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
Diodes and usamons
v_Enhance   47
N 12 minutes ago by EeEeRUT
Source: USA December TST for the 56th IMO, by Linus Hamilton
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?

Proposed by Linus Hamilton
47 replies
v_Enhance
Dec 17, 2014
EeEeRUT
12 minutes ago
IMO Genre Predictions
ohiorizzler1434   58
N 25 minutes ago by WLOGQED1729
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
58 replies
1 viewing
ohiorizzler1434
May 3, 2025
WLOGQED1729
25 minutes ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   2
N an hour ago by Tkn
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
2 replies
BR1F1SZ
Monday at 9:45 PM
Tkn
an hour ago
3-var inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
1 reply
sqing
an hour ago
sqing
an hour ago
IMO ShortList 2001, combinatorics problem 3
orl   37
N an hour ago by deduck
Source: IMO ShortList 2001, combinatorics problem 3, HK 2009 TST 2 Q.2
Define a $ k$-clique to be a set of $ k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.
37 replies
orl
Sep 30, 2004
deduck
an hour ago
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N an hour ago by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
an hour ago
Geo metry
TUAN2k8   3
N an hour ago by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
3 replies
TUAN2k8
Yesterday at 10:33 AM
TUAN2k8
an hour ago
Functional equation of nonzero reals
proglote   8
N 2 hours ago by jasperE3
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
8 replies
proglote
Oct 24, 2013
jasperE3
2 hours ago
Interesting inequalities
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^2-ab+b^2+a+b=3  $. Prove that
$$  \frac{39+\sqrt{13}}{78}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq  \frac{1}{2}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=3  $. Prove that
$$  \frac{19+\sqrt{10}}{39}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq    \frac{39+\sqrt{13}}{78}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=5  $. Prove that
$$  \frac{3}{5}> \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq     \frac{185+3\sqrt{21}}{402}$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
4-var inequality
sqing   2
N 3 hours ago by sqing
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
3 hours ago
Inspired by Bet667
sqing   2
N 3 hours ago by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
2 replies
sqing
Yesterday at 2:46 PM
sqing
3 hours ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   14
N 3 hours ago by MathLuis
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
14 replies
parmenides51
Jul 25, 2018
MathLuis
3 hours ago
GCD of terms in a sequence
BBNoDollar   1
N 3 hours ago by mashumaro
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
1 reply
BBNoDollar
Yesterday at 10:29 PM
mashumaro
3 hours ago
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N 3 hours ago by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
3 hours ago
Problem 4 (second day)
darij grinberg   93
N Apr 25, 2025 by Ilikeminecraft
Source: IMO 2004 Athens
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
93 replies
darij grinberg
Jul 13, 2004
Ilikeminecraft
Apr 25, 2025
Problem 4 (second day)
G H J
Source: IMO 2004 Athens
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
795 posts
#83
Y by
We first prove the case $n=3$. Assume FTSOC we have $(t_1,t_2,t_3) = (a,b,a+b+k)$, where $k \ge 0$. Then
\begin{align*}
3^2+1 &> (t_1+t_2+t_3)\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3}\right) \\
&\ge 5 + \frac{t_3}{t_1} + \frac{t_3}{t_2} + \frac{t_1}{t_2} + \frac{t_1}{t_3} \\
&= 7 + \frac{c+k}{b} + \frac{b+k}{c} + \frac{b+c}{b+c+k} \\
&\ge 10 + k\left(\frac 1b + \frac 1c - \frac{1}{b+c+k}\right) \\
&\ge 10,
\end{align*}
contradiction. We now show the case $n \ge 4$. Assume again FTSOC $t_1, t_2, t_3$ do not form the sides of a triangle. Then
\begin{align*}
n^2+1 &> \left(t_1 + t_2 + \ldots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_n} \right) \\
&= 10 + (n-3)^2 + \left(t_1+\ldots+t_3\right)\left(\frac{1}{t_4}+\ldots+\frac{1}{t_n}\right) + \left(t_4+\ldots+t_n\right) \left(\frac{1}{t_1}+\ldots+\frac{1}{t_3}\right) \\
&\ge 10 + (n-3)^2 + 2\sqrt{10(n-3)^2} \\
&= \left(n+\sqrt{10}-3\right)^2 \\
&\ge n^2+1,
\end{align*}
contradiction. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RedFireTruck
4223 posts
#84
Y by
Assume WLOG that $t_1\le t_2\le \dots\le t_n$.

