-->Z
My solution:
At a=0, $f(0)+2f(b)=f(f(b))$
Take t=f(b) to get $f(0)+2t=f(t)$
Therefore, f(x)=2x+n where n=f(0)
Could someone please clarify if this is right or wrong?
1 reply
prtoi
4 hours ago
KAME06
4 hours ago
can you solve this..?
Jackson0423   1
N 4 hours ago by GreekIdiot
Source: Own

Find the number of integer pairs \( (x, y) \) satisfying the equation
\[ 4x^2 - 3y^2 = 1 \]such that \( |x| \leq 2025 \).
1 reply
Jackson0423
May 8, 2025
GreekIdiot
4 hours ago
Gergonne point Harmonic quadrilateral
niwobin   0
4 hours ago
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
0 replies
niwobin
4 hours ago
0 replies
Combi that will make you question every choice in your life so far
blug   1
N 5 hours ago by HotSinglesInYourArea
$A$ and $B$ are standing in front of the room in which there is $C$. They know that there is a chessboard in the room and that on every square there is a coin. Every coin is black on one side and white on the other side and is flipped randomly. $A$ enters the room and then $C$ points at exactly one square on the chessboard. After that, $A$ must flip exactly one coin of his choice on the chessboard to the other side and leave. Finally, $B$ enters the room ($A$ and $B$ haven't met again after $A$ entered the room) and he has to guess which square did $C$ point at.
What strategy do $A$ and $B$ have that will make this happen every time?
1 reply
blug
Yesterday at 5:46 PM
HotSinglesInYourArea
5 hours ago
Functional equation
Pmshw   17
N 5 hours ago by arzhang2001
Source: Iran 2nd round 2022 P2
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for any real value of $x,y$ we have:
$$f(xf(y)+f(x)+y)=xy+f(x)+f(y)$$
17 replies
Pmshw
May 8, 2022
arzhang2001
5 hours ago
Hard Geometry
Jalil_Huseynov   3
N 5 hours ago by bin_sherlo
Source: DGO 2021, Individual stage, Day1 P3
Let triangle $ABC$ be a triangle with incenter $I$ and circumcircle $\Omega$ with circumcenter $O$. The incircle touches $CA, AB$ at $E, F$ respectively. $R$ is another intersection point of external bisector of $\angle BAC$ with $\Omega$, and $T$ is $\text{A-mixtillinear}$ incircle touch point to $\Omega$. Let $W, X, Z$ be points lie on $\Omega$. $RX$ intersect $AI$ at $Y$ . Assume that $R \ne X$. Suppose that $E, F, X, Y$ and $W, Z, E, F$ are concyclic, and $AZ, EF, RX$ are concurrent.
Prove that
$\bullet$ $AZ, RW, OI$ are concurrent.
$\bullet$ $\text{A-symmedian}$, tangent line to $\Omega$ at $T$ and $WZ$ are concurrent.

Proporsed by wassupevery1 and k12byda5h
3 replies
Jalil_Huseynov
Dec 26, 2021
bin_sherlo
5 hours ago
Mount Inequality erupts on a sequence :o
GrantStar   89
N Apr 24, 2025 by sangsidhya
Source: 2023 IMO P4
Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that
\[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$
89 replies
GrantStar
Jul 9, 2023
sangsidhya
Apr 24, 2025
Mount Inequality erupts on a sequence :o
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 IMO P4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sami1618
910 posts
#84 • 2 Y
Y by GeoKing, cubres
We show that we can not have $a_n=k$, $a_{n+1}=k+1$, and $a_{n+2}=k+2$, which is sufficient. Let $s=x_1+x_2+\dots+x_n$ and $t=\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}$. Then $$k^2=st$$$$(k+1)^2=(s+x_{n+1})(t+\frac{1}{x_{n+1}})\iff  2k=t\cdot x_{n+1}+s\cdot \frac{1}{x_{n+1}}$$$$(k+2)^2=(s+x_{n+1}+x_{n+2})(t+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})\iff 2k+2=(t+\frac{1}{x_{n+1}})\cdot x_{n+2}+(s+x_{n+1})\cdot \frac{1}{x_{n+1}}$$Applying AM-GM to the third equality we must have that $t\cdot x_{n+1}=s\cdot \frac{1}{x_{n+1}}\iff x_{n+1}=\frac{\sqrt{s}}{\sqrt{t}}$. Applying AM-GM to the fifth equality we must have that $(t+\frac{1}{x_{n+1}})\cdot x_{n+2}=$ $(s+x_{n+1})\cdot \frac{1}{x_{n+1}}\iff x_{n+2}=\frac{\sqrt{s}}{\sqrt{t}}$, a contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Sammy27
83 posts
#85 • 2 Y
Y by Eka01, cubres
Solution
This post has been edited 3 times. Last edited by Sammy27, Jul 17, 2024, 10:14 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dkedu
180 posts
#86 • 1 Y
Y by cubres
We will show that $a_{n+2} \ge a_n + 3$ which implies the result since $a_1 = 1$.