Claim: $a\ge b+c$ for $a,b,c>0$ means that $\frac{b+c}{a}+\frac{a}{b}+\frac{a}{c}\ge 5$.

Proof: Let $a=b+c+d$. Then, we wish to prove that $(1-\frac{d}{b+c+d})+2+\frac{b}{c}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\ge 5$. By AM-GM, $\frac{b}{c}+\frac{c}{b}\ge 2$ so it suffices to prove that $\frac{d}{b}+\frac{d}{c}\ge \frac{d}{b+c+d}$, which is true because $\frac1b+\frac1c\ge \frac1{b+c}$.

Notice that for $n=3$, we just need to prove that $t_3\ge t_1+t_2$ implies that $(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})\ge 10$. Expanding gives $$(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})=3+\frac{t_1}{t_2}+\frac{t_2}{t_1}+\frac{t_1+t_2}{t_3}+\frac{t_3}{t_1}+\frac{t_3}{t_2}\ge 3+2+5=10$$as desired.

Assume that for some $n-1\ge 3$, the assertion is true. We will now prove that the assertion is true for $n$.

If there exists $1\le i<j<k\le n-1$ such that $t_i$, $t_j$, $t_k$ don't form the side lengths of a triangle, then $$(t_1+t_2+\dots+t_{n-1})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n-1}})\ge (n-1)^2+1$$and $$t_n(\frac1{t_1}+\frac1{t_2}+\dots+\frac1{t_{n-1}})+(t_1+t_2+\dots+t_{n-1})\frac{1}{t_n}+1\ge 2n-1$$by AM-GM so $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n^2+1$$as desired.

Otherwise, assume that $t_n\ge t_1+t_2$. By AM-GM and our claim, $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n+2\binom{n}{2}-4+\frac{t_1+t_2}{t_n}+\frac{t_n}{t_1}+\frac{t_n}{t_2}\ge n^2+1$$as desired.

We are done by induction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blueprimes
352 posts
#86
Y by
For the sake of contradiction, assume that three of the $t_i$ are equal to $a$, $b$, $a + b + x$ where $a$, $b$ are positive reals and $x$ is nonnegative. By Cauchy-Schwarz, we have
$$(t_1 + t_2 + \dots + t_n) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) \ge \left(n - 3 + \sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} \right)^2.$$Now
$$\sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} = \sqrt{6 + \frac{2a}{b} + \frac{2b}{a} + x \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{a + b + x} \right) } \ge \sqrt{10}$$by AM-GM. To reach contradiction it is now sufficient to show $(n - 3 + \sqrt{10})^2 \ge n^2 + 1 \iff n \ge 3$ which is true. We are done.
This post has been edited 2 times. Last edited by blueprimes, Jun 24, 2024, 1:53 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dudade
139 posts
#88
Y by
FTSOC, suppose $t_1 > t_2 + t_3$. Therefore, we want to show
\begin{align*}
n^2 + 1 &> \sum_{1 \leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right) + n \\
&= \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \sum_{4\leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right)+n \\
&\geq \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \left(n^2 - n - 6\right) + n.
\end{align*}Thus, we want
\begin{align*}
7 > \left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)
\end{align*}if $t_1 > t_2 + t_3$. By AM-GM-HM, note that
\begin{align*}
\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right) &= \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right) + \left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right) + 2 \\
&\geq \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right)+ \dfrac{4}{\left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right)} + 2 = k + \tfrac{4}{k} + 2.
\end{align*}Since $k < 1$, then we must have that $k + \tfrac{4}{k} + 2 \leq 7$ which is clearly a contradiction. Thus, for all $i$, $j$, and $k$, there exists a triangle with side lengths $t_i$, $t_j$, and $t_k$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jolynefag
125 posts
#89
Y by
Elegant problem which is still being solved by people these days. So beautiful!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SomeonesPenguin
128 posts
#90 • 1 Y
Y by zzSpartan
storage
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
numbertheory97
42 posts
#91
Y by
Really nice problem! I think I accidentally ended up proving the stronger version.

Solution. Without loss of generality, suppose to the contrary that $t_1 \geq t_2 + t_3$. Then \[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]and \[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]by Cauchy-Schwarz, so we have \[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]But $f(x) = \left(x + \sqrt{10} - 3\right)^2 - x^2$ is strictly increasing, so $f(x) \geq f(3) = 1$, a contradiction. $\square$

Remark. (Motivation) Why $2/5$ and $3/5$? I originally tried using $1/2$'s, which was very close but broke slightly for $n = 3$. But if we use arbitrary coefficients $x$ and $1 - x$, it suffices to maximize the function \[f(x) = \sqrt x + 2\sqrt{2 - x},\]which by C-S or similar methods is greatest when $x = 2/5$. This fits perfectly when $n = 3$.
This post has been edited 1 time. Last edited by numbertheory97, Aug 23, 2024, 11:51 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ezpotd
1263 posts
#92
Y by
Sort $t_n$. Assume to the contrary that there exists some tuple $(t_1 \cdots t_n)$ such that the inequality is satisfied and $t_1 + t_2 \le t_n$. We show that if $n \ge 4$ we can form a similar tuple with $n -1$ elements. Of course, just remove $t_3$, the decrease on the left is $2n - 1$, the decrease on the right is $t_3 \frac{1}{t_3} + \sum_{i \neq 3} \frac{t_i}{t_3} + \frac{t_3}{t_i} \ge 2n - 1$ by AM-GM, so the inequality is preserved. Now we can just reduce to the $n = 3$ case, where we MUST have $10 > 3 + \sum_{sym} \frac{t_i}{t_j}$. Since $\frac{t_3}{t_1} + \frac{t_1}{t_3}, \frac{t_3}{t_2} + \frac{t_2}{t_3}$ are increasing for $t_3 > t_1, t_2$, if the inequality holds true for $t_3 > t_1 + t_2$, it must hold for $t_3 = t_1 + t_2$. Now plugging this value of $t_3$ gives $7 > 2(\frac{t_1}{t_2} + \frac{t_2}{t_1}) + 3$, which is NEVER true by AM-GM, so the original inequality can never hold in the conditions described.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GeorgeRP
130 posts
#93 • 1 Y
Y by Bumfuzzle
We will solve the problem by induction on n.

Base case: n=3. We have that:
$$10>(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \Leftrightarrow 7abc>\sum{a^2b}$$WLOG we can assume $a$ is the largest. Thus we need to show $a<b+c$. If we rewrite as a quadratic function in respect to $a$ (considering $b$ and $c$ as parameters) we have:
$$0>a^2(b+c)+a(b^2+c^2-7bc)+(b^2c+c^2b)=f(a)$$We however know:
$$ b+c>a \Leftrightarrow \begin{cases}
  f(b+c)\geq0 \\
  b+c>\frac{7bc-b^2-c^2}{2(b+c)} 
\end{cases} \Leftrightarrow \begin{cases}
  2(b+c)^3-8(b+c)bc\geq0\\
  3b^2-3bc+3c^2>0 
\end{cases} \text{ which are both true, finishing the base}$$
IH: for n-1