By Cauchy-Schwarz we have \[a_{n+2} = \sqrt{[(x_1+x_2+\dots+ x_n) + (x_{n+1} + x_{n+2})]\left[\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+ \frac 1{x_n} \right)+ \left(\frac{1}{x_{n+1}} + \frac{1}{x_{n+2}}\right)\right]} \ge \sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)} + \sqrt{\frac{(x_{n+1} + x_{n+2})^2 }{x_{n+1}x_{n+2}}} > a_n + 2\]where the final inequality is strict since $x_{n+1} \neq x_{n+2}$ so we have the result $a_{n+2} \ge a_n + 3$ since $a_i$ is an integer for all $i$ so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
brainfertilzer
1831 posts
#87 • 1 Y
Y by cubres
Note that $a_1 = 1$ so it suffices to show $a_{n+2}\ge a_n + 3$ for all $n$ and we'll be done by induction. Fix $n$ and let $S = \sum_{i\le n} x_i$ and $T = \sum_{i\le n}1/x_i$. Relabel $x_{n+1}, x_{n+2}$ as $p,q$ for convenience of typesetting. Then note that
\[ a_{n+2} = \sqrt{\left(S + p + q\right)\left(T + \frac{1}{p} + \frac{1}{q}\right)}\stackrel{\text{CS}}{\ge} \sqrt{ST} + \sqrt{p\cdot \frac{1}{p}} + \sqrt{q \cdot \frac{1}{q}} = a_n + 2.\]But note that equality only holds if $S/T = p/(1/p) = q/(1/q)$. But this means $p^2 = q^2\implies p = q$, impossible since the $x_i$ are distinct. Hence equality doesn't hold and we have $a_{n+2} > a_n + 2$. Since the $a_i$ are integers, this means $a_{n+2}\ge a_n + 3$ as needed.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
N3bula
277 posts
#88 • 1 Y
Y by cubres
Proceed by induction, clearly we have that $a_1=1$, I will now prove $a_{n+2}>a_{n}+2$ which gives $a_{n+2}\geq a_n+3$.
We get:
\[a_{n+2}^2=a_n^2+\left( \frac{1}{x_{n+1}} +\frac{1}{x_{n+2}} \right)(x_{n+1}+x_{n+2})+(x_{n+1}+x_{n+2})\left(\sum_{i=1}^{n}\frac{1}{x_i}\right)+\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)\left(\sum_{i=1}^{n}x_i\right)\]Now by two applications of am-gm and the fact $a_{n+1}\neq a_{n+2}$ we get:
\[a_{n+2}^2> a_n^2+4+2a_n\sqrt{\left( \frac{1}{x_{n+1}} +\frac{1}{x_{n+2}} \right)(x_{n+1}+x_{n+2})}\]A final am-gm application gives:
\[a_{n+2}^2>a_n^2+4a_n+4\]Which suffices.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bobthesmartypants
4337 posts
#89 • 1 Y
Y by cubres
Intimidating-looking problem that ends up being not so bad at all.

We use induction to show $a_{2n-1}\ge 3n-2$ and $a_{2n}\ge 3n$. Base cases $a_1=1$ is trivial, and $a_2\ge 3$ follows from Cauchy-Schwarz $$a_2= \sqrt{(x_1+x_2)\left(\frac{1}{x_1}+\frac{1}{x_2}\right)} > 2\implies a_2\ge 3$$noting that equality case $x_1=x_2$ cannot be obtained.

Now assuming $a_{2n-1}\ge 3n-2$ and $a_{2n}\ge 3n$, we first find by Cauchy Schwarz
\begin{align*}
a_{2n+1} 
&= \sqrt{(x_1+\cdots +x_{2n}+x_{2n+1})\left(\frac{1}{x_1}+\cdots + \frac{1}{x_{2n}}+\frac{1}{x_{2n+1}}\right)}\\
&= \sqrt{((x_1+\cdots +x_{2n})+x_{2n+1})\left(\frac{a_{2n}^2}{x_1+\cdots +x_{2n}}+\frac{1}{x_{2n+1}}\right)}\\
&\ge a_{2n}+1 \ge 3(n+1)-2
\end{align*}with equality case $x_{2n+1} = \frac{x_1+\cdots +x_{2n}}{a_{2n}}$, as desired.