IS: for n. Let $A=t_1+t_2+\cdots+t_{n-1}, B=\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{n-1}}$.WLOG we can assume $t_1\leq t_2 \leq \cdots \leq t_n$. We know that $n^2+1>(A+t_n)(B+\frac{1}{t_n}) \Leftrightarrow n^2-Bt_n-\frac{A}{t_n}>AB$. We will first show that $Bt_n+\frac{A}{t_n}\geq2n-2$. This follows from the fact:
$$Bt_n+\frac{A}{t_n}=\sum_{1\leq i \leq n-1}{\frac{t_n}{t_i}+\frac{t_i}{t_n}} \overset{AM-GM}{\geq} 2n-2$$From this it directly follows that $(n-1)^2+1>AB$, implying that $t_i, t_j, t_k$ are the sides of a triangle for $1\leq i<j<k\leq n-1$. We are now left to show that $t_n<t_1+t_2$ from which the desired will follow (because $t_1, t_2$ are the smallest). For simplicity let $t_1=x, t_2=y$. FTSOC let $t_n=x+y+k$ for some $k\geq0$. We know that by Cauchy-Schwarz $AB\geq(n-1)^2$. We will now show that $t_nB+\frac{A}{t_n}\geq 2n-1$.
$$t_nB+\frac{A}{t_n}= \frac{t_n}{t_1}+\frac{t_1}{t_n}+\frac{t_2}{t_n}+\frac{t_n}{t_2}+(\frac{t_3}{t_n}+\cdots+\frac{t_{n-1}}{t_n})+(\frac{t_n}{t_3}+\cdots+\frac{t_n}{t_{n-1}}) \overset{\text{AM-GM}}{\geq} $$$$ \geq \frac{a+b+k}{a}+\frac{a}{a+b+k}+\frac{a+b+k}{b}+\frac{b}{a+b+k}+2n-6 \geq \frac{a+b}{a}+\frac{a}{a+b}+\frac{a+b}{b}+\frac{b}{a+b}+2n-6=$$$$=\frac{(a+b)^2}{ab}+2n-5\overset{\text{AM-GM}}{\geq}2n-1$$From here it follows that:
$$n^2> AB+t_nB+\frac{A}{t_n}\geq n^2 \text{ \Large{ \lightning }} \text{ with which we are done}$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
N3bula
268 posts
#94
Y by
Expanding gives us:
\[\left(t_1 + t_2 + \dots + t_n\right)
  \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]Thus we get if for any $i$, $j$ and $k$ that $\sum_{sym}\frac{t_i}{t_j}\geq 7$ we get a contradiction,
thus it suffices to prove that if we have $a+b\leq c$ that $\sum_{sym}\frac{a}{b}\geq 7$, thus I will
prove that that sum is minimised when $a+b=c$, we get that proving that is equivalent to proving
\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]When $a+b=c$ and $k$ is a positive real. Thus the above is equivalent to:
\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]\[\frac{b+a}{ab}-\frac{1}{a+b}\]The above is minimised when $a=b$, thus its equivalent to:
\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]Which is clearly true.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eg4334
637 posts
#95
Y by
WLOG $t_1 \leq t_2 \leq \dots \leq t_n$, and FTSOC let $t_i + t_j < t_k$. The condition then gives by Cauchy that $$n^2 +1 \geq \left( n -3 + \sqrt{3+\sum \frac{t_i}{t_j} } \right)^2$$where the summation is taken over all ordered pairs in $t_i, t_j, t_k$. Now notice that $n^2 +1 \geq (n - 3 + \sqrt{10})^2$ has no integer solutions for $n \geq 3$, so we just need to prove that $$\frac{t_i}{t_j} + \frac{t_i}{t_k} + \frac{t_j}{t_i} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq 7 $$$$\frac{t_i}{t_k} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq5$$by AMGM. Also, $(\frac{1}{t_i} + \frac{1}{t_j})(t_i + t_j) \geq 4$ by Cauchy so this reduces into $$\frac{t_i+t_j}{t_k} + \frac{4t_k}{t_i+t_i} \geq 5$$which is obvious.
This post has been edited 2 times. Last edited by eg4334, Nov 29, 2024, 8:56 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#96
Y by
We show by induction on $n$ that if there exists at least one triple $(t_i, t_j, t_k)$ that are not the sides of a triangle, then $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$$\text{Base Case: }$ For $n=3,$ we must show that if $t_1, t_2, t_3$ cannot be the sides of a triangle, then $$10 \leq (t_1+t_2+t_3)\left( \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3} \right).$$WLOG, assume that $t_1 \leq t_2 \leq t_3,$ then we have that $t_3 > t_1+t_2.$ Also, because our inequality is homogenous, assume that $t_3=1.$ Letting $x=t_1+t_2,$ it suffices to show that $$7 \leq \frac{t_1}{t_2}+\frac{t_2}{t_1}+x+\frac{1}{t_1}+\frac{1}{t_2}.$$But by AM-GM, $\frac{t_1}{t_2}+\frac{t_2}{t_1} \geq 2,$ so it suffices to show that $$x+\frac{1}{t_1}+\frac{1}{t_2} \geq 5.$$But, $x \leq t_3=1 \implies (x-1)(x-4) \geq 0 \implies 5 \leq x+\frac{4}{x} \leq x+\frac{1}{t_1t_2} \leq x+\frac{1}{t_1}+\frac{1}{t_2}$ by $2$ applications of AM-GM at the end. Therefore, our base case is proven.