Similarly, we find $a_{2n+2}\ge a_{2n+1}+1$ with equality case $x_{2n+2} = \frac{x_1+\cdots +x_{2n+1}}{a_{2n+1}}$. Now, the key insight is to realize that these two equality cases cannot both be true. If they were, then $x_{2n+1}$ is the average of $x_1, \ldots, x_{2n}$ and $a_{2n}-2n$ zeros. Meanwhile, since for equality case $a_{2n+1} = a_{2n}+1$, then $x_{2n+2}$ is the average of $x_1, \ldots, x_{2n}$ and $a_{2n}-2n$ zeros and $x_{2n+1}$, the average of these numbers, which is still just the average of these numbers. In other words, $x_{2n+2} = x_{2n+1}$, which is disallowed. Thus, in actuality $a_{2n+2}\ge a_{2n+1}+2 \ge 3(n+1)$ as desired.

Bringing it back to the problem at hand, we get $a_{2023}\ge 3034$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
635 posts
#90 • 1 Y
Y by cubres
This is a really interesting problem, and I was pleasantly surprised when I managed to solve this in contest two years ago.

We proceed via induction to show that for all odd integers $n$ , $a_n \ge n + \frac{n-1}{2}$.

First note that
\[a_1 = \sqrt{x_1 \cdot \frac{1}{x_1}}=1\]
Now, assume the claim is true for some odd integer $m=2k+1 \ge1$. Throughout this proof, we will use the following key inequality,

Claim : For all positive real numbers $a$ and $b$,
\[(a+b)\left(\frac{1}{a}+\frac{1}{b}\right) \ge 4\]

Proof : This is simply Cauchy Schwarz. Note that,
\[(\sqrt{a}^2 + \sqrt{b}^2)\left(\frac{1}{\sqrt{a}^2}+\frac{1}{\sqrt{b}^2}\right) \ge (1+1)^2 = 4\]as desired.

Now, we are ready for the induction. Note,
\begin{align*}
a_{m+2} &=\sqrt{(x_1+x_2+\dots+x_{m+2})\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_{m+2}}\right)}\\
a_{m+2}^2 & = (x_1+x_2+\dots+x_{m+2})\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_{m+2}}\right)\\
a_{m+2}^2 &=  (x_1+x_2+\dots+x_m)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_m}\right) + \left(\frac{1}{x_{m+1}} +\frac{1}{x_{m+2}}\right)(x_{m+1}+x_{m+2})\\
& + (x_1+x_2 + \dots + x_m)\left(\frac{1}{x_{m+1}}+\frac{1}{x_{m+2}}\right) + \left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_m}\right)(x_{m+1}+x_{m+2})\\
& \ge (3k+1)^2 + 4 + \frac{(3k+1)^2(x_{m+1}+x_{m+2})}{x_1+x_2+\dots + x_m} + \frac{4(x_1+x_2+\dots + x_m)}{x_{m+1}+x_{m+2}}\\
& \ge (3k+1)^2 +4 + 2 \sqrt{\frac{4(3k+1)^2(x_{m+1}+x_{m+2})(x_1+x_2+\dots + x_m)}{(x_1+x_2+\dots + x_m)(x_{m+1}+x_{m+2})}}\\
& \ge (3k+1)^2 + 4(3k+1) + 4\\
& \ge (3k+3)^2\\
a_{m+2} & \ge 3k+3
\end{align*}But, if $a_{m+2}=3k+3$, equality must hold at each of the above inequalities. In particular we need
\[ \left(\frac{1}{x_{m+1}} +\frac{1}{x_{m+2}}\right)(x_{m+1}+x_{m+2})=4\]which requires $x_{m+1}=x_{m+2}$ which is impossible if $x_1,x_2,\dots,x_{m+2}$ are pairwise distinct positive reals. Thus equality does not hold and $a_{m+2}>3k+3$. Further since $a_{m+2}$ is an integer we must have $a_{m+2} \ge 3k+4$ so,
\[a_{m+2} \ge (m+2) + \frac{m+1}{2}\]which completes the induction. Thus,
\[a_{2023} \ge 2023 + \frac{2022}{2} = 3034\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8866 posts
#91 • 1 Y
Y by cubres
Observe that $a_1 = 1 < a_2 < \dots$. The crux of the problem is the following claim:

Claim: There does not exist an index $n$ such that $a_{n+1} - a_n = 1$ and $a_n - a_{n-1} = 1$ simultaneously.

Proof: Suppose such an index $n$ existed. Then observe that
\begin{align*}
a_n^2 &= a_{n-1}^2 + 1 + x_n\left(\frac 1{x_1} + \frac 1{x_2} + \cdots +\frac 1{x_{n-1}}\right) + \frac 1{x_n}\left(x_1+x_2+\cdots+x_n\right) \\
&\geq a_{n-1}^2 + 1 + 2\sqrt{(x_1+x_2+\cdots+x_n)\left(\frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}\right)} \\
&= (a_n+1)^2.
\end{align*}Importantly, equality only holds when $x_n = \sqrt{\frac{x_1+x_2+\cdots+x_{n-1}}{\frac 1{x_1} +\frac 1{x_2}+\cdots+\frac 1{x_{n-1}}}}$. Let $A$ and $B$ be the numerator and denominator of this quantity. The same argument applied to the index $n+1$ yields \[x_{n+1} = \sqrt{\frac{A+\sqrt{\frac AB}}{B+\sqrt{\frac BA}}} = \sqrt{\frac{A\sqrt{AB}+A}{B\sqrt{AB}+B}} = \sqrt{\frac AB} = x_n\]which is a contradiction. $\blacksquare$

So in particular, $a_{n+2} - a_n \geq 3$ for each $n$. It follows that $a_{2023} \geq 1 + 1011 \cdot 3 = 3034$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aiden-1089
295 posts
#92 • 1 Y
Y by cubres
Note that $a_1=1$. It suffices to show that $a_{n+2} \geq a_n+3$ for all $n$.
Noting that $x_{n+1}, x_{n+2}$ are distinct,
\begin{align*}
a_{n+2}^2 &=\left(\sum_{i=1}^{n+2} x_i \right)\left(\sum_{i=1}^{n+2} \frac{1}{x_i} \right) \\
&=a_n^2 + (x_{n+1}+x_{n+2})\left(\sum_{i=1}^n \frac{1}{x_i} \right) + \left(\frac{1}{x_{n+1}} + \frac{1}{x_{n+2}} \right)\left(\sum_{i=1}^n x_i \right) + (x_{n+1}+x_{n+2})\left(\frac{1}{x_{n+1}} + \frac{1}{x_{n+2}} \right) \\
&> a_n^2 + (x_{n+1}+x_{n+2})\left(\sum_{i=1}^n \frac{1}{x_i} \right) + \left(\frac{4}{x_{n+1}+x_{n+2}} \right)\left(\sum_{i=1}^n x_i \right) + 4 \\
&\geq a_n^2 + 4 \sqrt{\left(\sum_{i=1}^n \frac{1}{x_i} \right)\left(\sum_{i=1}^n x_i \right)} + 4 =(a_n+2)^2.
\end{align*}Since $a_{n+2}$ and $a_n$ are integers, $a_{n+2} > a_n+2 \implies a_{n+2} \geq a_n+3$ so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1008 posts
#93 • 1 Y
Y by cubres
Very beautiful problem, definitely one of the nicer algebra problems out there.
Claim 1: $a_{n+1} \ge a_n + 1$ with equality iff $x_{n+1} = \frac{x_1 + \dots + x_n}{a_n}$.
Proof: We have
$$a_{n+1} = \sqrt{a_n^2 + 1 + x_{n+1}  \left(\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}\right) +\frac{ x_1 + x_2 + ... + x_n}{x_{n+1}}}$$$$\stackrel{\textrm{AM-GM}}{\ge} \sqrt{a_n^2 + 1 + 2a_n} = a_n + 1.$$Equality holds iff equality holds in AM-GM, i. e.
$$x_{n+1}^2 = \frac{x_1 + x_2 + ... + x_n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}} = \frac{ x_1 + x_2 + ... + x_n}{\frac{a_n}{ x_1 + x_2 + ... + x_n}}$$$$\iff x_{n+1} = \frac{x_1 + \dots + x_n}{a_n}$$as desired. $\square$

Claim 2: We cannot simultaneously have $a_{n+2} = a_{n+1} + 1, a_{n+1} = a_n + 1$.
Proof: Note that for equality to hold in both cases, we need
$$x_{n+2} = \frac{x_1 + x_2 + \dots + x_n + x_{n+1}}{a_n + 1} = \frac{\frac{a_n+1}{a_n} \times x_1 + \dots + x_n}{a_n+1} = \frac{x_1 + x_2 + \dots + x_n}{a_n} = x_{n+1}.$$But the sequence is composed of distinct terms, so this is impossible. $\square$