$\text{Inductive Step: }$ Now, assume that for $n=k-1 \geq 3$ our proposition holds. Consider the sequence $t_1, t_2, \cdots t_k$ of positive real numbers. Then if there is at least one triple, say WLOG $(t_1, t_2, t_3),$ which are not the sides of a triangle then we have by our inductive hypothesis that
\begin{align*}
(t_1+t_2+\cdots+t_k)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_k} \right) &=(t_1+t_2+\cdots+t_{k-1})\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{k-1}} \right) + \sum_{i=1}^{k-1} \left( \frac{t_i}{t_k}+\frac{t_k}{t_i} \right)+1 \\
&\geq \left((k-1)^2+1\right)+(k-1)\cdot 2+1 \text{ by AM-GM} \\
&= k^2+1.
\end{align*}Therefore, our induction is complete. QED

Now, for the sake of a contradiction assume that there exists $t_i, t_j, t_k$ which are not the sides of a triangle, and $$n^2+1 > (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$But by our induction above, we have that $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right),$$a contradiction. Therefore, all $t_i, t_j, t_k$ are the sides of a triangle. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Marcus_Zhang
980 posts
#97
Y by
IMO P4
This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 23, 2025, 3:46 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cubres
119 posts
#98
Y by
Yapping
Storage - grinding IMO problems
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
616 posts
#99
Y by
Assume there exists a sequence $t_i$ such that $t_1 \geq t_2 + t_3.$ Then, we have that:
\begin{align*}
  & (t_1 + t_2 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \cdots + \frac1{t_n}\right) \\
  & \geq (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (t_1 + t_2 + t_3)\left(\frac1{t_4} + \cdots + \frac1{t_n}\right)\\
  & \qquad + (t_4 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (n - 3)^2 \\
  & \geq (n - 3)^2 + 2(3n - 9) + 3 + (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right)
\end{align*}Thus, all that is left is to prove that $$(t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) \geq 10$$Expanding, we have that $$3 + \frac{t_1}{t_2} + \frac{t_3}{t_2} + \frac{t_2 + t_3}{t_1} + \frac{t_1}{t_3} + \frac{t_2}{t_3}$$Now, differentiate w.r.t. $t_1.$ We get $f(t_1) = \frac1{t_2} + \frac1{t_3} - (t_2 + t_3)\frac1{t_1^2}.$ We claim that $t_1^2 \geq t_2t_3.$ However, this is clearly true because $t_1^2 \geq t_2^2 + t_3^2 + 2t_2t_3,$ while $t_2^2 + t_3^2 + t_2 t_3 \geq 0.$ Thus, $f(t_1)$ is increasing as $t_1$ increases. Thus, we can assume that $t_1 = t_2 + t_3.$ This clearly implies that $f(t_1) \geq 10.$ Hence, we are done.
Z K Y
N Quick Reply
G
H
=
a