Note that due to Claim 2, we must have $a_{n+2} \ge a_n + 3$. Therefore,
$$a_{2023} \ge a_{2021} + 3$$$$\ge a_{2019} + 6$$$$\ge a_{2017} + 9$$$$\vdots$$$$\ge a_1 + 3 \times 1011 = 3034$$as desired. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
complex2math
7 posts
#94 • 1 Y
Y by cubres
Let $S_n := x_1 + \cdots + x_n$ and $T_n := \frac{1}{x_1} + \cdots + \frac{1}{x_n}$, then
\[\begin{aligned}
a_{n + 2}^2 &= (S_n + x_{n + 1} + x_{n + 2})\left(T_n + \frac{1}{x_{n + 1}} + \frac{1}{x_{n + 2}}\right) \\
                    &= S_n T_n + \frac{S_n}{x_{n + 1}} + x_{n + 1}T_n + \frac{S_n}{x_{n + 2}} + x_{n + 2}T_n + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n + 1}}{x_{n+2}} + 2 \\
                    &\ge a_n^2 + 2a_n + 2a_n + 4 = (a_n + 2)^2
\end{aligned}\]
The equality cannot hold because $x_{n + 1} \neq x_{n + 2}$, thus $a_{n + 2} \ge a_n + 3$ and $a_{2n + 1} \ge 3n + 1$ follows easily by induction.

I am surprised that it is rated A3 on the shortlist, considering the difficulty of this problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pie854
243 posts
#95 • 1 Y
Y by cubres
Note that by CS: $$a_n=\sqrt{(x_1+\dots+x_{n-2}+x_{n-1}+x_n)(1/x_1+\dots+1/x_{n-2}+1/x_{n-1}+1/x_n)} \geq \sqrt{(x_1+\dots+x_{n-2})(1/x_1+\dots+1/x_{n-2})}+\sqrt{x_{n-1}\cdot \frac{1}{x_{n-1}}}+\sqrt{x_n\cdot \frac{1}{x_n}}=a_{n-2}+2.$$If equality were to hold, then we would have $\frac{x_n}{1/x_n}=\frac{x_{n-1}}{1/x_{n-1}}$ i.e. $x_n=x_{n-1}$, which is not possible. Thus, $$a_n>a_{n-2}+2\implies a_n\geq a_{n-2}+3.$$By induction it follows that $a_{2023}\geq 3033+a_1=3034$, as desired. 
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
656 posts
#97 • 1 Y
Y by cubres
We prove the following lower bound via induction:
Claim: $a_n\geq \frac{3n + 1}{2}$
Proof: We prove $a_n \geq a_{n - 2} + 2$ with induction. First, note that $a_1 = 1.$

Now, we prove the inductive step. Observe that
\begin{align*}
            a_n^2 & = a_{n - 2}^2  + \left(\frac1{x_n} + \frac1{x_{n - 1}}\right)(x_{n - 2} + x_{n - 3} + \dots + x_1) + (x_n + x_{n - 1})\left(\frac1{x_{n - 2}} + \frac1{x_{n - 3}} + \dots + \frac1{x_1}\right) + \left(\frac1{x_n} + \frac1{x_{n - 1}}\right)(x_n + x_{n - 1})\\
            & \geq a_{n- 2}^2 + 4 + 2\sqrt{(x_n + x_{n - 1})\left(\frac1{x_{n - 2}} + \frac1{x_{n - 3}} + \dots + \frac1{x_1}\right)\left(\frac1{x_n} + \frac1{x_{n - 1}}\right)(x_n + x_{n - 1})} \\
            & \geq a_{n - 2}^2 + 4 + 4a_{n - 2} = (a_{n - 2} + 2)^2
        \end{align*}
This finishes.
This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 11, 2025, 3:59 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Nari_Tom
117 posts
#98 • 1 Y
Y by cubres
Where is shortlist 2024 huh, some people in my country (including me) already have it and there is no reason to not post it in here cause it's potentially public now LOL
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sangsidhya
23 posts
#99
Y by
EthanWYX2009 wrote:
Motivation. It is obvious to see that $3034=2022\times\frac 32+1,$ which leads us to the lemma below$.$
Lemma. For $\forall n\in\mathbb Z_+,a_{n+2}\geq a_n+3.$
how does it intuitively lead to this?
Z K Y
N Quick Reply
G
H
=
a